if:
(2^n)-1 is prime
[b]prove that:[/b]
n is prime.

let us suppose that N is composed and d is a divisor strict of n
wa can write n=dd' and:
2^n-1=(2^d-1)(2^{d(d'-1)}+2^{d(d'-2)}.....+1)
the factor 2^d-1 is noncommonplace sow 2^n-1 is composed

[hide]Suppose that $ab=n$ ($a,b$ not equal to 1)

It is a well-known rule in number theory that $a-b|a^{n}-b^{n}$ for all integers $a,b$.

But $2^{n}=2^{ab}=2^{a^{b}}$
So $2^{a}-1|2^{a^{b}}-1^{b}=2^{n}-1$

So $2^{a}-1|2^{n}-1$
Since $2^{a}-1$ is neither 1 nor $2^{n}-1$ it follows that $2^{n}-1$ is composite.
[/hide]

edit: I see diane got there first, but anyway...

[hide]Contrapositive- If n is not prime, $2^{n}-1$ is not prime.
If n is not prime, it equals 1, in which case $2^{1}-1=0$ which is not prime, or it equals ab, where a and b are some positive integers not equal to 1. If n=ab, $2^{n}-1=(2^{a})^{b}-1=(2^{a}-1)(2^{ab-1}+2^{ab-2}+...+1)$. Since both terms are >1, $2^{n}-1$, is composite, so therefore is not prime. We have proved the contrapositive, so the original statement is true.[/hide]


I wasn't sure where to post this question but so i decided here.

I am currently in an argument to prove to my friend that n/0 is not infinity. I heard there are a few paradoxes that happen if n/0 is treated as infinity.

How would you explain why n/0 is NOT infinity?

Oh geez...
Try [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=203903]this[/url] if you dare...

It is undefined...
but it  can be indeterminable.

The limit as x approaches 0 for $ \frac{\sin{x}}{x}\equal{}1$

[quote="ashpotter"]I am currently in an argument to prove to my friend that n/0 is not infinity.[/quote]
Here's the proof: It's undefined.

There have been numerous topics about this; searching would have turned up some of them, like the one Brut3Forc3 pointed out.

Also, double post: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=208792 (probably accidental, but still).

sorry about the double post. What happened was that my browser lagged so i refreshed the page and clicked submit twice.


On the AMC 12, each correct answer is worth $6$ points, each incorrect answer is worth $0$ points, and each problem left unanswered is worth $2.5$ points. If Charlyn leaves $8$ of the $25$ problems unanswered, how many of the remaining problems must she answer correctly in order to score at least $100$?

$\text {(A)} 11 \qquad \text {(B)} 13 \qquad \text {(C)} 14 \qquad \text {(D)} 16\qquad \text {(E)}17$

[hide="Answer"]C.

We want the smallest integer $x$ such that $(2.5)(8)+x>100$.[/hide]

[quote="Apprentice"][hide]Blind luck: 17*6 = 102, so E[/hide][/quote]Read the question carefully. :)

[quote="Apprentice"][hide]Blind luck: 17*6 = 102, so E[/hide][/quote]
You're forgetting that she recieves points for those 8 unanswered questions.

EDIT: Oops. I posted like three seconds after you.  :oops:

[hide]Charlyn has 20 points from the unanswered questions, so she needs 80 more points. Each question is worth 6 points, so the smallest integer greater than $\frac{80}{6}$ is $14\implies\boxed C$[/hide]

Charlyn needs 80 points more. Then,

6X≥80

X≥80/6

X≥40/3

As X must be an integrer, X=14. So, Charlyn has to answer 14 answers correctly at least.

Then, the correct answer is C.


number theory
relatively prime
combinatorics unsolved
combinatorics

There are $n$ distinct points on a circumference. Choose one of the points. Connect this point and the $m$th point from the chosen point counterclockwise with a segment. Connect this $m$th point and the $m$th point from this $m$th point counterclockwise with a segment. Repeat such steps until no new segment is constructed. From the intersections of the segments, let the number of the intersections - which are in the circle - be $I$. Answer the following questions ($m$ and $n$ are positive integers that are relatively prime and they satisfy $6 \leq 2m < n$).

1) When the $n$ points take different positions, express the maximum value of $I$ in terms of $m$ and $n$.

2) Prove that $I \geq n$. Prove that there is a case, which is $I=n$, when $m=3$ and $n$ is arbitrary even number that satisfies the condition.

I'm pretty sure you have the additional condition that $m$ and $n$ are coprime (otherwise you can easily give a counterexample to (2)).

For the first part, it's easy to see that you can choose the points so that no three diagonals of the $n$-gon are concurrent, which means that $I\le (m-1)n$ elements: for each of the $n$ diagonals, there are $2(m-1)$ intersection points between that diagonal and other diagonals, and each point is counted twice.

For the second part, for each diagona $A_iA_{i+m}$ consider the point of intersection between it and $A_{i+m-1}A_{i+2m-1}$. It's easy to see that this gives an injection from the set of $n$ diagonals we have drawn to $I$, so $I\ge n$. As for the construction, start with a regular $\frac n2-$gon $X_0X_2\ldots X_{n-2}$, and take $X_1,X_3,\ldots,X_{n-1}$ to be the midpoints of $X_0X_2,X_2X_4,\ldots,X_{n-2}X_0$ respectively. After this, put $A_k=X_kX_{k+2}\cap X_{k+3}X_{k+5},\ \forall k\in\overline{0,n-1}$, where all the indices are modulo $n$.


"Question about a party?"
[hide][b]Anarchist[/b] - There should be no government; encompassing all anarchic movements even if they are at odds.
[b]Classical Conservative[/b] - Strong government, hierarchal.
[b]American Conservative[/b] - Domestic small-government, pro-military, relatively religious.
[b]Classical Liberal / Libertarian[/b] - There should be fewer laws and few restrictions on people or anything else by the government.
[b]American Liberal / New Deal Liberal /Neo-Liberal / Social Democrat[/b] - Big, economy-regulating equality-promoting socially-active program-full wealth-redistributing (usually minor) side.  Somewhere between Classical Liberals and Socialists.
[b]Democratic Socialist[/b] - Socialist as achieved by democracy (major government ownership of everything.)
[b]Revolutionary Socialist, Communist, Marxist[/b] - Communism or Socialist as achieved by revolution.
[b]Monarchy/Fascism/Authoritarianism[/b] - In all its forms.  One guy at the top.

If you feel that you aren't represented here, are at odds with the definitions provided, or have some other beef with the poll, go ahead and comment below and I may add your choice / remedy that.[/hide]

This has been done before...check the Polls section

i like how the majority of the people say they're communists

I am an anarchist, but I believe in its evolution through a communist revolution. :wink:

[quote="Dark_Shadow"]I don't see it.  Could you drop me a link, 7h3.demon.117?[/quote]

nvm it was a similar polticial question in the Round table...

Why isn't Potato on the list?

If there was a potato party, I'd totally vote for it. Potatoes are awesome.

You probably think I'm kidding, but most of my friends could tell you that I'm 100% serious. Let me illustrate some of the great things about potatoes:

1. They taste awesome.
2. They have most of your daily value of Vitamin C, potassium, and B6, and can act as a substitute for fiber. (Of course, wikipedia doesn't know all of the health benefits of potatoes.)
3. They are much cheaper than other food, in fact, I believe they could single-handedly wipe out hunger and poverty.
4. If there was a Potato party, America wouldn't be stuck with a duopoly controlling the government.
5. There's more, but I have to go cook some potatoes right now. Yes, I CAN smell your envy from here.

It is interesting how there are so many communists.


function
inequalities solved
inequalities

Let x, y, z be positive reals such that
   1/x + 1/y + 1/z = 1, x+y+z=10
Find maximal and minimal values of  A = x 2 + y 2 + z 2.

Source: Suggested by Namdung for School Team Selection Test, year ?

Could you please post your solution to this problem, Namdung? It is very interesting and I cannot find an elementary solution. Maybe we could study the function f(x)=(x^3-10x^2)/(1-x), but  I prefer ideas rather than computations. Thanks.

1) Firstly, we show that  2 <= x, y, z <= 5. Ideed:
    1-1/z = 1/x+1/y, 10-z=x+y so (1-1/z)(10-z) >= (x+y)(1/x+1/y) = 4
=> z^2 - 7z + 10 <= 0 => 2 <= z <= 5. Similarly with x, y.
2) Since x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx), we have to find max and min of xy+yz+zx.
     We have (x-2)(y-2)(z-2) >= 0 => xyz - 2(xy+yz+zx) + 4(x+y+z) - 8 >=0 => xy+yz+zx <= 32. The equality holds, for example when x=2, y=z=4.
     Similarly, (5-x)(5-y)(5-z)>=0 => 125 - 25(x+y+z) + 5(xy+yz+zx) - xyz >=0 => xy+yz+zx >= 125/4 (We use x+y+z=10 and xyz=xy+yz+zx).
The equality holds, for example when x=5, y=z=5/2.
3) Finally, we conclude that max = 75/2, min = 36.

Namdung

Super! Thank you very much, Namdung. It is absolutely splendid.


ARML
email
analytic geometry
graphing lines
slope

So the team list is [url=http://www.nycmathteam.com/2006-nyc-math-team/arml-selections/]up[/url].
I'm wondering if team members got the email too? I guess I can PM you a copy of the email (with password) if you don't have it already.

[quote="JBL"]Heh, they put the two Wensi's on the same team.[/quote]
...again. I noticed too.

[quote="JBL"]Also, is Kevin really not in the top 4 at Hunter?  I mean, it looks like he's fifth, since he appears to be captain of the team with the people who are almost on A, but still.[/quote]
Maybe not being at NYSML counted against him...

Ah, I didn't know he missed NYSML.

Breakdown of ARML members by school:
Stuyvesant	29
Brooklyn Tech	9
Hunter	8
Cardozo	6
Francis Lewis	3
Bowne	2
Newcomers	2
Bayside	2
HSMSE	2
Flushing	1
Saint Ann's	1
Trinity	      1
Bronx Sci	1
Midwood	1

[quote="JBL"]Heh, they put the two Wensi's on the same team.

Also, is Kevin really not in the top 4 at Hunter?  I mean, it looks like he's fifth, since he appears to be captain of the team with the people who are almost on A, but still.

I think team members did get the e-mail, although [i]I[/i] didn't.[/quote]

he's definitely in the top 4 at hunter...

EDITED due to satisfactory explanation of situation.

As Mr. Linker BCCs all the recepients, it's hard to tell who did't get emails, though i think he might have been emailing the people who went to NYSML

PS to joel: there was an email sent earlier (which you may or may not have gotten as well) about stating interest in going to ARML as coach- act upon that if you haven't gotten it, i can forward you stuff if you want (and if barry didn't already)

I'm posting a letter that I sent to the A Team Captains (slightly modified).

As of last night, several students have already expressed reservations with the published A team selections.  Having befriended most of the students on the A team, I understand the disappointment that they might be feeling.

This year, we've seen an unprecedented number of A team calibre students.  Unfortunately, the fact remains
that the A team ultimately comprises only 15 students and thus some very difficult choices had to be made.  During the committee's deliberations, we have taken into account all available scores as well as the input of no less than a dozen indviduals (including the two captains).  This process is imperfect, but necessary.  Based on those judgements, the committee reached a consensus that was recently published and we continue to stand by those decisions. Our hope is that everyone can now put this behind them and focus on the preparation of ARML.

Thanks Maria, I did get that one.

Anyhow, this just means that the N team is going to rock, too  :lol:

Not to keep being difficult, but while you're here, Jack:  Evelyn Israel from Bronx gets a 7 at NYSML, doesn't make the ARML team?  Ditto for Val Zakharevich, or however he spells that, at Murrow.

The Bronx Science prom is the night before ARML [u]again[/u]. She and another Bronx Science senior said they wouldn't go to ARML at NYSML.

Prom -- feh.  I figured it was probably something like that.

Val is going to some camp that weekend as well :(

Well that's odd...I think I received that e-mail, but my mind is always a slippery slope, so I'm not too sure about it.  Rest assured that I do want to come, though.

I was surprised a bit myself for the A team, but then again, if we are overflowing with high scorers, then some of our other teams stand a good shot of doing well as well.

The team I coached at NYSML remained relatively intact though....

--KC

I suppose someone will be doing some online relay practices sometime?
I'll be glad to host, or fill up a 2/3-full team, or just show up and watch.
My AIM is ManicBarry.

Agreed.  I shall volunteer as well.  I already have four sets of relays written up...

AIM: ixidor1987

P.S. What?  I need SOMETHING to do in World Cultures: China lectures!

--KC

.........Maybe i am just too weak to be on A team...... :(

It is getting more and more obvious that the nyc math team is only picking people from well-known school. It is ok to pick a lot of specialized high school students  'cause many of them are smart. (No offences to those students and schools) "A lot" is different from "most." It seems to me that "most" students picked came from those schools.
But is it fair to students from other schools?

I am in college right now and I still keep in touch with the math team coach in my high school. I am also helping in my school's math team. He told me that NYSML started to make changes in team selection. He told me that  NYSML would pick at least one student from each school. what I was thinking right away was, "Well, it seems that everyone can participate now."

When I look at the ARML team list now, it no longer looks like a nyc math team, but a "well-known schools math team. " And I refer back to the NYSML individual score, I notice that 2 people receive the same score have totally different fates. The person who come from those school can stay in the original team, while another person with the same score drops to reserve team. Someone from those school stays in the team, while the other person can't even get to ARML. ...more and more examples....

I understand that there are many different criteria on selection, besides NYSML score. But I personal don't think it is a good start...

Hi Simon, welcome to the forum!

First, some numbers:

Among 68 students on the ARML team, there are 14 high schools represented.  At NYSML, among 90 students, there were more than 20 schools represented.  From the seven schools which lost representation, Val Zakharevich was the highest individual scorer, with a 7, and was invited back to ARML but has other obligations.  Among the other students representing those schools, the highest individual score was a 3.
The Stuyvesant math team has, last I knew, more than 250 students.  That is, most probably, larger than the math teams of all other 13 schools sending students to ARML combined.
As the composition of the NYSML R team shows, there was a strong preference for choosing students from schools without large representation: 7 students were the lone representative from their school.  2 other students were one of two from their school to attend.  The 6 remaining students were from Tech, Hunter and Stuyvesant and accounted for the highest 4 scores on the team.  There were also at least three students who were the lone representative of their schools who were placed on a different NYC team.

Now, that all said, I have sympathy for the view that the NYSML and ARML team should incorporate students from as many high schools as possible, especially when the contest is downstate and so easily accessible to NYC.  My conversations with those responsible for choosing the NYSML and ARML teams, as well as the actual data (especially the composition of the NYSML R team) lead me to believe that "representing a non-Stuy/Hunter/Tech/Queens powerhouse school" is taken seriously into account in building the NYC math team.
When the number of students decreases by 25%, it should not be too suprising (even if it is disappointing) that the number of schools represented declines by even more than this.  I'm thinking about doing a more thorough analysis of the NYSML and ARML team lists, but I strongly suspect that NYSML score will be a much stronger predictor of ARML fate than school representation.

I just want to say that the Hunter girl who replaced me has the same NYSML individual score as mine.


[quote="manlio"]Let $p,q,r$ reals not equal $0$ and such that $p+q+r=1$and $a,b,c$ sides of a triangle. Prove that

$a^2pq+b^2qr+c^2rp \geq 0$[/quote]

Still not correct; take a = b = c = 1, p = -4, q = -3, r = 8.

  Darij

Hey, guys, there are still some nice and difficult problems unanswered. I think it would be more interesting to solve those ones than to try to repair some trivial inequalities. My opinion... hac...


I saw this in an incomplete paper, which only had the year:97.

We have n different primes:p_1,p_2,....p_n.and N=p_1p_2...p_n.
A and B write in turn composite divisors of N.with the rule that there should never be 2 numbers such that one of them divides the other.

Who can win?(A starts the game)

Could you give a little more explanation on how this game works?
When does a player "win"?

And you mean with N the product of all those primes? then I don't see how there can be 2 divisors that divide eachother...

excuse me I forgot to say that: the one who can't make a move will lose.
and that N is the product of those primes.

an explanation on the rule : if number p_1p_2p_3p_4 is written then no one can write p_1p_2 or p_1p_2p_3p_4p_5. because if so we may have :
p_1p_2 |  p_1p_2p_3p_4  or p_1p_2p_3p_4 |  p_1p_2p_3p_4p_5.

So, A writes N and win the game?

Pierre.

No......     don't put N.other wise it's not a problem.(surely not an olympiad one) by the way I found a solution:

A will put p_1p_2...p_(n-1) , for the next times till the end one must use p_n and a nonempty subset of
A={p_1 , ... p_(n-1)}which is not A. for the first move of B , he/ she has 2^(n-1)-2 chances. Now we calculate the number of possible positions that become impossible with each move.

if one writes p_1p_2...p_kp_n we name p_1p_2...p_k the A-part of his number.in this case all the 2^k -1 nonempty subsets of the A-part of his number can not be A-part for the next numbers. and also all the
2^(n-k-1)-1 nonempty subsets of A - {p_1 , p_2 ... p_n} plus his own choice that means he lessens an odd number of choices from the total of 2^(n-1)-2 possibilities.so he leaves an odd number of choices for A.and A lives an even number of choices for B and...

So ,that will be A that leaves 0 choices for B and A wins.

Is there any problem? :)

Maybe I see more simple way...
BTW, hossein11652 forgot to say, that players can't write 1.

I think, $B$ wins.
Strategy for $B$ is following: if $A$ writes $m$ then $B$ writes $N/m$.
It is clear if $A$ can make his turn, then $B$ can do it too.

actually putting N/m for B is not always a viable option-the numbers put down have to be composite,so if A chooses the product of n-1 primes,then the above move for B is violated.
in fact it looks like a player A game.in terms of sets ,one has the lattice if subsets of {1,2...n} except for the empty,the singletons and the complete set.and this game is equivalent to extending an antichain in this lattice.
A can start with the set {1,2...n-1},so any further choice has to be a set containing n,so A becomes player 2 in the variant of this same game on the set{1,2...n-1}(since n has to be part of every subsequent choice we can simply drop it) but with the singletons included in it.now in this poset,every element has its 'complement',so for any choice that B makes,A can choose the 'complement' choice,thus A always has a move.

thus this is just a rehash of the earlier solutions!

[quote="seshadri"]actually putting N/m for B is not always a viable option-the numbers put down have to be composite,so if A chooses the product of n-1 primes,then the above move for B is violated.[/quote]
 :blush: I didn't see that players must write composite divisors.


function
induction
number theory unsolved
number theory

[quote="creatorvn"]Find the largest positive integer $ x$ possible which is divisible by all the positive integers less than or equal to $ \sqrt[3]{x}$[/quote]

We are looking for pair $ x, y$ such that $ x$ is divisible by all natural numbers $ n$ with $ 1\le n \le y$, and such that $ (y+1)^{3}> x\ge y^{3}$.
Now look at the two functions
$ y^{3}$,
$ f(y) =$ the smallest integer which is divisible by all natural numbers $ n$ with $ 1\le n \le y$.
If we find greatest $ y$ such that $ (y+1)^{3}> f(y) \ge y^{3}$, the solution will be $ x=f(y)$. So let's find such $ y$.
We get that $ y=7$ works, so $ x=420$ works. There are no greater solutions because for $ y\ge 8$, $ f(y) > (y+1)^{3}$.

[quote="djuro"][quote="pco"][quote="djuro"]...There are no greater solutions because for $ y\ge 8$, $ f(y) > (y+1)^{3}$.[/quote]

That's true (and result 420 too) but it remains to give the proof.[/quote]

You're right. Well, it's not hard to see that $ f(y)$ grows much faster than $ (y+1)^{3}$, but I don't have a strict proof. Do you?[/quote]

I have.
Let $ p_{n}$ be the $ n^{th}$ prime number ($ p_{1}=2, p_{2}=3, p_{3}=5, p_{4}=7, p_{5}=11, \cdots$).

First, we can show with induction that $ \prod_{i=1}^{n}p_{i}>p_{n+1}^{3}$ $ \forall n\geq 5$ :
It's true for $ n=5$ : $ \prod_{i=1}^{5}p_{i}=2*3*5*7*11=2310>p_{6}^{3}=13^{3}=2197$
Then, if we have $ \prod_{i=1}^{n}p_{i}>p_{n+1}^{3}$, then $ \prod_{i=1}^{n+1}p_{i}>p_{n+1}^{4}$. But we have, with $ n\geq 5$, $ p_{n+1}>8$ and then $ p_{n+1}^{4}>(2p_{n+1})^{3}$. And we know that $ 2p_{n+1}>p_{n+2}$ (Bertrand - Chebyschev) and so $ \prod_{i=1}^{n+1}p_{i}>p_{n+2}^{3}$. Q.E.D.

Now, if $ x\geq 8^{3}$, $ x$ must be divisble by $ 2,3,4,5,6,7,8$ and so by $ 840>9^{3}$. So $ x$  must be divisble by $ 2,3,4,5,6,7,8,9$ and so by $ 2520>11^{3}=p_{5}^{3}$. But then, let $ k(x)\geq 5$ be the natural number such that $ p_{k(x)}^{3}\leq x<p_{k(x)+1}^{3}$.  $ x$must be divisible at least by $ \prod_{i=1}^{k(x)}p_{i}$ and so $ x>p_{k(x)+1}^{3}$ which is impossible. And so $ x$ can't be $ \geq 8^{3}$

Then, if $ 7^{3}\leq x<8^{3}$, $ x$ must be divisible by $ 1,2,3,4,5,6,7$ and so by $ 420$. So the greatest value is $ 420$


$ a$,$ b$,$ c$,$ A$,$ B$,$ C$ are positive numbers such that:
$ a\plus{}A\equal{}b\plus{}B\equal{}c\plus{}C\equal{}k$
prove that :
$ aB\plus{}bC\plus{}cA<k^2$

[quote="ghostchungho"]$ a$,$ b$,$ c$,$ A$,$ B$,$ C$ are positive numbers such that:
$ a \plus{} A \equal{} b \plus{} B \equal{} c \plus{} C \equal{} k$
prove that :
$ aB \plus{} bC \plus{} cA < k^2$[/quote]
I think, $ aB \plus{} bC \plus{} cA \leq\frac{3}{4}k^2$ is true too. :wink:

[quote="arqady"][quote="ghostchungho"]$ a$,$ b$,$ c$,$ A$,$ B$,$ C$ are positive numbers such that:
$ a \plus{} A \equal{} b \plus{} B \equal{} c \plus{} C \equal{} k$
prove that :
$ aB \plus{} bC \plus{} cA < k^2$[/quote]
I think, $ aB \plus{} bC \plus{} cA \leq\frac {3}{4}k^2$ is true too. :wink:[/quote]

 Try $ a\rightarrow k , b\rightarrow \frac{k}{2},   c\rightarrow 0$

a standard convexity (though here its more like linearity...) works :wink:

[hide="Solution(hope a nice one)"]We know that:
$ Abc\plus{}Bca\plus{}Cab\plus{}ABc\plus{}BCa\plus{}CAb<Abc\plus{}Bca\plus{}Cab\plus{}ABc\plus{}BCa\plus{}CAb\plus{}abc\plus{}ABC\equal{}(a\plus{}A)(b\plus{}B)(c\plus{}C)\equal{}k^3$, so
$ Abc\plus{}Bca\plus{}Cab\plus{}ABc\plus{}BCa\plus{}CAb<k^3$, so
$ aB(c\plus{}C)\plus{}bC(a\plus{}A)\plus{}cA(b\plus{}B)<k^3$, so
$ k(aB\plus{}bC\plus{}cA)<k^3$ hence
$ aB\plus{}bC\plus{}cA<k^2$.[/hide]

[quote="Marius Mainea"][quote="arqady"][quote="ghostchungho"]$ a$,$ b$,$ c$,$ A$,$ B$,$ C$ are positive numbers such that:
$ a \plus{} A \equal{} b \plus{} B \equal{} c \plus{} C \equal{} k$
prove that :
$ aB \plus{} bC \plus{} cA < k^2$[/quote]
I think, $ aB \plus{} bC \plus{} cA \leq\frac {3}{4}k^2$ is true too. :wink:[/quote]

 Try $ a\rightarrow k , b\rightarrow \frac {k}{2}, c\rightarrow 0$[/quote]
I have found my mistake.  :( Thank you!
It's also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106695

[quote="ghostchungho"]$ a$,$ b$,$ c$,$ A$,$ B$,$ C$ are positive numbers such that:
$ a \plus{} A \equal{} b \plus{} B \equal{} c \plus{} C \equal{} k$
prove that :
$ aB \plus{} bC \plus{} cA < k^2$[/quote]

  We have $ \sum {a(k\minus{}a)}\leq 0$  or  $ k^2\minus{}(a\plus{}b\plus{}c)k\plus{}ab\plus{}bc\plus{}ca\geq 0$   (*)

 Let $ a\leq b\leq c$

 Case 1)  $ c<a\plus{}b$  then $ \Delta\equal{}a^2\plus{}b^2\plus{}c^2\minus{}2(ab\plus{}bc\plus{}ca)<0$ and (*) is true.

 Case 2)  $ c\geq a\plus{}b$  ,then  if $ \Delta< 0$  is obvious.

 If $ \Delta\geq 0$ then $ k>c \geq\frac{a\plus{}b\plus{}c}{2}$  and $ f(c)\equal{}ab>0$ where $ f(x)\equal{}x^2\minus{}(a\plus{}b\plus{}c)x\plus{}ab\plus{}bc\plus{}ca$
so $ f(k)>0$

It is from the USSR national olympiad, year don't remember. 8 grade, if i'm not mistaken.

We want to show $ k^2\minus{}k(a\plus{}b\plus{}c)\plus{}ab\plus{}bc\plus{}ca>0$.
Sol1.
$ (k\minus{}a)(k\minus{}b)(k\minus{}c)\geq0$because$ 0<a,b,c<k$.
<=>$ k^3\minus{}k^2(a\plus{}b\plus{}c)\plus{}k(ab\plus{}bc\plus{}ca)\minus{}abc>0$=>$ k^2\minus{}k(a\plus{}b\plus{}c)\plus{}ab\plus{}bc\plus{}ca\geq0$ because $ abc>0$.
Sol2.
Let $ f(a,b,c)\equal{}k^2\minus{}k(a\plus{}b\plus{}c)\plus{}ab\plus{}bc\plus{}ca$.
let b,c is constant. so $ f(a,b,c)$ is a linear funtion. So we enough to think $ a\equal{}0,k$. similarly we enough to think $ a,b,c\equal{}0,k$..the next is easy.

We have
$ a(k \minus{} b) \plus{} b(k \minus{} c) \plus{} c(k \minus{} a) \le k^2 \implies k(a \plus{} b \plus{} c) \minus{} ab \minus{} bc \minus{} ac \le k^2$
But for a fixed a+b+c, ab+bc+ac is minimized when a=b=0, c equals the rest
But c must be less than k.


I was wondering, if I took the AMC 12 A, could I still take the 12 B?  If so how would I go about doing that because I don't think my school ordered it?  Thanks.

you might wanna look here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73356
it was posted just 2 hrs ago!


function
number theory unsolved
number theory

$ m,n\in Z \plus{} , m\neq n$
prove that $ \exists x \in \Re$
s.t. $ \frac {1}{3} \leq${$ {x(n)}$}$ \leq \frac {2}{3}$  and
$ \frac {1}{3} \leq${$ x(m)$}$ \leq \frac {2}{3}$
Here  {x} $ \equal{} x \minus{} [x]$

Any thoughts?
Please help me if you can solve


Prove that if a relation with more than one element is transitive, it cannot be a function. (or prove it wrong)

I'm not sure if this belongs here because they don't really teach this in middle school. Please PM me on where this should belong if it's in the wrong spot.

We didn't even know what elements [i]were[/i] in middle school. This definately belongs in HSB.

[quote="ashpotter"]Prove that if a relation with more than one element is transitive, it cannot be a function. (or prove it wrong)

[/quote]
I'm not sure...
[hide]If $ a$ gives $ b$ and $ b$ gives $ c$ then if it's transitive, $ a$ gives $ c$.  But then $ a$ gives $ b$ and $ a$ gives $ c$. Hence it cannot be a function.[/hide]

If the relation $ R$ is defined on $ S\times S$, and $ s$ is some fixed element of $ S$ then let $ R$ be such that $ aRb\Leftrightarrow b \equal{} s$. This seems to be a function  to me, and the relation is also transitive. I think this is a counterexample.

In the above argument, you didn't take into account the case $ b \equal{} c$.


[quote="emannes"]Put the digits 1 to 9 inclusive into a grid of nine squares (3 squares to a row, 3 rows) such that the sum of the digits in every row, column, and 3-square diagonal is 15.[/quote]
[hide]Sorry for the no picture.
Top left-4
Top Middle-3
Top Right-8
Middle Left-9
Middle Middle-5
Middle Left-1
Bottom left-2
Bottom Middle-7
Bottom Left-6[/hide]


function
symmetry
integration
calculus
real analysis
calculus computations

Show that if $ g$ is not integrable on $ [0,1]$, then the sequence of averages

$ \displaystyle\frac{1}{n}\displaystyle\sum_{k\equal{}1}^{n} g (\frac{k}{n})$

may or may not be convergent.

My question is why are we not asked to prove one way or the other.

Because you can't prove something that is false. (Well, I know some people who can, but let's not go into that.)

You need two functions $ g$: one such that the sequence of averages is convergent, the other such that the sequence is divergent. The second is quite easy: $ g(x) \equal{} 1/x$ will work. The first one is tricky, unless one takes an easy way out and redefines $ g(x) \equal{} 1/x$ at rational points. Let's find continuous (on $ (0,1)$) non-integrable $ g$ with this property...

if you show $ g(x)$ is not integrable with the Riemann - Stieltjes

is it enough to say that for every $ \epsilon$ and $ \xi \in [x_{k\minus{}1}, x_{k}]$ that

$ \sum_{k\equal{}1}^{n} g(\xi)[x_{k}\minus{}x_{k\minus{}1}]$ is not defined particularly when $ \xi \equal{} 0$, thereby $ g(x)$ not integrable?

Thanks for the help.

Can I use $ g(x) \equal{} \frac{1}{x^{2}}$ to make the sequence of averages convergent(by p-series test)

also can someone answer my question from the last post(using R-S to show NOT integrable)?

Um, no. If $ g(x) \equal{} \frac1{x^2},$ then

$ \frac1n\sum_{k \equal{} 1}^ng\left(\frac {k}{n}\right) \equal{} \frac1n\sum_{k \equal{} 1}^n\frac {n^2}{k^2} \equal{} n\sum_{k \equal{} 1}^n\frac1{k^2}.$

Yes, $ \sum_{k \equal{} 1}^{\infty}\frac1{k^2}$ converges. All that means is that as $ n\to\infty,$ $ \sum_{k \equal{} 1}^n\frac1{k^2} \equal{} L \plus{} o(1)$ for some specific positive $ L.$ (Yes, we know an exact value for $ L,$ but that's not needed for this argument.)

Then $ n\sum_{k \equal{} 1}^n\frac1{k^2} \equal{} Ln \plus{} o(n)$ and this tends to infinity as $ n\to\infty.$

All this gives us is a slightly more emphatic version of mlok's suggestion of $ g(x) \equal{} \frac1x,$ which would have given us $ \sum_{k \equal{} 1}^n\frac1k,$ which is $ \ln(n) \plus{} O(1)$ for large $ n.$ Both examples make the same point, and neither answers the question which was the last sentence in mlok's post.

Finding the answer to that last question [i]is[/i] tricky. I don't have it yet.

thanks for the explanation Kent.

how would i show that $ g(x) \equal{}\frac{1}{x}$ is NOT differentiable over $ [0,1]$(using Riemann -Stieltjes)?

thanks

[Responding to mlok's and Kent's question]

One way out is to use symmetry, as in a principal value; try $ f(x)\equal{}\frac1x\minus{}\frac1{1\minus{}x}\equal{}\frac{1\minus{}2x}{x(1\minus{}x)}$ on $ (0,1)$, and zero the values at the endpoints. The sum is always zero.

We could also try putting the singularity in the middle; $ f(x)\equal{}\begin{cases}\frac1{2x\minus{}1}&x\neq\frac12\\0&x\equal{}\frac12\end{cases}$ works.

This is risky, of course- it doesn't work for any other point $ c$ in place of $ \frac12$.

a big thank you jmerry, one last question:

does anyone have an idea how to show $ g(x) \equal{} \frac{1}{x}$ is not integrable using Riemann-Stieltjes?

thanks

Riemann-Stieljes can't be the right question without more information. To interpret $ \int_a^b f\,dg$ we need to know what $ g$ is.

As for the Riemann integral: any unbounded function must always fail to be Riemann integrable. Let's concentrate on $ \frac1x$ (defined to be zero at zero) on $ [0,1].$

Take any partition $ P.$ Take any $ M>0.$ The first interval of this partition has the form $ [0,a_1].$ Chose $ \xi_1$ so that $ a_1\cdot\frac1{\xi_1}>M.$ Then (since the rest of the terms are nonnegative), the Riemann sum on this partition is greater than $ M.$ Hence the Riemann sums cannot converge and the function is not integrable.

Could we have used $ g(x) \equal{} \begin{cases}1&x\in\mathbb{Q} \\0&x\in\mathbb{R\minus{}Q}\end{cases}$ for the case where $ g(x)$ is non integrable and the sequence of averages converges?

Yes, you could have, and mlok hinted at that. It's just that mlok decided to issue an extra challenge to get someone like jmerry to find the way.


Putnam
AMC
AIME
combinatorics proposed
combinatorics

[quote="Christine"]Find the largest two-digit prime that divides 200C100 (nCk denotes n chooses k)[/quote]

Let $ A \equal{} \binom{200}{100} \equal{} \frac {200!}{100!100!}$

Assume $ 100 > p\geq50$. Then $ p$ divides exactly twice the lower part of $ A$. If we want that $ p$ divides at least 3 times the upper part ($ 200!$), we need to have $ 3p\leq 200$ and so $ p\leq 66$.

And hence the result $ p \equal{} 61$

I've done this question before (and it's appeared on the forum before) -- what is it, an old Putnam or AIME or something?


nice ineq

multiplying the both sides by $ \prod{(2\plus{}a)}$ the ineq is equivalent to :
$ 8\prod{(4\minus{}a^2)} \geq \prod{(2\plus{}a)} \prod{(a\plus{}bc)}$

obviously $ \prod{(2\plus{}a)}  \leq  27$:
it sufficies to show that :

$ 8\prod{(4\minus{}a^2)} \geq 27\prod{(2\plus{}a)}$

but :\[ (\prod(4\minus{}a^2))^2\equal{}  (\prod(b^2\plus{}c^2\plus{}1))^2 \equal{} \prod(a^2\plus{}1\plus{}b^2)(a^2\plus{}1\plus{}c^2) \geq \prod(a^2\plus{}1\plus{}bc)^2 \geq \prod(2a\plus{}bc)^2\]  (Cauchy and am-gm)
hence it's enought to prove that: $ 8\prod(2a\plus{}bc) \geq 27\prod(a\plus{}bc)$ expanding and using $ \sum{x^2}\equal{}3$ it's equivalent to:
$ 4abc \plus{}5\sum{(ab)^2} \minus{}19(abc)^2 \geq 0$ but it's obvious since : $ abc\leq 1$ and $ \sum{(ab)^2} \geq 3(abc)^{\frac{4}{3}}$ :wink:

[quote="nguyenvanhuyen"]Given $ a,b,c > 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that:
$ 8(2 \minus{} a)(2 \minus{} b)(2 \minus{} c) \ge (a \plus{} bc)(b \plus{} ca)(c \plus{} ab)$
(Tran Quoc Luat)[/quote]

We have : $ 8(2 \minus{} a)(2 \minus{} b)(2 \minus{} c) \ge (a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2)$

We have to prove that : $ (a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2) \geq  (a \plus{} bc)(b \plus{} ca)(c \plus{} ab)$

It is easily by AM-GM ;)

[quote="anas"][quote="Materazzi"][quote="nguyenvanhuyen"]Given $ a,b,c > 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that:
$ 8(2 \minus{} a)(2 \minus{} b)(2 \minus{} c) \ge (a \plus{} bc)(b \plus{} ca)(c \plus{} ab)$
(Tran Quoc Luat)[/quote]

We have : $ 8(2 \minus{} a)(2 \minus{} b)(2 \minus{} c) \ge (a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)$

[/quote]

why? :wink:[/quote]
Because $ 4 \minus{} 2a \equal{} b^2 \plus{} c^2 \plus{} (a \minus{} 1)^2 \geq b^2 \plus{} c^2$.


rotation
conics
ellipse
linear algebra
matrix
quadratics
linear algebra unsolved

I know that the following equation represent an ellipse:
\[ \left( \frac{\minus{}2a\plus{}2b}{3} \right)^2 \plus{}
   \left(  \frac{a\minus{}4b}{6} \right)^2 \plus{}
 \left(  \frac{a\plus{}2b}{3}  \right)^2 \equal{} 1
\]

However, the major axis of this ellipse is not on the $ x$-axis. So I want to use some transformation to rotate it so that the major axis coincide with the $ x$-axis. How does one proceed for problems like this? Is there a way to use linear algebra to do this?

Eventually, I want it look like the standard equation for an ellipse:
\[ \frac{a^2}{m^2} \plus{} \frac{b^2}{n^2} \equal{} 1.
\]

Well, you've got three terms and that's bad. Expand everything out into a form $ pa^2 \plus{} 2qab \plus{} rb^2 \equal{} 1$.
Now, look at the matrix $ \begin{bmatrix}p & q \\
q & r\end{bmatrix}$ of this quadratic form. You want an orthogonal change of basis to diagonalize it.


function
integration
limit
Functional Analysis
advanced fields
advanced fields unsolved

$f: \mathbb{R}\to \mathbb{R}$ a continuous and bounded function, $\Omega$ a bounded open subset of $\mathbb{R}^{n}$. Show that there exists at least one function $u \in H^{1}_{0}(\Omega)$ such that
$\Delta u = f(u)$
Show that if $f$ is decreasing, this solution is unique.

We know that $\Delta: H_{0}^{1}(\Omega)\to H^{-1}(\Omega)$ is a bounded isomorphism with bounded inverse.  We construct a sequence of functions by setting $u_{i+1}=\Delta^{-1}f(u_{i})$.  The $f(u_{i})$'s are bounded in $L^{2}$ and consequently they are relatively compact in $H^{-1}$.  So the $u_{i}$'s are relatively compact in $H^{1}_{0}$.  Now take a convergenct subsequence to obtain the desired result.  (Or, you can use that other result you asked about regarding the fixed points of compact maps.)

The answer to the second question is totally independent.  Given two solutions $u_{1}$ and $u_{2}$, just apply the maximum principle to $u_{1}-u_{2}$ in the region where $u_{1}\geq u_{2}$.  Actually, the proof is more technical than this, since we only have $H^{1}_{0}$ rather than $C^{2}_{0}$.

Nice :)  I prefer your solution than mine (as you have guessed with fiexed point) because it is less involved, and shauder fixed point are not that easy to demonstrate  :lol:  Thank you

Here's another approach. Denote $U(x): =-\int_{0}^{x}f(t)dx$ then $u$ is a solution to our problem iff it is a critical point of the functional
\[\Phi(u)=\frac12\int_\Omega|\nabla u|^{2}dx+\int_\Omega U(u)dx. \]
Since $f$ is bounded $\Phi$ is weakly lower semi-continuous coercive (i.e. $\lim_{\|u\|\to\infty}\Phi(u)=+\infty$) functional on the reflexive Banach space $X: =H_{0}^{1}(\Omega)$. It remains to make use of the following statement:

[b]Proposition.[/b] Let $X$ be a reflexive Banach space, $\Phi\colon X\to\mathbf R$ a coercive weakly lower semi-continuous functional. Then $\Phi$ attains its minimum in $X$.


if m,n are natural numbers prove that
 $\frac{1}{m+n+1}-\frac{1}{(m+1)(n+1)}=< \frac{4}{45}$

Eh, there's a problem: $\frac{45}{4}\neq 10+\frac{1}{4}$, it's a 11!!
Now, I don't know how to solve "beautifully" the problem, but I hope that your proof still holds, even if I don't really understood why you were trying just a few values of m and n. Ok, you might say that after that the sum grows too much, but it would be better to demonstrate it, don't you think?  :wink:

[quote="nutz_for2.718281828"]I think you mean

$\frac{1}{m+n+1}-\frac{1}{mn+m+n+1}\leq\frac{4}{45}$

[hide="contradiction:"]

We can give the fractions common denominators:

$\frac{mn+m+n+1}{m^{2}n+mn^{2}+m^{2}+n^{2}+3mn+2m+2n+1}-\frac{m+n+1}{m^{2}n+mn^{2}+m^{2}+n^{2}+3mn+2m+2n+1}>\frac{4}{45}$

$\frac{mn}{m^{2}n+mn^{2}+m^{2}+n^{2}+3mn+2m+2n+1}>\frac{4}{45}$

$\frac{45}{4}>\frac{m^{2}n+mn^{2}+m^{2}+n^{2}+3mn+2m+2n+1}{mn}$

$\frac{45}{4}>m+n+\frac{m}{n}+\frac{n}{m}+3+\frac{2}{n}+\frac{2}{m}+\frac{1}{mn}=m+n+3+\frac{m+2}{n}+\frac{n+2}{m}+\frac{1}{nm}$

45/4 is 10+1/4.

$7+\frac{1}{4}>m+n+\frac{m+2}{n}+\frac{n+2}{m}+\frac{1}{nm}$

We now need to prove that, for all natural numbers m and n, that 7 and 1/4 is greater than all of that. If both m and n equal 1, then all of that equals 1+1+3+3+1, which is greater than 7.25. If m=1 and n=2: 1+2+3/2+4+1/2, which is again greater than 7.25. It doesn't matter which order we plug in values for m and n, so we don't need to find it if m=2 and n=1.

If m=3 and n=1: 3+1+5+blahblahblah, which is again greater than 7.25

It's the same thing with n=1, m=(4,5,6). We don't need to check any other pairs that have 1, because the first two terms added together will result in 8 or greater.

So let's try 2 and 2:

2+2+2+2+blahblahblah, which is greater than 7.

2 and 3:

2+3+4/3+5/2+blahblah, which is greater than 7.25

2 and 4:

2+4+1+blahblah, which is greater than 7.25

With 3, it's always greater than 7.25 Therefore, all of the above statements are false. That makes

$\frac{1}{m+n+1}-\frac{1}{mn+m+n+1}\leq\frac{4}{45}$ true.[/hide]

I WUV algebra!!! :rotfl:[/quote]

Ooh... pretty... </sarcasm>

Is that the only way to do it, or is there something... not as ugly?


$ \overline{AB}$ is the hypotenuse of a right triangle $ ABC$. Median $ \overline{AD}$ has length $ 7$ and median $ \overline{BE}$ has length $ 4$. The length of $ \overline{AB}$ is:

$ \textbf{(A)}\ 10\qquad
\textbf{(B)}\ 5\sqrt{3}\qquad
\textbf{(C)}\ 5\sqrt{2}\qquad
\textbf{(D)}\ 2\sqrt{13}\qquad
\textbf{(E)}\ 2\sqrt{15}$

[hide]

Let $ CE\equal{}x$ and $ CD\equal{}y$. By Pythagorean Theorem, $ x^2\plus{}4y^2\equal{}16$ and $ 4x^2\plus{}y^2\equal{}49$. Adding the two equations, $ 5x^2\plus{}5y^2\equal{}65$. Multiplying by $ \frac{4}{5}$ gets $ 4x^2\plus{}4y^2\equal{}52$, so $ AB\equal{}\sqrt{52}\equal{}2\sqrt{13}$. $ \boxed{\textbf{D}}$

[/hide]

[hide="Similar solution"]Join D to E

Let $ BD \equal{} CD \equal{} x$ and $ AE \equal{} CE \equal{} y$

As above we get two equations

$ 4x^2 \plus{} y^2 \equal{} 16\,\, .\, .\, .\, \left(1\right)$

$ 4y^2 \plus{} x^2 \equal{} 49\,\, .\, .\, .\, \left(2\right)$

Adding these two we get

$ 5\left(x^2 \plus{} y^2\right) \equal{} 65$

$ \implies x^2 \plus{} y^2 \equal{} 13$

In triangle $ CDE$,

$ x^2 \plus{} y^2 \equal{} DE^2$

$ \therefore DE^2 \equal{} 13$

$ \implies DE \equal{} \sqrt {13}$

It is known that the line joining two medians is both parallel to the third side and half its length.

$ \therefore AB \equal{} 2DE \equal{} 2\sqrt {13} \implies \boxed{D}$[/hide]


xyz=1 z,y,x>0
find the minimum value that can accept x^2+4*x*y+8*y^2+16*Z^2

Hmm...I move it into the inequalities forum...

Pierre.

I post my solution to it in pre-olympiad: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23742]www.mathlinks.ro/Forum/viewtopic.php?t=23742[/url]


Quadrilateral $ ABCD$ is circumscribed, rays $ BA$ and $ CD$ intersect in point $ E$, rays $ BC$ and $ AD$ intersect in point $ F$. The incircle of the triangle formed by lines $ AB$, $ CD$ and the bisector of angle $ B$, touches $ AB$ in point $ K$, and the incircle of the triangle formed by lines $ AD$, $ BC$ and the bisector of angle $ B$, touches $ BC$ in point $ L$. Prove that lines $ KL$, $ AC$ and $ EF$ concur.

[i]Author: I.Bogdanov[/i]

Let $ G \equiv AC \cap BD.$ Let $ (I)$ be incircle of quadrilateral $ ABCD,$ $ (X)$ incircle of triangle $ \triangle (AB, CD, BI)$ and $ (Y)$ incircle of triangle $ \triangle (BC, DA, BI).$ Let $ r, x, y$ be radii of $ (I), (X), (Y).$ Let $ (I)$ touch $ AB, BC, CD, DA$ at $ S, T, U, V.$ $ ST, UV, EF$ are polars WRT $ (I)$ of collinear points $ B, D, G,$ they are concurrent at the pole $ P$ of $ BD.$ By Pascal theorem for $ SSTUVV,$ $ A, G, P$ are collinear and by Pascal theorem for $ STTUUV,$ $ C, G, P$ are collinear $ \Longrightarrow$ $ P \in AC.$

$ \frac {\overline{KE}}{\overline{KB}} \cdot \frac {\overline{LB}}{\overline{LE}} \equal{} \frac {\overline{SE}}{\overline{SB}} \cdot \frac {\overline{TB}}{\overline{TE}}$ $ \Longleftrightarrow$ $ \frac {\overline{KE}}{\overline{SE}} \cdot \frac {\overline{TE}}{\overline{LE}} \equal{} \frac {\overline{KB}}{\overline{LB}} \cdot \frac {\overline{TB}}{\overline{SB}}$ $ \Longleftrightarrow$ $ \frac {x}{r} \cdot \frac {r}{y} \equal{} \frac {x}{y}.$

By Menelaus theorem for $ \triangle BEF$ cut by $ KL, ST,$ these 2 lines intersect at $ P \equiv EF \cap ST \cap AC.$

Let $I$ be the incenter of quadrilateral $ABCD$, let $P,Q,R,S$ be feet from $I$ to $\overline{AB},\overline{BC},\overline{CD},\overline{DA}$, respectively. Let $I_1,I_2$ be the incenters of triangles formed by lines $AB$, $CD$ and the bisector of angle $B$ and lines $AD$, $BC$ and the bisector of angle $B$, respectively.
Observe that by Brianchon's theorem, $\overline{AC},\overline{BD},\overline{PR},\overline{QS}$ are concurrent. Hence, by Pascal's theorem, $\overline{EF},\overline{AC},\overline{PQ},\overline{SR}$ are concurrent. Finally, it is sufficient to show $\frac{KE}{EB}\cdot \frac{BF}{FL}=\frac{KP}{PB}\cdot \frac{BQ}{QL}$ using Ceva/Menelaus, which is really equivalent to $\frac{KE}{KP}\cdot \frac{QL}{FL}=\frac{BE}{BF}$. Indeed, by angle bisector theorem, \begin{align*}
\frac{KE}{KP}\cdot \frac{QL}{FL}=\frac{I_1E}{I_1I}\cdot \frac{I_2I}{I_2F}=\frac{BE}{BI}\cdot \frac{BI}{BF}=\frac{BE}{BF},
\end{align*}as desired. $\blacksquare$

Apologize for reviving this post, but I can't construct the diagram. Can somebody show the diagram for me?  :(


The result of the Indian National Astronomy Olympiad is available at [url=http://www.hbcse.tifr.res.in/olympiads/inbo06_results/inao2006_result]INAO Result 2006[/url]

The site shows that the Camp will be from [color=red]May 1 - May 20, 2006[/color]...

Congrats and Best of luck!

Kindly continue all further discussion here! Thanks!

List of students selected on merit basis from the [color=darkblue][size=150]INAO examination[/size][/color]
for the International Astronomy Olympiad Training Cum Selection Camp 2006
(Note: The lists are in alphabatical order)

[size=150]Junior Group[/size]

Region       Name Of The Student            City
[color=red]Bhubneswar[/color] Soumyashant Nayak [color=red]Chandrasekharpur[/color]
[color=orange]Gulbarga[/color] Sharat S Ibrahimpur [color=orange]Gulbarga[/color]
[color=brown]Hyderabad A Krishna Sumanth Bagh Amberper
Hyderabad A. Mahathey Pavan Hyderabad
Hyderabad Adhokshaj Katarni Hyderabad
Hyderabad Arinjay Singh Parihar Hyderabad
Hyderabad C. V. K. Koundinya Hyderabad
Hyderabad E. Santhosh Kumar Hyderabad
Hyderabad S Mano Teja Reddy Hyderabad
Hyderabad Shruti Ranjan Satpathy Hyderabad
Hyderabad Srujan M.  Hyderabad
Hyderabad T. Chenna Reddy Hyderabad
Hyderabad T. Gowtham Kumar Rajamundry [/color]
[color=green]Jaipur Priyank Gupta  Jaipur[/color]
[color=olive]Mumbai  Abhishek Padmanabhan Thane
Mumbai  Adwait Vedant Mathkar  Mumbai
Mumbai  Kedar Shriram Tatwawadi  Nashik
Mumbai  Rahul Manoj Makhijani Mumbai
Mumbai  Rupan Ranjan Saharoy  Alibag, Raigad [/color]
[color=indigo]New Delhi Adarsh Prasad New Delhi
New Delhi Nikhil Kumar New Delhi [/color]
[color=violet]Pune Aniruddha Anand Bapat Pune
Pune Rohan Milind Khante Pune [/color]

[size=150]Senior Group [/size]

Region Name Of The Student City
[color=darkred]Bharuch Mohammad Fahim Bharuch [/color]
[color=red]Bhubaneswar Debajyoti Manadhata Bhubaneswar
Bhubaneswar Saswat Rath Bhubaneswar [/color]
[color=orange]Calicut Raziman T V Calicut[/color]
[color=brown]Hyderabad Jayakiran Akurati Hyderabad [/color]
[color=olive]Jaipur Gaurav Lahoti Udaipur
Jaipur Mehul Jain Jaipur
Jaipur Pranjal Bordia Udaipur
Jaipur Raman Sharma Jaipur [/color]
[color=green]Lucknow Manish Singh Rathaur Lucknow
Lucknow Udbhav K. Singh Allahabad [/color]
[color=olive]Mumbai Abhinav Sinha Mumbai
Mumbai Adtya Bhuvaneshwaran Mumbai
Mumbai Chiraag S Juvekar Mumbai
Mumbai Niranjan Aniruddha Viladkar Nashik
Mumbai Sivaramakrishnan Mumbai
Mumbai Tilak Sharma  Mumbai [/color]
[color=violet]Patna Abhay Kumar Patna
Pune Gund Ved Vishwas  Pune
Pune Prashant Sachdeva Pune
Pune Shreyas Ganesh Patankar Pune
Tirupati Prathyush Manchala Secunderabad [/color]

Anyone who qualified here ?? I didn't make it but then next year's always there...

[quote]Anyone who qualified here ?? [/quote]

i qualified  :lol:

udbhav

Hey udbhav,

Congratulations !!!!   :thumbup:

And just by the way,  do you have a copy of the senior question paper ?? Next year I guess I need to take up the senior astro-olympiad, so I'm pretty interested in getting this year's paper....

[quote="karthik_k"][quote="mysmartmouth"]Yeah, same here.  I think I was close to qualifying...lol.[/quote]

Hmm.......I thought you said on some other post that you've never been to India though you're Indian.....so how on earth did you write the INAO ?? Do they have something for non resident Indians as well ??  :?[/quote]

I guess I'm bad at sarcasm.  LOL.

Oh............sorry I didn't understand what you meant... :oops:

the question papers aren't circulated.i'll write down the question which i remember.

1) there is an aeroplane flying at the speed of 720km/h .it tries to complete a circular turn in the horizontal plane inclined at an angle of 15deg with the horizontal. find the radius of the circle in which it should turn.

2) four planets are placed on each of the corners of a square of side of lengh L.find the speed which each of the planets should have in order to be in cicular orbits in the same arrangement.

3) the distance of the earth from the sun is 1.5x10^11 m .the solar constant at earth is 1400 Wm^-2.s^-1.take radius of the earth as 600km.asuuming the sun to be a perfect  black body find the temperature of the surface of the sun.

4) by hubble's law it is known that v=hr where r is the distance of a galaxy or any such object and v is it's velocity and H is hubble's const.the wavelengh of a spectral line from a galaxy is found to be 4089 Angstrom(A) .it's laboratory wavelength is known to be 4081 A.find the distance of the galaxy from earth given that
1/H=15billion seconds.

Note- i have done my best to make sure that these questions resemble the original as much as possible.please point out if you find any mistakes with the questions.i can't remember the rest so i haven't written them down.

Udbhav

[quote]four planets are placed on each of the corners of a square of side of lengh L.find the speed which each of the planets should have in order to be in cicular orbits in the same arrangement.
[/quote]

Hey !!!!!!! We had that stuff on the junior paper too !!!!!!! Anyway, thanks a lot for those questions..... :lol:


OKAY GUYS
IGNORE THIS TOPIC PLZ
ALL INFO CONCERNING THIS MATCHUP CAN BE FOUND IN TEH OTHER TOPIC
thank you


Everyone will receive a unique piece. The game is divided up into two phases:

1) The Private, "taking" phase (by pm)
In this phase you download all files your target has. That is, they don't lose files (but don't gain either), and you get whatever they have (barring duplicates).

2) The Public, "giving" phase
You choose a person to "give" your files to (also a download). No person can be targeted more than once. For this phase you must post your target.

Then it repeats until the win condition is satisfied:

3) Get all the pieces to win.

Note that during Phase 1 I will tell you what files you already have. Also you will only know people's starting piece, not their entire collection.

Signups now  :)

Figurative sense.

So if player 1 has Pieces 1,3,8
and player 2 has Pieces 2,4,8

and player 1 downloads off player 2, player 1 now has Pieces 1,2,3,4,8.

as many as there are players, so in a game of 10 players there are 10 pieces.

I'll join too.

This reminds me of torrenting, for some reason.

1) algebra2000
2) AIME15
3) tennis123
4) yaofan
5) EggyLv.999
6) vahalla
7) budi713
8) Brut3Forc3.

This is enough, we will start now!
Everyone has the piece associated with their order of joining, so vahalla has piece 6.

PM your files to take, deadline on Wednesday.


I have started learning measure theory and i need some help (all i quite easy).

Let $X$ be an infinite set and let $\mathcal{A}$ be a set of collection of all subsets $A$ of $X$ such that either $A^{c}$ or $A$ is finite.$\mathcal{A}$ is not a $\sigma$-algebra.

Find the $\sigma$-algebra on $\mathbb{R}$ that is generated by the collection of all one point subsets of $\mathbb{R}$.

Find an example that the union of a collection o $\sigma$-algebras on a set $X$ fail to be a $\sigma$-algebra on X.

If its all very easy just give me a hint  :wink: .

1) Extract a sequence $(c_{n})$ out of $X$. All the singletons $\{c_{2n}\}$ are in script A, but the union is not (why not?).
2) At the very least, the generated set must contain all countable sets and the complements of these. This collection forms a $\sigma$-algebra, so you get the reverse inequality.
3) You want to take the union of some relatively "small" $\sigma$-algebras. This can be achieved by letting $B_{n}$ be the $\sigma$-algebra in the natural numbers generated by the first $n$ numbers.


geometry
trigonometry
rectangle
algebra proposed
algebra

Find a geometric solution for this: \[Arctan(\frac{1}{2})+Arctan(\frac{1}{3})=\frac{\pi}{4}\] (Iranian Olympiad)

If you want a geometrical solution, you should have posted it in the Geometry section.

Let $ACD$ be a triangle with $\angle{D}=90^{o}$ and $\tan{A}=\frac{1}{2}$, and $BC'D'$ be a triangle with $\angle{D'}=90^{o}$ and $\tan{B}=\frac{1}{3}$. If we let $CD=C'D'=1$, then we can use ACD and BC'D' to make a new triangle ACB such that $AC=\sqrt{5}$, $BC=\sqrt{10}$ and $AB=5$. By the law of cosine, we get $\cos{\angle{ACB}}=-\frac{\sqrt{2}}{2}$ which implies $\angle{ACB}=135^{o}$. Thus $\angle{A}+\angle{B}=45^{o}$.

Shobber,your solution is very nice. I give an other solution. I think this is simpler.

In the following rectangle, let $AB=BC=CD=FG=GH=HE=1$ and $AF=1.$
we have $\vartriangle BFD \sim \vartriangle BFC$ because $\frac{{FB}}{{BD}}= \frac{{BC}}{{FB}}.$Then \[\angle BCF = \angle BFD \Rightarrow \angle BDF+\angle BCF = \angle BDF+\angle BFD = \frac{\pi }{4}\]


!این جا برای مساله های جبر حالا دیگه لطف کنید سوالهاتون رو با همون سیستم بفرستید دیگه

mish lotfan yeki ye raah hale tamiz eraae bede :huh:
[url]http://www.mathlinks.ro/Forum/topic-59047.html[/url]

hatman mishe ye rahe halle tamiz erae dad . :D


If the perimeter of a square is $ 8 \sqrt{5}$ centimeters, what is the area of the square, in square centimeters?

\begin{align*}
4s=8\sqrt{5} &\Rightarrow s=2\sqrt{5} \\
&\Rightarrow s^2=\boxed{20}
\end{align*}


function
limit
induction
search
algebra proposed
algebra

This is a good one i believe...i find my solution approvably pretty.
This is mine of course but as you will see it's not really hard to come up with it...

Let f(x) be defined in R where x is not 0. Find all  functions f such that
f(x-c) - f(x) = f(x-c)*f(x/c)
for all real numbers x,c such that x-c,x,x/c not 0....

I have a solution if $f$ is continuous.

\[f(x-c)-f(x)=f(x-c)\cdot f\left(\frac x c\right)\quad(1)\]

Let $f(1)=a$. For $c=x-1$ one has
\[a-f(x)=a\cdot f\left(\frac x{x-1}\right)\quad(2)\]

Now, let $x\to\infty$. From $(2)$ we have
\[a-\lim_{t\to\infty}f(t)=a\cdot f(1)\]
\[\lim_{t\to\infty}f(t)=a-a^2\quad(3)\]

Hence the above limit exists. Knowing that, we could put $c$ fixed and $x\to\infty$ in $(1)$ and get the following:
\[\lim_{t\to\infty}f(t)-\lim_{t\to\infty}f(t)=(\lim_{t\to\infty}f(t))^2\]
hence
\[\lim_{t\to\infty}f(t)=0\]

Now, by $(3)$:
\[a-a^2=0\]

Now we have two cases:

1) $a=0$
It is obvious from $(2)$ that $f(x)=0$ for each $x$.

2) $a=1$.
We will prove by induction that $f(n)=\frac 1 n$ for each $n\in\mathbb{Z}$. Putting $c=1$ in $(1)$ follows:
\[f(x)=\frac{f(x-1)}{f(x-1)+1}\quad(4)\]
Hence, by induction hypothesis,
\[f(n)=\frac{\frac 1{n-1}}{\frac 1 {n-1}+1}=\frac 1 n\]
It is similar for negative numbers.

Let $x=p,c=q$. From $(1)$ follows:
\[f\left(\frac p q\right)=\frac{f(p-q)-f(p)}{f(p-q)}=\frac{\frac 1{p-q}-\frac 1 p}{\frac 1{p-q}}=\frac q p\]

Let $r\neq 0$ be real number, and let $(r)_i$ be a sequence of rationals such that $\lim_{n\to\infty}r_n=r$. Putting $x=r$ in $(4)$ one has:

\[\begin{array}{rl}
f(r)&=\displaystyle\frac{f(r-1)}{f(r-1)+1}=\frac{f(\lim_{n\to\infty}r_n-1)}{f(\lim_{n\to\infty}r_n-1)+1}=\lim_{n\to\infty}\frac{f(r_n-1)}{f(r_n-1)+1}=\\
&\displaystyle=\lim_{n\to\infty}\frac{\frac 1{r_n-1}}{\frac 1{r_n-1}+1}=\lim_{n\to\infty}\frac 1{r_n}=\frac 1 r\end{array}\]

Thus the problem is solved. There are two solutions: $f(x)\equiv 0$ and $f(x)=\frac 1 x$. It is easy to verify that these functions indeed satisfy the given equation.

HAHA, i just enter the forum to post this problem, (under the sourse "DOUVROPOULOS"...;-PPP) and i see that post.

Bojan Basic your solution seems ok, althgouht it can be improved.
The condition that f is continous is not needed to prove the result.
I have a solution, nice but highly non-trivial, in which i use only the conditions stated in top.
The guy , who post this problem, is Douvropoulos. (giati den ebales bre fisiologiko nickname ?), And his solution is very nice and cute. (exactly as his problem).

I really want to see more solutions, try it !

Well Bojab Basic your solution is good but actually not pretty because knowing a function is continuous gives you way to much info....
Try simple methods....I would like to see some more trying going on....
So I'll give a hint or so ...of my solution...:
Search for a substitution that will give us another relation combining which with the first we will be able to get rid of the term f(x-c) ...and have only x,c,  x/c , c/x in a new relation.....
That will be a good step i believe because doing so... having just x,c..... we can easily find symmetries......

To be even clearer:Instead of x,c using something else that will keep (x-c) as it is.Then combine the new relation with the first relation and find  f(x-c)....with respect to f(x),f(c),f(x/c),f(c/x).Then replace f(x-c) in the first relation..
Sorry if that sounds complicated

Give me what you got...The solver wins my gratitude..

....
Anto de nomizo na eipa tosa polla e???
Dikio exeis gia to nickname alla skeftika oti etsi opoios koitaei to nickname koitaei kai thn askisi....
ksereis an mporo na to allakso???
simeiosi:eides stergie...ego den evala simaia kai xazomares sto profile xaxaxaxaxaxaxa

pou na valo ekeinh thn askisi me thn anadromikh akolou8ia????

Quite nice. Consider the function $ g(x)=f(\frac{1}{x})$. The condition of the problem can also be written $ f(y)-f(x)=f(y)f(\frac{x}{x-y})$ for non-zero $x,y$ and different $x,y$. This becomes $ g(y)-g(x)=g(y)g(1-\frac{x}{y})$. Take $y=1$. Then for $x$ different from 0 and 1 we have $g(1)-g(x)=g(1)g(1-x)$. Thus if $ g(1)=0$ then we obtain only the zero function. Suppose not, then we have $ g(1-\frac{x}{y})=1-\frac{g(\frac{x}{y})}{ g(1)}$. Replace it in the last relation and you will find $ g(y)g(\frac{x}{y})=g(x)g(1)$ for all $x,y$ non-zero and different. Thus $ g(y)g(1-\frac{x}{y})= g(1) g(y-x)$ and thus we have for different $x,y$ non-zero the relations $ g(y)-g(x)=g(1) g(y-x)$ and $ g(xy)g(1)=g(x)g(y)$. Thus $ g(u+v)=g(u)+g(1) g(v)$ for $u,v$ non-zero with non-zero sum. Change them and you will find that if $g(1)$ is not 0 then $g$ is the zero function, false. Thus we must have $ g(1)=1$ and then it reduces to the classical problem of finding multiplicative and additive funtions.

[quote="Bojan Basic"][quote="harazi"]...and then it reduces to the classical problem of finding multiplicative and additive funtions.[/quote]
...for what, I believe, we must assume that the function is either continuous, or monotone, or bounded, or measurable (I'm not sure whether anything more is possible).

Really, I was thinking some time about this problem and couldn't solve it completely without an additional assumption. Any help?[/quote]

er, no. if you ONLY have that it's multiplicative, or ONLY that it's additive, you need additional assumptions. I believe harazi showed you have both.

You don't need any further assumption once you have shown they are both multiplicative and additive (since it follows immediately that they are increasing!), which is what I did.

Uh, sorry, I didn't know that. Thank you for clarifying.

Very nice!
The proposer's solution is similar to harazi's.

 :)

Well harazi.. first of all very elegant solution.. Anto is right mine is pretty similar but I find yours better…
Well here is mine anyway:
I kind of like it because it has a lot of nice substitutions….
\[ f(x-y) - f(x) = f( x - y )\cdot f\left (  \frac{x}{y} \right ) (1) \]
Now I set x -> ( -x )  ,   y-> ( -y) and we have
\[ f(x - y) - f( -y) = f( x - y )\cdot f\left (  \frac{y}{x} \right ) \]
That was what I meant as the first substitution….
Combining these two we have
\[  f(x-y)= \frac { f(x)-f(-y)}{ \displaystyle  f\left ( \frac{y}{x} \right ) - f\left ( \frac{x}{y} \right ) } \]
Now we replace what we found for $ f(x-y) $ in (1) and multiply out so we get:
\[ f(x)-f(-y)=f(x)\cdot f\left ( \frac{y}{x} \right ) - f(-y)\cdot f\left ( \frac{x}{y} \right ) (2)\]
This one is very nice I think…it has no $ (x-y) $ in it….
After having y=1 in the above relation and playing a bit with it i found:
\[ f(x)\cdot f\left ( \frac{1}{x} \right ) = 1 \]
So i used a similar technic to prove that f is multiplicative.....
(2) becomes:
\[ f(x) + f(-y)\cdot f\left ( \frac{x}{y} \right ) = f(x)\cdot f\left ( \frac{y}{x} \right ) + f(-y) \]
As we see the right part is the same if we have $ x $ or $ \frac{y}{x} $...So the first part will be the same
and we have..:
\[ f(x) + f(-y)\cdot f\left( \frac{x}{y} \right ) = f\left ( \frac{y}{x} \right ) + f(-y)\cdot f\left ( \frac{1}{x} \right ) \]
Using that and the fact that \[ f(x)\cdot f\left ( \frac{1}{x} \right ) = 1 \] we can easily deduce that:
\[ f(-y)=-f(x)\cdot f\left ( \frac{y}{x} \right ) \] which will easily become \[ f(y)=f(x)\cdot f\left ( \frac{y}{x} \right ) \]
that of course means \[ f(x)\cdot f(y) = f(xy) \]
I consider \[ g(x)=\frac{1}{f(x)} \]. $ g(x) $ is also multiplicative obviously but very easily it is additive..Use (1)....
So $ g(x)=x $ and thus $ f(x)=\frac{1}{x} $

There are some cases we have to check of course  but they are all alright and i am tired to write them here....


MATHCOUNTS
AMC
AIME
factorial
counting
distinguishability
number theory

I need some advice on how to write problems for mock tests, marathons, etc. All the problems I write are either ridiculously easy or have some fault in them that can't be fixed without, well, basically destroying the entire problem.

So this is why I want some advice in writing MATHCOUNTS-EASY AIME level problems  :D .

Easy AIME problems (like the first 2) generally require the solver to simply follow through.  They typically don't require too much ingenuity and as long as you know the concept, they're not too hard.  I've found that number theory has a lot of nice little formulas and theorems, that can be turned into problems-a few years ago, there was a problem involving the number of zeros in a factorial, which is a well-known theorem.  Trigonometric identities have a nice way of creating interesting problems.  Maybe use something well-known but put an alternative spin on it.

I've been the head of the problem-writing committee for the high school contest my university hosts for something like 6 or 7 years. All I can really say about writing questions: [b]It's not easy![/b]

I need the other committee members to bounce ideas off of, and they need to bounce their ideas off of me. Yes, we do steal ideas from other sources (no, I'm not going to reveal where), but there have to be some ideas that are original with us. My own specialty seems to be writing problems that 25%-50% of the competitors get right and that correlate pretty strongly with overall rank; I'm not so good at the questions that sort out the top handful and determine the winner.

But the process takes time. The questions have to checked and rechecked, and re-read to see if they make sense as written. Sometimes, questions are substantially edited; sometimes they're discarded. I we ever put out our first draft and called it the contest, we'd probably cause a riot.

(And if a really good idea for a question occurs to me ... I'm not going to tell you. I need it for my own purposes!)

Although not completely the same (for obvious reasons, as you'll see), when you take the SAT, the questions that appear on there were not written the night before.  They written several years back and field-tested to see if the question makes sense, if it's too difficult, too easy,...  In fact, on every exam, there is an extra section (indistinguishable from the others) that is not scored but instead used to test questions for future years.

You can't really do that for math contests, but what you can do is find some friends who you know will not be taking your exam and have them do your questions.  Get as much feedback as you can as that is the only way to get better.


calculus
integration
conics
parabola
geometry
analytic geometry
calculus computations

Given a line segment $ PQ$ moving on the parabola $ y \equal{} x^2$ with end points on the parabola. The area of the figure surrounded by $ PQ$ and the parabola is always equal to $ \frac {4}{3}$. Find the equation of the locus of the mid point $ M$ of $ PQ$.

I get $ M\left( t \plus{} 1,t^2 \plus{} t \plus{} {\scriptstyle\frac {1}{2}}\right)$ for real $ t$.

Let P and Q be the points $ (x_{2},y_{2})$ and $ (x_{3},y_{3})$ and let M be  $ (x_{1},y_{1})$.

Since we have to find equation of locus of M it means we have to find an algebraic equation which is satisfied by coordinates of every point on the path traced by M.

In other words we have to find an algebraic equation which relates $ x_{1}$ and $ y_{1}$.

Thus        $ x_{2}+x_{3}=2x_{1}$

               $ y_{2}+y_{3}=2y_{1}$

               $ x^2_{2}+x^2_{3}=2y_{1}$

               $ x_{2}x_{3}=2x^2_{1}-y_{1}$

               $ x_{3}-x_{2}=2\sqrt{y_{1}-x^2_{1}}$

Formulating equation for areas we get

  $ \frac{1}{2}(x_{3}-x_{2})(y_{2}+y_{3})-\int_{x_{2}}^{x_{3}}x^2 dx=\frac{4}{3}$

On simplifying,we get

$ y_{1}=1+x^2_{1}$

Thus the required locus is $ y=1+x^2$

That's correct answer. :lol:

Here is my solution:

Let $ \alpha ,\ \beta \ (\alpha < \beta)$ be the $ x$ coordinates of the points of intersection between the parabola $ y \equal{} x^2$ and the line $ PQ: y \equal{} l(x)$,

we can write $ x^2 \minus{} l(x)\equiv (x \minus{} \alpha)(x \minus{} \beta)$, so the area is $ \frac {4}{3} \equal{} \minus{} \int_{\alpha}^{\beta} (x \minus{} \alpha)(x \minus{} \beta)\ dx \equal{} \frac {1}{6}(\beta \minus{} \alpha)^3$, yielding $ \beta \minus{} \alpha \equal{} 2$. Let $ M(X,\ Y)$,

since $ P(\alpha ,\ \alpha ^2),\ Q(\beta ,\ \beta ^2)$ without loss of generality, we have $ X \equal{} \frac {\alpha \plus{} \beta}{2},\ Y \equal{} \frac {\alpha ^ 2 \plus{} \beta ^ 2}{2}\Longleftrightarrow \alpha \plus{} \beta \equal{} 2X,\ \alpha ^ 2 \plus{} \beta ^ 2 \equal{} 2Y$.

Thus use the equality : $ (\alpha \plus{} \beta )^2 \plus{} (\alpha \minus{} \beta)^2 \equal{} 2(\alpha ^ 2 \plus{} \beta ^ 2)$ to get $ (2X)^2 \plus{} 2^2 \equal{} (2Y)^2$, yielding $ Y \equal{} X^2 \plus{} 1$. The desired equation is $ y \equal{} x^2 \plus{} 1$.


for $ a,b,c > 0$ satisfying $ a^2 \plus{} b^2 \plus{} c^2\equal{}3$ prove inequality:
$ (2 \minus{} a)(2 \minus{} b)(2 \minus{} c) \ge \sqrt {abc}$

[quote="Dumel"]for $ a,b,c > 0$ satisfying $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$ prove inequality:
$ (2 \minus{} a)(2 \minus{} b)(2 \minus{} c) \ge \sqrt {abc}$[/quote]
Let $ a\plus{}b\plus{}c\equal{}3u,$ $ ab\plus{}ac\plus{}bc\equal{}3v^2$ and $ abc\equal{}w^3.$
Hence, your inequality is equivalent to $ f(w^3)\leq0,$ where $ f$ is increasing function.
But $ f$ gets a maximal value when $ w^3$ gets a maximal value,
which happens when two numbers from $ \{a,b,c\}$ are equal, which very easy to check.

Hi !

You can easily prove that :

$ (2 \minus{} a)(2 \minus{} b)\geq\frac {1 \plus{} c^{2}}{2} \equal{}\minus{} \frac {a^{2} \plus{} b^{2} \minus{} 4}{2}$

(Just expanding and SOS !  :wink: )

So we have :

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

But $ \frac {1 \plus{} c^{2}}{2} \geq c$ (and similarly for $ a,b$)

And we are done ! :P

[quote="Dumel"]for $ a,b,c > 0$ satisfying $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$ prove inequality:
$ (2 \minus{} a)(2 \minus{} b)(2 \minus{} c) \ge \sqrt {abc}$[/quote]
We have
$ 8(2\minus{}a)(2\minus{}b)(2\minus{}c) \ge (a\plus{}bc)(b\plus{}ca)(c\plus{}ab)$


Let $N_{n}$ denote the number of ordered $n$-tuples of positive integers $(a_{1},a_{2},\ldots,a_{n})$ such that \[1/a_{1}+1/a_{2}+\ldots+1/a_{n}=1.\] Determine whether $N_{10}$ is even or odd.

I'll show it is odd. Group equal $ a_i$ to obtain a sum of the form $ \sum \frac {n_i}{b_i}$ with $ \sum n_i \equal{} 10$ and $ \{a_1,...,a_{10}\} \equal{} \{b_1,...,b_r\}$. Then the number of such solutions is $ \frac {10!}{n_1!\cdots n_r!}$.

To check parity, we only need to consider arrangements where this number is odd. But $ \frac {10!}{n_1!\cdots n_r!}$ is odd only if $ n_1 \equal{} 2,n_2 \equal{} 8$, or $ n_1 \equal{} 10$. In the first we have $ 1 \equal{} \frac {2}{a} \plus{} \frac {8}{b} \Leftrightarrow (a \minus{} 2)(b \minus{} 8) \equal{} 16$, which has solutions $ (a,b)\in\{(3,24),(4,16),(6,12),(18,9)\}$, each appearing $ 45$ times, $ 4\cdot 45$ is even. In the latter case we have one solution. So the total is is odd.

So, can I ask you how many is $N_{10}$ exactly?


function
inequalities unsolved
inequalities

Find the lowest possible values from the function
\[ f(x) \equal{} x^{2008} \minus{} 2x^{2007} \plus{} 3x^{2006} \minus{} 4x^{2005} \plus{} 5x^{2004} \minus{} \cdots \minus{} 2006x^3 \plus{} 2007x^2 \minus{} 2008x \plus{} 2009\]
for any real numbers $ x$.

[quote="wangsacl"][hide="Solution"]Notice that the function can be factored into:
\[ f(x) \equal{} (x \minus{} 1)^2(x^{2006} \plus{} 2x^{2004} \plus{} 3x^{2002} \plus{} \ldots \plus{} 1004) \plus{} 1005\geq1005\]
Hence, the minimum value is $ 1005$. It is attained when $ x \equal{} 1$.[/hide][/quote]
Very Nice !  :o

Ans : 2009
 :rotfl:

for x < 0, all terms become positive.
for 0, f(n)=2009
for 1, f(n)=1-2+3.............+2009 which >2009..

concluded!! :P

Let $\forall n\in {\lbrace2,4,6,...\rbrace} f_n(x)=\sum_{i=1}^{n+1}(-1)^{i-1}ix^{n+1-i}$
We will prove that $\forall n\in {\lbrace0,2,4,6,...\rbrace} f_n(x)\ge \frac{n+2}{2}$ with equality iff $x=1$
It's clearly true for $n=2$ as $f_2(x)=(x_1)^2+2$. Now the induction step is:
$f_{n+2}(x)=x^2f_{n}(x)-(n+2)x+(n+3)\ge x^2\cdot \frac{n+2}{2}-(n+2)x+(n+3)=\frac{n+2}{2}(x-1)^2+\frac{n+4}{2}\ge\frac{(n+2)+2}{2} $
We see that equality is iff $((x=0\vee x=1)\wedge x=1)\iff x=1$
Hence $f(x)=f_{2008}(x)\ge \frac{2008+2}{2}=1005$ with equality iff $x=1$

Consider $f_{2n}(x)=x^{2n}-2x^{2n+1}+\cdots+2n+1$\\

$\textbf{\textcolor{red}{Claim:-}}$ $f_{2n}(x)$ has minimum at $x=1$ which is $n+1$\\

$\textbf{\textcolor{blue}{Pf:-}}$\\

$\textbf{Subclaim:-}$ $f_{2n}(x)=(x-1)^2(x^{2n-2}+2x^{2n-1}+\cdots+n)+n+1$\\

$\textbf{\textcolor{blue}{Pf:-}}$ $\textbf{\textcolor{magenta}{Base Case (n=1)}}$\\

$f_{2}(x)=x^2-2x+3=(x-1)^2(1)+2$ which makes base case true\\

now we consider that $\textbf{Subclaim}$ is true for case of $n$ , now we observe case of $n+1$\\

 we notice $f_{2n+2}(x)=x^2f_{2n}(x)-(2n+2)x+2n+3=x^2((x-1)^2(x^{2n-2}+2x^{2n-1}+\cdots+n)+n+1)=(x-1)^2(x^{2n}+2x^{2n-2}+\cdots+nx^2)+(n+1)(x-1)^2+n+2$\\

$f_{2n+2}=(x-1)^2(x^{2n}+2x^{2n-2}+\cdots+n)+n+2$ so $\textbf{Subclaim}$  is true from induction hypothesis $\square$\\

so clearly we have $f_{2n}(x) \geqslant n+1$ so minimum is at $x=n+1$ attainable at $x=1$ , so claim follows $\square$

so least possible value is $\boxed{1005}$


is it possible to boot from a bootable iso?

Hm, are you saying that you have an .iso file on your harddrive and want to boot from it? I seriously doubt that this is possible, the easiest way to do this is probaly just burn it to CD (or whatever kind of image it is).


Does exist function which satisfy this conditions:

1)be increasing monotomicaly in [a,b] or R.
2)has countable critical points .
3)in any interval like [c,d] which be a subset of  [a,b] or R we have at least one critical point.

Express a function if it exist.

A cheap example: $ f(x) \equal{} \sum_{n \equal{} 1}^{\infty}2^{ \minus{} n}\max\{0,x \minus{} q_n\}$, where $ \{q_n\}$ is some enumeration of rational numbers in $ [0,1]$. This function is differentiable everywhere except at rational points, which are therefore critical points for $ f$.

To make things more interesting, we should insist that $ f$ be differentiable - then a countable dense set of critical points is harder to create.

 :idea: How about this: $ g(x)\equal{} \sum_{n \equal{} 1}^{\infty}2^{ \minus{} n}(x \minus{} q_n)^{1/3}$ and $ f\equal{}g^{\minus{}1}$?


Suppose $ l_1$ and $ l_2$ are parralel lines and that the circle $ \Gamma$ touches both $ l_1$ and $ l_2,$ the circle $ \Gamma_1$ touches $ l_1$ and $ \Gamma$ externally in $ A$ and $ B,$ respectively, and that the circle $ \Gamma_2$ touches $ l_2$ in $ C,$ $ \Gamma$ externally in $ D$ and $ \Gamma_1$ externally in $ E.$ Prove that $ AD$ and $ BC$ intercect in the circumcenter of triangle $ \Delta BDE.$

A hint:

[hide]Try to show that the circumcenter of $ \Delta BDE$ is the intersection of the three common tangents of two of the circles, respectively.[/hide]


Please suggest books on Number theory and inequalities. Which are the most elementary ones?

I picked up "First Steps in Number Theory-A primer on Divisibility" by S.Shirali.Although it is a nice book,I find that it is a lot about explanations using a lot of English with little emphasis on proof writing based on mathematical language.As a result, I am sometimes forgetting what I learnt.I would like to have a book that is elementary,but uses more mathematical language and signs with good explanations.The" Challenges and Thrill" seemed to be a lot more advanced for a slow leaner like me.

And I am getting stuck in Inequalities."Challenges" was quite difficult for me.Can you suggest books that are very elementary?Although there are no shortcuts to labour,yet I felt the question would be best answered by the problem-solvers here.

Thank you in advance.

I think there is a book by Dmitri Fomin called mathematical circles or sth
I found that book very entertaining and basic.

but most of the problems there are quite below olympiad level.
you can try that one  :D  :D

@muks
Same with me when it comes to "challenges" inequalities.
Try out Hall and Knight for number theory.


inequalities
function
inequalities proposed

Let $ a, b, c, d$ be nonnegative real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 1$. If $ 0<k\le 2$, then

$ (1\minus{}a)^k\plus{}(1\minus{}b)^k\plus{}(1\minus{}c)^k\plus{}(1\minus{}d)^k \geq a^k\plus{}b^k\plus{}c^k\plus{}d^k$.

[hide]The inequality holds also for $ k\equal{}3$, $ k\equal{}4$ and $ k\equal{}5$.[/hide]

An idea:
Consider $ f(x) \equal{} (1\minus{}x)^k \minus{} x^k$
We can check that this function is decreasing, and has one inflection point.
These two conditions allow us to use a slightly modified version of half convex half concave theorem, then we know that
1. If $ k \in (0,1)$, then $ f$ is first convex then concave. Then we can assume some of $ a,b,c,d$ are 0, one is unknown and the rest are equal. By the given condition of $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1$, this is actually a one-variable inequality which can be done by calculus.

2. If $ k \in (1,2)$, then $ f$ is first concave then convex. We can do the same thing again.

For $ 0<k\le 2$ there is a nice solution using the AM-GM Inequality together with the Power Mean Inequality.


If the numbers of set A are all the odd single-digit positive integers, and the members of set B are all the prime integers, what is A ∩ B?

The only numbers needed are 1-9

A ∩ B means the set that contains all those elements that A and B have in common

A has 1, 3, 5, 7, 9

B has 1, 2, 3, 5, 7

9 is not prime - 3 * 3 = 9
8 is not prime - 2 * 4 = 8
6 is not prime - 2 * 3 = 6
4 is not prime - 2 * 2 = 4

The answer is 1, 3, 5, 7

Is there an equation for this problem?

[quote="pi^2/6"]1 isn't prime. And primes greater than 2 have to be odd.

So basically it's just primes greater than 2.[/quote]

3, 5, 7

???

Thank you


function
calculus
derivative
real analysis
real analysis unsolved

Show that a monotonous function is derivable almoust everywhere.

yes, it is very classical. you can find the proof in every real analysis book, i think.
they use Dini's Derivatives and Vitali's Covering Theorem.

Also appears in Walter Rudin's Book as a corollary, if I remember right, that every monotonic function is discontinuous at most at a countable set of points. As long as the 4 Dini numbers agree, and hence derivative, then techniques of basic calculus suffice.


Anyone have any ideas on the best ways to practice for relays?  My team has done all of the relays in the 1995-2003 ARML book and from [url]http://math.missouristate.edu/ARML/2007-03-17s.pdf[/url], and we still are looking for more problems.

like the thing is that there are two problems: either

a)people have difficulty solving the problems conceptually
or
b) people cannot do the computation fast enough

a) can be solved by just like doing lots of difficult math...i doubt this is a problem with zuming there :P
b) can be solved by just like do mathcounts sprint rounds till you pass out :lol:

my problem is b)...i am going to have to step my computation up...like on that stupid arithmetic game, i get like mid 20s where most people are getting like 40-50


MIT
college
Stanford
University of Chicago
geometry
ratio
Support

(Un)fortunately, I got in everywhere I applied:
MIT
UMich (honors math)
Stanford
Chicago
UCSD
UC Berkeley

I've sort of ruled out UCSD, because I want to get out of Southern California. Berkeley, Stanford, UMich will all be a similar price for me; MIT is a tad more expensive; U of Chicago is the most expensive.

Unfortunately, all I know about each of these schools is what is stated on their websites. So, all of them have 250 student groups and numerous a cappella groups, but that doesn't help me much.

I want to study mathematics, and I want to go on to grad school, but I don't care about "looking good" to grad schools.

I want to get involved in research. (I hear this may be a hard thing to do at Berkeley.)

It looks like I'd start off in each of these classes:
MIT-100B
Mich-295
Chicago-Honors Calc or Real Analysis--not sure
Stanford-51H
Berkeley-Advanced linear algebra-->intro abstract algebra

I will be going to MIT CPW and Stanford Admit Weekend. I've seen Michigan (and I really liked it, but there weren't any students there). I don't have time to visit any other schools.

I don't want to go to a "party" school (I guess all schools are party schools; I just want a group of more serious students and a reasonable amount of quiet in the dorms), but I don't think any of these schools fit into that category. (My sister went to Cal State San Diego, and it was apparently really bad.)

Anybody have any advise for narrowing down this list?

I have a mathematician friend who's very happy at UChicago, so I have a high opinion of it.  On the other hand, it's hard to beat MIT for math.  If you're curious, look at MIT's math website and look at the courses and professors to see if you're interested in the kind of stuff we do here.  (But of course I'm obligated to convince you to come here :P)  I should also mention that it's probably easier to do research as an undergraduate at MIT than at any of the other schools you listed.

If you want peace and quiet at MIT, you can definitely get peace and quiet.  Different dorms at MIT lead quite different lives; if you live in a particular part of campus you'll learn that MIT throws surprisingly good (and loud) parties, and if you live in another part of campus you may not ever hear any dance music ever.  But I think the best advertisement for MIT is the CPW experience you'll have, so see what happens after that.  Feel free to PM me if you want to ask me any questions or find me during CPW :)

I went from SoCal to Berkeley, so here's what I'd make of that choice:
Berkeley has all the benefits and drawbacks of being gigantic, both as a school and as a math department. There are tons of opportunities, and you can easily miss them all and fade into the bureaucracy. The dorms don't have enough space for everybody, so most students are out by the second year; one of the main perks of the Regents' scholarship is access for four years if you want it.
I didn't get into research there, but I did take some introductory graduate courses, and got a lot out of them.
I got plenty of peace and quiet there- too much, really. For a math major who wants it relatively quiet, the Foothill dorm is the place to start.

I'll also plug the benefits of medium distance: you can easily drive up to the Bay Area to move in, but you won't have parents visiting all the time. The Chicago/MIT/Michigan options all involve flying at the beginning and end of the school year, so moving bulky stuff is a big hassle.

Back in the day, my decision was mainly between UMich and MIT.  One of the main reasons I chose MIT was the people.  MIT is incredibly dense with the brightest minds in the nation.  Though, I have to say, it was more than just having a bunch of people that tool.  The work hard, play hard principle that is the norm at MIT is what really hooked me.  There's an amazing breadth of opportunities here and everyone finds a niche that they really enjoy.

Regarding the party school issue...yes, you'll find a thriving party scene at any of those schools.  At UMich, since you'd be in the honors dorm, it would probably be better than, say, being in a random dorm there.  As for MIT, tor gave you a pretty good description.  I'll add that everyone here is in the same boat in that they're extremely hosed during the week.  Most of the partying is done on Friday and Saturday night.  Most people head over to a fraternity to party.  Though there are a few dorms and sections of dorms that party as well (Baker, EC, senior house, certain sections of other dorms).  I lived in Simmons my freshman year which is definitely one of the quieter dorms, perhaps the quietest.  Basically, you should have no problem finding quiet space from Sunday-Thursday anywhere and on the weekend you have to seek out parties unless you live in a place that hosts parties.

That said, the nature of social life at MIT is quiet different from that almost anywhere else.  Before college I did very little socializing.  I got here and lived in Simmons for a term.  It was really quiet, which was nice, but lacked culture.  My good friends (one of which I knew from AoPS(!)) had joined a fraternity in the fall and would be moving out of Simmons sophomore year.  I really didn't understand fraternities and the value of socializing for a while (which is why I didn't rush in the fall).  After a semester in Simmons, I knew I had to find something else and thus explored the fraternity my friend was in (after teasing him for a whole semester since I thought fraternities were silly...)...and it was pretty amazing.  Sure they would party on Friday and/or Saturday night.  But the rest of the week was spent studying and doing psets.  When I got stuck, I could go ask others for help or just take my mind off of it for a while and engage in an interesting discussion.

Regarding CPW, it's a really awesome experience (which I missed for a track meet my year...).  You'll hopefully be able to understand the point I was trying to make in the previous paragraph by experiencing it first hand  :P  That said, you wont see quiet the usual MIT environment.  There will be way more things going on than usual.  In a few quick, sleepless days you'll be able to experience the vast diversity of things you can do at MIT.  In terms of the work hard, play hard spectrum, you'll definitely see us playing hard*...so keep that in mind, i.e. it's quieter at MIT normally.

* Good ol' clean fun, i.e. no alcohol.  CPW is officially dry and that policy is stuck to remarkably well, especially compared to other schools that have similar policies.

As much as I hate to say "school X is better than school Y," I would quickly narrow your choices down to MIT, Stanford, and Chicago, based on your stated priority of math over all else.  The other schools are great, too, but the undergraduate math majors, as a whole, will not be as strong, and this has a large effect on the difficulty of your undergraduate courses.

The main advantage of Chicago for you, as I see it, is that it has a reputation for having a serious student body.  However, the students you take classes with at MIT and Stanford will also be pretty serious about their studies, so this only has relevance for the overall environment (e.g. dorms).  So unless that's super-important to you, I'd rule out Chicago also, since it's more expensive, and again, the math majors (as a whole) are *probably* at a *slightly* lower level than the MIT-Stanford level.

That brings us to MIT vs Stanford.  At both of these schools, there are a ton of brilliant math students (and consequently a great peer group and challenging classes), so I don't think there's any point in doing a comparison there.  Obviously, both are top-notch in research.  Teaching quality is hard to compare, especially since the postdocs always change anyway.  Luckily, the two schools are located in two very different places with very different environments, so it shouldn't be hard to choose based on the non-math factors.  (I'll draw some caricatures of dubious accuracy.  MIT:  Cold place, urban, close to Boston, no riding in cars, sterile but cool campus, fewer women, reputation for hard work, high stress, and nerdiness, and you will be surrounded by engineers and linux.  Stanford:  Amazing weather, suburban but not in a crappy way, San Francisco is mostly irrelevant, beautiful sprawling Taco Bell-like campus, top-notch athletics, reputation for laid-back attitudes and quirkiness.)  Honestly, if I had to make your decision (for myself, not you), based on everything I know now, I think my head might explode.

The MIT people can correct me if I'm wrong, but isn't MIT one of the few places where you can make meaningful choices about where you'll live freshman year?  That may be attractive if you want quiet.  Because although many students at both Stanford and MIT love quiet, there are many more who do not.

P.S.  For the sake of others on this board, please avoid referring to your circumstances as "unfortunate" in any way.

[quote="yenlee"]MIT:  Cold place, urban, close to Boston, no riding in cars, sterile but cool campus, [b]fewer women[/b][/quote]
Do you know how many colleges there are in the Boston area?  :P

And yes, it's pretty cold in Boston in the winter, but I don't think no cars is a disadvantage - MIT is situated right between Boston and Cambridge and public transportation is great here, so you can get pretty much anywhere you want to go on foot or by bus and train.  It's quite refreshing, at least when the weather's nice.

[quote="yenlee"]isn't MIT one of the few places where you can make meaningful choices about where you'll live freshman year?[/quote]
Quite right.  While a lot of other schools lottery you into a freshman dorm, at MIT the only restriction to where you can live freshman year is that you have to live on campus, and you can still join a fraternity or other living group without moving in.  That gives each dorm a lot of power to shape its own unique culture and personality, which a lot of people wouldn't give up for anything.  One of the drawbacks to this system, though, is that it makes social life at MIT somewhat segregated.  It's easy to become insular.  Some people prefer the lottery system because it forces you to interact with a wide variety of people, and indeed some of the newer dorms are more diverse personality-wise since they haven't had as long to establish themselves, but I think you'll find that a lot of the quirkiest and most interesting people you'll meet will inevitably find themselves drawn to the quirkiest and most interesting dorms.

[quote="yenlee"]MIT:  Cold place, urban, close to Boston, no riding in cars, sterile but cool campus, fewer women, reputation for hard work, high stress, and nerdiness, and you will be surrounded by engineers and linux.  Stanford:  Amazing weather, suburban but not in a crappy way, San Francisco is mostly irrelevant, beautiful sprawling Taco Bell-like campus, top-notch athletics, reputation for laid-back attitudes and quirkiness.)  Honestly, if I had to make your decision (for myself, not you), based on everything I know now, I think my head might explode.[/quote]

based on [i]these factors alone[/i], this doesn't sound like a very difficult decision to me (i.e., Stanford and it isn't close?)

Urban vs. suburban and engineers vs. top-notch athletes are a matter of preference, which is what I think yenlee was getting at.  Although I don't like the implication that Stanford somehow has a quirkier reputation than ours or that MIT people aren't laid back :P

On that subject, MIT is one of the few schools that doesn't recruit athletes, nor does it accept legacies.  (We do accept "foundational admits" - e.g. for donating libraries - but what are you gonna do?)  I have friends at other colleges who complain that it's so obvious who the people at their school are who don't belong there, in the sense that their academics just aren't up to par, and I think you'll be hard-pressed to find people like that at MIT.  Our athletes are just as dedicated to their intellectual pursuits as anyone else.

[quote="t0rajir0u"]Urban vs. suburban and engineers vs. top-notch athletes are a matter of preference, [/quote]

Of course, in the same way that choosing a school with great weather versus a campus with cold weather, and choosing a school with a relatively even ratio of guys to girls or a school with mostly guys is a matter of preference.  I mean, I can think of (good) reasons to go to MIT over Stanford, but if I were deciding on that list alone I don't think my decision would take very long.

[quote="t0rajir0u"] I have friends at other colleges who complain that it's so obvious who the people at their school are who don't belong there, in the sense that their academics just aren't up to par, and I think you'll be hard-pressed to find people like that at MIT.  Our athletes are just as dedicated to their intellectual pursuits as anyone else.[/quote]

At one point in my life I probably would have worried about 'who belonged' at x school but it's ultimately a pretty ridiculous thing to do.  As a math major you aren't going to be interacting much in classes with dumb superstar athletes, and having a wider range of people on campus is almost certainly going to create a more interesting on-campus atmosphere (not to mention that division I college sports are pretty darn exciting).  I mean it's kind of hypocritical of me to say so, but nerdy math/physics/cs types aren't exactly who I want to spend most of my free time around, no matter how 'quirky' they might be.

I'll put in a few words regarding math at Stanford, although my experience is somewhat limited since I'm not a math major.

All of my math professors (four total) have been very good; if you choose the right professors to take classes with, you'll definitely get a lot out of it. I've heard that some math professors here are not great lecturers, so it's partially up to the student to try and figure out the best classes to take.

In particular, I took the 50H series last year (Leon Simon for the first 2 quarters, Simon Brendle for the 3rd). I think it is a very well-designed sequence of classes and both professors give good lectures. On the other hand, taking that sequence made me realize that I didn't really want to be a math major so make what you will of that. I will say that the classes are challenging and for the most part interesting; usually it got less interesting once I got too lost.

Personally, I chose Stanford over MIT for reasons related to what yenlee mentioned (e.g. weather, campus, distance, culture, etc). Without comparison to MIT (because I can't meaningfully make such a comparison), I think Stanford's math community is rather close-knit. There are a lot of awesome math students here, but a large number of them go into engineering (as I have), so the number of actual math majors is somewhat small.

Anyway, to sum it up, Stanford's math program is really good. And the school is just awesome in general. So you should come here :).

[quote="blahblahblah"]choosing a school with a relatively even ratio of guys to girls or a school with mostly guys is a matter of preference.[/quote]  Both MIT and Stanford freshman classes are majority male: at MIT, the ratio is [url=http://web.mit.edu/facts/enrollment.html]54.6:45.4[/url] while at Stanford it's [url=http://www.stanford.edu/about/facts/undergraduate.html]51.3:48.8[/url] -- a difference, but not a particularly large one, especially in light of the fact that there are several tens of thousands of non-MIT undergraduates in the immediate vicinity of MIT and 0 non-Stanford undergraduates in the immediate vicinity of Stanford.  Also, um, lots of people prefer to go to school in urban areas rather than suburban areas -- it's not even remotely comparable to the question of Boston vs. Palo Alto weather.

[quote="JBL"]Also, um, lots of people prefer to go to school in urban areas rather than suburban areas -- it's not even remotely comparable to the question of Boston vs. Palo Alto weather.[/quote]

Yeah.  I was being a bit sarcastic; while urban campuses versus suburban campuses definitely are a matter of personal preference, the other factors as listed by yenlee certainly seem to favor Stanford.  I have no particular bias towards or against urban campuses :)

[quote="yenlee"]As much as I hate to say "school X is better than school Y," I would quickly narrow your choices down to MIT, Stanford, and Chicago, based on your stated priority of math over all else.  The other schools are great, too, but the undergraduate math majors, as a whole, will not be as strong, and this has a large effect on the difficulty of your undergraduate courses.[/quote]

Just to play devil's advocate for a minute - Does "more difficult" necessarily mean "better educationally"?

And I think the upper level's of U. of Michigan's Honors Math program, and many of their upper level undergraduate/lower level graduate courses, will be comparable in difficulty to similar courses you'd find at MIT, Stanford, or U. of Chicago. They will also be populated by students who could have probably gone to one of those other schools.

The main difference academically may be the quality of some of the (non-honors) classes outside of the math department, and the variety of opportunities to learn mathematics outside of the classroom. Also at large Universities, the ability to get into the classes that you want can be a pain. This depends on the University though. The good news is that they usually have a large variety of classes, and they are offered frequently with a greater variety of professors to choose from.

Berkeley also has honors versions of the core upper-division classes. My experience with those was a bit uneven, but it does mean that you can find classes with other high-end students, and follow those up with graduate classes. Berkeley and Michigan come out fairly similarly on the academic options, so that choice really comes down to geography.

[quote="gauss202"]Also at large Universities, the ability to get into the classes that you want can sometimes be a pain.[/quote]This is not generally a problem for higher-level courses; it's the basic classes that fill, or force you into sections at odd times. Getting into Berkeley's Math H110 is easy, but Chem 1A will fill or come close. For advanced classes, having the class canceled for low enrollment is a much more relevant risk.

I don't think I care too much about the male-to-female ratio, unless it's something crazy like New Mexico Tech's. A few athletes probably won't bother me either.

I think the urban-suburban thing matter quite a bit. Surely there's got to be some sort of "college town" with restaurants, movie theaters, grocery stores, etc. within walking distance around Stanford (I'm not expecting anything quite like Ann Arbor); it can't just be suburbs.

[quote="mnop"]Surely there's got to be some sort of "college town" with restaurants, movie theaters, grocery stores, etc. within walking distance around Stanford (I'm not expecting anything quite like Ann Arbor); it can't just be suburbs.[/quote]
Clearly you haven't visited Palo Alto :P  Even [url=http://www.stanforddaily.com/cgi-bin/?p=493]the Stanford Daily[/url] admits it's not a college town.

haha t0rajir0u don't insult palo alto. palo alto is awesome. there are grocery stores, several great restaurants on university avenue, and numerous movie theaters in palo alto and in neighboring cities. downtown is very un-suburbia-ish.

Funny, that description sounds... exactly like every other affluent suburb in the US.

It is not possible to dislike Palo Alto unless you are a very ignorant or confused human being.

Also, I'm currently taking the 50H series at Stanford and they're really great classes. One of the best mathematical experiences I've had. Though echoing paladin8, among everyone I know in the class, I might be the only one who hasn't been completely put off the idea of a math major because of it.

How many students are in the Math 50H course? Are they all freshman, and are there any math majors who do not go through that course?

My year, 50H had about 30, I think.  I've heard it's significantly larger this year, because a lot fewer dropped out.  They're mostly frosh, but not all.  And there are plenty of math majors who don't go through it.  If you want to get your math degree with honors, though, it's required.

Actually, after I wrote that MIT had fewer women, I did google the actual ratio (which is really not that awful) and decided to edit out that comment.  I guess I forgot to do it.  So I agree that the male-female ratio is not such a big distinction.  But it's still true that people wear fewer clothes at Stanford.  (There, against my better judgment, I said it.)

@t0rajir0u, I'm sure that many people at MIT are laid back, but I think it's fair to say that, as a whole, MIT does not have a reputation for being laid back.  If anything, it has the opposite reputation (fairly or not).  And I'm definitely sure that MIT people can be quirky (see ZBT as Exhibit A), but Stanford does have the Stanford band, which is a significant component of Stanford culture.  (And I agree that not riding in cars is a good thing.  You seem to think that I came down on the side of Stanford, which is not quite true.)

@mnop, Palo Alto is not a college town.  It is a very wealthy and very nice suburb, with a cute little downtown area.  So yes, you can easily walk or take the Marguerite bus downtown and access lots of great restaurants and whatnot, although many of them are pricey.  There's also a Whole Foods and even a movie theater that plays classics.  (But undergraduates don't go grocery shopping, do they?)  It is definitely not the strip mall hell that you see in newer suburbs.  Also of note is the Stanford Shopping Center, which is actually a very pretty outdoor mall located on Stanford property.  And unlike most other suburbs, a lot of people *work* there, so it's crawling with software engineers, venture capitalists, lawyers, etc.

@gauss202, I do not equate "more difficult" with "better educationally."  But if you're serious about going into math, there's nothing more important than challenging yourself and having the support you need to do so.  The easiest path to challenging yourself is to take really hard classes (not that I say "hard" and not "advanced") with students of comparable ability.  A harder (but often superior) way to do it is to blaze your own trail of independent study, guided by helpful professors.  I guess I'm betraying my personal preference for the lazy path.  I also think it's nice to find the right peer group, which is easier to do at some places  than others (depending on who you are, of course).

Adding to Sly Si's comment:  If a student skips 50H at Stanford because he is too advanced, the requirement of taking 50H to get honors would surely be waived.

The class isn't much larger this year--I counted 36 scores in the 52H midterms, and five of those were not precisely from Stanford students.

@t0rajir0u - Do you know any of the athletes at MIT? I'm interested in playing basketball or tennis varsity, but I'm not sure if I will be able to do a sport along with studies at MIT. Can the people there manage to have a life and learn at the same time?

Of course.  MIT used to have 41 varsity sports and currently has 33 (unfortunately, we had to make some budget cuts), although you should be aware that we are a D-III school.  One of my fraternity brothers is on our crew team and another fences.  Generally, I think you'll find that plenty of MIT students have very active extracurricular lives.

I'm not even sure what prompted this question - you must be thinking of Caltech :P


Find $ x$ given $ 2^{2x\plus{}1}\minus{}2^{2x\minus{}1}\equal{}24$.