without using a calculator and with proof:

which is larger (60^(1/3)) or (2 + 7^(1/3))

Im just a beginner so I will try my best to answer this.

60^(1/3) = x
x^3 = 60

4^3 =64
3^3 = 27

So x is betweem 3 and 4, leaning closer to 4 b/c 60 is closer to 64 than it is to 27.

Lets approximate it as x = 3.8

y^(1/3) = 7
y^3 = 7

1^3 = 1
2^3 = 8

So y is between 1 and 2, leaning closer to 2 because 7 is closer to 8 than it is to 1.

Lets approximate it as: 1.9

y+2 = 3.9

3.8 and 3.9.....this is difficult.  I would guess that it is cuberoot 60 b/c

%change between 7 and 8 is 1/8 and % change between 60 and 64 is 4/60 = 2/30 = 1/15.  There is less change between 60 and 64 than there is between 7 and 8 so cuberoot 60 is closer to cuberoot 64 than cuberoot7 is to cuberoot8.

Probably wrong.

you got the right answer, but i was hoping for a more elegant solution,

here's a hint: [hide] prove the more general: which is larger? (4(x+y))^(1/3) or x^(1/3) + y^(1/3) (because if we let x = 8, y = 7 we get our case) for positive reals x,y.  Then let x=a^3, y=b^3 and try to see which is bigger.[/hide]

prove the more general: which is larger? (4(x+y))^(1/3) or x^(1/3) + y^(1/3) (because if we let x = 8, y = 7 we get our case) for positive reals x,y. Then let x=a^3, y=b^3 and try to see which is bigger.

(4x + 4y)^1/3 ? x^(1/3) + y^(1/3)

let x=a^3 and y=b^3

(4a^3 + 4b^3)^1/3 ? (a^3)^(1/3) + (b^3)^(1/3)
4^(1/3) * (a^3+b^3)^(1/3) ? a + b

4^1/3 * (a+b)^1/3 *(a^2 -ab +b^2)^1/3 ? a+b

8^1/3 =2
1^1/3 = 1
0 < 4^1/3 < 1

confused...

I guess if you know how a calculator does it then you could follow that method?

x=a^3, y=b^3

so (4(a :^3: +b :^3: ))^(1/3) compares to a+b
or  4(a :^3: +b :^3: ) compares to a :^3: +b :^3:+3a :^2: b+3b :^2: a

 3(a :^3: +b :^3: ) compares to 3(a :^2: b+b :^2: a)
   a :^3: +b :^3:  compares to a :^2: b+b :^2: a

Note that a :^3: +b :^3: -a :^2: b-b :^2: a :ge: 0
because it factos into (a-b)(a :^2: -b :^2: )
if a :ge: b, then the expression is positive
if a :le: b, then a-b is negative, and a :^2: -b :^2:  is negative, the product is also positive
so  a :^3: +b :^3:   :ge:  a :^2: b+b :^2: a

so (4(a :^3: +b :^3: ))^(1/3) >  a+b

this is hard, I probably couldn't have figure it out without the hint :oops:

That's very nice.  I would have got it without the hint, but it would have been very messy.  I cubed, combined terms, got a quadratic in the cube root of 7, found the roots of the quadratic, and was then going to end up comparing :sqrt:34 - 1 and the cube root of 7 which I can do by cubing, subtracting the constant to one side, and then squaring.  Very not pretty.


Find all integers a > 1 for which the least integer solution n of any congruence a^n = 1 (mod p) differs from 6 (p is any prime number).

[i]Mikls Schweitzer competition 1957 problem 8[/i]

We have a 6-1=(a-1)(a+1)(a 2-a+1)(a 2-a+1).

We must find all a such that if p|(a 2-a+1) then p|a-1, p|a+1 or p|(a 2+a+1). (hope you understand why)
Let p|(a 2-a+1)

If p|(a 2+a+1), then by substracting we get p|2a. Since (a 2-a+1) is odd, so is p, so p|a and from this p|1 (a contradiction)

If p|a-1 we get p|(a 2-2a+1) and p|(a 2-a+1), so p|a... contradiction.

If p|a+1, we get p|(a 2+2a+1), so p|3a. If p|a then we get a contradiction, so p=3.
So 3 is the only prime divisor of (a 2-a+1), so there is n such that
a 2-a+1=3 n. Multiply this by 4 and obtain
(2a-1) 2+3=4*3 n  [b](1)[/b]
If n=1, we get a=2.
If n>1 we get 3|2a-1, so 9|(2a-1) 2 and we get a contradiction in (1) mod 9.

So the only solution is a=2.


linear algebra
matrix
vector
induction
superior algebra
superior algebra solved

If A and B are square matrices with real entries and AB=BA, prove that
Spec(A+B) \leq Spec(A)+Spec(B). That is: each eigenvalue of A+B is the sum of an eigenvalue of A and an eigenvalue of B.

[quote="Moubinool"]
if they AB=BA then there P invertible such that

P^(-1)AP=T_A

P^(-1)BP=T_B[/quote]

Why?

Maybe you'll laugh, but what does "diagonalisable" mean?

It means that there exist a basis s.t the endomorphism associate to A can be written  like

(a 0............0)
(0 b0...........0)
.....................
(0...............0 z)

example

Consider p a projector in R^3, pop=p
R3^= kerp @ ker(p-I) , @ is direct sum
take a basis B of R^3 obtained a basis of  kerp and ker(p-I)
the matrix associate to p
will be like

(100)
(010)
(000)
or
(100)
(000)
(000)
or
I_3
or 0

[quote="Moubinool"]Theorem: A,B matrices in Mn(C), AB=BA they commute
then exist a common basis in wich they are simulatenously triangular superior.

You also have

Theorem: If A,B are diagonalisable matrices Mn(K) K=R or C
AB=BA then exist a common basis in wich they are diagonal.[/quote]

I really want a proof of that theorem, some hints or a link to a proof. I've never heard of such a result.

Theorem: A,B matrices in Mn(C), AB=BA they commute
then exist a common basis in wich they are simulatenously triangular superior.

proof:
a,b endormorphism of C^n associate to A and B
k eigenvalue of a
ker(a-kI) is stable by b there a vector y=/=0, y \in C^n
a(y)=ky, b(y)=qy
F subspace of C^n such that C^n=Cy@F where @ is direct sum
a', b' endomorphism of F defined by
a'(x)=p_F(a(x))
b'(x)=p_F(b(x)) where p_F is projector on F parallel to the line Cy
a',b' commute
consider block decomposition
matrix(a;{y} \cup (basis of F))=
(k *****************)
(0   matrix(a';basis of F))

matrix(b;{y} \cup (basis of F))=
(q****************)
(0 matrix(b';basis of F))

the commutativity of these matrices => mat(a') and mat(b') commute
a'ob'=b'oa'

by induction there is a basis f of F such that
mat(a';f)=
(s_1*****************)
(0 s_2***************)
.....................................
(00..........................s_(n-1))

mat(b';f)=
(w_1*****************)
(0 w_2 ***************)
.....................................
(00.............................w_(n-1))

in the basis {y} \cup f , a and b are simultanously triangular superior

Theorem: If A,B are diagonalisable matrices Mn(K) K= R or C
AB=BA then exist a common basis in wich they are diagonal

proof: Let E be the K-vector space K^n, a and b the endomorphism
associate to A,B
E= @_(k eigenvalues of a) Ker(a-kI)
the eigenspace ker(a-kI) is stable by b (commutativity)
Let b_k the restriction of b to ker(a-kI), b_k is diagonalisable
there is a basis B_k of ker(a-kI) in wich b_k is diagonalisable
 \cup _(k eigenvalue of a)B_k is a basis of E in wich a,b are simultanously
diagonalisable.

Thanks Moubinool for taking some time to explain these results.


linear algebra
matrix
vector
superior algebra
superior algebra unsolved

Prove that the symplectic group $ SP_{2n}(\mathbb{R})$ is path-connected, and conclude that every symplectic matrix has determinant 1.

$ Sp(V)$ ist generated by the transvections, i.e. maps of the form $ \sigma_{\lambda,u}: \equal{}v\mapsto v\minus{}\lambda\langle v,u\rangle u$ with a nonzero vector $ u$ and the symplectic form $ \langle\cdot,\cdot\rangle$.
The path $ [0,1]\to Sp(V): t\mapsto\sigma_{t\lambda,u}$ connects every transvection with the identity. So $ Sp(V)$ ist path connected. Now since every element $ A\in Sp(V)$ satisfies an equation of the form $ A^T JA\equal{}J$ with a skew-symmetric matrix $ J$, we have $ \det(A)\equal{}\pm 1$ and because of the path-connectedness $ \det(A)\equal{}\plus{}1$ for every $ A\in Sp(V)$.

(But you can have this one also from the fact, that the transvections generate $ Sp(V)$ because every transvection has determinant 1)

How can be proved that $ Sp(V)$ is generated by transvections? It isn't obvious for me.


Its my first time doing USAMTS and i was wondering what forms i need like permission or qualification? Also, can i scan my solutions and then email them, or do I have to use Latex? Finally, if i mail, i can mail on nov 21 right?
thx

You can read about how to send solutions here:

http://www.usamts.org/Problems/U_ProblemsSoln.php

The Entry Form is here:

http://www.usamts.org/MyUSAMTS/U_MyForms.php

You do not have to send in the Permission Form, but if you want your name to be listed among top scorers, we need that form completed.

You do not need to use LaTeX.  Fax or mail is fine.  Usually emailed scans don't come out too well, so we'd prefer fax or mail over emailed scans.  (Emailed PDF files which are not scans are fine).

Just out of curiousity.  Why do you say we.  Are you an organizer?

Mr. Rusczyk is one of the administrators of AoPS, and AoPS now runs USAMTS, from my understanding.

Mr. Rusczyk is the director of USAMTS. ;)


i am offended by this topic because it reminds me that i'm asian

I am offended by this topic because it is my topic and people are being offended by it and thus offending me.

This is not meant to be at all racist, and If needed I will ask for it to be locked but I have seen many comments around GFF about many people on here being Asian and I wanted to see what percentage of people here it makes up.  I apologize if it is offending to some people.

c'mon, how this have anything to do with race? this has absolutely nothing to do with race

[quote="junggi"]i heard that the race is very close
maybe that has sometihng to do with this thread?

i am asian[/quote]

YES@ THE RACE IS VERY CLOSE@

[quote="mdk"][quote="asmallgoat"]I am offended by this topic because it reminds me of how im not asian :noo:  :noo:  :noo:[/quote]

What are you trying to say? You wan't to be an Asian that bad?[/quote]

Can't blame her.
Being caucasian can be fun at times, true, but the thinking mind is sure to feel intellectually starved after a point.
I mean, try digesting the fact: "I am not asian"
First you'll think: "Who cares? This is more fun."
After some time you'll think: "Not that I can do anything about it."
And finally: "Oh daamn daaamn daaaaamn! :( "
(I have not had to pass through this agony by the way, because the very cause is missing in my case)

Yes

[size=9]The message is too small. Please make the message longer before submitting.[/size]

[quote="footballrocks41237"]Yes[/quote]

I'm honoured.

I might as well feel offended because so many people are.

[quote="funcia"][quote="mdk"][quote="asmallgoat"]I am offended by this topic because it reminds me of how im not asian :noo:  :noo:  :noo:[/quote]

What are you trying to say? You wan't to be an Asian that bad?[/quote]

asmallgoat has been told she wants to be Asian by an Asian person...[/quote]

Oh I see.
So ... ummm ... where's the contradiction?

[quote="bubka"][/quote][quote="funcia"][quote="mdk"][quote="asmallgoat"]I am offended by this topic because it reminds me of how im not asian :noo:  :noo:  :noo:[/quote]

What are you trying to say? You wan't to be an Asian that bad?[/quote]

asmallgoat has been told she wants to be Asian by an Asian person...[/quote][quote="bubka"]

Oh I see.
So ... ummm ... where's the contradiction?[/quote]
[url=http://www.spam.com/]here[/url]

anyway the topic of this thread is NOT "how many of you are offended by this topic?" but rather "how many of you are asian?" so LETS GET ON TOPIC PLEASE

asian as in azn as in chinese?

or asian as in from asia?

no. im friends w/most of the asians at my school though...

I don't understand how this is racist; it's just asking a question about nationality. He's not saying that being Asian is good or bad, he(or she)'s just curious. There's nothing wrong with people asking about ethnicity.

Anyways, I'm not Asian, but my friend is.

[quote="7h3.D3m0n.117"]I'm Asian.

Asian Pride![/quote]

That's exactly what I was going to post.

Asian Pride!

I love being Asian!!! I'm Chinese, but anything Asian is cool. I'm friends with most of the Chinese at my school, although there aren't that many in Utah Valley. Long Live Asia!

亚洲万岁！中国万岁！！！

WHITE BOY PRIDE!!!
                  \
not asian ____\ 8D

don't delete this one demon because if you do....

lol im not going to delete it....

over half my friends are white...and i have no problem with that.

i was born in the USA so that makes me...?

i am asian (indian)! most of my friends are chinese.. and one other indian (anaxerzia)

I'm Asian. My town has way to many Asians, we never meet people of other ethnicities. My school is like 60% Chinese.

cool, i'm asian too. it's not surprising that there's a high percentage of asians here. go china!

im asian...
chinese

rang ni bu ji ge ni ben dan!

I'm not stupid and you take that back!  :D

[quote="Cyrazeno"]i am asian (indian)! [/quote]

Woo go indians!

yup! i am ndn too.

most of my friends are chinese and ndn

i know this is really off topic, but if you can understand chinese, listen to the song Ugly Girl by Tie My Shoe. XD XD

[quote="pianoforte"]im not asian[/quote]
[quote="pianoforte"]im asian![/quote]

Nice switch there...
Anyway, I'm Indian


$ k$ is a natural number and $ n\equal{}2^{k}\plus{}1$.prove that $ n$ is a prime number if and only if $ 3^{\frac{n\minus{}1}{2}}\plus{}1$ is divisible by $ n$.

If $ n$ is prime and $ k>2$, then from $ (\frac{3}{n})\equal{}\minus{}1$ we get $ 3^{n\minus{}1}{2}\equal{}\minus{}1\mod n$.
If $ n\equal{}2^k\plus{}1$ and $ 3^{n\minus{}1}{2}\equal{}3^{2^{k\minus{}1}}\equal{}\minus{}1\mod n$, then for minimal period $ m$, suth that $ n|3^m\minus{}1$ we get $ m\equal{}2^kl$, l -odd. But $ m$ must be less than n. It give $ m\equal{}n\minus{}1$ or $ n$ is prime.


Find two five-digit numbers whose product is 3^12+ 15^6+ 625^3.

how would you do this problem?

[quote="mr. math"]Find two five-digit numbers whose product is $3^{1}2+15^{6}+625^{3}$.

how would you do this problem?[/quote]
[hide="hint"]notice that it is equal to $3^{12}+(3\cdot5)^{6}+5^{12}$ and compare it to the expansion of $(3^{6}+5^{6})^{2}$[/hide]


Using $\cot^{2}2t=\left({1-\tan^{2}t\over 2\tan t}\right)^{2}=\frac{(1-\tan^{2}t)^{2}}{4\tan^{2}t}$

and substituting $a=\tan^{2}x,b=\tan^{2}y,c=\tan^{2}z$, we get the system

$2a+b+{1\over b}=4$
$2b+c+{1\over c}=4$
$2c+a+{1\over a}=4$

Since $a,b,c$ are positive, we have $2a=4-b-{1\over b}\leqslant 4-2=2\iff a\leqslant 1$, similarly for the other variables.

Assume two of them are equal, WLOG $a=b$. Then $2c=4-a-{1\over a}=4-b-{1\over b}=2a\implies a=b=c$.

Therefore we'll assume $a\neq b\neq c\neq a$. WLOG assume $b=\min\{a,b,c\}$ or $b=\max\{a,b,c\}$. Subtracting the second equation from the first we get

$2(a-b)+(c-b)\left({1\over bc}-1\right)=0$

Since $b\neq c$, at least one of those two is strictly less than $1$, hence ${1\over bc}-1>0$. If $b=\min\{a,b,c\}$, then the LHS is strictly positive, and if $b=\max\{a,b,c\}$, then the LHS is strictly negative, hence in neither case can't be zero. Similar argument for the remaining combinations of $\max$ and $\min$.

Thus is proven that $a\neq b\neq c\neq a$ can't hold, which means that at least two are equal, which in turn means that all are equal. Then the first equation yields

$3a+{1\over a}=4\iff a\in\{1,{1\over 3}\}$

Hence $x=\arctan(\pm\sqrt{a})\in\{\pm{\pi\over 4},\pm{\pi\over 6}\}+k\mathbb{Z}$.

Therefore all the solutions are given by $(x,y,z)=(\pm{\pi\over 4}+k\pi,\pm{\pi\over 4}+l\pi,\pm{\pi\over 4}+m\pi)$ and $(x,y,z)=(\pm{\pi\over 6}+k\pi,\pm{\pi\over 6}+l\pi,\pm{\pi\over 6}+m\pi)$, where we can take any combination of $\pm$ signs, and $k,l,m$ are non-codependent integers.


Ok, we got this for extra credit today... its tricky here it is:
How many times in a 24 hour time span will the minute hand and the hour hand of an analog clock be collinear?????

I don't think it matters when it starts... it's a 24 hour span, so every possible time will appear twice.  That makes solving it easier.  As does remembering that wherever the hour/minute hand is, there are two place for the other hand to be to make them collinear.

wouldn't it just be [hide]sorry, but you need to do your own homework, otherwise you wouldn't get the full benefit of struggling on the problem and learning when you get the answer[/hide]

People, let this guy do his own homework.  That's not what this website is here for.

i remember a similar problem on the FMO two years ago...
but indeed: he should do his own homework  :P

Did anyone here who said "let him do his homework" first look at the marked gender for Abbs1307? :lol: Or is it just this monitor which is incorrect? :?  Or my vision is going...
Well think about this...when will the hands be collinear?  Is it a 12 hour or 24 hour clock?

[quote="seamusoboyle"]It will hit 12 three times if it starts on it.[/quote]
No, I think if it is only 24 hours, you wouldn't count the third time... because that would start another day.  And if that was not the case, then no matter what time you start on, you'd hit it three times.


The house numbers of the seven houses on Avni's side of the street are consecutive even numbers. The sum of those house numbers is 4522. Avni's house number is the median of the house numbers on her side of the street. What is Avni's house number?

just divide 4522 by 7, because she is the median of the group, resulting in [b]646[/b] as the answer.

This is a longer solution, but you could also do:
7x+42=4522 since the house numbers are consecutive even numbers. Then 7x = 4480 and so x=640. Therefore the first house number is 640, and since Avni's house is the median of the house numbers, we would need to add 6 to 640 to get 646.


a wooden block of mass m is kept immersed in a vessel containing water.a thread is tied to the block whose other end is fixed to the bottom of the vessel.the system is made to fall freely under gravity.find the tension in the thread?

Draw a free-body diagram and remember to keep track of all the forces-there is gravity, the tension, and the buoyant force, all acting on the block.


search
number theory
superior algebra
superior algebra solved

I am in the middle of a problem and can't seem to get the last part.  Any help would be appreciated.

For the equation x2 - 2y2 =  :pm:1, show that the list of integer solutions is infinite, find the pattern, and show the list of solutions is complete.

To show the list of solutions is infinite, I have used the idea that all integers that solve x2 - 2y2 =  :pm:1 is the same thing as all units in Z[:sqrt:2].  The first unit is 1+0 :sqrt:2, the second unit is 1+ :sqrt: 2, the third unit is 3+2 :sqrt: 2, etc. So any unit can be multipled by a preceeding unit to yield another unit. So the list is infinte.

The pattern is y[size=75]n+1[/size] [size=100]=[/size] x[size=75]n[/size] [size=100]+[/size] y[size=75]n[/size] and x[size=75]n+1[/size] [size=100]=[/size] x[size=75]n[/size] [size=100]+[/size] 2y[size=75]n[/size].

But how do I show this list is complete?

google search for pell equation will help you

where does $-1+\sqrt 2$ appear in your list?

Yes amfulger it is a standard pellian equation.
You can also see any standard number theory book.I prefer Burton or Niven & Zukerman. Its dealt in greater details there..
Good luck...


p is a prime
prove :
there exists n (n is a positive integer)p l n*n-n+3
iff
there exists m (m is a positive integer)p l m*m-m+25

if n^2 - n = -3 then (2n-1)^2 = -11. so the first condition is equivalent
to : -11 is a perfect residue.
second is equivalent to:-99 is a perfect residue.

suppose that -99 is perfect but -11 is not. then since -99=-11*9 , 9 is also not perfect. contradiction.

generalization:

(3,25) could be replaced with (n,m) were 4n-1 = x^2(4m-1)
-a

i used this lemma which is pretty simple and well-known:

a is a perfect residue iff a^(p-1)/2 is 1 mod p.


National Target Round #8:

A collection S of integers is defined by the following three rules: (I) 2 is in S ;(II) for every x in S, 3x and x+7 are also in S; (III) no integers except those defined by rules (I) and (II) are in S. What is the smallest integer greater than 2004 which is NOT in S?

Does anyone know how to do this w/o guess&check? (only way I know how to do this)

[hide]I got 2006. The way I did this is that if for every $x$ there is $x + 7$, and we can start with the number 6 which is $ 3 * 2 $, we get 6 + 7, which also has another 7 added to it because it is an integer in S. So basically, we have an equation $ 6 + 7x > 2004$ which simplifies to $ x > 286 \frac 37 $ making the least integer value that works greater than 2004, 2006. [/hide]

Sorry, the answers neither 2005 nor 2006. tarquin, I think u misread the problem and forgot that it wuz asking for the least integer NOT in set S.

Also, the problem's much harder than just multiplying 2 by 3 and continuously adding seven. You could multiply by 2 three times, then add 7 four times, then multiply twice, then add six times...well, u get my point.

I was trying to use mod to do this problem, but it didn't work out to well, since there's no number to mod out.

The only way I know how to do this is guess and check.

Well, the most numbers you will have to guess and check is 6... and I don't see anything wrong with that. However, I have a method that isn't guess and checking.

[hide]First, you find out that 2004 is 2 mod 7. Then you find what mods of 7 you can make using 2*3n, and making n less than 7 because 2*3^7 is bigger than 2004. 2*30=2 (mod 7), 2*31=6 (mod 7), 2*32=4 (mod 7), 2*33=5 (mod 7) then when n=4, it will be 1, when n=5, it will be 3, and when n=6, it will be 2 (mod 7). The only mod 7 that isn't included is 0 (mod 7), therefore the number must be divisible by 7, and the smallest one less than 2004 that is divisible by 7 is [b]2009[/b].[/hide]

Didn't Monkey day he tried to do it with mods?  Well, maybe he'll give the answer now.

Yeah, bob123, the answer is 2009. Good job.

You have to [hide]realize that no numbers in S can be divisible by 7[/hide].  I almost missed these critical 2 points last year :(.


(Alternative Question $3$) Let $a,b,c > 0$ and $ab+bc+ac = 1$. Prove the following general result holds true for any positive real numbers $t_{1},t_{2},t_{3}$:
$3abc(t_{1}+t_{2}+t_{3}) \leq \frac{2}{3}+a t_{1}^{3}+b t_{2}^{3}+c t_{3}^{3}$.

The ineq is equivalent to $\sum_{cyc}\frac{t_{1}^{3}}{bc}+\frac{2}{3abc}\geq \sum_{i=1}^{3}t_{i}$.Now from Holder  $\sum_{cyc}\frac{t_{1}^{3}}{bc}\geq \frac{(\sum_{i=1}^{3}t_{i})^{3}}{3\sum ab}$ and also $abc\leq \frac{1}{3\sqrt{3}}$ so we have to prove $2\sqrt{3}+\frac{(\sum_{i=1}^{3}t_{i})^{3}}{3}\geq 3\sum_{i=1}^{3}t_{i}$ which is a plain AM-GM  :D  :lol:


Can anyone find an appromaxiate using taylor expansion to estimate the following function.
b and w are constant.

Aren't Taylor expansions calculus?  If so, this is in the wrong forum...

Yeah, try "Calculus Computations and Tutorials" :)


Define $(b_{n})$ to be: $b_{0}=12$, $b_{1}=\frac{\sqrt{3}}{2}$ adn $b_{n+1}+b_{n-1}=b_{n}\cdot\sqrt{3}$. Find $b_{0}+b_{1}+\dots+b_{2007}$.

Note. Maybe this seems too easy, but I want to post all the problems...

[hide="Yes, it is easy"]Sequence is periodic [/hide]  :wink:


Hi everyone,

the success of Canadians at various math competitions is  inspiring, and of course is what many of us would like to emulate. It would nice if some of the more experienced Canadians began here a thread of advice - what you wish you knew when you were younger.

Along the lines of book recommendations, study methods, relaxtion tips, or
particular topics you found helpful for certain contests. Whatever has proved helpful to your progress. Many of us aren't fortunate to have a coach, or mentor who has the experience to guide. I hope this thread can act as something of subsitute.  :lol:

Thanks

Write contests.  Do as many as you can.  Get invited to camps as early as you can.  I didn't even do the Open until grade 11 because my school didn't offer it, and even then, I had to beg them to order one for me.  It's kinda disappointing.

Hoala!  I'm not among the elite of Canada, but have fun.  It can't be all work and no play.

And personally, experience has taught me that if your too worked up about it then you won't do as well as when you're relaxed or don't care.  Well, not caring is extreme, but you get the point.

From my petite connaissance (little knowledge)  :P , I believe that you have to actually work hard to achieve success. Do not procrastinate and finish your other devoirs (homework) first, which is obvious of course. Other than that, it's mainly practice, practice, practice... and not what my guitar teacher jokes: sleep, sleep, sleep!  :rotfl:

Homework is overrated.  The mathematics that I learned in high school outside of the curriculum has paid off immensely now that I am in university, whereas learning, say, French has had no benefit to me and will continue to have no benefit for me unless I ever choose to run for public office in Canada.

[quote="Sunny"]Hoala!  I'm not among the elite of Canada.[/quote]
Yes, you are. You were invited to winter camp last year, where only 15 top students get to go.

[quote="lightrhee"]Yes, you are. You were invited to winter camp last year, where only 15 top students get to go.[/quote]

Well, I've dropped out essentially.  I've stopped working on it


[quote="ehku"][color=blue][b]Problem. [/b] For three positive real numbers $ a, b, c$, prove that
\[ \frac {1}{(a \minus{} b)^2} \plus{} \frac {1}{(b \minus{} c)^2} \plus{} \frac {1}{(c \minus{} a)^2} \ge \frac {11 \plus{} 5\sqrt {5}}{2(a^2 \plus{} b^2 \plus{} c^2)}
\]
When does the equality hold ?[/color][/quote]
Since the inequality is symmetric we can assume that $ a \ge b \ge c$. If we fix $ a\minus{}b,b\minus{}c,c\minus{}a$, then the maximum of $ RHS$ is when $ c\equal{}0$.
So, $ x\equal{}a\minus{}b, y\equal{}b\minus{}c$ then the inequality is $ \frac{1}{x^{2}}\plus{}\frac{1}{y^{2}}\plus{}\frac{1}{(x\plus{}y)^{2}} \ge \frac{11\plus{}5\sqrt{5}}{2((x\plus{}y)^{2}\plus{}y^{2})}$.
Multiply $ x^{2}$ to this inequality and $ p\equal{}\frac{y}{x}$ then, the inequality is $ 1\plus{}\frac{1}{p^{2}}\plus{}\frac{1}{(1\plus{}p)^{2}} \ge \frac{11\plus{}5 \sqrt{5}}{2((1\plus{}p)^{2}\plus{}p^{2})}$ and when we multiply $ 2(p^{2}\plus{}(p\plus{}1)^{2})$ to this inequality, it is equivalent to $ \frac{4}{p(p\plus{}1)}\plus{}\frac{2(2p^{2}\plus{}2p\plus{}1)}{p^{2}(p\plus{}1)^{2}}\plus{}4p(p\plus{}1)\ge 1\plus{}\sqrt{5}$ . $ x\equal{}p(p\plus{}1)$ then $ \frac{8}{x}\plus{}\frac{2}{x^{2}}\plus{}4x\ge 1\plus{}5 \sqrt{5}$. If we differentiate $ \frac{8}{x}\plus{}\frac{2}{x^{2}}\plus{}4x$ to evaluate minimum, $ x^{3}\minus{}2x\plus{}1\equal{}0$ is the condition which is equal to $ (x^{2}\minus{}x\minus{}1)(x\plus{}1)\equal{}0$. Since $ x>0$, $ x^{2}\equal{}x\plus{}1$ and $ x\equal{}\frac{1\plus{}\sqrt{5}}{2}$. If we evaluate the value, minimum is when $ x\equal{}\frac{1\plus{}\sqrt{5}}{2}$.

The condition of equality is $ c\equal{}0$, and $ \frac{b}{a\minus{}b}(\frac{b}{a\minus{}b}\plus{}1)\equal{}\frac{ab}{(a\minus{}b)^{2}}\equal{}\frac{1\plus{}\sqrt{5}}{2}\iff a^{2}\minus{}2ab\plus{}b^{2}\equal{}\frac{\sqrt{5}\minus{}1}{2}ab \iff a^{2}\minus{}\frac{\sqrt{5}\plus{}3}{2}ab\plus{}b^{2}\equal{}0 \iff \frac{a}{b}$ or $ \frac{b}{a}\equal{} \frac{\frac{\sqrt{5}\plus{}3}{2}\plus{}\sqrt{\frac{3\sqrt{5}\minus{}1}{2}}}{2}$

You are really excellent[i] Msceok[/i]
      Very , very, very  nice solution
         Congratulation :lol:
  A beautiful problem and a extraordinary solution


function
trigonometry
search
real analysis
real analysis unsolved

if function $f$ and $g$ have the intermediate value property
can we prove that:$f+g$,$fg$,$f(g(x))$ also have the intermediate value property.

My meaning by intermediate value property is that:
if function $f$ has the median charater,then we have:
if $f(x_1)=a,f(x_2)=b$
then for $\any c \in (a,b)$ ,
there exist a $x \in (x_1,x_2)$,satisfy $f(x)=c$

It is called [i]intermediate value property[/i] or [i]Darboux property[/i] in English and the answer is "No" for the sum and the product and "Yes" for the composition.

thanks,and I have edited it.
And it is easy to prove for the composition.(I have done :) )
but can you give me the countexample for the sum and the product?
thanks

Let $h(x)=\frac1x\sin\frac1x$, $x>0$. Put $f(x)=\left\{\begin{aligned} h(-x),&\quad x<0; \\ 0,&\quad x=0; \\ h(x),&\quad x>0; \end{aligned}\right.$ and $g(x)=\left\{\begin{aligned} -h(-x),&\quad x<0; \\ 0,&\quad x=0; \\ 1-h(x),&\quad x>0. \end{aligned}\right.$
This gives an example for the sum. Can you figure out how to do the product case now?

thanks very much(and It is my honour to thank you ;) ),for product by your hint
$h(x)=|\frac 1 x \sin \frac 1 x|+1$ and
$f(x)=\{\begin{aligned} 1,&\quad x \le 0; \\ h(x),&\quad x>0,x; \end{aligned}$ and
$g(x)=\{\begin{aligned} 1,&\quad x \le 0; \\ \frac 2 {h(x)},&\quad x>0; \end{aligned}$

is this right?

and another problem:if $g$ has the stronger condtion that $g$ is continous,can we prove that $f+g$ has the intermediate value property?

[quote="zhaobin"] and another problem:if $g$ has the stronger condtion that $g$ is continous,can we prove that $f+g$ has the intermediate value property?[/quote]
No, even in this case we have a counterexample. Take $g(x)=x$. Now, take any function $h$ such that $h(I)=\mathbb R$ for every open interval $I$ (the problem of existence of such function has been discussed on this forum several times). Now take all points $x$ for which $h(x)=-x$ and change the value of $h$ at them to anything else. You'll get a function $f$ that also satisfies the condition  $f(I)=\mathbb R$ for every open interval $I$ (indeed, for every value $c$, the function $h$ attained that value infinitely many times on $I$ and only once, at $-c$, this value was changed when constructing $f$). Thus, $f$ has the intermediate value property. On the other hand, $f(x)+x$ is never $0$ but it takes arbitrarily large positive and negative values on every open interval, so it does not have the intermediate value property.

well,I don't know two facts
(1)can you show me a function $h$ such that $h(I)=R$ in any open  interval $I$?(I don't know how to search with such things,sorry)
(2)can you tell me that $f$ is also such that condtion(sorry,I'm poor in it)

 :( hehe,It take me two hours to prove that thing,

thanks  :)


abstract algebra
superior algebra
superior algebra unsolved

$ V\equal{}R^n$. The Clifford algebra $ Cl(V)$, also denoted $ Cl(n)$, is the quotient of the tensor algebra
$ \bigoplus _ {k>\equal{}0} V \otimes ... \otimes V$ by the two sided ideal generated by all the elements of the form $ v \otimes v \plus{}\|v\|^2$ for $ v \in V$.

Then how can I show that the multiplication rule for Clifford algebra $ Cl(V)$ is
$ vw\plus{}wv\equal{}\minus{}2<v,w>$?

Just use the definitions and everything's clear:
$ v^2\plus{}vw\plus{}wv\plus{}w^2\equal{}(v\plus{}w)^2\equal{}\minus{}\|v\plus{}w\|^2\equal{}\minus{}(\|v\|^2\plus{}2<v,w>\plus{}\|w\|^2)\equal{}v^2\minus{}2<v,w>\plus{}w^2 \Rightarrow vw\plus{}wv\equal{}\minus{}2<v,w>$.

[quote="Gockel"]Just use the definitions and everything's clear:
$ v^2 \plus{} vw \plus{} wv \plus{} w^2 \equal{} (v \plus{} w)^2 \equal{} \minus{} \|v \plus{} w\|^2 \equal{} \minus{} (\|v\|^2 \plus{} 2 < v,w > \plus{} \|w\|^2) \equal{} v^2 \minus{} 2 < v,w > \plus{} w^2 \Rightarrow vw \plus{} wv \equal{} \minus{} 2 < v,w >$.[/quote]

why $ (v \plus{} w)^2 \equal{} \minus{} \|v \plus{} w\|^2$ ?

why $ \minus{} (\|v\|^2 \plus{} 2 < v,w > \plus{} \|w\|^2) \equal{} v^2 \minus{} 2 < v,w > \plus{} w^2$?

because you mod out $ v^2 \plus{} \parallel{}v\parallel{}^2$


[b]a)[/b] Find four integers a, b, c, d such that the numbers $a^2+b^2$, $a^2+b^2+c^2$ and $a^2+b^2+c^2+d^2$ are perfect squares.

[b]b)[/b] Does there exist an infinite sequence whose members are perfect squares, and which has the property that for any natural n, the sum of the first n members of the sequence is a perfect square?

[b]PS.[/b] The original proposed solution (is Russian) can be found at

http://kvant.mccme.ru/1982/01/resheniya_zadachnika_kvanta_ma.htm

  Darij

a) An example would be $(a,b,c,d)=(30,16,288,816)$.

b) I think such a sequence exists: assume we have defined $a_i,\ 1\le i\le n$ s.t. $\sum_{i=1}^ka_i^2$ is a perfect square for all $1\le k\le n$. We want to define $a_{n+1}$.

Let $u=\sqrt{\sum_{i=1}^na_i^2}$. If $u$ can be written as $2pq$, then just take $a_{n+1}=p^2-q^2$. On the other hand, if $u$ cannot be written as $2pq$, this means that it's odd, so surely it can be written as $(p-q)(p+q)=p^2-q^2$, and then take $a_{n+1}=2pq$.

It's easy to see that in both cases we can select $p\ne q$. This is important because I assume we're not allowed to have sequences which become $0$ from some point on, because then the answer would be trivially "yes", since we can take a sequence formed only by $0$'s.

[quote="grobber"]This is important because I assume we're not allowed to have sequences which become $0$ from some point on, because then the answer would be trivially "yes", since we can take a sequence formed only by $0$'s.[/quote]

Indeed. The proposer of the problem seems to have forgotten this restriction...

Nice solution, by the way. It is very similar to the proposed solution, with the difference that the proposed solution (which is due to V. Sierpinski) doesn't have to distinguish two cases:

We will construct a sequence $\left( a_{i}\right) $ of natural numbers such that the for any n, the sum of the squares of the first n members of this sequence is an [i]odd[/i] square. We begin with defining $a_1=3$ and $a_2=4$. Now, if the first n members are already found, then we define the (n + 1)-th member as follows: We have $a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}=\left( 2k+1\right) ^{2}$ for some natural k (this is because the sum of the squares of the first n members of the sequence is an odd square); now, define $a_{n+1}=2k^{2}+2k$. Then, since $\left( 2k+1\right) ^{2}+\left( 2k^{2}+2k\right) ^{2}=\left( 2k^{2}+2k+1\right) ^{2}$, we have $a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}+a_{n+1}^{2}=\left( 2k^{2}+2k+1\right) ^{2}$, what is an odd square again. Hence we can construct our sequence $\left( a_{i}\right) $.

The first members of this sequence are 3, 4, 12, 84, 3612. In fact,

$3^2+4^2=5^2$;
$3^2+4^2+12^2=13^2$;
$3^2+4^2+12^2+84^2=85^2$;
$3^2+4^2+12^2+84^2+3612^2=3613^2$.

  Darij

It's quite funny because I once thought about [i]exactly[/i] the same problem when I went to a supermarket with my mother and my younger brother.  (My brother's also good at math so I was posing him some easier olympiad questions, then I got to thinking about this.)  It so happens that my solution is exactly the same as the proposed one.  I guess I never expected it to come out in Kvant (even 20 years ago) because it's really quite trivial anyway...


probability
counting
distinguishability
geometry
geometric transformation
rotation

7.
if n and m are whole numbers,39<n<81, and 19<m<41, whwats is the greatest value for n/m +m/n? express your answer as a mixed number.
i can get 4+1/4 but i'm not sure if it is the largest.

7(target). a driver approaches a toll booth and randomly selects two coins from his pocket. if the pocket contains 2 quarters, 2 dimes, and 2 nickels, what is the probability that the value of the two coings he selects will be at least enough to pay the 30-cent toll? express your answer as a common fraction.

7.(1992 lolz 7 again) in how many distinguishable ways can four indentical red chips and two indentical white chips be arranged in a circle?

2(target).to the nearest second, what is the first time between 4:00 and 5:00 that the minute and hour hands of a clock are perpendicular to each other? exress you answer in the form hr:min:sec.

i'm fairly sure that your answer for target 2 is incorrect. i remember that the answer was 4:05:something.

oops, it said perpendicular, I'll edit it.

[quote="Ihatepie"]7.basically try to find the greatest possible value for n/m since n is greater than m, so it is 4+1/4 which is your answer.

7.(target) so if he has 6 coins, there are 6C2 ways to choose them or 15.

so the ways it will be enough is QD (2*2=4 ways to do this) QQ, one way, or QN (2*2=4 ways to do this) 4+4+1=9

so 9/15 simplifies to 3/5.

7.(1992 lolz 7 again)Find ways to place the two white chips. They can be next to each other, have one red chip between them, or opposite. So 3. Rotations don't count I'm assuming.

2.(target)
so at 4:00 the degrees are the hour hand is 120 degrees ahead of the minute. The hour hand goes forward .5 degrees per minute and the minute 6, so it gains 5.5 degrees a minute. so 30/5.5 is 5 5/11 5/11=x/60
x is approximately 33, so it is

4:05:27[/quote]

kk, great. those were the answer i was getting too(for the most part.) i got the same answer as you did when i redid the clock problem....thanks for your help!!


function
inequalities
inequalities proposed

Let $ 1>a_1>a_2>\dots>a_{2k\minus{}1}>a_{2k}>0$ be real numbers.
Find the minimum value of the following expression:

$ \frac{1}{\sqrt{a_1\minus{}a_2}}\plus{}\frac{1}{\sqrt{a_3\minus{}a_4}}\plus{}\dots\plus{}\frac{1}{\sqrt{a_{2k\minus{}1}\minus{}a_{2k}}}$.

[quote="Erken"]Let $ 1 > a_1 > a_2 > \dots > a_{2k \minus{} 1} > a_{2k} > 0$ be real numbers.
Find the minimum value of the following expression:

$ \frac {1}{\sqrt {a_1 \minus{} a_2}} \plus{} \frac {1}{\sqrt {a_3 \minus{} a_4}} \plus{} \dots \plus{} \frac {1}{\sqrt {a_{2k \minus{} 1} \minus{} a_{2k}}}$.[/quote]

It seems that there is no minimum, but only an infimum.

Consider the modified problem where the constraints are
$ 1 \ge a_1 \ge a_2 \ge \cdots \ge a_{2k \minus{} 1} \ge a_{2k} \ge 0$.

Then we are minimizing a continuous function over a compact set, and the minimum is attained.
In the minimum $ a_{2n \plus{} 1} \equal{} a_{2n}$ for all $ n$; otherwise, increasing $ a_{2n \plus{} 1}$ a little
bit while keeping all other variables fixed would decrease the value of the sum.
Furthermore $ a_1 \equal{} 1$ and $ a_{2k} \equal{} 0$.

Then let $ b_n \equal{} a_{2n \minus{} 1} \minus{} a_{2n}$. These $ b_n$ are positive real numbers that add up to 1.
By Jensen, the sum is at least $ k\sqrt {k}$.

Since the minimum is only attained on the boundary of the feasible region of
the original problem, the answer is: The infimum is $ k\sqrt {k}$. There is no minimum.


uhhhh... plug into a calculator $\sin^{-1}{0.8503}$?  unless your calculator doesn't have inverse trigonometric functions...  :maybe:

If you don't have a calc on you, I'll say the answer:
[hide]58.244314038857297458986011662601[/hide]
Out to a TON of decimals.

[quote="jenny221"]Can anyone help with this problem?

Use a calculator to find the acute angle of sin q = 0.8503

q stands for theta

any idea?[/quote]
Uh, why are we doing your math homework?

[quote="jenny221"]Can anyone help with this problem?

Use a calculator to find the acute angle of sin q = 0.8503

q stands for theta

any idea?[/quote]

[hide]just plug in $sin^{-}1 .8503$ into your calculator (usually 2nd SIN) [/hide]


AMC AIME algebra polynomial trigonometry

I'm preparing for AMC/AIME, and the questions that I really fail at are tough AMC/easy-medium AIME algebra problems.  Is there any good book that focuses on this level of algebra?

I already have Intro to Algebra, and I'm pretty much done w/ it.  I also have the Intermediate Algebra book.

I agree with mathemonster.  The Interm Algebra book is what you're looking for.  I hear it's pretty good :)

[quote="soltixnj"]Try 101 Problems in Algebra by Andreescu/Feng, or Polynomials by Barbeau.[/quote]

Isn't this book more for hard AIME (like #12-15) and Olympiad?

[quote="soltixnj"]Try 101 Problems in Algebra by Andreescu/Feng, or Polynomials by Barbeau.[/quote]

+1 for polynomials, excellent book. However, you should get through intermediate algebra first.

[quote="Poincare"]101 problems of algebra is awesome. I really really really really wish I had it. But, its pretty hard to get.[/quote]

Yeah whats with that? Amazon always has 102 comb, 103 trig and 103 numb theory but not 101 algebra.


$ y$ varies inversely as the square of $ x$.  When $ y\equal{}16$, $ x\equal{}1$.  When $ x\equal{}8$, $ y$ equals:

$ \textbf{(A)}\ 2 \qquad
\textbf{(B)}\ 128 \qquad
\textbf{(C)}\ 64 \qquad
\textbf{(D)}\ \frac{1}{4} \qquad
\textbf{(E)}\ 1024$

[quote="AIME15"]$ y$ varies inversely as the square of $ x$.  When $ y \equal{} 16$, $ x \equal{} 1$.  When $ x \equal{} 8$, $ y$ equals:

$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 128 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ \frac {1}{4} \qquad \textbf{(E)}\ 1024$[/quote]
[hide]y varying inversely as the square of x translates into $ y\equal{}\frac{k}{x^{2}}$ where k is a constant. Solving when y=16 and x=1, we have $ k\equal{}16$. When x=8, we have the equation $ y\equal{}\frac{16}{8}\equal{}2$ A[/hide]

[quote="ckck"][quote="AIME15"]$ y$ varies inversely as the square of $ x$.  When $ y \equal{} 16$, $ x \equal{} 1$.  When $ x \equal{} 8$, $ y$ equals:

$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 128 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ \frac {1}{4} \qquad \textbf{(E)}\ 1024$[/quote]
[hide]y varying inversely as the square of x translates into $ y \equal{} \frac {k}{x^{2}}$ where k is a constant. Solving when y=16 and x=1, we have $ k \equal{} 16$. When x=8, we have the equation $ y \equal{} \frac {16}{8} \equal{} 2$ A[/hide][/quote]

You forgot to square the $ 8$ at the last part so the answer is actually $ \boxed{\text{(D)}}$.


geometry
conics
ellipse
geometry unsolved

[hide]$ S_{ABC}\equal{}pr\equal{}h_c.AB$ then we can draw the length of $ h_c$ easily.
$ C$ lies on two lines $ l,l'$ which parallel to $ AB$ and to be distant from $ AB$ the length of $ h_c$.
Since $ CA\plus{}CB\equal{}x\equal{}const$. We construct an ellipe which has two focuses $ A,B$ and the lengh of big axis is $ x$. It intersects $ l,l'$ at $ C$...[/hide]

Based on given data, the segments $ (p \minus{} c \ , p)$ can be constructed. Then the construction of  $ \triangle A'B'C'$ congruent to $ \triangle ABC$ follows. Incircle $ (I')$ touches $ A'C',B'C'$ at $ X,Y.$ The  quadrilateral $ C'XI'Y$ is then constructible. On the rays $C'X$ and $C'Y$ pick the points $ M,N,$ such that $ C'M \equal{} C'N \equal{} p$ $\Longrightarrow$ C'-excircle $ (I_c')$ is tangent to the sidelines $ C'X$ and $ C'Y$ at $ M,N$ $\Longrightarrow$ common internal tangents of $ (I'),(I_c')$ complete the two possible $ \triangle A'B'C'.$ From this we get the desired sides $ CA\equal{}C'A'$ and $ CB\equal{}C'B'.$ Thus  the construction of $ \triangle ABC$ follows straightforwardly.

how is the intersection l' with ellipse construction?

 :(


LaTeX
complex analysis
complex analysis unsolved

Apply the transformation f(z)=wo+t(z)^n to cosz with zo=0. Determine t(z) explicitly

this is the question, no specific transformation is mentioned.

What transformation are you speaking of¿
Can you please use $\LaTeX$¿


Simplify $\sin (x-y) \cos y + \cos (x-y) \sin y$.

$ \textbf{(A)}\ 1\qquad\textbf{(B)}\ \sin x\qquad\textbf{(C)}\ \cos x\qquad\textbf{(D)}\ \sin x \cos 2y\qquad\textbf{(E)}\ \cos x \cos 2y $

[hide="Answer"]$\sin (x-y)=\sin x\cos y-\sin y\cos x$, $\cos (x-y)=\cos x\cos y+\sin x\sin y$
$\sin (x-y)\cos y=\sin x\cos ^2y-\sin y\cos x\cos y=[1]$, $\cos (x-y)\sin y=\cos x\cos y\sin y+\sin x\sin ^2y=[2]$
$[1]+[2]=\sin x(\cos ^2y+\sin ^2y)=\sin x$[/hide]

Umm... something isn't right...

Oh Yeah, I notice something not right too. My answer is not one of five choices.

[hide]$sin(x-y)cosy + cos(x-y)siny$

$=\frac{1}{2}[sinx+sin(x-2y)] + \frac{1}{2}[sinx-sin(x-2y)]$

$=\frac{1}{2}[sinx+sin(x-2y)+sinx-sin(x-2y)]$

$=\frac{1}{2}(2sinx)$

$=sinx$
[/hide]

[hide]
we know that  $\sin(a+b)=\sin a \cos b + \cos a \sin b$.
then $\sin (x-y) \cos y + \cos (x-y) \sin y$.=$\sin(x-y+y)=\sin x$ [/hide]

Hmm, my answer isn't in the options too...

[hide]$\sin{(x-y)}\cos{y}+\cos{(x-y)}\sin{y}=\sin{((x-y)+y)}=\sin{x}$[/hide]

ya I also get the same thing , unless $x=y$ ....

[hide]Hmmm...what does that remind you of? *cough* sine addition formula *cough*

$\sin{[(x-y)+y]}=\sin{(x-y)}\cos{y}+\cos{(x-y)}\sin {y}$
$\sin{[(x-y)+y]}=\sin{x}$[/hide]
?? that's not one of the answer choices! :huh:

edit: :blush: that's what everyone else got too...

Hehe..  :blush:

You guys had the right answer. I put the wrong letter.  :P

sin(x-y)cos y+cos(x-y)sin x
=sin(x-y+y)=sin x

[quote="JesusFreak197"]Thought so. ;)

chess64, that's a very elegant solution right there. Good job! :)[/quote]

Thanks! :D

[quote="chess64"][quote="Fermat -Euler"][quote="JesusFreak197"]Thought so. ;)

chess64, that's a very elegant solution right there. Good job! :)[/quote]

i think that chess64's solution it's like my solution and prototypex24's solution  :gathering:[/quote]

Yeah it actually is pretty much the same. Maybe my presentation was better... :D[/quote]
i don't see that

Actually, I somehow missed that you two had the same solution. :huh:


Let be given a nonright-angled triangle $ABC$ with its altitudes $AA';BB';CC'$.Let $D;E;F$ be the centers of the escribed circles of the triangles $AB'C';BC'C';CA'B'$ opposite $A;A';B'$ respectively .The escribed circle of triangle $HBC$ opposute $A$ touches the lines $BC;CA: AB$ at $M;N;P$ respectively .Prove that the circumcenter of triangle $DEF$ is the orthocenter of triangle $MNP$

I think the problem is this:

[quote="chien than"]Let be given a nonright-angled triangle $ABC$ with its altitudes $AA';BB';CC'$.Let $D;E;F$ be the centers of the escribed circles of the triangles $AB'C';A'BC';A'B'C$ opposite $A;A';A'$ respectively .The escribed circle of triangle $ABC$ opposute $A$ touches the lines $BC;CA: AB$ at $M;N;P$ respectively .Prove that the circumcenter of triangle $DEF$ is the orthocenter of triangle $MNP$[/quote]

This problem won't be that hard if you know how to break it down.

[hide="Observations"]

Here are a few observations. Let's temporarily ignore $MNP$ and focus mainly on triangle $DEF$. Clearly if you call $I_{a}$ the $A$-excenter of $ABC$, you will have $I_{a}, B, E$ and $I_{a}, C, F$ are collinear triples. Now it seems that $D$ is the orthocenter of $I_{a}EF$ (I used computer to sketch, so it is much more accurate and easier. However, since there are many right angles in this problem here, the chance for this to be true isn't small and you may predict it after a while).

That helps a lot. Consider $MNP$ now. Notice that if our prediction is true, then the orthocenter of $DEF$ will be $I_{a}$ which is the circumcenter of $MNP$. Now, if the problem is right, then we will have an interesting reverse property: the orthocenter of $MNP$ is the circumcenter of $MNP$. Well, I haven't seen many pairs of triangle that have this property except when they are congruent or homothetic. With this direction, you can predict that these two triangles are congruent and from now on it's easy to do.

[/hide]

So this remains to proving $DEF$ and $MNP$ are equal. This is extremely easy if you notice $B'DC'$ and $B'CF$ are similar which gives you the similarity between $B'C'C$ and $B'DF$. It is easy now to calculate $DF$. From this you can also prove that $DF$ is parallel to the $B$ bisector of $ABC$. Do the same thing to the other side and you are done.


function
real analysis
real analysis unsolved

[quote="The Captain"]Let $X$ and $Y$ be two nonempty sets and $f$ be a bounded real valued function on $X \times Y$.
Prove or disprove:

$\sup_{x}\inf_{y}f(x,y)=\sup_{y}\inf_{x}f(x,y)$ .[/quote]
There are counterexamples with $|X|=|Y|=2$.

Here is a quantitative version of this problem: Let $X = \{1,\dots,m\}$ and $Y=\{1,\dots,n\}$, and let $f: X \times Y \to [0,1]$ be a random function with independent uniformly distributed values at all $(i,j) \in X \times Y$. Find $prob \left( \max_{x}\min_{y}f(x,y)=\max_{y}\min_{x}f(x,y)\right)$.


The Word Problem: A toy manufacturer is introducing two new dolls, My First Baby and My Real Baby. In one hour, the company can produce 8 First Babies or 20 Real Babies. Because of demand, the company produces at least twice as many First Babies as Real Babies. The company spends no more than 48 hours per week making these two dolls. The profit on each First Baby is 3.00 and the profit on each Real Baby is 7.50.
Find the number and type of dolls that should be produced to maximize profit.

First Step from the Answers Section in the back of the book:

Define the Variables: x= number of first babies y= number of real babies

Inequalities from the word problem: x ≥ 0, y ≥ 0, x ≥ 2y, 5x+2y≤ 1920

My Questions: Where did they get 5x+2y≤ 1920 from? How did they get it? Especially how did they get 1920 from that?

[quote]In one hour, the company can produce 8 First Babies or 20 Real Babies.[/quote]

That strange inequality comes from the above constraint.

Since the company can produce 8 First Babies in one hour, it can produce x First babies in $x/8$ hours. You can see this by using ratios.
8 babies : 1 hour :: x babies : t hours, implies that $t = x/8$ hours.

Likewise, if the company can produce 20 Real Babies in one hour, it can produce y Real Babies in $y/20$ hours.

So looking at the total time to make babies verses the total time available, you have $x/8+y/20 \leq 48$

Multiplying through by 40 gives $5x+2y \leq 1920$

By the way, I take it that you're in a Finite Math class. Where are you taking this course and what book are you using for it?

[quote="gauss202"]40 is the Least Common Multiple of 8 and 20. So its the smallest value you can multiply the inequality by to get rid of the fractions.[/quote]
It is not necessary to do that really.

From the fact that the company produces twice as many $x$ as it produces $y,$ we have:

$x-2y = 0$

Therefore,

\begin{eqnarray}\frac{x}{8}+\frac{y}{20}&\le & 48\\ x-2y &=& 0\end{eqnarray}

Substituting (2) into (1):

$\frac{y}{4}+\frac{y}{20}\le 48$

Multiplying both sides by 20:

$5y+y \le 960\\ \implies 6y \le 960 \\ \implies y \le 160$

$\therefore x \le 320$

$p = \$ 3.00 x+\$ 7.50 y \le \$ 3.00\left(320\right)+\$ 7.50\left(160\right) = \$ 2160$


Approximately $ 70\%$ of the earth's surface is covered with water.  Twenty-five percent of the remaining surface is green space.  What percent of the earth's surface is not covered with water and is not green space? Express your answer to the nearest tenth.

$ \frac{7}{10}\plus{}\frac{3}{40}\equal{}\frac{31}{40}$ of the earth is green space or water.

$ \frac{9}{40}\equal{}\frac{2.25}{10}\equal{}22.5 \%$.


vector
calculus
geometry
function
derivative
analytic geometry
complex numbers

Alright, so I am not quite sure of where to post this question, but since it's grade twelve vectors, I figure that here will be all good. I love math and everything, but vectors are seriously beginning to get on my nerves. We're currently learning how to find the distance from a point to a plane, which is all good, I can usually do it and all, but I much prefer to do math with a purpose. I cannot fathom why anyone would ever need to be able to write a single vector in terms of two others. If anyone could give a couple of reasons why vectors are...erm...useful, I would be eternally greatful. Even if it is some obscure, engineering-ish use that I likely wouldn't understand anyway, it would just help me to know that this is not completely without purpose (other than amusement, and impressing friends at parties, of course). I mean, even calculus has a purpose, if you ever need to know the rate that an inverted conical silo is filling up with grain...which you never really would. But if you ever needed to know...
I'm beginning to ramble (or maybe I've just been rambling all along) but this would really help me with my math. Thanks so much, everyone!
xox

Here's an analogy: Say you know basic complex numbers. You are given a problem: $ z\equal{}1\plus{}i$. Find $ z^{9}$. Obviously, you're not going to do out $ (1\plus{}i)^9$ - that would take forever. So you use polar form - $ z\equal{}\sqrt2e^{i\pi/2}$, so $ z^9\equal{}16\sqrt2e^{9i\pi/2}\equal{}16\sqrt2e^{i\pi/2}e^{4i\pi}\equal{}16\plus{}16i$. Putting the problem into different notation makes it bafflingly simple - vectors are just another way of looking at geometry problems. The distance from point to line formula is one example. It's incredibly, incredibly ugly to derive it with standard analytic geometry, but with vectors it's relatively docile.

Introductory courses can often seem useless and boring, especially if there is a lot of tedious computation involved. Often, the applications to such things are not immediately obvious. Don't be discouraged by this, though. Here are some areas where vectors prove useful:

[b]Physics, Chemistry, and Science in General:[/b] In the most fundamental sense, a vector is something with a direction and a magnitude. They came about at first from the study of basic physics (absolutely nothing obscure here) because physics has lots of properties just like that. Force, velocity, acceleration, these all have direction and magnitude. So basically, [u]whenever[/u] you are dealing with something with both [u]magnitude[/u] and [u]direction[/u], you have a vector.

[b]Data and statistics:[/b] Also, vectors are closely related to matrices, as you will find as you study linear transformations. Naturally, matrices are great for representing large sets of data, and thus vectors find their way into statistics as well.

Suppose you have some factor X, which varies with factors Y and Z (use any relevant example). Suppose you find a linear approximation to the data(of the form X=aY+bZ+c), and find that the data is approximated by a plane. Now you want to see how accurately this plane approximates a particular data point. Can you see why you may want to know the distance between a plane and a line.

[b]Functions of many variables:[/b] This brings us to another area of mathematics: multivariable calculus. Suppose you have some function f of two variables: f(x,y). You want to describe how f(x,y) is changing at a particular ordered pair; say (x,y)=(1,1). If it were a one-variable problem, you could be like, oh, easy, derivative! But when we are in many variables, what is the derivative? We no longer have a tangent line, but a tangent plane. One path on the plane might be going up, while another might be going to some side, etc. The answer is, the normal vector (aka. gradient), which is perpendicular to all lines contained within the tangent plane. For this reason, as well as others, the use of vectors is very important when studying functions of many variables. This is important because in real life, most things depend on more than one factor, and we must study how things vary with these many factors.

[b]Another Example:[/b]  I've included an example of a situation where it may help to know the distance between a point and a plane in the statistics section of this post. For your curiosity, I will address your other question - representing one vector in terms of others.

These are basically taking components. You see this all the time. Suppose you want to represent a point in space. How do you do it, usually? You define an origin, an x unit vector (which defines the x axis), a y unit vector (which defines the y axis), and a z unit vector. Then, you give your coordinates: say, (1,2,3). This means that this point is the x unit vector plus twice the y unit vector plus thrice the z unit vector. In other words, whenever you use Cartesian-style coordinates, you are representing a vector in terms of other vectors.

Sometimes it may help to use vectors other than the standard basis. This comes up in physics problems. For example, if a box is on a plane, how fast will it slide down? The weight of the box is a vector. You can break this into two components - one in the direction of the plane incline, and one in the direction perpendicular to the plane incline. You can use the former component to determine how much of the force of gravity is pulling the box down, and the latter to determine the force of friction.

[i]
-- I hope you find this helpful! [/i] :)

Thank you so much to both of you. I am fairly certain that I understand the first anology, though right now I would never be able to do it by myself, as unfortunately I do not know basic complex numbers. But it understand what you are saying and how the equation works, so thank you. Thanks very much, The Zuton Force, for the whole list of applications. You obviously spent quite a bit of time compiling it, and I greatly appreciate it. This does help me quite a bit...I suppose that it was possible that I was simply in need of reassurance. *sheepish grin* I am much better equipped, however to continue with the course...I am actually really excited because we will soon be starting calculus, which is much more enjoyable. I think that my main problem with vectors is that the problems that they give us do not make any sense. For example, the other day I was given point A (6, -5, 3) and was told that a force with the direction vector of (-4, 3, 5) moved the object from point A to point B (-2, 7, 4). Before beginning the problem, which I did know how to do, I drew out the points and the vector. I can imagine it in three space, and I do not understand how the force could possibly move an object from point A to point B. Maybe I just cannot fathom the complexity of three-space, but I asked my teacher and she was not able to alleviate my frustration either. She told me to just drop it and not to try and make sense of it, just to do it. But both of you were very helpful, and I thank you for that.

this will possibly give you the use of vectors.
 I  am also a class 12 student and i also enjoy doing math with a purpose.
 for eg.
  if you have to give a proof to the following prb.

  IF THE MID POINTS OF THE CONSECUTIVE SIDES OF ANY QUAD. ARE CONNECTED BY STRAIGHT LINES ,PROVE THAT THE RESULTING QUAD. IS A PARRALLELOGRAM.

 sol.  let ABCD be the given quad. and let P,Q,R,S be the mid points of the sides AB,BC,CD,DA RESPECTIVELY.IF a,b,c,d represent the vectors AB,BC,CD,DA respectively,
    then PQ=1/2(a+b),QR=1/2(b+c),RS =1/2(c+d) and SP=1/2(a+d).
   but  a+b+c+d=0.
  so, PQ=1/2(a+b)=-1/2(c+d)=SR
   AND  QR=1/2(b+c)=-1/2(d+a)=PS
 THUS OPP. SIDES ARE EQUAL AND PARRALLEL AND HENCE IT IS A PARALLELOGRAM. :roll:

 SO YPU CAN SEE THAT THE SOLUTION IS CONCISE.YOU CAN DO IT BY GEOMETRY ALSO IT MAY BE A LITTLE LONG AND ALSO IT IS BETTER TO HAVE MORE THAN ONE SOLUTION FOR A PROBLEM.

 ALSO IT IS SAID THAT ANY PROBLEM OF TWO DIMENSIONAL GEOMETRY CAN BE SOLVED BY USING CO-ORDINATE GEOMETRY,VECTORSAND COMPLEX NUMBERS SEPARATELY

BUT THE FUN LIES IN SORTING OUT THE SMALLES AND ELEGANT SOLUTION :roll:


inequalities
algebra
polynomial
inequalities proposed

Prove or disprove that for acute triangle with side lengths $a,b,c$

\[\frac1{(1+a^3)(1+b^3)}+\frac1{(1+c^3)(1+b^3)}+\frac1{(1+a^3)(1+c^3)}\leq\frac3{(1+abc)^2}\]

Your ineq... is wrong when a=b= 2/3c and lim(a--->oo)!!!
-_______________________
anh la SM day dao nay em co khoe ko?

[quote="Reyes"]Yes,you're right .But   this inequality is true :
$\sum\frac{1}{(1+a^3)(1+b^3)} \geq \frac{3}{(1+abc)^2}$. :D[/quote]
Are you sure? What will happen if $b=c=1$ and $a \to 0$?  :?

[quote="Ji Chen"]where the constant  $\frac{5-3\sqrt2+\sqrt{75-30\sqrt2}}{4}=1.6161718332405478802988\cdots$ is the best possible.[/quote]

[quote="Ji Chen"]where the best constant  $k_{0}$ is the maximal real root of the following irreducible polynomial:
$243k^{24}+14768k^{21}+88264k^{18}-1803816k^{15}+12124485k^{12}-43197624k^{9}$
$-505971198k^{6}-908764110k^{3}-387420489$.[/quote]

I know that's a little OT but....HOW CAN YOU GET THESE???


look at your book

also range of a [one-to-one] function is the same as the domain of the inverse of the function

[quote="machack"]f(x)= square root of ((3+x)/(2x-1))

Plz help me find the range of this and if possible the inverse and the domain/range of the inverse. (the range of the original function is the most important)[/quote]
Why don't you just graph the function?

$\begin{gathered}y^{2}= \frac{{3+x}}{{2x-1}}(y \geqslant 0) \Rightarrow x = \frac{{y^{2}+3}}{{2y^{2}-1}}\hfill \\ \Rightarrow reng(f) = [0,+\infty ]-\{ \sqrt{\frac{1}{2}}\}\hfill \\ \end{gathered}$


trigonometry
calculus
integration
logarithms
calculus computations

Answer the following question.

[1] Given $y=e^{-x}\sin x$, we have $y''+\boxed{\ }y'+\boxed{\ }y=0$

[2] $\frac{d}{dx}\frac{x+1}{x-1}$

[3] $\frac{d}{dx} x^3\sqrt{1+x^2}\ (x>0)$

[4] $\frac{d}{dx} x^2\sin (3x+5)$

[5] $\frac{d}{dx} 2^{\sin x}$

[quote="Jimmy"]1)

$y''+2y'+2y=0$

Which has a solution of $y=e^{-x}\left(C_1\cos x + C_2 \sin x\right)$.[/quote]

 :)

[quote="fedja"][quote="kunny"]Answer the following question.
[2] $\frac{d}{dx}\frac{x+1}{x-1}$
[/quote]
Hmmm, kunny, don't you think that it is [b]too[/b] simple for this forum? No offense meant and you are welcome to continue to post whatever you find appropriate (I also see that your posts are quite popular and some of your integrals were really good :) ) but I used to fail my business calculus students if they were not able to solve a problem like this... Anyway, this is just a remark. Thanks for your series. Do you plan to extend them to asymptotics and other more advanced topics? This would be really useful (in my humble opinion, of course).[/quote]

Thank you for your message, fedja.

Certainly [2] $\frac{d}{dx}\frac{x+1}{x-1}$ seems to be very easy.Probably almost  students use the differential of quotient.

kunny


Let $ A$ be a finite subset of $ \mathbb N$. $ A_2 \equal{} \{a \plus{} b|a;b \in A;a\neq b\}$.
Suppose that there exists two sets $ X;Y$ such that $ X_2 \equal{} Y_2$ and $ X\neq Y$.
Prove that $ |X| \equal{} |Y| \equal{} 2^k$
Note:$ \mathbb N$ means the set of positive integers!

[quote="mavropnevma"]Classical - elegant proof by Selfridge (and Strauss, I think). This type of problems are better exposed in a mini-lecture than launched as a challenge, in my opinion.[/quote]
Thank you for your opinion mavropnevma; but unluckily it is my homework.I worked very hard but receive nothing till now.
It seems very simple but I think, as mavropnevma, It is very hard!

The statement is clearly incorrect if $A_2$ is not understood as a multiset, e.g. $X=\{1,2,3,4,6,7,8\},$ $Y=\{1,2,3,5,6,7,8\}$ makes a counterexample. When $A_2$ is understood as a multiset it's the best to refer to [url="https://msp.org/pjm/1958/8-4/pjm-v8-n4-p17-s.pdf"]Selfridge and Strauss' paper[/url], but let me also note that [url="https://artofproblemsolving.com/community/c6h364143_china_tst_2010_problem_2"]China TST 2010 P2[/url] exactly reproduced the proof in the paper.


Two isosceles triangles with sidelengths $x,x,a$ and $x,x,b$ ($a \neq b$) have equal areas. Find $x$.

[hide="A little easy for Pre-Olympiad...? "]
For convenience, let $c = \frac{a}{2}, d = \frac{b}{2}$.

$K = \sqrt{x^2 - c^2} c = \sqrt{x^2 - d^2} d$
$c^2 (x^2 - c^2) = d^2 (x^2 - d^2)$

But $c \neq d$.  Therefore

$x^2 - c^2 = d^2$
$x = \boxed{ \frac{\sqrt{a^2 + b^2} }{2} }$.
[/hide]

I guess the problems are easy because it was the first year the olympiad was organized.

[quote=maraujo68]I found the answer: x = sqrt(a^2 + b^2)/2...[/quote]

Congrats on your first post!


How do you study for it on your own? and what resources do you use?

sorry, but...
what is USNACO?

USNCO or...USACO?

two things I did (i've tried twice and gone to camp twice):

1. read Peter Atkins's Chemical Principles.
be able to do all the problems in the book perfectly. basically.

2. do practice tests. repeatedly.
if anything, this is probably the best one can do.


YO GUYS!!!THIS IS MY 101st POST :thumbup:  :w00tb:  :clap2:

I'm making a special request on such an auspicious occasion( :rotfl: ).I am a learning kid who lives in India.Over here,we have limited amount of reference books.We do not have the privilege of learning from the awesome foreign books like the AOPS books and ACOPS.
So,friends i am requesting all of you who can help me and other users of this site to post some questions of your choice from AOPS,ACOPS, or any other sources over here.It will be of great help and you may also find some xtra useful methods to solve the same problem!!Any sum can be given,of any difficulty level(ofcourse of gs level) can be given.Important strategies to do certain type of questions can also be discussed which will be of great help to many users.

Feel free to post any mischellaneous question if u feel like.
Thank you all
 :lol:  :lol:  :lol:  :lol:  :lol:  :lol:  :lol:  :lol:  :lol:  :lol:

[quote="chess64"]Actually, we already post problems from AoPS and many other math books as well all over these forums. ;) Haven't you ever noticed? :)[/quote]

How is he supposed to know without actually having the books?  ;)

Plus, can't you buy the book online anyway?

[quote="jli"]congradulations about your 101th post.

you can go on amazon and i think they are selling the books.

if you go on ebay, they might be selling really cheap ones for like 5 or 6 bucks.[/quote]

I don't think used books are good because they have writing all over it (well, some cheap ones are new.)

(This is interesting: your post, jli was at 1:01. )

I bought Vol 2 for $\$$10.99$ off Half.com and it was very good. It seemed new. :)

I did think about buying books inline but it isnt safe,u know about someone hacking ur card no. and stuff.Anyway,it would still help if someone just posted a few questions frequently from here and there.Because i heard ACOPS AN AOPS books are good.

Can someone tell me a really good site that gives books at a really cheap rate?

can anyone who uses AoPS books tell me about them ? like how many volumes are there and what is the course in the various volumes...... . Whats the difficulty level like and other stuff ya know ........ :D


logarithms
trigonometry
calculus
derivative
function
calculus computations

Let $ S_k = \sum_{n=0}^{\infty} \frac{1}{(3n+k)^3}$. Then differentiating both sides of $ \log \Gamma(x) + \log \Gamma(1-x) = \log \left( \frac{\pi}{\sin \pi x} \right)$ three times, we obtain

$ \psi_2(x) - \psi_2 (1-x) = -2\pi^3 \frac{\cos \pi x}{\sin^3 \pi x}$

where $ \psi_2 (x) = \frac{d^3}{dx^3} \log \Gamma(x) = -2 \sum_{n=0}^{\infty} \frac{1}{(x+n)^3}$. Thus we have

\begin{eqnarray*}
S_1 + S_2 + S_3 & = & \zeta(3) \\
27(S_1 - S_2) & = & \pi^3 \frac{\cos \left( \frac{\pi}{3} \right)}{\sin^3 \left( \frac{\pi}{3} \right)} \\
S_3 & = & \frac{1}{27}\zeta(3)
\end{eqnarray*}

Solving this linear system gives the desired result.

I wonder if it is possible to evaluate the series $ \sum_{k\equal{}0}^{\infty}\frac{1}{(3k\plus{}1)^{2}}$.

It's a derivative of the digamma function, which I believe means "no" if you want to express it in terms of nice constants.

[quote="Timestopper_STG"]I wonder if it is possible to evaluate the series $ \sum_{k \equal{} 0}^{\infty}\frac {1}{(3k \plus{} 1)^{2}}$.[/quote]

But what [i]is[/i] possible to evaluate is the series $ \sum_{k \equal{} \minus{}\infty}^{\infty}\frac {1}{(3k \plus{} 1)^{2}}$


probability
expected value
AMC 12 B
algebra
system of equations
AMC

I was doing the 2003 AMC 12 B and these were the two questions I missed or didnt have a full solution

24.  Positive integers $ a,b,$ and $ c$ are chosen so that $ a<b<c$, and the system of equations

$ 2x+y=2003$ and $ y=|x-a|+|x-b|+|x-c|$has exactly one solution.  What is the minimum value of $ c$?

25.  Three points are chosen randomly and independently on a circle.  What is the probability that all three parawise distances between the points are less than the radius of the circle.

Umm for number 25, I just did this (and got the right answer but Im positive this isnt a correct way or traditional way)
I pick a random point, theres a 1 chance of that
The second point must be 60 degreees left or right of the first point so 120 degrees or 1/3 chance
The third point's probability depends on the second point.  If the second point was on the first point, then there is a $ \frac{1}{3}$ chance for the third point.  If the second point was the other extreme ($ 60$ degrees left or right of the first point), then theres only $ \frac{1}{6}$ chance of third point.  Since I dont know where the second point is going to be exaclty, ill take an expected value of $ \frac{1}{3}+\frac{1}{6}$ (the two extremes) and divide that by $ 2$ to get $ \frac{3}{12}$.  So $ 1*\frac{1}{3}*\frac{3}{12}=\frac{1}{12}$, which is the correct answer, but isnt there a better way to do this with linear programming methods.
And is that method I discribed even correct?

Hope that I am not wrong.

 For #24, by graphing the two equations you can see that the system have one solution if and only if the point (a, b+ c- 2a) lies on the line 2x + y = 2003, which implies b + c = 2003. Therefore the minimum value of c is 1002. The equality holds only when a <1001, b = 1001, c = 1002 . Then x = a and y = 2003 - 2a

 For #25, call the three points A, B, C and the circle O. Let x = ( OA, OB ), and y = ( OA, OC ). Then ( OB, OC ) = y - x. Without loss of generality we can assume that -180 < x, y, < 180. Now the conditions become:
    -60 < x < 60
    -60 < y < 60
    -60<y -x< 60 or y-x>300 or y-x<360
 By demonstrating these conditions on a coordinated plane, you can have that correct answer.

For [b]#24[/b] : The system of two equations has one solution for every value of a, b ,c . The minimum valye of c is 3 because a,b,c are positive integers and c is the biggest of them.

[color=red]edit:[/color] Of course I am wrong. I took only one branch of the second equation. Sorry.
 Second sight : I agree that the minimum value of c is 1002 .

If $ c=3$ then $ a=1$ and $ b=2$. Then the system of equations will have two solutions $ (x,y)$ which are $ (-1997, 5997 )$ and  $ (\frac{2009}{5}, \frac{5997}{5})$. This is contradictory to the condition that the system of equations has [b]exactly[/b] one solution.


calculus
integration
limit
trigonometry
real analysis
function
Support

Let $a$ be real number.
Find the following Limit value.

\[\lim_{n\to\infty} \int^{\frac{1}{n}}_{-\frac{1}{n}} (n-n^2|x|)\cos (a-x)dx\]

Change variable $y=nx$

$A_n=\int_{-1}^{1}(1-|y|)\cos(a-\frac{y}{n})dy=\int_{-1}^{1}f_n(y)dy$

$f_n(y)$ converge simply $(1-|y|)\cos a$

$|f_n(y)|\leq 1-|y|$

Lebesgue dominated convergence

$\lim A_n =\int_{-1}^{1}(1-|y|)\cos a dy$

$=\cos a$, of course.

This is a very standard "approximation to the identity" setup.

Suppose $\varphi(x)$ is an integrable function on $\mathbb{R}$ with integral 1. Let's make our life simple by assuming it has compact support. Define $\varphi_t(x)=\frac1t\varphi\left(\frac xt\right).$

Suppose g is a continuous function on $\mathbb{R}$.

Claim:
$\lim_{t\to0^+}\varphi_t*g(x)=\lim_{t\to0^+}\int_{\mathbb{R}} \varphi_t(x-y)g(y)dy = \lim_{t\to0^+}\int_{\mathbb{R}}g(x-y)\varphi_t(y)dy = g(x).$

Moubinool's argument works here. Use the last form of the convolution integral and make the substitution $y=tu.$ Then

$\int_{\mathbb{R}}g(x-y)\frac1t\varphi\left(\frac yt\right)dy = \int_{\mathbb{R}} g(x-tu)\varphi(u)du.$

Since I assumed $\varphi$ had compact support, this integral is taken over a bounded interval. As $t\to0$, $g(x-tu)$ tends (uniformly on that bounded interval) to $g(x).$ So the integral tends to $\int_{\mathbb{R}}g(x)\varphi(u)du=g(x).$

Of course, in kunny's problem, $\varphi$ is the "tent function": $\varphi(x)=\max(1-|x|,0).$

We can weaken many of these conditions and ask for other kinds of convergence (in norm, almost everywhere, etc., but the place to discuss things like that would be out on the main analysis forum, not in here.


Exactly what it sounds like. No correct answers or anything, just put some random stuff. No offensive sentences or "fills" like usual. When you fill something in, make sure you also place a sentence to be filled.

Example:

[quote="Person 1"]Please do not ____________.[/quote]
[quote="Person 2"]throw any cigarettes in the trash can

(next sentence)[/quote]

Be creative and stuff. Start:

A child: Why can't I ever _________?
Parent: Because you will ____________.

Child: Why can't I ever play?
Parent: Because you will fail your test if you don't.

My fill in:
Child: Mom, can I ________________________?
Mom: ___________________but be sure to________________.

[quote="k00lperson"]My fill in:
Child: Mom, can I have a baby?
Mom: Sure, but be sure to find a girl first.[/quote]

dragon96: I want my ________!
james4l: Well that's just too ________!

dragon96: I want my horoscope!
james4l: Well that's just too many in one week!

We should at least [i]try[/i] to make these humorous.  :no:

Teacher: How many ____________???
Student: Over ______!!!!!

This is pretty obvious.

[quote="1=2"]dragon96: I want my horoscope!
james4l: Well that's just too many in one week!

We should at least [i]try[/i] to make these humorous.  :no:

Teacher: How many ____children________???
Student: Over __1000000000000000000000____!!!!!

This is pretty obvious.[/quote]
mine is : 1=2 says k00lperson do u _______________________------
k00lperson ______________________BLAH!

1=2 says k00lperson do u like pi------
k00lperson: SNUGGLEFISTBLAH!

Minez:

Meewhee has an awesome __________
with a(n) _________,
and a(n) ___________.
Since she's so ______________.

Hehehehe...

1=2 says k00lperson do u like pi------
k00lperson: SNUGGLEFISTBLAH!

Minez:

Meewhee has an awesome _hairband_
with a(n) _zombie___,
and a(n) ___god_____.
Since she's so ___n00b_______.

Hehehehe...

Meewhee has an awesome little hairband that
with a ugly zombie it explodes,
and a n00by god n00bed her.
Since she's so ugly n00by god n00bed her.

NOOOOOOO!!! I'll ________ IT BACK YOU N00B!


Solve in real numbers the equation $(xy-9)^2+(xy+y^2)^2=18y^2$


search
function
geometry
similar triangles

[quote="abacadaea"]a square is inscribed inside triangle ABC such that one side of the square is on the hypotenuse BC. AB=6 and AC=8. find the side length of the square.[/quote]

when you mean hypotenuse, are you implying its a right triangle.

If its a right triangle, you look for two similar triangles.  its kinda hard to explain without a picture.

but the triangle to the left of the square is similar with the bigger one and so is the one to the right of the square so you have 3/4x+4/3x+x=10, then solve

The name of this thread pretty much destroys the point of solving the problem as it tells you which tool to use.  :P  (next time be a little more careful of what you name the thread, good problem though)

[hide]Notice that the other two verticies of the square will lie on $ AC$ and $ AB$ so if we let the side of the square be $ x$ and the square $ DEFG$ has verticies $ D$ and $ E$ on $ BC$ and $ F$ on AB and $ G$ on AC then by similar triangles (since they all have the same angles) DB=4x/3 and EC=3x/4 so adding them gives us 37x/12=10 so the side lenght of the square is just  120/37. [/hide]

note the info. it says help.

the future-yes it is a right triangle i forgot to say that.

sonny-i know the answer, im just confused on the procedure, all i know is that we need similar triangles.

[quote="the future"][quote="sonny"]The name of this thread pretty much destroys the point of solving the problem as it tells you which tool to use.  :P  (next time be a little more careful of what you name the thread, good problem though)

[hide]Notice that the other two verticies of the square will lie on $ AC$ and $ AB$ so if we let the side of the square be $ x$ and the square $ DEFG$ has verticies $ D$ and $ E$ on $ BC$ and $ F$ on AB and $ G$ on AC then by similar triangles (since they all have the same angles) DB=4x/3 and EC=3x/4 so adding them gives us 25x/12=10 so the side lenght of the square is just  24/5. [/hide][/quote]

Thats not right, its not just 4x/3+3x/4, you forgot the x of the square so the right answer should be 37x=120, x=120/37[/quote]

Ya you are correct, thanks for spotting that. EDITED!

abacadea-Ok well I hope you understand how you do it now.

Is it just me, or is this the same problem as #30 on the state sprint this year?

No it's not just you, it is #30 Sprint on this years state

yeah, i didnt like their solution. i couldnt understand it. so i asked u guys

With these recent contest problems, they are very likely to come up in the forum before you decide to post; please use the search function first.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=141801[/url]


trigonometry
geometry
perimeter
Science Olympiad

im trying to figure out that if u are given a certain perimeter what are the two regular polygons taht u can construct (not literally) that will give u the most and the least area
i figure i could use this info in a building competition in science olympiad (the middleschool one) although it turns out the the uses of this are more abstract then i thought  :( but if someone could just sove the maximum and minimum value of the following equation for me id be grateful :D
(cos(180/N))(N)(A^2)
4(sin(180/N))
this is supposed to be a fraction
A is the given peremiter so i dont know if its needed in the equation and N is the number of sides the formula is how to calculate the area of a polygon with the given peremiter and number of sides

Your question doesn't make much sense as stated.  We can't actually achieve a maximum because we can increase the number of sides indefinitely.  This gives us a sequence of polygons with larger and larger areas that approach the area of the circle with a given perimeter, which is the actual maximum.  On the other hand, conversely the minimum is achieved for an equilateral triangle.

doesnt a square have less area then an equilateral triangle with the same perimeter?

oh ur rite i got my thoughts confused with somehting else
the square has more area  :rotfl:  :blush: that was my thought when i first started this project

[quote="peregrinefalcon88"]$ \frac{NS^2}{4\tan{\frac{180}{N}}}$[/quote]
And that is the formula of a regular polygon with a side length S and N sides. Is there a coincidence that this is similar to the thing you had???

no i actually calulated that cause i was trying to create an equation


[quote]Let a,b,c >0 such at abc=1 . Prove that :
\[ \frac {1}{a \plus{} b^2 \plus{} c^3} \plus{} \frac {1}{b \plus{} c^2 \plus{} a^3} \plus{} \frac {1}{c \plus{} a^2 \plus{} b^3} \le 1\]
[/quote]

Use the CS ineq : $ (a \plus{} b^2 \plus{} c^3)(a \plus{} 1 \plus{} \frac {1}{c}) \ge (a \plus{} b \plus{} c)^2$

We need to prove that : $ \frac {\sum a \plus{} \sum \frac {1}{a} \plus{} 3}{(a \plus{} b \plus{} c)^2} \le 1$

$ \Longleftrightarrow \frac {\sum a \plus{} \sum ab \plus{} 3}{(a \plus{} b \plus{} c)^2} \le 1$

$ \Longleftrightarrow \frac {2}{3}.(a \plus{} b \plus{} c \minus{} 3)\left(a \plus{} b \plus{} c \plus{} \frac {3}{2}\right) \plus{} \frac {1}{3}.(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca) \ge 0$

@ you said that "very easy inequality" and posted it at box Unsolved...what a fun


limit
inequalities
calculus
calculus computations

1)
Let the sequence $a_n,b_n>0$ such that for $n\geq 1\in\mathbb{N}$ ,
$a_n=\frac{1}{2}(a_{n-1}+b_{n-1})$
$b_n=\sqrt{a_{n-1}b_{n-1}}$

Prove that $\lim_{n\to\infty}b_n$ exist .

2)
Let $a_n=\sum^n_{k=1}\frac{1}{n+k}$ , $n\in\mathbb{N}$. Prove that $a_n\uparrow$ has upper boundary

Can someone help me with these two question ?
By the way , what is $a_n\uparrow$ and $a_n\downarrow$ ? I saw this symbols but dont know the meaning ?

1)the difference between a and b becomes less than half in each step. so limit of difference is zero. AM-GM inequality gives b<a for all n.GM> smaller no. shows that b is an increasing series. similarly a is a decreasing series.b<a implies that the increasing b series is bounded above and a series is bounded below. an increasing series bounded above has to converge.so, limit of b exists.

2)each term of the summation is less than 1/n.there are n terms. so sum is bounded above by n/n = 1.
so the summation is bounded.[/code][/quote]

for (1) , i can understand $b_n\geq a_n$ for all $n$ but I still dont know how $b_n$ is increasing and $a_n$ is decreasing .Can you explain more detail ? thanks .

For (2) , it is ok . :)

[quote="Ramanujan"][quote="shyong"]for (1) , i can understand $b_n\geq a_n$ for all $n$ but I still dont know how $b_n$ is increasing and $a_n$ is decreasing .Can you explain more detail ? thanks .
[/quote]

You mean $a_{n} \geq b_{n}$ . It is $a_{n}-a_{n-1} =\frac{1}{2}(b_{n-1}-a_{n-1}) \leq 0$

and $b_n=\sqrt{a_{n-1}b_{n-1}} \geq \sqrt{b^{2}_{n-1}}=b_{n-1}$[/quote]

oh i see .Thanks you .

One last question , what does the symbol $\uparrow$ and $\downarrow$ means ? ( i do mention the question above , dont I?  :P )

[quote="fedja"]$\uparrow$ usually means "increases" and $\downarrow$ means "decreases". So, in (2), you are asked to prove that $a_{n+1}\ge a_n$ for all $n$. :)[/quote]

ok thanks for your help  :D

But I also notice that the symbol also appears like this

$\lim_{x\uparrow a}f(x)$

im guessing that have nothing to do with " increase" , "decrease" ?

"$x$ increases to $a$" or "$x$ approaches $a$ from the right." Similarly with decreasing. Often you can also write $x\to a^-$ or $x\to a^+$ for that. (As from below and from above, respectively).


Daca am un cub ABCDA'B'C'D' si planul $\alpha$ ai $AC'\perp \alpha$. Rotesc cubul in jurul lui AC' si astfel se genereaza un corp. Sa se afle volumul corpului obtinut.

[hide="numai pt luke999"]in ce clasa esti?[/hide]


inequalities
real analysis
real analysis unsolved

Let $ E$ be a uncountable subset of $ \mathbb{R}$, prove that there is a $ t\in \mathbb{R}$ st either $ E\cap ( \minus{} \infty, t)$ or $ E\cap (t, \plus{} \infty)$ is uncountable.

it seems clear for me when I have $ E \equal{} E\cap\mathbb{R} \equal{} (E\cap ( \minus{} \infty, t])\cup (E\cap (t, \plus{} \infty))$

Any idea is highly appreciated.

As stated, this is true of every $ t$; if both are countable, then $ E$ is countable.  Did you mean that [b]both[/b] intersections are uncountable?

t0rajir0u : can you give a prove that  both intersections are uncountable

thanks

One approach is to consider $ s\equal{}\sup\{t: E\cap (\minus{}\infty,t)\text{ is countable}\}$ and $ S\equal{}\inf\{t: E\cap (t,\plus{}\infty)\text{ is countable}\}$ and to show that $ s<S$ (just think a bit over what the other inequality would imply). Another approach is to note that every uncountable set $ E$ has an uncountable subset $ F$ whose intersection with every open interval is either empty or uncountable (the key here is that it suffices to consider open intervals with rational endpoints). See if you can figure out the details yourself :). Probably there are many other ways to proceed as well.

1234567a, I didn't look at your solution and I came up with basically the same thing.  Assume that no such $ t$ exists, so that $ E\cap (\minus{}\infty,t)$ and $ E\cap (t,\infty)$ are both countable for all $ t\in\mathbb{R}$.  Then $ (E\cap (\minus{}\infty,t))\cup (E\cap (t,\infty))\equal{}E\backslash\{t\}$ is countable, since a union of countable sets is countable.  But then $ E$ must be countable, a contradiction.

The argument seems fine unless I'm missing something.

The original assertion in the OP is quite trivially true.

We are, however, now trying to prove that we can find a $ t$ so that [b]both[/b] intersections are uncountable.

I am sorry when I checked it again ,  It must be as the following :

Let $ E$ be a uncountable subset of $ \mathbb{R}$, prove that there is a $ t\in \mathbb{R}$ st [b]BOTH [/b]$ E\cap ( \minus{} \infty, t)$ [b]AND[/b] $ E\cap (t, \plus{} \infty)$ is uncountable

fedja :  Your idea is so smart,  but I still don't get your idea fully

based on your idea I proved that $ s < S$ so I can find $ t : s < t < S$ and $ t$ is the exactly number I am looking for but:

If I let $ A \equal{} \{t: E\cap ( \minus{} \infty,t)\text{ is countable}\}$ and $ B \equal{} \{t: E\cap (t, \plus{} \infty)\text{ is countable}\}$,  I don't know why is $ A$  bounded from above and why is B bounded from below?
Could you or any one please point it out for me? and how can you know to define such $ s$ and $ S$? what is the motivation behind?  thanks

If $ A$ wasn't bounded above, then you could keep finding $ t$'s such that $ E\cap (\minus{}\infty,t)$ is countable, which would mean that $ E$ is countable which isn't true.


USAMTS geometry AMC AIME USA(J)MO USAMO search

I liked number four.  Two was not that difficult, and neither was 1.  Three was interesting ... I found it easy to discern the answer but difficult to prove it.  Five was a close second, but 4 was my favorite.

JB

I'm really not sure between 3 and 4. I liked 3 when I saw how simple it was, and I also liked the equation I got for 4. :P It was the second one that I figured out (after 2), although it took me a little longer to actually write an equation for it.

Five, because after working on it for so long, there was such a nice solution.

I liked #3 as there was a very nice way to solve it that wasn't so obvious.

[quote="4everwise"]We all like Mr. Rusczyk's geometry problems.  :P[/quote]
Yep. :)

I'm still waiting for the AoPS people to choose a problem 1 that isn't mostly tedious casework and straightforward counting.

[quote="Erin J. Schram"][quote="ZennyK"]I'm still waiting for the AoPS people to choose a problem 1 that isn't mostly tedious casework and straightforward counting.[/quote]
Under George Berzsenyi's system of writing the USAMTS, Problem 1 was alway a problem that did not require specialized mathematics. (Problem 2 was number theory, Problem 3 was algebra, analysis, or combinatorics, Problem 4 varied, and Problem 5 was geometry.) Plain casework fit the criterion.

The Art of Problem Solving people have written only two rounds so far. Give them time to get experience before they break tradition.

Erin Schram[/quote]
Ok, thanks. By the way, everyone at AoPS: you are all doing a marvelous job. Thank you and keep up the good work! :)

[quote="ZennyK"]By the way, everyone at AoPS: you are all doing a marvelous job. Thank you and keep up the good work! :)[/quote]
Ditto. I really enjoy doing the problems.

[quote="1234567890"]I agree. USAMTS is even funner to do, in my opinion, than an AIME.  The geometry problems are particuarly good. :)[/quote]

I'm not saying this applies to EVERYONE but USAMTS is more fun to do than AIME because it's totally different contest. In AIME, everyone is trying to score a certain point to make the goal called "USAMO." Hence, there are many pressures and given a 3-hour limit without calculator, you can easily make stupid mistakes and screw up. USAMTS is almost opposite of it. Although the problems require proofs, the level is not that much harder or easier. But since you're given about a month to do this, there is no pressure (as long as you work each day of course) and nothing like you get to qualify for USAMO. The best you can get is get AIME certification and maybe get a Gold Medal but it's nothing much compared to USAMO qualifier (no offense to USAMTS, this is a great contest and I like it very much --> Especially about being free and etc.). So, that might be why I say.

If people were given a same time to do AIME problems, they'll probably like AIME a lot too because some AIME problems are VERY nice and excellent as USAMTS problems (sometimes, better in some angle).

[quote="1234567890"]I agree. USAMTS is even funner to do, in my opinion, than an AIME.  The geometry problems are particuarly good. :)[/quote]
Argh... "funner" is not a word!  Anyway, I actually think the AIME is more fun than the USAMTS.  I really enjoy the proof-style contest without time pressure that the USAMTS provides, but I feel like the AIME is definitely more of a challenge given the time pressure and the fact that many of the problems rival or surpass the USAMTS in difficulty, and the thrill I get from a competition is often linked with the challenge I get from it (as long as it's not completely over my head to the point where I give up in frustration).

IMO, AIME is better than USAMTS. It probably takes me about the same amount of time for the five problems of USAMTS as for the 15 of AIME, if not more, and I've never really liked working a long time on hard problems as much as having problems with quicker, more elegant solutions, like the AIME. I'm not saying that this doesn't apply to the USAMTS, but AIME is always better from that regard. No matter how "simple" a solution is on USAMTS, it's going to require some work other than just writing it out.

I like the USAMTS a lot more. I don't think timing should be a factor in math. I think the difficulty of math should be in the math. When there's timing it adds the factor of a stress test. But in USAMTS you get to do more difficult problems, and just enjoy them on end. The USAMTS has less of a feel of competition, more of a feel of just doing math for fun (which is what I think the best thing). It's also good as it prepares people for what those deep in graduate math or mathematicians do, spend a long time solving difficult problems.

I also wonder why people seem to say here that the AIME is as difficult as the USAMTS, I'd think the USAMTS is harder. And to give an example I know someone who says the AIME isn't too hard, prolly will qualify for USAMO, but has no idea how to do USAMTS problems...the two can be a different skill though, solving harder problems versus solving them quickly. It's like the division between the fast and those who'll just do math on end but may be more able to solve the hardest problems.

PS/EDIT:Look below, I totally agree with K81o7.

I like the USAMTS more than the AIME because there is more time, less pressure.  You can really get into the problems and explore and extend them.  Rather than simply finding a solution and writing down the answer, with no credit even with a small error, you get to search through the problems and look at different approaches.  You can find the most elegant solution, and points are based on your thought process and ideas, rather than simply on your answer.  I learn a lot more from the USAMTS than from the AIME.

Of course, then again, they can't have people qualify for USAMO through it; not only do some people have more time than others to work on it, but the USAMTS is based on an honor system.  Anybody could just solve the problems and then give the solutions out to all of his/her friends, which would mean that a bunch of random schmoes go to USAMO.

[quote="zanttrang"][quote="1234567890"]I agree. USAMTS is even funner to do, in my opinion, than an AIME.  The geometry problems are particuarly good. :)[/quote]
Argh... "funner" is not a word!  Anyway, I actually think the AIME is more fun than the USAMTS.  I really enjoy the proof-style contest without time[/quote]

Eh.. I should watch out what I'm typing (while uploading problems..). I copied the phrase from 1234567890's post and didn't even know the mistake with "funner." Anyway, looks like many people have different opinions about USAMTS vs. AIME.

The AIME and the USAMTS involve two different styles of problem solving. Some people are better at more fast-paced problems, better at thinking under the gun, while others prefer the more research/exploratory style of the USAMTS. I think the AIME is more fun, but I'm better at the USAMTS.

[quote="4everwise"]Haha. Look at how number 5 is winning by like 45%!  :rotfl:

We all like Mr. Rusczyk's geometry problems.  :P[/quote]

I don't know how you guys do those geometry problems, but THEY ARE DIFFICULT!

The other problems are usually challenging and I love to do them.  I would rather do #1, #2, #3, and #4 together than do #5.

Any suggestions for improving my geometry skills?

(Also, I did #5 and got the correct answer, but my equations and calculations involved decimal, not exact, answers.  So, I found the degree measure to the nearest hundereth (it wound up being a whole number anyway).  Do you think I'll get a point deducted because I didn't show that my answer was exact?)

I think you'll probably get a point deducted.  Graders usually are less leniant on brute force solutions.  If you do brute force a problem, you have to be thorough to get full credit.  It depends on how generous the graders feel though.

[quote="calc rulz"]
I also wonder why people seem to say here that the AIME is as difficult as the USAMTS, I'd think the USAMTS is harder. And to give an example I know someone who says the AIME isn't too hard, prolly will qualify for USAMO, but has no idea how to do USAMTS problems...the two can be a different skill though, solving harder problems versus solving them quickly. It's like the division between the fast and those who'll just do math on end but may be more able to solve the hardest problems.
[/quote]

Well obviously first few AIME problems are easier than USAMTS. But the last few AIME problems are usually harder than USAMTS problems. Both AIME and USAMTS are good contests, but USAMO is the best of all :) !


There are three light switches turned off.  Every minute a person comes by and randomly flips exactly one of the switches.  On average, how many minutes will pass before all three light switches get turned on at the same time?

let $E_0,E_1,E_2,E_3$ be the expected value if $0,1,2,3$ lamps are switched on.

it is clear that $E_0= 1 + E_1$ (in minutes) , because only $1$ turned on lamp can result out of $0$ turned on lamps.
$E_1=1+ \frac13 E_0 + \frac23 E_2$ , because the probability of the on lamp turned off again is $\frac13$ and another lamp turned on is $\frac23$.
similar $E_2= 1+ \frac23 E_1+\frac13 E_3$.
but $E_3$ is $0$ because the $3$ lamps are turned on. so we got a lin. equation set:
\[ \begin{array}{rcl} E_0&=& 1 + E_1\\ E_1&=&1+ \frac13 E_0 + \frac23 E_2\\ E_2&=& 1+ \frac23 E_1 \end{array} \]

So we get
\[ E_1 = 1+ \frac13 \left(1 + E_1\right)+ \frac23 \left(1+ \frac23 E_1\right) \]
therefore $E_1 = 9$ and $E_0 = \boxed{10}$, which we were looking for

peeta


[quote="Utkirstudios"]If $x,y,z >0$, prove that

${\sum_{sym}\frac{1}{(x+y) \sqrt{xy+4z(x+y)}}\geq \frac{3}{2(xy+yz+xz) }}$[/quote]

Just stumbled upon this topic. The inequality is WRONG for x = y = 10 and z = 1.

  darij

[quote="darij grinberg"][quote="Utkirstudios"]If $x,y,z >0$, prove that

${\sum_{sym}\frac{1}{(x+y) \sqrt{xy+4z(x+y)}}\geq \frac{3}{2(xy+yz+xz) }}$[/quote]
 The inequality is WRONG for x = y = 10 and z = 1.

  darij[/quote]
darij, what do you think about  following inequality:
${\sum_{cyc}\frac{\sqrt{xy+4z(x+y)}}{x+y}\geq \frac{9}{2}}$ $?$ :wink:

This is really known: http://www.mathlinks.ro/Forum/viewtopic.php?t=28175

If you have anything like a nice proof, feel fre to post it. ;)

  darij

Thank you very much, darij, for your link. I didn't know it.
I think, this inequality is very hard.


modular arithmetic
induction
Euler
search
Congruences

Show that, for any fixed integer $\,n \geq 1,\,$ the sequence \[2, \; 2^{2}, \; 2^{2^{2}}, \; 2^{2^{2^{2}}}, \cdots \pmod{n}\] is eventually constant.

Write $a_n=\underbrace{2^{2^{...^2}}}_{n \text{ times}}$, then $a_{n+1}=2^{a_n}$.

Induction on $n$ (one could also use some straightforward way, but I think this way's nicer, and it kills topic 139, too):

We show that it is constant beginning from the $n-1$-th term (or earlier):

$n=1$ is clear.
For any $n>1$, we write $n=2^m n^\prime$ with $n^\prime$ odd.
The sequence $a_k$ clearly gets constantly $0 \mod 2^m$ for all $k \geq m \leq n-1$. So we are left to prove the same $\mod n^\prime$.

By induction, the sequence gets constant $\mod \tilde n := \varphi(n^\prime) < n^\prime \leq n$. Thus there is $c$ such that for all $k \geq \tilde n-1$ we have $a_k \equiv c \mod \varphi(n^\prime)$.
This gives $a_{k+1} = 2^{a_k} \equiv 2^c \mod n^\prime$ by the theorem of Euler-Fermat, meaning nothing else than constantness of the sequence for all $k>\bar n \leq n-1$, our result.

There is a relatively recent post, showing this is true for any positive integer $a$ instead of $2$. A search did identify it as [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=379220&hilit=constant[/url].

Yeah ; as mapronevma said ; there exists a stronger version of this porblem

   Here is it's solution ; with very clear writing :D

We will show the stronger bound that with $a_1 = 2, a_2 = 2^2, a_3 = 2^{2^{2}}, \ldots$, we already have that $a_i \equiv a_j \pmod{n}$ for all $i,j > \log_2(n)$. First note that if $a_k$ is congruent to $a_{k+1} = 2^{a_k}$ modulo $n$ for some $k$, then $a_k \equiv {a_j} \pmod{n}$, for all $j > k$ by induction. This means that we are done as soon as we can find two consecutive terms which are congruent modulo $n$. Now assume by induction that $a_k \equiv a_{k+1} \pmod{n}$ for all $n < 2^k$, which is trivially true for $k = 1$. We will prove that $a_{k+1} \equiv a_{k+2} \pmod{n}$ holds for all $n < 2^{k+1}$. So assume $2^k \le n < 2^{k+1}$ and write $n = l2^m$ with $l$ odd. \\

First we will deal with the even case; assume that $m \ge 1$. We actually intend to show that already $a_k \equiv a_{k+1} \pmod{n}$ in this case. First it is clear that $a_i \equiv a_j \equiv 0 \pmod{2^m}$ for all $i,j \ge k \ge m$, so it is sufficient to show $a_k \equiv a_{k+1} \pmod{l}$. But since $l < 2^k$, this already follows from the induction hypothesis.

So now let's move on to the odd case and assume $m = 0$. Since $\varphi(n)$ is even, we obtain (either by what we just proved or the induction hypothesis) that $a_{k+1} \equiv a_k \pmod{\varphi(n)}$. But by Euler's Theorem this implies $a_{k+1} = 2^{a_{k}} \equiv 2^{a_{k+1}} = a_{k+2} \pmod{n}$.


AMC
AIME
USA(J)MO
USAMO
AMC 10
AMC 10 A
AMC 12 A

Are you allowed to take both AMC 10 A and B AND Amc 12 A and B to qualify to AIME? So that means your best score would be taken in account with the AIME to justify USAMO qualification?

And you cant take the AIME twice? Why?

The two A tests are held on the same day, and the B tests. Unless you find a way to be in two places at once, no. I'm not sure if you can qualify through one A test and another B test, though...

I think this is in the FAQ.

[quote="gauss1181"]And you cant take the AIME twice? Why?[/quote]

The AIME is hosted twice because some students are not able to make one test date or another, not to give the lucky few who can make both two chances.  Many schools do not offer both test dates, so this would be an unfair advantage.

Wait...but both the Amc 10/12 A are on tje same day, and AMC 10/12 B are on the same day, right?

But why cant you take both AIME's if you could take both A and B of the AMC 10/12?

Please see Frequently Asked Questions AMC 10/12 #12, 14, 18, 20, and 21
and AIME #2, #3, and #7 at
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions


pigeonhole principle
set theory
superior algebra
superior algebra unsolved

Been working on this problem all night and have gotten no where. Maybe someone out there can show me the light :)

U = {1, 2, 3, .... , n, ....., 2n} for some natural number n
Let P be a subset of U such that |P| = n + 1
show that there exists x, y in P where x not equal y and x divides y or y divides x.

[quote="JBL"]This is an elementary pigeonhole argument, certainly not abstract algebra (which is the subject of the forum you've put it in).  Personally, I think this is a problem for which there aren't very many good hints, and you're better off struggling with it yourself.  The comments I will make are that (1) you need to deploy the pigeonhole principle, (2) the trick is in choosing the holes intelligently, and (3) note that $ n \plus{} 1$ is the smallest number that makes the question statement true.[/quote]

Sorry, I didn't know where to put this question since I did not see any forum sections titled set theory or discrete mathematics.

The problem I having is generalizing the choice of holes.

[quote="HouseUrMusic"]

The problem I having is generalizing the choice of holes.[/quote]

Any number in the set $ \mathbf{A} \equal{} \{1,2, \ldots, 2n \}$ is of the form $ 2^{p} q$ where $ q$ is an odd number in $ \mathbf{A}$.