I was invited... we should make a list

edit:  invited before comc, if you are wondering about comc invites

[quote="startogalaxy"]Just wondering, Alex, how do you get an invite for winter camp? Is it largely depended on Olymon?[/quote]
I was also wondering about this. The invite usually goes to
1) former IMO team members who are eligible to go again
2) people who did good on COMC
I know that they do look at Olymon, but I thought it was mostly COMC.

I've been thinking: is it hard to get "discovered" by the CMS that you're capable doing advanced mathematics? I know some people who are really good at mathematics but they don't have a clue about what the CMS does and the benefits. I guess it's always good to do some research (like I did) or be in a school where the elite are praised.  ;)  Of course, to all you people.  :)

Was talking to Adrian, they didn't invite on Olymon this year, just CMO results so far.  Then the rest will come from COMC results.

yes, i was invited based on CMO.  The rest will come from COMC as sarah said.

They're doing it differently this year, then.

Last year, they only invited one person strictly from COMC.  Of the top six COMC scorers, I think four were already invited before COMC, one (who hadn't done well on Olymon or CMO in the previous year) was invited based on COMC results (that was me, in case you're wondering), and the last one wasn't invited, even though he really deserved to be.

I think one more was a young female who'd done reasonably well on COMC, and the rest were from CMO, Olymon, and IMO.

[quote="Vatchern"]Winter camp? Where is this held?
I also have no idea what a CMS, CMO, COMC or Olymon is.
I am from Newfoundland, Grade 12, doing Advanced Math, can someone fill me in on the questions above?
thanks,
Nick.[/quote]
CMS stands for Canadian Mathematical Society. They try to promote mathematics in Canada. Visit [url=http://www.cms.math.ca/]here[/url].
CMO stands for Canadian Mathematical Olympiad. It is written in March, and you need to get invited to write this(around 100 people get invited, I think). Visit [url=http://www.cms.math.ca/Competitions/CMO/]here[/url].
COMC stands for Canadian Open Mathematics Challenge. It is written in November and any high school students can write it. Visit [url=http://www.cms.math.ca/Competitions/COMC/]here[/url].
Olymon, Mathematical Olympiads Correspondence Program, is for anyone who is interested in problem-solving. The problems appear [url=http://www.math.utoronto.ca/barbeau/]here[/url] every month, and you can submit complete or partial solution.

_________________________________________________________
ADDED:
Winter camp is for the top 15 high school mathematicians. Most of IMO team members are chosen from this camp. It's in January and has been at York University for few years now.

[quote="geoffreynham"]They're doing it differently this year, then.

[/quote]

Yep.  Because of the sadly low OLYMON participation.

So to get invited to winter camp this (academic) year, you need to do well in last year's CMO? That's a bit strange. So really winter camp is largely composed of the Honour Mentions and Champions of CMO last year, who haven't graduate. I've always thought it would solely depend on Olymon and COMC.

[quote="startogalaxy"] I've always thought it would solely depend on Olymon and COMC.[/quote]

It used to be.  Anyone remember that test they would send you if you did Olymon well?  They stopped using that a couple years ago, I'm not sure why, someone said it could be from cheating too much.  And now its because barely anyone does Olymon this year, so they invited on CMO instead.

[quote="lotus87"]Yea... it was called the Time-Limited Exam (TLE).  I wrote it a couple of times.  I think it was my school that complained to get rid of the exam because students were allowed to work on it for like a 2 week period, during which some cheated like crazy just to attend a single math camp...  :([/quote]

Ah ha that's what it was called.  I couldn't for the life of me remember what it was called, lol.  I hated that exam, I'm glad they got rid of it.

I'm coming too. I look forward to seeing you guys at winter camp.

(But first, must finish all those essays for application to U.S. universities. Due Jan 1!!!)

Looking forward to meet old and new friends!

does anyone know why boris isn't coming?  :?

[quote="carpo"]Did anyone see Richard Hoshino on the news teaching Sudoku to stranded travellers?

"The nine can't go here or here."[/quote]

ROFL.  Details? :)

[quote="TheDreamer"]ROFL.  Details? :)[/quote]This was on CTV NewsNet yesterday afternoon. Normally I don't watch CTV NewsNet, but I was on a plane.

http://www.ctv.ca/servlet/ArticleNews/story/CTVNews/20051227/atlantic_storm051227/20051228/

I wonder why Richard Hoshino was flying with Air Canada. I guess he booked late and the WestJet flights were full. ;)

Winter Camp starts tomorrow!!! :jump:
Don't forget to bring CanMoo :moose: , Richard!

[quote="chess64"]Tomorrow? Doesn't school start tomorrow? :huh:[/quote]

No, it starts January 9th in Ontario I believe.  I'm not sure about the rest of Canada.

And even if school does start tomorrow...so...what...

It's Winter Camp! :D

[quote="chess64"]Tomorrow? Doesn't school start tomorrow? :huh:[/quote]

School starts tomorrow for a lot of university kidz.

Did everyone see the pictures that Adrian took?
It was funny to see that CanMoo :moose:  was in most of the pictures. :D
BTW, am I the only one who cannot open the video files? :?
____________________________________
EDITED: 'file' is pluralized. :blush:

When I have time later this week, I'll be posting the photos that I took. Other campers that brought cameras should do the same.

Fantastic, and make sure that everyone will claim they copyrights of those photos at the same time :D  Plus, keep them all in a more math-wise order. :P

Now I shalt claim that I get the copyrights of the advice above :D

[quote="jpark"]where can I find the pictures?[/quote]
You mean you didn't get the mail from Adrian? Go [url=http://www.math.ucalgary.ca/~tang/WinterCamp2006]here[/url]. I PMed you the username and password.


algebra
polynomial
superior algebra
superior algebra unsolved

1)Let K be a field. what is $ Irr(x,K(x^n))$ $ n\in N^*$
2) Show that $ x\equal{}\frac{\sqrt{2}\plus{}\pi\sqrt{3}}{1\minus{}\pi^2}$is not a root for a polynom over Q.

1) if $ u \in K(x) \backslash K$, say $ u \equal{} f/g$ with coprime $ f,g \in K[x]$, then $ K(x)/K(u)$ is finite of degree $ max(deg(f),deg(g))$. this can be checked with the (unnormed) minimal polynomial $ g(t)*u \minus{} f(t) \in K(u)[t]$. in this case $ u \equal{} x^n$.

thanks. and for the second one anybody help?

If you know that some number $ \pi$ is transcendental, then any nonconstant rational $ f(x)$ with algebraic coefficients results again in a transcendental number.

The nature of transcendetal elements like $ \pi$ is that they imply an isomorphism $ \bar{\mathbb{Q}}(x) \cong \bar{\mathbb{Q}}(\pi)$ (as they define one on the polynomial rings). Thus it suffices to show that no nonconstant element of $ \bar{\mathbb{Q}}(x)$ is algebraic, being immediate as $ \bar{\mathbb{Q}}$ is already algebraic closed, thus the algebraic closure of $ \mathbb Q$ in $ \bar{\mathbb{Q}}(x)$.


calculus
integration
function
complex analysis
complex analysis unsolved

Calculate $ \iint_{a \le |z| \le b} (1\plus{}f(z)z) dx dy$, where f is analytic on the disc $ D(0,R), R>b>a>0$.

The average of any harmonic function on a circle/disk/annulus is its value at the center. Analytic functions are harmonic.

cool! so,
$ \iint_{a \le |z| \le b} (1\plus{}f(z)z) dx dy\equal{}$ $ \iint_{|z| \le b} (1\plus{}f(z)z) dx dy \minus{} \iint_{|z| \le a} (1\plus{}f(z)z) dx dy\equal{}$
$ \pi b^2 (1\plus{}f(0)*0)\minus{}\pi a^2 (1\plus{}f(0)*0)\equal{}\pi (b^2\minus{}a^2)$?
Is there a short proof of this theorem?


$ 6.$ Define a sequence $ < a_n > _{n\geq0}$ by $ a_0 \equal{} 0,$ $ a_1 \equal{} 1$ and
\[ a_n \equal{} 2a_{n \minus{} 1} \plus{} a_{n \minus{} 2},\]
for $ n\geq2.$
$ (a)$ For every $ m > 0$ and $ 0\leq j\leq m,$ prove that $ 2a_m$ divides
\[ a_{m \plus{} j} \plus{} ( \minus{} 1)^ja_{m \minus{} j}.\]
$ (b)$ Suppose $ 2^k$ divides $ n$ for some natural numbers $ n$ and $ k$. Prove that $ 2^k$ divides $ a_n.$

[hide] (a) Verifying the hypothesis for j =0 and j=1, we will prove that whenever the result hold for j=k-2 and j=k-1, it hold for j=k.

lets keep things easy by assuming that k is even

We want to prove for $ j \equal{} k \plus{} 1$

Then $ a_{m \plus{} k} \plus{} a_{m \minus{} k} \equal{} 2a_{m \plus{} k \minus{} 1} \plus{} a_{m \plus{} k \minus{} 2} \plus{} a_{m \minus{} k}$ But $ a_{m \minus{} k} \equal{} a_{m \minus{} k \plus{} 2} \minus{} 2a_{m \minus{} k \plus{} 1}$

Hence the expression becomes $ 2(a_{m \plus{} k \minus{} 1} \minus{} a_{m \minus{} k \plus{} 1}) \plus{} (a_{m \plus{} k \minus{} 2} \minus{} a_{m \minus{} k \plus{} 2})$ which from our assumption is divisible by $ 2a_m$

(b) is also easy by strong induction to prove that the sequence $ b_k \equal{} (p \plus{} q\sqrt 2)^{2^k} \plus{} (p \minus{} q\sqrt 2)^{2^k}$ is divisible by $ 2^k$[/hide]

For the second part, can obtain a_n by the recursion formula then replace n as m(2^k) and
factorize using (a^2 - b^2 ) and do  induction on k,and use binomial expansion.
unfortunately i didn't get it during the exam though.... :mad: [/code]

Part one follows by strong induction on $ j$.

Part two follows by plugging in $ j\equal{}m$ in part one followed by strong induction on $ v_2(n)$.


function
modular arithmetic
number theory
prime numbers
number theory unsolved

Determine all functions $ f$ defined on the natural numbers that take values among the natural numbers for which
\[ (f(n))^p \equiv n\quad {\rm mod}\; f(p)
\]
for all $ n \in {\bf N}$ and all prime numbers $ p$.

Fixed n .
From Fermat's theory we have :
$ f(n)^p\equiv f(n) (\mod p)$
So $ p|f(n)\minus{}n$ for all prime p .
Chose p is large enough such that $ p>|f(n)\minus{}n|$
So $ f(n)\equal{}n$
The equation has only onsolution $ f(n)\equiv n,\forall n\in N$

[quote="TTsphn"]
So $ p|f(n) \minus{} n$ for all prime p . [/quote]

Why?
The initial congruence is modulo $ f(p)$ not $ p$.

Pierre.

in the equation $ (f(n))^p\equiv n\pmod{f(p)}$ let $ n = p$ so we get that:

$ (f(p))^p\equiv p\pmod{f(p)}$

$ \Rightarrow f(p)\mid p$ but $ p$ is prime,thus we must have either $ f(p) = 1$ or $ f(p) = p$...

now note that as TTsphn said,there doesn't exist infinitely many primes with $ f(p) = p$,hence there are finitely many prime numbers $ \{p_1,p_2,\ldots ,p_k\}$ such that $ f(p_i) = p_i$ for $ 1\leq i\leq k$ and for every prime number $ p\not\in\{p_1,p_2,\ldots ,p_k\}$ we must have $ f(p) = 1$.

now note that for a fixed number $ n$ we have:

$ f(n)\equiv n\pmod{p_i}$ for $ 1\leq i\leq k$

thus:

$ p_i\mid f(n) - n$

$ \Rightarrow p_1p_2\ldots p_k\mid f(n) - n$

$ \Rightarrow f(n) = n'.p_1p_2\ldots p_k + n$ for every natural number $ n$ ($ n'\in\mathbb{N}_0$).

thus the function $ f$ must fulfill the following conditions:

is $ f(n) = n'.p_1p_2\ldots p_k + n$ and $ f(p_i) = p_i$ for $ 1\leq i\leq k$ and for $ p\not\in\{p_1,p_2,\ldots p_k\}$ we have $ f(p) = 1$.

now we'll show that every function $ f$ with the above conditions is a solution:

if $ p\not\in\{p_1,p_2,\ldots ,p_k\}$,i.e. $ f(p) = 1$ then we obviously have:

$ f(n)\equiv n\pmod 1$

now for $ i = 1,2,\ldots ,k$ we have $ f(p_i) = p_i$ thus:

$ (f(n))^{p_i}\equiv n\pmod{f(p_i)}$

$ \iff (n'.p_1p_2\ldots p_k + n)^{p_i}\equiv n\pmod{p_i}$

$ \iff n^{p_i}\equiv n\pmod{p_i}$

which is true by fermat theorem.

hence the function $ f$ is a solution iff there is a finite set of primes $ S = \{p_1,p_2,\ldots ,p_k\}$ such that $ f(p_i) = p_i$ for $ 1\leq i\leq k$ and for every prime number $ p\not\in S$ , $ f(p) = 1$ and also $ f(n) = n'.p_1p_2\ldots p_k + n$ where $ n'\in\mathbb{N}_0$.

I am sorry .
Take $ n\equal{}p$ then
$ f(p)|f(p)^p\minus{}p$ so $ f(p)|p$
Thus $ f(p)\in \{1,p\}$
If $ f(p)\equal{}1$ then  :
$ q|f(p)^q\minus{} p$ forall prime $ q$
$ \Rightarrow q |p\minus{}1$
It gives contradition .
So $ f(p)\equal{}p$
As above post we have $ f(n)\equiv n,\forall n\in N$

1)$ f(p) \equal{} 1$ or $ f(p) \equal{} p$ for all primes $ p$
2)if there exist infinitely $ p$ such that $ f(p) \equal{} p$ then $ f(n) \equal{} n$
3)Assume that the condition $ (2)$ is not true so we have only finitely $ p$ such that $ f(p) \equal{} p$
if there exist $ p > 2$ such that $ f(p) \equal{} p$ then for all primes $ q$ such that $ f(q) \equal{} 1$ we have :
$ p|q \minus{} 1$
but we know there exist infinitely primes in the form $ pk \minus{} 1$ , but we have only finitely $ p$ such that $ f(p) \equal{} p$ so for $ r$ (one of this primes) we have :
$ p|r \minus{} 1$ and then $ p|pk \minus{} 2$ which gives $ p \equal{} 2$ . Contradiction
so for all $ p > 2$ we have $ f(p) \equal{} 1$ and we have only two case :
1)$ f(2) \equal{} 1$ then every function such that for all primes $ p$ , $ f(p) \equal{} 1$ is solution.
2)$ f(2) \equal{} 2$ , in this case every function such that for all primes $ p > 2$ , $ f(p) \equal{} 1$ , $ f(2) \equal{} 2$ and $ 2|f(n) \minus{} n$ for all composite numbers $ n$ is solution

It is easy to conclude that $ f(p) \equal{} p$ for all primes.

Now consider prime $ q$ that does not divide $ n$.

Since $ f(q) \equal{} q$ divides $ f(n)^{q} \minus{} n$ it follows that q does not divide $ f(n)^{q}$ so:

$ f(n)^{q} \minus{} n \equal{} f(n)^{q \minus{} 1}f(n) \minus{} n \equal{} f(n)\minus{}n (\mod q)$

Hence if prime $ q$ does not divide $ n$ then $ q$ divides $ f(n) \minus{} n$ and since for each natural $ n$ we have infinitely many such primes $ q$ it follows that $ f(n) \equal{} n$ for all naturals $ n$.

Many of the solutions above are incorrect; there are non-identity functions that work.

Let $P(n, p)$ denote the statement $f(n)^p \equiv n \pmod{f(p)}$.

$P(p, p)$ gives $0 \equiv p \pmod{f(p)}$, so $f(p) \in \{1, p\}$ for every prime $p$. Now, we split into cases:
[list]
[*] If $f(p) = 1$ for every prime $p$, then $f$ evidently works.
[*] Else, suppose there is a prime $p_0$ with $f(p_0) = p_0$. $P(n, p_0)$ gives $f(n)^{p_0} \equiv n \pmod{p_0}$, so $f(n) \equiv n \pmod{p_0}$ for all $n$.
[list]
[*] If $p_0 \neq 2$, by Dirichlet theorem, there are infinitely many primes $q$ with $q \equiv -1 \pmod{p_0}$, so $f(q) \equiv q \pmod{p_0}$. Also $f(q) \in \{1, q\}$, so $f(q) = q$ for all these $q$, so $f(q) = q$ for infinitely many prime $q$.

Then, $P(n, q)$ gives $f(n) \equiv n \pmod{q}$ for infinitely many $q$, so $f$ is identity, which works.

[*] Else, $p_0 = 2$, so suppose $f(2) = 2$ while $f(p) = 1$ for odd prime $p$. Then $P(n, p)$ is satisfied for $p$ odd, and $P(n, 2)$ is $f(n) \equiv n \pmod{2}$; all such $f$ work.
[/list]
[/list]

The complete characterization of functions satisfying the condition is the following:
[list]
[*] $f$ identity.
[*] $f(p) = 1$ for all prime $p$.
[*] $f(p) = 1$ for all odd prime $p$, $f(2) = 2$, and $f(n) \equiv n \pmod{2}$ for all $n$.
[/list]

Let $P(n,p)$ denote the given assertion.

$P(p,p): f(p)^p\equiv p\pmod{f(p)}$, so $p\equiv 0\pmod{f(p)}\implies f(p)\in \{1,p\}$.

If $f(p)=1$ for all $p$, any such function works.

Now suppose there exists a prime $q$ with $f(q)=q$.

Note that $f(n)^q\equiv n\pmod q$, so $f(n)\equiv n\pmod q$ for all $n$.

Case 1: There exists a prime $q>2$ with $f(q)=q$.
Then by Dirichlets there are infinitely many primes $r$ with $r\not\equiv 1\pmod q$. For any such $r$, if $f(r)=1$, then $1\equiv r\pmod q$, not true. So $f(r)=r$ for infinitely many primes $r$. Thus, $f(n)^r\equiv n\pmod r$, so $r\mid f(n)-n$ for infinitely many $r$, so $f(n)=n$ for each $n$, which works.

Case 2: $f(2)=2$ and $f(q)=1$ for all primes $q>2$.
Then $f(n)\equiv n\pmod 2$. We can verify that anything of this form works.

So our final solution set is $\boxed{f(n)=n}$, $\boxed{\text{any function for which } f(p)=1 \text{ for all primes }p }$, $\boxed{\text{any function such that }f(2)=2, f(p)=1\forall p>2, f(n)\equiv n\pmod 2\forall n\in \mathbb{N}}$

Setting $n=p$ we get $f(p)\in \{p,1\}.$ If $f(p)=1,$ clearly this works. Otherwise $f(2)=2$ and $f(p)=1$ for others, then $f(n)\equiv n\pmod 2.$ This works. If $p>2,$ by Fermat's Little Theorem $f(n)\equiv n \pmod p.$ But Dirichlet's Theorem suggests $f(p)=p$ for infinitely many $p.$ Thus we make $p$ large so that $f$ is the identity. This works too.

We uploaded our solution [url]https://calimath.org/pdf/CMO2008-4.pdf[/url] on youtube [url]https://youtu.be/_Kdz1r18TwU[/url].


$1+w+w^2+w^3+w^4$=0 (i).

what i did is write it like $[(1+w)(1+w^2)(1+w^3)(1+w^4)]\cdot(1+w^2)(1+w^3)(1+w^4)\cdot(1+w^3)(1+w^4)\cdot(1+w^4)$.  i did that cause i was pretty sure the first bracket would come out to something nice that we could use (i) on... it comes out to 1.  Then you can continue to do the calculations, its not that hard, or you can write each consecutive factor as 1, $1/(1+w)$, $1/[(1+w)(1+w^2)]$ etc and multiply them but i dont know that thats any easier. i got -1 as answer.


geometry
trapezoid
trigonometry
geometry unsolved

let $ O$ be the centre of a circle. let $ OU,OV$ be perpendicular radii of  the circle.the chord $ PQ$ passes through the midpoint $ M$ of $ UV$.let $ W$ be a point such that $ PM\equal{}PW$,where $ U,V,M,W$ are collinear.let $ R$ be a point such that $ PR\equal{}MQ$,where $ R$ lies on the line$ PW$.prove that $ MR\equal{}UV$
ishfaq :maybe:

$ R$ has to be on the ray $ (PW,$ not on the opposite one.

Let $ (O, r)$ be the given circle. If $ S$ is a point on the ray $ (PQ \equiv (PM,$ such that $ PS \equal{} PR \equal{} MQ,$ then $ MWRS$ is isosceles trapezoid with equal diagonals $ MR \equal{} WS.$ Since $ PS \equal{} MQ,$ $ S$ is on a circle $ (O, \varrho)$ concentric with the given circle $ (O, r)$ and passing through $ M,$ tangent to $ UV$ at $ M,$ so that $ \varrho \equal{} OM \equal{} \frac {r}{\sqrt {2}}$ and

$ PS \equal{} \frac {PQ \plus{} MS}{2},\ \ PW \equal{} PM \equal{} \frac {PQ \minus{} MS}{2}.$

Denote $ \varphi \equal{} \angle OMS.$ Then

$ PQ \equal{} 2 \sqrt {r^2 \minus{} OM^2 \sin^2 \varphi} \equal{} r \sqrt {2} \sqrt {1 \plus{} \cos^2 \varphi}$

$ MS \equal{} 2 \varrho \cos \varphi \equal{} r \sqrt {2} \cos \varphi$

By the cosine theorem for the $ \triangle PWS$ with the angle $ \angle SPW \equiv \angle MPW \equal{} 2 \varphi,$

$ WS^2 \equal{} PS^2 \plus{} PW^2 \minus{} 2\ PS \cdot PW \cos 2\varphi \equal{}$

$ \equal{} \frac {PQ^2 \plus{} MS^2}{2} \minus{} \frac {PQ^2 \minus{} MS^2}{2}\ \cos 2\varphi \equal{}$

$ \equal{} r^2 (1 \plus{} 2 \cos^2 \varphi) \minus{} r^2 \cos 2 \varphi \equal{} 2r^2,$

so that $ MR \equal{} WS \equal{} r \sqrt {2} \equal{} UV.$


geometry
3D geometry
tetrahedron
inequalities
vector
AMC
USA(J)MO

Let $a, b, c, d$ be real positive numbers.

Prove that $a, b, c, d$ are the areas of the faces of a tetrahedron if and only if they verify the inequalities:
$a<b+c+d$, $b<a+c+d$, $c<a+b+d$, $d<a+b+c$

[quote]Prove that $a, b, c, d$ are the areas of the faces of a tetrahedron if and only if they verify the inequalities:

$a<b+c+d, b<a+c+d, c<a+b+d, d<a+b+c$[/quote]

One direction (areas of tetrahedral faces satisfies these inequalities):

[hide="solution"]
Put one vertex of a tetrahedron at the origin, and let $P$,$Q$,$R$ be the other vertices.

The magnitude of the cross product of two vectors is equal to twice the area of the triangle formed by the vectors.

Therefore, twice the areas of the 4 faces of the tetrahdron are:

$|P \times Q|, |Q \times R|, |R \times P|, |(P-Q) \times (P-R)|$

Expand the cross product inside the last expression and using the triangle inequality shows that it is less than the sum of the other three.
[/hide]

Yakko

===>
Project the tetrahedron onto the plane containing the face [tex]a[/tex] of largest area, the projection should be parallel to some line that intersects both the bottom face and the vertex opposite that face.
[tex]P(b)+P(c)+P(d) \ge P(a) = a[/tex], because the tetrahedron is a closed figure. But
[tex]b+c+d > P(b)+P(c)+P(d)[/tex].

I think the other direction can be done by letting [tex]a[/tex] be an equilateral triangle and [tex]b,c,d[/tex] be isosceles triangles.

Actually my idea with the equilateral and isosceles triangles is wrong.

BTW, would I be able to use something like this
[quote="schulmannerism"][tex]P(b)+P(c)+P(d) \ge P(a) = a[/tex], because the tetrahedron is a closed figure.[/quote] on the USAMO? It's intuitively obvious but it probably takes algebraic topology or something to prove.

I also managed to prove "=>" (it was easy), but "< =" seems very hard.

I think the other part can also be solved, but I am not sure of my example:
Let the 4 areas be $A,B,C,D$. Take 3 perpendicular axes $x,y,z$, so that $x\times y = A , y\times z = B, z\times x = C$. Now, by continuosly re-sizing the axes $x,y$ and respectively $z$, you can make sure that the other face is once very small, and another time almost equal to say the lowest one $A$. If we would take $A$ to be the biggest, by continuity, we once had area $D$.
I understand that it is a bit unclear, but the resizing is done like this: modify first $z$, and then modify $x,y$ so that they respect the conditions. When $z\to 0$ area of the other face $\to x\times y =A$. When $z\to \infty$ are of the other face tends to $0$.

i don't understand how can you obtain for certain area $D$ and in the same time let $A$, $B$, $C$ unchanged.

Sorry for some misunderstandings in the solution....
First of all, I have discussed the problem with a friend of mine and I am actually a bit wrong, we should take the axes $x,y,z$ and $60$ degrees each. The idea of the solution is to choose initially $x,y,z$ to have the areas we need, and then, by changing one of them, say $z$, to change also $x$ and $y$ to satisfy the area requierments, but to try to obtain different values for the $4^{th}$ area...
Just an aditional explanaition : if we modify $z'=z\epsilon$, then we modify $x'=\frac{x}{\epsilon}$ and $y'=\frac{y}{\epsilon}$
I hope the idea is at least clear, and maybe even correct :?  :?

[quote="DusT"]we should take the axes $x,y,z$ and $60$ degrees each.[/quote]

here you meant angle(x,y)=$60$ and so on?

In DusT's solution it seems that only two of the areas remain unchanged. The third changes .... Nooo :(

Do you have any ideas how to fix this ?


function
integration
inequalities
topology
Functional Analysis
real analysis
real analysis unsolved

Regarding the second question, I think $F$ is non-empty because the set $\{f\in C^1([0,1])|f(0)=f(1)=0,\ \int^1_0 f'^2(x)dx\le 1\}$ is compact wrt the $L^2$ metric on [$0,1$], and the application $f\mapsto \|f\|_2$ is then a continuous application from a compact metric space to the reals, meaning that it reaches its maximum (it also reaches its minimum, but that's trivially $0$ :)).

One problem might be the compactness: is that set compact or not? I think it is, because it is closed in the set $\{f\in C^1([0,1])|f(0)=f(1)=0,\|f\|_2\le T\}$ for some positive $T$ (it's easy to obtain a $T$ by applying Cauchy's inequality: $x\ge \int^x_0 f'^2(t)dt\cdot\int^x_0 1^2dt\ge \left(\int^x_0 f'(t)dt\right)^2=f^2(x)$). Another problem might be that we have extended $E$ to contain functions $f$ s.t. $\|f'\|_2<1$, but it's easy to see that the maximum for $\|f\|_2$ is still reached for functions $f$ which have $\|f'\|_2=1$ (by multiplying a function which does not have this property with some $\alpha$ with $|\alpha|>1$).

Finally, I don't know about the $|F|\le 2$ thing :). Of course, if $\|f\|_2=M$, then $\|-f\|_2=M$, so we must show that $f$ is essentially unique, i.e. only $f$ and $-f$ are Ok.

Is this correct? I'm a total newbie when it comes to functional analysis :?, so sorry if it's wrong.

Edit:

I missed the $C^2$ there, but i think the argument for the non-emptyness of $F$ can be modified to take this into account.


From $ 10$ to $ 4999$, how many integers which the sum of all digits s divisible by $ 4$ ?

 :nhl:

[quote="vinskman"]From $ 10$ to $ 4999$, how many integers which the sum of all digits are divisible by $ 4$ ?

 :nhl:[/quote]

Anybody has idea? :D :wink:

Lots of casework. Solving $ a\plus{}b\plus{}c\plus{}d\equal{}4k$ many times, with restrictions.

[quote="math154"]Lots of casework. Solving $ a \plus{} b \plus{} c \plus{} d \equal{} 4k$ many times, with restrictions.[/quote]

This direction to find I've tried.

Have u a solid way or more effective method to find ?

:bruce:

[quote="math154"]I haven't found a better way...[/quote]

I found a little bit of thread:

consider $ 1000$ to $ 4999$. Fiirst of all ,restricts $ 000\sim 999$ ,hundredth ,tenth and unit --these combinations totally are $ 1000$ integers.For each integer, we can find a suitable digit from 1 to 4, suppose the digit be a, such that $ a\plus{}b\plus{}c\plus{}d\equiv 0 \pmod 4$, for example b,c,d is $ 001$ ,we put $ a\equal{}3$,i.e.$ 3001$ can only be fixed  because $ a\equal{}1,2,3,4$  only.Thus , only $ 1000$ integers can be found for matching part of the question. Then ...

Do you agree?

I'm not exactly sure what you mean...could you explain more?

[quote="vinskman"][quote="math154"]I haven't found a better way...[/quote]

I found a little bit of thread:

consider $ 1000$ to $ 4999$. Fiirst of all ,restricts $ 000\sim 999$ ,hundredth ,tenth and unit --these combinations totally are $ 1000$ integers.For each integer, we can find a suitable digit from 1 to 4, suppose the digit be a, such that $ a \plus{} b \plus{} c \plus{} d\equiv 0 \pmod 4$, for example b,c,d is $ 001$ ,we put $ a \equal{} 3$,i.e.$ 3001$ can only be fixed  because $ a \equal{} 1,2,3,4$  only.Thus , only $ 1000$ integers can be found for matching part of the question. Then ...

Do you agree?[/quote]

For $ 1000$ to $ 4999$, this is correct.
Now, you need to find a way to count $ 10\sim999$.

The best way I've found is to look at the mod 4 sums for 0 to 40 , but that still isn't much prettier than casework.

0, 1, 2, 3, 0, 1, 2, 3, 0, 1, (0 - 9)
1, 2, 3, 0, 1, 2, 3, 0, 1, 2, (10 - 19)
2, 3, 0, 1, 2, 3, 0, 1, 2, 3, (20 - 29)
3, 0, 1, 2, 3, 0, 1, 2, 3, 0, (30 - 39)

It's easily seen that it balances itself for every 80 terms. It also balances itself for every 800 terms (because from 100 to 100 the pattern changes).
Conclusion -- it is balanced except for 800-999, or 880-899 and 980-999, where we get one 0, two 1, two 2 and one 3.
This can only mean that the distribution is $ (249, 251, 251, 249)$.
After excluding $ 0$, $ 4$, and $ 8$, we get a total $ \boxed{1246}$.

[hide="Solution with little casework"]From $ 1000$ to $ 4999$: For each terminating string of 3 digits, there is a unique leading digit to make the digits divisible by $ 4$. There are $ 4000$ numbers in the range, so there are $ 1000$ answers here.

From $ 10$ to $ 999$: Throw in $ 0$ through $ 9$; we'll subtract them out later. The generating function for each individual digit's residue mod 4 is $ f(x) \equal{} 3 \plus{} 3x \plus{} 2x^2 \plus{} 2x^3$. We want the value of all the terms with exponent divisible by $ 4$ in $ (f(x))^3$. We'll get this by the "roots of unity filter", taking $ \frac14 \left( (f(1))^3 \plus{} (f(i))^3 \plus{} (f(\minus{}1))^3 \plus{} (f(\minus{}i))^3 \right)$. This comes out to $ \frac{1}{4} ( 1000 \minus{} 2 \plus{} 2i \plus{} 0 \minus{} 2 \minus{} 2i ) \equal{} 249$. So there are $ 249$ answers here. Subtract out the $ 3$ we get from $ 0$ to $ 9$.

Final answer $ 1246$.[/hide]


calculus function logarithms integration algebra polynomial ratio

I have two questions regarding Calculus:

1)  Arrange the functions $f_1(x)=x^e,$ $f_2(x)=e^x,$ $f_3(x)=x^x,$ $f_4(x)=2^x,$ $f_5(x)=(\ln x)^{2x}$ in increasing order of growth.

2) The function $f(x) = \displaystyle \int^{x^2}_4 \sqrt{9+t^2} dt$ has an inverse.  Find $(f^{-1})' (0).$

Thanks in advance.

Exponentials kill polynomials; 1 is smallest.  Hence we also conclude 3 is greater than 5.  Take ratios, 2 is greater than 4.  Obviously 3 is greater than 2.  $\ln x$ is eventually greater than any constant, so we conclude 5 is greater than 2.

Putting everything together, 1< 4 < 2 < 5 < 3


A box contains some green marbles and exactly four red marbles. The probability of selecting a red marble is $ x\%$. If the number of green marbles is doubled, the probability of selecting one of the four red marbles from the box is $ (x \minus{} 15)\%$. How many green marbles are in the box before the number of green marbles is doubled?


analytic geometry
graphing lines
slope
algebra unsolved
algebra

Change y = x - 3 so that the graph of the new equation has a steeper, negative slope and the same y-intercept.

I really don't know how to solve this problem.  I've been working on it all night long.  Please help me.

1) this doesnt belong in algebra olympiad problems

2)

The graph of y=x-3 has a point at (0,-3) and slopes up 45 degrees.  Notice -3 is the y intercept of the line.

If we want to keep the y intercept but change the slope, then we change the coefficent about x, from 1, to say,.. -3.

So y = -3x - 3 is an example of a line with the same y intercept as y = x-3 but a steeper and negative slope.


If $a_{xy}= a_{x}a_{y}$ for all $x, y$ and $a_{x+y}= a_{x}+a_{y}$ for all $x, y$, either
$a)$ Prove that this must be the sequence $a_{n}= n$.
$b)$ Provide another sequence that works.

[hide]I think $a_{i}=0$ for all $i$ works too.  I believe these are the only two sequences.  My 'proof' first shows that that $a_{0}=0$ and either $a_{1}=1$ or all $a_{i}$ are $0$.  Then, the two sequences follow. [/hide]


[quote="Franz Joseph"]Does Cramer's Rule help?[/quote]

hmm i think it does. never tried with a 4x4 matrix thou

For $4 \times 4$ determinants, just lower it into a $3 \times 3$ and everything else is a piece of cake. But evaluating heaps of them take ages.

[quote="xxxxx"]can someone explain to me how I should do a problem like this?[/quote]
Well, when a mommy equation and a daddy equation love each other very much ... :P
My gut instinct says that there is a solution without actually finding w, x, y, and z, but if not you can use Cramer's Rule.
If you are uncomfortable with it and have time on your hands than isolate each variable and substitute it into the other equations until you have a quartic with which you can identify the sum of the roots easily with vietas...

[quote="xxxxx"]$x+5y-3z+6w=13$
 $2x+8y-2z+w=42$
 $3x-7y+11z-w=23$

add them together:
$6x+6y+6z+6w=78$
$x+y+z+w=13$

Is that it?[/quote]
As I said earlier, I'm sure that there is a better solution than solving for each variable. This is in general something you should look out for when asked for the sum or product of a group of items.

[quote="Ubemaya"][quote="xxxxx"]$x+5y-3z+6w=13$
 $2x+8y-2z+w=42$
 $3x-7y+11z-w=23$

add them together:
$6x+6y+6z+6w=78$
$x+y+z+w=13$

Is that it?[/quote]
Wait, how did you get those equations?[/quote]

what do u mean? all i did was pick three out of the four equations...


Find three vulgar fractions of the same value using all the digits 1 to 9 once only. Here is one solution of the problem :
$ \frac{3}{6} \equal{} \frac{7}{14}\equal{}\frac{29}{58}$

Sorry what is a vulgar fraction? Is it just one that hasn't been reduced?

[quote="chetjan"]Sorry what is a vulgar fraction? [/quote]

Read here http://en.wikipedia.org/wiki/Fraction_(mathematics)#Vulgar.2C_proper.2C_and_improper_fractions

Here's another 3 fractions that work.

$ \frac{2}{6} \equal{} \frac{3}{9} \equal{} \frac{58}{174}$.

How did you find that? Do you just have to guess and check?

Yeah, all I did was fiddle with the numbers a bit.

Check the third post (and follow the link).

Can anyone please share some trick, i do not get how to solve(yes i looked at the third post).

Like other then trial and error.


Let $h_a$, $h_b$ and $h_c$ be the heights and $r$ the inradius of a triangle.
Prove that the triangle is equilateral if and only if $h_a + h_b + h_c = 9r$.

Only a short addition: This was nearly the same problem as in the German Mathematical Olympiad 1998/99, grade 10, round 3.

Yeah, you're right. It was used in the Bundeswettbewerb Mathematik 1988 as second question in the first stage, for short BWM1988_1_2

ha+hb+HC=9r
we know, rs=∆
∆=(a/2)*ha
ha=(2∆)/a
Similarly,
hb=(2∆)/b
hc=(2∆)/c
Add,
2∆{(1/a)+(1/b)+(1/c)}=9r
2rs{(1/a)+(1/b)+(1/c)}=9r
2s{(1/a)+(1/b)+(1/c)}=9
(a+b+c){(1/a)+(1/b)+(1/c)}=9
So, after simply,
a²c+c²a+a²b+b²a+b²c+c²b=6abc
So,
(a√c-b√c)²+(c√a-b√a)²+(a√b-c√b)²=0
Sum of square of 3 numbers is 0 , so each must be equal to 0
So,
(a√c-b√c=0, a=b
c√a-b√a=0, c=b
a√b-c√b=0, a=c
so, a=b=c
Therefore triangle is equilateral.
#Krishijivi


So how was Shaastra 08? What all did you do? I guess par,mad and all must have owned the math contests :p

It was nice man. Obviously par and sowmya were expectionaly good and madness and chapli managed to get into the finals.
Apart from that the best event of shaastra was the shaastra nights laser show. It was truly amazing and ofc we had that NUS robotics show. The events were highly competitive.

@Rakesh
Is soumya in IIT Madras?
I thought he is in ISI or CMI.

he had dropped down to participate in the events he is in ISI B'lore

What about Bharghav? he is also there in IIT M rite as second year...
You both beat him???? :o  :o   :mad:

[quote="Aravind Srinivas L"]What about Bharghav? he is also there in IIT M rite as second year...
[b]You both beat him????[/b] :o  :o   :mad:[/quote]What does that mean actually, along with the smileys? :mad:
This time I've quoted you. So no point editing your post, right?

we didnt beat him up  :P
just that we got selected in math modelling he didn't.....
btw as akash asked what did u wanna say thru ur smiley's?

Nothng meant through the smileys ofcourse :D .
See, I wanted him to win thats why.
See Akashnil, don't think that I am posting everything with reason. I just wanted to use smileys  thts why.

And wat is math modelling actually about?

[quote="pardesi"]we didnt beat him up  :P [/quote]
I think he meant that you beat him in the competition, not beat him up and that's worse and certainly not a joke. He just looks at things that way and after reading some Metathread stuff, he thinks that the seniors of the forum are only good at those stuff, look at what he said to madness [url=http://www.mathlinks.ro/viewtopic.php?t=230664]here[/url] if you saw the unedited version of course...


If $ (X, |\cdot|)$ is a normed linear space, we say that the norm is strictly subadditive if $ |x \plus{} y| \equal{} |x| \plus{} |y|$ implies that $ x$ or $ y$ is a non-negative multiple of the other.

Why is this equivalent to the condition that $ |x| \equal{} |y| \equal{} 1$ and $ |x \plus{} y| \equal{} 2$ implies $ x \equal{} y$?

What I've attempted is the following: Suppose that $ |x\plus{}y| \equal{} |x|\plus{}|y|$ but that they're not non-negative multiples. Then let $ x' \equal{} |y|x$ and $ y' \equal{} |x|y$ which are of the same length, and $ x' \neq y'$. With $ c_1 \equal{} \frac{|x|}{|x|\plus{}|y|}$, $ c_2 \equal{}  \frac{|y|}{|x|\plus{}|y|}$

Then $ x\plus{}y \equal{} \frac{|x|\plus{}|y|}{|x\parallel{}y|}\left(c_1x' \plus{} c_2y'\right)$. Now we are given that the midpoint between $ x'$ and $ y'$ is inside the open ball $ B(0, |x\parallel{}y|)$, but if we could prove that the whole segment betwen $ x'$ and $ y'$ lies in the ball, then $ |x|\plus{}|y| \equal{} |x\plus{}y| \equal{}  \frac{|x|\plus{}|y|}{|x\parallel{}y|}\left|c_1x' \plus{} c_2y'\right| < |x|\plus{}|y|$ which is a contradiction.

I still don't know why this would be true though, and I've played with this forever :(

We shall prove that for every $ v \equal{} c_1x' \plus{} c_2y'$: $ |v| < N$, where $ N\equal{}|x\parallel{}y|\equal{}|x'|\equal{}|y'|$, $ c_1 \plus{} c_2 \equal{} 1$, $ c_1$, $ c_2$ are positive.

Assume that $ c_1 >1/2> c_2$. Then
\[ v\equal{} (c_1\minus{}c_2) x' \plus{} 2c_2\frac{x'\plus{}y'}{2} \leq (c_1\minus{}c_2) |x'| \plus{} 2c_2\left|\frac{x'\plus{}y'}{2}\right| <
(c_1\minus{}c_2) N \plus{} 2c_2 N \equal{} N.\]

(I hope I correctly understood what you are asking about.)

Yes, that's exactly what I meant. Thanks! I can't believe it was that simple (I was trying to keep halving midpoints and then apply continuity but couldn't really do it since I lacked compactness).

This problem was stated as a completely trivial fact in my book, but it's not so trivial to me. Are you aware of any better method of doing this?

Maybe, the following proof is a bit simpler. Note that if $ |x\plus{}y| \equal{} |x| \plus{} |y|$ then for every $ \alpha\in [0,1]$:
\[ |x\plus{}\alpha y| \geq |x\plus{}y| \minus{} (1\minus{}\alpha) |y| \equal{} |x| \plus{} |y| \minus{}  (1\minus{}\alpha) |y| \equal{} |x| \plus{} \alpha |y|.\]
and, clearly, $ |x\plus{}\alpha y| \equal{} |x| \plus{} \alpha |y|$.

Now, if $ |y| > |x|$, let $ \alpha \equal{} |x|/|y|$, and $ \hat y \equal{} \alpha y$, we have
\[ |x\plus{} \hat y| \equal{} |x| \plus{} \alpha |y| \equal{} |x| \plus{} |\hat y|,\]
and $ |x| \equal{} |\hat y|$, thus $ x\equal{}\hat y$.

I'm slightly confused. Isn't the definition of a subadditive that $ \| x\plus{}y \| \leq \| x \| \plus{} \| y \|$, with equality iff. $ \exists k \in \mathbb{R}^{\geq 0} \; : \; x\equal{}ky$

Yes TZF, that is equivalent to the definition Kalle is using plus the triangle inequality.


What is the least number of additional squares that must be drawn to make this figure have a vertical line of symmetry and a horizontal line of symmetry?  The figure may not be rotated.

[asy]size(3cm,3cm);

draw((0,1)--(0,3)--(1,3)--(1,2)--(2,2)--(2,1)--(3,1)--(3,0)--(2,0)--(2,1)--(0,1));
draw((0,2)--(1,2));
draw((1,1)--(1,2));[/asy]

you want a H shape, since that's the smallest figure that has the given squares & axes symmetry.
you add 1 on the bottom left, and 2 on the right. so that's 3 squares.

if u add a box to the right column in the center, top right corner, and bottom left corner, you form a letter H shape. thus the answer is [b]3[/b].


Describe each of the following graphs of Y. Give (a) its amplitude, its maximum and minimum and its period.

Y=3sin1/3x

y=tan2A

these two should give you plenty of work lol it sure did with me.

Y=3sin1/3x

Amp: 3
Max: 3
Min: -3
Period: 6pie

y=tan2A

Amp: 1
Max: infinite
Min: -infinite
Period: pie/2

This is easy if you understand the topic you're studying.  It's at a grade 10/11 level according to the school baords in Canada (which in my opinion has a low level in mathematics...soon to take out calculus in the Toronto area).  Try to refer to a textbook before posting such things here. :)

Masoud Zargar

[quote="seamusoboyle"]I think th first one is tricker than you give it credit...[/quote]

That depends...is the argument $(1/3)x$ or $\frac 1{3x}$?

id say if it was $\frac{1}{3}x$, he'd have written $\frac{x}{3}$


What is your opinion on arranged marriages?

They have no choice but if they *happened* to want to marry their match they agree, not necessarily do the parents count this but they agree.

It is important to distinguish between forced marriages, which you seem to be referring to, and ones that are merely arranged (not forced).

If the existing culture in a certain location/family has arranged marriages taking place most of the time, I think that continuing this is completely fine- as long as the potential spouses don't mind is fine. Forced marriages are a different matter.

[quote="bpms"][quote="JavaMan"]OK if both parties agree to it and want it. Bad if forced.[/quote]
Generally the children have no choice, the parents force them to. This is usually because the families are friends, and they want to merge the families.[/quote]

Um, that isn't the case for everyone that I know in my family.

[quote="7h3.D3m0n.117"]Arranged marriages are okay if, like what JavaMan said, both parties agree.[/quote]
Why arrange it at all if both parties agree to marriage?

[quote="solafidefarms"]Well it would provide a channel for one to divert all their romantic attentions to, though of course if one really disliked the desired spouse I suppose it wouldn't be helpful in that regard. And that way one wouldn't have to worry about the 50 million girls who might be the right one and would only have to deal with the one, reducing the romantic distraction.[/quote]
The problems are numerous:

a) He/she might not be the right one (I can think of some better reasons, but practically, personality might be one. Being gay could also put a damper on things.)
Also, consider if the other person does not like you?

b) What if one if attracted to to other chicks/dudes during the course of childhood?

c) Why must one have any romantic distraction at all? Why must marriage be required? Why must sex and marriage and children be broached at such a young age. I advocate cursory theoretical knowledge of the human body at a young age, but I do think some children aren't mature enough to appreciate the finer points, like what's entailed in raising a child.
(And from experience I can tell you: having an infant in the house sucks. Being the primary caregiver would suck more.)

d) What if one weds before one is ready, but only because culture and family dictates he or she must. I may be on friendly terms with quite a few girls in my grade (this forum needs [/brag] tags), but I wouldn't want to marry them, and certainly not raise children with them. Moreover, what if I find out when my hormones die down that marrying the girl with the biggest breasts and thinnest waist wasn't such a good idea after all if she can't do anything useful.

e) There's no "get out of jail free card". Marriage is a pretty big thing, and I would assume divorce is taboo or looked down upon in whatever society we are talking about.

[quote="solafidefarms"]Well it would provide a channel for one to divert all their romantic attentions to, though of course if one really disliked the desired spouse I suppose it wouldn't be helpful in that regard. And that way one wouldn't have to worry about the 50 million girls who might be the right one and would only have to deal with the one, reducing the romantic distraction.[/quote]

The whole point of love is trying to find someone you match with; if it takes 50 million tries, than so be it. Forcing someone into a marriage isn't going to work out just because it 'focuses their romantic attentions.'

[quote="Elemennop"]The whole point of love is trying to find someone you match with[/quote]

You're taking it as an axiom here that two people either "match" or "don't match."  Real life doesn't seem to bear this out: people who don't like each other at one point in time often find that their opinions change with time and increased familiarity, and people who believe they like each other very much often discover that positive feelings wear thin.  Not that I think this has very much bearing on the larger question, but you might consider trying to make your argument without leaning on that particular support.

Does arranged marriage mean the marriage arranged by others such as parents?

[quote="ifai"]Marriage should NOT be arranged by others, as it is a great affair in the life (although there are many divorced people nowadays). And the arranged marriages are ok if and only if both of the two people also want and agree.[/quote]

in real cases, the two younge people have to agree because they don't have a choice


calculus
integration
function
real analysis
real analysis unsolved

I want to know if the following is true (for another problem I am working on)

If $ f_n\to f$ a.e. in $ L^p(\Omega)$ with $ \|f_n\|_p,\|f\|\leq C$ for some uniform constant $ C$ and $ \mu(\Omega)<\infty$, does it follows that $ f_n\to f$ in $ L^p(\Omega)$?

I'm pretty sure it is (using Egoroff's theorem to get $ L^p$ convergence almost everywhere, and handling the remaining bit by noting that the integral of an $ L^p$ function over a set of very small measure cannot be very large), but I would like it if someone would confirm (not provide a proof) for me :)

What about $ f_n \equal{} n^{1/p}\,\mathbf{1}_{[0,1/n]}$ on $ [0,1]?$ Bounded $ p$ norm, bounded $ 1$ norm, tends to zero a.e. Doesn't tend to $ 0$ in the $ p$ norm.

Now if you add the hypothesis $ \|f\|_p\to \|f\|_p$ ...

I think I see the mistake in my 'proof', and I should have checked the three 'typical' examples (only two on a set of finite measure) where a.e. convergence doesn't imply L^p convergence.  I hope I can get my original problem now!


Same idea http://www.mathlinks.ro/Forum/viewtopic.php?highlight=root&t=134075  :maybe:


The integer $ x$ has 12 factors. The numbers 12 and 15 are
factors of $ x$. What is $ x$?

fine from the common multiples of 12 and 15.

L.C.M of 12 and 15 is 60.

and 60=2^2*3*5 so 60 has 12 factors.

...

and thats it.

x is 60.


floor function
number theory
number theory proposed

Let $ p$ be a prime number with $ p>5$. Consider the set $ X \equal{} \left\{p \minus{} n^2 \mid n\in \mathbb{N} ,\ n^2 < p\right\}$.
Prove that the set $ X$ has two distinct elements $ x$ and $ y$ such that $ x\neq 1$ and $ x\mid y$.

[i]Albania[/i]

We have $p - 1 \in X$ and $p - \lfloor \sqrt{p} \rfloor^2 \in X$. I believe $(p - \lfloor \sqrt{p} \rfloor^2 ) \mid p - 1$, for any $p$ prime (but I don't have any proof for the moment).
EDIT: And in fact for $p>5$ we have $(p - \lfloor \sqrt{p} \rfloor^2 )>1$, so the given numbers would meet the condition.

[quote="cyberjoe"]I believe $(p-\lfloor\sqrt{p}\rfloor^2)|(p-1)$, for any $p$ prime.[/quote]

Unfortunately, this result is not true.  The least counter-example is $p=23$, for which $(p-\lfloor\sqrt{p}\rfloor^2)=23-4^2=7\not|22$.

I'll chance offering my own proof.  Let $p=n^2+k$.  Clearly $0<k<2n$, and $k\ne n$.

Let $q=|n-k|$.  Clearly $0<q<n$, and $p-q^2=n^2-(n-k)^2+k=k(2n-k+1)$, so $p-q^2$ is a multiple of $p-n^2=k$, and $q<n$ means the two must be distinct.

BTW, isn't that the source of the problem : [i]Math. Magazine, Problem 1438, Proposed by David M. Bloom[/i] ?  :)
(And thanks Diarmuid, I strugled for some time trying to find a proof ;))

The source was another. After the solution I will tell it to you

Actually, just after I posted my attempt, I came across the source.

The solution given was the same as mine (up to notation), except that it took care of the case where $k=1$, which I forgot.

This is the link:
[url=http://www.kalva.demon.co.uk/balkan/bsoln/bsol962.html]www.kalva.demon.co.uk/balkan/bsoln/bsol962.html[/url]

My solution.
Let $1\in X$, so $p-n^2=1 \Longleftrightarrow p=n^2+1 $ with $n$ even number.
Now $ p-(n-1)^2=n^2+1-(n-1)^2=2n$ so the numbers $ y=p-(n-1)^2=2n $ and $x=p-1^2$  satisfy the conditions required.

Let $1\notin X$ and $n=[\sqrt p]$. Now $ n^2+1<p<(n+1)^2$.
If we put $x=p-n^2 $ we have $ p<n^2+2n $ and $p \neq n^2+n$.
So $ x-n\neq0 $ and $ 0<x<2n $; in other words, $ 0<\left|x-n\right|<n$.
Now we choose $ y=p-(x-n)^2\in X $, and we will have that $y=p-n^2+2nx-x^2=x(1+2n-x)$.
So $x$ divides $y$.

Let $p=r^2+q$ which $r=[\sqrt p], 1\le q\le2r-1$
if q=1 then r is even we have $p-(r-1)^2|p-1$ which equals to $2r|r^2$
In other case, we prove that $q|r^2+q-s^2$ holds for some s between 1 and r-1.
that's equal to $q|(r+s)(r-s)$  which can be easily made by letting one of the factors be q.

[hide="Solution"]
Let $m^2<p<(m+1)^2$ and $p=m^2+k$.  I claim that $\frac{p-a^2}{p-m^2}=\frac{m^2+k-a^2}{k}$ is an integer for $k\not=1$.  We have $\frac{m^2+k-a^2}{k}=1+\frac{m^2-a^2}{k}=\frac{(m-a)(m+a)}{k}$.

We know that $a<m$ and $k<2m+1$.  Thus, $m+a$ or $m-a$ can take on any values between $1$ and $2m-1$ excluding $m$.  We see that if $k=2m$, it would mean that $p=m^2+2m$ which is composite when $m>1$.  If $m=1$, then $p<5$.  If $k=m$, then $p=m^2+m$ is composite when $m>1$.  If $m=1$, then $p<5$.  Thus we can choose $a$ to be $|m-k|$, so we have $\frac{p-|m-k|^2}{p-m^2}\in\mathbb{Z}$.  However, we skip the case when $k=1$.

If $k=1$, then we have $p=m^2+1$.  Then I claim that $\frac{p-a^2}{p-(m-1)^2}=\frac{m^2+1-a^2}{2m}$ is an integer.  $m$ must be even since $p$ is prime.  Then let $a=1$, we have $\frac{m^2+1-a^2}{2m}=\frac{m}{2}$ which is an integer.  $\blacksquare$
[/hide]

Also, this should be moved to the Number Theory Unsolved Problems forum because it's from a contest.

The idea is pretty simple.

If $1\in X$ (i.e. $p=n^2+1$), take $x=p-(n-1)^2$ and $y=p-1^2$.

If $1 \not \in X$, then let $p=n^2+k$. Now take $x=p-n^2$ and $y=p-(k-n)^2$. (Since $p$ is a prime, so $0<k<2n\implies -n<k-n<n\implies y\in X$.)

And if the condition $n^2<p$ is removed, for every $x\in X$ we can find an appropriate $y$ such that $x|y$. (Using the same idea; for $x=p-m^2$, take $y=p-(p-m^2\pm m)^2$ )

Motivation can be from trying out small values.

Solution:

Let $k^2<p<(k+1)^2$
1. If $p-k^2 = 1$. Choose $x = p-(k-1)^2 = 2k$ and $y = p -1 = k^2$. It's obvious $2k|k^2$ since $k$ is even.
2. If $p-k^2 > 1$. Choose $ x = p-k^2$ and $y =p-(k-x)^2$. It's obvious $x|y$. It's enough to show that $y$ lies on $X$ (and $\neq x$) which happens if 1 $\leq (k-x)^2 < k^2$. The leftmost inequality, $1\leq (k-x)^2$, is true since if $k=x \implies p=x(x+1)$ would be a contradiction. The right most inequality is equivalent to $(k-x)^2 < k^2 \iff 2k> x>0$. Now if $x \ge 2k$, we have $(k+1)^2 > p=k^2+x \ge k^2+2k \implies p= (k+1)^2 -1 = k(k+2)$, contradiction.

You should [i]think twice[/i] before plagiarizing problems for an AoPS mock  ;) (if you know you know)

Delete any elements of $X$ that are equal to $1$, and let $p=a^2+b$ where $2 \leq b \leq 2a+2$, so $b=p-a^2$ is the smallest element of $X$. We want to find $y<a$ such that $p-a^2 \mid p-y^2 \iff p-a^2 \mid a^2-y^2=(a-y)(a+y)$. We now consider these cases:
[list]
[*] If $2 \leq b<a$, then let $y=a-b>0$.
[*] If $b=a$, then $p=a^2+a$, but since $a \geq 2$ as $p\geq 7$, $a \mid p$ while $a<p$, so $p$ is not prime. Hence we discard this case.
[*] If $a<b<2a$, then let $y=b-a>0$.
[*] If $b=2a$, we again have $a \mid p$ and $2\leq a<p$, so we discard this case.
[*] If $b=2a+1$, we have $p=(a+1)^2$ which is absurd, so we discard this case.
[*] If $b=2a+2$, then $p=a^2+2a+2$, hence $a$ is odd. Then take $y=1$, so $2 \mid a-y \implies 2(a+1) \mid (a-1)(a+1)$.
[/list]
This covers all cases, so we're done. $\blacksquare$


search
inequalities unsolved
inequalities

Try to search for it. It is very famous(Japan 1997).

[color=#FF0000][moderator edit: yes, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=54&t=146]here[/url], [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=31804]here[/url] or [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=28678]here[/url].] [/color]

it was also proposed in 1998 greece olympiad


function
trigonometry
real analysis
real analysis unsolved

Is there anyone who knows how to solve the following problem?

A billion thanks!!  :)

One note: the intial position is, in effect, discontinuous at $x=1.$ By using Fourier series techniques, we will find a "weak solution" to this PDE, but a classical solution - one with a twice-differentiable function - is too much to ask for. For this hyperbolic equation, some discontinuity should persist for all $t,$ although its location will migrate.

Hint: with that discontinuity, when we go to write

$u(x,t)=\sum_{k=1}^{\infty}b(k)\sin(\pi kx)\cos(\pi ckt),$

we should have that $b(k)=O\left(\frac1k\right)$ as $k\to\infty.$

Thanks Kent!!  :)
I come up with the following result after my computation. But I think there're some mistakes in there, esp for the n=0 case. Is there any way to include the n=0 case? Any other mistakes?  :huh:


Cute one, I like it :

Prove that for all a, b, c > 0 :

(a - ab + b)^(1/2) + (b - bc + c)^(1/2) >= (a + ac + c)^(1/2)

This has kind of an unexpected soln:

Let O, A, B, C be points in the plane. OA=a, OB=b, OC=c s.t. angles AOB and BOC are 60 deg. Then angle AOC = 120 deg. By applying cosine rule we find the left expression to be AB+BC and the third one to be AC, so AB+BC>=AC, triangle ineq. I like this one!

Indeed. It's a very nice one I think.

The idea of geometric interpretation is sometimes useful !


algebra
polynomial
group theory
abstract algebra
geometry
3D geometry
Galois Theory

Suppose that $ f\in\mathbb{Q}[x]$ , that $ f$ is irreducible over $ \mathbb{Q}$ and that $ \deg f=3$. Let $ L$ denote the splitting field for $ f$ over $ \mathbb{Q}$. What are the possible values $ [L: \mathbb{Q}]$? Give an example of a polynomial $ f$ to illustrate each possibility. Answer the same question if one omits the requirement that $ f$ be irreducible over $ \mathbb{Q}$.

This isn't too bad.  The key here is if $ f$ is irreducible, then one of the roots of $ f$, say $ \alpha$, has the property so that $ F(\alpha)$/$ F$ is a degree $ 3$ extension.  Thus because the Galois group must be a subgroup of $ S_{3}$, the only two possibilities are $ A_{3}$ and $ S_{3}$, as they both have size divisible by $ 3$. (Note: they also must be transitive, and these are the only two subgroups of $ S_{3}$ that are).  To show they are both possible, note that the splitting field of $ x^{3}-2$ of $ \mathbb{Q}$ is simply $ \mathbb{Q}(\omega, 2^{\frac{1}{3}})$, where $ \omega$ is a cube root of unity.  This extension is of degree $ 6$ and corresponds to the Galois group $ S_{3}$.  On the other hand, the splitting field of the polynomial $ x^{3}+x^{2}-2*x-1$ is irreducible and has Galois group $ A_{3}$, though this is less obvious.  The key to this being only a degree $ 3$ extension is that the discriminant
$ D=(ab)^{2}-4b^{3}-4a^{3}c-27c^{2}+18abc$
$ =(-2)^{2}-4(-2)^{3}-4(-1)-27(-1)^{2}+18(2)=49$ is a square.

If $ f$ does not need to be irreducible, then in addition to the extensions above, we can also have two more cases.  The first is trivial, where $ f$ has $ 3$ rational roots and so the extension is of degree 1.  The second is when $ f$ splits into a rational linear factor and an irreducible quadratic.  Hence the extension is of degree $ 2$, and the Galois group is simply $ \mathbb{Z}_{2}$.


so this is an article from the onion(btw, its not a real newspaper so dont get all offended)
but it's pure genius

http://www.theonion.com/news/index.php?issue=4133&n=2

LOL That was funny.....how did you find it?

just visit the onion website compulsively, i do

Do I believe in gravity now ?? :D
Hilarious website overall.... :rotfl:  :rotfl:

hahahha, that was totally rad!!! :lol:  :lol:

lolll...wow...is this linked from like bored.com or something?


trigonometry
algebra
polynomial
system of equations

I'm trying to learn how to do partial fraction. But my book (trig book) gives poor explanation and well, I'm okay with the simple ones by guess-and-check. But I don't know how to do the ones that are complicated.

Only one I did and got right..

$\frac {3}{x^2+x-2}$

[hide="My way"]

Since bottom factors into (x+2)(x-1) and x+2-(x-1) = x+2-x+1 = 3, we get:

$\frac {1}{x-1} + \frac {1}{x+2}$[/hide]

But my guess wouldn't work for ones like $\frac {x^2+12x+12}{x^3-4x}$ or even worse, $\frac {x^3-x+3}{x^2+x-2}$. (That is worse, right? :? )

Can anyone give me how to do each of them with some explanations? I noticed how useful knowing how to manipulate these can be useful but I'm not so confident on using them.. :(

I suggest [url=http://www.google.com]googling[/url] "Partial Fraction Decomposition".  Tell us if you still need help.

Just an edit in the one you got right it should be a minus sign.

I DID googling. ;)

I even looked at some threads but I practically don't get how to use on the real problems. That's only problem. :(

A couple things to note that I don't think came up in that other post:

Your second example really isn't all that bad.  Partial fraction decomposition is for when the degree of the polynomial in the numerator is less than that of the denominator.  Since that isn't the case, you can use polynomial long division and end up with $x-1+ \frac{2x+1}{x^2+x-2}$ and that is where the partial frac decompin comes in.  But that isn't all that daunting, now is it?

*edit: nevermind, taken care of  :lol:

Generally, you should try a system of equations if you know what the denominator factors into.

$\displaystyle \frac {3}{x^2+x-2}=\frac{A}{x+2}+\frac{B}{x-1}$
$\displaystyle 0x+3=Ax-A+Bx+2B$
$0=A+B$
$3=2B-A$
$A=-1, B=1$

$\displaystyle \frac {3}{x^2+x-2}=\frac{1}{x-1}-\frac{1}{x+2}$

Of course, there are special cases and whatnot...like if a factor is a square, you have to include it twice:
$\displaystyle\frac{Ax^2+Bx+C}{x^3-4x^2}=\frac{D}{x}+\frac{E}{x^2}+\frac{F}{x-4}$

My book explained it quite well... although my book might be your book.

Well here's an example:
$\frac{y}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}$
($y$ is just a variable for the numerator)  Now we just need to solve for A and B, which is relatively simple when we multiply both sides by $(x+1)(x-2)$
$y=A(x-2)+B(x+1)$
Now we sub in $x=2$ and $x=-1$ to get rid of the $A$ and $B$ terms, respectively, and solve for each of them.  So, say $y=2x+3$.  Then:
$A = \frac{-1}{3}$
$B = \frac{7}{3}$
So the new fraction is
$\frac {-1}{3(x-2)} + \frac{7}{3(x+1)}$

At least I think that's right.  I don't have my book with me to make sure, and I don't feel like verifying those fractions are equal.


I forgot to watch it. I was too busy preparing for my finals. :(

[quote="fireemblem13"][quote="mathnerd314"]So what did you think of Thursday's spelling bee?[/quote]
was this the national spelling bee?[/quote]
Yes.
The spelling bee was wicked.
I was hoping the Canadian girl would win.
That would of been funny if a Canadian won the national spelling bee.

:rotfl:

but how would a canadian get into the U.S. national spelling bee. i used to be good at spelling. :roll: i should have worked harder between then and middle school and i could've done something.

Last year they started admitting all English-speakers.

I did the spelling bee in fifth grade and I got 16th at state, but I never did it again because it was too nervewracking. Plus, the Rocky Mountain News bee has a stupid rule where if you get first place at state, you can't compete again. >_>

I wanted the Canadian girl to win as well.

I went from
$Spelling\longrightarrow Public Speaking\longrightarrow Math$

What does "Ursprache" in english mean and how do you pronounce it (couldn't find it in a dictionary)¿ This word exists exactly with that spelling in german and means something like the one original first language; kind of curious...

Edit: don't bother, I found it.

I probably would have written it correctly, except for the U , I would have written :
Ürsprache.
 :oops:
We don't have spelling bees
We do have New Spelling, which is incredibly cool, we just change our spelling every four or five years, and thus when you are a kid you learn two or three spellings at school (this year this is wrong but next year it won't). :huh:

Even newspapers and news broadcasters are now willing to boycot the latest version of the New Spelling, pledging allegiance to the renegade alternative White Spelling, their motivation is brilliantly expressed by this picture :

Text means :"White Spelling.  You can read and write with it." :D

[img]http://www.wittespelling.nl/imgs/AfficheFeb2006_A4.gif[/img]

I won my spelling bee at my school [size=59]in first grade though.[/size]

[quote="Captain Sugar"]I got 2nd in my state. i was close! :lol:  but i don't think i could have spelled about 4/5 of the words in that bee.[/quote]

Yeah, same here.

(Mathnerd314 continues to moan and groan about "dirndl".)

I would've spelled it wrong.

I didn't care about who was gonna win.

But, I sort of wanted a tie because both of the spellers were phenomenal.

Eh, I went from focusing on vocabulary (in middle school) to math (this year). My goal is to be second nationally in math one day. :rotfl:

As for the spelling bee, Katherine Close was back for her 5th year at NSB. I am convinced that Joe Shepherd, if his paper had let him, would have done will his 5th year too. But then, if Joe had been doing spelling this year, he might not have won Word Power. So I suppose it's all worderful. :)

(Unfortunately, I don't know Katherine, so I can't claim to know [i]all[/i] the middle school competition champions... :))

i think its easy to spell URSPRACHE, isnt it? ;)

I read all about it in the Economist. Kundalini? That's an easy one. Ursprache, not that easy.


inequalities
modular arithmetic
number theory unsolved
number theory

Let $\sigma (n)$ be the sum of divisors function. A number is called superperfect if $\sigma(\sigma(n))=2n$. Show that $p^2$, where $p$ is an odd prime, is not superperfect.

This problem has had me stumped for too long. Any and all input welcome.

[hide]
$s(p^2)=p+1$ but clearly $s(p+1)<\frac{p(p+1)}{2}$

the right-hand side of the inequality is $<2p^2$ for $p>3$ (by solving $3p^2-p>0$), so we have the proof
[/hide]

No, $s(p^2)=p^2+p+1$ since $p^2$ is also a divisor of $p^2$.

Bumping this one back up, as it has caused me a lot of stress... can anyone obtain a solution?

Excellent work, El_Doopa. A friend and I had worked on this problem for a decent amount of time, and our solution was along the same path as yours. But we couldn't eliminate case two, $p^2+p+1=q^{a_1}_1 q^{a_2}_2$. This is a nice proof, though... shows that sometimes you have to use brute force and follow it all the way through. Thanks so much for your help.

FWIW:   For all p of the from $6k+1$, $p^2+p+1$ is divisible by 3 and notdivisible by 9 so $s(p^2+p+1)$  has to be divisible by 4.
(For the other case when p is -1 mod 6,  looks like,  one has to resort to some argument  similart to the one given above)


Let $ a,b,c$ be positive real numbers. Prove that:
$ \frac {a}{9ab\plus{}(a\plus{}b\plus{}c)^2}\plus{}\frac {b}{9bc\plus{}(a\plus{}b\plus{}c)^2}\plus{}\frac {c}{9ca\plus{}(a\plus{}b\plus{}c)^2}\ge \frac {1}{2(a\plus{}b\plus{}c)}$

[quote="mathVNpro"]Let $ a,b,c$ be positive real numbers. Prove that:
$ \frac {a}{9ab \plus{} (a \plus{} b \plus{} c)^2} \plus{} \frac {b}{9bc \plus{} (a \plus{} b \plus{} c)^2} \plus{} \frac {c}{9ca \plus{} (a \plus{} b \plus{} c)^2}\ge \frac {1}{2(a \plus{} b \plus{} c)}$[/quote]

Let $ a \plus{} b \plus{} c \equal{} 3$,then the inequality becomes
\[ \frac {a}{ab \plus{} 1} \plus{} \frac {b}{bc \plus{} 1} \plus{} \frac {c}{ca \plus{} 1}\geq\frac {3}{2}\]
I thinks this is an old problem of Vasile.Just notice that
\[ \sum{\left(a \minus{} \frac {a}{ab \plus{} 1}\right)} \equal{} \sum{\frac {a^2b}{ab \plus{} 1}}\leq\frac {1}{2}\sum{a^{\frac {3}{2}}b^{\frac {1}{2}}}\leq\frac {1}{6}(a \plus{} b \plus{} c)^2 \equal{} \frac {3}{2}\]


Let $ a,b,c$ be non-nagetive real numbers satisfying $ a \plus{} b \plus{} c \equal{} 1$.
Prove that $ \sqrt {a \plus{} \frac {(b \minus{} c)^2}{4}} \plus{} \sqrt {b} \plus{} \sqrt {c}\le \sqrt {3}$

We have the same problem
[u]Problem 1:[/u]VQBC:
Let $ a,b,c$ be nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}1$.Prove the inequality:
$ \sqrt{a\plus{}\frac{(b\minus{}c)^2}{4}}\plus{}\sqrt{b\plus{}\frac{(c\minus{}a)^2}{4}}\plus{}\sqrt{c\plus{}\frac{(a\minus{}b)^2}{4}}\le \sqrt{3}\plus{}(1\minus{}\frac{\sqrt{3}}{2})(|a\minus{}b|\plus{}|b\minus{}c|\plus{}|c\minus{}a|)$
[u]Problem 2:[/u]PKH:
For all $ a,b,c\ge 0$ and $ a\plus{}b\plus{}c\equal{}1$ we have
$ \sqrt{a\plus{}\frac{(b\minus{}c)^2}{4}}\plus{}\sqrt{b\plus{}\frac{(c\minus{}a)^2}{4}}\plus{}\sqrt{c\plus{}\frac{(a\minus{}b)^2}{4}}\le 2$

this was also posted here [url]http://www.mathlinks.ro/Forum/weblog_entry.php?t=199502[/url]...

Hi thaithuan_GC, your problem is Chinese Girl's MO 2008


[quote="MCrawford"]The student council is selling school t-shirts for a fund raiser.  The number of t-shirts they can expect to sell $(t)$ is a function of the price, in dollars, charged $(p)$ with $t = 200 - 10p$.  How many t-shirts can they expect to sell at $\$7.50 each?[/quote]
[hide]
We know that the the number of t-shirts they can expect to sell is 200-10p. We also know that the price is $7.50. Plug the price into the equation.

200-10*7.5
=200-75
=125

They can expect to sell 125 t-shirts.
[/hide]

[quote="Raenef"]250-75=125[/quote]

You mean $200-75$ right?

[hide="to Raenef"] you should hide your answers and i think you made a careless mistake.  I think you mean $200-75=125$ :D [/hide]

[quote="MCrawford"]The student council is selling school t-shirts for a fund raiser.  The number of t-shirts they can expect to sell $(t)$ is a function of the price, in dollars, charged $(p)$ with $t = 200 - 10p$.  How many t-shirts can they expect to sell at $\$7.50 each?[/quote]
$t=200-10\dot 7.50=200-75=125$

[hide="answer"]t=200-10p and so t=200-75 solve for t and so t=125 :)[/hide]


kasi az zamane elame nataieje olampiad khabare daghigh dare?
avval az hame rizi badesham fizi va adabi va shimi?

elam natayej ke aslan, zamanesh ghati nist, hamoon tor ke sale pish
marhale 2 ye computer ye 10-12 roozi dirtar az hamishash elam shod.

valli... be ehtemale ziyad ta 26 miyad,

ye edeyi migan ke marhale 2, emsal khafan khahad bood o az in harfa,
avallan khob az koja midoonan, soalhayi ke tarh nashodan, sakhtan
dovoman chera bayad ba sal haye pish fargh kone?

nazaretoon chiye? :?:


Hi everyone, to all who have qualified Congrats, I'll be there too  :D .
Now to get to the point, does anbody know where we will be going on excursion?
I read somewhere that it is monsoon season in Vietnam during July, so that may restict us a bit.

Thanks in advance

You will have trip to Ha Long Bay, world's natural heritage, and Dam Long in the northern province of Ha Tay.

I wonder if the coodinators could also visit these places?


You could multiply them all out... and you'd have some cubic polynomial with imaginary coefficients...  (I think you're confusing a polynomial with a quadratic) Or... Is that problem an exact quote? I mean, could it have stated that it had real coefficients? In that case, you know that -2i is also a root so it would then be $(x^2+4)(x+3)(x-2)$

This problem is unfortunately from quote :(

[quote="chess64"][quote="heavymetal64"]Given the problem:

Write a polynomial equation with the following solutions: 2, -3, 2i

I understand this would give me (x-2), (x+3), (x-2i)

Now how would I get this in polynomial form? You cannot FOIL all three of these..[/quote]

We are looking for a polynomial that has degree at least 3. However, if we expand $(x-2)(x+3)(x-2i)$, we will get $i$ in the result, which isn't real. So it has to have at least degree 4. Thus $-2i$ must also be a solution. If we expand $(x-2)(x+3)(x-2i)(x+2i)$, we get our polynomial: \[ f(x) = x^3+x^3-2x^2+4x-24.  \][/quote]

It didn't say that the coefficients in the expanded form had to be real... therefore you can't assume that $-2i$ is also a solution, so I think that it should just be:
$(x^2+x-6)(x-2i)=x^3-(2i-1)\cdot{x^2}-(2i+6)\cdot{x}+12i$

It didn't say the coefficients had to be real... but it also didn't say the coefficients COULDN'T be real.  It also didn't say those three were the only solutions.  To keep the problem as clean as possilbe, I would follow chess64's strategy.  I believe the problem is meant to test your knowledge of polynomials (imaginary roots appear in conjugates) rather than just testing your ability to multiply.


LaTeX
trigonometry
inequalities
function
limit
calculus
calculus computations

Let {x(n)} be increasing seqences positive solutions of eqation tg x = x.
Find
lim {x(n)-x(n-1)} , n tends to infinity.

P.S. I hope that's clear. I will use TeX or LaTeX in future. And about problem: I think this must tends to pi.

Yes, your guess is correct.

Try to prove that $ x_n \minus{} \pi (n \minus{} 1/2)\to 0$, $ n\to\infty$.
[hide]For big value of $ n$, the value of $ \tan$ should be big as well.[/hide]

Use this $ x(n)\equal{}nPi \plus{}Pi/2\minus{}1/(nPi)\plus{}1/(2Pi*n^2)\plus{}o(1/n^2)$

Here is, I think, elementary solutiom of my posted problem. It is obviousus that
\[ n\pi < x_{n} < (n + 1/2)\pi\Rightarrow |x_{n} - x_{n - 1} - \pi| < \frac {\pi}{2},\quad n = 2,3,\dots\]
By inequality
\[ |x| < |\tan(x)|,\ \ \text {for}\quad |x| < \frac {\pi}{2}\]
periodicity of function $ \tan(x)$  , fact that  $ x_{n}$ are solutions of equation  $ \tan(x) = x$  and adition formula we have
\[ |x_{n} - x_{n - 1} - \pi| < |\tan(x_{n} - x_{n - 1} - \pi})| = |\tan(x_{n} - x_{n - 1})| = \big |\frac {\tan(x_{n}) - \tan(x_{n - 1})}{1 + tan(x_{n})\tan(x_{n - 1})}\big | = \big |\frac {x_{n} - x_{n - 1}}{1 + x_{n}x_{n - 1}}\big | = \frac {x_{n} - x_{n - 1}}{1 + x_{n}x_{n - 1}} < \frac {3\pi}{2[1 + n(n - 1)\pi^2]} < \varepsilon,\quad n\geq {n_{0}(\varepsilon)\]
follow requred
\[ \lim\limits_{n\to\infty}(x_{n} - x_{n - 1}) = \pi\]
[b] Remark[/b]

In fact, since

$ \tan(x_{n} - x_{n - 1} - \pi) = \tan(x_{n} - x_{n - 1}) = \frac {\tan(x_{n}) - \tan(x_{n - 1})}{1 + \tan(x_{n})\tan(x_{n - 1})} = \frac {x_{n} - x_{n - 1}}{1 + x_{n}x_{n - 1}} > \frac {\pi}{2[1 + (n^2 - \frac {1}{4})\pi^2]} > 0$

we see that must be  $ x_{n} - x_{n - 1} - \pi > 0$,  and hence
\[ \pi < x_{n} - x_{n - 1} < \frac {3\pi}{2}\]
And one more question!. While  $ \tan(x)$ is increasing function, together with previousus, is this sufficient to conclude that sequence

$ \{x_{n} - x_{n - 1}\}$

is decreasing?


if $m_a,m_b,m_c$ are medians and $r_a,r_b,r_c$ are exradios of triangle prove that :
$\frac{r_a {r_b}}{m_a{m_b}}+\frac{r_b{r_c}}{m_b{m_c}}+\frac{r_c{r_a}}{m_c{m_a}}\ge3$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=3485 and http://www.mathlinks.ro/Forum/viewtopic.php?t=16373 .

  darij


Find the $n$ th term of the sequence $\{a_{n}\}$ such that $a_{1}=1,\ na_{n}=(n-1)\sum_{k=1}^{n}a_{k}\ (n=2,\ 3,\ \cdots).$

[hide="Solution"]Denote $S_{n}= \sum_{k=1}^{n}a_{k}$. \[\begin{aligned}na_{n}= (n-1)S_{n}\ & \Longleftrightarrow\ S_{n}= \frac{na_{n}}{n-1}\\ (n+1)a_{n+1}= nS_{n+1}\ & \Longleftrightarrow\ (n+1)a_{n+1}= n\left(S_{n}+a_{n+1}\right) = n\left(\frac{na_{n}}{n-1}+a_{n+1}\right)\ \\ & \Longleftrightarrow\ (n^{2}-1)a_{n+1}= n\left(na_{n}+(n-1)a_{n+1}\right) = n^{2}a_{n}+(n^{2}-n)a_{n+1}\\ & \Longleftrightarrow\ (n-1)a_{n+1}= n^{2}a_{n},\end{aligned}\] for all $n\geq 2$. It's easy to see that $a_{2}= 1$, so \[\begin{aligned}a_{n}& = \frac{(n-1)^{2}a_{n-1}}{n-2}= \frac{(n-1)^{2}(n-2)^{2}a_{n-2}}{(n-2)(n-3)}= \cdots = \frac{\left(\left(n-1\right)!\right)^{2}a_{2}}{(n-2)!}\\ & = \frac{(n-1)!(n-1)!}{(n-2)!}= (n-1)\cdot\left(\left(n-1\right)!\right),\end{aligned}\] for all $n\geq 2$ and $a_{1}= 1$.[/hide]

I hate school; I'm never online when you post the problems, kunny  :P

[color=darkblue][b]Kunny[/b] (very nice your exercise !) and [b]Kurt Godel[/b] (very nice your proof !), here is a similar exercise.

Find the $n^{th}$- term of the following recurrent sequence : $\boxed{\ a_{1}=0\ ,\ a_{n+1}=\frac{1}{n(n+1)}\cdot \sum_{k=1}^{n}ka_{k}-\frac{1}{2}\ ,\ n\in N^{*}\ }\ .$
[hide="Answer."] $a_{n}=\frac{1-n}{n}\ .$[/hide][/color]


Prove that the numbers $ (b\minus{}c)(bc\minus{}a^2),(c\minus{}a)(ca\minus{}b^2),(a\minus{}b)(ab\minus{}c^2)$ cannot be positive simultaneously.

[quote="bigman"]Prove that the numbers $ (b-c)(bc-a^2),(c-a)(ca-b^2),(a-b)(ab-c^2)$ cannot be positive simultaneously.[/quote]

To put in $ \text{\LaTeX}$ just add two dollar signs, one at each end.

[quote="tonypr"][quote="bigman"]Prove that the numbers $ (b - c)(bc - a^2),(c - a)(ca - b^2),(a - b)(ab - c^2)$ cannot be positive simultaneously.[/quote]

To put in $ \text{\LaTeX}$ just add two dollar signs, one at each end.[/quote]

Replace $ a$ with $ b$ and replace $ b$ with $ a$, we still get the same numbers.

So WLOG , we can assume that $ a \ge b \ge c$ as it doesn't change the output numbers.

Assume that the numbers are positive simultaneously with $ (a,b,c)$. Then clearly $ a,b,c$ are pairwise distinct. So $ a>b>c$.

So $ \begin{cases}bc>a^2 .........(1) \\ ab>c^2 .........(2) \end{cases}$

[b]Case 1:[/b] $ a$ is positive. Then from $ (2)$, $ b$ is also positive. Then from $ (1)$, $ c$ is also positive. $ a>b>c>0$ gives $ bc<a^2$, which contradicts $ (1)$

[b]Case 2:[/b] $ a$ is negative, then $ 0>a>b>c$ gives $ ab<c^2$ , which contradicts $ (3)$.
[b]Case 3:[/b] $ a=0$, then from $ (2)$, $ 0=ab>c^2 \ge 0$, contradiction.

So it's impossible for any $ a$, done.

The sum of the three numbers is $ (b \minus{} c)(bc \minus{} a^2) \plus{} (c \minus{} a)(ca \minus{} b^2) \plus{} (a \minus{} b)(ab \minus{} c^2) \equal{} 0$.  Hence, not all three numbers can be positive.

I originally saw this problem on a Leningrad competition.


ΚΑΛΟ ΠΑΣΧΑ ΚΑΙ ΚΑΛΕΣ ΓΙΟΡΤΕΣ! :D  (Ξεκινω με αυτην) Λυστε την εξισωση:
$(x+y+z)^{2}=xyz$ οπου $x,y,z$ ακεραιοι.

προτείνω τρόπο λύσης κι αμα είναι σωστός βάζω λύσεις
εσείς θα μου πείτε
χ/ψ+ω
ψ/χ+ω
ω/χ+ψ
χψω/χ+ψ+ω

κι βρίσκουμε λύσεις για κάθε μια περίπτωση
χ/χ+ψ+ω (χωρίς βλάβη γενικότητας)
χψ/χ+ψ+ω(χωρίς βλάβη γενικότητας)
χψω/χ+ψ+ω δεν βγάζει λύσεις

Ωραια ακσηση! Θα ηθελα πολυ να δω την λυση...

[quote="silouan"]
Όμως $3(x+y)(y+z)(z+x)=3(x+y+z)(xy+yz+zx)-3$
[/quote]
 :?:

Η ταυτοτητα Euler ειναι αυτη [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/f932ed51cff3730d6c24514974d1ef7e.gif[/img] και πως προεκψε αυτο [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/1f044e53ed0546bfed402c1446c3de62.gif[/img] ?
Επισης το $|$ τι συμβολιζει?

Λάθος , συγχώρεσέ με  galeos .... :(

[quote="silouan"]Λάθος , συγχώρεσέ με  galeos .... :([/quote]
Sumbainoun auta :!:  :oops:

[quote="mailo"]Επισης το $|$ τι συμβολιζει?[/quote]
$a|b$ σημαίνει ότι το α διαιρεί το β τέλεια :wink:

[quote="silouan"]Wrong way .[/quote]

για ποιό λόγο?????? :o

Δεν θα γράψω ολοκληρωμένη λυση (είναι παρα πολύ μεγάλη) αλλα καποιες σκέψεις.
Η (9,9,9) είναι μια λυση.
Επισης αν η (χ,ψ,z) είναι μια λυση τοτε και η (ψz-2ψ-2z-χ, ψ ,z) είναι λυση. Το μονο που μενει είναι να δειξουμε οτι ολες οι λυσεις ανηκουν σε αυτη την ακολουθια λυσεων.
Αυτο γινεται με τη μεθοδο της επ απειρον καθοδου (infinite descent)  .
Αυτά μόνο για τους θετικούς ακεραίους .
Γιa την περιπτωση κάποιος από τους χ ,ψ και z να είναι μηδέν οι μόνες λύσεις είναι :
(α,0,-α) , (0,α ,-α) και (α,-α,0) .
Για tην περίπτωση καποιοι από τους χ , ψ , z να είναι αρνητικοί o τρόπος λύσης είναι παρόμοιος .

Καλή Ανάσταση σε όλους

Νίκος

[quote="nikos"]
Για tην περίπτωση καποιοι από τους χ , ψ , z να είναι αρνητικοί o τρόπος λύσης είναι παρόμοιος .

Καλή Ανάσταση σε όλους

Νίκος[/quote]

δηλαδή άπειρη κάθοδος???

[quote="nikos"].
Επισης αν η (χ,ψ,z) είναι μια λυση τοτε και η (ψz-2ψ-2z-χ, ψ ,z) είναι λυση. [/quote]

Εξαιρετικά Νίκο  :) ...

Α , δείτε και κάτι παρόμοιο

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=xyz%2A&t=91312

Re Silouane eksairetiko einai olo to ypoloipo!auto einai eukolo (mallon den prosekses tin askisi)
koita... vale ψζ-2ψ-2ζ=Α τοτε αν παρεις την αρχική γίνεται

χ^2 - Αχ + (ψ+ζ)^2=0

ε βάλε όπου χ το Α-Χ και θα δεις ότι είναι ισοδύναμες!

Πάντως εγώ με τη σειρά μου να πω ένα μπράβο στο Νίκο γι  αυτή τη λύση γιατι μου άρεσε πολυ (θα ήθελα βάβαια αν βρεις χρόνο να γράψεις λίγο πιο καθαρά τη σκέψη σου ρε Νick -des to sa xari....)!
ayta...

Kai kati allo....to link pou dinei o silouanos einai aplos katapliktiko!!prokeitai gia genikeusi tou parapano provlimatos mono pou ekei oi x,y,z anikoun sto N.

@ARXIMIDIS 6 mipos exeis omorfi lysi?

Έχεις δίκαιο Nick ,μετά το ποστάρισμα το κατάλαβα .
Δείτε και εδώ για μια άλλη λύση  :wink:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=abc%2A&t=113734

H βασικη μου ιδεα ειναι αν εχω 1 λυση (χ,ψ,z) με χ,ψ,z>9 προκυπτει ,αν χ ομεγαλυτερος, οτι 0<ψz-2ψ-2z-χ<χ
συνεχιζοντας ετσι φτανουμε σεμια λυση με min(x,ψ,z)<10. εκει εξεταζω καθε περιπτωση χωριστα που τελικα δεν ειναι τοσο ευκολο και ξεκουραστο..Παντως απο αυτες τις περιπτωσεις προκυπτουν ολες οι λυσεις.

ευχαριστώ πολυ τον Νίκο Ράπανο που off the record με βοηθησε
Καλή ανάσταση σε όλους μας

Φιλε Αμβροσιε καταρχην ειναι λαθος η παρατηρηση σου γιατι πρεπει $x|(y+z)^{2}$ και οχι $x|y+z$. Δεν ισχυει οτι αν $x|m^{2}$ τοτε $x|m$.

Δειτε κι εδω στην σελιδα 52. http://www.gallup.unm.edu/~smarandache/JozsefSandor2.pdf    :)

αν $z\geq y\geq x$ και $x^{2}-x(yz)+(y^{2}+z^{2})=0$ εχουμε οτι $(yz-x)^{2}-(yz-x)yz+(y^{2}+z^{2})=0$ ισοδυναμη με την $x(x-yz)+(y^{2}+z^{2})=0$ δηλαδη  αν $(x,y,z)$ μια λυση παιρνουμε  την $(y,z,yz-x)$, ομως $yz-x\geq z$ αρα οι λυσεις ειναι απειρες και παραγωνται με αυτο τον τροπο

Παρακαλω αν καποιος μπορει να βρει φορμουλα ολων των λυσεων και οχι μια αναδρομικη σχεση των λυσεων ας το κανει......Αν και πιστευω οτι ειναι αρκετα δυσκολο.....(δεν το εχω καταφερει) :( .


find the max of value of expression  :
         $A= 10a/(b+c) + 11b/(c+a) +12c/(a+b)$ where a,b,c $\in$ [1;2]

The given expression is convex in each of the variables, so you only have to check at the endpoints of the domain.

Pierre.


Bob gets 3.50 dollars allowance weekly. He donates 1/7 of it to charity and spends 1/7 of it on lunches. If he saves the rest, how many weeks will it take him to save 210 dollars?

[color=darkred]
[hide="solution o_o"]
He gets $\$3.50 a week, but he only saves 5/7 of it, which means $\$2.50 a week.

210/2.50 = 84, so 84 weeks.
[/hide]
[/color]

a quick tip: do not use the $\$ (dollar) sign inside a text with bbcode enabled. Use instead it's $\$\$\$ (if you want to use the hide tag, or click on the disable BBcode in this post checkbox, about the submit/preview buttons when you post).

[hide]3.50 * 5/7 = 2.50
210 / 2.50 = 84

84 weeks[/hide]

If I was him ,I wouldn't waste my allowence on school lunch.Just ask for his parents


Let $ABC$ be a right triangle with $\angle A = 90^\circ$.  Let $D$ be the foot of the altitude from $A$ to $BC$.  Let $E$ and $F$ be the incenters of $\triangle ABD$ and $\triangle ACD$.  Express $EF$ in terms of $r$, the inradius of $\triangle ABC$.

Denote $a=BC, b=AC, c=AB$ and the inradii of the small triangles by $r_E,r_F$. Let $E\in\triangle ABD$ and $F\in\triangle ACD.$

Let $T$ be the foot of the perpendicular from $F$ to $BC$ and $M$ be the foot of the perpendicular from $E$ to $FT$. Then $EMF$ is a right triangle with $EM=r_E+r_F$ and $MF=|r_E-r_F|$, hence $EF^2=EM^2+MF^2=2(r_E^2+r_F^2)$

Now $\triangle ABD\sim\triangle ABC\implies r_E: r=c: a\implies r_E=\frac{cr}{a}$, and $\triangle ACD\sim\triangle ABC\implies r_F: r=b: a\implies r_F=\frac{br}{a}.$

Therefore $EF^2=2\cdot\frac{c^2+b^2}{a^2}\cdot r^2=2r^2\implies EF=r\sqrt{2}$


Definition: An ordered set S is said to have the least-upper-bound property if the following holds:
If E is a subset of S, E is not empty, and E is bounded above, then sup E exists in S.

Let E be the set of all positive rationals p such that $ p^2 <2$. Doesn't E satisfy the said property and yet it can be shown that E has no least upper bound in Q?

Which tells you that $ \mathbb{Q}$ does not have the least upper bound property.


1.well in most books u will find the product when an alkyne is treated with 1 mol. hydrogen is the corresponding alkene why is the case of an alkene so formed not considered.
yes initially the rate of reaction of alkene is less but as it's conc.increases doesn't it react?

2.find the product and indiacte the major product(s) when 1 butyne is treated  with 1mol HCl?

1. You are right.

[quote="Pardesi"]well in most books u will find the product when an alkyne is treated with 1 mol. hydrogen is the corresponding alkene[/quote]

Actually this is only true with the use of a special catalyst (Lindlar catalyst for instance), otherwise a substantial amount of hydrogenated alkene is also formed.

[hide="Answer for 2"]The product is 2-chlorobut-1-ene. And what is the mechanism? How about if it were 2 eqs. of HCl?[/hide]

[quote="Pardesi"]for 2 first the vinylic carbocation is formed then why doesn't it rearrange to give a more stable allylic carbocation[/quote]

No, that is wrong. Some old textbooks also incorrectly described the mechanism that way. The mechanism is not that simply because vinylic carbocations are not stable. The actual mechanism (supported by the experimentally observed [i]third[/i] order kynetics) involves one molecule of alkyne and two of the hydrogen halide, and so it is named the "$ Ad_{E}3$" mechanism.

oh is it  :lol:
great can u describe a bit about that

"$ Ad_{E}3$" stands for "Addition electrophilic termolecular". In the transition state of this mechanism we will have the following arrangement,

[code]
             H-Cl
            /
          /
     R-C#C-H           The AdE3 mechanism
      /
    /
 H-Cl

[/code]

where the "/" lines represent nearly formed bonds and "#" represents the triple bond.


calculus
integration
symmetry
calculus computations

Let $ c$ be a constant number. Find the real value of $ x$ such that $ \int_0^c |t\minus{}x|\ dt\equal{}\int_0^x |t\minus{}c|\ dt$.

I am probably wrong, but it's quite obvious that $ x\equal{}c$. This way we get the same thing on both sides... Or?

if $ x \ge  c  \ge 0$ then

$ \int_0^c |t\minus{}x|\ dt \equal{} \int_0^c (x\minus{}t )dt \equal{} ( cx\minus{}\frac{c^2}{2})$

$ \int_0^x |t\minus{}c|\ dt\equal{} \int_0^c (c\minus{}t )dt \plus{} \int_c^x( t\minus{}c )dt \equal{}$
$ (c^2\minus{}\frac{c^2}{2})\plus{}(\frac{x^2}{2}\minus{}cx)\minus{}(\frac{c^2}{2}\minus{}c^2)\equal{}c^2\minus{}cx\plus{}\frac{x^2}{2}$

setting these equal we get

$ cx\minus{}\frac{c^2}{2} \equal{} c^2\minus{}cx\plus{}\frac{x^2}{2}$

or
$ 0\equal{}3c^2\minus{}4cx\plus{}x^2$ so $ x\equal{}\frac{4c \pm \sqrt{16c^2\minus{}12c^2}}{2}\equal{}2c \pm c \equal{} c, 3c$

if $ c \ge x \ge 0$ then similarly we get $ 0\equal{}c^2\minus{}4cx\plus{}3x^2$

and so $ x\equal{} \frac{4c \pm \sqrt{16c^2\minus{}12c^2}}{6}\equal{}\frac{2c}{3}\pm \frac{c}{3} \equal{} \frac{c}{3},c$

so if $ x\le c \le 0$

we have
$ \int_0^c |t\minus{}x|\ dt \equal{} \int_0^c (t\minus{}x) dt \equal{} ( \frac{c^2}{2}\minus{}cx)$

$ \int_0^x |t\minus{}c|\ dt\equal{} \int_0^c (t\minus{}c )dt \plus{} \int_c^x( c\minus{}t )dt \equal{}cx\minus{}c^2\minus{}\frac{x^2}{2}$

giving us the same solutions as the positive case, $ c \le x \le 0$ also.

now we know when we have $ xc \ge 0$ that the solutions are only $ x\equal{}\{\frac{c}{3},c,3c\}$

so what about when xc < 0 ?

let $ c>0, x<0$ then we have

$ \int_0^c |t\minus{}x|\ dt \equal{} \int_0^c (t\minus{}x )dt \equal{} ( \frac{c^2}{2}\minus{}cx)$

$ \int_0^x |t\minus{}c|\ dt\equal{} \int_0^x (c\minus{}t )dt \equal{} (cx\minus{}\frac{x^2}{2})$

setting these equal we get $ 0\equal{}x^2\minus{}4cx\plus{}c^2$
so $ x\equal{}(2 \pm \sqrt{3})c >0$ so there is no solution. and the same situation arises by symmetry when x>0 c<0.

therefore the complete set solutions is complete

That's right. The answer is $ x\equal{}\frac{1}{3}c, c, 3c$.


Let $n$ a natural number such that $(n,6)=1$
Prove that the sum of $n$ ''consecutive'' squares is multiple of $n$.


if [b]1+4(n!) 4[/b] divisible by [b]4n+1[/b]. prove that [b]n[/b] is a perfect square.

I'am interested...
How do you prove this  :?  ???

Where are you champions?  ;)
It's easy to prove that [b]4n+1[/b] is prime and that [b]4n+1[/b] is dividing [b]1+2n!+2(n!)2[/b] or [b]1-2n!+2(n!)2[/b].
It's not enough to solve this problem :(

[b]Robin Chapman's solution:[/b]
$4n+1$ must be prime (it has no factors $\leq n$, and so it is a prime or 3 times a prime, and the latter can't hold).
If $p = 4n+1$ is prime, $p = a^2 + b^2$ where:
$a = \frac{(2n)!}{2(n!)^2} (mod\ p)$ and $b = (2n)! a = \frac{-1}{2(n!)^2} (mod\ p)$.
If $p | (1 + 4(n!)^4)$ then $a^2 = \frac{(2n)!^2}{4(n!)^4} = -1/(-1) = 1 (mod\ p)$.
so $a=1(mod\ p)$ or $a=-1(mod\ p)$ and we have $a^2=1$.


inequalities
function
inequalities unsolved

[quote="Yosh..."]If $0<k\leq 1$ and $a_{i}\in R^{+}$

prove that $(\frac{a_{1}}{a_{2}+\cdots+a_{n}})^{k}+\cdots+(\frac{a_{n}}{a_{1}+\cdots+a_{n-1}})^{k}\geq \frac{n}{(n-1)^{k}}$[/quote]
the function $f(x)=\Bigr(\frac{x}{S-x}\Bigl)^{k}$ is convex, here $S=\sum_{i=1}^{n}a_{i}$. And use Jensen's inequality.

and yet again someone claims that
\[\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\geq \frac{3}{\sqrt{2}}\]
is true... :roll:

[quote="Sunjee"] the function $f(x)=\Bigr(\frac{x}{S-x}\Bigl)^{k}$ is convex...[/quote]
That's false if $0 < k < 1$ and $x < \frac{(1-k)S}2$ ...

[quote="Yosh..."]If $0<k\leq 1$ and $a_{i}\in R^{+}$

prove that $(\frac{a_{1}}{a_{2}+\cdots+a_{n}})^{k}+\cdots+(\frac{a_{n}}{a_{1}+\cdots+a_{n-1}})^{k}\geq \frac{n}{(n-1)^{k}}$[/quote]
That's manifestly false : take for example $\ n=3 ,\ k =\frac{1}{2}, a_{1}= a_{2}=1$ and  $a_{3}\rightarrow 0$ (near to $0$).
...
Of course the inequality holds if $k\geq 1$ because $f(x)=\Bigr(\frac{x}{S-x}\Bigl)^{k}$ becomes convex for $0 < x < S$.
Do you follow me?
 :cool:


用7種顏色塗在正六面體的6個面上, 不同面用不同顏色. 問可塗成多少種不同的正六面體?


geometry trigonometry quadratics algebra

THe area of a right triangle is 30. If the median to they hypotenuse is length 6, find b^2, where b is the length of the longer leg of the triangle.

thx ;)

[hide="I think..."]Let $a$ and $b$ be the legs, and $c$ the hypotenuse.  The area is $30$, so c must be $10$.  Also, $ab=60$.  We can also tell that $a^2 + b^2 = 100$.  Reducing & Combining gives us $a+yb=2\sqrt{55}$.  However, I'm stuck from there....[/hide]

[hide]The hypotenuse of a right triangle is twice the size of the corresponding median, hence $c=12$. Therefore $a^2+b^2=144.$ On the other hand, ${ab\over 2}=30\iff ab=60\iff a^2b^2=3600.$ Now we have two equations: $a^2+b^2=144,a^2b^2=3600.$ If we put $x=a^2,y=b^2$, we get $x+y=144, xy=3600$. The solutions to this system are the roots of the equation $t^2-144t+3600=0\iff t_{1,2}=\frac{144\pm\sqrt{144^2-4\cdot 3600}}{2}$

The discriminant can be factorized:
$144^2-4\cdot 3600=144^2-120^2=(144+120)(144-120)=\\=264\cdot 24=24\cdot 11\cdot 24=24^2\cdot 11$

Therefore $t_{1,2}=\frac{144\pm 24\sqrt{11}}{2}=72\pm 12\sqrt{11}$

Since we're asked for the square of the longer leg, we take $b^2=72+12\sqrt{11}$[/hide]

[hide]The length of the hypothenuse is twice the length of the median to it, so it's 12. The altitude to the hypothenuse is 5, since the area is given to be 30. Consider the acute angle to which side b is adjacent. The sine of this angle is 5/b and the cosine is b/12. So we have (5/b)^2+(b/12)^2=1. From here, you get a quadratic in b^2: (b^2)^2-144(b^2)+3600=0, which was solved by farenhajt above (I just completed the square).[/hide]

[hide]As Farenhajt wrote, we have$a^2+b^2=144$ and $ab=60$, so $(a+b)^2=a^2+2ab+b^2=264\Longleftrightarrow a+b=2\sqrt{66}$ ($>0$)

So $a$ and $b$ are the roots of the quadratic equation $x^2-2\sqrt{66}+60=0$.

$x^2-2\sqrt{66}+60=0 \iff x^2-2\sqrt{66}+66=6 \iff (x-\sqrt{66})^2=6$

The larger solution occurs when $x-\sqrt{66}=\sqrt{6}$, so $b=\sqrt{6}+\sqrt{66}$, which gives $b^2=72+12\sqrt{11}$.

(This approach is more valuable if we need to find $b$, not $b^2$...)[/hide]


Prove that, if [i]n[/i] is an integer, none of the numbers 7[i]n[/i] + 3, 7[i]n[/i] + 5, 7[i]n[/i] + 6 can be a perfect square.

The squares modulo 7 are 0,1,4,2,2,4,1 (corresponding to $ 0^2,1^2,2^2,\ldots 6^2$).  Since 3,5, and 6 are not in this list, we have that $ 7n \plus{} 3, 7n \plus{} 5, 7n \plus{} 6$ are not squares.

These numbers give $ 3$, $ 5$, $ 6$ as a reminder after dividing by $ 7$.

None of this can be a perfect square since perfect squares give

reminders $ {1,2,4}$ after dividing by $ 7$.


geometry
incenter
circumcircle
geometric transformation
reflection
power of a point
radical axis

Let $ ABC$ be a given triangle with the incenter $ I$, and denote by $ X$, $ Y$, $ Z$ the intersections of the lines $ AI$, $ BI$, $ CI$ with the sides $ BC$, $ CA$, and $ AB$, respectively. Consider $ \mathcal{K}_{a}$ the circle tangent simultanously to the sidelines $ AB$, $ AC$, and internally to the circumcircle $ \mathcal{C}(O)$ of $ ABC$, and let $ A^{\prime}$ be the tangency point of $ \mathcal{K}_{a}$ with $ \mathcal{C}$. Similarly, define $ B^{\prime}$, and $ C^{\prime}$. Prove that the circumcircles of triangles $ AXA^{\prime}$, $ BYB^{\prime}$, and $ CZC^{\prime}$ are coaxal (i.e. they two common points).

Vladimir Zajic (yetti on ML) and I have a neat synthetic proof by making use of inversion with respect to the incircle.

Let $\triangle I_aI_bI_c$ be the excentral triangle of $\triangle ABC.$ $I$ and $(O)$ become orthocenter and 9-point circle of $\triangle I_aI_bI_c.$ Thus $(O,R)$ cuts $\overline{I_bI_c}$ at its midpoint $M,$ which is also midpoint of the arc $BAC$ of $(O).$ According to problem 2.20 of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=334829]2005 mosp 2.20(g) 4.44(g)[/url], the points $M,I,A'$ are collinear. Inversion with center $I$ and power $\overline{IA} \cdot \overline{II_a}$ takes $(O)$ into the circumcircle $(U,2R)$ of $\triangle I_aI_bI_c$ and the line $BC$ into the circle $\odot(II_bI_c).$ Hence, ray $A'M$ cuts $(U)$ at the inverse $P$ of $A'$ and ray $IA$ cuts $\odot(II_bI_c)$ at the inverse $D$ of $X.$ Clearly, $P$ is the antipode of $I_a$ WRT $(U)$ and $D$ is the reflection of $I_a$ across $I_bI_c.$ Hence, the center $U_a$ of the inverse $\odot(I_aDP)$ of $\odot(AXA')$ is the intersection of $I_bI_c$ with the perpendicular to $UI_a$ though $U.$ Likewise, the centers $U_b,U_c$ of the inverses of $\odot(BYB'),$ $\odot(CZC')$ are the intersections of $I_cI_a,I_aI_b$ with the perpendiculars to $UI_b,UI_c$ through $U,$ respectively $\Longrightarrow$ $U_a,U_b,U_c$ are collinear on the orthotransversal of $U$ WRT $\triangle I_aI_bI_c.$ Since $U$ has power $-4R^2$ WRT $(U_a),(U_b),(U_c),$ then we deduce that $(U_a),(U_b),(U_c)$ are coaxal with common radical axis the perpendicular from $U$ to $\overline{U_aU_bU_c}.$ Since $(U_a),$ $(U_b),$ $(U_c)$ are coaxal, then their inverses $\odot(AXA'),$ $\odot(BYB'),$ $\odot(CZC')$ under the referred inversion are also coaxal.

Dear Mathlinkers,
for a metric proof, you can see
http://perso.orange.fr/jl.ayme   vol. 4 A new mixtilinear incircle adventure p. 36-43
Sincerely
Jean-Louis


for those of you who are taking AIME, which one are you gonna take?

It's hard to take AIME when you get a 78 on the AMC 10A!

YEAH!  I'm really embarassing myself here.

hopefully AIME II, but my proctors have yet to reply.. let alone let me know of my official score..

I'm beginning to suspect that they're doing this on purpose, since I don't attend their school anyway (and I probably got the school winner medal), but that's just my paranoia :ninja:

[quote="cognos599"]how did other people in your school do on the AMC 10?[/quote]
based on previous years' results, barely over 100

Well, I am definitely taking the AIMEFTS (American Invitational Mathematics Examination For The Smart).
I know that none of you guys are smart enough to be in league with me; only top 50 of AMC 10 and AMC 12 are taking it.  Yes, revel in the glory!

Being more serious, I'm taking the AIME I.  I really do not want to go back to school... (Spring Break has just ended for me)

hey do you guys know anyone elses scores on AIME?


ratio
geometry
circumcircle
trigonometry
perpendicular bisector
geometry proposed

Let $w$ be a fixed circle and two fixed points $A,B$ which belong to the exterior of the circle $w$. For a mobile circle $x,\ \{A,B\}\subset x$ I note $\{M,N\}\subset x\cap w$. Prove that the product $\frac{AM\cdot AN}{BM\cdot BN}$ is constant.

Let $x_1$ be a circle passing through the points A, B and tangent to the circle w at a point T. Let the common tangent t of the circles $x_1, w$ at the point T meet the line AB at a point P. Let l be any line through the point P and cutting the circle w at points M, N. The power of the point P to the circles $x_1, w$ is $PT^2 = PA \cdot PB = PM \cdot PN$. Since the point P on the line AB is fixed, the ratio $\frac{PA}{PB}$ is also fixed. Since $PA \cdot PB = PM \cdot PN$, the quadrilateral ABNM is cyclic and its circumcenter lies on the perpendicular bisector of the segment AB. Hence, the given family the circles x is identical with the family of the circumcircles of the quadrilaterals ABNM. Let D, E be the feet of normals dropped from the points A, B to the line MN. From similarity of the right angle triangles $\triangle PAD \sim \triangle PBE$, the ratio $\frac{AD}{BE} = \frac{PA}{PB}$ is fixed. AD, BE are the A-, B-altitudes of the triangles $\triangle AMN, \triangle BMN$ to their common side MN, hence, their areas are in the fixed ratio of their altitudes, $\frac{|\triangle AMN|}{|\triangle BMN|} = \frac{AD}{BE} = \frac{PA}{PB}$. Since the quadrilateral ABNM is cyclic, the angles $\angle MAN = \angle MBN$ are equal. It follows that the following ratios are all fixed:

$\frac{AM \cdot AN }{BM \cdot BN} = \frac{\frac 1 2 AM \cdot AN \sin \widehat{MAN}}{\frac 1 2 BM \cdot BN \sin \widehat{MBN}} = \frac{|\triangle AMN|}{|\triangle BMN|} = \frac{AD}{BE} = \frac{PA}{PB}$


Show that the equation for a line with nonzero x and y intercepts can be written as (x/a) + (y/b) = 1, where a is the x-intercept and b is the y-intercept. This is called the intercept form of the equation of a line.

well let $ y \equal{} nx \plus{} m$ we have an x intercept of $ x \equal{} \frac { \minus{} m}{n}$ and a y intercept of $ m$. Thus $ \frac {x}{\frac { \minus{} m}{n}} \plus{} \frac {y}{m} \equal{} 1$ is what you're trying to prove.

$ \frac { \minus{} nx}{m} \plus{} \frac {y}{m} \equal{} 1$

$ \minus{} nx \plus{} y \equal{} m$

$ y \equal{} nx \plus{} m$

$ \mathbb{Q.E.D}$


Suppose $M$ is a set of $2003$ numbers such that, for any distinct $a,b\in M$, the number $a^{2}+b\sqrt{2}$ is rational. Prove that $a\sqrt{2}$ is rational for all $a\in M$.

Let $a_{0}$ be any element in $M$. Then for any indices $i,j>0$ we have \[a_{0}^{2}+a_{i}\sqrt{2}\in Q\]  \[a_{0}^{2}+a_{j}\sqrt{2}\in Q\] substracting them we may obtain $(a_{i}-a_{j})\sqrt{2}\in Q$. Thus for every $i>0$ , $a_{i}$ has a form: \[a_{i}=a_{0}+q_{i}\sqrt{2}\] Where $q_{i}\in Q$.Now we nave for example \[a_{0}^{2}+a_{2}\sqrt{2}\in Q\]  \[a_{1}^{2}+a_{2}\sqrt{2}\in Q\] Again by substrcting we get: \[2q_{1}a_{0}\sqrt{2}\in Q.\] So \[a_{0}\sqrt{2}\in Q .\] Using our prevoius result for any $i\geq 0$, $a_{i}\sqrt{2}\in Q$.


My sister and I are going to make a test with problems that are around the AMC 12 level. We will not be posting the test until quite a bit later and are debating whether to make the test in AMC 10/12 format or to try something creative (still keeping the problems at AMC 12 level). Please post the format you prefer if you are interested in taking the test. Suggestions are welcome.

yeah i know, the one we are making is for waaaaaay later. in a couple months, so are you in or not?

I'd prefer the regular AMC format so that it's more comfortable when taking the test (unless your creative idea is exceptionally creative, unlike wearing mismatched socks for halloween and calling it "creative" *cough*cognos*cough*).
And, if you would so allow me, I'd like to be a proofreader of the test (and then, hopefully, I can get a perfect when I take it!).  Don't worry; I won't post the test all over the Internet when I get it to proofread...  And, yes, I realize that I won't be able to take the test if I'm a proofreader.

Note to others: I personally know cognos, so don't be surprised if I get accepted and if other people don't get to be proofreaders (now watch and laugh as I get rejected; I hate irony).

I'll tell you what, MathAndKnowledge, i will send you the test after i am done but before i post it. you take the test like an actual AMC 12 and send me your answers, i will grade your test and you can pm afterwards with any errors in the test. your score will be counted just like everyone else's and if there are any errors in the problems, i will give you the points for that problems*. for any questions or comments please pm, MathAndKnowledge.

* I will only give you points if you find a mistake in the actual problem/answer, not if you find a grammatical error or spelling error, etc.