trigonometry
vector
number theory unsolved
number theory

I obtained $ ad\minus{}bc\equal{}\pm 2\sqrt{3}, \ M\equal{}\frac 12 (ad\plus{}bc)$

Thanks.
But how can you find that? :(  :(  :(

\[ \boxed{\boxed{M \equal{} \pm \sqrt {3}}}\]

The key was kunny's calculation of ad - bc.  Think about how he got that, and you should get it.

actually I don't see why $ M = \pm \sqrt{3}$.  From our two equations, don't we get
\[ M \&= 1/2(ad + bc) \\
\&= 1/2(ad - bc + 2bc) \\
\&= 1/2(\pm 2 \sqrt{3} + 2bc) \\
\&= \pm \sqrt{3} + bc\]

[quote="kingofqcumber"]Oh I see!  So the key was to think of (a, b) and (c, d) as vectors on the radius 2 circle.  That's cool![/quote]

Yes. This is my first idea at first sight of the problem.

$ \left( \begin{array}{cc} a & b \\
c & d \end{array} \right) \left( \begin{array}{cc} a & c \\
b & d \end{array} \right) \equal{} \left( \begin{array}{cc} 4 & 2 \\
2 & 4 \end{array} \right)$

$ \left( \begin{array}{cc} a & c \\
b & d \end{array} \right) \left( \begin{array}{cc} a & b \\
c & d \end{array} \right) \equal{} \left( \begin{array}{cc} a^2 \plus{} c^2 & M \\
M & b^2 \plus{} d^2 \end{array} \right)$,

By the property of the determinants of matrices,

we have $ (a^2\plus{}c^2)(b^2\plus{}d^2)\minus{}M^2\equal{}4*4\minus{}2*2\equal{}12$.

Wait!  What about $ a \equal{} 1$, $ b \equal{} \sqrt{3}$, $ c \equal{} \minus{}1$, $ d \equal{} \sqrt{3}$?

That's also correct.  :|

We are to eaxmine several cases in detail back to the previous my post.

[quote="kunny"][quote="Yafmuzz"]Suppose that 4 real numbers a,b,c,d satisfy the conditions
$ a^{2} \plus{} b^{2} \equal{} c^{2} \plus{} d^{2} \equal{} 4$ and $ ac \plus{} bd \equal{} 2$
Find the set of all possible values the number $ M \equal{} ab \plus{} cd$ can take[/quote]

We can set $ a \equal{} 2\cos \alpha ,\ b \equal{} 2\sin \alpha ,\ c \equal{} 2\cos \beta ,\ d \equal{} 2\sin \beta$, thus $ ac \plus{} bd \equal{} 2\Longleftrightarrow \cos (\alpha \minus{} \beta) \equal{} \frac 12$, yielding $ \alpha \equal{} \beta \pm \frac {\pi}{3} \plus{} 2n\pi \ (n \equal{} 0,\ \pm 1,\ \pm 2,\ \cdots)$. W.L.O.G. we set $ (a,\ b) \equal{} (2,\ 0),\ (c,\ d) \equal{} (1,\ \pm \sqrt {3})$.
$ \therefore M \equal{} \pm \sqrt {3}$[/quote]

Because I calculated $ M \equal{} \pm \sqrt{3} \plus{} bc$, and $ \alpha$ and $ \beta$ are off by $ \pi/3$, it feels like the answer should just be $ M \equal{} \pm \sqrt{3} \plus{} 4 \sin \alpha \cos (\alpha \pm \pi/3)$.


integration
trigonometry
calculus
derivative
function
real analysis
real analysis unsolved

Let  $J_n (x)=\frac{1}{\pi}\int_0^{\pi} \cos (n\theta -x\sin \theta)d\theta\ (n=1,2,\cdots)$.

Prove that  $J_n (x)=\frac{x}{2n}\{J_{n-1}(x)+J_{n+1}(x)\}$.

by formula $\cos a+\cos b=2\cos \frac{a+b}2\cos \frac{a-b}2$ we have
\[
\pi (J_{n-1}+J_{n+1})=2\int_0^{\pi} cos(n\theta-x\sin\theta)\cos \theta d\theta,
\]

so
\[
-\frac{\pi x}2(J_{n-1}+J_{n+1})+\pi nJ_n=
\int_0^{\pi} cos(n\theta-x\sin \theta)(n-x\cos\theta)d\theta,
\]

we see that an expression under integral is a derivative of $f(\theta)=\sin (n\theta -x\sin\theta)$, so the integral equals $f(\pi)-f(0)=0$.

Thank you for your reply, Fedor Petrov.

To tell the truth, this problem was made an examination as follows.

Let  $J_n (x)=\frac{1}{\pi}\int_0^{\pi} \cos (n\theta -x\sin \theta)d\theta\ (n=1,2,\cdots)$.

(1) Express $J_n\ '(x)$ in terms of $J_{n-1}(x),J_n (x)$.

(2) Prove that  $J_n (x)=\frac{x}{2n}\{J_{n-1}(x)+J_{n+1}(x)\}$

(3) Prove that  $x^2 J_n\ ''(x)+J_n\ '(x)+(x^2-n^2)J_n (x)=0$

How do you solve for (1),(3)?

Thanks in advance.

kunny

$J_{n-1}$ is an integral of $cos((n\theta-x\sin\theta)-\theta)=\cos(n\theta-x\sin\theta)\cos \theta+\sin(n\theta-x\sin\theta)\sin\theta$. An integral of last summand equals $-J_n'(x)$. A first integral equals $\frac nx J_n$, as it is proved in my previous post. So, $xJ_n'=nJ_n-J_{n-1}$.

(3) follows from (1) and (2), just differentiate $xJ_n'$ one more and express $J_{n-1}$ and $J_{n-2}$ using (1) and (2).

Thank you for your replies, Fedor Petrov.

I can understand your solution which I wanted to know without using the generating function of the Bessel coefficients as follows.

\[e^{\frac{1}{2}z\left(t-\frac{1}{t}\right)}=\sum_{n=-\infty}^{\infty} t^n J_n (z)\]

kunny


Let $ a, b, c,k$ be positive real numbers.
Prove that the following inequality holds
${ \frac {a^4+k}{a^3b + k} + \frac {b^4+k}{b^3c + k} + \frac {c^4+k}{c^3a + k}}\geq3$

[quote="marin.bancos"]Let $ a, b, c,k$ be positive real numbers.
Prove that the following inequality holds
${ \frac {a^4 + k}{a^3b + k} + \frac {b^4 + k}{b^3c + k} + \frac {c^4 + k}{c^3a + k}}\geq3$[/quote]

From AM GM inequality we have

$ LHS \ge 3\sqrt [3]{\frac {(a^4 + k)(b^4 + k)(c^4 + k)}{(a^3b + k)(b^3c + k)(c^3a + k)}}$

Thus it remains to prove that :

$ \prod(a^4 + k) \ge \prod(a^3b + k)$

and From Holder inequality we have :

$ (a^4 + k)(a^4 + k)(a^4 + k)(b^4 + k) \ge (a^3b + k)^4$

$ (b^4 + k)(b^4 + k)(b^4 + k)(c^4 + k) \ge (b^3c + k)^4$

$ (c^4 + k)(c^4 + k)(c^4 + k)(a^4 + k) \ge (c^3a + k)^4$

Multiplying the three inequalities we obtain the desired result  :P

The second version is with Cauchy-Schwartz:
We have
\[{ \frac{a^{4}+k}{a^{3}b+k}+\frac{b^{4}+k}{b^{3}c+k}+\frac{c^{4}+k}{c^{3}a+k}}\geqslant \frac{(\sum a^2)^2+9k}{\sum a^3b+3k}\mbox{,}\]
and what remains to complete the proof is Vasc's famous inequality
\[ \left(a^2+b^2+c^2\right)^2\geqslant 3(a^3b+b^3c+c^3a)\mbox{.}\]
 :)

Both solutions are nice. Thanks for them. [b]Martin N.[/b]'s proof is the same as mine. [b]Abdek[/b]'s proof is a valuable one, because it avoids Vasc's inequality.


calculus
derivative
linear algebra
matrix
LaTeX
vector
real analysis

Could you please help me with this problem?

Let $ f(x) = (f_1(x), f_2(x))$ map $ R^2$ into itself where $ f_1$, $ f_2$ have continuous 1st/ 2nd partial derivatives in each variable. Assume that $ f$ maps origin to itself and that $ J_f(x)$ (Jacobian matrix) is an invertible 2x2 matrix for all $ x$. Put $ g(x) = x - (f'(x)^{-1})f(x)$

(i) Explicitly compute  $ J_g(x)$ using relation $ J_{f^{-1}}\cdot J_f =$ Identity matrix $ I_2$
(ii) Use (i) to prove there exists $ \delta>0$, $ M$ finite such that $ \|x\|<\delta \implies \|J_g(x)\| \le  M\|x\|$  and $ \|g(x)\| \le M\|x\|^2$

Thanks in advance!!!
[color=green]
Moderator edit: I $ \text{\LaTeX}$ified your post, which mainly was as simple as putting dollar signs around the formulae. See if you can use $ \text{\LaTeX}$ yourself next time :)[/color]

I do not understand: what is $ f'(x)^{\minus{}1}$ ? How do we take the Jacobian of g(x)?

$ f'(x)$ is the differential of $ f(x)$, i.e., the matrix conisting of 4 partial derivatives. $ f'(x)^{\minus{}1}$ is its inverse.

[quote="fedja"]$ f'(x)$ is the differential of $ f(x)$, i.e., the matrix conisting of 4 partial derivatives. $ f'(x)^{ - 1}$ is its inverse.[/quote]

I got       $ Jf = \begin{pmatrix} \partial_x f_1 & \partial_y f_1 \\
\partial_x f_2 & \partial_y f_2 \end{pmatrix}$

and

      $ Jf^{ - 1} = \frac {1}{\partial_x f_1 \partial_y f_2 - \partial_x f_2 \partial_y f_1} \begin{pmatrix} \partial_y f_2 & - \partial__y f_1 \\
- \partial_x f_2 & \partial_x f_1 \end{pmatrix}$

but I do not know how to get J_g? I do not think finding $ J_g$ by taking the partial derivatives is a good way to do it. It makes the answer look awfully ugly. I am curious if there is another way to do it?

The answer [b]is[/b] ugly if you want to write it down as an explicit 2 by 2 matrix :(
I honestly do not know what the author of (i) had in mind when asking such a question. You aren't doing anything wrong. Of course, you might play with the formal calculus a bit and get something like
\[ (J_g)h\equal{}[J_f^{\minus{}1}][(D J_f)(h)][J_f^{\minus{}1}]f\]
where $ DJ_f$ is the differential of $ J_f$ and $ h$ is any vector (so $ (DJ_f)(h)$ is a matrix for every $ h$ and both sides of the equation are vectors), but multiplying the matrices and writing everything down explicitly is still a nightmare even after that. The second part is already obvious from the above formula since $ f(x)\equal{}O(\|x\|)$ and the rest is $ O(\|h\|)$, but the first part is definitely something one should do not more than once in his life (preferrably never). So, alas, I cannot really help you here and I doubt anybody can. The only bright side here is that you weren't asked to do it in $ \mathbb R^3$.

Thanks for the suggestion. I will try to work on it and see where it gets me.


induction
number theory unsolved
number theory

More than two years have passed, no one is interested in it?

[quote="zhaoli"](1) Prove that there exist infinitely many positive integers $ n$ such that $ n$ divides $ 2^{n}+1$.
(2) Prove that there exist infinitely many positive integers $ n$ such that $ n$ divides $ 2^{n}+2$.

(1) and (2) are solved already. I saw them in a Chinese book. I conjecture that the following problem maybe correct.

(3) Let $ m$ be an arbitrary positive integer. Prove or disprove that there exist infinitely many positive integers $ n$ such that $ n$ divides $ 2^{n}+m$.[/quote]
1) $a_1=1$.
$a_n=2^{a_{n-1}}+1$.
If $a_{n-1}|2^{a_{n-1}}+1$ then $a_n=2^{a_{n-1}}+1|2^{a_n}+1$.
Hence there exist infinitely many positive integers $ n$ such that $ n$ divides $ 2^{n}+1$.

oh no.don't sad.can you show me  b solution?
Thank you.i think,someone interested in it but they can't solve your conjecture :D  :lol: as me and my friend-Thjch Ph4 Trjh
(the solution of Thjch Ph4 Trjh-i have read in my teacher book :D )

Let $ n>1 $ be an even integer, such that $ n \mid k = 2^n + 2 $.

We observe that $ \frac{k}{n} $ is odd, therefore: $ k-1 = 2^n + 1 \mid 2^k + 1 $

And, since $ \frac{2^k + 1}{k-1} $ is odd: $ 2^{k-1} + 1 \mid 2^{2^k + 1} + 1 \Rightarrow 2^k + 2 \mid 2^{2^k + 2} + 2 $

$ 2 \mid 2^2 + 2 = 6 $, hence the result of part (2).

An alternative proof of part (1):

$ 2^{3n} + 1 = (2^n + 1)(2^{2n} - 2^n + 1) $

$ 2^{2n} - 2^n + 1 \equiv 0 \mod 3 $,  for all odd positive integers $ n $.

Hence, by induction: $ 3^k \mid 2^{3^k} + 1 $, for all non-negative integers $ k $.


trigonometry
calculus
calculus computations

Use the trig identity: $ cos^2 x \equal{} \frac{1\plus{}cos{2x}}{2}$ and the power series for $ cos x$ to find a power series for $ f(x) \equal{} cos^2 x$


No one knows for sure. I don't know if the cutoff has ever climbed as high as a 72, but this year was probably one of the easiest ever. I know 10 people at our school scored 71 or above.

Well in 2001, 8 people got perfect on COMC, and about 70 people got 72+.

I don't think we will have 8 perfects this year, but I'm sure there's a handful of people who got 78-80. So I guess you can expect the cutoff to go be about 71/72. Which probably seems reasonable, and I would be surprised if the cutoff is less than 70 or over 73.

[quote="SM4RT"]
I don't think we will have 8 perfects this year, but I'm sure there's a handful of people who got 78-80. So I guess you can expect the cutoff to go be about 71/72. Which probably seems reasonable, and I would be surprised if the cutoff is less than 70 or over 73.[/quote]

Yes, this seems like a very reasonable estimate... I think that the cutoff will probably be about 72, because otherwise, too many people would make CMO...  :maybe:  :maybe:

Dude, I wish they could like send back like comments to see we lost points

It would be so useful.

But, yeah, I am still happy, and I think that the cutoff is at least 71, for I know 8 people with 71 or higher in my school as well, I believe.

i also got 71 on COMC and received a letter that says the committee would qualify more students by using 2008 CMO qualification repechage...

10 questions... COMC level...

i think the cutline is 72 (from group 1 to 3) and the group 4 (which have about 40 ppl) gets the Repechage and another chance to qualify!

I currently solved everything but one...  :(


Find all polynomials $ p(x)\in \mathbb{R}[x]$ such that $ p(x^{2008} \plus{} y^{2008}) \equal{} (p(x))^{2008} \plus{} (p(y))^{2008}$, for all real numbers $ x, y$.

seems no one want to solve :
put $ p(0) \equal{} c$
we have $ p(x^{2008}) \equal{} (p(x))^{2008} \plus{} c^{2008}$.so we have $ p(x^{2008} \plus{} y^{2008}) \equal{} p(x^{2008}) \plus{} p(y^{2008}) \minus{} 2c^{2008}$.
put $ g(x) \equal{} p(x) \minus{} 2c^{2008}$.so we have for every $ x,y\geq 0$ we have $ g(a \plus{} b) \equal{} g(a) \plus{} g(b)$
so we have $ p(x) \equal{} ax \plus{} b$ easily we have $ p(x) \equal{} x, \minus{} x$
indeed we only need that $ p$ is continuous(not polynomial),in this case we have $ p \equal{} x, \minus{} x,|x|, \minus{} |x|$

[quote="sumita"]
so we have $ p(x) \equal{} ax \plus{} b$ easily we have $ p(x) \equal{} x, \minus{} x$
indeed we only need that $ p$ is continuous(not polynomial),in this case we have $ p \equal{} x, \minus{} x,|x|, \minus{} |x|$[/quote]

How can you deduce this?

we have $ g(a\plus{}b)\equal{}g(a)\plus{}g(b)$ if $ p$ is continuous so $ g$ is also.so $ g$ is cauchy.this means that $ g(x)\equal{}ax(x\geq 0)$
so we have $ p(x)\equal{}ax\plus{}b$.from $ p(x^{2008})\equal{}(p(x))^{2008}\plus{}c^{2008}$ we have $ ax^{2008}\plus{}b\equal{}(ax\plus{}b)^{2008}\plus{}c^{2008}$ now easily we have $ b\equal{}c\equal{}0$and$ a\equal{}0,1,\minus{}1$


I'm [i]sure[/i] that this has been posted many times, but I cannot find any, so I'll have to ask:

Is there a site that has lots of Avatars? I kind of got mine just from copying someone else's, but I kind of want to have my own, and I'm just wondering if there is a link to an Avatar site. The ones that google gave me weren't too good...

Thanks

Just find a picture that you like and change it so that it fits the pixel requirements.  To change the picture I usually just use MS Paint.

I don't know of any such sites...well, there are buddy icon sites (those ones made basically for AIM) but they aren't very good.

http://www.avatarity.com
that's where a lot of people get theris.


modular arithmetic
number theory
prime factorization
Divisibility Theory

Prove that if the odd prime $p$ divides $a^{b}-1$, where $a$ and $b$ are positive integers, then $p$ appears to the same power in the prime factorization of $b(a^{d}-1)$, where $d=\gcd(b,p-1)$.

Let $ d \equal{} \gcd(b,p \minus{} 1) < p,b \equal{} p^kcd$, then $ \gcd(a,p) \equal{} 1$ and exist k,q, suth that $ bk \minus{} q(p \minus{} 1) \equal{} d$, therefore $ a^d \equal{} (a^b)^k(a^{ \minus{} 1})^{(p \minus{} 1)q}\equiv 1\pmod p.$
Let $ v_p(a^d \minus{} 1) \equal{} m\ge 1$, then $ a^{cd} \minus{} 1 \equal{} (a^d \minus{} 1)(a^{d(c \minus{} 1)} \plus{} a^{d(c \minus{} 2)} \plus{} ... \plus{} 1)$. Because $ a^{(c \minus{} 1)d} \plus{} a^{(c \minus{} 2)d} \plus{} ... \plus{} 1 \equiv c \not\equiv0\pmod p$ we get $ v_p(a^{cd} \minus{} 1) \equal{} v_p(a^d \minus{} 1) \equal{} m.$ If $ v_p(a^n \minus{} 1)\ge 1$, then $ v_p(a^{np} \minus{} 1) \equal{} v_p(a^d \minus{} 1) \plus{} 1$.
Therefore $ v_p(a^b \minus{} 1) \equal{} v_p(b) \plus{} v_p\left(a^{\gcd(b,p \minus{} 1)}\right)$.

[quote="Rust"]If $ v_p(a^n \minus{} 1)\ge 1$, then $ v_p(a^{np} \minus{} 1) \equal{} v_p(a^d \minus{} 1) \plus{} 1$.
Therefore $ v_p(a^b \minus{} 1) \equal{} v_p(b) \plus{} v_p\left(a^{\gcd(b,p \minus{} 1)}\right)$.[/quote]I understand all before this. But what does the first line mean and how does it imply the second line?

(assuming I corrected your typos correctly :?)

Edit: and I still can't follow the thing below. What is $ n$? Why is the identity valid? what with the extra factor $ c$?

We know $ v_p(a^{np} \minus{} 1) \equal{} v_p(a^d \minus{} 1) \plus{} 1$.
Therefore $ v_p(a^b \minus{} 1) \equal{} k \plus{} v_p(a^d \minus{} 1) \equal{} v_p(b) \plus{} v_p\left(a^{\gcd(b,p \minus{} 1)} \minus{} 1\right)$.

Does this solution work?

[hide=Solution]
We want to show that $v_p(a^b-1)=v_p(b)+v_p(a^d-1)$. Let $b=dn$. Now note that $ord_pa|d$ since $ord_pa|b$ and $ord_pa|p-1$. Thus $a^d \equiv 1\pmod p.$ Now, since $p$ is odd, and $\gcd(a,p)=1$, we can apply LTE and obtain that $v_p(a^b-1)=v_p(a^{dn}-1)=v_p(n)+v_p(a^d-1)$. Thus we just need to show that $v_p(b)=v_p(n)$. However, this is equivalent to showing that $v_p(d)=0$, which is true since $d<p$.
[/hide]


I know it has been a very long while:and my parents will probably email AoPS, but is it definitly too late too enroll in AMC 10? If I am willing to "miss" the first 5 lessons, and probably still pay the 150 dollars... (unsure)
Just wondering since I dont know if this is possible, as it has been a really long while.
Also, as a general question, if you enroll in a class late, would you still see the transcripts of when you weren't enrolled?

Also, would it be just better to buy AMC 10/AoPS V1/2 instead?

Yes, you can enroll by phone (619) 659 - 1612.  You will have access to the old transcripts.

Thanks - I may decide to go with the book instead, after looking at what the next 7 weeks have to offer...- still uncertain.:blush:
Ok, so old transripts can be accessed, thanks for answering that too

[quote="mathtyro"]Its been a couple weeks since i enrolled in the AMC 10 class, but i realized I dont have access to the class message board. does the AMC class not have one? or do I simply not have access to it?[/quote]You should be able to see it now.


geometry
circumcircle
incenter
geometry proposed

Let $ABC$ be a triangle. A circle is tangent to sides $AB, AC$ and to the circumcircle of $ABC$ (internally) at points $P, Q, R$ respectively. Let $S$ be the point where $AR$ meets $PQ$. Show that \[\angle{SBA}\equiv \angle{SCA}\]

A possible approach would be the following:
1. It is well-known that $I$ (the incenter) is the midpoint of $PQ$.
2. Since $RI$ bisects $\angle{BRC}$ we get that $RI$ passes through the midpoint of the arc $\widehat{BAC}$ of the circumcircle ($\Gamma$) of $\triangle{ABC}$.
3. Let $PQ\cap{BC}=\{T\}$ and let $V$ be the midpoint of the  $\widehat{BC}$ of $\Gamma$. Then $TIMV$ is cyclic, where $M$ is the midpoint of $BC$.
4. Let $RI\cap{BC}=\{L\}$, then $RTSL$ is cyclic too (from 2).
5. Observe that 4 is equivalent to the conclusion  :D .

I can't see why  $RI$ bisects $\angle{BRC}$ and why 4. is equivalent to the conclusion  :maybe:

Check [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17134]this[/url] out, it might be helpful.
Also note that $RA$ is a symmedian in $\triangle{PQR}$.

$RA$ meets circle $(O')$ at $D$
$RP$ meets circumcircle $(O)$ at $K$
Because $(O)$ and $(O')$ touches at $R$ so $O'P$//$OK$ so $OK \bot AB$
This means $RS$ is the bisector of $\angle BRA$
Also $\angle BPR= \angle PDR$ so Triangle $BPR$ and $PDR$ are similar
So $BP/PD=PR/RD$ $(2)$
Smilarly $CQ/QD=RQ/RD$ $(3)$
From (2) and (3) $BP/CQ=(PR*PD)/(QR*QD)$
Otherwise $DS$ is a symmedian in $\triangle{PDQ}$ so $PS/SQ=PD^{2}/QD^{2}$ and $PR/QR=PD/QD$
So $BP/CQ=PS/SQ$
Thus Triangle $BPS$ and $CQS$ are similar so $\angle SBA=\angle SCA$ $q.e.d$

See here

 http://www.mathlinks.ro/Forum/viewtopic.php?highlight=china&t=98968

Exactly the same problem with different angles  :D


Let ABC be a triangle, and let M, N be points on AB and AC, respectively, such that the tangents from B and C to the circumcircle of triangle AMN have the same length. Prove that the midpoints of AB, AC, BN, CM form a cyclic quad.

I used power of a point 2 times on each of B and C but didn't really know what to do.

This problem is rather simple.
Denote P,X,Y,Z,T be the midpoints of BC,AB,BN,AC,CM. Then P,X,Y and P,Y,Z are two lines
$ PX.PY\equal{}\frac{1}{4}CA.CN\equal{}\frac{1}{4}BA.BM\equal{}PZ.PT$, so X,Y,Z,T are concircle.

Can you explain how P,X,Y and P,T,Z (you said P,Y,Z but I believe you meant P,T,Z) are collinear?

Additionally, I wasn't sure if the circumcircle of AMN intersects BC, along with not being sure on which side of M X is on AB (same with N,Z on AC).
Is this information necessary?

First, P,X,Z are collinear because P is midpoint of BC, Z is midpoint of BN, X is midpoint of BA, and A,N,C are collinear. (It's the midline.) Similarly, P,T,Y are collinear. (T is midpoint of MC, Y is midpoint of AC.)

Now,

\[ (PZ)(PX) \equal{} \frac{CY}{2}\cdot\frac{CA}{2}\], because $ PX$ is the midline of triangle $ ABC$. Similarly,

\[ (PT)(PY) \equal{} \frac{BX}{2}\cdot\frac{BA}{2}\].

But by power of a point, the RHS of those two equations are equal, so we're done.

I see...
There is a small error: the two equations should be
\[ (PZ)(PX) \equal{} \frac {CN}{2}\cdot\frac {CA}{2}
\]
and
\[ (PT)(PY) \equal{} \frac {BM}{2}\cdot\frac {BA}{2}
\]
but otherwise it is right. Thanks for the clarity


Solve, for real x, y, z, the system:

$ 2x \plus{} 3y \plus{} z \equal{} 13$
$ 4x^2 \plus{} 9y^2 \plus{} z^2 \minus{} 2x \plus{} 15y \plus{} 3z \equal{} 82$

It is a problem that appears in some competitions and magazines before. It may be solved by using computer programs and 3D graphical interpretation of the equations...


inequalities
function
calculus
derivative
inequalities proposed

Let a,b,c>0 such that $ a\plus{}b\plus{}c\equal{}2$. Find the maximum of $ A\equal{}\sum \frac{ab}{\sqrt{2c\plus{}a\plus{}b}}$

[quote="natural"]Let a,b,c>0 such that $ a \plus{} b \plus{} c \equal{} 2$. Find the maximum of $ A \equal{} \sum \frac {ab}{\sqrt {2c \plus{} a \plus{} b}}$[/quote]
I think the maximum of $ A$ is $ \sqrt{\frac{2}{3}}.$
If $ a\equal{}b\equal{}c\equal{}\frac{2}{3}$ then, $ A\equal{}\sqrt{\frac{2}{3}}.$
We'll prove that $ A\leq\sqrt{\frac{2}{3}}.$
By Cauchy-Schwartz we obtain:
$ A^2\leq\sum_{cyc}\frac{1}{(2c\plus{}a\plus{}b)(7c\plus{}a\plus{}b)}\cdot\sum_{cyc}(7c\plus{}a\plus{}b)a^2b^2.$
Thus, it remains to prove that
$ \sum_{cyc}\frac{1}{(2c\plus{}a\plus{}b)(7c\plus{}a\plus{}b)}\cdot\sum_{cyc}(7c\plus{}a\plus{}b)a^2b^2\leq\frac{2}{3}\left(\frac{a\plus{}b\plus{}c}{2}\right)^3,$
which is true but very ugly.

[quote="natural"]Let a,b,c>0 such that $ a \plus{} b \plus{} c \equal{} 2$. Find the maximum of $ A \equal{} \sum \frac {ab}{\sqrt {2c \plus{} a \plus{} b}}$[/quote]
We see that
\[ \sum \frac {ab}{a \plus{} b \plus{} 2c} \le \frac {1}{4} \sum \left( \frac {ab}{a \plus{} c} \plus{} \frac {ab}{b \plus{} c}\right) \equal{} \frac {1}{4} \sum \left( \frac {ab}{b \plus{} c} \plus{} \frac {ca}{b \plus{} c}\right) \equal{} \frac {1}{4} \sum a \equal{} \frac {1}{2}
\]
Thus,
\[ \sum \frac {ab}{\sqrt {a \plus{} b \plus{} 2c}} \le \sqrt {\left( \sum ab\right) \left( \sum \frac {ab}{a \plus{} b \plus{} 2c}\right)} \le \sqrt{ \frac {4}{3} \cdot \frac {1}{2}} \equal{} \sqrt {\frac {2}{3}}
\]
Equality holds when $ a \equal{} b \equal{} c \equal{} \frac {2}{3}.$
:)

[quote="natural"]Let a,b,c>0 such that $ a \plus{} b \plus{} c \equal{} 2$. Find the maximum of $ A \equal{} \sum \frac {ab}{\sqrt {2c \plus{} a \plus{} b}}$[/quote]
[b]DELETED[/b]

Thanks Tchebytchev, for pointing out my mistake.

Back again :lol:

C-S: $ (\sum \frac {ab}{\sqrt {2 \plus{} c}})^2 \leq (\sum ab )( \sum \frac {ab}{2 \plus{} c})$

but $ \sum ab \le 4/3$
and
$ S \equal{} \sum \frac {ab}{2 \plus{} c} \equal{} \sum \frac {ab}{2} \frac {2}{2 \plus{} c} \equal{} \sum \frac {ab}{2} .(1 \minus{} \frac {c}{2 \plus{} c}) \equal{} \sum \frac {ab}{2} \minus{} \frac {abc}{2} (\sum \frac {1}{2 \plus{} c} )$
but
$ \sum \frac {1}{2 \plus{} c} \ge \frac {9}{8}$
thus
$ S \le \frac {1}{2}(\sum ab \minus{} \frac {9}{8} abc )$
but
$ \sum ab \minus{} \frac {9}{8} abc \equal{} ab(1 \minus{} 9/8c) \plus{} c(2 \minus{} c) \le (\frac {2 \minus{} c}{2} )^2 (1 \minus{} 9/8c) \plus{} c(2 \minus{} c) \le 1$
The last inequality follows after derivation of the function $ (\frac {2 \minus{} c}{2} )^2 (1 \minus{} 9/8c) \plus{} c(2 \minus{} c)$ on $ \mathbb{R^{*}_{ \plus{} }}$

[quote]$ \sum ab \minus{} \frac {9}{8} abc \equal{} ab(1 \minus{} 9/8c) \plus{} c(2 \minus{} c) \le (\frac {2 \minus{} c}{2} )^2 (1 \minus{} 9/8c) \plus{} c(2 \minus{} c) \le 1/2$

it's 1 and not 1/2 ;)[/quote]

it's ok now?

[quote="FOURRIER"][quote="FOURRIER"]Back again :lol:

C-S: (\sum \frac {ab}{\sqrt {2 + c}})^2 \leq (\sum ab )( \sum \frac {ab}{2 + c})

but \sum ab \le 4/3
and
S = \sum \frac {ab}{2 + c} = \sum \frac {ab}{2} \frac {2}{2 + c} = \sum \frac {ab}{2} .(1 - \frac {c}{2 + c}) = \sum \frac {ab}{2} - \frac {abc}{2} (\sum \frac {1}{2 + c} )
but
\sum \frac {1}{2 + c} \ge \frac {9}{8}
thus S \le \frac {1}{2}(\sum ab - \frac {9}{8} abc )
but
\sum ab - \frac {9}{8} abc = ab(1 - 9/8c) + c(2 - c) \le (\frac {2 - c}{2} )^2 (1 - 9/8c) + c(2 - c) \le [size=200]1/2[/size]
The last inequality follows after derivation of the function (\frac {2 - c}{2} )^2 (1 - 9/8c) + c(2 - c)$ on$\mathbb{R^{*}_{ + }}[/quote]

it's 1 and not 1/2 ;)

it's ok now?[/quote]
what?

[quote="tchebytchev"]FOURRIER wrote :
[quote] $ ab(1 \minus{} 9/8c) \plus{} c(2 \minus{} c) \le (\frac {2 \minus{} c}{2} )^2 (1 \minus{} 9/8c) \plus{} c(2 \minus{} c)$ [/quote]
why? choose $ 1 \minus{} 9/8c < 0$ (for example $ a \equal{} b \equal{} 1/2$ and $ c \equal{} 1$).[/quote]

ok I see now what you mean

if $ 1 \minus{} 9/8c < 0$

then it's easy

$ ab(1 \minus{} 9/8c) \plus{} c(2 \minus{} c) \le c(2 \minus{} c) \le 1$

ok?

[quote="shaaam"]

But the following function is concave in the given interval:

$ f(x) \equal{} \frac {(2 \minus{} x)^2}{4\sqrt {x \plus{} 2}}$
[/quote]
why this function is concave? , is Ithink that $ f''(x)\equal{}\frac{3x^2\plus{}20x\plus{}76}{16(x\plus{}2)^{\frac{5}{2}}}>0$.

[quote="tchebytchev"][quote="shaaam"]

But the following function is concave in the given interval:

$ f(x) \equal{} \frac {(2 \minus{} x)^2}{4\sqrt {x \plus{} 2}}$
[/quote]
why this function is concave? , is Ithink that $ f''(x) \equal{} \frac {3x^2 \plus{} 20x \plus{} 76}{16(x \plus{} 2)^{\frac {5}{2}}} > 0$.[/quote]

You are indeed right. The function is convex.  :oops:

P.S: I did a mistake because, i looked at the first derivative instead of second... (and first derivative is less than zero...)


Point E is chosen inside square ABCD such that AE+ED=$ \sqrt{3}$. How many such points are there? If a point is a reflection/rotation of another, then the points are considered distinct.

That depends a great deal on how large the square is ;)

The question is rather oddly phrased, though (or at least, I'm used to questions that begin "how many" having answers that are nonnegative integers) -- the points satisfying that condition (assuming you intended a unit square) form two arcs of an ellipse.

Unrelatedly, your signature ("$ \sqrt {12 \plus{} \sqrt {12 \plus{} \sqrt {12}}}\ldots \equal{} 4$") would look more like what you intend if you enclosed the $ \ldots$ withing the radical signs and added another +, i.e., $ \sqrt {12 \plus{} \sqrt {12 \plus{} \sqrt {12 \plus{} \ldots}}} \equal{} 4$.

I mistyped--I meant unit square

Also, according to your answer, is it infinity?

btw, the answer choices were
(A)2
(B)3
(C)4
(D)5
(E)more than 5


number theory
prime numbers
number theory proposed

Find all integers $k\geq0$ such that we have the maximum number of prime numbers in the following sequence:

$k+1,k+2,\cdots,k+100$

you can easily prove that all primes are in the form of : $30k+1 , 30k+7 , 30k+11 , 30k+13 , 30k+17 , 30k+19 , 30k+23 , 30k+29$
then by use mod 7 we can prove that all of numbers $30m+1 , 30m+7 , 30m+11 , 30m+13 , 30m+17 , 30m+19 , 30m+23 , 30m+29$ can't be prime because one of them is divisible by $7$(m>0) . after that we can prove that if k>7 then we have at most $25$ prime numbers , and we know that if k=1 we have $26$ primes . and we must check $k=0,1,2,3,4,5,6,7$.

Exactly...nice solution khashi :)
...and atlast after checking we'll get that the max number of primes is for $k=1$


modati hast ke hich soal e riyazi ye jaleb, kasi post nakardeh.  in soal e jalebi hast, vali kheyli assone (fekr konam, soal e #1 bood!).

for $m,n\in\mathbb{Z^+}$, prove:
$\gcd(m,n)=2\sum_{k=0}^{m-1} \lfloor{\frac{kn}{m}}\rfloor+m+n-mn$

Be darkhaast yeki az doostan in posto revive kardam :D

manzoor az $\left \| x \right \|$ baghimande $x$ bar $m'$ too chizi ke paeen neveshtam . manzoor az baghimande ham adad bein $0,1,\dots , m' -1$ e ;)

$d=(m,n) \implies m=dm' \, ,\,  n=dn' \, ,\,  (m' , n')=1 \\
\sum_{k=0}^{m-1} \left \lfloor \frac{kn}{m} \right \rfloor = \sum_{k=0}^{m-1} \left \lfloor \frac{kn'}{m'} \right \rfloor = \sum_{k=0}^{m-1} \frac{kn'}{m'}  - \frac{ \sum_{k=0}^{m-1} \left \| kn'  \right \|}{m'} \\
= \frac{1}{2} (\frac{n'(m)(m-1)}{m'} - \frac{\frac{m}{m'}(0+1+ \dots (m' -1))}{m'})
\\
\\
\implies 2 \sum_{k=0}^{m-1} \left \lfloor \frac{kn}{m} \right \rfloor = nm - n -m +d$


[quote="spanferkel"]very easy AM-GM:
$1=\prod_{cyc}\sqrt{a+b}\sqrt{a+b}=\prod_{cyc}\sqrt{(a+b)(b+c)}$
$\ge\prod_{cyc}\frac{a+2b+c}2=\frac1{16}\prod_{cyc}(2a+b+c)$
and $1=\prod_{cyc}(a+b)^{2}\ge \prod_{cyc}(4ab)=16^{2}(abcd)^{2}$
multiply both => done  :)[/quote]
Never underestimate a Chinese problem.
It should be $\sqrt{(a+b)(b+c)}\leq \frac{a+2b+c}{2}$.

:blush: oops I think your assumption is right. SO maybe this will now occupy my evening :roll:

I think you'v made a simple mistake..... :D :
$\prod_{cyc}\frac{a+2b+c}{2}=\frac{1}{16}\prod_{cyc}(2a+b+c)$
this step is also wrong..

how can you come up with this idea?it seems a bit magical for me,i'm not used to 4-variables ineq.... :P

It comes from the similar ineq for 3 varianbles, which is very wellknown!

Then, can u release the full solution...
That must be really great and helpful  :)

[quote="wupanruo"]lovaz,perhaps you mean:
$(a+b)(b+c)(c+a)\ge(a+b+c)(ab+bc+ca)$ :maybe:
a?this is not true,hava i understood sth?[/quote]
Multiply $\frac{8}{9}$ on right side and it works

[quote="Lovasz"]My solution is the inequality: \[(a+b)(b+c)(c+d)(d+a) \ge (a+b+c+d)(abc+bcd+cda+dab).\] [/quote]

How to continue?

Note the thesis $\prod_{\rm cyc}(a+b)=1$, and AM-GM. Is that OK?

then
$\prod{(2a+b+c)}\le(a+b+c+d)^{4}$
we need only to show:
$\frac{\sum{\frac{1}{a}}}{4}\ge(\frac{\sum\frac{1}{abc}}{6})^{1/3}$
and Maclaurin's ineq


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Sorry...Let A be a matrix over a field with characteristic polynomial
f_A(x)=det(xI-A).Which polynomials f_A(x) completely determine  the rank of A?

The characteristic polynomials that completely determine the rank of A are the polynomials that are not divisible by $x^2.$

If $x$ does not divide $f_A(x)$ then $A$ is invertible, hence of full rank.

If $x$ is a factor of $f_A(x)$ and $x^2$ is not, then $0$ is an eigenvalue. The geometric multiplicity of an eigenvalue is less than or equal to the algebraic multiplicity, but with the provision that the geometric multiplicity is at least $1.$ If $0$ is an eigenvalue of algebraic multiplicity $1$ then it is an eigenvalue of geometric multiplicity $1.$ That is, the null space of $A$ has dimension $1,$ which means that $\text{rank}(A)=n-1.$

However, if $x^k$ divides $f_A(x)$ for some $k>1$, the rank of $A$ could be anything between $n-k$ and $n-1.$ One could easily produce examples to show this.


In a number theory book I saw two statement seems to "contradict" each other, which one is right?

(1)... whether there are infinitely many integers n that $ n^2 \plus{} 1$ is a prime is unproved..

(2)there are infinitely many positive integers n such that $ 2^{2^n} \plus{} 1$ is prime.

it seems that if (2) is true, (1) will obviously be true(by taking n=$ 2^{2^{k \minus{} 1}}$ where k can be infinitely possible n in (2)).

(2) is far away from being proven. In fact, a lot of people think that it is prime for only finitely many $ n$.

(2) is an exercise from the book, originally stating "Prove that there are infinitely many prime by considering the sequence $ 2^{2^n} \plus{} 1$ where n is positive integer. Maybe I misunderstood?

You can prove the existence of infinitely many primes using that sequence without proving that any of these numbers is prime.
The keypoint is to prove that all these numbers are coprime, thus every such number must have at least (!) one new prime factor.


does anyone know where i can find a proof of this (monochromatic forced triangles)? thanks

Here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=goodman&t=5820

Pierre.


calculus
function
logarithms
number theory
relatively prime
algebra
polynomial

So, I am just making sure. Problems like X^X = 2 or any number or even X^(3^X) = 2 or any number can only be solved using advanced topics and calculus, right.

Calculus doesn't have much to do with it, but $x$ can't be found in terms of elementary functions at all.  Consult

http://mathworld.wolfram.com/LambertW-Function.html

[hide="Then the result is..."] That if $x^{x}= a$ then

$x \ln x = \ln a \implies$
$(\ln x) e^{ \ln x}= \ln a \implies$
$\ln x = W(\ln a) \implies$
$x = \boxed{ e^{ W(\ln a)}}$ [/hide]

Well, basically, problems like these do not have elementary solutions (at least not that I know of).

Follow-up problem: Let $x$ be the positive real number such that $x^{x}=2$. Prove $x$ irrational. Then, using the help of (*) theorem (whatever it's called), prove $x$ transcendental.

(*) if $a,b$ algebraic, $a$ is not 0 or 1, and $b$ irrational, then $a^{b}$ transcendental.

Oh yeah, and do not try to prove (*), its proof is probably way over all of our heads.

[hide="my solution"]
For the irrational part, suppose $x=\frac{p}{q}$ where $p,q$ relatively prime positive integers. Then we have
\[x^{p}-2^{q}=0\]
so by the Rational Root Theorem, $q=1$ and $x=p$, an integer. But now a contradiction is easy: either $p=1$ and $p^{p}=1$ or $p\geq 2$ and $p^{p}\geq 4$.

For the transcendental part, suppose $x$ algebraic. Then by (*), and using the fact that $x$ is irrational, $x^{x}$ would be transcendental. But of course 2 is algebraic, contradiction.
[/hide]

[quote="Clive123"]Problems like X^X = 2  can only be solved using advanced topics and calculus, right.[/quote]
To answer this question, we have to tell first, what does it mean that the equation can be solved?
The existence of a solution of $x^{x}=2$ follows from the Darboux property of continuous function $x\mapsto x^{x}$. In informal, everyday language, we often understand sentences like "equation $x^{x}=2$ can be solved" as "we are able to write the number $x$ such that $x^{x}=2$ as a 'combination' of well known numbers". And here we face an unsolvable problem: which numbers are 'well known' ?
Of course not all reals have their 'names'. In the sense mentioned above, $x^{x}=2$ cannot be solved.

The fact (*) in scorpius119's post is a very profound result of algebraic number theory: it's the Gelfond-Schneider theorem, proved in 1934.


Given a natural number $ k$, find the smallest natural number $ C$ such that

$ \frac{C}{n\plus{}k\plus{}1}\binom{2n}{n\plus{}k}$ is an integer for all integers $ n\geq k$.

[quote="easternlatincup"]Given a natural number $ k$, find the smallest natural number $ C$ such that

$ \frac {C}{n \plus{} k \plus{} 1}\binom{2n}{n \plus{} k}$ is an integer for all integers $ n\geq k$.[/quote]

For $ n \equal{} k$ the given expression takes the value $ \frac {C}{2k \plus{} 1}.$ So $ C$ must be at least a multiple of $ 2k \plus{} 1$. We'll prove that $ C \equal{} 2k \plus{} 1$ is enough!

In other words, we'll prove that $ \frac {2k \plus{} 1}{n \plus{} k \plus{} 1}\binom{2n}{n \plus{} k}$ is always an integer

Indeed, for any $ n>k$ we have:

$ \binom{2n}{n \plus{} k} \minus{} \binom{2n}{n \minus{} k \minus{} 1} \equal{}$

$ \frac {(2n)!}{(n \plus{} k)!(n \minus{} k)!} \minus{} \frac {(2n)!}{(n \plus{} k \plus{} 1)!(n \minus{} k \minus{} 1)!} \equal{}$

$ \frac {(n \plus{} k \plus{} 1)(2n)!}{(n \plus{} k \plus{} 1)!(n \minus{} k)!} \minus{} \frac {(n \minus{} k)(2n)!}{(n \plus{} k \plus{} 1)!(n \minus{} k)!} \equal{}$

$ \frac {(2k \plus{} 1)(2n)!}{(n \plus{} k \plus{} 1)!(n \minus{} k)!} \equal{}$

$ \frac {2k \plus{} 1}{n \plus{} k \plus{} 1}\cdot \frac {(2n)!}{(n \plus{} k)!(n \minus{} k)!} \equal{}$

$ \frac {2k \plus{} 1}{n \plus{} k \plus{} 1}\binom{2n}{n \plus{} k}$


induction
ratio
inequalities
algebra
binomial theorem

Using mathematical induction, prove that
n! < n^n; for n > or = 1

This is basically just using the ratio test on $ \frac{n!}{n^n}$

First of all it should read $ n!\leq n^n$

So for $ n=1$ this is true since $ 1!=1\leq 1^1$
Then if its true for $ n$ it implies its true for $ n+1$
$ (n+1)!\leq (n+1)^(^n^+^1^)$
$ (n+1)!=n!(n+1)\leq (n+1)^(^n^+^1^)$
Now divide by $ n+1$ and you get $ n! \leq (n+1)^n$
Since $ (n+1)^n \geq n^n \geq n!$ this is true.

[b]Follow-up:[/b]  Show that $ n! \le \left( \frac {n \plus{} 1}{2} \right)^n$.

[quote="Carbon57"]By AM-GM,

$ \sqrt [n]{n!}\le\frac {n \plus{} (n \minus{} 1) \plus{} ... \plus{} 1}{n}$

$ \sqrt [n]{n!}\le\frac {n \plus{} 1}{2}$

$ n! \le \left( \frac {n \plus{} 1}{2} \right)^n$[/quote]OK, now prove it using [u]induction[/u].

I think I got it, but might be missing some steps at the end  :(

for $ n = 1$ its clearly true.

we know $ 2^n n!\leq (n + 1)^n$
Then $ (n + 1)\leq (\frac {n + 2}{2})^(^n^ + ^1^) \Longleftrightarrow 2^n n! 2(n + 1)\leq (n + 2)^(n^ + ^1^)$

$ 2^n^ + ^1 n! (n + 1) 2\leq 2 (n + 1)^(^n^ + ^1^) \leq (n + 2)^(^n^ + ^1^)$
then $ \sqrt [n + 1]{2}(n + 1)\leq n + 2$
Which should be true for all $ n \geq 1$

Is the proof like this okay??? If not so (which it probably is), how should I do it?

(Anchor) n=2
2<4, which is correct

(Inductive hypothesis) n=k
k! < k^k

(Proof)
k! < k^k
k! * (k+1) < k^k * (k+1)
(k+1)! < k^(k+1) * k^k
Keeping in mind k!<k^k, we know that
(k+1)! < k^k+1 * k^k

[quote="NapoleonXIV"]
then $ \sqrt [n \plus{} 1]{2}(n \plus{} 1)\leq n \plus{} 2$
Which should be true for all $ n \geq 1$[/quote]

This statement is the essence of this problem. To make it more clear, you need to prove
[quote]
Claim: for all positive integers $ n$, there is $ \left(1 \plus{} \frac {1}{n}\right)^n \geq 2$.
[/quote]

theax

[quote](Proof)
k! < k^k
k! * (k+1) < k^k * (k+1)
(k+1)! < k^(k+1) * k^k
Keeping in mind k!<k^k, we know that
(k+1)! < k^k+1 * k^k
[/quote]

I believe you are incorrect in your conclusion because you want to show that (k+1)! < (k+1)^(k+1), not (k+1)! < k^k+1 * k^k

Can you do it? for that matter, can anybody else do it?

We are trying to get it into the form of:

(k+1)! < (k+1)^(k+1), right?
It's easy to get (k+1)! < (k+1)^k, but how do you get the exponent of k+1?

Yes, so assume $ k!<k^k$ and multiply by $ k\plus{}1$ on both sides to get

$ k!(k\plus{}1)\equal{}(k\plus{}1)!<k^k(k\plus{}1)<(k\plus{}1)^k(k\plus{}1)\equal{}(k\plus{}1)^{k\plus{}1}$

Somebody want to provide a full solution to t0rajir0u's follow-up question? :D

[u]Lemma:[/u] for all positive integers $ n$, there is $ \left(1 \plus{} \frac {1}{n}\right)^n \geq 2$.
[u]Proof:[/u] By Binomial Theorem
$ \left(1 \plus{} \frac {1}{n}\right)^n \equal{} \sum_{k \equal{} 0}^{n}{\binom{n}{k}\left(\frac {1}{n}\right)^k}$
Since $ n\geq 1$ and each term is non-negative,
$ \left(1 \plus{} \frac {1}{n}\right)^n \geq \sum_{k \equal{} 0}^{1}{\binom{n}{k}\left(\frac {1}{n}\right)^k} \equal{} 1 \plus{} \binom{n}{1}\frac {1}{n} \equal{} 2$. $ \blacksquare$

We'll proceed by induction now.
[u]Base:[/u] for $ n \equal{} 1$, $ 1!\leq \left(\frac {1 \plus{} 1}{2}\right)^1$.
[u]Inductive Hypothesis:[/u] suppose the inequality holds for some $ n\geq 1$, namely
$ n!\leq \left(\frac {n \plus{} 1}{2}\right)^n$ ..................(1)
By our lemma,
$ 2\leq \left(\frac {n \plus{} 2}{n \plus{} 1}\right)^{n \plus{} 1}$
$ n \plus{} 1 \leq \frac {(n \plus{} 2)^{n \plus{} 1}}{(n \plus{} 1)^n}\cdot \frac {1}{2}$ ..................(2)
Multiplying (1) and (2),
$ (n \plus{} 1)!\leq \left(\frac {n \plus{} 2}{2}\right)^{n \plus{} 1}$. QED

Wait, how does that solve the question?  :?:

It solves my follow-up question.  $ \left( \frac{n\plus{}1}{2} \right)^n$ is a much tighter bound, and much better approximates the real growth of $ n!$ (although we can do still better).


In a Triangle ABC, E and F are points on AC and AB respectively.
The Lines BE and CF intersect at P.
Area of BPF=5 sq. units
Area of PFAE=22 sq. units
Area of CPE=8 sq. units

Find the area of BPC....

I got the answer as 10 sq. units.

Can someone please confirm it or post the correct answer?

[hide="what I did"]I used area ratios but then I had two variables and two difficult equations to work with. Am I on the right track? [geogebra]059d0276ce2ca04b092e55933af1176a3ea769ff[/geogebra] [/hide]

I too don't know the answer. I just got the answer as 10 sq. units.

$ 10$ is correct  :)

[hide="solution"]
 [geogebra]f397c4182e60a10df725b297e6b16f3555843dba[/geogebra]

We look for triangles with the same height and collinear bases.
There are four such pairs.
Consider $ \bigtriangleup EPC$ and $ \bigtriangleup CPB$, with collinear bases $ a$ and $ b$ respectively

$ \Rightarrow \frac {5}{x}\equal{} \frac {[EPC]}{[CPB]}  \equal{} \frac {\frac {1}{2}ah}{\frac {1}{2}bh} \equal{} \frac {a}{b}\qquad(1)$

Setting up the other three equations;

$ \frac {a}{b} \equal{} \frac {22 \minus{} y}{y \plus{} 8} \qquad(2)$

$ \frac {c}{d} \equal{} \frac {27 \minus{} y}{y} \qquad (3)$

$ \frac {c}{d} \equal{} \frac {x}{8}\qquad(4)$

Now set $ (1) \equal{} (2)$ and $ (3) \equal{} (4)$, then find both in term of $ x$ and we get
$ \frac {216}{x \plus{} 8} \equal{} \frac {22x \minus{} 40}{5 \plus{} x}$

$ 22x^2 \minus{} 80x \minus{} 1400 \equal{} 0 \Leftrightarrow 2(x \minus{} 10)(11x \plus{} 70) \equal{} 0$

Now discarding the negative gives
$ \Rightarrow x \equal{} \boxed{10}$  :)
[/hide]

Neat Solution and thanks for confirming my answer.....

My solution only involved 1 variable but ended up with a cubic.....

I'm don't know how to create geometric diagrams in java at all but I'm giving you a JPG of how I did the problem with some explanation.....

 $ [CEQ\equal{}27]$ and $ [BFR]\equal{}30$. [Just use that area of triangles between the same parallels and on the same bases.] (Here $ [P]$ signifies area of region $ P$.)

The ratio $ \frac{|BE|}{|EQ|}\equal{}\frac{|a\plus{}8|}{|27|}$ as the ratio of triangles with same height is equivalent to the ratio of bases. Similarly, $ \frac{|CF|}{|FR|}\equal{}\frac{|a\plus{}5|}{|30|}$. Now, using the fact that $ \Delta BEC$ is similar to $ \Delta QEA$ and using that the ratio of there areas is square of the ratio of their sides, we find the area of  $ \Delta QEA$ in terms of $ a$. In a similar way, we get the area of $ \Delta RFA$ which is similar to $ \Delta CFB$.

Now, just use the fact that $ \Delta BPC$ is similar to $ \Delta QPR$ with ratio $ \frac{|BP|}{|PQ|}\equal{}\frac{|a|}{35}$. Equating the ratio (squared for areas), we have:

$ \frac{900}{a\plus{}5}\plus{}\frac{729}{a\plus{}8}\plus{}22\equal{}\frac{35^2}{a}\equal{}\frac{1225}{a}$.

Just check that $ a\equal{}10$ satisfies the above equation. I calculated the rest two roots, they were negative....

I HOPE YOU COULD PLEASE CHECK MY SOLUTION AND TELL ME IF THERE IS ANY ERROR.....

Let [EPF]=x, then by area ratios, [BPC]=40/x and [AEF]=22-x.
$ \frac{[AEF]}{[BEF]}\equal{}\frac{[AEC]}{[BEC]}$ by area ratios also, so
$ \frac{22\minus{}x}{x\plus{}5}\equal{}\frac{30}{5\plus{}\frac{40}{x}}$. Solving, x=4, so [BPC]=10

[quote="dgreenb801"]Let [EPF]=x, then by area ratios, [BPC]=40/x and [AEF]=22-x.
$ \frac {[AEF]}{[BEF]} \equal{} \frac {[AEC]}{[BEC]}$ by area ratios also, so
$ \frac {22 \minus{} x}{x \plus{} 5} \equal{} \frac {30}{5 \plus{} \frac {40}{x}}$. Solving, x=4, so [BPC]=10[/quote]

Amazing work!! Thanks a lot!!

[quote="Kondr"]Using Van Aubel theorem: 27/(a+8)+30/(a+5)=35/a

Nice problem.[/quote]

Thanks.. The solutions are getting better....

Thanks for introducing me to Van Aubel's Theorem as well. I'd never heard of it before....

[quote="dgreenb801"]Let [EPF]=x, then by area ratios, [BPC]=40/x and [AEF]=22-x.
$ \frac {[AEF]}{[BEF]} \equal{} \frac {[AEC]}{[BEC]}$ by area ratios also, so
$ \frac {22 \minus{} x}{x \plus{} 5} \equal{} \frac {30}{5 \plus{} \frac {40}{x}}$. Solving, x=4, so [BPC]=10[/quote]
couldnot get you????????????? :o
how can AEC be a triangle????
aec are collinear????????

u made a typo.......  :P
HOWEVER UR IDEA WAS NICE................. :)

Of course, I meant $ \frac{[AFC]}{[BFC]}$  :lol:


Hm, seems like that hoax actually got some people :P

That's weird, how a group would put together a 14-page newspaper just for a little prank.

They even re-made [url=http://www.nytimes-se.com/]The New York Times homepage.[/url] You can download the [url=http://www.nytimes-se.com/2009/07/04/todays-paper]"special edition" as .pdf here.[/url]

I liked the `Bush indicted on high treason charge'. Didn't get the `harvard business school closing' thing though.

How about the "Maximum Wage Law Succeeds"?  Nice twist.

With regards to MysticTerminator's comment, perhaps you remember a few years ago when Dennis Kucinich called for Dick Cheney to be brought up on similar charges.

[quote="TZF"]The PDF is no longer available. Could someone upload it to this thread?[/quote]

[url=http://www.nytimes-se.com/nytse/wp-content/uploads/NYTimes-SE_spreads.pdf]PDF with spreads[/url]

[url=http://www.nytimes-se.com/nytse/wp-content/uploads/NYTimes-SE.pdf]PDF without spreads[/url]

Hey, that is not so much a hoax. (the date is early...) see [url=http://www.reuters.com/article/vcCandidateFeed7/idUSTRE4AG25M20081118]this[/url]


Begin with the 200-digit number 9876543210987654321…543210, which repeats the digits 0-9 in reverse order. From the left, choose every third digit to form a new number. Repeat the same process with the new number. Continue the process repeatedly until the result is a two-digit number. What is that resulting two-digit number?

Two marbles are randomly selected without replacement from a bag containing blue and green marbles. The probability they are both blue is 1/6. If three marbles are randomly selected without replacement, the probability that all three are blue is 1/21. What is the fewest number of marbles that must have been in the bag before any were drawn?

[hide]1: If you start choosing every 3rd number, the digits of the new number, containing floor(200/3)=66 digits, have a pattern of 7418529630741852... Then, choosing every 3rd digit again, you get a 22 digit number (66/3) that has the digits 1234567890123... Repeat again to get a 7 digit number 3692581. And, once again to get the 2 digit number 98.

#2: call the probability of the 3rd marble chosen a/b. We know that the probability of choosing 3 blue marbles is (a+2)/(b+2)*(a+1)/(b+1)*a/b (if a is the number of blue marbles after 2 picks and b is the number of total marbles after 2 picks). Since we know that the product of the 1st 2 is 1/6, and the product of all 3 is 1/21, we get a/b by dividing the product of all 3, 1/21, by the product of the 1st 2, 1/6, and get a/b=3/7. This means that after the 2nd pick, the ratio of blue:total is 3:7. the least number of marbles total is therefore 7, and blue marbles 3. Adding 2 to get the total before choosing, we get 9 as the number of marbles we started with. Sorry if this was a little wordy. i wasn't sure how much detail i needed to go into.

[/hide]

#2:

[hide]The probability of getting a blue marble is b/(b+g); b is blue and g is green.  The probability of getting two consecutive blues is b/(b+g)*(b-1)/(b+g-1) which equals 1/6.  Then, the probability of getting three blues in b/(b+g)*(b-1)/(b+g-1)*(b-2)/(b+g-2) or 1/21.  Then we substitute the part in the 3rd equation which is found in the 2nd equation and you get: 1/6*(b-2)/(b+g-2)=1/21  (b-2)/(b+g-2)=2/7  7b-14=2b+2g-4  5b=2g+10.  This makes b even and g divisible by 5.  Then we guess and check for values.  Let's make b the smallest positive even integer, 2, because, of course, 0 wouldn't work because there has to be at least one blue marble.  Then with the 2 for b, that means g is equal to 0, which is impossible because the bag has to contain at least one green marble.  Then, let's try 4, the next even number, for b.  That makes g equal to 5, and that works.  Therefore, the smallest number of marbles needed is 4+5 or 9.[/hide]

A shorter solution for #2:

[hide]The probability that the first two selected are blue is 1/6.  The probability that a third marble selected (after two blue ones are removed) is (1/21)/(1/6)=2/7.  There must therefore be at least 7 after two have been removed, so 9 total.[/hide]

Here's a shorter solution for #1 (in spoiler)

[hide]Since you are taking every 3rd number, then every 3rd number of the 3rd numbers,etc, what you want are the powers of three.  With the 27th, 54th,81st,etc. you have more than two digits, but when you take the multiples of 81 (81,162) there are only two less than 200.  So you want the 81st and 162nd numbers.  Since they are in reverse order by tens, all that matters is the units digit of each.  So you take 200-81=139 and 200-162=38 and you take the last digit or each you get 98.[/hide]


Hi,

Many times in different contests' solutions, I see 'lemma's being introduced...

Honestly, I don't understand the use of it yet...
and I think I'll pretty much need it for future occasions...

Can anyone help me understand exactly what a lemma is and when & how it is used??

(Wikipedia doesn't give much help :huh: )

Generally, lemma is a proposition used in a larger proof. It is basically a theorem which is used to show the correctness of a more important theorem, or in solving a problem (proving its statement). Introducing lemmas makes proofs easier to read and analyze, and makes the proofs more systematic.

[color=green]Sorry for 'intruding' in Canadian national community  :lol: [/color]


show that there are infinitly many priems $p$ such that $2^{p-1} $is not equivalent to$ 1 (mod p^2)$.(Hint :use ABC conjecture).

LOL, using a conjecture to prove something is really cool :D

no other proof known for this fact?

Guys, are you sure this is not an open question? I mean, if we do not admit the ABC conjecture, after my knowledge this remains unsolved.

Yes, it is a conjecture. The "result" we have to prove is not that there are infinitely many non-Wieferich primes, but that the $abc$ conjecture implies that there are infinitely many non-Wieferich primes. :)

After sam-n posted this, I downloaded a paper on the subject, and it had a proof of this. I won't post it though, because I don't want to spoil the topic and I don't like to copy someone else's proofs :).

Yes, I understood that and I can prove it, but I did not know whether someone proved the conjecture till now (that one with the primes, not the abc conjecture).


topology
real analysis
real analysis unsolved

This is somewhat related to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126760]another problem I posted[/url].

Suppose $A$ and $B$ are compact subsets of $\mathbb R$ such that there exist continuous surjections $A\to B$ and $B\to A$. Does it follow that $A$ and $B$ are homeomorphic?

It does not.

Take $A$ to be the Cantor set, and take $B$ to be the disjoint union of a compact interval and the Cantor set. Since every compact metric space is a continuous image of the Cantor set, we easily get continuous maps from $A$ onto $B$ and from $B$ onto $A$ (for the latter, send the Cantor set in $B$ on $A$, and send the interval to an arbitrary point of $A$). However, $A$ is totally disconnected whereas $B$ is not.

Great, now I only have to become well-mannered. Actually, I first thought of the Lipschitz version:

Suppose $A$ and $B$ are compact subsets of $\mathbb R$ such that there exist Lipschitz surjections $A\to B$ and $B\to A$. Does it follow that $A$ and $B$ are bi-Lipschitz equivalent? (The latter means that there is a bijection $f\colon A\to B$ such that $f$ and $f^{-1}$ are both Lipschitz.)

[quote="mlok"]Great, now I only have to become well-mannered. [/quote]
This has nothing to do with mathematics, but can you explain why you use this phrase here? Why would you not be well-mannered?

Oh.. Voltaire. I didn't know the quote and didn't see it.

Taking out the middle fifth of each segment, I get a set of measure zero, because the measure is multiplied by $\frac45$ each time. But I think your construction works if $A$ is modified slightly, say, by removing intervals of length $\frac{1}{5^{n}}$ in each generation. (Which is probably what you meant.) The set will not be self-similar anymore, but it will be "self-bi-Lipschitz", which is good enough.

By the way, I asked this question because in another part of their book David and Semmes ask if the two Cantor sets in the topic linked above are bi-Lipschitz equivalent. Turns out, these are different problems (both open as far as I know.) EDIT: both answered in the affirmative by Rao, Ruan, and Xi (2006).


$(a+b)^2-2ab=2$
$(a+b)=\sqrt{2+2ab}$
$ab=x$
$2\sqrt{2+2x} + \frac 1x \ge 5$
$2x\sqrt{2+2x} - 5x+1 \ge 0$

That's not beautiful. And I don't even know how to complete it.

[quote="arqady"]Let $a>0,$ $b>0,$ $a^2+b^2=2.$
Prove that $2(a+b)+\frac{1}{ab}\geq5.$[/quote]

Maybe a trigonometric substitution?

[quote="edriv"]$(a+b)^2-2ab=2$
$(a+b)=\sqrt{2+2ab}$
$ab=x$
$2\sqrt{2+2x} + \frac 1x \ge 5$
$2x\sqrt{2+2x} - 5x+1 \ge 0$
[/quote]

It is not so ungly.It is very nice .
For $x\leq\frac{1}{5}$  the inequality is trivial .So let $x\geq\frac{1}{5}$

so  $2x\sqrt{2+2x} - 5x+1 \ge 0$  is equivalent to  $8x^3-17x^2+10x-1\geq 0$
which is equivalent to $(x-1)^2(8x-1)\geq 0$  which is true for $x\geq\frac{1}{5}$
and we are done

Ooh, I missed that factorization!  Nicely done.

edriv, silouan, very nice.
Try it:
Let $a>0,$ $b>0,$ $a^4+b^4=2.$
Prove that $4(a+b)+\frac{3}{ab}\geq11.$

I think that  $x+y=s$  and $xy=p$  would help .But the computations are monstrous

[quote="arqady"]Let $a>0,$ $b>0,$ $a^2+b^2=2.$
Prove that $2(a+b)+\frac{1}{ab}\geq5.$[/quote]

we first prove that

$ab+3\geq 2(a+b)$ . But after some sustitution from $a^2+b^2=2$ we will get $(ab-1)^2\geq 0$ .

Using this we have

$2(a+b)+\frac{1}{ab}\\\geq \frac{4(a+b)^2}{ab+3}+\frac{1}{ab}\\=\frac{8p^2+9p+3}{p(p+3)}\\=\frac{(3p^2+3)+5p^2+9p}{p(p+3)}\geq \frac{15p+5p^2}{p(p+3)}=5$

where $p=ab$

langtupanda, I think that your proof is very nice. :)
Try post #6. ;)

[quote="toetoe"]
$\frac{1}{ab}\geq\frac{1}{a^{2}b^{2}}$

[/quote]
This is not true. ;)

OK, langtupanda. Try the generalization:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88270


THere are infinite amounts of integers and real numbers. But there are more real numbers than integers. How is infinity bigger than infinity (or is infinity simply any value that has no end?)

You are a moderator of Basics Forum. You should be familiar with the 'Search' function. Please use search, this is discussed so many many many ... $(\text{many}^\text{many})$ ... times in MathLinks/AoPS.

I'm confused, what does the search function have to do with the problem?

[hide="My (probably wrong) answer, or IMO"]You cannot say that there are a greater number of real numbers than integers, because infinite isn't a number, it's a concept[/hide]

Yeah I'll bet this has been discussed alot.

There are more real numbers than integers since the set of integers cannot be put into bijection with the set of real numbers.

If you know set theory, they have different cardinalities.

[quote="10000th User"]You are a moderator of Basics Forum. You should be familiar with the 'Search' function. [/quote]

I have searchaphobia  :rotfl:

Okay, so what should I search for? (keywords?)

There is some wierd stuff in there, but I'll go with it  :D

If someone can find another link, good, but please pm me about it, I don't care too much about this anymore, so just lock this topic.


How to construct a two-dimensional unmeasurable set? Thank you.

A direct product of two nonmeasurable sets, of of a nonmeasurable set and one with positive measure.


I know I have posted a few times last year about this, but my coaches are still skeptical.

Any calculator that does not have a full QWERTY keyboard is allowed to be used in MC competitions. right?

I recently programmed a whole bunch of time-saving formulas on my calculator, but I'm not sure if I'm allowed to use them.

Thank you!!

[quote="kimmystar94"]Any calculator that does not have a full QWERTY keyboard is allowed to be used in MC competitions. right?[/quote]

That is correct.

[quote="kimmystar94"]I recently programmed a whole bunch of time-saving formulas on my calculator, but I'm not sure if I'm allowed to use them.[/quote]

Unless there is a new regulation (I'm sure MC officials will make that public and for all to know), then programs are allowed.

Im not quite sure but are graphical calculator's allowed?

Yes. They actually give those as an award for being in the national round.


Peter is never on AIM... it makes me wonder what happened to him.

also congratulations to keenan pepper from rickards high school.

Congrats to all of you, and especially Nick Wage from Wisconsin.

[quote="richpang"]:D  Especially Peter Diao from TJ  :D[/quote]

Very cool, I knew Peter from the Ross program this summer...congrats!

Thanks guys, this is so flattering lol.  :blush: Did Becca put you up to this Richard (I probably actually put her up to it)  ZSCool, you are Zachery right?    I'm afraid the book hasn't been in good use, might be coming back to Ross though after MOSP, either way, if I don't come Menyoung will return the book.  Linus, I lost the list sn's who are you?

I see 18-20 of our school's students on that list, and a couple dozen more AoPSers.  Congrats all!

The absence of certain names is certainly surprising.

5 from IL sounds like fun.

Yeah, but I was one of those people who got in the cheap way :blush: .  But MOP will be fun.

[quote="probability1.01"]Yeah, but I was one of those people who got in the cheap way:blush: .  But MOP will be fun.[/quote]
Even it is a cheap way:;) , it is a lot harder compared to the last year when all the 9th grade qualifiers were invited. I bet many of freshmen who got squeezed out are very jealous of you. And many sophomores wished that they were 9th graders. I believe you will have a great time with the MOP.

Congratulations to Shui Hu, from Minnesota, whom I saw in last year's Minnesota MATHCOUNTS countdown round.

Note that the last post in the thread (other than yours) was 2005...

I'm not trying to revive this thread after 11 years  :D .  My competition days ended long ago but I happened to see this page tonight. I wanted to leave a comment because I still remember reading this thread back in 2005 (note my screen name) ...

There was a time when I idolized the people mentioned in this thread.  After my Olympiad days ended, it took me many years to come to terms with my overall experience.  Making the IMO team was once my dream, but it never happened.  After high school, I went to college and then to work.  I even returned to academia for a while because I was still stuck in the past.  Only recently did I finally come to terms with my contest days.  2005 might seem like a lifetime ago for some people, but for me, it was yesterday.

奥数败将飘过

Well, it is nice to have closure. I'm glad you found it.

I hope a few years later, looking back, I will be proud of my contest days..... . But for that to happen I still need a lot of work!
:S

[quote]The list of MOPpers has been posted: http://www.unl.edu/amc/a-activities/a6-mosp/a6-1-mosparchives/2005-ma/05mospstate.html[/quote]
It says 404 error

Maybe it's because the link is over 10 years old.

[quote=adihaya]Wait, according to Google Translate the Chinese text you wrote meant [quote=Post #17 Chinese Text, translated]Mathematical Olympiad defeated drifting[/quote]
Do you mind sharing what it actually means? (just curious   :laugh:)[/quote]

奥数 means math olympiad.  败将 literally means "defeated general", but in this case it refers to anyone who failed at something.  飘过 means "to drift by/pass by."

This thread in particular, and AOPS in general, is like a living fossil.  As the years go by, many old competition records disappear from the internet, and the people who traded their childhood and teenage years for contest training move on with their life (usually). Most people who compete end up failing, while the few who succeed in their competition career are quickly forgotten by the next batch of ambitious contestants.  I think a significant portion of contestants believe during their contest days that what they are working for will matter in their lives.  However, looking at my own life and the lives of the people who did well (e.g. the high-scorers from any year),  the only major positive effect that math competitions had was on university admissions.  To be fair, that is a huge plus, but there is so much more to life, for which math competitions have either no effect or are even detrimental.  In hindsight, I see math competition as not much different from any other activity that a high schooler, middle schooler, or elementary schooler might engage in.

[quote=RaymondEP]mathlete_2000s, I just wanted to say that the message you're spreading is extremely valuable. So many kids spend their childhoods striving to reach the top, only to never get there. It can be devastating, but it's important to keep things in perspective. Math competitions offer something fun to get students doing more math outside the classroom, and probably encourage the development of some valuable skills, but it's only a small part of life.

Your post is valuable especially considering the timing. The AMC competitions are coming up in a week, and some kids are going to be devastated by their performance. They need to know that it's okay to fall short of their expectations sometimes. It's an inevitable part of life.[/quote]

What happens if you always fall short of your expectations?

That's a great question.  Falling short of expectations is a great chance to reflect:

1) From a technical standpoint, there is probably something that you can improve on.  For some contests, it may be that you need to avoid making mistakes, while for something like the USAMO, it may be that your approach to problem-solving is not optimal.  It could also be that you should handle stress during a test differently.  Whatever it is, the introspection is most valuable because it will allow you to improve in many areas of life later on.

2) As someone who fell short of my expectations in all but two contests over the course of eight years, I think the best way to handle disappointment is to realize that whatever it is that I am most passionate about does not matter nearly as much as I think it does.  Of course this is much easier said than done, especially when you are in the middle of competing.  If I could turn the clock back by 10-20 years and compete again, I would seek more balance by caring more about friendships, other subjects (I got As in everything but I didn't really care back then), growing as a person, and planning for the future after high school.  Having competitions (or anything for that matter) as only one among several top priorities instead of THE top priority would make failure much more soluble.

[quote]mathlete_2000s, I just wanted to say that the message you're spreading is extremely valuable. So many kids spend their childhoods striving to reach the top, only to never get there. It can be devastating, but it's important to keep things in perspective. Math competitions offer something fun to get students doing more math outside the classroom, and probably encourage the development of some valuable skills, but it's only a small part of life.

Your post is valuable especially considering the timing. The AMC competitions are coming up in a week, and some kids are going to be devastated by their performance. They need to know that it's okay to fall short of their expectations sometimes. It's an inevitable part of life.[/quote]

I personally am not going to be devastated because i know already i am going to do poorly on the AMC
*************************************************************************************************************************************
Its especially important to know where you stand in competitions. Otherwise kids can get seriously devastated by their performance
The thing with math competitions is just do the best you personally can.
If you do that then you have succeeded

[quote]What happens if you always fall short of your expectations?[/quote]
That would mean you would have to revise your expectations
Although many of us have great dreams of how we want to perform in competitions, often there is a difference between the reality and the dream.
And although its frustrating there are always going to be people much better than you, and its a fact we all have to accept in life.
The key is to really know where you stand in math competitions
and even if you think its not good if thats the best you can do then that is good

[quote]However, looking at my own life and the lives of the people who did well (e.g. the high-scorers from any year), the only major positive effect that math competitions had was on university admissions.
. . .
In hindsight, I see math competition as not much different from any other activity that a high schooler, middle schooler, or elementary schooler might engage in.[/quote]

Hmm.

I certainly agree with the second sentence, in that (a) many activities other than math contests provide similar long-term benefits, and (b) the math itself is not so important later on.  I also do agree that the *results* of math contests have essentially zero effect on life other than college admissions (and even then, there are much easier ways to impress college admissions officers than doing well on the USAMO :) ).

But I'm surprised that you think math contests (and by extension other HS activities) have few major positive effects at all.  For me there were at least two:

1. Teaching me how to work hard towards ambitious goals like USAMO/IMO.
Most of the useful skills I have now came from math contests instead of school.

2. Providing me with a peer group of other smart students with similar interests;
almost all the friends I have today are people I met through math contests in some way or another.
(I suppose it's possible that math contests were less social in 2005 than they are today; these days sites like AoPS, Facebook, etc. are much bigger and actually make it possible to keep in touch with everyone else in the world who does math contests. Certainly what I did in high school.)

Of course these same benefits can be obtained from many other activities; there is nothing unique to math contests here.


will everyone's acceptance/rejection letter be sent on the same day?

Probably not.

During the Early Application (App. A), I recieved my acceptance/rejection the night I sent them (I made pdf and sent them by e-mail).

Some people, it has been up to 5-7 days until their official acceptance/rejection.

Well, it's none of your worries since you did pretty well on Mathcounts.

(I was one of the students up on the stage at Texas Mathcounts.)


Hi Thanhliem,

If I understood correctly, then I guess there is some sort of error in your problem.

From the first eqn, $x_1 | 3 x_2$ $\Rightarrow$ $x_1 = 1, 3$ or $x_1 | x_2$

If $x_1 = 1$, then from the first eqn, $3 x_2$ is negative, which implies $x_2$ is negative. (Contradiction)

If $x_1 = 3$, then from the first eqn, $x_2 = 0$, and hence $x_3 = 0$ from the second eqn, and so on, upto $x_{2004} = 0$ and hence from the last eqn, $x_1 = 0$. (Contradiction)

So, let us assume $x_1 | x_2$, and $x_1 \neq 1, 3$.

Similarly, we can conclude $x_2 | x_3$ and so on. So,  $x_1 | x_{2004}$.
But from the last eqn, $x_{2004} | x_1$.
Hence, $x_1 = x_{2004}$  $\Rightarrow$ $x_1 = x_2 = x_3 = ..... = x_{2004}$.

So, puttin $x$ in place of $x_1$ and $x_2$ in first eqn, we have
$x^3 - 3 x^2 = 3 x$
$\Rightarrow x^2 - 3 x - 3 = 0$    as $x \neq 0$ (the integers are positive)

Roots are $\frac{3 \pm \sqrt{21}}{2}$ that are not integral.
Thus the only possible solution is $x_1 = x_2 = .... = x_{2004} = 0$, but this cannot happen as the integers should be positive.

Kindly clarify this.

**************************************************************************************
Sambit

I think the equations Thanhliem is trying to set up are:

[tex]x_1^3-3x_1^2=3x_2[/tex]

[tex]x_2^3-3x_2^2=3x_3[/tex]
.
.
.
[tex]x_{2004}^3-3x_{2004}^2=3x_1[/tex]

The first equation gives [tex]3|x_1[/tex], implying [tex]9|x_2[/tex], implying [tex]81|x_3[/tex], implying [tex]6561|x_4[/tex] and so on, leading nowhere fast.

Although far from vigorous, I believe I can safely say there exists no solutions in the positive integers.


hello, do you mean this inequality $ \sqrt{x\plus{}\frac{1}{x}}\plus{}\sqrt{1\plus{}\frac{1}{x}}>1\plus{}\frac{1}{x}$?
Sonnhard.

[quote="trannguyenquyND_VN"]solve:
  $ \sqrt {x \plus{} 1/x}$ +$ \sqrt {1 \plus{} 1/x}$ $ \succ$ (1+1/x)[/quote]
i think, you shouldn't post it in this forum :)


The fifth term of an arithmetic sequence is $ 11$. If the difference between two consecutive terms is $ 1$, what is the product of the first two terms?

The common difference is 1, so the first 4 terms are 10, 9, 8, and 7. The product of the first two terms is then $ \boxed{56}$.

If the common difference is 1 that means the first term is 4 less than the 5th so its 11-4=7 and the next term is just one more so the answer is 7*8=56.


inequalities
search
inequalities proposed

[quote="Allnames"]Let $ a,b,c > 0$ and $ abc \equal{} 1$  prove that
$ \frac {a}{\sqrt [3] {8 \plus{} b^3}} \plus{} \frac {b}{\sqrt [3] {8 \plus{} c^3}} \plus{} \frac {c}{\sqrt [3] {8 \plus{} a^3}}\ge \sqrt [3] {3}$[/quote]
We only need prove
$ (a\plus{}b\plus{}c)^4 \ge\ 24abc(a\plus{}b\plus{}c) \plus{} 3(ab^3\plus{}bc^3\plus{}ca^3)$
I think it is obvious  :wink:
Thanks! :lol:

[quote="Allnames"][quote="onlylove_math"][quote="Allnames"]Let $ a,b,c > 0$ and $ abc \equal{} 1$  prove that
$ \frac {a}{\sqrt [3] {8 \plus{} b^3}} \plus{} \frac {b}{\sqrt [3] {8 \plus{} c^3}} \plus{} \frac {c}{\sqrt [3] {8 \plus{} a^3}}\ge \sqrt [3] {3}$[/quote]
We only need prove
$ (a \plus{} b \plus{} c)^4 \ge\ 24abc(a \plus{} b \plus{} c) \plus{} 3(ab^3 \plus{} bc^3 \plus{} ca^3)$
I think it is obvious  :wink:
Thanks! :lol:[/quote]
Nice my friend,but the last is quite hard,can you detail it
My proof is quite nice[/quote]
Oh my friend , I think my proof is true. If we develop it , I think it is obvious.
And can U post your proof, pls? :wink:  I want see your nice proof
Thanks! :)

[quote="onlylove_math"][quote="Allnames"]Let $ a,b,c > 0$ and $ abc \equal{} 1$  prove that
$ \frac {a}{\sqrt [3] {8 \plus{} b^3}} \plus{} \frac {b}{\sqrt [3] {8 \plus{} c^3}} \plus{} \frac {c}{\sqrt [3] {8 \plus{} a^3}}\ge \sqrt [3] {3}$[/quote]
We only need prove
$ (a \plus{} b \plus{} c)^4 \ge\ 24abc(a \plus{} b \plus{} c) \plus{} 3(ab^3 \plus{} bc^3 \plus{} ca^3)$
I think it is obvious  :wink:
Thanks! :lol:[/quote]

Yes, you are right. I have proved that it is always correct for all $ a.b.c \in \mathbb{R}$
\[ (a \plus{} b \plus{} c)^4 \ge\ 24abc(a \plus{} b \plus{} c) \plus{} 3(ab^3 \plus{} bc^3 \plus{} ca^3),\forall a,b,c \in \mathbb{R}\]
But I don't think it is obvious. Please post your solution if you have  :wink:

[quote="zaizai-hoang"][quote="onlylove_math"][quote="Allnames"]Let $ a,b,c > 0$ and $ abc \equal{} 1$  prove that
$ \frac {a}{\sqrt [3] {8 \plus{} b^3}} \plus{} \frac {b}{\sqrt [3] {8 \plus{} c^3}} \plus{} \frac {c}{\sqrt [3] {8 \plus{} a^3}}\ge \sqrt [3] {3}$[/quote]
We only need prove
$ (a \plus{} b \plus{} c)^4 \ge\ 24abc(a \plus{} b \plus{} c) \plus{} 3(ab^3 \plus{} bc^3 \plus{} ca^3)$
I think it is obvious  :wink:
Thanks! :lol:[/quote]

Yes, you are right. I have proved that it is always correct for all $ a.b.c \in \mathbb{R}$
\[ (a \plus{} b \plus{} c)^4 \ge\ 24abc(a \plus{} b \plus{} c) \plus{} 3(ab^3 \plus{} bc^3 \plus{} ca^3),\forall a,b,c \in \mathbb{R}
\]
But I don't think it is obvious. Please post your solution if you have  :wink:[/quote]

Using Vasc's inequality we have
\[ \left( {a^2 \plus{} b^2 \plus{} c^2 } \right)^2 \ge 3\left( {a^3 b \plus{} b^3 c \plus{} c^3 a} \right)
\]
Thus it suffices us to prove the inequality
\[ \left( {a \plus{} b \plus{} c} \right)^4 \ge 24abc\left( {a \plus{} b \plus{} c} \right) \plus{} \left( {a^2 \plus{} b^2 \plus{} c^2 } \right)^2
\]

\[ \Leftrightarrow 4\sum {a^2 b^2 } \plus{} 4\sum {ab\left( {a^2 \plus{} b^2 } \right)} \ge 12abc\left( {a \plus{} b \plus{} c} \right)
\]
which obviously true. :)

[quote="RainL"][quote="zaizai-hoang"]

Yes, you are right. I have proved that it is always correct for all $ a.b.c \in \mathbb{R}$
\[ (a \plus{} b \plus{} c)^4 \ge\ 24abc(a \plus{} b \plus{} c) \plus{} 3(ab^3 \plus{} bc^3 \plus{} ca^3),\forall a,b,c \in \mathbb{R}
\]
[/quote]

\[ 4\sum {a^2 b^2 } \plus{} 4\sum {ab\left( {a^2 \plus{} b^2 } \right)} \ge 12abc\left( {a \plus{} b \plus{} c} \right)
\]
which obviously true. :)[/quote]
Try $ a\rightarrow\minus{}\infty,$ $ b\equal{}1$ and $ c\equal{}0.$ :wink:

[quote="arqady"][quote="RainL"][quote="zaizai-hoang"]

Yes, you are right. I have proved that it is always correct for all $ a.b.c \in \mathbb{R}$
\[ (a \plus{} b \plus{} c)^4 \ge\ 24abc(a \plus{} b \plus{} c) \plus{} 3(ab^3 \plus{} bc^3 \plus{} ca^3),\forall a,b,c \in \mathbb{R}
\]
[/quote]
\[ 4\sum {a^2 b^2 } \plus{} 4\sum {ab\left( {a^2 \plus{} b^2 } \right)} \ge 12abc\left( {a \plus{} b \plus{} c} \right)
\]
which obviously true. :)[/quote]
Try $ a\rightarrow \minus{} \infty,$ $ b \equal{} 1$ and $ c \equal{} 0.$ :wink:[/quote]

Thanks you for your notice, arqady.  :)

At least we got the solution for Allnames's problem.  :)

[quote="RainL"]

At least we got the solution for Allnames's problem.  :)[/quote]
Of cause!

[quote="zaizai-hoang"][quote="onlylove_math"][quote="Allnames"]Let $ a,b,c > 0$ and $ abc \equal{} 1$  prove that
$ \frac {a}{\sqrt [3] {8 \plus{} b^3}} \plus{} \frac {b}{\sqrt [3] {8 \plus{} c^3}} \plus{} \frac {c}{\sqrt [3] {8 \plus{} a^3}}\ge \sqrt [3] {3}$[/quote]
We only need prove
$ (a \plus{} b \plus{} c)^4 \ge\ 24abc(a \plus{} b \plus{} c) \plus{} 3(ab^3 \plus{} bc^3 \plus{} ca^3)$
I think it is obvious  :wink:
Thanks! :lol:[/quote]

Yes, you are right. I have proved that it is always correct for all $ a.b.c \in \mathbb{R}$
\[ (a \plus{} b \plus{} c)^4 \ge\ 24abc(a \plus{} b \plus{} c) \plus{} 3(ab^3 \plus{} bc^3 \plus{} ca^3),\forall a,b,c \in \mathbb{R}
\]
But I don't think it is obvious. Please post your solution if you have  :wink:[/quote]
Thank Zaizai ( and every body),can you post your proof for it,i reall want to know that
But,how about
$ \frac {a}{\sqrt [k] {8 \plus{} b^3}} \plus{} \frac {b}{\sqrt [k] {8 \plus{} c^3}} \plus{} \frac {c}{\sqrt [k] {8 \plus{} a^3}}\ge \sqrt [k] {3^{k \minus{} 2}}$ for all $ k\ge 2,k\in N$
Thank you

[quote="kunny"]$ \frac {e^x}{\sqrt [3]{8 \plus{} e^{3y}}} \plus{} \frac {e^y}{\sqrt [3]{8 \plus{} e^{3z}}} \plus{} \frac {e^z}{\sqrt [3]{8 \plus{} e^{3x}}}\geq \sqrt [3]{3}$ with $ x \plus{} y \plus{} z \equal{} 0$.[/quote]
Sorry kunny,but  the last ....
 :oops:

Sorry, I can't understand what you say, Allnames. :maybe:

[quote="Math pro"]Detail your solution,pls only,all names
My solution:
easy to see that:
$ \frac {a}{\sqrt [3] {8 \plus{} b^3}} \plus{} \frac {b}{\sqrt [3] {8 \plus{} c^3}} \plus{} \frac {c}{\sqrt [3] {8 \plus{} a^3}}$  $ \geq$
$ \frac {a}{\sqrt [3] {8 \plus{} a^3}} \plus{} \frac {b}{\sqrt [3] {8 \plus{} b^3}} \plus{} \frac {c}{\sqrt [3] {8 \plus{} c^3}}$
[hide]ap dung BDT hoan vj bac nao bjet' djch ho em cai(^_^)[/hide]
we need prove that
$ \frac {1}{\sqrt [3] {8 \plus{} x^3}} \plus{} \frac {1}{\sqrt [3] {8 \plus{} y^3}} \plus{} \frac {1}{\sqrt [3] {8 \plus{} z^3}}\geq\sqrt [3] {3}$
with xyz=1 (set 1/a=x...)
last ineq easy by mixing variable! :blush: (to esier you should put x^3=m,y^3=n,z^3=p and mnp=1)[/quote]
Are you sure about your proof??????

[quote="can_hang2007"][quote="Allnames"]But,how about
$ \frac {a}{\sqrt [k] {8 \plus{} b^3}} \plus{} \frac {b}{\sqrt [k] {8 \plus{} c^3}} \plus{} \frac {c}{\sqrt [k] {8 \plus{} a^3}}\ge \sqrt [k] {3^{k \minus{} 2}}$ for all $ k\ge 2,k\in N$
Thank you[/quote]
Allnames's statement is correct. By the way, the inequality holds of all $ k \ge 1$ and $ k$ is not neccessary integer.
After using weighted AM-GM, it suffices to prove the following inequality
\[ \sum \frac {a}{b^3 \plus{} k} \ge \frac {3}{k \plus{} 1}
\]
for all $ k \ge 0$ (notice that $ k$ in here and $ k$ in the original inequality are different)
Denote $ a \equal{} \frac {x}{y},b \equal{} \frac {z}{x},c \equal{} \frac {y}{z}$ where $ x,y,z > 0$, our inequality becomes
\[ \sum \frac {x^4}{y(z^3 \plus{} kx^3)} \ge \frac {3}{k \plus{} 1}
\]
which is obviously true by Cauchy Schwarz Inequality since
\[ \sum \frac {x^4}{y(z^3 \plus{} kx^3)} \ge \frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{\sum yz^3 \plus{} k\sum x^3y}
\]
and the Vasile's Inequality gives us
\[ \sum yz^3 \le \frac {1}{3}(x^2 \plus{} y^2 \plus{} z^2)^2
\]

\[ \sum x^3y \le \frac {1}{3}(x^2 \plus{} y^2 \plus{} z^2)^2
\]
The proof is completed.
We have done.
:)[/quote]
Nice,very nice,thank Can_hang,my proof is quite similar,but yours is stronger
with AMGM we have $ 3^{k \minus{} 2}(b \plus{} 2)(b^2 \minus{} 2b \plus{} 4)\le (\frac {b^2 \minus{} b \plus{} 3k}{k})^k$
so the ineq is
$ \sum \frac {ka}{b^2 \minus{} b \plus{} 3k} \ge 1$
then let $ a \equal{} \frac {x}{y},b \equal{} \frac {z}{x},c \equal{} \frac {y}{z}$ where $ x,y,z > 0$
we use CS,and Vasc ineq we have the proof (the last is quite similar to Can_hang 'one)

The following statements holds:

(a) Let $ a,b,c > 0$ and $ abc \equal{} 1$. If $ p > 0$, then
$ \frac {a}{\sqrt [3] {p \plus{} b^3}} \plus{} \frac {b}{\sqrt [3] {p \plus{} c^3}} \plus{} \frac {c}{\sqrt [3] {p \plus{} a^3}}\ge \frac 3{\sqrt [3] {p \plus{} 1}}$.

(b) Let $ a,b,c$ be real numbers such that $ a\plus{}b\plus{}c\equal{}3$. Then

$ a^3b\plus{}b^3c\plus{}c^3a \plus{}6abc \le 9$.

[quote="Vasc"]The following statement holds:
Let $ a,b,c > 0$ and $ abc \equal{} 1$. If $ p > 0$, then
$ \frac {a}{\sqrt [3] {p \plus{} b^3}} \plus{} \frac {b}{\sqrt [3] {p \plus{} c^3}} \plus{} \frac {c}{\sqrt [3] {p \plus{} a^3}}\ge \frac 3{\sqrt [3] {p \plus{} 1}}$.[/quote]
Yes, and I have showed before, VasC. :)
See my post above.

[quote="Vasc"]You are right, can_hang. :)
But this is an old inequality posted on this forum in 2006.[/quote]
Really????,vasc,because i am a new member so i dont know that,but i have found it by my self
Sorry because my title is stupid :blush:  :blush:

[quote="Vasc"]The following statements holds:

(b) Let $ a,b,c$ be real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Then

$ a^3b \plus{} b^3c \plus{} c^3a \plus{} 6abc \le 9$.[/quote]

See my solution here  :wink:  By the same way
http://www.mathlinks.ro/viewtopic.php?t=213886

[quote="zaizai-hoang"]Let $ a,b,c$ be real numbers such that $ a \plus{} b \plus{} c \equal{} 3.$Prove that:
\[ a^{3}b \plus{} b^{3}c \plus{} c^{3}a \plus{} 6abc\leq 9
\]
What do you think about the following solution  :wink:

The inequality is equivalent to:
\[ (a \plus{} b \plus{} c)^{4}\geq 18abc(a \plus{} b \plus{} c) \plus{} 9(a^{3}b \plus{} b^{3}c \plus{} c^{3}a)
\]

\[ \sum a^4 \plus{} 6\sum a^2b^2 \minus{} 5\sum a^3b \plus{} 4\sum ab^3 \minus{} 6abc \sum a \ge 0
\]

\[ \Leftrightarrow \frac {3\left( \sum a^{3} \minus{} 3 \sum a^{2}b \plus{} 6abc\right) ^{2} \plus{} \left( \sum a^{3} \plus{} 6\sum a^{2}c \minus{} 3\sum a^{2}b \minus{} 12abc\right) ^{2}}{ 18[(a \minus{} b)^{2} \plus{} (b \minus{} c)^{2} \plus{} (c \minus{} a)^{2}]}\geq 0
\]
The inequality was proved. The equality holds on $ a \equal{} b \equal{} c \equal{} 1$
[/quote]

@onlylove_math: are you sure what you posted is always correct? I don't think so. There are a lot of mistake in your post. For example:
[quote]$ f(k) \equal{} \frac {a}{\sqrt [k] {8 \plus{} b^3}} \plus{} \frac {b}{\sqrt [k] {8 \plus{} c^3}} \plus{} \frac {c}{\sqrt [k] {8 \plus{} a^3}} \minus{} 3^{\frac {k \minus{} 2}{k }}$
$ \rightarrow f '(k) \equal{} \sum \frac {\frac {1}{k^2}ln(8 \plus{} b^3)}{\sqrt [k] {8 \plus{} b^3}} \minus{} \sqrt [k] {3^{k \minus{} 2}}ln3.\frac {2}{k^2} \ge\ 0$ easy see :lol:[/quote]
What do you think if I say it is wrong? It should be:
 \[ f '(k) \equal{} \sum \frac {\frac {1}{k^2}aln(8 \plus{} b^3)}{\sqrt [k] {8 \plus{} b^3}} \minus{} \sqrt [k] {3^{k \minus{} 2}}ln3.\frac {2}{k^2} \ge\ 0\]
Easy to see? Really? Are you kidding or are you spamming? I am doubt that you have not ever solved it before you post into forum  :?:
And what do you think about the following topic
http://www.mathlinks.ro/viewtopic.php?p=1221396#1221396

Can you explain what you means in this one?
[quote="onlylove_math"][quote="Allnames"]$ Let,x,yz\ge 0$,$ xy \plus{} yz \plus{} zx \plus{} xyz \equal{} 2$
Find the best $ k > 0$ to this ineq is true for all $ x,y,z$

$ x \plus{} y \plus{} z\le \frac {1}{4xyz} \plus{} (12 \minus{} 8k)xyz \plus{} k \minus{} 2$
P\s>I think it is a nice ineq,with that condition,we have many other ineq :)[/quote]
It 'll be kill by CYH Put :$ a \plus{} b \plus{} c \equal{} p;ab \plus{} bc \plus{} ca \equal{} \frac {p^2 \minus{} q^2}{3} ;abc \equal{} r.$

Or put $ a \equal{} cosA;b \equal{} cosB;c \equal{} cosC$[/quote]
And
[quote="Allnames"][quote="Allnames"]$ Let,x,yz\ge 0$,$ xy \plus{} yz \plus{} zx \plus{} 2xyz \equal{} 1$
Find the best $ k > 0$ to this ineq is true for all $ x,y,z$

$ x \plus{} y \plus{} z\le \frac {1}{4xyz} \plus{} (12 \minus{} 8k)xyz \plus{} k \minus{} 2$
P\s>I think it is a nice ineq,with that condition,we have many other ineq :)[/quote]
with can_hang 's help,i have found some mistake in my problem  :blush:
but how abuot $ k \equal{} 3\sqrt 3 \minus{} 4$[/quote]

I just want to recommend you to do check carefully  before post any comment. I don't care do you listen to this thing? But let think of what I wrote here!

[quote="zaizai-hoang"]

@onlylove_math: are you sure what you posted is always correct? I don't think so. There are a lot of mistake in your post. For example:
[quote]$ f(k) \equal{} \frac {a}{\sqrt [k] {8 \plus{} b^3}} \plus{} \frac {b}{\sqrt [k] {8 \plus{} c^3}} \plus{} \frac {c}{\sqrt [k] {8 \plus{} a^3}} \minus{} 3^{\frac {k \minus{} 2}{k }}$
$ \rightarrow f '(k) \equal{} \sum \frac {\frac {1}{k^2}ln(8 \plus{} b^3)}{\sqrt [k] {8 \plus{} b^3}} \minus{} \sqrt [k] {3^{k \minus{} 2}}ln3.\frac {2}{k^2} \ge\ 0$ easy see :lol:[/quote]
What do you think if I say it is wrong? It should be:
\[ f '(k) \equal{} \sum \frac {\frac {1}{k^2}aln(8 \plus{} b^3)}{\sqrt [k] {8 \plus{} b^3}} \minus{} \sqrt [k] {3^{k \minus{} 2}}ln3.\frac {2}{k^2} \ge\ 0
\]
Easy to see? Really? Are you kidding or are you spamming? I am doubt that you have not ever solved it before you post into forum  :?: [/quote]
Yes, i agree with it . I have some mistake . And to be honest, I also have not ever solved it before you post into forum. Sorry for it. And I'm kidding or spamming , I think U shouldn't care it because It don't make influence for U . And U don't have power to say i am spamer  :oops:
[quote="zaizai-hoang"]And what do you think about the following topic
http://www.mathlinks.ro/viewtopic.php?p=1221396#1221396

Can you explain what you means in this one?
[quote="onlylove_math"][quote="Allnames"]$ Let,x,yz\ge 0$,$ xy \plus{} yz \plus{} zx \plus{} xyz \equal{} 2$
Find the best $ k > 0$ to this ineq is true for all $ x,y,z$

$ x \plus{} y \plus{} z\le \frac {1}{4xyz} \plus{} (12 \minus{} 8k)xyz \plus{} k \minus{} 2$
P\s>I think it is a nice ineq,with that condition,we have many other ineq :)[/quote]
It 'll be kill by CYH Put :$ a \plus{} b \plus{} c \equal{} p;ab \plus{} bc \plus{} ca \equal{} \frac {p^2 \minus{} q^2}{3} ;abc \equal{} r.$

Or put $ a \equal{} cosA;b \equal{} cosB;c \equal{} cosC$[/quote][/quote]
I think CYH for it is true and with put a=cosA... i don't sure  :roll: . And Can U prove Allnames 's [hide="problem"]http://www.mathlinks.ro/viewtopic.php?p=1221396#1221396[/hide] with $ k \equal{} 3\sqrt {3} \minus{} 4$ ? :roll:  I think it is quite hard at least for me . And I think U can't prove it by a easy way so U shouldn't say about it  :rotfl:
 I only say my emotion , really

[quote="onlylove_math"]
I think CYH for it is true and with put a=cosA... i don't sure  :roll:
[/quote]

Dear [b]onlylove_math[/b] and [b]Allnames[/b] ,

As I know [b]the CYH technique[/b] is a new technique about using the Cauchy Schwarz inequality of can_hang2007.

Thus when I read the message you post I have some ... that I cannot understand. ( So sorry for ... because I have no so much vocabulary )

So you know the CYH technique, don't you?

Regards,

RainL. :)

[quote="kunny"][quote="Allnames"]Let $ a,b,c > 0$ and $ abc \equal{} 1$  prove that
$ \frac {a}{\sqrt [3] {8 \plus{} b^3}} \plus{} \frac {b}{\sqrt [3] {8 \plus{} c^3}} \plus{} \frac {c}{\sqrt [3] {8 \plus{} a^3}}\ge \sqrt [3] {3}$[/quote]

Allnames, are you sure that the problem is too easy?

I don't think so at all. I would like to see the more basic solution, although there is no solution such like and I do want to know how these inequalities that seems beautiful or simple are made.[/quote]
Oh,kunny,I say it is easy because i found it, since Vasq ineq and $ \sum \frac {a}{b^2 \minus{} b \plus{} 4}\ge 1$ then i made it to be harder.Before i join Mathlink forum,i was heard that there are mant pro people with high knowlege here,and after sometime,i have found it is true ,for example vasc,arqady,can_hang,you,and more....So i think you can easyly to koll it (in the truth,i only want to send this problem to both my friend ,onlylovemath and math pro.Sorry every body
To vasc,i have seen your problem(my proof for it is quitr diffrent from),but i think the general'one is mine.Thank you
To RainL;Yes of course my dear,Can_hang has a friend who study in my school,so we (me and both friends) know it (i think the method which we know is quite fully) :lol: .Thank you

Sorry every body,because of me,the topic is too long

To zaizai-hoang :Are you mod here,yes i see,you arent,so why do you care my posts (and both friend ' one ) are spam,it is not suitable to you,May be, i think,you are also spamer :rotfl: (because in this to pic you have 5 posts and only 2 mention math.So my friends,dont care him,because he is similar to us :rotfl:

Exactly,all names :wink:
@zaizai:we should agure about math,not any thing because
with Math we are one! :wink:

Very nice Vasc, I will try to solve it as soon as possible  :)
To Allnames:
[quote]To zaizai-hoang :Are you mod here,yes i see,you arent,so why do you care my posts (and both friend ' one ) are spam,it is not suitable to you,May be, i think,you are also spamer  (because in this to pic you have 5 post and only 2 mention math.So my friends,dont care him,because he is similar to us [/quote]
Of course, I am not a mod in Mathlink Forum. I don't care about this and it is not the reason why I recommended to both of you. Because of a very simple reason. I am Vietnamese so I don't want to hear from foreigners that "oh Vietnamese are only professional spamers". I don't want to continue this thing. Stop at here.
PS: Your English (both of you) is really bad !!

[hide]2 chú coi lại các post mà xem. Anh chỉ khuyên như vậy thôi, ko nghe thì anh cũng bó tay. Ko gì anh mà ko ít người coi những post của các chú là spam, ko phải "toàn bộ" nhưng cách mà các chú post thì làm người ta khó chịu đấy. Hãy làm ra đã trước khi post bình luận của mình,  Và tốt nhất là đừng khích bác người khác, bài các chú giải đc thì anh cũng giải đc đấy. Anh vừa chat với allnames và đồng ý với allnames kết thúc chuyện này ở đây, ko nên cãi nhau nữa !!![/hide]

[quote="Vasc"][quote="zaizai-hoang"][quote="Vasc"]The following statements holds:

(b) Let $ a,b,c$ be real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Then

$ a^3b \plus{} b^3c \plus{} c^3a \plus{} 6abc \le 9$.[/quote]
See my solution here  :wink:  By the same way
http://www.mathlinks.ro/viewtopic.php?t=213886[/quote]
Indeed, you are right, zaizai-hoang. :)

A more general statement is the following:

Let $ a,b,c$ be real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. If $ p$ and $ q$ are real numbers such that $ 1 \plus{} pq\ge (p \plus{} q \minus{} 2)^2$, then

$ p\sum a^3b \plus{} q\sum ab^3 \plus{} 3(9 \minus{} p \minus{} q)abc \le 27$.[/quote]

The inequality is equalivent to
\[ \sum {a^4 }  \plus{} 6\sum {a^2 b^2 }  \plus{} \left( {4 \minus{} 3p} \right)\sum {a^3 b}  \plus{} \left( {4 \minus{} 3q} \right)\sum {ab^3 }  \plus{} \left( {3p \plus{} 3q \minus{} 15} \right)abc\left( {\sum a } \right) \ge 0
\]
\[ \Leftrightarrow \frac{1}{{18}}\sum {\left[ {3\left( {a^2  \minus{} b^2 } \right) \plus{} 3\left( {p \minus{} q} \right)ab \minus{} \left( {12 \minus{} 3p \minus{} 6q} \right)bc \plus{} \left( {12 \minus{} 6p \minus{} 3q} \right)ca} \right]^2 }
\]
\[ \plus{} \frac{{36\left( {p \plus{} q} \right) \minus{} 9\left( {p^2  \plus{} pq \plus{} q^2 } \right) \minus{} 27}}{{18\left( {48 \minus{} 36\left( {p \plus{} q} \right) \plus{} 9\left( {p^2  \plus{} pq \plus{} q^2 } \right)} \right)}}\sum {\left[ {3\left( {p \minus{} q} \right)ab \minus{} \left( {12 \minus{} 3p \minus{} 6q} \right)bc \plus{} \left( {12 \minus{} 6p \minus{} 3q} \right)ca} \right]^2 }  \ge 0
\]
true for all $ a,b,c \in \Re$.


Propose a synthesis of m-iodophenol starting from benzene.

Doubt
[hide]
Are we allowed to nitrate nitro benzene?  :)
[/hide]

i think it is allowed using excess $ \ce{HNO3}$
then we canselctive reduce one of them using $ \ce{(NH4)2S}$ and then diazotize it and heat that with water to m-nitro phenol...then diazotize it again then use $ \ce{KI}$

Of course we are, and by using the usual nitrating mixture (or other reagent), but not only HNO3.

yes using other but i meant an excess of $ \ce{HNO3}$ that doesn't rule out the possibility of usingother things

Excess nitric acid is not also good enough. A stronger acid (like sulphuric acid) must also be present in order to produce reasonable amounts of the electrophile (nitronium ion). With nitric acid only, that formation is slow.

yes that's what nitrating mixture is i suppose

Yes: the nitrating mixture is a mixture of concentrated nitric and sulphuric acids.

can you explain how you get iodo substituted prodeuct on treating the dioazonium chloride with [b]KI[/b]???
can you explain?
thanks....

Excuse this is supposed to be ArSN1 reaction which obviously takes place through the formation of phenyl carbocation(N2 is a good leaving group) and then attack by I-


wow, where would u even start? doesn't one have to be like 1,2,3, or maybe 4?

jorian

[quote="kyyuanmathcount"]I found this problem difficult without heavy guess/check.

There are six distinct ordered pairs of nonzero digits (b,x) such that the sum of the digits of the number $b^{x}$ is equal to the sum $b+x$.  What is the sum of all of the $b^{x}$ possibilities?[/quote]
How would you do this problem? Like jhredsox, I don't even know where to start. I only found one by brute force. $(2,2)$.

[quote="ckck"][quote="kyyuanmathcount"]I found this problem difficult without heavy guess/check.

There are six distinct ordered pairs of nonzero digits (b,x) such that the sum of the digits of the number $b^{x}$ is equal to the sum $b+x$.  What is the sum of all of the $b^{x}$ possibilities?[/quote]
How would you do this problem? Like jhredsox, I don't even know where to start. I only found two by brute force. $(2,2),(-1,2)$.[/quote]

When they say digits is it positive?

[quote="Ubemaya"]$-1$ isn't a digit.[/quote]
Oh my bad. Edited.

Dang, perfect beat me to it. But I gave a fuller solution...

This problem is actually very easy... it took me under 2 minutes I think. (I might be exaggerating... I have no sense of time).

There are only 81 possibilities. I guessed and checked starting systematically starting with lower exponents, because for high exponents the sum of the digits would be too high.

1: Obviously impossible,
2: 2, 5, and 8 work. They become 4, 25, and 64 respectively.
3: 5, 6, and 7 work. They become 125, 216, and 343 respectively.

The sum of 4, 25, 64, 125, 216, and 343 is 777... I think. I suck at adding in my head.

All i did was go through, starting w/2. i didn't think that it neccessary to say that. btw, ubemaya, $SOD({2}^{10})<SOD({2}^{9})$

Yeah, I know there are exceptions... but in general, the sum of the digits of higher powers are higher.

And you chose special numbers... 2^10 is just over a million so the sum of the digits is small...

Oops... for some reason I registered $2^{10}$ as $2^{20}$.

See, I'm stupid.


algebra
polynomial
linear algebra
matrix
geometry
3D geometry
sphere

Can someone explain to me how Eigenvalues are used to find the roots of polynomials? I've seen methods using companion matrices but I don't quite understand the technique. Any help is appreciated.

If you can link me to a webpage (maybe Wikipedia) that has the method you're interested in, I'd be glad to take a look :)

Eigenvalues are interesting and wicked useful.

For whoever is interested, I'd like to give an interpretation of an eigenvalues and eigenvectors.  Eigenvalues always come from a square matrix.  Every eigenvalue has an eigenvector buddy.

In linear algebra, you learn that every matrix actually makes a map from n-space to n-space.  Imagine the unit sphere in 3-space.  Take some 3x3 matrix and multiply every vector on the sphere with the matrix, and map them on a set of x-y-z coordinate axes.  What you get is a distorted version of the sphere.

Some of the vectors are lengthened, shortened and/or have their direction changed.  Eigenvectors are the vectors [i]whose direction didn't change[/i] but their length may be changed.  By "direction didn't change" I mean it's still parallel to its old version (so it may be pointing backward).  The eigenvalue that goes with the eigenvector tells you [i]how much the vector got stretched[/i] during the mapping process.

So for example, if a matrix has an eigenvalue of 2, that means there's some vector on the unit sphere which got doubled in length, but still points in the same direction.  An eigenvalue of -1 means that there's some eigenvector which got its direction reversed, but it didn't get stretched.

Thanks for you reply. The method I'm talking about involves writing a companion matrix and then finding its eigenvalues (I believe). If you have any knowledge of such a method (or any other similar method) please explain. I would also appreciate if you could talk about your interpretation of an eigenvalue/eigenvector. Thanks for the help.

[url=http://en.wikipedia.org/wiki/Companion_matrix]Companion matrices[/url] are something I'm seeing for the first time, but it looks like the way this method works is to use the fact that the characteristic polynomial of the matrix is the original polynomial you're considering. Note that the roots of a matrix's characteristic polynomial are the eigenvalues. So finding the eigenvalues of the companion matrix is the exact same thing as finding roots of the polynomial. The reason the former might be easier is because linear algebra is a very well understood topic and you have a ton of tools available to find those eigenvalues.

[url=http://www.amazon.com/Linear-Algebra-Right-Sheldon-Axler/dp/0387982582]This book[/url] has an excellent discussion of eigenvectors and their motivation. The general idea is similar to the one in the above reply.

Thanks for the help. Can someone explain some methods for finding eigenvalues without finding the roots of the corresponding characteristic polynomial directly? Thanks for all of the help.

Finding eigenvalues is a interesting, and typically hard to do.

Since most real life matrices are actually pretty messy, most eigenvalue finding algorithms have to approximate the eigenvalues, rather than compute them directly.  Usually, a good eigenvalue finding alg "factors" the original matrix first to make the process easier.

If you want to reasearch this, definitely look up the following topics:
"Power Iteration" and "QR algorithm" for finding eigenvalues
The Rayleigh Quotient
Hessenberg/Tridiagonal form

I'm digging these out of my Numerical Linear Algebra text by Trefethen and Bau
The above are fairly advanced topics, but I think on the way you'll learn a lot of interesting stuff.  Before tackling these, I highly recommend you check out the following topics:
Hermetian matrices, unitary matrices, singular values
QR factorization, LU factorization, SVD factorization

I found it really interesting, and it would be GREAT prep for learning those advanced topics.

Good luck

Power iteration is fairly straightforward.  The starting assumption is that there is a unique eigenvalue $ \lambda_{max}$ with maximal absolute value.  This means that when taking repeated powers of your matrix $ M$ multiplied by an arbitrary vector (that is, repeatedly applying $ \mathbf{M}$ to some vector), this eigenvalue will "dominate" the rest of the eigenvalues; in other words, it will be true asymptotically that

$ \mathbf{M}^k \mathbf{v} \approx \lambda_{max}^k \mathbf{v}_{max}$

where $ \mathbf{v}_{max}$ is the corresponding eigenvector.  By increasing $ k$ and computing the ratios $ \frac {\parallel{}\mathbf{M}^{k \plus{} 1} \mathbf{v}\parallel{}}{\parallel{}\mathbf{M}^k \mathbf{v}\parallel{}}$ it is therefore possible to get better and better approximations to $ \lambda_{max}$.  As I understand it, the problem of finding the rest of the eigenvalues boils down to eliminating the largest one and repeating this algorithm, although I am not sure if this is actually done by dividing out $ (x \minus{} \lambda_{max})$.

By the way, this isn't really Intermediate Topics.  It probably belongs [url=http://www.artofproblemsolving.com/Forum/index.php?f=346]here[/url].

Thanks for your reply and I appoligize for posting this in the wrong section. If possible, can someone telll me a fairly efficient way of deciding a good initial vector when using power iteration? All of the help is greatly appreciated.

Unless you have the bad luck to pick a vector orthogonal to the eigenvector, you should be fine.  I'm not sure what the accepted algorithm is, but it seems to me like computing $ \frac{\parallel{} \mathbf{Mv} \parallel{}}{\parallel{} \mathbf{v} \parallel{}}$ for a few orthogonal vectors $ \mathbf{v}$ and picking the largest one should be fine as a "first approximation."

Speaking of which, I think one way to eliminate the largest eigenvalue after it's found is to also compute the corresponding eigenvector (which is equally straightforward) and work in the subspace perpendicular to it, but I am not sure if numerical errors become too great with this technique.  (Possibly you want to continue establishing perpendicularity at each step.)


is there anybody who knows somthing or some links about history of numbers.(please forgive me if i shouldn't post this here(espesialy you Arne) but i don't know where should i post it.) :D

see this link [url=http://www.geocities.com/mathfair2002/school/arit/arithm1.htm]History of numbers[/url].

Here is a link.I hope this help you. :)
[url]http://www-groups.dcs.st-and.ac.uk/~history/Indexes/Number_Theory.html[/url]