On Sunday, Jan 2nd, I sent in my solutions, to solutions@usamts.org.

For both Round 1 and Round 2, I received a confirmation message very soon after I sent in my solutions. This time, I did not get one.

So just to be safe, I sent Mr. Rusczyk a message regarding whether I should send in my solutions again just to be safe.

The reason I am concerned is not that there is an overwhelming chance that the message did not send, but rather: IF it did not send, then there is no way for me to get credit for all the work I put into this Round (3), since the deadline is today, and I can't do anything past the deadline.

What should I do? Is there a way for me to reserve my right to re-submit my solutions even after the deadline if I promise that (1) I did not change a single letter in my solutions and (2) I promise that I did send it on Sunday (Jan 2nd)?

Or should I just calm down and not worry about it.

Thanks

First off, you probably /should/ calm down.  :)

Today is the postmark deadline; I believe (hope!) that sending in answers any time today is fine.  (So if you hear back today, no need to worry!)  Also, I forget where it's posted, but I thought there was some warning not to panic if you didn't get a confirmation message right away.  Am I just imagining that (does someone else know where that is?)

I'm no USAMTS official-type person, but I think it'd be reasonable to re-email a copy of your answers if you're that worried, just explaining that they are a copy to be used only in the event that the previous e-mail did not go through.  It's still a bit of a headache for the graders, though, especially considering that lots of people must have similar fears!  If you were snail-mailing your solutions, you'd be trusting the US Postal Service not to lose your answers.  Realistically, is the fear you have that your answers weren't sent greater than that you'd have if you had mailed paper copies?

Don't panic - we haven't yet marked the papers we've received (holiday weekend, eh).

[quote="sran"]So when would be a good time to start panicking? Say, Monday?[/quote]
Probably Monday at the latest.  They don't let us talk about solutions until a week after they're due, meaning it probably takes that long to enter all submissions into the database and deal with problems.  If they still haven't send you confirmation after that, it's likely that something is wrong.

Yeah, panic on Monday.  We'll mark off many, if not most, of the solutions as received tomorrow (for the ones we have received so far, of course).


Hi,

is it true that for any positive real numbers $x,y,z$ we have :

\[{{x}\over{\sqrt{x+y}}}+{{y}\over{\sqrt{y+z}}}+{{z}\over{\sqrt{z+x}}}\geq{{1}\over{\sqrt{2}}}(\sqrt{x}+\sqrt{y}+\sqrt{z}).\]

[quote="Michael"][quote="arqady"][quote="Michael"]Hi,

is it true that for any positive real numbers $x,y,z$ we have :
\[{{x}\over{\sqrt{x+y}}}+{{y}\over{\sqrt{y+z}}}+{{z}\over{\sqrt{z+x}}}\geq{{1}\over{\sqrt{2}}}(\sqrt{x}+\sqrt{y}+\sqrt{z}). \]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56071[/quote]

I think that the problem can be generalized to :
\[{{x}\over{(x+y)^{p}}}+{{y}\over{(y+z)^{p}}}+{{z}\over{(z+x)^{p}}}\geq{{1}\over{2^{p}}}(x^{1-p}+y^{1-p}+z^{1-p}) \]
for any $0<p<1$.[/quote]

try p very near to 1 and you see it is not true.
it is true for p<0.8 but i don't have the best constant


Protozoa from 3 different planets (A,B,C) were discovered by the space-probe harekrishna. Two protozoans of different types can merge together into one protozoan of the third type. After several such merges, only one protozoan remains in the test tube. What is its type if there 20 protozoans of type A, 21 of type B and 22 of type C.

Note: All facts mentioned are entirely ficticious.

[hide]In every step, the number of particular types of protozoa is changed by $ (\minus{}1,\minus{}1,\plus{}1)$ pattern (in any order). Hence $ |n_A \minus{} n_B|, |n_A \minus{} n_C|, |n_B \minus{} n_C|$ are invariant modulo $ 2$, because they can change either by $ 0$ or by $ 2$. Since in the initial state $ n_C \minus{} n_A \equal{} 2, n_B \minus{} n_A \equal{} 1, n_C \minus{} n_B \equal{} 1$, it's clear that the last two can never become zero, so the last protozoa standing will be of $ B$ type.[/hide]


limit
real analysis
real analysis unsolved

Discuss the below series:

$ a_{1}\minus{}a_{1}\plus{}a_{2}\minus{}a_{2}\plus{}...\plus{}a_{n}\minus{}a_{n}\plus{}...$

Is this series convergent or divergent :lol:

All of the even partial sums are zero; as for the odd partial sums, the $ 2k\minus{}1$st partial sum is $ a_k.$ Therefore this converges to zero if $ \lim a_k\equal{}0$ and diverges in all other cases.


[quote="darksaga"]εδω http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule#Proof_by_Cauchy.27s_mean_value_theorem λεει

Proof by Cauchy's mean value theorem

The most common proof of l'Hôpital's rule uses Cauchy's mean value theorem.

μπορει καποιος να το κανει λιγο πιο συγγεκριμενο αυτο?[/quote]
Υποθέτοντας τις $ f' ,g'$συνεχείς έχουμε: $ \frac{f(x)}{g(x)}=\frac{(x-x_{0})f'(k)}{(x-x_{0})g'(k)}$διότι $ f(x_{0})=g(x_{0})=0$ κ.λ.π

συγνώμη πού το γράφω εδώ για εξοικονόμηση τόπικ αλλά ρε παιδιά δεν με αφήνει να μπω το σιτε της ΕΜΕ μέσα δοκιμάστε κι εσείς μήπως είναι γενικό το πρόβλημα
ευχαριστώ

Γενικό το πρόβλημα .

chmod 777

 ...

ευχαριστώ θερμά τον mostel για την επιβεβαίωση
αν κάποιος δει διόρθωση του προβλήματος παρκαλώ να με ενημερώσει
ευχαριστώ

Το πρόβλημα δε θα διορθωθεί όσο δεν αλλάζουν mod σε 777.. θα τους στείλω ίσως ένα e-mail το απογευματάκι !


Find a formula for the maximum sections into which $N$ coplanar lines may divide a plane.

Is it 2N+3?
i just noticed that this is compltetly wrong. shoot!

[quote="Kamil Witek"]Is it 2N+3?[/quote]

$1$ line does not divide a plane into $5$ sections :wink:  .

[hide="Answer"] The proof is somewhat tedious, but a nice form is given by ${ N \choose 2}+{N \choose 1}+{N \choose 0}$.  This makes the generalization to $D$ dimensions clear. [/hide]

[quote]
The proof is somewhat tedious, but a nice form is given by ${ N \choose 2}+{N \choose 1}+{N \choose 0}$. This makes the generalization to D dimensions clear. [/quote]

I don't get how that works, can you explain? (if you don't want to, go ahead, I'm from middle school so I guess I have some time to figure this out...)

Consider the case in which the $k$th line intersects all $k-1$ previous lines for $k = 2, 3, \ldots, N$. You should get

[hide] $\frac{N(N+1)}{2}+1$ regions total, which can be rewritten as $\binom{N}{2}+\binom{N}{1}+\binom{N}{0}$. [/hide]

Alternatively, you could think of it as every nth line intersects the n-1 lines before it, causing the increase by n with each additional nth line.

...That isn't worded very well, is it?

[hide="Direct proof"] There are two types of "sections" into which $N$ lines may divide a plane.  Some of them have a "leftmost bottom point," and some do not (because they open up on the bottom).

Choosing any two of the lines produces an intersection point (in the maximal case, where none of the lines are collinear), which is the leftmost bottom point of some region; this produces ${n \choose 2}$ regions.

The sections that do not have a leftmost bottom point are not bounded below, so each of the $n$ lines contributes in dividing the rest of the plane into $n+1 ={n \choose 1}+{n \choose 0}$ sections. [/hide]
[b]Follow-up:[/b]  Can you generalize this proof to planes cutting a space?


A telecom company has a faulty system-- some times when you dial a 2-digit extensions, the system will transpose a pair of adjacent digits. e.g., a call to X318 will ring at X318, X138, or X381. while a call to X404 would be intended for extensions X404 or X044. The company do not replace the faulty system. A plan is used to correct the problem: a computer is used to detect the 3-digit extensions, whenever transposation of adjacent digits happen, the system wil reverse it to the correct dialing.
what is the greatest number of 3-digits extensions the company can assign under this plan.

For 3 digits, this is not very difficult: all of the extensions $aaa$ can be included.  Of the extensions $aab$, $aba$ and $baa$, any two are potentially confusable (if we include any two of them, a call to $aba$ is ambiguous).  The extension $abc$ excludes $acb$ and $bac$ directly, but then also excludes $cab$ and $bca$ (since if we include $cab$, for example, then a call to $acb$ becomes ambiguous), but allows for $cba$.  Thus, we can have one third of all extensions with at least 2 digits, and all of the extensions with only one digit.

Now, it'd be more interesting to see the general case for $n$-digit extensions.


For an integer $ k$ in base 10, let $ z(k)$ be the number of zeros that appear in the binary representation of $ k$. Let $ S_n \equal{} \sum^{n}_{k\equal{}1} z(k)$. Compute $ S_{256}$.

[hide]For every number $ a<128$, there exists a number $ b>127$ such that the digits of the binary representation of $ a$ are the exact opposite of those of $ b$.  For example, if $ a\equal{}17_{10}\equal{}00010001_2$, then $ b\equal{}238_{10}\equal{}11101110_2$.  Therefore we have a total of $ 8\cdot 128\equal{}1024$ zeroes.

However, we changed the representation of these numbers to make them 8 digits long, which we must correct.  We drop:
128 zeroes in the first position ([b]0[/b]...)
64 zeroes in the second position (0[b]0[/b]...)
32 zeroes in the third position (00[b]0[/b]...)
16 zeroes in the third position (000[b]0[/b]...)
8 zeroes in the third position (0000[b]0[/b]...)
4 zeroes in the third position (00000[b]0[/b]...)
2 zeroes in the third position (000000[b]0[/b]...)

However, this will give us $ \sum_{k\equal{}0}^{255}z(k)$.  To correct this, we add $ z(256)$ or 8 and subtract $ z(0)$ or 1.

The answer is actually $ 1024\minus{}(128\plus{}64\plus{}\cdots\plus{}2)\equal{}1024\minus{}254\plus{}8\minus{}1\equal{}\boxed{777}$.[/hide]

Let's represent each number up to 255 in the form ABCDEFGH, where each is either a 0 or a 1 (using leading 0's; that is, 3 would be 00000011).

For each such string, if we interchange the 0's and 1's, we get another number in the set {0,1, ... 255}, so exactly half of these numbers are 1's. There are thus (8*256)/2 = 1024 1's in this set. Removing 0 from the set and adding 256 to the set adds one 1, for a total of 1025 digit 1's.

Now we need to find the total number of digits without leading 0's.

There is one 1-digit number, two 2-digit numbers, ... 2^7 8-digit numbers, and one 9-digit number.

Thus, the total number of digits is $ 9\plus{}\sum_{j\equal{}1}^{8} j 2^{j\minus{}1} \equal{} 9\plus{}8(2^8)\minus{}2^8\plus{}1\plus{}9\equal{}1802$

Thus, there are $ 1802\minus{}1205\equal{}777$ 0's.

For a binary number with length $ n\plus{}1$, except for the leading $ 1$, all the other digits can be 0. Since there are $ \binom{n}{i}$ ways to choose i digits to be zeros, the total number of zeros to be had in this group of numbers is:
$ \sum_{i\equal{}1}^{n}{i\cdot \binom{n}{i}}$
$ \equal{}\sum_{i\equal{}1}^{n}{i\cdot \frac{n!}{i!\cdot (n\minus{}i)!}}$
$ \equal{}\sum_{i\equal{}1}^{n}{n\cdot \frac{(n\minus{}1)!}{(i\minus{}1)!(n\minus{}i)!}}$
$ \equal{}n\cdot \sum_{i\equal{}1}^{n}{\binom{n\minus{}1}{i\minus{}1}}$
$ \equal{}n\cdot 2^{n\minus{}1}$
Count the last number 256 separately, we get
$ S_{256}\equal{}\sum_{n\equal{}1}^{7}{n\cdot 2^{n\minus{}1}}\plus{}8\equal{}777$


function
combinatorics theorems
combinatorics

what methods are out there to solve recurrence relations? The only ones I currently know of are (1) generating functions (1a) Z-transform, and (2) characteristic equations (for linear constant coefficients).


inequalities
parameterization
inequalities proposed

[quote="thegod277"]
- if choose a=3, b=1, c=d=-3, we have $ x^4 \plus{} y^4 \plus{} z^4 \plus{} \frac {1}{3}(xy \plus{} yz \plus{} zx)^2\geq \sum_{sym}ab(a^2 \plus{} b^2)$
- ...
[/quote]
If $ a\equal{}3,$ $ b\equal{}1$ and $ c\equal{}d\equal{}\minus{}3$ I have obtained
$ x^4 \plus{} y^4 \plus{} z^4 \plus{} \frac {1}{3}(xy \plus{} yz \plus{} zx)^2\geq \sum_{cyc}xy(x^2 \plus{} y^2).$ :maybe:

oh, thanks you very much. I have edited. :)

[quote="thegod277"]This is my inequality:
Let a, b, c, d, e are real numbers such that $ a\geq 0, a \plus{} b\geq 0, 3a(a \plus{} b)\geq (c \plus{} d)^2$.
Prove that the inequality
$ a(x^4 \plus{} y^4 \plus{} z^4) \plus{} b(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2) \plus{} c(x^3y \plus{} y^3z \plus{} z^3x) \plus{} d(xy^3 \plus{} yz^3 \plus{} zx^3) \minus{} (a \plus{} b \plus{} c \plus{} d)xyz(x \plus{} y \plus{} z)\geq 0$. It is true for all x,y,z are real numbers.
Apply this inequality we will have
- if choose a=1, b=2, c=3 , d=0, we have Vasc-inequaltity:$ (x^2 \plus{} y^2 \plus{} z^2)^2\geq 3(x^3y \plus{} y^3z \plus{} z^3x)$
- if choose a=3, b=1, c=d=-3, we have $ x^4 \plus{} y^4 \plus{} z^4 \plus{} \frac {1}{3}(xy \plus{} yz \plus{} zx)^2\geq \sum_{sym}xy(x^2 \plus{} y^2)$
[/quote]
For $ d\equal{}0$, this inequality is equivalent to the following

$ \sum x^4\plus{}(3r^2\minus{}1)\sum x^2y^2 \plus{}3r(1\minus{}r)xyz\sum x \ge 3r\sum x^3y$,

which is in Algebraic Inequalities (page 67). For $ r\equal{}2/3$, we get
$ x^4 \plus{} y^4 \plus{} z^4 \plus{} \frac {1}{3}(xy \plus{} yz \plus{} zx)^2\geq 2\sum x^3y$.

Oh, I am sorry because I fogot a sign ||. It is $ 3a(a\plus{}b)\geq (|c|\plus{}|d|)^2$. :)

If this is really true, it's awful.
Unfortunately for the case [url]http://www.mathlinks.ro/viewtopic.php?t=185671[/url]
$ a\equal{}1, b\equal{}11, c\equal{}2, d\equal{}\minus{}6$
\[ 3a(a\plus{}b)<(|c|\plus{}|d|)^2.\]
I hoped it worked there.

[quote="thegod277"]I take very much time to prove the inequality. Finally, I have proofed it.:)
This is my inequality:
Let a, b, c, d are real numbers such that $ a\geq 0, a \plus{} b\geq 0, 3a(a \plus{} b)\geq (|c| \plus{} |d|)^2$.
Prove that the inequality
$ a(x^4 \plus{} y^4 \plus{} z^4) \plus{} b(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2) \plus{} c(x^3y \plus{} y^3z \plus{} z^3x) \plus{} d(xy^3 \plus{} yz^3 \plus{} zx^3) \minus{} (a \plus{} b \plus{} c \plus{} d)xyz(x \plus{} y \plus{} z)\geq 0$
is true for all x,y,z are real numbers.
Apply this inequality we will have
- if choose a=1, b=2, c=3 , d=0, we have Vasc-inequaltity:$ (x^2 \plus{} y^2 \plus{} z^2)^2\geq 3(x^3y \plus{} y^3z \plus{} z^3x)$
- if choose a=3, b=1, c=d=-3, we have $ x^4 \plus{} y^4 \plus{} z^4 \plus{} \frac {1}{3}(xy \plus{} yz \plus{} zx)^2\geq \sum_{sym}xy(x^2 \plus{} y^2)$
- ...
:D[/quote]
Don't want to ... spam in Inequalities Forum anymore, but...

If this inequality is true, I will [b]vote this as the greatest elementary inequality of this year[/b]!

Some months ago, I was also writing a paper to find a solution to this terrible kind of cyclic 3-variable inequalities when the equality case is terribly incredible! I've done by using EMV method (see EMV in Math reflections). This approach can deal with all fourth degree inequalities, but it lacks of generality, since it can handle specific problem easily, but can't deal with the general form (actually, it is applicable, but requires tons of computation).

[quote="Crux"]If this is really true, it's awful. [/quote]

Totally agree!

I think the following statement holds:

Let $ a, b, c$ be real numbers such that $ 3(a \plus{} 1)\ge b^2 \plus{} bc \plus{} c^2$. If $ x,y,z$ are real numbers, then

$ \sum x^4 \plus{} a\sum x^2y^2 \plus{} b\sum x^3y \plus{} c\sum xy^3 \ge (1 \plus{} a \plus{} b \plus{} c)xyz\sum x.$

The following statement is equivalent:

Let $ a, b, c$ be real numbers such that $ 3(a \plus{} 1)\ge b^2 \plus{} bc \plus{} c^2$. If $ x,y,z$ are real numbers, then

$ \sum x^4 \plus{} a\sum x^2y^2 \ge b\sum x^3y \plus{} c\sum xy^3 \plus{} (1 \plus{} a \minus{} b \minus{} c)xyz\sum x.$

for some particular cases I proved this using the sum of three squares. But it seems it fails for the general case and I can't understand how you derive the conditions on the parameters a,b,c,d.
I am waiting for a solution and hope it's not too ugly. Thanks in advance

Substituting $ a,b,c$ by $ \frac q{p}$, $ \frac {\minus{}r}{p}$, $ \frac {\minus{}s}{p}$, respectively, we get the statement:

Let $ x,y,z$ be real numbers. If $ p,q,r,s$ are real numbers such that $ p>0$ and $ 3p(p\plus{}q)\ge r^2\plus{}rs\plus{}s^2$, then

$ p\sum x^4\plus{}q\sum x^2y^2\plus{}r\sum x^3y\plus{}s\sum xy^3 \ge (p\plus{}q\plus{}r\plus{}s)xyz\sum x$.

It is interesting to write the statement as follows:

Let $ p, q, r$ be real numbers such that $ 3(1 \minus{} r)\ge p^2 \plus{} pq \plus{} q^2$. If $ x,y,z$ are real numbers, then

$ \sum x^4 \plus{} (p \plus{} q \plus{} r \minus{} 1)xyz\sum x \ge p\sum x^3y \plus{} q\sum xy^3 \plus{} r\sum x^2y^2.$

For $ r \equal{} 0$, we get the following nice generalization of Schur's inequality:

$ \sum x^4 \plus{} (p \plus{} q \minus{} 1)xyz\sum x \ge p\sum x^3y \plus{} q\sum xy^3$,

if $ p^2 \plus{} pq \plus{} q^2\le 3$ and  $ x,y,z$ are real numbers. Moreover, if $ p\plus{}q\equal{}1$, we obtain a known  :P inequality:

$ \sum x^4 \plus{} \sum x^3y \ge 2\sum xy^3$.

For $ r \equal{} 1 \minus{} p \minus{} q$, we get the inequality:

$ \sum x^4 \plus{} (p \plus{} q \minus{} 1)\sum x^2y^2 \ge p\sum x^3y \plus{} q\sum xy^3$,

if $ p^2 \plus{} pq \plus{} q^2\le 3(p \plus{} q)$ and  $ x,y,z$ are real numbers.

Thank you for your ideas! And that inequalities of Vasc is true, but on condition that $ ,3(a\plus{}1)\ge(|b|\plus{}|c|)^2$, not $ 3(a\plus{}1)\ge b^2\plus{}bc\plus{}c^2$. Any ideas? If not, I will post my solution after some hours. :)
Sorry, I don't good at English.
_________________________
Nguyen The Cuong

Great! :lol:  I have find a strongest condition for the inequality.  :lol: So, I will post my inequality again:
Let a, b, c, d are real numbers such that $ a\ge 0, 3a(a + b)\ge (min(|c| + |d|;|c| + |c - d|;|d| + |c - d|))^2$.
Prove that the inequality
$ a(x^4 + y^4 + z^4) + b(x^2y^2 + y^2z^2 + z^2x^2) + c(x^3y + y^3z + z^3x) + d(xy^3 + yz^3 + zx^3) - (a + b + c + d)xyz(x + y + z)\ge 0$
is true for all x,y,z are real numbers.
And this is my solution:
Rewrite the inequality in the form:
$ a(\sum x^4 - \sum x^2y^2) + c(\sum x^3y - \sum x^2yz) + d(\sum xy^3 - \sum x^2yz)$
$ + (a + b)(\sum x^2y^2 - \sum x^2yz)\ge 0$
And we have
$ (\sum x^4 - \sum x^2y^2) = \frac 12\sum (x^2 - y^2)^2 = \frac 12(p^2 + q^2 + r^2)$
$ \sum(x^3y - x^2yz) = \frac 13\sum(x^3y + 2y^3z + x^3z - xy^3 - 2x^2yz - xy^2z)$
$ = \frac 13\sum (x^2 - y^2)(xy + xz - 2yz) = \frac 13(pl + qm + rn)$
$ \sum(xy^3 - x^2yz) = \frac 13\sum(x^2 - y^2)(yz + yx - 2xz) = \frac 13(pm + qn + rl)$
$ \sum(x^2y^2 - \sum x^2yz = \frac 16\sum(x^2y^2 + y^2z^2 + 4z^2x^2 + 2xy^2z - 4x^2yz - 4xyz^2)$
$ = \frac 16\sum(xy + yz - 2xz)^2 = \frac 16(m^2 + n^2 + l^2)$
where
$ p = x^2 - y^2,q = y^2 - z^2,r = z^2 - x^2,l = xy + xz - 2yz,m = yz + yx - 2xz,n = zx + zy - 2yz$
Hence
$ \frac a2(p^2 + q^2 + r^2) + \frac {a + b}{6} (m^2 + n^2 + l^2) + \frac c3(pl + qm + rn) + \frac d3(pm + qn + rl)\ge 0$
We need have the inequality is true in cases:
1)$ 3a(a + b)\ge (|c| + |d|)^2$
Apply AM_GM and Cauchy-Schwarz inequalities we have
$ \frac a2(p^2 + q^2 + r^2) + \frac {a + b}{6}(m^2 + n^2 + l^2)\ge 2\sqrt {\frac {a(a + b)}{12} (p^2 + q^2 + r^2)(m^2 + n^2 + l^2)}$
$ \ge \sqrt {\frac {a(a + b)}{3}}max(|pl| + |qm| + |rn|;|pm| + |qn| + |rl|)$
$ \ge\frac {|c| + |d|}{3} max(|pl| + |qm| + |rn|;|pm + |qn| + |rl|)\ge  - \frac c3(pl + qm + rn) - \frac d3(pm + qn + rl)$.
Thus the inequality is true.
2)$ 3a(a + b)\ge (|c| + |c - d|)^2$
Because l+m+n=0, we have
$ \frac c3(pl + qm + rn) + \frac d3(pm + qn + rl) = \frac {c - d}{3}(pl + qm + rn) + \frac d3[p(m + l) + q(m + n) + r(n + l)] = \frac {c - d}{3}(pl + qm + rn) - \frac d3(pn + ql + rm)$
And similarly, apply AM_GM andCauchy_Schwarz inequality we have the inequality is true,too.
3)$ 3a(a + b)\ge (|d| + |c - d|)^2$
Prove sirmilarly.
Then the inequality is proved.  :D
I am very tired to write this post...
Any idea?

[quote="thegod277"]I don't spam in Inequalities Forum! I will post my solution, but not this time. Because I want see a solution of you. :)[/quote]

Have to say that your English is... I was to say "I don't want to spam in this forum anymore", but your problem is so nice to me  :lol: Congratulation!

Now, you make me to spam again... since

[quote="thegod277"]Great! Very Happy I have find a strongest condition for the inequality. Very Happy So, I will post my inequality again:
Let a, b, c, d are real numbers such that a\ge 0, 3a(a + b)\ge (min(|c| + |d|;|c| + |c - d|;|d| + |c - d|))^2.
Prove that the inequality
a(x^4 + y^4 + z^4) + b(x^2y^2 + y^2z^2 + z^2x^2) + c(x^3y + y^3z + z^3x) + d(xy^3 + yz^3 + zx^3) - (a + b + c + d)xyz(x + y + z)\ge 0
is true for all x,y,z are real numbers.
And this is my solution:
Rewrite the inequality in the form:
a(\sum x^4 - \sum x^2y^2) + c(\sum x^3y - \sum x^2yz) + d(\sum xy^3 - \sum x^2yz)
+ (a + b)(\sum x^2y^2 - \sum x^2yz)\ge 0
And we have
(\sum x^4 - \sum x^2y^2) = \frac 12\sum (x^2 - y^2)^2 = \frac 12(p^2 + q^2 + r^2)
\sum(x^3y - x^2yz) = \frac 13\sum(x^3y + 2y^3z + x^3z - xy^3 - 2x^2yz - xy^2z)
= \frac 13\sum (x^2 - y^2)(xy + xz - 2yz) = \frac 13(pl + qm + rn)
\sum(xy^3 - x^2yz) = \frac 13\sum(x^2 - y^2)(yz + yx - 2xz) = \frac 13(pm + qn + rl)
\sum(x^2y^2 - \sum x^2yz = \frac 16\sum(x^2y^2 + y^2z^2 + 4z^2x^2 + 2xy^2z - 4x^2yz - 4xyz^2)
= \frac 16\sum(xy + yz - 2xz)^2 = \frac 16(m^2 + n^2 + l^2)
where
p = x^2 - y^2,q = y^2 - z^2,r = z^2 - x^2,l = xy + xz - 2yz,m = yz + yx - 2xz,n = zx + zy - 2yz
Hence
\frac a2(p^2 + q^2 + r^2) + \frac {a + b}{6} (m^2 + n^2 + l^2) + \frac c3(pl + qm + rn) + \frac d3(pm + qn + rl)\ge 0
We need have the inequality is true in cases:
1)3a(a + b)\ge (|c| + |d|)^2
Apply AM_GM and Cauchy-Schwarz inequalities we have
\frac a2(p^2 + q^2 + r^2) + \frac {a + b}{6}(m^2 + n^2 + l^2)\ge 2\sqrt {\frac {a(a + b)}{12} (p^2 + q^2 + r^2)(m^2 + n^2 + l^2)
\ge \sqrt {\frac {a(a + b)}{3}}max(|pl| + |qm| + |rn|;|pm| + |qn| + |rl|)
\ge\frac {|c| + |d|}{3} max(|pl| + |qm| + |rn|;|pm + |qn| + |rl|)\ - \frac c3(pl + qm + rn) - \frac d3(pm + qn + rl).
Thus the inequality is true.
2)3a(a + b)\ge (|c| + |c - d|)^2
Because l+m+n=0, we have
\frac c3(pl + qm + rn) + \frac d3(pm + qn + rl) = \frac {c - d}{3}(pl + qm + rn) + \frac d3[p(m + l) + q(m + n) + r(n + l)] = \frac {c - d}{3}(pl + qm + rn) - \frac d3(pn + ql + rm)
And similarly, apply AM_GM andCauchy_Schwarz inequality we have the inequality is true,too.
3)3a(a + b)\ge (|d| + |c - d|)^2
Prove sirmilarly.
Then the inequality is proved. Mr. Green
I am very tired to write this post...
Any idea?[/quote]

This is indeed a very great, elegant, fantastic solution EVER! (I dont check it now, but pretty sure your method works well)

sorry, it is $ a\ge 0, a\plus{}b\plus{}c\plus{}d\plus{}e\ge 0, 3a(a \plus{} b)\ge a(a \plus{} b \plus{} c \plus{} d \plus{} e) \plus{} c^2 \plus{} cd \plus{} d^2$ :D
An we have $ 2a(2a\plus{}2b\minus{}c\minus{}d\minus{}e)\ge 3a(a\plus{}b\minus{}c\minus{}d\minus{}e)$, so this condition is stronger.

[quote="thegod277"]sorry, it is $ a\ge 0, a \plus{} b \plus{} c \plus{} d \plus{} e\ge 0, 3a(a \plus{} b)\ge a(a \plus{} b \plus{} c \plus{} d \plus{} e) \plus{} c^2 \plus{} cd \plus{} d^2$ :D
[/quote]
You are right, because $ \sum x^2y^2 \plus{}2xyz\sum x \ge 0$  :)
You are wrong, because the case $ a\equal{}c\equal{}d\equal{}0$, $ b\equal{}1$, $ e\equal{}3$ don't verify. :P

Hey, this is new result: Prove the inequaltiy
$ a\sum x^4 \plus{} 2b\sum x^2y^2 \plus{} c\sum x^2yz \ge (a \plus{} b)\sum xy(x^2 \plus{} y^2)$
is true for all real numbers x,y,z if a,b,c are [b]positive numbers[/b] such that $ b \plus{} 2c\ge 3a\ge 2c$
(Equality occurs [b]when[/b] $ (x,y,z) \equal{} (0,t,t) \equal{} (t,t,0) \equal{} (t,0,t)$ :)
A condition $ b \plus{} 2c\ge 3a\ge 2c$ is interesting, because coefficients is  1 :arrow: 2 :arrow: 3 :P
_________________
[i]I am The God.
The Cuong.[/i]

I think it is wrong, because for $ a\equal{}b\equal{}c\equal{}1$ we get an inequality better than Schur's inequality of four degree, which is not possible.

Oh no, you are right, Vasc. I am wrong.Forget and delete it. This is a true inequality
$ (p \plus{} 1)\sum x^4 \plus{} (p \plus{} 2q \plus{} 2)\sum x^2yz \plus{} 2(1 \plus{} 2q)\sum x^2y^2\ge 2(p \plus{} q \plus{} 1)\sum xy(x^2 \plus{} y^2)$
this inequality is true for all real numbers x,y,z, if p,q such that $ p\ge 1\ge q\ge \minus{} \frac 12$
Thanks for idea of you.
________________
[i]I am the god.
The Cuong.[/i]

Not any solution for it? It not, i will post my solution. :D
_____________
[i]The Cuong [/i]

[quote="thegod277"]Oh no, you are right, Vasc. I am wrong.Forget and delete it. This is a true inequality
$ (p \plus{} 1)\sum x^4 \plus{} (p \plus{} 2q \plus{} 2)\sum x^2yz \plus{} 2(1 \plus{} 2q)\sum x^2y^2\ge 2(p \plus{} q \plus{} 1)\sum xy(x^2 \plus{} y^2)$
this inequality is true for all real numbers x,y,z, if p,q such that $ p\ge 1\ge q\ge \minus{} \frac 12$[/quote]
There is a mistake, because for $ p\rightarrow \infty$, we must have
$ \sum x^4 \plus{} \sum x^2yz \ge 2\sum xy(x^2 \plus{} y^2)$,
which is not true.

The following statement holds:
Let $ b,c,d$ and $ x,y,z$ be real numbers. The inequality
\[ \sum x^4\plus{}b\sum x^2y^2\plus{}c\sum xy(x^2\plus{}y^2) \minus{}(1\plus{}b\plus{}2c)xyz\sum x \ge 0
\]
holds for all $ x,y,z \ge 0$ if and only if $ c\plus{}1 \ge 0$ and $ 2\plus{}b\plus{}2c \ge 0$ or $ c\plus{}1 \le 0$ and $ 1\plus{}b \ge c^2$.

[quote="ehku"]The following statement holds:
Let $ b,c,d$ and $ x,y,z$ be real numbers. The inequality
\[ \sum x^4 \plus{} b\sum x^2y^2 \plus{} c\sum xy(x^2 \plus{} y^2) \minus{} (1 \plus{} b \plus{} 2c)xyz\sum x \ge 0
\]
holds for all $ x,y,z \ge 0$ if and only if $ c \plus{} 1 \ge 0$ and $ 2 \plus{} b \plus{} 2c \ge 0$ or $ c \plus{} 1 \le 0$ and $ 1 \plus{} b \ge c^2$.[/quote]
We can reformulate the statement as follows:

Let $ b,c$ and $ x,y,z$ be real numbers. The inequality

\[ \sum x^4 \plus{} b\sum x^2y^2 \plus{} (2c\minus{}1 \minus{} b)xyz\sum x \ge c\sum xy(x^2 \plus{} y^2)
\]
holds for all $ x,y,z \ge 0$ if and only if $ b\ge (c\minus{}1)\max \{2,c \plus{} 1\}$.

It is great, ehku! I have just try to prove this inequality, but not well, I am lose. So, how do you proof it? ( I think by expand method $ y\equal{}x\plus{}p, z\equal{}x\plus{}q$ and $ p,q\ge 0$ is well, but very hard.) :)

Vasc, can you post a solution for a problem of ehku?
Oh, my problem is wrong! :( And this is right inequality
$ \sum x^4 \plus{} (1 \plus{} r)\sum x^2yz \plus{} 2r\sum x^2y^2\ge (1 \plus{} r)\sum xy(x^2 \plus{} y^2)$
This inequality is true for all real numbers x,y,z if $ r\in [0;3]$
A solution has posted in "secret 1 inequality", by me.

Ehku, a condition $ c\plus{}1\le 0$ and $ 1\plus{}b\ge c^2$ is wrong. It must be $ c\plus{}1\ge 0$ and $ 1\plus{}b\ge c^2$.
We can prove this inequality by that way
$ \sum x^4\plus{}b\sum x^2y^2\plus{}c\sum xy(x^2\plus{}y^2)\minus{}(1\plus{}b\plus{}2c)xyz\sum x$
$ \equal{}[\sum x^4\plus{}xyz\sum x\minus{}\sum xy(x^2\plus{}y^2)]\plus{}(c\plus{}1)[\sum xy(x^2\plus{}y^2)\minus{}2x^2y^2]$
$ \plus{}(c\plus{}1)^2(\sum x^2y^2\minus{}xyz\sum x)\plus{}(b\plus{}1\minus{}c^2)(\sum x^2y^2\minus{}xyz\sum x)\ge 0$
(true by Schur inequality) :wink:

[quote="thegod277"]Ehku, a condition $ c \plus{} 1\le 0$ and $ 1 \plus{} b\ge c^2$ is wrong. [/quote]
I think, this condition is correct.

[quote="thegod277"]It must be $ c \plus{} 1\ge 0$ and $ 1 \plus{} b\ge c^2$.[/quote]
This is a weak condition. The condition $ c \plus{} 1\ge 0$ and $ 2\plus{}b\plus{}2c \ge 0$ is better.

You are right, Vasc, since $ 1\plus{}b\ge c^2\ge \minus{}2c\minus{}1$.
We can prove this condition by similarly way .(You can see my solution to know more) :wink:


linear algebra
matrix
abstract algebra
geometry
geometric transformation
rotation

Prove the that [url=http://mathworld.wolfram.com/PauliMatrices.html]Pauli Matrices[/url] form a group under matrix multiplication.
[size=59][i]Dover Books - Matrices and Transformations[/i][/size]

The book has a different set of matrices than those on Wikipedia, MathWorld, and other sources online, though.
$ \begin{array} {ccc} \left( \begin{array} {ccc} 1 & 0 \\
0 & 1 \end{array} \right) & \left( \begin{array} {ccc} 0 & 1 \\
- 1 & 0 \end{array} \right) & \left( \begin{array} {ccc} 0 & - 1 \\
1 & 0 \end{array} \right) & \left( \begin{array} {ccc} - 1 & 0 \\
0 & - 1 \end{array} \right) \\
\left( \begin{array} {ccc} i & 0 \\
0 & - i \end{array} \right) & \left( \begin{array} {ccc} - i & 0 \\
0 & i \end{array} \right) & \left( \begin{array} {ccc} 0 & - i \\
- i & 0 \end{array} \right) & \left( \begin{array} {ccc} 0 & i \\
i & 0 \end{array} \right) \end{array}$

The work is a bit tedious but all you have to do is multiply every matrix by all the other matrices. If they do form a group, then every multiplication will yield another matrix from the set (this is assuming the group is abelian (communitive)). After that you must show an identitiy exists (which it clearly does). And then you must show that for every element $ A$ in the set, there exists an element $ B$ such that $ A*B\equal{}I$ where $ I$ denotes the identity.

Oh, pah.  The Pauli matrices are precisely the units in the matrix representation of the [url=http://en.wikipedia.org/wiki/Quaternion#Matrix_representations]quaternions[/url] as $ 2 \times 2$ complex matrices, and the units of any ring form a group.

Thanks to both of you.

[b]JZambrano201[/b] , I thought there might have been a trick I was missing so I did not have to multiply each of them together.

[b]t0rajir0u[/b] , thanks for that quaternions link; it was very interesting.  :)

However, can anyone tell me why there are a different set of Pauli matrices given by the book?  And, would not find the group involving [i]also[/i] involve multiplying more than two of those matrices, whether the same or different?
I [i]think[/i] that looking at them as sort of rotations in space (quaternions) would help.

It's likely that the complex representation of the quaternions is not unique.  What exactly is the set of matrices given by your book?

The matrices in my original post are from the book.  :)
In the MathWorld entry for the Pauli Matrices it lists three different matrices.


Solve in the reals for $ x$:

$ \sqrt{x \plus{} 2 \sqrt{x\minus{}1}} \plus{} 4 \sqrt{x \minus{} 2 \sqrt{x\minus{}1}} \equal{} 2 \sqrt{x\minus{}1} \plus{} 12$

[hide="Hint"] Let $ u \equal{} \sqrt{x \minus{} 1}$. [/hide]

[hide="different but probably wrong"]
Square both sides to start off.
$ x\plus{}2\sqrt{x\minus{}1}\plus{}16x\minus{}32\sqrt{x\minus{}1}\plus{}8\sqrt{x^2\minus{}4x\plus{}4}\equal{}4x\minus{}4\plus{}144\plus{}48\sqrt{x\minus{}1}$
Simplify
$ 25x\minus{}30\sqrt{x\minus{}1}\minus{}16\equal{}4x\plus{}140\plus{}48\sqrt{x\minus{}1}$
Simplify
$ 21x\minus{}156\equal{}78\sqrt{x\minus{}1}$
Square both sides one more time, solve, check for extraneous roots
[/hide]

[hide] by substituting $ y$ for $ 2\sqrt{x\minus{}1}$, I got $ 26$ for $ x$? [/hide]

Yep

[hide="Solution"]$ u : \equal{} \sqrt{x\minus{}1} \implies x \equal{} (x\minus{}1) \plus{} 1 \equal{} u^2 \plus{} 1$, giving:

$ \sqrt{u^2 \plus{} 2u \plus{} 1} \plus{} 4 \sqrt{u^2 \minus{} 2u \plus{} 1} \equal{} 2 u \plus{} 12$, or:

$ (u\plus{}1) \plus{} 4 (u\minus{}1) \equal{} 2u \plus{} 12 \iff 5u\minus{}3 \equal{} 2u\plus{}12 \iff u \equal{} 5 \implies x\equal{}26$[/hide]

This problem was on this year's WPI Invitational Math Competition, a local Massachusetts contest.


Χρόνια πολλά σε όλα τα μέλη του forum, χρόνια πολλά στους εορτάζοντες και εύχομαι η καινούρια χρονιά να είναι γεμάτη υγεία και πολλές πολλές ιδέες και προβλήματα! Το μόνο που δεν εύχομαι είναι η πρωτιά του subforum μας η οποία είναι ΔΕΔΟΜΕΝΗ με τον ρυθμό με τον οποίο προχωράμε! Να είστε καλά όλοι!

[color=red][size=150][b]ΚΑΛΗ ΧΡΟΝΙΑ - ΕΥΤΥΧΙΣΜΕΝΟ ΤΟ 2008[/b][/size][/color]

Αλέξανδρος Συγκελάκης

Καλή χρονιά με υγεία και ευτυχία σε όλους.

Και για να μην ξεχνιόμαστε μια γλυκιά γεωμετρία.

Έστω ένας κύκλος και μια διάμετρός του $ AB$ και μια χορδή $ CD$. Θεωρούμε ευθεία που διέρχεται από το κέντρο του κύκλου που τέμνει τις $ AC$ και $ BD$ στα $ R$ και $ S$ αντίστοιχα. Να αποδειχθεί η ισότητα $ <RCS\equal{}<RDS$.

Δεν ξέρω τι έχω πάθει και έγω γίνει πολύ τριγωνομετρικός τώρα τελευταία .Πάντως η τριγωνομετρία που χρησιμοποιώ είναι στα όρια του υποφερτού χωρίς πολλές  πράξεις .

;Estv $ O$  to perikentro .
Ως γνήσιος συμβολολάτρης θέτω $ \angle{DCR} \equal{} C_1$ , $ \angle{COR} \equal{} O_1$ , $ \angle{ROD} \equal{} O_2$
$ \angle{RCS} \equal{} x$ , $ \angle{SCO} \equal{} y$ , $ \angle{RDS} \equal{} x_1$ και $ \angle{RDC} \equal{} y_1$

Από trig Ceva στο CDO για το σημείο R έχουμε

$ \frac {\sin C_1}{\sin (x \plus{} y)}\cdot\frac {\sin O_1}{\sin O_2}\cdot\frac {\sin (x_1 \plus{} C_1)}{\sin y_1} \equal{} 1$

kai sto CDO gia to S exoume

$ \frac {\sin (C_1 \plus{} x)}{\sin y}\cdot\frac {\sin O_1}{\sin O_2}\cdot\frac {\sin C_1}{\sin (x_1 \plus{} y_1)} \equal{} 1$

Ισχύει $ x \plus{} y \equal{} x_1 \plus{} y_1$ άρα εξισώνοντας τα δεύτερα μέλη παίρνουμε

$ \frac {\sin (C_1 \plus{} x)}{\sin y} \equal{} \frac {\sin (C_1 \plus{} x_1)}{\sin y_1}$  άρα $ x_1 \equal{} x$ και $ y_1 \equal{} y$  , to ζητούμενο  :)

Πολύ ωραίο πρόβλημα Κώστα, και ωραία λύση Σιλουανέ κι ευχαριστώ πολύ για το σχήμα και για τη βοήθειά σου :)

Η δική μου λύση είναι η εξής:
Κατ' αρχάς θεωρώ ότι η $ CD$ τέμνει την $ AB$ (με όλες τις πιθανές διατάξεις βγαίνει απλά έτυχε να το σκεφτώ έτσι)
Εστω ότι η $ CS$ τέμνει τον κύκλο στο $ K$ και η $ DR$ τέμνει τον κύκλο στο $ L$.
Είναι εύκολο να αποδείξουμε με λίγο angle chasing ότι αρκεί η $ KL$ διάμετρος του κύκλου, δηλαδή αρκεί να διέρχεται από το $ O$ που είναι το περίκεντρο.
Αυτό το τελευταίο όμως προκύπτει άμεσα απ' το θεώρημα Pascal στο εξάγωνο $ BDLKCA$.

Αν υπάρχει κάπου απορία παρακαλώ να την πείτε γιατί η αλήθεια είναι ότι έχω κάνει πολύ περιληπτική τη λύση μου :blush:

To σχήμα στην λύση του mim .

Ας δούμε μία ισοδύναμη εκφώνηση για αυτή την όντως γλυκιά γεωμετρία.

[b][size=100][color=DarkBlue]Σε δοσμένο κύκλο $ (O)$ με διάμετρο $ AB,$ θεωρούμε τυχούσα χορδή $ AC$ και σταθερό σημείο $ R$ μεταξύ των $ A,$ $ C.$  Δια του $ B$ φέρνουμε μεταβλητή χορδή $ BD,$ η οποία τέμνει την ευθεία $ OR,$ στο σημείο έστω $ S.$ Αποδείξτε ότι $ \angle RCS \equal{} \angle RDS.$[/color][/size][/b]

Αντί για λύση βάζω μόνο το σχήμα και νομίζω ότι είναι αρκετό.

Καλή και δημιουργική χρονιά σε όλους, Κώστας Βήττας.

[size=100][color=Blue]EDIT: Δημήτρη και Σιλουανέ, με προλάβατε ως συνήθως...[/color][/size] :)


Find the least positive integer $ n$ such that for each prime $ p$, $ p^2\plus{}n$ is not a prime.

if $ p\equal{}2$, 4+n is not prime.

and if p was another prime number(which is an odd number), $ p^2$ is also an odd number, and to make $ p^2\plus{}n$ a non-prime number, it should be an odd number.

(for some p, i.e. 3(9+6 is not prime), even numbers can work, but we can't guarantee that it works for every other primes. but if n=odd, odd+odd becomes even, which is definitely not prime)

so 4+n is not prime and n is odd.

smallest n which satisfies these conditions is 5. (4+5=9 which is not prime, 5=odd)

answer : 5

Huh? I don't get it from the first sentence. Where did that claim come from? And some other claims are murky too...

[hide="maybe more clarity or formality"]
The answer is $ \fbox{5}$. We must prove a) that given $ n\equal{}5$, $ p^2\plus{}n$ is nonprime for all $ p$, and we must prove b) that $ 5$ is the smallest integer $ n$ that satisfies a).

To prove a), we let $ n\equal{}5$. All primes except $ 2$ are odd, and we consider the odd primes as a whole first. The parity of a number will not change if it is raised to a natural-number power, so all our $ p^2$ are odd. Since $ n\equal{}5$ is also odd, this gives entirely even $ p^2\plus{}n$. Since $ 2$, the only even prime, is also the smallest prime, we cannot produce any prime $ p^2\plus{}n$ from the odd $ p$.

If $ p^2\equal{}4$, $ p^2\equal{}4$, and $ p^2\plus{}n\equal{}9$. This is also nonprime.

But why couldn't we use smaller odd numbers? In proving b), we attempt to find simple counterexamples for $ n \in \{1,2,3,4\}$. If $ n\equal{}1$, we can let $ p\equal{}2$ and we produce $ p^2\plus{}n\equal{}5$. Similarly, a prime occurs with $ n\equal{}3$ and $ p$ remaining 2. Also, letting $ p\equal{}3$ disproves $ n\equal{}2$ or $ n\equal{}4$. Having exhausted all other possibilities, we have proved $ \fbox{n\equal{}5}$.
[/hide]

@FantasyLover: All the right ideas are in your proof, but you could organize it better.

@distracted523: Your proof is good. In a formal proof, it isn't necessary to contain certain comments like we look for small counter examples or whatever. You can just give them. I'd write something like:

$ (n,p,n\plus{}p^2)$: $ (1,2,5)$, $ (2,3,11)$, $ (3,2,7)$, $ (4,3,13)$ so $ n\ge 5$.

If $ n\equal{}5$, $ p\equal{}2$, then $ n\plus{}p^2\equal{}9$. If $ p>2$ then $ p$ is odd so $ 5\plus{}p^2>5$ is even and greater than $ 2$ so it is not prime.


The cubic polynomial $ f(x)$ is given by

$ f(x) \equal{} 2x^{3}\plus{}ax^{2}\plus{}bx \plus{}15$,

where $ a$ and $ b$ are constants. Given that $ (x\plus{}3)$ is a factor of $ f(x)$ and that, when $ f(x)$ is divided by $ (x\minus{}2)$, the remainder is $ 35$, find the values of $ a$ and $ b$.

Using these values of $ a$ and $ b$, divide $ f(x)$ by $ (x\plus{}3)$.

[quote="AndrewTom"]The cubic polynomial $ f(x)$ is given by

$ f(x) \equal{} 2x^{3} \plus{} ax^{2} \plus{} bx \plus{} 15$,

where $ a$ and $ b$ are constants. Given that $ (x \plus{} 3)$ is a factor of $ f(x)$ and that, when $ f(x)$ is divided by $ (x \minus{} 2)$, the remainder is $ 35$, find the values of $ a$ and $ b$.

Using these values of $ a$ and $ b$, divide $ f(x)$ by $ (x \plus{} 3)$.[/quote]
[hide]Since this polynomial is divisible by $ x\plus{}3$, $ f(\minus{}3)\equal{}0$, so we have $ 9a\minus{}3b\equal{}39$ or $ 3a\minus{}b\equal{}13$. We also have that $ f(2)\equal{}35$. This gives us $ 4a\plus{}2b\equal{}\minus{}31$. Solving gives $ a\equal{}\minus{}\frac{1}{2}$ and $ b\equal{}\minus{}\frac{29}{2}$. Therefore the cubic is $ f(x)\equal{}2x^3\minus{}\frac{1}{2}x^2\minus{}\frac{29}{2}x\plus{}15$, dividing by $ x\plus{}3$ results in $ 2x^2\minus{}\frac{13}{2}x\plus{}5$.[/hide]

Hi ckck. Thanks for your reply.

For $ f(2)\equal{}35$ I get $ 16\plus{}4a\plus{}2b\plus{}15\equal{}35$

$ 4a\plus{}2b\equal{}4$

$ 2a\plus{}b\equal{}2$

and so $ a\equal{}3, b\equal{}\minus{}4$

$ f(x)\equal{}2x^{3}\plus{}3x^{2}\minus{}4x\plus{}15 \equal{} (x\plus{}3)(2x^{2}\minus{}3x\plus{}5)$.

I agree with AndrewTom's answer;

@ckck: i think you forgot about the 35; it would be 4a+2b=35 instead of 4a+2b=-31.

By the condition, we can set $ f(x)\equal{}(x\plus{}3)(2x^2\plus{}cx\plus{}5)$, so
$ f(x)$ has residue of $ 35$ when $ f(x)$ is divided by $ x\minus{}2\Longleftrightarrow f(2)\equal{}5(2\cdot 2^2\plus{}c\cdot 2\plus{}5)\equal{}35\Longleftrightarrow c\equal{}\minus{}3$, yielding $ f(x)\equal{}(x\plus{}3)(2x^2\minus{}3x\plus{}5)$.


[size=167] Moderators, you can delete this post or move it to solved. I can sometimes just rush into getting helped instead of thinking. Sorry for that.[/size]

I got all of the points except for one that is mentioned in the book, but I couldn't figure out how to get it here is the problem.

Max the object function 100x+150y subjected to the constraint

$ x+3y\le120$
$ 35x+10y\le780$
$ x\le20$
$ x\ge0,y\ge0$

The points that I got is $ (0,40), (0,0), (0,20) and (12,36)$. I am missing $ (20,8)$
I got $ (12,36)$ by
$ y=\frac{-1}{3}x+40$
$ y=\frac{-35}{10}x+78$
Then, $ \frac{-1}{3}x+40=\frac{-35}{10}x+78$
So, $ x=12, y=36$

But how do I get $ (20,8)$?

Actually neveorrmind. I got it, it is that to get it I just need to plug in $ 20$ into $ y=\frac{-35}{10}+78$
There is $ 8$.

That's what I get for solving math at 1130pm.

Thus, plug all the number int the feasable set of equation to see whichever maximizes that.  And the answer will be $ (12, 36)$.

Sorry for wasting your time. :lol:


This is an easy problem I found in a Canadian Grade 9 Math Contest however I'm not sure how to approach this problem.  Can someone help me out and tell me how to solve it?

I have full solutions if you want them.  Please ask if you want them.

The question is uploaded as a picture.

For part (a), I suggest looking at the first 4 or so cases and finding the pattern.

1, 10, 100, 1000

First: One sum number - 1, 10, 100, 1000
Second: Two sums - 11, 101, 1001, 110, 1010, 1100
Third: Three sums: 111, 1101, 1110
Fourth: Four sums: 1111

Add them: 1111+ 3333+2322+1111 = 7877

Tell me if it's wrong

Even it might wrong, I suggest you begin with first sum, then second sum, third sum, etc pattern to do this problem. List the possible outcomes then add later.

Hope this helps!

PM me if you still have a questoin =)

lol...u were closed...u missed 1011.

i kinda now how to solve a but it's b that gives me problems.

Think about it this way. For a term like 10, you could form sums by either taking or not taking any of the remaining elements. So 2^3 sums will include 10. Same thing for all the other terms. So for the first part, we get 8(1000+100+10+1) or 8888. Try to apply the same reasoning to the second part.

The one easy way to work out this problem is to ask your question like this:

Since the number is either made up by 0 or 1, ask is 10 possible? or is 11 possible? Something like that and try to list the problem first.  I hope I can do it without listing but I prefer myself to be more accuracy, not speed.

In general,
The total possible number of subset of a set with n distinct element is 2^(n), because the total#of subset=nC0+nC1+nC2+nC3+...nCn= :Sigma: (nCk)    k=0 to k
Binomial Theorem (a+b)^n= :Sigma: nCk(a^(n-k))(b^(k)     k=0 to k
sub 1 for a and 1 for b, we have 2^n= :Sigma: nCk(1^(n-k))(1^k)
=:Sigma: (nCk)

that include empty set.so we subtract one =
2^(n)-1. (1,10,100,....10^n) is a set. Out of all subsets, The total # of subset has 10^n= 2^(n-1) because it's the same as number of subsets are there without 10^n. same thing for 1,10,...
the power sum is (1+10+...10^(n))*(2^(n-1))

Beta, where could people look at those forumlas you know?
Like do you just know them or know site to look at?
If there is, can you tell us? Because I love to see formulas..

[quote="beta"]AoPS book should have Binomial Theorem.[/quote]

definitely...

many of the more common fomulas used in math competitions are covered in AoPS (many with proofs as well)...

and most of the rest are covered in the classes...so it's all good...


function
trigonometry
complex analysis
real analysis
real analysis unsolved

Show that this formula defines a differentiable function on line $ \mathbb{R}$:

$ f(x)\equal{} \sum_{n\equal{}1}^{\infty} (x\minus{}nsin \frac{x}{n})$.

Thanks.

Let $ f_{n} \equal{} x \minus{} n\sin \left(\frac{x}{n}\right)$ be defined on an open interval I.

1. Is $ f'_{n}(x) \equal{} 1 \minus{} \cos \left(\frac{x}{n}\right)$ continuous on $ I$?

2. Does the sum        $ \sum_{n\equal{}1}^{\infty} f'_{n}(x)$                         converge uniformly on $ I$?

3. Does there exist some point $ a \in I$ for which the sum $ \sum_{n\equal{}1}^{\infty} f_{n}(a)$ converge?

Then the sum in question converges to a differentiable function in I and $ \left(\sum_{n\equal{}1}^{\infty} f_{n}(x)\right)' \equal{} \sum_{n\equal{}1}^{\infty} f'_{n}(x)$.

Expand out $ n\sin\left(\frac xn\right)$ as its power series in $ x$ centered at zero, and note that the lowest degree term cancels with the $ x$ we're subtracting it from. That gives us:

$ f(x)\equal{}\sum_{n\equal{}1}^{\infty}\sum_{k\equal{}1}^{\infty}\frac{(\minus{}1)^{k\plus{}1}x^{2k\plus{}1}}{n^{2k}(2k\plus{}1)!}.$

We can show that for each fixed $ x,$ this sum is absolutely convergent, since for large enough $ n,$ the sum is dominated by $ \sum_n\frac{x^3}{n^2}.$ (This convergence isn't uniform on $ \mathbb{R}$ but it is uniform on compact subsets of $ \mathbb{R}.$)

That absolute convergence then justifies interchanging the sums, leaving us with:

$ f(x)\equal{}\sum_{k\equal{}1}^{\infty}\frac{(\minus{}1)^{k\plus{}1}\zeta(2k)x^{2k\plus{}1}}{(2k\plus{}1)!}$

That is a power series with an infinite radius of convergence. (Note that $ \zeta(2k)\approx 1$ for large $ k.$) This is an entire function, so of course it is continuous and differentiable on $ \mathbb{R}.$

The first few terms of the power series written out:

$ f(x)\equal{}\frac{\pi^2x^3}{36}\minus{}\frac{\pi^4x^5}{10800}\plus{}\frac{\pi^6x^7}{945\cdot 5040}\minus{}\cdots$

[quote="Kent Merryfield"]This is an entire function, so of course it is continuous and differentiable on $ \mathbb{R}.$[/quote]
This is, of course, the key. Note that we do not need extensive computations to prove it. The simple observation that the terms of the series are dominated by $ n^{\minus{}2}$ on every compact subset of the complex plane is more than enough. ;)


logarithms
inequalities
inequalities unsolved
induction

Let $a_1, a_2, \dots$ be a sequence of real numbers satisfying $a_{i+j} \leq a_i+a_j$ for all $i,j=1,2,\dots$. Prove that
\[ a_1 + \frac{a_2}{2} + \frac{a_3}{3} + \cdots + \frac{a_n}{n} \geq a_n  \]
for each positive integer $n$.

The following is China MO 1993 Problem 6 (See [url=http://www.artofproblemsolving.com/Forum/topic-66332-20.html]CMO 1993[/url])

[quote]Let $f$ be a function from the positive real numbers to the positive real numbers such that $f(xy)\leq f(x)f(y)$ for any positive real numbers $x$ and $y$. Prove that for any positive real number $x$ and all positive integers $n$, $f(x^n)\leq f(x)f(x^2)^{\frac{1}{2}}\cdots f(x^n)^{\frac{1}{n}}$.[/quote]

If we let $a_i = \ln{f(x^i)}$, then $a_{i+j} \leq a_i+a_j$, then the CMO problems immediately follows from the above APMO problem.

I did not know the relation between these two problems because it was not me who translated CMO 1993......

[hide]We will prove this by induction. Note that the inequality holds for $ n\equal{}1$. Assume that the inequality holds for $ n\equal{}1,2,\ldots,k$, that is,
\[ a_1\ge a_1, \\ a_1\plus{}\frac{a_2}2\ge a_2, \\ a_1\plus{}\frac{a_2}2\plus{}\frac{a_3}3\ge a_3, \\ \vdots \\ a_1\plus{}\frac{a_2}3\plus{}\frac{a_3}3\plus{}\cdots\plus{}\frac{a_k}k\ge a_k.
\]
Sum them up:
\[ ka_1\plus{}(k\minus{}1)\frac{a_2}2a_2\plus{}\cdots\plus{}\frac{a_k}{k}\ge a_1\plus{}a_2\plus{}\cdots\plus{}a_k.
\]Add $ a_1\plus{}\ldots\plus{}a_k$ to both sides:
\[ (k\plus{}1)\left(a_1\plus{}\frac{a_2}2\plus{}\cdots\plus{}\frac{a_k}k\right)\ge (a_1\plus{}a_k)\plus{}(a_2\plus{}a_{k\minus{}1})\plus{}\cdots\plus{}(a_k\plus{}a_1)\ge ka_{k\plus{}1}.
\]
Divide both sides by $ k\plus{}1$:
\[ a_1\plus{}\frac{a_2}2\plus{}\cdots\plus{}\frac{a_k}k\ge\frac{ka_{k\plus{}1}}{k\plus{}1},
\] i.e.
\[ a_1 \plus{} \frac{a_2}{2} \plus{} \frac{a_3}{3} \plus{} \cdots \plus{} \frac{a_n}{n} \geq a_n.
\][/hide]

[quote]Let $f$ be a function from the positive real numbers to the positive real numbers such that $f(xy)\leq f(x)f(y)$ for any positive real numbers $x$ and $y$. Prove that for any positive real number $x$ and all positive integers $n$, $f(x^n)\leq f(x)f(x^2)^{\frac{1}{2}}\cdots f(x^n)^{\frac{1}{n}}$.[/quote]

have you a solution to this beautifull problem ? thanks

[quote=oty]
have you a solution to this beautifull problem ? thanks[/quote]
Well, as noted above by [b]fuzzylogic[/b], the CMO problem follows immediately from the APMO problem.

Sorry I didn't notice his post , thank you a lot Dear [b]Tintarn [/b]

@post#2 You showed, that there exists a sequence $a_i$ satisfying assumptions, that also satisfies thesis. But that's no proof, that every sequence $a_i$ satisfying condition $a_{i+j} \leq a_i+a_j$ gives $a_1 + \frac{a_2}{2} + \frac{a_3}{3} + \cdots + \frac{a_n}{n} \geq a_n $.
Your proof is incomplete then.

[quote=WolfusA]
Your proof is incomplete then.[/quote]
Sure. Read it properly. It's not even attempting to be a proof of the APMO problem.
It is just to show how the CMO problem follows from the APMO problem. And that is perfectly correct.

[quote=jgnr][hide]
Divide both sides by $ k\plus{}1$:
\[ a_1\plus{}\frac{a_2}2\plus{}\cdots\plus{}\frac{a_k}k\ge\frac{ka_{k\plus{}1}}{k\plus{}1},
\] i.e.
\[ a_1 \plus{} \frac{a_2}{2} \plus{} \frac{a_3}{3} \plus{} \cdots \plus{} \frac{a_n}{n} \geq a_n.
\][/hide][/quote]
Im sorry, how does this work?

[quote=Tintarn]I just read in Titu Andreescu's book "104 Number Theory Problems" where he gave another, combinatorial solution to this problem (ascribed to [i]Andreas Kaseorg[/i]) and I felt that this proof is too beautiful to not mention it here:
[hide=Solution]
We can extend the condition by induction to $a_{i_1+i_2+\dotsc+i_k} \le a_{i_1}+a_{i_2}+\dotsc+a_{i_k}$. We apply a combinatorial argument.

For any permutation $\pi$ in $S_n$, define $f(\pi,k)$ to be the number of $k$-cycles in $\pi$. Clearly,
\[1 \cdot f(\pi,1)+2 \cdot f(\pi,2)+\dotsc+n \cdot f(\pi,n)=n\]
Note also that $\sum_{\pi \in S_n} f(\pi,k)$ is the total number of $k$-cycles in all permutations of $n$ elements, which is $\binom{n}{k} (k-1)!(n-k)!=\frac{n!}{k}$.
Therefore, we have
\begin{align*}
a_1+\frac{a_2}{2}+\dotsc+\frac{a_n}{n}&=\frac{1}{n!} \sum_{\pi \in S_n} \left(f(\pi,1)a_1+f(\pi,2)a_2+\dotsc+f(\pi,n)a_n\right)\\
&\ge \frac{1}{n!} \sum_{\pi \in S_n} a_{1\cdot f(\pi,1)+2 \cdot f(\pi,2)+\dotsc+n \cdot f(\pi,n)}\\
&=\frac{1}{n!} \sum_{\pi \in S_n} a_n=a_n
\end{align*}
[/hide][/quote]

Very nice. I just wonder, why has Titu written a [i]combinatorial[/i] solution to an [i]algebra[/i] problem in his [i]number theory[/i] book? :P

[quote=DrYouKnowWho]Inductively, one can get \[a_k\geq \frac{ka_n}{n}\] Hence the conclusion easily follows.[/quote]
Unfortunately just wrong. Consider the sequence $(1,0,1,0,0,\dots)$ with $k=2, n=3$ as a direct counter-example.
As a general hint, it often helps to read previous solutions to a problem to judge its difficulty before claiming it to be "easy".


$ a_{n}$, $ b_{n}$, $ c_{n}$ reals sequences such that

$ a_{n}+b_{n}+c_{n}$ tends to $ 0$

and

$ e^{a_{n}}+e^{b_{n}}+e^{c_{n}}$ tends to $ 3$

Does the sequences $ a_{n}$, $ b_{n}$, $ c_{n}$ tends to $ 0$ ?

suppose the limits are independents
$ \lim_{n\to \alpha }\, a_{n}=a$
$ \lim_{n\to \alpha }\, b_{n}=b$
$ \lim_{n\to \alpha }\, c_{n}=c$

the system have infinitely complex solutions
$ a+b+c=0$
$ e^{a}+e^{b}+e^{c}=3$

one solution is all zero

but have non-zero solutions

But $ a,b,c\in \mathbb{R}$ because $ a_{n},b_{n},c_{n}\in\mathbb{R}\forall n$.

For all $ x$, $ e^{x}\ge 1+x$, with equality only at $ x=0$. Let $ f(x)=e^{x}-1-x$; we have $ f(a_{n})+f(b_{n})+f(c_{n})\to 0$, so $ f(a_{n})\to 0$ and similarly for the other two. This can only happen if each sequence tends to zero on its own.


graph theory
combinatorics proposed
combinatorics

this is kind of a puzzle, but i think graph pts vertices ... may be applicable here to prove or disprove the existence of a solution?

[img]http://i2.tinypic.com/2054ke9.jpg[/img]

i have been asked to reproduce the figure with the following conditions

1. pick up the pencil at max 5 times

2. no retrace

i doubt that this figure cannot be construct with these conditions. can anyone draw/disprove this? thank!

Do you want to use graph theory? Mmm... I don't know.

There obviously isn't any Eulerian path (a path that goes through all the edges without retracing) in this particular graph $G$, because there are more than two vertices with odd degree (like the Königsberg bridges).

In a general graph $G$, what happens if we can pick up the pencil $n$ times? That means that we can draw $n+1$ paths, doesn't it?
We must partition the graph $G$ into $n+1$ edge-induced subgraphs $G_{i}$ and each one must be a connected graph with an Eulerian path (no more than two vertices with odd degree), that can then be found with Fleury's algorithm.
Finding the minimum $n$ seems a difficult question.
But I think (please correct me if I'm wrong) each vertex with odd degree must be an initial or final point of an Eulerian path, so you need to partition the graph into at least $n+1=m/2$ subgraphs, where $m$ is the number of vertices with odd degree.
But is that enough? I'm not sure. I'll think about it.

I'll write down some properties of the graph in this particular case. Maybe they can help.
The graph $G$ has $12$ vertices:
- $8$ "outside" vertices $O$;
- $4$ "inside" vertices $I$.
We can see that
- every $O$ has degree $3$: $3$ neighbour vertices ($1$ $O$ and $2$ $I$);
- every $I$ has degree $7$: $7$ neigbour vertices ($4$ $O$ and $3$ $I$).
So, there are $\frac{8\times 3+4\times 7}{2}=26$ edges:
- $4$ edges $OO$;
- $16$ edges $OI$;
- $6$ edges $II$.
If we look at the line graph $LG$ of the graph $G$, we can see that
- every edge $OO$ has $(3-1)+(3-1)=4$ neighbour edges;
- every edge $OI$ has $(3-1)+(7-1)=8$ neighbour edges;
- every edge $II$ has $(7-1)+(7-1)=12$ neighbour edges.

to be more precise, condition (1) means you can draw at most 5 paths, not 5 + 1 paths

some claimed that they can draw this figure with 5 paths and no retrace. i doubt if this is possible

what is your conclusion after you have written down these properties lordWings?

Sorry, I didn't understand if you meant $5$ paths or $5$ interruptions ($6$ paths).

If you meant $5$ paths, then the answer is easier.
If we have $12$ vertices with odd degree, we need $6$ or more paths ($5$ are not enough) to draw it without retracing. Each vertex with odd degree should be the start or the endpoint of a path.

5 paths mean you can draw 5 continuous lines without retrace
e.g. you can draw the letter P in one path; you can draw the letter X in at least 2 paths, but not one path

since i know nothing about discrete mathematics, i need to ask you more questions

one can begin to draw by putting the pencil on the cross point of the "X's" - the intersections of the diagonal of the squares. so do we consider those points vertices? and why not?

If we have 12 vertices with odd degree, we need 6 or more paths ....
each vertex with odd degree should be the start or the endpoint of a path

is the latter statement a theorem in discrete math?

[quote="analysis"]one can begin to draw by putting the pencil on the cross point of the "X's" - the intersections of the diagonal of the squares. so do we consider those points vertices? and why not?[/quote]

You're right. I should have considered those points because you can change direction there when tracing a path. Probably I forgot them because the argument doesn't change much, since  these new vertices have an even degree (an even number of neighbour vertices).

[quote="analysis"]If we have 12 vertices with odd degree, we need 6 or more paths ....
each vertex with odd degree should be the start or the endpoint of a path

is the latter statement a theorem in discrete math?[/quote]
I'm not an expert in discrete math myself. I just had seen problems with Eulerian paths like the well known problem of the seven bridges in Königsberg. This problem is a situation I had never met before, but somewhat similar. Maybe there's a theorem, but it's easy to deduce it by yourself.

You can imagine the process of tracing the initial graph step by step, with $n$ paths. After each step (after each path has been traced), there remains a graph that still hasn't been traced. At the end, of course, one expects the remaining graph to be void (maybe it's easier to imagine paths of an eraser instead of a pencil, but you have to keep in mind that you can't go through an already erased line).

What's the difference between the remaining graph before and after each step?
In a path you must enter a vertex from an edge and leave by another, so the degree of each of those vertices is now an even number less than before ($2n$ if you have visited the vertex $n$ times during the path).
The only exceptions are the start and the endpoint of the path (because one time you don't enter or leave it); in this case obviously the degree of those vertices are now an odd number less than before ($2n+1$ if you have visited the vertex $n$ times during the path besides the beginning/end).

So, if there's a graph with $m$ vertices with odd degree, you'll need to trace at least $m/2$ paths so that these vertices are the start or the endpoint of some path. But maybe more, if you haven't been careful (or worst, if you haven't been able) to avoid remaining graphs that are not connected.

In this case, it's easy to trace the graph with only $6$ paths, so $6$ is the minimum.

Here is a simple solution - each arrow represents lifting the pencil between paths, and each step (except the last) adds the same five-line figure.

lordWings, your argument is beautiful by the fact that it is simple. i have no objection... i believe that you have proved there is no solution if we  are allow to draw 5 paths (unless i don't see a flaw somewhere). thank you!

Diarmuid, sorry for the unclear-posed problem. condition (1) suppose to mean you can draw only 5 continuous lines or paths (without retrace). lordWings have showed that the minimum is 6 paths, which you have constructed in your drawing. thank you!


If a,b,c >0, a+b+c>=27, prove that

(a+b+c)(ab+bc+ca)<=9(a+b+c)^2

If a b c are all very very big, then ineq will fail to hold.

If you meant $ a\plus{}b\plus{}c\equal{}27$ we get
$ ab\plus{}ac\plus{}bc \le 3^5$
$ \frac{ab\plus{}ac\plus{}bc}{3} \le 3^4$
$ \sqrt{\frac{ab\plus{}ac\plus{}bc}{3}} \le 3^2$
$ \sqrt{\frac{ab\plus{}ac\plus{}bc}{3}} \le \frac{a\plus{}b\plus{}c}{3}$

which is the maclaurins inequality

[quote="tkrass"]If you meant $ a \plus{} b \plus{} c \equal{} 27$ we get
$ ab \plus{} ac \plus{} bc \le 3^5$
$ \frac {ab \plus{} ac \plus{} bc}{3} \le 3^4$
$ \sqrt {\frac {ab \plus{} ac \plus{} bc}{3}} \le 3^2$
$ \sqrt {\frac {ab \plus{} ac \plus{} bc}{3}} \le \frac {a \plus{} b \plus{} c}{3}$

which is the maclaurins inequality[/quote]

You're complicating too much. If $ a\plus{}b\plus{}c\equal{}27$, than we just need to proove that $ 27(ab \plus{} ac \plus{} bc) \le 9*27^2$, or just $ ab \plus{} ac \plus{} bc \le 243$ what is trivial by well known inequality $ ab \plus{} ac \plus{} bc \le \frac {(a \plus{} c \plus{} c)^2}{3}$.

[quote="MathUniverse"][quote="tkrass"]If you meant $ a \plus{} b \plus{} c \equal{} 27$ we get
$ ab \plus{} ac \plus{} bc \le 3^5$
$ \frac {ab \plus{} ac \plus{} bc}{3} \le 3^4$
$ \sqrt {\frac {ab \plus{} ac \plus{} bc}{3}} \le 3^2$
$ \sqrt {\frac {ab \plus{} ac \plus{} bc}{3}} \le \frac {a \plus{} b \plus{} c}{3}$

which is the maclaurins inequality[/quote]

You're complicating too much. If $ a \plus{} b \plus{} c \equal{} 27$, than we just need to proove that $ 27(ab \plus{} ac \plus{} bc) \le 9*27^2$, or just $ ab \plus{} ac \plus{} bc \le 243$ what is trivial by well known inequality $ ab \plus{} ac \plus{} bc \le \frac {(a \plus{} c \plus{} c)^2}{3}$.[/quote]

It can't be *that* well known or that user would have used it, correct?
Don't presume/assume information on a fair percentage of users,
and just allow for different solutions without an attitude of dismissiveness.

Also, that fraction must be $ \frac {(a \plus{} b \plus{} c)^2}{3}$ instead of what you typed.


Does anybody know what the top score was for Indiana 2009 Mathcounts? (for individuals....was it like 46/46 or something insane??)Indiana doesn't have an official website for this type of thing. Also, can someone link me to the Calumet Chapter Mathcounts results(I can't find it, of course though I know the results. I somehow managed first place despite the fact I don't even know the quadratic formula. Probably lucky guesses.... I kinda wanted to know my score, but there was a computer bug and I never saw the results. I probably got somewhere in the ballpark of 25-28 out of 46, which makes me feel a little guilty for people who fail chapter with near perfect scores)

Please reply.

There's a reason that Indiana doesn't have a Mathcounts website....it's because no good competitors care about results.....they practice a lot, do their best and what's done during the contest is done
My advice to you is to work hard like all the national competitors and quit worrying about how you compare to other people

sorry im not from indiana, but...yeah

What sansuunoousama said is true; however, even hard workers want to see results, compare themselves with others, and smile at their achievements  :) . In terms of results, I forget if Mark Sellke told me he got a 45 or 46... (he won state).

I'm not a national competitor for one thing.  I phailed at state[63rd])  I understand that I shouldn't worry,  but I did want to know how I did in Calumet Chapter.  There was a computer bug when they were entering all the results, so they couldn't tell everybody how they did, just their places.  Probably because we had a new building at Valparaiso University.  They were supposed to mail the results, but they never did.  :( Is anybody else on AoPS from the Calumet Chapter?  Also, other top MC States, such as California, do have a separate site for Mathcounts FYI.

EDIT: This comment above is directed at sansuunoousama.  And thanks "the pink muskrat".  Well, if that's the top score, I'll have a tough time making nats.  But at least Mark Selkie is in 8th grade, and I won't have to worry about him!  I hope the 2010 MC tests are like the 2003 States test(If you've seen them they are about 30 times easier than other MC test).  And I think I have a right to at least know how I did in chapter as far as my score.  That way, I can improve for next chapter competition(everybody in my chapter didn't do so well.....so I know I can improve of it :))

California is a BIG state......just by population, they're going to have more talented kids.....and you realize Indiana has almost always outperformed California at nationals.....
the reason Indiana is always in the top five is because there are always at least four people that work their butts off to be the best....

your signature says you want to make Nats.....if that really is your goal, I'm just giving you advice......

I speak from experience.....

You mean you made Nats?

And thanks for the advice, it's just that I think I'm entitled to know my score for chapter.  After all, I took the test.  But oh well.

And I tried saying "Make MC Nats" being my signature, but I can't see it.  What
s going on?

So everybody else CAN see my signature, but I can't.

Anyways, I am practicing for  Mathcounts and hoping all goes for the best.  I have two more chances to make it, and after I order AoPS Volume 1 and finish reading it, I should be well prepared.

I made it and I've been working with three of the four that made it to Nationals from Indiana last year for the past few months......

ok....well have fun working then....

I worked as a runner at State last year so hopefully I can be there next year too..... see ya there

Yeah, well, I'll be actually practicing hard 1 month before states.........and can you answer my question about the signature dilemna?  I thought I'm supposed to see my own signature.

I cant see your signature....I had to look for it
go to edit profile....preferences.....posting messages.....always attach my signature should be marked yes if you want to see it when you post....k?

Thanks.  I've changed the settings, and it's working.  Will this also include my signature in all of my old posts too?

[quote="maybach"]I hope the 2010 MC tests are like the 2003 States test(If you've seen them they are about 30 times easier than other MC test).  [/quote]

You do realize that if the test is easier, everyone else does well too, right?

[quote="maybach"]Yes, but if I get 46/46(which is actually quite possible if it was like 2003) then it would be an unbreakable tie, and I'd go to nats no matter what.   :heli:[/quote]

Unless 4 or more other people do and then you have to do a tiebreaker round :heli:

Yeah.  After I posted this I realized about the tiebraker round(but I really don't know what it is)  Which is odd.  Because in my chapter competition 11 people went to states because there was an unbreakable tie for 10th........  How does a tiebraker round work,  anyway?

ah mathcounts.  Mathcounts was actually very beneficial to me because I did so poorly...blessing in disguise, taught me to work harder.
my advice: don't worry too much about it.  Do the competitions for fun and for the experience...it helps a lot not to be stressed

@sansuunoousama: hows your research going?  It's not a stretch to say that the 3 of you guys are some of the top math students in the entire nation, so I'm guessing it's going well.  Anyways good luck.

[quote="maybach"]I'm not a national competitor for one thing.  I phailed at state[63rd]) [/quote]
And no you did not "phail" at state.  Since you weren't even an 8th grader and you got 63rd, that's pretty good.  Just keeping working hard, and you'll reach your goal.

uhhh....yeah its going g.r.e.a.t?......
I was wondering what day Math Club was meeting this year.....

I don't know yet.  I'll get back to you once school starts; it has to be a good time for the sponsors.  I'll recommend Thursday, but I don't know what time they want it to be at.

Also, on a totally unrelated note from my previous post, but I guess back on topic...I just googled this and found it:
http://mathcounts.org/Document.Doc?id=325
Here is the list of the phone numbers and emails of Mathcounts Chapter sites.

To maybach: If you really want to see your results from last year, maybe you can email them to find out your score from last year.

Or perhaps your teacher at your school who sponsors Mathcounts will know your score; don't they usually get a score sheet back?

My mathcounts coach said they never mailed the results at all.  According to them, they had a "computer bug" and the way it sounds, they might not even have a record at all of my score.  Googling for my chapter did absolutely nothing.  Is anyone else on AoPS from the Calumet Chapter that participated in 2009 MC(perhaps from my school?)?

Sorry, but I am slightly confused.
From your previous response, it seems like you are implying that other students from your region should know your score...but I doubt that because you even said that no one from your school got scores back (since you said the teacher never got a sheet) and that apparently there was a "bug" for everyone in your chapter.

Anyways, they must have some record of your score because you even said in your original post that you got 1st.  They must have your score somewhere on a sheet of paper...otherwise how would they know that you got first (the initial grading from that day should be on paper-my mom was a grader once and she said that she recorded it on paper and got it checked by others)?  The website that I provided has the email of your chapter's director, so you can contact him/her if you really want.

Besides, my advice would be for you to not worry too much about it.  It was last year, and you have a brand new year this year.  Instead of worrying about what you got, you should try to improve your score so that you will even do better.

[quote="maybach"]
But at least Mark Selkie is in 8th grade, and I won't have to worry about him! )[/quote]

hope you mean going into 8th. you will have to worry about Mark, as well as the rest of the all 7th-grade national team from indiana. only state where no 8th graders were on the team.

The future is correct, your coach should have received a score sheet with your(and your teammates) score on it.  That score sheet would have been filled out before anything was entered in a computer.  The posters were also correct in that there were also at least 7 7th graders in the top 13 at the state competition.  No one in Indiana should be taking a trip to Nationals for granted.  It should be a fun year at the state competition.

It definitely will be  I think what happened though was(since this was a new building this year) according to what they said, was that they had the results, but they couldn't hand out scoresheets because a computer bug deleted the scores, so all they had were the results.  They promised to mail the results to the coaches, but it never happened.  But that isn't the point.

I wanted to know if there was somebody else from my area(or my school) in Art of Problem Solving.

Hmm, my first post...

Mark: Probably 1st again

David/Yushi: Probably 2nd/3rd

Vivian: Supposedly not doing MC this year

So, 1 slot up for grabs :(

You're going to need to practice a lot, lot, lot to make it.


quadratics
modular arithmetic
number theory proposed
number theory

prove that for every prime $p=3k+1$ there exsist natural $a$ such that:$a^2+a+1\equiv 0 (\mod p)$

$\left(\frac{-3}{p}\right) = \left(\frac{-1}{p}\right) \cdot \left(\frac{3}{p}\right) = (-1)^{\frac{p-1}{2}} \cdot (-1)^{\frac{p-1}{2} \cdot \frac{3-1}{2}} \cdot \left(\frac{p}{3}\right) = \left(\frac{3k+1}{3}\right) = \left(\frac{1}{3}\right) = 1$ so $-3$ is a quadratic residue mod $p$ and there exists $t$ such that $t^2 \equiv -3 \pmod{p}$. As $2 \not | p$ ($p=2$ is not $\equiv 1 \pmod{3}$), we know that $\exists x$ such that $2x \equiv 1 \pmod{p}$ so take $a = x(t-1)$. We then have $4(a^2 + a + 1) = (2a+1)^2 + 3 \equiv t^2 + 3 \equiv 0 \Rightarrow a^2 + a + 1 \equiv 0 \pmod{p}$, as desired.

Alternatively consider the equation $X^k-1$ modulo $p$.  By the factor and remainder theorem it cannot have more than $k$ roots.  So there is an non-root $h$ such that $h^k = a \not\equiv 1 \bmod p$ but $h^{3k} = a^3 \equiv 1 \bmod p$ by little Fermat.  So $(a-1)(a^2+a+1) \equiv 0 \bmod p$ but $a-1 \not\equiv 0 \bmod p$.  We have $a^2+a+1 \equiv 0 \bmod p$ as desired.

You can prove more:
For every prime $p \equiv 1 \mod 3$ exists $a,b$, such that $p=a^2+ab+b^2$.

You can prove a yet stronger result:
If p=1 mod 3 is prime, there exist x,y naturals such that p=a^2+3b^2.

Bomb

Eissenstein integers easily prove everything here since $a^2+ab+b^2$ is their norm.


linear algebra
matrix
inequalities
vector
linear algebra unsolved

Let $ Q$ be an $ n*k$ matrix, and $ rank(Q)\equal{}k$, prove $ I_n\minus{}Q(Q^*Q)^{\minus{}1}Q^* \geq 0$. where $ Q^*$ denotes conjugate transpose matrix of  Q, and $ \geq 0$ means being positive semi-define.

---------------------
This inequality is inside the proof of the matrix form of Cauchy-Schwarz inequality by Marshall and Olkin. (a article named "Matrix versions of the Cauchy and Kantorovich inequalities"). But I cann't find this article, (

What happens if we just stick a vector in there and try it?

Let $ A\equal{}I\minus{}Q(Q^*Q)^{\minus{}1}Q^*.$ What is $ x^*Ax?$

$ x^*Ax\equal{}x^*x\minus{}x^*Q(Q^*Q)^{\minus{}1}Q^*x$

It would help if we could write $ x\equal{}Qy$ for some $ y.$ Note that $ x\in \mathbb{C}^n$ while $ y$ would be in $ \mathbb{C}^k.$ We can't do that - since we could have $ n>k,$ then $ Q$ would not map $ \mathbb{C}^k$ onto $ \mathbb{C}^n.$

Even though we can't do that, let's pretend we can anyway. So assume $ x\equal{}Qy.$

$ x^*Ax\equal{}y^*Q^*Qy\minus{}y^*Q^*Q(Q^*Q)^{\minus{}1}Q^*Qy\equal{} y^*Q^*Qy\minus{}y^*Q^*Qy\equal{}0.$

All right, let's do this. Let $ V$ be the column space of $ Q.$ We can orthogonally decompose $ \mathbb{C}^n$ as $ V\plus{}V^{\perp}.$ We say that $ z\in V^{\perp}$ iff $ z^*Qx\equal{}0$ for all $ x\in\mathbb{C}^k.$

Write an arbitrary $ x\in\mathbb{C}^n$ as $ x\equal{}y\plus{}z$ where $ y\equal{}Qu$ and $ z^*Qv\equal{}0$ for all $ v.$

$ x^*Ax\equal{}(y\plus{}z)^*A(y\plus{}z)\equal{}y^*Ay\plus{}y^*Az\plus{}z^*Ay\plus{}z^*Az.$

We have shown that $ y^*Ay\equal{}0.$

$ z^*Az\equal{}z^*z\minus{}z^*Q(Q^*Q)^{\minus{}1}Q^*z\equal{}z^*z$ since $ v^*Q^*z\equal{}(z^*Qv)^*\equal{}0.$

By a very similar argument, $ y^*Az\equal{}y^*z.$ But $ y^*z\equal{}0.$ This was an orthogonal decomposition, after all. Similarly, $ z^*Ay\equal{}0.$

Hence $ x^*Ax\equal{}z^*z\ge0$ and $ A$ is positive semidefinite.

(Kind of a stream-of consciousness approach - I started writing it before I knew what the proof was. It could be cleaned up.)

A much more direct approach, from which we learn more anyway:

$ A^2\equal{}(I\minus{}Q(Q^*Q)^{\minus{}1}Q^*)^2$

$ \equal{}I\minus{}Q(Q^*Q)^{\minus{}1}Q^*\minus{}Q(Q^*Q)^{\minus{}1}Q^*\plus{} Q(Q^*Q)^{\minus{}1}Q^*Q(Q^*Q)^{\minus{}1}Q^*$

$ \equal{}I\minus{}2Q(Q^*Q)^{\minus{}1}Q^*\plus{}Q(Q^*Q)^{\minus{}1}Q^*\equal{} I\minus{}Q(Q^*Q)^{\minus{}1}Q^*\equal{}A.$

$ A$ is a Hermitian idempotent. Since it's Hermitian, it can be diagonalized. Since it's an idempotent (that is, a projection), its only eigenvalues are $ 0$ and $ 1.$ That makes it positive semidefinite.

(I realized this upon taking a second look at what I wrote above - that action I'm describing is clearly an orthogonal projection - a projection onto the orthogonal complement of the column space of $ Q.$)

And on that note, $ I\minus{}A\equal{}Q(Q^*Q)^{\minus{}1}Q^*$ is also an idempotent, and is a projection onto the column space of $ Q.$


Let $ a;b;c$ be positive real numbers.Prove that:
$ \frac{a\plus{}b\plus{}c}{3} \leq \frac{1}{4}.\sqrt[3]{\frac{(a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2}{abc}}$

[quote="chien than"]Let $ a;b;c$ be positive real numbers.Prove that:
$ \frac {a \plus{} b \plus{} c}{3} \leq \frac {1}{4}.\sqrt [3]{\frac {(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2}{abc}}$[/quote]
By Am-GM, we have
\[ (a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2 \ge \frac{64}{81}(ab\plus{}bc\plus{}ca)^2(a\plus{}b\plus{}c)^2\]
and
\[ (ab\plus{}bc\plus{}ca)^2 \ge 3abc(a\plus{}b\plus{}c)\]
Hence
\[ (a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2 \ge \frac{64}{27}abc(a\plus{}b\plus{}c)^3\]
and so
\[ \sqrt[3]{\frac{(a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2}{abc}} \ge \frac{4}{3}(a\plus{}b\plus{}c)\]
we have done. :)
Equality holds if and only if $ a\equal{}b\equal{}c.$


Let's say I want to evaluate the derivative of a function $ f(x)$ at $ x\equal{}10$. How do I write this in latext in this form:

$ D_{x}\left[f(x)\right]|_{x\equal{}10}$

The problem with the output above is that the vertical line bar is too small. Is there a command that I can use?

How about $ D_{x}\left[f(x)\right]\Big|_{x \equal{} 10}$
In order of size there's \big, \Big, \bigg, \Bigg which can be applied to all sorts of brackets ()[]{} etc etc


Let $ a;b;c\geq 0$ satisfying $ \sum a\plus{}abc\equal{}4$.Prove that:
       $ \sum a\geq\sum ab$

[quote="karis"]Let $ a;b;c\geq 0$ satisfying $ \sum a \plus{} abc \equal{} 4$.Prove that:
       $ \sum a\geq\sum ab$[/quote]
Use
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=83238


geometry
number theory
prime factorization

Let $S(n)$ denote the number of ways that $n$ can be expressed as the sum of two (not necessarily distinct) integer squares.  (Clarification:  $2^{2}+3^{2}, (-2)^{2}+3^{2}, 3^{2}+2^{2}$ are not distinct ways.)

Evaluate

$\sum_{k=1}^{2005}S(k)$

Heh...I wrote a paper related to this at PROMYS...

Well anyways, the value of $S(n)$ which would count those you listed as distinct is $4\tau(n)$, so you could just modify the sum a little bit, it seems...

[hide]
Well, basically any time we have $1 \le a^{2}+b^{2}\le 2005$, that adds one to the sum. So iterating through, we get

$a = 44$, $b = 0, 1, \ldots, 8$
$a = 43$, $b = 0, 1, \ldots, 12$

and so on. I can't think of a good way to add all of them together. Maybe interpret this as the number of lattice points in the first quadrant and inside the circle $x^{2}+y^{2}= 2005$? [/hide]

P.S. My post count refuses to go up...

Yes, those cases where it is $0$ have to be put into the formula, but it is still wrong as stated then.
Here an excerpt from the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=76610]formulary[/url]:

[b]2)[/b] Sum of two squares:
A positive integer $n$ can be represented as sum of two perfect squares iff all prime factors $p \equiv 3 \mod 4$ of $n$ occur an even number of times in the factorisation of $n$. $n$ can be written as sum of squares $\neq 0$ iff the previous condition holds and it has at least one prime factor $\equiv 1 \mod 4$ or $v_{2}(n)$ is odd.
There are exactly
\[\mathrm{r}_{2}(n) = 4 \cdot \sum_{\substack{ d \in \mathbb{N}\\ d|n \\ d \equiv 1 \mod 2}}(-1)^{\frac{d-1}{2}}= 4 \cdot \prod_{\substack{p \in \mathbb{P}\\ p\equiv 1 \mod 4}}(v_{p}(n)+1) \]
different solutions $(a,b) \in \mathbb{Z}^{2}$ to $n= a^{2}+b^{2}$.

[quote="t0rajir0u"][quote="K81o7"]the value of $S(n)$ which would count those you listed as distinct is $4\tau(n)$, so you could just modify the sum a little bit, it seems...[/quote]

Huh?  Even if I counted the cases I listed as distinct, $S(n) = 0$ for an awful lot of numbers.[/quote]

Sorry, what I meant was $4\tau(k)$ where $k$ is the largest divisor of $n$ whose prime factorization contains only 1mod4 primes.  And if, in the prime factorization of $n$, there is any odd power of a 3mod4 prime, it's 0.

So then you can take a list of possible 1mod4 prime combinations, and take multiples of their tau's (of course, dividing by 2 whenever you get a sum of two distinct squares)?

The value in question is, as paladin said, the number of lattice points in or on the 45 degree slice of that circle...so we can appproximate the value in question be taking that area (likening to Pick's theorem).


inequalities
function
combinatorics unsolved
combinatorics

Let $ (d_i)_{i\equal{}1}^n$ be the degree sequence of a planar graph $ G$. Then:
(i) if $ \delta(G)\geq 4$ prove that \[ \sum_{i\equal{}1}^nd_i^2<2(n\plus{}3)^2\minus{}62.\]
(ii) if $ n\geq 4$ prove that \[ \sum_{i\equal{}1}^nd_i^2\leq 2(n\plus{}3)^2\minus{}62.\] Show that equality can hold for every $ n$.
(iii) determine maximum of $ \sum_{i\equal{}1}^nd_i^2$ if $ G$ has girth at least $ 4$.

oho pierwszy raz zrobiłem coś czego Szymek nie zrobił :D ;)

II)
it's easy to prove the following inequality:
$ d_1 \plus{} d_2 \plus{} ... \plus{} d_n \le 6(n \minus{} 2) \equal{} 2 \cdot (n \minus{} 1) \plus{} (n \minus{} 4) \cdot 4 \plus{} 2 \cdot 3$
it's well-known fact. a proof in polish is here: http://www.ift.uni.wroc.pl/~cislo/md/mat4.pdf

let $ f(x) \equal{} x^2$
suppose that $ d_1,d_2,...d_n$ maximize $ f(d_1) \plus{} f(d_2) \plus{} ... \plus{} f(d_n)$. then $ d_i \ge 3$
$ f$ is increasing and convex function, so by Karamata inequality we have:
$ f(d_1) \plus{} f(d_2) \plus{} ... \plus{} f(d_n) \le 2f(3) \plus{} (n \minus{} 4)f(4) \plus{} 2f(n \minus{} 1) \equal{} 2(n \plus{} 3)^2 \minus{} 62$

the inequality truly holds for every $ n$ - here is an example: http://img709.imageshack.us/img709/4919/szampanskiegosylwestra.png


1)find all of the natural numbers like $n$ such that $7|2^n-1$                                                2) prove that is'nt natural number like $n$ such that  $7|2^n+1$


geometry
similar triangles
geometry solved

A quadr'l ABCD is insrcibed into a circle so that AB is a diameter. Suppose that AB and CD intersect at I, AD and BC intersect at J , AC and BD iinteresect at K and considwe a point N on AB. Prove that IK is perpendicular to JN if and only if N is the midpoint of AB.

Bomb

Even if $AB$ is not a diameter of $(O)$, you will always have $IK$ is perpendicular to $OJ$ (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5326]this problem[/url]). Hence since $O$ is the midpoint of $AB$ in this problem, $N$ has to be the midpoint of $AB$.

ooh , this is trivial , because $KI$ is the polar of $J$  ;)

Well this problem was the Q5 of the team training that i solved at December Vinh ..
K is orthocentre of the triangle ABJ ( AD perp to JB, BK perp to AJ ).
SO it is sufficent to prove MK.MJ = MI.MN, but MK.MJ=MA.MB (AKM,JMB similar triangles) so stp: MA.MB = MI.MN.
NA^2 - MN^2 = MI.MN
NA^2 / MN = MI + MN
NA^2 = MN*IN. ... (1)
But using Ceva and Menalaus ( with AD,BC,MJ concurrent and C,D,I collinear ) it follows:
BM/AM=BI/AI.
BM * AI = AM * BI
(r + MN)(NI - r ) = ( r - MN) ( r + NI)
-r^2 + MN * NI + r ( NI - MN) = r ^2 - MN*NI + r ( NI - MN)
2*MN*NI = 2 r^2
MN*NI = NA^2 as required. proving (1) and thus we are done.
Elaborating on treegoner's approach -->
Define a transformation wrt the circle diameter AB that maps points to lines and vica versa as follows:
X -> l s.t. the point X' is in L, with OX * OX' = r^2, and O,X,X' are collinear; and OX is perpendicular to l.
Then the point k is mapped to the line JI. To prove this consider the tangents from J to the circle meeting it at U and V.
We want to prove I,K,U,V are collinear (this is easily equivalant to proving the polarity result.)
Proving U,K,V is collinear is equivalant to that q3 you didnt solve on that hard ( i think s4? ) exam.

the zadgah of problem is ghotb o ghotbi.where did u find the problem?


why ? can you help me ?   :blush:    :D  :blush:

[quote="tranthanhnam"]let $a,b,c$be three positive real numbers   such that$abc \geq 1$ prove that :
$\sum\frac{a}{\sqrt{b+\sqrt{ac}}}\geq \frac{3}{\sqrt{2}}$[/quote]
By chauchy:
$\sum\frac{a}{\sqrt{b+\sqrt{ac}}}\geq \frac{(\sum \sqrt{a})^{2}}{\sum \sqrt{b+\sqrt{ac}}}\ge \frac{(\sum \sqrt{a})^{2}}{\sqrt{3\sum{b+\sqrt{ac}}}}$
$A=\sum a+\sum \sqrt{ac}\geq 9, B=\sum \sqrt{ac}\geq 3$
then$\sum\frac{a}{\sqrt{b+\sqrt{ac}}}\geq\frac{(\sum \sqrt{a})^{2}}{\sum \sqrt{b+\sqrt{ac}}}\ge$
$\frac{(\sum \sqrt{a})^{2}}{\sqrt{3\sum{b+\sqrt{ac}}}}$
$=\frac{\frac12 A+\frac12 A+B}{\sqrt{A}}$$\ge \frac{3\sqrt[3]{\frac14 A^{2}B}}{\sqrt{A}}\ge \frac{3}{\sqrt{2}}$

[quote="Hawk Tiger"][quote="tranthanhnam"]
$A=\sum a+\sum \sqrt{ac}\geq 9, B=\sum \sqrt{ac}\geq 3$ [/quote][/quote]

I'm sorry mate, but for $a=b=c=1$, $A=6$, so $6 \geq 9$ is a little bit far from truth, innit? :)

I am sorry for the typo.it should be $A\geq 6$

This problem come from Mathematics and Young (the contests not finish) :mad: Tranthanhnam, you are rapacious  :mad: (you used to post problems from magazines and test of forums, example VMEO III and MFC I  :mad: ).
I think you don't consider about problem  :( You will conclude this story  :)
I hope moderater will lock topic until the contests finish  :)

tranthanhnam: hi vọng tôi sẽ ko bao giờ phải xem những topic gian lận kiểu này của anh nữa. Anh phải biết rằng hành động của anh là rất tồi tệ và nó lập  đi lập lại ko ít lần. Nhưng dù sao tôi cũng hi vọng anh biết hối cãi  :)


I think we need to develop a poll in this forum to see what instruments are most commonly played by AOPSers.

[ [i]Moderator edit: I added the requested poll.[/i] ]

piano.
Why don't you make an actual poll?

I added the requested poll as a moderator edit to the first post in this thread. I didn't put in an option of "none" on the assumption that most people who visit this forum play SOME instrument, although in fact I don't myself. (I took flute lessons years ago, but all I remember of those is which side of my body to hold the flute on.)

I find this forum very interesting because my wife plays piano and flute and teaches piano as her occupation. We are getting heavily into jazz in recent months; I was playing a Dizzy Gillespie CD at home tonight.

P.S. Some of the poll choices were very broad; please specify your particular instrument in a reply to this thread, especially if it is a non-Western instrument.

Can you add an option for "More than one instrument"? Thanks.

[quote="jmerry"]I'm a violist, and slightly annoyed at being lumped with others.

Actually, math and the viola go together quite well in my experience; I was one of at least three violists at MOP 1999, out of 24 total students, and almost half of my viola section in college was associated with math, with the rest mostly music majors.

Of course, I understand the place of violins and the place of violas in the musical world. Here's one of my favorite jokes on the sublect:

What do you call the members of a string quartet?
[hide="answer"]A good violinist, a bad violinist, a former violinist, and someone who hates violinists.[/hide][/quote]

i play cello (10 years about) and unfortunately (sry jmerry) most of strings jokes i know are about violas lol. i cant think of any at the moment though....

I used to play clarinet, and did for three years, all through middle school. However, I dropped it in order to take a more advanced choir. So I voted for single reed, because I still play it sometimes. But, I would say that, now, I "play" the vocal cords. :P

[quote="sheepwarrior"]i play cello (10 years about) and unfortunately (sry jmerry) most of strings jokes i know are about violas lol. i cant think of any at the moment though....[/quote]
Don't worry - whatever the jokes are that you can't quite think of, I've heard them already. I played the violin from elementary school through college, stopped playing for about 15 years, than picked it back up and eventually changed to viola (partly jmerry's influence).

I play both piano and percussion, but am a far more avid Percussionist than pianist.  Therefore I voted percussion.
My specialty is in Keyboard percussion, but I can also play a lot of non-western stuff like djembe and steel pan.

I play both piano (6 or 7 years) and viola (4 or 5 years).

I play both piano and violin.  Played violin for about a year but piano for like... 6 or 7.

i play the piano, like most asian kids...

Oh yeah, thanks. I remember when he went to Ireland he found another man with a banjo like his and was really surprised.

I used to play the banjo. Was fun. Math +science took over

i play euphonium and a PVC didgeroo
heh heh, 2nd one doesn't count
im the only euphonium player here

[quote="cottontail"]I play piano.

And I just started taking erhu. (And I spelled it right, yes.)[/quote]

no you didn't
its 二胡
meh

[quote="eagle2"]violin is what i play and I tought myself to play piano.[/quote]

Juuuust about the exact opposite for me

I play the piano and I used to play guitar but i havent touched one in 5 years

Piano/Violin, but mainly piano lol so I chose that

My main instrument is piano, but I also play the harp and percussion.

[color=#0000FF]I play the violin (not very well though!!!!)... I think I have poor technique[/color] :|

I play the violin and used to play the piano, I wasn't very good at the piano either... :stink:

I play flute and have been for 5 years. I play a little bit of piano and a little bit of guitar, a little meaning I only know the basics (ex:chords)


calculus
derivative
integration
function
calculus computations

http://img257.imageshack.us/img257/4439/questionrf1.jpg

My main question is if the Differentiation under the Integral rule (http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign) is valid here, because that would make it easy.

The Wikipedia article is about differentiation with respect to a variable other than the variable of integration. In your case the derivative is taken with respect to $ t$, and then the result is integrated with respect to $ t$. Also long as $ f$ and $ g$ are $ C^1$-smooth, the derivative is continuous and the integral recovers the original function - this is the fundamental theorem of calculus. The integral is equal to $ f(g(x,\beta(x)))\minus{}f(g(x,\alpha(x)))$, etc.

How do we show that integral evaluates to what you said (and what i got, though I , obviously wrongly, used the Differentiation under the integral sign rule)? Is there another way to show the rule applies?


I'm trying to begin a study of analysis in order to prepare for other advanced topics and I was wondering what are the best books to use.

I'm familiar with the two standards: Rudin and Apostol. I've looked through Rudin and it's well [i]concise[/i]. The problems also seem to be quite difficult for someone who is just beginning to learn the subject. I have only glimpsed at the Table of Contents of Apostol and it already seems much more beginner friendly than Rudin. Yet, many colleges use Rudin as the main Analysis text.

So my question: If I use Apostol instead of Rudin will my understanding of some Analysis topics be sacrificed?
Or if you have any recommendations for other Analysis books please mention them.

There are a lot of books, and I've used several different ones when I teach the course. One that I've used fairly often is Kenneth Ross's [i]Elementary Analysis: the Theory of Calculus.[/i] That's a fairly thin book, and needs to be supplemented in some areas. (It doesn't go very deeply into metric spaces, and it  has nothing on multivariable analysis.)

I'd think it's better to find a book you can understand rather than claiming to go through something you don't understand. And self-study is difficult; there are advantages to having a teacher.

[quote="Kent Merryfield"]... there are advantages to having a teacher.[/quote]

True. Eventually I will have a teacher because I'm planning on taking this course come fall but there are some things I would like to understand before then.

[quote="Kent Merryfield"]... I'd think it's better to find a book you can understand.[/quote]

Thanks for the advice. I think I'll drop Rudin and explore some other books.

[quote="Kent Merryfield"]And self-study is difficult; there are advantages to having a teacher.[/quote]
Also advantages to having exams. Unless I am extremely interested in a topic there is no way I will learn things as carefully if I don't have to write an exam. I don't feel teaching is that important anymore; it feels like most classes are mainly self-study anyway.

In self study, there is no fear of failing to pass to motivate you to keep a good pace. ;)

In my opinion the better way to learn "high" level mathematics is in a small group of persons who do the self-study with orientation of an experienced person on the subject. Making discussions, and solving problems to each other on regular encounters. That way i learn things and don't forget.

[quote="HilbertThm90"]I'm completely opposite. I learn almost nothing in actual classes. This is because I know that I can get an A by memorizing the night before and regurgitating on a test. Then I promptly forget it all a few weeks later.

When I self-study it is all internally motivated. I have to learn it all, or else I would just be wasting my time going through the motions. I also only self-study things that really interest me.[/quote]

Your classes must be easy and/or you must be very smart. In my classes exams are in-class and so you are working mostly against a short time, which means you make a lot of silly errors which you dont have the option of going back to fix. Either way I haven't taken yet a math class where I could cram the night before and regurgitate.

I like to dialogue with my professors a lot. I often feel like its a one way conversation but it's better than nothing.

I can learn almost anything in any situation, but I don't enjoy it unless it's self taught.

[quote="HilbertThm90"][quote="§outh§tar"]
Your classes must be easy and/or you must be very smart. [/quote]

It is very sad, but my classes are easy. They have to be, since my school is mostly interested in training high school and middle school teachers. The majority of the class would fail if the traditional math majors were given tests that were actually challenging to us.[/quote]

Ditto for the college I will be attending for my freshman year.... what's worse though, is I had 10 of 15 minutes to sit and do nothing on a timed test in my AB calc class. My teacher then, gave a five minute extension to the class because no one else was done.

I got the [i]only[/i] perfect score.

I study independently of the class and I write my own notes. Works for me.


function
algebra
polynomial
real analysis
real analysis unsolved

Let $f: R\to R$ which verifies $f(x+y)=f(x)+f(y)$ and $P: R\to R$ a polynomial function(not constant). Show that if $Pof$ has primitives, then $f(x)=f(1)x$.
Show the same thing, if $e^f$ has primitives.

Already posted at least twice. See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32514[/url]


The number of ways in which n distinct objects can be arranged in a circle is (n-1)!

Four men and four women are forming a circle for a folk dance. In how many ways can this be done if we require that the men and women alternate positions.

If anyone can help with this problem, please do so.

The answer the the teacher has and even isn't sure of is 1152. Is that right?

You don't need to double post it.
Two answers are waiting for you in the other thread.


I am scheduled to leave for Japan tomorrow morning. Hope to meet some of you there.

I will be bringing with me some issues of the recently published "Romanian Mathematical Competitions 2003", where I am co-author. The book contains almost all meaningfull problems used in the Romanian Mathematical Competitions this year, including full solutions to the Team Selection Tests. It is written in English.

Anyone desiring this book contact me at the IMO, after the last contest day.

can i have the imo solutions of shortlist? I am rustamtm@yahoo.com


geometry
incenter
circumcircle
trigonometry
inequalities
inequalities proposed

[hide="Hint"][color=darkblue]Denote the incenter $ I$ and the second intersections $ A',\ B',\ C'$ of the circumcirle with the lines $ IA,\ IB,\ IC$ respectively. Then,

$ \sum (b+c)\sqrt{\frac{bc}{(b+c)^{2}-a^{2}}}\geq\frac{2}{\sqrt{3}}(a+b+c)$ $ \Longleftrightarrow$ $ \boxed{3\ge}\sum\cos\frac{B-C}{2}\ge \frac{2p}{R\sqrt 3}$ $ \Longleftrightarrow$ $ \boxed{6R\ge}\sum AA'\ge \frac{4p}{\sqrt 3}$. [size=117][b]Very interesting inequality ![/b][/size][/color][/hide]


logarithms probability function algebra domain

1. For each positive integer $ n$ define $ f(n)$ to be $ \log_8 n$ if $ \log _8 n$ is rational, or $ 0$ otherwise. Compute $ \sum_{1}^{2009} f(n)$.

2. In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Find x.
[code]
x | 19 | 96
1 | ?  | ?
? | ?  | ?
[/code]

3. Triangle $ ABC$ is a right triangle with $ \angle ABC$ a right angle, $ m\angle BAC \equal{} 60$ (typo said ACB), and $ BC \equal{}10$. Let $ E$ be randomly chosen point in $ \triangle ABC$, and extend $ AE$ to meet $ BC$ at $ D$. Find the probability that $ AD > 5\sqrt{2}$.

4. The function $ f$ has the property that for real number $ x$ in its domain, $ 1/x$ is also in its domain and $ x \equal{} f(x) \plus{}f(1/x)$. Find the largest set of real numbers that can be in its domain of $ f$.

5. What is the number of ordered pairs of real numbers $ (a,b)$ such that $ (a\plus{}bi)^{2009} \equal{} a\minus{}bi$.

(Rules were 40 minutes to write solutions.)

Darn darn darn pianoforte, you say trivial problems yet you post them? You should be ashamed of yourself :o

I think piano posted just for reference.


limit
Functional Analysis
advanced fields
advanced fields unsolved

Let 'H' be a Hilbert space and $(T_{n})$ be a sequence in BL(H).
Suppose that for each x,y $\in H$, the scalar sequence $(\langle T_{n}x,y\rangle)$
converges in  $\mathbf{K}$. SHOW THAT there exists
$T\in BL(H)$such
that $T_n\rightarrow T$ WEAKLY as $n\rightarrow \infty$

Suppose we prove that for every $x$ th sequence $T_nx$ converges weakly. Then we're almost done. Denote the weak limit of $(T_nx)_n$ by $Tx$. It's routine to check that $T$ is a linear operator on $H$. We should now prove that it's also bounded. Now, since $(T_nx,y)$ is bounded for fixed $x$ and $y$, it means that $T_nx$ is bounded (the uniform boundedness principle, applied to the linear transformations $y\mapsto (T_nx,y)$ as $n$). Again, by the uniform boundedness principle, this time applied to the transformations $T_n$, we conclude that there must be some $M>0$ s.t. $\|T_n\le M,\ \forall n$. Now, for fixed $x,y$, we have $(T_nx,y)\to(Tx,y)$, so for $y=Tx$ we get $(T_nx,Tx)\to(Tx,Tx)=\|Tx\|^2$. Since $(T_nx,Tx)\le\|T_nx\|\cdot\|Tx\|\le M\cdot\|x\|\cdot\|Tx\|,\ \forall n$, we have $\|Tx\|\le M\cdot\|x\|$, and since $x$ was arbitrary, we get $\|T\|\le M$, as desired.

All that's left now is to prove is that if $(T_nx,y)$ converges for every $x,y$, then $T_nx$ converges weakly for every $x$. We'll prove a general result from which this follows, namely that if $(x_n)_n\subset H$ is a sequence s.t. $(x_n,y)$ converges for every $y\in H$, then $x_n$ is weakly convergent, i.e. there is an $x\in H$ such that $(x_n,y)\to(x,y),\ \forall y\in H$.

We may as well assume our Hilbert space is separable and has an orthonormal basis $(e_i)_{i\ge 1}$, since given an orthonormal basis in the original Hilbert space, we’re only concerned with those members of this basis which appear in the Fourier expansion of the $x_n$’s. Let $x_n=\sum_{i\ge 1}\alpha_i^ne_i$, and let $\alpha=\lim_{n\to\infty}\alpha_i^n$ (this limit exists since the hypothesis tells us that the sequence $(\alpha_i^n)_n=[(x_n,e_i)]_n$ converges). If we prove that $\sum_{i\ge 1}|\alpha_i|^2<\infty$, then we can take $x=\sum_{i\ge 1}\alpha_ie_i$, and it will be easy to check that this is indeed the desired weak limit of $(x_n)_n$.

Now we apply that dreaded uniform boundedness principle again :) to conclude that $(x_n)_n$ is bounded. Take some $K>0$ s.t. $\|x_n\|\le K,\ \forall n$. For every $t\ge 1$ and every $n$ we have $\sum_{i=1}^t|\alpha_i^n|^2\le\|x_n\|^2\le K^2$, so $\sum_{i=1}^t|\alpha_i|^2=\lim_{n\to\infty}\left(\sum_{i=1}^t|\alpha_i^n|^2\right)\le K^2$. This works for all $t\ge 1$, so we get $\sum_{i\ge 1}|\alpha_i|^2\le K^2<\infty$, which is exactly what we wanted.


What are the possible numbers of common external tangents two congruent nonconcentric circles in the same plane can have?

[quote="SplashD"][hide]If a common external tangent is what I think it is, then the answer is $0,1,2,3,4$[/hide][/quote]I think that's impossible. Think about the symmetry and you know the answers must be even  :wink:

[hide="My guess:"]0, 2, 4[/hide]
Btw, Totally Zealous, what are "non-congruent circles"?  :roll:

Doesn't an external common tangent not intersect the segment connecting the centers of the 2 circles? How would there be 4?

What if a circle is internally tangent to a larger circle, then wouldn't there be 1?

Omg, you're right  :o How could I even forget they could touch internally. And you know, this means that 3 is also possible (if they touch externally!)

For the 4-case, look at the picture I added.

[quote="SplashD"]Congruent of nonconcentric means they can't be internally tangent though.[/quote]
Oh, yeah.

Kurt Godel, there are 2 external tangents and 2 [b]internal[/b] tangents in your picture. :maybe:

So is the answer [hide]$\boxed{2}$[/hide]?

[hide="this is the confusing solution AoPS volume one gave"]If the 2 circles intersect in two points, there are 2 tangents. If they are tangent, there are three tangents. If they don't intersect, there are four common tangents. Thus the possible numbers are 2,3,4.[/hide]

[quote="Totally Zealous"]common [size=200][b]external[/b][/size] tangent [/quote] is defined as a line, tangent to 2 circles not intersecting the segment connecting the centers of the 2 circles.

Yeah, you said external in the problem! :mad: