Nα βρεθεί το ελάχιστο του πολυωνυμου $ Q(x)$ με
$ \Large Q(x) \equal{} \begin{cases} \displaystyle \frac {(x \minus{} 4)Q(x \plus{} 1) \plus{} 20}{x},x\neq 0 \\
29, x \equal{} 0 \\
\end{cases}$

moderator edit : Thanks Dimitri !

Δεν ξερω αν λυνεται πιο απλα. Αλλα αν δεν εχω κανει λαθος καπου, εξισωνοντας μεγιστοβαθμιους συντελεστες βρησκουμε $ degQ \equal{} 4$ και αφου ευκολα εχουμε $ Q(3) \equal{} \minus{}5, Q(4) \equal{} Q(1) \equal{} 5, Q(\minus{}1) \equal{} 125, Q(0)\equal{}29$ μπορουμε να βρουμε το πολυωνυμο και μετα να βρουμε το ελαχιστο του με παραγωγους.

Eχεις ψιλοαριθμητικο το $ Q(3)\equal{}5$ και θελει πολλες πραξεις.....
Γραψε αναλυτικα πως βγαζεις τον βαθμο...
Λυνεται και χωρις γνωσεις συνεχειας -Δ.Λογισμου (Για τους μη εχοντες γνωσεις Γ Λυκειου)

Ειναι: $ x(Q(x) \minus{} Q(x \plus{} 1)) \equal{} \minus{} 4Q(x \plus{} 1) \plus{} 20$ για $ x$ διαφορο του μηδεν (αλλα και για 0, διοτι το πολυωνυμο ειναι συνεχης συναρτηση, μπορουμε ομως να μη γραψουμε τιποτα για συνεχεια και να πουμε απλα οτι ενα πολυωνυμο με απειρες ριζες ειναι μηδενικο). Θετουμε $ Q(x) \equal{} \sum_{k \equal{} 1}^n{a_kx^k}$ και αφου χρησημοποιησουμε το δυωνυμικο αναπτυγμα εξισωνουμε τους μεγιστοβαθμιους συντελεστες και λαμβανουμε: $ n \equal{} \binom{n}{1} \equal{} 4$

Βασικα μπορουμε μετα στην παραπανω σχεση να θεσουμε: $ Q(x) \equal{} ax^4 \plus{} bx^3 \plus{} cx^2 \plus{} dx \plus{} 29$ και μετα απο μερικες πραξεις να λυσουμε ενα σχετικα απλο συστημα και να βρουμε το πολυωνυμο... αυτο που δεν εχω κοιταξει και δεν μπορω να φανταστω πως θα μπορουσε να γινει ειναι το αν μπορουμε να βρουμε το ζητουμενο ελαχιστο χωρις να βρουμε τον τυπο του πολυωνυμου

Εχουμε την ιδια περιπου λυση

Σχεδον το ιδιο σκευτηκα πριν απο λιγο, και μπηκα να το γραψω αλλα ηρθα 2ος  :P
Απλα εγω ειπα:
Εχουμε $ degQ \equal{} 4$ (εχω γραψει πιο πανω απο που προκειπτει αυτο)
Ειναι τωρα $ Q(3) \equal{} Q(4) \equal{} Q(1) \equal{} Q(2) \equal{} 5 \Rightarrow Q(x) \minus{} 5 \equal{} k(x \minus{} 1)(x \minus{} 2)(x \minus{} 3)(x \minus{} 4)$ απο οπου λογο $ Q(0) \equal{} 29$ προκειπτει $ k \equal{} 1$ και
$ Q(x) \equal{} x^4 \minus{} 10x^3 \plus{} 35x^2 \minus{} 50x \plus{} 29 \equal{} (x^2 \plus{} 5)((5 \minus{} x)^2 \plus{} 5) \minus{} 121 \geq (5\sqrt {5})^2 \minus{} 121 \equal{} 125 \minus{} 121 \equal{} 4$
απο Cauchy - Swartz με ισοτητα για $ x\equal{}m\sqrt{5}, 5\minus{}x\equal{}\frac{\sqrt{5}}{m} \Rightarrow x^2 \minus{} 5x \plus{} 5 \equal{} 0$. Αρα το ελαχιστο ειναι ο αριθμος 4


[quote="orl"]Let $ (x_1,x_2,\cdots)$ be a sequence of positive numbers such that $ (8x_2 \minus{} 7x_1)x_1^7 \equal{} 8$ and
\[ x_{k \plus{} 1}x_{k \minus{} 1} \minus{} x_k^2 \equal{} \frac {x_{k \minus{} 1}^8 \minus{} x_k^8}{x_k^7x_{k \minus{} 1}^7} \text{ for }k \equal{} 2,3,\ldots
\]
Determine real number $ a$ such that if $ x_1 > a$, then the sequence is monotonically decreasing, and if $ 0 < x_1 < a$, then the sequence is not monotonic.[/quote]

We have $ x_{k \plus{} 1} \equal{} \frac{x_k^2}{x_{k\minus{}1}} \plus{} \frac {x_{k \minus{} 1}^8 \minus{} x_k^8}{x_k^7x_{k \minus{} 1}^8}$ and so $ 8x_{k \plus{} 1}x_k^7 \equal{} 8\frac{x_k^9}{x_{k\minus{}1}} \plus{} 8\frac {x_{k \minus{} 1}^8 \minus{} x_k^8}{x_{k \minus{} 1}^8}$

$ 8x_{k \plus{} 1}x_k^7 \minus{}7x_k^8 \equal{} 8\frac{x_k^9}{x_{k\minus{}1}} \minus{}7 x_k^8 \plus{} 8\frac {x_{k \minus{} 1}^8 \minus{} x_k^8}{x_{k \minus{} 1}^8}$

$ \implies$ $ 8x_{k \plus{} 1}x_k^7 \minus{}7x_k^8 \equal{} \frac{x_k^8}{x_{k\minus{}1}}(8x_k \minus{} 7x_{k\minus{}1}) \plus{} 8\minus{}8\frac { x_k^8}{x_{k \minus{} 1}^8}$

$ \implies$ $ (8x_{k \plus{} 1} \minus{}7x_k)x_k^7 \equal{} 8 \plus{} \frac{x_k^8}{x_{k\minus{}1}^8}((8x_k \minus{} 7x_{k\minus{}1})x_{k\minus{}1}^7  \minus{}8)$

$ \implies$ since $ (8x_2 \minus{} 7x_1)x_1^7 \equal{} 8$ : $ (8x_{k \plus{} 1} \minus{}7x_k)x_k^7 \equal{} 8$

And so $ \boxed{x_{k\plus{}1}\equal{}\frac{7}{8}x_k\plus{}\frac{1}{x_k^7}\text{  for }k>\equal{}1}$

Let $ f(x)\equal{}\frac{7}{8}x\plus{}\frac{1}{x^7}$ defined in $ \mathbb R^{\plus{}*}$. It's easy to see that :
$ f(x) > 8^{\frac{1}{8}}$ $ \forall x\neq 8^{\frac{1}{8}}$
$ f(x) < x$ $ \forall x>8^{\frac{1}{8}}$
$ f(x) > x$ $ \forall x<8^{\frac{1}{8}}$
$ f(8^{\frac{1}{8}})\equal{}8^{\frac{1}{8}}$

So :
If $ x_1\equal{}8^{\frac{1}{8}}$, $ x_n\equal{}x_1$ is a constant sequence.
If $ x_1 > 8^{\frac{1}{8}}$ $ x_n$ is a strictly decreasing sequence (always $ >8^{\frac{1}{8}}$)
If $ x_1 < 8^{\frac{1}{8}}$ $ x_2 > x_1$ and $ x_n$ becomes strictly decreasing from there and so $ x_n$ is not monotonic.

And so the answer is $ \boxed{a\equal{}8^{\frac{1}{8}}}$

[quote=orl]Let $ (x_1,x_2,\cdots)$ be a sequence of positive numbers such that $ (8x_2 \minus{} 7x_1)x_1^7 \equal{} 8$ and
\[ x_{k \plus{} 1}x_{k \minus{} 1} \minus{} x_k^2 \equal{} \frac {x_{k \minus{} 1}^8 \minus{} x_k^8}{x_k^7x_{k \minus{} 1}^7} \text{ for }k \equal{} 2,3,\ldots
\]

Determine real number $ a$ such that if $ x_1 > a$, then the sequence is monotonically decreasing, and if $ 0 < x_1 < a$, then the sequence is not monotonic.[/quote]
$x_{k+1}x_{k-1}-x_k^2=\dfrac{x_{k-1}^8-x_k^8}{x_k^7x_{k-1}^7}\Rightarrow \dfrac{x_{k+1}}{x_k}-\dfrac{x_k}{x_{k-1}}=\dfrac{1}{x_k^8}-\dfrac{1}{x_{k-1}^8}\Rightarrow \dfrac{x_{k+1}}{x_k}-\dfrac{1}{x_k^8}=\dfrac{x_k}{x_{k-1}}-\dfrac{1}{x_{k-1}^8}=\dfrac{x_2}{x_1}-\dfrac{1}{x_1^8}=\dfrac{7}{8}$
$\therefore x_{k+1}=x_k-\dfrac{1}{x_k^7}$. Let $f(x)=\dfrac{7}{8}x-\dfrac{1}{x^7}$, then $f(a)=a\Rightarrow a=8^{\frac{1}{8}}$, and the problem becomes easy.


floor function
number theory unsolved
number theory

Let two natural number $n > 1$ and $m$ be given. Find the least positive integer $k$ which has the following property: Among $k$ arbitrary integers $a_1, a_2, \ldots, a_k$ satisfying the condition $a_i - a_j$ ( $1 \leq i < j \leq k$) is not divided by $n$, there exist two numbers $a_p, a_s$ ($p \neq s$) such that  $m + a_p - a_s$ is divided by $n$.

If $d=(m,n)$, then the answer should be $d\cdot \left\lfloor \frac n{2d}\right\rfloor$, I believe.

Draw $n$ points around the circle, and connect each point to the $m$'th point in a counter-clockwise direction. We will get $d\ \frac nd$-sided polygons. No two points among the $k$ we choose can be consecutive in such a polygon, so from each polygon we can select at most $\left\lfloor \frac n{2d}\right\rfloor$, and the result follows.

See also : http://www.mathlinks.ro/Forum/viewtopic.php?t=6008 . :)


well, i notice that states are up they are posted on mathcounts.org

so what scores did u guys get? i thought personally that this year's test was harder than usual(maybe it was b/c i was so nervous).

i dont know my score since my "coach" forgot to ask for it....

I [i]think[/i] I got 35 (21/14), but I don't know for sure, since we didn't get scores back. I had so many stupid mistakes on sprint.

i got a 28/14 =42. just kidding i FAAAAAIIIIIIILLLLLLEEEDDD. id rather not say i did badly. do the mods mind if i make another copy of this post with a poll.

[quote="#H34N1"]GSB says he got like a 26/10[/quote]
Then he lies.

i though i got a 18+12=30 roughly, since i didnt get my scores but the answers for announced after(stupid mistakes on sprint and target, esp. for the donut prob.)

and somehow i made nats....

To anyone from Indiana: NO POSTING OF OUR SCORES AT ALL or Mr. Fish will REALLY kill you.  Thanks =P.

This year's test... I'm not exactly sure if it was harder or not... Mathcounts is relatively simple and perfects should be easy if not for stupid mistakes, etc.  Nerves are a big factor in this and cognos really did fail, just not as bad as me.  =P

got a 16/12 = 28...... and made nats.

OFF BY 1 ON  TARGET # 8!!!!!!!!!

I know alabama sucks....

33=21/12

First got 21/14
3rd got 20/12
4th got 20/12

My team had pretty bad scores-

33-me,26,21,16.

Although it was really fun because the guy who get second place on my team played me in the countdown finals. (I won.) I made a silly mistake on target.

ALSO: guessing got me to nats. just wanted to let it be known.

Are we allowed to tell this stuff, I thought the announcement lasted until Thursday 11:59p.m.....

Gah, I did so badly on Sprint I'm not saying my score, but I got 7 of the targets (missed the last one, of course.)
The last target was an interesting problem... I wasted my time checking numbers like 913. XD

[quote="distracted523"]Read the first post.[/quote]
Still it just says that the state is posted. Oh well, all you guys posted your scores so I will post mine. 19/14=33. Sprint was hard....Target was easy.

HAHAHAHA! I got 25/14 = 39. I got #3 wrong (the right angles in an octagon), #23 wrong (I forgot to multiply by 4 :blush: ), and something else wrong.

...But I guessed #21 and #27 correctly! (#21 was a total guess, by the way) :D

Oh yeah, and I got the pyramid one on the target wrong because I thought $ 5\sqrt{2}$ was $ \sqrt{20}$...

Anyway, I tied for second, but Jason beat me for some reason. MA is a really hard state...

i did really bad i got 25, but i'm a 6th grade

I believe I got 24/14=38

Missed last target, of course, and a bunch of sprint

[quote="l Bob l"][quote="Yongyi781"]HAHAHAHA! I got 25/14 = 39. I got #3 wrong (the right angles in an octagon), #23 wrong (I forgot to multiply by 4 :blush: ), and something else wrong.

...But I guessed #21 and #27 correctly! (#21 was a total guess, by the way) :D

Oh yeah, and I got the pyramid one on the target wrong because I thought $ 5\sqrt {2}$ was $ \sqrt {20}$...

Anyway, I tied for second, but Jason beat me for some reason. MA is a really hard state...[/quote]

If two students tie, the one with the most sprints wins (but i also got 25/14) so they determine the tiebreaker by focusing on which sprint questions that were answered wrong. Basically I think they average the question #'s of all the sprints you got wrong...and the person with the least average wins...like getting #3 wrong is better than getting #29 wrong. I hope this is clear :roll:

How do you remember which questions you got wrong?? I don't even recall a single one :blush:[/quote]

Simple. I just look at the questions on the website, compare their answers to my answers (that I somehow remember). (Consider it a gift that I have, as well as being slightly better than average at math.)

Also, as a side note, I got #3 wrong, and STILL GOT THIRD? That means all the questions you got wrong were in the first half of the competition...

Strange :huh: Before the award ceremony I predicted I would get a 42...but I got 3 less than that so ..probably means I did get some easy questions wrong

States was beastly hard this year compared to other years. Target was pretty easy, though I got #5 wrong  :blush: apparently included integers as 0 as an answer, ending up with 28

Did anyone else notice you could pwn two problems in target using balls and urns, doing the exact same thing (they even had the same answers)?

it was really ironic, right before the test my friend and i, were just talking about formulae, we talked about only balls and urns and distance from a point to a line and there were like 3 probs. on target which used these topics so when i saw the question about the shortest dist from a pt to a line, i almost screamed. i was thinking WOOOAHHHH that was ironic.

i got either 24/12 or 25/12. i never got my score back.

it all depends whether i actually did put 3*7=15  on that one problem or if i just think i did lol

wow...i kinda failed miserably, 19/8(???) i hate balls and urns but poned distance from a point to a line!!!

14th individual :-(  1st team

congrats andersonw who won in our state!

[quote="abacadaea"]lol. unimpossible, i did what you did on that one.  :rotfl:

first in our state was andersonw. he got something like 26/14, missed last one.

19/12 for me. i got really lucky(guessed for 3 points).[/quote]
I think i got around the same as andersonw...i know i got 14 on target...stupid binary.
But how did he get first? I only got 7th. and the cutoff was like, high 30's.
grr. i missed #8 on target. And to think that i found out how to make a binary program a week after states...

psh to all of you with ti-89s on target #8... =P

27/12- I don't know what I missed in Sprint, completely failed last page of Target. Forgot to halve the squares diagonal for pyramid question. My Sprint score was [b]2nd[/b] in my state!


solve this inequation : $ {x}^{4}\minus{}x\minus{}42 \leq 0$
thanks.

x^4 - x - 42 <= 0

Solving for (x^2 - 7)(x^2 + 6) = 0

x^2 = 7 or/ x^2 = -6 [Assuming we discount this because it yields complex number result]

x = +- sqrt7

Therefore, -sqrt7 <= x <= sqrt7

OOPS!, didn't notice that it was x^4 - x - 42 instead of x^4 - x^2 - 42!

[quote="gurmies"]x^4 - x - 42 <= 0

Solving for (x^2 - 7)(x^2 + 6) = 0

x^2 = 7 or/ x^2 = -6 [Assuming we discount this because it yields complex number result]

x = +- sqrt7

Therefore, -sqrt7 <= x <= sqrt7[/quote]

Can you explain to me, why should we solve $ (x^2 \minus{} 7)(x^2 \plus{} 6) \equal{} 0$?
Thx

gurmies is wrong. I think that the inequality has a typo.

If we could solve the inequality, we would seek some complex root of $ x^4\minus{}x\minus{}42\equal{}0$ by setting $ x \equal{} a \plus{} bi\ a,b\in{\mathbb{R}}$.

[quote="kunny"]gurmies is wrong. I think that the inequality has a typo.

If we could solve the inequality, we would seek some complex root of $ x^4 \minus{} x \minus{} 42 \equal{} 0$ by setting $ x \equal{} a \plus{} bi\ a,b\in{\mathbb{R}}$.[/quote]

the inequation is correct,
However, I inferred it from this one : $ x\left(x\minus{}1 \right)\leq \frac{42}{{x}^{2}\plus{}x\plus{}1}$
Maybe I should try other ways to sold the initial inequation  :(

[quote="kunny"]Sorry, certainly sketching the graphs of $ y \equal{} x^4$ and $ y \equal{} x \plus{} 42$ gives the solution of $ x^4 \minus{} x\leq 42$;$ \alpha \leq x\leq \beta$, where $ \alpha < \beta$ are rhe real roots of the equation ： $ x^4 \equal{} x \plus{} 42$. How would we find $ \alpha , \beta$? I tried to seek the real root by factor thorem, but couldn't. Finally we would use calculus, say, Newton-Raphson method.[/quote]

Newton-Raphson method!!! Would you please explain it more ??? I've never heard of it before. :blush:

I'm not sure how Newton-Raphson method works but I found my method to calculate pretty close values of x.

First we have $ x^4 \le x\plus{}42$.
Now we pick any real x, we only want $ x\plus{}42$ to be positive. Let's say $ x\equal{}10$
$ x^4 \le 10\plus{}42 \implies x \le 2.685349614$. Now we have to add 42 and take fourth root of the value and repeat until we find pretty close aproximation of x. That's the upper limit. To find lower limit, we just take negative solution of fourth root and we get $ x\in[\minus{}2.506861906\cdots,2,584013622\cdots]$

Newton-Raphson method is also known as Newton's method, and it is used to approximate roots of a function. Given an initial guess $ x_0$, a better guess is:

$ x_1 \equal{} x_0 \minus{} \frac {f(x_0)}{f'(x_0)}$

If the initial guess is close enough, this will converge into a root.

In our case, we have $ f(x) \equal{} x^4 \minus{} x \minus{} 42$ and $ f'(x) \equal{} 4x^3 \minus{} 1$.

If we use 2.5 as a initial guess (I got this guess by plugging random values for $ f(x)$), after four iterations we get $ x_4 \equal{} 2.584013573158372264678707697$, which should be close to a root.

Thank you all, especially ffao for explaining Newton-Raphson method.
This problem took me a lot of time. :(


calculus
derivative
trigonometry
LaTeX
trig identities
real analysis
real analysis unsolved

Hi everybody...
I'm rather stumped on how to prove this problem...
"Show that the curves y=xsinx and y=cosx cut at right angles"
I know that the derivatives of each are the inverse of each other, but I'm not sure how to show this... any help would be appreciated!
Thanks, Ryan.

uhm...
ok, just write the conditions: the two curves meet at points such that $x\sin x = \cos x$.
at these points, the product of derivatives is $(\sin x + x\cos x)\cdot(-\sin x)$.
you have to show that this product is $-1$, which is a little algebra... good luck ;)

Ok, that makes sense... so I set (-sinx)(sinx+xcosx)=0... I just can't seem to wrap my head around the algebra. Maybe my knowledge of trig identities is really rusty, but I can't find anything that would make it equal to -1.... I've worked it down to x+tanx and other similar things, but none of it seems to come close to the needed solution... Any additional help would be GREATLY appreciated! Thanks!

We are looking at this at the point where $x\sin x=\cos x$. Find a way to use that fact in your algebra.

Hello Canada !

(sin(x) + xcos(x))sin(x) = (sin(x) + cos(x)/sin(x)*cos(x))sin(x) = ((sin(x)^2 + (cos(x))^2)/sin(x)*sin(x) = 1

Sorry but I'm learning Latex now.(I have done cos(x)/sin(x) for x cause xsinx = cosx)

savoir attendre et douter - d'Alembert.


inequalities
vector
linear algebra
matrix
algebra
polynomial

[quote="Xevarion"]To avoid confusion I will use $[n, n, n]$ to represent symmetric sums and $(n, n, n)$ for cyclic in this post. Wikipedia uses square brackets for symmetric.

Now, Muirhead's inequality says that if the ordered $k$-tuple $\{n_{1}, n_{2}, ..., n_{k}\}$ majorizes $\{m_{1}, m_{2}, ..., m_{k}\}$, then $[n_{1}, ..., n_{k}] \ge [m_{1}, ..., m_{k}]$, right? I am wondering, what about cyclic sums? For example
$(6, 0, 0) \ge (5, 1, 0)$ by
$5a^{6}+b^{6}\ge 6a^{5}b$ AM-GM.
$(7, 3, 1, 0) \ge (5, 4, 1, 1)$ for example will not fall so easily (the system of equations has only the trivial solution).

Is there a general result?[/quote]

Good questions. Well, as you know Muirhead helps us to prove symmetric inequalities, not cyclic ones.

But it is not true, that $(t_{1},t_{2},t_{3})\ge(s_{1},s_{2},s_{3})$ holds generally. There is no Muirhead-like inequality for cyclic inequalities, because such inequality will be not valid.

[b]EDIT:[/b] I was thinking. Maybe it is valid. Because I am not able to find any counterexample.
I posted an example of prooving such inequality:
http://www.mathlinks.ro/Forum/viewtopic.php?p=584212

[b]EDIT2[/b]
Can be all such inequality (I mean like Muirhead but for cyclic sums) done using Rearrangement ineq?

conceive

[quote="Xevarion"]

Is there a general result?[/quote]Very very difficult I think.
I wonder if somebody has even an example of a cyclic sum ineq $(\alpha)\ge(\beta)$ (with $\alpha,\beta$ partitions of $n$) that holds but is [b]not [/b]provable by AM-GM after 'decomposing' $(\alpha)$ in a suitable way.
I had asked another, equally tough but not really related question about cyclic sums here:  [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=83869[/url]

Well it could be digressive, but I found a test for existence of way to prove inequalities with two cyclic sums by AM-GM inequality. (for only 3 variables)

Let's think about $\sum a^{x}b^{y}c^{z}\geq \sum a^{p}b^{q}c^{r}$. ($x+y+z=p+q+r$, and $x,y,z,p,q,r \geq 0$) First, we know that if $m=min(x,y,z,p,q,r)$, then we can divide both sides by $a^{m}b^{m}c^{m}$. There remains zero among $x,y,z,p,q,r$. If there is no zero among $x,y,z$, then applying $c=0$, we can get contradiction. So we can assume that $z=0$.

So we should look at $\sum a^{x}b^{y}\geq \sum a^{p}b^{q}c^{r}$. To prove this by AM-GM inequality, there should exist $\alpha,\beta,\gamma \geq 0$ such that \[\alpha a^{x}b^{y}+\beta b^{x}c^{y}+\gamma c^{x}a^{y}\geq a^{p}b^{q}c^{r}\alpha+\beta+\gamma = 1.\] This concludes with the following conditions: \[x \alpha+y \gamma = p\]  \[y \alpha+x \beta = q\]  \[y \beta+x \gamma = r\] Now solving the equalities, we get \[\alpha = \frac{px^{2}+qy^{2}-rxy}{x^{3}+y^{3}}\]  \[\beta = \frac{qx^{2}+ry^{2}-pxy}{x^{3}+y^{3}}\]  \[\gamma = \frac{rx^{2}+py^{2}-qxy}{x^{3}+y^{3}}.\] As $x^{3}+y^{3}>0$, the conditions are like \[px^{2}+qy^{2}-rxy, qx^{2}+ry^{2}-pxy, rx^{2}+py^{2}-qxy \geq 0\] For $px^{2}+qy^{2}-rxy \geq 0$, we get $r \leq p \frac{x}{y}+q \frac{y}{x}$. Adding $p+q$ at both sides, we get: \[x+y=p+q+r \leq p \frac{x+y}{y}+q \frac{x+y}{x}=(x+y)(\frac{p}{y}+\frac{q}{x})\] So it is equivalent that \[\frac{p}{y}+\frac{q}{x}, \frac{q}{y}+\frac{r}{x}, \frac{r}{y}+\frac{p}{x}\geq 1.\] [b]So that $\sum a^{x}b^{y}\geq \sum a^{p}b^{q}c^{r}$ can be proved by AM-GM is equivalent that $\frac{p}{y}+\frac{q}{x}\geq 1$, $\frac{q}{y}+\frac{r}{x}\geq 1$, and $\frac{r}{y}+\frac{p}{x}\geq 1$.[/b]

I tried to find an example that does not follow the conditions above (that means the inequality which cannot be proved by AM-GM) but can be proved by other way, but I couldn't. Moreover, I don't know if there are more than 3 variables..

Very nice formulation of the condition!
I've done about the same (without assuming $z=0$) and written the conditions as scalar products $v_{i}v\ge0$ ($i=1,2,3$) with vectors $v=(x^{2}-yz,y^{2}-xz,z^{2}-xy), v_{1}=(p,q,r), v_{2}=(q,r,p),v_{3}=(r,p,q)$, so a partition $(p,q,r)$ can be easily 'tested' for a given $(x,y,z).$ But your brillant idea of adding $p+q$ makes the result nicer. :)
Now from dabbling around I have the impression that these conditions might be not only necessary but sufficient... otherwise stated:
I [b]conjecture [/b]that [hide]at least in the case of three variables,[/hide] [b]the ineq is true iff it can be done by AM-GM[/b]. To prove that, it would of course be sufficient to "just" provide a counterexample if one of the conditions doesn't hold :lol:  . I'm working on it for some special cases. :P

Thanks, spanferkel.

[quote="spanferkel"]
I [b]conjecture [/b]that [hide]at least in the case of three variables,[/hide] [b]the ineq is true iff it can be done by AM-GM[/b].[/quote]

Actually, I conjectured the same thing, and I tried to find counterexamples. I found that $\sum a^{4}b^{3}\geq \sum a^{4}bc^{2}$ does not follow the conditions, but it seems solvable. (I believe it should be not solvable  :P )

How do you exactly define when it can be "proved by AM-GM"¿

Well, my approach for solving by AM-GM was similar to Sung-yoon Kim's, though the end result is a little different.
I'll do it for
$\sum_{cyc}a^{x_{1}}b^{x_{2}}\ge \sum_{cyc}a^{y_{1}}b^{y_{2}}c^{y_{3}}$
We need
$pa^{x_{1}}b^{x_{2}}+qb^{x_{1}}c^{x_{2}}+rc^{x_{1}}a^{x_{2}}+k\ge (p+q+r+k)a^{y_{1}}b^{y_{2}}c^{y_{3}}$
If we can find such $p, q, r, k$ we win.
So it is this system of equations:
$x_{1}p+x_{2}r = y_{1}(p+q+r+k)$
$x_{2}p+x_{1}q = y_{2}(p+q+r+k)$
$x_{2}q+x_{1}r = y_{3}(p+q+r+k)$
we can find nontrivial (i.e., useful) solution by row-reducing
$\left( \begin{array}{cccc}x_{1}-y_{1}&-y_{1}& x_{2}-y_{1}&-y_{1}k\\ x_{2}-y_{2}& x_{1}-y_{2}&-y_{2}&-y_{2}k\\-y_{3}& x_{2}-y_{3}& x_{1}-y_{3}&-y_{3}k \end{array}\right)$
Then we win if the solution allows $p, q, r > 0, k \ge 0$. Note that if $x_{1}+x_{2}= y_{1}+y_{2}+y_{3}$ then taking $k=0$ is necessary and allows us to always find a nontrivial solution for $p, q, r$ (though not necessarily a positive one).
In Sung-yoon Kim's case, it is
$4p+3r = 4(p+q+r)$
$3p+4q = p+q+r$
$3q+4r = 2(p+q+r)$
--
(1) is $4q+r = 0$ which means something will be negative.

Is my condition equivalent to Sung-yoon Kim's?

@ ZetaX: I guess for my approach, I would say it is solvable by AM-GM if my system of equations gives a solution for which $p, q, r > 0$. (Or more variables, if necessary -- this approach works for any number of variables.)

It seems very similar way to mine. Maybe equivalent, except I did not consider the case in which constant comes LHS. (But in that case homogenousity should be broken.)

Oh sorry, it is pretty much identical (except you did more ;)). I just wonder if there is a more direct way to check (such as from the determinant of the matrix?). I am thinking of the homogeneous case so there is no $k$.

@ZetaX: Well, I guess I should revise my statement to "when Sung-yoon Kim's approach fails, it is not AM-GMable." :)

[quote="Sung-yoon Kim"]... and I tried to find counterexamples. I found that $\sum a^{4}b^{3}\geq \sum a^{4}bc^{2}$ does not follow the conditions, but it seems solvable. (I believe it should be not solvable  :P )[/quote]

Counterexample of $\sum a^{4}b^{3}\geq \sum a^{4}bc^{2}$ : $a=6, b=1, c=2$!

Now I am gonna try to prove the conjecture that the conditions are sufficient and necessary. In other words, WTS : if $\frac{q}{x}+\frac{p}{y}< 1$, then $\sum a^{x}b^{y}\geq \sum a^{p}b^{q}c^{r}$ holds for [b]not any $a,b,c \geq 0$[/b].

[quote="Xevarion"]I just wonder if there is a more direct way to check (such as from the determinant of the matrix?).  :)[/quote]Well, the main determinant (in the general case) is $\Delta=3xyz-x^{3}-y^{3}-z^{3}$ which is always $<0$. So we need the determinants $\Delta_\alpha,\Delta_\beta,\Delta_\gamma\le0$ (by those I mean the ones that give $\alpha=\frac{\Delta_\alpha}\Delta$ etc.) and those are exactly my 'scalar products' multiplied by $-1$. I don't think we can get it more direct than that.

[quote="Sung-yoon Kim"]
Now I am gonna try to prove the conjecture that the conditions are sufficient and necessary. In other words, WTS : if $\frac{q}{x}+\frac{p}{y}< 1$, then $\sum a^{x}b^{y}\geq \sum a^{p}b^{q}c^{r}$ (*) [color=red]holds for [b]not any [/color]$a,b,c \geq 0$[/b].[/quote]I suggest a clearer wording: "[color=red]...doesn't hold [b]for all [/b]$a,b,c \geq 0$.[/color]" Do you agree?

Some special cases I've found counterexamples for: [hide="(easy to check)"]but I cannot guarantee they are all error-free :roll: [/hide]
Let $\epsilon>0$ smaller than, say, $\frac12$ and $N=\frac1\epsilon$. The following are cases such that $(x,y,z)\ge(p,q,r)$ holds if $\epsilon=0$ with at least one of Kim's three conditions being an equality.
$(x,y,z)=(4,2,0),(p,q,r)=(1-\epsilon,2,3+\epsilon)$ Then (*) is false for $(a,b,c)=(2^{-N},\sqrt{2},2^{N})$.
$(x,y,z)=(4,2,0),(p,q,r)=(1-\epsilon,2+\epsilon,3)$ or $(1,2-\epsilon,3+\epsilon)$. Then (*) is false for $(a,b,c)=(2^{-N},1,2^{N})$.
$(x,y,z)=(6,3,0),(p,q,r)=(1,4-\epsilon,4+\epsilon)$. Then (*) is again false for $(a,b,c)=(2^{-N},1,2^{N})$.
$(x,y,z)=(2k,2,0),(p,q,r)=(1-\epsilon,k+\epsilon,k+1)$ Then (*) is false for $(a,b,c)=(2^{-N},1,2^{\frac{N}{k-1}})$.
$(x,y,z)=(2k,2,0),(p,q,r)=(1-\epsilon,k,k+1+\epsilon)$ Then (*) is false for $(a,b,c)=(2^{-N},2^{\frac1{k(k+1)}},2^{\frac{N}{k-1}})$.  :P

Even more generally (and that's what we want :) ) for $(x,y,0)$ and diverse $(p,q,r)$ a good candidate for a counterexample may be $(a,b,c)=(2^{-N},2^{t},2^{\frac{y}{x-y}N})$. But it is becoming a bit messy to find suitable $t$ (independent of $N$ and hopefully small) for which this works.
Note that by homogenity we can assume wlog $x+y=1$ and furthermore, once we have chosen $p$, the [hide="constraints"] 1. $p+q+r=1$
2. at least one of Kim's conditions has equality[/hide] allow only four or less posssibilities for $q$ and $r$, and then we have for each triple $(p,q,r)$ two or three choices where to add/substract $\epsilon$.
This means there is only a finite number of cases to consider (12 or less).

[i]I think this can be done. [/i][b]It will be messy but maybe rewarding[/b] :lol:

[quote="spanferkel"][quote="Sung-yoon Kim"]
Now I am gonna try to prove the conjecture that the conditions are sufficient and necessary. In other words, WTS : if $\frac{q}{x}+\frac{p}{y}< 1$, then $\sum a^{x}b^{y}\geq \sum a^{p}b^{q}c^{r}$ (*) [color=red]holds for [b]not any [/color]$a,b,c \geq 0$[/b].[/quote]I suggest a clearer wording: "[color=red]...doesn't hold [b]for all [/b]$a,b,c \geq 0$.[/color]" Do you agree?

[/quote]

Clarification: "... should have counterexamples."  :)

Some comments:
--- From Sung-yoon Kim's conditions: Since $p+q+r = x+y$, write instead $x+p, q+r-x$ for $x, y$. Then we can rewrite the conditions as
$(0 \le ) (x-(-p+q+\frac{r}{2}))^{2}< \frac{p^{2}}{4}-qr$
and the other two by cycling. Hope this is interesting.

--- Noting some other cases: $n=1$ doesn't make sense. $n=2$ is trivially always doable by AM-GM. $n=3$ is done by S-yK above. $n=4$ is already much more complicated than $n=3$! The common denominator for the solutions is
$x^{4}-y^{4}+z^{4}-2x^{2}z^{2}+4xy^{2}z$
(I am too lazy/unskilled to analyze it for myself so it is posted here
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=588400#588400)
and the numerators are huge and I don't know how to make them into a nice condition as S-yK did with the earlier version:
$(x^{3}+y^{2}z)q+2xyzr+(xy^{2}+z^{3})s \ge x^{2}yp+xz^{2}q+y^{3}r+x^{2}zs$
and cycling. It seems the general case might be very hard... Maybe we should try to win for $n=3$ before moving on to generalization...

made it!! For the case of $\frac{q}{x}+\frac{p}{y}<1$, there is a counterexample: $a=m^{x-y}, b=m^{x}, c=1$ for big big enough $m$.  :lol:

So we can say the following statement:

[b]For positive real numbers $a,b,c$ and nonnegative real numbers $x,y,p,q,r$, $\sum_{cyc}a^{x}b^{y}\geq \sum_{cyc}a^{p}b^{q}c^{r}$ iff $\frac{q}{x}+\frac{p}{y}\geq 1$, $\frac{r}{x}+\frac{q}{y}\geq 1$, and $\frac{p}{x}+\frac{r}{y}\geq 1$.[/b]

Moreover, it is impossible that two of them are less than 1. Assume that two of them are less than 1. Then [b]wlog[/b] we can assume that $\frac{q}{x}+\frac{p}{y}< 1$ and $\frac{r}{x}+\frac{q}{y}< 1$, or $px^{2}+qy^{2}-rxy<0$ and $qx^{2}+ry^{2}-pxy<0$. So $p>q\frac{x}{y}+r\frac{y}{x}>q\frac{x}{y}+(p\frac{x}{y}+q\frac{y}{x})\frac{y}{x}=p+q\frac{x}{y}+q\frac{x^{2}}{y^{2}}>p$, contradiction. Similarly, it is impossible that one of them is equal to 1 and one of them is less than 1.

So we can paraphrase the statement above like this:

[b]For positive real numbers $a,b,c$ and nonnegative real numbers $x,y,p,q,r$, the followings are equivalent:
(1) $\sum_{cyc}a^{x}b^{y}\geq \sum_{cyc}a^{p}b^{q}c^{r}$,
(2) The inequality can be solved by AM-GM only,
(3) $\frac{q}{x}+\frac{p}{y}\geq 1$, $\frac{r}{x}+\frac{q}{y}\geq 1$, and $\frac{p}{x}+\frac{r}{y}\geq 1$,
(4) $K=(\frac{q}{x}+\frac{p}{y}-1)(\frac{r}{x}+\frac{q}{y}-1)(\frac{p}{x}+\frac{r}{y}-1)\geq 0$.[/b]  :)

Now let's try the case of four variables as Xevarion said.

Well, the day starts earlier in Korea I guess - Cuz I have found it too [hide][i]during the night  :ddr: [/i] [/hide] If $x,y\in\mathbb N$ (which we can assume wlog), $(a,b,c)=(2^{x},2^{y},2^{x+y})$ does the job. This means $m=2$ is sufficient. Note the form is symmetrical, and re-introducing $z$ we may even write it as $(2^{-z},2^{-y},2^{-x})$. Now, may [b]that[/b] be something to go for in four or more dimensions?
I have not the least numerical evidence and am not very strong in linear algebra, but maybe $(a_{1},...,a_{n})=(2^{-x_{n}},...,2^{-x_{1}})$ is the counterexample we are looking for (assuming $x_{i}\in\mathbb N$)? That would just be too nice. :roll:  :roll:

[quote="spanferkel"]If $x,y\in\mathbb N$ (which we can assume wlog), $(a,b,c)=(2^{x},2^{y},2^{x+y})$ does the job. This means $m=2$ is sufficient. [/quote]

Wouldn't that be $m=2^{y}$? I wouldn't be surprised if the general conditions and general counterexample are similar, but it doesn't seem nearly as easy to attack....

[quote="Xevarion"]
Wouldn't that be $m=2^{y}$? [/quote]No, everything is just multiplied by $2^{y}$. Note that my proof uses $x,y\in\mathbb N$ (and in fact $p,q,r\in\mathbb N$ which we can assume wlog too) as it uses the violated condition $\frac{q}{x}+\frac{p}{y}< 1$ in the form of $qy+px<xy\Rightarrow qy+px\le xy-1$.   :P

For your last sentence you are probably right  :huh:

I found that in case of four variables, the common denomiator can be zero or negative.. Let's consider $\sum a^{x}b^{y}c^{z}\geq \sum a^{p}b^{q}c^{r}d^{s}$. When $y=x+z$, it becomes zero, and the system has infinitely many solutions if $p+r=q+s$, or no solution otherwise.

I felt that this is like finding Ramsey's number, as we cannot generalize directly from smaller n. :( very messy.

Yes :( it seems to me that the system is solvable in general [b]iff $n$ is a prime number[/b]. Here is why:
We want the determinant $\Delta\ne0$, i.e. all eigenvalues of the matrix must be non-zero, unless of course $x_{1}=...=x_{n}$. Now it is easy to calculate those eigenvalues, and by the way that can be found [url=http://mathworld.wolfram.com/CirculantDeterminant.html]here[/url]. So we need the $n$-th roots of unity to be linearly independant over $\mathbb Z$, which of course is only the case iff $n$ is prime.

Having said that, let's see what happens if $\Delta=0$. Let's consider a small environment of $(x,y,z,0)$ in $\mathbb R^{4}$ and suppose AMGM works somewhere therein. Now $[0,1]^{4}$, where the solutions $(\lambda_{1},...,\lambda_{4})$ must sit, is a compact set. If we approach $(x,y,z,0)$, the $\lambda_{i}$ cannot go astray. So all the four determinants $\Delta_{p},...$ must be 0 as well, which is a contradiction.

This means, if $\Delta=0$, we cannot apply AMGM in a whole environment  anyway. :D

So we can still try to prove the general result. We might assume there is a solution to the system, i.e. an $n$-tuple of coefficients $(\lambda_{1},...,\lambda_{n})$, with at least one of them $<0$. And we would need to show that there exists then a certain set of variables, maybe my $(a_{1},...,a_{n})=(2^{-x_{n}},...,2^{-x_{1}})$, such that
$\sum_{cyc}a_{1}^{x_{1}}\cdots a_{n}^{x_{n}}< \sum_{cyc}a_{1}^{p_{1}}\cdots a_{n}^{p_{n}}$ with $p_{i}=\sum_{k=1}^{n}\lambda_{k}x_{k+i}$ (indices modulo $n$).
Even if we know that $\sum_{k=1}^{n}\lambda_{k}=1$, this is becoming really messy. Maybe somebody used to linear algebra wants to deal with those multiple sums by picking the leading terms on either side etc. etc. :P

Well.. I am thinking differnet way. I considered putting $x_{i}= x^{y}_{i}$ for $\sum_{cyc}x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{n}^{a_{n}}\geq \sum_{cyc}x_{1}^{b_{1}}x_{2}^{b_{2}}\cdots x_{n}^{b_{n}}$. Then the both sides are polynomials for $x$. So for given $a_{i}$'s and $b_{i}$'s, if there is some $y_{i}$'s such that the degree of RHS is bigger than the one of LHS, then the inequality doesn't hold, as there should be counterexamples for $x$ which is big enough. This means we may find some conditions in which the inequality does not hold, not the conditions in which the ineq holds. (Actually when I was finding counterexamples for $n=3$ and $\frac{q}{x}+\frac{p}{y}<1$, I used this and got proper $y_{i}$'s.)

You can see that by this way the problem can be changed into different problem treating maximums of two linear functions.

By this method I found that if the inequality $\sum a^{x}b^{y}c^{z}\geq \sum a^{p}b^{q}c^{r}d^{s}$ holds, then $px+qy+rz\geq \min (2xz,xy+yz)$, $qx+ry+sz\geq \min (2xz,xy+yz)$, $rx+sy+pz\geq \min (2xz,xy+yz)$, and $sx+py+qz\geq \min (2xz,xy+yz)$. This means that if at least one of them is violated, then the inequality does not hold. But, as you know, it is necessary condition, and it seems that these conditions are not so tough.

In 1987, I published an article in RMT Magazine,2, with the following statements.

[b]Theorem 1.[/b] Let $x,y,p,q,r$ be real numbers such that $x$ is nonzero and $x+y=p+q+r$. The inequality
$\sum_{cyc}a^{x}b^{y}\geq \sum_{cyc}a^{p}b^{q}c^{r}$ holds for any positive real numbers $a,b,c$, if and only if
$min(px+qy,qx+ry,rx+py) \geq xy$.

[b]Theorem 2.[/b] The above inequality holds if $p,q,r$ are positive numbers such that $4pqr \geq max(p^{3},q^{3},r^{3})$ and  $x+y=p+q+r$.

[b]Theorem 3.[/b] Let $x,y,p$ be positive numbers and let $q$ be a real numbers such that $x-y=p-q$ and $x>p$. Then the inequality
$\sum_{cyc}a_{1}^{x}a_{2}^{-y}\geq \sum_{cyc}a_{1}^{p}a_{2}^{-q}$ holds for any positive numbers $a_{1},a_{2},...,a_{n}$.

[b]Theorem 4.[/b] Let $x,y,p$ be positive numbers and let $q$ be a real numbers such that $x-y=p-q<>0$ and $py^{n-1}\geq qx^{n-1}$. Then the inequality
$\sum_{cyc}a_{1}^{x}a_{2}^{-y}\geq \sum_{cyc}a_{2}^{p}a_{1}^{-q}$ holds for any positive numbers $a_{1},a_{2},...,a_{n}$:

as much as 19 years ago...  :roll: Cool!
[size=75](in Theorem 1, it should say $\min(px+qy,qx+ry,rx+py) \geq xy.$)[/size]
What is RMT Magazine? Something Romanian I guess?
The condition $py^{n-1}\geq qx^{n-1}$ of your theorem 4 is funny. May you give a sketch of proof?

[quote="spanferkel"]What is RMT Magazine? Something Romanian I guess?
The condition $py^{n-1}\geq qx^{n-1}$ of your theorem 4 is funny. May you give a sketch of proof?[/quote]
RMT=a Romanian Magazine (Mathematical Review of Timisoara).
I will give an idea of proof, which helps to solve the general problem.
For the inequality $\sum_{cyc}a_{1}^{x}a_{2}^{y}\geq \sum_{cyc}a_{1}^{p_{1}}a_{2}^{p_{2}}...a_{n}^{p_{n}}$  with $x+y=p_{1}+p_{2}+...+p_{n}$,
the idea is to use the substitutions: $a_{i}^{x}a_{i+1}^{y}=x_{i}$ for $i=1,2,...,n$.
The inequality becomes
$x_{1}+x_{2}+...+x_{n}\geq \sum_{cyc}x_{1}^{q_{1}}x_{2}^{q_{2}}...x_{n}^{q_{n}}$, with $q_{1}+q_{2}+...+q_{n}=1$.
This inequality is true for any positive $x_{i}$ if and only if all $q_{i}\geq 0$.
I established in the RMT article "On some cyclic algebraic inequalities" only the results in Theorem 1...4 above.

[quote="Vasc"]
The inequality becomes
$x_{1}+x_{2}+...+x_{n}\geq \sum_{cyc}x_{1}^{q_{1}}x_{2}^{q_{2}}...x_{n}^{q_{n}}$, with $q_{1}+q_{2}+...+q_{n}=1$.
This inequality is true for any positive $x_{i}$ if and only if all $q_{i}\geq 0$.
[/quote]
THank you Vasc. Very nice! It seems to me this is (again!) exactly the case where the ineq can be solved by AM-GM. The $q_{i}$ are the coefficients of the decomposition.  :)

[quote="Vasc"]I will give an idea of proof, which helps to solve the general problem. [/quote]
Unfortunately, in the general case we can express the right - hand side using the substitution variables if and only if the cyclic matrix $\mathbf{M}\left(\vec{p}\right)$ (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=83869]here[/url]) is nonsingular.


Given two points A and B, and a line CD, find a point P (on the line CD) such that AP^2 + BP^2 is a minimum.

By the way, I have the answer, so I am posting this more as a challenge than for help.


search
combinatorics proposed
combinatorics

If permutations of the numbers $ 2,3,4,\ldots,102$ are denoted by
\[ a_1,a_2,a_3,\ldots,a_{101},
\]
find all such permutations in which $ a_k$ is divisible by $ k$ for all $ k$.

Not only has this been posted before, but it was posted before [i]by you[/i]: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1148361167&t=213461


[url=http://www.mathlinks.ro/resources.php?c=1&cid=114&year=2008]Here[/url] you can find the links to the problems of the Middle/Central European Mathematical Olympiad (MEMO) 2008. For your interest I als appened the official solutions for the single and team competitions. MEMO 2008 gold medal winner Bertram Arnold and member of the [url=http://www.mathlinks.ro/viewtopic.php?t=252051]TOP 17 Team for the IMO 2009, Bremen, Germany[/url] supplied the German report for the MEMO 2008.

[quote="Umut Varolgunes"]I can't open the file "TOP 17 Team for the IMO 2009, Bremen, Germany". Would you mind if I ask you what is it?[/quote]

Umut, that is not a file. That is only a link to another thread in the mathlinks forum.

[quote="Umut Varolgunes"]:blush: So, why I can't see it?[/quote]

I am not sure what the problem is. But the thing you are referring to is only a link to the following thread:

- http://www.mathlinks.ro/viewtopic.php?t=252051

which has the names of the German TOP 17 members for IMO 2009.

[code]Sorry, but only users granted special access can read topics in this forum.

Click Here to return to the Index
[/code]

It is probably in IMO Vorbereitung subforum (which is a restricted access area).

[quote="Erken"]Is that a top secret information?[/quote]

Yeah, absolutely. This forum is the only place that already has the problems for IMO 2009.  :D

Good that I am a mod there but do not know anything that it was restricted. Maybe people wanted to post the current German IMO TST problems there back then. I have no idea. Anyway I moved the thread to a different forum and you should be able to see it now.


algebra
polynomial
number theory proposed
number theory

Let ${a_n}$ be a sequence of integers such that for $n\geq 1$,
$(n-1)a_{n+1} = (n+1)a_n-2(n-1)$
If $2000$ divides $a_{1999}$, find the smallest $n\geq 2$ such that $2000$ divides $a_n$

We have $a_{n+1} = \frac{n+1}{n-1} a_n - 2$, so write $a_n = b_n(n-1)$.  Then $n b_{n+1} = (n+1) b_n - 2$.  Let's suppose that $b_n$ can be written as a polynomial or power series in $n$: in any case try $b_n = c_n n + d$.  Then
$n(n+1)c_{n+1} + dn = n(n+1)c_n +d(n+1) - 2$.  This suggests putting $d=2$and then $c_{n+1} = c_n$, hence $b_n = cn+2$ for constant $c$.  This works, and the sequence is determined by its first term $a_2$, so this is the unique solution and $c = b_2/2 - 1 = a_2/2-1$.  Now, if $2000 \mid a_{1998} = 1998(c.1999 + 2)$ then $c \equiv 2 \bmod 1000$.  So $a_n \cong ruent (2n+2)(n-1) \bmod 2000$ and this is $n^2 \equiv 1 \bmod 8$ and $\bmod 125$.  The first congruence simply says $n$ is odd, the second has solutions $\pm 1 \bmod 125$.  So the smallest $n \ge 2$ is 249.


search
number theory unsolved
number theory

1.Show that if x y z are positive integer ,then (xy+1)(xz+1)(yz+1) is a perfect square if and only if

xy+1,xz+1,yz+1 are all perfect squares

2.if a b c are positive integer such that  0<a^2+b^2-abc<=c

show that  a^2+b^2-abc is perfect square

search before posting:
1.http://www.mathlinks.ro/viewtopic.php?t=150369
2.http://www.mathlinks.ro/viewtopic.php?t=150372


I guess this is a n00bish question, but how do you install font packages? I'd like to be able to use the [url=http://www.tug.dk/FontCatalogue/calligraphicalfonts.html]Calligra font[/url]. How do I proceed?

That's a good page because they help you if you do some clicking. So suppose you want the Auriocus Kalligraphicus font. Click on this name and it leads you to http://www.tug.dk/FontCatalogue/aurical/index.html
To use this font it tells you that all you need is \usepackage{aurical} in your preamble. If you click on [i]LaTeX source of PDF sample[/i] then you'll get a sample tex file using this font.

Of course you have to download aurical first which the MiKTeX package manager will do for you. If it's a font MiKTeX doesn't know about , or you are using a different LaTeX distribution, then [url=http://tug.ctan.org/search.html#byDescription]CTAN Search[/url] (Use [i]Search the package descriptions You can search the descriptions of our packages from the Catalogue[/i] option) will get it for you.


Find pairs of integers x;y such that
y^2=1+x+x^2+x^3+x^4

maybe we use zip
if $ x > 0$  $ \implies$
$ 4y^{2} \equal{} 4x^{4} \plus{} 4x^{3} \plus{} 4x^{2} \plus{} 4x \plus{} 4$ $ \implies$
$ (2x^{2} \plus{} x)^{2} < 4y^{2} < (2x^{2} \plus{} x \plus{} 1)^{2}$  for $ x^{2} > 2x \plus{} 3$  $ \implies$
$ x\leq 3$  $ \implies$
$ x \equal{} 1$  $ \implies$  $ 5$ isn't perfect square
$ x \equal{} 2$  $ \implies$  $ 31$ isn't perfect square
$ x \equal{} 3$  $ \implies$  $ y \equal{} 11$ $ \implies$  $ (3,11)$ is a solution
similarly $ x < 0$  $ \implies$
$ 4x^{4} \plus{} 4x^{3} \plus{} x^{2} < 4x^{4} \plus{} 4x^{3} \plus{} 4x^{2} \plus{} 4x \plus{} 4$  for $ 3x^{2} \plus{} 4x \plus{} 4 > 0$  for $ x < 0$  .
$ 4x^{4} \plus{} 4x^{3} \plus{} 4x^{2} \plus{} 4x \plus{} 4 < (2x^{2} \plus{} x \plus{} 1)^{2}$  for $ x^{2} > 2x \plus{} 3$    for $ x < \minus{} 2$
only $ x \equal{} \minus{} 1$  $ \implies$  $ y \equal{} \pm 1$
if $ x \equal{} 0$  $ \implies$  $ y \equal{} \pm 1$   :wink:


If $ a,b,c$ are three positive real numbers , then

$ \frac {a}{(b\plus{}c)^2} \plus{} \frac {b}{(a\plus{}c)^2} \plus{} \frac{c}{(a\plus{}b)^2} \ge {\frac {9}{4(a\plus{}b\plus{}c)}}$

$ LHS\equal{}\frac{a^2}{a(b\plus{}c)^2}\plus{}\frac{b^2}{b(c\plus{}a)^2}\plus{}\frac{c^2}{c(a\plus{}b)^2}\geq\frac{(a\plus{}b\plus{}c)^2}{a(b\plus{}c)^2\plus{}b(c\plus{}a)^2\plus{}c(a\plus{}b)^2}$
We have to prove $ 4(a\plus{}b\plus{}c)^3\geq 9[a(b\plus{}c)^2\plus{}b(c\plus{}a)^2\plus{}c(a\plus{}b)^2]$
$ \Leftrightarrow 4(a^3\plus{}b^3\plus{}c^3)\plus{}3(a^2b\plus{}ab^2\plus{}b^2c\plus{}bc^2\plus{}c^2a\plus{}ca^2)\geq 30abc$. OK by AM-GM inequality!

oke!!
your solution is very nice!!!
I think we can let : a+b=2x;c+b=2y;a+c=2z
  the inequality becomes easily!!!

By Cauchy;\[ (a\plus{}b\plus{}c)(\frac {a}{(b\plus{}c)^2} \plus{} \frac {b}{(a\plus{}c)^2} \plus{} \frac{c}{(a\plus{}b)^2}) \geq (\sum_{cyc}\frac{a}{b\plus{}c})^2\]
\[ \geq \frac{9}{4}\] by nesbit


calculus
derivative
function
calculus computations

Let $ f(x) \equal{} (1 \plus{} x)\cdot\sqrt{(2 \plus{} x^2)}\cdot\sqrt[3]{(3 \plus{} x^3)}$. Determine $ f'(1)$.

[hide="simpler, but different answer"]$ f'(x)\equal{}\sqrt{2\plus{}x^2}\cdot\sqrt[3]{3\plus{}x^3}\plus{}g(x)$ where $ g(x)$ is a function with factor $ (x\plus{}1)$ so $ g(\minus{}1)\equal{}0$. Hence $ f'(\minus{}1)\equal{}\sqrt3\cdot\sqrt[3]2$[/hide]

@Johan : he wants $ f'(1)$ not $ f'(\minus{}1)$

I think it's a typo. See this: http://web.archive.org/web/20080126105837/www.kalva.demon.co.uk/vietnam/viet62.html


:D let x,y,z is positive reals and $\sum\limits_{cycle} xy = \frac{1}{6}$
and $F=\sum\limits_{cycle} (\sqrt{x^2+1}-x)$
Find min F?
 :lol:

Yes, I am pretty sure that is what he means.


[quote=bertram]Let $ ABCD$ be a (non-degenerate) quadrangle and $ N$ the intersection of $ AC$ and $ BD$. Denote by $ a,b,c,d$ the length of the altitudes from $ N$ to $ AB,BC,CD,DA$, respectively.

Prove that $ \frac{1}{a}\plus{}\frac{1}{c} \equal{} \frac{1}{b}\plus{}\frac{1}{d}$ if $ ABCD$ has an incircle.

Extension: Prove that the converse is true, too.

[If this has already been posted, I humbly apologize. A quick search turned up nothing.][/quote]

Suppose $AB,BC,CD,DA$ are tangent to the incircle at $E,F,G,H$. By Brianchon's (hexagon $AEBCGD$) the point $N$ lies on $EG$. By Brianchon's (hexagon $ABFCDH$) the point $N$ lies on $FH$.

Note that $\angle BEG=\angle EHG=\angle EGC$ and $\angle AHF=\angle HGF=\angle HFB$. By extended law of sines
\[
\frac{EN}{\sin\angle NGC}+\frac{NG}{\sin\angle BEN}=\frac{EG}{\sin\angle EHG}=2r=\frac{HF}{\sin\angle HGF}=\frac{HN}{\sin\angle NFB}+\frac{NF}{\sin\angle AHN}
\]
By intersecting chords theorem $EN\cdot NG=HN\cdot NF$. Thus
\[
\frac{1}{a}+\frac{1}{c}=\frac{1}{NG\cdot\sin\angle NGC}+\frac{1}{EN\cdot\sin\angle BEN}=\frac{1}{NF\cdot\sin\angle NFB}+\frac{1}{HN\cdot\sin\angle AHN}=\frac{1}{b}+\frac{1}{d}
\]


I don't think it is very easy...

[hide]$1$ in the first group, $4$ in the second group, $7$ in the third group (verify please) gives $12$ children as the final answer. [/hide]

[hide="Hmm..."]

Well since 28 30 and 31 are around 30, I divided 365 by 30 to get 12 R5.

Every time we have 31 walnuts per kid the total goes up by 1. Since we need 5 more than 365, there are 5 kids with 31 walnuts each, which leaves 7 kids with 30 walnuts each

So, 5+7=12 kids.[/hide]

[quote="davidlizeng"]some number of kids are split into 3 groups.
In the first group, each kid is given 28 walnuts.
In the second group, each kid is given 30 walnuts.
In the third group, each kid is given 31 walnuts.
In total, there are 365 walnuts.
Find the number of chidren.

p.s. the number of chidren in each group does not have to be equal.[/quote]

:rotfl:

[hide="BIG hint"]Map the number of walnuts per kid onto the number of days per month.[/hide]

Took me long enough...

oh lol. I can't believe that i didn't make that connection as listed in the hint. It seems so OBVIOUS. (oh no, i just swore in mathematical language.) It seems to make the problem that much easier. :blush:

Oh, that's a really cool problem, if you establish that no math is allowed.

yep, mathnerd314 is correct.
that's all you had to do.
 :lol:


[quote="allrighty"]Prove that if x1, x2, ..., xn are n real numbers in the closed interval (a,b), then
(x1 + x2 + ... + xn) / n is also in the closed interval (a,b).
Using induction,

Clearly, p(1) is true.
Now, assume p(k) is true, that is, if x1, x2, ..., xk are n real numbers in the closed interval (a,b), then (x1 + x2 + ... + xk) / k is also in the closed interval (a,b).

Then, how to show that (x1 + x2 + ... + x(k+1)) / (k+1) is in closed interval (a,b)?[/quote]
[hide="Inductive Proof"]
Like you said, the base  case is obviously true.  Now, we have that \[ a\le \frac{x_1\plus{}x_2\plus{}x_3...\plus{}x_n}{n}\le b\] \[ an\le x_1\plus{}x_2\plus{}x_3...\plus{}x_n\le bn\] \[ a\le x_{n\plus{}1}\le b\] Clearly, \[ an\plus{}a\le x_1\plus{}x_2...\plus{}x_{n\plus{}1}\le b\plus{}bn\] \[ a(n\plus{}1)\le x_1\plus{}x_2... \plus{}x_{n\plus{}1}\le b(n\plus{}1)\] \[ a\le \frac{x_1\plus{}x_2\plus{}x_3...\plus{}x_{n\plus{}1}}{n\plus{}1}\le b\] So, it is proven. [/hide]

[hide="non inductive proof"]
WLOG assume the largest number is $ x_k$, then the largest sum is $ k\cdot x_k$ (where all the numbers are the maximum), so the maximum average is $ x_k$

Seemed intuitive enough for me to mess up on it.
[/hide]


integration
logarithms
limit
calculus
calculus computations

Hello. My teacher gave each of us a review sheet to do over spring break to help us learn what will be over the test a week after the break. Here are a few questions that I cannot answer, so any help would be appreciated.

1) $ \displaystyle f(x) = \int_{2}^{e^x}\frac{1}{\sqrt{\ln t}}dt $

2) f(x) = 2^(3^x) and I'm supposed to find f ' (1).

I didn't know how to write this one using Latex. It kept putting the x next to the 3

3) $ \displaystyle \lim_{x\to 0}\left(\frac{1}{1+e^\frac{1}{x}}\right) $

Thanks again for any help.

1) I think the problem is that find $f'(x)$,if so, using fundamental theorem of calculus, we get $f'(x)=\frac{1}{\sqrt{\ln e^x}}\cdot (e^x)'=\frac{e^x}{\sqrt{x}}$.

2)$f(x)=2^{3^x}\Longleftrightarrow f'(x)=2^{3^x}\ln 2\cdot (3^x)'=2^{3^x}\ln 2\cdot 3^x\ln 3=$$2^{3^x}3^x
(\ln 2)(\ln 3)$,yielding $f'(1)=24(\ln 2)(\ln 3)$


AB = 360, BC = 507, CA = 780. M is the midpoint of AC, D is the point on AC such that BD bisects ∠ABC. F is the point on BC such that BD and DF are perpendicular. The lines FD and BM meet at E. Find DE/EF.

Source: AIME 2003

This is a pretty tedious solution involving a lot of calculation...some one can take the pain to check the calculations to verify the solution.....just incase.... :D....as I was in a hurry trying to solve this

From the right triangles BDF and BDE we have
DF/DE= 1+ EF/DE= [tan(B/2)/tan(<DBM)]...

Applying cosine formula in triangle ABC we have cos B and from which we have tan(B/2)= 2.023649504...

Applying Apollonius Theorem we have BM=203.038173..

Since angular bisector divides the opp side in the ratio of the other two sides we have AD from which we get....DM= 66.12456747404...

Applying cosine formula in triangles ABC & ABD for <A, we have BD= 186.52736642..

Now applying cosine formula in triangle BDM we have cos(DBM)...from which we have tan(<DBM)= .343110.......

Hence we have DE/EF value=.204166....

I don't think this is quite the way it's meant to be solved (elseway we'd have gotten it at school :D) but well I can't think of a better one either :(

you are right peter. this problem was from a contest where calculators are not allowed. so all these calculations would have been pointless. we need a solution that gives an exact answer.


absolute value
calculus
calculus computations

An object moving along a line that its position at time t(positive or negative) is given by s(x)=2x^3+9x^2-24x+6

1.on what intervals is moving to the right?
2.on what intervlas is moving to the left?
3.on what intervals its velocity increasing?
4.on what intervals its velocity decreasing?
5.on what intervals its speed increasing?

1. on what intervals is $ s'(x)$ positive?
2. on what intervlas is $ s'(x)$ negative?
3. on what intervals is $ s''(x)$ positive?
4. on what intervals is $ s''(x)$ negative?
5. on what intervals $ s''(x)$ has the same sign as $ s'(x)$?

I understand that velocity can be positive or negative, and speed is the magnitide (absolute value) of velocity.


linear algebra
matrix
induction
algebra unsolved
algebra

Let $ a$ and $ b$ be integers such $ x^2\minus{}x\minus{}1$ is a factor of $ ax^{17} \plus{} bx^{16} \plus{}1$. Find $ a$.

Let $ x^2\minus{}x\minus{}1|ax^{n\plus{}1}\plus{}bx^n\plus{}1\equal{}P(x), c\equal{}\frac{1\plus{}\sqrt 5 }{2},d\equal{}1\minus{}c$, then $ 0\equal{}P(c)\plus{}P(d)\equal{}aL_{n\plus{}1}\plus{}bL_n\plus{}2$ and $ 0\equal{}P(c)\minus{}P(d)\equal{}2\sqrt 5 (aF_{n\plus{}1}\minus{}bF_n).$
Therefore $ a(L_{n\plus{}1}F_n\minus{}L_nF_{n\plus{}1}\plus{}2F_n\equal{}0.$ But $ L_{n\plus{}1}F_n\minus{}L_nF_{n\plus{}1}\equal{}2(\minus{}1)^{n\plus{}1}$. Therefore $ a\equal{}F_n(\minus{}1)^n\equal{}F_{16}$.

Not all that different, but: let $ \alpha$ be a root of $ x^2\minus{}x\minus{}1$, i.e., $ \alpha^2\equal{}\alpha\plus{}1$.  Then $ \alpha^n \equal{} F_n\alpha \plus{} F_{n\minus{}1}$.

If $ x^2\minus{}x\minus{}1$ divides $ ax^{n\plus{}1}\plus{}bx^n\plus{}1$, then we must have $ 0\equal{}a\alpha^{n\plus{}1}\plus{}b\alpha^n\plus{}1\equal{}(aF_{n\plus{}1}\plus{}bF_n)\alpha \plus{} (aF_n\plus{}bF_{n\minus{}1}\plus{}1)$.

Since $ \alpha$ is irrational we get two equations in $ a$ and $ b$, which give $ a \equal{} \frac{F_n}{F_{n\minus{}1}F_{n\plus{}1}\minus{}F_n^2}\equal{}(\minus{}1)^n F_n$,
$ b\equal{}\frac{\minus{}F_{n\plus{}1}}{F_{n\minus{}1}F_{n\plus{}1}\minus{}F_n^2}\equal{}(\minus{}1)^{n\plus{}1} F_{n\plus{}1}$.

[quote="danilorj"]Eigenray, I didn't understand how did you simplify the expression of $ a$ to get to $ a \equal{} ( \minus{} 1)^{n}F_{n}$?[/quote]
Here is a useful trick when dealing with Fibonacci numbers: define the matrix
\[ F \equal{} \left(\begin{array}{cc}0&1\\1&1\end{array}\right).\]
By induction,
\[ F^n \equal{} \left(\begin{array}{cc}F_{n\minus{}1}&F_n\\F_n&F_{n\plus{}1}\end{array}\right).\]
In particular,
\[ F_{n\minus{}1}F_{n\plus{}1}\minus{}F_n^2 \equal{} \det F^n \equal{} (\minus{}1)^n.\]


Find all solutions in non-negative integers of the equation $(xy-7)^{2}=x^{2}+y^{2}$. I can find only four: (7,0), (0,7), (3,4), (4,3) by inspection.  I don't have a method.

Typical case bashing solution:

Suppose that $x \geq y$. We have:
$x^{2}y^{2}= x^{2}+y^{2}+14xy-49 \leq 26x^{2}-49 \leq 16x^{2}$
So we must have $y^{2}\leq 16$ or $y \leq 4$. It suffices to examine five easy cases, which should not be a problem for you.

Rather $y^{2}\leq 16$ than $y^{2}\leq 16x^{2}$. Thanks for pointing out the error, it eliminates one case.

Yeah, you're right prithwijit  :)

We can get more clearly solution by changing the form into
$(x+y)^{2}-(xy-6)^{2}=13$
$(x+y+xy-6)(x+y-xy+6)=13$

Let $A=x+y+xy-6, B=x+y-xy+6$, so $x+y=\frac{A+B}{2}, xy=\frac{A-B}{2}+6$

i) If $A=-13, B=-1$, we get $x+y=-7<0$

ii) If $A=-1, B=-13$, we get $x+y=-7<0$

iii) If $A=1, B=13$, we get $x+y=7, xy=0$
    $x=7,y=0$ and $x=0,y=7$

iv) If $A=13, B=1$, we get $x+y=7, xy=12$
    $x=3,y=4$ and $x=4,y=3$


hy my freind vive maroc . i want to ask if this problem is from morrocan tst 2007. it seem very easy.$cst$ function are obviously solution. :roll:

[quote="vietnam2007"]take the real function $f$ defined by $f(x)=x-1$ it's solution !!

For you Fermat 3     that problem isn't from the moroccan tst.I had created it recently.[/quote]
sorry,
i forget $a=-1$
then $x=0$ => $f(f(y))-1=-1+f(y)$
so, if $f(R)=R$ $f(x)=x-1$

[quote="vietnam2007"]Find all functions $f: R \to R$ such that :

$\forall (x,y) \in{R}^{2}$  $f(xf(y)+y)+f(x)=f(xy)+f(y)$[/quote]


You guys have been great help so far, thank you so much!

I got one more tricky problem to solve, if anyone could help, thanks!

I have to use Booth's Algorithm to find the product QM where

Q=11001 and M=01000

I have a table, 3 columns and 6 rows.

A  |  Q  |  Q-1
----------------
 0 | 11001  |  0
----------------
    |      |
----------------
    |      |
----------------
    |      |
----------------
    |      |
----------------
    |      |
----------------

The first row is filled with initialisation.  I have to basically fill out the values for each iteration!  I have been working at this for hours and just can't do it, I'm not clever enough!

Thanks for any help,

Answers are good, explanations are better!


I clicked on "Build current file" to compile my PDF.

The thing is, when I click on "View Output", it does not appear the content, it just appears a PDF window.

I also checked if the viewer path is correct. (I'm currently using LaTeX => PS => PDF.)

Thanks for the help.

Are you sure that your TeX file is error-free? Some errors are so bad that no output is produced at all. Also, look at the directory where your TeX file is. Do you see a pdf file with the same name there? If so, try to open it and see what the content is. If it is empty or if there is no pdf file at all, the problem is with creating the pdf file, not with viewing it. Go to Build->Define output profiles and look at what is there especially in the postprocessor section for the LaTeX=>PS=>PDF profile. Without additional information, it is hard to tell you more...

I just reinstalled TeXnicCenter and the problem was solved.

Thanks for your help fedja.


Let $ a_1,a_2,\cdots a_n \in \mathbb{R}^\plus{}$ such that $ \sum_{i\equal{}1}^{n}{\frac{1}{a_i\plus{}2}}\equal{}\frac{1}{2}$
Then find the minimum of $ \prod_{i\equal{}1}^{n}{a_i}$

Let $ \frac {1}{a_i \plus{} 2} \equal{} b_i$.
Then, $ 2 \sum_{i \equal{} 1}^n b_i \equal{} 1$, and $ a_i \equal{} \frac {1 \minus{} 2b_i}{b_i} \equal{} 2 \frac {P \minus{} b_i}{b_i}$ (where $ P \equal{} b_1 \plus{} .. \plus{} b_n$)

Multiplying all these equations, we get $ \prod_{i \equal{} 1}^{n} a_i \equal{} 2^n \frac {(P \minus{} b_1 )(P \minus{} b_2 ) .. (P \minus{} b_n )}{b_1 .. b_n} \geq 2^n (n \minus{} 1)^n$

($ \frac {(P \minus{} b_1 )(P \minus{} b_2 ) .. (P \minus{} b_n )}{b_1 .. b_n} \geq (n \minus{} 1)^n$ is trivial by AM-GM inequality)


Let $ a,b,c$ be nonnegative numbers such that $ ab\plus{}bc\plus{}ca\equal{}3$. Find the smallest positive number $ p$ such that

$ \frac {1}{a\plus{}p}\plus{}\frac {1}{b\plus{}p}\plus{}\frac {1}{c\plus{}p}\le\frac {3}{1\plus{}p}$.

Very interesting inequality, Vasc !

[color=red]EDIT: Sorry, I had a mistake I'll write a correct solution later ...[/color]

[quote="vardanch"]
Note that $ f(a, b, c)\minus{}f(a, b\plus{}c, 0) \equal{} 3abc\geq0,$ so we just need to prove:
$ f(a, b, 0)\geq 0$[/quote]
But $ ab\plus{}ac\plus{}bc\neq a(b\plus{}c)\plus{}a\cdot0\plus{}(b\plus{}c)\cdot0.$ :wink:
I think, the mixing variables method works here and the answer is $ 2\plus{}\sqrt3.$
Nice problem, Vasc!

[quote="arqady"]
But $ ab\plus{}ac\plus{}bc\neq a(b\plus{}c)\plus{}a\cdot0\plus{}(b\plus{}c)\cdot0.$ :wink:[/quote]

Yes ... , I knew that I had a mistake here ... (It must be $ f(a,b\plus{}c\plus{}bc/a, 0)$ ...  )

[quote="arqady"]
I think, the mixing variables method works here and the answer is $ 2\plus{}\sqrt3.$ [/quote]

Yes... However I have some problems while using mixing variables.

Look when $ a\plus{}b\plus{}c\geq 2\sqrt 3$ we have:

[hide]
....
\[ \Leftrightarrow\sum{a}\plus{}3k\leq 3abc\plus{}k\sum{a}\]
Let $ f(a,b,c) \equal{} (k\minus{}1)\sum{a}\plus{}3(abc\minus{}k)$. If $ k\leq 1$ then $ f(a,b\plus{}c\plus{}\frac{bc}{a},0) \equal{} (k\minus{}1)(\sum{a}\plus{}\frac{bc}{a})\minus{}3k\leq 0$, contradiction. Thus $ k > 1$, $ k\minus{}1\equiv L$ $ \Rightarrow$ $ L > 0$.

So, $ f(a,b,c) \equal{} L\sum{a}\plus{}3abc\minus{}3L\minus{}3$, and we need to find a positive real number $ L\rightarrow min$, such that $ f(a,b,c)\geq 0$ for every  nonnegative real numbers $ a,b,c$.

[color=blue]Let us prove that for every $ a,b,c$ (satisfying the problems condition) there is a positive real number $ x$, such that
$ f(a,b,c)\geq f(x,3/x,0)$ (note that the numbers (x,3/x,0) satisfy the problems condition)
<=> $ f(a,b,c)\minus{}f(x,3/x,0) \equal{} L(a\plus{}b\plus{}c\minus{}x\minus{}3/x)\plus{}3abc\geq 0$, $ a\plus{}b\plus{}c\equiv r$, so we just need to find $ x$, such that $ r \equal{} x\plus{}3/x$, and we are done.
On the other hand $ D \equal{} r^{2}\minus{}12$ and $ x \equal{} (\frac{r\pm\sqrt{r^{2}\minus{}12}}{2})$, which is solution for $ r\geq 2\sqrt{3}$ $ \Rightarrow$ we just need to prove
$ f(x,3/x,0)\geq 0$, which is obviously true (look in .pdf file) , and we easily find $ k$, then $ p$ ... [/color]
[color=red]
BUT, Arqady, what can I do in the case $ 3\leq r < 2\sqrt{3}$, in that case I have some doubts that there are numbers $ x,y$, such that always $ f(a,b,c)\geq f(x,y,0)$ (or maybe you wanted to say that MV with other constant (not zero )works...  :maybe: , Thank you ...
[/color]

[/hide]

[quote="vardanch"]
However I have some problems while using mixing variables.
[/quote]
I have meant the following reasoning:
if $ a = 0$ and $ b = c =\sqrt3$ then $ p\geq2+\sqrt3.$
Let $ f(a,b,c) =\frac{3}{1+p}-\frac{1}{a+p}-\frac{1}{b+p}-\frac{1}{c+p}.$
Hence, $ f(a,b,c)-f(a,\sqrt{(a+b)(a+c)}-a,\sqrt{(a+b)(a+c)}-a) =$
$ =\frac{2}{\sqrt{(a+b)(a+c)}-a+p}-\frac{1}{b+p}-\frac{1}{c+p}=$
${ =\frac{2bc+ab+ac-(b+c)\sqrt{(a+b)(a+c)}+p(\sqrt{a+b}-\sqrt{a+c})^{2}}{(\sqrt{(a+b)(a+c)}-a+p})(b+p)(c+p)}=$
${ =\frac{(\sqrt{a+b}-\sqrt{a+c})(b\sqrt{a+b}-c\sqrt{a+c})+p(\sqrt{a+b}-\sqrt{a+c})^{2}}{(\sqrt{(a+b)(a+c)}-a+p})(b+p)(c+p)}\geq0.$
Thus, it remains to check when
$ f(a,b,b)\geq0,$ where $ 2ab+b^{2}= 3$ id est, $ a =\frac{3-b^{2}}{2b}$ and $ 0 < b\leq\sqrt3.$
But $ f(\frac{3-b^{2}}{2b},b,b)\geq0\Leftrightarrow b^{3}-pb^{2}+(2p-3)b+2-p\leq0\Leftrightarrow$
$ \Leftrightarrow(b-1)^{2}(b+2-p)\leq0,$ which true for $ p = 2+\sqrt3.$

Yes, arqady, the best $ p$ is $ 2\plus{}\sqrt 3$.
I solved it using Schur's inequality: $ S^{3}\plus{}9abc\ge 12S$, where $ S\equal{}a\plus{}b\plus{}c$.

Wow Arqady ! Excellent solution ...

[quote="arqady"]
I have meant the following reasoning:...
 [/quote]

Now I understand what have you meant, thank you ...

Thank you,vardanch!
But I am wrong:
[quote="arqady"]
${ \frac{2bc+ab+ac-(b+c)\sqrt{(a+b)(a+c)}+p(\sqrt{a+b}-\sqrt{a+c})^{2}}{(\sqrt{(a+b)(a+c)}-a+p})(b+p)(c+p)}=$
${ =\frac{(\sqrt{a+b}-\sqrt{a+c})(b\sqrt{a+b}-c\sqrt{a+c})+p(\sqrt{a+b}-\sqrt{a+c})^{2}}{(\sqrt{(a+b)(a+c)}-a+p})(b+p)(c+p)}\geq0.$
[/quote]
It must be:${ \frac{2bc+ab+ac-(b+c)\sqrt{(a+b)(a+c)}+p(\sqrt{a+b}-\sqrt{a+c})^{2}}{(\sqrt{(a+b)(a+c)}-a+p})(b+p)(c+p)}=$
${ =\frac{(\sqrt{a+b}-\sqrt{a+c})(c\sqrt{a+b}-b\sqrt{a+c})+p(\sqrt{a+b}-\sqrt{a+c})^{2}}{(\sqrt{(a+b)(a+c)}-a+p})(b+p)(c+p)}.$
But ${ \frac{(\sqrt{a+b}-\sqrt{a+c})(c\sqrt{a+b}-b\sqrt{a+c})+p(\sqrt{a+b}-\sqrt{a+c})^{2}}{(\sqrt{(a+b)(a+c)}-a+p})(b+p)(c+p)}\geq0\Leftrightarrow$
$ \Leftrightarrow(\sqrt{a+b}-\sqrt{a+c})((c+p)^{2}(a+b)-(b+p)^{2}(a+c))\geq0\Leftrightarrow$
$ \Leftrightarrow(\sqrt{a+b}-\sqrt{a+c})(b-c)(p^{2}-2ap-3)\geq0,$
 which true if $ a=\min\{a,b,c\}$ and $ p\geq3.$ Done! :)

Can we prove that $ \frac{3(1\minus{}abc)}{a\plus{}b\plus{}c\minus{}3}\leq 3\plus{}2\sqrt{3}$? If this is possible, we can prove this problem basically. :maybe:


geometry
3D geometry
calculus
integration
number theory
prime numbers

[hide]Since they're prime numbers, we can't combine anything to make a perfect cube. Thus the smallest perfect cube with that as a factor is $p^{3}q^{3}r^{6}$.[/hide][/hide]

[quote="SplashD"]The question asks for the biggest, though it's probably a typo[/quote]

Oops! Yes, of course you are right :) I edited the original post to make sense now.

[quote="Hikaru79"]$p,q,r$ are three distinct primes. What is the smallest positive perfect cube that has $pq^{2}r^{4}$ as a factor?[/quote]

We should be able to assign the smallest primes to $p,q,r$ in the order of each variable's power, ie $(r,q,p)=(2,3,5)$.
Thus $p q^{2}r^{4}= 5 \cdot 3^{2}\cdot 2^{4}= (2^{3})(2 \cdot 3^{2}\cdot 5)$ and the smallest cube, $p^{3}q^{3}r^{6}= \boxed{60^{3}}$.

Yup, this is the same answer I got :)

Here's another question, a little more difficult, but which uses a similar idea.

Let $n$ be the smallest positive integer that is a multiple of $75$ and has exactly $75$ positive integral divisors (including 1 and itself). Find $\frac{n}{75}$

$n = p_{1}^{r_{1}}p_{2}^{r_{2}}\cdots$
$75 = (r_{1}+1)(r_{2}+1) \cdots$

To minimize the exponents, we should use 3, 5, and 5. And to minimize the bases, use primes 2,3, and 5 and take 2 and 3 to the highest powers. That makes $n = 2^{4}*3^{4}*5^{2}$, which is divisible by 75. Divide by 75 to get 432.


Prove that
$ {p \choose 1} \plus{} {p \choose 2} \plus{} ... \plus{} {p \choose k} \equiv 0$  $ (mod$ $ p^{2})$
($ k \equal{} [\frac {2p}{3}]$ and p is a odd prime.  )

sure $ [x]$ means laregest integer on $ x \minus{} 1 < [x] \leq x$

ex) [1]=1 [1.5]=1


Triangle ABC is acute and non-isosceles.
Points M and D on BC are such that AM is a median of the triangle and AD is the internal  bisector of angle A.
If the circle through A , M and D cuts AB at L and AC at N, prove that BL=CN.

[quote="me@home"][img]7048[/img]

Are you sure its right? It doesn't look like that from the image...[/quote]

[b]HINT[/b]

 Use :
a) The bisector theorem for $AD$
b) Powers from points $B,C$ to the circle($BD.BM = BL.BA$ ect). Then divide these (last) two relations and use the first.

Babis

I agree with me@home. The problem as stated is false.

Do you have an idea of what it should really say?

(Is it something silly like mislabeling or such?)

AD is an angle bisector not an altitude.My post has been edited.
I apologise for this mistake.

Now when the problem is corrected it is solved in [b]Sregiu's [/b]post (obviously he knew AD is bisector :lol: )

Ok, this problem is quite easy, but intriguing.

By the problem's hypothesis:

$BM=MC$ (1)

Applying bisectors' theorem on ABC, we have:

$\frac{BD}{DC}=\frac{AB}{AC}$. (2)

By the powers' theorem we also have:

$BD\cdot BM= BL\cdot AB$ (3) as well as $CM\cdot CD=CN\cdot AC$ (4)

Dividing (3) and (4) we get:

$\frac{BD}{DC}\cdot\frac{BC}{CM}=\frac{BL}{CN}\cdot\frac{AB}{AC}\stackrel{(1),(2)}{\implies}\ldots\implies\frac{BL}{CN}=1\implies BL=CN$ and our proof is complete.


integration
calculus
function
complex numbers
complex analysis
complex analysis unsolved

The following identity, adapted from a book, comes from the Schwarz's Theorem
\[ f\left( z \right) \equal{} \frac{1}
{{2\pi i}}\int\limits_{\left| \xi  \right| \equal{} R} {\Re \left\{ {f\left( \xi  \right)} \right\}\frac{{\xi  \plus{} z}}
{{\xi  \minus{} z}}\frac{{d\xi }}
{{i\xi }}}  \plus{} i\Im \left\{ {f\left( 0 \right)} \right\}\]when $ |z| < R$ where $ f$ an entire function. Could you explain a little bit?

What Schwartz theorem you are talking about? The thing is that if I understand your notations correctly then  the statement you posted is wrong (take $ f \equal{} 1$). :wink:

Well, this formula is due to Schwarz. But I think it's usually called the Herglotz representation of $ f$. I don't know what bookworm is asking, though. What do you want to be explained? Note that the integral defines an analytic function and that its real part is the poisson integral of $ \Re f$.

But there should be another coefficient for Herglotz formula (not $ i\xi$ in the denominator) :)


algebra
polynomial
inequalities
algebra proposed

Find all polynomials of degree 3, such that for each $x,y\geq 0$: \[p(x+y)\geq p(x)+p(y)\]

The answer is $p(x)=ax^{3}+bx^{2}+cx+d$
$a>0$ $d\leq \min\{0,\frac{8b^{3}}{243}\}$
and $a=0,b\geq 0$
Becuase $p(x+y)\geq p(x)+p(y)$
is equivlent to
$xy(3a(x+y)+2b)\geq d$
Let$x-\to \infty,y-\to \infty$ we get $a\geq 0$
$x=0$ we get that $d\leq 0$
if $b<0$
Let $x=y=-\frac{2b}{9a}$
then we get that $d\leq \frac{8b^{3}}{243}\}$
and if $a\geq 0$ $d\leq \min\{0,\frac{8b^{3}}{243}\}$
we have ${xy(3a(x+y)+2b)\geq xy(6a\sqrt{xy}+2b}\geq 0\geq d$ when $b\geq 0$
${xy(3a(x+y)+2b)\geq xy(6a\sqrt{xy}+2b}\geq \frac{8b^{3}}{243}\geq d$

[b]Hint:[/b]
If you denote $x+y=S, xy=P$, the problem it is equivalent with $P(3aS+2b)\ge d,\text{ }\forall S,P\ge 0$


[quote="kunny"]Let $x,y$ be real numbers such that $x\geq 0,\ y\geq 0.$ Find the minimum value of  $\frac{4(x^3+y^3)+5}{x+y+1}$ and the value of $x,y$ which give the minimum value.[/quote]
$\frac{4(x^3+y^3)+5}{x+y+1}\geq\frac{(x+y)^3+5}{x+y+1}.$
But $\frac{(x+y)^3+5}{x+y+1}\geq3\Leftrightarrow(x+y)^3+2\geq3(x+y),$ which true by Cauchy.
Hence, the minimum value is $3$ for $x=y=\frac{1}{2}.$

[quote="kunny"]Let $x,y$ be real numbers such that $x\geq 0,\ y\geq 0.$ Find the minimum value of  $\frac{4(x^3+y^3)+5}{x+y+1}$ and the value of $x,y$ which give the minimum value.[/quote]
Generalization
Let $x,y,n,\lambda $ be real numbers such that $x\ge 0,\ y\ge 0,n>1,\lambda >0.$ Find the minimum value of  $\frac{{{2}^{n-1}}({{x}^{n}}+{{y}^{n}})+n-1+n\lambda }{x+y+\lambda }$ and the value of $x,y$ which give the minimum value.


The sum of all but one of the interior angles of a convex polygon equals $2570^\circ$. The remaining angle is

$\textbf{(A)} \ 90^\circ \qquad \textbf{(B)} \ 105^\circ \qquad \textbf{(C)} \ 120^\circ \qquad \textbf{(D)} \ 130^\circ \qquad \textbf{(E)} \ 144^\circ$

[hide="Answer"]Since the polygon is convex, we have $180(n-2)=2570+x$, so $n=17$ and $x=130\Rightarrow \boxed{D}$.[/hide]

Because
$2570+180 > 180(n-2) \geq 2570$ since it is a convex polygon
$15.22 \geq n-2 \geq 14.22$
 n is an integer so n-2=15 therefore n=17


4.08g of a mixture of BaO and an unknown MC03
was heated strongly.The residue weighed 3.64g.This was dissolved in 100mL of 1N HCL. The excess acid required 16mL of 2.5N NaOH solution for complete neutralisation..Identify the metal M..

Plz rate the difficulty level of the problem...

I think for both HCl and NaOH, 1 normal is equivalent to 1 molar.

Metal carbonate decomposes to metal oxide and carbon dioxide.

Dissolved barium oxide will form bairum hydroxide, but it is neutralized by acid so you have barium chloride. (Each mole of BaO will neutralize 2 moles of HCl).

Obviously the amount of excess acid can be computed by computing the amount of NaOH solution, since NaOH neutralizes HCl on a one-to-one basis.

Subtract the moles of excess acid from the total moles of HCl at the beginning and dividing by two gives you the moles of BaO. Using standard molar masses we find out the grams of metal carbonate, say it's x grams.

Finally, the difference between 4.08g and 3.84g is accounted for by one mole of CO2 per mole of metal carbonate. Find the number of moles of CO2 in 0.24 grams, and that is the number of moles of metal carbonate per x grams.

Find the molar mass of metal carbonate. Subtract molar mass of a carbonate ion. This gives you the atomic weight of the metal. Done.

I'm not good in English..
But here's what I've tried..

[hide="My Solution"]
The carbon dioxide mass = $ 4.08 \minus{} 3.64 \equal{} 0.44$ grams.
The moles of carbon dioxide = $ \frac {0.44}{44} \equal{} 0.01$ moles.

$ \text{MCO}_{3(s)} \to \text{MO}_{(s)} \plus{} \text{CO}_{2 (g)}$
So, the moles of $ \text{MO}$ also 0.01 moles.

Moles of HCl = $ 0.1\times 1 \equal{} 0.1$ moles.
Beside that, the moles of excess HCl is the same as NaOH that used to neutralize, that is $ 0.016\times 2.5 \equal{} 0.04$ moles.
Then, the moles of HCl that used in the reaction is $ 0.1 \minus{} 0.04 \equal{} 0.06$ moles.

$ \text{MO}_{(s)} \plus{} 2\text{HCl}_{(aq)}\to \text{MCl}_{2(aq)} \plus{} \text{H}_2\text{O}_{(l)}$
The moles of HCl that used in this reaction is $ 2\times 0.01 \equal{} 0.02$ moles.

$ \text{BaO}_{(s)} \plus{} 2\text{HCl}_{(aq)}\to \text{BaCl}_{2(aq)} \plus{} \text{H}_2\text{O}_{(l)}$
The moles of HCl that used in this reaction is $ 0.06 \minus{} 0.02 \equal{} 0.04$ moles.
So, the moles of BaO : $ \frac {1}{2}\times 0.04 \equal{} 0.02$ moles.
The mass of BaO : $ 0.02 \times (137 \plus{} 16) \equal{} 3.06$ grams.

So, the mass of MO in the residue is $ 3.64 \minus{} 3.06 \equal{} 0.58$ grams.
Also, previously we know that the moles of MO is $ 0.01$ moles.
The molecular mass of MO is equal to $ \frac {0.58}{0.01} \equal{} 58$ grams per moles.
So, the atomic mass of M is $ 58 \minus{} 16 \equal{} 42$.
[/hide]

Do u have other stoichiometry problem that have a same level like this problem?
I need them for my exercise..
thanks so much.  :D


superior algebra
superior algebra unsolved

How to show that $ \bf{Q}_p$ is not algebraically closed for all primes $ p$. For some small primes, I could find an irreducible polynomial.

Is there an all-in-one proof?


If the mesophyll cells in C4 plants have chloroplasts, why don't they release oxygen?

Since mesophyll cells surround the leaf, they must release oxygen, whether it was produced in the bundle-sheath cells as usual, or in the mesophyll cells themselves, as you seem to suggest. The calvin cycle generally takes place in the bundle-sheath cells, though, so that is where the oxygen is produced and must then be released through the mesophyll cells. (I think...)

Since USABO is knocking at the door, let's clear this up. Oxygen is produced as a result of light-dependent reactions (namely the excitement of electrons in the chlorophyll a reaction center of photosystem II). Water is split to replace the electrons, leaving oxygen and a couple of protons. This is completely separate from the Calvin Cycle (in the bundle sheath cells in C4 plants) where carbon dioxide is fixed to produce carbohydrate.


There are 2 non intersecting circles inscribed  in <XOY. Let A be a point on OX. Suppose that B and C are on OY such that AB and AC are tangent to first respectively second circle and AB=AC. Prove that the sum of radi of the circles equals to the altitude from A in triangle ABC.

These are main steps of the solution:
r1/h = BO/(AO+BO+AB), r2/h = CO/(AO+CO-AC)
(AH = h and is perpendicular to BC)
To prove equality it is enough to prove
BO(AO+CO-AC)+CO/(AO+BO+AB) = (AO+CO-AC)(AO+BO+AB)
Using AC=AB and CO = BO+BC after direct opening parenthesis
we will get that that we need to prove:  BO*OC = AO2 - AB2
OB*OC = OH2  BH2 = AO2 - AB2 or AH2+BH2=AB2  q.e.d

Let $X,Y$ be the points where $AB,AC$ touch the respective circles. Let $U,S$ be the points where one of the circles (with radius $r_1$) touches $Ox,Oy$ respectively, and $V,T$ the points where the other circle (with radius $r_2$) touches $Ox,Oy$ respectively. I don't think it matters, but assume $B$ is between $C$ and $S$.

Let $h$ be the $a$-altitude of $ABC$. We have $PB+CQ=(r_1+r_2)\cdot\tan(\frac{\angle B}2)$. On the other hand, we have $PB+CQ=AB-\frac{BC}2=h\cdot\tan(\frac{\angle B}2)$, and from here we get the conclusion.

I just used the fact that AO1=AO2, which can easily be proved by angle chasing, and the rest is quite obvious.


Solve the equation $\cos^{n}{x}-\sin^{n}{x}=1$, where $n$ is a given positive integer.

[hide="Possible methods:"]
1) Euler's substitution.

2) Factoring + Pythagorean Identity (?).

Umm... somebody else think of more.  :P [/hide]

[quote="N.T.TUAN"]What is ''Euler's substitution'' ?  :maybe:[/quote]
It's this:  \[\frac{ e^{i \theta}+e^{-i \theta}}{2}= \cos \theta \]
\[\frac{ e^{i \theta}-e^{-i \theta}}{2i}= \sin \theta \]

Ok! And now, can you post your solution? :D

What's about the similar equation: $\cos^{m}x-\sin^{n}x=1$ where $m,n$ are positive numbers


I don't understand why this is true. Can someone show that it is?

Oh I got it. Please help me with this.

Show that \displaystyle\sum_{k=1}^{n}\((k-1)}=\displaystyle\sum_{k=0}^{n-1}\(k}.

Darn...it's not working.

[quote="airelemental135"] $\displaystyle\sum_{k=1}^{n}\((k-1)}=\displaystyle\sum_{k=0}^{n-1}\(k}.$[/quote]
Just add dollar signs. Also, \displaystyle is redundant here. If you want it not to be displaystyle, use \textstyle for that.

If we write it out, we are trying to show that
$(1-1)+(2-1)+\cdots+(n-1-1)+(n-1)=0+1+\cdots+(n-2)+(n-1)$. These sums match term by term, so it is clear. If you still don't see that, try induction on $n$.

[quote="airelemental135"]What do you mean by "add dollar signs"?[/quote]

He means put "$" characters around the math. So \displaystyle\sum_{k=1}^{n}\((k-1)}=\displaystyle\sum_{k=0}^{n-1}\(k} should be written as:
 $\displaystyle\sum_{k=1}^{n}\((k-1)}=\displaystyle\sum_{k=0}^{n-1}\(k}$
 Then it will work :)

[quote="airelemental135"]I am not so familiar with the use of LaTeX yet. But I am working on it. So just try to draw this out.

Summation Symbol. Index = "(k-1)", upper limit = "n", lower limit = "k=1"

Summation Symbol. Index = "k", upper limit = "n-1", lower limit = "k=0"

Show that these two are equal.[/quote]
Prove: $\sum_{k=1}^n(k-1)=\sum_{k=0}^{n-1}k$
$\sum_{k=1}^n(k-1)=\frac{n(n-1)}{2}$
$\sum_{k=0}^{n-1}k=\frac{0+(n-1)}{2}\cdot n=\frac{n(n-1)}{2}$

And the conclusion follows.

Masoud Zargar

That is not a particularly good way of proving it, however, since there are many sums which lack a closed form, but for which we can shift the summation as seen there.

[quote="airelemental135"]I am not so familiar with the use of LaTeX yet. But I am working on it. So just try to draw this out.

Summation Symbol. Index = "(k-1)", upper limit = "n", lower limit = "k=1"

Summation Symbol. Index = "k", upper limit = "n-1", lower limit = "k=0"

Show that these two are equal.[/quote]

$\sum_{k=1}^{n} (k-1) = \left( \sum_{k=1}^{n} k \right) - n$

Note that if we expand the term in parentheses, it is $1 + 2 + 3 + ... + (n-1) + n$, and subtracting $n$ we have $1 + 2 + 3 + ... + (n-1) = \sum_{k=0}^{n-1} k$.

Edit:  Really, there is no need for summation formulae here  :)

For teh first one, I just used Guass' summation formula on (n-1), since it began with 1.  For the second, I used teh formula $\frac{a+b}{2}\cdot n$, where a is the first term, b is teh lst term, and n is the number of terms.


geometry
circumcircle
geometric transformation
power of a point
radical axis
geometry proposed

Let $ K$ be the symmedian point of $ \triangle ABC$ and $ \triangle K_aK_bK_c$ the cevian triangle of $ K.$ The circle passing through $ A$ tangent to $ BC$ at $ K_a$ cuts the circumcircle $ (O)$ again at $ A'.$ Define cyclically $ B',C'$ and the following points: $ X \equiv CC' \cap BB',$ $ Y \equiv AA' \cap CC' $ and $Z \equiv AA' \cap BB'.$ Prove that $ AX,BY, CZ$ and the Brocard axis $ OK$ concur.

I don't know a synthetic solution of this problem, but it can be generalized substituting the symmedian point $K$ by any other point $P=(u:v:w)$. We'll get that the lines $AX$, $BY$ and $CZ$ always concur in the barycentric square $Q=(u^2:v^2:w^2)$ of $P$.

The locus of points $P$ such that $O$, $P$ and $Q$ are collinear is the cubic [url=http://bernard.gibert.pagesperso-orange.fr/Exemples/k168.html]K168[/url], through the symmedian point.

Thanks Francisco, I was already aware of the general configuration of cevians $AX,BY,CZ$ intersecting at the barycentric square of $P \ (\star).$ But I'm not able to prove synthetically that the barycentric square of the symmedian point (third power point), defined as the trilinear pole of the line passing through the exsimilicenters of the pairs of circles with  diameters $(BK_a, CK_a),$ $(CK_b,AK_b)$ and $(AK_c,BK_c),$ lies on the Brocard axis. Nevertheless, we can proof $(\star)$ prescinding from calculations with barycentrics.

Replace symmedian point $K$ by an arbitrary point $P$ and let $\triangle P_aP_bP_c$ be the cevian triangle of $P,$ the remaining notations are the same. Let lines $AA',BB'$ and $CC'$ cut $BC,CA$ and $AB$ at $M,N,L,$ respectively. $AM$ is radical axis of $(O)$ and the circle passing through $A$ and tangent to $BC$ at $P_a$ $\Longrightarrow$ $MA \cdot MA'=MB \cdot MC={MP_a}^2.$ Hence, inversion with center $M$ and radius $MP_a$ transforms the circles with diameters $BP_a$ and $CP_a$ into each other $\Longrightarrow$ $M$ is their exsimilicenter. Likewise, we have that $N,L$ are the exsimilicenters of the circles with diameters $CP_b,AP_b$ and $AP_c,BP_c$ $\Longrightarrow$ $MNL$ is the trilinear polar of the barycentric square $Q$ of $P.$ But since the pencil $A(L,N,X,M)$ is harmonic, it follows that $AX$ passes through $Q.$ Similarly, lines $BY,CZ$ pass through $Q.$


Let $f(x)$ has f"(x) is continous in $\mathbb R$ and satisfying:$f(0)=f(1)=a$.Prove that:
\[\max_{x\in [0,1]}\{f\"(x)\} \ge 8(a-b),b=\min_{x\in [0,1]}\{f(x)\}\]

first  I think it should be $\max_{x\in [0,1]}\{|f\"(x)|\} \ge 8(a-b),b=\min_{x\in [0,1]}\{f(x)\} $
if not $f(x)=x(1-x)+a$ will be a counterexample.
assume $f(x_0)=b$,$x_0=0,1$will be easy.
if $x_0 \in (0,1)$ we have $f'(x_0)=0$
use taylor formal in $x_0$ to $0$ and $1$
we have $a=b+f\"(c_1) \frac{x_0^2}{2}$
and $a=b+f\"(c_2) \frac{(1-x_0)^2}{2}$
$c_1 \in (0,x_0),c_2 \in (x_0,1)$
dispart with $x_0 \leq \frac{1}{2}$ and  $x_0 \geq \frac{1}{2}$
we are done :)

If we replaced $[0,1]$ by $[\alpha,\beta]$ ,we have:
\[\max_{x\in [\alpha,\beta]}\{f\"(x)\} \ge \frac {8(a-b)}{(\alpha - \beta)^2}\]


I have three questions.

1- I'm beginner in latex, so how can I download packages? (for windows xp users).
2- How can I write my own packages?
3- How can I draw picture like this
$ \setlength{\unitlength}{.5cm}\begin{picture}(2,2) \qbezier(2,1)(2,2)(1,2) \qbezier(1,2)(0,2)(0,1) \qbezier(0,1)(0,0)(1,0) \qbezier(1,0)(2,0)(2,1) \end{picture}$
or this
$ \setlength{\unitlength}{.5cm}\begin{picture}(5,4) \put(.5,.5){\line(5,0){5}} \put(.5,.5){\line(0,3){3}} \put(.5,3.5){\line(5,\minus{}3){5}} \put(1,.5){\line(0,1){0.5}} \put(.5,1){\line(1,0){0.5}} \put(\minus{}.1,3.5){A} \put(\minus{}.1,0){B} \put(5.5,0){C} \put(2.75,0){a} \put(3,2.1){b}  \put(0,1.75){c} \end{picture}$
??

[quote="stevem"]1. If you are using MiKTeX then use the Package Manager (MiKTeX, Browse Packages). If you are using a different system then you must say what it is.
2. Forget it, this is not for beginners.
3. Try [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_%28Vector_Graphics_Language%29]Asymptote[/url] which is easier to program and can be used in your tex documents. There is an [url=http://www.artofproblemsolving.com/Forum/index.php?f=519]AoPS Asymptote forum[/url] for help with producing diagrams.[/quote]
Thank you, stevem.

Hi mathson,

1. most important: [url=http://ctan.org]CTAN[/url],
2. have a look at [url=http://www.ctan.org/tex-archive/macros/latex/doc/clsguide.pdf]clsguide[/url] or at [url=http://www.ctan.org/tex-archive/help/Catalogue/bytopic.html#latexdocu]LaTeX Documentation by the LaTeX3 Project[/url] in the TeX Catalogue.
3. PSTricks would be an alternative, and TikZ. I'm preferring TikZ. Have a look at the [url=http://www.texample.net/tikz/examples/area/geometry/]geometry category[/url] and at others on texample.net. Further you could use external tools like Xfig or [url=http://texblog.net/latex-archive/graphics/jpgfdraw/]jpgfdraw[/url] ([url=http://texblog.net/latex-archive/maths/jpgfdraw-example/]example[/url]).

Stefan

That's a good example of the big advantage of using Asymptote - there is a wealth of expertise here ready and willing to help :)

Hi, thanks u all.  :blush:
But I cant download GSview 4.8. The link doesn't work!!  :?:

This link to [url=http://pages.cs.wisc.edu/~ghost/gsview/get49.htm]GSVIew 4.9[/url] works.


ABCDE is a regular pentagon,F is the mirror image of A across the line BE.Prove or disprove that the nonconvex pentagon BCDEF can tile the plane.


limit
algebra
polynomial
calculus
calculus computations

Evaluate \[ \lim_{n \rightarrow \infty} \frac{1^k + 2^k + \cdots + n^k}{n^k} - \frac{n}{k+1}  \]
  ;)

Yup  :D  Strange, You just shown the links to Kunny about Stolz Theorem, and this limit is just a particular case of the formula that Grobber metioned there. :rotfl:

Did you prove it with an elementary way?
 :roll:

No,I showed to kunny is not about the Stolz Theorem.There is another solution for his problem.
I agree with that your problem is a particular case of the formula that Grobber metioned there.  :)
And I think $k>0$ in your problem :?:
Anyway,I also use the Stolz Theorem to solve this one,I wonder if it is an elementary way?

Yes $k>0$.
I thinks i solved it too using Stolz theorem. But when i first see this problem, my solution use Bernoulli Polynomial/ Numbers (Maybe you can guess what?).  is it elementary or not ?
But this only when $k$ is integer.


We have 12 coins. One of them is not good (It may be heavier or lighter than the others). You have three times to use the scale to find it.

           \_____/            \_____/
                l______________l
                           l
             _________l___________

I hope you understand my scale.

Use code tags to do pictures that need spacing. :wink:

[hide="Hint"]Start by weighing 4 coins against 4 coins.[/hide]

sorry I don't know how to do use code tags.
and what are the next steps after that?
Remember we don't that the bad one is heavier or lighter than the others.
It's not easy.


function
analytic geometry
conics
parabola
calculus
calculus computations

Find the envelope of normals of $ y^2\equal{}4ax$.

Normal : $ F(x, y, t) = \pm \sqrt{\frac{t}{a}}(x-t) - y \mp 2\sqrt{at} = 0$.

Evolute : $ \{\begin{array}{l}F(x, y, t)=0 \\ F_t(x, y, t)=0\end{array} \iff 27ay^2=4(x-2a)^3$.

[quote="kan"]Can u explain me more.[/quote]

$ \{\begin{array}{l}F(x, y, t)=0 \\ F_t(x, y, t)=0\end{array} \iff \{\begin{array}{l}\pm \sqrt{\frac{t}{a}}(x-t) - y \mp 2\sqrt{at} = 0 \\ (\mp 2a \mp 3t \pm x)/(2\sqrt{at}) = 0\end{array}$

The envelope of normals is also known as the [url=http://mathworld.wolfram.com/Evolute.html]evolute[/url]. You can show that the two are equivalent by rigorous algebra then applying the definitions of first and second (maybe third? its been a while) derivatives. This isn't an easy subject, but its also not too heinous. You have to manage your variables correctly and keep in mind what you're trying to accomplish--there are a lot of steps. if you choose to find the locus as the envelope. Parametrize the equation then find a way to write the normal line at y=t as a function of t, so that you are left with the equation of a line in strictly x and y (for some constant t). Now find the coordinates of intersection of any given normal line at t with its neighbor at t+h for some real h. Take the limit as h approaches 0 and you will be left with the parametric equations for the rectangular eq's evolute. Kill t through substitution to get the rectangular form, then square some stuff to make it look pretty. Then you're done.

The evolute of a parabola is known as a [url=http://mathworld.wolfram.com/SemicubicalParabola.html]semicubical parabola[/url]. This curve is well documented so you should have no trouble filling in gaps using those links.  :)


trigonometry
function
calculus
integration
LaTeX
calculus computations

Determin the value of
\[-2e^{3}\sum_{n=1}^{\infty}\frac{}{}\cos{\left(\frac{9}{n\pi+\sqrt{n^{2}\pi^{2}-9}}\right)}\]
answer[hide] is $\frac{\pi^{2}}{6}$.[/hide]

[hide="A start..."] $\frac{9}{n \pi+\sqrt{n^{2}\pi^{2}-9}}= n \pi-\sqrt{n^{2}\pi^{2}-9}$

$\cos \left( n \pi-\sqrt{n^{2}\pi^{2}-9}\right) = \cos n \pi \cos \sqrt{n^{2}\pi^{2}-9}$

The sine term is identically zero, and $\cos n \pi = (-1)^{n}$, so our summation is

$2e^{3}\sum_{n=1}^{\infty}(-1)^{n-1}\cos \sqrt{n^{2}\pi^{2}-9}$ [/hide]

Can somebody find a period function satisfing the Dirichlet conditions such that is Fourier series is the Half-Range Cosine Series who coefficients inside are $\sqrt{\pi^{2}n^{2}-9}$ then to find the limit of summartion we evaluate the function at 1.

Since the answer does not contain $e$ maybe the function is something $e^{x}$ mutliplied be something else. Not sure.

It is very tempting to apply the residue theorem to something like $\frac{\cos\sqrt{z^{2}-9}}{\sin z}$ but the corresponding function does not decay at $\infty$, so one will have to be very careful. Also, some correction to the statement of the problem is necessary because, as written, the series, clearly, diverges (the general term tends to $1$, not to $0$ :P)

EDIT: I believe that the problem, as stated, is a very distorted version of
\[\sum_{n\ge 1}\left(\cos\left(\frac{1}{\pi n+\sqrt{\pi^{2}n^{2}-9}}\right)-1\right)=-\frac{e^{3}+e^{-3}-11}{4}\]

i don't know if the original problem stated is [b]correct [/b]in the first place,
but i am able to show (takes almost 3+ pages), using fredholm integral equation:

$\;-2\;\sum_{n=1}^{\infty}\;\frac{1}{n^{2}}\;\cos \left( \frac{9}{n\pi\;+\;\sqrt{n^{2}\pi^{2}-9}}\right) \;=\;\frac{\pi^{2}}{6}\cdot \frac{1}{e^{3}}$

r u sure you are not missing $\frac{1}{n^{2}}$ within the $\sum_{n=1}^{\infty}$ ?

sorry, but i just don't want to type out 3 pages worth of stuff if you are
not familiar with volterra/fredholm equations... i am sure there are
easier ways of finding the sum, but i can't seem to think of one other
than integral equations...  :rotfl:

Yes yore right! Sorry  :oops: .

[color=red]THE PROBLEM SHOULD HAVE BEEN:[/color]
\[-2e^{3}\sum_{n=1}^{\infty}\frac{1}{n^{2}}\cos{\left(\frac{9}{n\pi+\sqrt{n^{2}\pi^{2}-9}}\right)}\]

actually if you look at the $\LaTeX$-code I've put an \frac{}{}, but forgit to fill it  :blush: .

That makes it easy: as I said, use the residue theorem for $f(z)=\frac 1{z^{2}}\frac{\cos\sqrt{z^{2}-9}}{\sin z}$. The sum you need is the sum of all residues except that at $0$ (see t0rajir0u's post). It remains to compute the residue at $0$, which is a bit unpleasant but completely doable.


a nice one:
Let E a normed linear space of finite dimension. Let f an endomorphism of E and K a convex compact set of E. Prove that if f(K) is included in K then there is at least a fixed point of f in K.(the use of Brouwer theorem is not allowed  :D )

Just some ideas

f is continous, f is Lipchitzian
Compact probably Bolzano-Weierstrass
$F_n(x)=l_nf(a)+(1-l_n)f(x)$

Am I in good direction alekk ?

The sequence $x,Tx,T^2x,\ldots$ has a convergent subsequence ($Tx$ is the image of $x$, and $T^n$ means $T$ composed with itself $n$ times). This means that, if $\varepsilon>0$ is some constant, we can find $m,k$ s.t. $\|T^{m+k}x-T^mx\|<\varepsilon\ (*)$. Now put $x_{mk}=\frac{T^mx+T^{m+1}x+\ldots+T^{m+k-1}x}k$. From $(*)$ we get $\|Tx_{mk}-x_{mk}\|<\varepsilon$.

Since $K$ is compact, $\|Tx-x\|$ reaches its minimum on $K$, and since the previous paragraph shows that it can be as small as we want, it must be $0$ in some point of $K$.

What makes me doubt this is the fact that I don't seem to be using the finite dimension of $E$. Did I make any stupid mistakes? :?


induction
combinatorics unsolved
combinatorics

Several thin unit cardboard squares are put on a rectangular table with sides parallel to the sides of the table. The squares may overlap. Each square is colored with one of k colors. Given any k squares of different colors, we can find two that overlap. Show that for one of the colors we can nail all the squares of that color to the table with 2k-2 nails.

This one is a nice problem:

The idea is induction on the number of colors.

Pick the leftmost square, called it $a$, and  say it is of color 1. Let $A$ be the set of squares that overlap with $a$. Now lets ignore the squares in $A$ and $a$, and the rest of squares of color 1. For every $k-1$ squares among the remaining, there are two that overlap, because if there weren't, considering those $k-1$ plus square $a$ we would obtain a contradiction to the hipothesis. So we can apply inductive step here, so we get that for some color, all squarers in the remaining set can be pierced by $2k-4$ nails. But what about squares of this color in $A$?, easy!, we can pierce them with two nails in the right corners of $a$ so we use at most $2k-2$ nails to pierce all that color's squares and that's it.

For 2 colors, we do this: Pick the leftmost square, say its color is 1 and call it $a$, then $a$ intercepts every square of color 2, by the hipotesis, so we can nail everyone of color 2 with two nails in the riaght corners of $a$.

It has being a long time since i didnt post here, so I guess I am back!


An acute angle of a right triangle is 30°, and the hypotenuse is 40 units. What is
the area of the triangle? Express your answer as a decimal to the nearest tenth.

Ok, I know you have to use that sin., cos., tan. thing in this but I don't understand how.

No, use 30-60-90 triangle ratios.

The hypotenuse is 40 units so the shorter leg is 20 units, and the longer leg is $ 20\sqrt{3}$ units.  The area is approximately 346.4

If the side opposite the 30 degree angle is x

the side opposite 60 degree angle is $ x\sqrt{3}$

side opposite 90 degree angle (the hypotenuse) is $ 2x$.

Derive this using an equilateral triangle.

[quote="lifeisacircle"][quote="007math"]A more visual picture:

30-60-90:

[asy]pair A=(0,0), B=(10,0), C=(10,17.32);

draw(A--B--C--cycle);
label("$x$", A--B, S);
label("$x\sqrt{3}$", B--C, E);
label("$2x$", A--C, NW);
draw((8,0)--(8,2)--(10,2));[/asy][/quote]

Correction[/quote]

gah....thanks


always do the STR and MR for studying, as they cover general areas. although, if you're having trouble in a certain area go to guided practices and practices and ask as many questions as possible! math teachers want to help you :P - we hope

but seriously, try the above

-jorian

thanks for the info! btw...every week we have a chapter test..how should i study for that?

At the end of each chapter there is a "Chapter Summary and Review" , "Chapter Test" (50 percent chance its the test our teacher assigns), and "Maintaining Skills"

Also there are sections in each lessons..there are "Guided Practice", "Practice and Applications", "Standardized Test Practice", "Mixed Review", "Maintaining Skills", and a Quiz for the lesson...

I know you suggested STP and MR, if u want to suggest i should study then just reply

[hide="My Way Of Study"]I think that you should study everything, then when your done, you can just ask us aopsers for more questions![/hide]

[quote="jdmae86"]I have a book called McDougall Littell's Algebra 1 Concepts and Skills. In each chapter, the chapter has a "Chapter Summary and Review", and a "Chapter Test", sometimes my teacher doesn't use "Chapter Test" in the book and last "Maintaining Skills"

In each lesson, there is "Guided Practice" and "Practice", my class usually does this for homework. Then "Standardized Test Practice", and "Mixed Review".

If some1 can help me make a study planner for each lesson or chapter[/quote]

Well, I look at the lessons ahead and try to understand the concept. Then, to test if I truly get it, I would do the odd numbers pertaining to the stuff you learned (because the odd numbers show up on the answer keys in my math book). Now to study what you already know, just do some odd problems you've never seen before (and if you want, some evens but sometimes checking your answers (officially) might be hard) from that lesson. I never study for tests unless it is the year final.

Just use it to your advantage and absorb the material covered if you don't know it yet.  :lol:

(number of lessons)/(number of days) is my formula. Do the tests last.


Let $ a,b,c$ are positive real numbers. Find the best constant $ k$ that satisfy: $ a^{2}\plus{}b^{2}\plus{}c^{2}\leq k(ab\plus{}bc\plus{}ca)$ and there is an $ (a,b,c)$ make equality hold.

There isn't exist such constant $ k$ because,if we fix $ a$ and $ b\equal{}c\rightarrow0$ then $ k\geq\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{ab\plus{}bc\plus{}ca}\rightarrow\infty$

Do you mean $ (a,b,c)$ is length of a triagle.

OK, I see there is no constant for $ (a,b,c)$ are positive real numbers. So, let's $ (a,b,c)$ be the 3 length sides of a triangle. with this, is there any constant $ k$????

$ k \equal{} 2$  :wink:

Try $ a \equal{} 2,b \equal{} n,c \equal{} n\plus{}1$, then $ k >\frac{2n^{2}\plus{}2n\plus{}5}{n^{2}\plus{}5n\plus{}1}\equal{} 2\minus{}\frac{8n\minus{}1}{n^{2}\plus{}5n\plus{}2}$.

Let $ n\to\infty$ we have $ \frac{8n\minus{}1}{n^{2}\plus{}5n\plus{}2}\to 0$ then $ k \equal{} 2$ is best constant so that the condition . :)

Yes, I see. For $ k\equal{}2$, the equality can not hold. So that there is no constant satisfy the condition >.<