inequalities
induction
function
Cauchy Inequality
combinatorial geometry

Let a,b,c,d,e be positive reall numbers, show that:

$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+e}+\frac{d}{e+a}+\frac{e}{a+b}\ge \frac{5}{2}$

I've been trying to solve this problem for quite a while now...

At first, I thought I had to use Chebyshev's sum inequality:
http://en.wikipedia.org/wiki/Chebyshev's_sum_inequality

However, that dosnt seem to work.

I let:

$a \geq b \geq c \geq d \geq e$

Then:

$a+b \geq a+e \geq b+c \geq c+d \geq d+e$

Or:

$\frac{1}{d+e}\geq \frac{1}{c+d}\geq \frac{1}{b+c}\geq \frac{1}{a+e}\geq \frac{1}{a+b}$

But this order dosnt match up to what the hypotheis of Chebyshev's sum inequality...

Where am I going wrong?  :(

Well when you try to use Chebyshev the inequality $a+e \ge b+c$ does not necessarily hold, for example $10+1 \not\ge 9+8$. And I don't think Chebyshev is the best way to go for this inequality. I'm thinking

[hide]Cauchy.[/hide]

[hide="Solution?"]Cauchy-ize: then we can show
\[\frac{(a+b+c+d+e)^{2}}{\sum_{sym}ab}\ge \frac{5}{2}\]
Do some cancelling:
\[2(a^{2}+b^{2}+c^{2}+d^{2}+e^{2})\ge \sum_{sym}ab \]
Which is Rearrangement I think.

Equality at $a=b=c=d=e$?[/hide]

Or final step can go by applying am-gm to all ten possible pairs as follows; $a^{2}+b^{2}\ge 2ab$ then symmetric sum all those inequalities, cancel a 2 and you're done. I don't think rearrangement works because the symmetric combination on the right is trickier than the form on the left.

[quote="ch1n353ch3s54a1l"]
Do some cancelling:
\[2(a^{2}+b^{2}+c^{2}+d^{2}+e^{2})\ge \sum_{sym}ab \]
[/quote]
I think, $\sum_{sym}ab=12(ab+ac+ad+ae+bc+bd+be+cd+ce+de).$ :wink:

Really? Isn't $\sum_{cyc}ab=ab+bc+cd+de+ea$? Which isn't right either...

And I thought the symmetric sum was all the combinations of $ab$ with $a,b,c,d,e$?? Where did the 12 come from...*confused*

[quote="me@home"]Nice catch... just to clarify, $\sum_{sym}$ refers to every possible combination, which isn't at all what you meant to say! (since there are 5 terms, this means that the sum of the coefficients must account for $5!=120$ terms which is a huge overcount).

The proper notation is $\sum_{cyc}$ referring to the cyclic sum which is really what you meant to say[/quote]

Don't you mean $\binom{5}{2}=10$ instead of $5!$

Could you show me how did you apply Cauchy? I would be grateful for that :)

[quote="Je_nie_dieu"]Could you show me how did you apply Cauchy? [/quote]
$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+e}+\frac{d}{e+a}+\frac{e}{a+b}=$
$=\frac{a^{2}}{ab+ac}+\frac{b^{2}}{bc+bd}+\frac{c^{2}}{cd+ce}+\frac{d^{2}}{de+da}+\frac{e^{2}}{ae+be}\geq$
$\geq\frac{(a+b+c+d+e)^{2}}{ab+ac+ad+ae+bc+bd+bd+cd+ce+de}.$
But $\frac{(a+b+c+d+e)^{2}}{ab+ac+ad+ae+bc+bd+be+cd+ce+de}\geq\frac{5}{2}\Leftrightarrow$
$\Leftrightarrow(a-b)^{2}+(a-c)^{2}+(a-d)^{2}+(a-e)^{2}+(b-c)^{2}+$
$+(b-d)^{2}+(b-e)^{2}+(c-d)^{2}+(c-e)^{2}+(d-e)^{2}\geq0.$

For a generalization, take a look at Shapiro's Inequality.

Arqady, thank you! Well this is just beautiful.

[quote="arqady"][quote="Je_nie_dieu"]Could you show me how did you apply Cauchy? [/quote]
$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+e}+\frac{d}{e+a}+\frac{e}{a+b}=$
$=\frac{a^{2}}{ab+ac}+\frac{b^{2}}{bc+bd}+\frac{c^{2}}{cd+ce}+\frac{d^{2}}{de+da}+\frac{e^{2}}{ae+be}$
$\geq\frac{(a+b+c+d+e)^{2}}{ab+ac+ad+ae+bc+bd+bd+cd+ce+de}.$
But $\frac{(a+b+c+d+e)^{2}}{ab+ac+ad+ae+bc+bd+be+cd+ce+de}\geq\frac{5}{2}\Leftrightarrow$
$\Leftrightarrow(a-b)^{2}+(a-c)^{2}+(a-d)^{2}+(a-e)^{2}+(b-c)^{2}+$
$+(b-d)^{2}+(b-e)^{2}+(c-d)^{2}+(c-e)^{2}+(d-e)^{2}\geq0.$[/quote]

Sorry...i am not understanding this..can we simplyfy this further please?

I am coufused at:

$=\frac{a^{2}}{ab+ac}+\frac{b^{2}}{bc+bd}+\frac{c^{2}}{cd+ce}+\frac{d^{2}}{de+da}+\frac{e^{2}}{ae+be}$
$\geq\frac{(a+b+c+d+e)^{2}}{ab+ac+ad+ae+bc+bd+bd+cd+ce+de}.$

How did we arrive at this?
How do we know that the first part is greater than or equal to the second part?

Also, how do we know that this is true:

$\frac{(a+b+c+d+e)^{2}}{ab+ac+ad+ae+bc+bd+be+cd+ce+de}\geq\frac{5}{2}\Leftrightarrow$

The way I see it, if 5>3, and 5>x, x>3 is not nessissarily true.

Could we simplify this please?

[quote="arqady"][quote="problem_solver"]

$=\frac{a^{2}}{ab+ac}+\frac{b^{2}}{bc+bd}+\frac{c^{2}}{cd+ce}+\frac{d^{2}}{de+da}+\frac{e^{2}}{ae+be}$
$\geq\frac{(a+b+c+d+e)^{2}}{ab+ac+ad+ae+bc+bd+bd+cd+ce+de}.$

How did we arrive at this?[/quote]
I applied the Cauchy's inequality in Engel form:
Let $a_{1},$ $a_{2},...,$ $a_{n}$ are real numbers and $b_{1},$ $b_{2},...,$ $b_{n}$ are positive numbers. Then
$\frac{a_{1}^{2}}{b_{1}}+\frac{a_{2}^{2}}{b_{2}}+...+\frac{a_{n}^{2}}{b_{n}}\geq\frac{(a_{1}+a_{2}+...+a_{n})^{2}}{b_{1}+b_{2}+...+b_{n}}.$
It's very useful inequality.

[quote="problem_solver"]
Also, how do we know that this is true:

$\frac{(a+b+c+d+e)^{2}}{ab+ac+ad+ae+bc+bd+be+cd+ce+de}\geq\frac{5}{2}\Leftrightarrow$
[/quote]
Since, $\frac{(a+b+c+d+e)^{2}}{ab+ac+ad+ae+bc+bd+be+cd+ce+de}\geq\frac{5}{2}\Leftrightarrow (a-b)^{2}+(a-c)^{2}+$
$+(a-d)^{2}+(a-e)^{2}+(b-c)^{2}++(b-d)^{2}+(b-e)^{2}+(c-d)^{2}+(c-e)^{2}+(d-e)^{2}\geq0.$
[quote="problem_solver"]
The way I see it, if 5>3, and 5>x, x>3 is not nessissarily true.
[/quote]
You are right, it not necessarily true. But if $5>x$ and $x>3$ then $5>3$ is true. :)[/quote]

Thank you for explaining the Cauchy inequality part - I appreciate that.

However, I still do not quite understand the other part...

Since, $\frac{(a+b+c+d+e)^{2}}{ab+ac+ad+ae+bc+bd+be+cd+ce+de}\geq\frac{5}{2}\Leftrightarrow (a-b)^{2}+(a-c)^{2}+(a-d)^{2}+(a-e)^{2}+(b-c)^{2}+(b-d)^{2}+(b-e)^{2}+(c-d)^{2}+(c-e)^{2}+(d-e)^{2}\geq0.$

Where does this part come from:

$(a-b)^{2}+(a-c)^{2}+(a-d)^{2}+(a-e)^{2}+(b-c)^{2}++(b-d)^{2}+(b-e)^{2}+(c-d)^{2}+(c-e)^{2}+(d-e)^{2}\geq0.$

?

Squares are always positive.
Also, almost every inequality of this form can be reduced to a sum of squares... and arqady is a master at it!
To show the two sides are equal, notice that for each square you get two squared terms and a single mixed term of coefficient two. Thus when you add them all up, this balances out so each type of term has an equal coefficient.

By the way, the inequality which states that for real $a_{i}$ and positive real $x_{i}$ we have

$ \sum_{i=1}^{n}\frac{a_{i}^{2}}{x_{i}}\ge \frac{\left(\sum_{j=1}^{n}a_{i}\right)^{2}}{\sum_{k=1}^{n}x_{i}}$

is commonly known as Titu's Lemma.  It can be proved by induction or by using the Bunyikowski-Cauchy-Schwarz Inequality after multiplying through by $\sum_{k=1}^{n}x_{i}$.

As far as I know, it is [i]commonly[/i] known as Cauchy in Engel Form.

I should have edited this in, but whatever...

Inequalitities of this form (actually a slightly modified form, but whatever - I think they're equivalent) are known as Shapiro cyclic sum inequalities, after the Ashkenazi Jewish mathematician Harold Shapiro.  Vladimir Drinfeljd of the USSR did pioneering work on establishing the best constant $k$ such that the cyclic sum is greater than $kn$ for all $n$.  As it turns out, $k = \frac{1}{2}$ holds only for even $n \le 14$ and odd $n \le 23$ (approximately - I'm not completely sure).  Drinfeljd's work showed that the optimal constant was $\frac{1}{2}\chi(0)$, where $\chi$ was the function convex hull of $f_{1}(x) = e^{-x}$ and $f_{2}(x) = \frac{2}{e^{x}+e^{\frac{x}{2}}}$.  The constant's numerical value is approximately 0.4945668...

Oh - by the way, Drinfeljd won a Fields Medal for this work.


Sorry.
I really jumped to conclusions.

Please think my solution to be a value of a case of $f(1)=f(2)=\cdots=f(9)=0, f(10)=3628800$.


[quote="korammath"]Early, regular, or late? which should i do?[/quote]
Probably early, because it's cheaper if you get in to go.


Vise nisam aktivni takmicar, ali jos uvijek ne mogu bez takmicenja. Tako sam ove godine (bez ikakvih problema), dosao u posjed zadataka sa MMC prije nego sto se pomenuto takmicenje odrzalo u BiH i Hrvatskoj.

Svjestan sam ove pojave vec otprije. No, ove godine mi se cini da je tamicenje izgubilo svaki smisao. Korektan takmicar nece ni pogledati zadatke prije takmicenja, znam, ali draz takmicenja ubijaju upravo oni koji ce postupiti drugacije.

Ukinuti, ili reorganizovati.

Sta ostatak citateljstva misli o ovome?

Vidio sam (prije nego sto sam uopce vidio i ovaj post  :lol: ). I mogu ti reci da se slazem... Prosle godine npr. zadaci sa MMCa su se mogli naci na internetu (to sam, srecom otkrio tek poslije MMCa u BiH).

Zadaci su kako svi znamo bili izmijenjeni (u odnosu na ove objavljene na netu) a evo i (meni poznatih rezultata) za BiH

Bronzane medalje:  Admir Besirević, Zlatan Tucaković

Srebrne medalje: Haris Čausević i Vedran Karahodžić

Zlatna medalja: Salem Malikić :)

Za ostale ne znam mada pretopostavljam da ima nekoliko pohvala zbog one nejednakosti.

Kod nas je jedini riješen zadatak bila nejednakost, pa uglavnom je bilo tko ju je riješio dobio je nagradu.

Rezultati su bili drugačiji nego oni s IMO-a, iako sam ja i tada imao bod više od Žužića (on je imao 0 a ja 1 bod :D).

Saleme što si ti bio riješio?

mislim da je salem imao bodove 7-0-2-7, haris 7-0-1-7, ja 5-0-2-7, a ovi bronzani su imali po 9 bodova a pošto je negdje bilo i po 8 a nije bronza zaključujem da je cutoff za bronzu 9 a za zlato 16 (jer je slalem imao 16 a haris 15).....završavam mediteransku karijeru s 2 bronze i 1 srebrom  :D

BTW Da li je datum odrzavanja MMC strogo fiksiran i ko ga odredjuje?
Jer u zadnje nekolike godine, MMC se poklapa sa Velikom subotom, zbog cega ja nikad nisam ni ucestvovala na MMC.
Iako se i ja takodje pitam, koji je smisao MMC

Nije fixiran i određuje ga koordinator MMC-a za tu državu u ovom slučaju prof.dr. Šefket Arslanagić (molim one koji čitaju ovo da ustanu  :D  :D ) a koliko se ja sjećam kad sam bio prvi i drugi razred nije se poklapao s Velikom Subotom (posebno sam siguran kad sam drugi bio jer je tad Igor došao)....jednostavno treba neko doktoru u martu javiti kad je Velika Subota i neće biti problema ja mislim ...

[quote="delegat"]Zadaci su kako svi znamo bili izmijenjeni (u odnosu na ove objavljene na netu) a evo i (meni poznatih rezultata) za BiH

Bronzane medalje:  Admir Besirević, [i][b]Zlatan Tucaković[/b][/i]

Srebrne medalje: Haris Čausević i Vedran Karahodžić

Zlatna medalja: Salem Malikić :)

Za ostale ne znam mada pretopostavljam da ima nekoliko pohvala zbog one nejednakosti.[/quote]

Jel ovo sve 100% sigurno?

Mislio sam da cu uzeti pohvalu :D


ratio
SFFT
number theory
relatively prime
geometry
similar triangles

To find other problems and description, look here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=187874

17. In square ABCD, AD is 4 centimeters, and M is the midpoint of CD. Two lines are created in the square, BM, and AC with O being the intersection point. What is the ratio of OC to OA? Express your answer as a common fraction.

SOLVED

[hide] I have no clue how to do this but the answer key got 1/2[/hide]

19. The product of two positive integers minus their sum is 39. The integers are relatively prime, and each is less than 20. What is the sum of the two integers?

SOLVED

[hide] I got this by guess and check and I need a faster solution. The answer is 16.[/hide]

28. Given f(x)=1/(1-(1/(1-(1/(1-x))))), compute (f(f(-2)))^-2. Express your answer as a common fraction.

SOLVED
[hide]I got -1/4 as the answer to this but the answer key says 1/4.[/hide]

That's it. Thanks in advance.

[quote="xpmath"][hide="28"] For #28, I assume you got f(-2)=-2? SO you want to find $ f( - 2)^{ - 2}$, which is ${ \frac {1}{( - 2^2)}}$=1/4.[/hide]
[hide="#19"]SFFT (look in the Wiki) quickly gives (a-1)(b-1)=40, so a=9,b=6. I get 15?[/hide]

EDIT:[hide="#17"]Similar triangles quickly gives 1/2.[/hide]

I rushed, so some mistakes in there probably.[/quote]

about 28, I did get that as f(-2)

for 19, the two numbers have to be relatively prime.

Well, I don't know what you did.
Oops, it should be 11 and 5 then. The major step was just SFFT.


Okay guys, I guess we've been pretty lenient about this lately but please try to demonstrate self-control when trying to run up ur post count. Logging on and seeing the same name (*cough cough* no names mentioned) for the last 20 posts ought to tell you something...

Trying to stay away from just deleting posts by individuals that think they must post on every thread everytime anyone bothers to start a topic or reply to them...

Go do something productive...like some math problems...or helping me on my debate cases...or something...

Have a nice day. :-)

I agree that this is a serious problem, and appreciate you staying on top of it. Quality is more important than quantity.

What type of debate do you do? Ted Turner, Public Forum, Policy, LD??? I do LD, and I love it. Even though it's lots and lots of researching and hard work preparing a good solid case.

I used to do L-D, though I don't debate anymore.  Loved L-D, though I don't see where you're getting the thing about a lot of research.  Policy is far more labor-intensive (not to mention expensive) concerning the amount of research that is necessary.

Hey, Miths, I'd be happy to help with policy debate. I used to be a debate coach (of a winning team).

By the way, who is the new avatar?

[quote="MithsApprentice"]Okay guys, I guess we've been pretty lenient about this lately but please try to demonstrate self-control when trying to run up ur post count. Logging on and seeing the same name (*cough cough* no names mentioned) for the last 20 posts ought to tell you something...[/quote]

okay, okay! i'm sorry. i won't do that again.  by the way, i don't log on as often anymore.  i have better things to do. and its 29 posts, not 20 posts :D . i'm just trying to be accurate. :lol:

[quote="tokenadult"]Hey, Miths, I'd be happy to help with policy debate. I used to be a debate coach (of a winning team). [/quote]

That would be great.

I used to do CX. What I do now is policy debate in a "public forum"-like setting. The program is caled youth and government and basically you pretend like you're members of the state legislature (at the regional and state level) or Congress (at the national level) and you present bills and debate other ppl's bills. Bills can be on anything a normal senator or representative might present a bill on: education, taxes, etc. You follow all parliamentary procedures while you're on the floor the stuff like that.

However, I also help coach some LD and CX debate teams for a local high school, and that's where I could use some help.

Robinhood, LD can require large amounts of research depending on what format your particular organization/league uses. Contentions normally have at least some quotes to back them up.

Yes, but a few quotes and a couple of solid examples were usually all that were necessary for a decent argument.  At least, that was the way it worked when I did it.  Mind you, I wasn't very good, but I held ground that way.  And I was comparing LD to Policy, which requires tangible evidence of one's research.

sorry I haven't been helping moderate miths...I just got out of the hospital from breaking my arm...lol.

I thought he was talking about me for a second...


If, in the expression $ x^2 \minus{} 3$, $ x$ increases or decreases by a positive amount of $ a$, the expression changes by an amount:

$ \textbf{(A)}\ {\pm 2ax \plus{} a^2}\qquad
\textbf{(B)}\ {2ax \pm a^2}\qquad
\textbf{(C)}\ {\pm a^2 \minus{} 3} \qquad
\textbf{(D)}\ {(x \plus{} a)^2 \minus{} 3}\qquad\\
\textbf{(E)}\ {(x \minus{} a)^2 \minus{} 3}$

$ (x \pm a)^2 \minus{} 3 \equal{} x^2\pm2ax \plus{} a^2 \minus{} 3$

The difference between this and $ x^2 \minus{} 3$ is simply $ \pm2ax \plus{} a^2$, which is the answer.

Answer: A


geometry
geometric transformation
rotation
combinatorics unsolved
combinatorics

A circle is divided into $p$ equal sectors where $p$ is a prime number. In how many different ways can these $p$ sectors be colored with $n$ given colorings if two colors are considered different only when neither can be obtained from the other by rotating the circle?

$\frac{n^p-n}{p}+n$ from where Fermat theorem follows.

Using Burnised's lemma:
the identity has n^p fixed points, each of the p-1 rotations have n fixed points (since p is a prime, the only constalations that are the same after a rotation are one-colored). So the average number of fixed points is
$\frac{n^p+(p-1)n}p=\frac{n^p+np-n}p=\frac{n^p-n}p+n$.


What percent of $ 12\sqrt{12}$ is $ 3\sqrt3$? Express your answer to the nearest tenth.

We divide the larger number from the smaller one and then multiply by 100:

$ 100 \times \frac{3\sqrt {3}}{12\sqrt {12}}$

$ 100 \times \frac {3}{12\sqrt {4}}$

$ 100 \times \frac {1}{8}$

$ \boxed{12.5}$ percent

You have $12\sqrt{12}$ and $3\sqrt{3}$. $12\sqrt{12}$ is $24\sqrt{3}$ so you get $\frac{3}{24}$ which simplifies out to $\frac{1}{8}$.
$\frac{1}{8}$ is $ \boxed{12.5} $ percent


Εδώ θα παραθέτω την άσκηση της εβδομάδας και στο τέλος της εβδομάδας θα ανακοινώνονται οι λύτες και οι λύσεις τους .
Έχω ακόμα δύο ασκήσεις που μου έχετε στείλει .Θα προτιμούσα να μου στέλνετε για προτεινόμενες ασκήσεις , ασκήσεις όχι γνωστές από ΙΜΟ , ΒΜΟ κτλ και κατά προτίμηση σχετικά προτότυπες (αν δεν είναι προτότυπες ας μην είναι τουλάχιστον γνωστές ) . Τις λύσεις να μου τις στέλνετε σε PM . Ευχαριστώ ...
So let's begin

[color=red]ΑΣΚΗΣΗ 1[/color]

Να δειχθεί ότι δεν υπάρχουν πολυώνυμα $ P$ και $ Q$ ώστε
\[ \pi(x) \equal{} \frac {P(x)}{Q(x)}
\]
για κάθε  $ x\in\mathbb{N}$

όπου $ \pi(x)$ το πλήθος των πρώτων αριθμών $ \leq x$

Η άσκηση προτάθηκε από το Μιχάλη Σάββα

Πολύ μικρή η συμμετοχή σας στην άσκηση .... :(
Έλαβα μόνο 2 λύσεις και αυτές τηλεφωνικά από τους;Τάσο Βαφείδη και Ηλία Γεχάσκιελ .

Η λύση που προτείνουν και οι δύο πάνω κάτω είναι η ακόλουθη

Έστω ότι υπάρχουν πολυώνυμα $ P, Q \in \mathbb{C}[x]$ ώστε $ \pi(x) \equal{} \frac {P(x)}{Q(x)}$, για κάθε $ x$ $ x \in \mathbb{N} \equal{} \{1, 2, \ldots\}$.
και έστω $ m \equal{} \deg(P)$ και $ n \equal{} \deg(Q)$ και καθώς $ \lim_{x \to \plus{} \infty}\pi(x) \equal{} \plus{} \infty$, θα είναι $ m \ge n \plus{} 1$

 Τώρα έχουμε ότι $ \pi((m \plus{} 2)! \plus{} 2) \equal{} \pi((m \plus{} 2)! \plus{} 3) \equal{} ... \equal{} \pi((m \plus{} 2)! \plus{} m \plus{} 2) \equal{} t$  άρα σε  $ m \plus{} 1$ σημεία έχουμε ότι το πολυώνυμο  $ H(x) \equal{} P(x) \minus{} tQ(x)$ μηδενίζεται άρα είναι το μηδενικό πολυώνυμο άρα $ \frac {P(x)}{Q(x)} \equal{} t$ για κάθε $ x$ άρα $ \pi(x)$ σταθερή άτοπο άρα το ζητούμενο

***
Νέα άσκηση θα προταθεί μόνο εάν το θελήσετε (δηλαδή θελήσετε να συνεχιστεί ο θεσμός)

Ωραία άσκηση! Εμένα πάντως μου αρέσει σαν θεσμός και προτείνω να συνεχιστεί

και εγω θα ηθελα να συνεχιστει απλα στην συγκεκριμενη ασκηση δεν ειχα τις γνωσεις για να σκεφτω μια λυση,με ορια και μιγαδικους. :( παρ'ολα αυτα δεν παυει να ειναι μια προκληση. Ας μην σταματησει να υπαρχει,η αποψη μου.

Η ασκηση κατα την ταπεινη μου αποψη ειναι [b]απ τις  κομψοτερες[/b] που είδα τελευταια .
Συνδυάζει την ιδεα των τεραστιων κενων απο πρωτους στους φυσικους με το σταθερό σταθερο πολυωνυμο και αποδεικνυει οτι το πληθος των πρωτων μεχρι και το $ x$ δεν ειναι ρητη συνάρτηση του $ x$ ΄.
Για την  λυση της δεν χρειαζονται ουτε μιγαδικοι (το $ deg$ ειναι  βαθμος) ουτε ντε και καλα ορια .Το οριο που ορθα γραφει ο Sil (Βλεπετε εγινε πολυ φορμαλιστης τελευταία ή οι κακες παρεες με αναλυστες ......) το τελειωνει αμεσως αλλα αν θελεις παρακαμπτεται λεγοντας οτι αφου το πληθος των πρωτων ειναι απειρο πρεπει ο βαθμος του αριθ. να ειναι κατα ενα τουλ. μεγαλ.του βαθμου του παραν. γιατι αν δεν ηταν απο καποιο $ x$ και πανω το πηλικο των πολυωνυμων θα ηταν φραγμενη συναρτηση ατοπο γιατι το πρωτο μελος δεν ειναι.......

Ας συνεχίσουμε λοιπόν και βλέπουμε . :)

[color=red]ΑΣΚΗΣΗ 2[/color]
Η ακολουθία ($ x_n$) ικανοποιεί $ x_0\equal{}0$ , $ x_2\equal{}x_1\sqrt[3]{2}$  ,  $ x_3\in\mathbb{N}$  και

$ x_{n\plus{}1}\equal{}\dfrac{1}{\sqrt[3]{4}}x_n\plus{}\sqrt[3]{4}x_{n\minus{}1}\plus{}\dfrac{1}{2}x_{n\minus{}2}$ για κάθε $ n\geq{2}$

Να προσδιοριστεί ο ελάχιστος αριθμός ακεραίων όρων που πρέπει να έχει αυτη η ακολουθία

Η άσκηση προτάθηκε από τον Σμπώκο Δημήτρη (kritikos)

Giati stamathse h askhsh ths ebdomadas? Den ly8hke h askhsh h xemeiname apo askhseis? To deftero apokleietai pantws... ;)

Σιλουανέ βάλε τη λύση σου σε αυτήν την άσκηση για να συνεχιστεί η άσκηση της Εβδομάδας .


Hello, I am starting a Survivor competition, and if there is some body that would like to join, then you have to either say that you would like to join on this post, or youo can just private message me whether you would like to join or not. I am taking the first twenty contestants that would like to join.

  Here are the rules. I will divide the twenty contestants in to two teams, and each week I will post 3 problems and every body on each team has to give their answers to the three problems. You will get one point per question that you get correct, and I will post every bodies scores when everyone has done it. Based on the scores I will total them for each team, and the team that does the worst will be sent to trival council. (Supposedly). I will give the contestants on the team that had gone to trival council a certain number of days to discuss among them selves who they would like to vote out. Once you have made your descision you will private message me your vote, and the person that gets the most votes loses. (You want to vote the person that is causing your team the smallest benefit). You just have to say that you would like to join and then you are autimatically in.

I'll join....

(Hoping this game does not fail.)

I join.What is the difficulty level? Multiple choice, AIME style, or Proofs?

I guess I'll join.
I think proofs would take too long.

They're most likely gonna be mathcounts level...Note the only place he posts...

[quote="Wickedestjr"]why? it isn't going to be just mathcounts level, it is going to be a lot harder.[/quote]

Does it reach #10-15 on AIME? or just AMC and easy AIME.

I already got the problems. Are we allowed to post them?

1. What is the 27th Prime number not containing a 1 or a 3?

2. What is (1/2x-1/x)(1/2x+1/x)*

3. What is the sum of the first 30 positive perfect squares**

* I'm not sure if this reads $ (\frac{1}{2x}\minus{}\frac{1}{x})(\frac{1}{2x}\plus{}\frac{1}{x})$ or $ (\frac{1}{2}x\minus{}\frac{1}{x})(\frac{1}{2}x\plus{}\frac{1}{x})$, nor do I know if he wants us to answer in
terms of $ x$.

** The sequence being $ 1,4,9,16,\hdots$.

These are AIME level? There more like easy mathcounts level to me. :|

That is because dingzhou has changed to (^_^)  :rotfl:  :rotfl:

[quote="Wickedestjr"]Number 2 is the following;

2. (1/2x + 1/x)/(1/2x - 1/x)

And I was not able to get a hold of dingzou. He was not on the memberlist.

P.S. - Everybody except him should have the questions by now. If you don't please private message me.[/quote]

um...yeah...dingzhou does not exist anymore....but um....i never got any pm and i signed up for this.

[quote="pythag011"]Um... I do not think that these problems are olympiad level...

Probably medium AIME at most.[/quote]

um they're more like elementary aime level if I got the right questions

[quote="Wickedestjr"]Number 2 is the following;

2. (1/2x + 1/x)/(1/2x - 1/x)

[/quote]

...So is that $ \frac{1}{2}$ a coefficient or is the x part of the denominator?

And also, we pm you are answers right?

I do not think that we will be able to continue the Survivor because of all the problems in it, but I am still doing the Problem Solving Tournament, the other Problem Solving Tournament, and the Jackpot game.

I am planning on beginning the next survivor competition in the month of February.

If you would like to join the AoPS survivor, then you might want to join the one in February because the next one will be about a year after that one.


[quote="236factorial"]The base fo a triangle is 3 times the length of the altitude to that base. The area of the triangle is 96 square centimeters. Find the length of the base.[/quote]

[hide]Let $x$ be the base, and $y$ be the height.

$x=3y$

$\dfrac{xy}{2}=96$ which is $xy=192$

Subtituting,

$y(3y)=192$

$3y^2=192$

$y^2=64$

$y=8$

Solving for $x$, $x=24$[/hide]


inequalities
function
symmetry
inequalities proposed

Let $ a$, $ b$,$ c$ be positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
$ abc \plus{}\frac{12}{ab\plus{}bc\plus{}ac}\ge5$

[quote="ABRORBEK"]Let $ a$, $ b$,$ c$ be positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
$ abc \plus{} \frac {12}{ab \plus{} bc \plus{} ac}\ge5$[/quote]
trivial
$ abc \geq (b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)$ and then we have
$ abc \geq \frac {4}{3}\sum bc \minus{} 3$
and then you do by yourself

[quote="Mathias_DK"][quote="desmrhai"][quote="ABRORBEK"]Let $ a$, $ b$,$ c$ be positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
$ abc \plus{} \frac {12}{ab \plus{} bc \plus{} ac}\ge5$[/quote]
trivial
$ abc \geq (b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)$ and then we have
$ abc \geq \frac {4}{3}\sum bc \minus{} 3$[/quote]
I don't understand nothing of your solution.. How did you get $ abc \geq (b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)$ ??[/quote]
 :roll: chicken boy :rotfl:  :huh:
If $ (b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c) < \equal{} 0$ then Ine is true
If $ (b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c) > 0$ then exits $ \triangle ABC$ such that $ \sqrt{a} \equal{} BC;\sqrt{b} \equal{} CA;\sqrt{c} \equal{} AB$
we use $ cosA\equal{}\frac {b \plus{} c \minus{} a}{2\sqrt{bc}}$ and $ cosAcosBcosC \leq \frac {1}{8}$

[quote="desmrhai"][/quote][quote="Mathias_DK"][quote="desmrhai"][quote="ABRORBEK"]Let $ a$, $ b$,$ c$ be positive numbers such that $ a + b + c = 3$. Prove that
$ abc + \frac {12}{ab + bc + ac}\ge5$[/quote]
trivial
$ abc \geq (b + c - a)(c + a - b)(a + b - c)$ and then we have
$ abc \geq \frac {4}{3}\sum bc - 3$[/quote]
I don't understand nothing of your solution.. How did you get $ abc \geq (b + c - a)(c + a - b)(a + b - c)$ ??[/quote][quote="desmrhai"]
 :roll: chicken boy :rotfl:  :huh:
If $ (b + c - a)(c + a - b)(a + b - c) < = 0$ then Ine is true
If $ (b + c - a)(c + a - b)(a + b - c) > 0$ then exits $ \triangle ABC$ such that $ \sqrt {a} = BC;\sqrt {b} = CA;\sqrt {c} = AB$
we use $ cosA = \frac {b + c - a}{2\sqrt {bc}}$ and $ cosAcosBcosC \leq \frac {1}{8}$[/quote]
I don't know what chicken boy is supposed to mean, but i still don't get the point :S

The  mentioned inequality is trivial, indeed.
Just try Karamata for $ \{(a\plus{}b\minus{}c);(a\plus{}c\minus{}b);(b\plus{}c\minus{}a)\}\succ \{a;b;c\}$ and the concave function $ f(x)\equal{}ln (x)$.

If $ abc\geq \frac {4}{3}(ab \plus{} bc \plus{} ca) \minus{} 3$ is true indeed (i haven't verified it), then $ abc \plus{} \frac {12}{ab \plus{} bc \plus{} ca}\geq 4(\frac {ab \plus{} bc \plus{} ca}{3} \plus{} \frac {3}{ab \plus{} bc \plus{} ca}) \minus{} 3\geq 4.2 \minus{} 3 \equal{} 5$.
The first inequality may be equivalent to $ abv\geq (a \plus{} b \minus{} c)(c \plus{} a \minus{} b)(b \plus{} c \minus{} a)$ so the initial inequality should not be equivalent to it but a consequence.

[quote="Mathias_DK"][quote="mihai miculita"]$ b \plus{} c \minus{} a\ge 0, \ \ a \plus{} c \minus{} b\ge 0, \ \ a \plus{} b \minus{} c\ge 0 \Rightarrow \\
\Rightarrow (b \plus{} c \minus{} a).(a \plus{} c \minus{} b).(a \plus{} b \minus{} c) \equal{} \sqrt {(b \plus{} c \minus{} a)^2.(a \plus{} c \minus{} b)^2.(a \plus{} b \minus{} c)^2} \equal{} \sqrt {(b \plus{} c \minus{} a).(a \plus{} c \minus{} b)}.\sqrt {(b \plus{} c \minus{} a).(a \plus{} b \minus{} c)}.\sqrt {(a \plus{} c \minus{} b).(a \plus{} b \minus{} c)} \equal{} \sqrt {c^2 \minus{} (a \minus{} b)^2}.\sqrt {b^2 \minus{} (a \minus{} c)^2}.\sqrt {a^2 \minus{} (b \minus{} c)^2}\le abc.$[/quote]
The ineq was: $ abc \plus{} \frac {12}{ab \plus{} bc \plus{} ac} \ge5$. Are the two ineqs equivalent or what ?!?!?!?[/quote]
U don't understand?????
But BG Yoda understand. and then you know the mean of "chicken boy"

[quote="ABRORBEK"]Let $ a$, $ b$,$ c$ be positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
$ abc \plus{} \frac {12}{ab \plus{} bc \plus{} ac}\ge5$[/quote]
Let $ f(a,b,c) \equal{} abc \plus{} \frac {12}{ab \plus{} bc \plus{} ac} \minus{} 5$.
Then differentiate with respect to $ a$ twice: $ \frac {\Delta f(a,b,c)}{\Delta^2 a} \equal{} \frac {24(b \plus{} c)^2}{(ab \plus{} bc \plus{} ca)^3} > 0$. So $ f(a,b,c)$ is convex in $ a,b,c$.
Thus $ f(a,b,c) \ge f(\frac {a \plus{} b \plus{} c}{3},\frac {a \plus{} b \plus{} c}{3},\frac {a \plus{} b \plus{} c}{3}) \equal{} f(1,1,1) \equal{} 5$. QED.

[quote="matex (L)(L)(L)"]Does anybody know where can I find an article(or something like this) about Jensen's inequality or Karamata;s inequality in more than one variable.I mean what conditions should be fulfilled to use one of the below inequalities.

thanks for reply[/quote]
My solution is not valid as far as I'm concerned. This occured to me, as I searched this topic for more information.

Read it if you like, I don't know any papers about this :S

First the definition of convexity is as follows:
The function $ f : I \to \mathbb{R}$ is convex iff for all $ x,y \in I$ and $ \lambda \in [0;1]$:
\[ \lambda f(x) \plus{} (1 \minus{} \lambda)f(y) \ge f(\lambda x \plus{} (1 \minus{} \lambda)y)
\]
It is well known that it is enough that $ f''(x) \ge 0$. This is probably the hard thing to prove, so I'll leave it to you :P

First a proof that $ f(x) \plus{} f(y) \ge 2 f \left ( \frac{x\plus{}y}{2} \right )$ (When $ f: I \to \mathbb{R}$ is convex and $ x,y \in I$.)
Proof: It follows directly from the definition. Just let $ \lambda \equal{} \frac{1}{2}$.

$ f(x) \plus{} f(y) \plus{} f(z) \ge 3 f \left ( \frac{x\plus{}y\plus{}z}{3} \right )$ is a bit more complicated but not harder.
From the definition you get:
$ \frac{1}{3}f(x) \plus{} \frac{2}{3} f(y) \ge f \left ( \frac{x\plus{}2y}{3} \right )$
$ \frac{1}{3}f(x) \plus{} \frac{2}{3} f(z) \ge f \left ( \frac{x\plus{}2z}{3} \right )$
$ \frac{1}{2} f \left ( \frac{x\plus{}2y}{3} \right ) \plus{} \frac{1}{2} f \left ( \frac{x\plus{}2z}{3} \right ) \ge f \left ( \frac{x\plus{}y\plus{}z}{3} \right )$.

Addition yields:
$ f(x) \plus{} f(y) \plus{} f(z) \ge \frac{3}{2} f \left ( \frac{x\plus{}2y}{3} \right ) \plus{} \frac{3}{2} f \left ( \frac{x\plus{}2z}{3} \right ) \ge 3 f \left ( \frac{x\plus{}y\plus{}z}{3} \right )$

You can probably prove the general version in similar ways.

Now let us take a look at a two variable function. Let $ f : I \to \mathbb{R}$ be a convex in both variables and symmetric function. Then I'll prove:
$ 2f(a,b) \ge 4f(\frac{a\plus{}b}{2},\frac{a\plus{}b}{2}) \minus{} f(a,a) \minus{} f(b,b)$

First notice that $ \lambda f(x,y) \plus{} (1 \minus{} \lambda)f(x,z) \ge f(x, \lambda y \plus{} (1\minus{}\lambda)z)$ also holds here! (Just let $ g(y) \equal{} f(x,y)$ and it's obvious.)
Then by definition:
$ f(a,b) \plus{} f(a,a) \ge 2f \left (a,\frac{a\plus{}b}{2} \right )$
$ f(a,b) \plus{} f(b,b) \ge 2f \left (b,\frac{a\plus{}b}{2} \right )$
$ 2f \left ( a,\frac{a\plus{}b}{2} \right ) \plus{} 2f \left ( b,\frac{a\plus{}b}{2} \right ) \ge 4f \left ( \frac{a\plus{}b}{2},\frac{a\plus{}b}{2} \right )$
(Here we use the symmetry too)
Addition yields:
$ 2f(a,b) \plus{} f(a,a) \plus{} f(b,b) \ge 4f(\frac{a\plus{}b}{2},\frac{a\plus{}b}{2})$
The general $ f(a,b) \ge f \left (\frac{a\plus{}b}{2},\frac{a\plus{}b}{2} \right)$. Does not hold! Consider $ f(a,b) \equal{} a^2b^2$ which is clearly convex. Then it would mean that $ 16a^2b^2 \ge (a\plus{}b)^4$, which is clearly false!
But the true "theorem" on $ f(a,b) \equal{} a^2b^2$ yields:
$ 8a^2b^2 \plus{} 4a^4 \plus{} 4b^4 \ge (a\plus{}b)^4$

[quote="Mathias_DK"][quote="matex (L)(L)(L)"]Does anybody know where can I find an article(or something like this) about Jensen's inequality or Karamata;s inequality in more than one variable.I mean what conditions should be fulfilled to use one of the below inequalities.

thanks for reply[/quote]
My solution is not valid as far as I'm concerned. This occured to me, as I searched this topic for more information.

Read it if you like, I don't know any papers about this :S

First the definition of convexity is as follows:
The function $ f : I \to \mathbb{R}$ is convex iff for all $ x,y \in I$ and $ \lambda \in [0;1]$:
\[ \lambda f(x) \plus{} (1 \minus{} \lambda)f(y) \ge f(\lambda x \plus{} (1 \minus{} \lambda)y)
\]
It is well known that it is enough that $ f''(x) \ge 0$. This is probably the hard thing to prove, so I'll leave it to you :P

First a proof that $ f(x) \plus{} f(y) \ge 2 f \left ( \frac {x \plus{} y}{2} \right )$ (When $ f: I \to \mathbb{R}$ is convex and $ x,y \in I$.)
Proof: It follows directly from the definition. Just let $ \lambda \equal{} \frac {1}{2}$.

$ f(x) \plus{} f(y) \plus{} f(z) \ge 3 f \left ( \frac {x \plus{} y \plus{} z}{3} \right )$ is a bit more complicated but not harder.
From the definition you get:
$ \frac {1}{3}f(x) \plus{} \frac {2}{3} f(y) \ge f \left ( \frac {x \plus{} 2y}{3} \right )$
$ \frac {1}{3}f(x) \plus{} \frac {2}{3} f(z) \ge f \left ( \frac {x \plus{} 2z}{3} \right )$
$ \frac {1}{2} f \left ( \frac {x \plus{} 2y}{3} \right ) \plus{} \frac {1}{2} f \left ( \frac {x \plus{} 2z}{3} \right ) \ge f \left ( \frac {x \plus{} y \plus{} z}{3} \right )$.

Addition yields:
$ f(x) \plus{} f(y) \plus{} f(z) \ge \frac {3}{2} f \left ( \frac {x \plus{} 2y}{3} \right ) \plus{} \frac {3}{2} f \left ( \frac {x \plus{} 2z}{3} \right ) \ge 3 f \left ( \frac {x \plus{} y \plus{} z}{3} \right )$

You can probably prove the general version in similar ways.

Now let us take a look at a two variable function. Let $ f : I \to \mathbb{R}$ be a convex in both variables and symmetric function. Then I'll prove:
$ 2f(a,b) \ge 4f(\frac {a \plus{} b}{2},\frac {a \plus{} b}{2}) \minus{} f(a,a) \minus{} f(b,b)$

First notice that $ \lambda f(x,y) \plus{} (1 \minus{} \lambda)f(x,z) \ge f(x, \lambda y \plus{} (1 \minus{} \lambda)z)$ also holds here! (Just let $ g(y) \equal{} f(x,y)$ and it's obvious.)
Then by definition:
$ f(a,b) \plus{} f(a,a) \ge 2f \left (a,\frac {a \plus{} b}{2} \right )$
$ f(a,b) \plus{} f(b,b) \ge 2f \left (b,\frac {a \plus{} b}{2} \right )$
$ 2f \left ( a,\frac {a \plus{} b}{2} \right ) \plus{} 2f \left ( b,\frac {a \plus{} b}{2} \right ) \ge 4f \left ( \frac {a \plus{} b}{2},\frac {a \plus{} b}{2} \right )$
(Here we use the symmetry too)
Addition yields:
$ 2f(a,b) \plus{} f(a,a) \plus{} f(b,b) \ge 4f(\frac {a \plus{} b}{2},\frac {a \plus{} b}{2})$
The general $ f(a,b) \ge f \left (\frac {a \plus{} b}{2},\frac {a \plus{} b}{2} \right)$. Does not hold! Consider $ f(a,b) \equal{} a^2b^2$ which is clearly convex. Then it would mean that $ 16a^2b^2 \ge (a \plus{} b)^4$, which is clearly false!
But the true "theorem" on $ f(a,b) \equal{} a^2b^2$ yields:
$ 8a^2b^2 \plus{} 4a^4 \plus{} 4b^4 \ge (a \plus{} b)^4$[/quote]

I've seen the proof of $ f''(x)\geq 0 \Rightarrow f \minus{}convex$. It's something like that:
Denote $ g(x)\equal{}tf(x)\plus{}(1\minus{}t)f(y)\minus{}f(tx\plus{}(1\minus{}t)y)$ (I replace lambda with t). By differentiating $ g'(x)\equal{}tf(x)\minus{}tf(tx\plus{}(1\minus{}t)y)$. $ f''(x)\geq 0$ for all $ x\in [a,b]$ implies that f'(x) is increasing on [a,b] hence $ g'(x)\geq 0$ for $ x\geq y$ and $ g'(x)\leq 0$ if $ x\leq y$. That means $ g(x)\geq g(y)\equal{}0$.


[quote="verizonwires"]LOOK, I am from bangalore just heads up. I no Hebar Iyengar Tamil, and kannada.

anyway. How dare Dhoni drop Dravid, he is a great fielder which India needs.

AND y did that ool Dhoni drop Ganguly?

He always says " I believe the batsmen were at fault" but w8... RENT U A FRIKIN BATSMEN?????

I MEAN WaT the heck!!!!!

sorry guys but I dont like dhoni callin the shots.
---------------------------------------------------

What the heck is Rohit Shrma doin on the ODI team???[/quote]

You should hate Vengsarkar. He's the one who dunnit... But yes Dhoni has his fav and loyal players by his side.....

Well, Ganguly dropped and Dravid dropped . We all know who's next.

P.S: What do you mean by I no Hebar Tamil Iyengar?

Ganguly and Dravid might have been dropped, by Dhoni has been again and again proving himself by unconvention.. I was obviously disappointed by that decision, but now it doesnt hurt so much coz India is putting on a brave fight.. If you want ppl to stick on and finish games like Dravid, you've got ppl like Gambhir, Yuvraj and Dhoni himself who can stick to the wicket and as a bonus, add quick runs too.. Abt Dada, yeah I still feel he can open wit Sachin given India is not able to nurture really good openers, but frankly speaking, one fine day Dada will retire, Sachin will follow him in a year, and India will be left wit no good opener.. Look at Australia, they have nurtured Phil Jacques, and ever since he joined Aussie team, he has been rocking...

a completely useless topic. :ninja:
and i wonder why 12th ppl are posting here, go study for your boards you jobless ppl. :P

[quote="madness"]probably people are confused what instigates the guts and wat to do if u have low guts  :P
[hide]
Increase the content of non adrenaline  :rotfl:
[/hide][/quote]

Doesn't histamine smoothen the contraction of gut muscles??

God knws da leave it not so much nv and all  :D  ask shruthi the bio student  :P
P.S: This was meant to be a satire on all bio students very very much  :P

Kill yourself Rohit....

As for your question Shre, it does contract the smooth muscles, lining the gastrointestinal tract, what is otherwise known as the gut muscles....

[quote="shruthibaskaran"]Kill yourself Rohit....[/quote]

Expected that  :P  Neway ill do just that after march 3rd dontee worry  :D   :rotfl:

Im sure i will and no one can stop me  :P   :rotfl:   :(

Where is Verizon, Dhoni's greatest fan?? Probably he is gone to celebrate India's probable victory agst Australia??
Or did his comp crash as Pardesi's did??

And for all those who said Sachin can't chase, can't play in a big match, a final, he's given you'll a great reply today.... :coolspeak:  :first:
Adhu CENTURY.. HE ROCKZZZZZZZZZZZZZZZZZZZZZZZ....

It was superb. But it was a pity that Rohit Sharma was stumped the next ball :(

Shru, when someone bowls a ball and it hits your stump direct without going to any fielder in the way, by the official cricket terminology, it is called bowled and not stumped!!

Ah well, it doesnt matter anyway :D I thought stumped sounded much better in context :D And I like English more than cricket :p SO :p

@Verizon: What do you call this?? This is an amazing win... And as a great fan of Sachin, I tell u this... Its not just Pravin Kumar's bowling or Sachin's batting which won us the tournament... The major role was Dhoni's captaincy.. He has got the coolest customer you could get as captain, and he has got the guts to go by his decision, prove he has been right, has got the will power to finish off the game wit a victory.. So get it.. DHONI ROCKZZZ!!!!!

I too completely agree with what vignesh says... Dhoni led us brilliantly... with a cool head...  :D


Solve for dy/dx  using quadratic formula
Now put log x =p
u get
$ \frac{dp}{dx}\plus{}\frac{x}{2}\equal{}{\frac{\sqrt{x^2\plus{}4p}}{2}}$
wich can b solved

[quote="elastiboysai"]Solve for dy/dx  using quadratic formula
Now put log x =p
u get
$ \frac {dp}{dx} \plus{} \frac {x}{2} \equal{} {\frac {\sqrt {x^2 \plus{} 4p}}{2}}$
wich can b solved[/quote]

How do you solve it??

by putting 4p=x^2tan^2 theta
i think i have got the ans .
will post it tom.


$a,b,c$are three positive numbers,such that $a+b+c=1$
please prove that $\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\leq2(\frac{b}{a}+\frac{c}{b}+\frac{a}{c})$

now I have edited it. I think it is more clear now.
let us solve it~~~

grr... that assymetry of the RHS is annoying me.. that's probably why I failed the USAMO :D

Very nice problem, I would like to see your solution, [b]zhaobin[/b], so please post it (of course if you have it). :)

sorry.I don't have the solution.I have thought about for many many times.
hope somebody can post the solution.additional,it is from JMO.Is there someone japanese here can post the offical solution?

Then perhaps it would have been better to post it in the Olympiad section than in Pre-Olympiad?  (I also have been unable to solve it.)


$ 2\sin x\sin 80\equal{}\sin 50\sin (x\plus{}50)$

Does the exact value for x exist ?
If so , may I get the solution ?

Thank you  :)

hello, with the addition formulas you can write
$ 2\sin(x)\sin(\alpha) \equal{} \sin(\beta)(\sin(x)\cos(\beta) \plus{} \cos(x)\sin(\beta))$
with $ \alpha \equal{} 80 degrees$ and $ \beta \equal{} 50 degrees$.
From this we have
$ \sin(x)(2\sin(\alpha) \minus{} \sin(\beta)\cos(\beta)) \equal{} \cos(x)\sin^2(\beta))$
and
$ x \equal{} arctan\left(\frac {\sin^2(\beta)}{2\sin(\alpha) \minus{} \sin(\beta)\cos(\beta)}\right)$.
Sonnhard.

Can $ x \equal{} \arctan\left(\frac {\sin^2(\beta)}{2\sin(\alpha) \minus{} \sin(\beta)\cos(\beta)}\right)$ be simplified further?


I am trying to write up a lesson plan for 9/10th grade students on Geometry, so our school could be ready for Mu Alpha Theta. Any suggestions on what should be on it? (it should be Geometry/Algebra 2/Trig students-level)

I cant seem to think because I know too many things in Geometry that will be useless in Mu Alpha Theta.

I don't think there was much out of the normal curriculum on the geometry stuff.. I didn't really focus on any particular detail in studying, I just made sure I got through the curriculum. Sorry I can't be more of a help, but if anything pops into my mind, I'll be sure to post it.

The main thing is to practice practice practice. I started really studying in January (which is when the contests started-aka way late) and by March/April I got 10th at an invitational and 7th at states, and all I did was do practice test after practice test. The main key is to avoid mistakes and make sure you read the question properly!

Sadly I won't be going to nationals, I was the only one who placed at states from our school, and I really didn't want to spend my summer going over simple geometry problems to avoid easy mistakes and then pay for my plane ticket to go there. I hope you have a good time, and good luck to you!


Let $ a,b,x,y \in N$ $ a,b \neq 0$
and
$ 15a \plus{} 16b \equal{} x^2$
$ 16a \minus{} 15b \equal{} y^2$

$ min(min(x,y)) \equal{} ?$

[quote="bboypa"]0?  :huh:

$ 16a \equal{} 15b \implies a \equal{} 15A, b \equal{} 16B$

$ 15a \plus{} 16b \equal{} x^2 \implies 15^2A \plus{} 16^2B \equal{} x^2$

forexample

$ A \equal{} 16^2 , B \equal{} 20^2 \implies 15^2 16^2 \plus{} 16^2 20^2 \equal{} 16^2 25^2$[/quote]
I think it is not true!because $ y^2 \equal{} 16\times15\times 16^2 \minus{} 15\times16\times 20^2\neq0$

azz,scuse me :blush:

however min(min(x,y))=0 is "probably" 0 for a=15*481*K^2  and b=16*481*K^2 :roll:

BONUS: and what is the answer if $ a,b,x,y \neq 0$? :wink:


geometry
circumcircle
inradius
inequalities
trigonometry
inequalities proposed

If ABC is a triangle with circumradius R and inradius r, prove the inequality

$8R^2+4r^2\ge a^2+b^2+c^2$

some computations give the original ineq is equivalent to
$\frac{9}{4} + \cos^2 A + \cos^2 B + \cos ^2 C \ge \cos A + \cos B + \cos C$
which follows from
$\cos^2 A + \cos^2 B + \cos ^2 C \ge \frac{3}{4}$ and $\frac{3}{2} \ge \cos A + \cos B + \cos C$

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=18937 .

  Darij

darij, may i know how to prove the poland ineq mentioned in that thread?
Thanks.

I mean, a direct proof, instead of some variations like this one.

$8R^2+4r^2\ge a^2+b^2+c^2\Longleftrightarrow 8R^2+4r^2\ge 2p^2-2r^2-8Rr\Longleftrightarrow$

$p^2\le 4R^2+4Rr+3r^2\Longleftrightarrow IH^2=4R(R+r)+3r^2-p^2\ge 0$, q.e.d.

(I used the relation $ab+bc+ca=p^2+r^2+4Rr$).


[quote="Dumel"]a,b,c,d>0
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$
prove inequality
$ \sum_{}^{} \frac {1}{5 \minus{} a} \le 1$[/quote]
remember that:
$ \frac{1}{5\minus{}a} \leq \frac{1}{48}(a^2\minus{}1)\plus{}\frac{1}{48}(a\minus{}1)\plus{}\frac{1}{4}$
so :
$ \sum \frac{1}{5\minus{}a} \leq \sum(\frac{1}{48}(a^2\minus{}1)\plus{}\frac{1}{48}(a\minus{}1)\plus{}\frac{1}{4})\equal{}\frac{1}{48}(\sum a\minus{}4)\plus{}1\leq 1$

[quote="FelixD"]@tuandokim:
How did you find this estimate?[/quote]
oh i find k,l that satisfy this   :D
$ \frac {1}{5 \minus{} a} \leq k(a^2 \minus{} 1) \plus{} l(a \minus{} 1) \plus{} \frac {1}{4}$
and i use this $ a,b,c,d \leq 2$


A rectangular room measures 30 feet in length and 12 feet in height, and the ends are 12 feet in width. A spider rests at a point one foot up from the floor at the middle of one end. The center of the spider's web is located one foot down from the ceiling at the middle of the other end of the room. What is the shortest distance the spider can walk to get to her web?

you make a net and unfold it so it's a 32 and 24 legnth.  40 feet.

Originally in a 1905 brainteaser book.  Found it in the joy of mathematics.

[quote="nat mc"]Its one foot up from the floor and one down from the ceiling, so it is a right triangle with legs 24 and 30 that give us the shortest path of $6\sqrt{41}$[/quote]  that's wrong it's a 24,32 for the legs because you have to add 2 to 30 because of the ceiling and floor. 32 for the up and down feet.

There are different ways to unfold it:
24*30
23*31
13*41
12*42

I could draw it for you, but I don't know how to put it from Word to here.
 How do you do it?


LaTeX
absolute value
complex numbers
algebra unsolved
algebra

This is probably going to be simple for you, but I cannot arrive at an answer (although I know what it is).

How many sets (a,b) satisfy $(a+bi)^{2002}=(a-bi)$?

I was thinking about multiplying both sides by (a+bi) to get the following:

$(a+bi)^{2003}=a^2+b^2$

but I'm not sure what I would do with that.  By the way, this is an AMC-12 problem.  Any ideas?

Thank you.

first, consider the absolute value of complex numbers then we can see

abs(!z) = abs(z) = abs[(z)^2002] = abs(z)^2002

(z = a + bi, !z = a - bi, abs(z) = sqrt(a^2 + b^2))

so we can rewrite it as

abs(z)^2002 = abs(z)

abs(z) * [abs(z)^2001 - 1] = 0

so one solution is abs(z) = 0  --> a = 0, b = 0

other solution is abs(z)^2001 = 1  -->  abs(z) = 1

and then we can do as you say, multiply it by z

z^2003 = abs(z)^2 = 1

because z is raised to 2003 power, there is exactly 2003 solutions for z = 1

so 1 + 2003 = 2004 solutions

(soory, i dont know latex)

[quote="alexmcavoy@gmail.com"]Ok thanks for the explanation.  So basically the solutions are 1 multiplicity 2003 and 0, right?

Thanks again.[/quote]

No.  There are 2003 distinct solutions having modulus 1 (the 2003-th roots of unity)

Let $z=a+bi=\sqrt{a^2+b^2}e^{i\alpha}$ where $\alpha=Arg(z)$ then
the problem is $\sqrt{a^2+b^2}^{2002}e^{i2002\alpha}=\sqrt{a^2+b^2}e^{-i\alpha}$ so
$|z|=\sqrt{a^2+b^2}=1$ and $e^{i2003\alpha}=1$ and that's what blah*3 said.


Oh,It's a easy ineq in Mathlinks.
Assume that $a+b+c=1$
$(a+b)(b+c)(c+a)=(1-a)(1-b)(1-c) \le (\frac{3-a-b-c}{3})^3=\frac{8}{27}$
$a^2+b^2+c^2 \ge \frac{1}{3}$
And we are done!


Solve in the set of the real numbers :

$ 2a^{4x}\equal{}a^{2x^2}\plus{}a^8\ ,\ (\forall)a\in \mathbb{N}$

[hide]$ a \ge 1 \Longrightarrow a^{2x^2} \plus{} a^8 \ge 2\sqrt{a^{2x^2 \plus{} 8}} \equal{} 2a^{x^2 \plus{} 4} \ge 2a^{4x}$ by AM-GM, equality iff $ x\equal{}2$.[/hide]


i love both of them...
but i choosed france.

[quote="kostas"]Italy,I think they 'll win. but I like France too.[/quote]

remember the game betwin france and italy in 2000.

finally france won two - zero.

yes it was an amazing game, but Italy has younger players and an ambitious team. and the best defenders in the world !!! :D

Correct me if I am wrong, but I believe France won 2-1in Extra Time by Henry's Golden Goal. And that only increases the probablity of Italy winning this one by the law of averages.

But I voted France. I want Zidane to become a second Maradona.

[quote="bubka"]. I want Zidane to become a second Maradona.[/quote]

but i think zidan is the first zidan, not the second maradona. :D

[quote="Dave Vather"]Italy! There is no way to score them[/quote]

Yet USA, one of the weakest teams, did.

Nope, they didn't. The only team that has scored Italy is Italy

yes, that was an own goal

France for sure!!!!!! :lol:

That's still scoring against the defenders you know.

finally italy won the game,

congratualations... :D  :wink:


function
algebra
polynomial
algebra unsolved

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76767
[quote="peeta"]idk if it helps any but the only solution for
\[ f(f(n))+f(n)=2n+2001  \]
is \[ f(n) = n + 667  \]

this is, since, if $f(1)=a$ and $a_n=f^n(1) \implies a_0=1\text{ and }a_1 =a$ as well as $a_{n+2}= 2a_n+2001-a_{n+1}$ then
\[ a_n = \left(\frac{668-a}{3}\right)(-2)^n + 667n + \frac{a-665}{3}  \]
which is only positiv for all $n$ if $a=668$.

similary for all other $n=2,\dots$[/quote]

I found that posted . However I don't understand this solution inthen
\[ a_n = \left(\frac{668-a}{3}\right)(-2)^n + 667n + \frac{a-665}{3}  \]
.

Could anyone explain it clearier for me?

If i undrestood your question how to solve this kind of functions $f(f(n))+f(n)-2n=2001$ ?

Put $f(a_{n})=a_{n+1}$ and get this in the original equation gives:
$a_{n+2}+a_{n+1}-2a_{n}=2001$ take $l$ such that $l+l-2l=2001$ and put $x_{n}=a_{n}-l$ then $x_{n+2}+x_{n+1}-2x_{n}=0$
The caracteristic polynomial of this equation is:$r^2+r-2=0$ and its roots are $r_{1}=1$ & $r_{2}=-2$
Then in this case we have $x_{n}=\alpha(1)^n+\beta(-2)^n$ and then $a_{n}=\alpha(1)^n+\beta(-2)^n+l$
Now just solve the system  $a_{0}=\alpha(1)^0+\beta(-2)^0+l$      &     $a_{1}=\alpha(1)^1+\beta(-2)^1+l$
I think now its clear  :D

[quote="AYMANE"]take $l$ such that $l+l-2l=2001$[/quote]

What? :huh:

Ok :?  Is there any $l$ such that $l + l -2l = 2001$? (No,I think.)

And Anybody Help me please! More Clearier!!  :(

C'est vraiment affreux ce j'ai ecrit

Sorry !!!!!!!!!!!!!!!!! :blush:


mathnerd314 plays a game with 4 numbers:

He writes down 4 numbers a, b, c, and d.

In the next row he writes down 4 new numbers: |a-b|, |b-c|, |c-d|, |d-a|, and calls them e, f, g, and h, respectively.

In the next row he writes down 4 new numbers: |e-f|, |f-g|, |g-h|, |h-e|, and calls them i, j, k, and l, respectively.

He continues this process, using random symbols when he runs out of English letters :lol:, until the four numbers in the last row are all the same. What will the numbers all turn out to be? Prove your assertion.

[quote="xfactor"][hide]
Choose your four numbers as 0,x,2x,x for some x > 0.
Then the next, and last, row is x,x,x,x.
Since x can be chosen arbitrarily, we can't say exactly what the four numbers are, even though we may be want to prove how they are related algebraically.

[/hide]

Actually, according to the statement of the problem, you could just start by writing down x,x,x,x for any choice of x and end the process there since all 4 numbers are the same.  But I don't think this is what was meant.[/quote]
Yes, but what about any arbitrarily chosen a, b, c, and d?
I for one am stumped. I tried going thorough a few interations with symbols and nothing popped out at me.

What can you say about all four numbers after a few iterations? Prove.

[hide="hint"]
[hide="are you sure?"]
[hide="positive?"]
Infinite descent
[/hide][/hide][/hide]

[quote="mathnerd314"]mathnerd314 plays a game with 4 numbers:

He writes down 4 numbers a, b, c, and d.

In the next row he writes down 4 new numbers: |a-b|, |b-c|, |c-d|, |d-a|, and calls them e, f, g, and h, respectively.

In the next row he writes down 4 new numbers: |e-f|, |f-g|, |g-h|, |h-e|, and calls them i, j, k, and l, respectively.

He continues this process, using random symbols when he runs out of English letters :lol:, [b]until the four numbers in the last row are all the same.[/b] What will the numbers all turn out to be? Prove your assertion.[/quote]

[hide]Try the perfect squares.

1 4 9 16
3 5 7 15
2 2 8 12
0 6 4 10
6 2 6 10
4 4 4 4.

The number is 4 in this case.

Try the first few natural numbers.

1 2 3 4
1 1 1 3
0 0 2 2
0 2 0 2
2 2 2 2

Two different numbers?  :huuh: [/hide]

Whoops, my mistake. I guess I'll have to try that again...

$(m,n,o,p)=(56,42,88,102)$

$(q,r,s,t)=(14,46,14,46)$

$(u,v,w,x)=(32,32,32,32)$

It has something to do with the numbers originally chosen, but I'm not sure what.

[b]EDIT:[/b] Hmm, I just noticed that all of the numbers are powers of 2. Coincidence or not?

Well, it can be any number, because you just write down four numbers taht are the same.

[quote="PenguinIntegral"][quote="236factorial"]Let's see...

1 7 21 40
6 14 19 39
8 5 20 33
3 15 13 25
12 2 12 22
10 10 10 10

No. 10 isn't a power of two.[/quote]
Well, not an integer power........[/quote]
Well then you could say all e powers of two, it might have integer power or it might not

You never know if mathnerd314 used decimals.

Whoops, my mistake in the wording of the problem. What I meant was "until all four numbers are zero". I guess I was trying to be too cryptic in the wording :lol:. There. But you still need to prove that it smoothes to zero, or alternately, that eventually all the numbers will be the same in all cases.

Powers of 2 are on the right track but will get run over if they just stay there. :rotfl:


calculus
integration
calculus computations

Hi:
Can anyone help me with the indefinite integral of $ x(x^5\plus{}1)^{\minus{}1/3}$?
I've tried everything I know.
Thanks

The indefinite integral is not elementary.  Are you sure you have the problem right?

That is exactly how it's written in the problem set. It might be a misprint though. However near that one there are some similar ones which I've already solved (for example, the indefinite integral of $ x^{\minus{}2}(2\plus{}x^3)^{\minus{}5/3}$, which even though it seems non-elementary, it can be solved with algebra tricks (i.e. a tricky substitution).[hide]$ u\equal{}(2\plus{}x^3)/x^3$[/hide]
So I believe I might be missing "the trick" for the other one.
Thanks.

There is no trick.  [url=http://integrals.wolfram.com]Wolfram[/url] gives the indefinite integral in terms of hypergeometric functions.


A 1L balloon (balloon 1) contains 1.0 mol of atoms of an ideal gas, and another balloon with the same volume  (balloon 2) contains 1.0 mol of ethylene at 273 K, considered a Van der Waals gas.

(a) In which balloon the pressure is higher when the volume is increased?

(b) In which balloon there is more free space between atoms/molecules?

(c) In which balloon the pressure is higher when the volume is decreased?

For starters,

Ideal Gas Equation: PV=nRT
Van der Waals Equation: $ (P\plus{}\frac{n^2a}{V^2})(V\minus{}nb)\equal{}nRT$

[hide="Solution?"]
a. Say we increase the volume by a factor of k:
    Ideal: $ PkV\equal{}nRT \implies P\equal{}\frac{nRT}{kV}$
    Van Der Waals: $ P\equal{}\frac{nRT}{kV\minus{}nb}\minus{}\frac{n^2a}{k^2V^2}$
    for ethylene, a=4.612, b=0.0582
    so

ideal: $ P\equal{}\frac{1(0.082)(273)}{k}$
van der Waals: P= $ \frac{1\cdot0.082\cdot273}{k\minus{}0.0582}\minus{}\frac{1}{k^2}$

as k becomes larger, I think the bottom one becomes bigger as k-.0582 increases at a slower rate than K^2 does as k becomes larger.

b. more free space in the ideal, since there is a volume correction in the vdW equation

c. if we replace k by 1/k for large k in a., $ P\equal{}1(0.082)(273)k$ for the ideal and $ \frac{1\cdot0.082\cdot273}{\frac{1}{k}\minus{}0.0582}\minus{}k^2$

so the first one will increase faster I think because the -k^2 gets big really quickly making the second pressure lower.
[/hide]

This is my (bad?) attempt to use what I learned in class...

Okay
dp/dV=-nRT/V^2 for ideal and -nRT/(V-nb)^2+a(n/v)^2 for real
Using the data from atkins it appears that rate of fall of pressure with volume increase is less in 2.
Ideal Gas molecules are spaced more far apart as the molecules themselves are points
and  the last is using same logic as above 1


The first term of an arithmetic sequence is $ 2$. The sum of the third and the sixth terms is $ 25$. What is the fourth term?

hello, given that $ a_1\equal{}2$ and $ a_3\plus{}a_6\equal{}25$ then we get with
$ a_3\equal{}2\plus{}2d$
and
$ a_6\equal{}2\plus{}5d$
solving this we get
$ d\equal{}3$ and $ a_4\equal{}11$
Sonnhard.


$ S\subset\mathbb N$ is called a square set, iff for each $ x,y\in S$, $ xy\plus{}1$ is square of an integer.
a) Is $ S$ finite?
b) Find maximum number of elements of $ S$.

Do you mean x is not equal to y? Because if it were, $ x^2\plus{}1$ would be a square, and that is clearly impossible.

Of course you should consider $ x\neq y\in S$.

a)The answer is no.Assume for country that an infinite such set exists let $a,b,c$ be three distinct elements from the set then $(ax+1)(bx+1)(cx+1)$ is a square for infinity many $x$ which contradicts Siegel's Theorem.

as the user Taha sayd the first part can be done like that :

i will just leave my proof here if anyone wants an elementary proof

we claim the answer is $No$

for the sake of contradiction (god i love this sentence ) say we have infinite numbers on $S$

take 2 of them like $a , b$ with $ab=m^2 -1$ let $a+b=u , m^2 - 1 =w^2 $ remember that $u$ is integer but $w$ is not.

now look at this polynomial $P(x)= (ax+1)(bx+1)$ as we all know if $x$ is an element of $S$  then $P$ is the square of an integer

look more closely and get $ P(x)=(wx)^2 + ux + 1$

now lets construct 2 little polynomials $Q$ and $R$ such that

$Q(x)= (wx + u/(2w) ) ^2 , R(x) = (wx + u/(2w) -1) ^2 $

after a little work and using $AM-GM$ we get $Q>P>R$

let $x$ be an element of $S$ we have$ wx + u/2w > sqrt(P(x))>wx + u/2w -1$ with this we have $sqrt(P(x))=floor_of (wx + u/2w) $

i dont know how to draw floor in here so im just gana say $g(x)$ is the floor function

now let $S(x) = P(x) - (g(wx + u/2w) ^2) $

now let $S(x)=0$ and $S(x + 1/w) > 0 $ (we can get this by a little amount of work) so by inductive step we get $S(x + n/w) > S(x+ (n-1)/w) $  but since this should give us inifnite solutions as $S(x)=0$ is a contradiction

and we are done

.
.
.

by the way $bump$ for part $b$

[quote=Alphatrion]Doesn't the question asks for if - Every two elements product increased by $1$ is a perfect square ? There are infinitely many if set $S$ has only $3$ elements.[/quote]

The question asks among all possible sets $S$, what is the largest possible cardinality (Note that from $a)$ such maximum exist).  If you claim that the answer is $k$, you must give an example of such $S$ with $k$ elements and prove that no such $S$ with $k+1$ elements exist.

just to leave this here my proof for part $a$

also gives us that for part $b$  the $max$ cardinality is atmost 3 since if its 4 my proof gives by taking 2 of them and the smaller root of $S$ and using the inductive step i sayd we can get a contradiction so its atmost
3

and for an example $1,3,8$ work

This is a relatively well-known problem and very difficult. Such sets are known as Diophantine $m$-tuples. The answer for part $b$ is 4: the above example $1,3,8$ can be extended into $1,3,8,120$.
For reference, see [url = http://oeis.org/A030063]A030063[/url] on OEIS.

then can someone point out the misstake in my solution?

maybe the problem is with the number 1 it self? i dont know can some one give an example of 4 elements without number 1 in it?

@above
$\{n,n+2,4n+4,16n^3+48n^2+44n+12\}$ for  $n \in \mathbb{N}$ works and there is no 1.

and also [quote]after a little work and using $AM-GM$ we get $Q>P>R$

let $x$ be an element of $S$ we have$ wx + u/2w > \sqrt{P(x)}>wx + u/2w -1$ with this we have $\sqrt{P(x)}=\lfloor (wx + u/2w)\rfloor $

now let $S(x) = P(x) - (\lfloor(wx + u/2w)\rfloor ^2) $

now let $S(x)=0$ and $S(x + 1/w) > 0 $ (we can get this by a little amount of work) so by inductive step we get $S(x + n/w) > S(x+ (n-1)/w) $  but since this should give us inifnite solutions as $S(x)=0$ is a contradiction [/quote]
This is unclear to me
And also isn't it a bit too hard for Iran 3rd round  :stretcher:

yes actually i found out the hole in my proof it still works for part a with a little editing but not for part 2


Galois Theory
superior algebra
superior algebra theorems

Is it true that if P is irreducible over the finite field F with p^q elements (p prime, q>0) then the Galois group of P over F is cyclic of order the degree of P?

Yes.  It's generated by the Frobenius map $ x \mapsto x^{p^q}$.


Repair your PC!

http://www.technetguru.com/host/27/

  dg

Just because it's math I might do this to my PC :D .

did you know that this really a sport ?
you play it with a hammer...

super cool. they also must make one with laptops. :)


[quote="236factorial"]What is the next number in this sequence:

2, 3, 6, 11, 20, 37, 68, _

[hide="hint"]the first three numbers are just numbers to start the sequence[/hide][/quote]
[hide]
I didn't actually realize the pattern until I realized the Fibonacci Sequence.

I believe the pattern is that the number is the sum of the previous 3 terms...

So the next term should be $68+37+20=125$[/hide]

[hide]the answer is 20+37+68=125 :idea: our answer[/hide]

[hide]
_2 3 6 11 20 37 68      121!!!!
__1 3  5  9  17  31   (53)
___2 2  4  8   14  (22)
____0  2  4  6    (8)

work down finding differences between terms and then worked back up.

Would this sequence be a cubic then?

 [/hide]

[quote="4everwise"]The solution: [hide]Don't look at it until you've posted  :D [hide]The next term is the sum of the previous three terms.

$t_n=t_{(n-3)}+t_{(n-2)}+t_{(n-1)}$
Therefore, the next term is $68+37+20=125$[/hide] [/hide][/quote]

weird how I didn't think this way and everyone else did. :P

[hide]Th three numbers before it added together equal the next term so 20+37+68=[b]125[/b][/hide]

may be you are wrong because \[ 2+3+6+11+20+37+68=147\]
and \[5!=120 \rightarrow 5!-\frac{5!}{4}=90 \] and \[147+90=236\] so now we understand that his name is absolutly correct. :police:

[hide="no looky"]
this pattern took me a while but i figured it out.
The pattern is the next number is the sum of the previous three numbers.
So 20+37+68=125
125 is my answer[/hide]

HEHEHE Jensen, you messed up on your addition 147+90=[b]237[/b] :D

[quote="jhollenbeck"]how does this relate to 236???? ;) maybe you should change your title :D[/quote]

Look at the first three numbers of the sequence  :)

:rotfl: i thought it had to do with your handle's number :rotfl:its the 236 starting the equation  :blush:


Not Enough Information, I believe.  I mean there's nothing to work with!  It could be like anything!

Thats wat the problem is , i think its solveable

Where did you see this problem and what is the answer?

this problem is truly impossible, unless you're psychic...
there isnt anything to work with!

call the distance d, and t as time and r as change in rate...

d= 50t
d= (t+1/2)(50-r) (took him longer to get there, so he must be at a slower speed)

using second equation we get:
d = 50t -rt +25 - 1/2 r
50t = 50t -rt + 25 -1/2r
25 = r(t+1/2)

this might help...

it still doesnt matter because he could live a mile away from Easton, and it could have been a really big traffic jam, or he could have lived really far away and the traffic jam could have been spread out over a long time
but, thats a good start if there is an answer, in which i would look like a complete moron :blush:

Everyone is right, there is no one answer.

He could live 25 miles away, and the traffic jam could slow him down to 25 mph.

Also, he could live 50 miles away, and the traffic jam slows him to 33.33 mph on the way back.

There are infinite answers.

Warm up 6 of what year? every thing can be found except the rate that he came back at, so this problem is truly impossible.

There probably was a chart associated with this problem.. sometimes workouts and warmups have those...

[quote="GoBraves"][quote="kidwithshirt"]:D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :rotfl:

nice one guys, its on warm up 6.... maybe u guys know wat that is, from ur coaches[/quote]

There's no such thing on warm up 6??? Or atleast for this year there isnt.[/quote]

sry brave but its on warm 5...

[quote="kidwithshirt"]Tom drove to Easton from home at 50mph. His return trip took 30 min longer because of traffic jam. How many miles is Tom's home to Easton?

I got no idea how to solve it  :([/quote]

[hide]
I'd try to make a chart.

                R                     T             D
 Go          50                     x          50x
 Back       (50x)/(x+30)     x+30    50x

But that's obviously getting nowhere...and charts don't work on these forums!  :blush:
[/hide]


It's April 25, which is next Saturday. Figured there might as well be a thread about it. :D

:spam:

Norcross will have a team.  Yay.

[quote="yaofan"]i know this is a random question but do u guys know elkins pointe middle school?[/quote]

I do, John Xie from last year's national mathcounts team went there.

Also, is it a calculator test? I was looking over the test and some seemed computational-ish...o.0

Congrats to all who attended!

BTW, on a completely unrelated note, I may be late to practice on Mother's Day since my parents are taking me to Fogo de Chau for a graduation luncheon.

Also, I will not be able to come to the practice on the May 17 practice due to science olympiad.

Congrats to Miles for keeping 1st place as a high school students only award  :wink: , and to Walton for having an yet another great year.

@ azn6021023 and timwu:
People are usually allowed to miss one practice for personal reasons, but be sure to tell one of the coaches at or before the first practice.

Congrats to everyone who placed and the arml pplz! Also to Walton! :D

[quote="1076207"]Don't worry Santosh; he still has next year...  :lol:[/quote]

I'm not worrying, since I will have graduated by then.  :)

I can't come to the first practice, our school has grad bash that weekend.


trigonometry
geometry
rectangle
geometric transformation
homothety
USAMTS
number theory

In right triangle $ABC$ with right angle $C$, $CA=30$ and $CB=16$.  Its legs $\overline{CA}$ and $\overline{CB}$ are extended beyond $A$ and $B$.  Points $O_{1}$ and $O_{2}$ lie in the exterior of the triangle and are the centers of two circles with equal radii.  The circle with center $O_{1}$ is tangent to the hypotenuse and to the extension of leg CA, the circle with center $O_{2}$ is tangent to the hypotenuse and to the extension of leg CB, and the circles are externally tangent to each other.  The length of the radius of either circle can be expressed as $p/q$, where $p$ and $q$ are relatively prime positive integers.  Find $p+q$.

By Pythagoras, $AB = 34$.  Let $I_{C}$ be the $C$-excenter of triangle $ABC$.  Then the $C$-exradius $r_{C}$ is given by
\[r_{C}= \frac{K}{s-c}= \frac{240}{40-34}= 40.\]

[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/6cfdbce39cab426dc78fdd985c9fcca3.png[/img]

The circle with center $O_{1}$ is tangent to both $AB$ and $AC$, which means that $O_{1}$ lies on the external angle bisector of $\angle BAC$.  Therefore, $O_{1}$ lies on $AI_{C}$.  Similarly, $O_{2}$ lies on $BI_{C}$.

Let $r$ be the common radius of the circles with centers $O_{1}$ and $O_{2}$.  The distances from points $O_{1}$ and $O_{2}$ to $AB$ are both $r$, so $O_{1}O_{2}$ is parallel to $AB$, which means that triangles $I_{C}AB$ and $I_{C}O_{1}O_{2}$ are similar.

The distance from $I_{C}$ to $AB$ is $r_{C}= 40$, so the distance from $I_{C}$ to $O_{1}O_{2}$ is $40-r$.  Therefore,
\[\frac{40-r}{40}= \frac{O_{1}O_{2}}{AB}= \frac{2r}{34}\quad \Rightarrow \quad r = \frac{680}{57}.\]
Hence, the final answer is $680+57 = 737$.

Nice solution Naoki.  :D

Here's how I solved the problem during the test:

Note that the lines from the vertices of the triangle to the centers of the circles bisect the exterior angles. Additionally, the angles formed (after bisecting) are equal to half the supplement of the angles of the right triangle.

So... draw altitudes from the centers of the circles to the hypotenuse. This divides the hypotenuse into three segments. The middle segment has length $2r.$

Then use the tangent half angle formula to find relationships between $r$ and the remaining two segments. Solve for $r.$

So I guess my solutions was similar to the one by bpms, except you only need to solve a linear equation for r. Hope that's not too sketchy.  :roll:

I did it exactly like you 4everwise, but when I added $2r+\frac{r}{4}+\frac{3r}{5}$, I made it $\frac{37r}{20}$, rather than $\frac{57r}{20}$.  Grrrr...

Anyway, I thought this was a really nice problem, especially because I had just reviewed my trig identities the night before as part of getting ready...

[img]7896[/img]

Is my way of constructing the radius in sketchpad. Any new ideas of solutions from it? (note the homothecy to construct g...)

Start with a scaled 16-30-34 triangle.  Inscribe a circle.  The height, $h,$ and radius, $r,$ are found via $A=\frac{1}{2}\times 16s\times 30s=\frac{1}{2}\times 34s\times h=\frac{1}{2}\times rp,$ where $p$ is the perimeter.

Cut the figure through the circle and perpendicular to the hypotenuse.  Slide the two pieces in opposite directions along the hypotenuse until they are one diameter of the circle apart.  Complete the two partial circles.

The linear dimensions of the new triangle are $\frac{46s}{34s}=\frac{23}{17}$ times the size of the original.  The problem's 16-30-34 triangle sits above the circles.  Equate heights and solve for $r=6s$:

\begin{eqnarray}\frac{240s}{17}\times\frac{23}{17}& = & \frac{240}{17}+12s \\ 20s\times 23& = & 20\times 17+s\times 17\times 17 \\ s & = & \frac{340}{171}\\ r = 6s & = & \frac{680}{57}\end{eqnarray}

The answer is $\boxed{737}$

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=138340#138340]USAMTS Year 16, Round 2, Problem 5[/url] for a simpler application of this method.

[quote="nsato"]By Pythagoras, $ AB \equal{} 34$.  Let $ I_{C}$ be the $ C$-excenter of triangle $ ABC$.  Then the $ C$-exradius $ r_{C}$ is given by
\[ r_{C} \equal{} \frac {K}{s \minus{} c} \equal{} \frac {240}{40 \minus{} 34} \equal{} 40.
\]
[/quote]

Could someone tell me where does this formula come from??

$ [CI_{C}A] \equal{} \frac{1}{2}\cdot r_{C}\cdot b$
$ [CI_{C}B] \equal{} \frac{1}{2}\cdot r_{C}\cdot a$
$ [BI_{C}A] \equal{} \frac{1}{2}\cdot r_{C}\cdot c$

$ K \equal{} [ABC] \equal{} [CI_{C}A]\plus{}[CI_{C}B]\minus{}[BI_{C}A] \equal{}\frac{a\plus{}b\minus{}c}{2}\cdot r_{C}$.
$ \frac{a\plus{}b\minus{}c}{2}$ also equals $ s\minus{}c$. Thus, $ r_{c} \equal{} \frac{K}{s\minus{}c}$

First see this [hide="completely accurate, sophisticated, and to-scale Paint diagram"][img]http://postimage.org/image/5t09xi8rr/[/img][/hide]. To clarify, $S$ and $T$ are the feet of the radii to the intersections of tangent $\overline{AB}$ with $O_1$ and $O_2$ respectively. Let $r$ be the radius of each circle, so $ST=2r$.

$\overline{O_1B}$ and $\overline{O_2A}$ bisect exterior angles $\angle B$ and $\angle C$ respectively. Thus we have

$\tan{\angle{O_1BA}}=\tan{\left(\frac\pi 2-\frac{\angle{ABC}} 2\right)}=\cot{\frac{\angle{ABC}}2}=$
$\sqrt{\frac{1+\cos{\angle{ABC}}}{1-\cos{\angle{ABC}}}}=\sqrt{\frac{1+\frac{16}{34}}{1-\frac{16}{34}}=\frac53}$. Thus $BS=\frac35 r$.

Similarly,

$\tan{\angle{O_2AB}}=\tan{\left(\frac\pi 2-\frac{\angle{BAC}} 2\right)}=\cot{\frac{\angle{BAC}}2}=$
$\sqrt{\frac{1+\cos{\angle{BAC}}}{1-\cos{\angle{BAC}}}}=\sqrt{\frac{1+\frac{30}{34}}{1-\frac{30}{34}}=4}$. Thus $AT=\frac14 r$.

Solving $\frac35 r + 2r + \frac14 r=34$ gives $r=\frac{680}{57}$.

[quote=bpms][hide]Let the point where CB's extension hits the circle be D, and the point where the hypotenuse hits that circle be E. Clearly EB=DB. Let EB=x. Draw the two perpendicular radii to D and E. Now we have a cyclic quadrilateral. Let the radius be length r. We see that since the cosine of angle ABC is $\frac{15}{17}$ the cosine of angle EBD is $-\frac{15}{17}$. Since the measure of the angle opposite to EBD is the complement of this one, its cosine is $\frac{15}{17}$. Using the law of cosines, we see that $x^{2}+x^{2}+\frac{30x^{2}}{17}=r^{2}+r^{2}-\frac{30r^{2}}{17}$
This tells us that r=4x.

Now look at the other end of the hypotenuse. Call the point where CA hits the circle F and the point where the hypotenuse hits the circle G. Draw the radii to F and G and we have cyclic quadrilaterals once more.
Using the law of cosines again, we find that the length of our tangents is 2.4x. Note that if we connect the centers of the circles we have a rectangle with sidelengths 8x and 4x. So, 8x+2.4x+x=34. Solving we find that $4x=\frac{680}{57}$ so our answer is 737.[/hide][/quote]

Sorry for revive, can someone explain why cosine(ABC) is equal to 15/17 and not 8/17?

http://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_9

Here is wiki entry complete with diagram

Thanks

They use $\theta = \angle CAB$ so $\cos CAB = \frac{15}{17},$ which is also $\sin ABC$

[quote=nathantareep][quote=bpms][hide]Let the point where CB's extension hits the circle be D, and the point where the hypotenuse hits that circle be E. Clearly EB=DB. Let EB=x. Draw the two perpendicular radii to D and E. Now we have a cyclic quadrilateral. Let the radius be length r. We see that since the cosine of angle ABC is $\frac{15}{17}$ the cosine of angle EBD is $-\frac{15}{17}$. Since the measure of the angle opposite to EBD is the complement of this one, its cosine is $\frac{15}{17}$. Using the law of cosines, we see that $x^{2}+x^{2}+\frac{30x^{2}}{17}=r^{2}+r^{2}-\frac{30r^{2}}{17}$
This tells us that r=4x.

Now look at the other end of the hypotenuse. Call the point where CA hits the circle F and the point where the hypotenuse hits the circle G. Draw the radii to F and G and we have cyclic quadrilaterals once more.
Using the law of cosines again, we find that the length of our tangents is 2.4x. Note that if we connect the centers of the circles we have a rectangle with sidelengths 8x and 4x. So, 8x+2.4x+x=34. Solving we find that $4x=\frac{680}{57}$ so our answer is 737.[/hide][/quote]

I am sorry though your solution is right and you're offline but cos abc is equal to 8/17[/quote]

And the ratios will be r=5/3y and r=4x,so except your answer everything is wrong.

Let the circle with center $O_1$ meet line $AC$ at point $D$ and line $AB$ at point $F$, and let the circle with center $O_2$ meet line $BC$ at $E$ and meet $AB$ at point $G$. Extend $DO_1$ and $EO_2$ past $O_1$ and $O_2$, respectively, so that they meet at point $K$. Note that quadrilaterals $EO_2GB$ and $DO_1FA$ are cyclic, so by a simple angle chase we find that triangle $KO_1O_2$ is similar to triangle $ABC$.

We have $O_1O_2=2r$, so $KO_1=\tfrac{16r}{17}$ and $KO_2=\tfrac{30r}{17}$. Since $BG=BE=\tfrac{16r}{17}+r-16$ and $AF=AD=\tfrac{30r}{17}+r-30$, we find that $$\frac{30}{17}r+r-30+\frac{16}{17}r+r-16+2r=AB=34,$$ so $r=\tfrac{680}{57}$, and our answer is $\boxed{737}$.


analytic geometry
number theory
greatest common divisor
geometry
The Geometry of Numbers

Prove that if a lattice triangle has no lattice points on its boundary in addition to its vertices, and one point in its interior, then this interior point is its center of gravity.

Let $ \Delta OPQ$ be a triangle under the consideration.
Let one of the triangle's $ \Delta OPQ$ vertices be at the origin of the integer coordinates (say  $ O(0,0)$ ) , $ C$ be the point inside of $ \Delta OPQ$  and
 $ (x_1, y_1)$, $ (x_2, y_2)$ and $ (x_3,y_3)$ be the coordinates of $ P$,  $ Q$ and  $ C$, respectively.
Without loss of generality let's say $ 0 \leq x_i \leq y_i$ (it's the same the other way round).
Now, we observe that the point $ C$ is the common vertex of three small triangles  inside of  $ \Delta OPQ$: that is  $ \Delta OCP$, $ \Delta CPQ$ and $ \Delta CQO$.
  Firstly,   we notice that from the point  $ C$ the points  $ O$, $ P$ and $ Q$ are visible, since there are no other lattice points on  the sides $ CO$ , $ CP$ and $ CQ$.
 Secondly,  we observe that the points $ P$ and $ Q$ are both visible from the origin $ O$.Then $ gcd (x_i;y_i)\equal{}1$, and so  $ 0\leq \frac {x_i}{y_i} \leq1$ for $ i\equal{} 1,2,3$.
Therefore, $ \frac  {x_i}{ y_i}$ for any $ i$ ( $ i\equal{}1,2,3$)  belong to the Farey sequence.

     Hence, $ \Delta OCP$,  $ \Delta CPQ$ and $ \Delta CQO$  are the primitive lattice triangles.  The area  of any of these triangles  is the same, equal to $ \frac {1}{2}$.
     So, the  lattice point $ C$  is the center of gravity.

Using Pick's theorem, the area of $ ABC$ is $ \frac{3}{2}$, The area of each triangle $ \triangle{APC},\triangle{APB},\triangle{BPC}$ is $ \frac{1}{2}$ hence, point $ P$ divides the area in three equal parts, therefore it must be the baricenter.


How is the rule that $ a\equiv b \pmod{n}, a\equiv b \pmod{m} \Rightarrow a\equiv b \pmod{\text{lcm}{n}}$ follow from the Chinese remainder theorem? In some sense, it looks like it [i]is [/i]the ch. remainder theorem.

I assume you mean $ a\equiv b\pmod{\gcd(mn)}$ for the third one?

It is obvious that one solution to $ a\equiv b\pmod{m}, a\equiv b\pmod{n}$ is $ a\equiv b\pmod{\gcd(mn)}$, and CRT shows uniqueness.

:oops: I was confused by the notation.  But it is clear. Thanks!


Manu and Janani are playing a coin toss game with a fair penny. Manu gets a point if the penny lands on heads, and Janani gets a point if the penny lands on tails. The score is Janani 9, Manu 7, in a game to 10 points. What is the probability that Janani will win the game?

The only way Janani loses is if 3 straight flips land on tails.  So Janani wins with $ T$, $ HT$, or $ HHT$ and Manu wins only with $ HHH$.  Hence $ P(\text{Janani wins})\equal{}1\minus{}P(\text{Janani loses})\equal{}1\minus{}P(HHH)\equal{}1\minus{}\frac{1}{8}\equal{}\frac{7}{8}$.


geometry
3D geometry
linear algebra
matrix
number theory unsolved
number theory

Let $S$ be a set consisting of $43$ distinct positive integer not exceeding $100$. For each subset $X$ of $S$ let $t_{X}$ be the product of its elements. Prove that there exist two disjoint subsets $A$ and $B$ of $S$ such that $t_{A}t_{B}^{2}$ is the cube of a positive integer.

First note that $\pi(100)=25$ (i.e., there are 25 primes below 100).
Consider a matrix $M$ of size $43\times 25,$ where element $(i,j)$ equals the degree of $p_{j}$ (the $j$-th prime) in the factorization of the $i$-th element of $S$.

We will consider the matrix $M$ over the field $\mathbb{Z}_{3}=\{0,1,2\}.$ It is clear that the rank of $M$ does not exceed 25 (the number of columns). Since the number of rows in $M$ is greater than the number of columns, there exists a non-trivial linear combinations of rows of $M$ equal zero vector. In such a combination each row comes with a coefficient from $\mathbb{Z}_{3}.$ Let $A$ be a subset of $S,$ containing elements corresponding to rows coming with the coefficient 1; and let $B$ be a subset of $S,$ containing elements corresponding to rows coming with the coefficient 2. It is clear that $A$ and $B$ are disjoint, and that $t_{A}t_{B}^{2}$ is a cube of an integer.

P.S. The size of $S$ in this problem can be safely reduced to $\pi(100)+1=26$.


Was the above post cut off? Is there a quick solution to this kind of problem?

[hide]Label the columns 1-6 and the rows A and B. There is one path to B1, A1+B1=2 paths to A2, B1+A2=3 paths to B2, and so on. We we see that this pattern is the Fibonacci sequence, with square A1 being $F_1$. It takes $12$ steps moving alternately up and down diagonally to reach $B$, so there are $F_{12}=144$ ways to move from $A$ to $B$.
[/hide]


[u][i][b]Meet the Mod.[/b][/i][/u]
Hi. I'm meewhee if you did not know.
I attempted to mod an original game in the past, yet that did not succeed. Since I'm as active as ever. Why not?
Here's an entertaining little thing I made up from the top of my head....
[u][i][b]The Intro[/b][/i][/u]
3000 years ago, from this date...
a young princess wondered the halls of an abandoned cottage... hiding.
You see, the dictators from the other kingdom down south, have attempted to take over the kingdom out of anger.
The young, evil prince... Marius, of that kingdom has attempted many times to get the young princess, (Larissa) to marry him. Those many attempts have failed. Which then finally realizing the perfect plan of taking over the kingdom... and holding it hostage for the princesses love.
This plan would have worked perfectly you see. Yet the princess has fallen for a duke of another kingdom, Kristofer. Kristofer heard of the attack and worked with the kingdom to keep Larissa safe. That plan is somewhat failing. The guards from the kingdom of Pawn, (Marius' kingdom) have tracked down Larissa's placing, due to the trader on Larissa's and Kristofer's side. Oh, the heartache.
Larissa then had to somehow improvise so the guards would not find her. She threw her dress at a distant tree before making her way into the cottage, she was wearing her undergarments... which are not very comfortable in the Autumn.
She was pacing back and forth... hoping that Kristofer would save her.
2 hours have passed, and she was still alone.
It was both good and bad. The guards had no trace of her, but Kristofer and her kingdom had no trace or help either.
She was still pacing, it was dreadful. She didn't know how to feel.
As I have stated before, the cottage is abandoned, no furnishing at all. It took Larissa a while to realize there was a second floor of the house. Except the cottage did not have stairs, only a dusty old ladder.
She somehow managed to get up there, using the ladder and the panel which lead to the second floor. The first thing she noticed in the room, was a closet. It looked clean, an ivory cabinet.  She wondered... why the previous family leave a beautiful closet in an abandoned floor?
She worked her way in front of the door. Then attempted to open it. It was tightly sealed... but her days as a peasant somehow came in handy, it was just like carrying the old sickle.
When she opened it, a wisp flew out.
"Oh how joyous to leave the horrid closet!" The wisp sang while frolicking around the room.
Larissa looked at it in awe. A wisp? In a closet? Hmm. Peculiar.
The wisp then looked at Larissa.
"I sense a heartache. Is it you young one?"
Larissa cocked her head, "You can sense heartache? Strange power of a wisp."
The wisp's mood seemed serious, "Yes, yes... I can sense a lot. Who is this 'Kristofer' fellow?" The wisp asked.
Larissa then shocked said, "Mind reader?"
"Mhmm." The wisp then said.
Larissa felt lonely, she needed to speak to someone.
She then told the wisp her story.
The wisp... in shock said, "Being in the presence of a princess brings gratitude to the soul."
Larissa smiled. The first time in a while.
The wisp then had an epiphany, "I got it. I'll bring back knights from the future to help you. A quite nice quest, right?"
Larissa's smile grew wider, "That would be very nice wisp friend. I do need help in times like this."
Larissa crossed her arms and shrugged. "Hopefully they'll fulfill their duties."
"I'll make sure." The wisp said... before disappearing.
[b][i][u]Status. Rules & Sign-Ups[/u][/i][/b]
Status & Reputation;
Every miniature task within the quest will be rewarded with status points.
For example;
You have arrived in the kingdom, what would you like to do next?
a) Talk to the townspeople (+2 popularity)
b) Go study the kingdom in the local library (+2 knowledge, +1 happiness)
There are 5 different types of status:
- Popularity
- Strength
- Trust
- Knowledge
- Health
- Happiness
They interact with the storyline and at the end of the game, YAY REWARDS! :)

Action Items;
At some point in the game I will send you an individual choice.
For example:
The young boy in the shelter next door is hungry, what to do?
a) Steal food from team and give it to the boy (-3 Trust, +3 Popularity)
b) Ignore (+3 Trust, -3 Popularity)

[b]Do not speak of your action item if not public. Some will go through PM.[/b]
[b]This game is about strategy. Remember... think, think and think.[/b]
[b]Yeah, alliances are allowed.[/b]
[b]Remember: Activity. The princess' fate is now in your hands. Alert me if inactivity comes up, I hate having to find out myself.[/b]
[b][i][u]Sign-Ups[/u][/i][/b]
There will be 8 needed players. You will start out with 25 status points, please place them when signing up,
example sign-post;
"/in
Popularity - 5
Knowledge - 5
Strength - 5
Trust - 5
Health - 5"

Got it? Okay.

1. Chompy
2. EggyLv.999
3. monkeygirl13
4. westiepaw
5. spaceguy524
6. royalflower
7. bluecarneal
8. jjfun1

(Oh and updating the sign-up list is nice too)
(If still unclear, PM me, I'll try to explain)

hmmmmm...should i join.......
why not I have about 5 games running anyway

9 health
6 happiness
5 knowledge
1 strength
4 popularity

I vote we all alliance against ernie :)
no, jk.

Popularity - 1
Knowledge - 10
Strength - 3
Trust - 5
Health - 7

I'm in!

1. Chompy
2. EggyLv.999
3. monkeygirl13
4. westiepaw
5.
6.
7.
8.

Popularity-2
Knowledge-7
Strength-5
Trust-5
Health-6

3 more spots...
I wanna start tomorrow  :)

I'm in!
Popularity - 5
Knowledge - 5
Strength - 5
Trust - 5
Health - 5

How do I update the list?

Popularity - 2
Knowledge - 7
Strength - 8
Trust - 3
Health - 5

in

1. Chompy
2. EggyLv.999
3. monkeygirl13
4. westiepaw
5. spaceguy524
6. royalflower
7. bluecarneal
8.

Whew, last spot!
Popularity - 5
Knowledge - 5
Strength - 5
Trust - 5
Health - 5

1. Chompy
2. EggyLv.999
3. monkeygirl13
4. westiepaw
5. spaceguy524
6. royalflower
7. bluecarneal
8. jjfun1

The wisp set out.
It got lost in the space-time continuum. Joy.
It then finally arrived in the year 2009,
it was late at the time... 8 AoPS users just logged off to then go to sleep.
Right before they crawled under the blankets,
the wisp used it's power to transport them into the princess's cottage. It tagged along with them.
Before you knew it, the AoPS users arrived, with a thump to the ground following.
They sat before Princess Larissa herself,
when the wisp arrived,
everyone was in awe.
What was that speaking spot of light?
"Princess, according to what I have learned, the future predicts these to save the land."
The users looked at each other. Dumbstruck.
The princess and wisp then had a quiet conversation with each other.
The princess then looked at the AoPS users,
"You are on a quest, to protect me from the evil. I would like you to learn and defend myself. Somehow conquer Marius' wrath. Along the way bring Kristofer back to me... " she said dreamy eyed.
The wisp then said the rest.
"Your first task is to choose your next destination... "
Where do you want to go?
a) Colony of Maldor (specializes in mining and farming, independent, broken off from kingdoms)
b) Kingdom of Spark (Larissa's kingdom. Marius holding hostage)
c) Kingdom of Galamor (A kingdom specializing in knowledge and gossip)
d) Kingdom of Gaelwyn (Kristofer's kingdom, at the moment planning alliances and strategy)
e) Kingdom of Pawn (Marius' kingdom. Base of attack)

[img]http://i134.photobucket.com/albums/q93/candygirlmaddie/scan0001.jpg[/img]

*Notice 'Home Base' you will be redirected there if no decision is made soon.

[b]I give you till Friday.[/b]
[b]P.S. Do not comment on my mad drawing skillzz[/b]

I pick:
[b]c) Kingdom of Galamor (A kingdom specializing in knowledge and gossip) [/b]

BTW: You made a typo.[quote="meewhee009"]"Your first task is to choose [b]you[/b] next destination... " [/quote]

/picks up sharpened stick for a weapon. /walks over to the exit.

Butler and his stick go toward the exit.
Continue?

I run back to the others. I ask Marius, "Where would you suggest we go?"

Marius looks at Butler and says,
"Alas, I do not know what is best." He sighs,
"Not even the designated overseer can see what will happen in the future." He says, with a blank facial expression.

I run out of the building and look for any secret entrances etc.

I wake up and start to throw stones at a target on the floor.

Butler does so,
he sees two tunnels while exploring.
What to do? Carry on? Tell the others?

And spaceguy throws rocks at a target drawn on the sand.

Bluey's Optionos.
a) Discuss strategy
b) Go to the tunnels and decide
c) Talk to people in the fort
d) Ask Marius (something)


function
Putnam
real analysis
real analysis unsolved

I can't remember where I saw this problem; probably on a Putnam practice exam I found somewhere on the net.

Let $A$ be the set of all $a > 0$ such that for every continuous function $f$ with $f(0) = f(1) = 0$, there is an $x \in [0,1]$ such that $f(x) = f(x+a)$.  Find all numbers in $A$.

I know how to prove that $1/n \in A$ for every $n \in \mathbb{N}$, but I don't know how to prove no other numbers are in $A$ (which I believe is true).  Any thoughts?

Just wanted to note that it is true, and it's been mentioned before on the forum in several places. Here's one such thread, even though no solution is posted there: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30466[/url].


[color=darkblue]Let $ xyz \equal{} a\neq 0$, let $ x \plus{} y \plus{} z \equal{} \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 0$. Find $ A \equal{} \frac {x^6 \plus{} y^6 \plus{} z^6}{x^3 \plus{} y^3 \plus{} z^3}$ by $ a$?[/color]

its very easy :maybe:

[hide]1/x + (y+z)/yz =0 =>
1/x + -x / a/x =0 =>
x^3 =a[/hide]

[quote="kunny"]Since $ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx \equal{} 0$ and $ x^2 \plus{} y^2 \plus{} z^2 \equal{} (x \plus{} y \plus{} z)^2 \minus{} 2(xy \plus{} yz \plus{} zx) \equal{} 0$

$ \therefore x^3 \plus{} y^3 \plus{} z^3 \equal{} 3xyz \equal{} 3a,\ x^6 \plus{} y^6 \plus{} z^6 \equal{} (x^2)^3 \plus{} (y^2)^3 \plus{} (z^2)^3 \equal{} 3(xyz)^2 \equal{} 3a^2$.

$ \therefore \frac {x^6 \plus{} y^6 \plus{} z^6}{x^3 \plus{} y^3 \plus{} z^3} \equal{} \frac {3a^2}{3a} \equal{} a$.[/quote]

[color=darkblue]Thank Kunny very much. [/color]

You are welcome. :)

$ \boxed{A^3\plus{}B^3\plus{}C^3\minus{}3ABC\equal{}(A\plus{}B\plus{}C)(A^2\plus{}B^2\plus{}C^2\minus{}AB\minus{}BC\minus{}CA)}$

For $ A\plus{}B\plus{}C\equal{}0$,  we have $ A^3\plus{}B^3\plus{}C^3\equal{}3ABC$.

If $ x \plus{} y \plus{} z \equal{} 0 \equal{} xy \plus{} yz \plus{} zx,xyz \equal{} a\not \equal{} 0$, then $ x,y,z$ are 3 complex roots $ t^3 \equal{} a$, therefore $ x^3 \equal{} y^3 \equal{} z^3 \equal{} a,x^6 \equal{} y^6 \equal{} z^6 \equal{} a^2.$

To be honest, I imagined Vieta's formula, too.


Let $ a,b,c\ge 0$ such that $ a\plus{}b\plus{}c\equal{}1$. Prove that $ a^3\plus{}b^3\plus{}c^3\le 3(1\plus{}abc)$.

$ \sum a^3\equal{} (a\plus{}b\plus{}c)^3\minus{}3(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\plus{}3abc\equal{}1\minus{}3(ab\plus{}bc\plus{}ca)\plus{}3abc\leq 3(1\plus{}abc)$ $ \Leftrightarrow$
$ 0\leq 2\plus{}3(ab\plus{}bc\plus{}ca)$

Is this really the problem you wanted to post ?

[quote="nayel"]Let $ a,b,c\ge 0$ such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that $ a^3 \plus{} b^3 \plus{} c^3\le 3(1 \plus{} abc)$.[/quote]
  $ a^3 \plus{} b^3 \plus{} c^3 <3 \le 3(1 \plus{} abc)$.

Yes this was very easy. :blush:  Actually the origin was the following inequality:
\[ a^3\plus{}b^3\plus{}c^3\le \sqrt{3(a^3\plus{}b^3\plus{}c^3\minus{}3abc)}\plus{}3abc\]
However, the following inequality for $ a\plus{}b\plus{}c\equal{}1$ holds too:
\[ a^3\plus{}b^3\plus{}c^3\le \sqrt{3(a^3\plus{}b^3\plus{}c^3\minus{}a^2b\minus{}b^2c\minus{}c^2a)}\plus{}3abc\]

Nobody interested in the last inequality? :maybe:

[quote="nayel"]Yes this was very easy. :blush:  Actually the origin was the following inequality:
\[ a^3 \plus{} b^3 \plus{} c^3\le \sqrt {3(a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc)} \plus{} 3abc
\]
[/quote]
Let $ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}t$,we have $ t \leq \sqrt{3t}$ or $ t \leq 3$ ,not true for $ a,b,c \geq 0$ :wink:

[quote="nayel"][quote="dduclam"][quote="nayel"]Yes this was very easy. :blush:  Actually the origin was the following inequality:
\[ a^3 \plus{} b^3 \plus{} c^3\le \sqrt {3(a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc)} \plus{} 3abc
\]
[/quote]
Let $ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} t$,we have $ t \leq \sqrt {3t}$ or $ t \leq 3$ ,not true for $ a,b,c \geq 0$ :wink:[/quote]

If you read the original problem, it included $ a \plus{} b \plus{} c \equal{} 1$. And Svejk proved it too. That condition also applies to this problem. Please read above posts before posting something. :)[/quote]
$ a,b,c\geq0$? If $ a,b,c\geq0$and$ a\plus{}b\plus{}c\equal{}1$ we have $ t\leq3$ always true :wink:
I think you should write clearer in your post  :wink:

[quote="nayel"]Sorry if I made it difficult for you to understand, my bad. :oops:  The inequality you are talking about has been already proved to be easy. I was asking for this one:

For non-negative reals $ a,b,c$ such that $ a \plus{} b \plus{} c \equal{} 1$ prove that
\[ a^3 \plus{} b^3 \plus{} c^3\le \sqrt {3(a^3 \plus{} b^3 \plus{} c^3 \minus{} a^2b \minus{} b^2c \minus{} c^2a)} \plus{} 3abc
\]
[/quote]
ineq $ \Leftrightarrow (a^3\plus{}b^3\plus{}c^3\minus{}3abc)^2\le 3(a^3\plus{}b^3\plus{}c^3\minus{}a^2b\minus{}b^2c\minus{}c^2a)$
$ \Leftrightarrow (a\plus{}b\plus{}c)^2(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)^2 \le 3(a^3\plus{}b^3\plus{}c^3\minus{}a^2b\minus{}b^2c\minus{}c^2a)$
$ \Leftrightarrow (a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)^2 \le \sum_{cyc}(2b^3\plus{}c^3\minus{}3b^2c)\equal{}\sum_{cyc}(2b\plus{}c)(b\minus{}c)^2$
Note that $ 0 \le (a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)^2 \le a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca < 1$
We only need prove that $ a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca\equal{}\frac1{2}\sum_{cyc}(b\minus{}c)^2\le \sum_{cyc}(2b\plus{}c)(b\minus{}c)^2$
$ \Leftrightarrow \sum_{cyc}(2b\plus{}c\minus{}\frac1{2})(b\minus{}c)^2 \ge 0$ true by S.O.S $ \rightarrow (Q.E.D)$ :D


inequalities
LaTeX
inequalities unsolved

Can anyone find [b]nice solutions [/b]to these two inequalities I attached?  Thanks!

please write them down in separate topics. you can use LateX to make your post look nicer (a latex tutorial is located on the main page -> click on the [read-full] button :D)


AMC
AIME
AMC 10
trigonometry
AMC 12
Pythagorean Theorem
geometry
\/closed

I'm in 8th grade, and I average about a 125 on the AMC 10 and about 105 on the AMC 12. I am going to take the AIME problem series. Should I take one of the AMC 10/12 problem serieses as well?

I guess I'm in a similar situation.  I am taking the AMC 12 problem series but not the AMC10 one, because I can pretty much, assuming I don't make dumb mistakes (which doesn't happen much :oops: ), full score most of them.  I heard that the level of AMC 12 problems can be compared to the problems in mid AIME's.

[quote="Phelpedo"]Maybe 23,24,25, probably not anything lower (although the recent AIMEs have featured some very trivial early problems).[/quote]
You mean like problem one of AIME 2006 II (just use pythagorean theorem twice).

I think that was AIME I but AIME II was trivial too.

I am in the AMC 10 problem series and i want to know if this is the virtual classroom..

Let's just say that someone, cough *me* cough, sucks at trig. Would the AMC 12 prob. series still be good for me?

[quote="Phelpedo"]Probably. Hard to say if you give us such vague data.[/quote]
If $\tan x+\tan y=25$ and $\cot x+\cot y=30$, what is $\tan(x+y)$?
This problem was number 3 from an AIME in the early 80s. I drew a right triangle and used the info given to solve it, and spent 30 min.

[quote="Phelpedo"]Ok, so it looks like you might need more strength in basic identities and stuff (read the AoPS book, I suppose). At first glance it looks like the identity $\tan(x+y) = \frac{\tan x+\tan y}{1-\tan x \tan y}$ might be useful  :wink:

I'm quite sure that AMC12 Problem Series will be very helpful to you, even if it's too hard. The question is, is it worth the time and effort? Maybe you should sign up for it and see; you can always drop out after 3 classes if you find it too hard.[/quote]
I doubt it'll be too hard (I average a decent amount above what nsato said the minimum score to avoid being lost is). I'm just particularly weak in trig. I'm going to take the AIME problem series instead...

In my humble opinion, you should really take the AMC 12 problem series. Either that, or really brush up on your basic trig. ;)

I am getting first steps for Math Olympians instead...


If there is a prism and pyramid with the same base and height, and the tip of the pyramid is anywhere on the plane that contains the top of the prism, is the volume of the pyramid [b]always[/b] $ \frac{1}{3}$ the volume of the prism?

Is there a visual way to show this?

yes this is true.
for visully seeing try getting a pyramid of and prism of same base and hieght.(u can make by yourself also by paper).
fill pyramid with sand and u will observe that prism is filled after three fillings by the pyramid.


Probe that if $ n$ is natural, then:
\[ (x+1)^{2n+1}+x^{n+2}\equiv 0 (x^{2}+x+1)\]


Hi! I'm new and I have a problem. I don't think is a hard problem, but i just don't know how to solve it  :(

Proof that: $\log_{a}\frac{bc}{a}\cdot\log_{b}\frac{ca}{b}\cdot\log_{c}\frac{ab}{c}\leq1, \forall a,b,c\in(1,\infty)$

...by the way I used for the first time LaTeX and is't cool  :D

I dont know if im missing something rerally obvious, but i got it down to proving that this is true

$\displaystyle \leftrightarrow (x+y+1)(y-x+1)(x-y+1) \leq (1+x)(1+y)$ I might of seen a solution to this somewhere, not sure where.... Anyone else?

Hey! I've made some substitutinos but I didn't obtain that   :blush: . I'll check to see how you got there  :D

If you dont feel like it, that ruduces to proving that:
$x^3+x^2+y^2 x + y x \geq y^3+y^2+y x^2$

try this...

$\displaystyle \log_{a}\frac{bc}{a}\cdot\log_{b}\frac{ca}{b}\cdot\log_{c}\frac{ab}{c}\leq1, \forall a,b,c\in(1,\infty)$

$\displaystyle \leftrightarrow (\frac {\log{b}+log{c}}{log{a}}-1)(\frac {\log{a}+log{c}}{log{b}}-1)(\frac {\log{a}+log{b}}{log{c}}-1) \leq 1$

Just replacing a few variables, instead of $\log{a}$ we'll just replace it with $x$, or to make it easier for me, just with $a$ itself! (Ok i know its terrible form...)

$\displaystyle \leftrightarrow (\frac{b+c}{a}-1)(\frac{a+c}{b}-1)(\frac{a+b}{c}-1) \leq 1$

Ah crap have to go, see if you can finish!

I got it down to this

$(-a+b+c)(a-b+c)(a+b-c) \le abc$

but i dunno where to go from there

log a = a
log b = b
log c = c

Yes, that is what i got earlier, which i normalised down...

I'll finish it, since this is actually a fairly sophisticated problem:

As someone noted, it's equivalent to $(x+y-z)(y+z-x)(z+x-y)\leq xyz$

This is actually equivalent to Schur's inequality, and its proof can be found in many places on this forum.  To apply Schur, we need to know that $x\geq 0$; this is true because we have $a,b,c>1$.

Actually, I'll post a proof.
[hide]
Expanding, the original inequality is equivalent to:

$x^3+y^3+z^3+3xyz\geq xy(x+y)+xz(x+z)+yz(y+z)$, which is the form we'll prove it in.

Proof:  Because it is symmetric, we may assume $x\geq y\geq z\geq 0$.

Now, $x(x-y)(x-z)\geq y(x-y)(y-z)$ (compare termwise) and $z(z-x)(z-y)\geq 0$.  Sum these two inequalities and we get the desired inequality.[/hide]

Wow! I didn't knew about this inequality  :what?: . That's it then, but i still think there's another way to solve it.

Well, a generalised form (with solution) would be nice  ;)


Suppose,that $a$ and $b$ are natural numbers and $p=\frac{b}{4}\sqrt{\frac{2a-b}{2a+b}}$ is prime.
Find the maximum possible value of $p$.