ratio
number theory
relatively prime
greatest common divisor
geometric series

12. What is $\frac{1}{10}+\frac{1}{100}+\frac{2}{1000}+\frac{3}{10000}+\frac{5}{100000}+\frac{8}{1000000}+...$? (Yes, that is the Fibonacci sequence.)

19. How many 4-digit positive integers have no adjacent digits equal?

20. How many 4-digit positive integers have 4 distinct digits?

21. How many ordered pairs of integers (x,y) satisfy $x^{2}+y^{2}=65^{2}$?

23. What is the second smallest positive integer that leaves a remainder of 2 when divided by 5 and leaves a remainder of 3 when divided by 7?

Better yet, does anyone possess a PDF of the solutions? (Do they even publish that on the Internet? :ninja:

[hide="19"]There are 9 choices for the first digit (1-9), then 9 choices for the second (0-9 not including first digit), then 9 for the third, then 9 for the fourth. Thus $6561$ such numbers.
[/hide]
[hide="20"]There are 9 choices for the first digit (1-9), then 9 for the second (0-9 not including first digit), 8 for the second, 7 for the fourth. Thus $4536$ such numbers.
[/hide]
[hide="21"]We can just brute force to get $(\pm8,\pm1)$, $(\pm1,\pm8)$, $(\pm4,\pm7)$, $(\pm7,\pm4)$ as our solutions, there are 4 for each, so $16$ total solutions.
[/hide]
[hide="23"] Just list out the numbers congruent to 3 mod 7, then see which ones end in 2 or 7:
3,10,17,24,31,38,45,52
So $52$ is the second smallest.
[/hide]

[quote="13375P34K43V312"][hide="21"]We can just brute force to get $(\pm8,\pm1)$, $(\pm1,\pm8)$, $(\pm4,\pm7)$, $(\pm7,\pm4)$ as our solutions, there are 4 for each, so $16$ total solutions.
[/hide]
[/quote]

It's $65^{2}$, not $65$

anyway, for #12,

[hide]let the sum be $S$

$10S-S=1+\frac{1}{100}+\frac{1}{1000}+\frac{2}{10000}\cdots$
$9S=1+\frac{S}{10}$
$S=\frac{10}{89}$[/hide]

[quote="mathgeniuse^ln(x)"]Well, the ones I can think of for 65 are

(39,52)
(25,60)
(33,56)

I can't think of any more...

Of course there are eight for each one...

So 24...[/quote]

Actually, there are 36, as (16,63) also gives 8 solutions, and let's not forget (0,65)  :P .

[hide="how to do this problem"]
If $x$ and $y$ share a common divisor, it will divide 65. Suppose $x_{0}$ and $y_{0}$ are the integers that result from factoring out the greatest common divisor from $x$ and $y$. If the greatest common divsor is 65, we get $x_{0}^{2}+y_{0}^{2}=1$, giving (0,65). If it is 13, then $x_{0}^{2}+y_{0}^{2}=5^{2}$, and if it is 5, then $x_{0}^{2}+y_{0}^{2}=13^{2}$. These provide us with solutions (39,52) and (25,60) respectively.

Finally, if $x$ and $y$ are relatively prime, then we use a well-known formula: exactly one of $x$ or $y$ is even, and if $x$ is even, there exist relatively prime integers $m$ and $n$ such that $m^{2}+n^{2}=65$, $2mn=x$, and $m^{2}-n^{2}=y$. We see that (7,4) and (8,1) are the only possible values for $m$ and $n$ (other than negatives and permutations), leading to the solutions (56,33) and (16,63) for $x$ and $y$. [/hide]

On #23, could you use the Chinese Remainder Theorem or some variant of it?

I'm Chinese so... :D :harhar: No, I just don't like the idea of listing out possibilities unless it's faster.

[quote="mathnerd314"]On #23, could you use the Chinese Remainder Theorem or some variant of it?[/quote]

Yes.

[hide="Quick solution"] We have a unique solution $\bmod 35$.  It must be $\equiv 3 \bmod 7$, so cycling through, we have $3, 10, 17$, and $17 \equiv 2 \bmod 5$, so our solution $\bmod 35$ is $17$.  The second smallest positive integer in this equivalence class is $17+35 = \boxed{ 52}$. [/hide]

Well, $\phi^{n}$ is quite different from $F_{n}$ especially for the first few terms..

[quote="13375P34K43V312"]I came up with and idea for #12, but it doesn't seem to work. Could someone tell me why?

We know that $\frac{F_{n+1}}{F_{n}}$ converges to $\frac{1+\sqrt{5}}{2}$, so could we sum an infinite geometric series with first term $\frac{1}{10}$ and common ratio $\frac{1+\sqrt{5}}{20}$. This gives the sum as $\frac{2}{19-\sqrt{5}}$.[/quote]

Just because $\frac{F_{n+1}}{F_{n}}$ converges to the golden ratio doesn't mean $\frac{F_{n+1}}{F_{n}}$ is the golden ratio for all $n$. This clearly doesn't work in evaluating a sum like
\[\underbrace{1+1+1+\cdots+1}_{2006}+\left( \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots \right)\] If you know Binet's formula, you could sum two geometric series, but that's massive overkill (especially since a common way of deriving Binet's is to evaluate $F_{0}+F_{1}x+F_{2}x^{2}+\cdots$ by just using the basic formulas for $F$).

There was about a year long period where I did all summations with Fibonacci's by just plugging Binet's in and brute forcing it...I got really good at it. It was impressive how quick I could do it.

brute forcing...

shame shame shame  :fool:


inequalities
function
inequalities proposed

let$a,b,c>0$,$ab+bc+ac+2abc=1$.
Prove that a,$a+b+c\ge 3/2$.
                b,$\sum{\frac{1}{a}}\ge 4(a+b+c)$.
                c,$ab+bc+ac-2abc\le 1/2$

These 3 inequalities follow immediately from the substitution $a=\frac{x}{y+z}, b=\frac{y}{z+x}$ and $c=\frac{z}{x+y}$. The first one is Nesbitt and the second one is Levinson ;)

Silouan, this is an application of Levinson inequality for 3-convex function deduced by Levinson himself which states that: If $a, b, c>0$ then the following inequality holds:

                $\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\geq 4(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}).$   ;)

part a): by an application of the am-gm on the condition it follows that $\frac{1}{8}\geq abc$. now, $a+b+c\geq \frac{3}{2} \\\Leftrightarrow a^2+b^2+c^2+2abc\geq \frac{9}{4}-2(ab+ac+bc)+2abc=\frac{9}{4} -2(1-2abc)+2abc=\frac{1}{4}+6abc$,
but from the condition we have that $a^2+b^2+c^2+2abc\geq ab+ac+bc+2abc=1 \geq \frac{1}{4}+6abc$, so we're done!

part c): the inequality should be reversed. by $\frac{1}{8}\geq abc$ and the condition it follows that
$ab+ac+bc\geq \frac{3}{4} \geq \frac{1}{2} + 2abc \\\Rightarrow ab+ac+bc-2abc\geq \frac{1}{2}$


For a positive integer, $m$, answer the following questions.

1) Show that $2^{m+1}+1$ is a prime number, when $2^{m+1}+1$ is a factor of $3^{2^m}+1$.

2) Is converse of 1) true?

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=pepin%5C%27s+test&t=31158

Pierre.

The [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1852108#p1852108]problem[/url] and the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1852193#p1852193]converse[/url] are both true.


Does anyone here play Celtic music? If so, what are your favorite groups? I play flute and whistles, and I love Lunasa, and any of the less traditional Celtic music, but not to the extent of Celtic rock.

I've played "celtic" music a lot in orchestra, because for some reason, writing celtic music for middle school orchestras is the thing to do now.

I like traditional celtic music. Could you name any groups?

Lasairfhíona Ní Chonaola

she is the best

I just play celtic music on flute. i play aughrim and a lot of  anothers


Zadanie tego typu, że jak się nie wpadnie od razu to się długo myśli :) :

(i) $\{a_n\}$ - ciąg liczb rzeczywistych, ponadto jest ograniczony
(ii) dla pewnej liczby rzeczywistej $a \in (0,1)$ oraz dla każdego całkowitego $n \geq 2$ zachodzi $a_{n+1} \leq aa_n+(1-a)a_{n-1}$.

Dowieść, ze $\{a_n\}$ jest zbieżny.

Podlug zalecen staralem sie myslec krotko  ;) Mam "do polowy" jesli tak mozna to nazwac:

Rozwazmy ciag: $x_{n+1}=ax_n+(1-a)x_{n-1}$, gdzie $x_1=a_1$ oraz $x_2=a_2$. Latwo otrzymac ogolny wzor na ciag: $x_n=\alpha(a-1)^n +\beta$ gdzie $\alpha, \beta$ to pewne stale. Teraz jako, ze mamy $x_n \geq a_n$ dla kazdego $n$ wiec przechodzac do granicy otrzymujemy $\beta \geq \lim_{n \to \infty} a_n$. Problem w tym, ze nie wiem jak otrzymac oszacowanie $\lim_{n \to \infty} a_n \geq \beta$ co by zalatwialo sprawe.

Czy to dobra droga?  :)

rozwiazywanie dla ustalonego ciagu rekurencyjnego jest mylne bo niestety jest caly czas problem zwiazany z tym ze szacujesz w ten sposob z gory ale nie z dolu :( , wiec jedyne co mozesz wskazac to po prostu gorne oszacowanie granicy...nie zagladaj na org bo tam pewna osoba wkleila wzorcowy dowod (nie wiem po co)...a mnie zalezalo by sie ludzie sami tym pobawili...

ps. (twoje rozumowanie przychodzi powszechnie jako pierwsze go glowy :) - nie wiem czemu )

Megus, po za tym oszacowanie $\lim_{n \to \infty} a_n \geq \beta$ nie musi być prawdziwe.

_Szkic_ mojego _dowodu_:

Traktujemy ciąg ${a}$ jako funkcję $f(n) = a_n$.

Łatwo patrzeć na liczbę $aa_n + (1-a)a_{n-1}$ jako na punkt leżący na odcinku $(n-1,a_{n-1})$ , $(n,a_n)$ o współrzędnej $n-1 + a$.

Dodatkowo, jeżeli $a_n > a_{n+1}$ to wszystkie wyrazy ciągu po $a_{n+1}$ leżą pod tym punktem.

Ciąg $a_n$ jest ograniczony, więc musi mieć podciąg zbieżny $b_n$ do $g$. Trzeba udowodnić, że dla dowolnego $\epsilon$ wyrazy ciągu ${a}$ siedzą w epsilonowym otoczeniu $g$. Wiemy jednak, że od pewnego miejsca wyrazy ciągu ${b}$ siedzą w otoczeniu $\frac{\epsilon}{2}$ liczby $g$ (powiedzmy po wyrazie $b_m$).

Jeżeli $M$ jest liczbą ograniczającą ciąg ${a}$ od góry, to można udowodnić, że wszystkie wyrazy ciągu ${a}$ po $b_{m+c}$ będą mniejsze od $(1-a)^{c}(M - g - \frac{\epsilon}{2}) + g + \frac{\epsilon}{2}$. Wyrazy ciągu ${a}$ będa więc od pewnego momentu wpadały w epsilonowe otoczenie $g$.

Podaobnie, patrząc na liczby $aa_n + (1-a)a_{n-1}$ jak na punkty, też widać, że wyrazy $a$ nie mogą znajdować się daleko pod $g$, bo to przeczy zbieżności ciągu $b$ do $g$.

Dowód nie jest szczególnie ładny (jeżeli jest poprawny), pewnie Rafałowi chodziło o coś sprytniejszego :) ?

Pozdrawiam
Michał

Coś mi się widzi, że jest taki sprytny krótki sposobik... Nasza nierówność jest równoważna nst. nierówności: \[ a_{n+1}-a_n\geq (1-a)(a_n-a_{n-1}) \] jak się trochę z tymi kolejnymi różnicami pokombinuje, to wychodzi, że albo ten ciąg jest malejący, albo od pewnego miejsca będzie ciągle rosnący, a więc jako ciąg ograniczony będzie również zbieżny :lol:

Jastrząb, a ciąg wyznaczony przez Megusa

$\alpha (a-1)^n + \beta$ ?

On nie jest ani taki, ani taki.

Ano rzeczywiście.... troszkę mi się pomyliło przy przekształcaniu.... nasza nierówność to \[ a_{n+1}-a_n\leq (a-1)(a_n-a_{n-1}) \]

A wtedy już nie jest tak prosto... chociaż też wychodzą ciekawe rzeczy, bo na przykład \[ a_{n+1}-a_n\leq (a-1)^n(a_1-a_0) \]


In a quadrilateral ABCD ,it is given that AB is parallel to CD and the diagonals AC and BD are perpendicular to each other.Show that
1. AD.BC>=AB.CD
2. AD+BC>=AB+CD

First of all i would like to say that your problem misses a part. In your problem you don't specificate whether  $ AB \geq CD$ or otherwise. Still, the solution to the problem is :

 Using pithagora in AOB, COB, COD and AOD triangles you express AB, BC, CD or AD. so

$ \sqrt{BO^2\plus{}CO^2}\plus{}\sqrt{AO^2\plus{}DO^2}\geq \sqrt{BO^2\plus{}CO^2}\plus{} \sqrt{AO^2\plus{}DO^2}$

 $ 2\sqrt{(BO^2\plus{}CO^)(AO^2\plus{}DO^2)}\geq 2\sqrt{(BO^2\plus{}CO^2)(AO^2\plus{}DO^2)}$

 $ (BO^2\plus{}CO^2)(AO^2\plus{}DO^2)\geq (BO^2\plus{}CO^2)(AO^2\plus{}DO^2)$

 $ BO^2AO^2\plus{}BO^2OD^2\plus{}OC^2AO^2\plus{}OC^2OD^2\geq BO^2CO^2\plus{}BO^2DO^2\plus{}AO^2CO^2\plus{}AO^2OD^2$

 Then, you'll get:

$ AO^2(BO^2\minus{}OD^2)\geq CO^2(BO^2\minus{}OD^2)$

 $ AO^2\geq CO^2$

 $ AO\geq CO$
This isn't always true. Now you see why you need to know whether $ AB \geq CD$. If your problem would specify that  $ AB \geq CD$ , you're problem will be true. And, regarding the second part of the problem you just need to look at the first part, because it has already been done. I hope i didn't mess up the expresions to bad.:)

It is [url=http://www.isical.ac.in/~rmo/rmo97.pdf]Regional Mathematical Olympiad 1997; India; problem 4[/url]. :)
I posted it in Mathlinks a long time ago, and this solution is by ocha:
[b](RMO-1997, India) In quadr. ABCD; $ AB\parallel{}CD$ & $ AC\perp BD$. Show that :
(a)$ AD\cdot BC\ge AB\cdot CD$
(b)$ AD \plus{} BC\ge AB \plus{} CD$[/b]
[hide="Solution:(by OCHA)"]

[asy]draw((0,50)--(0,-100));
draw((-50,0)--(100,0));
draw((-50,0)--(0,50));
draw((0,-100)--(100,0));
draw((0,50)--(100,0));
draw((-50,0)--(0,-100));
label("$a$", (0,25), E);
label("$b$", (-25,0), N);
label("$ka$", (0,-50), E);
label("$kb$", (50,0), N);
label("$A$", (0,50), N);
label("$B$", (-50,0), W);
label("$C$", (0,-100), S);
label("$D$", (100,0), E);
label("$O$", (0,0), SW);[/asy]
Let ABCD be the quadr.
$ \triangle AOD\sim \triangle DOC$
So that $ \frac {OC}{AO} \equal{} \frac {OD}{BO} \equal{} k$ (say; where k is the scaling factor of these two triangles)
Letting $ AO \equal{} a, BO \equal{} b; \implies OC \equal{} ka, OD \equal{} kb$
We have some relations:
$ AB^2 \equal{} a^2 \plus{} b^2$
$ AD^2 \equal{} a^2 \plus{} k^2b^2$
$ CD^2 \equal{} k^2(a^2 \plus{} b^2)$
$ BC^2 \equal{} b^2 \plus{} k^2a^2$
Hence
$ AD^2\cdot BC^2 \equal{} a^2b^2 \plus{} k^2a^4 \plus{} k^2b^2 \plus{} k^4a^2b^2$
$ \equal{} a^2b^2(1 \plus{} k^4) \plus{} (ka^2)^2 \plus{} (kb^2)^2\ge 2a^2b^2k^2 \plus{} (ka^2)^2 \plus{} (kb^2)^2$
$ \equal{} k^2(a^2 \plus{} b^2)^2 \equal{} AB^2\cdot CD^2$
$ \therefore \boxed{AD\cdot BC\ge AB\cdot CD}$
Also; $ AD^2 \plus{} BC^2 \equal{} AB^2 \plus{} CD^2 \equal{} (a^2 \plus{} b^2)(1 \plus{} k^2)$
$ \implies AD^2 \plus{} BC^2 \plus{} 2AD\cdot BC\geq AB^2 \plus{} CD^2 \plus{} 2AB\cdot CD$
$ \implies (AD \plus{} BC)^2\ge (AB \plus{} CD)^2$
$ \boxed{\implies AD \plus{} BC\geq AB \plus{} CD}$
QED[/hide]
[hide]For more information on this useful method using similarities, you can visit [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=295946]Here[/url]  :blush:  . [/hide]


a)What is the maximum number of cycles in a graph of order $ n$ with girth $ d$?
b)what is the maximum number of length $ k$ cycles in a graph of order $ n$ with girth $ d$? (of particular interest is the case $ k\equal{}d$)

All right ... I came up with some (simple) esimations for this value. I put them all in a short research note found at: http://www.sfu.ca/~pvaladkh/main_8.html


Hey guys,

For various reasons, I have decided to "retire" from the position of moderator in this forum. Elyot has agreed to take over the role, and he will soon be your new mod.

I have enjoyed your posts, and I will still remain on MathLinks-AoPS in the future. Good luck  :D

Thank you very much for moderating, Yufei. :) Good luck at University!
Thanks, Elyot, for accepting the important role of moderating on this forum. :)

[quote="rem"][quote="rzsolt"]does this mean if I kill elyot, I get the job?[/quote]
no, because you will go to prison and there are no computers there.[/quote]

then I must also kill prison.

[quote="Elyot"]BWAHAHAH.  Now I'll finally have the power to corrupt all of you to the full extent of my abilities.  :D[/quote]

Aww.. but you still have to visit the Canadian forum once in a while! And.. are you sure you want Elyot of all people to moderate??

Elyot-I'm sure you'll be fine, but we'll all be pretty corrupted very very soon..

[quote="jpark"]

And.. are you sure you want Elyot of all people to moderate??
[/quote]

 :o You're so mean!! :(

NO FUDGE FOR JENN!

[quote="Elyot"][quote="jpark"]Aww.. no fudge for me?!!!

I'm of course disappointed, because I [i]really[/i] want to eat fudge every time we play Mao!  :P[/quote]

Well then the solution, of course, is to never play Mao again.  :lol:[/quote]

Don't say that, you'll make David cry!!  :D

[quote="rzsolt"][quote="WindingFunction"][quote="TheDreamer"]Bye Yufei, good luck at MIT, you will do wonders for math.[/quote]

Yufei at MIT means MIT starts winning Putnam again.   :([/quote]

remineds me of a question: Who are currently the best putnam students at UW? What about once half of the IMO's arive?[/quote]

According to last year's results, the best Putnams at Waterloo were me, Ralph Furmaniak, and Olena Bormaschenko, all former IMO.  I'm not sure how many of them will still be around... I think Olena might be graduating or something, but I'm not sure.

[quote="lightrhee"]Just a little question. Why does it still say "moderators: billzhao"?[/quote]
No idea. maybe the moderators have not changed yet. Like Yufei said, "Elyot has agreed to take over the role, and he will [b]soon[/b] be your new mod."
PS:
[quote="lightrhee"]I had no computer [b]excess[/b] for a week. :( [/quote]
lol
supposed to be "access" :)

[quote="lightrhee"]I had no computer excess for a week. :(
Just a little question. Why does it still say "moderators: billzhao"?[/quote]

Well, I already reminded Valentin. As soon as he remembers I guess ...

Yeah, the moderators will not get changed until Valentin returns from his vacation:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=601519#601519[/url]

DAAAAAAAAAAVVVVVVVIIIIIIDDDDDDDDD!!!!!!!!!

We haven't seen you around here for a while.   :lol:

[quote="jpark"]Whoa!!!!!!! It's DAVID!!

I was wondering why you were so quiet!  :D  When are you getting here?[/quote]
I was in Edmonton since August 13. I'm just busy packing + preparing to teach at Alberta math camp + getting a driver license. :juggle:

Hopefully there is no typo this time. :lol:


The natives of  P3 (aka. the Planet of  Periodic Prevericators) live under an absolute biological compulsion to either lie or tell the truth at various times. They have absolutely no choice in the matter - every time they begin making a precise true-or-false statement the compulsion takes over and they cannot choose which. Fortunately, it is periodic. The population is divided into strict classes, depending on when they lie or tell the truth:
Class 1: Every statement is a lie.
Class 2: Every second statement is a lie, the others are true.
Class 3: Every third statement is a lie, the others are true.
Class 4: Every fourth statement is a lie, the others are true.
Class 5: Every fifth statement is a lie, the others are true.
Etc.

Now it so happens that among the numerous secret societies on P3 there is one group that calls themselves the PsychoMaths. They create all sorts of numerical and logical puzzles and challenge one-another at their meetings. On of their favorite pastimes is playing the game of Number Match. A referee chooses 5 players, no two of which are of the same class, who play the game for 10 rounds. During each round each player secretly chooses and writes down a number from 1 to 5 and gives it to the referee. Once all numbers are in the referee exposes them all and delcares the winner, who is determined by the Match Rule: Any player whose chosen number is the same as that of any other player(s) is said to be matched; all other players are unmatched. The winner is the unmatched player having the highest number, and he scores points equal to his number; no other players score that round.  If all players are matched no one wins that round and no one scores.
(Examples: if the numbers are 4,5,5,3,4 then only 3 is unmatched and wins. If the numbers are 5,2,5,4,3 then 4, 3, 2 are unmatched and 4 wins.) After the 10 rounds are over the referee calculates the total scores for each player and then the winner is determined from these 5 scores by using the Match Rule again:  the player having the highest unmatched total score wins the game.

The game is played in secret and after the game is over each of the 5 players in turn makes 6 statements about the game which the referee carefully records on paper, in the exact order stated. During this recording session absolutely no other statements are made by the players. For the written statements the players are assigned letters A,B,C,D,E in order to remain anonymous. Then the statements are presented to the full membership of the society as a puzzle, and they attempt to deduce who was the winner.

Below are the statements of a recent game, listed in exact order:

A1: I won the second round.
A2: I won the last 2 rounds with equal scores.
A3: No one's total score was higher than mine.
A4: I never won with 5.
A5: I won the fifth round.
A6: I did not win the game.

B1: Three of the players never chose 4.
B2: I won the third round.
B3: My total score was lower than any of the others.
B4: I chose 2 in every round.
B5: I won the seventh round.
B6: I won the sixth round.

C1: I did not win the eighth round.
C2: My total score was 14.
C3: I chose 4 in the first round.
C4: I won the first two rounds.
C5: I did not win the sixth round.
C6: I chose a different number in each round I won.

D1: No one won 2 consecutive rounds.
D2: I chose 4 in every round.
D3: I won the first round.
D4: No one won more than 3 rounds.
D5: I won 2 of the first three rounds.
D6: I lost every even-numbered round.

E1: I did not win the first round.
E2: No one ever chose less than 3.
E3: No one ever won with 3.
E4: I won the sixth and seventh rounds.
E5: I won the seventh and eighth rounds.
E6: I did not win the third round.

Can you determine the winner?

Remember: each player is of a different class. All 5 total scores can be determined, but not the exact number choices of the losing players of each round.

Also note that when a player makes his first statement you do not know at what point in his lie-cycle he is at, so his first statement could be true or false depending on the last time he lied as well as his class.

Can someone be of class 6 or something more than 5?

Yes, somebody can be a class 6, 7, or 8 etc.

This is a nice and [b][size=200]HARD[/size][/b] problem.

!!!!!!!!!I GOT IT!!!!!!!!! :wow:

I spent loads of hours doing it, and I was almost crazy because I got the conclusion that it was impossible for 2 times. Luckily, I could find my mistakes. Finding our own mistakes is the hardest thing, you know. I'm going to die now. Very tired. Hope my answer is correct.

[hide="Answer"]B was the winner.[/hide]
Where did you get it from? It's very hard problem, and also the answer is unqiue. I wonder how it was invented.

Great job :D ! It is a very tough problem! (You were correct)

I got it from an incredible logic puzzles site, which I will choose not to release (for now) for purposes that I plan to post more of their puzzles  :evilgrin: . Here is a "full solution" if you are interested:

[hide="Full Solution"](0) Note that if any player makes 2 consecutive false statements then he must be of class 1. Also if any player is not of class 1 then he cannot make 2 consecutive false statements.

(1) Assume D1 is true:  then E4 and E5 must both be false, and so E is of class 1 and so E1 is false. This means E won the first round. So D3 is false. Now if D5 is true then D must have won rounds 2 and 3 since he did not win 1. But this contradicts D1. So D5 is false. But then since D3 and D5 are both false, D must be of class 2 (class 1 being taken by E). But this would make D1 false also, a contradiction. So the assumption is false and D1 must be false.

(2) Assume D3 is true: then since by (1) D1 is false, D2 must be true (else all D are false by (0)). So by D2 and D3, D won the first round with 4. So D must have been unmatched in round 1 and so C3 is false. Also C4 is false since D won the first round. So C is of class 1 and C1 and C5 are false. This means C won round 6 and 8, so E4 and E5 are both false. This means E is of class 1, which contradicts C being class 1. So the assumption is false and D3 must be false. And since D1 is also false by (1), then D must be of class 1 or 2, and in either case D5 must be false also.

(3) Assume D4 is true: then by (2) D is of class 2 and D2 is true also. Now by D4, C must have won 3 rounds or less. If C2 is true, C must have won 3 rounds with scores of 5,5,4 (in some order) since no other combination of 3 or fewer numbers adds up to 14. This means C won with 4 in some round and so must have been unmatched. But this contradicts D2 being true. So C2 is false. Now if C4 is true then, since D2 is true, C3 must be false. But then C2 and C3 are both false and so C is class 1 which contradicts C4 being true. So C4 is false. Now since C2 and C4 are false, C must be of class 1 or 2. But since D is of class 2, C is of class 1 and so C1 and C5 are false. This means C won round 6 and 8, so E4 and E5 are both false. This means E is of class 1, which contradicts C being class 1. So the assumption is false and D4 must be false. Then, since D3 is false also by (2), D must be of class 1 and all of D's statements are false.

(4) Assume B5 is true: then E4 and E5 are both false which makes E class1. But this contradicts D being class 1 by (3). So the assumption is false and B5 must be false. And since B cannot be of class 1, then B cannot have 2 consecutive false statements, so B4 and B6 are both true. This means E2 and E4 are both false. And since E cannot be class 1, then E must be of class 2. So E1, E3, E5 are all true and E6 is false. And by E6 (false) E won round 3, so B2 is false. So, since B5 is also false, B is of class 3. This means B1 and B3 are both true.

(5) Since B6 is true then C5 is also true. And since E5 is true, so is C1. Now observe that C could not win with 2 since, by B4,  he would be matched by B. Also C could not win with 3, by E3. So C could only win with 1,4,5. This means that either C's total score is 10 or less or C won more than once with some number. So C2 and C6 cannot both be true. So one or both are false and since C cannot be of class 1,2,3 (already taken) then C3 and C4 must both be true in any case. Since C4 is true, then A1 is false. And since A cannot be of class 1,2,3 (already taken) then A2, A3, A4 must all be true.

(6) Now by C4, C won rounds 1 and 2. By E6 (false) E won round 3. By B6, B won round 6. By E5, E won rounds 7 and 8. And by A2, A won rounds 9 and 10. This leaves only rounds 4 and 5 open. Now by  D6 being false, D must have won an even-numbered round, and only 4 is open, so D won round 4. And by D4 being false, someone won more than 3 rounds. The only way this is still possible is if E won round 5.

(7) Since E won round 5, A5 is false. This makes A class 4 and A6 is true. Since A only won 2 rounds, the highest his total score could be is 10, so by  A3, no one had a higher total score so C2 is false. And C cannot be class 4 since A is, so C6 must be true. (C's class is 5 or more but it can't be determined.)

(8) Since B won only round 6, then by B4, B's total score was 2. So by B3 every other player had a total score of 3 or more. Also, by B4, no other player could win with 2 (since he would be matched by B), and by E3, no player won with 3. So winning numbers for players other than B can only be 1,4,5, with a total more than 2.

(9) Now A only won the last 2 rounds, and by A2 his 2 scores were equal. They could not be both 1 since his total must be more than 2. They could not be both 5 by A4. So they were both 4 and A's total score was 8. Also, C only won the first 2 rounds, with a first round score of 4, by C3. And by C6, C's other score cannot also be 4. So it must be 1 or 5. But if it is 5 then C's total score would be 9, which is higher than A's 8, which contradicts A3. So C's second round score is 1, for a total of 5. Now since A and C both won with 4, then by B1, B,D,E never chose 4, so D and E could only have won with 1 or 5. Since D won only 1 round, and his total score cannot be less than 2, it cannot be 1, so D's score and total must be 5. Now E won 4 rounds, and by A3 his total cannot be higher than 8. So E's scores mut either be 1,1,1,1 for a total of 4, or 1,1,1,5 for a total of 8. If E had a total score of 4, A would have an unmatched score of 8 with all others lower, and so would win the game. But this contradicts A6. So E had a total score of 8.

(10) Now the total scores are  A=8, B=2, C=5, D=5, E=8. So by the Match Rule the game was won by B.[/hide]


I already know what a few peoples users are...but I'm curious as to everyone else. If you went to mathzoom this summer plz post. :D

btw, I'm Sarah

DG.  :D

Aww... The message is too small. Please make the message longer before submitting.

I think you fall under the catagory of ppl whose usernames I know.  :D


algebra
polynomial
number theory unsolved
number theory

Players 1 and 2 pick digits in turn. Player 1 start with digit $b_{0}$, then player 2 picks digit $b_{1}$, etc ... Player 1 wins the game iff $0.b_{0}b_{1}\dots$ is transcendental. Who wins ?

Can someone explain what a transcendental number is? I've never had a definition.

A transcendental number is one that is not the solution of any polynomial with integer coefficients. (i.e. is not algebraic). It would seem player 1 would win by choosing digits of one or more transcendental numbers.

A complex number $t$ is called transcedental iff there doesnt exist a polynomial $p(x)$ with rational coefficients such that $p(x)=0.$

There are more than countable many possibilities for player $1$ to choose his numbers. Assume that player $1$ will always give a list of the numbers he will take to player $2$ at the beginning. Now assume that player $2$ has an answer to all of them, yielding that we get different reals for different sequences (not exactly since it is not unique, but there are at most two possibilities for each real to be represented that way), thus more than countable real algebraic numbers, contradiction.

@clinkmath: no, player 1 would not win for sure by this.

[quote="ZetaX"]A complex number $t$ is called transcedental iff there doesnt exist a polynomial $p(x)$ with rational coefficients such that $p(x)=0.$[/quote]

You mean $p(t)$, right? Also, does that mean all transcendental reals are irrational?

Yes, $x \to t$.
And also yes, all transcedentals (also the complex ones) are irrational.

I think Player One will win because if Player One picks digit as following sequence: $12345678910111213141516171819\cdots$

Then we can prove $0.1*2*3*4*5*6*7*8*9*1*0*1*1*1*2*1*3*1*4*1*5*\cdots$ is a transcendental number!

Think about it. It is a bit hard to prove.

@ ZetaX: This actually means that Player 1 can win ... how to explicit a method ?

@ Libra: this would give a method for Player 1 to win ... how do you prove this result ?

I actually had Libra's example in mind but it didn't seem clear to me how to prove that it is transcendental (even though, well, it seems very intuitive).

The method I would use if I were player 1: I would put a bijection (defining an order, as for the usual bijection $\mathbb{N}^{2}\rightarrow \mathbb{N}$) between $E_{n,m}$ and $\mathbb{N}$, where $E(n,m)$ is the set of the solutions of the polynomial equations with integer coefficients $\leq m$ and degree $n$; then, for each element of $E_{n,m}$, I would play a digit such that $0.b_{0}\dots$ does not equal that element. This way, every algebraic number will be forbidden at one point.

Sorry :( , I found that I just prove it is irrational. I think we should construct a number which is a Liouville number. Maybe it has this form $X=0.1a1b0c0d0e1f0g0h0i0j0k0l0m0n0o0p0q0r0s0t0u0v0w1x0y0z0\cdots$

I have proven that if $0.0a0b0c0d0e0f0g0h0i0j0k0\cdots$ is a algebraic number, then $X$ will be a Liouville number.

I have a Note about transcendental number. Maybe it will inspire you. :oops:


trigonometry
real analysis
real analysis unsolved

oh! you are right... sorry, that was so silly of me...   :(

i will [u]edit[/u] it now...  :maybe:


Princeton
college
Harvard
geometry
function
topology
complex analysis

Hi!

I am reading about the general exams at Princeton. A very interesting collection of previous student's impression of them are here: http://www.math.princeton.edu/generals/

What kind of preparation would you have to have to pass these? At first glance they seem pretty easy ('what is the definition of the riemann mapping theorem?'), but then some questions could be very difficult? Are these taken by first year Ph.D's? It is funny that Terry Tao's is there, where he says that he doesn't know Harmonic analysis  :o  That makes me think they are terrifically difficult!

Thanks in advance,

Los aus Finnland

The Princeton grad program is very unusual in that there isn't a lot of formal grading done. It's such a small community that the professors pretty much "know" how much progress the students are making. Also, the emphasis is much less about "book learning" and more about getting students ready for original research.

Students usually take their General Exam (an oral examination) in the middle/end of their second year. Some do it in their first year. It must be passed before a student can begin work on their PhD dissertation. The students take them very seriously, even though, I think, nearly all pass. Part of the stress is the fact that it's an oral exam, so you have to think on your feet and you are face-to-face with your executioners! (Undergraduates also have to take an oral examination, but it is very stress free). You can get a sense for the kind of questions that are asked in the general exams by reading the accounts of them posted by the students, and by looking at the list of often-asked questions.

I get the impression that the more stringent test for the students, however, is the [url=http://gradschool.princeton.edu/academics/processes/#comp000044bf59cc000000137a3c1d]annual re-enrollment process[/url], and of course, the dissertation itself.

I wonder if the gernerals here are any harder than the preliminary exams at other schools? I do think they are.

Where are you in school now?

I haven't seen the preliminary exams at very many schools, but I know the [url=http://math.berkeley.edu/graduate_prelim.html]Berkley ones[/url] are pretty difficult (at least imo), and the [url=http://www.math.cam.ac.uk/teaching/pastpapers/]Cambridge Tripos[/url] exams (which are for undergraduate/masters degree candidates there, but are simlar to american preliminary exams) are notoriously difficult.

The Harvard qualifying exams sound pretty brutal to me -- written, not oral, but on a fairly broad range of subjects (differential geometry  :o ) over a two-day period.  MIT's quals, by contrast, are oral exams with no fixed syllabus, and so the difficulty is largely a function of which professors you ask to examine you.

Oral quals and written quals are completely different beasts.  I don't think it really makes sense to compare them.  Here's a bit of my impressions about quals at various places, but I make no warranties of accuracy.  In any case, you can find most schools' written quals online, so you can form your own opinions.

Princeton.  Oral.  Everything you could possibly want to know about Princeton's quals is contained in Los's link.  Almost everyone passes on the first attempt.

Harvard.  Written.  Probably the broadest of all quals (covering algebra, algebraic geometry, algebraic topology, complex analysis, differential geometry, and real analysis), but consequently the questions don't go particularly deep.  They seem to want to make sure you know the basics of a LOT of topics.  I don't know anything about the pass rate.

Berkeley.  Written.  It has a reputation for being difficult because a lot of people flunk it  However, Berkeley used to be easier to get into than the other top tier schools, and there was an expectation that some admits would flunk out.  (I'm not sure about how things are now.)  Consequently, I don't think Berkeley's quals are as hard as they reputed to be.  I think there is also an oral exam later in the PhD program.

MIT.  Oral.  Like all oral exams, difficulty depends strongly on your examiners.  I imagine that it's similar to Princeton's, but it seems to have less of a "reputation."  My vague impression is that almost everyone passes on the first attempt.

Stanford.  Written.  The topics are pretty narrow, but the problems can be very challenging.  Most of the students would flunk the exam horribly if they didn't study the old exams, which tend to have similar problems.  So the emphasis seems to be on demonstrating that you can master certain tools and techniques.  Several years ago they eliminated 1 of the 3 exams, so now there's now only algebra and real analysis.  (So it's kind of like the opposite of Harvard.)  Typically, a few students fail at least one of the two exams on the first attempt.  There is also an oral qual later in the PhD program.

Chicago.  I'm a little fuzzy here, but I think that the finals for the first-year course curriculum (which is a big deal at Chicago) effectively act as written quals, with an oral qual later in the PhD program.

A word on orals:  Almost everyone has them, either in the earlier or later stages of the PhD program, and from what I understand, they're easy to "pass" everywhere.  But one should keep in mind that students do not study for orals in order to merely "pass."  Your oral committee is usually made up of the professors who know you (and will probably one day write you a recommendation), and almost always includes your thesis advisor.  The goal is usually to not look stupid, and that is often difficult to achieve.


Using mathematical induction prove the following:

6 | 7^n - 1   For all n that belongs to Z*

hello, hm, induction is here sennseless for me, it is easy with mod-calculation, a one-line proof:
$ 7\equiv 1 \mod 6$ therefore $ 7^n \minus{} 1\equiv 0 \mod 6$.
Sonnhard.

I agree with Sonnhard (except that he left out $ \equiv$ and meant to write $ 7^{n}\minus{}1 \equiv 0 \mod 6$,) but:

Basis step: For $ n \equal{} 1$, $ 7^{n} \minus{} 1 \equal{} 7 \minus{} 1 \equal{} 6$ which is divisible by $ 6$.

Inductive step: Assume true for $ n \equal{} k$, that is $ 7^{k} \minus{} 1 \equal{} 6a$, say. Then $ 7^{k \plus{} 1} \minus{} 1 \equal{} 7(7^{k} \minus{} 1) \plus{} 6 \equal{} 7\cdot 6a \plus{} 6 \equal{} 6(7a \plus{} 1)$.

hello, thank you for your hint it was a typo in my $ \text{\LaTeX}$ code.
Sonnhard.


Find all possitive integer (x,y) such that:
$x^3=y^2-4$

hum as i known in the book "Elementary Theory of Numbers" written by Waclaw Sierpinski, 1963, Publisher: Wacsaw, Poland[in English] he gave the full proof for some kind of equation $x^3-y^2=a$. The proof is elementary but long!

[quote="mtphaovibul"]No solution:
since $x^3=y^2-4$ we can factor as $x*x^2=(y-2)(y+2)$
Case 1:
let $y-2=x$ and $y+2=x^2$ by subtracting second equation from first, we obtain:
$-4=x-x^2$ which have irrational roots----no solution
Case 2:
let $y-2=x^2$ and $y+2=x$ by subtracting second equation from first, we obtain:
$-4=x^2-x$ which have complex roots ----no solution[/quote]

It's wrong.
x is not a prime.

oh yeah, I forgot about that
but I still belive it's has no solution
and let my old post that x isn't a prime


inequalities
AMC
USA(J)MO
USAMO
inequalities open

we know $H.M \geq  A.M$
hence taking h.m of $\frac{1}{1+a+b}+\frac{1}{1+b+c }+\frac{1}{ 1+c+a}$

we get $\frac{3}{\frac{1}{1+a+b}+\frac{1}{1+b+c }+\frac{1}{ 1+c+a}}\geq  \frac{3+2(a+b+c)}{3}$
we can rearrange it
we get $\frac{9}{3+2(a+b+c)}\geq  \frac{1}{1+a+b}+\frac{1}{1+b+c }+\frac{1}{ 1+c+a}$...................(i)

Now taking A.M >= G.M in a,b,c (because a,b,c>0)
hence
$\frac{a+b+c}{3}\geq  \sqrt[3]{abc}$
and $\sqrt[3]{abc}\geq  1$ (Given)
hence $\frac{a+b+c}{3}\geq  1$
$a+b+c \geq  3$
substituting in (i)
we get $\frac{1}{1+a+b}+\frac{1}{1+b+c }+\frac{1}{ 1+c+a}\leq  1$

For m,n,p > 0 the following definitions are valid:
AM(m,n,p) = (m+n+p)/3
GM(m,n,p) = (mnp)^(1/3)
HM(m,n,p) = 3/(1/m + 1/n + 1/p)
One can verify that: AM >= GM >= HM.

So then your proof has an error. You said HM >= AM while HM <= AM !

the inequality is probably wrong (the sign must be reversed)

Cauchy - Swarz Inuality : $(\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a})(3+2(a+b+c))\geq 9$   (*)

by Am-Gm Inequality we know that : $3+2(a+b+c)\geq 3+2\times3\sqrt[3]{abc}\geq3+6=9$   (**)

Now implying (**) to (*) we get that desired sum is greater or equal to 1

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

davron

[quote="Davron"]the inequality is probably wrong (the sign must be reversed)

Cauchy - Swarz Inuality : $(\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a})(3+2(a+b+c))\geq 9$   (*)

by Am-Gm Inequality we know that : $3+2(a+b+c)\geq 3+2\times3\sqrt[3]{abc}\geq3+6=9$   (**)

Now implying (**) to (*) we get that desired sum is greater or equal to 1

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

@ozodstudios: please edit your post!

davron[/quote]

this was a problem from the tournament of towns 1997 and the sign is not reversed...

counter example for your proof davron take $(2,\frac{1}{2},1)$ then LHS=$\frac{131}{140}<1$

Davron's proof is obviously wrong as well - if $ab\geq 9$ and $b\geq 9$, then it doesn't necessarily follow that $a\geq 1$!

Anyways, this is just that USAMO inequality written in a different form.  Put $a=x^{3},b=y^{3},c=z^{3}$, so that $xyz=1$, and then the left hand side becomes \[\sum \frac{1}{a^{3}+b^{3}+abc}\] which is awfully familiar.

[quote="Davron"]the inequality is probably wrong (the sign must be reversed)

davron[/quote]

Nope. If we take a=b=c=n which is increasing infinitely, the LHS would converge to 0, which is smaller than 1. Your prrof is wrong as (*) and (**) don't mean the last.

Now put $a=x^{3},b=y^{3},c=z^{3}$, and we get $xyz \geq 1$.

So $\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}$
$\leq \frac{1}{xyz+x^{3}+y^{3}}+\frac{1}{xyz+y^{3}+z^{3}}+\frac{1}{xyz+z^{3}+x^{3}}$
$\leq \frac{1}{xyz+x^{2}y+xy^{2}}+\cdots$, as for positive real $a$ and $b$, $a^{3}+b^{3}\geq a^{2}b+ab^{2}$ by AM-GM ineq.

By $\frac{1}{xyz+x^{2}y+xy^{2}}+\cdots$
$= \frac{1}{xy}\cdot \frac{1}{x+y+z}+\cdots$
$= (\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx})\frac{1}{x+y+z}$
$= \frac{1}{xyz}\leq 1$, we can get the ineq.

PS. Actually this is an old USA problem.

Sure, sorry for my rush and do you know where I did mistake?
it is in taking the min value, for ex TM>N if M>P then we can't say that T>N/P.
@guys you are fast, I observed that I did mistake and then I look back and here are many proofs :-)
thanks

davron

[quote="Sung-yoon Kim"]
Now put $a=x^{3},b=y^{3},c=z^{3}$, and we get $xyz \geq 1$.
So $\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}$
$\leq \frac{1}{xyz+x^{3}+y^{3}}+\frac{1}{xyz+y^{3}+z^{3}}+\frac{1}{xyz+z^{3}+x^{3}}$
[/quote]

The inequality
$\frac{1}{1+a+b}\leq \frac{1}{xyz+x^{3}+y^{3}}$
is wrong, because $xyz \geq 1$ implies
$\frac{1}{1+a+b}\geq \frac{1}{xyz+x^{3}+y^{3}}$

So guys, how do you prove the inequality \[\frac{1}{{1+a+b}}+\frac{1}{{1+b+c}}+\frac{1}{{1+a+c}}\le 1\] when $abc > 1$ ?

Put $a=\ell x$, $b=\ell y$, $c=\ell z$, where $\ell>1$ and $xyz=1$.

Then we have \[\sum \frac{1}{1+\ell(x+y)}\leq \sum \frac{1}{1+x+y}\] and we proceed as before.

[quote="Leon"]
So guys, how do you prove the inequality \[\frac{1}{{1+a+b}}+\frac{1}{{1+b+c}}+\frac{1}{{1+a+c}}\le 1\] when $abc > 1$ ?[/quote]

Clear the denominator, the inequality is equivalent to:

$a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)+2abc \geq 2+2(a+b+c)$
Since $abc \geq 1$ so $a+b+c \geq 3$ and $2abc \geq 2$
It remains to prove that $a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b) \geq 2(a+b+c)$
It isn't hard since
$\sum (a^{2}b+a^{2}c+1)-3 \geq \sum 3 \sqrt[3]{a^{4}bc}-3\geq \sum 3\sqrt[3]{a^{3}}-3 =3\sum a-3 \geq 2(a+b+c)+(a+b+c-3) \geq 2(a+b+c)$
 :)


Given an acute angles triangles $ABC$ and $H$ is its orthocentre(markaze ertefai). The external bisector of the angle $BHC$ meets the sides $AB$ and $AC$ at the points $D$ and $E$ respectively. The internal bisector of the angle $BAC$ meets the circumcircle of the triangle $ADE$ again at the point $K$. Prove that $HK$ is through the midpoint of the side $BC$.

farz konid $F$  mahale barkhorde nimsaze $A$ ba $MH$ bashad , midanim ke $MH$ az rooberoo ghotriye $A$ migozarad pas dar mosalase $AMH$ darim:

$AF=\frac{4R^2\cos A.\cos(\frac{B-C}{2})}{2R(1+\cos A)}=\frac{2R\cos A.\cos(\frac{B-C}{2})}{\cos\frac A2^2}$
hamchenin darim (G mahale barkhord nimsaze A ba DE ast):

$AG=2R\cos A.\cos(\frac{B-C}{2})$

$AE=\frac{2R\cos A.\cos(\frac{B-C}{2})}{\cos\frac{A}{2}}$

ke rabete zir barayeshan bargharar ast:
$AE^2=AG.AF$
pas FE va FD bar AB va AC be tartib amood hastand.

[quote="amin-nn"]nimsaz [b]A[/b] ,[b]MH[/b] ro dar [b]X[/b] va dayere mohiti ro dar [b]Y[/b] ghat mikone ke [b]M[/b] vasat [b]BC[/b] .[b]Z[/b] paye amood [b]M[/b] bar [b]AB[/b] va [b]CP[/b] ertefa .
chon [b]AE[/b]=[b]AD[/b]. kafie sabetkonim [b]DX[/b] bar [b]AB[/b] amood ast .ya [b]HX[/b]/[b]XM[/b]=[b]PD[/b]/[b]DZ[/b] (1).
[b]HX/XM=AH/MY[/b] ,[b]PD=PB-DB[/b] .agar  in ha ro dar (1) gharar bedim ,be dast miad .[/quote]

baghye rahe hal ro mishe befresti? ;)


Nα βρεθουν οι ρητες λυσεις του συστηματος

$a^{2}+c^{2}+17(b^{2}+d^{2})=21$
$ab+cd=2$

(Made in Iran...)
 :D  :D  :D

Ράλλη μπορείς να βάλλεις λύση ?

Olokliromeni???mpa...An exis to imo compendium,pane sel263 to 1992...(longlist)

 :D

An den to exei (o silouan 'h opoiosdipote allos) tha to vrei sti gnwsti istoselida!! Einai to teleutaio vivlio BIBLOS pou exw valei sto tmima "Diagwnismoi"

Alexandros

Το έχω το βιβλίο αλλά δεν έχει λύση ........ :o

λύση δόθηκε 4 χρόνια μετά  [url=http://www.mathematica.gr/forum/viewtopic.php?f=111&t=6843]εδώ[/url]


geometry
perimeter
square
dissection
triangulation
IMO Shortlist
IMO Longlist

The square $ABCD$ has to be decomposed into $n$ triangles (which are not overlapping) and which have all angles acute. Find the smallest integer $n$ for which there exist a solution of that problem and for such $n$ construct at least one decomposition. Answer whether it is possible to ask moreover that (at least) one of these triangles has the perimeter less than an arbitrarily given positive number.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)


上次的不等式指什么？
CAUCHY是什么意思？


induction
geometry
geometric transformation
reflection
email
AMC
USA(J)MO

Okay so I was thinking that we could be efficient by putting questions that we have for the program cooridnators in one thread. That way, it is easier for them to answer their questions, and there is a higher likelyhood that our questions will be answered. Please number your questions.

Q1 What is the deal with the courses/tracks/etc?

Q2 To what extent does the difficulty of the Induction packet reflect the difficulty of the problems that we will face at the camp?

Q3 To what extent should we expect to improve? [question from penguinintegral]

Q4

I've been told there is flooding in Dallas; will camp be affected?

*cue SRV's "Texas Flood"*

Q5

Will pianos be available? Chess boards? Decks of playing cards?

To what extent will we have access to math books? If I need to review something in ACoPS or 101 Algebra problems, could I?

Q6

Roughly how many problems total do you expect we will do at camp?

Q7

Will we have private tutoring to correct weak areas?

[quote="PenguinIntegral"]Q4

I've been told there is flooding in Dallas; will camp be affected?

*cue SRV's "Texas Flood"*[/quote]
LOL

[quote="PenguinIntegral"]Q4

I've been told there is flooding in Dallas; will camp be affected?

*cue SRV's "Texas Flood"*
[/quote]

We can use boats to go from classroom to classroom

But then we would spend too much time trying to prove our boat routes and parking patterns were optimal.

 :(

I have a better idea.  Why don't we just prove that University of Texas at Dallas cannot be flooded, and be done with it? :P

LOL...  :P

Anyway, if you don't know the answer to a question, don't clutter the thread with uninformed speculation. Divert the chatter to another topic.

I just got an email about the track system.

Q8

How broadly do you define combinatorics? Will it be broad and competition mathy enough to help with, say, problem 3 on day 1 and problem 1 on day 2 of USAMO 2007?

Q9 Could you give two sample problems to illustrate the difference between the hard and not hard tracks? Or if we really want to take on combination that is 1-3 and we screwed up the AIME, should we still take the harder course?

(note to self: Do not leave AIME testing after two hours having solved 9 problems because you are "pretty sure you didn't make any errors"  :wink: )

Q10
What will be doing on weekends?
Q11
Where are we going on those field trips?
Are there any plans to go to the theater?
if we're gonna watch harry potter 5 over there, i might restrain myself to watching it before camp

Q12
Is Six Flags on the last day before we leave (August 5)? Because I have to leave August 4 in the morning... :(

Q13: Has Zac become so overloaded in preparing for AwesomeMath that he doesn't have time to check this forum?

Q14 is beer good

Q15 why aren't tracks made so that we can cover all 4 areas, not just 2? is this a scam to make us come twice?

Q16: Can we get the handouts/psets for tracks that we are not in?
Q17: Is there a copier readily available to students?
Q18: Is zuming scary?
Q19: Will these questions get answered?
Edit:
Q20: What exactly is "Algebra"?

Q20 are guitar amps available?
Q21 can i bring a suicidal cat?
Q22 will tennis balls be available or should we bring our own?

[quote="ccy"]
Q18: Is zuming scary?[/quote]
He's intimidating if you misbehave.

Question whatever number we are on: How many people can do each track? It seems like a lot of people want to do track 3, are you going to try to put the same amount of people in each track, or will you be flexible?

[quote="PenguinIntegral"]It implies that in the packet, Mustafa.[/quote]

I don't wanna read the packet

Also:

How will we get meals? Will there be like cereal in the apartments, or will we go to a dining hall? What is going to happen for food, and is it good food?

Instead of paying money to copy hundreds of pages of math problems, I would assume you have them all Latexed up and could probably just email everything to us, or at least burn a CD of them. Is this plan acceptable to you?

I would assume there is a dining hall, that's how it is at most college campuses. The food is usually decent, you'll get your daily nourishment (to be specific they probably will have cereal), but it probably won't be gourmet and will get repetitive (especially breakfast i imagine)

I know that there are 6 people + mentor in the room, does each apartment have its own bathroom or are there bathrooms with like 10 showers in them that are shared by everyone on a floor?

Guys, stop worrying so much! The accommodations won't be like home, but I'm sure it'll be comfortable and you'll be fed well.

As for the tracks, regardless of what you choose, you will be challenged and you won't be disappointed. (The first thing you'll learn when you get to the camp --- you know a lot less than you think).

[quote="SplashD"]There is AC, right?[/quote]

I would assume so, because it's like 90 degrees at 8 AM...

[quote="mustafa"]How much money should we bring, if any?[/quote]
I think it said 150 bucks on the instruction sheet.

to quote the information packet:

"Base your spending money on these options, not to exceed $ \$200.00$"

[quote="sapphyre571"]I looked again at the brochure and it said that we have classes Tuesdays through Saturdays, so what happens on Monday? is it just a free day or what?[/quote]

we take trips

[quote="Nawahd Werdna"]Question: Will there be computer access... and if so for how long... is there wifi or free ethernet or something??? I'm debating whether to bring a laptop/pda....  Mainly because my primary communication with my parents is through email, and my cellphone doesn't work in the US....  :maybe:  :maybe:[/quote]

There will be limited wireless internet access, but laptop use will be moderately restricted.

I'd only recommend bringing a laptop if you need it for communication, not for recreation.

(edited for clarity)


calculus
integration
inequalities
algebra
polynomial
function
inequalities proposed

[b]Let  ${\mathcal M}$  be the set of all polynomial functions $p(x)=ax^3+bx^2+cx+d$ which are non-negative on $[0,1] .$
 Find the least  number $\lambda$ such that \[ \int\limits_{0}^{1}(\lambda-x)p(x)\; dx \ge 0\; \; \; \; \; \; \mbox{for \ all}\; \; p\in {\mathcal M} \; .  \] [/b]

$\lambda _{min} = \frac {4}{5}$

 :shock:

The answer is $\displaystyle  \lambda_{min}=\frac{6+\sqrt{6}}{10}\; . $ Alex

Try to find the real coefficients $ c_0,c_1,c_2 $ and the points $ x_1,x_2\in [0,1] $
such that
\[ \int\limits_{0}^1 f(x)\; dx = c_0f(0)+c_1f(x_1)+c_2f(x_2)
\]
for all polynomials $ f $ of degree $\le 4 $. There is only one solution, moreover with all coefficients $ >0 .$ Then apply above equality when  \[ f(x):=(\lambda -x)p(x) \; .\]


Prove that with each pair of reals a,b ;
Always have $c\in(0,1)$ satisfy $|ac+b+\frac{1}{c+1}|>\frac{1}{24}$

What about a=0, b=$-\frac{2}{3}$, c=$\frac{1}{2}$?
Perhaps you meant there exists a c such that...  In that case, I don't know.

Wumbate exactly!!! exists a 'c' satisfy.

Hey, nothings coming to my head?? :huh:  :help:  :noo:  :surrender:  :read:  :stretcher:

Rushil, one month bumps are a bit ridiculous, especially when you're not adding anything in this and some of your other posts. You're not obligated to post in every topic that you can.

Would Dirichlet's approximation theorem be useful? The form of the theorem is almost exactly the same.

well Rushil, at first you were an average user of this forum but lately i see you are spamming more often. perhaps you want me to introduce DarkKnight's video to you? it's really cool if you got some spamming problems...

hey people why are you after me. I am just seeing the problems to check if something can be of use to me. I found this question interesting but couldn't understand the solution. That's why I posted here!!!


Show that equation $ \frac {A^2}{x\minus{}a} \plus{}\frac {B^2}{x\minus{}b}\plus{}...\plus{}\frac {H^2}{x\minus{}h}\equal{}x\plus{}\lambda$ cannot have real roots where A,B,C..... and a,b,c are real numbers

[quote="jackie-chan01"]Show that equation $ \frac {A^2}{x \minus{} a} \plus{} \frac {B^2}{x \minus{} b} \plus{} ... \plus{} \frac {H^2}{x \minus{} h} \equal{} x \plus{} \lambda$ cannot have real roots where A,B,C..... and a,b,c are real numbers[/quote]

Huh ?

Choose $ \lambda \equal{} \frac {A^2}{u \minus{} a} \plus{} \frac {B^2}{u \minus{} b} \plus{} ... \plus{} \frac {H^2}{u \minus{} h} \minus{} u$  and the equation has at least a real root $ x \equal{} u$  :)

[quote="jackie-chan01"]So you are saying question is wrong  ?[/quote]

What I said was that there obviously exists $ A,B,C, ...,a,b,c,...,\lambda$ such that the required equation has at least one real root.

Btw, it's elementary to show that if $ a < b < c < ... < h$ are $ n$ distinct numbers, then the equation has exactly $ n \plus{} 1$ real roots (one in $ (\minus{}\infty,a)$, one in $ (a,b)$, one in $ (b,c)$, ... , one in $ (g,h)$ and one in $ (h,\plus{}\infty)$)

And so, yes, I'm saying question is wrong.


A long time ago, you could buy eight hens for a dollar or one sheep for a dollar, and cows were ten dollars each. A farmer buying animals of each type bought a hundred animals for a hundred dollars. What animals did he buy ?

[hide] Set up a system of equations:

$ 8x\plus{}y\plus{}\frac{z}{10}\equal{}100$

$ x\plus{}y\plus{}z\equal{}100$

Subtracting, $ 7x\equal{}\frac{9z}{10}$. Solving this equation, we see that $ x\equal{}9$ and $ y\equal{}70$. This means the farmer bought $ 72$ hens, $ 21$ sheep, and $ 7$ cows.[/hide]


What are the four distinct positive integers with the smallest sum that can be placed on the sides of two identical tetrahedral dice so that the sum of the two bottom numbers has ten distinct possibilities?

I think the answer is......
[hide]
1,2,5,and 7

The ten distinct possibilities for the sum of the rolled dice are 2,4,10,3,6,7,8,9,12, and 14. I don't think that there is a combination of four such numbers with a lower sum, since if you start with two as one of the numbers, you get 2,3,5,9 which has a higher sum than 1,2,5,7.

I used guess and check.  :D Can anybody think of a smarter solution?[/hide]

Same thing...
[hide]I used guess and check but i only looked at prime numbers...[/hide]


What is the greatest possible value of $ x\plus{}y$ such that $ x^{2} \plus{} y^{2} \equal{}90$ and $ xy\equal{}27$.

$ x\plus{}y\equal{}\sqrt{x^2\plus{}2xy\plus{}y^2}\equal{}\pm12$, so the greatest value is $ 12$


limit
function
inequalities
IMC
college contests

Let $A_n$ be the set of all the sums $\displaystyle \sum_{k=1}^n \arcsin x_k $, where $n\geq 2$, $x_k \in [0,1]$, and $\displaystyle \sum^n_{k=1} x_k = 1$.

a) Prove that $A_n$ is an interval.

b) Let $a_n$ be the length of the interval  $A_n$. Compute $\displaystyle \lim_{n\to \infty} a_n$.

This problem already appeared in Crux some years ago.

Pierre.

Can you post the solution the Crux ?
thx

I do not have the exacte reference  but it was something like that :

Let $y_i = arcsin(x_i)$ so that $x_i = sin(y_i)$ and $y_i \in [0, \frac {\pi} 2$

From the well-known inequality $sin(x) \geq \frac {2x} {\pi}$ for $x \in [0; \frac {\pi} 2$ (using the concavity of $sin$), we deduce that :
$\frac {2} {\pi} \sum_{i=1}^n y_i \leq \sum_{i=1}^n sin(y_i) = 1$

In another hand, using the concavity of $sin$ and the fact that $arcin$ is increasing, we have :
$\frac 1 n = \frac 1 n \cdot \sum_{i=1}^n x_i = \frac 1 n \cdot \sum_{i=1}^n sin(y_i)\leq sin( \frac 1 n \cdot \sum_{i=1}^n y_i)$
so that $n \cdot arcsin( \frac 1 n ) leq \sum_{i=1}^n y_i$

It follows that $A_n \subset [n \cdot arcsin( \frac 1 n ); \frac {\pi} 2]$ and these two bounds are reached, for $y_i = arcsin(\frac 1 n )$ for all $i$, and for $y_1 = \frac {\pi} 2 $, and $y_i = 0$ for $i \geq 2$.

Moreover, let $ \varphi$ be the function defined on $[0,1]^n$ onto $R$ such that $\varphi (x_1,...,x_n) = arcsin(x_1) + ... arcsin(x_n)$.
Then $\varphi$ is continuous on $X = \{ (x_1,...,x_n) / 0 \leq x_i \ 1$ and $x_1+ \cdots + x_n = 1 \}$. Note that $X$ is connexe thus $\varphi(X)$ is a connexe in $R$, which means that it is an interval.
Thus $A_n= [n \cdot arcsin( \frac 1 n ); \frac {\pi} 2]$ .

Let $l_n$ be its length.
We have $l_n = \frac {\pi} 2 - n \cdot arcsin( \frac 1 n)$
Since $ x \cdot arcsin( \frac 1 x) = \frac {arcsin (\frac 1 x) - arcsin(0)} { \frac 1 x - 0}$ it follows that the limit of $x \cdot arcsin( \frac 1 x ) = arcsin'(0) = 1$ as $x$ tends to $+ \infty$.
Thus the desired limit is $\frac {\pi} 2 - 1$.

Pierre.

[quote="grobber"]Since it's a continuous function from $\mathbb R^n$ to $\mathbb R$, (a) follows. As for (b), $\arcsin$ is convex, so the value of the sum reaches its minimum when all the variables are equal, and this minimum tends to $1$ as $n\to \infty$. On the other hand, by showing that $f:[0,a\le 1]\to \mathbb R,\ f(x)=\arcsin(a-x)+\arcsin(a+x)$ is increasing, we can show that the sum reaches its maximum when one variable is $1$ and the others are $0$. This maximum is $\frac \pi 2$, so $\displaystyle \lim_{n\to \infty}a_n=\frac \pi 2-1$.[/quote]that's the solution I gave :)

If you didn't know that the image of a connected set by a continuous function is a connected set (and that the only connected subsets of $\mathbb{R}$ are intervals), here is another way to prove Part (a).

Once you have realized the sum is maximized when $(x_1,x_2,...x_n)=A=(1,0,...,0)$ and minimized when  $(x_1,x_2,...x_n)=B=\big(\frac{1}{n},\frac{1}{n},...,\frac{1}{n}\big)$,  we can just move $(x_1,x_2,...x_n)$ from $A$ to $B$ in such a way that the sum is always decreasing.

Consider the points $A_0=A$,$A_1=\big(1-\frac{1}{n},\frac{1}{n},0,...,0\big)$, $A_2=\big(1-\frac{2}{n},\frac{1}{n},\frac{1}{n},...,0\big)$, ...$A_{n-1}=B$.

Move from $A_k$ to $A_{k+1}$ by changing only two variables simultaneously, along the path $\big(1-\frac{k}{n}-t,\frac{1}{n},..\frac{1}{n},t,0,...,0 \big)$ where   t varies from $0$ to $\frac{1}{n}$. The sum decreases along this path because $\arcsin(a-t)+\arcsin(t)$ is decreasing for $t \le \frac{a}{2}$. By the intermediate value theorem, it takes all of the values between its value at $A_k$ and its value at $A_{k+1}$.


summer program
Mathcamp
MathPath
geometry
HMMT
Duke
college

So, who's going to MathCamp 2009?

Me, for one

Actually, I should probably ask people to write something about their selves, but since is the first thing I've done of this kind (Damn you Britain), I have no idea what to suggest. Any suggestions?

[b]Which grade are you in?[/b]

People have already received their decisions?!?! O.o
my anxiety level just got tripled

i dont think so, i haven't gotten anything yet

neither have i.

i think some ppl who sent their apps in early have gotten theirs back.

I think they aim to have them done by the 15th or so. I got mine back particularly early for some reason

Hopefully, they won't be dragged out until the 15th. The suspense is quite unnerving XP

If it's like previous years, they will probably send out most decisions by the end of this week. However, this is just a guess, and it would likely depend how many applications they received and such. I believe I got my acceptance about a week and a half after the deadline...I don't think they'll drag it out till the 15th.

Oh, and I should post an introduction, since I'm going. Heh, I'm totally copying and pasting this from last year. Again. Although I edited more this time. But, anyway, that's why I have a lot more than my grade here :P .

I'm Laura Zehender, from near Richmond, VA (I have lived here my whole life). I am 17 and will be a senior next year (eeek, I'm old :maybe: ), and have been homeschooled since 2nd grade. This will be my third year at Mathcamp :D (went in '07 and '08), and I went to MathPath in '06. Randomly, I own eight camp t-shirts. My interests/activities include....
Math (duh): especially combinatorics, geometry, and knot theory, thanks to camp last year :) .
Dance: for regular dancing, ballet and tap now, jazz and pointe previously--I've danced for 13 years now. Last summer at MC, I got into partner dancing, thanks to all of the awesome people there who taught us/brought us to local dances. I can remember a decent amount of contra and swing, and a little of waltz and salsa. I hope there's a lot of partner dancing again, it's awesome.
Classics: Latin (which I've done for way too long), Mythology (mainly Greek), and Ancient history in general.
Music: I play piano well enough, although I'm horrible in comparison to people at Mathcamp (I can't wait to hear all of the amazing musicians at MC again :laugh: .) I especially like playing Tchaikovsky music (if you see a book of Tchaikovsky music lying around camp, that'd probably be mine.) I mainly listen to country, although I don't object to listening to almost any kind of music. I've heard lots of genres in dance classes over the years.
Shooting: Rifle and bow shooting are my favorites, although I also do shotgun (and only at targets...yeah.) I've been in a 4-H shooting club for two years now, and I placed 4th in rifle at the 4-H state meet last fall :-). Oh, and I also enjoy shooting with a camera (and in this case, I do shoot people :wink: .) Prepare for a ridiculous amount of photos on Facebook after camp...
Arts&Crafts: I waste way too much time on these. I like modeling clay (especially to make miniatures, which I sorta collect), pencil or pen drawing (also, prepare for some of those many photos turning up as drawings on Facebook later), beading, making (sometimes mathematical) jewelry, photography and photo editing, embroidery, and knitting (I learned how to knit a mobius strip at MC '08; it's pretty fun.) And now painting with acrylics.
Cubing: Although I'm not very good at it, it's fun, and I know too many people (from Mathcamp) who are really into it for some interest in it to not have rubbed off on me..

I also am a bookworm (I read mostly fantasy, but often science fiction and historical fiction, too, and random other books. And classics. As in, really old classics from ancient history [like, Plato's Dialogues, the Odyssey...]) Um, let me stop before I list books I like, because that could get out of hand :wink: . To cover a couple topics that came up in the previous introduction threads, I'm a Libertarian, an ENFJ (Myers-Briggs type). And I'm a chocoholic--although just dark chocolate. Like, [i]really[/i] dark chocolate :D .

Oh, and I also am excited to hopefully try some more random board games at Mathcamp; there are always tons of obscure but fun ones there. Just don't try to make me join a Mao game. People have [i]tried[/i] :roll: .

This year I've gone to PUMaC, HMMT, and Duke math meets, and I'm going to ARML (anyone else going to Penn State?) This is only my second year of going to math meets, and they're super fun. This year I went to everything on the Chesapeake team (with which I'm going to ARML again...yay for there finally being a team to cover non-TJ Virginia :lol: ), except for Duke. I sneaked on the GA team for Duke math meet for the second time, despite not being from GA....

Technically my name is James Koppel, but Mathcamp last year made me accept that I will probably always be "Jimmy." I'm a 17 year-old senior from St. Louis, MO. Next year I'll be going to Carnegie-Mellon, unless I get off the MIT waitlist.

Even more so than math, I am extremely interested in computer science, especially in programming languages (if you're curious what a context-free language with Hindley-Milner type inference which supports anonymous functions but not first class functions would look like and want a BNF syntax for one, ask me). My favorites are Ruby and Lisp (currently trying to learn Clojure), but I have done medium to large projects in about a half dozen others. I am a huge fan of Paul Graham, and a member of news.ycombinator.com.

In math, I have grown enamored of number theory. I know little about topology, but it sounds really cool, and I am quite keen on studying it. And, of course, I like theoretical computer science. I'm currently trying to self-study the lambda calculus, and hope to teach a class on it during Week 5.

Now, to try to think of my interests that aren't primarily math or CS...

Music: I will try to evangelize my new favorite, E.S. Posthumus. That probably won't be too hard, considering how surprised I was to find that Mathcampers mostly like the same kind of music I do (classical music, video game themes, and Broadway themes were all I heard for most of camp).

Chess: I have a decent collection of trophies, but I suck by Mathcamp standards (my rating's been hovering around 1100).

Languages: I love German.

Books: I've only recently (6 months ago) gotten back into a constant reading habit. Since then, I've read The Fountainhead by Ayn Rand, Thus Spake Zarathustra by Friedrich Nietzsche, The Selfish Gene by Richard Dawkins, and am currently reading a mix of Atlas Shrugged by Ayn Rand and Goedel, Escher, Bach by Douglas Hofstadter.

Politics: If you didn't already guess from the dose of Rand, I am an ardent libertarian (Ron Paul 2012!). I didn't know Billy was libertarian until he mentioned it in the yearbook, and I didn't know Laura was until just now, so it seems we have to shout that out a bit more.

Martial Arts: I've been doing classical fencing casually for 6 years, and will have a 2nd degree black belt in Tae Kwon Do by the time camp starts.

Video games: I'll play a good game from any genre, but I'm mainly interested in TBSs. I'm a huge fan of the Heroes of Might and Magic series, but I also like 4X games. I also like RTSs and RPGs a lot. (I'm currently playing the original Baldur's Gate.)

* also copies-and-pastes-and-modifies *

Hi, I'm Billy, homeschooled soon-to-be-not-senior-anymore (MIT next year!), and my favorite word is 'poecilonym'. I do absolutely love COUNTRY MUSIC! and singers thereof (particularly Sara and Brad and Danielle and Martina and Eric, in no order). I'd like to imagine I can carry on an intelligent although divagating conversation about politics, religion, philosophy, history, vocabulary, farming, &c. I've been to HMMT and Pumac x2 and DMM x4 and ARML x4. I also organized the first and so far only homeschool-run math meet in GA as a junior.

Oh yes, I do math, and computer science, and sesquipedaliophilia; I dabble in biology and history. (I also talk archaicly when I am either uncomfortable with what I'm saying or am trying to be humorous) I've done work both in encryption and in simulation, I'm also a fan of programming languages and have studied lambda calculus some amount, to reply to Jimmy. I shotgun, I enjoy Attack! (risk++), hearts, stratego, fluxx, mao, mao, mao, and I run slowly. I've never solved a cube.  And I'm sure I forgot at least something else about me. Oh yes, I enjoy country music greatly. :)

Politics and religion: It is not actually true that all returning Mathcampers are libertarian. It is actually true that I can only think of one other returning camper (and two not-returning campers) who are also libertarian. I will say, though, that libertarianness must be modified by religion: I'm a Christian. Joel Salatin, who appeared on the front of USA Today for Earth Day 2009, has said he is a Christian libertarian environmentalist farmer. I really look up to Mr. Salatin, and so will happily also claim this. Though I haven't sold at a farmer's market in some years now.

Reading: Just finished Watership Down after the two days of the USAMO. Bunnies are aliens. Reading Hitchhiker's Guide when it's juxtaposed with me and I have time. Just discovered the joys of Coleridge. Have read the Complete Works of O. Henry in the last year. Love love love Pascal, the KJV Bible, the Silmarillion and to a lesser extent Lord of the Rings. Enjoy the magazines [i]First Things[/i], [i]Reason[/i], and old [i]Mother Earth News[/i] (new ones are all citified). Get my news from a combination of those magazines, the Green Left Weekly, Dr. Laura's blog, news.ycombinator.com, Phoronix, and Schneier's blog. I'm not especially well-versed in US news since the last election: I can't tell you what I think of Obama's presidency. But I [i]can[/i] tell you that right before MIT CPW I went to a Tax Day Tea Party and walked the 3 miles back to MIT holding up my "Free Markets not Free Loaders" sign.

I'll be a two-time camper after this summer - I'm semi-fluent in Food Tongue and hope I meet more awesome people and do less stupid things this summer than last. (Apple cauliflower Sweet Tea.) (Sweet Tea would be an interest of mine that I didn't mention :P. Sweet tea! Yee-haw!)

(Based on the current flu situation, I move that we stick to only verbalized nosebops, without physical nosebops. * nosebop * )

By the way:

[b]Theorem[/b]: Mao is the most fun game
[b]Proof[/b]: Suppose there were a game G that was more fun than Mao. You could then institute a rule in Mao that called for playing G as one of its effects. Then Mao would be at least as fun as G (likely more fun, since the rule could call for interactions between the "embedded" game G and the host game of Mao).  Contradiction!

[quote="Darmani"][b]Theorem[/b]: Mao is the most fun game
[b]Proof[/b]: Suppose there were a game G that was more fun than G. You could then institute a rule in Mao that called for playing G as one of its effects. Then Mao would be at least as fun as G (likely more fun, since the rule could call for interactions between the "embedded" game G and the host game of Mao).  Contradiction![/quote]

I'm sure there's an error in that proof, I just haven't found it yet :P .

And I didn't know you were Libertarian until this thread, either. Knew Billy was, as I knew him before Mathcamp.

Oh, and I also read Fountainhead and Atlas Shrugged this year! Good job if you get through all of Atlas Shrugged, I read quickly and it took me ages to get through. I nearly got stuck on the 55-page speech :o (Ayn Rand obviously wants to make absolutely sure readers understand her philosophy, both of those book have huge speeches by the main character as a climax.)

And @Billy, our sets of favorite country music artists do not seem to intersect...Dierks Bentley, Sugarland, Taylor Swift, and Carolyn Dawn Johnson are way better :yup: .

EDIT: Also, if offline, my nose function can only be abused...erm, [i]activated[/i] by actual nosebops  :no: . (Are you serious about no real nosebops? I think if someone had flu at camp it'd get around pretty quickly anyway, since we're around each other all day :huh: . Anyway, let's stop confusing new campers who haven't heard of nose functions.)

[quote="Darmani"]
[b]Theorem[/b]: Mao is the most fun game
[b]Proof[/b]: Suppose there were a game G that was more fun than G. You could then institute a rule in Mao that called for playing G as one of its effects. Then Mao would be at least as fun as G (likely more fun, since the rule could call for interactions between the "embedded" game G and the host game of Mao).  Contradiction![/quote]

this proof does not comply with negative values of fun, generally classified as the long, boring intervals in between fun, which the other aspects of Mao could bring along.

btw, what is mao?

I guess if you mean "real chess" as opposed to "virtual chess" or "solitaire chess" rather than as opposed to "nonstandard chess," Mathcamp would be a good place to get your share.

[quote="Darmani"]I just went to last year's yearbook and counted. I got 89 boys to 27 girls, for a boy:girl ratio of about 3.3:1.[/quote]

Wait, really?  :| It seemed a lot higher when I was there, for some reason.

Before I made the count, I was going to post and say it was closer to 5:1 (i.e.: something like 100 boys, 20 girls), so feel free to grab a yearbook and recount.

I actually meant to say "lower" in my last post, and it came out as "higher" somehow. =/

Your count looks right, though.

So, I remember reading somewhere that the schedule of possible classes isn't usually decided until at least a week before camp, but is it ever put up online beforehand? Or do we find out when we get there?

Don't worry, you'll have plenty of time to figure out which classes you're interested in, and people (campers from previous years and staff members) are always willing to help you decide.

You can also find schedules from previous years at the main Mathcamp website, as I'm sure you're aware.

Hi guys,
@mathcamp alumni: How many classes, on average, did you take per day?
In general, considering the amount of homework etc., how many classes should one take?

Btw I'm Aniruddha (or Ani) from India. I'll be attending mathcamp this year.

There are 4 classes and 1 colloquim offered on most days during the first 4 weeks of camp. Usually, people take 3 classes though each day because 4 classes are very hard to handle if you want to get the most out of every class. In addition to 3 classes, some campers, like me, opt to do a project starting during the second or third week of camp. I would highly recommend doing this as I have loved my projects each of the past two years.

Just to point this out: it looks like we won't have wifi in the dorms after all. It says we will on the FAQ and such, but according to the precamp info .pdf we won't have it (and I'd wager it's more likely that the pdf is right, since it's more recent). There's still ethernet though, so it doesn't really matter :P

Don't give up so easily, Zeph.  Anyone have a router?

I should also add that people often take too many / too difficult classes at the beginning, but eventually learn. So be cautious (but not too cautious).

Besides, we have these things called 'academic advisors' that act as a buffer to warn you before you take 4 4-star classes or 1 1-star class.

[quote="ycz9000"]And you can always drop out of a class if it's too hard/easy/boring/interesting for you. While joining a class after it's already started is tough, there are usually a few classes that don't last the entire week, so dropping actually isn't too damaging to your schedule.[/quote]

too interesting? I can totally see that happening.


inequalities
Cauchy Inequality
inequalities unsolved

For $ a,b,c > 0$ and such that $ abc \equal{} 1$ . Prove that:
          $ (a^2  \plus{} b^2  \plus{} c^2 )^2  \ge (a^3  \plus{} b^3  \plus{} c^3 ) \plus{} ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a)$
I use Schur and p,q,r
 $ r \equal{} 1.$
       $ p^4  \minus{} 4p^2 .q \minus{} p^3  \plus{} 4.q^2  \plus{} 2p.q \ge 0$

This problem in my opinion is quite trivial
We have
$ (a^3 \plus{} b^3 \plus{} c^3 ) \plus{} ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a)\equal{}(a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)$
It suffices to prove that
$ a^2\plus{}b^2\plus{}c^2 \ge a\plus{}b\plus{}c$
which is easy to prove by Cauchy inequality :)


inequalities
calculus
derivative
inequalities proposed

Could anyone prove that for all positive x, y and z

(x + y)^z + (y + z)^x + (z + x)^y > 2

[quote="Jean-Louis"]Could anyone prove that for all positive $x, y$ and $z$

$(x+y)^{z}+(y+z)^{x}+(z+x)^{y}> 2$[/quote]
$a^{b}>\frac{a}{a+b}$ for $a>0$ and $0<b<1.$
If $x\geq1$ or $y\geq1$ or $z\geq1$ your inequality holds, while
if $\{x,y,z\}\subset(0,1)$ then
$\sum_{cyc}(x+y)^{z}>\sum_{cyc}\frac{x+y}{x+y+z}=2.$ :wink:
It's a very old inequality.

Thanks arqady,

What does 'cyc' notation stand for ?

Cyclic sum means add up the cyclic permutations of the variables (with n variables there are n cyclic permutations)

How do you prove $a^{b}>\frac{a}{a+b}$?

Is this correct :

If you derivate the inequality :

(x+y)^z + (x+z)^y + (y+z)^x > 2

you get,

{ z * [(x+y)^(z-1] } + { y * [(x+z)^(y-1)] } + { x * [ (y+z) ^ (x-1) } > 0

and take it to :

[z * 2^(z-1) ] + [ y * 2^(y-1) ] + [ x * 2^(x-1) ] > 0

 Since we know that the derivative of X, Y, and Z.. is basically 1 )


Try to prove these identities without fully expanding both sides

$ (a\plus{}b\plus{}c)^3\equal{}a^3\plus{}b^3\plus{}c^3\plus{}3(a\plus{}b)(b\plus{}c)(c\plus{}a)$

$ (a\plus{}b\plus{}c)^5\equal{}a^5\plus{}b^5\plus{}c^5\plus{}5(a\plus{}b)(b\plus{}c)(c\plus{}a)(a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca)$

$ 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)$

$ \boxed{(x \plus{} y)^3 \equal{} x^3 \plus{} y^3 \plus{} 3xy(x \plus{} y)}$,

$ (a \plus{} b \plus{} c)^3 \equal{} a^3 \plus{} (b \plus{} c)^3 \plus{} 3a(b \plus{} c)(a \plus{} b \plus{} c)$
$ \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} 3bc(b \plus{} c) \plus{} 3a(b \plus{} c)(a \plus{} b \plus{} c)$
$ \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} 3(b \plus{} c)\{bc \plus{} a(a \plus{} b \plus{} c)\}$
$ \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} 3(b \plus{} c)\{a^2 \plus{} (b \plus{} c)a \plus{} bc\}$
$ \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} 3(b \plus{} c)(a \plus{} b)(a \plus{} c)$
$ \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} 3(a \plus{} b)(b \plus{} c)(c \plus{} a)$

$ \boxed{x^3\plus{}y^3\equal{}(x\plus{}y)(x^2\minus{}xy\plus{}y^2)}$

$ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}(a\plus{}b)^3\minus{}3ab(a\plus{}b)\plus{}c^3\minus{}3abc$
$ \equal{}(a\plus{}b)^3\plus{}c^3\minus{}3ab(a\plus{}b\plus{}c)$
$ \equal{}(a\plus{}b\plus{}c)\{(a\plus{}b)^2\minus{}(a\plus{}b)c\plus{}c^2\minus{}3ab\}$
$ \equal{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)$

Alternative Solution for $ (a\plus{}b\plus{}c)^3\equal{}a^3\plus{}b^3\plus{}c^3\plus{}3(a\plus{}b)(b\plus{}c)(c\plus{}a)$:

[hide]
Assume that $ a,b,c$ are the roots of $ P(x)\equal{}x^3\plus{}a_2x^2\plus{}a_1x\plus{}a_0\equal{}(x\minus{}a)(x\minus{}b)(x\minus{}c)$, such that the LHS is $ (\minus{}a_2)^3\equal{}\minus{}a_2^3$, and the RHS is $ S_3\plus{}3(\minus{}a_2\minus{}a)(\minus{}a_2\minus{}b)(\minus{}a_2\minus{}c)\equal{}S_3\plus{}3P(\minus{}a_2)$.

By [url=http://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_sums]Newton's Sums[/url], we eventually get $ S_3\equal{}3a_1a_2\minus{}a_2^3\minus{}3a_0$, and since $ P(\minus{}a_2)\equal{}\minus{}a_2^3\plus{}a_2^3\minus{}a_1a_2\plus{}a_0\equal{}\minus{}a_1a_2\plus{}a_0$, we have

$ S_3\plus{}3P(\minus{}a_2)\equal{}\minus{}a_2^3$, as desired.
[/hide]

For the other two, the argument is similar to above: we use Vieta's and Newton's to express everything in terms of the coefficients of $ P(x)\equal{}x^3\plus{}a_2x^2\plus{}a_1x\plus{}a_0\equal{}(x\minus{}a)(x\minus{}b)(x\minus{}c)$.

Another proof for these should be in the site itself. The diagram at the left showed $ (x \plus{} y)^2$ geometrically. From there you might get something. (I don't know how to put diagrams here.)

Other solution for $ 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{} ac \minus{} bc)$:

[hide]Calculate the determinant $ \begin{vmatrix} a & b & c \\  c & a & b \\ b & c & a \end{vmatrix}$ and you get LHS.

Using determinant proprieties, we have:

$ \begin{vmatrix} a & b & c \\  c & a & b \\ b & c & a \end{vmatrix} \equal{}  \begin{vmatrix} a\plus{}b\plus{}c & a\plus{}b\plus{}c & a\plus{}b\plus{}c \\  c & a & b \\ b & c & a \end{vmatrix} \equal{} (a \plus{} b \plus{} c)\begin{vmatrix} 1 & 1 & 1 \\  c & a & b \\ b & c & a \end{vmatrix}$

And you get RHS.[/hide]

How would you prove this property (I know this property but I don't know the proof to it)?
$ \begin{vmatrix} a & b & c \\
c & a & b \\
b & c & a \end{vmatrix} \equal{} \begin{vmatrix} a \plus{} b \plus{} c & a \plus{} b \plus{} c & a \plus{} b \plus{} c \\
c & a & b \\
b & c & a \end{vmatrix}$

By the way, nice proof using determinant!


algebra
polynomial
vector
linear algebra
matrix
inequalities
search

Prove or disprove and salvage if possible:

Let $ u$, $ v$ be linearly independent $ 3 \times 1$ matrices with real entries.  Then $ u u^T \minus{} v v^T$ has a positive eigenvalue, a negative eigenvalue, and an eigenvalue of $ 0$.

(The cross-product of $ u$ and $ v$ will give you an eigenvector with eigenvalue $ 0$.)

the characteristic polynomial of $ A \equal{} uu^T \minus{} vv^T$

is $ X(X^2 \minus{} (\parallel{}u\parallel{}^2 \minus{} \parallel{}v\parallel{}^2)X \plus{} (u|v)^2 \minus{} \parallel{}u\parallel{}^2\parallel{}v\parallel{}^2)$

with $ (u|v)^2 \minus{} \parallel{}u\parallel{}^2\parallel{}v\parallel{}^2 \leq 0$

one eigenvalues is positive the other negative

This would work with $ u$ and $ v$ being $ n\times 1$ matrices, with $ n\ge 3.$ In that case, $ 0$ is an eigenvalue of multiplicity $ n-2,$ and there is a positive eigenvalue, and there is a negative eigenvalue.

Let $ w$ be any vector in the orthogonal complement of the span of $ u$ and $ v.$ The reason for saying $ n\ge 3$ is so this orthogonal complement will be nontrivial. Then $ u^Tw=0$ and $ v^Tw=0.$ As a result, $ (uu^T-vv^T)w=0.$ That's the eigenspace corresponding to $ \lambda=0.$

Now, let's ask if it is possible to find eigenvectors of the form $ \alpha u+\beta v.$

$ (uu^T-vv^T)(\alpha u+\beta v)=\lambda(\alpha u+\beta v)$

$ \alpha\|u\|^2u+\beta\langle u,v\rangle u-\alpha\langle u,v\rangle v-\beta\|v\|^2v=\lambda\alpha u+\lambda\beta v$

Since $ u$ and $ v$ were given as independent, we may conclude that $ \alpha$ and $ \beta$ satisfy
\begin{align*}\alpha\|u\|^2+\beta\langle u,v\rangle&=\lambda\alpha\\-\alpha\langle u,v\rangle-\beta\|v\|^2&=\lambda\beta\end{align*}
This means that $ \lambda$ is an eigenvalue of the matrix
\[ B=\begin{pmatrix}\|u\|^2&\langle u,v\rangle\\-\langle u,v\rangle&-\|v\|^2\end{pmatrix}\]
Since $ \det B=\langle u,v\rangle^2-\|u\|^2\|v\|^2,$ which is negative (for independent $ u,v$) by the Cauchy-Schwarz inequality, then $ B$ has one positive and one negative eigenvalue. Furthermore, the steps are reversible, and by finding $ \alpha$ and $ \beta,$ we have enough information to construct the eigenvectors for $ A.$

See the notion of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1332522724&t=159717]congruence of matrices.[/url]

Essentially what I have shown is that $ A$ is congruent to $ \begin{pmatrix}B & 0 \\
0 & 0\end{pmatrix},$ where $ B$ is the $ 2\times 2$ block shown in my post above. The matrix $ P$ such that $ P^TAP \equal{} \begin{pmatrix}B & 0 \\
0 & 0\end{pmatrix}$ is simply the matrix with $ u$ and $ v$ as the first two columns, finished out with a basis for the orthogonal complement of $ \text{span}\{u,v\}.$

In the same way, you could show that $ uu^T \plus{} vv^T$ has exactly two positive eigenvalues (distinct unless $ u$ and $ v$ are orthogonal and the same size), with the rest of its eigenvalues being zero.

Correction: I think I showed similarity, not congruence. Doing the congruence described above seems to get me an upper left $ 2\times 2$ block of
\[ \begin{pmatrix}\|u\|^2\minus{}\langle u,v\rangle^2&(\|u\|^2\minus{}\|v\|^2)\langle u,v\rangle\\ (\|u\|^2\minus{}\|v\|^2)\langle u,v\rangle&\langle u,v\rangle^2\minus{}\|v\|^2\end{pmatrix}\]

[quote="Singularitarian"]Thanks for the reply.  Is there a non-brute force way to come up with that characteristic polynomial?  Or do you just have to write out $ det(XI \minus{} (uu^T \minus{} vv^t))$ in terms of the components of $ u$ and $ v$, and simplify.[/quote]

You can do it without components of $ U$, $ V$

$ rank(A)\leq rank(U.U^{t}) \plus{} rank(V.V^{t}) \equal{} 2$ then $ det(A) \equal{} 0$

for $ A$ matrix in $ M(3,R)$ the caracteristic polynomial is

$ P_{A}(X) \equal{} X^3 \minus{} tr(A)X^2 \plus{} \frac {1}{2}((tr(A))^2 \minus{} tr(A^2))X \plus{} det(A)$

$ tr(A) \equal{} \parallel{}U\parallel{}^2 \minus{} \parallel{}V\parallel{}^2$

compute $ A^2$ then $ tr(A^2) \equal{} \parallel{}U\parallel{}^4 \plus{} \parallel{}V\parallel{}^4 \minus{} 2(U|V)^2$

now you have all materials to get $ P_{A}(X)$
_____________________________________________________________

Others questions :

1/ $ A$ is trigonalisable ?

2/Study the diagonalisation of $ A$ ?

A useful notion here: tensor rank. The tensor rank of a matrix $ A$ is the minimum $ k$ such that $ A$ can be written as the sum of $ k$ matrices of the form $ uv^T$ with $ u,v$ $ n\times 1$ vectors. This generalizes to $ m$-linear forms by considering sums of operators $ f_1(x_1)f_2(x_2)\cdots f_m(x_m)$.

For ordinary matrices, the tensor rank is always equal to the usual rank.
By the structure we are given, $ A$ has tensor rank at most 2. Our calculations show that this rank is actually equal to 2, and give the signature of the quadratic form.

Thanks for the excellent replies, everyone!  That was very enlightening.

[quote=Kent Merryfield]
Correction: I think I showed similarity, not congruence. Doing the congruence described above seems to get me an upper left $ 2\times 2$ block of
\[ \begin{pmatrix}\|u\|^2\minus{}\langle u,v\rangle^2&(\|u\|^2\minus{}\|v\|^2)\langle u,v\rangle\\ (\|u\|^2\minus{}\|v\|^2)\langle u,v\rangle&\langle u,v\rangle^2\minus{}\|v\|^2\end{pmatrix}\][/quote]

One minor correction here: the correct matrix obtained via congruence is
\[ \begin{pmatrix}\|u\|^4\minus{}\langle u,v\rangle^2&(\|u\|^2\minus{}\|v\|^2)\langle u,v\rangle\\ (\|u\|^2\minus{}\|v\|^2)\langle u,v\rangle&\langle u,v\rangle^2\minus{}\|v\|^4\end{pmatrix}\]

(which, btw, has a negative determinant).


Βάζω και εγώ μια άσκηση γιατί έχω καιρό να βάλω...[u]αλλά δεν ξέρω λύση σε αυτήν[/u]  :oops:

Αν $ a,b,c$ πλευρές τριγώνου να δείξετε ότι:

$ (abc)^2\geq\bigg(\frac {4S}{\sqrt {3}}\bigg)^3$

Με λίγο algebraic manipulation έχουμε ισοδύναμα:

$ 3\sqrt3*(R^3)\geq(abc)$

Με νόμο ημιτόνων καταλήγουμε σε:

$ \frac {3\sqrt3}{8}\geq sinAsinBsinC$

η οποία είναι γνωστή και εύκολο να αποδειχθεί.

Σόρρυ για την περιληπτικότητα... Γράφω Ιστορία... :(

Άλλος τρόπος:
Επειδή $ abc\equal{}4RS,$ αρκεί να αποδείξουμε ότι $ 3\sqrt{3}R^2\geq4S\equal{}4sr.$ Όμως αυτή προκύπτει με πολλαπλασιασμό των: $ R\geq2r$ και $ 3\sqrt{3}R\geq2s.$ Η τελευταία προκύπτει από την $ \sum sinA\leq\frac{3\sqrt{3}}{2}.$

:blush: πράγματι δεν ήταν τόσο δύσκολο... αλλά αν δεις την πηγή απ'την οποία πήρα το πρόβλημα, δεν θα μπορούσες να φανταστείς ότι λύνεται τόσο εύκολα.

Ποια ήταν η πηγή Δημήτρη? :?:


Hello!

I want to make sure that I get all the facts straight.

[b]1.[/b] First I need to take the TOEFL and GRE General+Subject tests. And the results will be sent for free(*) to four departments/universities.

[b]2.[/b] I fill in the papers that each department require and send them in.

And some questions.

[b]Q1.[/b] I've heard that the departments charge a fee for each application. Is this fee related to (*)? Do I still have to pay this fee (if it exists, I could be misinformed) to the four departments that I've chosen the results (GRE&TOEFL) be sent to at (*)?

[b]Q2.[/b] Say I take the TOEFL this year, GRE General test next year and the GRE Subject test in two years and wish to apply to graduate school in three years. According to (1) the results will be sent to four departments each time. But the TOEFL result will be sent three years prior to the date I'm actually applying to grad school. How does this work? Will this be a problem somehow?

[b]Q3.[/b] If I get an external scholarship from back home (but faaaaar from what I really need) will this affect my chances of getting support (e.g. RA and TA) at the college I am applying to?

[b]Q4.[/b] I've hard that apartments could often be found cheaper outside campus. Any good site out there you recommend?

Thats all I can think of! But if there is something I am missing and [i]should [/i]know, please don't hesitate in telling me!

Thank you!

You will also need letters of recommendation (about 3?), usually sent directly by your references. It may help if one of them comes from someone who is known by someone at the department of your choice. (This is not necessary, of course).

Q1: the application fee is completely different from the fees associated with TOEFL and GRE. I think that the application fee is typically charged by the Graduate School rather than the department itself, and is supposed to cover the expenses of handling your application. The amount of this fee varies from one university to another, but it's strictly positive almost everywhere.

Q2: I would not recommend taking TOEFL three years in advance. Universities want to see relatively recent test scores, since foreign language skills can deteriorate over time. Some may not even keep test scores files for that long (I'm not sure about this.) Taking TOEFL later will also allow you more time to polish your English if needed. [I applied one year later than initially planned, mostly because of TOEFL. And even then I only got the minimum score that was accepted by my target school. :blush: ]

Q3: I don't think so.

Q4: Websites like rent.com or apartments.com do exist, but apartment search is local by nature. Go to your target school's website and look for "off-campus housing". [url=http://studentlife.tamu.edu/agoss/]Here is an example[/url]. BTW, it has some interesting options in the roommate search:
[quote] Alcohol: Never/Rarely/Sometimes/Often
 Study Preference: Quiet/Background noise/Lots of distractions
 Study Habits: Library/Home/Both/[b]Not at all[/b][/quote]

[quote="kahsi"][b]Q2.[/b] Say I take the TOEFL this year, GRE General test next year and the GRE Subject test in two years and wish to apply to graduate school in three years. According to (1) the results will be sent to four departments each time. But the TOEFL result will be sent three years prior to the date I'm actually applying to grad school. How does this work? Will this be a problem somehow?[/quote]
I don't know anything about the TOEFL, but with respect to the GRE: when you take the GRE, you have the option of selecting 4 schools to which to send your scores; it is not mandatory.  If you then take the other (subject instead of general, or vice versa) exam, you may choose to have both scores sent to your schools together.  It may also be the case (but you should check with the ETS, because I do not know, I am just guessing) that you don't need to use the 4 reports immediately and can still get them free at a later date.

Thanks mlok and JBL.

I've sent a mail to ETS and get back to you when they mail me back.

I have a little question on the letter of statement.
http://owl.english.purdue.edu/handouts/pw/p_perstate.html

[quote]I am highly aware of the superb reputation of your school, and my conversations with several of your alumni have served to deepen my interest in attending. I know that, in addition to your excellent faculty, your computer facilities are among the best in the state. I hope you will give me the privilege of continuing my studies at your fine institution.[/quote]

Is this a good thing to put in it? It feels like but-kissing ( :blush: ) but maybe the faculty like this type of text  :lol:

Also,

[quote]Are there any gaps or discrepancies in your academic record that you should explain (great grades but mediocre LSAT or GRE scores, for example, or a distinct upward pattern to your GPA if it was only average in the beginning)?[/quote]

[quote]They also fail to explain errors or weaknesses in their background. Even though we might wish to admit a student, sometimes we can't in view of a weakness that they haven't made any effort to explain. For example, perhaps they haven't told us that they were ill on the day that they took the LSAT or had an automobile accident on the way. Such things are legitimate reasons for poor performance. I mean, we understand that life is tough sometimes. We need to know what happened, for example, to cause a sudden drop in the GPA.[/quote]

I have got fives (top grade) on all courses except for two which I only got fours. The reason I got fours was that both the exams were(100% of the grades depended on the exams) at the same time and that 10h (with a 1h lunch in between) of constant writing was too much both mentally and physically which lead to stupid mistakes (but also that I couldn't prepare for each exam as I would like to). Would that be a place to mention it or would they look at me as a student who tries to blame his bad results at other things than himself. Because, clearly it wasn't as serious as an "automobile accident".

[quote="kahsi"][quote]I am highly aware of the superb reputation of your school, and my conversations with several of your alumni have served to deepen my interest in attending. I know that, in addition to your excellent faculty, your computer facilities are among the best in the state. I hope you will give me the privilege of continuing my studies at your fine institution.[/quote]

Is this a good thing to put in it? It feels like but-kissing ( :blush: ) but maybe the faculty like this type of text  :lol:
[/quote]
It does feel VERY over the top. If I was the one reading that text I would think:
1) The person is BSing
2) The person is immature, not used to doing things in a social context.
3) The person doesn't really know why he wants to go

Maybe that was a little harsh, but I do think it's a very bad thing to put in an application. My intention is not at all to call you immature, but to stop you from writing that. Instead, I think you should write why you really want to go, and mention the actual reasons why you think it's a great institution, without stating this explicitly. That is, while you should write reasons that you are eager to go to their school you should mention specific reasons and not vague things like "excellent faculty". Don't explicitly compliment them, it just looks forced.

Agreed! But it's strange that they (Purdue University) bring it up as a "successful statement". As you say it sounds forced ("krystat"). And the person reading it might also think "doesn't the applicant have anything else to write about himself/herself and feels he/she has to put this gibberish in order to get slightly more volume" as they do when a person sends in letters of recommendations written by non-faculty people (but by relatives, friends etc etc).

But, ahhh. It's not my style and I wouldn't write such things even if it increased my chances.

[size=59]Correcting a spello' it should say "butt-kissing" and not "but-kissing"[/size]

Edit: 100 posts!  :)

One of the purposes of the personal letter is to convince the person reading it that you really have looked at his university and you really are interested in studying there.  Direct compliments are standard and reasonable.

is it possible to send GRE scores to more than 4 schools and if it is then about how much does it cost for each additional school?

[quote]Additional score reports (ASRs) can be ordered by mail or by phone.
    * Download and complete the Additional Score Report Form (PDF) or use the form found in the center of the Bulletin.
    * Mail it with the fee to the address on the form.
    * The fee is U.S. $\$$15 per score recipient.
    * Requests are processed within 10 working days after receipt.[/quote]
Source: [url=http://www.ets.org/]http://www.ets.org/[/url]


arithmetic sequence
algebra
system of equations

1) A series has sum to n terms given by: Sn = 2n + 3n^2

[color=darkred]a) Find the first 3 terms of the series.
b) By evaluating Sn-S(n-1) find a formula for the nth term of the series.
c) Hence show it is an arithmetic progression. [/color]

2) In an arithmetic progression, the sum of the first 10 terms is 170 and the 8th term is 32. Find the first term (a)  and the common difference (d).

Could you explain it please? In a few words..

OK.
Because $S_n=a_1+a_2+\cdots+a_n$ and $S_{n+1}=a_1+a_2+\cdots+a_n+a_{n+1}=S_n+a_{n+1}$, so we have: $a_{n+1}=S_{n+1}-S_n$. As long as we know $S_n$ in terms of $n$, we can find a formula for $a_n$.
(This also implies the mistake I made in my solution.)

The definition of arithmetic sequence is $a_{n+1}-a_n=\text{Constant}$, so...

:huh: Let's come back to that one..next century.

2) In an arithmetic progression, the sum of the first 10 terms is 170 and the 8th term is 32. Find the first term (a) and the common difference (d).

(I have a feeling this is a bit easier..)

We can use the arithmetic sequence sum formula: $\frac{n}{2}(t_1+(t_1+(n-1)d))$
Where $d$ is the common difference and $t_1$ is the first term
So writing the sum of the first 10 terms is such that:

$\frac{n}{2}(t_1+(t_1+(9)d)=$
$(2t_1+9d)(5)=$
$10t_1+45d=170$

The nth term in an arithmetic sequence is given by $t_n=t_1+(n-1)d$

Thus we write it as:
$t_1+7d=32$

So now we have a system of equations in $t_1$ and $d$.

$10t_1+45d=170$
$t_1+7d=32$

Solving by multiplying the second equation by 10.. and subtracting from the first we get...
$10t_1+45d=170$
$-(10t_1+70d=320)$
$= -25d=-150$
$d=6$

plug into $7(6)+t_1=32$
$t_1=-10$

$t_1=-10$
$d=6$


modular arithmetic
combinatorics unsolved
combinatorics

Determine the number of subset $ S$ of the set $ T\equal{}(1,2,....,2005)$ such that the sum of elements in $ S$ is congruent to $ 2006\pmod {2048}$.

[quote="Wickedestjr"]Well, there are 2005^2 subsets of T total, and that might help anybody else to solve the problem, but other than that, I've got nothing else.[/quote]

I think there are $ 2^{2005}\minus{}1$ subsets of T total.
But how to solve this problem?... :o

[quote="Bachukas"][quote="Wickedestjr"]Well, there are 2005^2 subsets of T total, and that might help anybody else to solve the problem, but other than that, I've got nothing else.[/quote]

I think there are $ 2^{2005} \minus{} 1$ subsets of T total.
But how to solve this problem?... :o[/quote]

It is 2^2005. I just learned all about sets and subsets and all that stuff, and I got 100% on the test.

Let $ T_1\equal{}\{1, 2, 4, 8, 16, \dots , 1024\}$, and let $ T_2$ be the remaining elements.  For every subset of $ T_2$, there is exactly one subset of $ T_1$ that can be combined with it to give you a total sum of $ 2006$ modulo $ 2048$.