hi guys.
I like to play drums but I don't know drum is better or electric guitar.
what's your opinion??

Depends on which YOU like better? Many people in bands have at least a superficial ability to play drums, electric/rhythm guitar, and bass.


algebra
polynomial
quadratics
number theory
greatest common divisor

Find all rational roots of the given polynomial $ x^2 \plus{} y^2 \plus{} z^2 \plus{} x \plus{} y \plus{} z \equal{} 1$ .

[hide="Hidden by mod"]$ (2x \plus{} 1)^2 \plus{} (2y \plus{} 1)^2 \plus{} (2z \plus{} 1)^2 \equal{} 7$

Let 2x+1=a, 2y+1=b, 2z+1=c.

$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 7$

Multiply through by the common denominator to get $ a_1^2 \plus{} b_1^2 \plus{} c_1^2 \equal{} 7d_1^2$

with $ a_1,b_1,c_1,d_1$ integers, with gcd(a_1,b_1,c_1,d_1)=1 (the only extraneous case is when they are all equal to 0, which cannot happen).

Mod 8, the quadratic residues are 0,1,4 with 4 if and only if it's equal to 2.

If none of $ a_1,b_1,c_1$ is equal to 2, then we readily see that they are all congruent to 0 mod 8, contradicting the gcd condition.

Therefore, WLOG, $ a_1 \equal{} 2$ (we'll permute the final solution if there exists one).

Mod 8, we have now $ 4 \plus{} b_1^2 \plus{} c_1^2 \equiv 7d_1^2$.
Now, if $ d_1 \equal{} 2$ then we have $ b_1^2 \plus{} c_1^2 \equal{} 24$ which has no solutions in the integers, hence $ d_1 \ne 2$ from this.
We also see that if $ b_1,c_1 \ne 2$, then the L.H.S. can only take on only the values 4,5,6 mod 8 while the R.H.S. can only take on the values 0,7 mod 8, so therefore, one of $ b_1, c_1 \equal{} 2$, WLOG, $ b_1 \equal{} 2$

$ c_1^2 \equiv 7d_1^2$ mod 8, which happens only when $ c_1^2 \equiv d_1^2 \equiv 4$, or
$ c_1^2\equiv d_1^2 \equiv 0$ so all of a_1,b_1,c_1,d_1 are even, hence their gcd is not 1, contradiction!

Hence, no solution.[/hide]

[hide="Nicer way to finish"]From rofler's solution, we get $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{}7d^{2}$, where $ (a,b,c,d)\equal{}1$.  If $ d$ is even then the right side is 0 modulo 4, and since squares are 0 or 1 modulo 4, $ a,b,c$ must all be even also, a contradiction.  Therefore $ d$ is odd.  Then the right side is 3 modulo 4, so $ a,b,c$ must all be odd.  But odd squares are 1 modulo 8, so the left side is 3 modulo 8, while the right side is 7 modulo 8, contradiction.[/hide]


inequalities
search
inequalities unsolved

Let $a_{1\rightarrow n},b_{1\rightarrow n},c_{1\rightarrow n}\ge 0$.
We have that:

   $\sum_{cyc}^{a_1,...,a_n} a_1^{b_1}a_2^{b_2}...a_n^{b_n} \ge \sum_{cyc}^{a_1,...,a_n} a_1^{c_1}a_2^{c_2}...a_n^{c_n}$

and                    $\sum_{i=1}^n b_i=\sum_{i=1}^n c_i$

Now,what can we say about the $(b_1,...,b_n)$ and $(c_1,...,c_n)$?

[hide]I think $(b_1,...,b_n)$ contains less zeros than $(c_1,...,c_n)$
but i dont know anything when the numbers of the zeros be equal.[/hide]

Your question is related to so called Murhead inequality ("Bunching"). And your conjecture seems to be wrong. For example, if b = (1,1,1) c = (2,1,0) then

        a^b <= a^c

But if  b = (3, 0, 0), c = (2, 1, 0) then we have other direction of ineq.

So, we cannot says anything about number of zeros.

About Murhead, you can search in internet many information.

Namdung

Thanks Namdung,I found the Murhead.Very interesting inequality .
I was facing lots of problems where are the parametires were subsituide inder the contestant powers and i just wondered that we have a theorem for it or not . Right now i am finding a very use full tool due  :lol: .Thanks agian :lol:

About those zeros , If you do not know the murhead then it is at least a good guess isnt it ? :)


Consider the set $ S$ of all numbers of the form $ a(n)\equal{}n^2\plus{}n\plus{}1, n \in \mathbb{N}.$ Show that the product $ a(n)a(n\plus{}1)$ is in $ S$ for all $ n \in \mathbb{N}$ and give an example of two elements $ s,t$ of $ S$ such that $ s,t \notin S$.

[quote="moldovan"]Consider the set $ S$ of all numbers of the form $ a(n) \equal{} n^2 \plus{} n \plus{} 1, n \in \mathbb{N}.$ Show that the product $ a(n)a(n \plus{} 1)$ is in $ S$ for all $ n \in \mathbb{N}$ and give an example of two elements $ s,t$ of $ S$ such that $ s,t \notin S$.[/quote]

1) $ a(n)a(n\plus{}1)\equal{}a((n\plus{}1)^2)\in S$

2) $ 5^2\plus{}5\plus{}1<(1^2\plus{}1\plus{}1)(3^2\plus{}3\plus{}1)<6^2\plus{}6\plus{}1$ and so $ a(1)a(3)\equal{}39\notin S$


abstract algebra
function
trigonometry
absolute value
group theory
superior algebra
superior algebra unsolved

im looking at this factor group $R/Z$ for about 2 days now.
am trying to understand what it means
i know that the left coset should be equal to the right coset also and
then be equal with the facotr group so is the following right ?
$R/Z=\bigcup {\hat{x}}$.
i think we can define this class of equivalence x like this
$\hat{x}=\left\{y\in R|y^{-1}x \in Z \right\}=\left\{y\in R| \frac{y}{x} \in Z\right\}$
but i dont know exactly when $\frac{y}{x} \in R$.
im not sure if this is right please help

It is not right. $(\mathbb R,\cdot)$ is not a group, and even if you take $\mathbb R^*$, this does not have $(\mathbb Z^*,\cdot)$ as a subgroup.
The group to consider is $(\mathbb R,+)$.

You can prove there is an isomorphism $\mathbb R/\mathbb Z\simeq S^1$, where $S^1$ is the group of the elements of absolute value $1$ in $\mathbb C$.
I hope the group operation is clear in $\mathbb S^1$.

To construct the morphism, take $\varphi: \mathbb R\to S^1$ given by $\varphi(x)=e^{2\pi ix}$.
Verify that it is well defined i.e $\varphi(x)\in S^1$ for all $x\in \mathbb R$, that it is a group homomorphism, that it is surjective and that its kernel is $\mathbb Z$.

amfulger
you are right,i was looking at an example of the use of first isomorphism
theorem on groups.
it was saying that $R/Z \simeq C^{*}$
because there is an morphism $f: R\to C^{*}$
with $f(x)=cos\( 2 \pi x\) + i\cdot sin\( 2 \pi x\)$.
now for this f we have $Ker\(f\)=Z$
and also $Im\(f\)=C^{*}$ so that we can apply first
theorem of isomorphism.
probably your function $\varphi(x)$ is the same as the one in
the example i am reading.
but please help me understand what is $R/Z$ this is what interests me now

If $\mathbb C^*=\mathbb C\setminus\{0\}$, then they are wrong because their $f$ is not surjective.
Indeed $e^{ix}=\cos x+i\sin x$.

You can think of $\mathbb R/\mathbb Z$ as the interval $[0,1)$ made into a group with the operation given by addition "modulo $1$".
This group operation $\oplus$ is explicitly given by:
\[ a\oplus b=\left\{\begin{array}{cc}a+b,&\mbox{if }a+b<1\\a+b-1,&\mbox{otherwise}\end{array}\right. \]

amfulger i understand that this is a rather trivial problem for you
but please help me.
amfulger could you please follow a more non-intuitive path to
proove that $R/Z$ is as you say it is ?
because i was thinking about that too..but then again...i
want to solve allot of exercices and tough i read the theory
i thought i should have understood what $R/Z$ means
exactly,but i don't.
so please follow a theoretically-based explanation as to
what exactly $R/Z$ is please.

[quote="spx2"]follow a theoretically-based explanation as to
what exactly $R/Z$ is please.[/quote]

From theoretical point of view,  $\mathbb R/\mathbb Z$ is the factor group of $\mathbb R$ by $\mathbb Z$. Nothing more than that, no better way to look at it.
$\mathbb R/\mathbb Z=\{\hat x|x\in\mathbb R,\ \hat x=\{y\in\mathbb R|y-x\in\mathbb Z\}\ \}$.
I was giving you nice ways to look at it via isomorphisms with $S^1$ and $([0,1),\oplus)$.


Yesterday; was the last time I was online, I was on mobile a while ago but well, I thought it was a glitch.
I was a Riemann Hypothesis (2 and a half stars) then it suddenly dropped to 2 stars, I had about 400 and something posts, and you can even check that I have 35-36 pages of posts if you look at my profile. Is this a bug? Or was it done for a purpose? Thanks.

This is due to many Games Forum games being moved to the Fun Factory, like Scrabble, The Word Game, etc.


Find all natural numbers $n$ for which every natural number whose decimal representation has $n-1$ digits $1$ and one digit $7$ is prime.

It is clear that $n=1$ satisfies the requirement. We will show that there are no other such $n$. Assume that $n>1.$

Number whose decimal representation has n-1 digits 1 and one digit 7 is of the form $\frac{10^n-1}{9} + 6\cdot 10^k$ where $0\leq k\leq n-1.$ We need to find such $n$ that $\frac{10^n-1}{9} + 6\cdot 10^k$ is prime for every $k=0,1,\dots,n-1.$

Note that unless $\frac{10^n-1}{9}$ is divisible by 7, there exists $k_0$ such that $\frac{10^n-1}{9} + 6\cdot 10^{k_0}$ is divisible by 7 (thus, is not prime).
On the other hand, if $\frac{10^n-1}{9}$ is divisible by 7 then $n$ is divisible by 6, implying that $\frac{10^n-1}{9} + 6\cdot 10^k$ is divisible by $3$ (for any $k$). Therefore, there are no other $n$ for which $\frac{10^n-1}{9} + 6\cdot 10^k$ is prime for every $k=0,1,\dots,n-1.$

[quote="maxal"]It is clear that $ n=1$ satisfies the requirement. We will show that there are no other such $ n$. Assume that $ n>1.$

Number whose decimal representation has n-1 digits 1 and one digit 7 is of the form $ \frac{10^{n}-1}{9}+6\cdot 10^{k}$ where $ 0\leq k\leq n-1.$ We need to find such $ n$ that $ \frac{10^{n}-1}{9}+6\cdot 10^{k}$ is prime for every $ k=0,1,\dots,n-1.$

Note that unless $ \frac{10^{n}-1}{9}$ is divisible by 7, there exists $ k_{0}$ such that $ \frac{10^{n}-1}{9}+6\cdot 10^{k_{0}}$ is divisible by 7 (thus, is not prime).
On the other hand, if $ \frac{10^{n}-1}{9}$ is divisible by 7 then $ n$ is divisible by 6, implying that $ \frac{10^{n}-1}{9}+6\cdot 10^{k}$ is divisible by $ 3$ (for any $ k$). Therefore, there are no other $ n$ for which $ \frac{10^{n}-1}{9}+6\cdot 10^{k}$ is prime for every $ k=0,1,\dots,n-1.$[/quote]
Something is wrong with this, as $ 17$ and $ 71$ are both primes. And obviously the problem is that $ 3 \nmid \frac{10^{n}-1}{9}$, unless $ 3 \mid n$.

My original proof does not work if $ k_0\geq n$, in which case $ \frac {10^{n} \minus{} 1}{9} \plus{} 6\cdot 10^{k_{0}}$ contains $ k_0 \plus{} 1 > n$ digits.
By the definition of $ k_0$, we have $ k_0 < 7$, implying that cases $ n < 7$ have to be verified separately:
for $ n \equal{} 2$, $ k_0 \equal{} 4$ and here we have primes $ 17$ and $ 71$ as pointed out by [b]s.tringali[/b];
for $ n \equal{} 3$, $ k_0 \equal{} 3$ but here $ 117$ is not prime;
for $ n \equal{} 4$, $ k_0 \equal{} 5$ but here $ 1711$ is not prime;
for $ n \equal{} 5$, $ k_0 \equal{} 2$ and the original proof works here;
for $ n\equal{}6$, $ \frac{10^n \minus{} 1}{9}$ is divisible by 7 and the second part of my proof applies.

So, the corrected answer to the problem is $ n \equal{} 1$ and $ n \equal{} 2$.


topology
trigonometry
advanced fields
advanced fields unsolved

I don't understand this problem I was given. Let $A\subset \mathbb{R}^{2}$ be a topologist's sine curve, $A=\left \{\left (x,\text{sin }\frac{1}{x}\right )|0<x\leq 1\right \}$ and $a$ a point in the closure $\bar{A}$ of $A$. Let $b=(1,\text{sin }1)\in A$. Prove that $a$ and $b$ can't be connected by a path in $\bar{A}$.

In my opinion, every $a\in A,a\not \in \bar{A}\setminus A$ can be connected to $b$ since $A$ is rectifiable. What I am doing wrong?

[quote]$a$ a point in the closure $\bar{A}$ of $A$.[/quote]
This was meant to be "$a\in \bar{A}\setminus A$".


the source http://www.mathhelpforum.com/math-help/showthread.php?t=3399
thanks a lot to	ThePerfectHacker
Consider the following sets.
$S_{1}=\{a_{1}n^{2}+b_{1}n+c_{2}|n\in \mathbb{Z}\}$
$S_{2}=\{a_{2}n^{2}+b_{2}n+c_{2}|n\in \mathbb{Z}\}$
.....
$S_{k}=\{a_{k}n^{2}+b_{k}n+c_{k}|n\in \mathbb{Z}\}$
Where, $a_{1},a_{2},...,a_{k}\not = 0$

Prove that,
$\bigcup_{j=1}^{k}S_{j}\not = \mathbb{Z}$
For any $k\geq 1$
-----
What I am trying to show, informally, that a set of integers cannot be placed in classes of quadradic progressions.

For example, with linear progressions it is always possible. Consider
$3k,3k+1,3k+2$-they contain all the numbers.
good fourm BTW

Here's a hint:
[hide]Look at the common differences.[/hide]
And here's the non-rigorous application of the hint:
[hide]"Eventually," each quadratic will have gaps of at least $2k+1$ between consecutive terms.  Since we have only $2k$ "branches" at most to work with, beyond that "eventually" point we must be missing 1 out of every $2k+1$ numbers.[/hide]

The details aren't hard to work out, but my keyboard is broken so I can't type parentheses and so I'm not going to do it for you :-]

Ah, while I'm here: on the linear equations, I noticed that all three of your equations have coefficient of $x$ equal to 3, and that $\frac13+\frac13+\frac13 = 1$.  Also, the obvious solutions for any $k$ are to do exactly the same thing, $kx, kx+1, \ldots$.  So, here's my question: are there other solutions?  You certainly are going to need that $\sum \frac 1{a_{k}}\geq 1$ [Prove this!].  But suppose I give you some set like $a_{1}= 2, a_{2}= 3, a_{3}= 6$.  Can you find a solution in this case?  In every such case?


hey, me and my friends are planning to make the biggest rube goldberg machine sometime next year. Its still in the early planning stages. if you have any idea on what to make, please share.

also, would it be okay to ask for donations on this forum? a 150 step rube goldberg machine might cost alot of money...

I made a 34 step machine last year for the C Division "Mission Possible" event in Science Olympiad. If you want, I can e-mail you the ETL (energy transfer list). I can tell you that mechanical energy transfers are a) the least expensive, b) the easiest to troubleshoot, and c) usually quite reliable. However, chemical transfers are the coolest (Don't mind me, I'm just extremely biased towards anything chemistry-based.) and electrical transfers are just plain annoying.

150 steps is a lot. Do you know what the current world record is?

What's a Rube Goldberg machine? I want to build one of those things if it's not too expensive.

A Rube GOldberg machine is basically a machine that uses a lot of levers and pulleys and stuff to make a simple action, such as pressing a button, or pouring out milk.

Apparently the world record is at 113 steps right now. If I give a donation, will you give me credit if you get a new record? :D

isn't a rube goldberg machine something that transfers energy through lots of different types of processes?

well..  i dunno.. depends how much you donate i guess.. im thinking about like making t shirts and telling big companies to donate in exchange for their ad on our backs.. but if u do donate.. some kind of acreditation is surely required.. thanks... ;)

:)

at our school, there is every a project called "The Incredible Machine" during the project week where we build such rube goldberg machine. it's always quite funny there, i did that project 3 years ago.

Peter

You seem to be trying for a lot of world records.


Soooo, how did everyone do on the AIME II?

7, see poll (6 stupid mistakes)

Will a 7 make it with the floor or a 213 with the index?

Will a AIME score of 8 would be enough to go to USAMO if still in middle scholl.

I don't believe there's a discrepency between middle school and high school cutoffs, except that your index does not matter for grades 7-10.

[quote="fordbhaskar"]Of course! How do you know?[/quote]

Hmm, I think there's a stereotype that people from Michigan tend to over-estimate the USAMO index and floor.

[quote="dbell0971"]Do you think an AIME II score of 7 will make it?[/quote]

Based on the difficulty level of the AIME II, I would say probably not, but a floor of 7 is still within reason. I would say the floor will be somewhere 7-9.

The floor should be the same for both AIMEs.  I predict a floor of 8.  Certainly not 9, since we have the USAMO Expansion.

The AIME I looks too easy for a 7. In my opinion, they are about the same difficulty.

I thought AIME II was about the same as AIME I.  My completely innacurate guess would be 8 floor and an index about ten less than last year (220-226ish)

Stereotypes are often false. That said, I predict 230.

[quote]Hmm, I think there's a stereotype that people from Michigan tend to over-estimate the USAMO index and floor. [/quote]

Not really fair; no one knows the score yet, so we can't possibly overestimate. And it's just this year - many of us were convinced that the floor would not go over 8 last year.

[quote="nat mc"][quote="solafidefarms"]Personally I think the floor shall be 7 for the I and 9 for the II, just judging from how I think I'd do on the II. The II looks like it was soooo easy.[/quote]
No way, the AIME I was childs play. They were so simple, and it was almost impossible to make a mistake. On the II, there were many computation and reading thingies that could have screwed you up.[/quote]

i wouldnt call it a "childs play" ;)

every one is capable of making mistakes, on any test
and i believe the tests were both around the same level of difficulty

[quote="Xantos C. Guin"][quote="nat mc"][quote="solafidefarms"]Personally I think the floor shall be 7 for the I and 9 for the II, just judging from how I think I'd do on the II. The II looks like it was soooo easy.[/quote]
No way, the AIME I was childs play. They were so simple, and it was almost impossible to make a mistake. On the II, there were many computation and reading thingies that could have screwed you up.[/quote]

I messed up on problems 6, 9, and 10.
It is very possible to make a mistake.
Even the people that write the AIME probably made 1 or 2 mistakes max.[/quote]

Yep, i made stupid mistakes on....8, 9, 10, 11, 12, 15 :(  :(
And the point of the aime was to find the smart way, not to do lots of computations and the Aime I is more difficult in that respect.

[quote="Phelpedo"]Well if you try to find the "smart" way, the AIME II is the same. I guess the test writers just didn't expect people would actually attempt to add up a 3-term fibonacci-type sequence all the way to its 28th sum.[/quote]

That plus the elegant way was harder to find for the I.

But all this doesn't really matter since the list is already made, so whatev.......


I've just found a very nice PDF Printer. It's freeware, so no watermarks or annoying text appears when priting out PDFs. It's easy to install and use.

Package is for Windows: http://sourceforge.net/project/showfiles.php?group_id=57796&package_id=53473&release_id=228013

Project home: http://sourceforge.net/projects/pdfcreator/

A lot of commercial pdf printers are based on Ghostscript, the backend of that printer.   :)

can you put the link in the TeX guide too?


Ich hatte während eines 2-wöchigen Auslandaufenthaltes Zeit, die vier Aufgaben zu lösen.

I. Wie ausführlich muss man seine Lösungen darlegen, wie ausführlich darf man sie darlegen (es gibt Abzug, wenn die Lösungen zu detailliert dargelegt werden)?
D.h., müsste man noch begründen, warum sich die Höhen eines spitzwinkligen Dreiecks innerhalb des Dreieckes schneiden oder darf man das als evident vorraussetzen?

II. Wie läuft das "Kolloquium" ab? Die Beschreibung, die auf der offiziellen Seite gegeben ist, erscheint mir da etwas zu vaage... Einige hier waren ja auch schon bei einem solchen Kolloquium dabei, die könnten bitte auch Beispielfragen schreiben, wenn möglich!

Danke für die Antworten

Blender

I. Die Höhen denk ich können vorausgesetzt werden. Ansonsten lieber zu ausführlich als zu knapp solange man es nicht übertreibt [sorry wenn dies so gut wie die Nullaussage ist].

II. Das hat schon seinen Grund dass es nur vage Angaben gibt: das ist individuell stark verschieden.
Ablauf:
- zuerst ein Gespräch über sonstwas (= Familie, Hobbies, Vereine, ...), so etwa 10 Minuten (Länge kann stark variieren)
- dann der mathematische Teil (Rest der 60 Minuten): die Themenschwerpunkte kann man selber vorgeben (allerdings wird verschieden stark davon abgewichen), die Professoren werden einem dann nach diesen zugewiesen; als Tip könnte man vielleicht geben: nicht vom Thema abbringen lassen dass man gerade will.

Hmm... Beispielfragen...
Was mir noch von meinem Gespräch (Februar 2005) einfällt an Sachen:
Die Themen liefen auf (quadratische) Zahlkörper, Funktionentee und (doofen) Fargen zum Sekantenverfahren für diophantische Gleichungen hinaus; der 3. von mir genannte Schwerpunkt Algebra wurde komplett weggelassen (ok es gibt nur wenige die so dumm sind gleich drei anzugeben).
Fragen zu Zahlentee (alles nur inhaltlich wiedergegeben):
- Lässt sich dies auch auf rationale Punkte auf der Sphäre im dreidimensionalen Raum verallgemeinern¿ (es ging um rationale Punkte auf dem Einheitskreis per Sekantenverfahren)
- warum betrachtet man Die Ringe $\mathbb{Z}[\tau]$ definiert durch $\tau = \sqrt{n}$ falls $n \equiv 2,3 \mod 4$ und $\tau = \frac{1+\sqrt{n}}{2}$¿ (es ging um quadratische Zahlringe und deren Ordnungen)
- noch so einiges das ich nichtmehr weiß...
Funktee:
- weshalb ist die Cauchy'sche Integralformel so eine starke Aussage¿ (Christian H. kann zu dieser und allen anderen Fragen wohl auch was beisteuern)
- beweise den Satz von Liouville für holomorphe Funktionen

okay, danke, das reicht schon
jetzt kann ich mir relativ viel darunter vorstellen...

Ich habe nur das Pech, dass ich zum Beispiel die Cauchy'sche Integralformel gar nicht kenne... hat man dann gleich verloren oder sagen die einem was das ist?^^

Es ist auch so, dass es bei mir oft an der Stelle schwierig wird, wo in den Lösungen steht: "it is a well-known fact that..."
Woher erfährt man denn diese "well-known facts"? einfach durch vielfaches Lesen anderer Lösungen und wikipedia-durchforsten, oder gibt es dafür spezielle seiten, wie z.b. die theorems&formulas-foren?

Nach der Integralformel fragen die auch nur wenn du sowas (=Funktee) als Thema angibst. Mit Thema Kombinatorik kommt dann eben z.B. wohl eine Frage zu Graphen oder weiß der Geier was. Wenn ihnen sonst nichts mehr einfällt (gar nicht so selten) fragen sie scheinbar Schulspam.
Übrigens hab ich ziemlich viel Schmarrn gelabert (u.a. hab ich 3 Ansätze für Liouville gebraucht).

Ich glaub man nennt etwas "well known" wenn man keine Lust hat den Beweis hinzuschreiben und hofft das er irgendwo schon steht  :D
Nachschauen sowie ungezieltes Herumlesen in Wikipedia wie auch MathWorld kann ich nichtempfehlen, beide kann man zum Anlesen vin "Grundtheoremen" vergessen.
Ich empfehle Bücher zum Thema ProblemSolving oder einfach MathLinks (letzteres kist mit Vorsicht zu genießen weil es eben nicht schön sortiert etc. ist, dafür ist es "allgemeiner").
Und es gibt noch:
http://www.mathlinks.ro/Forum/viewtopic.php?t=76610
*Werbung mach ;)*

Okay, danke, dann werde ich wohl diesen Beitrag und diverse Bücher darüber lesen...

Woher bekommt man die? Auch im normalen Buchhandel, oder nur bei Amazon u.s.w.?

Edit:
[quote]- Two squares: $(a^{2}+b^{2})(c^{2}+b^{2}) = (ac-bd)^{2}+(ad+bc)^{2}$[/quote]

auf der linken seite der gleichung steht 2 mal b, da müsste aber einmal b stehen, da hast du dich wohl vertippt.

Edit2:
Ah, auf aops.com gibt es ja eine Liste von Büchern. Gibt es da auch gute deutsche Bücher oder sind die nicht zu empfehlen?

Was genau ist eigentlich dein Ziel¿
Denn: für die 3. Runde BWM ist meiner Meinung nach (Spezial)Wissen in einigen Bereichen (viel¿) besser als gute ProblemSolvingSkills.

Aber von Büchern für ProblemSolving habe ich nur den Engel, und ich weiß nicht ob der so toll ist (Zahlentee ist dort z.B. ein Witz).

An "Spezialbüchern" kann ich empfehlen:
Jürgen Wolfart: Einführung in die Zahlentheorie und Algebra
Königsberger: Analysis 1 (und 2)
Bosch: Algebra (benötigt minimale Kenntnisse lineare Algebra)
Freitag, Busam: Funktionentheorie 1 (benötigt Kenntnisse Analysis, aber kann wenn man zwei drei Sätze einfach hinnimmt auch mit Schulwissen gelesen werden)
Harte aber schöne Kost: Hatcher: Algebraic Topology (online: http://www.math.cornell.edu/~hatcher/AT/ATpage.html )
[hab versucht sie nach Schwere zu ordnen]

Diese wie auch jedes andere Buch würde ich mir aber anschauen bevor ich es mir kaufe (bei Hatcher erübrigt sich das Kaufen ;) ); aber dabei nicht den Fehler machen in der Mitte aufzuschlagen und sich zu wundern dass man nichts versteht...
Vorteil slcher Bücher ist: man hat etwas auf Dauer davon und  kann man es oft auch in Aufgaben gebrauchen (z.B die Dinge aus Wolfarts Buch).

PS: Danke@Hinweis, werds mal korrigieren.

Danke für die Hilfe!
Mein Ziel... naja, das letzte bezog sich eigentlich darauf, dass ich allgemein besser werden wollte, das war dann zwar nicht mehr über die 3. RUnde vom BWM, aber da der Thread schon mal offen war...

Kann man sowas wie $3/5x+7 = 78 \implies x = (Ergebnis)$ direkt hinschreiben oder muss man da auch noch Zwischenschritte machen?

Ich denke ein Zwischenschritt schadet auf jeden Fall nicht.

Hallo $\pi J$  :)

@auchnefrage: netter nickname :D

(sorry fürs spammen)

Wie schreibt ihr eure Lösungen denn auf? Meine Trial-VErsion von Mathtype ist leider abgelaufen, deshalb weiß ich jetzt nicht, was man da benutzen könnte... Gibt es da was kostenloses? Kann man das auch als Wordplugin benutzen? Oder gibt es da auch einen solchen kostenlosen Formeleditor, mit dem man auch Text schreiben kann???

Ich benutze dazu $\LaTeX$.  Obwohl ich nicht allzu geübt im Umgang mit $\LaTeX$ bin, klappt das ganz gut.

Danke! An Latex habe ich auch schon gedacht, kann man da denn auch Zeichnungen einbinden, z.B. für Aufgabe 1 oder Aufg. 3?

Ich glaube du musst die Zeichnungen mit einem anderen Programm erstellen und später in das $\text{\LaTeX}$-Programm einbinden. Laut der $\sqrt{ \textnormal{WURZEL}}$, geht das mit dem Paket graphicx und dem Befehl \includegraphics. Zum Erstellen dieser Datei empfiehlt die $\sqrt{ \textnormal{WURZEL}}$  $\textnormal{METAPOST}$.
Ich persönlich habe dies noch nie ausprobiert und nutze die einfachste Variante: Ich zeichne mit Skizzen per Hand.  :)

Wie soll man eigentlich als Normalsterblicher bei der 3.Runde eine Chance haben?  :(
Gut ich nehme dieses Jahr zwar das erste Mal am BWM teil aber trotzdem läuft es gut.
Aber von den meisten Sachen die ZetaX genannt hat habe ich noch nie was gehört.(ich hätte Zahlentee auch als Schwerpunkt genommen)

Bsp:
Ich hab schon mal was über Ringe gelesen aber ich wüsste das dann nicht aus dem Stegreif auch hab ich keine Ahnung  was diese Cauchy´sche Integralformel ist.

Was ich schon schaffen würde wär sowas wie der kleine Fermat aber das andere!!??

Darf man denn z.b. die Vandermondesche Identität oder die Formel für die Partailsummen einer geometrischen Reihe voraussetzen oder soll man das extra beweisen???


Is it possible to pay the class fee by using Paypal?

No.  Since we are able to take credit cards directly, we don't need to use Paypal.


logarithms
calculus
integration
function
calculus computations

Try the condensation method

$S_n=\sum_{k = 2}^\infty \frac{1}{\sqrt{k} \cdot \left(\ln k\right)^n}=$
$\sum_{k = 1}^\infty 2^k \frac{1}{\sqrt{2^k} \cdot \left(\ln 2^k\right)^n} =$
$\sum_{k = 1}^\infty \frac{\sqrt{2^k} }{\left(\frac{k}{2}\right)^n \cdot (2 \ln 2)n}=$
$\frac{1}{(2 \ln 2)^n} \sum_{k = 1}^\infty \frac{2^{\frac{k}{2}} }{\left(\frac{k}{2}\right)^n}$
But there exists $N$ we will have $\left(\frac{k}{2}\right)^n<2^{\frac{k}{4}},\forall k>N$.
So, $(2 \ln 2)^nS_n>\sum_{k=N}^{\infty} \sqrt{4}^k \to \infty$. So $S_n \to \infty$.

Correct me if I'm wrong. I never used this method before.

The answer is immediate to someone who knows what to look for. But just for grins, I'll try out the condensation test on it:

If $a_k>0,\,a_k\to0$ as $k\to\infty$ and $a_k$ is monotone decreasing for large enough $k$

$\sum_ka_k$ converges if and only if $\sum_m2^ma_{2^m}$ converges.

So we compare our series to

$\sum_m\frac{2^m}{\sqrt{2^m}(\ln(2^m))^n}=C\sum_m\frac{2^{m/2}}{m^n}$

Since the exponential grows faster than any power of $m,$ the terms tend to infinity and the series diverges.

[I was composing this while xirti posted underneath me; I'll post it anyway.]

I never heard of this condensation method :o

What do you mean by "The answer is immediate to someone who knows what to look for"?

If $(a_n)\subset (0,\infty)$ is monotone we have that $\sum 2^ia_i^2 \le \sum a_i \le \sum 2^ia_{i+1}\le \sum 2^{i+1}a_{i+1}$. From this we have that $\sum a_n$ is convergent iff $\sum 2^ia_i^2$ is convergent. It is already discussed on the forum.

Arne, try to prove that these two series $\sum_{n\geq 1} \frac{1}{n}$ and $\sum_{n\geq 2} \frac{1}{n\ln n}$ are divergent by applying this useful method. ;)

[quote="Arne"]What do you mean by "The answer is immediate to someone who knows what to look for"?[/quote]
For large $k$ and positive $\epsilon$, $(\ln k)^n$ is much smaller than $n^{\epsilon}$. Therefore, the series diverges by comparison to $\sum_k\frac1{k^{\frac12+\epsilon}}$ if we choose $\epsilon\le\frac12$.

[quote="Arne"]Study the convergence of \[ \sum_{k = 2}^\infty \frac{1}{\sqrt{k} \cdot \left(\ln k\right)^n}  \] where $n$ is a positive integer.[/quote]

I didn't even know that trick had a name...

But I think it can be done much shorter, at normal class level. For every n there is a K such that for all $k>K$, we have $(\ln k)^n < \sqrt{k}$. Which means that for all k>K, the terms of $S_n$ are larger than those of the harmonic series, thus $S_n$ diverges.

Nice idea, Peter, but that's just what jmerry wrote.

By the Cauchy Integral Critera this series is convergent iff $\int_2^{\infty}\frac{dx}{\sqrt{x}(\ln x)^n} = \int_{\ln 2}^{\infty}\frac{e^{y/2}}{y^n}dy$ is convergent, and the second integral is quite obviously divergent since the integrand tends to infinity.

While the method feels a bit "childish" since the Integral Criteria is almost never needed, it does transform almost every series involing complicated things with $\ln$ into integrals that are easy to study.

it's general , we call it , Bertrand critera ,  1/(x^a*ln(x)^b ) is integrable over [2,+00[ iff (a>1 and b a real ) or ( a=1 and b>1) it's very easy to show using usuel comparaisons between polynoms and logarithm functions .

[quote="xirti"][quote="Kalle"]By the Cauchy Integral Critera this series is convergent iff $\int_2^{\infty}\frac{dx}{\sqrt{x}(\ln x)^n}$ is convergent.[/quote]

Why? Could you give a proof for "Cauchy Integral Criteria "[/quote]

Let $f(x) = \frac{1}{\sqrt{x}(\ln x)^n}$. $f$ is a decreasing and positive function for $x > 1$, so by integral estimation we have that

$\int_2^{n+1}f(x)dx \le \sum_{k = 2}^nf(k) \le f(2) + \int_2^nf(x)dx, \quad \forall n \ge 2$

Since the sum is majored by the integral and vice versa we have that the series is convergent if and only if the integral is convergent.

This is normally called the integral test in english literature, I think.

Indeed Kalle, that's how people most often call it :)


$ x,y,z$ are positive real numbers such that : $ xy+yz+zx=3$ and $ x\leq y\leq z^$  prove that :

$ \frac{x}{x^{2}+y^{2}+1}+\frac{y}{x^{2}+z^{2}+1}+\frac{z}{z^{2}+y^{2}+1}\leq \frac{1}{xyz}$

Try $ x=z=\sqrt{3}-\delta$, $ y=2\delta$.  As $ \delta$ gets small, the lhs goes to $ \frac{\sqrt{3}}{2}$ and the right to $ \frac{1}{3}$.  So it's false.  You could even just let $ y$ get very large and the other two very small to get the inequality to be false.  There also appear to be examples when it's true though: try $ y, z$ very small, $ x$ very large.

The sum is neither cyclic nor symmetric.  Are you sure it shouldn't be, maybe $ \sum\frac{x}{y^{2}+z^{2}+1}$? (though I think that doesn't hold either).  Or maybe a cyclic sum?

Edit: Okay, I see.  Nice inequality!

I have to think about it...it would suffice to show it's $ \le \frac{x+y+z}{x^{2}+y^{2}+z^{2}}$, but I'm not sure that approach works...

now the inequality is true so try to prove it !


function
real analysis
real analysis solved

Let f:R->R the function defined as
f(x)=sum from 1 to infinity from 1/n^x.
a.Show that limf(x)=1 for x->oo
b.Find the image of the function.

http://mathworld.wolfram.com/RiemannZetaFunction.html

For b) ]1;+oo[


conics
ellipse
geometry proposed
geometry

$AD$ is the altitude on side $BC$ of triangle $ABC$.
If $AB+AC = BC+AD$, and $BC=k$ (constant), find the range of $AD$.

When AD = 0, then the vertex A of ABC, may be any point on segment line BC, between B, C. So, we have AB + AC = BC + 0 = k

     When AD = n, then we can construct the vertex A, as follows:

     We draw an ellipse (e) with focuses the points B, C, main axis 2a = k + n  and secondary axis 2b = [n( n + 2k )]½

     We draw two lines (p), (q), parallel to BC, from both sides of it, in distance d = n, which intersect (e) at four points with the property as we want. ( AB + AC = BC + AD = k + n ).

     So, the range of altitude AD of ABC, is all the rates from 0 to infinity.

     Kostas Vittas.

It is a nice method, but you made an arithmetic mistake and the conclusion is absurd. If $AD \rightarrow \infty,$ then k = BC is negligible and $AB+AC \sim 2AD,$ so that the desired equality cannot hold. The ellipse minor half-axis is $b = \frac{\sqrt{n(n+2k)}}{2},$ which is not always greater than n. n = AD is maximum when the lines $p, q \parallel BC$ at the distance n = AD from the line BC touch the ellipse, hence $n \le b = \frac{\sqrt{n(n+2k)}}{2},$ $AD = n \le \frac{2}{3}k = \frac{2}{3}BC.$

Dear all.

     I apologize for my blindness and yetti, thank you very much for your advices.

     I thought that when the vertex A of ABC go to infinity, then the relation AB + AC = 2AD, is OK ( because BC is negligible ), but of course it is impossible for the altitude AD, to be taken twice. Also there is no meaning in the relation, if we consider BC = ∞.

     Sorry again.
     Kostas Vittas.


I want to have a more precise view of what people think the index will be in poll form- the other thread has way too wide ranges for the choices

I think the index will be about 203 (voted for 203-204.5)

after the aime ii it will be significantly lower because im taking it. :P

[quote="Klebian"]I think the index will be about 203 (voted for 203-204.5)[/quote]
I certainly hope so. :(

I've already accepted the fact that I haven't qualified for USAMO, and I'm thinking about next year.  Shooting for 240+ index for next year :)

[quote="Yongyi781"]Lol. Look at how much better life is for one who is not at the borderline of 200-210 :D[/quote]

above or below?

I assume above, lol.

Life must be the best at around 280-ish.

Interviewers and the like at perfect 300, and slightly below that (280-295.5) they would be disappointed of "barely missing perfect".

Shooting for 12-14 next year.


Presumably everyone's heard of $E=mc^2$. In what units is $c$ expressed? (meters per second, kilometers per second, etc.)

It depends on the units of $E$ (or rather, it determines it).  $c$ in $\text{m s}^{-1}$ and $m$ in $\text{kg}$ gives $E$ in $\frac{\text{kg}\cdot\text{m}^2}{\text{s}^2}=\text{N}\cdot\text{m}=\text{J}$.  Different units for $c$ give different units for $E$.


Let $n\geq 3$ and $A_1,A_2,\ldots,A_n$ be points on a circle. Find the largest number of acute triangles that can be considered with vertices in these points.

[i]G. Eckstein[/i]

Solution. (based on the author's idea)

We can assume the points $A_1,A_2,\ldots,A_n$ are ordered counterclockwise on the circle, indexed in $\mathbb{Z}_n$ (thus $A_{mn+k}=A_k$). Denote by $A_iA_j$ the arc of the circle starting from $A_i$ and ending in $A_j$ (in counterclockwise direction). We call an arc {\it non-acute} if $m(A_iA_j)\geq 180^\circ$. Let $x_s$ be the number of non-acute arcs $A_iA_{i+s}$ (having $s-1$ points in the interior). Clearly, $m(A_iA_j)+m(A_jA_i)=360^\circ$, hence at least one of the arcs $A_iA_j$ and $A_jA_i$ is non-acute, and thus we can infer that
\[(*)\qquad\qquad x_s+x_{n-s}\geq n\]
for every $s=1,2,\ldots,n-1$. Equality holds if and only if there are no diametrically opposite points $A_i,A_{i+s}$.

We will count the number of non-acute triangles $A_iA_jA_k$. Clearly, such a triangle has exactly one angle corresponding to a non-acute arc. For each non-acute arc $A_iA_j$ having $s-1$ points in the interior there are $n-s-1$ non-acute triangles $A_iA_jA_k$, for which $\angle A_k $ corresponds to the arc $A_{i}A_j$, namely those with $A_k$ in the interior of the arc $A_jA_i$. It follows the number of non-acute triangles is
\[ N=x_1\cdot (n-2)+x_2\cdot (n-3)+\cdots + x_{n-2}\cdot 1+x_{n-1}\cdot 0.\]

By regrouping terms and using (*) we obtain for $n$ odd
\[N\geq 0\cdot(x_{n-1}+x_1)+1\cdot(x_{n-2}+x_2)+\cdots +\dfrac{n-3}{2}\cdot(x_{\frac{n+1}{2}}+x_{\frac{n-1}{2}})\geq\]
\[\geq n\left(1+2+\cdots +\dfrac{n-3}2\right)=\dfrac{n(n-3)(n-1)}8.\]
Thus the number of acute triangles is at most
\[\dbinom {n}{3} - \dfrac{n(n-3)(n-1)}8 = \dfrac{n(n-1)(n+1)}{24}\]
and it can be reached, for example, for a regular $n$-gon.

By regrouping terms and using (*) we obtain for $n$ even
\[N\geq 0\cdot(x_{n-1}+x_1)+1\cdot(x_{n-2}+x_2)+\cdots +\dfrac{n-4}{2}\cdot(x_{\frac{n+2}{2}}+x_{\frac{n-2}{2}})+\dfrac{n-2}{2}\cdot x_{\frac{n}{2}}\geq\]
\[\geq n\left(1+2+\cdots +\dfrac{n-4}2 +\dfrac{n-2}4 \right)=\dfrac{n(n-2)^2}8.\]
Thus the number of acute triangles is at most
\[\dbinom {n}{3} - \dfrac{n(n-2)^2}8 = \dfrac{n(n-2)(n+2)}{24}\]
and it can be reached, for example, for an almost regular $n$-gon (with no diametrically opposite vertices).

Very nice solution ! Thank you very much mavropnevma.

Another approach. The idea is first to determine the conditions the extremal configuration should satisfy and then to calculate the number of acute triangles it contains.
Assume $A_1,\ldots, A_n$ is an extremal configuration with maximal number of acute triangles. Let's denote $M=\{A_1,A_2,\ldots,A_n\}$ and for $A\in M $, $A'$ be the antipodal point of $A$. Then $A'\not\in M$, since otherwise we can slightly move $A'$ increasing the number of the acute triangles. We are going to prove that the number of points in each of the two semicircles formed by $A, A'$ are almost the same. More precisely, if $AB$ denotes the arc of the circle starting from $A$ and ending in $B$ in counterclockwise direction, we have:
\[ |\,\#\{X\mid X\in M, X\in AA'\}- \#\{X\mid X\in M, X\in A'A\}\,|\le 1 \]
for every $A\in M$.
Assume on the contrary it's not true and let $\#\{X\mid X\in M, X\in A'A\}> \#\{X\mid X\in M, X\in AA'\}+1$. Let $B_1 \in M$ be the next to $A'$ (counterclockwise) and $B\in M$ next to $A$. Suppose $m(A'B_1) < m(AB) $ Then if we take $B_1$ and place it slightly next to $A'$ in clockwise direction, we will get a new configuration with more acute triangles, contradiction. Consider now the case $m(A'B_1) \ge m(AB) $. But then $\#\{X\mid X\in M, X\in B'B\}> \#\{X\mid X\in M, X\in BB'\}+1$ will also hold, and proceeding, we eventually will get a contradiction.
It remains a routine calculation of the number of the acute triangles.


function
algebra
polynomial
algebra proposed

From http://www.mathlinks.ro/viewtopic.php?t=241570 , I have a problem:
Let a positive integer $ n$. Find all functions $ f: \mathbb R\rightarrow\mathbb R$ such that $ f(x^n\plus{}f(y))\equal{}y\plus{}(f(x))^n$
A friend of mine have already solven it but his solution is rather long (4 pages), the answer is f(x) = x or f(x)=-x. I expect a shorter solution from Mathlinkers who like functional equations.

Let $ a\equal{}f(0)$.
(1) $ P(0,y): f(f(y))\equal{}y\plus{}a^n$
We see $ f$ is both surjective and injective. Plug (1) back,
$ f(x^n\plus{}f(y))\equal{}f(f(y))\minus{}a^n\plus{}(f(x))^n$, or
(2) $ f(x^n\plus{}y)\equal{}f(y)\minus{}a^n\plus{}(f(x))^n$.
Let $ y\equal{}0$ in (2), we get
(3) $ f(x^n)\equal{}a\minus{}a^n\plus{}(f(x))^n$.
Then plug this back to (2), we have
$ f(x^n\plus{}y)\equal{}f(x^n)\plus{}f(y)\minus{}a$.
No matter $ n$ is even or odd, we can show for any $ x$ and $ y$,
(4) $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)\minus{}a$.
Hence, for any integer $ r$, we have $ f(r)\equal{}a\plus{}br$ where $ b\equal{}f(1)\minus{}a\neq0$.
Looking back (3),
$ a\plus{}br^n\equal{}a\minus{}a^n\plus{}(a\plus{}br)^n$.
When $ n>1$, we treat both sides as polynomial of $ r$ and will have $ a\equal{}0$ from coeffcient-matching.
If $ n\equal{}1$, $ y\equal{}0$ gives $ f(a)\equal{}a$, then $ y\equal{}a$ gives $ f(a)\equal{}2a$. Hence $ a\equal{}0$ for $ n\equal{}1$.
Hence, we always have $ a\equal{}f(0)\equal{}0$.

Now (1) (3) and (4) turn into
(5) $ f(f(x))\equal{}x$
(6) $ f(x^n)\equal{}(f(x))^n$
(7) $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$.
For $ n>1$, the problem can be solved as in
http://www.mathlinks.ro/viewtopic.php?t=241570 .

For $ n\equal{}1$, we still need some more work with $ f(f(x))\equal{}x$ and $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$.


analytic geometry
geometry
incenter
geometric transformation
homothety
geometry solved

Its given a triangle ABC. Incycle of that triangle(with center O) touches the sides BC, AC, AB, at points D, E, F respectively. The midpoints of AO, BO, CO are G, H, I respectively. Prove that lines DG, EH and IF have common point.

[color=blue][i]Solution in barycentric coordinates[/i][/color]

With respect to the reference triangle $\triangle ABC$ the homogeneous coordinates of $A,B,C,O,D,E,F,G,H,I$ are:

$A(1:0:0)$, $B(0:1:0)$, $C(0:0:1)$, $O(a:b:c)$

$D(0:a+b-c:a-b+c)$

$E(a+b-c:0:-a+b+c)$

$F(a-b+c:-a+b+c:0)$

$G(2a+b+c:b:c)$

$H(a:a+2b+c:c)$

$I(a:b:a+b+2c)$

The equations of the lines $GD$, $EH$, $IF$ are:

$GD:$  $(a-b-c)(b-c)x+(-2a^2+b^2+ab-3ac-c^2)y+(2a^2+3ab+b^2-ac-c^2)z=0$

$EH:$   $(a^2+ab-2b^2-3bc-3c^2)x+(c-a)(a-b+c)y+(a^2+3ab+2b^2-bc-c^2)z=0$

$IF:$    $(a^2-b^2+ac-3bc-2c^2)x+(a^2-b^2+3ac-bc+2c^2)y+(b-a)(a+b-c)z=0$

The lines $GD,EH,IF$ are concurrent (i.e. the triangles $\triangle GHI$ and $\triangle DEF$  are prospectifs) since:

$
\left| {\begin{array}{*{20}c}
   {(a - b - c)(b - c)} & { - 2a^2  + b^2  + ab - 3ac - c^2 } & {2a^2  + 3ab + b^2  - ac - c^2 }  \\
   {a^2  + ab - 2b^2  - 3bc - 3c^2 } & {(c - a)(a - b + c)} & {(a^2  + 3ab + 2b^2  - bc - c^2 )}  \\
   {(a^2  - b^2  + ac - 3bc - 2c^2 )} & {(a^2  - b^2  + 3ac - bc + 2c^2 )} & {(b - a)(a + b - c)}  \\
\end{array}} \right| = 0
$

The perspector is the point $P((b+c)(-2a^3+2ab^2+2ac^2-4abc-a^2b-a^2c+b^3+c^3-b^2c-bc^2:...:...)$

Leon

Hi. I hate coordinates
If you look at the picture ,you can understand what i am going to do.
I use Ceva's theorem in GHI
GI is parallel to AC ===>  MHN ~ HIG ===> GP/PI  =  ME/EN
AO is bisector of A and GM is parallel to AB ===> GM=AM and with this way NI = NC
G and I are midpoint of AO and OC ===> GI = b/2 and wit this way GH=c/2 ,  HI=a/2
MHN ~ HIG ===>(c/2)/(c/2 +AM) = (b/2)/MN = (a/2)/(a/2 + NC)  ==> (b/2)/MN = (c/2 +b/2+a/2)/(b+a/2+c/2)
===> we can find AM and NC
we know that AE=(b+c-a)/2 and CE=(b+a-c)/2 ===> we can find ME and NE ===> we can find GP/PI
we can do this with other side of GHI  then use Ceva's theorem :lol:
if you can't understand i can help you. :roll:  :D

[quote="Naser_S"]Hi. I hate coordinates [/quote]

Your solution is nice ! :)

With my solution I have liked only explain how the barycentric coordinates can be used in some kinds of problems.  ;)

Leon

Nares halal a qo kerac kombiker@  :lol: . That was my solution of that problem.

Cossack wrote:
Nares halal a qo kerac kombiker@ :lol:
i can't understand what is your mean?

I change the notations:

[color=blue][b]Problem.[/b] Let I be the incenter of a triangle ABC, and let D, E, F be the points where the incircle of triangle ABC touches the sides BC, CA, AB. Let U, V, W be the midpoints of the segments AI, BI, CI. Prove that the lines UD, VE, WF concur at one point.[/color]

[i]Solution.[/i] Consider the homothety with center I and factor 2. This homothety maps the points U, V, W (being the midpoints of the segments AI, BI, CI) to the points A, B, C, respectively. On the other hand, this homothety maps the points D, E, F to three points X, Y, Z on the lines ID, IE, IF, and these points satisfy $IX=2\cdot ID$, $IY=2\cdot IE$, $IZ=2\cdot IF$, so that we have IX = IY = IZ (since ID = IE = IF, as the points D, E, F lie on the incircle of triangle ABC). Hence, after the Kariya theorem ( http://www.mathlinks.ro/Forum/viewtopic.php?t=30114 post #2), the lines AX, BY, CZ concur at one point. But since the points A and X are the images of the points U and D under our homothety with center I and factor 2, the line AX is the image of the line UD under this homothety; similarly, the lines BY and CZ are the images of the lines VE and WF. Hence, the lines UD, VE, WF concur at one point, since their images AX, BY, CY under our homothety concur at one point. The problem is solved.

  Darij


geometry
perimeter
pigeonhole principle
function
area of a triangle
Heron\u0027s formula

Please post reply any formulas or problem solving tricks.

Using Heron's formula, we get $ \sqrt {9(9 \minus{} 5)(9 \minus{} 6)(9 \minus{} 7)} \equal{} 6\sqrt {6}$.

EDIT: Blah right. Thanks mz.

In case you didn't understand the formula, it is
\[ \sqrt {(s)(s \minus{} a)(s \minus{} b)(s \minus{} c)}
\]
\[ s\equal{}\frac{perimeter}{2}\]\[ a,b,c\equal{}sides\]

(Revised) Pigeonhole Principle: If there are $ n$ pigeons, and $ n\plus{}1$ holes, then some pigeon must have at least two holes...

Clearly this is a very useful problem solving technique.

[quote="saszs"]All you geniuses passed.[/quote]

[sarcasm]your cool[/sarcasm]

What? [sarcasm]??? is there a function???

this is epic spam but no there isnt i jsut use that since sarcasm doesnt travel well over cyberspace


The notation isn't entirely clear- how many entries of 5, and how many of 19?

Thanks you !!!
[b]I remember !!!I remember !!!I remember !!![/b]

You didn't answer the question.

In general, your problems are hard to understand- you haven't done a very good job of translating into English, and there are other clarity issues like this one. Repetitive titles are also bad- you have another thread with the same title on the first page here. Please try to choose descriptive titles, and edit your questions for understandability.


what do you think of the US policy in iraq?
i have a debate soon on it.

I have not much idea, except that it appears not to be the most altruistic policy in the history of mankind. Of course whatever US is doing there is mostly for the Iraqis, for whom the US hearts are daily bleeding so, and in doing it they are making some magnanimous sacrifices themselves, yesm even considering all these noble intentions, maybe, just maybe, an American or two has a quarter of an eye on some hydrocarbons, no doubt from an entirely scholarly point of view. And the ungrateful sewer rats from the third world, who are upto their collars with US aids, and who wouldn't have had a strong enough voice had it not been for the latter's benevolence, dare to attach an unholy implication to these hydrocarbons. It is perhaps a furter testimony, if any such thing is still needed, of the generosity of the greatest country on planet earth, that in spite of this ingratitude, they have [i]still[/i] not withdrawn from Iraq, which would have left the poor things to die for themselves, and would have served them right!

i'm not a fan of bush and all his policies, but i think the media too often blame it all on bush
yes, he did lead us to an unseccesful war, but didn't 60 percent of americans supported it at the beginning?
yes, they need improvement in many strategies, but it is very hard to catch people who after firing a RPG, can just lose the weapon and blend in with the rest of crowd, or jump off the building and onto the us soldiers and detonate the bomb on his back.
and even if the public doesn't like this, i think they should increase tax to fund the war to provide better equipment, and this is the first war that the us has fought that hasn't increased the taxes at all, well that's just making the nation less aware of the soldiers dying half way across the world.

It was a mistake to begin with and it is our responsibility to leave as fast as physically possible and safe (which most experts agree is about 6 months). The war was based on lies, as the multitude of evidence that we know the intelligence agencies had (and that experts, current inspectors such as Hans Blix, and previous inspectors such as Scott Ritter were telling us) that no WMD existed in Iraq was ignored by the administration, and as we know that any connection between Iraq and al-Qaeda was completely fabricated given the presence of a Pentagon intelligence report at the time cleanly stating that no connection existed and that in fact Saddam and al-Qaeda hated each other. Ever since then, we have faced the chief question of how to get out, as thousands of our soldiers, not to mention hundreds of thousands of innocent Iraqis, have died, and the presence of a US occupation in the heart of the Arab world has attracted recruits for al-Qaeda in droves. It is clear now that our presence has given rise to and is currently only aggravating an Iraqi civil war. Some argue that the country will descend into genocide and a terrorist haven if we leave. However, the fact is that as long as US troops stay, Iraqis will have a civil war anyway, and the situation would be worse than if US troops leave, due to the presence of an additional nationalistic insurgency and a magnet for al-Qaeda. The most important concept to remember is that the United States is not involved in a war, but in an occupation, and there is no such thing as "winning" an occupation.

I suspect that you have already heard most of these arguments. However, this serves as a refresher on why this war is illegitimate and why we need to withdraw (or as they phrase it today, "redeploy").

If we do pull out of Iraq, we should go after the nations that are providing money for these terrorists and attack them.  And we'd have to be ruthless.  I don't understand why this hasn't been done yet.

Perhaps because
(1) the American army is too busy being blown up in Iraq to be blown up elsewhere right now, and
(2) point (1) is true not because other countries are evilly funding our enemies, defeating the competent and righteous efforts of the American government and military, but rather because we started an impossible war with bad motivations and have run it incompetently from start to finish.

Among the lessons of the Iraq war, one will surely [i]not[/i] find that "they are bad guys" is a good reason to start a war, nor that more wars are a good way to ensure security.

Well, I don't know if "occupation" is the right term, I mean : there have been elections.
But still, I think this war is catastrophic.  It is a new Vietnam.  Exactly what people predicted has happened : a fairly swift and easy victory for the coalition troops, followed by a never-ending guerrilla war. :(

I don't see the solution :huh: .  We are talking about a region that as far as I know has never been democratic, where different languages and religions exist:

[img]http://upload.wikimedia.org/wikipedia/commons/f/f6/Iraq_demography.jpg[/img]

Make something like that democratic and you are bound for civil war and a break-up. The scariest thing is that there are no clear borders, and everyone will be trying to make those borders as favorable as possible for his own party.


factorial
function
algebra
floor function

Generally, when we count the number of zero's in a certain factorial, we use the formula with the floor function and division, and count the number of 5's. My question is, how do you count the number of a power of a prime? For example, if I want to count to number of 4's in a given factorial (as in, a problem where you are asked to count the number of zero's in base 4).

You can also use the greatest integer function with those too:

For instance, how many $2$s are there in $100!$, we can see through the greatest integer function that there are $97$ twos in $100!$ and as your question states that we need to found out how many $4$s, we simply divide by $2$ and get $48$.

[quote="hqin2007"]count the number of 5's[/quote]You do it the same way to count the 5's.  After all, 5 is a prime, and there's nothing that would make counting other primes different than 5's.

I don't know If I am correct or wrong.
But here It is what I do

100 divide by 2 = 50
100 divide by 4 = 25
100 divide by 8 = 12
100 divide by 16 = 6
100 divide by 32 = 3
100 divide by 64 = 1

So 50 + 25 + 12 + 6 + 3 + 1 = 97
there are 2^97 in 100!


I ran a program to test this for values up to 2000 and it still remains true.  But can anyone actually [i]prove[/i] it?

if $\sqrt{a^2+b^2}=$a whole number then,
$(\sqrt{a}+\sqrt{b})^2=$not whole number.
but $a$ and $b$ must be natural numbers

If the square of a sum of square roots is whole then the square root of the sum of squares is not.

The second one means that you have $2\sqrt{ab}$ being an integer (just factor it out to see).
So $4ab=m^2$ must be a square. Since in both conditions we can factour out common factors, we may assume that $a$ and $b$ are coprime, thus $a=x^2$ and $b=y^2$ are perfect squares.
Now the first conditions says that $a^2+b^2=z^2$ is a perfect square too and with the above we get $x^4+y^4=z^2$, which is known to have no solutions except $xy=0$ which is excluded here (excluded since $0$ is not considered natural here; I don't want to see an answer in this thread saying anything else, it's clear that $0$ isn't seen as natural by the author).


geometry
rectangle
power of a point
radical axis
geometry solved

[quote="kalyanram_a"]Prove that the difference of the squares of the tangents to two coplanar circles, form any point P in their plane, varies as the perpendicular from P on their radical axis.[/quote]

Well, the square of the tangent from a point to a circle is the power of the point with respect two the circle.

Hence, we can restate your problem as follows:

Given two circles k and l and any point P in their plane, show that the difference of the powers of P with respect to k and with respect to l is proportional to the distance from P to the radical axis of k and l.

[i]Proof.[/i] If X and Y are the centers of the circles k and l, and x and y are their respective radii, then we have to show that the expression

  (PX2 - x2) - (PY2 - y2)

is proportional to SP, where S is the orthogonal projection of P on the radical axis of the circles k and l.

We will use directed segments.

Let P' and S' be the projections of P and S on the line XY. Then, the quadrilateral PP'S'S is a rectangle; we have SP = S'P', and also PX2 - PY2 = P'X2 - P'Y2, since PP' is perpendicular to XY.

Hence,

  (PX2 - x2) - (PY2 - y2) = (PX2 - PY2) - (x2 - y2) = (P'X2 - P'Y2) - (x2 - y2) = (XP' 2 - P'Y2) - (x2 - y2)

  = (XP' + P'Y)(XP' - P'Y) - (x2 - y2) = XY * (XP' - P'Y) - (x2 - y2) = XY * (XS' + S'P' - P'S' - S'Y) - (x2 - y2)

  = XY * (XS' - S'Y + 2 S'P') - (x2 - y2) = XY * (XS' - S'Y) + 2 * XY * S'P' - (x2 - y2)

  = (XS' + S'Y)(XS' - S'Y) + 2 * XY * S'P' - (x2 - y2) = (XS' 2 - S'Y2) + 2 * XY * S'P' - (x2 - y2)

  = (S'X2 - S'Y2) + 2 * XY * S'P' - (x2 - y2) = (S'X2 - x2) - (S'Y2 - y2) + 2 * XY * S'P'.

But, since S' lies on the radical axis of k and l, we have S'X2 - x2 = S'Y2 - y2, and hence

  (PX2 - x2) - (PY2 - y2) = 2 * XY * S'P' = 2 * XY * SP,

proving that the difference of the powers of P with respect to the circles k and l is proportional to the distance SP of the point P to the radical axis of k and l. Qed..

Strange that nobody had solved this problem yet: it was an easy and straightforward calculation.

  Darij Grinberg


trigonometry
ratio
geometry
inequalities
circumcircle
perpendicular bisector
angle bisector

let ABC be a triangle .there exist a point P in the interior of it,inwich:

angle PAB=10 ,  angle PBA=20 ,  angle PCA=30 ,  angle PAC=40

prove that:

               ABC is isoceles.

We know the angle $\angle A = \angle PAB + \angle PAC = 10^\circ + 40^\circ = 50^\circ$. Let M be the midpoint of the side AC and let CP meet the perpendicular bisector of the side CA at a point O. The triangle $\triangle AOC$ is isosceles, hence $\angle OAC = \angle OCA \equiv PCA = 30^\circ$ and $\angle OAB = \angle A - \angle OAC = 50^\circ - 30^\circ = 20^\circ$. Since $\angle PAB = 10^\circ$, AP bisects the angle $\angle OAB$. Let CP meet that side AB at a point Q. The remaining angle of the triangle $\triangle AQC$ is $\angle AQC = 180^\circ - (\angle A + \angle QCA) = 180^\circ - (50^\circ + 30^\circ) = 100^\circ$. Using the sine theorem for this triangle,

$\frac{AQ}{CA} = \frac{\sin \widehat{QCA}}{\sin \widehat{AQC}} = \frac{\sin 30^\circ}{\sin 100^\circ} = \frac{1}{2 \sin 80^\circ}$

From the isosceles triangle $\triangle AOC$ with $\angle AOC = 180^\circ - 2 \angle OCA = 180^\circ - 60^\circ = 120^\circ$,

$\frac{AO}{CA} = \frac{\sin \widehat{OCA}}{\sin \widehat{AOC}} = \frac{\sin 30^\circ}{\sin 120^\circ} = \frac{1}{2 \sin 60^\circ}$

and combining with the fact that the angle bisector AP cuts QO in the ratio of the adjacent triangle sides,

$\frac{PQ}{PO} = \frac{AQ}{AO} = \frac{\sin 60^\circ}{\sin 80^\circ}$

Let a normal to the side CA from the vertex B meet CQ at a point R. Then $\angle QBR = \angle AQR = 90^\circ - \angle A = 40^\circ$. Since $\angle PBA = 20^\circ$, BP bisects the angle $\angle QBR$. The remaining angles of the triangle $\triangle BQR$ are $\angle BQR \equiv \angle BQC = 180^\circ - \angle AQC = 180^\circ - 100^\circ = 80^\circ$ and $\angle QRB = 180^\circ - (\angle QBR + \angle QRB) = 180^\circ - (40^\circ + 80^\circ) = 60^\circ$. Using the sine theorem for the triangle $\triangle BQR$ and combinig with the fact that the bisector BP cuts QR in the ratio of the adjacent triangle sides,

$\frac{PQ}{PR} = \frac{BQ}{BR} = \frac{\sin 60^\circ}{\sin 80^\circ}$

Thus we obtained $\frac{PQ}{PR} = \frac{PQ}{PO}$, which means that the points R, O are identical. Since both BR and MO are perpendicular to CA, these 2 lines passing through the point $R \equiv O$ are identical. As a result, the vertex B lies on the perpendicular bisector MO of the side CA, which means that the triangle $\triangle ABC$ is isosceles.

trig ceva; working in degrees.

So $\sin30 \sin(40+x) \sin10 = \sin(40-x) \sin20 \sin40.$

Now
$\sin(x+y)\sin(x-y) = [\sin x \cos y]^2 - [\cos x \sin y]^2$

$= \sin^2 x (1 - \sin^2 y) - \sin^2 y (1 - \sin^2 x) = \sin^2 x - \sin^2 y$

So we have $\frac{\sin(40-x) }{\sin(40+x)} = \frac{sin^2 20 - sin^2 10}{sin^2 30 - \sin^2 10}$

Also, $\sin^2 10 = \frac{1 - \cos 20}{2}$.  Put $c,s = \cos 20, \sin 20$.  Then

 $\frac{sin^2 20 - sin^2 10}{sin^2 30 - \sin^2 10} = \frac{s^2 - \frac{1-c}{2}}{\frac{1}{4} - \frac{1-c}{2}} = \frac{-4c^2 + 2c + 2}{2c-1}$

and $\frac{-4c^2 + 2c + 2}{2c-1} = \frac{-8c^3 + 6c + 2}{4c^2 - 1} = \frac{1}{4c^2 - 1}$ using triple angle cosine formula.

It remains to see what value of x satisfies $\sin(40+x) = (4c^2 - 1)(\sin 40-x)$

After grunting, and dividing by some non-zero quantities, it boils down to $(s\cos x - c\sin x) = 0$. we must have $\sin20\cosh - \cos20\sinh = 0 \Rightarrow \sin(20-x) = 0$.  So x = 20 is the only acceptable answer  (we need |x| < 80)

thanks to yetti and singular...

but i think these easy problems are for theother people :P

Then you should clearly state "This problem is for the other people.", otherwise, it is for anybody :P. Anyway, I just clean up geometry problems in the pre-olympiad section, which go or are about to go out of scope without any replies.

Yetti.

i think you are so cool in geometry yetti.your solution is always the best. :D

jensen,i thaught about u solution.but if u post it here,it will be better.

I am working in degrees and I note $m(\angle PBC)=x$.
Thus, $m(\angle PCB)=80-x$ and $\sin (80-x)=\cos (10+x)$.
From the trigonometric appearance of the Ceva's theorem results:
$\sin 10\cdot \sin x\cdot \sin 30=\sin 40\cdot \sin 20\cdot \cos (10+x)\Longleftrightarrow$
$\sin 10\cdot \sin x=(\cos 20 -\cos 60)\cdot \cos (10+x)\Longleftrightarrow$
$\cos (10-x)-\cos (10+x)=\cos (30+x)+\cos (10-x)-\cos (10+x)\Longleftrightarrow$
$\cos (30+x)=0\Longleftrightarrow 30+x=90\Longleftrightarrow x=60\Longleftrightarrow A=C\ (a=c).$

A proof without trigo, although the problem have appeared many times on many sites meanwhile:
If $D$ is the circumcenter of $\triangle APC$, then $\triangle APD$ is equilateral, from $\angle ACP=30^\circ$; consequently $\angle DAC=\angle ACD=20^\circ$. Since $\angle PAB+\angle ABP=30^\circ, PB$ is perpendicular bisector of $AD$, hence $BD=AB, m(\widehat{ABD})=40^\circ=2\cdot m(\widehat{ACD})$ and $B$ is the circumcenter of $\triangle ACD$, with $AB=BC$, done.

Best regards,
sunken rock


For questions 1 and 2 below, let f(x) = x^2, g(x) = 3x and
h(x) = (sqrt{x}) + 1.  Express each function as a composite of f, g and/or h.

(1) G(x) = 3x^2

(2) R(x) = 9x

these are quite easy questions.....
1. $fog$
2. $gog$

Okay...easy for you but how did you get the answers?

Can you explain that?

Thanks!

The $\circ$ notation doesn't seem to me to be consistent in its usage, so I won't use it.

I think the simplest way to think of composition of functions is to think of each function as a set of instructions.

For example, $f(x) = x^{2}$ can be represented by the single instruction "I square my input."  And $g(x)$ can be represented by the single instruction "I multiply my input by $3$."

Now, how can we put $G(x)$ into words?  Well, first it squares its input and then it multiplies by $3$.  In other words, first we did the same instruction as $f$, and then we did the same instruction as $g$, so

$G(x) = g(f(x)) = g( x^{2}) = 3x^{2}$

Similarly, one way to think about $R(x)$ is that we multiply our input by $3$ twice.  In other words, we applied the instruction from $g$ twice, so

$R(x) = g(g(x)) = g(3x) = 9x$

Edit:  You ask a lot of questions on the forum that are obviously from a math class.  Do you feel that your teacher is not doing an adequate job preparing you for your homework?  You might want to consider getting a tutor rather than cluttering the forum with homework.

The second reply is truly internet math tutoring.

I thank you.


Since we are at least two to like topology, a short one which, I think, deserve a second thought :

Is the closure of the open disk ({x in E, d(a,x) < r)}, r > 0, always the closed disk ({x, d(a,x) <= r)} with same center and radius ??

There is not hypothesis on E (just a metric space).

Are there any restrictions on r? Consider when r = 0. The first set is empty, but the second set is {a}.

If you assume r > 0 and that you're in the standard metric topology - then yes. It is not hard to prove, but it is a good exercise. If you have some other topology then not necessarily.

Gauss202, are you sure ? So try to prove it. I did assume we were in IR^n.

[hide]

Consider IN + |x|, compare the closure of B_opened(0,1) = {0} and B_closed(0,1) ={0,1}  [/hide]

In a general normed space, it's definitely false that the closure of the open unit ball is the closed unit ball. (I'm trying to come up with a counterexample now.)

Haha that was painless enough. There you go.


quadratics
function
algebra
polynomial
algebra unsolved

Determine g(x) = (px+q)^2(2x^2-5x-3) where p, q are constants,
given that, on division by         (x - 2)^2, the remainder is -39-3x.

I have a fairly long solution but there must be efficient methods for remainders on division by functions higher than linear.
What are they?

[hide="First method (sort of cheap)"] $(px+q)^{2}(2x^{2}-5x-3) = f(x) (x-2)^{2}-39-3x$

Letting $x = 2$,

$(2p+q)^{2}(-5) =-39-6 =-45 \implies (2p+q)^{2}= 9$

Now, I'm going to cheat and differentiate both sides.

$2p (px+q) (2x^{2}-5x-3)+(px+q)^{2}(4x-5) = f'(x) (x-2)^{2}+2f(x) (x-2)^{2}-3$

And plug in $x = 2$ again.

$2p (\pm 3) (-5)+9 (3) =-3$

Some casework is necessary.

[b]Case:[/b]  $2p+q = 3$.  Then

$-30p =-30 \implies \boxed{ p = 1, q = 1 }$

[b]Case:[/b]  $2p+q =-3$.  Then

$30p =-30 \implies \boxed{ p =-1, q =-1 }$ [/hide]
Edit:

[hide="Second method"] Make the substitution $t = x-2$.  Then $x = t+2$ and so

$(px+q)^{2}(2x^{2}-5x-3) = (pt+q+2p)^{2}(2t^{2}+8t+8-5t-10-3) = (p^{2}t^{2}+2p(q+2p)t+(q+2p)^{2})( 2t^{2}+3t-5)$

We want to find the remainder upon division by $t^{2}$; in other words, the linear and constant terms.  This means we can ignore all of the quadratic terms (this is basically doing polynomial arithmetic $\bmod (x-2)^{2}$).  So,

$( 2p(q+2p)t+(q+2p)^{2})(3t-5) \equiv ( 3(q+2p)^{2}-10p(q+2p) ) t-5(q+2p)^{2}\equiv-39-3x =-45-3t$

So we have the system

$-5(2p+q)^{2}=-45 \implies (2p+q)^{2}= 9$
$(q+2p)(3q-4p) =-3$

The same casework as the first method applies here.  [/hide]


If $n!5!=6!$, then what is the value of $n$?

[hide=" Solution"]n!5! = 6!
n!    = 6!/5!
n!    = 6
[color=blue]n     = 3[/color][/hide]

[hide]i used guess and check and i got 3[/hide]

[hide=" :D Click me"]

$n!5! = 6!$

$n! = \frac{6!}{5!}$

$n! = \frac{720}{120}$

$n! = 6$

$\boxed{ \text{I know by heart that }3! \text{ is } 6}$

$\boxed{ \boxed{ \boxed{ \boxed{ \boxed{n = 6}}}}}$

[/hide]

kinda late, but this looks like a fun problem...

[hide]n!5!=6!

divide the 5! from both sides, and simplify

6X5!/5!

6=n!

n=3[/hide]

[hide]$n!$ must equal 6, and 3!=6, so n=3.[/hide]


[i] Χθες εκεί που ξεφύλλιζα ένα ρουμάνικο βιβλίο γεωμετρίας του 1983 για τάξεις 9-10 βρήκα και αυτή τη βασική άσκηση:[/i]

 Σε τρίγωνο $ ABC$ είναι  $ \angle A \equal{} 60$.Να αποδειχθεί ότι $ IO\equal{}IH$ , όπου $ I,O,H$ είναι το έγκεντρο , το περίκεντρο και το ορθόκεντρο του τριγώνου.

[hide]Eukola prokiptei oti ta $ B,H,I,O,C$ einai omokiklika kai i kiklos autos exei meso to N pou einai to meso tou toksou $ BC$ (pou den periexei to $ A$). Episis, $ AH// \equal{} 2OM \equal{} ON$, opou $ M$ to meso tis $ BC$. H teleutaia prokuptei kai apo to oti i $ \angle A \equal{} 60$. Opote to $ HNOA$ einai rombos opote $ \angle HNI \equal{} \angle INO$, apo opou exoume to zitimeno.

Fusika iparxei kai pio apli lisi (xwris tous kiklous etc.) arkei mono na doume oti $ AH \equal{} AO$. Episis, kseroume oti oi $ AH,AO$ einai isogwnies ws pros tis $ AB,AC$ opote exoume to zitoumeno.

Apla i prwti lusi einai pio endiaferousa apo theoria (oso afora tous mikrous...)[/hide]


You could refer to many different sources for speed calculation. Know the tricks; in this case, know that multiplying by 36 is just multiplying by 40 then subtracting multiplying by 4. Once you have one of those, you can figure out the other. Know that multiplying by 4 is doubling twice.

Thanks!

sorry if this is considered spam

Ah. I went to that site too..I think it has improved my speed as well..:o

Try looking at this attached document. I found this from Mr. Frost (the MA National Mathcounts coach from many past years)'s Math Team website when googling "mathcounts speed."

Yes, I know that page.  :wink:
I think it is very help ful for beginning MATHCOUNTS participants.

[quote="bluecarneal"]Sometimes if there are mental calculations that would take unreasonably long 20+sec....
You could use a calculator.[/quote]

Err, but you do realize that FTW takes most of its problems from non-calculator stuff like MC Sprint/Countdown, right? FTW's main purpose is meant for you to practice speed math [i]without[/i] a calculator.

[quote="AIME15"]http://www.zetamac.com/arithmetic has a good arithmetic speed game you can practice with.[/quote]

When I go there, when I click start game, my download box pops up and it asks me if I wish to save a file.  Is that supposed to happen?   :huh:

No. This website leads to the "Arithmetic Game". I dont think anything is supposed to pop up.  :maybe:

Never mind, it worked this time.  Weird.   :huh:

Probably something with your computer...

[quote="batteredbutnotdefeated"]some people may know me on FTW. I'm the guy with the second lowest rating...

Anyways, after an infinite amount of losses, I have concluded that I mostly lose because I'm unable to do calculations such as 27*36 in like 10 seconds. How can I improve my speed? If I could learn just a couple simple tricks, my rating would shoot up, and I would have a shot at winning next year's chapter CD.

I'm might even make nats. :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:
I'm not that good...[/quote]

Use tricks. 27*36=81*12=162+810=972

Or, 27*36=27^2+9*27=3^5+3^6=243+729=972

that's < 3 seconds.

What is your "second lowest" rating? just curious....


function
search
algebra unsolved
algebra

Find all reals $ a$ such that there exists a nonconst function $ f : (0;1]\to R$ and :
$ a \plus{} f(x \plus{} y \minus{} xy) \plus{} f(x)f(y)\leq f(x) \plus{} f(y)$ for any $ 0 < x,y\leq 1$


[quote="ComplexZeta"]Has anyone used Musix$\text{\TeX}$? I installed the packages this morning and started using it, but it seems somewhat more complicated than $\text{\LaTeX}$. Can anyone recommend a book on it? Thanks.[/quote]

I haven't seen a book on it, but I did find this:

http://www.tug.org/tex-archive/macros/musixtex/taupin/musixdoc.pdf

Yeah, that's what I was using as a reference before, but I don't think it's really that great.

Try this:

http://www.lilypond.org

It's not MusiXTeX, but it typesets music and can be used in tex/latex documents.


If $ b_1,b_2,...,b_n$ are scalene permutation of positive numbers $ a_1,a_2,...,a_n$ , prove that:

$ \frac{a_1}{b_1}\plus{}\frac{a_2}{b_2}\plus{}...\plus{}\frac{a_n}{b_n}\ge n^2$

What's a scalene permutation? (Google's showing no results.)

same Query as Zhero's ... [i]Wolfram Mathworld[/i] is also showing no result for [b]scalene permutation[/b]  :!:

if it meant just ordinary permutation, doesn't $ A.M \geq G.M$ on the fractions

$ \frac{a_1}{b_1}, \frac{a_2}{b_2}, ... \frac{a_n}{b_n}$

give you

$ \frac{\frac{a_1}{b_1} \plus{} \frac{a_2}{b_2} \plus{} ... \plus{} \frac{a_n}{b_n}}{n} \geq (\frac{a_1}{b_1}\frac{a_2}{b_2}...\frac{a_n}{b_n})^{\frac{1}{n}}$

Since $ b_i$ s are a permutation of  $ a_i$ s,

$ a_1.a_2 ... a_n \equal{} b_1.b_2.... b_n$

So,

$ \frac{a_1}{b_1} \plus{} \frac{a_2}{b_2} \plus{} ... \plus{} \frac{a_n}{b_n} \geq n$

and not $ n^2$ on RHS  ??

[b]Kindly clear my doubt [/b] :what?:

Can someone please answer my doubt above ??

 :trampoline:

Well, take $ a_i\equal{}b_i$ for all $ i\equal{}1,\ldots,n$. Now, LHS is $ n$. Since $ n\leq n^2$ for integers, the correct formulation of this inequality is $ \sum_{i\equal{}1}^{n}\frac{a_i}{b_i}\geq n$.


integration
trigonometry
logarithms
calculus
calculus computations

How to solve this one

$ \int \frac{dx}{\cos x}$

It has been post several time:
Suppose that $ \sec x\plus{} \tan x \not\equal{}0 \Rightarrow \int \sec xdx\equal{}\int \frac {\sec x(\sec x\plus{} \tan x)}{(\sec x \plus{}\tan x)}\equal{}Ln \mid \sec x\plus{} \tan x \mid \plus{}c$

Might be the same thing, but I think it is easier to multiply by $ \frac{\cos x}{\cos x}$, rewrite $ \cos^2 x \equal{} 1 \minus{} \sin^2 x$ and put $ t \equal{} \sin x$

[quote="0714446459923"]the easiest way of all of course is to just have it already memorized[/quote]
This is precisely what I tell my class. The "method" that ehsan2004 gave is essentially what you would get if you differentiated the answer and then simply reversed the steps. That's a lot easier to accomplish when you already know the answer. Just let it be a case of accidental or serendipitous discovery, and accept it as something to remember.

$ \int \frac{1}{\sin x}\ dx\equal{}\int \frac{1}{2\sin \frac{x}{2}\cos \frac{x}{2}}\ dx\equal{}\int \frac{\frac{1}{\cos ^ 2 \frac{x}{2}}}{2\tan \frac{x}{2}}\ dx$

$ \equal{}\int \frac{(\tan \frac{x}{2})'}{\tan \frac{x}{2}}\ dx\equal{}\ln \left|\tan \frac{x}{2}\right|\plus{}Const.$

$ \therefore \int \frac{1}{\cos x}\ dx\equal{}\ln\left| \tan \left(\frac{x}{2}\plus{}\frac{\pi}{4}\right)\right|\plus{}Const.$


Let $ I$ be the incenter of a triangle $ ABC$ with $ \angle BAC \equal{} 60^{\circ}$. A line through $ I$ parallel to $ AC$ intersects $ AB$ at $ F$. Let $ P$ be the point on the side $ BC$ such that $ 3BP \equal{} BC$.

Prove that $ 2\angle BFP \equal{} \angle ABC$


function
conics
ellipse
complex numbers
complex analysis
topology
complex analysis unsolved

If $f(z)$ in an entire function such that $f(z) \in \mathbb{R}$ for all $z: |z-1|+|z-2|=2$ prove that $f$ is constant .

the imaginary part of $f$ is an harmonic function that is null on the ellipse $|z-1|+|z-2|=0$. Thus it's null inside the ellipse. This shows that $f$ is real inside the ellipse, and it's easy to conclude from here because an entire function that is real on a set with nonempty interior is constant.

Hint: Consider the set $E = \{z: |z-1|+|z-2| < 2\}$, and its boundary $\partial{E}= \{z: |z-1|+|z-2| = 2\}$ (which is an ellipse). Look at the image of $E$. It should be an open set since every entire function maps an open set to an open set. On the other hand, the boundary of $f(E)$ is a subset of $f(\partial E)\subset{\mathbb R}$. But a subset of $\mathbb R$ is not a boundary of any bounded open set.


After 16 problems!

The highest score:

captcha000 --> 9 points

Next ones:

RC-7th --> 2
LynnelleYe --> 2
towersfreak2006 --> 2
shake9991 --> 2
h_s_potter_2002 --> 2

People with one point
dts
white_horse_king88
Wumbate
Optimosis
themonster
Rep123max

Zero
Lucky707

Negatives
Snowstorm
juggle7


My ugly proof:
$\frac{a^2+ac}{b^2+bc}+\frac{b^2+ba}{c^2+ca}+\frac{c^2+cb}{a^2+ab}\geq 2(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b})\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}((a^5c-a^4bc)+2(a^4c^2-a^3b^2c)+(a^3b^3-a^2b^2c^2)+(a^3b^2c-a^2b^2c^2))\geq0,$
which obviously true.

there is a nicer solution to this inequality .
Try to solve the stronger one :
$\sum_{cyc} \frac{a^2+ac}{b^2+bc}\geq \sqrt[3]{\prod_{cyc} \frac{(a+b)}{c}} \left( \sum_{cyc} \frac{a}{b+c}\right)$
 :)

That seems expansion is the most enjoyable method of arqady.

My solution:

[b]Problem:[/b]
Prove that for all positive reals $a,b,c$ we have:
\[ \frac{ab+b^2}{ac+c^2}+\frac{bc+c^2}{ba+a^2}+\frac{ca+a^2}{cb+b^2} \ge 2\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right).  \]

[b]Solution:[/b]
Applying AM-GM, we obtain:
\[ \left[\frac{ab+b^2}{ac+c^2} \equiv \frac{b(a+b)}{c(a+c)}\right]+\frac{4bc}{(a+b)(a+c)} \ge 4\frac{b}{c+a},  \]
Take sum of three inequalities,
\[ LHS+\sum_{cyc}\frac{4bc}{(a+b)(a+c)} \ge 4\sum_{cyc}\frac{a}{b+c}  \]
Moreover,
\begin{eqnarray*} LHS \ge 4\sum_{cyc}\frac{a}{b+c}-4\sum_{cyc}\frac{bc}{(a+b)(a+c)} &\ge& 2\sum_{cyc}\frac{a}{b+c} \\ \iff 2\sum_{cyc}\frac{bc}{(a+b)(a+c)} &\le& \sum_{cyc}\frac{a}{b+c} \\ \iff a^3+b^3+c^3+3abc &\ge& \sum_{sym}a^2b, \end{eqnarray*}
Reducing to Schur’s.
Done.

[quote=Albanian Eagle]Show that for any positive real numbers $a,b,c$ we have:
$\frac{a^2+ac}{b^2+bc}+\frac{b^2+ba}{c^2+ca}+\frac{c^2+cb}{a^2+ab}\geq 2(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b})$[/quote]

[img]http://s9.sinaimg.cn/middle/006ptkjAzy76rDO7XXq18&690[/img]

[quote=ductrung][quote="Albanian Eagle"]also another way of proving it is:
$ LHS\geq \sqrt{\left(\sum_{cyc}\frac{a}{b+c}\right)\left(\sum_{cyc}\frac{a+b}{c}\right)}$
and then use Levinson's inequality  :)[/quote]
Yes, such soution is posted here [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=102719[/url] together with a stronger version.[/quote]
The following inequality is also true
[img]http://s9.sinaimg.cn/middle/006ptkjAzy76rJcf4b628&690[/img]


function
induction
algebra unsolved
algebra

Let $f$ be a function from the nonnegative integers to the nonnegative integers.

Determine all possible functions $f$ if

\[f(x^2 + y^2) = xf(x) + yf(y).\]

The same nice identity that was used to solve the french equation $ f(x^2+y^2)=f^2(x)+f^2(y)$ works here too.

Pierre has posted a solution for the french problem in which he uses a nice identity. Try to find the solution.

I'm sorry but I can't find any posts by Pierre :(

I think you can use induction and identities like $(5n + 1)^2 + 2^2 = (4k + 2)^2 + (3k - 1)^2$... is that what you are talking about? I remember solving the problem about $f(m^2 + n^2) = f(m)^2 + f(n)^2$ using these identities. (I think it was me who posted them and not Pierre but I am not sure :P)


conics
parabola
calculus
calculus computations

The figure below (attachment) shows a road running in the shape of a parabola from the bottom of a hill at A to point B. At B it changes to a line and continues on to C. The equation of the road is represented by R(x), also shown below (in the attachment).

B is 1000 feet horrizontally from A and 100 feet higher. Since the road is smooth, R'(x) is continuous. What is the value of b?

Multiple Choice:
A---> 0.2
B---> 0.02
C---> 0.002
D---> 0.0002
E---> 0.00002

When given a problem of this type, you should divide the vertical distance by horizontal and multiply the result by two. So, $ b \equal{} \frac {100}{1000}\cdot 2 \equal{} 0.2$.

[i][I suppose this is how calculus is taught at high school level.][/i]

could somone check this for me.

anyone have any other methods for this?


AMC
AIME
complex numbers
algebra
polynomial
Rational Root Theorem

[quote]w and z are complex numbers such that w2 + z2 = 7, w3 + z3 = 10. What is the largest possible real value of w + z?

from AIME 1983[/quote]

In order for w + z to be a real number, their complex parts must cancel out.  So the complex part of one must be the negative version of the other.

Let w = a + ci and z = b - ci.

[tex]w^2 = (a+ci)^2 = a^2 + 2aci + c^2*i^2 = a^2 - c^2 + 2aci[/tex]
[tex]z^2 = (b-ci)^2 = b^2 - 2bci + c^2i^2 =  b^2 - c^2 - 2bci[/tex]

So [tex]w^2 + z^2 = 7[/tex] becomes
[tex]a^2 - c^2 + 2aci + b^2 - c^2 - 2bci = 7 + 0i[/tex]

So
[tex]a^2 + b^2 - 2c^2 = 7[/tex]
and
[tex]2aci - 2bci = 0i[/tex]
[tex]a = b[/tex]

Knowing that, we rewrite z = b - ci as a - ci.  Our other equaiton, [tex]a^2 + b^2 - 2c^2 = 7[/tex], becomes [tex]2a^2 - 2c^2 = 7[/tex].  We can then rewrite this:

[tex]2c^2 = 2a^2 - 7[/tex]
[tex]c^2 = a^2 - 7/2[/tex]

Now for the other one:
[tex]w^3 = (a+ci)^3 = a^3 + 3a^2ci + 3ac^2i^2 + c^3i^3[/tex] [tex]= a^3 - 3ac^2 + 3a^2ci - c^3i[/tex]
[tex]z^3 = (a-ci)^3 = a^3 - 3a^2ci + 3ac^2i^2 - c^3i^3 = a^3 - 3ac^2 + c^3i - 3a^2ci[/tex]

So [tex]w^3 + z^3 = 10[/tex] becomes
[tex]a^3 - 3ac^2 + 3a^2ci - c^3i + a^3 - 3ac^2 + c^3i - 3a^2ci = 10[/tex]
[tex]2a^3 - 6ac^2 = 10[/tex]
[tex]a^3 - 3ac^2 = 5[/tex]

Before, we found that [tex]c^2 = a^2 - 7/2[/tex].  Substituting this in:
[tex]a^3 - 3a(a^2 - 7/2) = 5[/tex]
[tex]a^3 - 3a^3 + 10.5a = 5[/tex]
[tex]2a^3 - 10.5a + 5 = 0[/tex]
[tex]4a^3 - 21a + 10 = 0[/tex]

Now use the rational root theorem to find the roots are a = 2, 1/2, or -5/2.    The maximum sum is a + b = 2a = 4.   We need to check that a=2 is possible, though.  When a = 2, c = sqrt(2)/2, which is a real number.  So a = 2 checks.  The answer is then [b]004[/b].

It's posted here before. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11980&highlight=]here[/url]


Putting 2n numbers, $a_{1}.......a_{n}$and $b_{1}......b_{n}$ they are all different into a table with a $nxn$ squares on it.
The number put into i row, j column =$a_{i}+b_{j}$
Prove if the products of the numbers in each row are equal, then the products of the numbers in each column are also equal.
i.e. when the product of the numbers in 1st row=2nd row=3 rd row=....=nth row,
then 1st column= 2nd column=3 rd column=...=nth row

The problem is from USSR 1991.
Consider the following polynomial \[f(x)=\prod_{i=1}^{n}(x+a_{i})-\prod_{j=1}^{n}(x-b_{j})\] that its degree is less than $n$. According to problem hypothesis $f(b_{j})=\prod_{i=1}^{n}(a_{i}+b_{j})=c$ for $j\in\{1,2,\dots,n\}$. Polynomial $f(x)-c$ has at least $n$ roots. Thus $f(x)=c$. Then $c=f(-a_{i})=-\prod_{j=1}^{n}(-a_{i}-b_{j})=(-1)^{n+1}\prod_{j=1}^{n}(a_{i}+b_{j})$, and we are done.

[quote="Omid Hatami"]The problem is from USSR 1991.
Consider the following polynomial \[f(x)=\prod_{i=1}^{n}(x+a_{i})-\prod_{j=1}^{n}(x-b_{j})\] that its degree is less than $n$. According to problem hypothesis $f(b_{j})=\prod_{i=1}^{n}(a_{i}+b_{j})=c$ for $j\in\{1,2,\dots,n\}$. Polynomial $f(x)-c$ has at least $n$ roots. Thus $f(x)=c$. Then $c=f(-a_{i})=-\prod_{j=1}^{n}(-a_{i}-b_{j})=(-1)^{n+1}$, and we are done.[/quote]
 $c=f(-a_{i})=-\prod_{j=1}^{n}(-a_{i}-b_{j})=(-1)^{n+1}$
Isn't it equal to the originial one when n is odd?
giving the $(-1)^{n+1}=1$

Excuse me, I missed something in my solution and I corrected.


calculus
integration
logarithms
real analysis
real analysis unsolved

$\int_{0}^{+\infty}\frac1{\((log(x+2))^{p}}\, dx$ for any p>0

I tried numericaly integrate it in Maple, I am convinced that it diverges for all p>0.
But can someone give a analytic proof?

From  $\log x \leq x-1$ we have

$\\ \int_{0}^{+\infty}\frac1{x+1}\, dx \leq \int_{0}^{+\infty}\frac1{\log(x+2)}\, dx$

and  $\int_{0}^{+\infty}\frac1{x+1}\, dx =+\infty$

thy this

$\ln(x+2)\leq (1+x)$ hence

$\int_{0}^{\infty}\frac{dx}{\ln^{p}(2+x)}\geq \int_{0}^{\infty}\frac{dx}{(x+1)^{p}}=\infty$ for $0<p\leq 1$
thus the integral diverges for $p\in (0,1]$.

Well $p$ was not there from the beginning  :roll:

Do this

If $x+2=t$ then the integral equals

$\int_{2}^{\infty}\frac{dt}{\ln^{p}t}$

now if $t=e^{y}$ then we get that

$I=\int_{\ln 2}^{\infty}\frac{e^{y}}{y^{p}}dy$, and this integral equals $\infty$ for any $p>0$.

There's an obvious connection between this request and the suggestion jmerry made [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=121329]here[/url].


number theory
greatest common divisor
number theory solved

Prove that \[ \gcd{\left({n \choose 1},{n \choose 2},\dots,{n \choose {n - 1}}\right)} \] is a prime if $n$ is a power of a prime, and 1 otherwise.

if $n=p^a$, with p prime, then $p \mid {n \choose k}$ : there is never enough 'p' in the denominator to 'kill' every p in the numerator, ${n \choose 1 } = n$ so gcd is a power of p. ${p^a \choose p^{a-1}}$ is not divisible by $p^2$. All this proves $gcd = p$

if $n = N * p^a$, with p prime and $p \nmid N$, then ${n \choose p^a}$ is not divisible by p  (all power of p are 'killed'). Since ${n \choose 1 } = n$ the gcd divides n but is not divisible by any prime divisor of n. It must be 1.


Mathcamp will award $200 scholarships to the top 30 females and the top 30 males in the USAMTS this year, contingent upon the awardees earning admittance to Mathcamp.  Winning the scholarships does not guarantee admission to Mathcamp; winners must apply for Mathcamp and earn admission just like everyone else.

[b]USAMTS participants should login to the USAMTS pages and complete the Gender field on the My USAMTS Profile page.  Only students who have completed this field are eligible for the scholarships.[/b]

Will a list of eligible students be posted, or should we count from the top and guess people's gender?

Also, if one (or more) of the top 30 of each gender does not attend Mathcamp, do the 31st place people get the potential scholarship?


[quote="zhoraster"]$ 2^4\times$ the number of circular arrangements of $ 5$ people $ \equal{} 384$.[/quote]Not quite. The problem only states that $ A$ must sit opposite $ B$ and $ C$ must sit opposite $ D$. Just those two pairs, nothing else.

I think that the correct answer would be 4 times the circular arrangements of 8 people: $ 4 \times 7! \equal{} 20160$, but I'm not too sure.

[quote="tabletenniser"]How many different arrangements are there if 10 people are seated around a round table, and A must sit opposite to B, and C must sit opposite to D?[/quote]
[color=blue]
Fix $ \ A$ somewhere on the circle. Then position opposite to  $ \ A$ will be occupied by $ \ B\ .$

Now there are $ \ 8$ positions left, $ \ C$ can be placed in $ \ 8$ ways and position opposite to  $ \ C$ will be occupied by $ \ D\ .$

And finally there are $ \ 6$ positions and  $ \ 6$ persons left. They can be arranged in $ \ 6!\equal{}720.$

Hence total ways $ \boxed{\boxed{\ \equal{} 8\times 6!\equal{} 5760\ }}$
[/color]

Thank you makar, I think your solution is correct.  :D

thank u so much for helping, i think the answer should be 5760...


Let $ m\equal{}\frac{(4^{p}\minus{}1)}{3}$, where $ p$ is a prime number exceeding $ 3$. Prove that $ 2^{m\minus{}1}$ has remainder $ 1$ when divided by $ m$.

\[ m\minus{}1\equal{}\frac{4}{3}(4^{p\minus{}1}\minus{}1)\equal{}4pk, k\equal{}\frac{4^{p\minus{}1}\minus{}1}{3p}\in N\]
and $ 4^p\equal{}3m\plus{}1$, therefore \[ 2^{m\minus{}1}\equal{}4^{2pk}\equal{}(3m\plus{}1)^{2k}\equal{}1\mod m.\]


Hello all
I am a student without learning anything about this topic. So I want to ask that anyone can give me a brief introduction on this topic.
Thank you!!!!!!

Well, in layman's term, matrix is an arrangement of items into labeled rows and columns. It shows the relationship between two categories of features that are relevant to the items in the matrix.

And in mathematical terms, matrix is an array of numbers in rectangular form. Those numbers constituting the matrix are what we called as entries or elements. Or in general, we use variables with subscripts to denote these entries..Look at the figure below:

The operations between matrices do not follow as the conventional arithmetic operations of numbers..You may refer to a basic linear algebra book to learn all of these or perhaps algebra w/ trigo books, i'm sure these books have ample discussions about matrices... :lol:


1.What integer $ n$ has the property that $ 5^{96}$ is greater than $ n^{72}$ and $ 5^{96}$ is less than $ (n\plus{}1)^{72}$?
[hide="Answer"]8[/hide]
2.The sum of two numbers is two morethan their positive differece. What is the product of the least two positive integers that will satisfy these conditions?
[hide="Answer"]2[/hide]
3.If the square of Winslow's age is added to Abby's age, the sum is 209. If the square of Abby's age is added to Winslow's age, the sum is 183. What is the sum of their age?
[hide="Answer"]27[/hide]

2.[hide]

so two numbers x, and y

x+y=2+x-y
2y=2
y=1
x=2

so 1*2=2.[/hide]

3.[hide]
w^2+a=209
a^2+w=183.

so find a square that is close to those two numbers. You find 196. So assume Winslow is 14, so abby is 13. And it all works out.
14+13=27.[/hide]

1.[hide]Take the 12th root of everything to get:
$ 5^4,n^3,5^4,(n \plus{} 1)^3$[/hide]

I hope this is a target, cause I wouldn't want to be multiplying everything. Anyways, hope this helps.

Edit: Oh yeah, thanks, when I first saw 96, 72 I first realized that they were both divisible by 12.

actually, like jvenezuela said on another topic, you should take the 24th root of all expressions and you end up with $ n^3, 625, (n\plus{}1)^3$ and this becomes alot easier

1.

[hide]Using what stevenmeow said, we can simplify the equatiosn to:

$ 5^4, n^3, (n\plus{}1)^3$

Now we could use trial error.  First, you know that a number less than five won't be greater than $ 5^4$, so we take 6 and go up.

$ 6\rightarrow5^4>6^3, 5^4>7^3$  6 can't work.
$ 7\rightarrow5^4>7^3, 5^4>8^3$ 7 can't work.
$ 8\rightarrow5^4>8^3, 5^4<9^3$  8 works.

So our answer is $ \boxed{8}$[/hide]

2:

[hide]Lets say a is one number and b is another:

$ a \plus{} b \equal{} 2 \plus{} a \minus{} b\rightarrow2b \equal{} 2\rightarrow2 \equal{} b$

b is 1.  NOw we just find a number a which will satisfy the equation, which is 2.  So:

$ 1\cdot2 \equal{} \boxed{2}$[/hide]

3.

[hide]Set up equations:

$ a \plus{} w^2 \equal{} 209$
$ a^2 \plus{} w \equal{} 183$

Solving the equation probably isn't the best idea since you are taking a timed test.  You coould use trial and error which would get the answer faster.  First, you would try the closest square to those numbers :13 and 14.  Now we try it out if a=13 and w=14:

$ 13 \plus{} 14^2 \equal{} 209$
$ 13^2 \plus{} 14 \equal{} 183$

Seeing these work, we do 13+14 and get $ \boxed{27}$[/hide]


I used the following $ \textrm{\text{\LaTeX}{}\sin}$

in Texer. Can someone please explain the result.

Also when I move my mouse over a TEX sample in the forum there's no pop-up showing the tex code any ideas.

THX

Remove \textrm so you have $ \text{\LaTeX}{}\sin$. See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1260818#1260818 (and the rest of that thread) for an explanation of what is happening, which is not a TeXer bug.

There is no popup in Firefox as it doesn't recognise the alt tag. You would have to persuade Valentin to add a title tag. Meanwhile click on the output to get a new window.

Interesting read.  So Latex's attempt to fix a bug creates the repitition.  I got the same result using other editors.

Firfox use to give me mouse over pop-ups on Latex in this site but I have updated Firefox a few times.

So is it a specific version of Firefox or has the structure of the Aop site changed or both?

[quote="Johnny"]Firfox use to give me mouse over pop-ups on Latex in this site but I have updated Firefox a few times.
So is it a specific version of Firefox or has the structure of the Aop site changed or both?[/quote]As far as I'm aware Firefox has never supported the alt tag tooltip which is only supported by IE. AoPS hasn't changed. See also [url=http://gadgetopia.com/post/3206]Firefox, ALT Tags, and Tooltips[/url]. The w3schools.com site says:
[quote]The alt attribute is meant to be used as an alternative text if the image is not available, not as a mouse-over text. To show a mouse-over text on images or image-maps, use the title attribute, like this: <img src="angry.gif" alt="Angry face" title="Angry face" />[/quote]

Personally, Chrome has served best for me in LaTeX dunno why. IE is really slow w/ LaTeX (at least w/ me). But I really IE don't get me wrong.  :)


calculus
ratio
limit
function
geometric series
geometric sequence
algebra

i have been given the problem:
                       ∞
    the series     ∑   ((2n)!/(n!)^2) * (1/4)^n
                     n=0

i need to figure out a test that can be applied that shows that this is divergent.
this divergence test is said to be found in an advanced calculus book, but i have not been successful in finding it so far.
any help would be greatly appreciated..

~civil97~

This is a positive term series, so it either converges or sums to infinity, and some comparison test should be used to decide which. Our first instinct is to try to compare it to a geometric series. For that purpose, we use the ratio test. But we find that the limit of the ratio is 1. This means that we cannot usefully compare this series to a geometric series; that the terms grow or shrink slower than any geometric sequence with ratio different than 1.

What item from more advanced calculus could we be missing? Something having to do with the factorial. Specifically, we could use [b]Stirling's approximation:[/b]

As $n\to\infty,\,n!\sim \sqrt{2\pi n}\,n^ne^{-n}.$

The specific meaning we have in mind for the symbol "$\sim$" is this:

$\lim_{n\to\infty}\frac{n!}{\sqrt{2\pi n}\,n^ne^{-n}}=1.$

So here we have

$\frac{(2n)!}{(n!)^24^n}\sim\frac{\sqrt{4\pi n}\,(2n)^{2n}e^{-2n}} {(2\pi n)n^{2n}e^{-2n}4^n}=\frac1{\sqrt{\pi n}}.$

So the series diverges by limit comparison to $\sum_n\frac1{\sqrt{n}}.$

$\Sigma {2n \choose n}x^n = 1/\sqrt{1 - 4x}$ increases to $\infty$ as $x$ increases to $1/4$, so the sum is certainly infinite at $x = 1/4$.

Of course, you won't see that series unless you know what you're looking for; it's easily proved with Newton's binomial theorem, but finding it in the first place is subtle.

It does come up if you've looked closely at Catalan numbers before.

Yes, I meant if the generating function for $2n \choose n$ is known in advance, to deduce from it the divergence.    It would be interesting to compute the asymptotics of the binomial coefficients directly from the generating function rather than using Stirling's formula.


Prove that the polynomial:
$\displaystyle  x^2(x^2-y^2)(x^2-1)+y^2(y^2-1)(y^2-x^2)+(1-x^2)(1-y^2)$
cannot be written as the sum of squares for any number of polynomials.

Well known?

It is very similar to example by R.M. Robinson (1973).

Who want to prove $S(x,y)\geq 0$? I will suppose it is proved.

Suppose $S(x,y)=\sum f_{i}^{2}(x,y)$. Since $S(x,y)=0$ for points $(\pm 1,\pm 1)$, $(\pm 1,0)$ and $(0,\pm 1)$, it means that all $f_{i}$ are zeros in these points too.
We will prove $f_{i}(0,0)=0$. Indeed, let points $(\pm 1,\pm 1)$ have weight 1, points $(\pm 1,0)$ and $(0,\pm 1)$ have weight -2, and point $(0,0)$ has weight 4. It is clear $\deg f\leq 3$, so $f$ consists of summands $x^{p}y^{q}$, $p+q\leq 3$. Consider the sum of values $f$ in mentioned points accordingly to their weight. If $pq=0$ then monom $x^{p}y^{q}$ gives 0 in the whole sum. If $p,q>0$ then nonzero summands come only from points $(\pm 1,\pm 1)$, and their sum is nonzero iff $p$ and $q$ are even, but it is impossible since $p+q\leq 3$. Therefore, the weighted sum of $f$ in these 9 points is always 0. It follows that if $f=0$ in 8 of 9 points then it is zero in the nine-th point too!
Thus $f_{i}(0,0)=0$, but $S(0,0)=1$. Contradiction.


How come there is a file size limit to the avatars we have. I can't load something because it is just a fex kilobites over.

its kinda anoying for me, but its pretty simple use a nice paint program like Adobe Photoshop to resize the image.


Let $ a,b,c>0$ such that $ \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}\equal{}1$. Prove that

$ \frac{b(a\plus{}c^2)\plus{}c^3}{(a\plus{}c)(b\plus{}c)}\plus{}\frac{c(b\plus{}a^2)\plus{}a^3}{(b\plus{}a)(c\plus{}a)}\plus{}\frac{a(c\plus{}b^2)\plus{}b^3}{(c\plus{}b)(a\plus{}b)}\geq\frac{5}{4}$

We have ∑▒ab/((a+c)(b+c))≥ 3⁄4 and
 ∑▒c^2/((a+c))≥1/2 with    ‘’’’’Cauchy Schwartz’’’’’


Its wrong?  (cant have 2 boxes)

Consider:

First box:

size 1,2,3,4 of all colors (32 balls)

Second box:

size 1,2,3,4 of all colors (32 balls)
plus size 4 of yellow (+1 ball)

total 65 balls; since only 4 sizes we are guaranteed to get 2 balls the same size if we take 5 the same color, but in a box there is no 3 balls with the same size and color.

----

If there is just one box, then atleast one color has 9 balls among it, and we can only have 4 different sizes among these 9 (else we pick all 5 diff. sizes and it contradicts the size criteria we have) and now there are 3 balls with the same size.

It's easy: here's my solution.
By the pigeonhole principle, you know that there are 4 different sizes. Thus, with 4 different colours and 4 different sizes, you know that there are 4*4, or 16 different types of balls.Once again, using the pigeonhole principle, you know that the maximum number of balls any of the boxes can hold so that there are no 3 of the same colour and same size is 32, i.e., 2 of each variety. However, the other box would have 65-32=33 balls and the pigeonhole principle tells us that there exist 3 balls of the same size and colour. Thus, no matter how the balls are divided, there will always be 3 balls of the same size and same colour in the same colour of the same box.

QED

there are 2 boxes.. and 4 colors.. Now as out of 5 balls at least 2 are of the same size by converse of $PHP$  we have that the balls can have 4 different sizes... so there are $2.4.4=32$choices in total..As there are total 65 balls by $P.H.P$ we have that at least 3 balls with the same color and radius must lie in the same box.
QED

[quote=k2s][quote=matt276eagles][hide]By the Pigeonhole Principle, one box contains at least 33 balls, and some nine of these balls are the same color.  Suppose for the sake of contradiction that no three of these balls are the same size.  Then there must be five different sizes among these nine balls, contradicting the fact that of any five balls of the same color, at least two of them are the same size.  Therefore, three of these balls are the same size, and we're done.[/hide][/quote]

DOUBT:why would there be 5 different sizes among the 9 balls of the same color?[/quote]

Actually it should be 4 different sizes it's a typo

[quote=Rushil]Two boxes contain between them 65 balls of several different sizes. Each ball is white, black, red or yellow. If you take any five balls of the same colour, at least two of them will always be of the same size(radius). Prove that there are at least three ball which lie in the same box have the same colour and have the same size(radius).[/quote]

By $PHP,$ there are $atleast$ $33$ balls in one box(let's say $A$)
By $PHP,$ there are $atleast$ $9$ balls of the $same$ colour in $A$.
By given condition and $PHP$, there are $3$ balls of the $same$ $size$ among those.

By Pigeon Hole Principle, we know that one of these boxes must contain at least 33 balls.

By Pigeon Hole Principle, we know that the must be at least one colour with 9 balls in same box.

By Pigeon Hole Principle, we know there will be maximum 4 sizes for each colour. If it exceeds 5 or is 5, then there will not be at least 2 out of 5 of same size and colour.

By Pigeon Hole Principle, the colour with 9 balls, will have at least 3 of the same size.

This means that there are atleast 3 balls which lie in the same box have the same colour and size.


advanced fields
advanced fields unsolved

prove that if $A\sim B ,B\sim C$ then$A\sim C$

I don't understand the question (or maybe I do). What are $A$, $B$ and $C$? What is $\sim$ in this case?


How do I incorporate normal text into an array of equations?
I want to give descriptions, such as which identity I used, while solving this huge trigonometry equation.
Same thing for log stuff.

Thanks.

With, \text{what you want to say}, like this:

$ f(x)\equal{}\begin{cases}\minus{}x&\minus{}1<x<1\\0&\text{otherwise}\end{cases}$

If you want a line of explanation inside a list of equations use \intertext - for example
\begin{align}
n = 1 \\
\intertext{but sometimes we have}
n = 0
\end{align}


calculus algebra function domain trigonometry

hi i'll be taking calculus ab next year and i have this summer packet to do and i ran into a few problems that i have no idea what to do with.

Solve the following trigonometric equations for the given domain:

sinx = cos(x)          on[0, 2pi]

and i have a few more problems like that, but i have no idea what the on[... means, could someone please walk me through this problem?

also this is a problem that i dont really know how to do:

Find the exact value:

sec^2(11pi/6)

Thank You in Advance

For the second one

secant is equal to reciprocal cosine $\sec{x}= \frac{1}{\cos{x}}$
You can use the unit circle (http://members.aol.com/williamgunther/math/ref/unitcircle.gif)
to find the value of $\cos{\frac{11\pi}{6}}$, which is in the 4th quadrant, and equals $\frac{\sqrt(3)}{2}$

Since we know that $\sec{x}= \frac{1}{\cos{x}}$ we have

$\sec^{2}\frac{11pi}{6}= \frac{1}{(\frac{\sqrt{3}}{2})^{2}}$
or
$\frac{4}{3}$.

the "on $[,]$" part means find all the solutions within that interval, so for the problem you gave, you want to find all $0 \leq x \leq 2\pi$ that satisfy the equation

For the second problem...do you know how to find the exact value of $\cos \frac{11\pi}{6}$?  If so, then use the identity $\cos x= \frac{1}{\sec x}$ to find $\sec \frac{11\pi}{6}$ and square that to get your answer.


logarithms
calculus
calculus computations

Study the serie $\sum_{k=1}^n ln(1+\frac{(-1)^k}{k^a(ln(k))^b})$

We'd better start the series at $k=2$ rather than $k=1.$ I'll assume that.

$\ln\left(1+\frac{(-1)^k}{k^a(\ln k)^b}\right) = \frac{(-1)^k}{k^a(\ln k)^b} - \frac1{2k^{2a}(\ln k)^{2b}}+o(k^{-3a})$

It will turn out below that the fate of $\sum_{k=2}^{\infty}o(k^{-3a})$ will never be decisive.

The series $\sum_{k=2}^{\infty}\frac{(-1)^k}{k^a(\ln k)^b}$ will converge as an alternating series whenever $a>0.$

Thus the decisive factor will be the convergence of the all-negative-term series

$-\sum_{k=2}^{\infty}\frac1{2k^{2a}(\ln k)^{2b}}$

When we're done examining that, our overall result for the original series is:

For $a>1,$ or for $a=1,\,b>1,$ the series converges absolutely.

Otherwise, and for $a>\frac12$ or for $a=\frac12,\,b>\frac12$, the series converges conditionally.

Otherwise (meaning $a<\frac12,$ or $a=\frac12,\,b\le \frac12,$) the series diverges.