summer program
Mathcamp
MIT
college
probability
email
PROMYS

Can anyone that's been there tell me how it is? (I'm contemplating an application for next year...)

It's really good.

Mostly what I like is that you can do absolutely anything. There are only 4 rules:

1. Be excellent to each other.
2. Don't do stupid stuff.
3. No fires.
4. Don't divide by 0 except under staff supervision.

Some times rule 0, "don't get the hosting college mad" is added. Also, you have to sign in between 8 and 10, after which you're not supposed to leave the immediate vicinity of the dorms, but if you forget to sign in or go off somewhere else and they find out about it they usually don't get too mad.

There is no curfew and you go to whichever class you want or whatever trip if you want to. You can wake up whenever you want to. You can go walk around the town (you have to sign out and be in groups of at least 2, and you have to be with a staff if going out after 10).

The trips are really good too. Last year I went hiking, biking, amusement park, laser tag, rock climbing, Seattle, animal reserve, etc. There was also a good LARP (Live Action Role Play: basically a role playing scenario except you're in an actual building and you pick up actual objects and you enter actual rooms). I was also in the Contrapositones, a very good a capella group there. They also organised a space combat game over telnet so that was fun.

The math stuff: There were so many interesting classes and guest speakers. They varied in length from a day to four weeks. Also, I liked that you don't have to sign up for classes. You just go, and if you don't like (that rarely happens) or it or it's too hard or too easy, you can always go to a different one the next day.

After classes there's other cool stuff. Other than the card/boardgames, the guest speakers sometimes talk about interesting non-math topics, and there's FUN (you'll have to go there to find out what it is), and people often play soccer or frisbee. They organise a convenient wall with the days of the week on it so people can easily put up the time and place of things they've organised and so other people can see what's going on each day.

Overall point: If you've got those 5 weeks free and you like to be able to do whatever you want with your time, you should definitely most certainly go to Mathcamp.

If you have any other questions about Mathcamp I will certainly try to answer them

[quote="Yasha"]Some times rule 0, "don't get the hosting college mad" is added.[/quote]

We (read: some people who were messy with Zome) got Colby College (well, the cleaning people) mad in 2001.  The commons room in one (maybe more?) of the dorms got shut for a few days.  It's rather interesting they're letting Mathcamp come back there.  (I went to 2001 and 2002, and I'm coming back this year.)

I'd definately vouch for Mathcamp - I really got a lot out of it in 2001 and 2002 mathwise, and it was great fun in many other ways too.  The people are also really cool.

So that's why they were so picky about everyone cleaning up their zome last year. . .

[quote="brian216"]im a bit bitter because I applied having solved 9 of the problems, and I was waitlisted. I understand that they have a lot of qualified people, and some had to be turned away, but what upset me was that I know of two people who got in only having solved 8 problems. I am happy for them and all, but I think you get my meaning.

oh well, those of you who are going, have a good time![/quote]

Were your solutions complete and well written?

Well first make sure your solutions to 9 problems are all correct.  If so, then.. well..

We got the Puget Sound staff mad too - remember Mira saying they took [i]photos[/i] of the Seward common room on 3rd floor? That was just after the sleepover when Chris VL dragged over his mattress, and there were pizza boxes all over the place. The trashcan was overflowing. Hey, why did they open a new pizza place if they didn't want people buying there... ;)

But I concur. MC is THE destination for summer. People are incredible (especially the visiting faculty, minus perhaps Serge Lang), classes are incredible, and the events are even more incredible. My favourite was the puzzle hunt and its associated lack of sleep. I really got into this puzzle solving thing afterwards; I participated in MC alum's team for the MIT puzzle hunt '04.

Go! By all means, go!

I didn't have any, and I doubt anyone did. In particular, your roommate wouldn't be very happy about it. ;) The common room closes at midnight, so you can't stay in there (legitimately, anyhow) all night.

But I can't vouch for the MOORE METHOD people, who from what I've heard were given insane amounts of brainsplitting work. Read:

Dave Savitt: "Did anyone do the homework problem I assigned?"
(Dan Gulotta raises his hand.)
Dave: "You brave, brave soul."

My problem sets came from Group Theory, Linear Algebra, and Beginner and Hardcore Problem Solving. None of them were [i]required[/i]; however, if you did them, the lecturers were able to provide comments and suggestions, which were very, very useful... but I didn't go as far as not sleeping to complete them.

Good to hear.  About what % of the assigned problems did you do?

[quote="cats..."]how many sleepless nights were there because of work on problem sets or something (and not cause of activities like the puzzle hunt)?[/quote]

Making HW (or going to classes) obligatory would conflict with Mathcamp's basic principle of letting you do whatever you want. You go to classes and do HW not because someone makes you but because you want to get something out of the five weeks. Also, the classes that give you homework are clearly labeled HW. Usually it's not a lot of HW, except for Moore Method classes (those are really hard anyways) and Problem Solving. Still, though, there's never near enough HW to make you stay up all night, and if there was, you could just not do it. No one's going to look at you like you're a failure if you don't (except for maybe Moore Method classes, where doing the HW is necessary for the class to proceed). I never spent more than half an hour on HW, and I only stayed up for the Puzzle Hunt and the All Nighter at the end, as well as for the tea.

The main point is that Mathcamp is not a very structured camp, unlike many others. You can go to as many classes and do as much HW and go on as many trips as you want.

btw Cats, are you going to Mathcamp this year or are you thinking of applying for next year?

Oh yeah, I will actually be at MIT for the puzzle hunt next winter, not just virtually as last time. Yay!

all of my solutions were extremely well written, and I can guarantee that 8.5 of them were correct. One was a two parter where the second part I was not 100% on, but it was still most likely correct. Anyway, what can you do? Not much I guess. I'm not really bitter, just confused and disoriented, and mostly disappointed.

May I ask when you applied?  Since I applied late as well and got in. :D

Osiris, or others who have been there before, do you have other recommendations for classes?

Also, is there some size limit to the classes?  I imagine a lot of people here are planning on taking Mr. Rusczyk's class.

i am arriving late evening (close to midnight), so afternoon doesn't sound too good.

I'm arriving late afternoon.  Hope I don't miss the BBQ lol...

Last year the BBQ was held around 6 PM.

I'll just briefly outline my schedule:

[b]Got much out of:[/b]

Dorin:
Beginning Problem Solving, weeks 1 through 5, */**
Hardcore Problem Solving, weeks 1 through 4, ***/****

Mark / Chris: Group Theory, weeks 1 and 2, **

Mira: Linear Algebra, weeks 1 through 3, **

Jeremy: Cryptography, *

I devoted weeks 4 and 5 to my project (artificial intelligence), and hence wasn't able to spend that much time on classes.

[b]Went to but understood little:[/b] (not that they're not good - I think I'm not good enough for them ;) )

Kenny's First Order Logic, ***
Mr. Soibelman's P-and-q numbers, ****

[b]Visiting faculty:[/b] (ranked according to my personal opinion, from high to low)

Blair Kelly
John Conway
Sanjoy Majahan
Josh Tenenbaum
Woody Dudley
Titu Andresscu
Serge Lang
Richard Guy

Note: what I've written above is by no means a complete listing of Mathcamp life. The yearbook, the projects, the trips, the YELLOW PIG DAY (ooh icecream), ultimate frisbee, cracks about UPS caf food, cracks about other camps, and most of all, PUZZLE HUNT, they went completely unmentioned. Experience them all!

Does anyone know how Beginning Problem Solving is? I don't know whether to take Hardcore Problem Solving or take both.

Osiris, I think you mixed up the stars for Dorin's classes...  Wouldn't the "Hardcore" Problem Solving have been the **** one?  At least, I mean, I [i]hope[/i] it would be. ;)

I don't know if its changed since I've been there (I attended in 2001 and 2002), but if you don't have much experience with Olympiad-level problem solving, prepare to be very intimidated by the problems in Dorin's harder class.  I know [i]I[/i] was in 2001.  He uses a number of problems from old Olympiads.

I [i]am[/i] going to give a HUGE recommendation for Dorin's classes (at least the harder one, which I attended when I was there in 2001 and 2002 - and will again this year).

You didn't mention the LARP!?!  Did they not have a LARP last year?!?  I hope they have one this year :).  A messed-up, interesting one that doesn't go as planned, like in 2002.

Also, if you are a glutton for [i]hard[/i] math (who isn't?  right?), you can take a Moore-method class.  Such classes involve the students being given the statements of all of the results, but having to prove (and present) them themselves.  In 2001 the Moore-method class was Ring Theory, in 2002 it was Abel's Theorem.  I heard from someone (is this correct?) that there were multiple Moore-method classes in 2003, the most notable being Topology.

cats... - AFAIK, there is no size limit, although they usually like to keep certain classes (like the Abel's theorem class in 2002) small to make sure they have the people who actually are working to keep up.  For the more popular guest-lectures, though, they usually use a large lecture hall that will fit pretty much all of Mathcamp.

I'm not quite sure why I'm writing all of this, as those to whom it is relevant will be getting the information when they get there....

See you all in Maine, in less than 24 hours (I hope...)!

My bad.  :D

I didn't attend the MOORE method classes or the LARP, so I ... yeah ... forgot. I don't know how I could have missed the signup for LARP...  :blush:

It sounds like the we (AoPsers) are arriving all throughout the day tomorrow...maybe sometime on Monday would be better to try to all meet so that we will all be there?

sry to bump this, but what grade(aops level) would mathcamp be good for?

IRL Grade: High School (for maturity reasons) - however middle schoolers can do it if they're mature ig

MathCamp is more research math oriented from what I've heard, so we really cant pinpoint it to your "aops level"

Is there like a specific math level I need to be? For example, do I need to qualify USAMO or smthn

oh geez sry about 18 year bump i didnt look

[quote=Yasha]Mostly what I like is that you can do absolutely anything. There are only 4 rules:

1. Be excellent to each other.
2. Don't do stupid stuff.
3. No fires.
4. Don't divide by 0 except under staff supervision.[/quote]

richard rusczyk said that in a video once ig


way to spam guys.

on the other hand, I heard that too, resurrection.

dude this is so not spam this is definitely relevant to the florida forum.

and ya. this lyndon kid is like sucky at everything apparently. i hear he didn't even pass the amc 10  :rotfl:


Prove these:

Theorem :"A convex quadrilateral is circumscripted at a circle of center $O$ iff it has four points satisfying the tangent property ( that is there are 4 points $X,Y,Z,W$ on its sides $DA,AB,BC,CD$ satisfying $AX=AY,BY=BZ,CZ=CW,DW=DX$) "

Lemma:"If the four points $X,Y,Z,W$ satisfy the tangent property then they are concyclic on a circle with the same center $O$"

But we know that a quadrilateral has an incenter iff teh sums of opposite sides are equal. This is somewhat stronger, I think (actually, the harder implication is stronger in this statement).

Ohp!!!!! Sorry , it was only a Lemma . :blush:

I will correct it.

I don't really understand what is now the stronger and what the weaker implication, but actually the whole problem is quite trivial:

We have to prove the following two assertions:

[b]Assertion 1.[/b] If ABCD is a quadrilateral and there exist four points X, Y, Z, W on its sides DA, AB, BC, CD such that AX = AY, BY = BZ, CZ = CW and DW = DX, then the quadrilateral ABCD has an incircle. If O is the center of this incircle, then the points X, Y, Z, W lie on one circle centered at the point O.

[b]Assertion 2.[/b] If ABCD is a circumscribed quadrilateral and X, Y, Z, W are four points on its sides DA, AB, BC, CD such that AX = AY, BY = BZ and CZ = CW, then DW = DX. And, of course, we can thus apply Assertion 1 to get the other properties of the configuration.

[hide="Proof of Assertion 1"]
[i]Proof of Assertion 1.[/i] Since AX = AY, BY = BZ, CZ = CW and DW = DX, we have AY + BY + CW + DW = AX + BZ + CZ + DX. In other words, (AY + BY) + (CW + DW) = (BZ + CZ) + (DX + AX), so that AB + CD = BC + DA. By a well-known theorem, this yields that the quadrilateral ABCD has an incircle. If O is the center of this incircle, and X', Y', Z', W' are the points where this incircle touches the quadrilateral's sides DA, AB, BC, CD, then we have AX' = AY' since the two tangents from a point to a circle are equal; since we also have AX = AY, we thus get XX' = AX' - AX = AY' - AY = YY'. Also, we have OX' = OY' (since the points X' and Y' both lie on the incircle of the quadrilateral ABCD, and the point O is the center of this incircle), and < OX'X = < OY'Y (since < OX'X = 90° and < OY'Y = 90°). Thus, the triangles OX'X and OY'Y are congruent, so that OX = OY. Similarly, OY = OZ and OZ = OW, so that OX = OY = OZ = OW, and it follows that the points X, Y, Z, W lie on one circle centered at the point O. This completes the proof of Assertion 1.[/hide]

[hide="Proof of Assertion 2"]
[i]Proof of Assertion 2.[/i] Since the quadrilateral ABCD has an incircle, a well-known theorem yields AB + CD = BC + DA. In other words, (AY + BY) + (CW + DW) = (BZ + CZ) + (DX + AX), or, equivalently, AY + BY + CW + DW = AX + BZ + CZ + DX. This, together with AX = AY, BY = BZ and CZ = CW, implies DW = DX, and Assertion 2 is proven.[/hide]

  Darij


trigonometry
inequalities
inequalities unsolved

[quote="hien"]Let $A,B,C$ be three angles of triangle $ABC$. Prove that:
$\left(\frac{\sqrt{3}}{\sin A}-1\right).\left(\frac{\sqrt{3}}{\sin B}-1\right).\left(\frac{\sqrt{3}}{\sin C}-1\right)\geq 1$[/quote]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=64246 and feel free to post a non-Jensen proof. ;)

  darij

Thanks, but what a pity! It's a nice problem but a solution is so much complicated! :?:


For a,b,c > 0 prove that $ \frac{1}{a^{2}}\plus{}\frac{1}{b^{2}}\plus{}\frac{1}{c^{2}}\ge \frac{a\plus{}b\plus{}c}{abc}$

[quote="Elio (n)"]For a,b,c > 0 prove that $ \frac {1}{a^{2}} \plus{} \frac {1}{b^{2}} \plus{} \frac {1}{c^{2}}\ge \frac {a \plus{} b \plus{} c}{abc}$[/quote]
It is homogeneous so we can asume that $ abc\equal{}1$
The $ LHS\geq \frac{1}{ab}\plus{}\frac{1}{bc}\plus{}\frac{1}{ca}\equal{}a\plus{}b\plus{}c$  :)

Expand the right side so that the statement to be proven reads as:

$ \frac{1}{a^2}\plus{}\frac{1}{b^2} \plus{} \frac{1}{c^2} \geq \frac{1}{ab}\plus{}\frac{1}{bc} \plus{} \frac{1}{ca}$

Let $ x \equal{} \frac{1}{a}$, $ y \equal{} \frac{1}{b}$, $ z \equal{} \frac{1}{c}$, so that the statement reads as:

$ x^2\plus{}y^2\plus{}z^2 \geq xy\plus{}yz\plus{}zx$
$ \iff 2x^2\plus{}2y^2\plus{}2z^2 \minus{} xy\minus{}yz\minus{}zx \geq 0$
$ \iff (x\minus{}y)^2 \plus{} (y\minus{}z)^2 \plus{} (z\minus{}x)^2 \geq 0$


This is the last question in the competition and probably the hardest one:

It is a surprising fact that 1x2x3x4x5x6=8x9x10. It is even more surprising that 8x9x10x11x12x13x14 is equal to another such product of consecutive whole numbers. What is the smallest number in this other product?

Can anyone help?

 :?:

$\begin{array}{|c|c|c|c|c|c|c|}\hline &&&&&& \\ 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline &&&&&& \\ 2^{3}& & 2 & & 2^{2}& & 2 \\ \hline &&&&&& \\ & 3^{2}& & & 3 & & \\ \hline &&&&&& \\ & & 5 & & && \\ \hline &&&&&& \\ &&&&&&7 \\ \hline &&&&&& \\ &&&11&&& \\ \hline &&&&&& \\ &&&&&13& \\ \hline \end{array}$

So $8 \cdot 9 \cdot 10 \cdot 11 \cdot 12 \cdot 13 \cdot 14=2^{7}3^{3}5^{1}7^{1}11^{1}13^{1}$

So we have to find a product of consecutive numbers with these prime divisors, not others. This means that the product shouldn't contain the number 17,19 etc

We take the multiples of 11 and the multiples of 13:
$11,13,22,26,33,39,44,52,55,65,66,77,78,88,91,99,...$

We want to find two successive terms in this sequence such that there is not between them a prime $>13$ (or a product of such a prime)
What I mean:

Between $13$ and $22$ there is the prime $17$. [color=red]Rejected[/color]
Between $22$ and $26$ there is the prime $23$. [color=red]Rejected[/color]

Continue this search and you will find that
Between $65$ and $66$ there is not any prime (not even any number). [color=blue]Accepted[/color]

So we can use the numbers 65,66 to make a sequence of consecutive numbers. But we cannot go to 67 because this is a prime $>13$

Finally we find that $63 \cdot 64 \cdot 65 \cdot 66 = (3^{2}7^{1})(2^{3})(5^{1}13^{1})(2^{1}3^{1}11^{1}) = 2^{7}3^{3}5^{1}7^{1}11^{1}13^{1}$

Many thanks. Very clever solution indeed.


1. isobutyl chloride + $ \ce{AgNO3}$ in ethanol/water
2. neopentyl bromide + $ \ce{AgNO3}$ in ethanol/water
3. [i]t[/i]-butyl bromide heated in ethanol

Is this a problem for us to solve, but that you already know how to solve, or is a problem in which you have some doubts? In the second case, please show some of your own thoughts first.

It is a problem for people to solve, and I have the answer.

[hide="Answer"]In the first two reactions, silver nitrate promotes the departure of chloride ion, leaving behind a primary carbocation, which then rearranges to a secondary (case of 1.) or a tertiary (case of 2.) carbocation. The only bases/nucleophiles present are ethanol and water, which are weak bases. Therefore, mixtures of products will be obtained: ethers (from nucleophilic attack of ethanol), alcohols (from nucleophilic attack of water), and alkenes (elimination).

For 3. you have solvolysis, and so, even with heat, substitution and elimination products are expected.[/hide]


I want to know what everybody thinks of imaginary numbers. I think they should be allowed in the country, so long as they don't take our jobs. What do you think?

Billy

I think we need a mathnic cleansing; all imaginery are to be executed.

I think that imagination is ok, as long as people don't take it too seriously. On the other hand, I think we should put irrational numbers in mental institutions, because they are a danger to us and to our society.

Imaginary numbers are awesome, they make great friends!

I'm sorry to say that that comment has forced me to label you irrational.

Off to the asylum you go.

But I wasn't even talking about irrational numbers.  You like imaginary numbers too.

Yes, and that is why I'm in a strait-jacket.

Well, I'm not going to the asylum, because I don't want to (imaginary numbers are still cool).

You [b]are[/b], whether you like it or not. So there!

Hmm... are complex numbers proof of intelligent design? Especially those irreducible complex numbers, like i+1...

I think imaginary numbers should be put in jail to make math and life easier.

As a representative of the imaginary numbers, I believe that REAL numbers should be put in jail to make life easier. No more calculations involving numbers like 1 which don't have any imaginary life significance!

No one answered my question, which could be of major scientific relevance.

Does the existence of irreducible complex numbers (such as 1+i) imply Intelligent Design? I think it does. Furthermore, I think that since Intelligent Design requires God to be rational (hence the qualifier intelligent), we must live in a universe where God is a rational number. Thus, anyone who believes in irrational or imaginary numbers should be burned, because they are in league with the Devil! Burn them!!!! Muahahahaha!!!!!!!!!!

But if imaginary numbers don't exist, then complex numbers don't exist, so they cannot be used as a proof of Intelligent Design. Hence ID is not true, meaning there is no God. So there is no life after death, meaning all of life is pointless, so I might as well kill myself.

*commits suicide*

See, that's what happens when you ignore imaginary numbers.

I'm not saying that we should ignore them, I'm saying we should declare war on them! (Big difference, you know what I mean?)

maybe we could just execute abs() on the imaginary space to make it real? :?

Yes! But we should take it one step further: abs(dictatorship)!!!!!!!!!!!!!!!!!!!!!!!!11!!1one!!!won@!@@##!@2!~~`122@21

or we can just start a game of mafia here until the next game is launching :)

What about quaternions? I think we should let them into the country.

Why would you want to declare war on imaginary numbers?  They are AWESOME!!!


The measure of a given angle is 180degrees/n , where n is a positive integer
not divisible by 3. Prove that this angle can be trisected by Eucliden means
(straightedge and compass).

guys i just fixed a typo...take a look at it again. thanks.

Em, what about $ n \equal{} 2$? I think you mean when $ 0 \not\equiv \frac {180}{n} \pmod 3$

What about it?  A 90 degree angle is trisectable.

Sorry, I read it as "can't" be trisected.


complex numbers
real analysis
real analysis unsolved

$u_{n}\in l^{1}(\mathbb{N})$ a sequence of complex numbers such that $\sum_{N}u_{n}^{k}= 0$ for every $k \in \mathbb{N}^{*}$. Does it follow that the sequence is null ? And if one does not assume that $\sum |u_{n}| < \infty$ ?

The first one was already discussed in a thread by Omid Hatami. There exist sequences that are not zero if we do not assume they are in $l^{1}(R)$, but I will let the people thinking before posting a proof.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=entire&t=101156]This[/url] is the topic harazi mentioned (there's only a proof for the first part there).


logarithms
inequalities
calculus
calculus computations

Is it possible to prove that the sequence

$a_{n}=\ln(1+a_{n-1})$, $a_{1}=10$

converges to 0?

Well, $0$ is the only fixed point of $f(x) = \ln (1+x)$.  Moreover, the sequence is clearly monotonically decreasing (Bernoulli's inequality or what have you) and bounded below.


[url=http://artofproblemsolving.com/Community/AoPS_Y_IndBlog.php?blog_id=1233]Complex Zeta's Blog[/url]

A four page proof of the four color theorem?  Hoorah!

I would like to see such a proof. I am not specialist in Graph Theory, but Thomas' proof of the four-color problem is not that small (and far from trivial) but it is the shortest one so far (that hasn't been disproved :P )

Every year there are a few new (flawed) hand proofs of 4CT.  Perhaps ComplexZeta will post a bit more detail, such as which equivalent problem was supposedly solved, and such?

By the way, Roberton/Seymour/Thomas are working on (or will soon be working on) a hand proof of 4CT; they have an NSF collaborative grant to do so.  I think that's a more promising project.


Ron has eight sticks, each having an integer length. He observes that he cannot form a triangle using any three of these sticks as side length. The shortest possible length of the longest of the eight sticks is:

(A) 20     (B)21     (C)22     (D)23     (E)24

Not too hard but can't find good process to achieve the answer

[hide]In a triangle, the sum of two sides has to be greater than the third, so you can use the Fibonacci sequence:

1,1,2,3,5,8,13,21

So the answer is B[/hide]


geometry
conics
hyperbola
Asymptote
geometric transformation
reflection
trigonometry

An acute angle is given (with vertex $ A$). Construct the $ ABC$ triangle with the smallest area where $ P\in\overline{BC}$, $ B$ is on one of the arms of the angle, $ C$ is on the other one.

A tangent to a hyperbola with the given asymptotes $ Ax,\ Ay$ (rays the $ \angle xAy$) cuts its asymptotes at $ B,\ C.$ For any tangent to a fixed hyperbola, the area of the $ \triangle ABC$ is constant. For the area of the $ \triangle ABC$ to be minimum, there must be only one tangent from the point $ P,$ i.e., $ P$ actually lies on the hyperbola and it is the tangency point.

[i]1st solution:[/i] The tangency point $ P$ is the midpoint of the segment $ BC$ cut on the hyperbola asymptotes $ Ax,\ Ay.$ Using a parallel projection, project the ray $ Ay$ into $ Ay',$ such that the $ \angle xAy' \equal{} 90^\circ$ is right and the projection $ P'$ of $ P$ is on the angle bisector of this right angle. The ray $ Ax$ stays in place in this projection. To find direction of this parallel projection, draw a parallel to $ Ax$ through $ P$ intersecting $ Ay,\ Ay'$ at $ Q,\ S$ and reflect $ Q$ into $ R$ in the midpoint of $ PS,$ yielding $ PR \equal{} QS.$ Perpendicular to $ Ax$ through $ R$ meets angle bisector of the right $ \angle xAy'$ at $ P'$ and $ PP'$ is the direction of the parallel projection. The projected problem is trivial: the $ \triangle ABC'$ with $ P' \in BC'$ and the minimum area is isosceles right, yielding $ B, C'$ ($ Ax$ stays fixed !). A parallel to $ PP'$ through $ C'$ (ray of the parallel projection) cuts the original ray $ Ay$ at $ C$ (and $ P$ is of course the midpoint of $ BC$).

[i]2nd solution:[/i] Putting the x-axis into the bisector of the $ \angle xAy,$ the hyperbola $ \mathcal H$ equation is $ \frac {x^2}{p^2} \minus{} \frac {y^2}{p^2 \tan^2 \frac {A}{2}} \equal{} 1.$ Since $ P \in \mathcal H,$ $ p^2 \equal{} x_P^2 \minus{} \frac {y_P^2}{\tan^2 \frac {A}{2}}.$ Drop a normal $ y_P \equal{} PR$ from $ P$ to the x-axis (bisector of the $ \angle A$) and draw a parallel to the ray $ Ax$ intersecting the x-axis at $ Q,$ $ RQ \equal{} \frac {y_P}{\tan \frac {A}{2}}.$ A circle with diameter $ x_P \equal{} AR$ intersects a circle centered at $ R$ and with radius $ RQ$ at $ S,\ S',$ giving $ p \equal{} AS,$ the hyperbola semimajor axis. A circle $ (A)$ centered at $ A$ and with radius $ p \equal{} AS$ is the hyperbola pedal circle, intersecting the x-axis at $ U,\ V,$ the hyperbola vertices. A normal to the x-axis at $ U$ intersects the rays $ Ax,\ Ay$ at $ W,\ W'$ and a circle centered at $ A$ and with radius $ AW \equal{} AW'$ intersects the x-axis at $ E,\ F,$ the hyperbola focal points. A circle with diameter $ PF$ (where $ F$ is the hyperbola focus more distant from $ P$) touches the pedal circle $ (A)$ at $ T,$ because $ BC$ is a hyperbola tangent at $ P$ and the $ \angle PTF \equal{} 90^\circ$ is right. $ PT$ meets the hyperbola asymptotes $ Ax,\ Ay$ at $ B,\ C$ and again, $ P$ is the midpoint of $ BC.$

Look here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=171914

EDIT: Oh, I'm sorry, [b]yetti[/b]!

That was a different problem, you were supposed just to [b]prove[/b] that P bisects BC, not to [b]construct[/b] the minimum area triangle. Once you realize that BC is tangent to a hyperbola with the given asymptotes Ax, Ay at P, it is always possible to project the $ \triangle ABC$ by a parallel projection (which preserves segment ratios on a line) to an isosceles or even an isosceles right triangle with $ AB \equal{} AC',$ which puts the projected tangency point P' at the projected hyperbola vertex, and P' being the midpoint of BC' follows just by symmetry.

I think that we are overcomplicating things...

We can just reflect $ A$ in $ P$ to $ A'$ and draw lines parallel to the rays through $ A'$ to make a parallelogram.

[color=darkred][b][u]Modified enunciation.[/u][/b] Given are an acute angle $ \widehat {XOY}$ and an its interior fixed point $ F$. Ascertain the positions of the points $ \|\begin{array}{c} M\in (OX \\
\ N\in (OY\end{array}$ for which $ F\in MN$ and the area of the triangle $ MON$ is minimum.[/color]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ \|\begin{array}{c} A\in (OX\ ,\ FA\parallel OY \\
 \\
B\in (OY\ ,\ FB\parallel OX\end{array}$ and $ \|\begin{array}{c} [OAFB] = 2a \\
\ [FAM] = x \\
\ [FBN] = y\end{array}$

Observe that $ [FOA] = [FOB] = a$ and $ \triangle FAM\sim\triangle NBF$.

Thus, $ \frac {[FOM]}{[FON]} = \frac {FM}{FN} = \sqrt {\frac {[FAM]}{[FBN]}}$ $ \implies$ $ \frac {x + a}{y + a} = \sqrt {\frac xy}$ $ \Longleftrightarrow$

$ (x + a)\sqrt y = (y + a)\sqrt x$ $ \Longleftrightarrow$ $ (\sqrt x - \sqrt y)(a - \sqrt {xy}) = 0$ $ \Longleftrightarrow$

$ x = y$ or $ xy = a^2$ ([b]constant[/b]). Therefore, $ [MON]$ is minimum $ \Longleftrightarrow$

$ x + y$ is minimum $ \Longleftrightarrow$ $ x = y = a$ $ \Longleftrightarrow$ $ FM = FN$, i.e. $ \|\begin{array}{c} AM = OA \\
\ BN = OB\end{array}$ .[/color]


function
calculus
derivative
integration
Functional Analysis
real analysis
real analysis unsolved

Let $ C^{0,1/2} \equal{} \{f: \mathbb{R}\to\mathbb{R} | \sup_{x\in\mathbb{R}}|f(x)| \plus{} \sup_{x\neq y}\frac {|f(x) \minus{} f(y)|}{|x \minus{} y|^{1/2}} < \infty \}$.

Show that $ f\in C^{0, 1/2}$ iff there exists $ C > 0$ such that for all $ \varepsilon > 0$ there exists bounded $ \phi\in C^\infty(\mathbb{R})$ satisfying:

$ \sup_{x\in\mathbb{R}}|f(x) \minus{} \phi(x)|\leq\sqrt {\varepsilon}$ and $ \sup_x \sqrt {\varepsilon}|\phi ' (x)|\leq C$.

It seems to me that what is being called $ C^{0,1/2}$ is actually the same as $ Lip(1/2)$ which makes it a Banach space with the thing that is less than infinity as the norm on the space. This line of reasoning didn't provide fruitful.

The other thing I noticed was how suspiciously similar $ \sup_{x\neq y}\frac {|f(x) \minus{} f(y)|}{|x \minus{} y|^{1/2}} \equal{} M < \infty$ is to $ \sup_{x\in\mathbb{R}}|f(x) \minus{} \phi(x)|\leq\sqrt {\varepsilon}$ making me think that $ \phi$ has is going to look something like: $ \phi(x) \equal{} \sup_{y\in B_\varepsilon(x)}|f(x) \minus{} f(y)|$ (with the ball a deleted neighborhood) since clearly we have that both individual sup's being finite. Unless this is just going to be brute force, that wasn't leading anywhere either.

I've seen that condition (with various exponents) called Holder continuity.

The usual way to generate $ C^{\infty}$ approximations is to convolve with a bump. What does that do here? Assuming uniform continuity, if $ |f(x) \minus{} f(y)| < \delta$ when $ |x \minus{} y| < \epsilon(\delta)$ and the bump $ \theta$ is supported in $ [ \minus{} \delta,\delta]$, $ |\theta*f(x) \minus{} f(x)| < \epsilon$ for all $ x$. Assuming boundedness ($ |f(x)|le M$), $ (\theta*f)' \equal{} \theta'*f$ and $ (\theta*f)'(x)\le M\|\theta'\|_1$. In practice, this $ \theta$ is $ \theta(x) \equal{} \frac1{\delta}\theta_1\left(\frac {x}{\delta}\right)$ - a scaled version of a standard bump. This makes $ |\theta'|_1 \equal{} \frac1{\delta}|\theta_1|_1$

We want our $ \phi$ to be $ f*\theta$. What do the bounds look like? We have $ |f \minus{} \phi|_\infty\le \epsilon(\delta)$ and $ |\phi'|_\infty\le \frac {K}{\delta}$ for some constant $ K$. For this estimate to scale right, we would need $ \epsilon(\delta)\le A\delta$, a Lipschitz condition. That's far too strong.
It is at least a start- we can easily show that boundedness and uniform continuity are necessary, and the above shows that Lipschitz is sufficient.

Now, how can we improve the estimate? There's not much we can do for the estimate of $ |f \minus{} \phi|$, but the derivatives have a lot of room to work with. Since $ \theta'$ takes on both large negative and large positive values, we can expect lots of cancellation. In particular, $ \int \theta' \equal{} 0$, so adding a constant to $ f$ doesn't change $ \phi'$. If $ g(x) \equal{} f(x) \minus{} f(x_0)$, $ (g*\theta)' \equal{} (f*\theta)'$, and $ (g*\theta)'(x_0)\le \epsilon|\theta'|_1$, estimating $ g$ by its supremum in the relevant region.

This gives $ |f \minus{} \phi|_\infty\le \epsilon(\delta)$ and $ |\phi'|_\infty\le \epsilon(\delta)\frac {K}{\delta}$; if $ \epsilon(\delta)\le A\sqrt {\delta}$, this estimate is good enough. The uniform Holder condition is thus sufficient.

There's another part: showing that the uniform Holder condition is necessary. I'll leave that for now. [Edit- OK, it's easy. I'll leave it anyway.]


Let A,B,C be angles of a Triangle, solve this equation:
$sin^2A+sin^2B=\sqrt[2006]{sinC}$

 :)  :)

[quote="MrWise"]$sin^2A+sin^2B=\sqrt[2006]{sinC}$

 :)  :)[/quote]

well.....you didn't state the relation between $A, B$ and $C$
But i guess they're angles of a triangle, no?

Absolutely!
A,B,C are angles of a triagle.

[quote="MrWise"]So, what is the solution?   :?:[/quote]

But you didn't state your problem clearly!

[quote="MrWise"]Let A,B,C be angles of a Triangle, solve this equation:
$sin^2A+sin^2B=\sqrt[2006]{sinC}$[/quote]

Is it clear?  ;)

[b]Partial solution[/b]

When $C=\frac \pi 2$
The equation becomes $\sin^2 A + \sin^2 B = 1$ with $A+B = \frac \pi 2$, $A,B>0$.
The equality holds for all (A,B) with $A+B = \frac \pi 2$ and  $A,B>0$.

When $C < \frac \pi 2$
Then $\frac \pi 2 < A + B < \pi$.
$\sin C < 1 \implies$ RHS < 1.
Suppose $A > B$. Then $B < \frac \pi 2$.
Let $A+B = \frac \pi 2 + \delta$.
LHS
 $=\sin^2 A + \sin^2 B = \sin^2(\frac \pi 2 -B + \delta)+\sin^2 B$
$> \sin^2( \frac \pi 2 -B)+\sin^2 B = \cos^2 B + \sin^2 B = 1$
There are no solutions in this case.

When $C> \frac \pi 2$
I have no conclusion yet.  :P

[quote="MrWise"]$\sqrt[2006]{SinC}$
$\Rightarrow 0 < SinC \le 1$
$\Rightarrow 0 < C \le \frac{\pi}{2}$

So we have conclustion:
$C=\frac{\pi}{2}$
$0<A,B<\frac{\pi}{2}$ and $A+B=\frac{\pi}{2}$[/quote]

Hard to understand.
WHAT IF $\sin C = \frac 1 2$ ? I think $C=150^0$ is also possible????

I don't think so. $A,B,C$ are angles of a triangle and so if $C=150^o$ we'll get RHS greater than LHS. ;)

I don't mean that.
Look at the implication $0< \sin C \leq 1 \implies 0< C\leq \frac \pi 2$
This is not true in general.
At least there should be some explanations for this line. :)

[quote="luimichael"][b]Partial solution[/b]
$\sin^2(\frac \pi 2 -B + \delta)+\sin^2 B$
$> \sin^2( \frac \pi 2 -B)+\sin^2 B$ [/quote]
That's not always true: if $sin^2(\frac \pi 2 -B + \delta)$ and $\sin^2( \frac \pi 2 -B)$ are both greater than $\frac\pi 2$ then we'll get the opposite result.

You are right.
The statement only holds for a certain range of values of $\delta$.
But I am wondering why the author of this thread used that title ?
********************************************************************
For the case $C > \frac \pi 2$
[hide]It can be shown that $\sin^2A+\sin^2B-\sin^2C=2 \sin A \sin B \cos C$ (by Sine Rule and Cosine Rule.)
The equation can be re-written as $2\sin A\sin B\cos C = \sqrt[4012]{\sin^2 C} -\sin^2C$.
In this case, simply note that LHS < 0 and RHS > 0.
Therefore, there are no solutions in this case.[/hide]


[quote][color=darkred][b][u]"Very strong" extension (last hour).[/u][/b] Let $ ABCD$ be a convex quadrilateral. Define

$ \{\begin{array}{ccc} E\in AB\ \ \ \ ,\ \ \ \ F\in DC & \mathrm {\ \ so\ that\ \ } & \frac {\overline {EA}}{\overline {EB}} = \frac {\overline {FD}}{\overline {FC}} \\
 \\
G\in AD\ ,\ I\in EF\ ,\ H\in BC & \mathrm {\ \ so\ that\ \ } & \frac {\overline {GA}}{\overline {GD}} = \frac {\overline {IE}}{\overline {IF}} = \frac {\overline {HB}}{\overline {HC}}\end{array}$

and $ \{\begin{array}{c} U\in DE\cap BG \\
 \\
V\in BF\cap DH\end{array}$ . Prove that  $ \{\begin{array}{c} I\in AV\cap CU\cap GH \\
 \\
AU\cap CV\cap BD\ne\emptyset\end{array}$ .[/color][/quote]
[color=darkblue][b][u]Remark I.[/u][/b] If denote $ X\in AU\cap CV\cap BD$ , then $ Y\in EG\cap HF\cap BD$ and

$ X$ , $ Y$ are (evidently, I say !) harmonical conjugate w.r.t. the pair of points $ \{B,D\}$ .

[b][u]Remark II.[/u][/b] See the my "easy (old)" and "strong (new)" extensions from [/color]

http://www.mathlinks.ro/Forum/viewtopic.php?t=182893

This time, I post at first the schema and the solution later.

I think that the difficult (but well known) part of this nice problem, is the collinearity of the points $ G,$ $ I,$ $ H.$

Two elementary proofs of this result, have already been posted before in this forum.

Kostas Vittas.

[color=darkblue][b]Thank you, Vittasko,[/b] for the your nice remark $ I\in GH$ and the upper drawing of the figure.

I added the your remark in the enunciation of the my problem. Now it is more nicely ![/color]

[quote][color=darkred]Let $ ABCD$ be a convex quadrilateral and the points

$ \begin{array}{ccc} E\in AB\ \ \ \ ,\ \ \ \ F\in DC & \mathrm {\ \ so\ that\ \ } & \frac {\overline {EA}}{\overline {EB}} \equal{} \frac {\overline {FD}}{\overline {FC}}\ \ ; \\
 \\
G\in AD\ ,\ I\in EF\ ,\ H\in BC & \mathrm {\ \ so\ that\ \ } & \frac {\overline {GA}}{\overline {GD}} \equal{} \frac {\overline {IE}}{\overline {IF}} \equal{} \frac {\overline {HB}}{\overline {HC}}\ \ .\end{array}$

Define the points  $ \begin{array}{c} U\in DE\cap BG \\
 \\
V\in BF\cap DH\end{array}$ . Prove that  $ I\in AV\cap CU\cap GH$ and $ AU\cap CV\cap BD\ne\emptyset$ .[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b]

$ 1\blacktriangleright$ Let $ J$ , $ K$ be the points for which the quadrilaterals $ ADJE$ , $ BCKE$ are parallelograms.

Denote $ \begin{array}{c} L\in EJ\ ,\ GL\parallel AE \\
 \\
M\in EK\ ,\ HM\parallel BE\end{array}$ $ \implies$ $ \boxed {\begin{array}{ccccc} DJ\parallel KC & , & \frac {FD}{FC} \equal{} \frac {DJ}{KC} & \implies & F\in JK \\
 \\
\frac {IE}{IF} \equal{} \frac {LE}{LJ} & , & \frac {IE}{IF} \equal{} \frac {ME}{MK} & \implies & \begin{array}{c} I\in LM \\
 \\
\overline {LIM}\parallel JK\end{array} \\
 \\
GL\parallel MH & , & \frac {GL}{MH} \equal{} \frac {IL}{IM} & \implies & \boxed {I\in GH}\end{array}}\ .$

$ 2\blacktriangleright$ Aplly the  [b]Menelaus' theorem[/b] to the transversals :

$ \boxed {\begin{array}{c} \overline {BGU}\ \mathrm {si}\ \triangle ADE\ \ : \ \frac {\overline {BE}}{\overline {BA}}\cdot\frac {\overline {GA}}{\overline {GD}}\cdot \frac {\overline {UD}}{\overline {UE}} \equal{} 1\Longrightarrow \frac {\overline {CF}}{\overline {CD}}\cdot\frac {\overline {UD}}{\overline {UE}}\cdot\frac {\overline {IE}}{\overline {IF}} \equal{} 1\Longrightarrow I\in CU \\
 \\
\overline {DHV}\ \mathrm {si}\ \triangle BCF\ \ : \ \frac {\overline {DF}}{\overline {DC}}\cdot\frac {\overline {HC}}{\overline {HB}}\cdot \frac {\overline {VB}}{\overline {VF}} \equal{} 1\Longrightarrow \frac {\overline {AE}}{\overline {AB}}\cdot\frac {\overline {VB}}{\overline {VF}}\cdot\frac {\overline {IF}}{\overline {IE}} \equal{} 1\Longrightarrow I\in AV\end{array}\Longrightarrow \boxed {I\in AV\cap CU}\ }\ .$

$ 3\blacktriangleright$ Denote $ \begin{array}{c} X\in AU\cap BD \\
 \\
Y\in CV\cap BD\end{array}$ .Aplly the [b]Ceva's theorem[/b] to the points :

$ \boxed {\begin{array}{c} U\ \ \mathrm {and}\ \ \triangle ABD\ : \ \frac {\overline {XB}}{\overline {XD}}\cdot\frac {\overline {GD}}{\overline {GA}}\cdot\frac {\overline {EA}}{\overline {EB}} \equal{} \minus{} 1 \\
 \\
V\ \ \mathrm {and}\ \ \triangle BCD\ : \ \frac {\overline {YB}}{\overline {YD}}\cdot\frac {\overline {FD}}{\overline {FC}}\cdot\frac {\overline {HC}}{\overline {HB}} \equal{} \minus{} 1\end{array}\Longrightarrow \frac {\overline {XB}}{\overline {XD}} \equal{} \frac {\overline {YB}}{\overline {YD}}\Longrightarrow X\equiv Y\Longrightarrow \boxed {AU\cap CV\cap BD\ne\emptyset}\ }\ .$[/color]


LaTeX
inequalities unsolved
inequalities

[quote="Extremal"]http://ivanisvili.narod.ru/Extr1.doc[/quote]

ok people. try to find when p=2, and n=2,3.

rewrite by latex:
$\frac{\sum_{j=1}^{m}(\sum_{i=1}^{n}a_{ji}x_{i})^{p}}{|\sum_{i=1}^{n}x_{i}^{p}|}\to max,min$
$x_{i}$ is not equal to $0$


http://www.youtube.com/watch?v=a81YvrV7Vv8&feature=related

 :huh:


geometry
trapezoid
incenter
rectangle
trigonometry
angle bisector
perpendicular bisector

Let $ABC$ be an isosceles triangle with $AB=AC$. Consider a variable point $P$ on the extension of the segment $BC$ beyound $B$ (in other words, $P$ lies on the line $BC$ such that the point $B$ lies inside the segment $PC$). Let $r_{1}$ be the radius of the incircle of the triangle $APB$, and let $r_{2}$ be the radius of the $P$-excircle of the triangle $APC$. Prove that the sum $r_{1}+r_{2}$ of these two radii remains constant when the point $P$ varies.

[i]Remark.[/i] The $P$-excircle of the triangle $APC$ is defined as the circle which touches the side $AC$ and the [i]extensions[/i] of the sides $AP$ and $CP$.

Strangely enough, nobody posted a solution to this one until now, so let me do:

[color=blue][b]Problem.[/b] Let ABC be an isosceles triangle with AB = AC, and let T be the midpoint of its base BC. Let P be a point on the extension of the segment BC beyound B. Let $r_{1}$ be the radius of the incircle of triangle APB, and let $r_{2}$ be the radius of the excircle of triangle APC opposite to the vertex P. Prove that $r_{1}+r_{2}=AT$.[/color]

[i]Solution.[/i] Let the incircle of triangle APB have the center $O_{1}$ and touch the side PB of this triangle at $Q_{1}$. Then, since $r_{1}$ is the radius of this incircle, we have $O_{1}Q_{1}=r_{1}$ and $O_{1}Q_{1}\perp PB$. Let the excircle of triangle APC opposite to the vertex P have the center $O_{2}$ and touch the extended side PC of this triangle at $Q_{2}$. Then, since $r_{2}$ is the radius of this excircle, we have $O_{2}Q_{2}=r_{2}$ and $O_{2}Q_{2}\perp PC$.

The relations $O_{1}Q_{1}\perp PB$ and $O_{2}Q_{2}\perp PC$ rewrite as $O_{1}Q_{1}\perp BC$ and $O_{2}Q_{2}\perp BC$, so that $O_{1}Q_{1}\parallel O_{2}Q_{2}$. Thus, the quadrilateral $O_{1}O_{2}Q_{2}Q_{1}$ is a trapezoid with its bases $O_{1}Q_{1}$ and $O_{2}Q_{2}$ being perpendicular to the line BC. Hence, if M and N are the midpoints of the legs $O_{1}O_{2}$ and $Q_{1}Q_{2}$ of this trapezoid, then the line MN is parallel to its bases $O_{1}Q_{1}$ and $O_{2}Q_{2}$, and thus perpendicular to the line BC, and we have $MN=\frac{O_{1}Q_{1}+O_{2}Q_{2}}{2}$. In other words, $MN=\frac{r_{1}+r_{2}}{2}$, so that $r_{1}+r_{2}=2\cdot MN$.

Being the incenter of triangle APB, the point $O_{1}$ lies on the angle bisector of the angle APB. Being the excenter of triangle APC opposite to the vertex P, the point $O_{2}$ lies on the angle bisector of the angle APC. Actually, these two angle bisectors are one and the same line - the angle bisector of the angle formed by the rays PA and PB = PC. Since the points $O_{1}$ and $O_{2}$ both lie on this angle bisector, the midpoint M of the segment $O_{1}O_{2}$ must also lie on this angle bisector; hence, < APM = < TPM.

Now, we can describe our situation as follows: The points A, B, C are chosen on the sides of an angle with the vertex P such that the point B lies between the points P and C, and the point A lies on the other side of the angle. The point $O_{1}$ is the incenter of triangle APB, and the point $O_{2}$ is the excenter of triangle APC opposite to the vertex P. Then, according to the Remark at the end of [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=190381#p190381]http://www.mathlinks.ro/Forum/viewtopic.php?t=30389 post #2[/url], since AB = AC, we have $O_{1}A=O_{2}A$. Thus, the triangle $O_{1}AO_{2}$ is isosceles, and the midpoint M of its base $O_{1}O_{2}$ must be the foot of its altitude from its apex A; in other words, $AM\perp O_{1}O_{2}$, so that < AMP = 90°. On the other hand, the triangle ABC is isosceles with AB = AC; hence, the midpoint T of its base BC is the foot of its altitude from its apex A; thus, $AT\perp BC$, so that < ATP = 90°. From < AMP = 90° and < ATP = 90°, it follows that the points M and T lie on the circle with diameter AP. Now, since < APM = < TPM, the chords AM and TM have equal chordal angles in this circle. Hence, these chords must have equal length: AM = TM. Thus, the triangle AMT is isosceles. Hence, the midpoint W of the base AT of this triangle is the foot of its altitude from its apex M; in other words, $MW\perp AT$.

Now, the quadrilateral MNTW has three right angles: < MNT = 90° (since $MN\perp BC$), < NTW = 90° (since $AT\perp BC$) and < TWM = 90° (since $MW\perp AT$). Thus, it is a rectangle, so that MN = WT. But since the point W is the midpoint of the segment AT, we have $2\cdot WT=AT$.

Hence, $r_{1}+r_{2}=2\cdot MN=2\cdot WT=AT$, and the problem is solved.

Note that the problem is equivalent to the one discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=18930 .

  Darij

here is my solution:

  Every time an isosceles triangle is mentioned without a shape limitation,

  it becomes our international responsibility to degenerate it to a segment

  by reducing the base under it to zero.  Once this is done here, we see

  that the circles have orthogonal internal and external tangents, and with

  this acknowlegement comes the realization that the sum of the radii in

  question stops being questionable and  becomes fixed.

  Thank you.

   M.T.

 @ Darij:   T is for   Tolouse-or-not-to-loose-a-traction.

Ι  have a trig solution.

Let $k=\widehat{APC}$
From the triangle $ADP$ we  have that  $PA=\frac{AD}{sin2k}$.

Also $PD=\frac{ADcos2k}{sin2k}$

So after some algebra we have that
$PA+PD=AD\frac{ 1}{tgk}$

So $r_{1}+r_{2}=(PA+PD)cotk=tgk\cdot AD\cdot \frac{1}{tgk}=AD=ct$

Coming at Darij's solution, with his notations, to see that $M$ lies on the perpendicular bisector of $AT$, there is an easier way: $APTM$ is cyclic, $PM$ the angle bisector of $\angle APT\implies AM=MT$, q.e.d.

Best regards,
sunken rock

[u]See[/u] [b]PP6[/b] from [url=http://www.artofproblemsolving.com/blog/56429][color=red][b]here[/b][/color][/url].


Is  there  any proof of this :

If $ p_1,p_2,p_3,...........p_n$ are prime than $ p_1 p_2 p_3..........p_n \plus{} 1$ is also prime?

Uh I'm not sure this is necessarily true.
For example if u took four primes, 5,7,11,13, 5(7)(11)(13) + 1 = 5006 which is not prime. In fact, for any primes p1, p2, p3... pn where none of the primes are even, p1p2p3...pn + 1 will always be composite. This is because that product will be odd so that odd product plus one will be even and greater than 2, obviously not prime.
Even if 2 is one of the primes it's not necessarily true either: 2*7 + 1 = 15.

Unless you are referring to that proof for the existence of an infinite amount of primes. Then it is like this, assuming the set of prime numbers is finite, take the product of all the primes and add one. However, this number p1p2p3...pn + 1 cannot be divided by any of those primes (it will always be 1 mod pi), and by the definition of a prime number p1p2p3...pn + 1 is a prime number. Therefore there are an infinite amount of primes.


I wanted to study Thermodynamics and Fluid mechanics, but i didn't have a good books to study, could you please recomended me the books, I needed it to prepare my national physics olympiad

Well,
If you want to prepare for Physics olympiad. Better check books used at undergraduate level for the concepts.
Then do problem books like Irodov or Krotov. I dont know whether they will be available at your place.
Most important of all is to solve as many of previous papers as possible.

I do think [url=http://cid-2b244a38c6f3a633.skydrive.live.com/browse.aspx/.Public/Thermodyn%20%5E0%20Fluid%20mechanics]these e-books[/url] will help you.

Thanks, for your help :)  , btw do you follow the international physics olympiad?

I think [url=http://cid-2b244a38c6f3a633.skydrive.live.com/self.aspx/.Public/Upgrade%5E_Your%5E_Physics.pdf]this[/url] might also be useful.  :roll:

I think it suffices to refer to general physics books used by undergarduates...the thermodynamics and fluid mechanics concepts there are enough for the olympiad..it's just that the books on thermo and FM would just take your time...and i don't think hardcore topics like VLE, solution thermo, mixing, etc would appear in the exam...but if you insist, i recommend those thermo and FM books used by engineering students especially ChE's..

What about a collection of problems on these 2 areas? Are there any really good books?

Helpful Texts:
Problems and Solutions on Thermodynamics and Statistical Mechanics by Yung-Kuo Lim
Schaum's Outline of Thermodynamics With Chemical Applications
Schaum's Outline of Thermodynamics for Engineers
Schaum's Outline of Fluid Mechanics and Hydraulics
Schaum's Outline of Fluid Dynamics


geometry
rectangle
analytic geometry
modular arithmetic
linear algebra
matrix
induction

We are given an $m\times n$ array filled with numbers s.t. no matter which two rows and columns we choose, the sum of the two numbers located at opposite vertices of the rectangle formed is equal to the sum of the other two numbers (also at opposite vertices). Some numbers are erased, but from what's left we can still tell what the numbers initially were. What's the minimal number of cells which are non-empty?

This is not original statement ;)
Where do you take "All-Soviet-Union" problems?

From [url=http://members.tripod.com/~PertselV/RusMath.html]here[/url]. There was another point, a), stating the same problem for $(m,n)=(2,10)$, but I skipped over that :).

In original statement you is asked to prove that minimal number of non-empty cells is m+n-1.

Oh.. Well, if you browsed that website you must have seen that the statement there doesn't mention $m+n-1$, so it's not my fault that this wasn't accurate :).

Taking the entries with at least one of coordinates $1$ we achieve $m+n-1$.
Now we prove $m+n-2$ don't suffice.
Cal an array satisfying the conditions of the problem a good array.
Now reduce the arrays mod 2 (we reduce the addition mod 2 too).
 I claim there are exactly $2^{m+n-1}$ good arrays.
Indeed every good array is bijectively defined by the entries which have at least one coordinate 1.
Now if there are at most $m+n-2$ entries from which we can recover the array there would be at most $2^{m+n-2}$ good arrays mod 2 since there are at most $m+n-2$ entries defining all the table
QED

And again I am ready to say nonsense, but...  :blush:

You assume that
1) array contains integer numbers
2) array can be recovered only from cells positions regardless numbers in them.

Maybe there are some particular m+n-2 real numbers s.t. array can be recovered from m+n-2 cells.

I can see where you're going, Iura, but the numbers need not be integers, so I don't know what $\pmod 2$ means :). Anyway, here's my argument:

The condition resembles that of matrices of rank $1$, except it's for addition instead of multiplication. Just like a matrix of rank $1$ has the property that there are $m$ numbers $x_i$ and $n$ numbers $y_j$ s.t. $a_{ij}=x_iy_j,\ \forall (i,j)\in (\overline{1,m},\overline{1,n})$, our matrix has the property that there are $m$ numbers $u_i$ and $n$ numbers $v_j$ s.t. $a_{ij}=u_i+v_j$. This would need a proof in a contest, but it's not hard to prove.

Assume WLOG $a_{11}=u_1+v_1$ is given. From the given numbers we need to find $u_i+v_j$ for all pairs $(i,j)$, so we need to find all $u_i-u_1,\ v_j-v_1,\ (i,j)\in(\overline{2,m},\overline{2,n})$. This means that we need at least $m+n-2$ linear equations of the type $u_i-u_1+v_j-v_1=k$, which means that we need at least $m+n-2$ other numbers (except for $u_1+v_1=a_{11}$).

Is this Ok? I'm not very sure about it.

Why don't you both accept my solution?
I mean: suppose we have $m+n-2$ cells $c_1 \cdots c_{m+n-2}$ such that the number from any cell can be expressed (linearly) by using the numbers in these cells.
Then if we consider all good arrays mod 2 there are at most $2^{m+n-2}$ arrays since every good array is defined by the sequence of numbers written in
$c_1 \cdot c_{m+n-2}$ and there are $2^{m+n-2}$ different sequences of
$m+n-2$ numbers $mod$ $2$.
Contradition, since as I've shown before there are exactly $2^{m+n-1}$ good arrays qith entries mod 2.

Iura! I can accept your solution with small uncertainty. ;)
Official solution uses induction on (n+m), but it is not absolutely clear for me too.


Find all rational solutions to $ x^4 \plus{} y^4 \equal{} 17$

We will try to find one solution.
Let $ x = \frac{p_1}{q_1}$ and $ y = \frac{p_2}{q_2}$ and let $ p_1, p_2 \in \mathbb{Z}$ and $ q_1,q_2 \in \mathbb{N}$
Then we have:
$ p_1^4q_2^4 + p_2^4q_1^4 = 17q_1^4q_2^4$
This means that $ q_1 \mid q_2$ and $ q_2 \mid q_1$. And since $ q_1, q_2 > 0$ we have $ q_1 = q_2 = q$

So it is reduced to $ p_1^4 + p_2^4 = 17q^4$
Keep in mind that $ ( p_1, p_2, q ) = 1$
It is obvius that $ p_1+p_2$ is odd, so say that $ p_1, q$ is odd and $ p_2$ is even. In addition to that we'll say that both $ p_1$ and $ p_2$ are nonnegative.

If $ p \mid q$ where p is a prime, we have to prove that $ p \mid p_1, p_2$. Then we are done and we can set $ q=1$ and $ (x,y) = (\pm 1, \pm 2)$ or its permutation.

In other words $ a^4 + b^4 = 0 \bmod p^4 \Rightarrow p \mid a,b$

Please help..


Darn it you beat me to it.  It messed up my geometry diagram fro a problem I was solving...

5.6 on the Richter sclae... it was my first earthquake.  You guys all right? 5.6 =Can cause major damage to poorly constructed buildings over small regions. At most slight damage to well-designed buildings.

[quote="CheeseIsGood"]I'm scared right now... there was just an earthquake. I hope the big one doesn't come...[/quote]
Dude... I felt it too...

That was pretty creepy...  :(

All good though.  How's everyone?

[quote="CheeseIsGood"]I'm scared right now... there was just an earthquake. I hope the big one doesn't come...[/quote]

what do you mean "the big one"
don't tell me theres more

There have already been three aftershocks which were too small for us to feel.

There may be more... 60% chance of a strong earthquake in the next 30 years or something.

http://earthquake.usgs.gov/eqcenter/recenteqsww/Quakes/nc40204628.php

That's today's earthquake.

[quote="Ubemaya"]Yeah, I felt it too. It was around 8:04 I think. I was playing SSB64 :( and my Jigglypuff died :'( .
It felt really really small though.

This has been the second earthquake I felt.[/quote]
how does your jigglypuff dying have anything to do with the earthquake. It doesn't even shake your tv and stuff.

ya my home was shaking it was pretty scary though. There was a little damage in SJ

My 4*4 Rubik's cube on my 2*2 stand fell off.
But I wasn't there! I was in Palo Alto, and I was playing violin, and I didn't even notice it. The lights were shaking a bit though.
About my 3rd earthquake. My first major one, though.
Thank god this wasn't the big one.

PS Was anyone able to get through with their cell phones immediately after the earthquake?

Looking at that USGS map, the location is uncomfortably close to right on the Hayward Fault. When Bay Area people aren't worrying about the San Andreas, the other big fault to worry about is the Hayward. Further north, it runs right through the UC Berkeley campus. (Where I live in Southern California, the fault to worry about is the Newport-Inglewood, which I'm right on top of, rather than the San Andreas, which is 50 miles away.)

[quote="Piggypi314"]Guys you have no idea how relieved I was to realize it was an earthquake... I thought it might have been my parents mad at me again.[/quote]
LOL dude me too

I was in play rehearsal.. We have a live band, and at first I thought it was just the drummer playing really loudly.
Then he stopped playing, and the shaking kept going.. and I was like OMGWTFBBQ!

I felt it

It was weird, cuz my sister was screaming, so I thought was just being annoying again

my house is pretty darn awesome, nothing fall at all (lamp was swinging a bit though)

Random earthquake story: I was just coming out of a shower and there was an earthquake. And you know how they tell you to get under the nearest table? Yeah...

I think I must be totally oblivious to earthquakes or something.  I missed both this one and the one last spring.

Felt it in Millbrae, I was eating at a restaurant. And everyone felt it, but they were like "meh, whatever" and kept on eating.

I suck at piano. I'm taking level 4 CMs this year, after 5 years of playing.

[quote="xscapezaer"][quote="vishalarul"][quote="Ubemaya"]Since davidlizeng doesn't go to AoPS anymore, here's a quote from him:

me: Did you feel the earthquake?
him: i was playing pathetique [that's a piano piece where you play really loud chords]
i banged a chord and then the whole floor shook and i was like wtf, i cant be that strong[/quote]

sonata pathetique by beethoven?[/quote]

@ubemaya & vishalarul: Eh, u all play piano?

I felt the quake. It wasn't that bad, though. The ground shook. Things attached to the ceiling were shaking pretty bad, tho.[/quote]

really
is it just my house then? barely anything moved. curtains shaked a little.

[quote="moogra"][/quote][quote="xscapezaer"][quote="vishalarul"][quote="Ubemaya"]Since davidlizeng doesn't go to AoPS anymore, here's a quote from him:

me: Did you feel the earthquake?
him: i was playing pathetique [that's a piano piece where you play really loud chords]
i banged a chord and then the whole floor shook and i was like wtf, i cant be that strong[/quote]

sonata pathetique by beethoven?[/quote]

@ubemaya & vishalarul: Eh, u all play piano?

I felt the quake. It wasn't that bad, though. The ground shook. Things attached to the ceiling were shaking pretty bad, tho.[/quote][quote="moogra"]

really
is it just my house then? barely anything moved. curtains shaked a little.[/quote]
My walls were moving back and forth.   :o

I think it depends on how close you were to the earthquake

I live pretty far away from you guys
therefore i did not feel the earthquake

glad to hear that you guys are ok (however I am not glad to hear that anirudh is ok).


geometry Asymptote AoPSwiki Alcumus \/closed

How do you post heavily geometric problems that involve diagrams? For that matter, how do you even post geometric problems that consist of only a small diagram? How do you post a diagram?

You can either learn Asymptote (from the AoPSWiki) or use Geogebra.  Next time you are writing a post, click the Geogebra button above the input box and see what happens.

if this is coming from alcumus you can click on the diagrsm, wait, copy the code given to you, paste it on your post, and continue writing. When your post comes out it will have the diagram on it.


A generic hurt and heal game. This one's with smilies though.
Remember: Hurt= -2, Heal= +1 You may not post consecutive posts, and you can only vote once per hour.
Once a smiley reaches 0, it's dead and you cannot hurt or heal it.

 :D :5
 :P :5
 :rotfl: :5
 :ninja: :5
 :play_ball: :5
 :heli: :5

READY!!!!!!!!!!!!!!!!!!!!!!!!!!! GO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Also, update the list every time you vote!
I'll start:

Heal Mr. Green (YAY!)
Hurt Razz (yeah right izzy)

Mr. Green: 6
Razz: 3
Rotfl: 5
Ninja: 5
PlayBall: 5
Helitongue: 5

Didn't the mod here say no more hurt and heals??

[quote="gauss1181"]A generic hurt and heal game. This one's with smilies though.
Remember: Hurt= -2, Heal= +1 You may not post consecutive posts, and you can only vote once per hour.
Once a smiley reaches 0, it's dead and you cannot hurt or heal it.

 :D :7
 :P :1
 :rotfl: :5
 :ninja: :5
 :play_ball: :5
 :heli: :5

READY!!!!!!!!!!!!!!!!!!!!!!!!!!! GO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Also, update the list every time you vote!
I'll start:

Heal Mr. Green (YAY!)
Hurt Razz (yeah right izzy)

Mr. Green: 7
Razz: 1
Rotfl: 5
Ninja: 5
PlayBall: 5
Helitongue: 5[/quote]

Heal Razz (sorry guass but I like razz too!)
Hurt Helitongue

Mr. Green: 7
Razz: 2
Rotfl: 5
Ninja: 5
PlayBall: 5
Helitongue: 3

Mr. Green: 8
Razz: Out
Rotfl: 5
Ninja: 5
PlayBall: 5
Helitongue: 3

Heal Mr. Green
Hurt Ninja

Mr. Green: 9
Rotfl: 5
Ninja: 3
PlayBall: 5
Helitongue: 3
RAWR RAWR RAWR RAZZ IS DEAD :D

Mr. Green: 10
Rotfl: 5
PlayBall: 5
Helitongue: 3
Ninja: 1

6th place: Razz

(Edited to switch order)

Mr. Green: 11
Rotfl: 5
PlayBall: 5
Helitongue: 1
Ninja: 1

6th place: Razz

Ninja is dead.

Mr. Green: 12
Rotfl: 5
PlayBall: 5
Helitongue: 1
5th place: Ninja
6th place: Razz

Mr. Green: 13
Rotfl: 5
PlayBall: 5
4th place: Helitongue
5th place: Ninja
6th place: Razz

Mr. Green: 11
Rotfl: 6
PlayBall: 5
4th place: Helitongue
5th place: Ninja
6th place: Razz

Mr. Green: 12
Rotfl: 6
PlayBall: 3
4th place: Helitongue
5th place: Ninja
6th place: Razz

Mr. Green: 10
Rotfl: 7
PlayBall: 3
4th place: Helitongue
5th place: Ninja
6th place: Razz

Mr. Green: 11
Rotfl: 5
PlayBall: 3
4th place: Helitongue
5th place: Ninja
6th place: Razz

Mr. Green: 9
Rotfl: 6
PlayBall: 3
4th place: Helitongue
5th place: Ninja
6th place: Razz

Audition 1 Mr. Green wins
Audition 2 Smile wins
Audition 3 Wink wins
Audition 4: Idea wins
Audition 5: Rotfl wins

Audition 6:
No: PWNT
Love: PWNT
Wacko: 4
What?: 3

Audition 7:

Alien: 4
Cool: 4
First Prize: 5
Dry: PWNT

Audition 1 Mr. Green wins
Audition 2 Smile wins
Audition 3 Wink wins
Audition 4: Idea wins
Audition 5: Rotfl wins

Audition 6:
No: PWNT
Love: PWNT
Wacko: 5
What?: 1

Audition 7:

Alien: 5
Cool: 2
First Prize: 5
Dry: PWNT

Audition 1 Mr. Green wins
Audition 2 Smile wins
Audition 3 Wink wins
Audition 4: Idea wins
Audition 5: Rotfl wins
Audition 6: Wacko wins

Audition 7:

Alien: 4
Cool: 3
First Prize: 4
Dry: PWNT

Audition 8:
Laugh : 8
 :coolspeak: : 6
 :oops_sign: : 6
 :whistling: :7

Alien: 5
Cool: 1
First Prize: 4
Dry: PWNT

Audition 8:
Laugh : 8
Cool (speak) : 4
Oops! : 6
Whistling :8

Audition 1 Mr. Green wins
Audition 2 Smile wins
Audition 3 Wink wins
Audition 4: Idea wins
Audition 5: Rotfl wins
Audition 6: Wacko wins

Audition 7:
Alien: 3
Cool: 2
First Prize: 4
Dry: PWNT

Audition 8:
Laugh : 8
Cool (speak) : 4
Oops! : 4
Whistling :9

Audition 1 Mr. Green wins
Audition 2 Smile wins
Audition 3 Wink wins
Audition 4: Idea wins
Audition 5: Rotfl wins
Audition 6: Wacko wins

Audition 7:
Alien: 4
Cool: 2
First Prize: 2
Dry: PWNT

Audition 8:
Laugh : 8
Cool (speak) : 2
Oops! : 4
Whistling :10

Audition 7:
Alien: 4
Cool: PWNT
First Prize: 3
Dry: PWNT

Audition 8:
Laugh : 9
Cool (speak) : PWNT
Oops! : 4
Whistling :10

Audition 1 Mr. Green wins
Audition 2 Smile wins
Audition 3 Wink wins
Audition 4: Idea wins
Audition 5: Rotfl wins
Audition 6: Wacko wins

Audition 7:
Alien: 5
Cool: PWNT
First Prize: 1
Dry: PWNT

Audition 8:
Laugh : 8
Cool (speak) : PWNT
Oops! : 5
Whistling :9

Audition 1 Mr. Green wins
Audition 2 Smile wins
Audition 3 Wink wins
Audition 4: Idea wins
Audition 5: Rotfl wins
Audition 6: Wacko wins
Audition 7: Alien wins

Audition 8:
Laugh : 9
Cool (speak) : PWNT
Oops! : 4
Whistling :8

Audition 9:
Yingyang: 6
W00T?: 8
Pilot: 6
Play Ball: 7

Mr. Green: 9
Rotfl: 2
PlayBall: 1
4th place: Helitongue
5th place: Ninja
6th place: Razz

oops

Audition 8:
Laugh : 9
Cool (speak) : PWNT
Oops! : 2
Whistling :9

Audition 9:
Yingyang: 6
W00T?: 8
Pilot: 4
Play Ball: 8

Audition 1 Mr. Green wins
Audition 2 Smile wins
Audition 3 Wink wins
Audition 4: Idea wins
Audition 5: Rotfl wins
Audition 6: Wacko wins
Audition 7: Alien wins

Audition 8:
Laugh : 10
Cool (speak) : PWNT
Oops! : PWNT
Whistling :9

Audition 9:
Yingyang: 4
W00T?: 8
Pilot: 5
Play Ball: 8

Audition 1 Mr. Green wins
Audition 2 Smile wins
Audition 3 Wink wins
Audition 4: Idea wins
Audition 5: Rotfl wins
Audition 6: Wacko wins
Audition 7: Alien wins

Audition 8:
Laugh : 8
Cool (speak) : PWNT
Oops! : PWNT
Whistling :10

Audition 9:
Yingyang: 4
W00T?: 8
Pilot: 3
Play Ball: 9

Audition 1 Mr. Green wins
Audition 2 Smile wins
Audition 3 Wink wins
Audition 4: Idea wins
Audition 5: Rotfl wins
Audition 6: Wacko wins
Audition 7: Alien wins

Audition 8:
Laugh : 9
Cool (speak) : PWNT
Oops! : PWNT
Whistling :8

Audition 9:
Yingyang: 4
W00T?: 8
Pilot: 1
Play Ball: 10


Sides AB, BC and CD of (simple) quadrilateral ABCD have lengths 4,5 and 20, respectively. If vertex angles B and C are obtuse and $\sin C=-\cos B=\frac{3}{5}$, then side AD has length

(A) 24
(B) 24.5
(C) 24.6
(D) 24.8
(E) 25

It means "whose opposite sides don't intersect mutually" i.e. convex.

Can we use the identity $\cos (\theta)=-\cos (\pi-\theta)$?

And I must ammend myself: Concave quadrilateral can still be simple, for example an arrowhead shape.

Who can prove that...

[hide="SPOILER"]
$\angle ABC = 2(\angle CDA)$ ?[/hide]

[hide]
Extend $\overline{AB}$ and $\overline{CD}$ to intersect at $X$. We have $\cos \angle XBC=\sin \angle XCB=\frac{3}{5}$ which means
\[\text{YAY!!!!! }\triangle CBX\text{ IS A 3-4-5 TRIANGLE!!!!!}\]
which makes $AD$ equal to $\sqrt{(3+4)^{2}+(4+20)^{2}}=\boxed{25}$.
[/hide]

[quote="Go Around the Tree"]dang dude.

bpms what course are you taking? you are eligible for math counts and are like taking extremely advanced math  :lol:[/quote]
Well I'm taking Alg. 2, but I haven't learned anything in it, it's only been review of Alg. 1 so far :(. I haven't learned anything in school in math since alg. 1, I learn all the advanced math I know on my own, which is why I suck at math...

[quote="bpms"][quote="DanK"]Can you please explain to me how you got that $\cos\angle XBC=\sin\angle XBC=\frac{3}{5}$?  I don't see how the new triangle implies that...[/quote]
Use the most basic definition of sin... Opposite over hypotenuse...[/quote]
  It is easy to prove that $\cos{\angle XBC}=\frac{3}{5}$ algebraically, but but how do we know that triangle  :maybe: XBC is right?  Also, assuming that it IS right, the fact $\cos{\angle XBC}=\sin{XBC}$ $(\frac{3}{5}=\frac{4}{5}???)$ doesn't make any sense.   :maybe: I have a feeling that I'm missing something important but it still makes no sense

i made a typo, it was supposed to be $\sin \angle XCB$. Edited.

I still don't get how your solution makes sense :maybe:

Okay, I'll try to explain it. Sorry for the lack of LaTeX, if you don't understand have someone else put it into latex.

Draw the quadrilateral ABCD, and make it so that side AB and side CD are "slanted" toward side BC. Now extend AB and DC toward BC, and call their intersection X.

Angle XBC is 180 minus angle B and angle XCB is 180 minus angle  C. For all angles X, cos(180-X) = -cos(X) and sin(180-X) = sin(X) (Use a circle diagram to see why this is true). So cos(XBC) is equal to -cos B, and sin(XCB) is equal to sin C, which are both equal to 3/5.

Using the right-triangle model, cos(XBC) is XB/BC (by the way, angle X is supposed to be a right angle). We know that BC is 5, so XB must be 3, and XC must be 4 by Pythagorean's Theorem.

Now we have a right triangle AXD with right angle X, and legs 4+3=7 and 20+4=24. Using Pythagorean's Theorem, the length of AD is $25$.


1. Find the center and the radius of the circle
x2 + y2 – 2x + 2y + 2 = 16

2. Find the equation of the inverse relation y = 2x - 8

I think you meant
1. Find the center and the radius of the circle
$x^{2}+y^{2}-2x+2y+2=16$

2. Find the equation of the inverse relation $y=2x-8$

Answer is
C(1,-1) r=4
Since the circle equation is
$(x-x_{0})^{2}+(y-y_{0})^{2}=r^{2}$

Thanks for the help.  Do you know the answer to the second one?

Inverse?

Sub in y for x and x for y.

x = 2y - 8
y = (x+8)/2 = x/2 + 4


손찬웅 >이재동 > 김택용
에서 2009 EVER 온게임넷 스타리그!!

[quote="ZerAn"]손찬웅 >이재동 > 김택용
에서 2009 EVER 온게임넷 스타리그!![/quote]

I DIED!


calculus
derivative
function
analytic geometry
graphing lines
slope
calculus computations

[quote="ArmiAldi"]can i have this step by step please:
Find the equation of the line tangent to the curve at the point
xy + y = 2 ; (1, 1)
Thanks[/quote]
Do you know implicit differentiation?

[b]Step 1:[/b]
Differentiate the function given, so you can find the slope.

[hide]
$ xy \plus{} y \equal{} 2$

$ y \plus{} xy' \plus{} y' \equal{} 0$

$ xy' \plus{} y' \equal{} \minus{} y$

$ y'(x \plus{} 1) \equal{} \minus{} y$

$ y' \equal{} \frac { \minus{} y}{x \plus{} 1}$
[/hide]

[b]Step 2:[/b]
Plug in the points given into $ y'$

[hide]
$ y' \equal{} \frac { \minus{} y}{x \plus{} 1}$

$ y' \equal{} \frac { \minus{} 1}{2}$

So the slope ($ m$) is $ \frac { \minus{} 1}{2}$ or $ \minus{} .5$
[/hide]

[b]Step 3:[/b]
Using the original points, and the slope, write the equation for the tangent line.

[hide]
$ y \minus{} y_{1} \equal{} m(x \minus{} x_{1})$
$ y \minus{} 1 \equal{} \minus{} .5(x \minus{} 1)$
And then simplify from there. (If it were an AP exam, I believe you can leave it like that as your final answer)
[/hide]


Let $a,b,c$ and $x,y,z$ be positive numbers such that $a+b\geq c+z, b+c\geq a+x, c+a\geq b+y$. Prove or disprove that

\[\Bigl(\frac ax+\frac by+\frac cz\Bigl)\Bigl(\frac{a+x}{b+y}+\frac{b+y}{c+z}+\frac{c+z}{a+x}\Bigl)\geq \Bigl(\frac{a+b}{x+y}+\frac{b+c}{y+z}+\frac{c+a}{z+x}\Bigl)^2\]

This is not ture, for example $a=b=c=x=2,y=z=1$:

$\Bigl(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\Bigl)\Bigl(\frac{a+x}{b+y}+\frac{b+y}{c+z}+\frac{c+z}{a+x}\Bigl)=\frac{185}{12}$;

$\Bigl(\frac{a+b}{x+y}+\frac{b+c}{y+z}+\frac{c+a}{z+x}\Bigl)^{2}=\frac{196}{9}$.


This is probably from Japan 2001. :maybe:

From CBS inequality
\[ \left(\sum x_i^2\right)\left(\sum y_i^2\right)\ge \left(\sum x_iy_i\right)^2 \plus{} (x_1y_2 \minus{} x_2y_1)^2 \\
\iff (x_1y_2 \minus{} x_2y_1)^2\le 1 \minus{} \left(\sum x_iy_i\right)^2
\]
Let $ \sum x_iy_i \equal{} x$. Then it remains to show that $ 1 \minus{} x^2\le 2|1 \minus{} x|$

Verify that from CBS inequality $ x^2\le 1\implies \minus{} 1\le x\le 1\implies 0\le 1 \minus{} x\le 2$. Therefore $ |1 \minus{} x| \equal{} 1 \minus{} x$ and we get $ 1 \minus{} x^2\le 2(1 \minus{} x)\iff (x \minus{} 1)^2\ge 0$, which is obviously true.

[quote="nayel"]

From CBS inequality

$ \left(\sum x_i^2\right)\left(\sum y_i^2\right)\ge \left(\sum x_iy_i\right)^2 \plus{} (x_1y_2 \minus{} x_2y_1)^2$[/quote]

Sorry, but why is this? I can't see how it's CBS (you mean Cauchy, right?)

Yes, CBS inequality is the Cauchy-Scwarz inequality, or more precisely, the [url=http://en.wikipedia.org/wiki/CBS_inequality]Cauchy-Bunyakovski-Schwarz inequality[/url]. The inequality I used follows from the identity on which Cauchy's inequality is based
\[ \left(\sum x_i^2\right)\left(\sum y_i^2\right) \equal{} \left(\sum x_iy_i\right)^2 \plus{} \sum_{i,j \equal{} 1}^n(x_iy_j \minus{} x_jy_i)^2
\]

This problem is from Korea National Mathematics Olympiad (known as Final KMO or FKMO in Korea) 2000.


function
number theory theorems
number theory

Dear group members,

I believe that the formula

(x^2/zeta(2)+1+O(xlogx))/2

is a better approximation for the partial sum (from 1
to [x]) of the Euler's totient function. I tried
numerically and see that it gives better
approximations than the formula in Apostol's book
(Theorem 3.7). Is this formula known?

Sincerely Yours,
Murat Aygen

Sorry. I learned that this is already known. Please remove this topic.


wouldn't preparing for AIME be more useful then preparing for the AMC 12 because if you do that then you would have the AMC 12 covered as well?

You could make the same argument. Why not just prepare for the USAMO? Or the IMO? Or the Chinese TST? Prepare at the highest level that you can.

Please keep your posts to a single topic. Hit reply instead of new topic when you want to reply to someone's posts. Thanks.

well the aime is harder, and the amc 10 is easier. do the best one you can

If you're comfortable doing most of the AMC-10 problems, try this:

1) Go through the last few problems on each AMC-10 and the first 5 problems on every AIME.
2) Solve problems 6-10 on every AIME.
3) That should we more than enough.

 :)


calculus
integration
logarithms
calculus computations

hello, setting $ t\equal{}\sqrt{1\plus{}x}$ we get $ x\equal{}t^2\minus{}1$ and $ dx\equal{}2t\,dt$ so your integral is
$ 2\int\frac{1}{2\minus{}t^2}\,dt\equal{}\frac{\sqrt{2}}{2}artanh(\frac{\sqrt{2}}{2}t)\plus{}C$.
Sonnhard.

Thanks Sonnhard.

I have:

Let $ y^{2} \equal{} 1\plus{}x$. Then

$ I\equal{} \int \frac{2y}{(2\minus{}y^{2})y}\ dy \equal{} \int \frac{2}{2\minus{}y^{2}}\ dy$

$ \equal{}\frac{1}{\sqrt{2}} \int(\frac{1}{y\plus{}\sqrt{2}} \minus{} \frac{1}{y\minus{}\sqrt{2}})\ dy$

$ \equal{}\frac{1}{\sqrt{2}} \ln(\frac{y\plus{}\sqrt{2}}{y\minus{}\sqrt{2}})$

$ \equal{} \frac{1}{\sqrt{2}} \ln(\frac{\sqrt{1\plus{}x}\plus{}\sqrt{2}}{\sqrt{1\plus{}x} \minus{} \sqrt{2}}) \plus{} c$.


Suppose that you do not solve a problem on USAMTS, but you give some ideas as to how you could approach the problem, and go somewhat in depth (so if you tried a method and it didn't work). Could you get at least one point for presenting your ideas?
If you are very close to the answer, but do not realize it, could you possibly get 2 or 3 points?

Thanks.

It all depends on the grading criteria. For some problems, any nontrivial work gains at least 1 point. For other problems, like this year's round 2 #5, any solution that claims the maximum is 2.5 receives no credit, no matter how long it is.

It is my impression that a score correlates most directly to how much the submitted work contributes to actually solving the problem, regardless of how nontrivial or interesting it is. USAMTS is pretty generous with this though...


Is there a positive integer with at most $4$ digits whose value is increased by exactly $60\%$ when the first digit is moved to the end of the number?
Prove your conclusion

Suppose you have such a number of exactly 4 digits $a,b,c,d$, say $X$.  Then $X = 1000a + 100b + 10c + d$.  The result of moving the digit gives $b,c,d,a$ which represents $Y = 1000b + 100c + 10b + a$.  So $Y-a = 10(X-1000a)$.  You're also told $Y = 1.6 X$.  So $(42/5)X = 999a$.  There are now only a few cases to try.


1)Να βρεθουν οι ακεραιοι x,y ωστε $ 2x^{4} \plus{} 1 \equal{} y^{2}$
2)Να βρεθουν ολοι οι ακεραιοι n ωστε το $ P(x) \equal{} x^{5} \minus{} nx \minus{} n \minus{} 2$ να μπορει να γραφει ως γινομενο δυο μη σταθερων πολυωνυμων του χ με ακεραιους συντελεστες
(εχω+αλλες αλλα δν μπορω να τς γραψω σε latex...)
ΡΑΛΛΗΣ :D

Για τη δεύτερη κοιτάξτε εδώ
http://www.mathlinks.ro/Forum/viewtopic.php?t=49879

Ράλλη καλή επιτυχία στην ΙΜΟ :)

Ευχαριστο πολυ!Ειδα τις λυσεις που δινουν αλλα υπαρχουν+αλλες(δν τις θυμαμαι ολες τωρα π.χν=-19 +αλλες..)παντως κανεις δεν εκμεταλευτηκε το γεγονος οτι P(-1)=-3 για καθε ν εZ....το οποιο οδηγει πολυ πιο ευκολα στις λυσεις.(Το θεμα ειναι απο μια μεσογειαδα,δν θυμαμαι ακριβως χρονια,αλλα ηταν δευτερο θεμα)

ΡΑΛΛΗΣ :D


function
geometry
geometric transformation
reflection
geometry proposed

Let $\mathcal O_{1}$ and $\mathcal O_{2}$ two exterior circles. Let $A$, $B$, $C$ be points on $\mathcal O_{1}$ and $D$, $E$, $F$ points on $\mathcal O_{1}$ such that $AD$ and $BE$ are the common exterior tangents to these two circles and $CF$ is one of the interior tangents to these two circles, and such that $C$, $F$ are in the interior of the quadrilateral $ABED$. If $CO_{1}\cap AB=\{M\}$ and $FO_{2}\cap DE=\{N\}$ then prove that $MN$ passes through the middle of $CF$.

It can be easily computed trigonometrically. Denote $L=BD \cap CF$, $K=CF\cap AE$ and prove that $BL=LC=KE=KF$. Now if you denote $\angle ABC =\alpha$ and $\angle CAB =\gamma$ you can express everything needed in terms of trigonometric functions with $\alpha$ and $\gamma$. That's a boring solution, anything more interesting?

Hello! Thanks for the trigo solution, I need to learn to do trigo like that :) Just thought I'd point out that this problem seems quite related to a lemma that appeared in the forums before:
[b]Lemma:[/b]  Let the incircle of triangle $ABC$ have centre $I$ and meet $BC, CA, AB$ at $D,E,F$, and $M$ be the midpoint of $BC$. Then $ID, EF$ and $AM$ are concurrent. (Similarly, if $I_{A}$ is the $A-$excentre and is tangent to $BC, CA, AB$ at $D_{A}, E_{A}, F_{A}$ then $I_{A}D_{A}, E_{A}F_{A}$, and $AM$ are concurrent. )
For the incircle case, refer to Virgil Nicula's proof in [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=80872[/url].
The excircle case can be done in the same way. :P
For this problem, just extend $AD$ and $BE$ to intersect, say at $X$, and let $FC$ meet $AD$ and $BE$ at $Y$ and $Z$ respectively. The two circles are an incircle and excircle of triangle $XYZ$. Using the lemma , the median of the triangle $XYZ$ passes through $N$ and $M$. Also $FY=CZ=s-y$, so so the median also passes through the midpoint of $CF$. :)

[quote="Megus"]and prove that $BL=LC=KE=KF$. [/quote]

Are you sure? In my diagram it is clear that $BL\neq LC$, $KE\neq KF$!

Here is my complete solution:
[img]8497[/img]

$X,Y$ are the points where internal and external common tangents meet. The parallel to $XY$ through $N$ meets $PX,PY$ at $X',Y'$.
The quadrilaterals $O_{2}NX'D$ and $O_{2}NEY'$, having two opposite right angles, are cyclic, and the triangle $O_{2}DE$, two of its sides being the radii, is isosceles. Therefore $\angle O_{2}X'N = \angle O_{2}DN = \angle O_{2}EN = \angle O_{2}Y'N \Rightarrow O_{2}X'Y' \mbox{isosceles}$.
$O_{2}N$, being height, is also median: $NX' = NY'$, but then K, the intersection of the line $PN$ and $XY$, is the midpoint of $XY$.
It is known that the midpoint of the side of a triangle is the midpoint of the points of contact of incircle and excircle with that side. Therefore $KF = KC$. Now consider the triangles $KFN, KCM$: they have both a right angle and $\angle FKN = \angle CKM$, and $KF = KC$. Then they are congruent.

$KM = KN$.  :)

Oh-hoo...oo, my solution same as @bryan_hooi's  solution.

My solution:

Let $ X=AD \cap BE, Y=CF \cap AD, Z=BE \cap CF $ .
Let $ T $ be the reflection of $ F $ in $ O_2 $ .

It's well known that $ CY=ZF $ .

Since $ AC \parallel DT , BC \parallel ET $ ,
so figure $ \triangle ABC \cap O_1 \cap M $ and figure $ \triangle DET \cap O_2 \cap N $ are homothetic with center $ X $ ,
hence we get $ X, M, N $ are collinear . ... $ ( \& ) $

Since $ XM $ pass through the midpoint of $ YZ $ (see [url=http://yufeizhao.com/olympiad/geolemmas.pdf]lemma 7[/url] ) ,
so combine with $ ( \& ) $ we get $ MN $ pass through the midpoint of $ CF $ .

Q.E.D

A little typo in question, D, E, F are on O2, not O1.

Firsly, denote $L=AD \cap BE$.
Let $X$ be the second intersection of the diameter $CM$ with the circle $(O_1)$. $Y$ is analogous.

Because $CF$ is the common internal tangent, $CX \perp CF$ and $YF \perp CF$,so $CX || YF$.

We know that the homotethy that takes $(O_1)$ to $(O_2)$ is centered at L, and because diameters $CX$ and $YF$ are parallel, then that homotethy takes $CX$ to $YF$.This shows that lines $CY$, $XF$ and $MN$ all pas through $L$
.
By intersecting $CF$ with $LA$ and $LB$ at $S$ and $T$, a well known lemma gives that $LM$ bisects $ST$, and because in the triangle $LST$ , $O_1$ and $O_2$ are the incircle and excircle,the conclusion follows.


probability
combinatorics proposed
combinatorics
coins

For every natural number $ t$, $ f(t)$ is the probability that if a fair coin is tossed $ t$ times, the number of times we get heads is 2008 more than the number of tails. What is the value of $ t$ for which $ f(t)$ attains its maximum? (if there is more than one, describe all of them)

heads x
tails y
then
x+y=t
x-y=2008
so x=$\frac{t+2008}{2}$ and y=$\frac{t-2008}{2}$
f（t）=$\frac{1}{2}^t\binom{t}{\frac{t+2008}{2}}$
at last maybe t=2012


Solve

$ x\plus{}y\plus{}z\equal{}a \\ x^2\plus{}y^2\plus{}z^2\equal{}b^2 \\ xy\equal{}z^2$

where $ a$ and $ b$ are constants.

Give the conditions that $ a$ and $ b$ must satisfy so that $ x,y,z$ are distinct positive integers.

This is [url=http://www.mathlinks.ro/viewtopic.php?p=343297#343297]IMO 1961[/url]


Line $ l$ is a straight line. What is the value of $ x$ in the diagram?
[asy]size(150); defaultpen(linewidth(0.8)); import geometry_dev;
draw(origin--(10,0),Arrows);
pair X = (5,0) + 5*dir(80);
draw((5,0)--X,EndArrow);
markangle(Label("$x^{\circ}$",align=NE),X,(5,0),(10,0),radius=10);
markangle(Label("$x + 20^{\circ}$",align=NW),X,(5,0),(0,0),n = 2,radius=10);[/asy]

The angles form a linear pair, so they are supplementary.

$ (x \plus{} 20) \plus{} x \equal{} 180$

$ 2x \plus{} 20 \equal{} 180$

$ 2x \equal{} 160$

$ \boxed{x \equal{} 80^{\circ}}$


1.    One day, a person went to horse racing area, Instead of counting the number of human and horses, he instead counted 74 heads and 196 legs. Yet he knew the number of humans and horses there. How did he do it, and how many humans and horses are there?  :D

This should be in Classroom Math or MATHCOUNTS, not Pre-Olympiad.

[hide]$x+y=74,\,2x+4y=196$

$x=50\text{ men},\,y=24\text{ horses}$[/hide]

74 heads and 196 legs
[hide]
x+y=74
2x+4y=196

it's x+y because both humans and horses have one head.
It's 2x+4y because humans have 2 legs and horses have 4 legs.

from the above equation, youcan do many ways but the fastest would be to find 2y first.

2x+4y=196
2x+2y=148

therefore, 2y=48
y=24
now you can subsitue y for either equation
to find x, which turns out to be 50 [/hide]

no this should be in the puzzles and brainteasers section


There is a carnival game. There is 3 different types of balls.  Red, blue and  green. There are 3 red, 3 blue, and 3 green balls. These balls are all put into one bag. You get 3 chances, and the goal is to pick 3 green balls in a row. Once you take a ball out, you don't put it back in. What is the probability of winning the game?

Edit: 6 balls (3 green and 3 blue) and 4 chances to get 3 in a row

$1-\left(\frac{1}{2}\cdot\frac{3}{5}\cdot\frac{3}{4}\cdot1\right)=\frac{31}{40}$

I think

[quote="ashpotter"][hide]$\frac{1}{3}*\frac{1}{4}*\frac{1}{7}=\frac{1}{84}$[/hide][/quote]

Seeing as we all have different answers, just putting down an answer without an explanation doesn't really help.  Please explain your thinking.

[quote="philB"][quote=hunter34][hide]1-\left(\frac{1}{2}\cdot\frac{3}{5}\cdot\frac{3}{4}\cdot1\right)=\frac{31}{40}[/hide]
i think[/quote]
I think this seems a little high...

[hide="6 balls of each color with three balls?"]
4 * (3*2*1*15) * 1/18 *1/17 * 1/16 * 1/15 = $\boxed{1/204}$[/hide]

[hide="or 3 colors with 6 balls?"]
4 * (6*5*4*15) * 1/18 *1/17 * 1/16 * 1/15 = $\boxed{4/51}$[/hide][/quote]
How'd you get these? It's 3 colors with three balls of each color, so there are 9 balls in all, not 18.

[quote="Ihatepie"]now there is 6 balls, and 4 chances to get 3 in a row[/quote]

I was (attempting :D) to answer this question Temperal.


Find all $ a\in\mathbb{R\plus{}}$ such that :
Always exist $ k\in\mathbb{R\plus{}}$ and function $ f: R\rightarrow R$ such that: $ \frac{f(x)\plus{}f(y)}{2}\geq f(\frac{x\plus{}y}{2})\plus{}k|x\minus{}y|^{a}$ for all $ x,y\in\mathbb{R}$ and $ x$ isn't equal $ y$.

let $ S \equal{}\{a > 0|\ \exists k > 0,\exists f: R\to R:\ f(\frac{x\plus{}y}{2})\ge\frac{f(x)\plus{}f(y)}{2}\plus{}k|x\minus{}y|^{a}\}$
and take $ for\ a\in S:\ F_{a}\equal{}\{f: R\to R| f(\frac{x\plus{}y}{2})\ge\frac{f(x)\plus{}f(y)}{2}\plus{}k|x\minus{}y|^{a}\}$
if $ a\in S$ then $ ]0,a]\subset S$ (because $ a > h$ gives $ F_{a}\subset F_{h}$)
then $ \exists b\in\mathbb{R}^{\plus{}}\cup\{\plus{}\infty\}:$ such that $ S \equal{} ]0,b]\ or ]0,b[$


trigonometry
calculus
calculus computations

Does $\sum_{k=0}^{m-1} \sqrt{n^2+kn+1} \; cos(\frac{2k\pi}{m})$ (where $m$ is a natural number, not equal to 0)  converge ?

What are you asking? For fixed $m,$ does this have a limit as $n\to\infty$? Is that the question?

Yes, for fixed $m$ as $n\to \infty$. Sorry !

$\sqrt{n^2+kn+1}=n+\frac k2+O(n^{-1})$ as $n\to\infty$

$\sum_{k=0}^{m-1}\sqrt{n^2+kn+1}\cos\left(\frac{2k\pi}m\right)=$

$n\sum_{k=0}^{m-1}\cos\left(\frac{2k\pi}m\right)+ \frac12\sum_{k=0}^{m-1}k\cos\left(\frac{2k\pi}m\right)+O(n^{-1})$

Now, let $\omega=e^{2\pi i/m}.$

$\sum_{k=0}^{m-1}\cos\left(\frac{2k\pi}m\right)= \Re\left(\sum_{k=0}^{m-1}\omega^k\right) =\Re\left(\frac{1-\omega^m}{1-\omega}\right)=0$

Thus as $n\to\infty,$ the original expression converges, and it converges to

$\frac12\sum_{k=0}^{m-1}k\cos\left(\frac{2k\pi}m\right)$

(Which could be explicitly computed.)


Let $ABC$ be a triangle and $P$ a point in its plane.

Let $R$ be the radius of the circumcircle of triangle $ABC$.

Prove that

$\frac{PA}{BC^{2}}+\frac{PB}{CA^{2}}+\frac{PC}{AB^{2}}\geq\frac{1}{R}$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=6073 (in that topic, the problem was posed only for P inside triangle ABC, but I solved it for arbitrary P).

  darij

A question just appears in my mind:
If ABC is fixed, then where is P when $\sum\frac{PA}{a^2}$ is minimized? Is there any geometric interpretation?


One throws randomly 120 squares of the size 1x1 in a 20x25 rectangle. Prove that one can still place in the rectangle a circle of the diameter equal to 1 in such a way that it does not have common points with any of the squares.

Pigeon hole principle ^^
20 x 25 = 500 and 1 square can cover 4 places. 4 x 120 = 480
500 < 480..

*More gasp* $ 500 < 480$? I never knew that! :P


A set of cards consists of 8 red and 7 black cards. Three cards are dealt at random without replacement. What is the probability that the three cards dealt are the same color? Express your answer as a common fraction.

We have two possibilities: the three cards are all red or the three cards are all black. Let's consider the first case, all red.

The probability of such an event occurring is $ \frac {8}{15} \cdot \frac {7}{14} \cdot \frac {6}{13} \equal{} \frac {8}{65}$.

Our second case is then all black.

The probability of such an event occurring is $ \frac {7}{15} \cdot \frac {6}{14} \cdot \frac {5}{13} \equal{} \frac {1}{13}$.

We add these two fractions together to get a result of $ \boxed{\frac {1}{5}}$.


Everyone in my school except me understands what they are. i only know a few examples like
you suck
thats what she said

i am confused.

Um, what does it sound like it means? It really isn't difficult at all to figure it out...

Carl (I mean....Catalyst!), don't go around poisoning the minds of innocent rising 8th graders! :wink:

For example:

Its not big enough. That's what she said.

I recall a funny lecture by my art teacher, she was explaining the size of one of the projects we were to work one:

"It's about this big."

and someone (could be me  :roll: ) said "That's what she said" in an undertone. Ironically everyone but the teacher heard it.

Hmm, during my lunch period, my social outcast friends, who I'm forced to hang out with during that period b/c in my opinion the rest of the period is [size=42]retarded[/size] mentally challenged, try to be cool by making that's what she said jokes, and I'm like, "Yeah, you're really bad at this."  Obviously, they're like "that's what she said!"  And I'm like...  Darn, that actually made sense... :|

lol sheesh somebody just explain instead of giving examples that you clearly wouldn't get if you didn't understand the concept

basically it is a sexual innuendo with an implied "in bed" appended to the "that's what she said"

Dood i love these i make them all the time at skool

My all  time favorite:

I'm thirsty and I need 75 cents.

The only one I got was at CTY. Cross-dressing day on July 4th?  :wacko:

I was wacking one of my classmates with a bra. I don't know why. He said "Stop wacking me with women's clothes." The rest of the story you know.

Hahahah.
Professor: ...and that is the Big Bang Theory.
Student: (nonchalantly) Yes, that's what she said. (urbandictionary)  :P

Wow, I didn't know that so many people on AoPS were as perverted as Arvind...

[quote="ZhangPeijin"]Band teacher:
It's not that hard to play trombone, all you have to do is put your mouth on it, and blow.
Some kid in the back row: THAT'S WHAT SHE SAID![/quote]
Uh, trombone is hard to play though...(well not for people who don't play well, then they think it's easy)

[quote="Flame"]no offense but you Americans suck at making jokes... if you want to prove me otherwise then tell me some really good joke, not this crap...[/quote]

Exactly what I intended to say...  :rotfl:

[quote="Flame"]no offense but you Americans suck at making jokes... if you want to prove me otherwise then tell me some really good joke, not this crap...[/quote]

What's the point of putting in "no offense" if you're going to call our jokes "crap?"

And also, how about the inverse of your second statement? How can you prove Americans have no sense of humor if you can't prove your [country] has a better one?

thats easy... I speak English and you don't speak Croatian :rotfl:
so basically I can't prove my humor isn't crap but you can prove yours isn't... which you are not doing  :D
if there is American who can speak some Balkan language (Croatian, Bosnian, Serbian...) please say so, so I could tell you a joke  :wink:

NOTE: you must agree that jokes cannot be translated

True, not all jokes are funny in more than one language, so you might also say that not every Balkan joke will be considered "humorous" by Americans.

But your whole concept of joke is different. It's more laughing [i]at[/i] someone then laughing [i]with[/i] someone. I mean you put examples yourselves. This can be funny to you, but is it funny to person you are laughing to? Don't get me wrong, we have that too, but its not a real joke. Jokes are short, entertaining (to everyone) stories. This "that's what she said" fits only in short.
And knock knock jokes? Never been funny to me.

And I know from experience Czechs have similar sense of humor. Also some Germans I know.

This topic is starting to stray off topic.  The perverted but funny topic of "that's what she said" jokes has transformed into a mathematical/philosophical debate about the essence of humor :) .  I agree that these kinds of jokes are funny in a very stupid, raw way, but there's no need to debate about it...

[quote="Flame"]But your whole concept of joke is different. It's more laughing [i]at[/i] someone then laughing [i]with[/i] someone.[/quote]

Ah, I understand where you're coming from. The "that's what she said jokes" and "yo momma jokes" aren't necessarily meant for poking fun at someone. In fact, most of them are used as a humerous debate. For example, my friend the other day was annoying everyone by saying "I did your mom last night in bed" (with varients). I got annoyed enough to say "No, my mom is still waiting for you to hit puberty." No he wasn't offended or anything, since these jokes [b]are meant jokingly, and not taken seriously[/b], and he simply said "Oh dang, you got me good there."

@Math Geek: We're not getting off topic, since we're kind of delving into the purpose of the "that's what she said" jokes.  :P

an example. that rollercoaster ride was fun. Thats what she said.

this is one of the last places i would expect this kind of topics...

[quote="Flame"]But your whole concept of joke is different. It's more laughing at someone then laughing with someone. I mean you put examples yourselves. This can be funny to you, but is it funny to person you are laughing to? [/quote]

yes it is, since most of the time no one takes these seriously

[quote]
And knock knock jokes? Never been funny to me. [/quote]

knock knock jokes are used to entertain 4 year olds...but sometimes they are so lame that they are funny...

After solving a hard math problem,

"Hey, that's clever!"
"That's what she said."

After someone does something awful,

"I want you to think long and hard."
"Thats what she said"

[quote="Flame"]But your whole concept of joke is different. It's more laughing [i]at[/i] someone then laughing [i]with[/i] someone.[/quote]

One of my teammates points at me when he laughs. He is even older than me! Well, two days but that's not the point. I only laugh. Not point. I only point when it's not taken seriously.

little boy: Are you a man?
Dad: Well, yes I am a man.

You know the story. :)

Personally, I use "That's what she said!" only in the most grammatically ambiguous situations.  For example: "He doubts she could've done what they claimed she did."
[url=http://xkcd.com/174/]I just couldn't resist.[/url]

the xkcd comic of friday also had this phrase in it

[url]http://xkcd.com/436/[/url]