Ni hao!

I appologize for not typing in Chinese. I have just re-formatted my computer, and I have not yet installed a Chinese type-setting program yet. However, my computer can read Chinese, so I welcome all responses, in both Chinese and English! (It would actually be easier for me to find the books if the titles are typed in Chinese.)

I was wondering what good math books are out there that are written in Chinese. I'm not really looking for textbooks; I'm more interested in books on recreational mathematics, mathematicians (anecdotes and biographies), the philosophy of mathematics, the development of mathematics, and the like. Any suggestions?

不知道你能不能买到[b]张景中[/b]先生的书. 他写的一系列丛书:漫话数学,新概念几何,数学与哲学,从数学教育到教育数学......都是很值得一读的好书!

谢谢大家的回信！

我昨天买了几本张景中的书。他的书在书店都非常受欢迎。虽然他的书是写给中学生的，我相信他的书会很好看！

可惜，我没有机会找到《数学和数学家的故事》。

你也可以多利用图书馆.那里的书不仅免费,而且往往藏着一些经典的东西!

海南出版社<数学的故事>

《美丽心灵——纳什传》一部很好的书，介绍了一个非常传奇的人。
如果你想看数学的发展，你可以去买《当代数学精英——菲尔兹得主及其建树》。
有时间的话可以买几部数学史方面的书来看。

[quote="dingdongdog"]《美丽心灵——纳什传》一部很好的书，介绍了一个非常传奇的人。

看过，好书。好像是《普林斯顿的幽灵－纳什传》，
介绍了纳什在博弈论中的贡献，但不幸患分裂症，最后与病魔斗争终于康复，并获1994年诺贝尔经济奖。

李学数的书不错，一系列数学和数学家的故事，入门的书。
另外，《几何学的故事》《无穷之旅》都不错。


2000 points are given on a tree.  Is it possible that their pairwise distances are 1, 2, 3, ..., C(2000, 2)? (i.e. all the distances are distinct and take on those values.)

It is an old problem which comes from the USSR (with 1986 instead of 2000).
We used it for our correspondance program in France in 2001...and it can be proposed each year until 2025...

First note that a tree is a connected graph with no cycle, thus between two arbitrary points there is one and only one path.
Let $A$ be one of the points.
Let $x$ be the number of points $B$ (including $A$ itself) such that the length of the path $AB$ is even. These points will be called even.
Let $y$ be the number of points $C$ such that the length of the path $AC$ is odd. These points will be called odd.

Let $n = x+y$ (here $n = 2000$). (1)

There are exactly $xy$ pairs of points $\{ B,C \} $ such that $B$ is even and $C$ is odd.
Moreover for such a pair, the length of the path $BC$ is odd.

Conversely, if the length of the path $BC$ is odd, then exactly one of the points is odd. Thus, since the lengths of the paths are exactly $1,2,..., \frac {n(n-1)} {2}$ it follows that among these numbers exactly $xy$ are odd.
- If $n = 4k$ or $n = 4k+1$, then $ \frac {n(n-1)} {2}$ is even, then $xy = \frac {n(n-1) }{4}$. Thus from (1), we have $n = n^2 - 4xy = (x-y)^2$.
- If $n = 4k+2$ or $n = 4k+3$ then $ \frac {n(n-1)} {2}$ is odd, thus $xy = \frac {1} {2} ( \frac {n(n-1)} {2} + 1)$, from which we deduce that $n-2 = (x-y)^2$.

Then, if such a configuration is possible, $n$ is a perfect square or $n-2$ is a perfect square. And this is not a property of $n = 2000$.

Pierre.

Maybe I don't get the problem right, but I think it has the folloing generalisation:
Let ai be any numbers, same thing for their difference.
In a Chinese ST this problem is present.


How many units are in the sum of the lengths of the three altitudes in a triangle with sides 7, 24, and 25?  Express yoru answer as a decimal to the nearest hundredth.

[hide]The altitude times the opposite side has to equal 168.  We can just divide 168 by 7, 24, and 25 to get 24, 7, and 6.72 respectively.  Then you just add them up to get 37.72.[/hide]


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Wow, the IMO seemed such a short length of time. I wonder if the Canadian IMO members have been online lately.  :huh: I can't wait until it's my turn for a crack at the IMO.  :ninja:

[quote="TheDreamer"][quote="Sunny"]Heheheh we have an awesome team this year.  I sure hope they did good...Jenn says the problems were somewhat easy this year, so I'm hoping for the best![/quote]

Heh.  My prediction, although optimistic, is that they got at least 3 golds :D[/quote]
I'm more pessimistic than you'd honestly think.  :oops:

[quote="TheDreamer"]Well these forums have been inactive for a while; and IMO is now over; so cmon guys, let us know!

Did Elyot get gold or spray paint?

Was David injured at opening ceremonies?  :D

We want to know!

GOGO team canada[/quote]

Whoa... Don't speak of any injuries. I cannot even fathom what might have happened due to the hurricane (if it materialized).  :ninja:

Glad someone found a computer :D

Sounds like you guys did really well, given the number of medals.

What did the rest of the ppl do while Elyot spun glow sticks?  :bruce:  :mad:  :mad:  :mad:  :mad:  :mad:

"persuation by intimidation  during coordination. "  Whaaa?  ;)

Haha, yeah it'll be nice staying a few extra days lol.  Neways GOOD JOB GUYS, you did us proud!   :)

[quote="Richard Peng"]Hi guys, this is Richard at the IMO (it took some efforts to find a computer)
Most of the marking of the contest are done. We´re expecting a gold for Yufei, silvers for the two penguins, bronzes for Elyot and Dave.
Elyot spun glow sticks on stage during the opening ceremony, which was great. :lol:  Also, the Newzelanders sent flying kiwi birds into the crowd. :rotfl:
CAN0 (the team mascot, CanMoo) also did great......it was really nice to use a moose for persuation by intimidation :mad:  during coordination.
I´ll try to keep you guys posted in the next few days. The hurricane is gonna to be awesome.[/quote]

OoOoOooOoo... So many medals.  :D CONGRATULATIONS!   :lol: BTW, does anyone know what is Canada's ranking? XD (<--- sideways smile) I've been reading a chapter about exponential growth and decay in a mathematics book that I found at the public library. I don't think they include information in the AoPS books about how to solve these problems. Is this chapter important for my preparation for different math-writing competitions? It's proved to be interesting and something that I haven't heard about before but it would be nice to hear some comments.  :lol:

[quote="Richard Peng"]I´ll try to keep you guys posted in the next few days. The hurricane is gonna to be awesome.[/quote]I was wondering why no Canadians post in the IMO 2005 forums  :lol: Congrats to y'all for the results!

[quote="d.lam.86"]I have a pessimistic guess but I'll keep that to myself.  ;) I don't want to upset anyone but don't get me wrong, I support Canada's IMO team.  :lol: Whoa, did you guys hear about the hurricane in Merida, Mexico? That's just scary.  :ninja:[/quote]

Heh.  Unless you know them,  you can't really have a guess....especially a pessimisic one!

Congratulations Team Canada!!!!

You all did amazing!!

Cheers,
Richard.

who is CAN3 and who's CAN5? My prediction is that Peng is CAN3 and Richard CAN5, since I don't see Peng mess up the 1st question :) also I see richard come up with right ideas for tough questions and wasn't able to finish it ;)

good job, canadian team! hopefully, Canada can get into top 10 next year in IMO

Wow, we've got a fan! Thanks Elyot, we liked you as well, we really made some fun. I can agree with all the things I know about that you wrote in you're post-imo-post. I especially recognized myself in this:

[quote="Elyot"]Still... nobody's funnier than the Dutch.  They ended up taking the tape off their windows and putting it back on again to spell the word HELP.  Crazy.[/quote]
When our leader, which was located in the hotel across the street, saw this, he had no doubt which room was ours (Johan and I lived in that room). He wanted to make a picture the next day but then we already changed the text to: "VIVA LA MEXICO CAMERONAS". Camarones means something like losers/*ssh*l*s.

Although, one of our most craziest plans didn't even work. We wanted to have some elections for the greatest people of the imo, in the categories "Most beautifull girl", "Best dressed guy", "Nicest guide" and "Most idiotic couple". Of course Johan and I wanted to win that last one ourselves :P.

Too bad I can't participate in next year's imo, but well, I'm going to university, which must be like some kind of full-year-imo.

Yours sincerely, Sjoerd (NLD1)

[quote="billzhao"]Due to popular demand, I've just uploaded the few hundred pictures that I took over the past month. You may find them at [url]http://photos.yahoo.com/z_yufei[/url]

If you want any picture in high-res, please email or PM me.

[quote="Elyot"] Where are all those cute pictures of IMO girls with the moose?  Did we get enough to make a calendar?[/quote]

When did this happen?

And Elyot, how's that IMO essay (which you volunteered to write) coming along?

For the first time, I was interviewed by an English newspaper (albeit a very small and local one) regarding IMO. Here's the [url=http://www.insidetoronto.ca/to/northy/story/2927519p-3391963c.html]article[/url].

Look how they spelled "combinator[b]ix[/b]." Also, "Calling high school math '[b]trivia[/b],' " Only Olympians will get what Lin actually said.[/quote]

Actually... nobody has said anything to me about the essay yet... I'm not actually sure what I'm supposed to write or who I'm supposed to send it to.

I remember taking some IMO moose shots... I think maybe some were on (Richard's??) camera.  I don't really remember though.

Those pics are awesome.  Sooo many memories.  Thanks!

[quote="billzhao"]Due to popular demand, I've just uploaded the few hundred pictures that I took over the past month. You may find them at [url]http://photos.yahoo.com/z_yufei[/url]

If you want any picture in high-res, please email or PM me.

[quote="Elyot"] Where are all those cute pictures of IMO girls with the moose?  Did we get enough to make a calendar?[/quote]

When did this happen?

And Elyot, how's that IMO essay (which you volunteered to write) coming along?

For the first time, I was interviewed by an English newspaper (albeit a very small and local one) regarding IMO. Here's the [url=http://www.insidetoronto.ca/to/northy/story/2927519p-3391963c.html]article[/url].

Look how they spelled "combinator[b]ix[/b]." Also, "Calling high school math '[b]trivia[/b],' " Only Olympians will get what Lin actually said.[/quote]

Lmao, "high school math is trivia".  But actually, if you think about it, it means the same thing as "trivial" in this case.

Aww Lin should've also said "random" and "ownage"; along with "trivial" those seem to be his favourite words :D

Maybe you guys should've also said "there's lots of mandos and cantos here", isn't that DMCI slang?

Another fond moment at IMO...

At the Mexican club, the DJ actually played NUMA NUMA!!!!  I was sitting near the bar with some of the girls from the dutch team and it came on, and I looked at them and I said "you know what this is, right?" and they said "yea..." and I said "you know we gotta go down there and dance."  So EVERYONE went down to the dancefloor and went nuts to NUMA NUMA!

Just imagine 30 IMO PEOPLE under the lights and laser beams cheering: "something breshna numa numa yay!  numa numa yay!  numa numa numa yay!"  "Maya he!  Maya ho!  Maya ho!  Maya ha-oh!"

or something to that effect!  It was memorable.

[quote="Valentin Vornicu"][quote="Elyot"]Just imagine 30 IMO PEOPLE under the lights and laser beams cheering: "something breshna numa numa yay!  numa numa yay!  numa numa numa yay!"  "Maya he!  Maya ho!  Maya ho!  Maya ha-oh!"[/quote]This sounds pretty much as "Dragostea din tei", a cool Romanian song :)[/quote]

Yes, that's the one.  It got very famous on the internet as "Numa Numa" though.  There are about 50 fanimutations of it and various other things... some actually quite funny.  Everybody knows the song... nobody actually knows what it really is (except for some of my Romanian friends...)

Just a interesting fact. On the trip back (the Merida->Mexico City), I took a photo of this volcano emitting smoke:
http://ca.pg.photos.yahoo.com/ph/peng_ri1/detail?.dir=e802&.dnm=4725.jpg&.src=ph
I just saw on the news that it blew up spectacularly yesterday. http://msnbc.msn.com/id/8111182/ :rotfl:

Also, for all you guys interested about the moose, I have it, muhahahaha. :moose:  The plan is to bring it to Poland as the Canadian mascot for the IOI in August. I'll return it to David's custody in winter camp next year (one of my favorite moments of the IMO was David casually telling a couple of New Zealanders who wanted to burn the moose that he has a black belt)

[quote="Richard Peng"]Just a interesting fact. On the trip back (the Merida->Mexico City), I took a photo of this volcano emitting smoke:
http://ca.pg.photos.yahoo.com/ph/peng_ri1/detail?.dir=e802&.dnm=4725.jpg&.src=ph
I just saw on the news that it blew up spectacularly yesterday. http://msnbc.msn.com/id/8111182/ :rotfl:
[/quote]

Richard... There's NO WAY that's the same volcano.  The MSNBC article says:

The volcano, which straddles the line between Colima and Jalisco states [b]430 miles west of Mexico City[/b], has had six spectacular eruptions in the past three weeks.

We were flying from Merida, which is EAST of Mexico City.  There is no way we could have passed by this volcano, which is 430 miles WEST of Mexico City.

[b]PWNED!!![/b]

Richard dont feel too bad, but mexico is a volcanic region.    :)  There are volcanos all over the place.  Besides scientists can predict when a volcanos about to errupt even if they cant guess the time or day, the volcano u passed by is no where near erruption, u probly saw clouds.   ;) If it was about to blow airplanes wouldnt be allowed near it anyway.  Honest and naive mistake is all.  :P

[quote="Richard Peng"] I'll return it to David's custody in winter camp next year (one of my favorite moments of the IMO was David casually telling a couple of New Zealanders who wanted to burn the moose that he has a black belt)[/quote]

Ah hahahaha, good on ya David!!

Double post, I know, but Dorette arrived in Calgary once again last night.  Hehehe I'm thrilled, she brought me an IMO t-shirt, hat and pin from mexico...

[quote="Valentin Vornicu"]I was wondering why no Canadians post in the IMO 2005 forums  :lol: Congrats to y'all for the results![/quote]

I did I think!  but might have been after this message.
btw congrats all.

[quote="bugzpodder"]oh btw nice avatar valentin, its the same as my teacher's avatar when he posts on a different forum :P[/quote]:D :yoda: thanks :P


Evaluate:
\[ \left( \binom{11}{1} \minus{} \binom{10}{1} \right) \plus{}
\left( \binom{11}{2} \minus{} \binom{10}{2} \right) \plus{} \cdots \plus{}
\left( \binom{11}{10} \minus{} \binom{10}{10} \right)\]


probability
inequalities
geometry
calculus
analytic geometry
triangle inequality

Something very close to your problem: choose a number between 1 and 2.

What is the probabilty that you chose 1.5 ? Answer : 0, because there is a infinity of numbers between 1 and 2.

For your stick, i think there is a number of way to break it which tends to infinity...

[quote="Ragingg"]If I have a stick and I break it into 3 pieces, what is the probability these three pieces form a triangle?[/quote]

1/4.  This problem is well-known, and has also been well-discussed around these parts.

If it's so well known, why don't you write the proof?

My solution is that the two cuts must be on the opposite sides of the midpoint and there cannot be more than half the length of the stick between the two cuts.  Thus it is 1/2*1/2=1/4!  :D

[quote="Wumbate"]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13138
rcv's solution is pretty clear.
Although, if someone has a clearer solution than that, I'd like to hear it.

Mine goes something like this: assume the stick goes from 0 to 1, and is broken at points a and b such that 0>x>y>1.  Then the lengths are x, y-x, and 1-y.  By the triangle inequality, you get y>.5, x<.5, 1+2x>2y.  Plotting all these inequalities gives you an area of 1/8, out of the total area of 1.  I must've done something wrong.  It may be that I assumed x>y, when it could also be x<y, in which case you'd multiply the 1/8 probability by 2 to account for the other possibility and come out with the answer of 1/4, but I'm not sure if that's where my mistake is.[/quote]

I looked at my earlier post and I have a simple solution without calculus:

WOLOG, the stick has length 1.

Consider

-----C1--|------C2----  as before.

Clearly, the two Cs have to be on opposite sides of the middle.  This has probability 1/2.

Now, let the midpoint have coordinate M, and let k < M

Assume C1 has coordinate M-k.  We pair up this coordinate with C1 --> k and consider the points such that C2 - C1 = 1/2

Since we can pair up all of the k's with an M-k such that their probabilities average out to 1/2, we conclude that the probability of having C2-C1 <= 1/2, given c1, c2 are on opposite sides of the midpoint is 1/2.

Multiplying, we get 1/2*1/2 = 1/4


algebra
polynomial
superior algebra
superior algebra solved

looks nasty
I knew this method too but I didn't manage to make de computations... I hope that one will find a nicer solution.

[quote="Kuba"]Let $P \in R[X]$ be a polynomial which satysfies for every $n\in N$ the following equality:$\displaystyle P(n) = \sum\limits_{k = 1}^n {k(k^2  - 1)(k^2  - 4)(k^2  - 9),n = 1,2,3,...} $
Find $P(-n)$.[/quote]

$\displaystyle P(n) = \sum_{k = 1}^n {(k-3)(k-2)(k-1)k(k+1)(k+2)(k+3) = 7!\sum_{k = 1}^n  {k+3\choose 7} = 7! {n+4\choose 8}}$, we are done!


calculus
integration
trigonometry
function
real analysis
real analysis solved

[u][b]The author of this posting is : prowler[/b][/u]
____________________________________________________________________

I am interesting in solution because I am beginner in Calculus.

$\int_0^{\infty} \frac{dx}{x^n+1} = \frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}$

[u][b]The author of this posting is : Kent Merryfield[/b][/u]
____________________________________________________________________

Note that we must have $n>1$ for the integral to converge and for this formula to be valid. However, $n$ need not be an integer.

The "simplest" method I know involves some advanced knowledge, namely the residue theorem from complex variables. Integrate $\frac{1}{z^n+1}$ along a closed path in the plane that includes the interval $\left[0,R\right]$ on the real axis, an arc of the circle of radius $R$, then the segment $z=te^{2\pi i/n}$ for $t\in \left[0,R\right]$, traced backwards. This integral can be calculated by the residue theorem from the one simple pole at $z=e^{\pi i/n}$. Letting $R\to \infty$ leaves us with the integral along the two rays, and those are specific constant multiples of each other. Do the appropriate algebra, and you get the result above.

Does anyone have a good method that doesn't use such advanced results?

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

This problem was posted and solved about 1 year ago. Though my method was the same as Kent's one, I don't remember whether another approaches were proposed.
Note that formula is true for all real $n>1$, not only integer.

[u][b]The author of this posting is : eugene[/b][/u]
____________________________________________________________________

If you don't like Kent solution you can use replace $x=t^{\frac{1}{n}}$ and then you wil obtain that our integral is equal to the $\frac{1}{n} B\left(\frac{1}{n},1-\frac{1}{n}\right)=\frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}$ where $B\left(p,q\right)$ is betta function

but the prove of $B\left(\frac{1}{n},1-\frac{1}{n}\right)=\frac{\pi}{\sin\frac{\pi}{n}}$ is not very simple... You need Taylor series...


The numbers on my watch are equally spaced in a circle.  It is now 2 o'clock.  What is the measure of the smaller angle between the hour and minute hands on my watch?

[hide]if you divide 360 degrees by 12 u see there 30 degre between each number. so theres 60 degrees between the 12 and 2.
[/hide]

divide 360 by 2 wich makes 6 hours and which is 180 degrees...

divide it by 2 again which is 90degrees..

3 : 90 = 2 : x

x = (90 * 2) / 3 = 60

60 degrees


Solve integer  equation:
$ abc\equal{}(a\plus{}b)(b\plus{}c)(c\plus{}a)$
 where  $ a,b,c$  are  integer numbers

AM>GM
$ abc \equal{} (a \plus{} b)(b \plus{} c)(a \plus{} c)\geqslant 2\sqrt {ab}*2\sqrt {bc}*2\sqrt {ac} \equal{} 8abc$
So we see that the equation has only solutions when one of a,b,c=0 and
 $ a \equal{} \minus{} b$ or
$ b \equal{} \minus{} c$ or
$ a \equal{} \minus{} c$ or
$ a \equal{} b \equal{} c \equal{} 0$

edit: oh right Ali, sorry :oops: ..

[quote="Pablo09"]AM>GM
$ abc \equal{} (a \plus{} b)(b \plus{} c)(a \plus{} c)\geqslant 2\sqrt {ab}*2\sqrt {bc}*2\sqrt {ac} \equal{} 8abc$
So we see that the equation has only solutions when one of a,b,c=0 and
 $ a \equal{} \minus{} b$ or
$ b \equal{} \minus{} c$ or
$ a \equal{} \minus{} c$ or
$ a \equal{} b \equal{} c \equal{} 0$[/quote]
you cannt use AM-GM,because $ a,b,c$ arent nessasarily NONEGATIVE...


Given real numbers $a, b, c, d$ with $a + b + c + d = 0$ prove that $a^3 + b^3 + c^3 + d^3 > 0$ if and only if $a^5 + b^5 + c^5 + d^5 > 0$.


What is the fastest way to do this. I want a way in less than 30 seconds.

The sum of three numbers $a$, $b$, and $c$ is $99$. If we increase $a$ by $6$, decrease $b$ by $6$, and multiply $c$ by $5$, the three resulting numbers are equal. What is the value of $b$?

if you had paper, just make a and c in terms of b  and solve the equation.  that should be around 20-30 seconds if you're fast.

I want to do it mentally anyway. I saw my friend do it mentally in like 30 seconds :|

[hide="Really easy (but only because of a special case in this particular problem)...dont peek until you got it!"]
$(a+6)+(b-6)=10c=a+b$... do you get it now? (i'd say that solves for $c$ in less than 5 s for sure)[/hide]

Well, if u want REALLY quick, use logic to find a plausible value for c. First number I tried was 10, gave me 56 and 44 for and a and b, which I knew were too large. Then I tried 9, and voila, 39,9,51. b is 51.

is this how you won the cd in our chapter? complete guessing?
  :clap:   :thumbup:

i know, he won our chapter cd by a longshot and we have the strongest chapter in Indiana.

[quote="#H34N1"]I want a mental way. What if this was a state cd[/quote]

Too bad this is a [b]slight[/b] modification of a 2002 state sprint problem.

didn't the actual one say 88? Wait... if it said 77, would the answer be 7?
EDIT:for c?

I'd hardly call 3-2, 3-1, 3-2 a longshot :P.

but to be fair, on 14 questions I answered first on 13 of them. Stupid mistakes X_X.


This is just my conjecture!

Suppose that $ A\subset \mathbf{R}$ is a Lebesgue measurable set but not Borel measurable. Then $ A$ can be decomposed as $ A\equal{}B\cup C$, where $ B\cap C\equal{}\varnothing$, $ B$ is Borel measurable and $ C$ is a Lebesgue measurable null set

Have you tried proving it? If so, what do your attempts look like?

I wonder why you add the requirement that $ A$ is not a Borel set.

As a direct result from the definition, any Lebesgue measurable set is the union of an $ F_\sigma$ set and a null set. Note that an $ F_\sigma$ set is always Borel.


graph theory
combinatorics proposed
combinatorics

Let be given a table 2nx2n. Find the least number k such that there is a way to paint k squares so that when leave out n arbitrary lines and n arbitrary columns, there is still at least a square painted.

I don't think it works...choose $n$ lines with at least $n$ marked points and delete them. Then choose $n$ colummns with the remaining marked point (if any) and you delete all the marked points.

Pierre.

Yup, I seem to have misunderstood: I thought we had to delete the intersections of $n$ lines and $n$ columns. I have no idea why I understood it like that though: there's nothing in the text suggesting such an approach :D. $k$ must clearly be $>2n$, as you have already mentioned (you've erased that now :)).

Sorry..

[Edit: another guess: $4n+1$ if $n$ is even and $4n$ if $n$ is odd; it's probably bogus (:D), but it comes from looking at what happens if we mark the two diagonals, which, intuitively, seems like the "worst case"]

Everything i did was wrong.....So i deleted it!

considere a bipartite graph, of $2n$ vertex in each "side", and an edge between vertex i on the right and vertex $j$ on the left if the intersection of line $i$ and colmun $j$ is painted.

Look taht if we select $n$ vertex from the left, we need they be joint to at least $n+1$ vertex in the right. This is equivalent to the condition the problem asks.

If we create a $2n+2$ cycle and a simple marriage between the $2n-2$ vertex not in the cycle, we get a bipartition with $2n+2+n-1=3n+1$ edges.

So i think the answer is $3n+1$, but to prove it, take $n$ vertex from the left, they are joint with $n+1$ from the right, so one of them has degree $>1$. doing this we find there are $n+1$ vertex of degree at least 2. taking the other $n-1$ vertex with one of the vertex of degree at least 2, we find there must be at least 3n+1 edges, it is easy computations.

Can anyone check it?

squars painted is a star*.
the maximal numbers of remove stars from leave out n line is 2n.

because if this value less than 2n,when one line is exists in that line it has one star.and out  of that n line one line exists then
it has 2 star.

now swap two lines l1to l 2  with continue  2n stars deleted.
     now maximum numbers of stars after leavs n line is n
                                becuase  3n -2n=n.
maximum sit  those stars down in n columns thus remove these
n columns.

considere a bipartite graph, of  vertex in each "side", and an edge between vertex i on the right and vertex  on the left if the intersection of line  and colmun  is painted.

with inductive proving if number of star sit stars is  3n+1   we cant remove n line and n column and we win .

with our graph with  2 vertex in each side is trivial (base case).

for n true  we determine for n +1 so true.
 2 vertex in each side with add  a,b in x-side and a'  b' in y side and conect a to a' and b to b' and a to b'  with  3 edges.

now minimum of numbers  for leave  is  2 in this comptanet and complete inductive becuse  3 edge add and  2 vertex. t is a minimum for number for 2n-1*2n-1 for delete stars.
t>2n-2
t+2>2n

proof is complete.


invariant
System
algorithm
polynomial
combinatorics
IMO
IMO 1993

Let $n > 1$ be an integer. In a circular arrangement of $n$ lamps $L_0, \ldots, L_{n-1},$ each of of which can either ON or OFF, we start with the situation where all lamps are ON, and then carry out a sequence of steps, $Step_0, Step_1, \ldots .$  If $L_{j-1}$ ($j$ is taken mod $n$) is ON then $Step_j$ changes the state of $L_j$ (it goes from ON to OFF or from OFF to ON) but does not change the state of any of the other lamps. If $L_{j-1}$ is OFF then $Step_j$ does not change anything at all. Show that:

(i) There is a positive integer $M(n)$ such that after $M(n)$ steps all lamps are ON again,

(ii) If $n$ has the form $2^k$ then all the lamps are ON after $n^2-1$ steps,

(iii) If $n$ has the form $2^k + 1$ then all lamps are ON after $n^2 - n + 1$ steps.

For (a), observe that lights can never be all off. Suppose lights are all off at Stepj, L(j-1) must be on, contradiction.
Denotes the situation of lights at Stepj as Sj. The number of pairs (j mod n, Sj) is finite, smaller than or equal to n*2^n. Take N > n*2^n, there exist i < j < N  s.t.  (i mod n, Si) = (j mod n, Sj).
Note each Sk is uniquely determined by any previous Sl (l < k), conversely, known Sk can determine each previous Sl. By Si = Sj, we can claim S(j-i) = S(i-i) = S0, which is the situation that all lights are on. M(n) = j - i.

At each position, assign light number $ 1$ if it is on, $ 0$ if it is off.
(a) First, the sequence cannot terminate: it can only terminate if it reaches configuration with all lamps off, which is not possible, since if after some step $ S_{k}$ the configuration is all lamps off, that must have been the configuration before step $ S_{k}$, so backtracking, the initial configuration must have been all off, which isn't true. Consider the infinite set of configurations after $ nt$ steps for some integer $ t$. Since there are only finitely many different configurations, some two must coincide. Choose the first such pair, after say $ i$ and $ j$ steps. So after $ j$ steps, the configurations just cycle. Now consider backtracking from step $ i$ to step $ 1$ (by backtracking, replace the relavent pair of switches as follows: (0,0) and (0,1) are unchanged, (1,0) goes to (1,1) and (1,1) goes to (1,0) ) : consider the sequence of steps taking step $ i$ to step $ 1$. Now applying the same sequence of steps to step $ j$, we can see by backtracking, after exactly $ i$ steps we will reach a configuration with all lamps on.

(b) Let's looking at the configurations $ K_{i}$, where $ K_{i}$ is the configuration obtained after $ ni$ steps. Denote $ K_{i,j}$ be the number assigned to light $ j$ in configuration $ K_{i}$.It's easy to see that, working in $ (mod$ $ 2)$, $ K_{i\plus{}1, 1} \equal{} K_{i,0} \plus{} K_{i, 1}$. (This is true, since two adjacent numbers (1,1), (1,0), (0,0), (0,1) are changed to (1,0), (1,1), (0,0), (0,1) respectively). From there, an induction gives us, that for $ 1 \leq j \leq n$, $ K_{i\plus{}1, j} \equal{} \sum_{l\equal{}0}^{j} K_{i,l} ... (1)$.
Looking at Pascal's triangle, for $ i\equal{}1,2$ we notice that the sequence: $ K_{i,0}, K_{i,1} ... K_{i,n\minus{}1}$ is identical to $ \binom {i} {i} , \binom {i\plus{}1} {i},  ... \binom {i\plus{}n\minus{}1} {i}$. Using $ (1)$, and well-known properties of Pascal triangle (i.e.: $ \binom {n\plus{}1}{k\plus{}1} \equal{} \binom {n}{k} \plus{} \binom {n}{k\plus{}1}$), and induction, we can see that this pattern is true $ i\equal{}1,2 ... n\minus{}1$. In particular, let's look at configuration $ K_{n\minus{}1}$: for $ 1 \leq i \leq n\minus{}1$, we have $ K_{n\minus{}1, i} \equal{} \binom {i\plus{}n\minus{}1}{i} \equal{} \binom {2^{k} \minus{} 1 \plus{} i}{2^{k} \minus{} 1}$. But using Lucas's Theorem ( http://planetmath.org/encyclopedia/LucassTheorem.html ), since the binary representation of $ 2^{k} \minus{} 1 \plus{} i$ contains at least one $ 0$, and $ 2^{k} \minus{} 1$ contains only $ 1$-s, it follows that, working in $ (mod$ $ 2)$, $ K_{n\minus{}1, i} \equal{} 0$ for $ i\equal{}1,2 ... n\minus{}1$ - so after $ n(n\minus{}1)$ steps we end up with a configuration with a $ 1$ for light # $ 0$, and a $ 0$ for every  other light. From there, after $ n\minus{}1$ steps, since each following steps toggles one light, we will have configuration with all lights on.

(c) Let's keep the notation from proof of (b) (i.e., the $ K_{i}$ notation), and let $ K'_{i}$ denote the arrangement obtained after $ ni$ turns (but here $ n \equal{} 2^{k} \plus{} 1$, and not $ n\equal{}2^{k}$). By looking at case $ k\equal{}2$ and comparing $ K_{i}$ with $ K'_{i}$, the following pattern becomes obvious: for $ i\equal{}1, 2, ... 2^{k}$, configuration $ K'_{i}$ is obtained from $ K_{i}$, by shifting $ K_{i}$ in cyclic direction twice, and adding in a $ 0$ for the lamp in position # $ 1$. We can prove this by using induction, and our equation $ (1)$ which tells us how to get configuration $ K'_{i\plus{}1}$ from $ K'_{i}$. Using this, we can see that $ K'_{n\minus{}2}$, every lamp except for lamp #$ 2$ is a $ 0$. Using equation $ (1)$, $ K'_{n\minus{}1}$ consists of all lamps except lamp # $ 1$ being a $ 1$. Thus, after one more step, all the lamps become $ 1$-s - so in $ n(n\minus{}1)\plus{}1 \equal{} n^2 \minus{}n \plus{}1$ steps, all lamps are on again.

We'll work in $\mathbb{Z}/2\mathbb{Z}$. Let $f(i)=1$ denote the initial state of lamp $L_i$ for $0\le i\le n-1$, and $f(i+kn)$ denote the state of lamp $L_i$ after it is modified $k$ times.

Then $S_j$, for $j\ge0$, gives us $f(j+n)= f(j+n-1)+f(j)$ and thus
\[G(x)=\frac{1}{1-x-x^n}=\sum_{r\ge0}f(r)x^r.\](We are working in $\mathbb{Z}_2[x]$.)

(i) Considering $n$-tuples $(f(t+1),f(t+2),\ldots,f(t+n))$, the periodicity is clear by the fact that at most $2^n$ such $n$-tuples can exist and a fixed $n$-tuple uniquely determines the entire sequence ($f(x)=f(x-1)+f(x-n)$ and $f(x-n)=f(x)-f(x-1)$).

(ii) If $n=2^k$, then it's equivalent to show that
\[\frac{x^{n^2-1}-1}{1-x-x^n}=(x^{n^2-1}-1)G(x)=P(x)\]for some polynomial $P\in\mathbb{Z}_2[x]$. But
\[x^{n^2}+x=(x^n)^n+x\equiv(x+1)^{2^k}+x\equiv x^{2^k}+1+x\equiv0\pmod{x^n+x+1},\]as desired (note that $\gcd(x,x^n+x+1)=1$).

(iii) If $n=2^k+1$, then it's equivalent to show that
\[\frac{x^{n^2-n+1}-1}{1-x-x^n}=P(x)\]for some polynomial $P\in\mathbb{Z}_2[x]$. But
\[x^{n^2-n+1}+1=(x^n)^{n-1}x+1\equiv(x+1)^{2^k}x+1\equiv(x^{2^k}+1)x+1\equiv0\pmod{x^n+x+1},\]as desired.

Solution from [i]Twitch Solves ISL[/i]:

For part (i), the sequence of states (and time indices modulo $n$) is finite, so it is certainly eventually periodic; however, because the operation is invertible, it is totally periodic.
Throughout the problem we always work modulo $2$. By a [i]round[/i] we mean a sequence of $n$ consecutive operations. Note that a round has the following effect: \begin{align*} 	(a_1, \dots, a_{n-1}, a_0) 	\mapsto (&a_2+a_3+\dots+a_0, \\ 	& a_1 + a_2, \; a_1 + a_2 + a_3, \; 		\dots, \; a_1 + a_2 + a_3 + \dots + a_0). \end{align*} (We prefer to list $L_1$ first, rather than $L_0$.)

For (ii), we give a description of the process for $n = 2^{k+1}$ in terms of the process for $n = 2^k$. We write our states as binary strings of length $n$, with the $i$th bit corresponding to $L_i$ (the last bit corresponds to $L_n = L_0$). The description goes as follows: [list] 	[*]We split the string into $L_1 L_2 \dots L_{n/2} \mid 		L_{n/2+1} \dots L_n$. 	[*]There are two copies of the process for $n=2^k$ side by side 	until we reach the string $100\dots \mid 1000$. 	[*]After the next round, we get $11\dots11 \mid 00\dots00$. 	[*]Then, the left half plays another copy of the $n=2^k$ process, 	with the right half being all zero, 	until we reach $1000 \mid 0000$ 	[*]Then $n-1$ turns later we get $111\dots111$ as needed. [/list]

For concreteness, if $n=4$ is given by \[ 1111 \xmapsto{4\text{ moves}} 1010 	\xmapsto{4\text{ moves}} 1100 	\xmapsto{4\text{ moves}} 1000 	\xmapsto{3\text{ moves}} 1111 \] in terms of the rounds, then $n=8$ is given by \begin{align*} 	1111\mid1111 	&\xmapsto{8\text{ moves}} 1010\mid1010 	\xmapsto{8\text{ moves}} 1100\mid1100 \\ 	&\xmapsto{8\text{ moves}} 1000\mid1000 	\xmapsto{8\text{ moves}} 1111\mid0000 \\ 	&\xmapsto{8\text{ moves}} 1010\mid0000 	\xmapsto{8\text{ moves}} 1100\mid0000 \xmapsto{8\text{ moves}} 	1000\mid0000. \end{align*} and seven moves later we return to all $1$'s. One can now check by a straightforward induction that the number of moves is $n^2-1$ exactly.

For (iii), we give a description for $n=2^k+1$ in terms of that for $n=2^k$. All that needs to be observed is that the situation after the $i$th round of $n=2^k+1$ is the same as the $i$th round of $n=2^k$, except with last bit moved to the first, and one extra $0$ appended, up until we reach the special state $0010000\dots$. For example, for $n = 9$ we have \begin{align*} 	111111111 	&\xmapsto{9\text{ moves}} 001010101 	\xmapsto{9\text{ moves}} 001100110 \\ 	&\xmapsto{9\text{ moves}} 001000100 	\xmapsto{9\text{ moves}} 001111000 \\ 	&\xmapsto{9\text{ moves}} 001010000 	\xmapsto{9\text{ moves}} 001100000 \xmapsto{9\text{ moves}} 001000000 \end{align*} and then after an additional $n+1$ moves we return to all $1$'s. The total move count can be computed to be the desired $n^2-n+1$.

This is a notation problem.

We break the problem into blocks of $n$ moves. Let the initial configuration be $C_0$, the configuration after $n$ moves be $C_1$, etc.

(i) Some $C_i$ and $C_j$ are the same for $i<j$ by the pigeonhole principle. Since the procedure can be done in reverse, $C_{j-i}$ and $C_0$ agree as desired; $M(n) = (j-i)n$ works.

(ii) For ease of writing, let $A$ denote ON and $B$ denote OFF, so the starting configuration is $\underbrace{AAA\cdots A}_{2^k}$; the element at index $z$ corresponds to $L_z$ with $0$-indexing. Define the following sequence of sequences of $A$ and $B$ called $W_1,W_2,\dots$ as follows. Let $W_{2^q} = \underbrace{AA\cdots A}_{2^q} \underbrace{BB\cdots B}_{2^q}$. For $1\le t < 2^q$, let $\lfloor \log_2 t\rfloor + 1 = p$, then define $W_{2^q+t}$ as $\underbrace{W_tW_t\cdots W_t}_{2^{q-p}}\underbrace{BB\cdots B}_{2^q}$. Note that this new sequence has exactly $2^{q+1}$ total instances of $A$ or $B$ when expanded. We claim that after $sn-1$ steps for $1\le s < 2^k$, the sequence of $A$s and $B$s is $\underbrace{W_sW_s\cdots W_s}_{2^{k-q-1}}$ where $q=\lfloor \log_2 s\rfloor$, shifted so that the initial $A$ of this sequence corresponds to $L_{n-1}$. The result is clear at $s=1$. Now, we show the result by induction for $s=a$ with $2\le a < 2^k$. Let $a=2^q+t$ so that $q$ is a positive integer and $0\le t\le 2^q-1$. If $t < 2^q-1$, we have two cases: either there exists some $1\le p < q$ such that $t=2^p-1$, or there does not. In the former case, note that each $W_tW_t$ sequence starting with the first metamorphosizes into $W_{t+1}$ by the result for $s=t+1$, whereas the $B$ terms in the second half of $W_a$ are unaffected. In the latter case, the result is trivial because each $W_t$ is replaced by $W_{t+1}$. Now suppose $t=2^q-1$. It is clear by induction that $W_a = W_{2^{q+1}-1} = A\underbrace{BB\cdots B}_{2^{q+1}-1}$; the result is clear at $q=0$, and each subsequent case follows from the last by the definition of $W_a$. Then $n$ moves transmute the first instance of $W_a$ into $\underbrace{AA\cdots A}_{2^{q+1}}$, the second instance into $\underbrace{BB\cdots B}_{2^{q+1}}$, and so on. This means that the new sequence is $\underbrace{W_{a+1}\cdots W_{a+1}}_{2^{k-q-2}}$, as desired.

Thus it suffices to observe that $n$ moves from $A\underbrace{BB\cdots B}_{2^k-1}$ transforms it into $\underbrace{AA\cdots A}_{2^k}$; we are done.

(iii) The idea is the same as that of (ii), but instead we claim that for $1\le s < 2^k$ and $q=\lfloor \log_2 s\rfloor$, the expression is $B\underbrace{W_sW_s\cdots W_s}_{2^{k-q-1}}$ at step $sn+1$ by induction on $s$. The result is trivial at $s=1$, and for $s\ge 2$, note that the manipulations of (ii) work exactly the same way because lamp $L_{n-1}$ is always assigned to $B$, so it does not flip. Then to go from $(n-2)n+1$ to $(n-1)n+1$ flips all the lamps $L_i$ with $i=0$ or $2\le i\le n-1$ back to $A$, so we are done.


I have a stack of coin whose probability $ p$ of coming up heads is uniformly distributed.  I takes a coin, flip it, and it comes up heads; what's the expected value of $ p$?


number theory
relatively prime
greatest common divisor

Can fractions be relatively prime? If so, how are they prime?
For example, is 1/2 and 1/3 relatively prime?

Two integers are said to be relatively prime if their greatest common divisor is $ 1$.  But if we are working with fractions, everything divides everything else, so it doesn't make sense to talk about greatest common divisors.

so relatively prime only applies to integers?


inequalities vector trigonometry calculus integration function

Let $ z_1$, $ z_2$, $ z_3$ be three complex numbers. Prove the inequality

$ \left|z_1 \plus{} z_2\right| \plus{} \left|z_2 \plus{} z_3\right| \plus{} \left|z_3 \plus{} z_1\right| \leq \left|z_1\right| \plus{} \left|z_2\right| \plus{} \left|z_3\right| \plus{} \left|z_1 \plus{} z_2 \plus{} z_3\right|$.

i like the following theorem very much(sometimes known as the leningrad method):
suppose we have vectors $ a_1, ..., a_n$ and $ b_1, ..., b_m$ in the plane and for every line through the origin(or some other point, it doesn't matter), let the projection of the vectors onto the line be $ A_1, ..., A_n$, $ B_1, ..., B_m$. suppose we always have $ |A_1| \plus{} ... \plus{} |A_n|\geq |B_1| \plus{} ... \plus{} |B_m|$. then $ |a_1| \plus{} ... \plus{} |a_n|\geq |b_1| \plus{} ... \plus{} |b_m|$.
proof: parametrise the lines through the origin by the angle $ \alpha$. integrate our inequality from $ 0$ to $ 2\pi$. the projection of a vector $ x$ to a line has length $ \left|x\right|\left|\cos \phi\right|$ where $ \phi$ is the angle between the vector and the line. as the line turns from $ 0$ to $ 2\pi$, the angle $ \phi$ takes all values from $ 0$ to $ 2\pi$ exactly once, thus the integral is proportional to the sum of the length of the vectors(with positive factor). that gives the result.

now, we just have to check that our result holds for vectors which lie on one line. but that is very easy using some case analysis.

maybe there's a more elementary proof, but i wanted to give an example how this theorem helps to kill such inequalities.

Peter

[quote="Peter Scholze"]i like the following theorem very much(sometimes known as the leningrad method):[/quote]

I know one of the Russian participants at the IMO 2004 who always uses this method with the attribute <<х*ёвый>> (ca. <<f*cking>>), and actually I am quite of the same opinion... you see, I am not converted to calculus yet. I actually knew this solution well enough (you probably remember we were solving exactly this problem in exactly this way during the opening ceremony of this IMO  :D ). I wanted to post it, too, but then I saw that Manlio asked for a nice solution...  :D  BTW, I wouldn't hesitate to use this method on a mathematical olympiad, as I used continuity in a geo problem of the DeMO 2002 or limits in the IMO 2004, but I am really not satisfied with such solutions  :(

For those who don't know, here is something about the source of the problem:

One of the 11th grade problems on the Allunion Olympiad 1976 was to prove that for any four vectors $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$, $\overrightarrow{d}$ on the plane satisfying the equation $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}=\overrightarrow{0}$, the inequality

$\left| \overrightarrow{a}\right| +\left| \overrightarrow{b}\right| +\left| \overrightarrow{c}\right| +\left| \overrightarrow{d}\right| \geq \left| \overrightarrow{a}+\overrightarrow{d}\right| +\left| \overrightarrow{b}+\overrightarrow{d}\right| +\left| \overrightarrow{c}+\overrightarrow{d}\right| $

holds.

Of course, since $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}=\overrightarrow{0}$, we have $\overrightarrow{d}=-\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right) $, and hence this inequality transforms into

$\left| \overrightarrow{a}\right| +\left| \overrightarrow{b}\right| +\left| \overrightarrow{c}\right| +\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right| \geq \left| \overrightarrow{b}+\overrightarrow{c}\right| +\left| \overrightarrow{c}+\overrightarrow{a}\right| +\left| \overrightarrow{a}+\overrightarrow{b}\right| $

for three arbitrary vectors $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ in plane. This is exactly the same as Manlio's inequality, since the complex numbers he uses are vectors. Actually I do have a Russian book containing the solutions of the Allunion Olympiad problems, but this particular one is not there (I see only a reference to another book which I don't have). So I would like to second Manlio's question...

  Darij

[quote="harazi"]Or from Popoviciu's inequality.[/quote]

I had this idea, too. But is Popoviciu's inequality also valid for functions from $ \mathbb{R}^{n}$ to $ \mathbb{R}$ ? (EDIT: It isn't!) I wonder because the only proof of Popoviciu's inequality I have ever seen uses Karamata, and I don't think that Karamata works for vector functions.

  Darij

I know there are three different proofs of Hlawka inequality published in these three articles of Gazeta Matematika A :

1) Gazeta Matematika A    7/1984,   pag 241     ----- author: Costantinescu

2) Gazeta Matematika A    1-2/1985, pag 48-50 ---- author : Beneze

3) Gazeta Matematika  A    ?/1985,   pag. 145-147----author:???????

I want to tell you the story about the ??????????????? in the third reference.

Three days ago I went by train to Parma, an University Town 400 km far from my house, to search those three articles about Hlawka ineq.

In fact, this is the only town in Italy that has some loose numbers of Gazeta Matematika A .

I found only the third article and I was so happy of that :lol: .

But I was so in a hurry that when I made a photostat of it I missed the first two pages, so now I have only the last page pag. 147 in which I can read only the first two references.

And I missed also the paper where I wrote the exact text of reference number 3, so I cannot remember the author and the month (in the articles of GM A the name of the author is in the first page).

When I was on train I realized of that  :(  .

A long trip for nothing. :?

Sorry, if you have wasted your time to read my story but I told it only to justify those terrible ????????????????????.

[quote]However, the proof mentioned by Darij is in the third reference.  [/quote]

[quote="Peter Scholze"]i like the following theorem very much(sometimes known as the leningrad method)[/quote]

Does this generalize to n dimensions?

In one-dimensional case, the proof given by  Pierre Bornsztein for  Hlakva's inequality is similar to a proof given in [3] in more general case,i.e. for Popoviciu's inequality.[ Compare with case when f(x)=|x| ].
 Hlawka's inequality holds in any pre-Hlibert (unitary) space.
It was conjectured in [7] that Hlawka's inequality is true in every real normed space. An answer to this question is given in [2] . Hlawka inequality does'nt holds for complex spaces. For other informations see [1],[3],[4]-[6] .

[b]   REFERENCES [/b]
[1] Hlawka Edmund (?), (quoted in :
[1-a] Hornich H.,[i] Eine Ungleichung fuer die Vektorlaengen , [/i]  Math. Z. 48 (1942) 268-274.)
[2]Kelly L.M. , Smiley ,D.M. and Smiley M.F. ,[i] Two dimensional  spaces are quadrilateral spaces,[/i] Amer.Math.Monthly      72 (1965) 753-754.
[3]Lupas A., [i]Asupra unei inegalitati pentru functii convexe [/i],(Romanian), Gazeta     Matematica ,Seia A , 1982 , Nr.1-2 ,49-52.
[4]Mitrinovic D.S.,[i] Analytic Inequalities [/i],Springer Verlag, 1970
[5]Pecaric J.E., Proschan F. and Tong Y.L.,[i] Convex Functions, Partial Orderings and Statistical Applications[/i] , Academic Press, 1992.
[6]Popoviciu T., [i]Sur certaines inequalities qui caracterisent les                  fonctions convexes, [/i]  An.Sti.Univ. Al.I.Cuza Iasi Sect.I-a , Mat.(N.S.)               11-B , (1965) 155-164.
[7] Smiley ,D.M. and Smiley M.F., [i]The polygonal Inequalities,[/i]                 Amer.Math.Monthly, 71 (1964) 755-760.

[quote="pbornsztein"]Hlawka's inequality follows immediately from the triangle inequality and the following identity (from Hlawka...) :
$(|a|+|b|+|c|-|b+c|-|c+a|-|a+b|+|a+b+c|) \cdot (|a|+|b|+|c|+|a+b+c|) = (|b|+|c|-|b+c|) \cdot (|a|-|b+c| + |a+b+c|) + (|c|+|a|-|c+a|) \cdot (|b|-|c+a|+|a+b+c|) + (|a|+|b| - |a+b|) \cdot (|c|-|a+b|+|a+b+c|).$

Pierre.[/quote]

How to think of such identity?


If a+b+c=1
Please prove: sigma(1/(a^2+b))>=6.75

I'm sorry for my poor English.

[quote="idontthinkso"]If a+b+c=1
Please prove: sigma(1/(a^2+b))>=6.75
[/quote]
It's wrong. See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88792
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=160017

Thank you. My classmate give me this wrong inequality.I don't think it's right ,but I couldn't find why it's wrong.Thanks for your answer and links.

My English isn't very good.I hope you can understand me.


Find the sum of all integers $ k$ such that $ \binom{23}{4} \plus{} \binom{23}{5} \equal{} \binom{24}{k}$.

Use Pascal's Identity: $ \binom{23}4\plus{}\binom{23}5\equal{}\binom{24}5\equal{}\binom{24}{24\minus{}5}$, and since $ 5\ne24\minus{}5$, the sum is $ \boxed{24}$.

*One might think this is just a trick question and the answer must be $ k\plus{}(24\minus{}k)\equal{}24$, but what if $ k\equal{}12$?

Then all that needs to be done is test k = 12, if that doesn't solve the equation, then the answer is 24.

[quote=rockets8128][quote=ArsenalFC][hide=Alcumus Solution]
By Pascal's Identity, we have $\binom{23}{4} + \binom{23}{5} = \binom{24}{5}$. However, we also have $\binom{24}{5} = \binom{24}{24-5} = \binom{24}{19}$. There are no other values of $k$ such that $\binom{24}{5} = \binom{24}{k}$, so the sum of all integers that satisfy the problem is $5+19 = \boxed{24}$.
[/hide]
[hide=Alcumus Challenge]Challenge: Is it a coincidence that the answer is 24?[/hide]
Seriously...[/quote]

lol just copy paste the alcumus sol lol[/quote]

A little bit hypocritical there

[hide=Solution]
For any value of $k$, we have $\binom{24}{k}=\binom{24}{24-k}$. Therefore, our answer is $k+(24-k)=\boxed{24}$.

Yes, that was the "coincidence" it was talking about.

Sorry for bumping this, but I wanted to do the challenge.
For any $r$ and $n$, $r+n-r=n+r-r=n$, because combining like terms. It's that simple.
I just realized pianoman24 said the same thing, sorry.

[quote=yofro]What a troll! The two sols will be $\binom{24}{k}$ and $\binom{24}{24-k}$, we don't even need to find k because it is just $k+24-k=24$[/quote]

Yeah algebra solves the problem

[quote=fucongcong][quote=awin][quote=yofro]What a troll! The two sols will be $\binom{24}{k}$ and $\binom{24}{24-k}$, we don't even need to find k because it is just $k+24-k=24$[/quote]

unless k=12, but it's not in this case :)[/quote]

how do you its not? :huh:[/quote]

It's clearly too big in this situation

[hide=e e e e e e e e e e e e]so i was being really stupid and proceeding to count 6 numbers from the right (24, 23, 22, 21, 20, 19) then right after that i just realized n C r is the same as n C n-r

hehe[/hide]

it is not a coincidence that the answer is [hide=Answer]24[/hide].

we just need to test [color=#888]k[/color] = [color=#f00]1[/color][color=#ff9a00]2[/color], and if no, then the answer is [color=#ff9a00]2[/color][color=#0f0]4[/color].

[quote=sw993]The factorial numbers add to 24 since the top number is 24.[/quote]

by pascal's identity, we have 24 choose (I don't know latex) k, we also have 24 choose 24-k. k+24-k=24. one-second answer.

This question is rather simple, simply realize that the values of $k$ that will work will be flipped around the numbers $12,12$ in a list of numbers $1$ to $24$. Because of this, it will definitely sum up to 24.

[quote=Lucasyao][quote=sw993]The factorial numbers add to 24 since the top number is 24.[/quote]

by pascal's identity, we have $\left(24 \choose k \right)$, we also have $24$ choose $24-k$. $k+24-k=24$. one-second answer.[/quote] here's yours but latexed

[quote=awin][quote=yofro]What a troll! The two sols will be $\binom{24}{k}$ and $\binom{24}{24-k}$, we don't even need to find k because it is just $k+24-k=24$[/quote]

unless k=12, but it's not in this case :)[/quote]

if k were 12 then just guess 24 and if its not then guess 12, its guaranteed to be either of those (lol trade accuracy xp for resistance xp)


probability
number theory
greatest common divisor
prime numbers

How many unique sets of 4 distinct prime numbers exist for which the sum of the members of the set is 38?

Please post a solution along with an answer.

Yeah, it's really tedious. Can you post your solution as well?

the answer is four, and I can't do anything with the wording since it's an existing problem.

[hide]Do the problem now with 34 instead of 38 (solution too)[/hide]

[hide]You should immediately eliminate two because of the parity.  For the rest, you just list and brute force like you have never brute forced before.  It might help to write down the prime list so its a little easier.[/hide]

You're not taking something into account. Brute force is boring, long, and very easy to make a mistake in, which is why I'm trying to see if anyone else can do it any other way.

Brute force accounts for ten of my points on the Sprint and twelve of my points on the Target at Nationals :roll: ... there are more advanced ways to brute force involving organization, parity, and other stuff. Of course all of the prime numbers have to be odd, and if we start the list with groups that start with 3 and 5 and work our way forwards, we can do it with a much lower chance of error and much higher speed.

(it's the only way I've learned, and the only way I will learn)

Oh, yeah. I forgot another thing about brute force. It takes a whole lot of time.

For example, try to brute force #10 on this year's national team round.

I'm guessing my definition of brute force is different from yours :p .

My idea of brute forcing involves making a list while organizing it and trying to do as little work as possible. It's kinda contradictory, but it's still not elegant, which is why I hate typing solutions >_____>

My definition of brute force is to drawing a tree diagram for 2^100 or rigorously trying to prove a complex theorem.

The wording IS strange.   :?
It asks for "unique sets" which simply means that you can't count {3,5,7,23} twice.  It does not say "distinct prime numbers", so the answer should be much larger and more tedious; for example, {2,2,5,29} should be included.  My guess is that "How many sets of distinct prime numbers.." was intended.

Here's another brute force question: :P

[b]Two different integers are randomly selected from the set of integers greater than 0 and less than 11 . What is the probability that their GCD is not a prime number? Express your answer as a common fraction.[/b]

[quote="Treething"]Wait, is four correct? The wording was slightly weird, so it could be four or some other answer.

I just made an organized list.

3 - 5 - 7 - 23
3 - 5 - 11 - 19
3 - 5 - 13 - 17
3 - 7 - 11 - 17[/quote]

Yeah, that is what I did, and that is how I got the answer of 4.

[quote="pianoforte"][hide]1          9
2          4
3          2
4          1
5          1

17/(10C2)=17/45 ?[/hide][/quote]

No, that's not right.  :P Not sure what you did though. :huuh:

[hide="My solution"]

I drew a chart and found that there were 32 possibilities out of 45.  Too many to list.

So the answer is 32/45.

$\frac{32}{45}$.[/hide]


ANyone here play MMORPGS?
I play Runescape- username Kevin1000

hey, i play runescape too- RS name lionkiller

18mil+ in bank ;)

I used to play runescape, but then just quit for some reason.

^same here. was "y2kbug" or maybe it was "y2kbug2".

I still play the multiplayer games on Jagex, "Hexaditidom" as always.

ragnarok online.. and maybe world of warcraft.. but i dont want to get myself addicted to WoW at this time.. heard its the best game ever tho

My Runescape name is fr4nkthetank.  For some stupid reason, Frankthetank wasn't available, so I substituted 4 for a.

I used to be addicted to online games like runescape and neopets.  But now that school is starting and my neopets account got hacked I don't play those kinds of things anymore.

wowowowowowowow
finally another 14 year bump
5th one yay

stop reviving ur not supposed revive old threads


Sorry I don't remember the exact question, but heres what the questions are asking:
#1:
you have the following 50  b  c
                               d   e  f
                               g   h  2
The product of the diagonals, the rows, and the column are the same. Find the sum of all possible values of g.

#2:
You choose three numbers without replacement from the set (1...1000), say a, b, and c. Then you choose three numbers from the remaining 997 numbers, say e, f and g.
The probability that a box of dimension a*b*c can be put into a box of dimension e*f*g is a fraction, find the sum of the numerator and denominator.

Thats all. By the way, does anyone know where to get all the AMC 10s and AMC 12s with solutions and explanations?

For the second question, I don't know if this is right but here goes.

[hide]
In order for one box to fit into another, all the dimensions $a,b,c$ must be less than each variable $d,e,f$. So we have \[a<b<c<d<e<f\] Since the variables $a,b,c$ and $d,e,f$ can be interchanged to get the same box, the variable placement does not matter for each of those groups. There are $\binom{1000}{6}$ ways to choose 6 increasing numbers for the variables in the inequality and $1000 \cdot 998 \cdot 997 \cdot 996 \cdot 995$ possible ways to choose the 6 variables. Therefore, the probability is \[\frac{\binom{1000}{6}}{1000 \cdot 998 \cdot 997 \cdot 996 \cdot 995}=\frac{1}{720}\] So the answer is \[1+720=721\][/hide]

[quote="NeverOddOrEven"]In order for one box to fit into another, all the dimensions $a,b,c$ must be less than each variable $d,e,f$. [/quote]

I wouldn't be so sure about this.  For example, a box with dimension $1, 2, 20$ clearly fits into a box with dimensions $2, 3, 21$, even though it is not true that $20 < 2$.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=462570#p462570]here[/url] for the second problem.

does anyone hav a solution to the first one? and for the second one, does first picking 3 numbers, then picking another 3 matters?


[img]http://i33.tinypic.com/212762f.jpg[/img]

is this correct?

[hide][img]http://i34.tinypic.com/168wfio.jpg[/img][/hide]

Yes, this should be correct. All equipotential points are correctly marked as same nodes.


geometry
circumcircle
geometric transformation
reflection
parallelogram
cyclic quadrilateral
geometry proposed

In acute triangle $ ABC$, let $ M$ be the midpoint of $ BC$, and let $ B_1,C_1$ be points on $ AC,AB$ such that $ B_1M\equal{}C_1M$. Let $ X\equal{}BB_1\cap CC_1$. Suppose that $ AB_1XC_1$ is cyclic.

Prove that $ X$ is the orthocenter of triangle $ ABC$.

[quote="massnet"]In acute triangle $ ABC$, let $ M$ be the midpoint of $ BC$, and let $ B_1,C_1$ be points on $ AC,AB$ such that $ B_1M \equal{} C_1M$. Let $ X \equal{} BB_1\cap CC_1$. Suppose that $ AB_1XC_1$ is cyclic.

Prove that $ X$ is the orthocenter of triangle $ ABC$.[/quote]
The exception is when $ AB \equal{} AC$.  Anyways, here's the solution otherwise:
[hide="Solution"]
Note: This solution only discusses the case for which $ AB\ne AC$, or else $ B_1$ and $ C_1$ can lie on $ AC$ and $ AB$ respectively such that $ B_1C_1\parallel BC$.

Let $ O$ be the circumcenter of cyclic quadrilateral $ B_1C_1BC$ and assume that $ O\ne M$.  Furthermore, let $ N$ be the midpoint of $ B_1C_1$.  Since $ M$ is the midpoint of $ BC$, $ OM\perp BC$.  Also, $ M$ and $ O$ lie on the perpindicular bisector of $ B_1C_1$ since $ OB_1 \equal{} OC_1$ and $ MC_1 \equal{} MB_1$.  Therefore, $ MN\perp B_1C_1$ and $ ON\perp B_1C_1$.  This means that $ M$, $ O$, and $ N$ lie on a line and this line is perpindicular to $ B_1C_1$.  Therefore, $ OM\perp B_1C_1\implies B_1C_1\parallel BC$.  Hence, $ AM$ meets $ B_1C_1$ at its midpoint, so $ A$, $ N$, and $ M$ are collinear.  Since $ MN\perp B_1C_1\implies AM\perp B_1C_1\implies AM\perp BC$, so $ A$ lies on the perpindicular bisector of $ BC$, which means that $ AB \equal{} AC$, which is a contradiction.  Thus, $ M \equal{} O$, meaning that $ MB \equal{} MC \equal{} MC_1 \equal{} MB_1\implies \angle CC_1B \equal{} 90 \equal{} \angle BB_1C$, so $ X$ is the orthocenter of $ \triangle ABC$, as desired. [/hide]

EDIT: Sorry, [b]highlanderI[/b]; I misread the problem.

Why is $ B_1C_1BC$ cyclic? and if its cyclic then you do not need such a long solution! it comes imediatly that $ \hat{AC_1X}\equal{}\hat{AB_1X}\equal{}90^o$.

Here's a solution that responds to the actual problem now, I hope it is correct:
[hide="Solution"]
Reflect $ B_1$ across $ M$ to form $ P$.  Thus, since $ B_1M \equal{} MP$ and $ MB \equal{} MC$, we have that $ B_1BPC$ is a parallelogram.  Also, $ \angle AC_1X \plus{} \angle AB_1X \equal{} 180$ since $ AB_1XC_1$ is cyclic.  Hence,
\[ 180 \equal{} 360 \minus{} \angle AC_1X \minus{} \angle AB_1X \equal{} (180 \minus{} \angle AC_1X) \plus{} (180 \minus{} \angle AB_1X) \equal{} \angle CC_1B \plus{} \angle BB_1C\]
Now, since $ B_1CPB$ is a parallelogram, we have that $ \angle BB_1C \equal{} \angle BPC\implies \angle BPC \plus{} \angle BC_1C \equal{} 180$, so $ C_1CPB$ is cyclic.  Now, we have that $ MC_1 \equal{} MB_1 \equal{} MP$, so $ M$ lies on the perpindicular bisector of $ C_1P$.  Since $ MB \equal{} MC$, we have that $ M$ also lies on the perpindicular bisector of $ BC$.  Thus, $ M$ is the intersection between the perpindicular bisectors of $ BC$ and $ C_1P$, so $ M$ is the circumcenter of cyclic quadrilateral $ C_1CPB$.  Hence, $ MB \equal{} MC \equal{} MC_1$, which means that $ \angle BC_1C \equal{} 90$ and similarly, $ \angle BB_1C \equal{} 90$, so $ X$ is the orthocenter of $ \triangle ABC$, as desired. [/hide]


probability
function
algebra
floor function

This is a nice example of a problem in which it might appear as if some detailed computation is necessary, when it isn't.

Let $ x$ be a randomly selected integer from the set $ {10000, 10001, \ldots , 99999}$. What is the probability that $ x^2$ and $ (x\plus{}10000)^2$ have the same number of digits?

Sorry, but that is incorrect.

And you also missed the trick.

[hide]It is unnecessary to calculate the $ \sqrt{10^9}$.[/hide]

lol the hint makes it waaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa...y clear sorry too much sugar today....  :wow:  :wow:

If they both have 9 digits, then $ 10000\le x < \sqrt{10}10000$ and $ 10000 \le x+10000 < \sqrt{10}10000$

$ 10000 \le x < \(sqrt{10}-1)10000$

Now, if they both have 10 digits, $ 10000\sqrt{10} \le x < 100000$ and $ 10000\sqrt{10} \le x+10000 < 100000$

$ 10000\sqrt{10} \le x < 90000$

If they both have 11 digits..., well they just won't. Therefore, letting $ \sqrt{10}10000=k$ we get a nice solution ;)

you mean

both $ x^2$ and $ (x \plus{} 10000)^2$ have [color=red]9[/color] digits:

x = 10000 ~ [color=red]21622[/color] = [color=red]11623 [/color]ways

both $ x^2$ and $ (x \plus{} 10000)^2$ have [color=red]10[/color] digits:

x = 31623 ~ [color=red]89999[/color] = [color=red]58377[/color] ways

[color=red]58377 + 11623 = 70000[/color]

[color=red]answer : 7/9[/color]

note: [x] represents the floor function too lazy to look it up
lol indeed you do not need to take any square root
let $ k\equal{}\sqrt{10^9}$
note that $ 20000<k<90000$
which is all we need to know about k (and that it is not an integer) both are easily verified
for both $ x^2$ and $ (x\plus{}10000)^2$ having 9 digits, $ 10000 \le x \le [k]\minus{}10000, [k]\minus{}19999 \text{numbers}$
for obth squares having 10 digits, $ [k]<x \le 89999, 89999\minus{}[k] \text{possibilities}$
total = $ [k]\minus{}19999\plus{}89999\minus{}[k]\equal{}70000$
there are 90000 5 digit numbers
our answer is $ \frac{70000}{90000}\equal{}\frac79$


Fie $x_{n+1}=arctgx_{n}$ .Demonstrati ca daca $x_0>0$ avem :
                       $ \mathop {\lim }\limits_{n \to \infty }nx^2_{n}= \frac{3}{2}.$

Se stie ca $x$ >= $arctgx$ deci $x_n$ e descrescator. Fiind marginit, e convergent, $\mathop {\lim }\limits_{n \to \infty }x_n = 0$. Avem $\frac{1}{x^2_{n}}$ strict crescator si nemarginit, deci putem aplica Cesaro-Stolz.

$ \mathop {\lim }\limits_{n \to \infty }nx^2_{n}= \mathop{\lim }\limits_{n \to \infty}\frac{n}{\frac{1}{x^2_{n}}}=
  \mathop {\lim }\limits_{n \to \infty}\frac{(n+1) - n}{ \frac{1}{x^2_{n+1}} - \frac{1}{x^2_{n}}}=
  \mathop {\lim }\limits_{x \to 0 }\frac{1}{\frac{1}{arctg^2{x}}-\frac{1}{x^2}}
$, care cu cateva artificii si un l'Hospital iese 3/2.

Ai dreptate Frozen dar eu nu stiu deocamdata artificile respective .
Poate sa posteze si cineva restul solutiei.
                                                                                Multumesc

Continuam:
$\mathop {\lim }\limits_{n \to \infty } nx^2_{n} = \mathop {\lim }\limits_{x \to 0 }\frac{arctg^2{x}*x^2}{x^2-arctg^2{x}} = \frac{x^2}{{(\frac{x}{arctg{x}}})^2-1}=\mathop {\lim }\limits_{x \to 0 } \frac{x^2 * {\frac{x-arctg{x}}{arctg{x}}}*{\frac{arctg{x}}{x-arctg{x}}}}{(1+\frac{x-arctg{x}}{arctg{x}})^2-1}=\frac{1}{2} * \mathop {\lim }\limits_{x \to 0 } x^2 * \frac{arctg{x}}{x-arctg{x}}=
\frac{1}{2} \mathop {\lim }\limits_{x \to 0 } \frac{x^3*\frac{arctg{x}}{x}}{x-arctg{x}} =
\frac{1}{2} \mathop {\lim }\limits_{x \to 0 } \frac{x^3}{x-arctg{x}}
$
Aplici l'Hospital, obtii $\frac{1}{2} \mathop {\lim }\limits_{x \to 0 } \frac{3x^2}{1-\frac{1}{1+x^2}}=
\frac{3}{2} \mathop {\lim }\limits_{x \to 0 }\frac{x^2(1+x^2)}{x^2} = \frac{3}{2}$

Mai sus am folosit faptul ca $\mathop {\lim }\limits_{x \to 0 } \frac{(1+f(x))^r - 1}{f(x)} = r$ cand  $\mathop {\lim }\limits_{x \to 0 } f(x) = 0$


geometry
circumcircle
perpendicular bisector
geometry proposed

[quote="tchebytchev"]Two circles $k_1$ and $k_2$ with centers $O_1$ and $O_2$ respectively touch externally at $A$. Let $M$ be a point inside $k_2$ and outside the line $O_1O_2$. Find a line $d$ through $M$ which intersects $k_1$ and $k_2$ again at $B$ and $C$ respectively so that the circumcircle of $\Delta ABC$ is tangent to $O_1O_2$.[/quote]

Finde a point $X$ diferent from $A$ in $O_1O_2$ so that $\frac{O_2X}{XO_1}= \frac{O_2A}{AO_1}$.
$XB$ intersects $k_1$ and $k_2$ at $B_2$ and $C_2$ respectively.
$O_1B_1$ intersects $O_2C_2$ in Y
By Menelao:
$\frac{O_2X}{XO_1}\frac{O_1B_1}{B_1Y}\frac{YC_2}{C_2O_2}=1$
then
 $\frac{O_2A}{AO_1}\frac{O_1B_1}{B_1Y}\frac{YC_2}{C_2O_2}=1$
$O_1A=O_1B_1$
$O_2A=O_2C_2$
then $YB_1=YC_2$
so the incircle of triangle $O_1O_2Y$ is tangent in $A,B_1,C_2$ so the circuncircle of $AB_1C_2$ is tangent to $O_1O_2$
QED

Invert al the image with center A, let M' by the inversion of M and $L_1,L_2$ the inversion of $K_1, K_2$ respectively.

Let $L_3$ by a line parallel to $L_1$ in the middle of $L_1,L_2$.
The perpendicular bisector of AM' cuts $L_3$ in O.
the circle $K_3$ with center O an radius OM'=OA
cuts $L_1$ in A' and A'' and $L_2$ in B' and B''.

Is easy to see that C'B' and C''A'' are parallels and perpendicular to $L_1,L_2$.
so $K_3$ is the image of a line that passes through M and cuts $L_1$ and $L_2$ in B and C.
The line B'C' (parallel to $O_1O_2$) is the image of a circle tangent to  $O_1O_2$ in A and passes through  B and C


analytic geometry
geometry
modular arithmetic
trapezoid
induction
trigonometry
number theory proposed

Find all positive integers $ n$ having the following properties:in two-dimensional Cartesian coordinates, there exists a convex $ n$ lattice polygon whose lengths of all sides are odd numbers, and unequal to each other. (where lattice polygon is defined as polygon whose coordinates of all vertices are integers in Cartesian coordinates.)

Beautiful problem - fits into the category of [i]combinatorial geometry of numbers:-)[/i]

Any even $ n\ge4$ works, and any odd $ n$ fails.

Let $ A_i(x_i,y_i)$, $ i \equal{} \overline{1,n \plus{} 1}$, $ A_{n \plus{} 1} \equal{} A_1$ be the sequence of vertices of our polygon. Since $ A_iA_{i \plus{} 1}$ is odd, $ x_{i \plus{} 1} \plus{} y_{i \plus{} 1}\not\equiv x_i \plus{} y_i\pmod2$. Then if $ n$ is odd, from one side $ x_1 \plus{} y_1\equiv x_n \plus{} y_n$. On the other side $ A_n$ and $ A_1$ are consecutive vertices, so the above incongruency must hold.

Let now $ n$ be even. I assume there are several ways...

We will construct an $ n$-gon $ A_1A_2\ldots A_n$, so that $ Ox\|A_1A_n\|A_2A_{n \minus{} 1}\|\ldots$ and $ A_1A_n\cap A_2A_{n \minus{} 1}\cap\ldots\neq\emptyset$. We use induction. Assume we have constructed such a convex polygon $ A_1A_2\ldots A_{2k \minus{} 2}$, where $ n \equal{} 2k$. Let $ M$ be a lattice point whose $ Ox$ coordinate lies both on $ A_1A_n$ and $ A_{k \minus{} 1}A_k$ (such a point exists, since $ A_1A_n\cap A_2A_{n \minus{} 1}\cap\ldots\neq\emptyset$. Consider the line $ d$ through $ M$ parallel to $ Oy$. Clearly, we can pull the right side of the polygon (wrt $ d$) by any number even number of units, and obtain an also good polygon. Call this transformation a $ T$-extension if after applying it the side $ A_{k \minus{} 1}A_k$ becomes equal to $ T$, where $ T$ is obviously odd.

So having constructed a convex $ A_1A_2\ldots A_{2k \minus{} 2}$, let $ \alpha_i \equal{} \angle(A_{i \plus{} 1}A_i,Ox)$ for $ i \equal{} 2,\ldots,k \minus{} 1$ and $ \beta_i \equal{} \angle(A_iA_{i \plus{} 1},Ox)$ where $ i \equal{}k, k \plus{} 1,\ldots,n \minus{} 1$ and all angles are taken acute. From the convexity of our polygon $ \alpha_2 > \alpha_3 > \ldots > \alpha_{k \minus{} 1}$ and $ \beta_{n \minus{} 1} > \beta_{n \minus{} 2} > \ldots > \beta_k$. Define $ \theta \equal{} \min\{\beta_k,\alpha_{k \minus{} 1}\}$.

We will show that there is a $ T$-extension and two points $ X$ and $ Y$ so that $ A_{k \minus{} 1}XYA_k$ is a trapezium with $ XY$, $ A_{k \minus{} 1}X$ and $ YA_k$ all odd and $ \max(\angle(XA_{k \minus{} 1},Ox),\angle(YA_k,Ox)) < \theta$. The last condition will ensure the convexity of $ A_1\ldots A_{k \minus{} 1}XYA_k\ldots A_{2k \minus{} 2}$.

Let $ A \equal{} A_{k \minus{} 1}$, $ B \equal{} A_k$. A $ T$-estension makes $ AB$ of length $ T$, where $ T$ is odd, arbitrarily large. Let $ m \equal{} 6k \plus{} 3$ for some $ k$. Note that $ (6k \plus{} 3)^2 \plus{} (18k^2 \plus{} 18k \plus{} 4)^2 \equal{} (18k^2 \plus{} 18k \plus{} 5)^2$ and $ (6k \plus{} 3)^2 \plus{} (6k^2 \plus{} 6k)^2 \equal{} (6k^2 \plus{} 6k \plus{} 3)^2$.
So it is worth taking the altitude of the trapezium $ AXYB$ equal to $ 6k \plus{} 3$. Clearly $ AX \equal{} 18k^2 \plus{} 18k \plus{} 5$ and $ YB \equal{} 6k^2 \plus{} 6k \plus{} 3$, which are odd numbers. Note also that $ XY \equal{} AB \minus{} (18k^2 \plus{} 18k \plus{} 4) \minus{} (6k^2 \plus{} 6k)$, so it is odd. Moreover this proves the base of the induction - the case $ n \equal{} 4$.

Since $ \tan\angle XAB \equal{} \frac {6k \plus{} 3}{18k^2 \plus{} 18k \plus{} 4}$ and $ \tan\angle{YBA} \equal{} \frac {6k \plus{} 3}{6k^2 \plus{} 6k \plus{} 2}$, for sufficiently large $ k$, we'll have that these values are less than $ \tan\theta$, ensuring the angles are less than $ \theta$. It is quite clear they are different hence $ AX\neq BY$. Moreover, for big enough $ k$, $ AX,BY >$ all other sides except $ XY$ and $ A_1A_{2k \minus{} 2}$ which, by means of a sufficiently large extension (enough large to cover all our requirements), can become the largest sides of $ A_1A_2\ldots A_{k \minus{} 1}XYA_kA_{k \plus{} 1}\ldots A_{2k \minus{} 2}$.

Actually this problem is very easy(even to the degree of trivial). It was also on one of the quizzes of the 2008 USA Black MOSP and was solved by almost all people.

Similarly as above we show that $n$ is even and greater then 2. I do the construction in different way. At first we prove following lemma:

[b]Lemma "I said that"[/b] : [i]For every even $n\ge4$, even positive number $s$ and odd positive numbers $l_1<l_2,<dots<l_{n-2}$ there are odd positive numbers $K,k_1,k_2,\dots,k_{n-2}$ such that  numbers $l_1k_1,l_2k_2,\dots,l_{n-2}k_{n-2}$, $Ks^2-1$ and are all pairwise different and[/i] \[sK=l_1k_1+l_2k_2+\dots+l_{n-2}k_{n-2}.\]

[b]Proof of the lemma[/b] : Let $L=l_2+l_2+\dots+l_{n-2}$. Notice that $L$ is odd. Then just put $k_i=l_1$ for all $1<i\le n-2$ and $K=K'l_1$. Then we'll get $sK'=L+k_1$. Obviously for any choice of $K'$ we get unique odd $k_1$. Now just use odd $K'$ which is so high, that $Ks^2-1$ and $l_1k_1=l_1(sK'-L)$ don't have the same value (it's possible, because $s\ne0\neK$) and they are neither equal to any of $k_il_i$. Two of $k_il_i$-s obviously don't have the same value, because $l_{i+1}>l_i$. $\square$

Back to the problem. We'll just take $n-1$ rightangle triangles with sides $4a,4a^2-1$ and $4a^2+1$. First one we want to put on $x$-axis by its odd side. The rest we want to put from top of this triangle whole the way down, one after another, with even side horizontal. And in order to make it work, we just simply find odd coefficients to multiply all the triangles, which exist because [i]I said that[/i].

Q.E.D.

[hide=comedy central]
We show the answer is even $n \geq 4$. This is clearly necessary, so we show sufficiency.

Note the convex condition is basically irrelevant because we can keep on "flipping" reflex angles until we get a convex polygon. Thus, it suffices to prove there exists any polygon.

In addition, if we have two polygons, one with $a$ edges and one with $b$ edges, then any nonnegative linear combination of $a$ and $b$ works, because we can just stick the two polygons together at a vertex and scale one up by a large odd factor so the conditions are preserved. Thus, it suffices to show that $n=4$ and $n=6$ are possible.

For the former, take the vectors $[3,4] \cdot A$, $[3, 4] \cdot -3A$, $[8, 15] \cdot 17 \cdot 40$, and $[20, 21] \cdot 13 \cdot 42$, where $A = 17 \cdot 40 + 13 \cdot 42$. For the latter, take the vectors $-[1000137, 0]$, $[1000001, 0]$, $[12, 5]$, $[24, -7]$, $[40, -9]$, $[60, 11]$. Both sets of vectors should sum to $0$ (i might have messed the first one up but the idea is clear), so we are done.

The vectors have commas because latex can't do spaces
[/hide]

[hide=wait]
i'm not sure if the above solution actually works because when you get two parallel sides in a row stuff dies. I think the second construction is probably generalizable though to all $2 \bmod 4$, so it still should work out, albeit with more words.
[/hide]

[quote=khina]
The vectors have commas because latex can't do spaces
[/quote]

FYI, $\LaTeX$ can do spaces, although its implementation is a bit clunky.
\[
(3\,4)\qquad (3\; 4)\qquad (3\;\;4)\qquad(3\;\;\;4)
\]

I believe latex can also do spaces if you insert a tilde, as such: $3~4~5$
(Compare this to the same latex without spaces, which is $345$)


geometry
calculus
integration
search
calculus computations

this is a fun question i think! :D

$\textbf{(i)}$ there is a circular patch of radius $\textbf R$, and a dog is
tied firmly to a point on the [i]inside [/i]of the boundary
of the patch using a rope of length $\textbf L$. :P

what would be the value of $\textbf L$  such that the dog can
cover an area [u]exactly[/u] $\frac{1}{2}$ of the patch ?

[ PS: you can set up an appropriate integral and then
proceed to solve that ]

$\textbf{(ii)}$ now consider tying the dog to a point on the [i]outside [/i]
of the boundary. let $\textbf L<\pi\textbf R$ such that the dog is unable
to reach around any further than the point exactly opposite
to where it is tied.

what would be the value of $\textbf L$ now such that the dog can
cover an area [u]exactly[/u] $\frac{1}{2}$ of the patch ?

what happens when $\textbf L=\pi\textbf R$ ? does the shape of the
covered area remind you of a familiar curve ?  :|

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=739520161&t=24514]A problem[/url] which is related to one part of this.

that's cool.  :D

i personally like this problem, too...... i've always wondered
if the curve traced out is actually a [b]cardioid[/b] when $ \textbf L=\pi\textbf R$.

It's not a cardioid. In particular, a large portion of the graph is merely a semicircle.


algebra
polynomial
inequalities
limit
quadratics
inequalities unsolved

I have been trying to find a lower bound $f(a)$, a polynomial, for the quantity $\frac{a}{a+1}$. It would be used in a problem where $a+b+c=1, a,b,c\geq 0$ and equality at $a=b=c=\frac13$. Naturally, I want: \[\frac{a}{a+1}\geq f(a) \iff (3a-1)^{2}(1-a)k\geq 0 \] or such (I have proven that such an inequality needs at least that degree, if not one more I believe). I realize this is equivalent to finding $g(a)\geq \frac{1}{a}$ and have tried infinite series bounds. However, all of my methods fail and I only get upper bounds on $\frac{a}{a+1}$. Can someone who is more seasoned help me?

I don't understand.  Do you mean a positive lower bound?  But $\lim_{a \to \infty}\frac{a}{a+1}= 1$ which implies that $f(a)$ must be a constant polynomial.

A positive lower bound on $(0,1)$ I mean. In fact, it seems as if the opposite holds for $a>1$. But of course we are only considering a from 0 to 1.

Ah, right, sorry.

Could you tell us the entire problem?  This might not be the best way to do it.

(One way I can see is that you want equality at $a = \frac{1}{3}$, so $f( \frac{1}{3}) = \frac{1}{4}$.  Determine negative values of $f(0)$ and $f(1)$ that make $f(a)$ a likely quadratic lower bound and solve for it...)

Hmmm there are a lot of very experienced inequality problem solvers here, I'm really curios about their input...

What do you think I have left to explain, I thought it was rather clear?

Yes, but I mean could you tell us the original inequality?  It doesn't seem to me like this is the best approach...

Sorry. No original inequality :|

But you could easily fit it to one:

Just recently someone posted this one:
$\sum \frac{a^{2}}{(a+1)^{2}}\geq \frac{3}{16}, a+b+c=1$


Find all primes $ p,q$ such that $ p|q\plus{}6$ and $ q|p\plus{}7$

$ \frac {q \plus{} 6}{p}$ and $ \frac {p \plus{} 7}{q}$ should both be integers. so in particular $ q \minus{} 7 \leq p \leq q \plus{} 6$. this means $ \frac {q \plus{} 6}{p} \leq \frac {q \plus{} 6}{q \minus{} 7}$. but if $ q > 20$, $ \frac {q \plus{} 6}{q \minus{} 7}< 2$, which means then that we need $ \frac {q \plus{} 6}{p} \equal{} 1$, that is, $ p \equal{} q \plus{} 6$. then $ \frac {q \plus{} 13}{q}$ must be an integer, and the only solution in this case is $ (p,q) \equal{} (19,13)$. so there are no solutions for $ q > 20$. and an easy check shows that the only solution with $ q < 20$ is $ (19,13)$.


A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains $ 100$ cans, how many rows does it contain?

$ \textbf{(A)}\ 5\qquad
\textbf{(B)}\ 8\qquad
\textbf{(C)}\ 9\qquad
\textbf{(D)}\ 10\qquad
\textbf{(E)}\ 11$

The sum of first n odd numbers is $ n^2$ so n=10.
D

We have
\[ 1\plus{}3\plus{}\cdots \plus{} \left(2n\minus{}1\right)\equal{}100.\]
The LHS is the sum of the first $ n$ odd positive integers, which is equivalent to $ n^2$. Then $ n^2\equal{}100 \implies n\equal{}10 \rightarrow \boxed{\textbf{D}}$.


Look here.... http://www.kalva.demon.co.uk/putnam/putn76.html

problem B6

I am asking you, is there any mistakes in that solution...
I don't think that there is such number that $\sigma(n)=2n+1$ because
[hide]
You se that it can't have two or three diferent divisors, so suposse that it has more then 3 divisors $1=d_1<d_2<...<d_{k-1}<d_k=n$ so $d_1+d_2+...+d_{k-1}+d_k=2n+1$ and from that $d_2+...+d_{k-1}=n$ but $d_2d_{k-1}=n$ so $d_2=2$ and $d_{k-1}=2$ so $contradiction$ [/hide]

That doesn't change anything as you can see I made typo
I will edit it now...
But do you agree that this is correct and that solution (folow the link) is wrong because there is no such number

[quote="M@re"] $d_2d_{k-1}=n$ so $d_2=2$ and $d_{k-1}=2$ so $contradiction$[/quote]
I do not understand. What do you mean?

Try to understand this
I was very sleepy when I posted this...:) :) :) :) so I talk foolishness
Everything is OK, except me :D

Forget about it...

Bye


geometry
trapezoid
function
algebra
domain
national olympiad

I just found Finnish mathematical competition problems in the Internet. Here comes the most advanced level problems (1st round). Have fun!

1. The harmonic mean of two non-zero numbers $a$ and $b$ is number $h$ such that
\[ \frac{1}{h}=\frac{1}{2}\left (\frac{1}{a}+\frac{1}{b}\right ).  \]
Let $ABCD$ be a trapezoid with $AB||CD$. Let $s$ be the line segment parallel to $AB$ and $CD$ going through the intersection point of the diagonals $AC$ and $BD$. Prove that the length of $s$ is the harmonic mean of $AB$ and $CD$.

2. Solve the equation
\[ \sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0.  \]

3. Determine all function $f$ whose domain is $\mathbb{R}\setminus\{0\}$ satisfying
\[ f(x)+8f\left (\frac{1}{x}\right )=-63x  \]
for all $x\not =0$.

4. Let $k$ be the number of right angles of a given $n$-gon.
1) Prove that $k<\frac{2}{3}(n+2)$.
2) Assume that $n$ is divisible by 3. Prove that the case $k=\frac{2}{3}n$ is possible.


Gauss
geometry
parallelogram
geometry proposed

Let P be an interior point of a triangle ABC. Cevians AP, BP, CP intersect sides
in points A, B, C. Prove that:
1) Lines that join midpoints of segments BC, AC, AB
with midpoints of sides BC, AC, AB respectively concur in a point T.
2) T, P, G where G  gravity center, are collinear.

I will denote with M(XY) the midpoint of an arbitrary segment XY.

Let T be the centroid of the four points A, B, C, P. Then, I will prove that this point T lies on the lines M(B'C')M(BC), M(C'A')M(CA) and M(A'B')M(AB), and that the points T, P, G are collinear. This will obviously solve the problem.

By the Gauss-Newton theorem, applied to the complete quadrilateral formed by the lines AB, CC', CA, BB', the midpoints of the segments BC, B'C' and AP are collinear. Hence, the line M(B'C')M(BC) coincides with the line M(BC)M(AP). But from the fundamental properties of centroids, the centroid T of the four points A, B, C, P lies on the line M(BC)M(AP). Hence, we have proven that the point T lies on the line M(B'C')M(BC). Similarly, we see that it also lies on the lines M(C'A')M(CA) and M(A'B')M(AB).

Now it remains to show that the points T, P, G are collinear. But this is again a consequence of properties of centroids, since the point G is the centroid of the points A, B, C, and the point T is the centroid of the points A, B, C, P. Moreover, we have PT : TG = 3 (with directed segments).

  Darij

[quote="hucht"]Whe you say T is de centroid of ABCO Do you mean that M is the centroid of ABCO's Wittenbauer paralelogram?[/quote]

I don't say that T is the centroid of ABCP (not ABCO, by the way, there is no O in my post). I say that T is the centroid of the four points A, B, C, P, i. e. the midpoint of the segments $M_{AB}M_{CP}$, $M_{BC}M_{AP}$, $M_{CA}M_{BP}$, where $M_{XY}$ denotes the midpoint of any segment XY. This centroid is the center of the Varignon parallelogram (the parallelogram $M_{AB}M_{BC}M_{CD}M_{DA}$. The center of the Wittenbauer parallelogram is the "area centroid" of the quadrilateral ABCP and something completely different.

  Darij


[color=darkred]$C_{k}\blacktriangleright$ Se dau punctele $A$ , $B$ , doua segmente de lungimi $x\ ,\ y\ ,\ x\ne y$ si numarul natural $k\in\overline{1,3}$. Sa se construiaca cu linia si compasul pozitia punctului $P\in (AB)$ pentru care $\frac{PA}{PB}=\left(\frac{x}{y}\right)^{k}$.
$C\blacktriangleright$ Se dau punctele $A$ , $B$ si patru segmente de lungimi diferite doua cate doua $x\ ,\ y\ ,\ z\ ,\ t$. Sa se construiasca pozitia punctului $P\in (AB)$ pentru care $\frac{PA}{PB}=\frac{x}{y}\cdot\frac{z}{t}$.

[b]Remarca.[/b] In primul rand este esential sa avem cel putin o constructie. In al doilea rand, vom aprecia impreuna constructia cu cele mai putine constructii elementare si cu utilizarea cercului cat mai putin. Cu alte cuvinte, o problema de optim constructiv.[/color]

[quote="Virgil Nicula"][color=darkred]$C_{k}\blacktriangleright$ Se dau punctele $A$ , $B$ , doua segmente de lungimi $x\ ,\ y\ ,\ x\ne y$ si numarul natural $k\in\overline{1,3}$. Sa se construiaca cu linia si compasul pozitia punctului $P\in (AB)$ pentru care $\frac{PA}{PB}=\left(\frac{x}{y}\right)^{k}$.
[/color][/quote]

Pt $C_{1}$ luam segmentele de lungimi $x,y$ respectiv in punctele $A,B$, paralele , in semiplane diferite fatza de $AB$., unind cealelalte extremitati ale fiecarui segment , se obtine punctul $P$ prin intersectia cu $AB$, prin thales.

Cum m-am gandit eu mai departe, este asa: construim acum conjugatul armonic al lui $P$ in raport cu $A,B$ : luam un trunghi oarecare cu 2 varfuri in $A,B$, fie al treilea $C$, apoi 2 drepte din $A$, respectiv $B$ concurente cu $CP$ in $Q$. Notam intersectile lui $AQ,BQ$ cu $BC,AC$ cu $M,N$. acum prelungind $MN$ pana taie $AB$ obtinem conjugatul armonic $P'$ al lui $P$ in raport cu $A,B$.

Pasul urmator : ducem cercul de diametru $PP'$. si mai luam un punct arbitrar $C'$ pe el. acum am obtinut un triunghi cu raportul laturilor $\frac{x}{y}$

Am efectuat aceasta constructia pt ca sa demonstrez ca putem pleca fara a restrange generalitatea de la faptul ca se poate construi un triunghi cu "baza $(AB)$ cu laturile "de sus" in raport $\frac{x}{y}$

pentru $C_{2}$ acum ne este suficient sa construim simediana din $C$ care taie $AB$ intr-un punct $P_{2}$ ce imparte $(AB)$ in raportul $(\frac{x}{y})^{2}$ (teorema lui steiner)

La restul, o sa scriu maine. :P


Find the smallest perfect square whose decimal representation begins and ends with three 4's

The number must end with a $2$ or an $8$. Throwing around some numbers shows that the only numbers under $1000$ whose squares are $\equiv 444 (\text{mod}\;1000)$ are $038$, $538$, $462$, and $962$.

We can expect the number to start with either a $66$ or a $21$ based on their squares ($4365$ and $441$ respectively).

For ending $038$, I'm getting $667038^{2}$ to be the minimum.
For ending $538$, I'm getting $2107568^{2}$.
For ending $462$, I'm getting $2107462^{2}$.
For ending $962$ I'm getting $2107962^{2}$.

Of these four, the minimum is $667038^{2}=\boxed{444939693444}$, assuming I didn't miss any possibilities.


integration
real analysis
real analysis solved

how do i solve this ?

$(0<m\leq n)\\
\\
\\
\int{\frac{x^m}{(x-a_1)\cdot(x-a_2)\cdot(x-a_3)\cdot ... \cdot (x-a_n)}dx}$

solution is to brake the fraction into rational fractions or somethin and then the result is somethin like
$\sum{c_i \cdot ln(x-a_i)}$

yeah, that's the idea right there...

write

$\frac{x^m}{(x-a_1)\cdot(x-a_2)\cdot(x-a_3)\cdot ... \cdot (x-a_n)} = \frac{c_1}{x-a_1}+...+\frac{c_n}{x-a_n}$.

then $c_i = \frac{a_i^m}{\prod_{j \neq i} (a_i-a_j)}$

please explain why you chose $c_i$ of that value

That was a calculation of the value of $c_i$. Multiply both sides by $x-a_i$ and then evaluate at $x=a_i$.

well you want

$ \frac{x^m}{(x-a_1)\cdot(x-a_2)\cdot(x-a_3)\cdot ... \cdot (x-a_n)} = \frac{c_1}{x-a_1}+...+\frac{c_n}{x-a_n} $

multiply both sides of this by $(x-a_1)\cdot(x-a_2)\cdot(x-a_3)\cdot ... \cdot (x-a_n)$... so that what you want is

$x^m = \sum_{i=1}^n c_i \prod_{j \neq i}(x-a_j)$

now, if you let $ x = a_i $, the LHS is $a_i^m$ while on the RHS every term is 0 except $c_i \prod_{j \neq i}(a_i-a_j)$.  so $c_i$ must be $ \frac{a_i^m}{\prod_{j \neq i} (a_i-a_j)}$

yeah, i see jmerry already responded.  heh.

oops, there was an error in my original post.  i switched $a_j$ and $a_i$.  it's fixed now.

so that everything cancels except the term with $c_i$, which you want to compute.


geometry
inequalities
incenter
trigonometry
angle bisector

Let  $l_a, l_b, l_c$ be the angle bisectors of a triangle $ABC$ and $r$ its inradious.

Find the angles of this triangle knowing that
$\frac{1}{ l_a}+\frac{1}{ l_b}+\frac{1}{ l_c}=\frac{1}{r}$

 :)

[quote="silouan"]It is easy.From a very well known ineq we have that
$\frac{1}{ l_a}+\frac{1}{ l_b}+\frac{1}{ l_c}\leq\frac{1}{r}$
The equality holds when the triangle is equilateral. So the angles are all 60 degrees[/quote]
I don't know this inequality before, but after I saw your post, I found out that I've just proved it!

Let $I$ be the incentre of the $\triangle{ABC}$ and $D$ be the intersection of the angle bisector of $\angle{A}$ and $BC$. Hence:
\[\frac{ID}{AD}=\frac{\triangle{BIC}}{\triangle{ABC}}\]

Let $\angle{A}=2x$, $\angle{B}=2y$, $\angle{C}=2z$, we will have: $ID=\frac{r}{\sin{(x+2y)}}$. Sum up:

\[\sum \frac{ID}{AD}=\frac{\triangle{BIC}+\triangle{BIA}+\triangle{CIA}}{\triangle{ABC}}\]

\[\sum \frac{r}{l_a} \cdot \frac{1}{\sin{(x+2y)}} = 1\]

\[\frac{1}{r} = \sum \frac{1}{l_a} \cdot \frac{1}{\sin{(x+2y)}} \geq \sum \frac{1}{l_a}\]

With equality holds if $x+2y=y+2z=z+2x=90^{o}$, which gives: $x=y=z=30^{o}$.

Good ineq ye? And also a good proof from you.


function
algebra
polynomial
floor function

Let $ f$ be a polynomial of degree 2005 with the leading coefficient 2006 such that for $ n\equal{} 1, 2, 3, ..., 2005$, $ f(n)\equal{}n$.  Find the number of 0's at the end of $ f(2006)\minus{}2006$.

[hide]
Let $ g(n) \equal{} f(n) \minus{} n$, so $ g(n)$ is also a polynomial of degree $ 2005$. We also know that $ 2005$ roots of $ g(x)$: $ g(n) \equal{} 0$ for $ n \equal{} 1,2,3, \ldots, 2005$. Thus $ g(n) \equal{} f(n) \minus{} n \equal{} 2006(n\minus{}1)(n\minus{}2)(n\minus{}3)\cdots(n\minus{}2005)$.

We want $ f(2006)\minus{}2006 \equal{} 2006(2006\minus{}1)(2006\minus{}2) \cdots(2006\minus{}2005) \equal{} 2006!$. The number of zeroes at the end of $ 2006!$ is given by the highest degree of $ 5$ dividing into this number (since there will be more 2s), and is $ \left\lfloor\frac{2006}{5}\right\rfloor \plus{} \left\lfloor\frac{2006}{25}\right\rfloor  \plus{} \left\lfloor\frac{2006}{125}\right\rfloor  \plus{} \left\lfloor\frac{2006}{625}\right\rfloor$ or something along those lines.
[/hide]


A stock tank that measures 10 feet by 1 yard by 1 yard is being filled by a garden hose and the water is
rising 3 inches per hour. How long will it take to fill the tank?

A) 3.3 hours B) 10 hours C) 12 hours D) 30 hours E) 40 hours

We have to know which of the measurements is the height.


Joe is a businessman who flies from city to city. His home is Albuquerque. On his next trip, he needs to visit Baltimore, Cleveland, Denver, and Evanston, in no particular order. The airline he uses has direct flights between each of these cities. In how many different ways can he fly from Albuquerque, travel to each of the other cities, and return to Albuquerque, without visiting any of the other four cities more than once?

in long countdown questions you should skip to the end of the problem.
In how many [b]different ways[/b] can he fly...
visiting any of the other [b]four[/b] cities more than once?
he needs to visit 4 cities, how many different ways you get
$ 4!$
which is 24


The rules of a certain game is as follows: a coin is repeatedly flipped, and each time it lands on heads 1 point is added and each time it lands on tails 2 points are deducted. The player's initial score was 0, and the game is over once the score reaches 100. What is the probability that within n flips of the coin, the player has won?


is the answer... [hide]171?[/hide] I'm pretty sure it's wrong.

The problem was written incorrectly in the book.

The real problem at nats was something different, and I forgot what it was.

Is the answer
[hide]
$86$ $?$
[/hide]
By the way, real problem was same except first mean was 76.

Answer in key is

[hide]93[/hide]

But I have no idea how they get  - I got 84

The key phrase is that in the original it says [b]no integer is less than 70[/b]
[hide]
Otherwise, you could have the answer as 100.
Data Set = 59,60,61,66,100,100,100
Method: It is easy to figure out that sum of last two tests = 166. To max out the mode, only one mode in last two tests.
sum of first five tests = 380. Two modes in first set. To max out, First set is 2x+3n+3=380, or 2x+3n=377(x=mode). Max score=100, so n>_59
n also is above or equal to 70. n has to be odd. From here you can test and figure out

[/hide]

[quote="guyinPA"]Reid took 7 tests. (First of all, who is named "Reid" - ok, that was an editorial - sorry -- anyway, On the first five tests that he took, his mean score is 76 points.  His mean score for all seven tests is 78 pts.  If the mode is unique and has a frequency of three, what is the greatest possible value of the mode for his seven test scores? (I have ans. but I need solution, please!) Clearly, I'm missing something...[/quote]
Reid Barton...

[quote="life1"]I don't get where you get the 3 from in 2x +3n = 3 ???[/quote]
[hide]The $2x$ is the the two modes in the first five tests, and the $3n$ $+$ $3$ is that to maximize the mode, the three extra numbers should be consecutive. $n$ $+$ $(n+1)$ $(n+2)$ $=$ $3n$ $+$ $3$[/hide]


Set $a_{1},a_{2},a_{3},\cdots,a_{n}>0$ , and $f(x)=\sqrt[x]{\sum_{i=1}^{n}a_{i}^{x}}$

Ask for the monotone interval and the range of $f(x)$ .

MAybe you meant
$f(x)=\sqrt[x]{\frac{1}{n}\sum_{i=1}^{n}a_{i}^{x}}$,
which is strictly increasing for all $x\in\mathbb{R}$, and
$f(\mathbb{R})\equiv (\min\{a_{1},a_{2},a_{3},\cdots,a_{n}\},\max\{a_{1},a_{2},a_{3},\cdots,a_{n}\})$,
unless all the $a_{i}$ are equal, and then $f$ is constant.
[url]http://en.wikipedia.org/wiki/Generalized_mean[/url]


geometry
circumcircle
analytic geometry
linear algebra
matrix
incenter
angle bisector

Given triangle $ ABC$ with circumcircle $ (O)$, centroid $ G$. Angle bisector $ AA', BB', CC'$ intersect at $ I.$ $ B'C' \cap (O)\equal{}\{A_1,A_2\},$ $ A'C'\cap (O)\equal{}\{B_1,B_2\},$ $ A'B'\cap (O)\equal{}\{C_1,C_2\}.$ Prove that the radical center of $ (A'A_1A_2), (B'B_1B_2)$ and $ (C'C_1C_2)$ lies on the line $ IG.$

can someone explain the meaning of RADICAL CENTER?plz :blush:

Radical center of 3 circles is the point which has same powers respective to all 3 circles, it's the intersection of 3 radical axes of those 3 circles taken 2 at a time, under condition that those radical axes aren't parallel.

For convenience rename $\omega_a,\omega_b,\omega_c$  the subject circles. Let $B'C'$ cut $BC$ at $D$ and let $M$ be the midpoint of $BC.$ From $(B,C,A',D)=-1,$ it follows that $\overline{DB} \cdot \overline{DC}=\overline{DA'} \cdot \overline{DM},$  but $p(D,\omega_a)=\overline{DA_1} \cdot \overline{DA_2}=\overline{DB} \cdot \overline{DC}$ $\Longrightarrow$ $p(D,\omega_a)=\overline{DA'} \cdot \overline{DM}$ $\Longrightarrow$ $M \in \omega_a.$  Therefore

$p(B,\omega_a)=\overline{BM} \cdot \overline{BA'}=\frac{ca^2}{2(b+c)} \ , \ p(C,\omega_a)=\overline{CM} \cdot \overline{CA'}=\frac{ba^2}{2(b+c)} $

On the other hand, let $AI$ cut $(O) ,\omega_a$ again at $S,P$ and $B'C'$ at $U.$ Then

$p(A,\omega_a)=-\overline{AP} \cdot \overline{AA'}=-(\overline{PU}-\overline{AU}) \cdot \overline{AA'}=-\left ( \frac{p(U,(O))}{\overline{UA'}}-\overline{AU} \right ) \cdot \overline{AA'}$

$p(A,\omega_a)=-\frac{\overline{AA'} \cdot ( \overline{AU} \cdot \overline{US}-\overline{AU} \cdot \overline{UA'})}{\overline{UA'}}=-\frac{\overline{BA'} \cdot \overline{A'C} \cdot \overline{AU}}{\overline{UA'}}$

$p(A,\omega_a)=-\frac{ac}{b+c} \cdot \frac{ab}{b+c} \cdot \frac{b+c}{2a}=-\frac{abc}{2(b+c)}$

Together with the powers of $B,C$ WRT $\omega_a$ found at the beginning, the barycentric equation of the circle $\omega_a$ is

$\omega_a \equiv a^2yz+b^2zx+c^2xy-\frac{x+y+z}{2(b+c)}(-abcx+a^2cy+a^2bz)=0$

By cyclic permutation of $a,b,c$ the equations of $\omega_b,\omega_c$ are given by

$\omega_b \equiv a^2yz+b^2zx+c^2xy-\frac{x+y+z}{2(a+c)}(b^2cx-abcy+b^2abz)=0$

$\omega_c \equiv a^2yz+b^2zx+c^2xy-\frac{x+y+z}{2(a+b)}(c^2bx+c^2ay-abcz)=0$

The barycentric coordinates of the radical axes of the pairs $\omega_a,\omega_b$ and $\omega_b,\omega_c$ are respectively

$[-abc(a+c)-b^2c(b+c) \ , a^2c(a+c)+abc(b+c) \ , a^2b(a+c)-b^2a(b+c)]$

$[b^2c(a+b)-c^2b(a+c) \ , -abc(a+b)-c^2a(a+c) \ , abc(a+c)+b^2a(a+b)]$

These two lines intersect at the radical center of $\omega_a,\omega_b,\omega_c$  with coordinates

$(a^2(b^2+c^2+ab+ac):b^2(a^2+c^2+ba+bc):c^2(b^2+a^2+ca+cb)),$

which is $X_{1193},$ i.e. the third Saragossa point of $I$ lying on the line connecting $I(a:b:c)$ and $G(1:1:1).$ Therefore, we have an additional observation: Denote the midpoints of the arcs $AC,AB$ of $(O)$ not containing $B,C$ as $T,V,$ $A_0 \equiv TC' \cap VB'.$ Define analogously $B_0,C_0,$ keeping in mind that the midpoint of the arc $BC$ not containg $A$ is $S,$ then the lines $SA_0,TB_0,VC_0$ concur at the radical center of $\omega_a,\omega_b,\omega_c.$

[quote="luisgeometria"]$p(A,\omega_a), \,p(B,\omega_a), \,p(C,\omega_a)$ [/quote]
Dear Luis,
since you are using coordinates, you will never be admitted in the Walhalla of the [i]synthetic[/i] geometers. Therefore, you can boldly use a more [i]natural[/i] method than all these powers. Triangle $A'B'C'$ is $[ \begin {array}{ccc} 0&a&a\\b&0&b\\c&c&0\end {array} ] $. The corresponding trigone is described by the adjoint matrix $[ \begin {array}{ccc} -bc&ac&ab\\bc&-ac&ab\\bc&ac&-ab\end {array} ] $. By hypothesis, line $\Delta(x,y,z)\doteq -bcx+cay+abz=0$ is the radical axis between $\Gamma$ and $\omega_a$. Thus, equation of the required circle is \[
\omega_a(x,y,z)=\Gamma(x,y,z)-k\,(x+y+z)\,\Delta(x,y,z)
\] Constant $k$ is obtained from $A'\in \omega_a$ and your $k=a/2/(b+c)$ is reobtained.

In addition to your last remark, triangles $A',B',C'$ and $A_0,B_0,C_0$ are perspective too (yet another cevian nest), the perspector being X(386).

Best regards, Pierre.

It's easy to generalize the problem for any cevian triangle. Keeping the same notation of my previous post, let $\triangle A'B'C'$ be the cevian triangle of the point $P(u:v:w).$ Circle $\omega_a$ passes through $A'(0:v:w),$  $M_a (0:1:1)$ and the radical axis of $(O),\omega_a$ is $B'C' \equiv -vwx+uwy+uvz=0.$  Hence the equation of $\omega_a$ is given by

$\omega_a \equiv a^2yz+b^2xz+c^2xy-\frac{a^2(x+y+z)}{2u(v+w)}(-vwx+uwy+uvz)=0$

By cyclic exchange of $a,b,c$ and $u,v,w$ we get the equations of $\omega_b, \omega_c$ as:

$\omega_b \equiv a^2yz+b^2xz+c^2xy-\frac{b^2(x+y+z)}{2v(u+w)}(vwx-uwy+uvz)=0$

$\omega_c \equiv a^2yz+b^2xz+c^2xy-\frac{c^2(x+y+z)}{2w(u+v)}(vwx+uwy-uvz)=0$

Radical center $R_0 (x_0:y_0:z_0)$ of $\omega_a,\omega_b,\omega_c$ has coordinates:

$x_0=ua^2[vw(b^2+c^2)+u(c^2v+b^2w)]$

$y_0=vb^2[uw(a^2+c^2)+v(a^2w+c^2u)]$

$z_0=wc^2[uv(a^2+b^2)+w(a^2v+b^2u)]$

We deduce straightforwardly some particular cases, such as:

If $P \equiv I$ incenter of $\triangle ABC,$ then $R_0 \equiv X_{1193}.$  Third Saragossa point of $I.$
If $P \equiv G$ centroid of $\triangle ABC,$ then $R_0 \equiv X_{39}.$  Brocard's midpoint.
If $P \equiv G_e$ Gergonne point of $\triangle ABC,$ then $R_0 \equiv X_{65}.$ Orthocenter of intouch triangle.
If $P \equiv N_a$ Nagel point of $\triangle ABC,$ then $R_0 \equiv X_{3057}.$
If $P \equiv H$ Orthocenter of $\triangle ABC,$ then $R_0 \equiv X_4 \equiv H.$

As a final remark, if $P$ runs through the Lucas cubic of $\triangle ABC,$ then its isogonal conjugate $P^{-1}$  WRT $\triangle ABC$ lies on the line connecting the radical center $R_0$ with the circumcenter of $\triangle ABC.$ For a proof see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=297530]Pedal, Cevian and radical center[/url].


[quote="HTA"]Determine the way to analysed the natural number n into the sum of k numbers where the value of P which is the product of that k numbers is max[/quote]
[hide="hint"]Let $n=a_{1}+a_{2}+...+a_{k}$
Then, there exists a analysis so that $P=a_{1}a_{2}...a_{k}$ max
We have three lemma:
-First, we easily have non of the $a_{i}$ is equal to $1$
-Second, we have non of the $a_{i}\ge 5$
If $a_{1}=t \ge 5$
But we also have $t<2(t-2)$ then we can replace $t$ by $2$ and $t-2$
Thus, we obtain $2(t-2)a_{2}...a_{n}>ta_{1}...a_{n}$. CONTRADICTION!
-Third, we see that:
$4=2+2=2.2$
Then, we can replace $4$ by $2$ and $2$.
From the three lemma, we clearly see that $a_{i}\in 2;3$
But,
${\begin{array}{l}2+2+2=3+3\\2.2.2<3.3\end{array}}$
Thus, we can replace three numbers $2$ by two numbers $3$ to get the bigger product.
Hence, we should use $a_{i}\in 2;3$ and the amount of $a_{i}$ which is equal to $2$ is $\le 2$
Then, we find out what is the remainder when n is divided by $3$ to decide to use the number $2$
For example, $1000=3k+1=3(k-1)+4=3(k-1)+2+2$
QED[/hide]

[quote="tunganh"]Prove that for any ten integers we always have the algebraic sum of four of them is divided by 25[/quote]
there is some hint
[hide="some hint for this one"]the lemma , for any four integers numbers , we always have the algebraic sum of 2 of them is divided by 5
[hide]dear tunganh , i think if you have another problem doesnt concern about the first problem , you should make its own topics where it can be given more attentions

your friend [/hide][/hide]