quadratics
pigeonhole principle
graph theory

What is the largest group of people you could have so that both the following were true:
Among any 3 of them, at least 2 were acquainted.
No four of them were mutually acquainted?

You should know that Ramsey theory has some of the most difficult open problems in mathematics. For example, even the very simple-sounding question ``How many people does it take so that there must exist a collection of five people all of whom are acquainted with each other people or all of whom are strangers to each other?'' has not been solved.

Yeah. But the two I've posted are quite solvable.

Maybe I'm missing something...but this doesn't seem solvable...

There could be any number of ppl...since u haven't placed any restraints as far as how big of a group mass we're talking about or how many ppl they each know or stuff like that...

maybe im just not thinking though...

I agree. With no other conditional this problem can't be solved.

Don't be reassured just yet, for I could (and most likely) stand confuted by confuted...I mean Simon :) </jk>

hmm...well 3M over there said it was solvable...

It is solvable.. the first condition is telling you that there are quite a few acquaintances and the second one is telling you there aren't that many, sort of. Those two conditions will give a limit on the number of people..

yet...there's no set number os exactly how many ppl there are...

[color=cyan]because that's what you're trying to figure out.

In other words, the conditions cannot be met with, say, 1000000 people.  In fact, it is unlikely they could be met with a smaller big number, like say 100 people  (I'm just guessing here).  The question is, what is the biggest number of people for which both conditions can be met?[/color]

So you're working with a group mass thats not set?

lol...yea...but still...its kinda...i dunno...b/c its juss too hard for me to tackle...

will someone let me know what they have a solution?

[color=cyan]Okay, I can outline a frame for the solution, I believe:

The first condition tells you that there are at least 2*C(n, 3)/k(n) edges, at least.  k(n) is a correction factor that I'm not sure of having not worked on this very hard -- basically, it is to divide out the multiple overcounts you get for an edge.  For example, when we consider ABC we know we have two edges, and when we consider ABD we know we have two edges.  However, AB might be an edge in both cases, and would have been extracounted somehow.  There's something that must be divided by to account for this.

The second condition tells you that there are a maximum number of edges: I'm not so sure how to quantify this yet.  I'll keep working on it.    It should be linear in n, I hope, since if it were quadratic it would tend to be larger than our lower bound, which won't give us any information.  You should be able to put the top and bottom condition together to find a maximal value on the number n.

Feel free to fill in the (very large) gaps.[/color]

Heres a hint.

Find an arrangement of [hide]8[/hide] people that works.

Prove that you cannot have [hide]9[/hide] people.

To prove that it doesn't work for 9 people:

[hide] This is equivalent to coloring a complete graph of 9 vertices with 2 different colors such that there are no cycles of 3 or more of one color (say wlog this is 2) and no cycles of 4 or more of the other (say color 1).

Then pick any vertex. By pigeonhole, at least 4 of the 8 lines from that vertex to the others must be of one color. Say it is color 2. Then we cannot connect the 4 vertices with lines of color 2, and they must thus be connected with edges of color 1, forming a 4-cycle, an impossibility.

Now say 4 of the edges extending from that vertex we picked are all of color 1. The 4 vertices the edges extend to must have at least 2 edges of color 1, and this makes a 4-cycle.

Hope this isn't too confusing![/hide]

I think this works to prove that 8 is Ok:

We draw a regular 8-gon (so it's easier to se the configuration). We number the vertices from 1 to 8 going around the 8-gon, then we connect the pairs of edges (x, x+3) and (x, x+2) where x is taken modulo 8. I'm not sure it works. Could smb please check it?

I don't think the second condition is satisfied; there are some 4-cycles (1,3,5,7 are all connected)

A solution with 8 people is an octagon with 2 inscribed squares (ie connect x+1 and x+2.)


calculus geometry algebra polynomial parameterization Putnam

Hello. Who do you think real math Olympiad genius is. Ciprian Manolescu
 Reid Barton , Christian Reiher, Eugenia  Mallinikova or Lazslo Lovasz.Post your comments also. THANKS

Nikolai Nikolov from Bulgaria has 3 gold 1 silver and he get a special prize for his solution of problem 6 IMO 1995 using prime root of unity.

didn't knew that! another thumb up for the Bulgarians!

basically the Balkan Rulz! :D

Indeed the Balkan rullz! Perhaps it's the air we breeth! :)
It seems that more people are better here than in any other place of the world, I am also thinking about the number of people here when I am talking about this! Lets think of how many people live in China or India and then think about Romania and Bulgaria...

I think its more to do with the Maths culture thats there in the East European countries and Russia than anything else. That is, the level of 'awareness' among the public, the popularity of mathematics competitions among the schoolgoing populace and the overall historical and sociological environment that makes the  'nurture' component of mathematical development [i]very effective[/i] (compared to India for instance, where such awareness is far lower, though it is perhaps better than several other countries). I trust that no one in a serious mood will disagree that the 'nature' component(that is genes, and the 'air' you breathe) does not vary significantly from country to country.

hey,i think the eastern european countries have an highly advanced culture in mathematics and that is not wot we find in india,where the awareness is far less,in my state it's nil,the ppl are so obsessed with typical bookwork problems(which are damn easy) and they dont take ne interest  in nething far more challenging.(but the attitude ic changing,i guess)
so there is no doubt balkan rulz (and russia as well,petersburg olympiad is so challenging!!!).but hey!!sumthing's happnd to the balkan math olympiad.it's becoming easier.dearth of problems,huh!!(can't believe it.but this years geometry problem was coool)
and i think ciprian manolescu is the real math olympiad genius.and btw where is christian reiher from?
regards

that's a very accurate and also wise oppinion and I totaly agree with it. we should try to make better use of this potential in our own countries, although I might prove extremely hard.

about the this year's geometry problem from the Balkan Mathematical Olympiad - it's only fair (for me :D) to say that I have proposed it, and it's probably one of the simplest in BMO history - I really didn't thought for a minute that they will actually chose it :D

Well it's not that simple( the problem you-Valentin Vornicu- proposed for the BMO), perhaps it seems simple from your point of view cause you already know a solution.. or  because you are very good :D.. but it is kinda hard.. for a geometry problem! not like the one's from IMO 2003 but for a BMO contest it seems hard enough!
I think more Romanian people should try to propose problems for international contests because we have the talent to make up pretty hard ones! :D

bye!

Hello. Can any one post information about Dan Voiculescu and his performance at IMO.
                                                  THANKS

i noticed that ciprian manolescu is winning the poll. Ive never heard of him or most of these people though. Can someone tell me about manolescu and perhaps why most people seem to be voting for him. (Not that they shouldn't be voting for him, i have no opinion after all because I am not familiar with most of these guys)

He must be pretty good, since he's beating Reid Barton (Reid taught himself calculus and got a 5 on the AP Calculus BC exam when he was 10 years old, for just one example  ;) ).

Edit: Ciprian Manolescu won three gold medals at the IMO. As an undergraduate in college, he got first place on the Putnam in 1997, 1998, and 2000. Search for him on Google to learn more.

It's not the he won 3 gold medals, it's that he scored 3 perfect scores in IMO (world record, never broken since then 1995-1996-1997). In 1996 he was the only perfect scorer, and also won the first place with the Romanian Team.

He also received the Morgan Undergraduate prize for exceptional research in Mathematics.

PS not that I don't think Reid is a very good mathematician, but I've also taught myself calculus, so that really doesn't say much. Ciprian basically taught himself everything ! ;)

I should've done my homework before I voted, because Ciprian definitely gets my vote now. :)

p.s. Okay, but did you teach yourself calculus when you were nine?

Nop. I did that when I was 14. I didn't like too much maths before I was 13 years old :D ...

Anyway, what I tried to say is not that I am as good as Reid or something, just that teaching yourself something (like calculus) is not a genius criterion :D

[quote="orl"][quote="esti"]
I think T.Tao is still leader on the list by the age parameter performance.
(Silver at 11,gold at 12-13 y.old.)
No? [/quote]

Indeed no. Look at my post on page three of this thread to see you are not right about his medal history.  ;)[/quote]
Ok, I won't go in details,but I can't resist not to ask [b]who else[/b] ,except Tao ,has won IMO gold before turning 13th birthday?  ;)

[quote="esti"][quote="chess64"]Answering the original question: Alex Zhai :D[/quote]
I would think the difficulty of winning IMO gold in preteen age is in the same ballpark as becoming a chess GM in that age.
What is Sergey Karjakin in chess world ,T.Tao is in math world :);)[/quote]

Why did you quote me? :?

[quote="saeid"]Maryam mirzakhani is the most Genius in the world from iran.[/quote]

There were many smart Iranians at the IMO, but they are still relatively young and at the beginning of their careers as scientists.   Mirzakhani has certainly progressed to bigger and better things since the IMO.  She gave a very natural (or more natural) proof, using dynamical systems, of Kontsevich theorem on moduli spaces of curves.  From the older generation (pre-Khomeini), Cumrun Vafa is probably the most famous from Iran, and extremely well respected by mathematicians.

[quote="Arne"]I don't know how he did it. He must be 21 or 22 years old now :)

Edit: he became fourth in the Norvegian mathematical olympiad 1997, so... :)[/quote]

and he has obtained his PhD? I remember you gotta have the degree to be leader. Is it true? :o

[quote="Soarer"]
and he has obtained his PhD? I remember you gotta have the degree to be leader. Is it true? :o[/quote]

I don't think this is true. Can anyone confirm?

You do not have to have a PhD to become Leader (or Deputy for that matter) of an IMO team. For example the former dutch team leader didn't have a PhD. If you look at the results page on the IMO2004 website you will also find a list of leaders with their titles. A good dozen of the leaders are not either Dr. or Pr. , and hence didn't have a PhD.

Anyone winning the IMO is a genius.  Some are natural like Reid probably, and some it took much more practice.  I don't think i could win a medal and i already have a masters in math

[quote="neelnal"]How many gold medalists at the IMO have won a fields medal?[/quote]

Perelman is the [i]ne plus ultra[/i] of the difference between an excellent IMO record and important mathematical research.  His IMO score was an early sign of talent, obviously, but his now-legendary papers are incredibly impressive.  It's hard to think of anything that spectacular on any one problem in a long, long time (even more so than Wiles, in my opinion).

[quote="Tiks"]Hey I think you should add to the list another new big genius
[b]Iurie Boreico[/b](Moldova)
I think next year he will beat Ciprian's record with tree perfect scores,also he got a special prise for outstanding solution for IMO 2005 ,Q.3.
Another reasone is that from all 7(if I am not mistaken) partisipants of IMO 2006,who had perfect scores last yere,he is the only hwo again get 42 points this year!And at last he is a very good person :) .[/quote]

Yeah, he's really good. He was a mentor at AwesomeMath (woot!)


calculus
integration
limit
real analysis
real analysis unsolved

Suppose f in R on [0,A] for all A < infinity, and f(x) -> 1 as x -> +infinity. Prove that
\[\mathop{\lim }_{t \to 0}t\int_{0}^\infty{e^{-tx}f(x)dx = 1}\]
(t > 0).


geometry
circumcircle
parallelogram
geometry proposed

The circle S passes through vertices A and C of a triangle ABC, and intersects de sides AB and BC in D respectively E. Let F be the point of intersection of the lines CD and AE. The circumcircle of the triangle BEF intersects S in the points E and G. Prove that GD bisects BF.

This problem is cool  :)  I made a wrong direction at first. Here is my solution
     Let $P$ and $Q$ be the intersections of $CD$ and $GB$, $BD$ and $GF$. We have the two following triangles are similar $GFB$ and $GAC$ as $\measuredangle{GBF}= \measuredangle{GEF}=\measuredangle{GCA}$ and $\measuredangle{GFB}=\measuredangle{GEB}=\measuredangle{GAC}$. Hence $\measuredangle{BGF} = \measuredangle{CGA}$.
     Notice that $\measuredangle{PGQ} = \measuredangle{BGF} = \measuredangle{CGA} = \measuredangle{CDA} = \measuredangle{FDQ}$. This implies that $PDQG$ is cyclic. Hence $\measuredangle{GPQ} = \measuredangle{GDQ}=\measuredangle{GDA}=\measuredangle{GCA}=\measuredangle{GBF}$. Therefore $PQ//BF$. By applying the Ceva 's theorem to triangle $BGF$ with the point $D$, we obtain that $GD$ passes through the midpoint of $BF$.

wow.. nice solution treegoner...
my solution
M is the intersection of BF with GD... then [it's not hard I think]
$\triangle MDB\sim \triangle MBG$ and $\triangle MDF\sim\triangle MFG$
u find then that
$MB^2=MD\cdot MG$ and
$MF^2=MD\cdot MG$
hence the conclusion.

This reminds me somewhat of a BWM problem of last year. Here's my solution:
Take [tex]H[/tex] to be the intersection of line [tex]GD[/tex] and circle [tex]BEF[/tex].
Then we've got:
[tex]\angle DHB=\pi-\angle GDA=\angle AEG=\angle FEG =\angle FHD [/tex]
Thus [tex]FH//DB[/tex]
By a similar angle chase we get [tex]HB//DF[/tex]
So [tex]DBHF[/tex] is a parallelogram and with bisecting diagonals

[quote="schulmannerism"]How do you prove those triangle similarities, Kuba?[/quote]

Well, let me just rewrite Kuba's solution with more details.

[color=blue][b]Problem.[/b] Let a circle S pass through the vertices A and C of a triangle ABC, and intersect the sides AB and BC in the points D and E, respectively (apart from the points A and C). Let F be the point of intersection of the lines CD and AE. The circumcircle of the triangle BEF intersects the circle S in the point G (apart from E). Prove that the line GD bisects the segment BF.[/color]

[i]Solution.[/i] We will use directed angles modulo 180°.

Let the line GD meet the line BF at a point M. Then,

  < MFD = < (BF; CF) = < (BF; BC) + < (BC; CF) = < FBE + < ECD.

Since the points F, B, E and G lie on one circle, we have < FBE = < FGE. Since the points E, C, D and G lie on one circle (namely, on the circle S), we have < ECD = < EGD. Thus, < MFD = < FGE + < EGD = < FGD = - < MGF. Also, it is clear that < FMD = - < GMF. Thus, the triangles MFD and MGF are inversely similar, and it follows that MD : MF = MF : MG, so that $MD\cdot MG=MF^2$.

On the other hand,

  < MBD = < (BF; AB) = < (BF; AF) + < (AF; AB) = < BFE + < EAD.

Since the points B, F, E and G lie on one circle, we have < BFE = < BGE. Since the points E, A, D and G lie on one circle (namely, on the circle S), we have < EAD = < EGD. Thus, < MBD = < BGE + < EGD = < BGD = - < MGB. Also, trivially < BMD = - < GMB. Hence, the triangles MBD and MGB are inversely similar, what yields MD : MB = MB : MG, so that $MD\cdot MG=MB^2$.

Combining $MD\cdot MG=MF^2$ and $MD\cdot MG=MB^2$, we see that $MF^2=MB^2$, and thus MF = MB. In other words, the point M is the midpoint of the segment BF. Hence, the line GD, passing through this point M, bisects the segment BF. Problem solved.

[b]PS.[/b] The problem is from the Russian IMO TST 2002, and it's Russia 98 = geo problem 167 in the op02 project.

  Darij


easy application of the pigeon hole principle- atleast 2 of the 4 integers have same parity and atleast 2 of them have same residue mod3. result follows

you have all 6 possible differences among the 4 $a_i$. 2 of the $a_i$ must be congruent mod 3, so their difference is divisible by 3. if 3 of the $a_i$ have the same parity, then 3 of the differences are even, so the product is divisible by 8. otherwise 2 of the $a_i$ are odd and 2 are even. in that case 2 of the differences are even, and the product is divisible by 4. either way, the product is divisible by $3\cdot 4 = 12$.

edit: looks like someone got here before me

Pvael Kozlov had solved this ... so this one is not new

Davron

oh, obviously it's not new. it's too basic to be new.

This and it's generalisation were posted about 11 times before ;)


I was just wondering how strict the USAMTS graders are on showing work. For example, could one simply make the following statement:

...
Substituting, we have (2x^2 + 1) + (y - 1) = (5x^2 - y) + 7. Solving, we have y = (3x^2 + 7)/2.

Does one actually need to show the work to get from (2x^2 + 1) + (y - 1) = (5x^2 - y) + 7 to y = (3x^2 + 7)/2?

Fierytycoon[/code]

That will be fine. If you state something without proof, then you may lose points. If you state something without doing two steps of algebra and say that that was what you did, you won't run into any difficulty.

[quote]two steps of algebra[/quote]

How about, say, solving a quadratic without any work? Such as giving (2a+b)x^2 - a^2x + 7 and then solvin for x in terms of a, b without work?

Fierytycoon

[quote]Anything that would be an early exercise in a school textbook you can do without showing any work.[/quote]

...But that depends how you define "early". Perhaps for some ppl, solving partial diffential equations are simple mental exercises.  8-)

Fierytycoon

By early I meant that in a school textbook it will be one of the first exercises at the end of a section. In general those are easier than the ones at the end of a section. If you think a problem looks like something at the end of a section, I would advise writing out a few intermediate steps.

[color=cyan]The simple way to deal with the problem of not knowing how much work to put is to put all of it.  This is also true for any other proof-based contest, in particular USAMO.  More work cannot possibly be worse, unless it's actually wrong.[/color]

Yes, but putting lines and lines of pure algebraic work isn't exactly what ppl call...suave.  8-)

Fierytycoon

Sure, but I'm not usually too thrilled when I read a paper or a book, and they say that getting from one line to the next is ``trivial'' or ``obvious,'' and then it takes me 20 minutes to figure out why it's true and also completely nonobvious (to me).

That happened to me when I tried to read some old-a** Number theory book I picked up...

As to the original topic of the post, I would just show as much work as needed. It's all about gettin' the points dude. No one else is gonna see this except your grader and you so there ain't no need to make yourself look "suave" by elimminiating what would otherwise corroborate your work.

My two Rupee.
IW.OAV

I read a olympiad-tips thing somewhere, and the one thing I advise you to do is this: [i]never use the words "clearly", "obviously", etc.[/i], since it may not seem so clear to your grader (or any reader) and he may be irritated by this.

Yeah I heard that somewhere. Though 'trivial' is a real fun word to use :D..

From MathWorld (on Trivial):

Related to or being the mathematically most simple case. More generally, the word "trivial" is used to describe any result which requires little or no effort to derive or prove. The word originates from the Latin trivium, which was the lower division of the seven liberal arts in medieval universities (cf. quadrivium).

According to the Nobel Prize-winning physicist Richard Feynman  (Feynman 1997), mathematicians designate any theorem as "trivial" once a proof has been obtained--no matter how difficult the theorem was to prove in the first place. There are therefore exactly two types of true mathematical propositions: trivial ones, and those which have not yet been proven.

The opposite of a trivial theorem is a "deep theorem."

That's right - in [i]Surely You're Joking, Mr. Feynman[/i], Feynman jokes that mathematicians "can only prove trivial things."

Sure. After a result has been proven, it is technically ``trivial.'' Feynman also didn't really like mathematicians very much it seems if I remember correctly from SYJMF.

True. Feynman wasn't a fan of mathematicians, though he did say the following,

"To those who do not know Mathematics it is difficult to get across a real feeling as to the beauty, the deepest beauty of nature. ... If you want to learn about nature, to appreciate nature, it is necessary to understand the language that she speaks in."

He liked mathematics, but not mathematicians. There's a difference.

Je sais. I agree with him on the mathematics, and I don't know mathematicians.

[quote="Lord Venom"]there are a lot of things others would find trivial that I would not (even though I am of an adequate skill level in competition mathematics [not brilliant, but not a doofus]).[/quote]

But the word trivial is very different from a word like "obvious" or "easy."  All of those words are based on your opinion.  By contrast, trivial refers to anything that has already been proven.  As soon as you prove something, it becomes trivial.  That's what Feynman was referring to.

I see. Well, I suppose a vast proportion of modern mathematics is "trivial" in that regard. However, there are many things that I have considerable trouble understanding. I don't have a direct example right now, but one example would be the physics concept of tension. It just doesn't connect with me.


Is it possible to draw overlapping circles in LaTeX, like a Venn Diagram? If possible, how would one go about doing it? Thanks.

Either see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=211639]How to draw Venn diagrams using Latex[/url] or use Asymptote.

Hi,

you could also try TikZ, have a look at the [url=http://www.texample.net/tikz/examples/venn-diagram/]Venn diagram example[/url] on texample.net.

Stefan


Hello friends, there is something to talk about The Indian Oil Cup (ODI Tri Series) starting on the 30th of July. After a long gap India will be on the cricket pitch again facing the Lankans and the in-form Windies in the three nation ODI tournament starting on July 30th .Sri Lanka has never been a happy hunting ground for either India or the Windies as both of them lag behind the hosts on its homeground. In Sri Lanka, India has won only 9 out of 30 ODIs played and Windies managed to win 2 out of 7 played between them and the Lankans.The hosts will definitely hold the upper hand in the beginning of the Tri Series but both India and West Indies will be giving them the toughest fight possible with the likes of Sehwag, Ganguly, Dravid. I got the latest of informations about The Indian Oil Cup from http://cricket.123india.com/SERIES/TRISERIES2005/ . Chanderpaul and his pretty young and inexperienced team is ready to set the ground on fire. It is an eagerly awaited series and possibly sets the tune for the three teams for the long cricket season ahead. Though Sachin Tendulkar is not playing in this series I’ll go for the Indians to clinch this series.

[quote="Aarav"]A captain is as good as his players are. Don't really think, Dravid alone can inspire Indian cricket to new heights.[/quote]
So you mean to say Ganguly is rubbish?
Dravid is a good batsman but not a good batsman when he bats.
You probably know that his average falls rapidly when he is a captain.
I think India will be last in the series.

[quote="riddler"]Dravid is a good batsman but not a good batsman when he bats.[/quote]

That doesn't make much sense...maybe you meant, "Dravid is a good batsman but not a good batsman when he's captain."? :roll:  ;)

[quote="chess64"]That doesn't make much sense...maybe you meant, "Dravid is a good batsman but not a good batsman when he's captain."?  [/quote]
Oops :blush:


Let $S$ be the set of all partitions of $2000$ (in a sum of positive integers). For every such partition $p$, we deﬁne $f (p)$ to be the sum of the number of summands in $p$ and the maximal summand in $p$. Compute the minimum of $f (p)$ when $p \in S .$

$x$ is the maximal element and $y$ is the number of the summands in $p$. $x\cdot{y}\ge2000$ then from am gm $x+y\ge2\sqrt{x\cdot{y}}\ge2\sqrt{2000}>89$.  $x$ and $y$ are integers so $x+y\ge90$. also $40\cdot50=2000$ hence $90$ works.


Here is the problem:

Wanda the witch agrees to trade one of her magic broomsticks to Gaspar the ghost in exchange for one of his gold chains. Gaspar is somewhat skeptical that the broomstick is in working order and insists on a guarantee equal to the number of links in the gold chain. As a guarantee he insists on paying by the installment plan, one gold link per day until the end of 63-day period, with the balance to be forfeitif the broomstick malfunctions during the guarantee period. Wanda agrees to this arrangement, but insists that the installment payment be effected by cutting no more than three links in the gold chain. Can this cutting be done, and, if so, what links in the chain should be cut? The chain initially consists of 63 closed gold links arranged in a simple linear order (not closed into a circle).
 :roll:

[Moderator: I don't like bold text with size 18 coloured in maroon]

How can A give B, 63 chain links in 63 days, one each day, by cutting only three or less chain links.  The chain initially consists of 63 closed links arranged in a simple linear order (not closed into a circle).


hy everybody I really need your help I need a writing about coming subject
There are many other examples from different fields of science. Physics, chemistry, communications or biology.
How should we deal with new technology?

* Should we accept every new technology with "open arms", the best thing ever!!
* Or should we be more conservative and be very suspicious of new technology?
* Or do we decide what happens?

so please I reaaly want him and if possible in a few hours coming before tomorrow afternoon

Unfortunately, people have better things to do than do other people's homework.  :wink:


probability
geometry
3D geometry
analytic geometry
combinatorics unsolved
combinatorics

three points are chosen from the interval [0,1].Fİnd the probability of choosing these points such that the longest distance betweeen two points is  <=  1/2

I think I have an idea Take the cube with one corner $(0,0,0)$ and another $(1,1,1)$ with side $1$, parralel to axes. Now any point in this cube correspond to a choice of 3 points on the segment $[0,1]$. Draw the regions for which these points are at a distace less than $\frac12$. It is easied if you will study first the 2-D case :D

sorry,but i could not understand,can u explain it in other words

His referring to the fact that you can treat it geometrically. Represent the outcome by a point with coordinates $0\le x,y,z\le 1$ in the cube $[0,1]\times[0,1]\times[0,1]$, where $x,y,z$ are the lengths of the segments you get.


trigonometry
function
calculus
derivative
real analysis
real analysis unsolved

Is f uniformly continuous on $ x\in\mathbb{R}$ ??

$ f(x)\equal{}x^2*sin(1/x)  ,x\neq 0$
 $ f(x)\equal{}0,x\equal{}0$

thank you!

$ \vert f'\vert \le 3$ since $ \vert f'(x)\vert \equal{} \vert 2x\sin(1/x) \minus{} \cos(1/x)\vert \le 2 \plus{} 1$ and $ f'(0) \equal{} 0.$ Now by Lagrange
$ \vert f(x)\minus{}f(y)\vert \equal{} \vert f'(c)\vert \vert x\minus{}y\vert \le 3 \vert x \minus{} y \vert$

Let me take away the tool that Arrigo-Sacchi just used.

Consider: $ g(x)=\begin{cases}x^2\sin(\frac1{x^3})&x\ne 0\\0&x=0)\end{cases}.$

I claim that $ g(x)$ is uniformly continuous on $ \mathbb{R}.$

Prove that.

We observe that the function is continuous.
The derivative is $ f'(x) \equal{} 2x\sin {1\over x^3} \minus{} {3\over x}\cos {1\over x^3}$ and $ f'(0) \equal{} 0.$ Break the reals as
$ ( \minus{} \infty, \minus{} a]\cup [ \minus{} a,a]\cup [a, \plus{} \infty).$ On $ [ \minus{} a,a]$ we invoke the Heine-Borel theorem although the derivative is unbounded. On $ [a, \plus{} \infty)$ the derivative is uniformly bounded thanks to $ \vert \sin(1/x^3)\vert \le 1/x^3.$   As for  $ ( \minus{} \infty, \minus{} a]$ the function is odd. Combining all these arguments we have the uniformly continuity.

[quote="Kent Merryfield"]Let me take away...[/quote]


[quote="pvthuan"]I want to post the following inequality which has a nice and simple solution in Solved Problems but I do not know why this is impossible.

Let $a,b,c,d$ be nonegative real numbers such that \[a^{2}+b^{2}+c^{2}+d^{2}=1,\] prove that \[\frac{a^{3}}{1-bcd}+\frac{b^{3}}{1-cda}+\frac{c^{3}}{1-dab}+\frac{d^{3}}{1-abc}\geq\frac47.\] [/quote]

[hide]By Cauchy in Engel form, you have

$\frac{a^{3}}{1-bcd}+\frac{b^{3}}{1-cda}+\frac{c^{3}}{1-dab}+\frac{d^{3}}{1-abc}\ge \frac{(a^{2}+b^{2}+c^{2}+d^{2})^{2}}{(a+b+c+d-4abcd)}=\frac{1}{a+b+c+d-4abcd}$

So it remains to prove that $7\ge 4(a+b+c+d)-16abcd$

but I don't know how to go on from here.[/hide]

Come on, you are on the right track. :lol:

I just can't get around the $abcd$ term :wallbash:

SplashD use the GM-AM inequality and a^2+b^2+c^2+d^2=1 to get to 16abcd term!!!!I dont post the solution because I let you to resolve it!!It was your idea to use the Engel inequality!!!

[quote="SplashD"]Can I have a hint?  :wallbash_red:[/quote]
Before you employ Cauchy, you need an approriate grouping of the terms, and finally, use AM-GM inequality. I hope you can get that done.

[quote="SplashD"][hide]By Cauchy in Engel form, you have

$\frac{a^{3}}{1-bcd}+\frac{b^{3}}{1-cda}+\frac{c^{3}}{1-dab}+\frac{d^{3}}{1-abc}\ge \frac{(a^{2}+b^{2}+c^{2}+d^{2})^{2}}{(a+b+c+d-4abcd)}=\frac{1}{a+b+c+d-4abcd}$

So it remains to prove that $7\ge 4(a+b+c+d)-16abcd$

but I don't know how to go on from here.[/hide][/quote]

[hide="continued"]$7\ge 4(a+b+c+d)-16abcd$

By AM-GM, we have that
$1\ge 16abcd$

Adding these two inequalities gives us
$8\ge 4(a+b+c+d)$
$\Leftrightarrow 2\ge a+b+c+d$

which is true by Cauchy. [/hide]

Here is one more alternative. \[(a+b+c+d-4abcd)^{2}=[a(1-bcd)+b(1-cda)+c(1-dab)+d(1-abc)]^{2}\leq\sum_{\rm cyclic}(1-bcd)^{2}.\] Then we prove that \[\sum_{\rm cyclic}b^{2}c^{2}d^{2}+\frac{15}{16}\geq 2\sum_{\rm cyclic}bcd.\] By AM-GM \[b^{2}c^{2}d^{2}+\frac1{64}\geq\frac14 bcd.\] Adding similar inequalities, we finish by proving that \[abc+bcd+cda+dab\leq\frac12.\] Anyway, your proof is better.

[quote="SplashD"][hide]I once grouped it as
$4\sum (a-\frac12)^{2}+16abcd\ge1$

By AM-GM,
$1\ge 16abcd$

Adding, we get

$4\sum (a-\frac12)^{2}\ge 0$ which is true by the trivial inequality[/hide][/quote]
I am afraid that you have a mistake :D I think your mistake for what we should have to prove and what we know.

[quote="pvthuan"]
Anyway, your proof is better.[/quote]
Yes,his all two solutions is better,but not right. :D

Anyway,your solution is nice,Thuan :)

Oh, sorry, I have not checked his solution carefully. It's wrong since he reversed the sign.   :maybe:

My solution again uses the technique of grouping. This is to make sure that the equality can hold in all applications.

[quote="SplashD"][quote="pvthuan"]It's wrong since he reversed the sign.   :maybe:
[/quote]

So, how would I fix it?[/quote]

This is unrepairable. Mine is an alternative.

[quote="pvthuan"]I have discovered that my solution is WRONG. :blush:[/quote]
Why?I haven't seen it :blush:

[quote]Then we prove that :
$\sum_{cyclic}b^{2}c^{2}d^{2}+\frac{15}{16}\geq 2\sum_{cyclic}bcd$
[/quote]

unfortunately you need this inequality to hold with reversed sign...

[quote="pvthuan"]It's wrong since he reversed the sign.[/quote]

My problem is worst. I can't see where he has reversed the sign. :|

[quote="SplashD"]

[hide="continued"]$7\ge 4(a+b+c+d)-16abcd$

By AM-GM, we have that
$1\ge 16abcd$

[b]Adding these two inequalities [/b]gives us
$8\ge 4(a+b+c+d)$
$\Leftrightarrow 2\ge a+b+c+d$

which is true by Cauchy. [/hide][/quote]

  This is a [color=red]logical mistake [/color]in inequalities.Zhaobin pointed it out , too.

 [u]Babis[/u]


Find the ten-digit number which uses each of the digits 0 to 9 once and has the following properties:
the first digit from the left is divisible by 1,
the number formed by the first two digits from the left is divisible by 2,
the number formed by the first three digits from the left is divisible by 3,
the number formed by the first four digits from the left is divisible by 4,
the number formed by the first five digits from the left is divisible by 5,
the number formed by the first six digits from the left is divisible by 6,
the number formed by the first seven digits from the left is divisible by 7,
the number formed by the first eight digits from the left is divisible by 8,
the number fromed by the first nine digits from the left is divisible by 9,
the number itself is divisible by 10.

3816547290

I use brute force after noticing properties like all even positions from left must have even numbers, the fifth position from left is 5, the last position is 0...

[quote="AndrewTom"]Thanks greentreeroad.

Is the solution unique?[/quote]I know this to be a fact, and I think it's pretty well-known. You can prove it to yourself by brute forcing cases.


geometry trapezoid trigonometry inequalities

Each side of a quadrialteral is tangent to a circle of radius 1. Prove that the area of the quadrilateral is at least 4.

Are we assuming that the circles are non-overlapping or something?

There is only one circle, and it is inscribed in the quadrilateral

See the attached diagram.

$AM=\tan\alpha\implies [OMA]={1\over 2}\tan\alpha\implies [OMAQ]=\tan\alpha$

Therefore $[ABCD]=\tan\alpha+\tan\beta+\tan\gamma+\tan\delta$. Also, $\alpha+\beta+\gamma+\delta=\pi$. Since $\tan x$ is convex on $\left(0,{\pi\over 2}\right)$, by Jensen we have

${1\over 4}\tan\alpha+{1\over 4}\tan\beta+{1\over 4}\tan\gamma+{1\over 4}\tan\delta\geqslant\tan\left({1\over 4}\alpha+{1\over 4}\beta+{1\over 4}\gamma+{1\over 4}\delta\right)=1$, which yields

$[ABCD]\geqslant 4$

The equality is satisfied for and only for $\alpha=\beta=\gamma=\delta={\pi\over 4}$, i.e. for a square.

APPENDIX: I see there's "trapezoid" in the title and "quadrilateral" in the text. If a trapezoid is indeed in question, the solution can be greatly simplified.

mustafa, please clear the dilemma.

Sorry, the original question was for a trapezoid, but I thought that it would be more fun for a quadrilateral, and it would still be true, so I changed it. If you want, you can prove it for a trapezoid also.

[quote="mustafa"]If you want, you can prove it for a trapezoid also.[/quote]

he already did..

He meant the simplified proof for trapezoid only.

[hide="trapezoid"]

Notice that the median of the trapezoid is $\frac{b_{1}+b_{2}}{2}\ge 2$, with equality when the trapezoid is a square (I thought trapezoids meant ONLY one pair of sides parallel...guess not) because the distances between the midpoints of the legs of the trapezoid is at least the diameter of the circle. (we can use geometry theorems to make a more rigorous proof, but I'll let someone else try it :P ).

Area of trapezoid  = $\frac{(b_{1}+b_{2})h}{2}$

Since $h=2$ (h=2r), the area of the trapezoid $\ge 4$[/hide]


If $ b^3 \equal{} .25$, what is the value of $ b^{\minus{}3}$?

b^3 and b^-3 are reciprocals, since they multiply to be 1.  1/.25=4.

If $ a^x \equal{} b$, then $ a^{\minus{}x} \equal{} \frac{1}{b}$.

[quote="ernie"]If $ a^x \equal{} b$, then $ a^{ \minus{} x} \equal{} \frac {1}{b}$.[/quote]

Tsk tsk erniepoo. Stop being like izzy. Does that only apply for reals? What if b=0?


integration
real analysis
real analysis unsolved

[quote="misan"]using [u][b]only[/b][/u] the [i]following [/i]2 definitions, i.e.

$ \zeta{(s)}\;=\;\sum_{n=1}^\infty\;\frac{1}{n^{s}}\;\;\;\quad\text{and}\;\;\;\quad\Gamma{(s)}\;=\;\int_{0}^\infty\;x^{s-1}\cdot e^{-x}\;\;\;\textbf dx$

prove that

$ \zeta{(s)}\cdot\Gamma{(s)}\;=\;\boxed{\int_{0}^\infty\;\frac{y^{s-1}}{e^{y}-1}\;\;\;\textbf dy}$[/quote]

Substitute $ x=nt$ to get:
$ n^{-s}\Gamma (s) = \int_{0}^{\infty}e^{-nt}t^{s-1}dt$
Take the sum,
$ \sum_{n=1}^{\infty}n^{-s}\Gamma (s) = \sum_{n=1}^{\infty}\left( \int_{0}^{\infty}e^{-nt}t^{s-1}dt \right)$
Assuming [b]uniform convergence[/b],
$ \zeta (s) \Gamma(s) = \int_{0}^{\infty}\left( \sum_{n=1}^{\infty}e^{-nt}t^{s-1}\right) dt$
So,
$ \zeta (s) \Gamma (s) = \int_{0}^{\infty}\frac{e^{-t}t^{s-1}}{1-e^{-t}}dt$
Da taa,
$ \zeta (s) \Gamma (s) = \int_{0}^{\infty}\frac{t^{s-1}}{e^{t}-1}dt$


floor function
number theory unsolved
number theory

[color=darkblue]Let $ n\in N, k\in Z$. Find $ S\equal{}\displaystyle \left [ \frac{n\plus{}1}{2}\right] \plus{} \displaystyle \left [ \frac{n\plus{}2}{2^2}\right]\plus{}...\plus{} \displaystyle \left [ \frac{n\plus{}2^k}{2^{k\plus{}1}}\right]$[/color]

[quote="t0rajir0u"]Has been posted several times.  I assume $ \left[ x \right]$ means the floor function.

Let the sum be $ S_n$.  We claim that $ S_n \equal{} n$ and prove by induction.  $ n \equal{} 0$ is clear.  When we replace $ n$ by $ n \plus{} 1$ a particular term increases if and only if $ \frac {n \plus{} 1 \plus{} 2^k}{2^{k \plus{} 1}}$ is an integer; in other words, $ n \plus{} 1 \plus{} 2^k \equal{} a \cdot 2^{k \plus{} 1} \Leftrightarrow 2^k | n \plus{} 1, 2^{k \plus{} 1} \not | n \plus{} 1$.  This $ k$ is unique.

Alternately, [url=http://en.wikipedia.org/wiki/Hermite%27s_identity]Hermite's identity[/url] for $ n \equal{} 2$ gives $ \left\lfloor x \right\rfloor \plus{} \left\lfloor x \plus{} \frac {1}{2} \right\rfloor \equal{} \lfloor 2x \rfloor$.  Define $ R_n \equal{} \left\lfloor \frac {n}{2} \right\rfloor \plus{} \left\lfloor \frac {n}{2^2} \right\rfloor \plus{} ...$.  Then

$ S_n \plus{} R_n \equal{} \sum_{k \equal{} 0} \left\lfloor \frac {n}{2^k} \right\rfloor \equal{} n \plus{} R_n \Leftrightarrow$
$ S_n \equal{} n$.[/quote]

Thank you very much.


There's a right triangle ABC, where C is a right angle.  Circle O is inscribed in ABC, where o is the center.  O is tangent to AC and BC at D and E respectivley.  Prove that quadrilateral CDOE is a square.

This is probably en easy problem, but I put it in this forum, since proofs shouldn't really be in Getting Started.  So try not to answer this if this is too easy for you.

if somebody wishes to solve this again, i just did a quick proof

[hide]well, the tangents are rt angles, and rt angle of triangle, since 3 of 4 angles right, the last one right, therefore it is a rectangle, and since radii congruent, it is rhombus, and since it is a rectangle and a rhombus, it is a square[/hide]

yes, E should be on CB not AB, i assumed that...

[quote="behemoth571"][quote="h_s_potter2002"]There's a right triangle ABC, where C is a right angle.  Circle O is inscribed in ABC, where o is the center.  O is tangent to AC and AB at D and E respectivley.  Prove that quadrilateral CDOE is a square.[/quote]

Shouldn't the question state AC and BC, not AC and AB?[/quote]

THat's changed.  Sorry about that.


One problem mentioned a net, and referred to a diagram, that would be folded into a cube. The goal was to calculate the maximum "lateral product" of the faces not including the top and bottom.

Yeah, I encountered that problem as well. I figured it'd be a die, and so I said 120, so when it told me I got it wrong, I was kinda confused.

well the answer is 144 because 1 and 5 are opposite so u just have 2 ask levans to fix that problem.

Well Lateral faces are all faces except for the top and bottom so the greatest product would be 120 because:

You could have 1&6 on the top and bottom=120=4x3x2x5    2,3,4,5 would be lateral faces
2&5 on the top and bottom=1x3x4x6=72    1,3,4,6 would be lateral faces
Or
3&4 on the top and bottom= 1x2x5x6=60  1,2,5,6 would be lateral faces

 :thumbup:

It isn't necessarily a die. Did you read any of the previous posts?

well you can assume its a die because what other cube has numbered sides?

There are many problems with this problem, simply go to the review

and click on "Report error".

And yeah, you can't assume its a die. Of course, you might want

to assume that the numbers are 1, 2, 3, 4, 5, and 6, but you can't

assume that its a die


function
algebra
polynomial
real analysis
real analysis unsolved

Denote by $ \mathcal{H}$ a set of sequences $ S\equal{}\{s_n\}_{n\equal{}1}^{\infty}$ of real numbers having the following properties:

(i)     If $ S\equal{}\{s_n\}_{n\equal{}1}^{\infty}\in \mathcal{H}$, then $ S'\equal{}\{s_n\}_{n\equal{}2}^{\infty}\in \mathcal{H}$;

(ii)    If $ S\equal{}\{s_n\}_{n\equal{}1}^{\infty}\in \mathcal{H}$ and $ T\equal{}\{t_n\}_{n\equal{}1}^{\infty}$, then
              $ S\plus{}T\equal{}\{s_n\plus{}t_n\}_{n\equal{}1}^{\infty}\in \mathcal{H}$ and $ ST\equal{}\{s_nt_n\}_{n\equal{}1}^{\infty}\in \mathcal{H}$;

(iii)   $ \{\minus{}1,\minus{}1,\dots,\minus{}1,\dots\}\in \mathcal{H}$.

A real valued function $ f(S)$ defined on $ \mathcal{H}$ is called a quasi-limit of $ S$ if it has the following properties:

If $ S\equal{}{c,c,\dots,c,\dots}$, then $ f(S)\equal{}c$;
If $ s_i\geq 0$, then $ f(S)\geq 0$;
$ f(S\plus{}T)\equal{}f(S)\plus{}f(T)$;
$ f(ST)\equal{}f(S)f(T)$,
$ f(S')\equal{}f(S)$

Prove that for every $ S$, the quasi-limit $ f(S)$ is an accumulation point of $ S$.

We proceed by contradiction, assuming $f(\{s_n\})$ is not an accumulation point of $\{s_n\}\in\mathcal{H}$.
Then there is a polinomial with integer coefficients $P(x)$ such that $P(f(\{s_n\}))<0$ but $P(s_i)>0, \forall i$.
This is a contradiction because $\{P(s_n)\}\in\mathcal{H}$ and $f(\{P(s_n)\})=P(f(\{s_n\}))$.

[quote="ociretsih"]Then there is a polinomial with integer coefficients $P(x)$ such that $P(f(\{s_n\}))<0$ but $P(s_i)>0, \forall i$.[/quote]  Well, this is false as written, but it can be rescued by applying (i) enough times to kill off the (finitely many) $s_i$ inside some $\varepsilon$-ball around $f(\{s_n\}_n)$.  (These are just the copies of the value $f(\{s_n\}_n)$ that belong to the sequence.) Nice.

[quote="JBL"][quote="ociretsih"]Then there is a polinomial with integer coefficients $P(x)$ such that $P(f(\{s_n\}))<0$ but $P(s_i)>0, \forall i$.[/quote]  Well, this is false as written, but it can be rescued by applying (i) enough times to kill off the $s_i$ inside some $\varepsilon$-ball around $f(\{s_n\}_n)$.  Nice.[/quote]

Good point! I completely forget the point! Thank you for corrrecting me!


Hurray, it's my birthday! Sing happy birthday to me.  :D
What sucks is my birthday is the same day as Thanksgiving this year, so I only get one celebration.


What is the Exact Value of the sum of
1/3 + 2/21 + 3/91 +... 99/96069403

I have figured out that the numbers follow this pattern
1/(1 + 1^2 + 1^4) + 2/(1 + 2^2 + 2^4) + 3/(1 + 3^2 + 3^4) + ...99/(1 + 99^2 + 99^4)

Some help on this question would be appreciated.
Thanks

Does the person who gave you this problem  know the solution? If not, there is probably no simple solution to this problem.

Bill: Yes My teacher knows the answer.

t0rajir0u: Thank you for the hint. However could you show me how to use that to calculate for exact value? that is my main problem.

[quote="paladin8"][quote="V4Vendetta"][quote="rofler"]Try expanding each term by the hyper hint, and cancelling out equal terms.[/quote]
I fail to see how that will give me the sum of the 99 terms.

 Perhaps someone can show me the solution? That would be extremely helpful as this is a challenge question for my level, and I would really love to understand how this works.[/quote]

[hide] The sum is

$\frac{1}{2}\left[\left(\frac{1}{1^{2}-1+1}-\frac{1}{2^{2}-2+1}\right)+\left(\frac{1}{2^{2}-2+1}-\frac{1}{3^{2}-3+1}\right)+\cdots+\left(\frac{1}{99^{2}-99+1}-\frac{1}{100^{2}-100+1}\right) \right]$.

Can you simplify it now? [/hide][/quote]

Yes i understand now. However I just have one final question How did t0rajir0u go from his middle step to his final step? I apologize if its obvious, but I can't see it.

[quote="V4Vendetta"][/quote][quote="paladin8"][quote="V4Vendetta"][quote="rofler"]Try expanding each term by the hyper hint, and cancelling out equal terms.[/quote]
I fail to see how that will give me the sum of the 99 terms.

 Perhaps someone can show me the solution? That would be extremely helpful as this is a challenge question for my level, and I would really love to understand how this works.[/quote]

[hide] The sum is

$\frac{1}{2}\left[\left(\frac{1}{1^{2}-1+1}-\frac{1}{2^{2}-2+1}\right)+\left(\frac{1}{2^{2}-2+1}-\frac{1}{3^{2}-3+1}\right)+\cdots+\left(\frac{1}{99^{2}-99+1}-\frac{1}{100^{2}-100+1}\right) \right]$.

Can you simplify it now? [/hide][/quote][quote="V4Vendetta"]

Yes i understand now. However I just have one final question How did t0rajir0u go from his middle step to his final step? I apologize if its obvious, but I can't see it.[/quote]
It's far from obvious if you've never seen it before. It's something called fraction decomposition that's used a lot in telescoping. I'm not sure how to explain it, but maybe if you google it you can find some simpler examples.

The simplest example is

$\frac{1}{n(n+1)}= \frac{1}{n}-\frac{1}{n+1}$

From this decomposition we can evaluate

$\sum_{k=1}^{n}\frac{1}{k(k+1)}= 1-\frac{1}{n+1}$

you assume there is a nice solution for the problem; the first tactic you always try on these types of problems [i.e. rational functions] is to do a partial fraction decomposition, then other things like comparing to integrals/telescoping are the next step


geometry
analytic geometry
Pythagorean Theorem

A point inside a square is positioned so that the distances to four vertices are 27, 21, 6, and [i]x[/i] units. If [i]x[/i] is a whole number, what is the value of [i]x[/i]?

[quote="Art of Owna"]Whoa, that thing has a name?  I thought it was just like one of those things you can do using analytic geometry and distance formula[/quote]
It can be proven with Pythagorean Theorem in two steps.

sorry if this sounds stupid:

why can't it be like

27^2+21^2=6^2+X^2

It can, but then x is not a whole number


geometry
3D geometry
counting
distinguishability
rotation
group theory

Each face of a cube is to be painted with one of 6 colours such that every 2 adjacent faces has 2 different colours. In how many ways may these faces be painted?

[hide="Assuming rotationally indistinguishable"]
Equivalently, only the opposite faces of the cube may have the same color.

If all six faces are distinct, then we fix one face (say $ \color{red}{R}$), and there are $ 5!$ ways to color the other faces. However, there are four ways to rotate the cube about the $ \color{red}{R}$ face, so we have $ \frac {5!}{4} \equal{} 30$ ways.

If one color is repeated, there are $ 6$ ways to determine the repeated color. There are $ {5\choose 4} \equal{} 5$ ways to determine the four remaining colors to be used, and there are $ \frac{(4 \minus{} 1)!}2 \equal{} 3$ ways to place these four colors (since these four colors lie on a circular band and are reflectionally indistinguishable). Thus we have $ 6 \times 5 \times 3 \equal{} 90$ ways.

If two colors are repeated, then there are $ {6 \choose 2} \equal{} 15$ ways to pick the two repeated colors, and $ {4\choose 2} \equal{} 6$ ways to pick the remaining two colors. Given these four colors all the cubes formed are rotationally indistinguishable, so there are $ 15 \times 6 \equal{} 90$ ways.

If three colors are repeated, then given any three colors all of the cubes are rotationally indistinguishable, so there are $ {6\choose 3} \equal{} 20$ ways.

These total $ 30 \plus{} 90\plus{} 90 \plus{} 20 \equal{} 230$ ways?
[/hide]

[hide="Interesting problem..."]First, lets arbitrarily choose a side to start with.  There are 6 possible colorings for this side.  Now lets arbitrarily choose one of the adjacent sides, where there are 5 possible colorings.  Circling around, the next side has 4 colorings, the next one 4, and the final has 3.  So far we have 6*5*4*4*3 possibilities.

The last side is where it gets tricky.  Depending on the colorings of the previous 4, we have from 2 (the starting color and the only unused color) to 4 possible colorings (where opposing sides of the previous 4 are colored the same).  There are 5*4 ways for the opposing sides to be the same colors, 5*4 times 2(choose one of the final two sides to be different than its opposite) times 2 (only two colors left to choose from) ways that we can have 3 color choices for the final side, and 5*4*3*2 ways to have two color choices left.  So we have 6(5*4*4+5*4*2*2*3+5*4*3*2*2) = 6(80+240+240)=3360 ways disregarding rotational indistinguishability.[/hide]

[quote="facis"]Circling around, the next side has 4 colorings, the next one 4, and the final has 3.  So far we have 6*5*4*4*3 possibilities.[/quote]  No: if the first and third sides you colored have the same color then there are 4 choices for the fourth face, not 3.

Indeed the answer is 230. The official solution assumes rotational indistinguishability. However, it seems quite easy to miscount in this problem! Maybe group theory can simplify things a bit

[quote="JBL"]No: if the first and third sides you colored have the same color then there are 4 choices for the fourth face, not 3.[/quote]

Good point, but it makes no difference to my solution since I break it down into the different cases after that.

As for the rotational indistinguishability...  probably easier to do with combinations as azjps showed.


floor function
combinatorics unsolved
combinatorics

There are $k$ heape of stones, the total number of stones being $n^2$ ($k,n \in N$). It is known that no matter how one divided any heap into two smaller non-empty ones, there would be two heaps with the same number of stones among $(k+1)$ heape obtained. Find the smallest possible value of $k$.

Is there the requirement that each heap should contain at least 2 stones initially?
Otherwise, taking $k=n^2-1$ with one heap of 2 stones and $k-1$ heaps of 1 stone obviously maximizes $k$.

And if each heap must contain 2 or more stones, then $k= \lfloor n^2/2 \rfloor$ should be the maximum of $k$, since each heap must contain 2 or more stones.

Sorry, but you have to find the MINIMUM of $k$.

I see... I must say I'm quite sorry.

For $n=2$ the minimum is obviously $2$, for $3|n$ it is 3 (just divide the stones into 3 equal heaps) and for $n >2 it is probably 4 (unless heaps of 1 stone are allowed).


calculus
derivative
quadratics
function
algebra
polynomial
quadratic formula

find the largest disk abt the origin whose image under the mapping $ f(z) \equal{} z^{2}\plus{}z$ is one-one. obviously its going to be one-one in the disc of radius half since the derivative doesnt vanish there. i also found that on the circle of radius $ \frac{1}{\sqrt 2}$ its not 1-1. any ideas on how to solve this problem?

You've got the directions wrong- it can't be injective on any disk with a zero derivative in the interior, but injectiveness isn't guaranteed just by having nonvanishing derivative.

It is in fact 1-1 on the disk of radius $ \frac{1}2$; the image of the boundary has a cusp but doesn't cross itself. That's as large as possible.

can you explain a little more jmerry? what is the idea behind looking at the boundary?

i thought that non vanishing derivative guarantess injectiveness in a neighbourhood. is that false? can i have a counterexample?

How about $ e^{z}$ on any tall enough strip, or even all of $ \mathbb{C}$?

The boundary idea is based on looking at winding numbers- if it hits a point twice, it winds around that point twice. A loop which winds twice around a point crosses itself.

A simpler method: Construct an inverse by the quadratic formula, with a half-plane including the disk of radius $ \frac{1}2$ as its image.

By the Noshiro-Warschawski Theorem, an analytic function g(z) = z + ... (normalized so that g(0) = 0 and g'(0) = 1) is univalent in |z| < 1 if Re(g'(z)) > 0 (in |z| < 1).   A simple calculation finds that g(z) = z + 0.5z^2 is the second-degree polynomial with the largest second real coefficient that satisfies the hypotheses of the theorem.   By noting that g'(z) = 1 + z and g'(-1) = 0, we also conclude that 0.5 is the largest second coefficient possible for univalence.  If h(z) = z + bz^2 for b>0.5, h'(z) = 0 for some value of z in |z| < 1 and hence h(z) is not univalent in |z| < 1.

Now, letting f(z) = 0.5*g(2z) = z + z^2, the univalence of g(z) in |z| < 1 is equivalent to the univalence of f(z) in |z| < 0.5.


I've got my solution, but I'm looking for another solutions, since my solution seems unnatural even for me :(

If $ a,b,c$ are sides of a triangle, then

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

[quote="dyd"]I've got my solution, but I'm looking for another solutions, since my solution seems unnatural even for me :(

If $ a,b,c$ are sides of a triangle, then

$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} \frac {3}{2}\left( a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b)\right)\geq a^4 \plus{} b^4 \plus{} c^4 \plus{} 3abc(a \plus{} b \plus{} c)$[/quote]

substitution$ a\equal{}x\plus{}y,b\equal{}y\plus{}z,c\equal{}z\plus{}x$

The following inequality is equivalent to $ 2\sum{x^2(x\minus{}y)(x\minus{}z)}\plus{}3\sum{x(y\plus{}z)(x\minus{}y)(x\minus{}z)}\geq 0$

[quote="dyd"]Hmm,, i c... the first sum is schur... what is the second one anyway?

EDITED : oh i c.. both sum is schur's... Thx  :lol:[/quote]

$ \sum{x(y \plus{} z)(x \minus{} y)(x \minus{} z)} \equal{} \sum{x^3y} \plus{} \sum{xy^3} \minus{} 2\sum{x^2y^2}\geq 0$

from AM-GM the inequality holds

you are right :lol:


I was just wondering if any of you had read these books? I am reading Angels and Demons right now, and besides some basic scientific fallacies (they basically dumbed down antimatter to be "something from nothing" which i dont think is a very good explanation, and there is also the Mach 15 aircraft... hm... most fighter jets operate between mach 1 and 2....), its a very good book. I was wondering, do any of the religion vs. science, religion and science topics discussed in this book mean anything to any of you? I have always believed that science, in a sense, explains religion, and that the scientific impossibilities in the Bible- such as the creation of the world in seven days- are merely an old school mortal's representation of God's divine power. My proof for the existence of a God? For one, the periodic table, the fact that there can be order in a seemingly random world screams to me of a higher power. Other demonstrations of order within chaos throughout the mathematical and scientific worlds I believe "prove God" if you will.
On a related note, my middle school religion workbook (I went to a Catholic middle school) had a blurb in it on how religion and science do not have to disagree with each other. It is one of the few things I still remember from those workbooks, because it had such a huge impact on my beliefs.
There are also the messages in the Bible that are seemingly at war with one another that I wanted to discuss (this is more in The Da Vinci Code than in Angels and Demons, now). Why does the Bible say "love one another" if people use (or used) the same book to demonize homosexuals and cast down African Americans (my history book says that the Bible's message of following the Lord was reinterpreted to basically mean "do what Master says")? I think that the problem is that people don't read the whole Bible. To me, the Bible is a message, a holy book, yes, but still written by mortals. Mortal men decided which books would constitute the Gospels. Men, not God, wrote those words, and men, not God, translated the texts into numerous languages. Sometimes inaccurately. It pains me to hear people that will turn their backs on new discoveries in Christianity (the Gnostic Gospels anyone?) because it does not agree with their tradition.
What is religion?

I am just wondering if anyone had any thoughts on these topics.

[i]So dark the con of man....[/i]

I think the errors you spotted in the book you are reading are typical of that author's pattern of writing books chock full of error.

I've read all of his:

Digital Fortress
Angels and Demons
Da Vinci Code
Deception Point

I can't really offer you anything about religion since I was never exposed to it. My parents are non-religious Chinese, and they've never exactly described religion to me. All I know is what I learn classes... which isn't too much..

But Dan Brown is an awesome writer. And the connections between science and religion he makes are absolutely amazing... especially in Angels and Demons.

what do you mean by that, tokenadult? i havent done as much research on angels & demons as i have on the da vinci code, but the man seems quite up on his religious and art history. granted, i think that he is going to take some author's license when it comes to science, but still from what i could tell, the meat of the stories (the history) is as accurate as history can be.

I think tokenadult is referring to false "facts" of the sort mentioned in the critique of [i]The Da Vinci Code[/i] found here: [url]http://www.tektonics.org/davincicrude.htm[/url].  Dan Brown does not get much of his history right, either.

If you would like to have some serious discussion on some of the more philosophical/religious things you brought up, I would suggest posting on the "Religion Discussion" thread in this forum where we can have more in-depth and involved discussions without deviating from the thread topic.

And that is where one gets confused... who to believe? The internet? A fiction author? or... no one?

[quote="JBL"][quote="Setofallsets"]Dan Brown does not get much of his history right, either.[/quote]

Okay, here's a question: if this is a novel, why does anyone question the quality of his history?  (If it's relevant, I have not read any of his books.)[/quote]

because these books plainly state historical FACT on the front pages. many consider it reputable.

Well, having just gone to Amazon and looked at all of the covers of his books, I saw no such labelling.  Additionally, several of his books are labelled "A Novel."  Whether people choose to believe that fiction is fact is none of my business, and I don't see why it should be a fiction-writer's burden to make sure people don't take his books too seriously.

Apparently labeling the books novels, which they indisputably are, is a way of disclaiming responsibility for their internal claims to be factual accounts of history. In other words, a lot of people take the Dan Brown books as historical novels, by genre, something like Johnny Tremaine, when they are really fantasy novels by genre. The books are not being presented to the public in a completely honest fashion.

[quote="Brown"]All descriptions of artwork, architecture, documents, and secret rituals in this novel are accurate.[/quote]

This is a claim of Dan Brown in the preface material of the Da Vinci Code. So this is no longer just fiction or fantasy. The author himself makes a direct claim to the historical basis of his work being completely accurate.

This distinction between history and fiction has been controvertial for a very long time. It has crept up recently in a lot of movies; namely JFK by Oliver Stone, Amistad by Stephen Spielberg, and The Passion of the Christ by Mel Gibson. The question becomes how historically rigorous does a movie or book have to be - and to what degree does how it is marketed affect the standard of historical rigor by which it is judged?

I, personally feel, that the burden of scrutiny lies with the individual reader/audience member. They have to realize that NO histories are completely accurate - and historical fictions are, by necessity, even less accurate. They are usually not written by people with formal training in history, and their primary aim is not to detail historical facts. Like all pieces of art (which is that they are), their primary aim to convey general themes, emotions, or ideas.

So the audience has to be the ultimate judge of the credentials of the person writing or making the product; and they should not trust an artistic rendition as a primary or authoritative source of information on history (no matter what they may say on the cover or in advertising).

I'm sure that most authors of historical fiction try fairly hard to get certain of their details historically accurate, but the fact remains that that is not their area of training or their ultimate goal. Historical fictions cannot be judged in the same way as histories - they can only be compared with other historical fictions, and with other fictions in general.

Speaking of historical fictions - Has anyone read [u]The Sandreckoner[/u]? It's a pretty good historical fiction about Archimedes.

Yes, I have, and highly recommend it to the others here.

I also wrote a review (http://www.livejournal.com/users/landofnowhere/5766.html) of it for my online journal a year ago.

The issue of historical accuracy in novels is interesting.  Some historical novelists pride themselves on their accuracy of detail, and it would be legitimate to criticize such a novelist for doing sloppy research.  Generally it's more important to stay true to the spirit of an era than its facts... and the worst historical inaccuracies are of that type.  (Have you ever noted the disproportionate number of modern-thinking feminists in historical novels, especially those written for preteenage girls?)  But then some authors merely use history as a springboard for their imagination, and in that case historical accuracy isn't the issue there.  I haven't read the Da Vinci code, so I can't make a call there.

(P.S.  It is also standard practice in historical fiction to include a postscript that explains which parts of the book are based on historical fact, and which are fiction.)

[quote="gauss202"]
Speaking of historical fictions - Has anyone read [u]The Sandreckoner[/u]? It's a pretty good historical fiction about Archimedes.[/quote]

blah!!!! I also fight with myself on whether or not I should post in any "religious" related threads because my views are highly radical and extreme and I think I might inadvertantly (or quite purposely) offend someone...

So I'll just say that my friend took yesterday off work and read The Da Vinci Code and said "If your religious, it's not that great... If you're not, you'll love it." Needless to say, as soon as I finish this Sword of Truth series by Terry Goodkind (most awesome fantasy series EVER!), I'll be reading the Da Vinci Code and some other books by this author.

Actually, that is highly inaccurate. My Catholic friend loved this book. So did another kid at school I know, who is Jewish. While it may infuriate the highly devout, again we have to point out that it is FICTION. Fiction, dear. I suppose one must tell oneself that it is for pure entertainment and leave it at that, but again, its not exactly presented honestly.
Also, if you are highly anti-Christian/anti-Catholic/anti-religious in general, you may enjoy [i]Angels and Demons[/i] more. It is the prequel to [i]The Da Vinci Code[/i] and is, in my opinion, more anti-Catholic.
On the same note, I think that you would benefit from seeing the movie Saved!. It has that Home Alone kid in it, whose name I can't spell to say my life. But that's beside the point. It is anti-Evangelist, in a sense. But not really anti-Christian. It just questions the intolerance and the hatred that people try to defend with their religions. As on Yahoo!Movies:
[quote]The first thing teenagers figure out is that it is enticingly easy to make fun of believers in any category. What's nice about this movie is that it does so while still entirely respectful of belief. It begins as a satire of new age-y holier than thou people who spend more time worrying about the appearance of Christianity than the values. But it concludes with a renewed commitment to a faith that engages the mind and heart. You could even call it grace. [/quote]

First, where did that quote come from? I don't seem to find it in this thread?

[quote]Actually, that is highly inaccurate. My Catholic friend loved this book. So did another kid at school I know, who is Jewish. While it may infuriate the highly devout, again we have to point out that it is FICTION. Fiction, dear. I suppose one must tell oneself that it is for pure entertainment and leave it at that, but again, its not exactly presented honestly.
[/quote]

Typically, in books that are pure fiction, there is a note that goes along these lines ... "All persons, events, etc... in this book are purely fictional and only related to real evens and people...."

However, in The Da Vinci Code we get a message saying that all rituals, art, documents, societies, etc... (see my earlier post) is accurate. So this book isn't just pure fiction. Its a book that claims to have a firm footing in history. So for me, it doesn't anger me because I am highly devout. It angers me because the author is lying when he claims that all details are accurate.

Actually, it appears that he doesn't say that ALL details are accurate, just that [i]certain[/i] details are accurate (all rituals, art, documents, societies, etc.). But actual events, conversations, characters, motives, etc. are not necessarily included in that claim.

All the books are awsome.  As many of you know I am a very strong Catholic and the books have not offended me in anyway.  I have read Deception Point, Da vinci Code, Digital Fortress,  and Angels and Demons at least 3 times each.  I didn't like Da vinci Code as much as I did the others but I found the controvercy that it displayed between the Catholic religion and science very interesting and spellbinding.  The only thing I have noticed about the history of the books is that all of the religion history is correct.  They used different popes but thats understandable.

Actually, some of the stuff was accurate, while some of it differed directly from what my English teacher has told us (my English teacher is very learned on some of these subjects). Also, has anyone checked out the Mach15 plane? I found a picture on his website but I'm still not sure how accurate that is, I mean, it is [i]his[/i] website. I believe it is called an X-33. On there, it says that they forecast it being a means of public transportation by 2006. Can anyone dispute that? It seems a little soon for me. I mean, I don't believe there are any commercial planes that even go the speed of [i]sound[/i], let alone 15x that.
However... On the more accurate note, the basic theme that the Catholic church began rather shadily... I think that is pretty well proven, is it not? That they got together and picked which books would go in the Bible and whatnot?

I've only read [i]The DaVinci Code[/i] so far but I'll probably be reading some more books like it soon because I really liked it...maybe some of that arises from the fact that I'm not Christian so I didn't find anything particularly offensive or disagreeable about the book...though I must admit the Catholic Church was portrayed in a rather bad light...

Whether the history is completely accurate or not, though, I think most people agree it's a good read, and I personally find stuff about secret codes and secret knowledge fun to read in general.  Also, it was very interesting to go look up the paintings in the book and try to find the details mentioned...

Mach 15 please!  Thats never going to happen.  Its impossible.  A body cannot handle that pressure even way up high.  At least not on Earth inside the atomosphere.  The fast planes go right now is like 2 or maybe 3 Mach at the most.  And thats mainly fighter jets.  Physics won't allow a Mach 15 in the atomosphere.  Maybe in space but not here.

The Space shuttle reaches to about mach23 while still in the earths atmosphere... why isn't the body able to handle it? You made a statement based on something you have no idea about.

Just letting you know... doesn't matter to me either way...

I do know about it spoon.  I have read about it and asked a lot of people about it they all give me the same answer.

glad someone could clarify this for me. by the way, for anyone interested: http://www.danbrown.com/secrets/angels_demons/plane.html
For pictures of the plane. Also on his website:
[quote]NEW YORK TO TOKYO IN AN HOUR
The United States now has an aircraft that can travel at Mach 15 (over 10,000 mph.) The X-33 space plane cruises at an altitude of almost 100 miles high, on the edge of space. Although the X-33 is no longer classified, specifications of the X-34 remain secret. Aerospace specialists predict civilian passengers will be riding in planes like this by 2006.[/quote]

It's the projection of this technology being available by 2006 that blows my mind. I think that is a little out there. But again, like spoon pointed out, I'm talking about something I know very little about.

[quote="Celeborn"][quote="BHorseMath"]Mach 15 please!  Thats never going to happen.  Its impossible.  A body cannot handle that pressure even way up high.  [/quote]

It's acceleration that a body may not be able to handle, not speed.  In fact, there is no way to feel what speed you're going.  If you travel in a straight line at the same speed, it feels as if you're staying still.[/quote]

Thats what I was saying.  forgive me if it came across wrong.

[quote]The only thing I have noticed about the history of the books is that all of the religion history is correct. They used different popes but thats understandable.[/quote]

[quote]On the more accurate note, the basic theme that the Catholic church began rather shadily... I think that is pretty well proven, is it not? That they got together and picked which books would go in the Bible and whatnot?[/quote]

This is the problem - not all of the religious history is correct. Brown takes certain historical pieces of information, ignores others, and eventually twists his way into a "shady" description. Couple examples: He maintains that the roman emperor Constantine was the person who picked the gospels that would form the Bible, whereas this is completely false. He did sponsor this task, but certainly took no part in it himself. The idea that there were over 80 gospels is unproven - there are less than 50 known of, and most of those are written far later than the four we have today, which are considered more accurate, datewise. Also, Brown claims that more gospels have been found in the Dead Sea Scrolls (where he gets the date when the scrolls were found wrong) which is false.

I wouldn't have a problem with this book at all, but it seems that the majority of its readers put concrete trust into the historical information that the author writes, which, lightly put, can be very misleading.

EDIT - Its not accurate to say that the foundation of the Catholic Church took place in 300 AD, since before the schism there was no separate Catholic / Orthodox Church.


quadratics
logarithms
algebra
special factorizations

Hello,

I have a problem which states :

e^x - e^-x = 3   ;   solve for x (real numbers)

I know that multiplying by a common component (e^x) can allow you to use quadratic form and equation.  But, how would you solve this question algebraically without the use of quadratic equation?  I know my question is rather vague, but is there a way to solve this algebraically without the quadratic method?

[hide]Let $ e^x\equal{}y$, so $ x\equal{}\ln y$ Thus,
$ y\minus{}\frac{1} {y}\equal{}3$.
$ y^2\minus{}1\equal{}3y$.
$ y^2\minus{}3y\equal{}1$
$ y^2\minus{}3y\plus{}(\frac{3} {2})^2\equal{}1\plus{}(\frac{3} {2})^2$
$ (y\minus{}\frac{3} {2})^2\equal{}\frac{13} {4}$
$ y\minus{}\frac{3} {2}\equal{}\pm\frac{\sqrt{13}} {2}$
$ y\equal{}\frac{3 \pm \sqrt{13}} {2}$
Note that when we take the log of any base, the thing we are taking the log of must be positive, so $ x\equal{}\ln (\frac{3\plus{}\sqrt{13}} {2})$

This is called [url=http://www.artofproblemsolving.com/Wiki/index.php/Completing_the_square]completing the square[/url][/hide]

The quadratic formula you wish to avoid is nothing more than the general result of the procedure above (completing the square).


Let $ a$, $ b$ and $ n$ be positive integers. Prove that $ n!$ divides \[ b^{n\minus{}1}a(a\plus{}b)(a\plus{}2b)\cdots(a\plus{}(n\minus{}1)b).\]

you mean this?
$ b^{n\minus{}1}a(a\plus{}b)(a\plus{}2b)\cdots(a\plus{}(n\minus{}1)b)|n!$

or you mean this ?
$ n!| b^{n\minus{}1}a(a\plus{}b)(a\plus{}2b)\cdots(a\plus{}(n\minus{}1)b)$


geometry
calculus
email
inequalities
number theory
graph theory
set theory

[b]I want to learn the most important theorems used in middle school.Could anyone help me?I mean, some theorems, if you colud send to me....I like learning theorems, I know many theorems, but I want more and more.. :) [/b]

Well, I can give you all the theorems you want but you probably understand half of those. I want to know what kind of math background (i.e. algebra, geometry, advanced algebra, calculus, number theory, graph theory, set theory??) you have so that the theorems I provide would be relevent.

-Swapnil

Thank you very much!I want to know all the theorems at geometry.If you know a lot, please send me them to my email adress(mate2006slatina@k.ro).Also, if you can send me some useful inequalities(I know the inequalities of Holder,Cauchy,Cebishev,Bernoulli,Murhead,Hardy,Kurliciunka...)... :D

Go to http://www.mathworld.com and read as much as you'd like, but make sure you know how they're derived.  They'll be of even more use.

Thanks! That is usually my problem with mathworld...I can hardly know how it is derived....actually i'll post one I don't get from wiki...

Can somebody post any link to a book where i can find some proves of useful theorems in algebra?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that if H is a subgroup of a finite group G, then the number or right cosets of H in G equals the number of left cosets of H in G.

PLEASE HELP!! Any hints???

Also one more!!!

Prove that if H and K are subgroups of a finite group G, and K is a subgroup of H, the [G:H][H:K].

THANK YOU SO MUCH!!!

i will give some hints but I don't think it is good for you if i give everything
work on it and if you fail, come back

map every coset gH onto $H g^{-1}$  (so map left cosets onto right cosets)
now YOU check that :

this map is well defined (i used an element g in my coset, show that my image $H g^{-1}$ does not depend on which g)
it is injective
it is surjective

for the second, when G is finite , don't you have a formula for [G:H] in terms of G and H ?  Use this and you find the result

I think it works too( anyone else?) when G is not finite?

[G:H]= [G] = [H] * K is what i got as a defintion for index from my book... is that what is goign to help me

Could you elaborate?  I find that weird?

for the second one all i have to do it say o(G)/ o(H) * o(H)/o(K) and those cross out and i will get the o(G)/o(K).. but what i dont understand is when do i use the fact that K is a subgroup of H?


1. For all the value, x,y,z, are positive real numbers and all the natural numbers n, prove that :
$ \frac {x^n}{y\plus{}z} \plus{} \frac {y^n}{z\plus{}x} \plus{} \frac {z^n}{x\plus{}y} > \frac {{(x\plus{}y\plus{}z)}^{n\minus{}1}}{2.{3}^{n\minus{}2}}$

From rearrangement,
\[ \sum\frac{x^n}{y\plus{}z}\ge\frac{3(x^n\plus{}y^n\plus{}z^n)}{2(x\plus{}y\plus{}z)}\]
But by Chebyshev's inequality we have \[ 3^{n\minus{}1}(x^n\plus{}y^n\plus{}z^n)\ge (x\plus{}y\plus{}z)^n\]
Thus \[ \sum\frac{x^n}{y\plus{}z}\ge\frac{(x\plus{}y\plus{}z)^{n\minus{}1}}{2\cdot 3^{n\minus{}2}}\]


hi
which is the diferrence betwen  shortlisted and longlisted problems from the imo

Read this: http://www.mathlinks.ro/viewtopic.php?t=15623
In short notes:
Longlist contains all the proposed problems from all countries which have sent their proposals
Shortlist is a list of problems, chosen from the Longlist by the Problem Selection Committee (20-30 problems) from which the Jury chooses 6 problems for IMO.


calculus
integration
analytic geometry
real analysis
real analysis unsolved

Show that

$\int_{-\infty}^{\infty}e^{-x^{2}}\,dx=\sqrt{\pi}$

classic trick:
compute $\int_{\mathbb{R}^{2}}e^{-(x^{2}+y^{2})}dx dy = (\int_{\mathbb{R}}e^{-u^{2}}du)^{2}$ with polar coordinate ..

What happens when you try to compute $\int_{\mathbb{R}^{3}}e^{-(x^{2}+y^{2}+z^{2})}\,dx\,dy\,dz$ in spherical coordinates?

For that matter, what happens when you try to compute

$\int_{\mathbb{R}^{n}}e^{-(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2})}\,dx_{1}\,dx_{2}\dots dx_{n}$

in the $n$-dimensional analogue of spherical coordinates?

[hide="What I am trying to do."]This is about one approach to finding the $n$-dimensional measure of the unit ball (or any ball) in $\mathbb{R}^{n}.$[/hide]

$\int_{R^{n}}e^{-(x_{1}^{2}+\ldots+x_{n}^{2})}dx_{1}\ldots dx_{n}= \int_{\mathbb{S}^{n}}\int_{0}^{\infty}e^{-r^{2}}r^{n-1}dr dS=H^{n-1}(\mathbb{S}^{n}) \int_{0}^{\infty}r^{n-1}e^{-r^{2}}dr=I^{n}$
where $I=\int_{\mathbb R}e^{-r^{2}}dr$. Hence this is a nice way to find $H^{n-1}(\mathbb{S}^{n})$, with integration by part.

This has probably been posted before. Anyway it is well known so
please look [url=http://en.wikipedia.org/wiki/Gaussian_integral]here[/url] for example.


trigonometry
geometry
angle bisector
trig identities
Law of Sines

Use the law of sines to prove the angle bisector theorem.

Let the angle bisector from $A$ meet $BC$ at $D$.
Then $\sin \angle BAD = \sin \angle CAD = t$, say
Also $\sin \angle ADB = \sin \angle ADC = u$ (since they are supplementary)

Considering the Sine Rule on $\Delta ABD$ and $\Delta ACD$,

$\frac{\frac{AB}{u}}{\frac{BD}{t}}=\frac{\frac{AC}{u}}{\frac{CD}{t}}=1$
$\frac{AB}{BD}=\frac{AC}{CD}$

[hide="Not hard"]
WLOG let $\angle B$ be bisected. In $\triangle ABC$, let $\angle B=2\theta$, $\angle A = \alpha$, and let the intersection of the angle bisector and side $AC$ be $D$. Then, $\angle ADB=\pi-(\theta+\alpha)$ and $\angle BDC=\theta+\alpha$.

By the law of sines, $\frac{CB}{CD}=\frac{\sin(\theta+\alpha)}{\sin \theta}$

$\frac{AB}{AD}=\frac{\sin[\pi-(\theta+\alpha)]}{\sin \theta}=\frac{\sin(\theta+\alpha)}{\sin \theta}$.

Set them equal and we're done.[/hide]

another proof:
let $D\in AB$ be such that $\angle ACD=\angle DCB$

$\frac{AD}{BD}=\frac{[CAD]}{[CBD]}=\frac{ \frac{1}{2}\cdot CD\cdot AC\cdot\sin\angle ACD}{ \frac{1}{2}\cdot CD\cdot BC\cdot\sin\angle BCD}=\frac{AC}{BC}$

the synthetic proof is nice too...


Does there exist a sequence of 2005 consecutive positive integers that contains exactly 25 primes?

Yes.
Let's consider $f(n) = \pi (n+2005)-\pi (n)$, where $\pi (x)$ is the number of primes less than or equal to $x$.
Thus $f(n)$ is the number of primes among $n+1,n+2,\cdots, n+2005$.

It is easy to verify that according to $n+1$ and/or $n+2006$ is a prime or not, we have $f((n+1)-f(n) =-1,0,1$.
Since $f(1) > 25$ and since it is known that there are arbitraly large string of consecutive integers containing no primes, which ensures that there exists $n$ such that $f(n)=0$, it follows that there exists $m$ such that $f(m) = 25.$

Pierre.

Posted before at
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100038[/url].

Edit: Pierre beat me to it.


inequalities
function
inequalities unsolved

prove that:
if a>1,and k,n$ \in N$,k<n
then:
$ \sqrt[k]{a^k\minus{}1} < \sqrt[n]{a^n\minus{}1}$

Consider the function $ f(x) \equal{} (a^x \minus{} 1)^{\frac 1 x}$.

Then $ \frac{df}{dx} \equal{} \frac{1}{x^2}(a^x \minus{} 1)^{\frac 1 x} \left( \frac{a^x \ln(a^x)}{a^x\minus{}1}\minus{}\ln(a^x\minus{}1) \right) > 0 \forall x$.

[quote="mathangel"]Consider the function $ f(x) \equal{} (a^x \minus{} 1)^{\frac 1 x}$.

Then $ \frac {df}{dx} \equal{} \frac {1}{x^2}(a^x \minus{} 1)^{\frac 1 x} \left( \frac {a^x \ln(a^x)}{a^x \minus{} 1} \minus{} \ln(a^x \minus{} 1) \right) > 0 \forall x$.[/quote]
can you tell me,what is this:$ \frac{df}{dx}$?

[quote="tuandokim"][quote="mathangel"]Consider the function $ f(x) \equal{} (a^x \minus{} 1)^{\frac 1 x}$.

Then $ \frac {df}{dx} \equal{} \frac {1}{x^2}(a^x \minus{} 1)^{\frac 1 x} \left( \frac {a^x \ln(a^x)}{a^x \minus{} 1} \minus{} \ln(a^x \minus{} 1) \right) > 0 \forall x$.[/quote]
can you tell me,what is this:$ \frac {df}{dx}$?[/quote]
The function $ f(x)$ differentiated over $ x$. You probably know it as $ f'(x)$.

[quote="mathangel"]Consider the function $ f(x) \equal{} (a^x \minus{} 1)^{\frac 1 x}$.

Then $ \frac {df}{dx} \equal{} \frac {1}{x^2}(a^x \minus{} 1)^{\frac 1 x} \left( \frac {a^x \ln(a^x)}{a^x \minus{} 1} \minus{} \ln(a^x \minus{} 1) \right) > 0 \forall x$.[/quote]
I don't think so:
$ \frac {df}{dx} \equal{} (a^x \minus{} 1)^{1/x}*( \frac{\minus{} ln(a^x \minus{} 1)}{x^2} \plus{} \frac{a^x*ln(a)}{(x*(a^x \minus{} 1)})$

[quote="Mathias_DK"][quote="tuandokim"][quote="mathangel"]Consider the function $ f(x) \equal{} (a^x \minus{} 1)^{\frac 1 x}$.

Then $ \frac {df}{dx} \equal{} \frac {1}{x^2}(a^x \minus{} 1)^{\frac 1 x} \left( \frac {a^x \ln(a^x)}{a^x \minus{} 1} \minus{} \ln(a^x \minus{} 1) \right) > 0 \forall x$.[/quote]
I don't think so:
$ \frac {df}{dx} \equal{} (a^x \minus{} 1)^{1/x}*( \frac { \minus{} ln(a^x \minus{} 1)}{x^2} \plus{} \frac {a^x*ln(a)}{(x*(a^x \minus{} 1)})$[/quote]
That's the exact same thing!
$ \frac {1}{x^2}(a^x \minus{} 1)^{\frac 1 x} \left( \frac {a^x \ln(a^x)}{a^x \minus{} 1} \minus{} \ln(a^x \minus{} 1) \right)$ $ \equal{} (a^x \minus{} 1)^{1/x}*( \frac { \minus{} ln(a^x \minus{} 1)}{x^2} \plus{} \frac {a^x*ln(a)}{(x*(a^x \minus{} 1)})$.......[/quote]
Are you sure?

[quote="tuandokim"][/quote][quote="Mathias_DK"][quote="tuandokim"][quote="mathangel"]Consider the function $ f(x) = (a^x - 1)^{\frac 1 x}$.

Then $ \frac {df}{dx} = \frac {1}{x^2}(a^x - 1)^{\frac 1 x} \left( \frac {a^x \ln(a^x)}{a^x - 1} - \ln(a^x - 1) \right) > 0 \forall x$.[/quote]
I don't think so:
$ \frac {df}{dx} = (a^x - 1)^{1/x}*( \frac { - ln(a^x - 1)}{x^2} + \frac {a^x*ln(a)}{(x*(a^x - 1)})$[/quote]
That's the exact same thing!
$ \frac {1}{x^2}(a^x - 1)^{\frac 1 x} \left( \frac {a^x \ln(a^x)}{a^x - 1} - \ln(a^x - 1) \right)$ $ = (a^x - 1)^{1/x}*( \frac { - ln(a^x - 1)}{x^2} + \frac {a^x*ln(a)}{(x*(a^x - 1)})$.......[/quote][quote="tuandokim"]
Are you sure?[/quote]
Yes


calculus
integration
calculus computations

the integral after u substitution is E, right?

So we want to make the substiution $ u \equal{} \sqrt {x \minus{} 1}$ on $ \int_2^5 \frac {\sqrt {x \minus{} 1}}{x}~dx$.

So if $ u \equal{} \sqrt {x \minus{} 1}$, then $ du \equal{} \frac {1}{2} \cdot (x \minus{} 1)^{ \minus{} \frac {1}{2}}\cdot dx \equal{} \frac {1}{2\sqrt {x \minus{} 1}} \cdot dx \equal{} \frac {1}{2u} \cdot dx$.  Thus $ du \equal{} \frac {dx}{2u} \iff dx \equal{} 2u \cdot du$.  And we also know that $ x \equal{} u^2 \plus{} 1$.  So we have the following:

$ u \equal{} \sqrt {x \minus{} 1}$
$ x \equal{} u^2 \plus{} 1$
$ dx \equal{} 2u \cdot dx$

And the bounds are going to be:

$ x \equal{} 2 \implies u \equal{} \sqrt {x \minus{} 1} \implies u \equal{} \sqrt {2 \minus{} 1} \implies u \equal{} 1$
$ x \equal{} 5 \implies u \equal{} \sqrt {x \minus{} 1} \implies u \equal{} \sqrt {5 \minus{} 1} \implies u \equal{} 2$

Thus:

$ \int_2^5 \frac {\sqrt {x \minus{} 1}}{x} \equal{} \int_1^2 \frac {u}{u^2 \plus{} 1} \cdot 2u \cdot du \equal{} \boxed{E. ~~\int_1^2 \frac {2u^2}{u^2 \plus{} 1}~du}$


Let a,b,c be positive real numbers and $ \sqrt{1987\plus{}a}\plus{}\sqrt{1987\plus{}b}\equal{}2\sqrt{1987\plus{}c}$
Prove that:
$ \frac{1}{2}(a\plus{}b)\geq c$

Let $ t\equal{}1987$. Replace $ a$, $ b$, and $ c$ with $ x$, $ y$, and $ z$. Then we can easily solve for $ x$; $ z\equal{}\frac{\sqrt{t^{2}\plus{}t(x\plus{}y)\plus{}yz}}{2}\minus{}\frac{t}{2}\plus{}\frac{x\plus{}y}{4}$.

$ 2x\plus{}2y\ge 4z\iff$
$ 2x\plus{}2y\ge 2\sqrt{(t\plus{}x)(t\plus{}y)}\minus{}2t\plus{}x\plus{}y\iff$
$ (t\plus{}x)\plus{}(t\plus{}y)\ge 2\sqrt{(t\plus{}x)(t\plus{}y)}$

which is true by the am-gm inequality.

[quote="galois01"]Let a,b,c be positive real numbers and $ \sqrt{1987\plus{}a}\plus{}\sqrt{1987\plus{}b}\equal{} 2\sqrt{1987\plus{}c}$
Prove that:
$ \frac{1}{2}(a\plus{}b)\geq c$[/quote]Here is a slightly different solution.
By the Jensen Inequality, we have $ \frac{\sqrt{1987\plus{}a}\plus{}\sqrt{1987\plus{}b}}{2}\leq\sqrt{\frac{1987\plus{}a\plus{}1987\plus{}b}{2}}\equal{}\sqrt{1987\plus{}\frac{a\plus{}b}{2}}$. Replace $ \frac{\sqrt{1987\plus{}a}\plus{}\sqrt{1987\plus{}b}}{2}$ with $ \frac{2\sqrt{1987\plus{}c}}{2}\equal{}\sqrt{1987\plus{}c}$ so we have $ \sqrt{1987\plus{}c}\leq\sqrt{1987\plus{}\frac{a\plus{}b}{2}}\implies c\leq\frac{a\plus{}b}{2}$.


Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that
\[ \sqrt[3]{4(a^3\plus{}b^3)}\plus{}\sqrt[3]{4(a^3\plus{}c^3)}\plus{}\sqrt[3]{4(b^3\plus{}c^3)}\geq\sqrt[3]{9\left((a\plus{}b)^3\plus{}(a\plus{}c)^3\plus{}(b\plus{}c)^3\right)}\]


Hi Guys,
There is a problem in the Rice University Annual Mathematics Test which I do not understand. Please explain it to me.
What is the largest six-digit number whose digits are all different such that the sum of the squares of the digits in the even-numbered positions equals the sum of the squares of the digits in the odd-numbered positions?
Ans. 986702

Thanks,

You could let the number be represented by $ ABCDEF$ where each letter represents a distinct digit and $ A \not\equal{} 0$.

Then you have to find a solution to the following equation:

$ A^2 \plus{} C^2 \plus{} E^2 \equal{} B^2 \plus{} D^2 \plus{} F^2$

You could probably find the answer through some trial and error.

also, you may assume that
$ A>\equal{}B$, $ A>\equal{}C>\equal{}E$ , and that $ B>\equal{}D>\equal{}F$ (i don't know how to latex greaterthan or equals)

it may help to write the equation as

$ (A\minus{}B)(A\plus{}B) \plus{} (C\minus{}D)(C\plus{}D) \plus{} (E\minus{}F)(E\plus{}F) \equal{} 0$

[quote=">Aver<"]The question is you to pick the greatest number which satisfies the 1st, 3rd and 5th digits' squares sum (ODDS) equal to 2nd, 4th and 6th (EVENS). Consider this number to be $ ABCDEF$ as the others said.

So if we write up the equation,

$ A^2 \plus{} C^2 \plus{} E^2 \equal{} B^2 \plus{} D^2 \plus{} F^2 \implies (A \minus{} B)(A \plus{} B) \plus{} (C \minus{} D)(C \plus{} D) \plus{} (E \plus{} F)(E \minus{} F) \equal{} 0$ ~ which mathking123 already said.

Then the answer comes from when $ A \equal{} B, C \equal{} D, E \equal{} F$ because we can't pick the negative situation (greatest number). So the greatest we can write is $ 998877$[/quote]

However, all digits must be distinct...

Refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=245790]this[/url].


Γειά σας. Σας γράφω δύο ασκήσεις από το υπέροχο βιβλίο "Ακολουθίες Πραγματικών Αριθμών" του Μαρίνου Ζήβα. Είναι πολύ ευκολότερες από τις ασκήσεις με τις οποίες καταπιάνεστε εδώ αλλά νομίζω ότι τα μαθηματικά είναι όμορφα είτε είναι εύκολα είτε είναι δύσκολα.
[b]1)[/b] Να δείξετε ότι κάθε όρος της ακολουθίας:
α(ν)= 1/6ν^3 + 1/2ν^2 - 2/3ν + 1    είναι ακέραιος.

[b]2)[/b] Να βρεθεί η ακολουθία για την οποία ισχύει ότι α(ν)=1 για ν:άρτιο και
α(ν)=0 για ν:περιττό.

Gia tin 1...
[hide]exume $a_{n}=\frac{n^{3}+3n^{2}-4n+6}{6}$
opote arki ndo $n^{3}+3n^{2}-4n+6$ isoipolipo $0 (mod6)$
dilladi oti $n^{3}+3n^{2}-4n=n(n-1)(n+4)$ ine pol 6 gia ka8e n fisiko..
to opio ine profanes..pernontas $n=6k+u$ me $u=0,...5$[/hide]
Gia tin 2....
[hide]endiaferon 8a exi an oles i akolu8ies me ayti tin idiotita anikun se mia euriteri ikogenia akolu8ion..
mia pu vrika ego ine i $a_{n}=|sin(n+1)\frac{\pi}{2}|$[/hide]

 :D

[quote="RAKAR"]Gia tin 2....
[hide]endiaferon 8a exi an oles i akolu8ies me ayti tin idiotita anikun se mia euriteri ikogenia akolu8ion..
mia pu vrika ego ine i $a_{n}=|sin(n+1)\frac{\pi}{2}|$[/hide][/quote]
ωραία σκέψη  :wink:
Βάζω δύο ακόμη (στοιχειώδεις):

[hide="2"]
$b_{n}= (n+1) \mod 2$

$c_{n}= \frac{1+(-1)^{n}}{2}$[/hide]

Rakar, μπράβο θαυμάσια ιδέα για την  δεύτερη. Να προτείνω και εγώ μία που σκέφτηκα.
α(ν)= [ (-1)^ν + 1]/2
Νομίζω ότι πρέπει να την δουλέψουμε αυτήν την δεύτερη γιατί η υποψία του Rakar ότι ανήκουν σε ευρύτερη οικογένεια είναι νομίζω σωστή. Ευπρόσδεκτη κάθε ιδέα.

Παιδιά συγνώμη, δεν πρόσεξα καλά αυτές γράψατε τελευταία και έγραψα την ίδια. Συγνώμη.

Να αποδειχθεί ότι η ακολουθία 49, 4489, 444889, 44448889, 4444488889, ...,  αποτελείται αποκλειστικά από τετράγωνα ακεραίων.  :)
(από το βιβλίο του Paul Zeitz, The Art and Craft of Problem Solving)


function
blogs
geometry
geometric transformation
rotation
complex numbers
algebra

How can generating functions help in combinatorial problem solving?? Could you lease give me an example of a problem like that and its sollution/??/
Thank you in advance....

See [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=184510]a blog post of mine[/url].  I also highly recommend [url=http://www.math.upenn.edu/~wilf/DownldGF.html]generatingfunctionology[/url].

I am also doubtful about the use of generating function in solving problems in combinatorics. It seems to me that any problem solved by using generating functions can also be solved with more elementary methods. Can anyone give some examples that generating function technique will give an advantage to more elementary techniques?

In how many ways can we fill a bag with 50 fruits subject to the following conditions:

- The number of apples must be even.
- The number of bananas must be a multiple of 5.
- There can be at most four oranges.
- There can be at most one pear.

(Albert Meyer, Mathematics for Computer Science Course Notes)

This problem is pretty easy with generating functions, but not so easy otherwise.

Some subtleties need to be addressed here.  First, sometimes "solving" a problem by generating functions does not involve giving the same answer as solving it by other means.  Often we can compute the generating function of some interesting combinatorial sequence without being able to give a closed formula for that sequence itself.

Second, the reason generating functions are useful is that they automatically encode the [url=http://en.wikipedia.org/wiki/Rule_of_product]rule of product[/url] in simple algebraic manipulations.  The simplification is analogous to studying rotation using complex numbers instead of rotation matrices; what it lets us do is simplify and combine otherwise abstruse computations in an elegant format.

Consider [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=184510]the following proof[/url].  The balls and urns problem has a well-known and straightforward counting solution (by dividers), but what is interesting about this proof is how a combinatorial problem is solved using only the general binomial theorem, an [b]algebraic[/b] theorem.  Similarly, it is much easier to prove that

$ (1 \minus{} 1)^n \equal{} \sum_{k\equal{}0}^{n} (\minus{}1)^k {n \choose k} \equal{} 0$

by inspecting the generating function than by a counting argument.

On the other hand, one might argue that generating function solutions provide less insight into the nature of a problem, and this might be true.  Try reading the book I linked to.


There are n elements in a circle. We should assign a color to each element out of 7 colors, so that each two adjacent elements will get a different color.
Q(n) is the sulution... Now I should solve it with recursion somehow.
I know that for Q(2) = 6*7 and Q(3) = 6*7*5 are trivial.
But for Q(4) we have to consider already two cases.. where element 1 and 3 has the same color (and then element 4 can have 6 choices) or a different color (and then element 4 can have 5 choices)

Thanx for you help.


Prove if $ a,b,c>0$ then

$ \sum_{cyc}\frac{1}{2a^2\plus{}bc}\ge\frac{1}{ab\plus{}bc\plus{}ca}\plus{}\frac{2}{a^2\plus{}b^2\plus{}c^2}$


Given triangle $ ABC$.Circle passing through $ B$ and $ C$ intersects  $ AB$ and $ AC$ at $ C_1$ and $ B_1$ respectively.Let $ BB_1$ and $ CC_1$ intersects at $ S$.Prove that the orthocenters of $ \bigtriangleup ABC$,$ \bigtriangleup AB_1C_1$ and $ S$ are collinear.

Dear Mathlinkers,
1. let C'1, B'1 the feet of the C1, B1-altitudes of AB1C1.
2. BC and B1C1 being antiparallels wrt AB, AC,   B'1C'1// BC.
3. Let H, H1 be the orthocenters wrt ABC, AB1C1.
4. HBC and H1C'1B'1 being homothetic, we are done...
Sincerely
Jean-Louis

Let $ H_1$ be the orthocenter of $ \triangle AB_1C_1$, $ H_2$ be the orthocenter of $ \triangle ABC$. Let $ C_2,B_2$ be the projections of $ H$ onto $ AB_1,AC_1$ respectively. Let $ C_3,B_3$ be projections of $ H_2$ onto $ AB,AC$ respectively. Denote $ \omega_b,\omega_c$ by the circle with diameter $ BB_1,CC_1$, respectively. We have $ \overline {H_1C_1}\cdot \overline {H_1C_2} \equal{} \overline {H_1B_1}\cdot \overline {H_1B_2}$, which implies that $ \mathcal {P}_{H_1/\omega_b} \equal{} \mathcal {P}_{H_2/\omega_c}$. With the same argument for $ H_2$ and $ S$. Hence, we conclude that $ H_1,H_2,S$ lie on the radical axes of $ \omega_b,\omega_c$. Which leads to the fact that $ H_1,H_2,S$ are collinear. Our proof is completed then.

considering HBC and H'B'C' are homothetic
the correspondent points H,H' and S are collinear


inequalities
algebra
system of equations

Say we are to prove an inequality in a, b, and c with the constraint that abc=1. Why are we allowed to make a substitution like a=(x^2)/(yz), b = (y^2)/(zx), c = (z^2)/(xy)? I can see that after substituting those expressions into abc=1, the equality would still hold, but I don't understand how we can be sure that such x, y, and z necessarily exist. This may seem like a very simple question, but I'd appreciate if someone could clarify this point for me. Thanks.

[quote="JBL"]Try solving the equations for x, y, z in terms of a, b, c.  If you can, then such x, y, z exist.[/quote]

What if the subtitution is a big complicated, involving a few variables, such that the system of equations is not easy to solve (even here, with only 6 variables, the process is not easy).

This is a pretty easy one to solve for -- it's symmetric, after all.  Anyhow, I can't think of circumstances under which big complicated substitutions come up, for exactly the reasons that you're talking about.  However, I haven't seen a whole lot of these types of problems compared to other people.  If you can show me an example of a problem with a more-complicated-than-this substitution, I'll think about it, but I'm not in a position to say anything more general and have it be useful.

[quote]This is a pretty easy one to solve for -- it's symmetric, after all.[/quote]

*shuffles feet* ...

Hmm, so whats the easiest way to solve for x, y, z in this example?

Sorry I took so long getting back to you.  My earlier advice on this was wrong.  You can't really solve those equations for a single solution x, y, z.  In fact, we see that if we have a solution, multiplying x, y and z by a constant will yield another valid solution.  So the best you can do is solve it up to a ratio.  The simplest way to do this is write them as
a = x^3/(xyz), b = y^3/(xyz), c = z^3/(xyz), and we see that x^3:y^3:z^3 = a:b:c, so in particular, one solution is x = a^(1/3), y = b^(1/3), z = c^(1/3).  Then you can clear by a constant if you like to make them nice numbers.  Anyhow, the idea behind coming up with such an expression relies upon the fact abc = 1; you need to find other variables to substitute in that will make that true.

Let me emphasize something I didn't: variables that will make it true regardless of their values.  (Of course, here we're constrained to non-zero, but you know what I mean.)  So, you note that if we multiply the numbers x^2/(yz) together cyclicly, we get 1; so that is the reason we thought to make those substitutions for a, b, c to begin with.


[quote="GameBot"]A 100-foot rope is cut into three pieces. The first piece is three times
as long as the second, and the third piece is 20 feet shorter than the
second. How many feet long is the shortest piece?[/quote]

Let the second piece have length $ x$. We then have $ 3x \plus{} x \plus{} x \minus{} 20 \equal{} 100 \implies 5x \minus{} 20 \equal{} 100 \implies x \equal{} 24$. However, we want to find the shortest piece, so our answer is $ 24 \minus{} 20 \equal{} \boxed{4}$.


Find the number of finite sequences $ \{a_1,a_2,...,a_{2n\plus{}1}\}$,formed with nonnegative integers,for which $ a_1\equal{}a_{2n\plus{}1}\equal{}0$ and $ |a_k\minus{}a_{k\plus{}1}|\equal{}1 \forall k \in \{1,2,...,2n\}$.

This is a standard presentation of Dyck paths, one of the most canonical objects counted by the Catalan numbers.


Not sure if this is in the right place.

Express $\prod_{k=1}^{n}\frac{2^{k}}{2k}$ in closed form.

[quote="SplashD"][quote="i_like_pie"][quote="Treething"][hide="answer"]
$\frac{2^{\frac{n(n-1)}{2}}}{n!}$

... I think.
[/hide][/quote]
You're close. :)[/quote]

I get the same answer as well, are you sure it's wrong?[/quote]
[hide="My solution"]$\prod_{k=1}^{n}\frac{2^{k}}{2k}=\frac{\prod_{k=1}^{n}2^{k}}{n!2^{n}}=\frac{2^{\frac{n(n+1)}{2}}}{n!2^{n}}$

I forgot to simplify. I stopped at the last step.[/hide]

[quote="Tim_Lou"]they are the same thing.[/quote]

Yes, just with the $2^{n}$ canceled


Let ABC be a triangle. Prove that:

 \sum (a 2 +b 2 )/(a^4+b^4) \leq 1/2Rr