Your cat Fermat wasn't at the door to greet you when you came home. After a little while, you start to get worried about him. So you decide to look all over your apartment to find him.

[hide="Living Room"]You look around your living room. You decide that he must be either behind the TV, in, around, or under the couch, behind the bookshelf, or in the Big Box O Movies.

[hide="TV"]You look all around the TV, but you can't find the little theorem. Now what?

[hide="Watch TV"]Congrats, you just lowered your IQ by two points.[/hide]

[hide="Take pride in your Plasma TV"]Oh what an awesome TV! High-def, 40-inch, 10 grand that could have been used to pay off the phone bill down the drain. You feel like an idiot.[/hide]

[hide="Start doing the Soulja Boy Dance"]You have lost 50% of your AoPS Popularity.[/hide][/hide]

---------- Line of Different Thing ----------

[hide="Couch"]You look around, under, and in your couch. No Fermat. Now what?

[hide="Sleep"]You fall asleep. You wake up in the bathtub. Weirdo...[/hide]

[hide="Play your Wii"]When you stop, it's 11P.M. You just wasted 6 hours of time that you could have used to find your cat.[/hide]

[hide="DIE, EMO BOY!!!"]You hold your breath, but you give up before you have the chance to pass out. You try a few more times, but you just aren't emo enough. Too bad. :noo: [/hide][/hide]

---------- Line of Different Thing ----------

[hide="Bookshelf"]You look around your bookshelf. You even lift it up at one point. Only to have it come crashing down on your toe.

[hide="Cry"]I would too; heavy bookshelf!!![/hide]

[hide="Read a book"]You have just gained 2 IQ points![/hide]

[hide="Smash your finger with a hammer"]You forget about your toe, but your finger is killing you. Emo boy.[/hide][/hide]

---------- Line of Different Thing ----------

[hide="Big Box O Movies"]You see the box top is not fully on the box, so you are hopeful. You lift the top. Nothing but movies.

[hide="Look through your movies"]Dude your movie taste sucks!!![/hide]

[hide="Cry"]What, again?![/hide]

[hide="Watch movies"]You see that your movie taste sucks.[/hide][/hide][/hide]

[color=red]-------------------------------- Line of Different Room --------------------------------[/color]

[hide="Kitchen"]After examining the kitchen, you realize that Fermat can be in one of four places: in the oven(emo cat), in the cabinet, in the fridge, or in the freezer.

[hide="Oven"]You look in the oven. You get blasted by the heat, and you see charred remains. When you take it out, you see that you burned your chicken. Good Job! :D

[hide="Pout"]Wait, what? You pout???[/hide]

[hide="Cry"]Wow you are sensitive![/hide]

[hide="Take a huge bite"]Tastes like chicken. You vomit.[/hide][/hide]

---------- Line of Different Thing ----------

[hide="Cabinet"]You look in the Cabinet. You see lots of human food, and spilled cat food. You also hear a meow in the closet. There is your cat! At least he wasn't charred or frozen.

Now if you find this, PM me following number:
152[i]586[/i]48103541128765125795324984.3141592653589793208462
I will be making another maze tomorrow. I will PM it to you, so only the people who find this number will see it, and you won't be able to quote me.[/hide]

---------- Line of Different Thing ----------

[hide="Fridge"]You find that the fridge is off for some reason, and you frantically turn it back on so that your food doesn't spoil.

[hide="Forgive me"]but I can't think of anything for you to do here.[/hide][/hide]

---------- Line of Different Thing ----------

[hide="Freezer"]You look in the freezer. There is absolutely no place where your cat can fit. But there is room in the fridge, maybe you should try there...[/hide][/hide]

[color=red]-------------------------------- Line of Different Room --------------------------------[/color]

[hide="Bathroom"]Now there are two places he could be: in the bathtub, or under the sink.

[hide="Sink"]You see a leaky pipe and nothing more.

[hide="Hit pipe with wrench"]Well [b]that's[/b] over-reacting if I ever saw over-reaction.[/hide]

[hide="Shave"]Good idea.[/hide]

[hide="Wash hands"]What a good little child![/hide][/hide]

---------- Line of Different Thing ----------

[hide="Bathtub"]He ain't in there, but a naked lady is! This isn't your apartment!!! :rotfl:

[hide="Run away"]What I would do.[/hide]

[hide="Stare"]You get slapped. I am lol-ing right now.[/hide]

[hide="Get in"]Wow you are a bad person![/hide][/hide][/hide]

[color=red]-------------------------------- Line of Different Room --------------------------------[/color]

[hide="Office"]There are four places your cat could be: near the computer, the futon, the Big Box O Goodies, or the closet.

[hide="Futon"]You don't find him near the futon.

[hide="Sleep"]G'night.[/hide]

[hide="I CANNOT TAKE IT ANYMORE!!!"]Jumping out of windows is fun.[/hide]

[hide="Moo"]Ha ha you lost.[/hide]
[color=white][color=orange][/color]and you have found the forbidden text. Handle this like a man and don't quote again.[/color]
[hide="OMG"]YOU FOUND A PENNY! :D[/hide][/hide]

---------- Line of Different Thing --------

[hide="Big Box of Goodies!!!"]You find every kind of goodie you know of; but you don't find what you're looking for: a cat.

[hide="WHAT A RIP-OFF!!!"]I agree.[/hide]

[hide="Eat the fish"]Now THAT is disgusting.[/hide]

[hide="THE BOX IS POSSESSED!!!"]It starts singing Macarena and you are driven insane.[/hide][/hide]

---------- Line of Different Thing ----------

[hide="Computer"]Not there.

[hide="Go to AoPS"]Now THAT'S the spirit! :D[/hide]

[hide="Play WoW"]Umm... What???[/hide]

[hide="Desk"]You slam your hand on the desk in frustration, and you get cut by a rusty nail. Soon you see a line going up your arm and you pass out. Dead.[/hide][/hide]

---------- Line of Different Thing ----------

[hide="Closet"]You see the closet.

[hide="Examine"]You examine it and you see that it is in fact a closet.

I did it. I examined it.[/hide]

[hide="Go into closet"]You come out the other side in your mom's house. You go back through and get to your dad's house. And you cycle through.[/hide]

[hide="Scared of the dark??? :P"]You say "What?! I'm not scared of the dark!" You prove it by turning off the lights and going into the closet. You make mocking noises, and you come out saying "There isn't anything in the closet," and that's when you are grabbed by the mean nasty bunny. Eaten alive.[/hide][/hide][/hide]

[color=red]-------------------------------- Line of Different Room --------------------------------[/color]

[hide="Hallway"]Nothing there.

[hide="Scream"]You see your cat jumping out of a window. Nice job you scared it away.[/hide]

[hide="Yell"]You get a noise complaint from all surrounding neighbors.[/hide]

[hide="Sob"]What did that get you???[/hide][/hide]

Note: This maze is quote-proof. :evil:

GRAAR!!! I was just told that my quote-proofing thing doesn't work. :( I'm gonna have to call this off.

DO NOT PRIVATE MESSAGE ME THE NUMBER! IF YOU FIND IT, JUST POST HERE!!!

The 5 right before the 8 is italicized as well...

[quote="1=2"]
Now if you find this, [b][i][u]PM[/u][/i][/b] me following number:
[/quote]

Erhem. I already PMed it to you

Haha....I did something that made it UN-QUOTE-PROOFABLE....


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In the last time a sadly large number of people have posted problems from ongoing (!) competitions on MathLinks in the hope to obtain solutions to these problems "for free". Particularly, problems from the Vietnamese Mathematics&Youth magazine seem to be popular among these cheaters.

Here: http://www.nxbgd.com.vn/toanhoctuoitre/?p=7&id=6&ReportID=289&ph=25 you can find the latest few problems of Mathematics&Youth. The deadline for them is 15 October 2007. Please don't help anyone with any of these problems until that date!

Don't believe everyone who tells you that "the deadline is over". The user nmtvn has made false claims of this kind.

There has also been some cheating going on with problems from the Polish MO and with the Indian IMO TSTs. It would be nice if someone would post a link to the problems of such competitions in time before cheaters post the problems here. It would also be great if someone posts a link to the respective Mathematics&Youth problems when they come out so that cheating can be prevented in future.

To everyone who posted solutions to ongoing contest problems: when I see such solutions, I move them into a hidden forum in order to move them back. Some other moderators delete them instead. To all such moderators: Please, if you delete them, save them on your computer so that you can repost them later when the deadline is over! Otherwise, the authors of the solutions (who often didn't know that they were solving a problem from a running contest) get their posts deleted for apparently no reason, what is not quite a motivating experience.

  Darij

Thanks Darij
the users who have been posting the indian postal test problems are
[url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=32407]chessfreak[/url] ,[url=http://www.mathlinks.ro/search.php?search_author=mnrc_2]mnrc_2[/url]
and another one  is suppose whom i am not getting
who has been warned sufficiently many times .
so please don't answer any of their questions before proper clarifications.

Another nickname of probably the same Vietnam cheater: [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=33669]pavarotti[/url].

  dg

another one is [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=32861]hoasua[/url]

They are young kids, and actually, the "contest" in MY magazine of Vietnam is not so serious... so I think they will think back. Don't treat them too seriously too, since we are mature.

I am pretty sad because all the mentions here are about Vietnamese members. Hope that such a thing ends.

Some different members used to post M&Y problems : toan hoc , friendlist , Harry Potter...  :mad:

[quote="hungkhtn"]I am pretty sad because all the mentions here are about Vietnamese members. Hope that such a thing ends.[/quote]

more participants + contests more important for people $ \to$ more cheaters. Noone says that your countrymen are particularly inclined to cheat.

  darij

[b]GUIDELINE FOR POLISH MO PROBLEMS[/b]

Polish MO problems: http://www.om.edu.pl/index2.php?poziom=9&main=zadania.php
Current Polish MO: http://www.om.edu.pl/zadania/om/om59_1.pdf

Even without knowledge of the language, you can identify most questions by the mathematical formulas included. You can translate the names of the months using the Polish wikipedia. The first series (I seria) is over now (deadline was 8 October 2007). I have moved some of the topics back. I am starting to delete some cheater topics which have received no reply.

[b]RECOGNIZING CHEATERS[/b]

There is no common rule to recognize cheaters, but here are some observations I have done on some of them.

Cheaters don't hesitate to post the same problem over and over, opening up a new topic each time the older one is locked or deleted.

Sometimes they post their problems in regional fores where they hope that nobody will detect their cheating: Kisiu111 posted a Polish MO problem in the Georgia forum ( http://www.mathlinks.ro/Forum/viewtopic.php?t=167330 ; please delete the topic). Stupidly enough for them, problems are seldomly solved on regional fores, while there are still many watchful eyes.

Sometimes they claim false sources for their problems (Rand0m claims his problem comes from "USA california math classes", http://www.mathlinks.ro/Forum/viewtopic.php?t=165238 ; Hoasua claimed his - ongoing M&Y - problem was from Uzbekistan MO 1998, http://www.mathlinks.ro/Forum/viewtopic.php?t=169275 ).

What I found particularly cool was Kisiu111 [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=167675]posting a running contest problem[/url], getting an answer from Loup Blanc and then [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=169411]starting another topic[/url] where he presented that answer as his own and asked for a proof.

If you see a problem posted in JPEG format rather than text, it is pretty likely that it is from an ongoing Mathematics&Youth problem sections. Some cheaters are even too lazy to write the problem in LaTeX and just cut it out of the JPEG file of the magazine.

Deleting one's post after it has been recognized as cheating has been practiced, too (I have just deleted such a topic).

Most of the cheaters seem to have a very poor understanding of mathematics and scrutinize every not completely trivial step in proofs ( http://www.mathlinks.ro/Forum/viewtopic.php?t=167995 ).

Cheaters are impatient because they have a deadline. If they don't get the help they want in time, or they can't understand the proofs because they are not detailed enough, they start crying around ("Can anybody help me, please?", "can anybody help me with this problem, please? I cannot do it anyway....", "So what about the answer?", "Noone can do it?", "For all: Do you have a more simple solution for this problem? The solution of mszew is very long and very difficult to understand!"), bumping up their topics.

Some get aggressive when their posts are revealed as cheating ( http://www.mathlinks.ro/Forum/viewtopic.php?t=165235 ), claiming nonsense like that the contest is open and anyone is allowed to help, but also attacking others who rightfully claim that the topic has to be locked ("You're very impolite? You think that you're good, do you? I don't need your solution!" - http://www.mathlinks.ro/Forum/viewtopic.php?t=168943 ).

Some claim that the deadline is over and thus the problem can be discussed when it is not so.

Some completely ignore posts revealing their problems as ongoing contest problems. Nmtvn has made himself a name here: Being told several times that his problem is from a current magazine contest, he continued reposting it answering "But there isn't any true and full solution in that topic! All of solutions are false or not full!" and "Where's the solution for your problem? Don't you have it, do you?". ( http://www.mathlinks.ro/Forum/viewtopic.php?t=167965 )

Almost as good: Knowing exactly why one's old topic was locked and then "returning" it. ( http://www.mathlinks.ro/Forum/viewtopic.php?t=168712 )

Don't expect cheaters to post exactly the same problem they got in the contest. Some will try to generalize the problem, often yielding complete nonsense; others will change notations (Kisiu111 posted the same Polish MO geometry question with 3 different ways to label the points).

  Darij

[url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=33670]thesunflower[/url] is another M&Y cheater. Thanks ZetaX for the info.

  darij

Can some Vietnamese here post a link to the official problem pages of Maths&Youth (the link in my first post only gives one issue) AND tell us the deadlines for the issues?

  darij

Hello,

I need your help:

From which sources (contests or journals) do the following problems come, and when are the respective deadlines? I have to decide whether to unhide them (they are currently in the hidden forum).

[quote="kisiu111"]We have angle with vertex P and point A lies inside this angle. Find points X and Y on another sides angle, that |PX|=|PY| and sum |AX|+|AY| is the smallest. I know that I must rotate A, but I don't know how, can anyone help me?[/quote]

[quote="thesunflower"]Write $ [x]$ is the largest integer not exceeding x and {x} = x - [x].
Find $ lim\limits_{n \to \infty} {(7 + 4\sqrt {3})^n}$[/quote]

[quote="thesunflower"]$ n$ is a positive integer number. Consider $ (2n + 2)$ real numbers $ a,b,a_1,a_2,...,a_n,b_1,b_2,...,b_n$
such that $ a_i \neq 0, i = 1,2...,n$ and the function $ F(x) = \sum\limits_{i = 1}^n \sqrt {a_ix + b_i} - (ax + b)$ satisfies:
There exists distinct real numbers $ m,n$ such that $ F(m) = F(p) = 0$.
Prove that $ m$ and $ p$ are the only real solutions of $ F(x)$[/quote]

[quote="chien than"]Find $ x;y \in N$ such that $ x^3 + x = y^5$[/quote]

[quote="thesunflower"]Consider a sequence $ (u_n)$, $ (u_1 = 1)$, $ u_{n + 1} = \frac {16u^3_n + 27u_n}{48u^2_n + 9}$.
Find the largest integer wich is smaller the sum S of 2008 summands
$ S = \frac {1}{4u_1 + 3} + \frac {1}{4u_2 + 3} + ... + \frac {1}{4u_{2008} + 3}.$[/quote]

Please provide sources (URLs of the competitions or journals; they can be in Polish, Vietnamese or any other language). Thanks!

  darij

There are problems from the newest Mathematics and Youth Magazine (7/2008)

[b]FOR LOWER SECONDARY SCHOOLS[/b]

[b]T1/373.[/b] [b](For $ 6^{th}$ grade)[/b] Write the numbers $ 1$, $ 2$, $ 3$, $ \ldots$, $ 2007$ in an arbitrary order and let $ A$ be the resulting number. Can $ A \plus{} 2008^{2007} \plus{} 2009$ be a perfect square?
[b]
T2/373.[/b] [b](For $ 7^{th}$ grade)[/b] Consider the following two polynomials $ f(x) \equal{} (x \minus{} 2)^{2008} \plus{} (2x \minus{} 3)^{2007} \plus{} 2006x$ and $ g(y) \equal{} y^{2009} \minus{} 2007y^{2008} \plus{} 2005y^{2007}$. Let $ s$ be denote the sum of all the coefficients of $ f(x)$ (after expansion). Find $ s$, and the value of $ g(s)$.

[b]T3/373.[/b] Find all positive integer solutions of the following system of two equations
\[ \begin{cases} x \plus{} y \plus{} z \equal{} 15 \\
x^3 \plus{} y^3 \plus{} z^3 \equal{} 495\end{cases}
\]
[b]T4/373.[/b] Let $ a$, $ b$, $ c$ be non-negative real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. Find the maximum value of the expression:
\[ (a \plus{} b \plus{} c)^3 \plus{} a(2bc \minus{} 1) \plus{} b(2ca \minus{} 1) \plus{} c(2ab \minus{} 1)
\]
[b]T5/373.[/b] Given a triangle $ ABC$ where $ \angle ABC$ is not a right angle. Let $ AH$ and $ AM$ denote, respectively, the altitude and the median through vertex $ A$. Choose a point $ E$ on the ray $ AB$ and $ F$ on the ray $ AC$ such that $ ME \equal{} MF \equal{} MA$. Let $ K$ be reflection point of $ H$ over $ M$. Prove that four points $ E$, $ M$, $ K$ and $ F$ lie on a single circle.

[b]FOR UPPER SECONDARY SCHOOLS[/b]

[b]T6/373.[/b] Solve for $ x$
\[ \sqrt {x \plus{} \sqrt {x^2 \minus{} 1}} \equal{} \frac {9\sqrt 2}{4}(x \minus{} 1)\sqrt {x \minus{} 1}
\]
[b]T7/373.[/b] Prove that in any acute triangle $ ABC$, the following inequality holds
\[ \frac {\tan A}{\tan B} \plus{} \frac {\tan B}{\tan C} \plus{} \frac {\tan C}{\tan A}\ge\frac {\sin 2A}{\sin 2B} \plus{} \frac {\sin 2B}{\sin 2C} \plus{} \frac {\sin 2C}{\sin 2A}
\]
[b]T8/373.[/b] The incircle of a triangle $ ABC$ meets $ BC$, $ CA$, $ AB$ respectively at $ A_1$, $ B_1$, $ C_1$. Let $ p$, $ S$, $ R$ be respectively, half of the perimeter, the area and the circumradius of $ \triangle ABC$. Let $ p_1$ be half of perimeter of $ \triangle A_1B_1C_1$. Prove the inequality $ p_1^2\le\frac {pS}{2R}$. When does equality occur?

[b]TOWARDS MATHEMATICAL OLYMPIAD[/b]

[b]T9/373.[/b] Let $ A$ ($ A\subset \mathbb{N}$) be a non-empty set satisfying the condition: If $ a\in A$ then $ 4a$ and $ [\sqrt a]$ are also in $ A$ (where $ [x]$ denotes the integer part of $ x$). Prove that $ A \equal{} \mathbb{N}$.

[b]T10/373.[/b] Let $ a$ be a natural number which is greater than $ 3$ and consider the sequence $ (u_n)$ ($ n \equal{} 1, 2, \ldots$) defined inductively by $ u_1 \equal{} a$ and $ u_{n \plus{} 1} \equal{} u_n \minus{} \left[\frac {u_n}{2}\right] \plus{} 1$ for all $ n \equal{} 1, 2, \ldots$ (where $ \left[x\right]$ denotes the greatest integer that is $ \leq x$). Prove that there exists $ k\in\mathbb{N}^{*}$ such that $ u_n \equal{} u_k$ for all $ n\ge k$.

[b]T11/373.[/b] Find all polynomials with real coefficients $ P(x)$, $ Q(x)$ and $ R(x)$ such that $ \sqrt {P(x)} \minus{} \sqrt {Q(x)} \equal{} R(x)$ for all $ x\in\mathbb{R}$.

[b]T12/373.[/b] Let $ ABCD$ be a tetrahedron with the centroid $ G$ and the circumradius $ R$. Prove that
\[ GA \plus{} GB \plus{} GC \plus{} GD \plus{} 4R\ge\frac {2}{\sqrt 6}(AB \plus{} AC \plus{} AD \plus{} BC \plus{} CD \plus{} DB)
\]
And the dealine for submitting solutions is 15 September, 2008.

[url]http://www.nxbgd.com.vn/toanhoctuoitre/?p=7&id=6&ReportID=413&ph=37[/url]
the latest problems ^^. The deadline is two months from the day they were published.
It's not good for Vietnamese guys because this fact. I feel so sorry because i'm a Vietnamese student, too.

T12 \379 http://www.mathlinks.ro/viewtopic.php?p=1384060#1384060
And warn lhplhp,please

Not always do the contest makers invent their own problems ........users may come to know of the problems from a different source.So banning them does not seem right.But there are also contest-cheaters who would intentionally cheat. So I think it is the mods' responsibility to unlock the problems after the contest is over .

[quote="N.N.Trung"][quote="Potla"]A new contest cheater:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=254483
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=254481

Name is N.N. Trung from Vietnam.
Moderators, please warn him.
http://www.artofproblemsolving.com/Forum/search.php?search_author=N.N.Trung

This is N.N. Trung's posts. profile link:
http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=55355[/quote]
Oh, sorry, I am a new member and I do not know that posting problems from continous contest is banned on Mathlinks. I don't participate in this contest, but if you want, I will never post similar problems. Thanks so much.[/quote]

I am very sorry that I was so rude to you.
But you should have at least that little bit of intelligence to guess that other people participate in M&Y too. In fact, I do not participate in the M&Y either. It is natural of people to think that you are a cheater if you post ongoing contest problems. After the contest is over, you can tell us for solutions to the previous issues.

At last, sorry for my inconvenience.

For uper secondary schools

T6/380.Prove that in any triangle ABC, the following inequality holds:
$ \frac{{\cot A.\cot B.\cot C}}{{\sin A.\sin B.\sin C}} \le \left( {\frac{2}{3}} \right)^3$

T7/380.There are 17 ornament betel-nut trees around a circular pond. How many ways are there to chopped pss 4 trees with the condition that no two consecutive trees be removed ?

T8/380.Solve the following system equations with parameter a
$ \left\{ \begin{array}{l}
 2z(x^2  \plus{} a^2 ) \equal{} x(x^2  \plus{} 9a^2 ) \\
 2y(z^2  \plus{} a^2 ) \equal{} z(z^2  \plus{} 9a^2 ) \\
 2x(y^2  \plus{} a^2 ) \equal{} y(y^2  \plus{} 9a^2 ) \\
 \end{array} \right.$

Well I remeber T6/380 being posted in the pre olympiad section. Well, maybe, it is another well known problem.

I've some questions on contest cheaters.

Why they could post these topics while they were contesting?

In a contest including Olympiad, when they were contesting, any kinds of communicating tools, such as computers and mobile phones, were not allowed to take in.

In IMO or some similiar contests, even we can't take the calculator in! So I'm amazed that the cheater had been able to post their problems before the deadtime.

Could someone answer my questions?

These are magazine contests - a magazine publishes series of problems for solving, and gives a deadline for the solvers before the next edition is published.

frankvista: There are (here in Germany, at least) lots of homework contests, i. e. contests were you are to solve problems at home until a given deadline.
And yes, there are magazine contests as well.

thank you darij.
I hate cheaters. they spoil the right of the people who are really the best ones.

Hi, sorry to revive an old topic, but for the second time in a row someone has cheated in the RMO using AOPS. Is there any way to report cheaters or file a complaint about them? Also, is there a way in which AOPS administrators can track the person's IP address if the user name is given, so that the person can be caught?

The user name of the cheater is solo1729 (who claims to be from California, though that is highly unlikely).

I think that we must REPORT the topics which we are sure of being taken from a contest. Just posting an apparently berating comment there would not only anger the 'cheater' but also sometimes pull you into seemingly inevitable arguments and I suppose we ought not to fight unnecessarily here at a solemn place which has its own environment of mathematical competitiveness.

If one posts a comment as such, then if the topic is locked, it demoralises the person who posted the problem, but if we just report the thing, even the cheater wouldn't be at doubts with us and the work would be done.

We recently had a user (sorry for reviving it) who posted 8 problems from the current issue of the Mathematical Reflections and it was very unlucky that 2 of them had got understandable solutions posted :
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=590877
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=590878
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=590879
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=590880
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=590881
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=590885
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=590886
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=590883

Good enough they were locked, I do wonder how many might have had sent solutions to those problems. You see that it also disappoints the proposers as they might have had their hardwork on it.  :roll:


calculus
integration
inequalities
function
limit
calculus computations

Let $a,b,c$ be positive numbers such that $abc=\frac{1}{16}.$
Prove the following inequality.

\[ \int_0^{\infty} \frac{x^2}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}\ dx\leq \pi. \]

From Holder's inequality $(x^2+a^2)(x^2+b^2)(x^2+c^2) \geq (x^2 +(abc)^{ \frac{2}{3}})^3 =\left (x^2 +\left ( \frac{1}{16}\right)^{ \frac{2}{3}} \right)^3$

Therefore

$\\ \int_0^{\infty} \frac{x^2}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}\ dx\ \leq \int_0^{\infty} \frac{x^2}{\left (x^2 +\left ( \frac{1}{16}\right)^{ \frac{2}{3}} \right)^3 }\, dx = \pi$

If you want i'll show the computations  :)

The first step is this
$\\ \bullet \, \frac{x^2}{(x^2+a)^3} = \frac{1}{(x^2+a)^2}-\frac{a}{(x^2+a)^3}$

The second step is this
$\\ \bullet$ Let $I_{n}= \int { \frac{1}{(x^2+a)^n} dx}$. Then $I_{n+1} = \frac{2n+1}{2na} \, I_{n} + \frac{x}{2na (x^2+a)^n }$

Now it is almost solved...  :D

[quote="kunny"]I see, thank you, Ramanujan. You can calculate directly the integral in terms of $a,b,c.$[/quote]

NOOOO sorry the above post is for $a=\left ( \frac{1}{16}\right)^{ \frac{2}{3}}$ or for any other constant.

Not for $a,b,c$ in your post. :)

Nice :)  !

We can find this also without complex integrals
$\\ \int \frac{x^2}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}\ dx = \frac{a(c^2 - b^2) arctan \left ( \frac{x}{a} \right ) +b(a^2 - c^2) arctan \left ( \frac{x}{b} \right ) + c(b^2 - a^2) arctan \left ( \frac{x}{c} \right ) }{(b^2 - a^2)(c^2 - b^2)(a^2 - c^2)}$ .

(If my calculations are correct :D )

 :)

Thank you for your replies, anik_andrew, Ramanujan.

To anik_andrew: From your calculation, we have $\int_0^ \infty f(x)dx=\frac{\pi}{2}\cdot \frac{1}{\sigma(a,b)\ \sigma(b,c)\ \sigma(c,a)}$ for some expression $\sigma(x,y).$

To Ramanujan: Your calculation is also correct, because if we differentiate your result, we can obtain the integrand.

this calculation is good to the practice of factrization.

kunny


geometry
incenter
circumcircle
cyclic quadrilateral
similar triangles
angle bisector
geometry solved

1) Let ABCD be a cyclic quadrilateral such that AD+BC=AB.  Prove that the bisectors of the angles ADC and BCD meet on the line AB.

2) In triangle ABC, with AB perpendicular to AC, let D be the foot of the altitude from A, and E, F the projections of D onto AB and AC, respectively.  Prove that (BE/CF)=(AB/AC)^3

1) For convenience, I rewrite the problem as below:

Let BCXY be a cyclic quadrilateral such that BY+CX=XY. Prove that the bisectors of the angles YBC and XCB meet on the line XY.

Extend BY and CX. Let the intersection point be A. Let I be the incenter of ABC and D,E,F the feet the of perpendicular to BC,CA,AB. We have to show that I lies on XY. Suppose $X'\in AC$ and $Y'\in AB$ are the points such that X'Y'//XY. Clearly the value of X'Y'-BY-CX is continuous increasing when X' moves nearer to C. Thus there exists unique X'Y' s.t. X'Y'-BY-CX=0. Therefore it suffices to prove the converse of the statement only(that is, if I lies on XY, then XY=BY+CX).
By similar triangles, $XY+FY+EX=IX+IY+\frac{r}{h_a}a=\frac{r}{h_a}(a+b+c)=a=BF+CE$. The result follows then.

(one may notice I have missed a case in the solution, but it can be done by considering the excenter wrt to A instead :) )

2) is pretty straightforward by trigonometry.

Here are different (and, I think, simpler) solutions to both of the problems:

[b]Solution to Problem 1).[/b]

Since AD + BC = AB, we can divide the segment AB into two parts of lengths AD and BC, respectively; in other words, we can find a point S on the segment AB such that AS = AD and BS = BC.

Let the circumcircle of triangle CSD meet the line AB at a point T (different from the point S, of course). WLOG assume that the point S lies between the points A and T on the line AB (else, we can simply rename A and D into B and C). Then, since the quadrilateral CTSD is cyclic, we have < TCD = 180° - < TSD. But 180° - < TSD = < ASD. Thus, < TCD = < ASD.

Since AS = AD, the triangle SAD is isosceles, so that its base angle equals $\measuredangle ASD=\frac{180^{\circ}-\measuredangle DAS}{2}$. Hence, $\measuredangle TCD=\measuredangle ASD=\frac{180^{\circ}-\measuredangle DAS}{2}=\frac{180^{\circ}-\measuredangle DAB}{2}$. But since the quadrilateral ABCD is cyclic, we have < BCD = 180° - < DAB; thus, we get $\measuredangle TCD=\frac{\measuredangle BCD}{2}$. It follows that the line CT is the angle bisector of the angle BCD. Similarly, the line DT is the angle bisector of the angle ADC. Hence, the point T is the point of intersection of the angle bisectors of the angles ADC and BCD. But since the point T lies on the line AB, we can therefore state that the angle bisectors of the angles ADC and BCD intersect on the line AB. Problem [b]1)[/b] is solved.

[b]Solution to Problem 2).[/b]

By the catet theorem, applied to the right-angled triangle ADB with its altitude DE, we have $BE\cdot AB=BD^2$.

By the catet theorem, applied to the right-angled triangle ADC with its altitude DF, we have $CF\cdot AC=CD^2$.

By the catet theorem, applied to the right-angled triangle ABC with its altitude AD, we have $BD\cdot BC=AB^2$ and $CD\cdot BC=AC^2$.

Thus,

$\frac{BE}{CF}=\frac{BE\cdot AB}{CF\cdot AC}\cdot\frac{AC}{AB}=\frac{BD^2}{CD^2}\cdot\frac{AC}{AB}=\left(\frac{BD}{CD}\right)^2\cdot\frac{AC}{AB}$
$=\left(\frac{BD\cdot BC}{CD\cdot BC}\right)^2\cdot\frac{AC}{AB}=\left(\frac{AB^2}{AC^2}\right)^2\cdot\frac{AC}{AB}=\frac{AB^4}{AC^4}\cdot\frac{AC}{AB}=\frac{AB^3}{AC^3}=\left(\frac{AB}{AC}\right)^3$.

This solves Problem [b]2)[/b].

  darij

[quote="pestich"]The angle bisectors of <ADC and  <BCD  intersect in O and
      continue to intersect mid-perpendiculars of AC and BD  on the circumcircle
      of ABCD.   Construct circle center O tangent to AD,DC,CB.
      Draw the another tangents from points A  intersecting bisector of <C
      in M and from point B intersecting bisector of <D in N.  MN is also tangent to
      circle O.  By Brianchon  AB concurrent with angle bisectors of <C and <D.

      Col.  Pestich[/quote]Could you detail this solution? How did you use $AD+BC = AB$?  :?  :? I don't think it's correct.

Hi Valentin,

  What I tried to do was an attempt to show that a circumscripting hexagon could
  be constructed that has line AB as a common angle bisector for both A and B
  and the condition AB+DC=AB is used here.  Without it the bisector is a broken line.

   It is is not a complete solution, not as solid and without reproach as Darij's,
   and as usual I do not claim it to be that.  I just wanted to mention Brianchon's
   name, for it is unfairly neglected in many ways in favour of its dual  - Pascal,
   whom we invariably imagine as a 16 year old fella.

    Maj.  Pestich

I tried reconstructing your solution and I ended up solving the problem instead :P :)

I'm glad that I could provide to you this small favour.

    Are we going to see your solution in this forum?

   Maj.  Pestich

I ment I tried to solve it your way, but did not succeede. I solved the problem as Darij did (almost) in the end :)


How many odd positive integers are factors of 480?

Listing them:

1
3
5
15

So, our answer is $ \boxed{4}$.

Also we can factor and we get $ 2^{5}\cdot{3\cdot{5}}$ which means there is 1 possibility for the number of 2's in the factor, adn 2 for 3 and 5, so $ 1\cdot{2\cdot{2}}$


Hi!

I'm starting with Latex and how can I write this Formula:
[img]http://mathworld.wolfram.com/l1img1635.gif[/img]

I've tryied with this code:
[code]\[
    \left[
        \begin{array}{cc}
            n & $\displaystyle\sum_{i=1}^n x_i$ \\
            $\displaystyle\sum_{i=1}^n x_i$ & $\displaystyle\sum_{i=1}^n x_i^2$
        \end{array}
    \right]
    \left[
        \begin{array}{c}
            a\\
            b
        \end{array}
    \right]
    &=&
    \left[
        \begin{array}{c}
            $\displaystyle\sum_{i=1}^n y_i$ \\
            $\displaystyle\sum_{i=1}^n x_iy_i$
        \end{array}
    \right]
\],[/code]
But it isn't working... :(

Best regards,
[PT]Devilishly

\[
    \left[
        \begin{array}{cc}
            n & \displaystyle\sum_{i=1}^n x_i \\
            \displaystyle\sum_{i=1}^n x_i & \displaystyle\sum_{i=1}^n x_i^2
        \end{array}
    \right]
    \left[
        \begin{array}{c}
            a\\
            b
        \end{array}
    \right]
    =
    \left[
        \begin{array}{c}
            \displaystyle\sum_{i=1}^n y_i \\
            \displaystyle\sum_{i=1}^n x_iy_i
        \end{array}
    \right]
\]

[code]
\[
    \left[
        \begin{array}{cc}
            n & \displaystyle\sum_{i=1}^n x_i \\
            \displaystyle\sum_{i=1}^n x_i & \displaystyle\sum_{i=1}^n x_i^2
        \end{array}
    \right]
    \left[
        \begin{array}{c}
            a\\
            b
        \end{array}
    \right]
    =
    \left[
        \begin{array}{c}
            \displaystyle\sum_{i=1}^n y_i \\
            \displaystyle\sum_{i=1}^n x_iy_i
        \end{array}
    \right]
\]
[/code]

Don't put dollar signs around the sums inside the array (since the entire array is already in math mode), and don't have ampersands around the equal sign in the middle.


Hi,

Does anyone know of a source for past IMO results (I'm looking for a very extensive results page, including results from IMO 1959)?

I've visited http://www.srcf.ucam.org/~jsm28/imo-scores/ but it's earliest entry is for IMO 1977 and it's also missing some years in between.

Any help would be greatly appreciated.

Thanks!

[url]http://www.imo.org.yu[/url]
Click there on the "Past IMO Results" button on the left :wink: .


geometric transformation
ratio
geometry proposed
geometry

Let $S$ be exsimilicenter of the excircles $(I_{b}),(I_{c})$ of $ABC (AC>AB)$. Drawn the small circle $(O)$ tangent to the excircles $(I_{b}),(I_{c})$ and $BC$ at $T$. If $H$ and $M$ are the foot of the altitude from $A$ and the midpoint of $BC$  respectively, prove that: $\frac{1}{ST}=\frac{1}{TH}-\frac{1}{TM}$

Let $T_{b},\ T_{c}$ be tangency points of $(I_{a}),\ (I_{b})$ with BC, $r_{b},\ r_{c}$ their radii, and $s$ the triangle semiperimeter. Since $T_{b}B = CT_{c}= s-a,$ M is also the midpoint of $T_{b}T_{c}.$ A is the internal similarity center of $(I_{b}),\ (I_{c}),$ the intersection of their common internal tangents.

$T_{b}T_{c}= T_{b}C+CT_{c}= s-a+s = b+c$

$T_{b}M = \frac{T_{b}T_{c}}{2}= \frac{b+c}{2}$

$T_{b}H = T_{b}T_{c}\cdot \frac{r_{b}}{r_{b}+r_{c}}= (b+c) \cdot \frac{r_{b}}{r_{b}+r_{c}}$

$\frac{r_{b}}{r_{c}}= \frac{T_{b}S}{T_{c}S}= \frac{T_{b}T_{c}+T_{c}S}{T_{c}S}= \frac{b+c}{T_{c}S}+1$

$\frac{r_{c}}{r_{b}}= \frac{T_{c}S}{T_{b}S}= \frac{T_{b}S-T_{b}T_{c}}{T_{b}S}= 1-\frac{b+c}{T_{b}S}$

$T_{c}S = (b+c) \cdot \frac{r_{c}}{r_{b}-r_{c}},\ \ T_{b}S = (b+c) \cdot \frac{r_{b}}{r_{b}-r_{c}}$

Invert the 3 tangent circles $(I_{b}),\ (I_{c}),\ (O)$ in a circle (T) with arbitrary radius $\rho$ and centered at the tangency point T of (O) with BC. The line BC passing through the inversion center T is carried into itself, the circle (O) tangent to BC at the inversion center T goes into a line $o \parallel BC$ and the excircles $(I_{b}),\ (I_{c})$ into 2 congruent circles $(I_{b}'),\ (I_{c}'),$ both tangent to the parallel lines $o \parallel BC,$ $r_{b}' = r_{c}'$ being thair radii and $T_{b}',\ T_{c}'$ their tangency points with BC. The inversion center T is a similarity center of $(I_{b}') \sim (I_{b})$ and $(I_{c}') \sim (I_{c}),$ with similarity coefficients equal to the power of inversion $\rho^{2}$ divided by the power of the inversion center T to the original circle, respectively.

$\frac{r_{b}'}{r_{b}}= \frac{TT_{b}'}{TT_{b}}= \frac{\rho^{2}}{TT_{b}^{2}},\ \ \frac{r_{c}'}{r_{c}}= \frac{TT_{c}'}{TT_{c}}= \frac{\rho^{2}}{TT_{c}^{2}},\ \ r_{b}' = r_{c}'\ \Longrightarrow\ \frac{TT_{b}}{TT_{c}}= \sqrt{\frac{r_{b}}{r_{c}}}$

In addition, $TT_{b}+TT_{c}= T_{b}T_{c}= b+c.$ Calculating $TT_{b},\ TT_{c}$ from their ratio and sum:

$TT_{b}= (b+c) \cdot \frac{\sqrt{r_{b}}}{\sqrt{r_{b}}+\sqrt{r_{c}}},\ \ TT_{c}= (b+c) \cdot \frac{\sqrt{r_{c}}}{\sqrt{r_{b}}+\sqrt{r_{c}}}$

Calculating the segment lengths TS, TM, TH:

$TS = T_{b}S-TT_{b}= (b+c) \left(\frac{r_{b}}{r_{b}-r_{c}}-\frac{\sqrt{r_{b}}}{\sqrt{r_{b}}+\sqrt{r_{c}}}\right)= (b+c) \cdot \frac{\sqrt{r_{b}r_{c}}}{r_{b}-r_{c}}$

$TM = TT_{b}-T_{b}M = (b+c) \left(\frac{\sqrt{r_{b}}}{\sqrt{r_{b}}+\sqrt{r_{c}}}-\frac{1}{2}\right)= \frac{b+c}{2}\cdot \frac{\sqrt{r_{b}}-\sqrt{r_{c}}}{\sqrt{r_{b}}+\sqrt{r_{c}}}$

$TH = T_{b}H-TT_{b}= (b+c) \left(\frac{r_{b}}{r_{b}+r_{c}}-\frac{\sqrt{r_{b}}}{\sqrt{r_{b}}+\sqrt{r_{c}}}\right)=$

$=(b+c) \cdot \frac{\sqrt{r_{b}r_{c}}}{r_{b}+r_{c}}\cdot \frac{\sqrt{r_{b}}-\sqrt{r_{c}}}{\sqrt{r_{b}}+\sqrt{r_{c}}}$

Subtracting the reciprocal segment lengths:

$\frac{1}{TH}-\frac{1}{TM}= \frac{1}{b+c}\cdot \frac{\sqrt{r_{b}}+\sqrt{r_{c}}}{\sqrt{r_{b}}-\sqrt{r_{c}}}\cdot \left(\frac{r_{b}+r_{c}}{\sqrt{r_{b}r_{c}}}-2\right) =$

$= \frac{1}{b+c}\cdot \frac{(\sqrt{r_{b}}+\sqrt{r_{c}})(\sqrt{r_{b}}-\sqrt{r_{c}})}{\sqrt{r_{b}r_{c}}}= \frac{1}{b+c}\cdot \frac{r_{b}-r_{c}}{\sqrt{r_{b}r_{c}}}= \frac{1}{TS}$

If $X,Y$ are the touch point of the small circle with $(Ic),(Ib)$ respectively, now: $\angle HXT= \angle TYM$

Exsimilicenter $S$ of $ (I_c) \sim (I_b)$ is also center of the positive inversion $\mathcal{I}$ taking $(I_c),(I_b)$ into each other. If $ P,Q$ denote the tangency points of $(O)$ with $ (I_c)$ and $ (I_b),$ then $ PQ$ passes through $S,$ since the set of  circles externally/internally tangent  to $(I_b),(I_c)$ transforms into itself under $\mathcal{I}.$ The 9-point circle of $\triangle ABC$ belows to this set indeed

$ SQ \cdot SP \equal{} ST^2 \equal{} SM \cdot SH \Longrightarrow \ ST^2 \equal{} (ST \plus{} TM)(ST \minus{} TH)$

$ ST^2 \equal{} ST^2 \minus{} ST \cdot TH \plus{} ST \cdot TM \minus{} TH \cdot TM$

$ ST \equal{} \frac {TM \cdot TH}{TM \minus{} TH} \Longrightarrow \ \frac {1}{ST} \equal{} \frac {1}{TH} \minus{} \frac {1}{TM}$


geometry
geometric transformation
homothety
power of a point
radical axis

A problem I just came across seemed like a fun one for this group:

Given two circles, and their four common tangent segments (two external, two internal).

Prove that the midpoints of the four tangent segments all lie on one line.

--Joshua Zucker

[hide="Hint"]Consider the radical axis of the two circles; prove that the four midpoints lie on it.[/hide]

The locus of points which have equal power w.r.t. both circles is a line, the radical axis. You need to prove that the midpoints have equal powers w.r.t. both circles.

[hide="Radical Axis Solution"]
Well, since we have the midpoints of the common tangents, the power with respect to the first circle for any of the midpoints $M$ is $(MT_1)^2 = (MT_2)^2$, the power with respect to the second circle, with tangent points $T_i$. Then it is on the radical axis of the two circles, done.
[/hide]
[hide="Hint for Less Technical Solution"]
Homothety
[/hide]


Let $\displaystyle1=d_{1}<d_{2}<...<d_{k}=n$ be the divisor of $n$.Find all positive integer $n$ such that:
$d_{7}^2+d_{10}^2=(\frac {n}{d_{22}})^2$

[i]Note: I didn't check the answer, you should check if the value of n below is valid.[/i]

For any positive integer n, let d(n) denote the number of positive divisors of n (including 1 and n itself).

n/d_22 is a positive integer and is a divisor of n, let d_i=n/d_22
obviously i>=11 and d(n)=i+22-1

In the equation d_7^2 + d_10^2 = d_i^2,
(i) d_7 or d_10 is divisible by 4
(ii) d_7 or d_10 is divisible by 3
(iii) d_7,d_10 or d_i is divisible by 5
(to prove this, write d_7,d_10,d_i in the form k(2uv),k(u^2-v^2),k(u^2+v^2), gcd(u,v)=1)

therefore 3x4x5=60|n

list down the divisors: 1,2,3,4,5,6,10,...
it's obvious that d_7=7 or 8 or 9 or 10

factorize d_7^ 2= d_i^2 - d_10^2, we get the following solutions:

(d_7,d_10,d_i)=(7,24,25),(8,6,10),(8,15,17),(9,40,41),(9,12,15),(10,24,26)

(8,6,1) is discarded because d_10>d_7

(7,24,25) is discarded because if we list down the divisors : 1,2,3,4,5,6,7,8,10,12,14,15,...,24, it's obvious that d_10$\neq$24, similary discard (9,40,41),(10,24,26)

[u]if d_7=8, d_10=15, d_i=17[/u]
then i=11, d(n)=11+22-1=32
n is divisible by 8,15,17
n is not divisible by an odd prime other than 3,5,17, otherwise d(n)>32
therefore n is of the form 2^a.3^b.5^c.17^d (a>=3, b,c,d>=1)
d(n)=(a+1)(b+1)(c+1)(d+1)=32
the only solution is a=3,b=c=d=1, so [u]n=8*3*5*17[/u]

(if i=12, then d_12*d_22=n ==> d(n)=12+22-1=33, but 2,3,5,17|n, therefore n is divisible by 2^a.3^b.5^c.17^d (a,b,c,d>=1) ==> d(n) is divisible by (a+1)(b+1)(c+1)(d+1), this contradicts with d(n)=33 which cannot be divided by (a+1)(b+1)(c+1)(d+1))

[u]if d_7=9, d_10=12, d_i=15[/u]
then i=11, d(n)=32
n cannot be divided by a prime p other than 2,3,5
suppose n is divisible by odd prime p>5, then n is divisible by 2^a.3^b.5^c.p^d (a,b>=2, c,d>=1)
therefore d(n)>=(2+1)(2+1)(1+1)(1+1)=36 contradiction

therefore n is of the form 2^a.3^b.5^c, a>=2, b>=2, c>=1
(a+1)(b+1)(c+1)=d(n)=32
the only solution is a=3, b=3, c=1, [u]n=8*27*5[/u]

(note i$\neq$12 and i$\neq$13, proof is similar to above)


Problem 2 (Group I). Industrialy, aluminium, Al(l), can be obtained by electrolysis of the oxide Al2O3 dissolved in cryolite, using carbon as one of the electrodes. This electrolitic reaction can be described by the chemical equation

$ 2Al_{2}O_{3}(l)+3C(s) \longrightarrow 4Al(l)+3CO_{2}(g)$.

2.1 Regarding the above chemical equation, select from the alternatives A - D the correct one:

A - One of the half-reactions is described by the equation $ C(s)+2O^{2-}(l)+4e^{-}\longrightarrow CO_{2}(g)$.
B - The variation in oxidation number of Al and C is -6 and +6, respectively.
C - There is the transfer of 3 mol of electrons by each mol of Al(l) formed.
D - In the positive electrode occurs the formation of Al(l).

2.2 Compute the mass of Al(l) obtained by electrolysis when there is the formation of 200 L of CO2(g), at 298.15 K at the pressure of 101 kPa.

i think its' C but i can't undesrtand the statement of C
for part 2
let $ m$ mass of $ Al$ formed then
$ m=\frac{4M_{Al}PV}{3RT}$

Yes, your answers are correct. Why don't you understand the statement of C?

whether 3 moles of Al is formed by transfer of 3 moles or is it the other way round

Each mole of aluminum requires the transfer of 3 moles of electrons.


LaTeX
algebra
polynomial
difference of squares
special factorizations

Please refer to the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10857]Getting Started Series Challenge Problems Introduction[/url]for an explanation of my objectives for this series of messages.

This is an easy problem, in that it involves only beginning algebra, but it is a hard problem in that it is intricate. The key to this problem is following the smart procedure rather than the tedious procedure, so SHOW YOUR WORK (in hidden text) so we can discuss ways to attack this kind of problem. The source of this problem is the Chinese-language self-study guide [i]Chuzhongsheng Wu Shi Zi Tong Congshu Shuxue Pian[/i] [[i]Self-Study without a Teacher Series for Junior High Students: Mathematics Volume[/i]] edited by Liu Yunlei (Beijing: Beijing Kexue Jishu Chubanshe, 1998) (ISBN 7-5304-2158-1) page 64.

Simplify:

[tex]\displaystyle\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}[/tex]

You've got to love the [url=http://www.artofproblemsolving.com/Forum/viewforum.php?f=123]LaTeX capabilities[/url] of this site.

Please show your answer and your steps in spoiler text so that others can try this out. This is meant to be a challenge problem for [b]Getting Started[/b] problem-solvers.

Ok here goes

[hide]

first notice that (a+b)(a-b)=(a^2-b^2)

Second notice that the numerator of each fraction is the same as the degree of the polynomial of the denomonator.

This paticular sequance does somthing really cute, in that the x's cancel out on the top and the bottom is an expanded version of the difference of squares, so adding the first two you get 2/(1-x^2) adding that to the next one, you get 4/(1-x^4) then 8/(1-x^8) then 16/(1-x^16)[/hide]

Yup I got [hide] 16 / 1 - x^16 [/hide] also. Nice problem. Much like the ones on Aops book.


Give an example of a martingale $ X_n$ with $ \sup_n |X_n| < \infty$ a.s. and $ \exists a, \ b: \ P(X_n \equal{} a\ inf. often) \equal{} 1, \ P(X_n \equal{} b \ i.o.) \equal{} 1$, that is bounded a.s. divergent martingale.

Here's a rough idea on how you might want to make your construction.

We're going to start at $ X_0\equal{}0$, and ideally what we'd like to do is to jump back and forth between 1 and -1 at each step afterwards.  However, we can't do that because of the martingale requirement.  So what we do is insert a very small failure probability $ p_n$ at each step.  Most of the time we'll just be jumping back and forth between $ \minus{}1$ and $ 1$, but at each step there'll be some small chance that $ X_{n\plus{}1}$ is something else.

We really want three things to happen:

(1) $ X_{n}$ better be finite with probability 1 for each $ n$, or we're already stuck.

(2) If $ X_{n\plus{}1}$ is a failure ($ X_{n\plus{}1} \neq \pm 1$), then we want (with probability 1) to eventually get back to our $ 1, \minus{}1, 1, \minus{}1$ ideal.

The first two are enough to guarantee that $ X_n\equal{}1$ and $ X_n\equal{}\minus{}1$ infinitely often, but not to guarantee that $ X_n$ is bounded, as we could have a infinite collection  of failures that send $ X_n$ farther away from 0 each time.  So we'll add a third condition:

(3) With probability 1 the number of failures is finite.

We now turn to actually constructing our martingale.  $ X_1$ will be equally likely to be $ \pm1$ (just to get ourselves started on our path).  If $ X_n$ is positive, then $ X_{n\plus{}1}$ will equal $ \minus{}1$ with probability $ 1\minus{}p_n$ (where we'll eventually pick $ p_n$ to be very small).  Otherwise, $ X_{n\plus{}1}$ will equal $ a$, where $ a$ is some number dependent on $ p_n$ and $ X_n$.  If $ X_n$ is negative, then $ X_{n\plus{}1}$ will equal $ 1$ with probability $ 1\minus{}p_n$, and will otherwise equal $ \minus{}a$.

Condition (1) is satisfied by construction.  For condition (2), at each step we'll get back to our ideal sequence with probability $ 1\minus{}p_n$, so, for small $ p_n$, we will almost surely get back in finite time.

 It remains to check condition (3) and that $ X_n$ can be made a martingale.  To check condition (3), note that at each step we fail with probability $ p_n$.  How can we choose $ p_n$ to guarantee the number of failures to be finite a.s.?

We still haven't used $ a$, so once you choose $ p_n$ you should be able to just pick $ a$ to make it a martingale.


function
linear algebra
linear algebra unsolved

I want to show that if the map q: V x V -->W is both bilinear and linear, then q must be the zero map. I'm started with the following, but I don't know the argument to proceed from this:

b(λ1v1+λ2v2,λ1v1+λ2v2)=λ1b(v1,v1)+λ1λ2b(v1,v2)+λ1λ2b(v2,v1)+λ2λ2b(v2,v2)=...

thnx

What field are we working over? For the sake of the argument, I'll assume $ \mathbb{R}.$ (I have the vague sense that characteristic 2 might cause trouble.)

How about this: if $ q$ is not the zero mapping, then there is some $ (v,w)\in V\times V$ such that $ q(v,w)\ne 0.$

Now, what can you say about $ q(tv,tw)$ for arbitrary $ t\in\mathbb{R}?$

By the bilinearity of q I suggest to write:

q(tv,tw)=tq(v,0)+tq(0,w)=0

Which implies that q must be the zero mapping. I don't see why this only true if q is linear, how is the linearity for q defined when it is a function of two arguments?

Bilinearity doesn't work that way. That's linearity: $ q(a\plus{}b)\equal{}q(a)\plus{}q(b)$, where $ a\equal{}(v,0)$ and $ b\equal{}(0,w)$.
Then bilinearity says that $ q(v,0)\equal{}q(v,0\cdot w)\equal{}0\cdot q(v,w)\equal{}0$.

Kent's argument is based on a weaker version of linearity; just the constant multiplication rule $ q(t(v,w))\equal{}tq(v,w)$.


I just took the Mathcounts test in math class

and it was [color=blue]kyyuanmathcount: of some unknown difficulty :P.  [/color]

If you're talking about the school test, this topic should be locked/deleted.

Also there's really no point in this.
Delete please.

If you see a spam thread, please do not just post more spam.  Locked.


If $ \left(r \plus{} \frac {1}{r}\right)^2 \equal{} 3$, then $ r^3 \plus{} \frac {1}{r^3}$ equals

$ \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6$

[hide="Click for solution"]
$ r^3\plus{}\frac{1}{r^3}\equal{} \left (r\plus{}\frac{1}{r} \right ) \left (r^2\minus{}1\plus{}\frac{1}{r^2} \right )$.  From $ \left ( r\plus{}\frac{1}{r} \right )^2\equal{}r^2\plus{}2\plus{}\frac{1}{r^2}\equal{}3 \implies r^3\plus{}\frac{1}{r^2}\equal{}1 \implies r^3\plus{}\frac{1}{r^3}\equal{}0$, or $ \boxed{\textbf{(C)}}$.
[/hide]


Korleis gjekk det med dokke på nordisk? fekk dokke til masse? eg fekk til oppgåve 1 trur eg, og så svarte eg på 2 og 4. får kanskje 5-6-7 poeng. Syntest 1 var litt rar. 3 og 4 var vanskelige.

Erik her.

Gikk ikke så veldig bra, svarte litt på noen, men vet ikke om jeg fikk til noe...

Geir her.

Klarte oppgave tre ved modulobetraktning. Skrev noen forutsetninger for løsningene av oppgave 1 og 2. Er jeg heldig får jeg 7-8 poeng.

Huff, ditta va vanskelig! Va de noken som fekk noken andre verdia for k enn -1 og 1 på oppgåve 2? Og skulle m vere 9 i oppgåve 3? Det var veldig vanskelig i bevise noke som helst...

Fikk ikke til noe som helst nesten jeg.
Er det noen som klarte enern som kan poste beviset.

Og treern.

Knut

Ja, er det noen som kan poste fulle løsninger for noen av oppgavene hvis dere har klart dem?

Geir, kan du skrive løsningen din på oppgave 3 på forumet?

Noen som klarte 2ern. Kan den legge ut svaret

ingen som har fullstendig løysing på oppgåve 2?

Grunde her

Jeg klarte å komme fram til et (tilsynelatende) vanntett bevis ved Reductio ad absurdum på oppgave 4. Maks 5 poeng, altså.


A 0.4508 gram sample of iron is dissolved in a small amount of concentrated nitric acid forming Fe3+ ions in solution and is diluted to a total volume of 500.0 mL.  Calculate the molarity of Fe3+.

Do you simply divide 0.4508 grams of Fe by the molar mass of Fe (55.85), or is there some mole ratio step that I'm leaving out?
I know that after you convert it to moles, you divide by .5000 L of solution.

Please help.

[quote="RHSLilSweetie07"]Do you simply divide 0.4508 grams of Fe by the molar mass of Fe (55.85)[/quote]

Yes.

[quote="thatin"]well typically you have to write the reaction, find the molarity of Fe and then say that the molarity of Fe is equal to the molarity of Fe3+[/quote]

this is only correct way. you can't do anything without reaction.


Which integers have the following property? If the final digit is deleted, the integer is divisible by the new number.

Call this integer is $ n$
Then $ n=10A+b$ where $ b\in\{0,1,..,9\}$
If we delete the final digit then we has number $ A$
Ipply that $ A|10A+B\Leftrightarrow A|b$  [color=blue](1)[/color]
 [u]Case 1:[/u] $ b=0$
Then (1) true for all $ A>0$
Imply that solution is $ n=10A$ where $ A\in N$
[u]Case 2[/u] $ b$ is different from 0.
Then $ A|b$
For $ b=1$ we have $ n\in\{11\}$
For $ b=2$ we have $ n\in\{12,22\}$
For $ b=3$ we have $ n\in \{13,33\}$
For $ b=4$ we have $ n\in \{14,24,44\}$
For $ b=5$ we have $ n\in\{15,55\}$
For $ b=6$ we have $ n\in\{16,26,36,66\}$
For $ b=7$ we have $ n\in\{17,77\}$
For $ b=8$ we have $ n\in\{18,28,48,88\}$
For $ b=9$ we have $ n\in {\19,39,99\}$
An easy problem. :)


For all positive real numbers $a,b,c$ the following inequality holds
$(a^{7}+b^{7}+c^{7})^{3}\ge 9a^{3}b^{3}c^{3}(a^{12}+b^{12}+c^{12})$

This is not a right solution. To ehku: Do you use expanding to solve this problem?


What is the minimum value of $ (\sin 2007x)(\cos 2007x)$?

I believe that these problems are taken from the Math League contests

[quote="13375P34K43V312"][quote="AstroPhys"][hide]
As the sine value approaches $ -1$ the cosine approaches zero and as the cosine approaches $ -1$ sine approaches zero, so that method will not work.
To get around this try and find where $ | \sin x | = | \cos x |$ so it yeilds the largest value when they are multplied and that one is negative so it reaches the lowest product, this at $ \frac{3 \pi }{4}+\pi k$ , $ k \in \mathbb{Z}$, for both of the functions.  For this problem it is specifically $ x= \frac{\frac{3 \pi }{4}+\pi k}{2007}$ .
The answer then becomes $ \boxed{ \left( \frac{-\sqrt{2}}{2}\right) \left( \frac{ \sqrt{2}}{2}\right) = \frac{-1}{2}}$ .
[/hide]

Also, where did you get these problems from? They seem very similar to ones from a recent math competition.[/quote]

That's not rigorous...the double angle thing is the way to go here. Or...

[hide]calculus[/hide][/quote]

What do you mean double angle?  Inst it intuition it must be -1/2. its when cos and sin are equla and negative of that.

it may be intuitive, but to prove that it is minimum, you would want to use the double angle formula for sine :)


Need some help with the following problems please:

The roots of $x^2+ax+b=0$ are $a$ and $b$ where $a$ and $b$ do not equal to 0. Compute $(a,b)$.

Compute the sum of the digits of the first 100 positive odd integers

Given that $\sqrt[3]{26+15\sqrt{3}} = a+b\sqrt{3}$, where $a$ and $b$ are positive integers, compute $(a,b)$

No idea there...maybe one of the more expericend members can help you out  :?

For the first one...
[hide]$x^2+ax+b=(x-a)(x-b)$
$x^2+ax+b=x^2-(a+b)x+ab$
Therefore
-(a+b)=a
And
b=ab
So
1=a
Which means
a+b=-a
1+b=-1
b=-2
So the original quadratic was $x^2+x-2$.
[/hide]

How did you factor $x^2+ax+b$ as $(x-a)(x-b)$? I thought it wasn't possible to factor the equation  :?

Anyone want to explain this?

[quote="Mario_KaRt"]How did you factor $x^2+ax+b$ as $(x-a)(x-b)$? I thought it wasn't possible to factor the equation  :?[/quote]
I didn't factor it. We know it from the fact that a and b are the roots of the equation. We want x=a and x=b, which would only result from x-a=0 and x-b=0.
Oh yeah, and I got the third one:
[hide]$\sqrt[3]{26+15\sqrt{3}}=a+b\sqrt{3}$
Cube both sides.
$26+15\sqrt{3}=(a+b\sqrt{3})^3=a^3+3a^2b\sqrt{3}+9ab^2+3b^3\sqrt{3}$
We can assume that 26 is equal to the two terms that don't have a $\sqrt{3}$ in them.
$26=a^3+9ab^2$
Since 26 can be factored as 2*13, and the RHS can be factored as $a(a^2+9b^2)$, we guess that a is 2 and see if it works.
$2^2+9b^2=13$
$9b^2=9$
$b^2=b=1$.
We might as well check our work by substituting a=2 and b=1 in $15\sqrt{3}=3a^2b\sqrt{3}+3b^3\sqrt{3}$. It checks, therefore $\sqrt[3]{26+15\sqrt{3}}=2+\sqrt{3}$.[/hide]

That is typo.

For the first one, I recognize this problem from one of the AMC-10's(Now, I did too much that I even memorize the problem) :D

One pair that I become sure is such that:

$x^2+x-2$ where a=1 and b=-2

Factor this out, we'll get $(x+2)(x-1)$ which has roots 1 and -2, just like what they asked you too.

happy holiday.

hope this helps!

[quote]Given that $\sqrt[3]{26 + 15\sqrt{3}} = a + b\sqrt{3}$,
where  $a$ and $b$ are positive integers, compute $(a.b)$[/quote]
[hide]
Cube both sides:  $26 + 15\sqrt{3} \;= \;(a + b\sqrt{3})^3 \;= \;(a^3 + 9ab^2) + (3a^2b + 3b^3)\sqrt{3}$

Then: $a(a^2 + 9b^2) = 26$ ... and: $3b(a^2 + b^2) = 15\;\;\Rightarrow\;\;b(a^2+b^2) = 5$

Since $a$ and $b$ are positive integers, we find (by inspection) that: $a = 2,\;b = 1.$[/hide]

[quote="Silverfalcon"]For the second, how many odd numbers have one digit? How many have two digits? And is there any odd number with three digits that is less than 100?

That's a big help. 1,3,5,7,9 are only ones with one digit. And if you can figure out how many odd two digit numbers are there and multiply by 2, and add 5 to it, you'll get the answer.[/quote]

How did you get that Silverfalcon? Can you explain a bit further?

This is the break down for the first 100 positive odd integers:

5 single digit integers (1-9 odd)
45 double digit integers (11-99 odd)
50 triple digit integers (99-199 odd)

However, there are no triple digit integers less than 100...

The first positive odd integer is 1, and the 100th odd integer is 199. As per your advice, this is what we get:

$(45*2)+5 = 95$

I am pretty sure that is not the correct answer  :?

[quote="Mario_KaRt"]So...does anyone have a difinitve answer?[/quote]
If $\overline{abc}$ is an odd integer from 1 to 199, then ${c\in\{1,3,5,7,9}$ and $\overline{ab}$ can be arbitrary number from 0 to 19 (20 numbers total). Thus we have following groups of odd numers: $\overline{ab1}$, $\overline{ab3}$, $\overline{ab5}$, $\overline{ab7}$, $\overline{ab9}$; each group contain 20 numbers.
Let's sum up sum of digits in first group:
(0+0+1)+(0+1+1)+...+(0+9+1)+(1+0+1)+(1+1+1)+...+(1+9+1)=10+2.45+20=120;
for second group: (0+0+3)+(0+1+3)+...+(0+9+3)+(1+0+3)+(1+1+3)+...+(1+9+3)=10+2.45+3.20=160;
and so on.
Final answer is: 120+160+200+240+280=1000.
Hope it is correct.

[quote="splintercell990"]For some reason, I am getting a 1000 as the answer :?[/quote]
I miscalculated.

[quote="splintercell990"]For some reason, I am getting a 1000 as the answer :?[/quote]

Could you explain how you got that answer?

I think splintercell990 did it in the same way as Myth.

Yep, as jen7 said, I did it the same way as Myth (actually, I must confess, all I did was notice a small mistake in Myth's solution, and informed him about it)


function
trigonometry
real analysis
ceiling function
real analysis unsolved

For which functions $f$ defined on nonnegative reals with positive values can we find a positive sequence $a_n$ converging to 0 and such that $\sum {a_n}$ diverges but $\sum{f(a_n)}$ converges?

I think you want to restrict this to monotone functions $f.$ Without that restriction, I could find some, well, peculiar examples.

Like this: let $f(x)=x^2+\frac1x\sin^2\left(\frac{\pi}x\right).$

Then $\sum_nf\left(\frac1n\right)$ converges.

If you do restrict to monotone functions, the condition $f(x)=o(x)$ as $x\to0$ is worth investigating - but I already have my doubts that it's both necessary and sufficient.

On second thought, we don't need $f$ to be monotone, and $o(x)$ was the wrong condition.

We can find a positive sequence $a_n$ tending to zero such that $\sum_na_n=\infty$ and $\sum_nf(a_n)<\infty$ precisely when

$\liminf_{x\to 0^+}\frac{f(x)}x=0.$

Note that for the "peculiar" example in my previous post, this liminf is indeed zero.

Proof:

First, if the liminf is not zero, then $\exists\,\delta>0$ and $c>0$ such that for $x\in(0,\delta),\,f(x)>cx.$

Now suppose $a_n\to0.$ Then $\exists\,N$ such that for $n>N,\,a_n<\delta.$

Then $\sum_{n>N}f(a_n)>c\sum_{n>N}a_n.$ Thus if $\sum a_n$ diverges, $\sum f(a_n)$ must also diverge.

Next, the converse. Suppose that $\liminf_{x\to 0^+}\frac{f(x)}x=0.$

Then there exists a sequence $x_k$ such that $x_k\to0$ and $\frac{f(x_k)}{x_k}\le2^{-k.}.$ Assume also that each $x_k<1.$

Let $m_k=\lceil\frac1{x_k}\rceil$ Then $1\le m_kx_k<1+x_k<2.$

Construct the sequence $a_n$ by repeating the term $x_k$ $m_k$ times.

Then $\sum_{n=1}^{\infty}a_n=\sum_{k=1}^{\infty}m_kx_k\ge\sum_{k=1}^{\infty}1=\infty$

but $\sum_{n=1}^{\infty}f(a_n)=\sum_{k=1}^{\infty}m_kf(x_k)\le\sum_{k=1}^{\infty}2^{-k}m_kx_k\le2\sum_{k=1}^{\infty}2^{-k}=2<\infty$


function
real analysis
real analysis unsolved

Prove or disprove:
f,g:[a,b]->[a,b] are continuous. And f(g(x))=g(f(x)) for all x in [a,b]. Then there exist y s.t. f(y)=g(y)

[hide]If $ f(y) > g(y)$ for all $ y\in [a,b]$, then the function $ f \minus{} g$ has strictly positive minimum $ m$ on $ [a,b]$. It follows that $ f([a,b])\subseteq [a \plus{} m,b]$ and $ g([a,b])\subseteq [a,m \minus{} b]$. Let $ I_1 \equal{} f([a,b])\cap g([a,b])\subseteq [a \plus{} m,b \minus{} m]$. The assumption $ f\circ g \equal{} g\circ f$ implies $ f(I_1)\subseteq I_1$ and $ g(I_1)\subseteq I_1$. This process can be continued: $ I_2 \equal{} f(I_1)\cap g(I_1)$ is a subset of $ [a \plus{} 2m,b \minus{} 2m]$ and so on. Since the interval $ [a,b]$ is finite, we have a contradiction.[/hide]

The special case $ g(x)\equiv x$ is simply the one-dimensional [url=http://en.wikipedia.org/wiki/Brouwer_fixed_point_theorem]Brouwer fixed point theorem[/url]. Is the $ n$-dimensional version true? More precisely:
Let $ B^n$ be the closed unit ball in $ \mathbb R^n$. If $ f,g\colon B^n\to B^n$ are continuous maps such that $ f\circ g \equal{} g\circ f$, is it true that $ f \equal{} g$ at some point?


I'm trying to use Mathematica to solve a certain two-variable equation in integers, does anyone know how to do  that? Or at least could I tell it to check all values from 1 to 1000 and tell me if there are any solutions or something like that? The equation is around the complexity level of $b(289-a^{2}) = (b-a)(b+a)a^{2}$ (that is not the actual equation though).

Also, if this is not the right forum, please move it to the best place.

FindInstance[{
b*(289 - a^2) == (b - a)*(b + a)*a^2,
 (a | b) ∈ Integers,
  a > 0, b > 0
  }, {a, b}, 1][[1]]

{a -> 17,b->17}

mathematica only find one solution

If you try 2 it give error.

FindInstance[{
b*(289 - a^2) == (b - a)*(b + a)*a^2,
 (a | b) ∈ Integers,
  a > 0, b > 0
  }, {a, b}, 2][[1]]


1) What is the AIME qualification score for 12B/10B 2008?

2) Has anyone received their official 12B/10B 2008 score?

Thanks!

Hey cognos, if you're still online, go to FTW so we can talk.

It usually takes around three weeks for the A test qualification scores to be announced, and a little less than that for the B tests.. so it should be announced anytime now.


Let $ ABC$ be an acute-angled triangle.Let $ P$ be a point on $ OH$.Prove that:

$ PA\plus{}PB\plus{}PC\leq3R$, where $ R$ is the usual notation for triangle.

[quote="Inequalities Master"]Let $ ABC$ be an acute-angled triangle.Let $ P$ be a point on $ OH$.Prove that:

$ PA \plus{} PB \plus{} PC\leq3R$, where $ R$ is the usual notation for triangle.[/quote]

This problem was discussed at http://www.mathlinks.ro/viewtopic.php?t=19316 . Of course, $ OH$ must mean "the segment $ OH$" here, but triangle $ ABC$ needs not be acute as long as $ P$ lies inside triangle $ ABC$.

  darij


Suppose $ a^5 \plus{} b^5 \equal{} 2$. Prove that $ a \plus{} b\leq2$.

Sketch the graphs of $ a^5\plus{}b^5\equal{}2,\ a\plus{}b\equal{}k$ for $ a>o,\ b>0.$

[quote="kunny"]Sketch the graphs of $ a^5 \plus{} b^5 \equal{} 2,\ a \plus{} b \equal{} k$ for $ a > o,\ b > 0.$[/quote]
Graphing is sometimes trustless and unacceptable. Could you please me a mathematical proof?

[quote="t0rajir0u"]After scaling by $ a \plus{} b$, the problem can be restated as follows:  suppose $ a \plus{} b \equal{} 2$.  Prove that $ a^5 \plus{} b^5 \ge 2$.[/quote]
The following may be better: Suppose $ a\plus{}b>2$, then $ a^5\plus{}b^5>2$. But how to prove it?

[quote="ifai"][quote="kunny"]Sketch the graphs of $ a^5 \plus{} b^5 \equal{} 2,\ a \plus{} b \equal{} k$ for $ a > o,\ b > 0.$[/quote]
Graphing is sometimes trustless and unacceptable. Could you please me a mathematical proof?

[quote="t0rajir0u"]After scaling by $ a \plus{} b$, the problem can be restated as follows:  suppose $ a \plus{} b \equal{} 2$.  Prove that $ a^5 \plus{} b^5 \ge 2$.[/quote]
The following may be better: Suppose $ a \plus{} b > 2$, then $ a^5 \plus{} b^5 > 2$. But how to prove it?[/quote]

generally, $ \sqrt[5]{\frac{a^5\plus{}b^5}{2}}\ge\frac{a\plus{}b}{2}.$ This is the power mean inequality, and plugging $ a\plus{}b\equal{}2$ in yields $ a^5\plus{}b^5\ge2\cdot 1^5 \equal{} 2.$

the power means inequality is for non-negative numbers, so a little explanation is required.

for the case where one of $ a, b$ is negative (both can't be negative) we may argue as follows. let $ a, b \geq 0$, such that $ a^5 \minus{} b^5 \equal{} 2$. we need to prove $ a \minus{} b \leq 2$. now,

$ a^5 \minus{} b^5 \equal{} (a \minus{} b)(a^4 \plus{} a^3b \plus{} a^2b^2 \plus{} ab^3 \plus{} b^4)$

note that the second factor is at least 1, since all the summands are non-negative, and clearly $ a > 1$. so if $ a \minus{} b > 2$, $ a^5 \minus{} b^5 > 2\cdot 1 \equal{} 2$, a contradiction


superior algebra
superior algebra unsolved

Let $ A$ be a ring (not necessarily commutative) and let $ N$ be a nilpotent ideal in A.
We know that $ \forall _{n \in N} a^n \minus{}b^n \in N$ where $ a, b \in A$
Is it truth that then $ a\equal{}b$?

May be $ a*n \equal{}  b*n$, then it is not true.

But what is $ n$? If n can be 1 it is truth

I think [b]you[/b] should know what $ n$ is...

  darij

???
if a , b are all in N??? a does not equal  to b.??


linear algebra
matrix
algebra proposed
algebra

A is the 4 x 4 matrix with a11 = 1 + x1, a22 = 1 + x2, a33 = 1 + x3, a44 = 1 + x4, and all other entries 1, where xi are the roots of x4 - x + 1. Find det A.


Let $ a,b\in \mathbb{C}$ such that $ |a|\equal{}|b|\equal{}1$ and $ a^2\plus{}b^2\equal{}\minus{}1$ . Prove that $ a\equal{}\pm\left(\cos \frac{\pi}{3}\plus{}i\sin \frac{\pi}{3}\right)$ and $ b\equal{}\pm\left(\cos \frac{2\pi}{3}\plus{}i\sin \frac{2\pi}{3}\right)$ .

We have $ 1\equal{}|b^2|^2\equal{}|\minus{}1\minus{}a^2|^2$, but $ |1\plus{}a^2|^2 \equal{} 1\plus{}a^2\plus{}\overline{a}^2 \plus{}|a|^4$, so $ a^2 \plus{} \overline{a}^2 \plus{} 1\equal{}0$. Now, $ \overline{a} \equal{} 1/a$, so we have $ a^2\plus{}1/a^2\plus{}1 \equal{} 0$, i.e. $ a^4\plus{}a^2\plus{}1\equal{}0$, yielding exactly the required conclusion.


Let $P$ be an interior point of a triangle $ABC$ and let $BP$ and $CP$ meet $AC$ and $AB$ in $E$ and $F$ respectively. IF $S_{BPF} = 4$,$S_{BPC} = 8$ and $S_{CPE} = 13$, find $S_{AFPE}.$

[quote="Singular"]AP hits BC at D.  Put areas AEP, AFP as x, y.

FP:PC = 4:8 = y/(x+13)
EP:PB = 13:8 = x/(y+13)

It's two var two eqn, solve easily for x and y , thus x+y.[/quote]There is a slight error in Singular's answer

The ratio EP:PB should be 13:8 = x/(y+4)

[hide="Here is a short and complete solution."] Denote $u=[AEF],\ v=[PEF],\ x=[AEPF]\ (u+v=x)$.
$(1)\ \ \boxed {\ [BPF]\cdot [CPE]=[PEF]\cdot [PBC]\ }\Longrightarrow 4\cdot 13=8\cdot v\Longrightarrow \boxed {\ v=\frac{13}{2}\ }\ .$
$\frac{[BAE]}{[BCE]}=\frac{EA}{EC}=\frac{[AEF]}{[CEF]}\Longrightarrow \frac{4+\frac{13}{2}+u}{8+13}=$ ${\frac{u}{\frac{13}{2}+13}}=\frac{4+\frac{13}{2}}{8-\frac{13}{2}}=\frac{21}{3}=7\Longrightarrow$
$\boxed {\ u=21\cdot \frac{13}{2}}\Longrightarrow x=u+v=\frac{13}{2}\cdot (1+21)=13\cdot 11=143\Longrightarrow \boxed {\ x=143\ }\ .$
[color=red][b]Remark.[/b][/color]
[color=darkred]The relation $(1)$ is well-known in a convex quadrilateral $BCEF$, where $P\in BE\cap CF\ !$[/color][/hide]

Let $D$ be $AP \cap BC$. Let $P$ be the center of mass for $\triangle ABC$
Since
\[\frac{[FBP]}{[BPC]} = \frac{FP}{PC} = \frac{1}{2}\]
\[\frac{[ECP]}{[BPC]} = \frac{EP}{PB} = \frac{13}{8}\]
We assign weights $13$, $7$, $8$, and $14$ to $B$, $C$, $E$, and $F$ respectively.
This implies than $A$ has a weight of $1$ and $D$ has a weight of $20$.
$\frac{13}{8} = \frac{BP}{PC} \Rightarrow [BDP] = \frac{104}{20}$ and $[CDP] = \frac{56}{20}$
$\frac{AP}{PD} = \frac{[APC]}{[PCD]} = \frac{[BAP]}{[BPD]} = \frac{20}{1} \Rightarrow [AEP] + [PEC] = 56$ and $[ABP] + [BPD] = 104 \Rightarrow [AFPE] = [AFP] + [AEP] = 143$

I think I was the only one thinking those quantities were perimeters of their respective triangles. Lol and sorry for the long bump


MIT
college
calculus
geometric series
geometric sequence
real analysis
real analysis unsolved

mit calculus final practice exam:
find, with proof, all values of $x$ such that

$\displaystyle\sum_{i=1}^{\infty}ne^{-nx}$

converges.

I think you meant $ie^{-ix}$.

Let $r = e^{x}$.  Then the summation becomes

$\sum_{i=1}^{\infty} \frac{i}{r^i}$

Because $r > 0$, it follows (from sum of a geometric series; easy to prove, I won't bother) that the sum converges iff $r > 1$, that is,

$\boxed{ x > 0 }$.

wait, you can look at it as a geometric series despite having the i term in the numerator?

When I wrote what I said about looking at it as a geometric sequence, I meant the following identity:

Let $S = \sum_{i=1}^{\infty} \frac{i}{r^i}$.

Now, $\frac{S}{r} = \sum_{i=1}^{\infty} \frac{i}{r^{i+1}} = \sum_{i=1}^{\infty} \frac{i-1}{r^i}$

Then $S - \frac{S}{r} = \sum_{i=1}^{\infty} \frac{1}{r^{i+1}}$
$S = \frac{r}{r-1} \left( \sum_{i=1}^{\infty} \frac{1}{r^{i+1}} \right)$

From which convergence follows.  :)

Alternately:

Let $s$ be a real such that $1 < s < r$.  Then

$\exists M : \forall m > M, \frac{m}{r^m} < \frac{1}{s^m}$

Then convergence also follows.


This is a very easy question.
How are you going to explain that the centroid of a triangle is always inside the triangle.

Of course, we know the proof. But i want to explain it in laymen's term. Any suggestion?

Depends on what definition of centroid you're taking.  If it's the more-or-less intuitive notion as center of mass (that is, the "average" of the points of the triangle), then it is true for any convex shape by definition.

Well, not becoming really mathematical or anything, you could say that the medians, which go from a vertex of a triangle to the midpoint of the opposite edge, is always contained within the triangle. Therefore, the centroid (the intersection of the medians) is obviously inside the triangle.


In how many ways we can choose 3 non empty and non intersecting subsets from $ (1,2,\ldots,2008)$.

[quote="nicetry007"]The answer is $ 4^{2008} \minus{} \binom{3}{1} 3^{2008} \plus{} \binom{3}{2} 2^{2008} \minus{} \binom{3}{3} 1^{2008}$. This follows from an inclusion exclusion argument.[/quote]
Can you explain more clearly?

Here's an intermediate question: How many ways are there to select $ 3$ non-intersecting subsets of $ \{1, \dots n\}$?  The way to think about it is to instead think about, for each number, choosing whether that number is in set $ A$, set $ B$, set $ C$, or none of the sets.  This gives you $ 4^n$.

Now apply inclusion exclusion, with the conditions you're excluding as $ c_1 \equal{}$"Set $ A$ is empty", $ c_2 \equal{}$"Set $ B$ is empty" and $ c_3 \equal{}$"Set $ C$ is empty".

Note that we're assuming here that the order of set $ A$ vs. set $ B$ vs. set $ C$ matters.  If this is not the case, you have to divide by $ 6$ to account for this at the end.

Can't we just use stars and bars? Wouldn't it then just be 2007 c 2?

No, because you don't have to contain all of the elements of the set.


linear algebra
matrix
search
linear algebra unsolved

Prove that any $n \times n$complex matrix is a product of two complex symmetric matrices.
There are similarities between this and my previous post on this forum,which get no any reply.Maybe this is a little more difficult than that one.As well,it is more gorgeous.

It has already been discuss on the Forum, and I think that Harazi gave a solution. (you just have to prove it for Jordan black, which is not that hard ..)

a link http://www.mathlinks.ro/Forum/viewtopic.php?search_id=390394483&t=67449 ,but no solution was provided  :lol:


function
integration
real analysis
real analysis unsolved

Let $ \varphi\in L^1(R^s)$, and define $ f$ by equation $ f(x) \equal{} \int_{R^s}\varphi(x \plus{} y)\overline{\varphi(y)}dy$   $ \hspace{2mm}(x\in R^s)$

Prove that $ f$ is positive definite.

Positive definite in what sense? In the first one that comes to mind (letting $ f$ generate an inner product on some subspace of $ L^2$), it isn't.

If in the sense that $ g(x,y) \equal{} f(x\minus{}y)$ is nonnegative definite, then it's straightforward from the definition.


[quote="Beginner"]An ancient Sicilian legend says that the barber in a remote townwho can be reached only by
traveling a dangerous mountain road shaves those people, and only those people who do not
shave themselves. Can there be such a barber?[/quote]
If he never shaves, then yes.


trigonometry
geometry
inradius
vector
circumcircle
geometric transformation
reflection

Prove that in a triangle, $AH=2R\cos(A)$ where H is the orthocenter and R is the circumradius.

Prove that in a triangle, $1+\frac{r}{R}=\cos(A)+\cos(B)+\cos(C)$, where $r$ denotes the inradius and $R$ denotes the circumradius.

[quote="farbod_mp2002"]oh my...the second identity:

1 + r/R = CosA + CosB + CosC

will give the simplest solution to my last question:

"prove that there are no triangle satisfying CosA + CosB + CosC = 1 " ...

how nice :)

(im sorry it was'nt a solution at all! )[/quote]

It also proves $1 < \sum \cos A \leq \frac32$.  :D

[hide="vectors for the first one"] Place the circumcenter at the origin as usual. Since $H = A+B+C$, we have $AH = (A+B+C)-A = B+C$. Reflect $B$ across the origin to $B'$. Then $B' =-B$ so $B'C = C-(-B) = B+C$, i.e. $AH = B'C \Rightarrow |AH| = |B'C|$.

Now consider triangle $BB'C$. Since $|BB'| = 2R$, $|BC| = 2R\sin{A}$ by the Law of Sines, and $\angle BCB' = \frac{\pi}{2}$ since $BB'$ is a diameter, we know

$|B'C|^{2}= |BB'|^{2}-|BC|^{2}= 4R^{2}-4R^{2}\sin^{2}{A}= 4R^{2}\cos^{2}{A}$.

Thus $|AH| = |B'C| = 2R \cos{A}$. Note that this quantity is negative when $\angle A > \frac{\pi}{2}$, indicating that $H$ lies outside of the triangle. [/hide]

How do you know $H=A+B+C$? That's a cool vector property of triangles, though.

If you simply show that $AH$ and $BC$ are perpendicular, then by the uniqueness of the orthocenter and symmetry the result follows.

[hide="hint"] Consider $(H-A) \cdot (C-B)$. [/hide]

[hide="you can also do this for the first one"]
Let $O$ be circumcenter of $\triangle ABC$ and $M$ be midpoint of $\overline{BC}$, then $\angle OBC=90^\circ-A$ so $OM=R\cos A$.

Now showing $AH=2R\cos A$ follows from similar triangles! Letting $G$ be the centroid of $\triangle ABC$, the dilation that takes $A$ to $M$ takes $\triangle ABC$ to the triangle formed by its midpoints. Then $H$ is taken to the orthocenter of the midpoint triangle, which happens to be $O$.

Therefore $\overline{AH}$ is sent to $\overline{MO}$ so $\triangle AGH\sim MGO$. Since we know $AG=2GM$, we get $AH=2OM$, finishing the problem.
[/hide]

[hide="the second one is basically a trig bash"]
We have
\[\frac{r}{R}=\frac{K}{Rs}=\frac{R(\sin 2A+\sin 2B+\sin 2C)}{2s}=\frac{\sin 2A+\sin 2B+\sin 2C}{2(\sin A+\sin B+\sin C)}\]
This follows from the area formulas $rs=K=\frac{1}{2}R^{2}(\sin 2A+\sin 2B+\sin 2C)$ (by dissecting the area into three triangles with vertices at the incenter for the first one, and the circumcenter for the second one), and the Extended Law of Sines.

Now we can verify the identity because
\[2(\sin A+\sin B+\sin C)(\cos A+\cos B+\cos C)\]
\[=(2\sin A\cos A+2\sin B\cos B+2\sin C\cos C)+2\sum_\text{cyc}(\sin A\cos B+\cos A\sin B)\]
\[=\sin 2A+\sin 2B+\sin 2C+2\left( \sin (A+B)+\sin (A+C)+\sin (B+C)\right)\]
\[=\sin 2A+\sin 2B+\sin 2C+2(\sin A+\sin B+\sin C)\]
\[\Rightarrow \cos A+\cos B+\cos C=1+\frac{\sin 2A+\sin 2B+\sin 2C}{2(\sin A+\sin B+\sin C)}=1+\frac{r}{R}\]
as desired.
[/hide]

[quote="Xaero"][color=darkred]Prove that in a triangle, $1+\frac{r}{R}=\cos A+\cos B+\cos C,$ where $r$ denotes the inradius and $R$ denotes the circumradius[/color].[/quote]
[color=darkblue][b]Proof.[/b] [/color]
\begin{eqnarray*}r&=&(p-a)\tan\frac{A}{2}=R(\sin B+\sin C-\sin A)\cdot\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}}\\ &=&2R\left(\sin\frac{B+C}{2}\cos\frac{B-C}{2}-\sin\frac{A}{2}\cos\frac{A}{2}\right)\cdot\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}}\\ &=&2R\left(\cos\frac{B-C}{2}-\cos\frac{B+C}{2}\right)\sin\frac{A}{2}\\ &=&4R\prod\sin\frac{A}{2}\\ \Longrightarrow 1+\frac{r}{R}&=&1+4\prod\sin\frac{A}{2}\\ \Longrightarrow 1+\frac{r}{R}&=&\sum\cos A\end{eqnarray*}

[quote="Bibobeo"]$\sum a(b^{2}+c^{2}-a^{2})=2abc+\prod (b+c-a)$[/quote][quote="scorpius119"]$\cos A+\cos B+\cos C=1+\frac{\sin 2A+\sin 2B+\sin 2C}{2(\sin A+\sin B+\sin C)}=1+\frac{r}{R}$[/quote][size=150][color=darkred][b]

Very nice ! I like these identities. Thank you ![/b][/color][/size]

[color=darkblue][b]Another proof for an acute triangle.[/b] Denote the circumcircle $C(O,R)$ and the middlepoints $M,N,P$ of the sides $[BC],[CA],[AB]$ respectively.

$\{\begin{array}{c}OD=R\cdot \cos A\mathrm{\ \ \ a.s.o.}\\\\ \sum [BOC]=[ABC]\end{array}\|$ $\implies$ $\sum \frac{a\cdot R\cos A}{2}=pr$ $\implies$ $\boxed{\ \sum a\cdot\cos A=\frac{r}{R}\cdot (a+b+c)\ }$.

Add thie relations  $\{\begin{array}{c}a\cos A+b\cos B+c\cos C=\frac{r}{R}\cdot (a+b+c)\\\\ a\cos B+b\cos A=c\\\\ b\cos C+c\cos B=a\\\\ c\cos A+a\cos C=b\end{array}$ $\implies$ $\cos A+\cos B+\cos C=1+\frac{r}{R}$.

[b]Remark.[/b] The [u]Euler's relation[/u] in an acute triangle $ABC$ : $\boxed{\ OD+OE+OF=R+r\ }$.[/color]

What?  :( Why do you rate my post as poor?
If you don't believe, just expand and you'll get what I wrote is true, then just use The Cos Law ...  :wink:

The rating system is pretty biased right now, since about 90% of people do not rate things (I think). So you probably got one or two bad ratings and then your post will look like spam when it isn't. I thought my posts were at least average, but who knows.


function
integration
calculus
derivative
calculus computations

Let f be a differentiable function and g continuous one, defined on R. If they satisfying $ \displaystyle \int\limits_0^{f\left( x \right)} {f\left( t \right)g\left( t \right)dt}  \equal{} g\left( {f\left( x \right)} \right) \minus{} 1$, find g

I'm not sure if this is correct or not.

[hide]Define a function $ A(u)$ such that $ A'(u) \equal{} f(u)g(u)$ and $ A(0) \equal{} 1$.

Then, $ \int\limits_{0}^{f\left( x\right)}{f\left( t\right)g\left( t\right)dt}\equal{} A(f(x)) \minus{} A(0) \equal{} g(f(x)) \minus{} 1$

Thus, $ A(x) \equal{} g(x)$ for all $ x \in f(\mathbb{R})$.

So, we have: $ g'(x) \equal{} f(x)g(x)$.

Therefore: $ \dfrac{\,dg}{\,dx} \equal{} f \cdot g \Longleftrightarrow \dfrac{\,dg}{g} \equal{} f \,dx \Longleftrightarrow \int \dfrac{\,dg}{g} \equal{} \int f \,dx \Longleftrightarrow \ln(g) \equal{} F \Longrightarrow g \equal{} e^F$

Thus, $ g(x) \equal{} e^{F(x)}$ where $ F(x)$ is an anti-derivative of $ f(x)$ such that $ g(0) \equal{} e^{F(0)} \equal{} 1 \Rightarrow F(0) \equal{} 0$.[/hide]

I cant see anything wrong in your proof except the step you solving the equation $ g'(x) \equal{} f(x)g(x)$
We can myltiply by $ e^{F(x)}$ and get $ g(x) \equal{} c e^{F(x)}$ with $ c \equal{} e^{\minus{}F(0)}$


vector calculus function algebra polynomial quadratics domain

hi, i wonder which textbook(s) would prepare me well for Waterloo's math 145,147 (if anyone's taken it before)? the courses generally more focus on proofs than applications of theories and formulas. also, i've completed AP cal AB and some basic vectors, so any suggestions of textbook(s) suitable for me?

What does that course cover?  Is it multivariable calculus, linear algebra, differential equations,...?  If you tell us the subject matter, we can recommend books.

Calculus:
Review of the Fundamental Theorem. Methods of integration. Further applications of the integral. Improper integrals. Linear and separable differential equations and applications. Vector (parametric) curves in R2. Convergence of sequences and series. Tests for convergence. Functions defined as power series. Taylor polynomials, Taylor's Theorem, and polynomial approximation. Taylor series. Suitable topics are illustrated using computer software.
Algebra:
A study of the basic algebraic systems of mathematics: the integers, the integers modulo n, the rational numbers, the real numbers, the complex numbers and polynomials.

and i would like textbooks that place their focuses on proofs :)

Having taken 145 and 147 (and currently taking their successor courses) I can (hopefully!) steer you in the right direction:

145 introduces elementary number theory and some aspects of abstract algebra. We started with stuff like the fundamental theorem of arithmetic, linear diophantine equations, fermat's little theorem, the chinese remainder theorem, euler's theorem and the totient function, RSA cryptography, some primality tests. Then moved on to an intro. to rings and fields, Quadratic number domains, the Gaussian integers, Polynomial rings (with concepts of divisibility and factorization throughout all these things), Gauss' lemma, Eisenstein's criterion. I also remember a detour into complex numbers somewhere along the way that contained an elementary proof of the fundamental theorem of algebra.

147 is a fairly rigorous introduction to single variable calculus. In my year we started with the least upper bound property and expanded on to sequences and limits of sequences, the bolzano weierstrass theorem, then limits of functions and their sequential characterization, the intermediate and extreme value theorems, the derivative (heavy emphasis on the derivative as a linear approximation), the mean value theorem and some applications, and we finished with a section on Taylor polynomials, Taylor's theorem, and big O.

Keep in mind both of these courses emphasize the theoretical aspects of these topics.

This should help people in recommending books. In my opinion the profs. are given pretty free reign in these courses since only one section is offered a semester and they aren't watched too closely by the "higher ups" so content differs a bit from year to year. For the algebra course, we followed a set of course notes called "Integers, Polynomials, and Finite Fields" but they're published by the university and so it might be tough to get your hands on them. For the calculus course, Spivak is the standard reference book (keep in mind it's a reference book, we cover the topics in a bit different order and in a little bit different a way). Some profs put "Introduction to Analysis" by Wade on their list as a reference book as well.

EDIT: The descriptions you gave were for the 13X stream. The first one you gave is the calc 2 outline (you will not see an integral in your first semester if you take the 14X stream  :P) and the second on you gave is for 135.

Since this 147 course seems to be advanced single-variable calculus, I recommend Spivak's "Calculus".  Terrific, terrific book.

Is there a specific edition worth investing into? Or are they all worth it.

ALso, welcome fellow waterlooers ! (I go there too and start my 2nd year in 1 week :))

Usually, content does not change too much from edition to edition, but I always go for the most recent one.

Mhmm I am currently studying from James Stewart's Calculus - Early Transcendentals.

Is Spivak's book much more useful than this one? Is the one I am currently studying from (it was suggested for the courses I am in) good enough for a very good background?

I believe Stewart's book is more plug and chug.  Spivak is a rigorous development from scratch.

I would disagree somewhat with that characterization. Stewart is certainly a more traditional introductory calculus book, but I wouldn't go so far as to call it "plug and chug". While many of the problems are meant as fundamental "skill building" exercises of the type - take this derivative, compute this area, etc., it does have some "problem solving" type of problems in it as well. And the exercises in the book are good for most students who are getting a first introduction to calculus.

The biggest difference between Stewart and Spivak is the level of maturity and ability to handle abstraction possessed by the reader. Examples are the main body of Stewart's book, whereas proofs form a significant body of Spivak's text. Ideally, a student should be exposed to both books - Stewart (or some similar book) for the intuition and the "how", and Spivak (or some similar book) for the deeper understanding and the "why".

i was wondering if spivak's book was rigorous enough to be considered an analysis book...

Rigorous enough - yes. Comprehensive enough - no, not in the modern sense of real analysis.


Isabella's house has 3 bedrooms. Each bedroom is 12 feet long, 10
feet wide, and 8 feet high. Isabella must paint the walls of all the
bedrooms. Doorways and windows, which will not be painted, occupy 60
square feet in each bedroom. How many square feet of walls must be
painted?

The walls take up $ 2(12 \cdot 8 \plus{} 10 \cdot 8)\minus{}60$ in a room, so 3 rooms is $ 876$.


algebra
polynomial
combinatorics unsolved
combinatorics

Hi, everyone,

I'm trying to get a reality check on some problems I saw posted on the Web

http://www.ams.sunysb.edu/~tucker/ams301-3tests.html

for a third course examination in a university combinatorics course. How hard are these problems? Are these too hard to fit in this Advanced Section, or are they at about the right level for this forum? Are you readily able to solve them? How long does that take?

Thanks for any comments you have.

I don't think they are difficult...
Anyone who learnt some Introdutory Combinatorics is able to solve them.
But it might take some time on computation...

[quote="liyi"]Anyone who learnt some Introdutory Combinatorics is able to solve them.
But it might take some time on computation...[/quote]

How much time, would you estimate?

Thanks for any comments you have.

eh...i think half an hour is enough...  :maybe:

eum... just wondering why time really matters to post here:
no need to care for OUR time we waste in this :D

Anyway liyi, I didn't try them but I'd think I'd need at least an hour for each contest edition...

[quote="Peter VDD"]Anyway liyi, I didn't try them but I'd think I'd need at least an hour for each contest edition...[/quote]
what do you mean by "each contest edition" here?

never mind :P

they're easy indeed but I had no clue what the rook polynomial was :)

1. Find the rook polynomial and give an expression for the number of matchings of the 5 men (rows) and 5 women (columns) given the following 8 conflicting paris (M1,W4), (M2,W2), (M3,W1), (M3.W3), (M4,W3), (M4,W5), (M5,W2), (M5.W4).

So I thought if I have to figure out the theorem for each problem -- luckily that ain't necessary for the rest...


Juniors (Grades up to 10)

[b]International Mathematics
TOURNAMENT OF THE TOWNS[/b]

[b]O-Level Paper, Spring 2004[/b]

1. [3] In triangle ABC the bisector of angle A, the perpendicular to side AB from its midpoint, and the altitude from vertex B, intersect in the same point. Prove that the bisector of angle A, the perpendicular to side AC from its midpoint, and the altitude from vertex C also intersect in the same point.

2. [3] Find all possible values of n :ge: 1 for which there exist n consecutive positive integers whose sum is a prime number.

3. Bucket A contains 3 litres of syrup. Bucket B contains n litres of water. Bucket C is empty. We can perform any combination of the following operations:

- Pour away the entire amount in bucket X,
- Pour the entire amount in bucket X into bucket Y,
- Pour from bucket X into bucket Y until buckets Y and Z contain the same amount.

(a) [3] How can we obtain 10 litres of 30% syrup if n = 20?
(b) [2] Determine all possible values of n for which the task in (a) is possible.

4. [5] A positive integer a > 1 is given (in decimal notation). We copy it twice and obtain a number b = [u]aa[/u] which happened to be a multiple of a2. Find all possible values of b/a2.

(e.g. a = 47, b = 4747)

5. [6] Two 10-digits integers are called neighbours if they differ in exactly one digit (for example, integers 1234567890 and 1234067890 are neighbours). Find the maximal number of elements in the set of 10-digit integers with no two integers being neighbours.

I think that X, Y, and Z can denote any of the three buckets.

I was [b]very[/b] faithful in transcribing the problems. I even left the grammatical mistakes intact. ;)

The only personal addition is the e.g. part for #4 Junior / #2 Senior.

#2 seems a bit easy: [hide]There arent any -- if n is odd then the average of the consecutive numbers is an integer that is greater than one, and if n is even the average always has a fractional part greater than 1/2 and an integer part greater than one.  In the first case the sum is divisible by n, and in the second the sum is divisible by n/2.  Is that right?[/hide]

That looks ok, except those calculations should give you the two trivial solutions :P

BUMP -- come on guys, you can do these.  The first 3 aren't too bad.

#5) [hide]If we take  a set of 10 digit numbers that are equivilent (mod 11) then each pair differs by a multiple of 11.  There are no multiples of 11 that have only one non-zero digit because (11, 10) = 1.  No two numbers in the set can be neighbors because they differ by at least two digits.  The maximum number of elements in a set with numbers that are all equivalent mod 11 is ceiling(9000000000/11) = 818181819.

We can see that there is no set with more members because in such a set two numbers would have to differ by less than 11, meaning they would be neighbors.[/hide]

I think at the end you meant more members, not less.

You haven't shown that such a set exists, though.

Oh yes -- I did mean more members, ill edit that.  Im not sure what you mean about showing that the set exists.  Do you mean that I did not specify a specific remainder mod 11?

Sorry ... my previous comment was wrong.

But I still don't think it's perfect.  For example, the numbers 102 and 111 don't differ by 11, but they also aren't neighbors.  So you could potentially fit many more numbers in then you're allowing for.

818181819 was my answer ('twas also a question on the senior paper), but apparently it's wrong - a friend of mine got 9e8.

Joel: 100 and 111 have the same parity mod 11, and 100 and 102 are neighbours. One's argument to show that the set is of maximal size is by showing that you can't add a number, b, (j mod 11) to the set (k mod 11), j != k, without the new number becoming neighbours with another number. One considers b :pm: 1 * 10n with n an integer ranging from 0 to 8. (Or is it 0 to 9?)

My proof, due to the time constraint, wasn't very rigourous. Though I doubt I would be able to make it rigourous even if I had enough time...

We'll see...

(P.S. this question is extremely similar a question [I don't know which] on 1990 USAMO)

9e8?

Any theory you have should apply equally well to 3- or 4-digit base-2 numbers, which are small enough that you could actually calculate it by hand (or write a slick computer program, probably, but that's definitely beyond my ability).

Osiris:  if you can convince me it works on a scale of numbers that I can actually do it out by hand for, I'll believe you.  That means base 2, no more than 4 digits, or base 3, no more than 3 digits, or any other base, no more than 2 digits.  I'm skeptical both of your answer and of your friend's, but I'd like to see the theory applied in a case where it's actually verifiable.

Also, what if I include numbers with leading 0's?  (That is, count 002 as a 3-digit number, so that we can add it to any set of 3-digit numbers without having neighbors.   What happens to your answer then?


$ a, n$ are integers

Prove that:

$ n \plus{} 1 \equal{} a^2 \implies n \plus{} 1 | 2(n \minus{} 1)!$

[hide]Given $ n,a \in \mathbb{Z}$ (assumed to be positive only), and $ n\plus{}1\equal{}a^2$, $ n\plus{}1|2(n\minus{}1)! \Rightarrow a^2|2(a^2\minus{}2)$
For $ a\equal{}2$, we have $ 4|4$.
For $ a\geq3$, all numbers from $ 1$ to $ a^2$ include $ a$ multiples of $ a$. Thus, $ a^a|a^2!$. Taking away $ a$ and $ a\minus{}1$ will remove one multiple of $ a$, so $ a^{a\minus{}1}|(a\minus{}2)!$. Since $ a\geq3$, we have $ a^2|(a\minus{}2)!$.[/hide]

[hide]

let's write the

$ n \plus{} 1 | 2(n \minus{} 1)!$   as $ a^2|2(a^2 \minus{} 2)! \equal{} (2a^2 \minus{} 4)!$

if $ 2a^2 \minus{} 4 > a^2$ then the above must be true ...

and that is true exepct for $ a \equal{} 1$ or $ a \equal{} 2$. for them we can check that te first statement is true and we finish the problem

[/hide]

$ 2(a^2 \minus{} 2)! \equal{} (2a^2 \minus{} 4)!$

I'm not so sure about that...

I just noticed it ... was it meant

$ 2*((n\minus{}1)!)$ or $ (2(n\minus{}1))!$?

It doesn't really matter, as both $ 2[(n \minus{} 1)!]$ and $ [2(n \minus{} 1)]!$ contain the term $ 2n \minus{} 2 \equal{} 2a^{2} \minus{} 4$, so all you would have to prove is that $ 2a^{2} \minus{} 4 \geq a^{2}$, for $ a^{2} > 1$.

[b]Brut3Forc3[/b] — your last statement is false: $ a^{2} > (a \minus{} 2)!\ \forall a \geq 2, \ \therefore a^{2} \nmid (a \minus{} 2)!$

[hide]
When $ a\equal{}2$, we have $ n\equal{}3$ and $ 4|4$.

Now consider when $ a\ge 3$. Note that since $ 2a<a^{2}\minus{}2$, $ (a^{2}\minus{}2)$ will have a factor of $ a$ and $ 2a$, or $ a^{2}$. Therefore, $ a^{2}|2(a^{2}\minus{}2)!$.[/hide]


geometry
3D geometry
sphere
search
set theory

Can a sphere be cut into finite pieces and reconstructed into two congurent spheres? If it's possible, can you prove it.

Actually the Banach-Tarski Paradox states this problem. Can anyone show a proof for it?

I may be mistaken, but I was under the impression that this proof required very extensive advanced mathematics -- not the kind of thing that you can just whip out on a forum in a couple of minutes.

just by looking, I don't think so.  The outer curved part of your sphere won't have a match with any other sphere

The answer to this question was once the source of some dispute in the mathematical world, and is one of the most bizarre and counter-intuitive results in mathematics. This is the famous Banach-Tarski paradox, so named after two famous Polish mathematicians who constructed it. The proof relies on an, at the time, easily believable axiom of set theory called the Axiom of Choice.

The Axiom of Choice, formulated in 1904 by Zormelo (the father of modern Set Theory), states that given any set of non-empty sets, you can pick one member of each of those sets and form a new set with them. Seems harmless enough right?

Well that's what most mathematicians thought until 1924 when the Banach-Tarski paradox was revealed. After the B-T paradox there was some dispute as to whether the AC was "true".

Well it was shown by Godel in 1940 that the AC was consistent with the rest of mathematics - meaning that it didn't lead to any contradictions with things that had already been proven. And in 1963 it was shown by Cohen that the AC is also indpendent of the rest of mathematics - meaning that you could choose whether you wanted to believe in it or not!!

But lots of other interesting (and seemingly "true" mathematics) relies on the AC, so nowadays most mathematicians have come to accept the AC, and hence the B-T paradox, as true.

Keep in mind that the decomposition and reconstruction of the sphere from one into two is not in large "chunks" but by collections of by "points." Making it difficult to really visualize.

The proof of the Banach-Tarski paradox is not elementary, but here is a website that presents a proof of it just in case you want to see it: http://abel.math.umu.se/~frankw/articles/bt/Banach-Tarski.html

You can also find lots of information about it by doing a search or also by looking at the MathWorld site.


trigonometry
real analysis
real analysis unsolved

[b][u]without[/u][/b] the aid of fourier's techniques, find the following 2 sums...
defined for $ \mathbf{Re}\;(x)\in\left[\;0\;,2\pi\;\right]$

$ \textbf{(i)}\;\;\sum_{n=1}^\infty\;\;\frac{\cos\;nx}{n^{2}}$

$ \textbf{(ii)}\;\;\sum_{n=1}^\infty\;\;\frac{\sin\;nx}{n^{2}}$

for $ \textbf{(i)}$, i get $ \boxed{\zeta{(2)}-\frac{\pi\cdot x}{2}+\frac{x^{2}}{4}}$,
but for $ \textbf{(ii)}$, i don't even think it is possible to obtain a sum at all...
(even if attempting with fourier's series or any other means!)

is that right?  :(


logarithms
calculus
trigonometry
complex numbers

What is 8 to the power of (log base 4 of 6)?

Please explain.

hello, we have $ \log_4 6\equal{}\frac{\ln(6)}{\ln(4)}\equal{}\frac{1}{2}\frac{\ln(6)}{\ln(2)}$  we have
$ 8^{\frac{1}{2}\frac{\ln(6)}{\ln(2)}}\equal{}x$, taking the logarithm of both sides we get
$ \frac{1}{2}\frac{\ln(6)}{\ln(2)}\cdot 3\ln(2)\equal{}\ln(x)$
$ \frac{3}{2}\ln(6)\equal{}\ln(x)$
$ \ln(6^{\frac{3}{2}})\equal{}\ln(x)$
$ \sqrt{6^3}\equal{}x$
$ 6\cdot\sqrt{6}\equal{}x$.
This is te desired result.
Sonnhard.

Didn't you already ask this in another thread? It's $ \log_e$, where $ e$ is Euler's number.

No, I haven't asked that yet. But thanks.  :)

[quote="Wickedestjr"]No, I haven't asked that yet.[/quote]

http://www.mathlinks.ro/viewtopic.php?p=1275375#1275375

 :lol:

:blush: Oh, I don't remember that. Sorry isabella.

When you get from [ln(6)]/[ln(4)] to [1 ln (6)]/[2 ln (2)], is that because 6^1 = 6, and 2^2 = 4?, or is that just a coincidence?

Yes -- you may want to familiarize yourself with the basic algebraic [url=http://en.wikipedia.org/wiki/List_of_logarithmic_identities#Algebraic_identities]properties of logarithms[/url].

Alright, I think that I am sort of getting the hang of it.

When it says that e^(pi)(i) + 1 = 0, is the (i) the square root of negative one?

that would be true unless you are reading a very very unused text.

If you're dealing with only real numbers, don't use that last property of logarithms. Let's take a look at what happens when you use that:

$ \ln x \equal{} \ln ( \minus{} ( \minus{} x)) \equal{} \ln ( \minus{} x) \plus{} \pi i \equal{} \ln x \plus{} 2 \pi i$
$ \therefore \ln x \equal{} \ln x \plus{} 2 k \pi i \; \; \forall k$

In other words, the complex logarithm is multivalued.

When we are taking the logarithm of positive reals, we generally use the unique real value, so that the logarithm is actually a function.

Of course, in elementary algebra, $ e^{i \pi}$ is nonsense. But in calculus, you will learn that this can be transformed into an infinite series and is equivalent to the polar/trigonometric form of complex numbers (this is the cis thing, you know).

Don't be bothered, just wait and you'll experience the magnificence of calculus!

not necessarily calculus, it also comes in trigonometry. (i think)

It comes in trigonometry, but the method on how $ e^{i\pi}$ is obtained is by infinite series (usually taught after some topics of calculus).

yeah, that's true.... (for 7h3.d3m0n.117: I'm not trying to act like a smarty pants, i'm just saying that calculus may be used there)