Here is a nifty little trick: take any three-digit number, $ABC$, where $A\neC$.  Note: $A$, $B$, and $C$ are digits.  Reverse the digits of $ABC$ and find the positive difference, which is represented by $XYZ$ (again, $X$, $Y$, and $Z$ are digits).  Reverse the digits of $XYZ$, find the sum of the two numbers, and you'll always get $1089$.  Every time.

Can anyone prove this?  Be aware that I am talking base 10.

[hide="Hint 1 (to give you a jump start)"]What are all the possible values for $|ABC-CBA|$?[/hide]

[hide="Hint 2 (may practically give it away)"]$(100a+10b+c)-(100c+10b+a)$[/hide]

Edit: thanks for showing me how to do a $\ne$ sign.

[quote="DanK"]
Shouldn't that be 198, 297, 396, 495, 594, 693, 792, 891?[/quote]

Yes you're right i was talking on the phone while i type it up i'm not surprised there were typos  :oops:

Posted and proven: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=80469[/url]

Didn't see that topic.  But here's how I'd prove it:
[hide]Instead of three-digit number $ABC$, let us use $100a+10b+c$, where $A=a$, $B=b$, and $C=c$.  So we find $|(100a+10b+c)-(100c+10b+a)|$ and we get $99*|a-c|$.  The possible values for $|a-c|$ are 1, 2, 3, 4, 5, 6, 7, 8, and 9.  So $ABC-CBA$ must be 99, 198, 297, 396, 495, 594, 693, 792, or 891.  Note that no matter what, $Y=9$ and $Z=9-X$ (getting that from $X+Z=9$).  So now, $X=x$, $Y=y$, and $Z=z$ to get the following: $XYZ+ZYX=(100x+10y+z)+(100z+10y+x)=101x+20y+101z=101(x+z)+20y=101(9)+20(9)=1089$.[/hide]
It's actually a pretty interesting trick, if you ask me.


advanced fields
advanced fields unsolved

[b]Let p(x),q(x),andr(x) be open statements relative to the set S.Then ~(exists x∈S)[(p(x)∨q(x))∧r(x)] is equivalent to[/b]
(A) (∨x∈S){[(~p(x))∨(~q(x))]∨[~r(x)]}
(B) (∨x∈S){[(~p(x))∧(~q(x))]∨[~r(x)]}
(C) (∨x∈S){[(~p(x))∧(~q(x))]∨[r(x)]}
(D) (∨x∈S){[(~p(x))∨(~q(x))]∧[~r(x)]}
(E) (∨x∈S){[(~p(x))∨(~q(x))]∧[r(x)]}

\forall is the code for $ \forall$ and \exists is the code for $ \exists$.

Remember that $ \neg (\exists x\in S)(P(x))$ is equivalent to $ (\forall x\in S)(\neg P(x))$.

So carefully negating the statement is done by DeMorgan first to get $ [\neg(p(x)\vee q(x))\vee\neg r(x)]$ now just DeMorgan a second time through the first parenthesis to get $ [(\neg p(x)\wedge\neg q(x))\vee\neg r(x)]$. If you do these one step at a time they aren't bad. So the answer is B.

Do I smell GRE practicing with these two questions?


Can someone help me finding the official solution of TST Romania 2006, Russia MO 2006

RUS NMO 2006: http://www.mathlinks.ro/Forum/viewtopic.php?t=86556
ROM TST 2006: http://www.mathlinks.ro/Forum/resources.php?c=142&cid=26&year=2006


If you choose to go through B, then there are 2 x 3 = 6 paths. If you choose to go through C, then there are 3 x 4 = 12 paths. Thus, there are a total of 6 + 12 = 18 paths.


Let $ a,b,c$ be positive real numbers. Prove that
$ \left( \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a}\right)^2 \ge 3\left(\frac{a^3}{b}\plus{}\frac{b^3}{c}\plus{}\frac{c^3}{a}\right).$

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers. Prove that
$ \left( \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}\right)^2 \ge 3\left(\frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a}\right).$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153560


Each point of a circle is painted black or white.

(a) Prove that there exists a quadrilateral inscribed in this circle, with two parallel sides (i.e. a trapezoid or a rectangle), such that all its vertices are painted the same color.

(b) Can one claim that there exists an inscribed rectangle such that all its vertices are painted the same color?

(c) Can one claim that there exists an inscribed trapezoid such that all its vertices are painted the same color (a rectangle is not a trapezoid  :) )?

Nice problem, solution:

a)
[hide]
We know that on the circle there are an infinite number of points. In this condition, with $ 2$ colors, let's call an "arc" a strand of connected points on a circle that have the same color. There must be an "arc" with infinite number of points:

Denote $ P_n$ as the number of points in each "arc".

$ P_1 \plus{} P_2 \plus{} P_3 \plus{} P_4.. \plus{} P_n \equal{} \infty$, so it follows that that at least one of $ P_n$ has to be $ \infty$.

With an "arc" with an infinite number of points, a [b]trapezoid[/b] can be inscribed, and if the arc extends over the length of half the circumference, a [b]rectangle[/b] can be inscribed.
[/hide]
b)
[hide]
Consider splitting a circle in half through its center. Color one half black and the other white. A rectangle is symmetrical on both sides therefore has two verticies on each side. Contradiction. So the answer is [b]no[/b].
[/hide]
c)
explained in part (a

Your argument is completely bogus.  To see the bogosity, choose a circle of circumference 1 and choose a fixed point $ P$ on the circle.  For any point $ Q$, if the arclength $ PQ$ is rational, color $ Q$ white, and if it's irrational, color $ Q$ black.  Every point is arbitrarily close to points of both colors, i.e. there is no continuous stretch of points of any given color.


trigonometry
geometry
incenter
geometry solved

The angle bisectors of a triangle are equal.  Prove the triangle is isosceles.

well, say r/sin(A/2) = r/sin(B/2), but A+B<360, so A=B.  :)

This question is can be done using trig or cyclic quadrilaterals. But it is in the euceledian geometry section in by book. So is their a method to do it using simple euceledian geometry.*

[size=59]Excuse me for any spelling mistakes[/size]

Sure, suppose AI = BI, then triangle AIB is isoceles, so <BAI = <ABI, so <A = <B.  ;)

I believe the initial problem was something different from what you are solving. Not the segments AI, BI, CI, where I is the incenter of triangle ABC, were supposed to be equal, but the segments AD, BE, CF, where D, E, F are the points of intersection of the angle bisectors of the angles CAB, ABC, BCA with the lines BC, CA, AB. The problem is then to show that if AD = BE = CF, then the triangle ABC is equilateral. In fact, this is an easy problem: By the Steiner-Lehmus theorem, AD = BE yields BC = CA, and BE = CF yields CA = AB, so that we get BC = CA = AB, and thus the triangle ABC is equilateral, and the problem is solved. The Steiner-Lehmus theorem can be found with proofs in http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/steiner-lehmus , http://www.math.niu.edu/~rusin/known-math/99/steiner-lehmus , http://www.kalva.demon.co.uk/short/soln/sh9012.html , http://groups.yahoo.com/group/Hyacinthos/messages/1309?xm=1&m=e&viscount=4 and many other sources.

  Darij

Actually, the question is to prove Steiner-Lehmus' theorem it seems.

Bomb


$(a+b)^{2}$ - $ab$ $\equiv$0 (mod $7^{3}$)

Find any $(a,b)$  where a and b are integers.

Jamshed: sorry everyone I forgot to note that

$((a+b),7)=(a,7)=(b,7)=1$

Jimbo


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

I am really desperate on this question, I've spent lot of time but still dont know, please help if you can

Let f: R ---> S be a ring homomorphism for rings R and S
Suppose that S of M is a S-module. Define an action of R on (M,+) by r*m = rm, where r is in R, m belongs to M. Show that R is a R-module.

This question is not hard, we just use definition of R-module to verify. But my problem is that I dont know how to define the action of R on M because we are given f is ring homomorphism

This is my attempt:

If I define f: R ---->S by f(r) = s
then I have a problem because I dont know how to define the action: R x M ---> M by f(r)m = rm. Is this right?

Do you know what I mean?. If you dont undrstand what I write, please ask me.

Thank you so much for helping me.[/hide]

That must be what's intended. Take that $ f(r)\cdot m$ as the definition of $ R$'s action.


Prove that for $n \in \mathbb{N}$ the equation $x^2 + y^2 = 4^n$ has no solutions in $(x,y) \in \mathbb{Z}^2$.

if $x^2+y^2$ is divisible by 4, then $x$ and $y$ are both even, so clearing 4, we may assume that the right side is not divisible by 4, so n=0. but then one of $x$, $y$ must be 0, contradiction(at least if we assume that $x,y\neq 0$, else there were solutions).


Euler
calculus
derivative
geometry
3D geometry
pyramid
videos

"I agree with those people who just dont care for imo, since if someone just solves problems all day, he will lose his imagination and just become a problemsolvingmachine without any thinking further, about new things.
The other thing is the time, who cares if one can solve 999 problems in x hours, when he becomes a mathematician there will be no time limit.
So what the imo organizers should do, is not a contest where you have to solve problems in a given time, but like find something new discover a new proof for a theorem etc."

What all you have to say on this ? Learning and investigate deep and advanced math or solve lot of problems and clear all the olympiads and examinations . ????

[color=green]Moderator says: there is never a good reason to block-quote the entire post immediately preceding your own.[/color]

I don't agree with your assertion that you would lose you imagination if you keep doing IMO problems. Most of the IMO problems require the true work of a mathematician, you begin to play around with the problem, then you make some hypotheses, you test them and look at the problems in new ways and then try other things until you either brute force an answer or you have a mini-eureka moment and solve the problem, this is true with most national and international mathematical Olympiads.

One of the most important qualities a mathematician can have is the ability to problem solve, intuition and dedication. Although its true that mathematicians have no time limit but its fairly important to solve a problem in a short amount of time, a mathematician that can solve a certain problem in 2 days that another mathematician can solve in 5 months will do better and be more notable. Take Euler, in 1 year he wrote 54 papers, which is about a paper a weak (and the majoring of his work was significant, not derivative restatements). Or, more contemporary, Terrence tao in a year publishes between 26 and 60 papers. Its their ability in problem solving that allows them to see this faster than other mathematicians.

Also, consider the fact that since the IMO has been running, of the 40 mathematicians that have recieved the fields medal, 8 of them have done the IMO. And in 2006, of the 4 recipients, 2 had been to the IMO! So its clear that the IMO is a key feature that gives a way for gifted high school students the possibility to use their imagination and solve difficult problems.

But i am intrigued by you idea of being given a "research" problem and finding a new proof or discovering something new from it. I dont think this would work. Consider how the IMO problems are strictly non calculus, this is because most high schools around the world do not go past elementary calculus, and as such cannot pose problems that require knowledge of a topic in a higher range such as lie algebra or algebraic topology. If they did construct such a competition that required knowledge of undergraduate mathematics for high school students it would only appeal to the geniuses which havent already started their degree. And so there would not be that many people who fit the requirements.

Although i am intrigued about an undergraduate competition where you are given a difficult research problem and have to finish first. That would be interesting.

[quote="srinath.r"]So what the imo organizers should do, is not a contest where you have to solve problems in a given time, but like find something new discover a new proof for a theorem etc."[/quote]

Welcome to the real world?

@kurt math
Actually those words were not mine ,I just copy pasted it from somewhere . Anyway thats immaterial
Even I disagree that doing problems and problems will lose your imagination skill . But I insist that there is no point in doing problem after problem after problem after problem. Usually people (there are exceptions) just solve problems and that  according to them that  becomes math for them . See it is like this
,"ok today take this X country sheet in the Yth year " after few hours "thank god I solved this sheet entirely and where is the Y+1th year's sheet "
A person liking math ,will not do like this . See it just becomes routine and monotonous and as I said he is one typical problem solving machine . When the mind is tuned in that way , they think only when some book or a problem demands ,say for example writing down powers of 2 and getting some nice conjecture may be interesting and also investigating the fibonacci numbers and pascal triangles ,but people who are typical problem solving machine will just think about them only when the religious problem solving book they use gives a problem demanding "Prove that nth fibonacci number is ............" or . That too he will leave the problem once he finishes solving it and go to the next problem without even extending the dynamics and boundary of the problem . If these are all there ,how can he become a research mathematician where self and automatic investigation is important .
I am not  telling that problem solving is a disadvantage for a mathematician . I am telling problem solving is not a pre-requisite ....problem solving helps but without it  still one can make a very good mathematician .

@Lezarus,sorry but I am not able to understand what you mean ?

I think you're suggesting that the sole purpose of math competitions is to prepare people to do research math.  I don't think this could be much farther from the truth.  I'm a freshman in college who spent a lot of time during high school doing olympiad math, and I don't think I'd be well-suited at all to doing research math.  But still, I enjoyed solving olympiad math problems and taking the olympiads themselves more than I can describe, and I value my experience with math competitions more than anything else I did academically in high school.  Olympiad math taught me how to solve problems.  It kind of taught me how to do math, but more importantly it taught me how to solve problems.  I'm definitely not planning to go into research math, and I'm not even planning to major in math, but the math contest experience was still incredibly valuable for me.

Maybe you'd still argue that the purpose of math competitions SHOULD be to prepare people to do research math.  But if this were true, then the target audience of math competitions would be much, much smaller than it is now, and other people like me wouldn't benefit from them.  The people who really want to prepare for research math can just do math research.  Problem solved.  :)

@matt eagles
If that 'you'  in your post refers 'me' ,I am sorry but I never meant olympiad mathematics trains people for research math . This statement is absolutely wrong as far as I am  concerned . But since you are in college now ,are you saying that having a very good flair in olympiad type problems ,you have an exceptional scope in research mathematics .?

@JBL ,thank you very much for giving the link ,very apt in fact  :clap2:

[quote="srinath.r"]I never meant olympiad mathematics trains people for research math . [i]This statement is absolutely wrong[/i] as far as I am  concerned .[/quote]
Hmmm... I would say that math. olympiads amount for about one half of my training as a researcher (the other half being university courses, books, etc.). I mean, not only I learnt a lot of useful ideas there earlier than I would see them anywhere else, but, most importantly, my brain acquired quite a lot of flexibility, which is really hard to develop later in life. I'm not at the top as a researcher, of course, but I fancy I'm pretty far from the bottom too. So, I felt like I should object to the italicised phrase.

In my humble opinion, when you are between 9 and 15 years old, nothing trains you better. The important part at that age range is not learning some "deep theories" but just learning to think properly.

The time constraint many people complain of is also a rather good thing really: you need to learn to think "fast" if you want to get anywhere. Fast thinking is not about some frantic efforts but about deep concentration and ability to cut the unpromising approaches at early stages, put the likely to succeed solution scheme together, gloss quickly over the parts that you are sure will work, find counterexamples to wrong steps, rearrange the scheme accordingly, etc., all done in real time and without losing control over your ideas. It is much more about being able to organize your thinking process efficiently than about anything else and one of the best ways to learn efficient organization is to do things under time pressure.

At last, olympiad math. teaches you to play the game fairly: you either get a solution or not in the end. No sidestepping, $ \varepsilon$-advances, and other things that are responsible for a huge amount of junk research are allowed. You learn to take the "partial progress" for what it really is: a failure that may possibly be turned into a success sometimes later.

Of course, the current MO format (which includes not only competitions themselves but also problem solving oriented math. circles, etc.) may not be the only possible way to train talented kids. Everybody is welcome to try something else if he thinks it'll work better. I just do not see much value in hypothetical considerations of what [i]might[/i] be better from an abstract standpoint. If you are really sure that you know how to do things properly, take a bunch of local kids interested in math. (very few places do not have them available for such experiments) and try your approach on them. If you manage to keep them in your class for more than one day, you have something to share with others. If not, then your ideas went wrong somewhere. So far, the overwhelming majority of successful experiments I know of have had solving olympiad type problems as a baseline (though other ingredients were often present as well).

As to ITYM, I wish the organizers good luck but until they get their own full pyramid of similar research type math. competitions, research-oriented math. clubs, etc. at all levels, I would be very reluctant to declare the project a success.

@fedja ,You have got a point . It is mainly to do problems and be quite familar solving olympiad problems (having some tricks under the sleeves) ,so how does it matter when (age) you try problems . You said that learning deep theories at age between 9-15 cant train you well or solving problems and problems only will train you well at that age . This may be wrong ,what if we go about a theory centered problem solving approach, there is no point in absolutely learning just theory ,what if we think deeply about those things ,frame problems and try solving them .This is also a kind of "problem solving" right ,where you use the theory to frame problems and try solving them .This is also something like a research . Why cant young people just pick some stuffs from air and start working on them . The main point is to keep yourself touch with problem solving right ? So how does it matter when and where we solve problems and it doesnt matter from which 'olympiad book' we solve a problem  ,we are still solving problems .Framing theories and problems will develop a more creative thinking isnt it ?Because most people are mad after olympiad problems and they solve lots and lots of problems and they think about some beautiful things only when a problem or statement or a book  demands it,this hinders creativity and research thinking isnt it ? I think both learning theories and some good familiarity with problem solving even at a young age is important and essential for a research mathematician .
What do you say on this?

[quote="fedja"]At last, olympiad math. teaches you to play the game fairly: you either get a solution or not in the end. No sidestepping, $ \varepsilon$-advances, and other things that are responsible for a huge amount of junk research are allowed. You learn to take the "partial progress" for what it really is: a failure that may possibly be turned into a success sometimes later.[/quote]
I actually see this as a drawback of olympiad math :  it's too open-and-shut.  Many "problems" in the works right now (e.g., the problem of classifying finite groups) are too monumental to expect anything but partial progress in a lifetime.

On the other hand, I don't see the time limit as a bad thing necessarily either.  After all, each of us only gets a finite amount of time in this life to do things. ;)

I see olympiad math as a mathematical analogue to hard technical exercises in the musical world.  In music, technical exercises help for "real" music (large-scale works), but there's much more to the big works, and you don't have to perfect the hardest etudes in order to play concert pieces well.  The exercises generally lack the "vision" and scope of "real" pieces (in the way that olympiad problems often lack the grand purpose of unified theories), but they have their moments, and some of them are [url=http://www.youtube.com/watch?v=939Yl9w9LBs]quite beautiful[/url] in their own right.  Some of the etudes do teach certain specific techniques, but more than that, they teach you how to learn technically difficult music.  It's not a perfect analogy (I think etudes are more necessary for musicians than olympiads are for mathematicians), but it has some merit.

Math contest jingoists claim that contests are invaluable because they improve problem solving skills. This is tautologically true, but no one can know how well these math contest skills transfer to other problems, and consequently we cannot know the actual value of this skill set relative to the value of doing some other educational activity. Opinions on this issue break down something like this: Adults who enjoyed math contests and succeeded in them are more likely to think that these skills are valuable, while successful adults who did not have such experiences are more likely to see these skills as unimportant.  (It's only human nature. The same goes for other questions such as, "Does learning music make you better at math, or vice versa," or, "Do video game skills help cognitive development, or are they totally useless?" Anyone who thinks highly of themselves will approve of most aspects of the way they were brought up.)

[quote="yenlee"]no one can know how well these math contest skills transfer to other problems, and consequently we cannot know the actual value of this skill set relative to the value of doing some other educational activity.[/quote]
I choose to humbly disagree if by "other problems" you mean "other mathematical problems". Of course, the ability to solve mathematical problems may have nothing to do with the ability to solve other types of problems, but within mathematics, the correlation seems to be quite strong. Also, whatever mathematical educational activity you choose, it will include solving some kind of problems on your own. The big advantage of contest problems is that they do not require learning some complicated concepts to just be able to understand what is being asked. You learn to think in abstract way about already familiar objects. Then, when you start learning abstract concepts, you'll just discover that you are just being told the names that you were missing all the time before.

I still remember how I was first told the $ \varepsilon\minus{}\delta$ definition of continuity during my first year at the university. My reaction was "But, yes, of course, that is what it should be! How stupid of me not to figure it out earlier by myself!". The reason was, of course that I already had no trouble with quantifiers and had quite a clear idea of what a continuous function should be and how it should and should not behave. Where did it come from? It came from solving numerous problems when the solution was that some integer quantity could change only by $ 1$ at each step and the point was to show that it was $ 0$ somewhere if it was negative in the beginning and positive at the end. All I needed to do now was to replace the word "minimum" by the word "infimum" and all the theory of continuity just clicked in place effortlessly. Seeing how my calculus students struggle with the very same definition now and how my colleagues discuss whether to introduce it at all in low-level classes, I cannot help seeing quite a lot of value in olympiad problem solving for university education.

Since yenlee talks about not absolute value, but "value relative to other activities", I just want to humbly ask "What activities?". In my case, as a child, I spent quite a lot of time reading magazines like Kvant but I still viewed the "theoretical" articles there as explanations of solutions to some problems I could understand. If I failed to understand the question the article was devoted to, I usually skipped over it. If I understood the question, I usually spent some time trying to figure out the answer myself and then compared my thoughts to what was written. So, again, it was "problem solving" rather than "concept learning".

In the 10th grade I was shown some theoretical books like van-der-Vaerden's "Algebra" but I still strongly preferred to read something like Khinchin's "Three pearls of number theory". Later I found some balance between learning "theory" and learning "theorems" but I should confess that I still prefer a good theorem to a good theory and believe that a theory is just nothing but a huge tree of theorems and tools for proving them, that its only value comes from the wide range of problems it allows to solve, and that the best contribution you can make to it is to solve a question in it that was unsolved by a method that has never been used before (in other words, to make a really important contribution to a theory, you need to break its established rules and pathways of thinking and to do something that goes against the current, something for which the knowledge of the current theory is useful only to understand what is asked but not to arrive at the answer)

[quote]Anyone who thinks highly of themselves will approve of most aspects of the way they were brought up.[/quote]
100% agree. So, when somebody tells you what his, say, mathematical, upbringing was and recommends you to follow a similar path, the first thing you should do is to evaluate the result that is right in front of your eyes, and then to decide whether you would like to get to the same point or to avoid that point by all costs. In other words, if you are not an expert yourself, you'd better somehow make your own judgement about the speaker's competence in the subject before you listen to what he says about it. I just thought that such things go without mentioning :).

[quote]what if we go about a theory centered problem solving approach, there is no point in absolutely learning just theory ,what if we think deeply about those things ,frame problems and try solving them[/quote]
The problem with "learning theory" and "thinking about it deeply" is that it pre-conditions your thinking. If it were possible to create some theory that embraces all mathematics and teach it to schoolchildren, I would see nothing wrong with that but I do not see how it is feasible. Narrow specialization is one of the worst nightmares in modern mathematics and to introduce it early is the surest way to design a narrowly-thinking brain. This is, of course, something we can argue about but I believe that before the age of 15 or so it is more useful to go broad in education than to go deep. Olympiad problems have enough versatility. They do not give you a deep knowledge of any particular subject but they allow you to learn the "language" of all of them (not the buzzwords, of course, but most of the elementary moves and constructions).

Any particular theory is very restricted. I remember that in the 10th grade I went to a seminar in Geometry of Numbers. It was fascinating in the beginning but after the first 5 lectures, it became very repetitive: there was one main idea (Minkowski theorem) and we just applied it to everything that we could and tried to find the critical determinant of whatever star-shaped body was there. Interestingly enough, it all resulted in my first research paper (on the local minimality of the Reinhardt octagon) but the key to the solution was an area formula that I learnt a year before when solving an olympiad problem in geometry.

@fedja thank you very much for the detailed post.
See,by me telling "learning theories " ,I do not mean that spend your time entirely just by book reading and knowing some 1000 facts . You said that learning theories and thinking about it deeply pre-conditions your thinking. But I have one question to ask ,for example imagine a situation like this ,you are interested in learning some abstract algebra and you pick some book and start reading it .You  learn what is a group ,next what I am telling is once you are introduced to this why not think deeply about it "Oh,this is a group ,so what relation and  what significance this has " . How does it limit your thoughts .What you say is correct if I just get the book -see definitons ,see lemmas,see proofs and if I am not putting any own effort apart from understanding it . By seeing some definitions and getting some insights how does it limit my thinking if I am thinking about it deeply ?By you telling me to do more problems when I am young without learning theories as doing just the latter will limits my thoughts ,I can invent group theory? It is practically not possible for everyone who keeps on trying problems after problems to  reinvent the entire mathematics . At one point of time you will be introduced or learn some theory and it can be at any time.Concept learning and problem solving are related when the concept is within the scope of a particular problem ,so I am telling once you sit on a problem and you manage to solve it and if it is interesting why not go deep about it -explore somethings and look for resources which have already dealt with that concept and improve yourself? This definitely contributes for becoming a research mathematician right ? I am mainly telling there is no point in solving 1000 problems only one after an other.
Please forgive if I have misunderstood what you mean to say.
Unrelatedly ,you had known so much compared to me when you were in 10th ( I am in 10th now :) ),maybe your education system is very good that way .In our country education system is horrible and they teach fundamental theorem of arithmetic in school only in 10th .

[quote="fedja"]I should confess that I still prefer a good theorem to a good theory and believe that a theory is just nothing but a huge tree of theorems and tools for proving them[/quote]
Hm . . . this sounds to me like, "A novel is nothing but a huge string of sentences, cleverly arranged."  I think that often theories have some sort of unifying theme.  I agree that ultimately a theory should solve large numbers of problems (or at least a few difficult problems), but I like to think that usually there is some sort of larger focus than individual problems, and sometimes the focus of the theory is most naturally stated in a way that does not involve "problems" explicitly.  For instance, a main idea of category theory seems to be, "Study classes of objects by focusing on their morphisms (structure-preserving functions)."  This isn't a "problem" really, but it clarifies a lot of similarities between different areas of math and, at its best, saves a lot of effort.  (For instance, if you take the snake lemma for modules and turn it into the more general form about objects in an arbitrary abelian category, your proof becomes about twice as short because of duality.)

[quote]The problem with "learning theory" and "thinking about it deeply" is that it pre-conditions your thinking.[/quote]
Is this necessarily a bad thing?  After all, many theories took generations' worth of the best efforts of brilliant mathematicians to reach their present state.  Might it not be good to consider the approaches of theirs that seem to have been the most fruitful?  Besides, couldn't you say that olympiad problems pre-condition you to think in terms of small steps achieved by "elementary" means (i.e., methods that disregard possibly useful existing knowledge about the subject)?

I actually think of olympiad problems as things that go pretty deep into a subject called "contest math" without getting a very broad scope of mathematics. :)   There is a specific vocabulary of inequalities and an array of geometric techniques, and an set of commonly used combinatorial things, but vast areas of mathematics stay untouched---algebra is essentially confined to elementary number theory and to clever manipulation of real or complex functions, analysis is usually avoided except for the inequalities I mentioned, and "geometry" for the most part exists only in a very strict, Euclidean sense of the word.

I found that my olympiad training helped me when I started learning theories, too.  But I feel like I merely got low-level, technical fluency, and not much insight into how areas of math "work" and interact on larger scales.  (Of course, the low-level fluency is in some ways a prerequisite for broader knowledge.)

To bring in my tuppence of wisdom - how about the sheer satisfaction of getting a problem right? Nowadays everybody is so very much result-oriented and questions everything on the basis of the value-added and return-on-investment, that we forget the essence of just being there and able-bodied and able-minded to enjoy an abstract challenge. I direct you to (re)read Herman Hesse's "Magister Ludi or The Glass Bead Game". It is also true that I am somewhat of an eccentric, so I do not want to seem prone on proselytizing ...

[quote="Boy Soprano II"]
Is this necessarily a bad thing?  After all, many theories took generations' worth of the best efforts of brilliant mathematicians to reach their present state.  Might it not be good to consider the approaches of theirs that seem to have been the most fruitful?  Besides, couldn't you say that olympiad problems pre-condition you to think in terms of small steps achieved by "elementary" means (i.e., methods that disregard possibly useful existing knowledge about the subject)?[/quote]
Very much true ,I agree 100%  :lol: . This is the problem with people who just keep on solving hundreds of problems. Once again I say ,they think only when a problem demands and they do a problem only if there is  an exam (They say like this ,oh god I have to clear this exam and make it to the team this year ...thats their motive)
[quote]
I actually think of olympiad problems as things that go pretty deep into a subject called "contest math" without getting a very broad scope of mathematics. :)   There is a specific vocabulary of inequalities and an array of geometric techniques, and an set of commonly used combinatorial things, but vast areas of mathematics stay untouched---algebra is essentially confined to elementary number theory and to clever manipulation of real or complex functions, analysis is usually avoided except for the inequalities I mentioned, and "geometry" for the most part exists only in a very strict, Euclidean sense of the word.[/quote]
Even this is correct (atleast to me )

So Boy Soprano and Fedja what are your final conclusions ,I seem to have forgot whats the result :D  :rotfl:


summer program
Mathcamp
AMC
USA(J)MO
USAMO

Hi everyone,

The application deadline for Mathcamp 2009 has been [b]extended[/b] to this [b]Monday[/b], April 27th, at midnight CDT.  (That is, by the end of Monday.)  This is to help accommodate anyone who received a late brochure.  Good luck!

[quote="crazypianist1116"]Will all applications up to that time be considered equally, or will applications received before that time be considered first? Also, when are recommendations due?[/quote]

look on their webpage

It says on the website that "The application deadline has been extended to 11:59pm CDT on Monday, April 27, 2009. (All applications received by that time will receive full consideration.)". Does this mean that applications received after that date and time will receive some consideration? How much consideration?  :oops:

I think if they have spots open after the applications from before the deadline then they consider late applications? I think you should trust DanZ more than me, but... also, you're applying to Mathcamp too? :D I wonder if you remember me >_> I have the same name as you, and I went to AMSP too!

whoa. leoxnlin you're applying too!!! sweet! =]]

[quote="DanZ"]All applications received up through Monday night will indeed be considered equally!

As to late applications: we do accept them.  We set up a waitlist following admissions, and late applications are competing for spots on that waitlist.[/quote]

This year I'm traveling - but want to apply next year! Where's next year's camp going to be?

[quote="crazypianist1116"]Timezones get me confused. Are there 4 more hours or 3 more hours?[/quote]

You  doing it last minute too?
Which time zone do you live in, eastern i assume?

Haha, yeah isn't everybody? And yes, I do live in Eastern.

I remember reading somewhere it is due at 11:59 p.m. central time
I think that means 12:59 a.m. eastern time.

Yep yep, I was finishing mine at the last minute too. 10 minutes or so 'til deadline  :D

lol. same boat as you Zeph... turned in.... 11:58 cdt. -.- scary. scanner was wayyyy too slow. ><

approximately when will we know whether we've gotten in or not?

Last year, notices were emailed exactly two weeks after the application due date.

Are we free to discuss the qualification quiz problems now that the deadline has passed?

Noooo. At least judging from other years, no. Mathcamp often gets late applications (I think they go on the waitlist, and I know people who have gotten in from a late application. For example, my brother, who finished the application on time but didn't realize there was a special "submit" button :P ...) So don't discuss the QQ problems for a while, some people might still be working on them.

Just to reply to a couple of remaining questions:

1) Yes, please don't talk about solutions yet --- we do want to wait until this year's camp is totally filled so that those doing late applications don't get any advantage.

2) The camp moves each year and we're not sure where we'll be yet for certain.  If we stick to our usual schedule, we'll be at Colby College in Waterville, ME, which is an absolutely fantastic campus.

Congratulations to everyone who got their applications in on-time --- if barely so! --- and good luck in the process.  We'll be grading pretty intensely and will try to get responses out ASAP.

This is terrible... I just submitted my application. :oops: USAMO shall be my excuse. I wonder what my chances are of getting in.

[quote="leoxnlin"]USAMO shall be my excuse.[/quote]

Pssst, don't tell Dan that!


parameterization
function
logarithms
Galois Theory
real analysis
real analysis unsolved

1. solve  $ \frac {x}{y} \equal{} \frac {logy}{logx}$
2. solve  $ \frac {x}{logx} \equal{} y$
which means, represent x,y in a parameter "t", and both x(t),y(t) shall be elementary function, or else show it's impossible.

3. solve y'(x)=y(ax) (y can be viewed as generalized function. y shall be as smooth as possible. )

I do not really understand #2: what's wtong with the trivial solution $ x\equal{}t>0$, $ y\equal{}\frac t{\log t}$?

In #1 there is a trivial solution $ x\equal{}y\equal{}t>0$ but there is also another one, which I suspect to be "non-elementary" but I have no idea how to prove non-elementarity in such a strong sense.

Now, for #3, the power series method is your friend if $ |a|\le 1$. Indeed, if $ y\equal{}\sum_{k\ge 0}c_k x^k$, then the equation is equivalent to $ kc_k\equal{}a^{k\minus{}1}c_{k\minus{}1}$ from where it is not hard to determine all the coefficients given $ c_0$. The classical proof of Picard's existence and uniqueness theorem shows that those are the only continuously differentiable solutions defined everywhere in this case. When $ |a|>1$, I believe that the only continuously differentiable solution defined of the entire real line is $ y\equal{}0$. This seems like a decent problem, indeed (neither trivial, nor unsolvable).

The last remark: please, post [color=red][b]one[/b][/color] problem per thread (or, at most, a couple of related questions). In this case #1 and #2 belong more or less to the same category, but #3 has nothing to do with any of them and should be posted in a separate thread.

For #1, I think there is a "well-known" parameterization $ x \equal{} t^{t/(1\minus{}t)}$ and $ y \equal{} t^{1/ (1 \minus{} t)}$.  Or something like that.  One trick to derive it is to set $ t \equal{} y/x$ and then get rid of $ y$.

You are right, Ravi. :) So, we are now left with #3 for $ |a|>1$.

sorry for the ambiguous #2. It means to present x by an elementary function of y, that is, the inverse function. I think it dosen't exist, if it has to be elementary.

the #3 may can be solved by Fourier transform.

thanks for your answer.

It's known that Lambert's W function is not elementary.  From there, we can show that the solution to #2 for $ x$ is not an elementary function of $ y$.

Can you give some explanation how to prove this "W" isn't elementary?

See the link I referenced in the following thread:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=186268[/url].

Haven't thought this topic has been studied so much. But still, it seems no one ever considered composing. Since, once composing is allowed, one has to concern the specific propert of a particular class of functions, thus the theory will no longer be "context free"——free to function's "own"  properties.

It's been a long time that i try to prove y''+p(x)y'+q(x)y=0 has no solution on extension of adding into the function "field" step by step solution of diff eq as y'+u(x)y+v(x)=0, which means the "first“ u(x) and v(x) is solutions of y'+P(x)y+Q=0 where P and Q are "usual" (elementary) function of p and q. Is it possible to prove it (or more, even travial) via so-called differential Galois theory? (if composing is not "allowed")

By the way, maybe it's in the same way to prove Gamma function isn't elementary, but what about ques: is there an elementary function coinside with Gamma function on every integer points? Any one has some ideas?


the tissues of a leaf from the upper to the lower surface are:

1 epidermis, mesophyll layer, epidermis,

2 cuticle, epidermis, spongy mesophyll, palisade mesophyll, epidermis, cuticle

3 epiderrmis, palisade layer, spongy layer, epidermis

4 cuticle, epidermis, palisade mesophyll, spongy mesophyll, epidermis, cuticle

5 all are wrong

please i need some help, in different texts books i find different answers, soes someone has a clue?????

i would say that the 4the one is correct

the cuticle and epidermis are on both sides...

the palisade mesophyll cells must be on top, since they are the cells that perform most photosynthesis, and light is mostly on top of the leaf.
the spongy mesophyll cells must be on the bottom because of the stomata guard cells which are mostly on the bottom.


[quote="joejia"]What is the vlaue of $(2^3)^{\frac{4}{3}}$?[/quote]

  Is there really any point in posting excersies rather than actual problems.

Remember this is Basics forum, if you think these are too easy, don't do them, let others learn, and you should move on ;)

However, other students learn when they can study solutions of other problem solvers, besides that the basics forum is probably the most active forum right now, and has the most problems to solve.

I think you are looking for more challenging problems. Can you try MATHCOUNTS-High School forums? From most of the solutions I see you post, you are more advanced than most MATHCOUNTS---Getting Started level forums, just like me :) I think [b]joejia[/b] is doing very good job because everyone start with easy problems. As they go along, they can advance and try more challenging problems.

[quote="G-UNIT"]lol, by mathcounts-highschool, i think you mean "getting started", and mathcounts getting started, "classroom math", but the real reason i think mose people are here, that includes you and I, is beacause the mathcounts forum has been rather inactive lately and the only new problems can be found in these forums, the "less challenging ones".[/quote]
Haha, no need to worry, I got message from RCV:

[size=75][color=darkblue]<Moderator RCV's note: 10000th:  Please read the post titled "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=148301#p148301]Using this Forum[/url]".  Quoting Mr. Rusczyk:  "The Basics forum is for [b]middle school students[/b] who are [b]just beginning[/b] with math contests.">[/color][/size]

This is my last post in The Basics. Bye  :bye:


induction
function
algebra proposed
algebra

Let $f$ be a solution.
Notice that for each $n$ we then have $f(f(n)) \leq \max\{n,f(n)\}$, with equality if and only if $f(n)=n$.
Let $f^k$ denote the $k^{th}$ iterate of $f$.

Suppose that there exists $a$ such that $a > f(a).$
Then $f^2(a) < a$ and by an easy induction $f^k(a) < a$ for all $k>0$ (1).
Since there is a finite number of integers in $\{1,2,...,a-1\}$ it follows that there exist $0<i<j$ such that $f^i(a) = f^j(a) = f^i(f^{j-i}(a))$ and using injectivity of $f$ it leads to $f^{j-i}(a) = a$ which contradicts (1).

Thus, $a \leq f(a)$ for each $a>0$.
Then $f(a) \leq f(f(a))$ for each $a$.
On the other hand, from the initial remark we also have $f(f(a)) \leq f(a)$.
Thus $f(a) = f(f(a))$ for each $a$, and using injectivity of $f$ we deduce that $f(a) = a$.

Conversely, it is easy to verify that $f(x) = x$ is indeed a solution.

Pierre.

Can you spare us some easy questions, pbornsztein?  :D  :D  :D
BTW, very nice solution (and typeset).

Sorry.....But I left two days. :D

Pierre.

I also solved it with your solution in 45 minutes

As injective so either monotonically increasing or decreasing.If it is a decreasing function then it makes a contradiction as suppose $f(a)=t$ then $f(a+t-1)=1$ and after that it gets no value.So must be a monotonic increasing function.
So,$f(1)\geq 1$ and likewise $f(n)\geq n$.
Now see $f(n)\geq \frac{n+f(n)}{2}\geq f(f(n))\geq f(n)$ so,all are equal which gives $f(n)=n$.

No, injectivity doesn't mean monotonicity. The function $f(1) = 2, f(2) = 1$ and $f(x) = x$ otherwise is injective (bijective actually) but not monotonic.

My first solution is same as pbornsztein's solution

2nd solution : Theorem :
If $a_n$ be a sequence of non negative numbers such that $2a_{n+1}\leq a_n+a_{n-1}$ then sequence is bounded.
Now choosing $a_k=f^k(n)$ we are done.

But your theorem is almost precisely what pbornsztein has done. If $2a_{n+1} \leq a_n + a_{n-1}$, then $a_{n+1} \leq \max\{a_n, a_{n-1}\}$. And since then clearly $\max\{a_{n+1}, a_{n}\} \leq \max\{\max\{a_n, a_{n-1}\}, a_{n}\} \leq \max\{a_n, a_{n-1}\}$, we're done.

[hide="solution"]
Note that $f(f(n))\le \dfrac{n+f(n)}{2}\le \text{max}\{n, f(n)\}$. Similarly, $f(f(f(n)))\le \text{max}\{f(n), f(f(n))\}\le \text{max}\{n, f(n)\}$, and it is easy to apply induction to prove $f^k(n)\le \text{max}\{n, f(n)\}$. Thus, the sequence $n, f(n), f(f(n)), \ldots$ is a bounded function, and therefore eventually periodic. Now suppose that the fundamental period $p$ of this sequence is greater than $1$. Then for big enough $n_0$, for all $k> n_0$, $f^{k+p}(n)=f^k(n)$. Now consider $\max\limits_{1\le i \le p} f^{k+i}(n)=f^{k+i_0}(n)$ for some $1\le i_0\le p $; we have $2f^{k+i_0}(n) > f^{k+i_0-1}(n)+f^{k+i_0-2}(n)\implies f^{k+i_0}(n) > \dfrac{ f^{k+i_0-1}(n)+f^{k+i_0-2}(n)}{2}$ and the inequality is strict because $p\ge 2$; but this gives a contradiction with the problem condition. Thus, eventually $f^k(n)$ becomes constant. Thus, for $k\ge n_0$ for big enough $n_0$, $f^{k+1}(n)=f^k(n)$. However, since $f$ is bijective, it follows that $f(n)=n$ which is our only solution. [/hide]

[quote=Valentin Vornicu]Find all one-to-one mappings $f:\mathbb{N}\to\mathbb{N}$ such that for all positive integers $n$ the following relation holds:
\[ f(f(n)) \leq \frac {n+f(n)} 2 . \][/quote]

 We claim $f(n) = n$ are the only solutions. It is enough to show that $f(n) \leq n \quad \forall n \in \mathbb{N}$. Suppose $f(n) < n$ for some $n \in \mathbb{N}$.
Note that by induction $f^k(n) < f(n) \quad \forall n \in \mathbb{N}$.
Let $(a_k)_{k \geq 0}$ be a sequence defined by $a_k = f^k(n) \quad \forall k \in \mathbb{N} , \quad a_0 = n$. Note that the sequence is bounded and discrete. So, there exist $k, l \in \mathbb{N}$ such that $a_k = a_l$. By the injectivity of $f$, we have $a_{k-l} = a_0$ which implies that $a_{k-l-1} = a_1$ which leads to a contradiction. $\blacksquare$

[hide=Nope, unfortunately] I think this one can be solved without using injectivity.

Fix $n$ and let $a_m = f(f(\cdots f(n) \cdots))$ (the number of iterations of $f$ is $m$, treating $a_0 = n$) for $m\geq 0$. Substituting $n$ with $a_m$ in the given inequality, we get $a_{m+2} \leq \frac{a_{m+1} + a_m}{2}$, i.e. $2a_{m+2} \leq a_{m+1} + a_m$. If this was an equality we would be able to solve this linear recurrence - this motivates to try to prove (by strong induction on $m\geq 0$) the inequality $a_m \leq \frac{a_0+2a_1}{3} + \frac{2}{3}(a_0-a_1)(-\frac{1}{2})^m$.

For $m=0$ and $m=1$ equality holds. Now if we assume that the claim holds for all indices not exceeding some $m\geq 1$, then
$$a_{m+1} \leq \frac{a_{m} + a_{m-1}}{2} \leq \frac{a_0+2a_1}{3} + \frac{1}{3}(a_0-a_1)\left(-\frac{1}{2}\right)^{m-1}\left(1 + \left(-\frac{1}{2}\right)\right) = \frac{a_0+2a_1}{3} + \frac{2}{3}(a_0-a_1)\left(-\frac{1}{2}\right)^{m+1} $$
which completes the induction.

On the other hand,  $a_{m} > 0$ and so $0 < \frac{a_0+2a_1}{3} + \frac{2}{3}(a_0-a_1)\left(-\frac{1}{2}\right)^{m}$, i.e. $2(a_1-a_0)(-\frac{1}{2})^m < a_0 + 2a_1$. Pick $m$ to be odd, then (as $a_0 = n > 0$ and $a_1 = f(n) > 0$) after multiplying with $(-2)^m = -2^m < 0$ we get $2a_1 - 2a_0 > -2^ma_0 - 2^{m+1}a_1 \Leftrightarrow \frac{a_1}{a_0} > \frac{2^m+2}{2^{m+1}+2}$. The right-hand side of the latter can attain values arbitrarily close to $1$ and so choosing $m$ to be sufficiently large yields $a_1 \geq a_0$, i.e. $f(n) \geq n$.

Hence $f(n) \geq n$ for all $n$. Now it is obvious how to finish. We deduce $f(f(n)) \geq f(n)$ and now the given condition implies $f(n) \leq \frac{n+f(n)}{2}$, i.e. $f(n) \leq n$. Therefore $f(n) = n$ and direct verification shows that this function works. [/hide]


All of my notes happen to say different things, so I'd appreciate some help on this: do centrioles replicate during cytokinesis or during the telophase?

Centrioles replicate before a cell divides, so I guess during interphase, which could be either S or G2 growth stages, but definitely before prophase of mitosis and way before cytokinesis or telophase.

Centrioles replicate JUST before mitosis at the end of the G2 stage.

They replicate in interphase, when replication occurs..


calculus
geometry
3D geometry
derivative
integration
Gauss
calculus computations

I'm new to calculus (just spent today learning it). Looking at the other topics in this forum like differential equations which is so much more advanced than what I'm asking, I'm really not sure whether this topic is in the right place.

Here's a few problems I couldn't figure out:

[hide="1"] A cube is expanding in such a way that its sides are changing at a rate of 2 cm/s. Find the rate of change of the total surface area when its volume is 125 cubic centimetres. [/hide]

[hide="2"] A spherical ball is being inflated at a rate of 20 cubic centimetres per second. Find the rate of increase of its radius when its surface area is $ 100 \pi$ square centimetres. [/hide]

[hide="3"] Variables $ x$ and $ y$ are related by the equation $ 2xy \equal{} 195$. If $ x$ is increasing at the rate of $ \displaystyle{\frac {3}{2}}$ units per second, find the rate of change of $ y$ when $ x \equal{} 15$ units. [/hide]

1) Call the sides $ x$.  Then the volume at any time is $ V=x^{3}$ and the surface area is $ S=6x^{2}$.  We are given that $ \frac{dx}{dt}=2\mbox{cm}/\mbox{s}$.  We are asked to find how fast the surface area is changing when $ V=125\mbox{cm^{3}}$.  Then
$ \frac{dS}{dt}=12x\frac{dx}{dt}=12x(2)=24x$
When the volume is 125, the $ x=5$ so
$ \frac{dS}{dt}=24\times 5=120\mbox{cm^{2}/s}$

[quote="JRav"][hide]1) Call the sides $ x$.  Then the volume at any time is $ V = x^{3}$ and the surface area is $ S = 6x^{2}$.  We are given that $ \frac {dx}{dt} = 2\mbox{cm}/\mbox{s}$.  We are asked to find how fast the surface area is changing when $ V = 125\mbox{cm^{3}}$.  Then
$ \frac {dS}{dt} = 12x\frac {dx}{dt} = 12x(2) = 24x$
When the volume is 125, the $ x = 5$ so
$ \frac {dS}{dt} = 24\times 5 = 120\mbox{cm^{2}/s}$[/hide][/quote]

Oh, I made an error by saying that $ \frac {dS}{dt} = 6x^2\frac {dx}{dt}$ (or a similar error) by forgetting to differentiate $ 6x^2$ with respect to $ x$.

How is the surface area of a solid in three-dimensional space related to its volume, if it even is? I noted that $ \frac {dV}{dt} = 3x^2$, which is apparently half of $ 6x^2$. Could it be a coincidence?

Well for a circle with radius $ r$, the area equals the derivative of the circumference/perimeter with respect to $ r$. Is there any similar relationship with other polygons? (Prove it)

Surface area is computed using surface integrals, and volume by triple integrals.  Surface integrals and triple integrals are connected using Gauss' Divergence Theorem.


So if you want to change $ 111101011_2$ into base 10, would it be $ 2^8\plus{}2^7\plus{}2^6\plus{}2^5\plus{}2^3\plus{}2^1\plus{}2^0$ right??

And if you want to change 234 into base 2, you start off with finding the largest number, (i.e $ 2^7$, and go from there, right??

Can anyone post some questions about base?

AHHH, I'M SO SORRY!

Here are some basic practice questions.

Change these to base 10

11100_2
1001001_2
102_3
423_6
2301_4
34_7
1006_8

Change these base 10 numbers

20 to base 2
381 to base 2
20 to base 4
381 to base 4
129 to base 3
999 to base 3
2500 to base 8

Check your work on some of them.

You make some error, and I think, "He still doesn't get it," but then you don't make the error in the next problem and get it right. So it must be carelessness. That's what's killing you.

I'm not telling you which ones are wrong because I want you to look yourself.

[quote="Chinaboy"]Check your work on some of them.

You make some error, and I think, "He still doesn't get it," but then you don't make the error in the next problem and get it right. So it must be carelessness. That's what's killing you.

[/quote]

So true... I am ashamed...

I caught 2 of them..so far.
I'll try to look out for some more if there are more.

$ 11100_2\equal{} \ 2^4\plus{}2^3\plus{}2^2\equal{}\boxed{28}$

$ 102_3 \equal{} \ 3^2\plus{}2(3^0)\equal{}\boxed{11}$

I only caught one more..

129 to base 3 $ \rightarrow 3^4 \plus{} 3^3 \plus{} 2(3^2) \plus{} 3^1 \equal{} \boxed{11210}$


limit
function
real analysis
real analysis unsolved

Let [tex]f:I\to I[/tex], where [tex]I\subset\mathbb{R}[/tex] is an interval, such that [tex]f[/tex] is continous and for all [tex]x\in I[/tex] there exists [tex]n_x\in\mathbb{N}^{*}[/tex] such that [tex]f^{n_x}(x)=x[/tex].

a) Prove that if [tex]I=[a,\infty)[/tex], then [tex]f(x)=x[/tex], for all [tex]x\in I[/tex].

b) Prove that if [tex]f\neq 1_I[/tex], then for [tex]I=(a,\infty)[/tex], we have  [tex]\lim_{x\to\infty}f(x)=a[/tex].

I saw this before. It was proposed for the Romanian NMO, but I never knew one thing: is $f^{n_x}$ supposed to be a power or a composition of functions? Sorry if it's a stupid question (which I'm sure is the case :)).

Well, it's not hard to show that $f$ is injective, since we can partition $\mathbb R$ in cycles. We construct an equivalence relation $\equiv$ on $\mathbb R$, where $x\equiv y\iff$ there is an $n_x(y)$ s.t. $f^{n_x(y)}(x)=y$. The cycles I'm referring to are the equivalence classes corresponding to this relation.

$f$ is thus strictly monotonic. In the first case we must have $f(a)=a\Rightarrow f$ is strictly increasing and from here the conclusion easily follows. In the second case, if $f\ne 1_I$, then $f$ must be strictly decreasing, and, again, we get the conclusion.


Suppose that by flipping fair coin nine times,  we have head "H" . What is the probability of having tail "T" next time?
(

[Edit] I didn't know you were asking for help.

Since each case is independent, the past nine flips don't matter. Therefore, the last flip has 1/2 chance for head or tail.

just 50%

because the to see the result, you don't need the previous coins

Excellent answers ...
Key word here is[b]  fair[/b] coin .. and  this by [b]definition[/b] means that the probablity is 50%  of getting a head in [i]any [/i] throw.

I was freaking out when I first saw the problem because I was wondering how you would do a problem like that, but after looking at everyone's post, I agree that it's 1/2

[hide="the answer is one-half"]$\frac{1}{2}$[/hide]

1/2 of course! ha ha! That's funny!

(ahem)

Anyway, the result of the heads before doesn't affect the next toss.

nonono
50% if will be 50%
the other 50% is that 25%

yeah, unless the coin can see into its own past and then change its flip to even things out, then 1/2. But watch out, coins are sneaky, what if it lands on a edge?

what if it doesn't? ........ :D  :lol:  :P

i doubt that will ever happen.  it happened in some old movie though.  i forget what it was called.  it was about these two guys who live in a house with a really smart rat.

no it was the movie Mouse Hunt <- funny movie

actually my old science teacher had someone do it once in her class
she took like 3 pictures of it and made this big deal about it

o yeah and the probability is [hide]1/2[/hide]

hey dinoman that's the movie i was talking about!
thanks! :D

This discussion is going to nowhere involving no math...

try to avoid this kind of discussions

Because each flip is an independent clause, the probablilty is 1/2 on each throw!

yes, now that we know the next flip is 1/2 chance, here's a little twist:
imagine, we went back to before the first flip was made. what is the probability that all ten flips are heads?

[hide] 1/2 [/hide]

The twist: [hide] 1/2^10 or 1/ 1024 [/hide]


A uniform hoop rolls without slipping down a $ 15^{\circ}$ inclined plane. What is the acceleration of the hoop's center of mass?

My work:
$ a_{CM}\equal{}r\alpha$
$ \equal{}\frac{r\tau_{net}}{I}$
$ \equal{}\frac{r*r_{perpendicular}*F}{mr^{2}}$
$ \equal{}\frac{r^{2}(\cos 15)(mg\sin 15)}{mr^{2}}$
$ \equal{}2.45 m \slash s^{2}$

Is this correct?

Er, looking back and thinking again, is the correct answer 9.8 m/s^2?

since it is pure rolling no work is done by firction so
By Conservation of energy let it move a distance $ \text{x}$ along the incline then we have
$ \text{mgx}\sin \theta \equal{} \frac {\text{mv}^{2}}{2} \plus{} \frac {\text{I}\omega^{2}}{2}$
differentiating both sides with time we get
$ \text{mgv}\sin \theta \equal{} \text{mva}_{cm} \plus{} \text{I}\frac {\text{va}_{cm}}{\text{R}^{2}}$
thus we have
$ \boxed{\text{a}_{cm} \equal{} \frac {\text{g}\sin\theta}{1 \plus{} \frac {\text{I}}{\text{mR}^{2}}}}$
note this holds for any object pure rolling down an incline of inclination $ \theta$


for the last one, is n assumed to be a power of 2?

I believe the answers you gave are correct, but you may want to give solutions and use spoiler. If you don't know how to use spoiler, umm... someone's signature used to give a good explanation. Dunno where it went. But basically, just put the tags [ spoiler ] and [/ spoiler ] (without the spaces) before and after the text you want hidden

Yup yup, the first three are so famous it's not that difficult to find the solutions.

The explanation was on my signature when I was (secretely) doing the "AoPS Tip of the Week" series (which wasn't recognised by anyone :( )
Anyways I put it back up since I realized that was the most helpful out of the 6 (yes, I patiently did it for 1 and a half months or so :lol: )

Yes, you can assume that is a power of 2. If you have read the question closely it said [quote="question"]and n is a number in the pattern[/quote], but oh well, now you know.

Actually, I did notice the "Tip of the Week."  I thought it was a rather good idea.  I just didn't comment on it.

(Your efforts do not go unnoticed :) )

I like number 4.

(Thanks Joel :D)

number 4's actually my original:
The first three I picked up common ones used in MATHCOUNTS a lot
The following 5 I just made it up except for the middle one (third one) which was on my school's Problem of the Week and is also very popular
The last three is also original but it involves either a combination of addition and subtraction or multiplication/division

This was really interesting too...here's what I got...I think a few of them are wrong...Tare posts pretty good problems! I think I'll revive all the unsolved ones... :D

1. $ \frac{n(n\plus{}1)}{2}$

2. $ \frac{n(n\plus{}1)(5n\plus{}1)}{12}$

3. $ (\frac{n\plus{}1}{2})^2$

4. $ (\frac{n\plus{}1}{2})^2\plus{}(\frac{n\minus{}3}{2})^2\minus{}4$

5. $ \frac{(n\plus{}1)(n\plus{}2)}{6}$

6. $ \frac{2^n\minus{}1}{2^n}$

7. $ 2^{n\plus{}1}\minus{}1$

8. $ \frac{5(3^n\minus{}1)}{2}$

9. $ \frac{n}{2}\plus{}4$

10. $ \frac{1}{n}$

11. $ 2^{\frac{n(n\minus{}1)}{2}}$

This was 7 years ago. But since it's not solved, I guess it qualifies as an "okay" to revive.

The thing with these is, Tare didn't really specify what he wanted the equations as. Sometimes the equations work better in terms of n, while sometimes they work better in terms of x (in the sense that n is the xth term). So I'm gonna respond to Barbie's considering that.

@BarbieRocks:
1, 3, 5, 7, 9, 10: yes
4: yes but it can be simplified
6, 8, 11: that's the equation using x
2: no either way

warning: 2 and 11 are messy in terms of n

you realize something like $ 2^{\log_{2}n}$ is very easily simplified?
but yes, everything you said was technically correct, even if they aren't as pretty as possible :P


I came across the following series
$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{{\pi}^2} {6}$ but I don't know why.
Can anyone show me how to prove this?

Have you read Fourier Series? If yes then expand f(x) = x .

Sorry, I don't know about Fourier Series.
Is there any other method?

Thanks a lot for the notes.
I can understand one of the proofs which makes use of the series expansion of arc sine. :)
Best Regards!
As a further question, is it easy to evaluate the value of a series like
$\sum_{n=1}^{\infty}\frac{1}{n^2+n+1}?$


Let $ n$ be a positive integer and $ n >\equal{} 2$ and $ 0 < a_1 < a_2 < ... < a_{2n \plus{} 1}$ be positive real numbers. Prove that :
$ \sqrt [n]{}a_{1} \minus{} \sqrt [n]{a_{2}} \plus{} \sqrt [n]{a_{3}} \minus{} ... \minus{} \sqrt [n]{a_{2n}} \plus{} \sqrt [n]{a_{2n \plus{} 1}} < \sqrt [n]{a_{1} \minus{} a_{2} \plus{} a_{3} \minus{} ... \minus{} a_{2n} \plus{} a_{2n \plus{} 1}}$ :)

why "infinite numbers"??

anyway, i think Jensen will take care of it.


Two questions:

1. How do i increase my rating?

2. When will i be able to rate the posts?

I think you should be able to rate posts just about now. Your rating depends on the ratings the users of the site give to your posts. The more and better, nice and concise posts you will have the higher rating you will get.

Thanks.

When I tried to rate your post, it said that i need at least 25 posts to be able to do that.

Well then you need 25 posts :)
Also remember that your ratings will matter as much as your current rating is. So basically if you just got 25 posts, but all of them are rated 1, then your rating will be low, so whenever you rate something it won't matter that much.


8. The second and fourth terms of a geometric sequence are 2 and 6. Which of the
following is a possible first term?

$ (A) \minus{}\sqrt{3} \qquad (B) \minus{}\frac{2\sqrt{3}}{3} \qquad (C) \minus{}\frac{\sqrt{3}}{3} \qquad (D) \sqrt{3} \qquad (E) 3$

Let r be the ratio.
$ 2r^2 \equal{} 6; r^2 \equal{} 3; r \equal{} \pm \sqrt{3}$
So the first term is $ \frac{2}{\pm\sqrt{3}} \equal{} \pm\frac{2\sqrt{3}}{3}$
which could be B.

The sequence is thus $ \minus{}\frac{2\sqrt{3}}{3}, 2,  \minus{}2\sqrt{3}, 6, ...$


You use (1+a)^2/(b+c)=(1+2a+a^2)/(1-a)=-a+1+4a/(1-a)=-a+1+4a/(b+c).
Now yuo have LeftSide=sum(-a+1+4a/(b+c))=2+4(sum(a/(b+c))). With Nesbitt (sum a/(b+c)=>3/2) we get: LeftSide=>2+4*3/2=8 as we wanted to proove.

Misha

[quote="manlio"]For $a,b,c$ positive reals such that $a+b+c=1$ prove that

$\frac{(1+a)^2}{(b+c)}+\frac{(1+b)^2}{(c+a)}+\frac{(1+c)^2}{(a+b)} \geq 8$[/quote]

Nesbitt accounts for 2*3/2 = 3
Sum[a^2/(b+c)] >= (a+b+c)/2 by rearrangement = 1/2

So we must have Sum[1/(b+c)]>= 9/2

But this greater than or equal to 3^2/2(a+b+c) by cauchy, hence we are done.

Another easy sol: we have $0<a,b,c<1$ and for $x<1$ we have that $f(x)=\frac{(1+x)^2}{1-x}$ is convex, so Jensen gives us:

$f(a)+f(b)+f(c) \ge 3f\left(\frac{a+b+c}{3}\right)=3f\left(\frac13\right) = 8$ :)

By ineq Cauchy or Andreescu(tell it as you want) we have

$\frac{(1+a)^2}{(b+c)}+\frac{(1+b)^2}{(c+a)}+\frac{(1+c)^2}{(a+b)} \geq\frac{(1+a+1+b+1+c)^2}{2(a+b+c)}=8 $and we are done

Look, you can give it many names, but not Andreescu. Titu Andreescu doesn't have anything to do with this inequality.

in many books in Greece we told this ineq as Andreescu. but it doesnt matter. My solution is true sure?


Provide a synthesis of hexan-1,4-diol using propene as the only organic starting material.

1. Reaction with NBS to give CH2(Br)-CH=CH2
2. React this with (CH3-CH=CH)2-CuLi to give CH3-CH=CH-CH2-CH=CH2
3. React with BH3,THF and H2O2 and NaOH (hydroboration-oxidation)

[quote="Hell_ever"]2. React this with (CH3-CH=CH)2-CuLi to give CH3-CH=CH-CH2-CH=CH2[/quote]

From where did that Gilman reagente came from?

[quote="Hell_ever"]3. React with BH3,THF and H2O2 and NaOH (hydroboration-oxidation)[/quote]

Reaction of borane (or equivalent) with a diene gives...

I think this works, though I haven't studied the NBS reaction yet (just looked it up on Wikipedia so far) so I may be wrong.

[hide="click text"]
Take one equivalent of propene and react with NBS to get 3-bromo-1-propene
Take another equivalent of propene, carry out hydroboration/oxidation, then oxidize using PCC to get propanal

React the 3-bromo-1-propene with Mg in the presence of the propanal and follow with acidic workup to get 4-hydroxy-1-hexene, then carry out hydroboration/oxidation again to yield the desired product[/hide]

I think NBS acts as a slow-releasing source of Br_2, so adding this to propene would give you 2-bromopropane (or 1-bromopropane if you added in anti-Markovnikov fashion such as with peroxides)

I found this site: [url]http://www.organic-chemistry.org/namedreactions/wohl-ziegler-reaction.shtm[/url].  It looks like the reaction would be carried out in peroxides with low concentrations of NBS, HBr and BR2.  The low concentrations are supposed to ensure that substitution at the allylic position occurs rather than addition of HBr or BR2 across the double bond.

Yes, the low concentration of molecular bromine provided by NBS favors allylic substitution at the expense of double bond addition, since this last path is reversible.

this was my idea; I don't know much about protecting groups but I will try to check to see if this makes any sense later

Start with propene, and add molecular bromine in a basic solution (lots of hydroxide). The bromonium ion forms, and the hydroxide attacks in S_N2 at the less hindered end to give 2-bromopropan-1-ol. Protect the alcohol with a protecting group (benzyl ether, maybe?) by benzyl bromide and sodium hydride. I think this is sufficiently bulky that if we now try to eliminate the bromine with LDA in an E2, it eliminates on the other side to give us essentially a propene molecule with this benzyl ether on the third carbon. Hydroborate (BH_3 in THF, then H_2O_2 and NaOH), tosylate with TsCl, attack with Br-, insert Mg, and attack on a molecule of propanal, formed by taking another equivalent of propene, hydroborating, and oxidating with CrO_3. Acid quench and remove the protecting group with palladium bonded to charcoal.

[quote="Hideaway"]I'm not sure about protecting groups, but I did spot one other thing in your synthesis: you use Jones' reagent to oxidize the 1-propanol to propanal.  This would yield a carboxylic acid since 1-propanol is a primary alcohol and the solvent is aqueous.[/quote]

I dont think so, it only oxidises till the aldehyde stage as far as i know :maybe:

Well, it does react to give the aldehyde initially.  However, since the reaction is carried out in water, the hydrate of the aldehyde will be present in solution.  And since the hydrate has two -OH groups, it reacts further with CrO3 (irreversibly) to give the carboxylic acid.

You can get around this problem by using pyridinium chlorochromate (PCC) in CH2Cl2, which is an aprotic solvent.  This way, since there isn't any water present, the hydrate can't form and the reaction stops at the aldehyde.

[quote="Hideaway"]You can get around this problem by using pyridinium chlorochromate (PCC) in CH2Cl2, which is an aprotic solvent.[/quote]

That is Sarret's reagent i think :)
And also Collins reagent can be used! (Or am i confusing the names? :D )

I'm sorry; I meant to use Collins reagent. (we learned Collins oxidizes to aldehyde, Jones oxidizes to carboxyl). I might have gotten the reagents mixed up when writing down the chemical name.


geometry
3D geometry
linear algebra
matrix
algebra
polynomial

How would one describe $ \{ A \in M_{2x2}(\mathbb{R}) : A^2\plus{}A\plus{}I \equal{} O\}$?

What are the eigenvalues of such a matrix?

And, because the characteristic polynomial is real, we must have both members of that conjugate pair.

We could attack this directly: we know what the characteristic polynomial must be. From that we get that $ \operatorname{tr}A\equal{}\minus{}1,\det A\equal{}1.$ That is, if

$ A\equal{}\begin{pmatrix}a&b\\c&d\end{pmatrix},$ then $ a\plus{}d\equal{}\minus{}1$ and $ ad\minus{}bc\equal{}1.$

There are many solutions to that. One nice one is $ \begin{pmatrix}0&1\\\minus{}1&\minus{}1\end{pmatrix}.$


Bjorn's height is 160 cm plus one-sixth of his height. What is the number of centimeters in Bjorn's height?

Since 160 + 1/6 of his height is his height, 160 is 5/6 of his height. Therefore, his height is 192 cm.

As an equation (clarification).

$ 160 \plus{} \frac{1}{6} x \equal{} x$

$ 160 \equal{} \frac{5}{6} x$

$ 160 \cdot \frac{6}{5} \equal{} x$

$ \boxed{192} \equal{} x$


hello  :D

show that

$ \sum_{n\equal{}0}^\infty\sum_{m\equal{}0}^\infty\dfrac{1}{m\cdot n\cdot (m\plus{}n)^{2p}}\;\equal{}\;\boxed{\dfrac{a}{b}\cdot \pi^{6p}}$

where $ a,b\;>0\quad\text{and}\quad a,b\in\mathbb N \qquad ( \text{for}\quad p\equal{}1,2,3,\ldots\; )$

[quote="JRav"]I thought you would stop with these. :lol:[/quote]

 :rotfl:

it is fairly easy, you can use poisson summation to get at the answer.  :lol:

Maybe sums must starts from $ m \equal{} 1$ and $ n \equal{} 1$, but not from $ m \equal{} 0$ and $ n \equal{} 0$.

I don't know Poisson Summation...  :oops:

First, denote:

$ F(x,y,z)\equal{}\sum^{\infty}_{n\equal{}1}\sum^{\infty}_{m\equal{}1}\frac{x^n y^m z^{m\plus{}n}}{m \cdot n \cdot (m\plus{}n)^{2p}}$

Now, we have:

$ \frac{\partial ^2 F}{\partial x \partial y}(x,y,z)\equal{}f(x,y,z)\equal{}\sum^{\infty}_{n\equal{}1}\sum^{\infty}_{m\equal{}1}\frac{x^n y^m z^{m\plus{}n}}{(m\plus{}n)^{2p}}\equal{}\sum^{\infty}_{n\equal{}1}\frac{z^{n\plus{}1}}{(n\plus{}1)^{2p}}\sum^{n\minus{}1}_{i\equal{}0}x^i y^{n\minus{}i\minus{}1}$

Now:

$ F(x,y,z)\equal{}\sum^{\infty}_{n\equal{}1}\frac{z^{n\plus{}1}}{(n\plus{}1)^{2p}}\sum^{n\minus{}1}_{i\equal{}0}\frac{x^{i\plus{}1} y^{n\minus{}i}}{(i\plus{}1)(n\minus{}i)}$

Furthermore:

$ F(1,1,z)\equal{}\sum^{\infty}_{n\equal{}1}\frac{z^{n\plus{}1}}{(n\plus{}1)^{2p}}\sum^{n\minus{}1}_{i\equal{}0}\frac{1}{(i\plus{}1)(n\minus{}i)}$

Now, we know that:

$ \frac{1}{(i\plus{}1)(n\minus{}i)}\equal{}\frac{1}{1\plus{}n}\left(\frac{1}{1\plus{}i}\plus{}\frac{1}{n\minus{}i}\right)$

And this gives us:

$ \sum^{n\minus{}1}_{i\equal{}0}\frac{1}{(i\plus{}1)(n\minus{}i)}\equal{}\frac{1}{1\plus{}n}\left(\sum^{n\minus{}1}_{i\equal{}0}\frac{1}{1\plus{}i}\plus{}\sum^{n\minus{}1}_{i\equal{}0}\frac{1}{n\minus{}i}\right)\equal{} 2\frac{H_n}{n\plus{}1}$

And this makes:

$ F(1,1,z)\Big|_{z\equal{}1}\equal{}\sum^{\infty}_{n\equal{}1}\sum^{\infty}_{m\equal{}1}\frac{1}{m \cdot n \cdot (m\plus{}n)^{2p}}\equal{}2\sum^{\infty}_{n\equal{}1}\frac{H_n}{(n\plus{}1)^{2p\plus{}1}} \neq \frac{a}{b}\pi^{6 p}$

For example, if we set $ p\equal{}2$ we get:

$ \sum^{\infty}_{n\equal{}1}\sum^{\infty}_{m\equal{}1}\frac{1}{m \cdot n \cdot (m\plus{}n)^4}\equal{}\frac{\pi^6}{630}\minus{}\zeta^2(3)$

and that $ \neq\frac{a}{b}\pi^{12}$ unless some huge brake through has occurred regarding $ \zeta(3)$.

[quote="milin"]$ \frac {\partial ^2 F}{\partial x \partial y}(x,y,z) \equal{} \sum^{\infty}_{n \equal{} 1}\sum^{\infty}_{m \equal{} 1}\frac {x^n y^m z^{m \plus{} n}}{(m \plus{} n)^{2p}}$[/quote]

This is not quite right (given your definition of $ F$), and that may account for the eventual discrepancy.


algorithm
modular arithmetic
number theory
prime factorization

I'm implementing the pollard-rho algorithm, which requires that one finds $ x^2 \plus{} 1\pmod{N}$ repeatedly, which after a certain number of iterations leads to a factor of $ N$, where $ N \subset \mathbb{Z}$, and $ x$ is usually a small (arbitrary) number. For large $ N$ (like $ N$=10266001) my implementation fails after a few iterations. Of course for small numbers (like 3337) it works. I don't see the problem, since taking the modulus of a number shouldn't produce a larger (i.e. whose mantissa is greater than 24 bits) number, but a smaller one!

Have you considered how you are calculating x^2 + 1 when x = 10266000? (x^2 won't fit in an int; it would fit in a 64 bit number).

no. But now that I think of it, i can't find a reason why $ x^2\plus{}1\pmod{N}$ [i]wouldn't [/i]be large.

After a few iterations, you're going to run into a number $ x < N$ such that $ x^2 > N$ - in particular, $ x^2$ will be larger than the size of an int.  That's why your implementation is failing.

Either you need to find a way of calculating $ x^2 \plus{} 1 \bmod N$ that does not involve calculating $ x^2$ (which is difficult without knowing the prime factorization of $ N$) or you need to use smaller $ N$.

Just like TripleM says, your $ x$ is always in $ [0,N)$ but the computer needs to be able to store $ x^2$ in its memory because it first squares it, then adds $ 1$ and only then divides the result by $ N$ to find the reminder. You should be fine if you use 64-bit variables.

10266001 is not large  :wink:
It would probably be much better to use some arbitrary-precission integers. There surely is some library fitting to your language, and even if not, it would not be harder to write one than writting all that "big-number-avoidance" (as long as you don't want really fast multiplication of big numbers).

With numbers of this size, naive multiplication is actually fastest :)

However, I don't see what's wrong with repeated doubling here - it has the same runtime as naive arbitrary precision multiplication, requires less storage, and keeps the numbers small. (processor math is faster, of course)

There is a simple fact that is wrong: whats the purpose in writing a factoring algorithm for numbers with 32 bit or less¿  :wink:

Practice? Or maybe there's a need for the factorization, but there isn't any code written in the particular manner required?

[quote="JBL"]You could also implement a version of multiplication-mod-$ N$ that wouldn't require you to ever deal with numbers larger than $ 2(N \minus{} 1)$ by doing repeated doubling.[/quote]

Couldn't you change it so that if it is larger than 1/2 of N, you put it into the -ve form? Then you could reduce the addition to larger than N-2

A signed integer requires an extra bit anyway...

All pretty pointless since you can simply use long long and you're done.

So anyway, for arbitrary precision mathematics is [b]bc[/b] the easiest way to go.


Given are $\triangle ABC$, area $S$ with any $M$ on plane prove that
$bMA+cMB+aMC\ge 4S$
a nice corollary when $M\equiv G$(centroid):
$\frac{m_{a}}{h_{b}}+\frac{m_{b}}{h_{c}}+\frac{m_{c}}{h_{a}}\ge 3$ :P

I 've posted smth similar [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=geometric&t=76453]here[/url] some time ago .
I would like to see a solution of that too. :wink:

It looks like this one:With point M in triangle ABC,area S,we have
      aMA+bMB+cMC $\ge$ 4S

yes Albanian Eagle your inequality is really hard , I saw a solution in AMM of Garder but it is very long, my inequality (with median) is only corollry of yours, but it is nice if we see it as a special case of my inequality (with any $M$)! :)


Hey ppl...according to u..wats the world's [b]best[/b] website??
See my favourite-[url]http://home.comcast.net/~wolfand/[/url]
It is sooo cool
Please post ur views about this website!!

According to me:www.mathlinks.ro. Very cool!People can discuss problems and have a lot of fun  :lol:.They can meet participants from IMO's,teachers, they can meet a lot of nice people, they learn modern cool mathematics!It's a great site!You should definately try it!It is genial!
Thanks to Valentin Vornicu for creating it, it is the best!!!

[quote="funcia"]How do I get it away?

EDIT: I did but had to Alt Ctrl Delete everything! Is that going to spam me?[/quote]
 nope- it's a completely harmless website...u just have to continue until u get to 150! Isn't it funny?

oh yeah- my other fav websites-mathlinks.....i tried it and it rox! :lol: and the website which we r currently on,duh! (and google obviously)

Is that the 150 questions link? I loved it, but i won't click it again!

[quote="funcia"]http://www.nanonet.go.jp/english/

AND

http://ihatecalculus.com/?[/quote]

<-- Look at username  :P

[url]http://www.math.utah.edu/~cherk/mathjokes.html[/url] owns.

hmm I always thought the 'calc' in your username meant 'calculator' :ninja:

i did and i meant that website as a spoof..

[quote="Klebian"]I just held down enter, and it went through all the boxes in about 10 seconds.[/quote]
cool!
i never tried that!


Let $m=AB$ and $n=BC$, where $A,B,C$ are digits and the numbers are written to the base $10$. Find all such two-digit $m,n$ such that $\frac{m}{n}=\frac{AB}{BC}=\frac{A}{C}$.

Show that there are no such $m$ and $n$ when written to the base $p$ for some prime $p$, unless $A=B=C$.

My solution for the first part is tedious. I haven't attempted the second part yet. (Edit: turns out the second part is pretty straight-forward..)

[quote="xxreddevilzxx"][hide="Part 1"]
$\frac{10A+B}{10B+C}=\frac{A}{C}$
$10AC+BC=10AB+AC$
$9AC+BC=10AB$
$C(9A+B)=10AB$

Case 1: They are all distinct.

Then $C=AB$, because C cannot equal 10A, 10B, or 10. so then matching, $9A+B=10$. The only solution possible here is $(A,B,C)=(1,1,1)$

Case 2: $C=A$. Then $10A = 9A+B$, $A=B$, so then it becomes $A=B=C$ and the solutions are $(A,B,C)=(1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6), (7,7,7), (8,8,8), (9,9,9)$. The solution would come out the same if we set $C=B$

Case 3: $A=B$.

$10AC=10A^2$
$AC=A^2$
$A=C$, which instantly leads to $A=B=C$, giving the same solution as case 2.

So in any cases, $A=B=C$. That leads to $m=n=11, 22, 33, 44, 55, 66, 77, 88, 99$
[/hide]

[hide="Part 2"]
$\frac{pA+B}{pB+C}=\frac{A}{C}$
$pAC+BC=pAB+AC$
$(p-1)AC+BC=pAB$
$C((p-1)A+B)=pAB$

Since $p$ does not divide $C$, we know that $p|(p-1)A+B$. This is only possible if $B \equiv -A mod{p}$, meaning $B=A$. Then going back and substituting,
$(p-1)AC+AC=pA^2$
$pAC=pA^2$
$A=C$

thus the numbers $m$ and $n$ exist in some base $p$ where $p$ is a prime if $A=B=C$
[/hide][/quote]There's something wrong with part 1, because there are other solutions than $A=B=C$. [hide="partial solution"]$\frac{m}{n}=\frac{10A+B}{10B+C}=\frac{A}{C}\Longrightarrow10A(C-B)=C(A-B)$. Clearly $A$ and $C$ are both nonzero (from the conditions). Either $C-B=A-B=0\Longrightarrow{A}=B=C$ and the trivial solutions are \[\frac{11}{11}=\frac11,\frac{22}{22}=\frac22,\ldots,\frac{99}{99}=\frac99.\] Otherwise, the LHS and RHS are nonzero, and we have $10\mid{C}(A-B)$. Both factors are clearly less than $10$. Thus we can divide into the following cases:

$C=5$
$A-B=\pm2,\pm4,\pm6,\pm8$

or

$C=2,4,6,8$
$A-B=\pm5$.

I checked all sixteen cases and found the four solutions \[\frac{16}{64}=\frac14,\frac{49}{98}=\frac48,\frac{26}{65}=\frac25,\frac{19}{95}=\frac15.\][/hide]
Part 2 looks good. Here's how I did it. [hide="part 2"]$\frac{pA+B}{pB+C}=\frac{A}{C}\Longrightarrow{p}A(C-B)=C(A-B).$

Of course $C<p$ and $A-B<p$. Either the LHS and RHS are both $0$ and we have the same trivial solutions as before, or $p\mid{C}(A-B)$, a contradiction.[/hide]

[hide]$\frac{10a+b}{10b+c}=\frac{a}{c}$
$(10a+b)c=(10b+c)a$
$10ac+bc=10ab+ac$
$9ac=b(10a-c)$
Either (a) $b=9$ or (b) $10a-c \equiv 0 \pmod {9}$, or (c) both $b$ and $10a-c$ are divisible by 3.

(a) $b=9$
$ac=10a-c$
$a(c+1)=10a$
$c=9 \Rightarrow a=9$

(b) $10a \equiv c \pmod {9} \Rightarrow a \equiv c \pmod {9} \Rightarrow a=c$.

Therefore $b=\frac{9ac}{10a-c}=\frac{9a^2}{9a}=a$, so $a=b=c$.

(c) both $b$ and $10a-c$ are divisible by 3

$b \equiv 0 \pmod {3}$
$10a \equiv c \pmod {3}$
$a \equiv c \pmod {3}$
$(a,\{ c \})=(3,\{ 3, 6, 9 \} ), (6,\{ 3, 6, 9 \} ), (9,\{ 3, 6, 9 \} )$

but $b=\frac{9ac}{10a-c} \equiv 0 \pmod {3}$, so $\frac{ac}{10a-c}$ is an integer.

Hmm...maybe somebody else will be able to finish this for me. i have to go[/hide]


[quote="kingof21penguins"]this one is not to bad
[hide]1st digit -4 possibilities
2nd digit -3 possibilities
3rd digit-3 possibilities
4th digit-2 possibilities
5th digit-4 possibilities
4*3*3*2*4=288 possibilities
i suggest making the problem so that 3 and 4 cannot be the same as 1 and 2[/hide][/quote]

How do you know the first and fourth digit are different? They could both be 7 for all you know, so that means that there aren't 4 possibilities for the 5th digit.

Pretty easy. The last part is clever though.

[hide]4 possible primes
3 remaining primes
3 digits greater than 6
2 remaining ""
[b]2 to 4[/b] above numbers.

How many ways are there to get 2 as the last number? There aren't any since there is only 1 prime greater than 6. How about 3? 1*3*1*2 (7 must be chosen for each)=6. Therefore, we have 6*3+(4*3*3*2-6)*4=[b]282[/b][/hide]

I think that it is just 4*3*3*2*4, because it never said that all the digits had to be different.

Oh wait, on yes, the last digit...

[hide]Digit primes are: 2,3,5,7
1st digit: 4 choices
2nd digit: 3 choices
3rd digit: 3
4th digit: 2
5th digit: 4
4*3*3*2*4=288
BUT...
There are 6 situations where there are only [b]3[/b] choices for the 5th digit, so we must subtract 6*1
>>282<<[/hide]

$4 \times 3 \times 3 \times 2 \times 4 = 288$, but this doesn't work at least for the last digit because there might only be 3 different choices, depending on whether any of the first four digits are the same.

WHAT? How'd you get 6 different situations where you have only three digits to choose from? I got 24, because you could match the 1st digit with the 3rd, 1st with 4th, 2nd with 3rd, or 2nd with 4th, and 6x4=24. Plus, you forgot to multiply by 3 for the last digit. You can't forget that.

In that case the answer is [hide]288-(24x3)=216. And by the way, jbo5, you were right about counting the number of ways the 5th digit could be any of 2-4 values, but I think you made a little error in the end.

But don't take my word for it. [/hide]

I still have't seen a right answer, or at least one I believe to be right.

You need to break this into two cases:

1) If 7 is chosen as one of the two primes and one of the two >6 numbers

2) 7 is not chosen twice

If seven is chosen twice, there are only three possibilities for the last digit
In seven not chosen twice, there are four possibilities

I'll post a solution if I don't see a right answer soon.

[hide="new answer"] 4 x 3 x 3 x 2 x 7 = 504 [/hide]

[quote="frost13"]I gave this to one of my classes today as a simple counting problem, but when I started working towards the answer I realized it wasn't as simple as I had intended :)

How many 5 digit ATM passwords can be formed using these rules :

1st digit must be prime
2nd digit must be a different prime
3rd digit must be greater than 6
4th digit must be greater than 6 but different than 3rd digit
5th digit must match one of the first four digits[/quote]

A lot of casework <_< .

Definitely more than 72.

Ok. I changed my mind :). That problem was very tricky and I'm still not sure I got it right, but I think I'm on the right track this time.

[hide]Assuming that there are three different numbers:
1*3*1*2*3
However, the 7 (represented by 1) could be in two different places for each double so multiply by 4 to get 72.

Assuming that there are four different numbers and one of the first two is 7:
1*3*2*1*4
Double because the seven could be in either spot to get 48.

Assuming that there are four different numbers and one of the last two is 7:
3*2*1*2*4
Double because the seven could be in either spot to get 96.

Assuming that there are four different numbers and none is 7:
3*2*2*1*4 = 48

I think the answer is 264.[/hide]

I'm probably not half as good at maths as many people on this site but I'll "guess" anyway since noone apparently has found the solution yet. [hide="My guess"]I say 324.[/hide]

But now when I think about it jb05's latest answer is very reasonable.

ok, jb05 has it, but it a totally different way then I did it.  I'll add my solution too, good job jb05.

[hide]  I started by looking for the probability that 7 would be used in the first AND second pair of numbers.  I did that by actually looking for the probability that a 7 wouldn't be used.

In the first two numbers, there is a $\frac{3}{4}$ chance of not using 7 in the first digit, and then $\frac{2}{3}$ chance of still not using 7 in the second chance.  THe probability of not using 7 either time is the $\frac{3}{4}$ x $\frac{2}{3} = \frac{1}{2}$  So there is a $\frac{1}{2}$ chance of not using and of using a 7 in the first pair.

If you use the same process for the second pair, there is a
$\frac{2}{3}$ x $\frac{1}{2} = \frac{1}{3}$ chance of not using a 7 in spots 3 & 4, so a $\frac{2}{3}$ chance of using a 7 in the second pair.

The probability of using 7 in each pair(once in 1st two digits and once in 2nd two digits) is  $\frac{1}{2}$ x $\frac{2}{3} = \frac{1}{3}$.  That also mean there is a $\frac{2}{3}$ chance of not using two 7s and ensuring there are 4 different digits used in the first 4.

So our two cases become :

Using 7 twice and only having 3 choices for last digit: $\frac{1}{3}$(4)(3)(3)(2)(3) = 72

Using 7 once of not at all giving 4 choices for the last digit : $\frac{2}{3}$(4)(3)(3)(2)(4)= 192

For a total of 72 + 192 = 264 passwords

[/hide]


Who here made the ARML team this year?  (Myself and Alex [altheman] were the only ppl from our school, I think.)

And to everyone, congrats!!  :)

WOOOOOOO 1010101010!!!!!!

i want more problems...lol....the only one left to do is #9...the number of ways from two lattice points is easy...i just have to find a nice way to break up the cases...or i can be lame and calc brute force with a summation

Gah! I got 12 at tryouts! One point from the cut off!  I cant beleive some of the mistakes I made,  sqrt(148) = 2 sqrt(37)  NOT 4 sqrt(37), honestly how did I mess that up.  If I could add, subract, multiply, and reduce radicals correctly, I would have had at least 17. bah!

yea...blah blah...i could have gotten 24/24 if i was not being stupid...

i made a computation error on the d=rt problem
the domain one i found the number of integers, not the sum
the system i dropped k=-5 for no reason...
x^2+1=13k i was being stupid...i have proposed harder problems along the same vein...i was just being stupid and got confused about finding all solutions vs a subset of the solutions

[quote="Altheman"][quote="cincodemayo5590"][quote="djshowdown2"]Ooh ooh I made it, 22.  Similar mistake nateharman1234, $\sqrt{\frac{15}{2}\cdot\frac{1}{2}\cdot\frac{5}{2}\cdot\frac{9}{2}}= \frac{15\sqrt{3}}{4}$ NOT $\frac{15\sqrt{3}}{2}$ so I got $\frac{49\sqrt{3}}{16}$, but I [i]guess[/i] I'm content.

Quien eres cincodemayo?[/quote]

Me llamo Sam Bloom, y mi cuenta en el concurso fue 13. (Sooo many stupid mistakes, lol.)

(Yea, Alex I was "lame" on that hw question and just did calc brute force. Yay!!)[/quote]

that last sentence does not make sense

what is this calc brute force you speak of...i don't have said brute force[/quote]

[quote="Altheman"]WOOOOOOO 1010101010!!!!!!

i want more problems...lol....the only one left to do is #9...the number of ways from two lattice points is easy...i just have to find a nice way to break up the cases...[b]or i can be lame and calc brute force with a summation[/b][/quote]

Lol. And the last sentence of yours doesn't make sense (grammatically) either.

[quote="bpms"][quote="Altheman"]ono but i used a theorem that is quite "tasty"....so to speak...so the proof is not so messy now...thats why i was saying that your sentence did not make sense... since i have a "tasty" solution...involving a certain baked good...apple flavored, in fact

also...i believe you should taking something that starts with an A-, and ends with a -MIE...and just happens to be linked in my sig[/quote]
What's an A-...-MIE?[/quote]

it could not possibly be an AIME could it? and that is not what is in my sig...OF COURSE NOT

lol...sarcasm sucks on forums

woo arml

who was like OMG 9b is annoying...yes i found the summation along the diagonals...but like...we have so little space to write stuff...

woo arml

grrr I want my big prize.  Did he say anything to you altheman?

i was like hey, whats the prize

and he was like...did you really...then i busted out the sheets

and he was like...next practice...blah..."I will have something"

which is like...hmm...nothing special


geometry
rhombus
conics
ellipse
trigonometry
analytic geometry
incenter

[color=darkblue]Let $ABCD$ be a rhombus and let $\Xi$ be the ellipse which is tangent to the sides of the rhombus in the its middlepoints. For a mobile point $M\in\Xi$ denote the projections $A'$ , $B'$ , $C'$ , $D'$ of the vertices of the rhombus to the tangent $MM$ in the point $M$ at the ellipse $\Xi\ .$ Prove that $AA'\cdot CC'=BB'\cdot DD'\ .$

[b]Particular case.[/b] If $ABCD$ is a square then the ellipse $\Xi$ becomes the its incircle. Prove at least this case.[/color]

[color=darkblue]I promised for a long time when I"ll come home, I"ll extend the problem w.r.t. the incircle in a square (which appeared at this section - the down particular case). I am waiting at least a proof.[/color]

Let $A(2a,0), \, B(0,2b), \, C(-2a,0), \, D(0,-2b)$. The ellipse must pass through $\left( \pm a, \pm b \right)$, so it has the equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 2$.

The parametric equation is $r(t) = \left( a \sqrt 2 \cos t, b \sqrt 2 \sin t \right)$, so that $\dot r (t) = \left(-a \sqrt 2 \sin t, b \sqrt 2 \cos t \right)$.
Thus, if $M(x,y) \in \xi$, then the tangent at $M$ has the equation $\frac{x-a \sqrt 2 \cos t}{y-b \sqrt 2 \sin t}= \frac{-a \sqrt 2 \sin t}{b \sqrt 2 \cos t}$, i.e.
\[xb \cos t+ya \sin t-ab \sqrt 2 = 0 . \]
Giving $t$ the values $\frac{k \pi}4, \, k \in \left\{ 1, 3, 5, 7 \right\}$, we see that my ellipse is the right one.

Now, for an arbitrary $t$, we have that $A A^\prime = d \left( MM, A \right) = \frac{\left| 2 a b \cos t-ab \sqrt 2 \right|}{b^{2}\cos^{2}t+a^{2}\sin^{2}t}, \, C C^\prime = \frac{\left|-2 a b \cos t-ab \sqrt 2 \right|}{b^{2}\cos^{2}t+a^{2}\sin^{2}t}$, so
\[A A^\prime \cdot C C^\prime = \frac{\left| 4 a^{2}b^{2}\cos^{2}t-2 a^{2}b^{2}\right|}{\left( b^{2}\cos^{2}t+a^{2}\sin^{2}t \right)^{2}}. \]
Similarly,
\[B B^\prime \cdot D D^\prime = \frac{\left| 4 a^{2}b^{2}\sin^{2}t-2 a^{2}b^{2}\right|}{\left( b^{2}\cos^{2}t+a^{2}\sin^{2}t \right)^{2}}. \]
Hence, the conclusion is equivalent to $\left| 2 \cos^{2}t-1 \right| = \left| 2 \sin^{2}t-1 \right|$, which is clearly true.

[quote="Virgil Nicula"][color=darkred]Let $ABCD$ be a rhombus and let $\Xi$ be the ellipse which is tangent to the sides of the rhombus in the its middlepoints. For a mobile point $T\in\Xi$ denote the projections $A'$ , $B'$ , $C'$ , $D'$ of the vertices of the rhombus to the tangent $TT$ in the point $T$ at the ellipse $\Xi\ .$ Prove that $AA'\cdot CC'=BB'\cdot DD'\ .$
[b]Particular case.[/b] If $ABCD$ is a square then the ellipse $\Xi$ becomes the its incircle. Prove at least this case.[/color][/quote]

[color=darkblue][b]Proof (synthetic) of the particular case.[/b] Suppose that the tangent $t$ in the point $T$ to the incircle $w=C(I,r)$ cuts the lines $CD$ , $AB$ in the points $M$ , $N$ respectively so that $M\in (CD)$ and $B\in (AN)\ .$ Denote the middlepoints $P$ , $R$ of the sides $[AB]$ , $[CD]$ respectively. Observe that $\triangle IPN\sim\triangle MRI\ ,$ i.e. $\frac{IP}{PN}=\frac{MR}{RI}$ and $\frac{AA'}{BB'}=\frac{AN}{BN}=$ $\frac{AP+PN}{PN-PB}=$ $\frac{IP+PN}{PN-IP}=$ $\frac{1+\frac{IP}{PN}}{1-\frac{IP}{PN}}=$ $\frac{1+\frac{MR}{RI}}{1-\frac{MR}{RI}}=$ $\frac{RI+RM}{RI-RM}=$ $\frac{MR+RD}{CR-MR}=$ $\frac{MD}{MC}=\frac{DD'}{CC'}\ ,$ i.e. $\frac{AA'}{BB'}=\frac{DD'}{CC'}\ ,$ what means $AA'\cdot CC'=BB'\cdot DD'\ .$

[b]Remark.[/b] The conditions $AA',BB',CC',DD'\perp t$ can weaken at $AA'\parallel BB'\parallel CC'\parallel DD'\ ,$ where $\{A',B',C',D'\}\subset t\ .$

[b]Proof (analytical) of the proposed problem.[/b] Choose the system of the coordinates so that $A(a,0)$ , $B(0,b)$ , $C(-a,0)$ , $D(0,-b)\ ,$ where $a>0$ and $b>0\ .$ Thus, the ellipse $\Xi$ has the equation $2\left(b^{2}x^{2}+a^{2}y^{2}\right)=a^{2}b^{2}$ and is tangent in the point $E\left(\frac{a}{2},\frac{b}{2}\right)$ to the line $AB$ with the equation $bx+ay=ab\ .$ For a mobile point $T\left(\frac{au}{\sqrt 2},\frac{bv}{\sqrt 2}\right)\in\Xi$ where $u^{2}+v^{2}=1\ ,$ the equation of the tangent $t\equiv TT$ in this point $T$ to the ellipse $\Xi$ is $\sqrt 2(bux+avy)-ab=0\ .$ Observe that $AA'\cdot CC'=BB'\cdot DD'$ $\Longleftrightarrow$ $\left|2u^{2}-1\right|=\left|2v^{2}-1\right|\ ,$ what is truly because $\left(2u^{2}-1\right)+\left(2v^{2}-1\right)=$ $2\left(u^{2}+v^{2}\right)-2=0\ .$[/color]

[b][size=117]1. – Alternative proof of the particular case.[/size][/b] $($ In my drawing, it has been considered a cartecian system at point $I$ $($ as the incenter of the square $ABCD ),$ the points $A(0, 6),$ $B(6, 0),$ $C(0,-6),$ $D(-6, 0),$ $T(-1.8, ...),$ where $T,$ is the tangency point of the line $(t),$ to the incircle $(w)$ of $ABCD,$ with $IT = r.$

We denote as $C'',$ $D'',$ the intersection points of the circumcircle $(I),$ of $ABCD$ with $IA = R,$ from the segment lines $CC',$ $DD',$ respectively. Also we denote as $E,$ $F,$ the midpoints of the segments $AC'',$ $DD'',$ respectively.

Because of $BD''\perp DD''$ $\Longrightarrow$ $BD''\parallel (t)$ $\Longrightarrow$ $BD''\perp CC''.$ So the right triangles $\bigtriangleup AC''C,$ $\bigtriangleup BD''D,$ are congruent and hence, we have that $AC'' = DD''$ $\Longrightarrow$ $C''E = DF.$

It is easy to prove that $TA' = TC',$ $TB' = TD'$ and also that $AA' = CC'',$ $BB'' = D'D''.$ So we have that:

$(AA')\cdot (CC') = (C'C'')\cdot (CC') = (IA)^{2}-(IC')^{2}= R^{2}-(IC')^{2}$ $,(1)$

$(BB')\cdot (DD') = (DD')\cdot (D'D'') = (ID')^{2}-(IA)^{2}= (ID')^{2}-R^{2}$ $,(2)$

From $(1),$ $(2),$ we conclude that it is enough to prove that $(IC')^{2}+(ID')^{2}= 2R^{2}$ $,(3)$

But $(IC')^{2}= (IT)^{2}+(C'T)^{2}= r^{2}+(C'T)^{2}$ $,(4)$ and $(ID')^{2}= (IT)^{2}+(D'T)^{2}= r^{2}+(D'T)^{2}$ $,(5)$

From $(4),$ $(5)$ $\Longrightarrow$ $(IC')^{2}+(ID')^{2}= 2r^{2}+(C'T)^{2}+(D'T)^{2}= 2r^{2}+(DF)^{2}+(IF)^{2}= 2r^{2}+R^{2}$ $= 2R^{2}$

$($ Because of $C'T = C''E = DF,$ $D'T = IF,$ $IA = ID = R$ and $2r^{2}= R^{2}$ $)$ and the proof is completed.

Kostas Vittas.

PS. I will post here next time, the synthetic proof I have in mind , for the proposed problem.

[b][size=117]2. – Synthetic proof of the proposed problem.[/size][/b] $($ In my drawing $IA = IC = 3.7,$ $IB = ID = 6.0,$ where $I,$ is the intersection point of diagonals of the given rhombus $ABCD$ $).$ Let $EF,$ the major axis of the inellipse $(\Xi)$ of $ABCD,$ tangents to it’s sidelines at their midpoints and let $KL$ be, the minor one $($ $E,$ between $D,$ $I$ and $K,$ between $A,$ $I$ $).$

We draw the circle $(I'),$ centered at the point $I,$ with radius $R' =IE = IF$ and we denote as $A_{1},$ $C_{1},$ the intersection points of the segment line $AC$ $($ $A,$ between $A_{1},$ $I$ $),$ from the tangent lines to $(I'),$ through the point $D.$ Because of the inellipse $(\Xi),$ tangent to $AD$ at it's midpoint, we conclude that the circle $(I'),$ tangents to $A_{1}D$ at it's midpoint also. Similarly the circle $(I'),$ tangent to $A_{1}B,$ $BC_{1},$ $C_{1}D,$ at their midpoints and so, the quadrilateral $A_{1}BC_{1}D,$ is a square, with the circle $(I),$ centerd at $I$ and radius $R = IB = ID,$ as it's circumcircle.

Also we denote as $S,$ the intersection point of the segment line $BD,$ from the line $(t),$ which tangents to inellipse $(\Xi),$ at a $($ mobile $)$ point $T$ and let $M,$ $N,$ be, the intersection points of the segment lines $CD,$ $AB$ respectively, from the line $(t).$

We will prove that $(AA')\cdot (CC') = (BB')\cdot (DD'),$ where $A',$ $B',$ $C',$ $D',$ are the orthogonal projections of $A,$ $B,$ $C,$ $D$ respectively, on the line $(t).$

$\bullet$ Through the point $S,$ we draw the line $(t'),$ tangent to the circle $(I'),$ at the point so be it $T',$ which intersects the segment lines $C_{1}D,$ $A_{1}B,$ at points $M',$ $N',$ respectively. It is well known that $MM'\parallel TT'\parallel AC\parallel NN'$ and it has already been proved that $(A_{1}A'_{1})\cdot (CC'_{1}) = (BB'_{1})\cdot (DD'_{1})$ $,(1)$ where $A'_{1},$ $B'_{1},$ $C'_{1},$ $D'_{1},$ are the orthogonal projections of $A_{1},$ $B,$ $C_{1},$ $D$ respectively, on the line $(t').$

From $(1)$ $\Longrightarrow$ $\frac{A_{1}A'_{1}}{BB'_{1}}= \frac{DD'_{1}}{C_{1}C'_{1}}$ $,(2)$ but $\frac{A_{1}A'_{1}}{BB'_{1}}= \frac{N'A'_{1}}{N'B}$ $,(3)$ and $\frac{DD'_{1}}{C_{1}C'_{1}}= \frac{M'D}{M'C_{1}}$ $,(4)$

$($ Because of $A_{1}A'_{1}\parallel BB'_{1}$ and $DD'_{1}\parallel C_{1}C'_{1}$ $).$

From $NN'\parallel AA_{1}$ and $MM'\parallel CC_{1}$ $\Longrightarrow$ $\frac{N'A_{1}}{N'B}= \frac{NA}{NB}$ $,(5)$ and $\frac{M'D}{M'C_{1}}= \frac{MD}{MC}$ $,(6)$

From $(2),$ $(3),$ $(4),$ $(5),$ $(6)$ $\Longrightarrow$ $\frac{NA}{NB}= \frac{MD}{MC}$ $,(7)$

From $DD'\parallel CC'$ and $AA'\parallel BB'$ $\Longrightarrow$ $\frac{NA}{NB}= \frac{AA'}{BB'}$ $,(8)$ and $\frac{MD}{MC}= \frac{DD'}{CC'}$ $,(9)$

From $(7),$ $(8),$ $(9)$ $\Longrightarrow$ $\frac{AA'}{BB'}= \frac{DD'}{CC'}$ $,(10)$ $\Longrightarrow$ $(AA')\cdot (CC') = (BB')\cdot (DD')$ and the proof is completed.

Kostas Vittas.