Let $ a,b,c\ge \frac{1}{5}$ such that $ a\plus{}b\plus{}c\equal{}1$. Show that:

\[ 27\ge \frac{2}{abc}\minus{}\left(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}\right)\ge 25\]

[quote="bobbym"]Hi;

For a = 1/2 and b= 1/4 and c = 1/4 we have:

$ \frac {2}{abc} \minus{} \left(\frac {1}{a^{2}} \plus{} \frac {1}{b^{2}} \plus{} \frac {1}{c^{2}}\right) \equal{} 28$

which is greater than 27, violating the constarints on the inequality.[/quote]

Yes, i'm sorry the upper bound is wrong, but the lower bound is still correct.

[quote="Pain rinnegan"][quote="fjwxcsl"]

     $ {\frac {275}{9}} \equal{} F(\frac {1}{5},\frac {1}{5},\frac {3}{5})\ge{F(a,b,c)}\ge{F(\frac {1}{5},\frac {2}{5},\frac {2}{5})} \equal{} 25.$[/quote]

Can you explain a bit more what you used ?(i hope not the computer :lol: )[/quote]

Let

$ a \equal{} \frac {1}{5} \plus{} \frac {2}{5}{\frac {x}{x \plus{} y \plus{} z}},b \equal{} \frac {1}{5} \plus{} \frac {2}{5}{\frac {y}{x \plus{} y \plus{} z}},c \equal{} \frac {1}{5} \plus{} \frac {2}{5}{\frac {z}{x \plus{} y \plus{} z}},{x,y,z}\ge{0}.$

[quote="Pain rinnegan"]Let $ a,b,c\ge \frac {1}{5}$ such that $ a \plus{} b \plus{} c \equal{} 1$. Show that:
\[ 27\ge \frac {2}{abc} \minus{} \left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right)\ge 25\]
[/quote]
The conclusion should be revised as follows,

\[ \frac{275}{9}\ge \frac {2}{abc} \minus{} \left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right)\ge 25.\]

Let $ a,b,c$ be  positive real numbers such that $a+b+c=1.$ Show that $$
abc(1+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\geq \frac{28}{27}$$
[color=#f00]Janous,Klamkin[/color]
Where?

Let $a_1,a_2,\cdots,a_n  (n\ge 2)$ be positive real numbers such that $a_1+a_2+\cdots+a_n=1.$ Prove that$$a_1a_2\cdots  a_n\left(1+\frac{1}{a^2_1}+\frac{1}{a^2_2}+\cdots+\frac{1}{a^2_n}\right)\geq\frac{n^3+1}{n^n}.$$

Let $ a,b$ be positive real numbers such that $a+b=1.$ Show that $$
ab(1+\frac{1}{a^2}+\frac{1}{b^2})\geq \frac{9}{4}$$
[b]Proof:[/b] $ab\leq \frac{1}{4}.$
$$ab(1+\frac{1}{a^2}+\frac{1}{b^2})\geq \frac{9}{4}\iff a^2b^2+1\geq \frac{17}{4}ab \iff (ab-\frac{1}{4})(ab-4)\geq 0.$$

leaving the upper bound part as it is wrong we can solve it by lagrangian multipliers our constrain function is a+b+c=1 so by differentiating with respect to x,y,z and finding whether the solution is concave or convex we can find the local minim and maxima.


7 kids are playing tag. Player 1 starts as 'it', and can tag anyone except him/herself. There is also the no tagbacks rule meaning that a player can't tag the person who tagged him/her last. Suppose all players are of equal skill and speed.

After 7 tags, what is the probability that player 1 will be 'it' again?

[hide]I got $856/3125$ as an answer... Is it correct?
If correct, i'll explain how i got it, if not, i'll need to look at it again :) [/hide]

[hide]
Let P(n) be the probability that the Player 1 is it after the nth tag.
Clearly, P(1)=P(2)=0.
Also, P(n)=(1/5)*(1-P(n-1)-P(n-2)) since there is 1/5 probability of being tagged if Player 1 isn't it, or wasn't it the last time.

P(1)=0
P(2)=0
P(3)=1/5
P(4)=4/25
P(5)=16/125
P(6)=89/625
P(7)=456/3125
[/hide]

[hide]Hm from P(3) to P(4) why did you multiply by $\frac{4}{5}$ Doesn't he still have a $\frac{1}{5}$ chance of being it?

Ah, nvm. You have to take into account that he could have been it the time before. Ok! Cool stuff.[/hide]


By Cauchy, $\left( 1+\frac{a_{1}^{2}}{a_{2}}\right) \left( 1+a_{2}\right) \geq \left( 1+a_{1}\right)^{2}$ etc. Now you just multiply them.


MATHCOUNTS
summer program
Mathcamp
\/closed

I'm interested in becoming a moderator in this site possibly somewhere around the middle school/getting started/intermediate level.  How are moderators selected and how would one go about becoming one?

We usually "recruit" moderators from among those that have proven to be active and reliable members of the community, and of course if they want to (there have been cases of people refusing :) ). At this point those forums you mention are literally full with moderators :) so you'll have to wait untill the forums become too populated to be easily moderated by the moderators there.

I'll volunteer too. I think by the time we need new moderators, it'll be about the 10000 user mark. I think I would do well in the Mathcounts forum or the Getting Started High School forum.

Billy

I am available if I am one day needed.  I think it would not be for a while, but with that new Arts & Sciences forum, we might need a few more moderators.

[quote="Valentin Vornicu"]Your application is noted. :)

Just as a side note: in general we would like to make mods from old, trusted members of the site (my idea of old is being at least 1 year old) since mods have the power to delete posts and topics. Of course exceptions appear (like ztbb's moderatorship in the MathCamp forum, etc.) or in cases in which a "young" member has already proven himself worthy (djmenez for example, moderator of the Spanish forums).[/quote]

I have plenty of time to prove myself...have I proven myself yet? :D

Can i be a administrator? :P
Or can i be a moderator?

[quote="Silverfalcon"]Hey, Valentin, just wondering were you the only administrator when MathLinks and AoPS were separate sites?

I knew we had two administrators at first in AoPS (rrusczyk and MCrawford) and I think DPatrick joined little bit later.

But I have no idea for MathLinks. :P Except it was pretty large growing community for Olympiad levels before it combined with AoPS. (I was going to join MathLinks but I forgot for couple weeks and when I came back, AoPS and MathLinks were one site.  :D )[/quote]Yes, I was the only Admin.

[quote="colts18"][quote="Valentin Vornicu"][quote="math92"]Can i be a administrator? [/quote]It is impossible.[/quote]

What do you mean???(Why Not?)[/quote] :huh: Because it just is impossible.

[quote="Valentin Vornicu"][/quote][quote="colts18"][quote="Valentin Vornicu"][quote="math92"]Can i be a administrator? [/quote]It is impossible.[/quote]

What do you mean???(Why Not?)[/quote][quote="Valentin Vornicu"] :huh: Because it just is impossible.[/quote]

But what if they make IMO and get a college degree and whatnot and AoPS hires them.  Then they'd be admins.  :)

[quote="bubala"]But what if they make IMO and get a college degree and whatnot and AoPS hires them.  Then they'd be admins.  :)[/quote]That's a totally different approach :) :D But for now math92 must be careful with his spam warnings ;)

Topic is officially closed.


Support symmetry inequalities probability counting distinguishability parameterization

I have always wondered why many esteemed professors of physics have included chapters entitled "The Arrow of Time" (or something similar) in their popular science books.  They seem to purport it as a "great mystery" why time only ever seems to move forward.  They state that all their physical law equations fully support "time reversal symmetry", which is obviously true.  Then, based on this (and also on observation of the world in general), they ask questions like "Why does a scrambled egg never unscramble itself?" and "Why do the ashes from a bonfire never burst into flame and turn back into a tall pile of wood?" etc.  OK, it's a "popular science" book for the non-physicist, so questions like these are interesting and worth answering.  But this is where I find it annoying, i.e. the way they answer these questions.  The esteemed professors always seem to answer these questions by using simple thermodynamic equations and talking about "entropy" etc.  But these "thermodynamic/entropic" explanations are not actually primary explanations at all, and they give the layman reader ABSOLUTELY NO EXPLANATION of what time actually is.  If the esteemed professors explained what time ACTUALLY WAS, then it would be BLATENTLY OBVIOUS why time only ever seems to move forward.  And it would also make these "thermodynamic/entropic" explanations totally unnecessary.  In fact, it would show these "thermodynamic/entropic" explanations to be nothing more than explanations based on "emergent properties".  And, consequently, it would show that these "emergent property thermodynamic/entropic" explanations do not actually "explain" anything at all.  Why don't these professors just explain the real nature of time, instead of giving explanations which don't actually "explain" anything?  To be quite honest, I have my suspicions, but I really don't know for certain.  It's definitely not because the "real" explanations would be too difficult for the general reader to understand.  The "real" explanations are actually a lot easier to understand that the "thermodynamic/entropic" non-explanations that are presently being put forward.  I can only think of two possible reasons for the behaviour of these professors: (1) There is some kind of "giant conspiracy" going on which has a goal that is being kept from us.  (But I think this is so unlikely that it's not even worth contemplating).  And: (2) These professors have actually managed to become professors of physics without ever understanding what time actually is THEMSELVES.  So, consequently, they give out the "thermodynamic/entropic" non-explanations because they are the only "explanations" they have to give.  And, amazing though it seems, I have actually managed to gather a significant amount of statistical evidence that suggests this second reason is actually correct.  A while ago, I sent a batch e-mail to 200 physics professors in Universities all over the world.  In this e-mail I asked them of they would please supply me with an exact definition of what time actually is.  Out of 200 sent, I actually received 11 replies.  I could not believe the nonsense that I read in some of these replies.  No two "definitions" were the same, and none were precise.  One was obviously just meant to "humour" me because it said: " Time is what a clock measures" and nothing else.  And some of the "definitions" actually blatently contradicted some of the other "definitions".  But the most amazing thing was this: they were ALL either incomplete or just dead wrong.  I know that this is not a statistically valid sample, but it amazed me none the less.  I am not saying that ALL physics professors are blissfully ignorant of what time actually is.  But I am saying that I have not come across one yet who does fully understand what time actually is.  Also, I found it HORRIFYING that two PROFESSORS of physics could actually give CONTRADICTORY DEFINITIONS of a concept as fundamental as time.

There is a PRECISE definition of what time actually is.  And when you hear it, the "real" explanations to several "mysteries of physics" will become totally obvious to you almost immediately.  If you actually manage to read to the end of this extremely long post without falling asleep, you will discover exactly which "mysteries of physics" I am talking about.

Well, you've read this far and, if you don't mind, I'd like to make an educated guess as to what you are thinking right now.  I suspect that you are thinking something along these lines: "This guy is obviously a total crank!  How could he possibly know something that professors of physics are not aware of?  I bet he's going to start spouting some mystical nonsense about "God" or "The Creator" or something like that in a minute."  Actually, if you are thinking along these lines, I wouldn't blame you.  It does seem a bit incredible, doesn't it?  But I would like to assure you that neither "God", "The Creator", "Aliens", "UFOs" or any other sort of mystical crap plays any part in my "real explanations".  All I have used to arrive at my conclusions is pure logic.  And if anybody can find fault with anything I say, I will be very happy to hear it!

Anyway, if anyone has found what I have written vaguely interesting, and wants me to give the "real" explanation as to why time only ever seems to move forward, then please post a message saying so.  If I get any interest, I will post the "real" explanation, and I will also post the "real" explanation as to why time seems to "slow down" under the influence of acceleration or a "gravitational field".  I have also written an explanation of a basic philosophical error which Einstein made in his theories of special and general relativity.  As you probably know, Einstein's relativity theories seemed to back up the results of the Michelson/Morley experiment by dispensing with the need for a "luminiferous aether" as a "carrier" for the propagation of light etc.  But I will explain that Einstein's philosophical error means that his theories of relativity actually logically support the EXISTENCE of a "carrier medium/aether", whilst Einstein unfortunately believed that his theories supported the exact opposite of this!  Modern physics is now starting to seriously explore this "luminiferous ether" theory again, but it's not called the "luminiferous ether" any more.  It's now called the "quantum foam" or the "string background" or something similar.  I actually have totally logical explanations for all of the above "things", and they are so simple that a child could probably understand the logic behind them.

I am also attempting to extrapolate my "real explanations" to a level where I can use them to explain other "mysteries" like the results of the "two slit experiment" etc.  My "two slit real explanation" should be finished soon, it just needs a bit of serious rewording to cut out the unnecessary "bloat".  So I will post it when it is completed, if people are actually interested in it.

I don't think that my "real explanations" would be of any use in attempting to refute "Bell's Inequality" or similar quantum non-locality "stuff" etc.  I actually believe that "Bell's Inequality" will actually soon be refuted by someone identifying a problem with timing constraints, detector efficency or simply discovering that an important "assumption" is not necessarily valid.

For example, see here:

[code]http://math.ucr.edu/home/baez/physics/Quantum/bells_inequality.html[/code]

OK, this post really is getting too long now.  As I said earlier, if anyone is interested in hearing my "real explanations", please put up a post saying so.  If I get any interest, I will post them within the next few days.

Okay I'll bite.

What have you got for a definition/explanation of time?

This of course (whether or not it is a mystery) is certainly a worthy subject of discussion.

As to Bell's theorem, it has good mathematical credentials.  Personally, I think there are several possibilities for accounting for the 'paradoxes'.  But that seems to be another thread.  Expound time my good man, expound time.

Thanks for replying Rogue Physicist.

I'm glad that someone actually replied, because I have almost finished removing the "bloat" from my "real explanations" and I will be posting them here either tomorrow or the next day.  I hope you enjoy reading them, and I will be very interested in reading any comments that you may make.

I would just like to reiterate that I am not some kind of "mystical crank".  I have a B.Sc. (Hons) in Mathematics, and I am also a total athiest.  So, any sort of mystical crap is as abhorrent to me as it is to most? other "apprentice mathematicians".

Fortunately (or unfortunately, depending on how you look at it) my "real explanations" require no understanding of mathematics whatsoever.  They just require the ability to follow a few simple non-mathematical logical steps.

Anyway, I won't say any more about them now.  I'll be posting them here shortly, so I'll just let them speak for themselves after I've posted them.

Regarding Bell's theorum, I totally agree with you that it has good mathematical credentials.  As you obviously know, the fundamantal reasoning behind it is based on simple probability theory.  If someone can understand that 5 chances out of 9 is a probability which is greater than 50%, then they are not very far away from understanding the underpinnings of Bell's Theorum.  So if the underlying mathematical reasoning could be faulted, I think it would have been faulted by now!

But, even though the mathematics is pretty unshakeable, I still think that Bell's Theorum will eventually be proved to be faulty.  Possibly through one of the non-mathematical reasons I mentioned in my earlier post.

OK, I'm off now.  My next post will contain my precise definition of "time" and the "real" explanations that I mentioned previously.  I hope you enjoy reading them.

[quote="doctortrag"] If the esteemed professors explained what time ACTUALLY WAS, then it would be BLATENTLY OBVIOUS why time only ever seems to move forward. ... These professors have actually managed to become professors of physics without ever understanding what time actually is THEMSELVES.  ...  Out of 200 [e-mails] sent, I actually received 11 replies.  I could not believe the nonsense that I read in some of these replies.  No two "definitions" were the same, and none were precise.  One was obviously just meant to "humour" me because it said: " Time is what a clock measures" and nothing else.  And some of the "definitions" actually blatently contradicted some of the other "definitions".  But the most amazing thing was this: they were ALL either incomplete or just dead wrong.  I know that this is not a statistically valid sample, but it amazed me none the less.  I am not saying that ALL physics professors are blissfully ignorant of what time actually is.  But I am saying that I have not come across one yet who does fully understand what time actually is.  Also, I found it HORRIFYING that two PROFESSORS of physics could actually give CONTRADICTORY DEFINITIONS of a concept as fundamental as time.

There is a PRECISE definition of what time actually is.  And when you hear it, the "real" explanations to several "mysteries of physics" will become totally obvious to you almost immediately.  If you actually manage to read to the end of this extremely long post without falling asleep, you will discover exactly which "mysteries of physics" I am talking about.[/quote]

[b]Time[/b] A dimension that enables two otherwise identical events that occur at the same point in space to be distinguished.
     -- [b] A Dictionary of Physics (ISBN: 0-19-280030-2)[/b]

So much for the "mysteries of physics."  ;)

[quote="doctortrag"]Time is a dimension that enables two otherwise identical events that occur at the same point in space to be distinguished.
-- A Dictionary of Physics (ISBN: 0-19-280030-2)

I wonder why none of the professors who bothered to reply to my e-mails didn't quote this "official" definition of "time" to me?

Perhaps, even though they don't seem to realise what time actually is, they DO realise that this "official" definition is completely meaningless and that it doesn't "define" or "explain" anything at all really.

If anybody bothers to read my next post, it will soon become very clear to them how unsatisfactory this "official" definition of "time" is.[/quote]

I wonder why you consider such a definition to be meaningless. It is as logically meaningful as a definition of anything else would be.  Though I suspect you're more interested in displaying your own peculiar genius than learning anything about physics.

Time is a dimension; i.e., a fundamental physical measure, as such it cannot be broken down into anything more basic. As for what the definition I gave explains, it explains the human ability to discriminate between one event and another. In other words, as far as we are concerned if A and B occur simultaneously at the same location then A and B are identical.

[quote]Time is a dimension that enables two otherwise identical events that occur at the same point in space to be distinguished.
-- A Dictionary of Physics (ISBN: 0-19-280030-2) [/quote]I too find this 'definition' disturbingly circular.  It invites one to postulate that a point in space can extend along another dimension, 'time' and remain a unique point.

It makes assumptions that one could make measurements of some kind that validly measure and identify 'points' in a space that have a kind of permanence of their own extending through time.

It seems to suggest that one could via some valid measuring process arrive at an identity between two 'points' at different times, which remain indistinguishable by that process.  The points can then only be distinguished by an associated parameter 'time', yet somehow we should accept the measuring process as valid.

Looked at differently, we are to embrace a system of measurement of 'space' independant of time, yet in the same breath accept 'time' as a dimension of equal validity and type as the other (three?) dimensions of space, at least in so far as classing 'time' as a dimension.   Presumably, we identify 'time' as a (true?) dimension in type, while admitting its uniqueness as an instance of dimension(s).

We are asked to embrace a philosophical system and outlook which treats 'time' as a dimension, presumably to manipulate in a manner analogous to spacial dimensions, while acknowledging its uniqueness, and (especially in the the assumed next step, namely the Minkowskian Spacetime of Special Relativity) peculiar difference from the other three dimensions, in the context of a system of 4-dimensional geometry and topology.

But there is no reason to treat 'time' as a dimension at all, at least in the sense of giving it an existance similar to spacial dimensions.  Since 'time' has no spacial dimension of its own, except on graphs and charts we have artificially constructed, it remains something entirely alien to true 'dimensions' as we experience them.

[quote="Magnara"]There is no true "definition" of time in any sense.  Nor is there of length, or mass for that matter.  We can describe all of these fine, we just have no idea what we're describing.  As a math person rather than a physics person, I imagine that we're in the same place as the 17th/18th century mathematicians who described continuity and limits, but without definitions of what they really were.  Unfortunately, those definitions were easier to come by than any definition for time/mass/length.[/quote]

Magnara,

I believe that you are close to the heart of the issue. However, to my mind, it is not as complicated as all that. Time is a dimension: a subspace of the real numbers. It is also an observable in the quantum mechanical sense, a coordinate of a position vector in $\mathbb{R}^4$.

The question is not so much what is time? we know the answer to that question well enough already. The question really is why can't we move backward along the time axis in the same way that we can move backward along any of the three other dimensions of space? Of course to me this seems a stupid question because if time is that thing which prevents multiple events from occuring simultaneously at the same point in space, then we cannot move backward in time by definition.

[quote]if time is that thing which prevents multiple events from occuring simultaneously at the same point in space, then we cannot move backward in time by definition.[/quote]Unfortunately, this argument if it had any force would work for forward motion in time also.  It obviously doesn't, and so has no force in either direction.

Instead, it points toward the real question, that is, whether or not time is a continuum or a quantized variable.

[quote="Rogue Physicist"][quote]Time is a dimension that enables two otherwise identical events that occur at the same point in space to be distinguished.
-- A Dictionary of Physics (ISBN: 0-19-280030-2) [/quote]I too find this 'definition' disturbingly circular.  It invites one to postulate that a point in space can extend along another dimension, 'time' and remain a unique point.[/quote] A point in space can and does extend along the dimension of time, and is yet unique (in space). Time is not space; space and time are different.

[quote="Rogue Physicist"]It makes assumptions that one could make measurements of some kind that validly measure and identify 'points' in a space that have a kind of permanence of their own extending through time.[/quote]You're suggesting they don't? I'm sitting at my computer right now, it is the very same machine it was yesterday in exactly the same position as yesterday, and yesterday the same was true of the day before it.

[quote="Rogue Physicist"]It seems to suggest that one could via some valid measuring process arrive at an identity between two 'points' at different times, which remain indistinguishable by that process.  The points can then only be distinguished by an associated parameter 'time', yet somehow we should accept the measuring process as valid.[/quote] Yep, it's called trusting your wrist watch.

[quote="Rogue Physicist"]Looked at differently, we are to embrace a system of measurement of 'space' independant of time, yet in the same breath accept 'time' as a dimension of equal validity and type as the other (three?) dimensions of space, at least in so far as classing 'time' as a dimension.   Presumably, we identify 'time' as a (true?) dimension in type, while admitting its uniqueness as an instance of dimension(s).[/quote]A dimension, according to dictionary.com, is a fundamental physical measure. There is no reason to insist that such a thing necessarily has a "space-like" quality.

[quote="Rogue Physicist"]We are asked to embrace a philosophical system and outlook which treats 'time' as a dimension, presumably to manipulate in a manner analogous to spacial dimensions, while acknowledging its uniqueness, and (especially in the the assumed next step, namely the Minkowskian Spacetime of Special Relativity) peculiar difference from the other three dimensions, in the context of a system of 4-dimensional geometry and topology.[/quote] Mechanically speaking time is the canonical conjugate of energy. Thus it is a physical observable. Including it as a dimension of a vector space $\in \mathbb{R}^4$ works. Because it works, arguing about it amounts to sound and fury signifying nothing. And with that, sir, I bid you good day.

Well I'm still waiting for the original poster to explain time his new way.

Presumably this will revolutionize our brains as we know them.

[quote="doctortrag"]Having a PhD in Physics seems to rob some people of their common sense. Remember two things: (1) Time cannot run backwards, and (2) "Bell's Inequality" is absolute nonsense; whether you have a PhD in Physics or not.[/quote]

I'm sorry, I cannot just let this one pass. As to (1) I completely agree. However, as to (2) anyone who believes Bell's inequality is nonsense doesn't have the slightest clue about whereof they speak.

Bell's inequality involves the measure of sets; it hasn't got anything to do with physics necessarily. It's real analysis; i.e. math; therefore, anyone who says it is "absolute nonsense" is an ignoramus.

The interesting part, and I'll give you the benefit of the doubt here as I believe this is what you were actually referring to, is Bell's theorem. Bell's theorem says that: 1) If logic holds in the quantum world, and 2) hidden variables exist (real but unmeasureable physical quantities) and 3) the actions of hidden variables are local then Bell's inequality follows.

The problem comes because every quantum experiment done to date has shown Bell's inequality is false.  That Bell's inequality is false implies that one or more of the three premises of Bell's theorem is also false. The falsification of each premise presents its own peculiar philosophical problems.

Any world where logic doesn't hold is one in which a given statement maybe both true and false simultaneously.

If hidden variables do not exist then nothing has any intrinsic reality; i.e., each of us owes his or her existence to the notion that someone else is observing him or her.

If the actions of hidden variables are not local then all physical actions are interconnected; i.e. entangled by a universal Butterfly effect.

[quote="-Job-"]Maybe time can be broken down as a function f(x, y, z). I can imagine that we could define time as being the difference between the x-coordinate of a particle A sitting still and of a particle B moving along the x-axis.
Of course for this to work we would need a definition of movement that is not dependent on time. Maybe one solution is to define the x-coordinate of the particle B recursively, say B.x = B.x+1 for a linear trajectory. Then the x-coordinate of the particle A which is not moving along the x-axis, is A.x = A.x. Hence we can let time t=B.x-A.x
In essence, maybe we can let x, y, or z act as a time dimension. This is a very naive approach, the intent is to imply that maybe time is a construct, or an observed relation, rather than something more. If humans had not a short or long term memory than it wouldn't be possible to diferentiate between present and past and there would be no perceived time, we would have to explain things differently, on different terms, but not necessarily incorrectly.[/quote]Nice attempt at unconventional thinking; however, if this were true then one should age according to one's spatial position. There is no evidence that this is the case. The most workable hypothesis thus far is that time is a dimension of a four dimensional manifold, but one along which regress is impossible.

It doesn't necessarily follow that one should age according to one's spatial position. Suppose that the two particles A and B generate time for other functions to use. Let's say that you age according to a function f(t). If f(t) uses the t obtained by the two particles above (t=B.x-A.x) then moving around does not necessarily make one older, though moving particle B about the x-axis might  ;) .
It would probably be more "realistic" to say a particle's behavior is a function of the time produced by neighboring particles.
I like the idea of defining movement recursively. If we have a particle A whose coordinates are defined as:
A.x = A.x + f(...)
A.y = A.y + g(...)
A.z = A.z + h(...)
Then we may identify the function f as the momentum or "energy" of the particle A along the x-axis. The difference between the functions f of a particle A and a particle B give the velocity of A relative to B, along the x-axis. It seems nice.


A lamppost in a park can illuminate an area of radius 40 feet. What is the minimum number of lampposts that must be installed in the park below so that the entire region is illuminated?
[asy]draw((0,0)--(0,3)--(5,3)--(5,0)--(0,0));
label("$200\sqrt2$",(0,3)--(5,3),N);
label("$120\sqrt2$",(5,3) --(5,0),E);[/asy]

I won't be rigorous on this, but every lamppost can illuminate a square of side 40rt(2), and there are 15 squares of such size, so 15 lampposts.


Let $ a,b$ and $ c$ be nonnegative real numbers with $ \frac{1}{a^2 \plus{} 1}\plus{}\frac{1}{b^2 \plus{} 1}\plus{}\frac{1}{c^2 \plus{} 1}\equal{}2$. Prove that: $ ab \plus{} bc \plus{} ca\leq\frac32$

Let $ x=\frac{1}{2a^2+2},y=\frac{1}{2b^2+2},z=\frac{1}{2c^2+2}$. Thus $ a=\sqrt{\frac{1-2x}{2x}},b=\sqrt{\frac{1-2y}{2y}},c=\sqrt{\frac{1-2z}{2z}}$. Then from the given condition $ x+y+z=1$ and we need to prove that \[ \sum_{cyc}\sqrt{\frac{(1-2x)(1-2y)}{2x\cdot 2y}}\le \frac 32\]
Or \[ \sum_{cyc}\sqrt{\frac{(y+z-x)(z+x-y)}{xy}}\le 3\]
But we have $ \sum_{cyc}2\sqrt{\frac{(y+z-x)(z+x-y)}{xy}}\le \sum_{cyc}\left(\frac{y+z-x}{y}+\frac{z+x-y}{x}\right)=6$. Hence we are done.

The inequality is also true when $ a,b,c$ are real numbers.

$ \frac{1}{a^2 \plus{} 1}\plus{}\frac{1}{b^2 \plus{} 1}\plus{}\frac{1}{c^2 \plus{} 1}\equal{}2$
$ \rightarrow \frac{a^2}{a^2 \plus{} 1}\plus{}\frac{b^2}{b^2 \plus{} 1}\plus{}\frac{c^2}{c^2 \plus{} 1}\equal{}1$
by Cauchy-Schwarz; \[ \sum_{cyc}\frac{a^2}{a^2 \plus{} 1} \geq \frac{(a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2\plus{}3}\]
$ \rightarrow a^2\plus{}b^2\plus{}c^2\plus{}3 \geq (a\plus{}b\plus{}c)^2$
$ \rightarrow ab\plus{}bc\plus{}ca \leq \frac{3}{2}$

[quote="tdl"]Let $ a,b$ and $ c$ be nonnegative real numbers with $ \frac {1}{a^2 \plus{} 1} \plus{} \frac {1}{b^2 \plus{} 1} \plus{} \frac {1}{c^2 \plus{} 1} \equal{} 2$. Prove that: $ ab \plus{} bc \plus{} ca\leq\frac32$[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103762


function
probability
integration
expected value
calculus
calculus computations

Hi guys,

I got a problem here that I hope you could help me with.

Assume that we have function f(t) and int(-inf,f(t)*t^p,+inf) is zero for p's up to M where M>=0. This is the property of a wavelet function with M vanishing moments.

In a specific problem, I need to calculate this:

[i]a[/i]=int(-inf, t*f(t) , +inf)/int(-inf, f(t) , +inf).

But since both the nominator and denominator are equal zero in case M>=1, I can't find [i]a[/i].

so, I was wondering if you could find a solution to this problem.

Thank you in advance

I'm pretty sure the answer is that you're asking the wrong question. I could say "not a number", but that's not very satisfying.
This looks like an expected value- and that isn't a particularly meaningful concept when we apply it to something with negative density.

why should it be wrong?! [i]a[/i] is the  mean value (centroid) of f(t). And this is how it's defined based on what I see in papers. Basically, I am trying to find the mean value of a wavelet function which has M vanishing moments. And having M vanishing moments means that the int(-inf , f(t)*t^p , inf) for all p<=M is zero. This property of wavelet functions makes it hard to use this definition of mean value.

I hope I made it a bit clearer, and I hope you could show an alternative for finding the mean value of a function with such a property.

Thank you.

When I see a mean value, I expect that it's with respect to a probability distribution. $ |f|^2$ might be a probability density, but $ f$ certainly isn't. I don't see how it ever makes sense to divide by $ \int_{\mathbb{R}}f$.

I'm not that good in math. This is how the mean value is defined in this paper:

J. Flusser and T. Suk, “Invariants for Recognition of Degraded 1-D Digital Signals,” Proc. 13th Int’l Conf. Pattern Recognition, vol. 2,
pp. 389-393, Vienna, Austria, 1996.

You can find the definition I gave before, right after equation 2 in this paper.


Please vote.Do express your views  here too.

[quote="1=2"]Canada is so gonna win.

What's Euro 2008 anyway???[/quote]
[url=http://www.google.com/search?hl=en&q=Euro+2008]Google it?[/url]
I didn't know what it was either... :P

I don't know why, but I'm a fan of the French team.


modular arithmetic
vector
geometry
geometric transformation
algebra
polynomial
floor function

Iranian National Olympiad (3rd Round) 2006 began an hour ago.

[b]First Day: Number Theory Exam[/b]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=612540#612540]Problem 1.[/url]
$n$ is a natural number. $d$ is the least natural number that for each $a$ that $gcd(a,n)=1$ we know $a^{d}\equiv\pmod{n}$. Prove that there exist a natural number that $\mbox{ord}_{n}b=d$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=612541#612541]Problem 2[/url]
$n$ is a natural number that $\frac{x^{n}+1}{x+1}$ is irreducible over $\mathbb Z_{2}[x]$. Consider a vector in $\mathbb Z_{2}^{n}$ that it has odd number of $1$'s (as entries) and at least one of its entries are $0$. Prove that these vector and its translations are a basis for $\mathbb Z_{2}^{n}$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=612542#612542]Problem 3[/url]
$L$ is a fullrank lattice in $\mathbb R^{2}$ and $K$ is a sub-lattice of $L$, that $\frac{A(K)}{A(L)}=m$. If $m$ is the least number that for each $x\in L$, $mx$ is in $K$. Prove that there exists a basis $\{x_{1},x_{2}\}$ for $L$ that $\{x_{1},mx_{2}\}$ is a basis for $K$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=612543#612543]Problem 4[/url]
$a,b,c,t$ are antural numbers and $k=c^{t}$ and $n=a^{k}-b^{k}$.
a) Prove that if $k$ has at least $q$ different prime divisors, then $n$ has at least $qt$ different prime divisors.
b)Prove that $\varphi(n)$ id divisible by $2^{\frac{t}{2}}$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=612544#612544]Problem 5[/url]
For each $n$, suppsoe $L(n)$ is the number of natural numbers $1\leq a\leq n$ that $n|a^{n}-1$. If $p_{1},p_{2},\dots,p_{k}$ are prime divisors of $n$, define $T(n)$ as $(p_{1}-1)(p_{2}-1)\dots(p_{k}-1)$.
a) Prove that for each $n\in\mathbb N$ \[n|L(n)T(n)\] b)Prove that if $gcd(n,T(n))=1$ then $\varphi(n)=L(n)T(n)$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=612545#612545]Problem 6[/url]
a) $P(x),Q(x)$ are polynomials with rational coefficients and $P(x)$ is not the zero polynomial. Prove that there exist a non-zero polynomial $Q(x)\in\mathbb Q[x]$ that \[P(x)|Q(R(x))\] b) $P,Q$ are polynomial with integer coefficients and $P$ is monic. Prove that there exist a monic polynomial $Q(x)\in\mathbb Z[x]$ that \[P(x)|Q(R(x))\]

5.   n | L(n)T(n)  ?

 6.  what is R(x)  ?

I may have changed the problem statements, not the context! Please tell me if anything is wrong!
In all problems, $S_{n}= \{1,2,\ldots,n\}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110415][b]Problem one[/b][/url]
Let $A$ be a family of subsets of $S_{n}$ such that no member of $A$ is contained in another. Sperner’s Theorem states that $|A|\leq{n\choose{\lfloor\frac{n}{2}\rfloor}}$. Find all the families for which the equality holds.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110416][b]Problem two[/b][/url]
Let $B$ be a subset of $\mathbb{Z}_{3}^{n}$ with the property that for every two distinct members $(a_{1},\ldots,a_{n})$ and $(b_{1},\ldots,b_{n})$ of $B$ there exist $1\leq i\leq n$ such that $a_{i}\equiv{b_{i}+1}\pmod{3}$. Prove that $|B| \leq 2^{n}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110417][b]Problem three[/b][/url]
Let $C$ be a (probably infinite) family of subsets of $\mathbb{N}$ such that for every chain $C_{1}\subset C_{2}\subset \ldots$ of members of $C$, there is a member of $C$ containing all of them. Show that there is a member of $C$ that is not a subset of any other member.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110419][b]Problem four[/b][/url]
Let $D$ be a family of $s$-element subsets of $S_{n}$ such that every $k$ members of $D$ have non-empty intersection. Denote by $D(n,s,k)$ the maximum cardinality of such a family.
a)	Find $D(n,s,4)$.
b)	Find $D(n,s,3)$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110420][b]Problem five[/b][/url]
Let $E$ be a family of subsets of $S_{n}$ with the property that for each $A\subset S_{n}$ there exist $B\in F$ such that $\frac{n-d}2\leq |A \bigtriangleup B| \leq \frac{n+d}2$. (where $A \bigtriangleup B = (A\setminus B) \cup (B\setminus A)$ is the symmetric difference of $A$ and $B$). Denote by $f(n,d)$ the minimum cardinality of such a family.
a)	Prove that if $n$ is even then $f(n,0)\leq n$.
b)	Prove that if $n-d$ is even then $f(n,d)\leq \lceil \frac n{d+1}\rceil$.
c)	Prove that if $n$ is even then $f(n,0) = n$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110421][b]Problem six[/b][/url]
The National Foundation of Happiness (NFoH) wants to estimate the happiness of people of country. NFoH selected $n$ random persons, and on every morning asked from each of them whether she is happy or not. On any two distinct days, exactly half of the persons gave the same answer. Show that after $k$ days, there were at most $n-\frac{n}{k}$ persons whose “yes” answers equals their “no” answers.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110422][b]Problem seven[/b][/url]
a) Prove that for any two natural numbers $n$ and $k$, there is a number $n(k,r)$ with the following property:
If we color the members of $S_{N}$($N>n(k,r)$) with $r$ colors, there exist numbers $a_{1}<a_{2}<\ldots<a_{k}$ whose sum doesn’t exceed $N$, such that for any $I,J\subset S_{k}$ whose maximums are equal, $\sum_{i\in I}a_{i}$ and $\sum_{j\in J}a_{j}$ are colored with the same color.
b) Suppose that we’ve colored the natural numbers with finite colors. Show that for any arbitrary large $k$, there exist $k$ numbers $a_{1}<a_{2}<\ldots<a_{k}$ such that the set $G = \{\sum_{i\in I}a_{i}| I\subset S_{k}\}$ is monochromatic.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=632795#632795]Problem 1[/url]
For positive numbers $x_{1},x_{2},\dots,x_{n}$, we know that $\prod_{i=1}^{s}x_{k}=1$. Prove that for each $m\geq n$ \[\sum_{k=1}^{s}x_{k}^{m}\geq\sum_{k=1}^{s}x_{k}^{n}\] [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=632796#632796]Problem 2[/url]
Find all real polynomials that \[p(x+p(x))=p(x)+p(p(x))\] [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=632797#632797]Problem 3[/url]
Find all real $x,y,z$ that \[\left\{\begin{array}{c}x+y+zx=\frac12\\ y+z+xy=\frac12\\ z+x+yz=\frac12 \end{array}\right.\] [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=632798#632798]Problem 4[/url]
$p(x)$ is a real polynomial that for each $x\geq 0$, $p(x)\geq 0$. Prove that there are real polynomials $A(x),B(x)$ that $p(x)=A(x)^{2}+xB(x)^{2}$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=632799#632799]Problem 5[/url]
Find the biggest real number $k$ that for each right-angled triangle with sides $a,b,c$: \[a^{3}+b^{3}+c^{3}\geq k(a+b+c)^{3}\] [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=632800#632800]Problem 6[/url]
$P,Q,R$ are non-zero polynomials that for each $z\in\mathbb C$, $P(z)Q(\bar z)=R(z)$.
a) If $P,Q,R\in\mathbb R[x]$ prove that $Q$ is constant polynomial.
b) IS the above statement correct for $P,Q,R\in\mathbb C[x]$?

1.CHOOMBAM
   we call a convex polyhedron,orderly if its sides be regular polygons.
   we call a orderly polyhedron,CHOOMBAM if no side be triangle.

	1.show that for every CHOOMBAM there exist an sphere that include all of  the vertices.

	2.show that utmost 3 polygons used in a CHOOMABAM.

	3.prove that there exist just one kind of a CHOOMBAM that the sides is pentagon & hexagon.(every two polygon must be                               	   used)

	4.for every $n>3$ a prism that the upper & inferior sides be two regular polygon and the lateral sides be $n$ square is a 	  CHOOMBAM.
	  prove that there is finite CHOOMBAM except these CHOOMBAMs.

2.thine tube
water flow in an infinite thin tube(we can let the tube be a line).we know that if a mulecule be in a poine with lenthe $x$,after $t$ second
it will be in a point with lenght $P(t,x)$.
prove that if $P$ be a polynomial with two variable then celerity of molecules are constante and equale.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=634184#634184]Problem 1[/url]
Suppose that $A\in\mathcal M_{n}(\mathbb R)$ with $\text{Rank}(A)=k$. Prove that $A$ is sum of $k$ matrices $X_{1},\dots,X_{k}$ with $\text{Rank}(X_{i})=1$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=634185#634185]Problem 2[/url]
$f: \mathbb R^{n}\longrightarrow\mathbb R^{m}$ is a non-zero linear map. Prove that there is a base $\{v_{1},\dots,v_{n}m\}$ for $\mathbb R^{n}$ that  the set $\{f(v_{1}),\dots,f(v_{n})\}$ is linearly independent, after ommitting Repetitive elements.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=634186#634186]Problem 3[/url]
Suppose $(u,v)$ is an inner product on $\mathbb R^{n}$ and $f: \mathbb R^{n}\longrightarrow\mathbb R^{n}$ is an isometry, that $f(0)=0$.
1) Prove that for each $u,v$ we have $(u,v)=(f(u),f(v)$
2) Prove that $f$ is linear

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=634188#634188]Problem 4[/url]
$f: \mathbb R^{n}\longrightarrow\mathbb R^{n}$ is a bijective map, that Image of every $n-1$-dimensional affine space is a $n-1$-dimensional affine space.
1) Prove that Image of every line is a line.
2) Prove that $f$ is an affine map. (i.e. $f=goh$ that $g$ is a translation and $h$ is a linear map.)

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=634196#634196]Problem 1[/url]
Prove that in triangle $ABC$, radical center of its excircles lies on line $GI$, which $G$ is Centroid of triangle $ABC$, and $I$ is the incenter.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=634197#634197]Problem 2[/url]
$ABC$ is a triangle and $R,Q,P$ are midpoints of $AB,AC,BC$. Line $AP$ intersects $RQ$ in $E$ and circumcircle of $ABC$ in $F$. $T,S$ are on $RP,PQ$ such that $ES\perp PQ,ET\perp RP$. $F'$ is on circumcircle of $ABC$ that $FF'$ is diameter. The point of intersection of $AF'$ and $BC$ is $E'$. $S',T'$ are on $AB,AC$ that $E'S'\perp AB,E'T'\perp AC$. Prove that $TS$ and $T'S'$ are perpendicular.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=634198#634198]Problem 3[/url]
In triangle $ABC$, if $L,M,N$ are midpoints of $AB,AC,BC$. And $H$ is orthogonal center of triangle $ABC$, then prove that \[LH^{2}+MH^{2}+NH^{2}\leq\frac14(AB^{2}+AC^{2}+BC^{2})\] [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=634199#634199]Problem 4[/url]
Circle $\Omega(O,R)$ and its chord $AB$ is given. Suppose $C$ is midpoint of arc $AB$. $X$ is an arbitrary point on the cirlce. Perpendicular from $B$ to $CX$ intersects circle again in $D$. Perpendicular from $C$ to $DX$ intersects circle again in $E$. We draw three lines $\ell_{1},\ell_{2},\ell_{3}$ from $A,B,E$ parralell to $OX,OD,OC$. Prove that these lines are concurrent and find locus of concurrncy point.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=634201#634201]Problem 5[/url]
$M$ is midpoint of side $BC$ of triangle $ABC$, and $I$ is incenter of triangle $ABC$, and $T$ is midpoint of arc $BC$, that does not contain $A$. Prove that \[\cos B+\cos C=1\Longleftrightarrow MI=MT\]

And the last one. This is the final exam.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111388]Problem 1 : Choombam[/url]
A regular polyhedron is a polyhedron that is convex and all of its faces are regular polygons. We call a regular polhedron a "[i]Choombam[/i]" iff none of its faces are triangles.
a) prove that each choombam can be inscribed in a sphere.
b) Prove that faces of each choombam are polygons of at most 3 kinds. (i.e. there is a set $\{m,n,q\}$ that each face of a choombam is $n$-gon or $m$-gon or $q$-gon.)
c) Prove that there is only one choombam that its faces are pentagon and hexagon. (Soccer ball)
[img]http://aycu08.webshots.com/image/5367/2001362702285797426_rs.jpg[/img]
d) For $n>3$, a prism that its faces are 2 regular $n$-gons and $n$ squares, is a choombam. Prove that except these choombams there are finitely many choombams.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110688]Problem 2 : Infinite pipe[/url]
A liquid is moving in an infinite pipe. For each molecule if it is at point with coordinate $x$ then after $t$ seconds it will be at a point of $p(t,x)$. Prove that if $p(t,x)$ is a polynomial of $t,x$ then speed of all molecules are equal and constant.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110694]Problem 3 : Cantor set in Z[/url]
For $A\subset\mathbb Z$ and $a,b\in\mathbb Z$. We define $aA+b: =\{ax+b|x\in A\}$. If $a\neq0$ then we calll $aA+b$ and $A$ to similar sets. In this question the Cantor set $C$ is the number of non-negative integers that in their base-3 representation there is no $1$ digit. You see \[C=(3C)\dot\cup(3C+2)\ \ \ \ \ \ (1)\] (i.e. $C$ is partitioned to sets $3C$ and $3C+2$). We give another example $C=(3C)\dot\cup(9C+6)\dot\cup(3C+2)$.

A representation of $C$ is a partition of $C$ to some similiar sets. i.e. \[C=\bigcup_{i=1}^{n}C_{i}\ \ \ \ \ \ (2)\] and $C_{i}=a_{i}C+b_{i}$ are similar to $C$.
We call a representation of $C$ a primitive representation iff union of some of $C_{i}$ is not a set similar and not equal to $C$.
Consider a primitive representation of Cantor set. Prove that
a) $a_{i}>1$.
b) $a_{i}$ are powers of 3.
c) $a_{i}>b_{i}$
d) (1) is the only primitive representation of $C$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110687]Problem 4 : Tiling of space[/url]
The image shown below is a cross with length 2. If length of a cross of length $k$ it is called a $k$-cross. (Each $k$-cross ahs $6k+1$ squares.)
[img]http://aycu08.webshots.com/image/4127/2003057947601864020_th.jpg[/img]
a) Prove that space can be tiled with $1$-crosses.
b) Prove that space can be tiled with $2$-crosses.
c) Prove that for $k\geq5$ space can not be tiled with $k$-crosses.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111394]Problem 5 : Calculating ruler[/url]
A caclculating ruler is a ruler for doing algebric calculations. This ruler has three arms, two of them are sationary and one can move freely right and left. Each of arms is gradient. Gradation of each arm depends on the algebric operation ruler does. For eaxample the ruler below is designed for multiplying two numbers. Gradations are logarithmic.
[img]http://aycu05.webshots.com/image/5604/2000468517162383885_rs.jpg[/img]
For working with ruler, (e.g for calculating $x.y$) we must move the middle arm that the arrow at the beginning of its gradation locate above the $x$ in the lower arm. We find $y$ in the middle arm, and we will read the number on the upper arm. The number written on the ruler is the answer.
1) Design a ruler for calculating $x^{y}$. Grade first arm ($x$) and ($y$) from 1 to 10.
2) Find all rulers that do the multiplication in the interval $[1,10]$.
3) Prove that there is not a ruler for calculating $x^{2}+xy+y^{2}$, that its first and second arm are grade from 0 to 10.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110681]Problem 6 : Translations[/url]
Assume that $C$ is a convex subset of $\mathbb R^{d}$. Suppose that $C_{1},C_{2},\dots,C_{n}$ are translations of $C$ that $C_{i}\cap C\neq\emptyset$ but $C_{i}\cap C_{j}=\emptyset$. Prove that \[n\leq 3^{d}-1\] Prove that $3^{d}-1$ is the best bound.
P.S. In the exam problem was given for $n=3$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111389]Problem 7 : Kearting[/url]
We have finite number of distinct shapes in plane. A "[i]convex Kearting[/i]" of these shapes is covering plane with convex sets, that each set consists exactly one of the shapes, and sets intersect at most in border.
[img]http://aycu30.webshots.com/image/4109/2003791140004582959_th.jpg[/img]
In which case Convex kearting is possible?
1) Finite distinct points
2) Finite distinct segments
3) Finite distinct circles

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110690]Problem 8 : Embedding in Q^n[/url]
We mean a traingle in $\mathbb Q^{n}$, 3 points that are not collinear in $\mathbb Q^{n}$

a) Suppose that $ABC$ is triangle in $\mathbb Q^{n}$. Prove that there is a triangle $A'B'C'$ in $\mathbb Q^{5}$ that $\angle B'A'C'=\angle BAC$.
b) Find a natural $m$ that for each traingle that can be embedded in $\mathbb Q^{n}$ it can be embedded in $\mathbb Q^{m}$.
c) Find a triangle that can be embedded in $\mathbb Q^{n}$ and no triangle similar to it can be embedded in $\mathbb Q^{3}$.
d) Find a natural $m'$ that for each traingle that can be embedded in $\mathbb Q^{n}$ then there is a triangle similar to it, that can be embedded in $\mathbb Q^{m}$.
You must prove the problem for $m=9$ and $m'=6$ to get complete mark. (Better results leads to additional mark.)

Thank you Mohammad Hossein.

And my special thanks to Omid Hatami for publishing all of the nice problems in a nice manner :)


induction
number theory open
number theory

Let $a_{1},a_{,}..,a_{k}$ are positive integers $((a_{i},a_{j})=1 \forall i \neq j)$
Find maximum $n \in N (n \neq 0)$ such that equation
$\sum_{i=1}^{k}( \frac{\prod_{j=1}^{k}a_{j}}{a_{i}}x_{i})=n$
have no positive integer root

There are no sulution ,isn't there ?  :(  I think it very nice  :)

The answer is:

$(k-1) \prod_{j=1}^{k}a_{j}-\sum_{i=1}^{k}( \frac{\prod_{j=1}^{k}a_{j}}{a_{i}})$.

We will prove this using induction.  For $k = 1$, the above expression is $-1$, which is indeed the smallest number that cannot be expressed as $x_{1}$ with $x_{1}\geq 0$.  Now assume the above is the correct answer for $k$.  Then

$a_{k+1}((k-1) \prod_{j=1}^{k}a_{j}-\sum_{i=1}^{k}( \frac{\prod_{j=1}^{k}a_{j}}{a_{i}}))$

is not expressible as $\sum_{i=1}^{k}( \frac{\prod_{j=1}^{k+1}a_{j}}{a_{i}}x_{i})$, and any larger multiple of $a_{k+1}$ is.  It follows that

$(\prod_{j=1}^{k}a_{j}) (a_{k+1}-1)+a_{k+1}((k-1) \prod_{j=1}^{k}a_{j}-\sum_{i=1}^{k}( \frac{\prod_{j=1}^{k}a_{j}}{a_{i}}))$
$=(k ) \prod_{j=1}^{k+1}a_{j}-\sum_{i=1}^{k}( \frac{\prod_{j=1}^{k+1}a_{j}}{a_{i}}) = n$

is not expessible as $\sum_{i=1}^{k+1}( \frac{\prod_{j=1}^{k+1}a_{j}}{a_{i}}x_{i})$.  But any integer $m$ larger than $n$ is.  Choose $i$ so that $m \equiv i\prod_{j=1}^{k}a_{j}(\mod a_{k+1})$.  Then $m-i\prod_{j=1}^{k}a_{j}$ is divisible by $a_{k+1}$ and larger than $a_{k+1}((k-1) \prod_{j=1}^{k}a_{j}-\sum_{i=1}^{k}( \frac{\prod_{j=1}^{k}a_{j}}{a_{i}}))$, so it is by induction it is expressible as $\sum_{i=1}^{k}( \frac{\prod_{j=1}^{k+1}a_{j}}{a_{i}}x_{i})$, as desired.


The real sequence $\{ x_n \}$ is defined by:
\[ x_1=a>0  \]
\[ x_{n+1} = 1-\frac{1}{x_n+2}  \]
Find $\lim_{n\to\infty}$.

It´s not difficult to prove by induction that $x_{n}>0$ for all natural $n$.
And so it´s clear that $0<x_{n}<1$. And now it´s just analyze the monotonic of this sequence.
Hence we can conclude that $x_{n}$ is bounded and monotonic, and so it´s a convergence sequence, and the value of it´s limit is $\frac{\sqrt{5}-1}{2}$.

[quote="blahblahblah"][quote="Altheman"]note that once you show that it is convergent, you can say for large n that $x_n=x_{n+1}$, and that is how he got the limit[/quote]

No you can't.  But if there is a limit, it has to satisfy the same relation as the members of the sequence.[/quote]

huh? i don't follow, it seems to me that is the same as saying that $L=a_n$ and $L=a_{n+1}$, which becomes $a_n=a_{n+1}$

That doesn't mean that $x_n=x_{n+1}$ for large $n$.

I think we must classify some cases of value $a$.

I think Altheman is right. For significantly large $n$, $x_n=x_{n+1}$. You just first have to prove that it does have a limit


Let a, b, c and x, y, z be non-negative numbers such that
a + b + c = x + y + z.
Prove that
ax(a + x) + by(b + y) + cz(c + z) ≥ 3(abc + xyz).

Apply CBS $ (a^2x\plus{}b^2y\plus{}c^2z)(yz\plus{}zx\plus{}xy)\geq xyz(a\plus{}b\plus{}c)^2$
 But $ (a\plus{}b\plus{}c)^2\equal{}(x\plus{}y\plus{}z)^2\geq 3(xy\plus{}yz\plus{}zx)$ therefore  $ a^2x\plus{}b^2y\plus{}c^2z\geq 3xyz$

 Similarly, $ ax^2\plus{}by^2\plus{}cz^2\geq 3abc$
 Adding up these inequalities yields the desired rezult.

  PS This inequqlity is propose from Vasile Cartoaje in MS 2005


i was just wondering are there any olympiad level books (dealing with problems/training i.e The USSR Olympiad Problem Book---> very good book) that people would reccommend

Some books that I am not sure of are:

1.) Mathematical Olympiad Challenges- [b]Titu Andreescu and Bogdan Enescu[/b]
2.) Mathematical Olympiad Treasures- [b]Titu Andreescu  and Razvan Gelca[/b]

if anyone owns these, or knows of any other good books post them here

-cheers

I own no. 2 which is a moderately good book. Dunno about no. 1, however I heard that it should be pretty good.

hmm..they should be available, but as i am in the US, i don't know for sure  :)  Anyway, valentin how hard is the subject matter of the first one..what level would you classify it as (very hard, hard, easy)

Nr. 1 isn't very hard, and nr. 2 isn't hard either.

nice...then maybe i will buy them. btw do u own them arne?

I do not own them but they were used for our IMO training last year, and in fact they were even too easy for us, and we are only the belgian team :D

oh wow...then i probably shouldn't use them as a primary resource for training, but i will buy them just the same---> any other books u would reccommend arne or anyone else?

try to solve all the previous solved questions on the forum :)

the advice that valentin gave is the best piece of advice u wud get ,splintercell .coz the problems posted on this forum would fill many books easily.
and btw,this site is BETTER than any book,coz i dont have any books but still managed a top 60 finish in this year's olympiad (scored only 60 out of 120  :D tho,the highest was 88,that's coz i bungled in two question i cud easily have done had there been no other things on my mind :D  :D (i.e physics olympiad practicals :) ).either way,i am not going for the training this year coz as the present list for selected person shows only 5 students selected from 12th grade.sorry abt boring the readers abt my exploits :D

BUT this site just makes my day (no other place where u get to read what some of the best brains have got to say(i.e solve :D )

okay then, now I know what I need to do :D


inequalities
search
inequalities proposed

Let $ a,b,c\geq 0$.Prove that:
$ a^4(b\plus{}c)\plus{}b^4(c\plus{}a)\plus{}c^4(a\plus{}b)\leq \frac{1}{12}(a\plus{}b\plus{}c)^5$

see for more http://www.mathlinks.ro/viewtopic.php?search_id=1705705862&t=30732

[quote="Inequalities Master"]Let $ a,b,c\geq 0$.Prove that:
$ a^4(b \plus{} c) \plus{} b^4(c \plus{} a) \plus{} c^4(a \plus{} b)\leq \frac {1}{12}(a \plus{} b \plus{} c)^5$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=82525


inequalities
function
inequalities proposed

Suppose $x,y,z$ are positive reals such that $x+y+z=1.$ Prove that \[(4+3x^{2})(4+3y^{2})(4+3z^{2})\geq81.\]


trigonometry
limit
calculus
calculus computations

[quote="bos1234"][u]Hence[/u] find the value of $ y \equal{} tan^{ \minus{} 1}x \plus{} tan^{ \minus{} 1}\frac {1}{x}$ for $ x > 0$, and determine the value of y when $ x < 0$
[/quote]
since $ y \equal{} \tan^{ \minus{} 1}x \plus{} \cot^{ \minus{} 1}x$, then for $ x > 0$, $ y\equal{}\frac{\pi}{2}$, and for $ x < 0$, $ y\equal{}\minus{}\frac{\pi}{2}$

But how did you determine that $ y \equal{} \pm\frac{\pi}{2}$? and why?

Yes, [b]Misan[/b], i.e. for any $ x\ne 0$,

$ f(x)\equiv\arctan x \plus{} \arctan\frac 1x \equal{} \frac {\pi}{2}\cdot\mathrm {sgn}\ x$ and $ f'(x) \equal{} 0$ a.s.o.

sgn means sign, so $ \textrm{sgn} ( \minus{} 7) \equal{} \minus{} 1$, $ \textrm{sgn} (\pi) \equal{} 1$ and $ \textrm{sgn}(0) \equal{} 0$.  Equivalently, $ \textrm{sgn}(x) \equal{} \frac{x}{|x|}$ for $ x \neq 0$, $ \textrm{sgn}(0) \equal{} 0$.  "a.s.o" means "and so on."

[quote="JBL"]"a.s.o" means "and so on."[/quote]
The standard English abbreviation is Latin rather than English: [i]etc.[/i]. I know that German uses its own resources rather than Latin to create the standard abbreviation [i]usw[/i]. If I were to try to render that German [i]usw.[/i] into idiomatic English, I'd probably use "asf." for "and so forth."

To really respect the Latin, one should use [i]etc.[/i] for things and [i]et al.[/i] for people - not a distinction that strictly adhered to by many English speakers. if "asf." were ever to catch on (unlikely), there would be no such distinction.

[quote="Virgil Nicula"]For any $ x\ne 0$,

$ f(x)\equiv\arctan x + \arctan\frac 1x = \frac {\pi}{2}\cdot\mathrm {sgn}\ x$ and $ f'(x) = 0$ a.s.o.[/quote]
[color=darkblue]$ f'(x) = 0$ for any $ x\in \mathcal R^*$ $ \implies$

exists $ \{k_1,k_2\}\subset \mathcal R$ so that  $ f(x) = \{\begin{array}{ccc} k_1 & \mathrm {if} & x < 0 \\
 \\
k_2 & \mathrm {if} & x > 0\end{array}$.

For ascertain these constants we can proceed such :

$ 1\blacktriangleright\lim_{x\to\ - \infty}f(x) = - \frac {\pi}{2}$ [b]or[/b] $ \lim_{x\nearrow 0}f(x) = - \frac {\pi}{2}$ [b]or[/b] $ f( - 1) = - \frac {\pi}{2}$ $ \implies k_1 = - \frac {\pi}{2}\ .$

$ 2\blacktriangleright\lim_{x\to\infty}f(x) = \frac {\pi}{2}$ [b]or[/b] $ \lim_{x\searrow 0}f(x) = \frac {\pi}{2}$ [b]or[/b] $ f(1) = \frac {\pi}{2}$ $ \implies k_1 = \frac {\pi}{2}\ .$

In conclusion, $ f(x) = \{\begin{array}{ccc} - \frac {\pi}{2} & \mathrm {if} & x < 0 \\
 \\
\frac {\pi}{2} & \mathrm {if} & x > 0\end{array}$ $ \implies$

for any $ x\in \mathcal R^*$, $ \arctan x + \arctan\frac 1x = \frac {\pi}{2}\cdot\mathrm {sgn}\ x\ .$[/color]


Please help me with the following trig question....it looks terribly simple, but I keep getting stuck!!

Prove tan10+tan70=tan50+tan60

I wonder, as a previous part of this question involved proving tan 50+tan60+tan70=tan 80, if you can get anywhere by using tan50+tan60-tan70 to try and get tan10?

Thanks.

uhh i proved it by using the fact that $ \tan{50}\plus{}\tan{60}\plus{}\tan{70}\equal{}\tan {80} ...a$.

lets say $ \tan{10}\plus{}\tan{70}\equal{}\tan{50}\plus{}\tan{60}$, then $ \tan{50}\plus{}\tan{60}\minus{}\tan{10}\equal{}\tan{70} ...b$.

$ a\minus{}b\equal{}2\tan{70}\equal{}\tan{80}\minus{}\tan{70}$.

$ \tan{70}\equal{}\tan{(80\minus{}10)}$

$ \tan(a\plus{}b)\equal{}\frac{\tan{a}\plus{}\tan{b}}{1\minus{}\tan{a}\tan{b}}$

so $ \tan(80\minus{}10)\equal{}\frac{\tan{80}\minus{}\tan{10}}{1\plus{}\tan{80}\tan{10}}$

$ \tan{80}\tan{10}\equal{}1$

so we have $ 2*\frac{\tan{80}\minus{}\tan{10}}{2}\equal{}\tan{80}\minus{}\tan{10}$

left hand side = right hand side

therefore, $ \tan{10}\plus{}\tan{70}\equal{}\tan{50}\plus{}\tan{60}$


A diver goes rapidly to the surface of the water from a depth of 4.08 m without expelling air from the lungs. By what factor increases the volume of his lungs when he gets to the surface?

The pressure decrease when the diver rises is equal to the pressure of water above him.
P=F/A

The force in this case is the weight of the water above the diver.
P=mg/A= Vgd/A

Here, d is density and V is volume.  Since V=Ah we can rewrite this as
P=gdh= (9.81 m/s^2)(1000 kg/m^3)(4.08 m)
 =4.00 x 10^4 Pa
Adding in atmospheric pressure, we see that the pressure drops from 1.413 x 10^5 Pa to 1.013 x 10^5 Pa.
Since volume is inversely proportional to pressure, the volume of the gas increases by the quotient of the two pressures, 1.40 x increase.


geometry
circumcircle
trigonometry
quadratics
Pythagorean Theorem
algebra
linear equation

[quote="Fang-jh"]Let circle O be the circumcircle of triangle ABC. Circle O1 touches the sides AB, AC and circle O at the points D,E,F, respectively. Denote by M the midpoint of line segment DE. AM intersects circle O at N. Prove that MF is perpendicular to NF.[/quote]

I'm sorry, i may not be understanding this correctly  :blush:  ....but do you mean "Circle $O_{1}$ touches..." = Circle $O_{1}$ is  [i]tangent to...[/i]?

Assume that $(O),\ (O_{1})$ are tangent internally. Then $(O_{1})$ is the mixtilinear A-incircle of the triangle $\triangle ABC$ with radius $r_{1}= \frac{r}{\cos^{2}\frac{A}{2}},$ where r is the triangle inradius. This was proved in the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=46418]Important circles[/url] using Casey's (generalized Ptolemy's) theorem. It is almost trivial to get it by trigonometry.

[hide="trigonometric calculation"]
Let C' be the midpoint of AB, $OC' = R \cos C,$ where R is the triangle circumradius. Let a parallel to AB through O meet the radius $O_{1}D$ of $(O_{1})$ at P. Since $O_{1}$ is on the internal bisector of $\angle A,$ we have $AD = \frac{r_{1}}{\tan \frac{A}{2}}.$ Then

$OO_{1}= R-r_{1}$

$O_{1}P = O_{1}D-PD = O_{1}D-OC' = r_{1}-R \cos C,$

$OP = C'D = AD-AC' = \frac{r_{1}}{\tan \frac{A}{2}}-\frac{c}{2}$

where c = AB, $\frac{c}{2}= R \sin C.$ Using Pythagorean theorem for the right angle triangle $\triangle OPO_{1},$

$OO_{1}^{2}= O_{1}P^{2}+OP^{2}$

$(R-r_{1})^{2}= (r_{1}-R \cos C)^{2}+\left(\frac{r_{1}}{\tan \frac{A}{2}}-R \sin C\right)^{2}$

Multiplying things out, we see that the absolute term of this quadratic equation for $r_{1}$ is zero, hence we get a linear equation for $r_{1}.$

$R^{2}-2Rr_{1}+r_{1}^{2}= r_{1}^{2}-2Rr_{1}\cos C+R^{2}\cos^{2}C+\frac{r_{1}^{2}}{\tan^{2}\frac{A}{2}}-\frac{2Rr_{1}\sin C}{\tan \frac{A}{2}}+R^{2}\sin^{2}C$

$r_{1}\left[r_{1}+2R(1-\cos C) \tan^{2}\frac{A}{2}-2R \sin C \tan \frac{A}{2}\right] = 0$

Discarding the root $r_{1}= 0$ and using some half angle trigonometric formulas,

$r_{1}= \frac{R[\sin C \sin A-(1-\cos C)(1-\cos A)]}{\cos^{2}\frac{A}{2}}= \frac{R[\cos A-\cos (C+A)+\cos C-1]}{\cos^{2}\frac{A}{2}}=$

$= \frac{R(\cos A+\cos B+\cos C-1)}{\cos^{2}\frac{A}{2}}= \frac{r}{\cos^{2}\frac{A}{2}}$
[/hide]

DE is perpendicular to the internal bisector of $\angle A,$ which cuts it at its midpoint M. Let T be the foot of a normal from M to AB.

$MT = AM \sin \frac{A}{2}= AD \cos \frac{A}{2}\sin \frac{A}{2}= \frac{O_{1}D}{\tan \frac{A}{2}}\cos \frac{A}{2}\sin \frac{A}{2}= r_{1}\cos^{2}\frac{A}{2}= r$

Thus the midpoint M of DE is the triangle incenter. Now

$\frac{O_{1}F}{O_{1}A}= \frac{r_{1}\sin \frac{A}{2}}{r_{1}}= \sin \frac{A}{2}$

$\frac{O_{1}M}{O_{1}F}= \frac{AO_{1}-AM}{O_{1}F}= \frac{r_{1}-r}{r_{1}\sin \frac{A}{2}}= \frac{1-\cos^{2}\frac{A}{2}}{\sin \frac{A}{2}}= \sin \frac{A}{2}$

Consequently, the triangles $\triangle AO_{1}F \sim \triangle FO_{1}M$ are similar by SAS and the angles $\angle FAN \equiv \angle FAO_{1}= \angle MFO_{1}\equiv \angle MFO$ are equal. Let the circumcircle tangent at F meet the internal bisector $AM \equiv AN$ of $\angle A$ at K (this is just for notation purpose). Then $\angle NFK = \angle FAN = \angle MFO.$ Since $OF \perp KF,$ it follows that $MF \perp NF$ as well.

If we assumed that $(O),\ (O_{1})$ are tangent externally, then $(O_{1})$ would be the mixtilinear A-excircle of the triangle $\triangle ABC$ with radius $r_{1}= \frac{r_{a}}{\cos^{2}\frac{A}{2}},$ where $r_{a}$ is the triangle A-exradius. We could then show in exactly the same way that M is the triangle A-excenter and that $MF \perp NF.$


polynomial
modular arithmetic
superior algebra
superior algebra unsolved

Let L = (3+sqrt(-3))/2 in the field Q[sqrt(-3)]. Assume L is prime in Z[sqrt(-3)].

Given that x = 1 mod L, prove that x^3 = 1 mod L^3.
Do the same thing again, this time with the 1's replaced with -1's.

I moved this to the superior algebra forum (under college playground) so please reply there.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=904947#904947[/url]

[quote="aidan"]What does the field $ \mathbb{Q}[\sqrt{-3}]$ mean?[/quote]
This question doesnt belong here. It should be in the Superior Algebra forum.

But since you asked I tell you. There are integers called "Gaussian Integers" and they have the form $ \{a+bi | a,b\in \mathbb{Z}\}$ and some nice properties. What he posted is called a "Quadaric Number Field" these are $ \{a+bi\sqrt{3}|a,b\in \mathbb{Q}\}$ they are similar to Gaussian integers instead you replace the $ \mathbb{Z}$ by $ \mathbb{Q}$ basically. (They are really nice for solving some diophantine equations).

Right idea, but I think you have some names incorrect.

$ \mathbb{Z}[i]$ is the gaussian integers.

$ \mathbb{Z}[i\sqrt{3}]$ are the eisenstein integers.

If you replace the Z with a Q, potentially you could replace integers with rationals. But I am not sure if that switch produces standard terminology.

[quote="Altheman"]
$ \mathbb{Z}[i\sqrt{3}]$ are the eisenstein integers.
[/quote]

Actually, I believe $ \mathbb{Z}[\omega]$ are the eisentein integers

Haha yea I think that seems more right. But I swear that I saw somebody calling those the eisenstein integers...

$ \rm{HINT}\quad X^{3}\minus{}a\equiv (X\minus{}a)^{3}\pmod{3}$


ratio
trigonometry
floor function
integration
calculus
complex numbers
geometric series

Let $x=\frac{\displaystyle\sum_{n=1}^{44} \cos n^\circ}{\displaystyle \sum_{n=1}^{44} \sin n^\circ}.$  What is the greatest integer that does not exceed $100x$?

The denominator is approximentaly 16.43.
The numerator is approximentaly 39.66.
So, the fraction is approx. 2.414.

Thus, the answer is 241.

[hide]Consider the sum $\sum_{n=1}^{44} \text{cis } n^\circ$. The fraction is given by the real part divided by the imaginary part.

The sum can be written $-1+\sum_{n=0}^{44} \text{cis } n^\circ=-1+\frac{\text{cis } 45^\circ-1}{\text{cis } 1^\circ-1}$ (by DeMoivre with geometric series)

$=-1+\frac{\frac{\sqrt{2}}{2}-1+\frac{i \sqrt{2}}{2}}{\text{cis } 1^\circ-1}=-1+\frac{\left( \frac{\sqrt{2}}{2}-1+\frac{i \sqrt{2}}{2} \right) (\text{cis } (-1^\circ)-1)}{(\cos 1^\circ -1)^2+\sin^2 1^\circ}$ (after multiplying by conjugate)

$=-1+\frac{\left( \frac{\sqrt{2}}{2}-1 \right) (\cos 1^\circ -1)+\frac{\sqrt{2}}{2}\sin 1^\circ +i\left( \left(1-\frac{\sqrt{2}}{2} \right) \sin 1^\circ +\frac{\sqrt{2}}{2} (\cos 1^\circ -1)\right)}{2(1-\cos 1^\circ)}$

$=-\frac{1}{2}-\frac{\sqrt{2}}{4}-\frac{i\sqrt{2}}{4}+\frac{\sin 1^\circ \left( \frac{\sqrt{2}}{2}+i\left( 1-\frac{\sqrt{2}}{2} \right) \right)}{2(1-\cos 1^\circ)}$

Using the half-angle tangent formula, this becomes $\left( -\frac{1}{2}+\frac{\sqrt{2}}{4}[\cot (1/2^\circ)-1] \right) +i\left( \frac{1}{2}\cot (1/2^\circ)-\frac{\sqrt{2}}{4}[\cot (1/2^\circ)+1] \right)$.

Dividing the two parts and multiplying each part by 4, the fraction is $\frac{-2+\sqrt{2}[\cot (1/2^\circ)-1]}{2\cot (1/2^\circ)-\sqrt{2}[\cot (1/2^\circ)+1]}$.

Although an exact value for $\cot (1/2^\circ)$ in terms of radicals will be difficult, this is easily known: it is REALLY BIG!

So treat it as though it were $\infty$. The fraction is approximated by $\frac{\sqrt{2}}{2-\sqrt{2}}=\frac{\sqrt{2}(2+\sqrt{2})}{2}=1+\sqrt{2}\Rightarrow \boxed{241}$.[/hide]

$\frac{\cos{x}+\cos{(45-x)}}{\sin{x}+\sin{(45-x)}}=1+\sqrt{2}$
...everything cancels, so $\lfloor 100(1+\sqrt{2}) \rfloor = \boxed{241}$

Sorry for bringing up a dead topic, but could you (altheman) explain the solution?

just expand that, and things cancel, i cannot really give a good reason why i would have thought of that, but it is a variation of pairing terms, so i suppose that is why i did that

well, there's one trig identity that states $\sin{A} + \sin{B} = 2\sin{\frac{A+B}{2}}\cos{\frac{A-B}{2}}$, and $\cos{A} + \cos{B} = 2\cos{\frac{A+B}{2}}\cos{\frac{A-B}{2}}$. use it with $A = x$ and $B = 45-x$, then everything cancels and we are left with $\cot{\frac{45}{2}}$.  Now just use half angle and we can find the solution.

btw, those two follow from angle add and angle subtract formula.  Plug in $A = x+y$ and $B = x-y$ and you'll see

Transform sum in products or rewrite the denominator sum of cosines; since is $x=45$, $\cos x-\cos(90-x)=\frac{\sqrt{2}}{\sqrt{2}}(\cos x-\sin x)=\sqrt{2}\cos(x+45)$(iii). writing the fraction as $\displaystyle\frac{\sum_{i=1}^{44} \cos i}{\sum_ {i=1}^{44} \sin i}-1+1=$$\frac{\sum_{i=1}^{44} \cos (i+45)}{\sum_ {i=46}^{89} \cos i}+1$ and using (iii) and rewriting (i haven't considerated $\sin 45$ and $\sin 90$ in the formulas  because i'm in a hurry :D )  we get $\sqrt{2}+1$. the answer is $100\cdot (1+1/41)=241$

[quote="seraph283"]$x=\frac{\displaystyle\sum_{n=1}^{44} \cos n^\circ}{\displaystyle \sum_{n=1}^{44} \sin n^\circ}=\frac{\cos 1 + \cos 2 + \dots+\cos 44}{\sin 1+\sin 2+\dots+\sin 44}=\frac{\cos (45-1)+\cos(45-2)+\dots+\cos(45-44)}{\sin 1+\sin 2+\dots+\sin 44} =\left(\frac{1}{\sqrt{2}}\right)\left(\frac{(\cos 1+\cos2+\dots+\cos44)+(\sin1+\sin2+\dots+\sin44)}{\sin1+\sin2+\dots+\sin44}\right) = \left(\frac{1}{\sqrt{2}}\right)\left(1+\frac{\cos 1 + \cos 2 + \dots+\cos 44}{\sin 1+\sin 2+\dots+\sin 44}\right)$

Therefore, $x=\left(\frac{1}{\sqrt{2}}\right)\left(1+x\right)$
$x\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)=\frac{1}{\sqrt{2}}$
$x=\frac{1}{\sqrt{2}-1} = 1+\sqrt{2}$
$\lfloor 100x \rfloor = \lfloor 100(1+\sqrt{2}) \rfloor = \boxed{241}$[/quote]
This is the best solution which explains itself, that's how I solved it. These sums of trig stuff with degrees shouts "easy work up ahead if you use pairing!!!"

How do you pull the $\frac{1}{\sqrt{2}}$ term out?

$\cos (a-b)=\cos a\cos b+sin a\sin b$ and $\cos 45=\sin 45=\frac{1}{\sqrt{2}}$.  He then factored $\frac{1}{\sqrt{2}}$ out.

I can't believe this is a number 11.

Anyone who doesn't feel like working through trigonometry could just throw up a prayer and integrate, and it would come out right. And that would be a free point in 30 seconds.

If you look at my post, you will see that that integral follows immediately...

[quote="azjps"]$ \frac {\displaystyle \sum_{n \equal{} 1}^{44} \cos n^{\circ}}{\sum_{n \equal{} 1}^{44} \sin n^{\circ}} \approx \frac {\int_{0}^{45} \cos n^{\circ}\, dn}{\int_{0}^{45} \sin n^{\circ}\, dn} \equal{} \sqrt {2} \plus{} 1$, how lucky.[/quote]

Could you please explain how you got the limits of integration?  Thanks!

I probably should have written that as $ \sum_{n \equal{} 1}^{44}\cos \left(\frac {\pi}{180} \cdot n\right) \approx \int_0^{45} \cos \left(\frac {\pi}{180} \cdot x\right)\, dx$.

The bounds were simply picked to be nice numbers; Altheman gave an explanation above as to why such an integration works as long as you pick bounds symmetric about $ 45/2^{\circ}$.

I don't really get Altheman's post.  Which post are you referring to?

Why is there a pi/180 in your integral?

Thanks!

Converting to radians (the integral is still the same thing; the goal is to have the limits of the integral to approximately equal to the limits of the sum), though of course after integrating, the factors of $ 180/\pi$ cancel out in the fraction.

I'm referring to post #4, where Altheman found that $ (*) \cos x \plus{} \cos (45 \minus{} x) \equal{} (\sqrt{2}\plus{}1)(\sin x \plus{} \sin (45 \minus{} x))$ where $ x$ is in degrees, and therefore that $ \int_0^{45} \cos\left(\frac{\pi}{180}\cdot x\right)$ $ \equal{} \int_0^{45/2} \cos\left(\frac{\pi}{180}\cdot x\right) \plus{} \int_0^{45/2} \cos\left(\frac{\pi}{180}\cdot (45 \minus{} x)\right)$ $ \equal{} (\sqrt{2} \plus{} 1) \int_0^{45/2} \sin\left(\frac{\pi}{180}\cdot x\right) \plus{} \sin\left(\frac{\pi}{180}\cdot (45 \minus{} x)\right)$ $ \equal{} (\sqrt{2}\plus{}1)\int_0^{45} \sin\left(\frac{\pi}{180}\cdot x\right)$, though if one had found the identity $ (*)$ first, the answer to the problem is obvious anyway.

p8: First we simplify $\cos k + cos(45-k) = 2 \cos \frac{45}{2} \cos \frac{45-2k}{2}$ and $\sin k + \sin(45-k) = 2 \sin \frac{45}{2} \cos \frac{45-2k}{2}$. But if we divide these, we get $\frac{\cos \frac{45}{2}}{\sin \frac{45}{2}} = 1+\sqrt2$. Thus $x = 1+\sqrt2 \implies \boxed{241}$.

remark: Wow this cancellation was pretty surprising!


[quote="236factorial"]In a polygon of nine sides, the length of each side is 8. What is the apothem to the nearest integer?[/quote]
[hide]
If you divide the 9-gon into 9 congruent isosceles triangles, each with their "top angle" equal to $\frac{360}{9}=40^\circ$ Making each of the base angles equal to $70^\circ$

Take just one of the isosceles triangles. The apothem would be the height of the triangle. If you draw the height, you have a right triangle, with angle measures of 90, 20, and 70 degrees. The side of the triangle opposite of the 20 degrees is equal to 4, and the side opposite the 70 degree angle is the apothem or height of this triangle.

So, if we make the apothem equal to $x$, then

$sin(20^\circ)=\frac{4}{x}$

[size=75][color=green]mod edit: \sin {blah} for sine![/color][/size]

$\sin {20^{\circ}}=\frac{4}{x}$

$x=\frac{4}{\sin {20^\circ}}$

$x\approx\frac{4}{0.342}$

$x\approx11.6959\approx12$[/hide]

[hide]
divide into 9 triangles. Becuase the base angles are congruent, these are isosceles triangles. 360/9=40. The top angle is 40 and the other 2 are 180-40 /2=70.

$\sin{20}=\frac{4}{b}\\b=\frac{4}{\sin{20}}$
b is appoxomately 12.

[/hide]


A 24-foot by 72-foot rectangular dance floor is completely
tiled with 1-foot by 1-foot square tiles. Two opposite corners
of the dance floor are connected by a diagonal. This diagonal
passes through the interior of exactly how many tiles?

I am thinking (once again, not sure) that the answer of a problem like this would just be the number of tiles on the longer side, maybe if the two rows of tiles are both even?  So would the answer be [b]72[/b]?

You use the Pythagorean Theorem.

$ 24$ and $ 72$ are the lengths of the legs, and we're looking for the length of the hypotenuse.

$ a^2 \plus{} b^2 \equal{} c^2$

$ 24^2 \plus{} 72^2 \equal{} c^2$

$ 5760 \equal{} c^2$

$ \sqrt{5760} \equal{} \sqrt{c^2}$

$ 24\sqrt{10} \equal{} c$

$ 75.9 \approx c$

Hmm...I'm getting an irrational number...what am I doing wrong?

izzy, ur finding  the length of the diagonal, when ur supposed to find how many tiles the diagonal passes through.

I believe you are having a super-stupid moment.

Well, I know it's not 72!!!

So YOU explain how to solve it.  :roll:

[quote="isabella2296"]But each tile is 1 foot by 1 foot, so...the length of the diagonal should be the same as the number of tiles, shouldn't it? Unless I'm having a super stupid moment... :huh:[/quote]

this method used only applies for squares, not rectangles.

Ah-ha!

[color=blue]* ernie found a way[/color]

You know how when you draw a diagonal of a 1 x 3 rectangle, you get a diagonal that passes through the three unit "boxes"?

You use that here, only blow up the proportions by multiplying by 24.

Thus, you have a 24 x 72 rectangle, with a diagonal that passes through [b]72 unit boxes[/b].
____________________________________________________________________________________________

This is my 250th post.  I thank all the AoPSers who are reading this response, as it is a turning point in my AoPS career.  I would like to thank the whole AoPS community for such a wonderful math atmosphere and for giving me the opportunity to complete this achievement.  And most of all, I would like to thank the fans, for all their support in my journey to reach these 250 posts.

Now wait for my ultra-speech at 500 posts!  :D  :D  :D


Find $ \sum_{r\equal{}1}^{n} r(r\plus{}1)(r\minus{}2)$, expressing your answer in a fully factorised form.

[quote="kunny"]$ \equal{} \sum_{r \equal{} 1}^n (r \minus{} 1)r(r \plus{} 1) \minus{} \sum_{r \equal{} 1}^n r(r \plus{} 1)$

$ \equal{} \frac {(n \minus{} 1)n(n \plus{} 1)(n \plus{} 2)}{4} \minus{} \frac {n(n \plus{} 1)(n \plus{} 2)}{3}$
[/quote]Is there a pattern in this? My way of getting the closed-form would be writing the whole expressions as a linear combination of $ \sum r^3,\sum r^2,\sum r,\sum1$. But is there anything special about the closed form when the summand is a product of consecutive numbers (like in $ (r\minus{}1)(r)(r\plus{}1)$ and $ r(r\plus{}1)$?)

Think Hockey-Stick Identity (incidentally one of the typical ways of proving the sum of consecutive powers formulas).


Find the polynomials $ f(x)$ having the following properties:

(i) $ f(0) \equal{} 1$, $ f'(0) \equal{} f''(0) \equal{} \cdots \equal{} f^{(n)}(0) \equal{} 0$
(ii) $ f(1) \equal{} f'(1) \equal{} f''(1) \equal{} \cdots \equal{} f^{(m)}(1) \equal{} 0$

From the hypotheses, it is easy to deduce that $ f(x) \equal{} x^n g(x) \plus{} 1 \equal{} (x\minus{}1)^m k(x)$

Equivalently, $ (x\minus{}1)^m k(x) \minus{} x^n g(x) \equal{} 1$. [1]

Since $ \gcd ( (x\minus{}1)^m , x^n ) \equal{} 1$ for all $ m,n$, we can find canonical polynomials $ K(x)$ and $ G(x)$ such that [1] holds.

Then the general solution must be of form:

$ k(x) \equal{} K(x) \plus{} h(x) x^n$ and $ g(x) \equal{} G(x) \plus{} h(x) (x\minus{}1)^m$ for some $ h(x) \in \mathbb{R}[x]$

So, all the polynomials to be found must be of form

$ f(x) \equal{} K(x) \plus{} h(x) x^n$ for any $ h(x) \in \mathbb{R}[x]$.

See a very much related question [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=463940[/url], with much more insight in the discussions following.


ABCD is a square side 1. Points A', B', C', D' are taken on the sides AB, BC, CD, DA respectively so that AA'/AB = BB'/BC = CC'/CD = DD'/DA = 1/n. The strip bounded by the lines AC' and A'C meets the strip bounded by the lines BD' and B'D in a square with an area 1/1985. Find n.

---Taken from the Kalva site---

Hey, I was just about to post this one too. I have a question with Kalva's solution:

[hide][quote]
$AP = \sqrt{1-x^2}$, so by similar triangles, $x/1 = \sqrt{1-x^2}/(1 - 1/n)$. Hence $x = 1/\sqrt{1 + (1 - 1/n)^2}$, [color=green]so the side of the small square is[/color] $x/n = 1/\sqrt{n^2 + (n-1)^2}$. It is easy to check that $n^2 + (n-1)^2 = 1985$ for $n = 32$.
[/quote]
How is the side of the small square $x/n$? :huh:
[/hide]

[hide] Let $C'E$ be a line perpendicular to $A'C$.
Let $F$, $G$, $H$, and $I$ be the vertices of the smaller square (clockwise order starting from the top of the smaller square).
We are given that $CC' = \frac{1}{n}$
Then $DC'=1-\frac{1}{n}$
Since the trinagle $C'CE$ and $DGC'$ are similar,
$1-\frac{1}{n}: \frac{1}{n}=GD: \frac{1}{\sqrt{1985}}$
Thus $GD=\frac{n-1}{\sqrt{1985}}$.
The triangle $B'DC$ and $BGC'$ are also similar,
So, $GD$:$DC'$=$GC'$:$B'C$.
Substitute $GD=\frac{n-1}{\sqrt{1985}}$, $DC'=1$,  and
$B'C=DC'=1-\frac{1}{n}$.
Hence, $GC'=\frac{(n-1)^2}{n\sqrt{1985}}$.
Now $GD^2$+$GC'^2$=$DC'^2$ by the Pythagoras theorem.
Then solve for n.

[/hide]


limit
trigonometry
function
calculus
calculus computations

find these 3 limits:  :D

$ \textbf{(i)}\;\lim_{y\to 0}\;\left( \frac{\sin y}{y}\right)^{\frac{1}{y}}$

$ \textbf{(ii)}\;\lim_{y\to 0}\;\left( \frac{\sin y}{y}\right)^{\frac{1}{y^{2}}}$

$ \textbf{(iii)}\;\lim_{y\to 0}\;\left( \frac{\sin y}{y}\right)^{\frac{1}{y^{3}}}$

Suppose $ f$ is a real-valued function which is of $ C^{n+1}$ class at some neighborhood of zero with $ f(0) = 1$, $ f^{(1)}(0) = 0, \ \cdots , f^{(n-1)}(0) = 0$.

Then $ \lim_{h\to0}\ \{ f(h) \}^{1/h^{n}}= \exp \left( \frac{f^{(n)}(0)}{n!}\right)$

Proof of this statement is quite easy. By Taylor's theorem, $ f(x) = 1+\frac{f^{(n)}(0)}{n!}x^{n}+o(x^{n})$ and hence $ \frac{\lnf(x)}{x^{n}}= \frac{f^{(n)}(0)}{n!}+o(1)$.

Applying this,

$ \textbf{(i)}\;\lim_{y\to 0}\;\left( \frac{\sin y}{y}\right)^{\frac{1}{y}}= 1$

$ \textbf{(ii)}\;\lim_{y\to 0}\;\left( \frac{\sin y}{y}\right)^{\frac{1}{y^{2}}}= e^{-\frac{1}{6}}$

$ \textbf{(ii)}$ means that there is $ 0<r<1$ such that $ \left( \frac{\sin y}{y}\right)^{\frac{1}{y^{2}}}$ lies between 0 and $ r$  for all sufficiently small $ x$. So

$ \textbf{(iii)}\;\lim_{y\to 0}\;\left( \frac{\sin y}{y}\right)^{\frac{1}{y^{3}}}= 0$

sweet. :coolspeak:

how 'bout these 3 limits ?  :D

$ \textbf{(i)}\;\lim_{y\to 0}\;\left( \frac{\tan y}{y}\right)^{\frac{1}{y}}$

$ \textbf{(ii)}\;\lim_{y\to 0}\;\left( \frac{\tan y}{y}\right)^{\frac{1}{y^{2}}}$

$ \textbf{(iii)}\;\lim_{y\to 0}\;\left( \frac{\tan y}{y}\right)^{\frac{1}{y^{3}}}$

The same arguments apply, but this time, $ \frac{\tan y}{y}=1+\frac13y^{2}+O(y^{4}),$

Hence,

[b](i)[/b] $ 1.$

[b](ii)[/b] $ e^{\frac13}$

[b](iii)[/b] $ \infty$, if it's $ \lim_{y\to0^{+}}$

(Note that sos440 was implicitly assuming a limit from the right in his third answer.)


A circle of radius e passes through the center of a square with side  :pi: . Find the difference of the areas of the non-overlapping regions.

Please don't do it if you've seen it before...

Won't it just be :pi:e:^2:  - :pi::^2: = :pi:(e:^2: - :pi:)?

Subtract the areas of both figures, the difference between the two regions will be the same for any area of overlap.  Consider the circular region A and the square region B.  Now let's call the overlapping area C.  We're trying to find the difference in areas of the two non-overlapping regions so (A-C) - (B-C) = A - B.  It can be seen that just finding the difference between the areas of the square and the circle will be our answer.

Good job.

This problem took my friend half an hour to figure out :p


Here is an easy problem for you guys to try out - see how long it takes you (or just do it if you want), relating to palindromes...

The Problem:

Sam Driver is a salesman. Last Monday, Sam left his hometown of Twin Forks and drove 112 miles due north to Painted Rock. On Tuesday, he left Painted Rock and drove 180 miles due east to Wabash. On Wednesday, he left Wabash and traveled in a southwesterly direction directly for home.

Sam kept a running total of the number of miles he traveled. He noted that when he left home on Monday, the last number recorded in his travel log was 606 miles. When he returned home on Wednesday, he said to himself, "This is remarkable! When I left home on Monday, the number of miles was a palindromic number. When I arrived in Wabash, I recorded another palindromic number in my travel log. And now, the number of miles recorded in my travel log is one less than a palindromic number."

So the question is, how many palindromes are there between the number Sam recorded in his travel log when he left home on Monday, and the number he recorded in the log when he arrived home on Wednesday night?

Try it out, it's not that bad - explain how you got your answer if you post a solution...

[hide]

On his trip, Sam travels $112+180+\sqrt{112^{2}+180^{2}}=504$ miles, so when he gets home his log reads $606+504=1110$.

I'm assuming you mean "between" exclusively, so the palindromes between 606 and 1110 are $6a6$ for any digit $a$ other than 0, $7a7$, $8a8$, and $9a9$ for any digit $a$, and 1001.  So $9+10+10+10+1=40$ palindromes.

   [/hide]


[quote="linzk3"]Let y=log2(sqrt(t)) and x=log8(t) where t<100.
Eliminate the variable t and rewrite the relationship so that y is a function of x.  Give the domain and range.

*in y=log2(sqrt(t)) its log base 2 and in x=log8(t) its log base 8[/quote]

[hide]We have the parametric equations:
$ x\equal{}\log_{8}{t}$
$ y\equal{}\log_{2}{\sqrt{t}}$
And $ t>0$

By the definition of logs, we have:
$ 8^{x}\equal{}t$
$ 2^{y}\equal{}\sqrt{t}\Rightarrow2^{2y}\equal{}t$.
Thus $ 2^{2y}\equal{}8^{x}\equal{}2^{3x}$.
And thus $ 2y\equal{}3x\Rightarrow{}y\equal{}\frac{3x}{2}$.

I think the domain is all real numbers?  :maybe:
[/hide]


calculus
derivative
integration
function
calculus computations

If f is a function that is continuous for all real numbers, then what is d( Integral of f(t)dt from 0 to x^2)/dx
thank you

\frac{d}{dx}\{F(x^2)-F(0)\}=2xF(x^2)=2xf(x^2).

I don't understand how you got that...you came from 2xF(x^2) to 2xf(x^2)

In addition how can you assume intergral of f(x)dx=F(x)+C

[quote="Kle"]In addition how can you assume intergral of f(x)dx=F(x)+C[/quote]
That wasn't an assumption, that was a definition. By the Fundamental Theorem of Calculus, every continuous function has an antiderivative (primitive). kunny was simply defining a new function $F$ to be that antiderivative. There's only one property of $F$ that you really know, and that's that $F'(x)=f(x).$ And that explain's kunny's last step.

Let me take another stab at explaining it. Let me define the following function:

$G(u)=\int_0^uf(t)\,dt.$

Then by the Fundamental Theorem of Calculus, we know that $G$ is differentiable and that $G'(u)=f(u).$

The problem then asks for $\frac{d}{dx}G(x^2).$ By the chain rule:

$\frac{d}{dx}G(x^2)=G'(x^2)\cdot\frac{d}{dx}(x^2)=2xf(x^2).$

RaduSorici's mistake was ignoring the chain rule.

At its core, my explanation is not really any different than kunny's, I'm just giving you some alternate notation.

Here's an exercise to see if you understand:

Let $h(x)=\int_{-\sqrt{x}}^{3\sqrt{x}}e^{-t^2}\,dt.$ For $x>0,$ compute $h'(x).$

First substitute $u=x^2$, then use the fundamental thm, then the chain rule.  Which is, now that I look at it, exactly what Kent Merryfield said.


How do I center latexed stuff? like
$ asasdf $
in my USAMTS file?

If you want to write an equation that's centred, write [code]\[x = y\][/code] to get
\[x = y\]

Otherwise, use the center environment:
\begin{center}
blah
\end{center}

For some reason it didn't work.  Am I doing this incorrectly, or does TexNic not recognize it?
\[ x=y \]

See what's going on over there?

Or you can write it by using displaystyle.
[code]$$x=y$$[/code]
Gives

\[x=y\]


ratio
LaTeX
analytic geometry
AoPSwiki
geometry
AMC
AIME

In $ \triangle ABC$ points $ D$ and $ E$ lie on $ \overline{BC}$ and $ \overline{AC}$, respectively. If $ \overline{AD}$ and $ \overline{BE}$ intersect at $ T$ so that $ AT/DT \equal{} 3$ and $ BT/ET \equal{} 4$, what is $ CD/BD$?
[asy]unitsize(2cm);
defaultpen(linewidth(.8pt));

pair A = (0,0);
pair C = (2,0);
pair B = dir(57.5)*2;
pair E = waypoint(C--A,0.25);
pair D = waypoint(C--B,0.25);
pair T = intersectionpoint(D--A,E--B);

label("$B$",B,NW);label("$A$",A,SW);label("$C$",C,SE);label("$D$",D,NE);label("$E$",E,S);label("$T$",T,2*W+N);

draw(A--B--C--cycle);
draw(A--D);
draw(B--E);[/asy]$ \textbf{(A)}\ \frac {1}{8}\qquad \textbf{(B)}\ \frac {2}{9}\qquad \textbf{(C)}\ \frac {3}{10}\qquad \textbf{(D)}\ \frac {4}{11}\qquad \textbf{(E)}\ \frac {5}{12}$

[hide="Solution"]
The above diagram is obtained from triangle similarities, where we start by letting $ a = [ATE]$ and $ x = [TEC]$ to get that

$ \frac {[ATB]}a = 4\implies[ATB] = 4a$, $ \frac {4a}{[BTD]} = 3\implies[BTD] = \frac {4a}3$, and $ \frac {a + x}{[TCD]} = 3\implies[TCD] = \frac {a + x}3$. Also,

$ \frac {\frac {4a}3 + \frac {a + x}3 + x}x = 5\implies\frac {5a + 4x}3 = 5x\implies x = \frac5{11}a$.

Now, our desired ratio is $ \frac {CD}{BD} = \frac {[ACD]}{[ADB]} = \frac {a + \frac5{11}a + \frac {a + \frac5{11}a}3}{4a + \frac43a} = \frac {\frac {16}{11}\left(1 + \frac13\right)}{4\left(1 + \frac13\right)} = \boxed{\textbf{(D)}\ \frac4{11}}$.
[/hide]

Edit: $ \text{\LaTeX}$ errors fixed.

I used double Menelaus. By Menelaus, we have
$ \frac{BD}{BC} \cdot \frac{CE}{EA}\equal{}\frac{1}{3}$
$ \frac{AE}{EC} \cdot \frac{CD}{DB} \equal{}\frac{1}{4}$
Let $ \frac{CD}{DB}\equal{}x$, then $ \frac{BD}{BC}\equal{}\frac{1}{x\plus{}1}$. So we have
$ \frac{1}{x\plus{}1} \cdot y \equal{}\frac{1}{3}$
$ \frac{1}{y\plus{}1} \cdot x\equal{}\frac{1}{4}$
Solving, $ x\equal{}\frac{4}{11}$

Hm... I just solved this problem with mass points; it was significantly easier than the coordinate solution on AoPSWiki; in fact, I might edit it in.

Anyway, the problem only gives some ratios, but the problem is general enough to consider a special case which still preserves the ratio. Therefore, an affine transformation can be applied which preserves the ratios, so you can assume that the lines are congruent.

[b]EDIT:[/b] I inserted my mass points solution into the wiki page. If anyone sees any errors, feel free to post or PM me.

What's "mass point", and how to use it?  :huh:

[quote="dgreenb801"]I used double Menelaus. By Menelaus, we have
$ \frac {BD}{BC} \cdot \frac {CE}{EA} \equal{} \frac {1}{3}$
$ \frac {AE}{EC} \cdot \frac {CD}{DB} \equal{} \frac {1}{4}$
Let $ \frac {CD}{DB} \equal{} x$, then $ \frac {BD}{BC} \equal{} \frac {1}{x \plus{} 1}$. So we have
$ \frac {1}{x \plus{} 1} \cdot y \equal{} \frac {1}{3}$
$ \frac {1}{y \plus{} 1} \cdot x \equal{} \frac {1}{4}$
Solving, $ x \equal{} \frac {4}{11}$[/quote]
We can alternately multiply the first two equations to get $ \frac {BC}{CD} \equal{} \frac {BD \plus{} CD}{CD} \equal{} 1 \plus{} \frac {BD}{CD} \equal{} 12.$

EDIT; Ah, never mind... I didn't catch the typo - it should have been $ AC$ instead of $ EC$, which multiplying the first two equations won't give a nice cancelation anymore...

Mass point geometry is a way of using the concept of the lever to solve geometric problems involving cevians and ratios. It usually tends to yield faster solutions, but is somewhat of a hard concept to understand. Click  [url=http://sanjosemathcircle.org/handouts/2007-2008/20071114.pdf]here[/url] for more information.

So far, I am able to apply it. I am currently at the process of understanding how and why it works. After you are comfortable, try AIME I 2009 Problem 5 using mass points.

It's actually not that hard to understand...
If you have a line, $ AC$, with a point $ B$ between $ A$ and $ C$, such that $ AB/BC \equal{} x/y$, then we may say that $ AB\equal{}xk$ and $ BC\equal{}yk$ for some real $ k$.

Alternatively, we can pretend $ AC$ is a "rod", and $ B$ is a "fulcrum" such that $ AC$ is sitting on $ B$ like a see-saw. If we place a weight (i.e. "mass") on $ A$, of $ y$, for example, we must place a weight of $ x$ on $ C$ to preserve the ratio (this is a definition). Now $ B$ must have a mass of the sums of whatever it's balancing (again, you can treat this as a definition if you don't understand the physics behind it) so the mass of $ B$ is $ x\plus{}y$ to make $ AC$ balance on $ B$.

It's harder to explain than to understand...

[hide="Faster Solution?"]
Scale the diagram so that $ BT \equal{} 4$ and $ TE \equal{} 1$. Draw a line parallel to $ \overline{BE}$ from $ D$ that hits $ \overline{AC}$ at $ F$. $ \frac {TE}{DF} \equal{} \frac1{DF} \equal{} \frac {AT}{AT \plus{} TD} \equal{} \frac34\implies DF \equal{} \frac43$. Thus, $ \frac {CD}{BD} \equal{} \frac {CD}{BC \minus{} CD} \equal{} \frac {DF}{BE \minus{} DF} \equal{} \frac {\frac43}{5 \minus{} \frac43} \equal{} \boxed{\textbf{(D)}\,\frac4{11}}$. I like this technique since it's usually a few seconds faster than mass points. :P
[/hide]

[hide="Solution Using Area"]Draw $CT$, and without loss of generality, let $[ECT]=1$. Then, $[BTC]=4$. If we let $CD/BD=r$, we find that $[TDB]=\frac{4}{r+1}$, and $[TDC]=\frac{4r}{r+1}$. Also, $[ABT]=3[TDB]=\frac{12}{r+1}$ and $[AET]=\frac{1}{4}[ABT]=\frac{3}{r+1}$. Since $[ATC]=3[TDC]$, we have: \[\frac{3}{r+1}+1=\frac{12}{r+1}\] Multiplying both sides by $r+1$, we find: \[3+r+1=12r\]\[11r=4\]\[r=\boxed{\frac{4}{11}}\].[/hide]

From NICE's Areas document.

[hide=Solution]Applying Menelaus's Theorem gives $1=\frac{DA}{AT} \cdot \frac{TE}{EB} \cdot \frac{BC}{CD}$. We have $\frac{DA}{AT}=\frac{4}{3}$ and $\frac{TE}{EB}=\frac{1}{5}$, so $\frac{BC}{CD}=\frac{15}{4}$. The answer is hence $\frac{4}{15-4}=\frac{4}{11} \implies \boxed{\text{(D)}}$.[/hide]

[hide=Sol Using Mass Points]

We see that since $AT/DT=3$ and $BT/ET=4$, we get that we can let the mass of $T$ be $20$. That way, we get that the mass of $B$ is $4$ and the mass of $E$ is $16$. Similarly, since the mass of $T$ is 20, we get that the mass of $A$ is $5$ and the mass of $D$ is $15$. Using this, we see that since the mass of $D$ is $15$ and the mass of $B$ is $4$, we get that the mass of $C$ is $11$. Using this, we get that $BD: CD=11:4$. This means that $$CD/BD=4/11.$$


calculus
integration
LaTeX
calculus computations

Please use Latex correctly, Are you referring to

$ \int\frac{1}{(3^{x}\plus{}1)(x^{2}\plus{}1)}\,dx$

or to

$ \int\frac{x^{2}\plus{}1}{3^{x}\plus{}1}\,dx$?

[quote="the_wonder"]$ \int{\frac{1}{(3^{x}\plus{}1)(x^{2}\plus{}1)}}dx$

Sorry :)[/quote]

hmmm......i don't think this integral will be doable ! you can easily do their [i]sum [/i]but not their [i]product[/i]... i don't think it is possible !  :maybe:


function
real analysis
number theory solved
number theory

A multiplicative function is one that satisfies:
$f(m\cdot n)=f(m)\cdot f(n)\mid gcd(m,n)=1$

Can anyone prove that $\displaystyle\sum_{1=1}^{n}gcd(i,n)$ is multiplicative?

$\sum_{i=1}^{n}gcd(i,n)=\sum_{d|n}d\varphi(\frac{n}{d})$ (since number of such $1\leq i\leq n$, that $gcd(i,n)=d$ is $\varphi(\frac{n}{d})$ ) $=\sum_{d|n}\frac{n}{d}\varphi(d)=n\sum_{d|n}\frac{\varphi(d)}{d}$.
And since $\varphi(n)$ is multiplicative, $\frac{\varphi(n)}{n}$ also multiplicative, and using known fact, that if $f(n)$ is multiplicative, then $\sum_{d|n}f(d)$ is multiplicative, we get, that $g(n)=\sum_{d|n}\frac{\varphi(d)}{d}$ is multiplicative and $\sum_{i=1}^{n}gcd(i,n)=n*g(n)$ is also multiplicative, q.e.d.

This is one part of problem from IMO Shortlist 2004

And the other thing: don't post problems into Solved Section  :!:

pavel_kozlov: Even shorter - the convolution of two multiplicative function is multiplicative. Hence your first equality suffice.


A coin is flipped eight times, and the sequence of heads and tails occurring is recorded. How many distinct sequences are possible?

This is just $ 2 \times 2 \times 2 \times \ldots \times 2\equal{}2^8\equal{}\boxed{256}$.


Given $ x,y,z$ real numbers show the inequality:

$ 3(x^2y \plus{} y^2z \plus{} z^2x)(xy^2 \plus{} yz^2 \plus{} zx^2) \ge xyz(x \plus{} y \plus{} z)^3$

We don't need expanding.  :)

I will prove the [b]equivalent problem[/b]:

Given that $ a, b, c$ are any real numbers, prove that;

$ 3(x^4y^2 \plus{} y^4z^2 \plus{} z^4x^2)(x^2y^4 \plus{} y^2z^4 \plus{} z^2x^4) \ge x^2y^2z^2(x^2 \plus{} y^2 \plus{} z^2)^3$

[b]Solution.[/b]

Using Cauchy - Schwarz Inequality:

$ 3(x^4y^2 \plus{} y^4z^2 \plus{} z^4x^2)(x^2y^4 \plus{} y^2z^4 \plus{} z^2x^4) \equal{}$

$ \equal{}3(x^4y^2 \plus{} y^4z^2 \plus{} z^4x^2)(x^4z^2 \plus{} y^4x^2 \plus{} z^4y^2)$

$ \geq 3(x^4yz \plus{} y^4zx \plus{} z^4xy)^2 \equal{} 3x^2y^2z^2(x^3 \plus{} y^3 \plus{} z^3)^2$

It is enough to prove that $ 3x^2y^2z^2(x^3 \plus{} y^3 \plus{} z^3)^2 \geq x^2y^2z^2(x^2 \plus{} y^2 \plus{} z^2)^3$

which follows from Holder's inequality:

$ (1 \plus{} 1 \plus{} 1)(x^3 \plus{} y^3 \plus{} z^3)(x^3 \plus{} y^3 \plus{} z^3) \geq (x^2 \plus{} y^2 \plus{} z^2)^3$

Proof is complete.

but it don't hold for all reals anyway, try a=1, b=0, c=-1 gives $ \minus{}3>0$  :)

[quote="great math"]Personally, Shaam. your solution is right although I found that you proved for positive real numbers.  :wink: . Anyway, there is a nice note on your solution.( only use Cauchy- Schwartz)[/quote]
Shaam's solution is right, but the question is wrong: it should be $ \forall x, y, z\geq0$. It is funny that Shaam assumed $ x: \equal{}x^2,......$ which meant that he assumed that $ x, y, z\geq0$. :lol:


geometry
trapezoid
geometric transformation
reflection
perpendicular bisector

Consider $ \triangle ABC$ with $ AB\equal{}AC$, and points $ M$ on $ AB$ and $ N$ on $ AC$.
If $ P \equal{} BN \cap CM$, show that $ MN \parallel{} BC \iff \angle APM \equal{} \angle APN$.

[b]First part: if $ MN\|BC$, then $ A\hat{P}N=A\hat{P}M$[/b]

$ BCNM$ is an isosceles trapezium, so $ PB=PC$ (to see why, draw perpendiculars to $ BC$ from $ M$ and $ N$ and look for congruent triangles). We also have $ AB=AC$, so $ AP$ is the perpendicular bisector of $ BC$, and hence the bisector of $ B\hat{A}C$. Now the result becomes obvious (just angle chase).

[b]Second part: if $ A\hat{P}N=A\hat{P}M$, then $ MN\|BC$[/b]

Reflect everybody across $ AP$. Clearly line $ \line{BN}$ reflects to line $ \line{CM}$. If $ AP$ is not the bisector of $ B\hat{A}C$, then the reflection of $ C$ will be in ray $ \ray{NB}$ and will be different from $ B$, a contradiction ($ AC'=AB$, angle chase again). So $ AP$ is the bisector of $ B\hat{A}C$. From this, we see thet $ M$ is the reflection of $ N$, so $ AP$ is perpendicular bisector of $ MN$. But it is already the perpendicular bisector of $ BC$, so we are done.


what are the integer solutions to

x^2 + 2 = y^3

?

See http://www.mathlinks.ro/Forum/viewtopic.php?t=15408 or http://www.mathlinks.ro/viewtopic.php?t=84491 .


linear algebra
matrix
geometry
3D geometry
sphere
vector
real analysis

I think that it is really good exercise ;)
consider the set of all $2*2$ matrices $X\in \cal{M}$ of rank $1$.
a)show that in a neighborhood of matrix$X_0=\left(\begin{array}{cc}1&0\\\ 0&0\end{array}\right)$ S is diffemorphic two a two dimensional disc :)
b)is this true locally for all matrices in S?
c)Describe S globally.(how many connected component does it have?is it closed ?if not what are limit points  and how does S approach them?what is intersection of S with the unit sphere in $\cal{M}$)

$f: \mathbb R^4-{0}\to\mathbb R$, $f(a,b,c,d)=ad-bc$ has $\nabla f(a,b,c,d)=(d,-c,-b,a)\neq 0$.
so $f^{-1}(0)$ is quite nice, but i'm not sure about the dimension ...

Global description: A matrix of rank 1 is a product $uv^t$ for two nonzero vectors $u,v$. This product is not unique, since we can multiply $u$ by a constant and $v$ by the inverse of that constant. Thus $S$ is the direct product of two projective lines and a copy of the field with the origin removed. If the field is $\mathbb{R}$, this is the direct product of the standard torus and $\mathbb{R}^*$. That's locally 3-dimensional everywhere- the original question is just wrong.