geometry
incenter
ratio
inradius
AMC
AIME
number theory

Triangle $ABC$ has $AB=21$, $AC=22$, and $BC=20$.  Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$, respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$.  Then $DE=m/n$, where $m$ and $n$ are relatively prime positive integers.  Find $m+n$.

i got different...

[hide="solution"]

Clearly $\triangle ADE \sim \triangle ABC$, draw the perpendicular from $A$ to $\overline{BC}$, it intersects $\overline{DE}$ at $X$ and $\overline{BC}$ at $Y$

so we have $\frac{DE}{BC}=\frac{AX}{AY}$

so we find the altitude by $A=rs=AY*\frac{1}{2}*20 \implies AY=\frac{rs}{10}$

then we note that since $DE$ passes through the incenter, we find the distance between those lines to be $r$

so we have

$\frac{DE}{20}=\frac{ \frac{rs}{10}-r }{\frac{rs}{10}}=\frac{rs-10r}{rs}=1-\frac{10}{s}$

so $DE=20*\left( 1- \frac{20}{63}\right)=\frac{20*43}{63}=\frac{860}{63} \implies m+n=\boxed{923}$[/hide]

[hide="Answer"]The altitude from $A$ to $BC$ is perpendicular to $DE$, and we call the altitude $AF$ and the point where $DE$ and $AF$ intersect $G$. $A=\sqrt{(31.5)(11.5)(10.5)(9.5)}=\frac{21\sqrt{1311}}{4}$, and we have $r=\frac{A}{s}=\frac{\sqrt{1311}}{6}$, while the altitude has length $\frac{21\sqrt{1311}}{40}$, so the ratio $\frac{AG}{AF}=\frac{20}{63}$, and therefore $\frac{DE}{BC}=\frac{DE}{20}=1-\frac{20}{63}\Rightarrow DE=\frac{860}{63}\Rightarrow 860+63=\boxed{923}$, since they increase inversely.[/hide]

[b]EDIT:[/b] Fixed a small error in my solution (didn't say what I wanted to correctly).

[hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=444[/img]

Clearly $\triangle ADE \sim ABC$. Since we don't know any of the ratios of their sides, we need something else--let's try their altitudes, since the altitude of $\triangle ADE$ is equal to the altitude $h$ minus the inradius $r$ of $\triangle ABC$, or $h-r$.

But before we do that, we need to establish a relationship between $h$ and $r$. Since the semiperimeter $s = \frac{63}{2}$, we have \[ [ABC] = \frac{63r}{2},  \] since $K=rs$. Also, since $K=\frac{1}{2} \cdot (20) \cdot h$, we have \[ \frac{63r}{2} = 10h,  \] so $\frac{r}{h} = \frac{20}{63}$.

Now we can use the ratios between sides and altitudes to determine $DE$. Specifically, \[ \frac{DE}{20} = \frac{h-r}{h} = 1- \frac{r}{h} = \frac{43}{63},  \] hence $DE = 20 \cdot \frac{43}{63} = \frac{860}{63}$, and our answer is $\boxed {923}$.
[/hide]

Good solutions... I wasted time actually finding the area before realizing that I didn't need to. :P

[quote="vvr1590"]So does the altitude to BC contain the incenter? How do you know that?[/quote]
No, it doesn't.

yeah, i just understood that it doesn't need to, only the distance is important. thanks.

Sorry, I don't follow...if the altitude doesn't contain the incenter, wouldn't the distance between DE and BC be less than r...? Then you can't take the ratio $ \frac{h\minus{}r}{h}$...

[quote="sonny"]Well the pependicular distance between two parallel lines is always the same.  since the altitude is perpendicular to BC it will be perpendicular to DE and the distance between the points where it hits DE and BC will also be r . therefore we can subtract the two.

and this is a really old topic.[/quote]

Yeah, but...if you draw the radius from the incenter to BC, it's not perpendicular...if you drew the perpendicular from the incenter to BC and the radius from the incenter to BC you get some kind of funny right triangle with r as the hypotenuse...

[quote="ch1n353ch3s54a1l"]

Yeah, but...if you draw the radius from the incenter to BC, it's not perpendicular[/quote]
[url]http://mathworld.wolfram.com/Incircle.html[/url]

Look at the picture, the radius from the incenter to a side in a triangle is always perpendicular.

Oh, WOW, right, all I could think was that the line connecting the vertex to the incenter was an angle bisector...and wasn't perpendicular to BC...clearly they're not the same line...all right, thanks, makes sense now o_O

(Sorry for reviving but I have aquestion).
Understand conventionel Heron's/inradius/similar-triangle method but don't get alternative "mass point" formula. Whats that and can you explain how to apply to this problem (AIME 2001 1 #7). thanks!

Yup . Use the angle  bisector formula to obtain $\frac{860}{63}$

Dude. Don't 9-year bump a post thats been answered alread y

Distance between DE @nd BC = inradius , that's done

First, we need to calculate the area and inradius of this triangle. The area is $\frac{21\sqrt{1311}}{4}$, making the heigh from $A$ to $BC$ equal to  and the inradius is $\frac{\sqrt{1311}}{6}$, so the distance from $A$ to $BC$ is equal to $\frac{21\sqrt{1311}}{40}$, so the distance from $A$ to $BC$ is $\frac{21\sqrt{1311}}{40} - \frac{\sqrt{1311}}{6} = \frac{43\sqrt{1311}}{120}$. Thus, the ratio of the height from $A$ to $G$ to the height from $A$ to $F$ is $\frac{43}{63}$, which is equal to the ratio of $DE$ to $BC$ or equivalently $DE$ to $20$. Thus, $DE = \frac{860}{63} \implies \boxed{923}$.

Same as AMC 12A 2011 #13, just different numbers

Oh yeah, I just noticed that haha. I've been going through old AIMEs, sometimes it helps more than you realize ;)

[b]Solution:[/b] Notice that we just want to find the altitude from $A$ to $\overline{BC}$ and $r$. This is because $\triangle{ADE}$ is similar to $\triangle{ABC}$. $A=rs$ gives us that $A = r \cdot \frac{63}{2} \Rightarrow r = \frac{2A}{63}$. Also, it is clear that the altitude is $\frac{A}{10}$ by $A = \frac{1}{2}bh$. Now, the altitude from $A$ to $\overline{DE}$ is just $\frac{A}{10} - \frac{2A}{63} = \frac{43A}{630}$. Notice that this is $\frac{43}{63}$ of the value of the length of the altitude from $A$ to $\overline{BC}$. Hence, $\frac{43}{63}$ is also the similarity ratio. Therefore, our final answer is $\frac{43}{63} \cdot 20 = \frac{860}{63} \Rightarrow \boxed{923}$.


find the number of solution of the eguation x+y+z=2007 and x<y<z. :maybe:

je sais que c est un simple exercice a l occasion    c est un exercice de tkil  :mad:


According to my proctor, I got a perfect score on the AMC 10B... will they re-examine me because this is the first time my school is sitting for the AMC and this is the first time I sat for it as well? And can I sit for the AIME if I'm not an American citizen or permanent resident?

No.  I believe the only circumstance they might consider retesting is if there was strong evidence that you cheated.. which I am sure there is not.
And yes I believe you would be able to take the AIME.

You are absolutely allowed to take the AIME. It's the USAMO that you aren't (and even that doesn't always hold true...)

[quote]It's the USAMO that you aren't (and even that doesn't always hold true...)[/quote]

Really? Under what conditions is sitting for the USAMO permitted?

Wait, I thought non-US citizens or permenant residents are still allowed to take the USAMO. Aren't Canadians allowed to take it?

I think that you might be able to take it, provided your score at AIME is high enough, and your school makes some sort of official request to the AMC. However I am pretty sure your score will not be published (or even graded by the AMC official graders). It would be like taking a mock test, just for preparation ...  :?

[quote]U. S. citizens and students legally residing in the United States and Canada (with qualifyng scores) are eligible to take the USAMO.[/quote]
This is from the [url=http://www.unl.edu/amc/]AMC web site[/url]...

I'm not sure what they exactly mean by "students legally residing in the United States and Canada"... Do they mean permanent residents? :?

(By the way, I'm not a U.S. citizen either, and I was still allowed to take the AIME. So... as some people said, you're certainly eligible to take the AIME.)

Well, you could be illegally residing in the United States or Canada, in which case you would probably not be allowed to take the USAMO.

Having a Visa means legal residence. So can we take the USAMO if we aren't citizens but have a Visa?

if you're illegally residing, why the heck would you be taking the test anyway?

So does that mean that Chess is an illegal resident? :P

What? No, I'm a legal resident, obviously... I said I would be screwed if they didn't allow non-citizens to take it. I'm not a citizen but I am a legal resident. ;)

I think what it's saying is that essentially, if you go to a school in the US, or are a US citizen, you can take the USAMO.

How exactly would the AMC check to see whether test-takers are legal residents?

I remember that a few years ago, the rule was US citizens and permanent residents, but a concern was that many kids didn't really know what permanent residency was.

[quote="chess64"]What the heck are you talking about, 123s?

Anyway, I think you can take the test even if you're not a citizen, as long as you're not an illegal resident. At least I hope that's the case, or I'd be screwed. :stretcher:[/quote]

Oh you know what he's talking about.  :D

Do they ask for your social security number?

I remember this same thing discussed in AoPS while back. Maybe joml88 or tetrahedr0n can confirm this. I think you're probably be able to take it if you contact Mr.Dunbar. As for counting as a score, I don't see why your paper should not be graded. But I do see that you will not be sent to MOP if you're not U.S. citizen. Obviously, U.S won't put non-U.S. citizen in their Olympiad camp, right? :)

No MOP? Darn. :mad:

Anyway, for AIME eligibility:

[quote="2006 AIME Teacher's Manual"]Any student who is officially enrolled in high school (or below) and is taking at least one course at the school, and has not graduated, is eligible to take the AIME (with qualifying score). US and Canadian Citizens and students residing in the United States (with qualifying scores) are eligible to take the USAMO.[/quote]

And USAMO:

[quote="2006 USAMO Teacher's Manual"]U. S. citizens and students legally residing in the United States and Canada (with qualifying scores) are eligible to take the USAMO.[/quote]

Can't find anything about MOP though... :(

[quote] But I do see that you will not be sent to MOP if you're not U.S. citizen. Obviously, U.S won't put non-U.S. citizen in their Olympiad camp, right?  [/quote]

I think it's OK if you are just a permanent resident. Not 100% sure.

To answer the original question, i am not citizen or green card, and i took USAMO for the past 2 years because i live in us.  if you are on visa but living in US or Canada, then yo can take the usamo.  otherwise yo can't.

[quote="Silverfalcon"]Obviously, U.S won't put non-U.S. citizen in their Olympiad camp, right? :)[/quote]

Wrong. Mr. Dunbar has told me that you don't have to be a citizen to attend MOSP. *phew*

chess, there have been people in the past who have gone to IMO without citizenship, so you shouldn't be that surprised.


Um no it is not. Autoclickers are a good way to crash the site.

[quote="ZetaX"]The better question is: why do you even ask¿ Doing such a thing means:
- you are a lousy cheater.
- you just want to show off.
- you care for that meaningless stupid number.
Get a life  :wink:[/quote]Support.

Not only that but if people keep doing that I will simply shut down that option or just make it a random number (a 3-digit one) and/or ban your IP address.

[quote="Valentin Vornicu"]Not only that but if people keep doing that I will simply shut down that option or just make it a random number (a 3-digit one) and/or ban your IP address.[/quote]

How do you identify the cheaters? I link my blog through a bunch of other forums and that's how I accumulate my views - but how is that different than those who autoclick for small boosts?

See it this way: a false positive does no harm  :wink:

Sigh...when false positives increase, false negatives decrease, which is exactly what we want...haha we just learned this in AP Statistics yesterday. But I see your point.

When you autoclick you send multiple requests for the same page from the same IP over and over and over again. Puts useless stress on the server, and we will ban you. It's as simple as that.

AIME15:

Exactly why would you want to do that? Do you actually think it is even worth your time to cause others such an inconvenience? If you do, at least do it in a non-educational website.

[quote="AIME15"]Just wondering, is it allowed to use an autoclicker to get more blog views?[/quote]

We are banning people who do this.

I think they only ban the IP address. (It has happened to me before)

[quote="shtsxc12"][quote="rrusczyk"]
We are banning people who do this.[/quote]

I assume that you are going to ban the ip address/loginID that was running the autoclick, not the owner of the blog, right ?

Somebody may do this to other person's blog, without the knowledge of the owner, such as the case of stevenmeow's blog. It is not fair to reset stevenmeow's "Total Visit".[/quote]

We ban the IP of the autoclicker, not the owner of the target blog.

As for stevenmeow's Total Visit count, that was very clearly highly inflated for a long time.  We don't want to encourage that sort of thing.

I can't think of a possible reason someone would care about their blog's view count...Even a really pathetic one...

It's pretty obvious: it's used as a status of popularity. The traditional view goes that if your blog has more views, it's more popular.


geometry
3D geometry
tetrahedron
geometry proposed

Three faces of a tetrahedron are right triangles, and the fourth face is a non-obtuse triangle. Prove that:

[b]a.)[/b] The fourth face is a right triangle if and only if there exists (at least) one vertex of the tetrahedron such that two and only two of the three planar angles at this vertex are right.

[b]b.)[/b] If all faces of the tetrahedron are right triangles, then the volume of the tetrahedron equals $\frac{1}{6}$ of the product of the three shortest edges not belonging to one and the same face.

[b]Note by Darij:[/b] A planar angle at a vertex of a tetrahedron is the angle between two edges. The term ''three shortest edges not belonging to one and the same face'' is a bit obscure. The meaning is: We take the two shortest edges $a,$ $b$ of the tetrahedron. Then we consider the set $S$ of all edges of the tetrahedron [i]not belonging to the same face as the first two edges[/i] $%
a,$ $b.$ (In particular, $a$ and $b$ themselves are not in $S.$) The shortest of the edges from $S$ is called $c.$ Then, the edges $a,$ $b,$ $c$ are referred to as the ''three shortest edges not belonging to one and the same face''. The solution states that the volume of the tetrahedron is $\frac{1}{6}\cdot AC\cdot BC\cdot
AD,$ if we assume that the right angles in the tetrahedron are $90  =\measuredangle ACB=\measuredangle CAD=\measuredangle BAD=\measuredangle BCD. $

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)


A bag contains 3 red marbles and 2 black marbles.  Two person, A and B, take turns drawing a marble from the bag without replacing any of the marbles drawn.  The first person to draw a black marble wins.  What is the probability that person A, who draws first, wins?

How would you solve this problem without making a probabltiliy tree?

Thanks!

[hide]okay, well there can only be a max of 4 turns, so it shouldn't be that hard. I'm not sure if this counts as a probability tree... o-0

Turn 1: Prob that A wins = 2/5, loses = 3/5
And if we get that far...
Turn 2: Prob that B wins = 2/4, B loses = 2/4
Turn 3: Prob that A wins = 2/3, A loses = 1/3
Turn 4: Prob that B wins: 1, B can't lose

So... for A: he can win @ turn 1 with a  2/5 chance and he can win @ turn 3 with a 3/5*2/4*2/3 chance.

2/5 + 3/5*2/4*2/3 = 2/5+1/5 = 3/5 that A wins, leaving B with a 2/5 chance.

(to check, B can win at turn 2 with a 3/5*2/4 chance and at turn 4 witha 3/5*2/4*1/3*1 chance, which equals 2/5).[/hide]

Well, as superwizard's saying, I think the best way to solve this is indeed to draw a probability tree. The number of marbles is changing every time so you need to do something similar to a tree (even if you don't draw one, ultimately you're going to have to do something paramount to drawing a diagram that shows all the different probabilities at each eventuality).


We are given an acute triangle  $ABC , AB < AC$  , the altitudes $AD , BE , CZ$  and the point $N$ , where line $EZ$  meets line $CB$ .A line which comes through  $D$  and  is parallel  to  $ZE$  meets lines  $AB , AC$ at $K , L$ respectively .Prove that the circumcircle of triangle $NKL$  bisects side $BC$.

 [u]Babis[/u]

See

http://www.mathlinks.ro/Forum/viewtopic.php?t=54020
http://www.mathlinks.ro/Forum/viewtopic.php?t=29208
http://www.mathlinks.ro/Forum/viewtopic.php?t=56754

  darij


Hi all!

I am interested in silk road math olympiad.
does anyone know something or has some stuff about it??

any help appreciated.
Thanks

 :)

[quote="Omid Hatami"]The Silk Road Mathematical Olympiad began in 2002 in countries Iran ,Turey, China, Uzbekstan,... by the sugestion of Turkey. Each year the Iranian students participate in this contet before the IMO. I have posted some of the questions in this forum[/quote]

Thanks for your reply.

But I have another couple of questions about it...
Where can I find the problems as well as info about the contest?
I'll search for those problems...Is there the source in that??

Thanks, have success in everything!
 :)


THERE ARE PRESENTLY FOUR INDIANS IN THIS FORUM hell_ever Shreyas Pardesi and me
HOPE THE NEW YEAR BRINGS IN MORE POSTS AND INDIANS TO THIS SITE

AND MOST IMPORTANTLY HAPPY NEW YEAR  :D     :)

Happy new year everyone
thanks madness  :D

HAPPY NEW YEAR EVERYONE!!!! :)  :ninja:  :)

[b][i]HAPPY NEW YEAR EVERYONE :thumbup:  :thumbup: [/i][/b]

Don't fool yourself Rohit there are plenty of Indians around. However, when there's no fruitful discussion they refrain from taking part.

Ya but someone has to take the initiative right?    :D   :P   :wink:

Wishing all of you a happy and prosperous new Year! Let 2008 be you year!

Hey baby,
I get the relavance of ur topic.. A forum shud always b kept active n i gess in the absence of activity it dusn matter gettin into topics not that fruitful.. :D

Po da loose see the private msg i sent you
You can see it on the top


오... 5/4에 입대하라고 오늘 새벽에 친히 메일이 왔군요. 입대 11일 전에 보내다니, 요런 상큼한 병무청같으니라고... =_=;;

자세한 건 제 블로그에서...

[url]http://sos440.tistory.com/167[/url]

I got an e-mail from the Military Manpower Administration at dawn today, informing me of the fact that they imposed the duty for the conscription service starting at May 4. I was nonplussed at how 'early' they inform such a critical thing. Only 11 days are left for me to enjoy this everyday life! Even more, it's depressing that enjoying mathematics during this period of conscription is quite hard. =.=

[quote="undefined117"]That's unfortunate... will you still post on AoPS? I guess you could look at it positively as a character-building experience, but two years out of our prime? :(

I am also a bit worried about my conscription service coming up in... two years? Six? I'll be in the U.S., but I don't know too much about immigration policies.[/quote]

Now I'm bracing myself up for the upcoming service, trying to look at positive things. Since I'm supposed to be working at electronic & networking department, there would be some rest time for my own study if I'm sufficiently accustomed to the military service.

But I'll recommend you to try all the possible things to shun the conscription service.  :lol:


ratio geometry calculus Euler Gauss probability function

If math is all based on assumptions, then why does it work so well with our nature/universe?

Not all mathematics is based on assumptions, that is why there are proofs.  This pure and dsicrete mathematics is what is then used to explain the universe.
If you are talking about the mathematics behind the major theories used for explaining the universe, none of them are written in stone and can be overturned with a valid contradiction.  This then leads to the reason to why they "work so well" in explaining things, because the theories have not been disproven, they appear to be that way.  An example of this is with Newtonian Mechanics not being completely efficent in explaining physical interactions, besides the well known Quantum Mechanics showing this, there are others such the dark matter-energy controversy that are known to not be explained by Newtonian Mechnics alone.  Another example, is String Theory, which has yet to be proven or disproven, whose mathematics, besides being complex, is theoretical.  This, though, does not prevent String Theorists from trying to explain the universe based on their theory.

why does pi and the golden ratio appear in a bunch of places in math? and what math is not based on assumptions. I think every math I know has some axioms or postulates that they then branch off of, even arithmetic. why does the Pythagorean theorem work if we made up math? I know that  mathematicians say that math and reality have no definite relationship, but I would just like to attempt to understand what the relationship they do have is.

$\pi$ and the golden ratio appear in math because they were created [i]by[/i] math. Pi is the ratio of a diameter of a circle to its circumference, and so it appears. Therefore, it has a tendency to appear in physics relating to movement described by a circle, like pendulum periods and spring periods. (Why it appears in some quantum mechanical formulas I have no idea, ask someone else :roll: ) The golden ratio is simply the ratio of a+b:a as a:b (if I did my notation correctly). So I think the question you were really meaning to ask is why pi and the golden ratio appear in nature. Now, I don't have a major in biology, but I do know that many animal's bodies have the golden ratio somewhere in them. Why? No clue, perhaps that specific ratio is just the most biologically efficient (and therefore more likely to have evolved).

The pythagorean theorem works exactly because we made up math. The Pythagorean theorem is a result of math that we made up, so it must follow that if the math we made up works, then so to should the pythagorean theorem.

Otherwise, perhaps this topic should be moved to another forum?

This is the wrong forum for this, but it'll get moved eventually. Also, you guys use the word "invent" a lot, I suggest reading up on mathematical realism for a different viewpoint.

EDIT: actually, the world invent was only used once... but look up mathematical realism anyways  :P.

Mathematical philosophy is something that interests me quite a bit and although I haven't had the chance to take any such course on it (yet), I'll offer my insights as to how I perceive it. I'll use the context of mathematical formalism, which to my understanding frames mathematics as the investigation of systems based on axioms. As a consequence, the real-world implications have nothing to do with the significance of mathematics [i]within[/i] mathematics itself: We choose the mathematical system which best represents our understanding of the universe (or, a specific piece of the universe). I think the best example is geometry and the parallel postulate, and how our interpretation of this postulate allows us to "rebuild" geometry in a sense, to suit different situations in the real world (around a black-hole, for instance) or even to "rebuild" geometry in ways that has no apparent use to our reality (for now).

Sorry if this should be in a different forum but I didnt see a forum for this topic. I was just puzzled at why it works in our universe. I said to someone the same thing you said to me (they work because we made it) and they said that it is just stupid. Thanks for your answer anyways.

I think it's more complex than, "it works because we made it" (and, in a sense, I don't even fully agree with the "we made it" part). There are topics in mathematics that do not have any immediate applications to the real world, we choose the mathematical systems that best suit our understanding of the world, this is why it "works"; because it's all relative. Also, there really is no wrong answer to this question; it's a philosophical discussion.

i have a quick question, while on my quest of this answer, i would like to know some different forms of math (not calculus, pysics, etc) but how non-euclidian geometry is to just normal geometry, i would like to know what maths there are to this math. i know finitism is one.

[quote="stuck"]If the world were not based around math, it would have to be based upon randomness. (Right? I'm assuming NP problems are still mathematical in nature) Therefore, if the world were to be based on randomness, then the possibility of surviving in such a universe would be extremely small, moreso since the conditions of such a universe would change every moment. Then, apply the Anthropic Principle, and you're done![/quote]

The world is based more on randomness than on maths. What you missed is this: the chances of our dying is extremely small, so the Anthropic principle makes us wonder why people do die after all, and even at a regular basis.

@linksterz: nice to see this inquisitiveness, but you are going along the wrong lines, as pointed out by stuck. theories in maths are meant to work well in maths, what's puzziling you perhaps is why it should work in nature. I'll tell you a story.

Archimedes, Eudoxus and Euclid brought rigour into maths. 2000 years later, people like Weierstrass were putting their finishing touches on a rigorous system of mathematical analysis, which had been moulded over years by Euler, Gauss, Cauchy, Lagrange, Riemann and the like. In the beginning of the 20th century, the outstanding mathematician David Hilbert decided to go a bit further. He declared a programme to formalize maths once and for all, and develop a system of axioms and rules such that anything consistent with the system could be proved rigorously, while the rest could be disproved rigorously.

So far so good. Hordes of mathematical enthusiasts joined Hilbert's quest for the unassailable mathematical question. Its fortunate though that some didn't, for otherwise the brilliant young Austrian Kurt Godel couldn't have dealt the crushing blow he did to Hilbert's dreams, by proving that in any system of axioms there are statements which can be seen to be true, but which can't be proved to be so withing the system.

Godel's incompleteness theorem, as it came to be called had fascinating implications. Follow this closely: the statements were [i]seen[/i] to be true, but they couldn't be [i]proved[/i]. This is where mathematical insight comes in, as against formal mathematical proofs. This came as a shock to formalists. They got round by saying there is nothing called absolute truth, something can only be provable or not, but this didn't impress the others much.

The other school of thought is based on something called Platonic reality, which answers many of your questions. These people believe, as Plato did, that the fundamental mathematical truths are just there, waiting to be stumbled upon by man. Maths and nature have remained mingled for a long time, and man have discovered (not thought is up, see, but [i]discovered[/i]) this slowly. If another civilization springs up on some distant planet, they too, will form the natural numbers, draw a circle, notice the constant ratio of its circumference and its diameter, and will much later be intensely astonished to see that this ratio on squaring and dividing by six equals the infinite sum of harmonic series of the squared integers. [i]That[/i] is the essence of Platonic reality.

Of course we are too young to adopt any viewpoint: formal or Platonic, but your questions are essentially Platonic in nature. I've learnt the above things from chapters 2, 3, 4 of the book [i]The Emperor's New Mind[/i] by Roger Penrose, and I'd highly recommend it, if you like to make your brain go crazy and do sommersaults.

As for the third school of thought, finitism, i don't have mcuh idea, because Mr Penrose did dwell on it but briefly. I only know that it refuses to recognise the concept of infinity. But its not a very widespread cult, and you don't have to worry about it now.

isn't the golden ration basically x when $x=\frac{1}{x}+1$.
In English, the receprical of the golden ratio is the same as the golden ration plusminus 1

[quote="JBL"][quote="stuck"]If the world were not based around math, it would have to be based upon randomness. (Right? I'm assuming NP problems are still mathematical in nature) [/quote]

This doesn't even make any sense.  Mathematics can be used to study randomness, just as it can be used to study anything which can be "modeled."  (Consider probability theory, or any branch of mathematics with the word "stochastic" in it.)  The universe isn't "based around" mathematics -- mathematics is an abstract formal system which happens to be extremely flexible and so is very useful for describing the way things behave.
The fact that we have good mathematical descriptions for many things is a property of [i]mathematics[/i], not a property of the universe.[/quote]

I agree that the world behaves randomly, but my contention is that it is an [i]interpretable[/i] randomness, like you pointed out. A randomness where one Planck second we exist in three dimensions and another we live in five (and variations on the same theme) are not able to be approximated by mathematical reasoning (nor would we be able to survive at all in these worlds, let alone atoms). The randomness has sense and meaning.

EDIT: Ignore this paragraph, I though about it, and don't agree.
I'm not sure if by randomness you mean that from one state to another the outcome is random, or the functions (so to speak) that define state transitions are random. If it is the latter, than I believe our views agree.[/i]


trigonometry
geometry
geometric transformation
homothety
geometry unsolved

Give two circle $ (O), (O')$. $ I\in (O)\cap (O')$( I is a fixed point) the line d passing throught I and cut (O),(O') at N,N' respectively.Find the maximum of $ IN.IN'$ when d moving

I will change the notations:
[b]let circles $ C_1,C_2$ with radius $ r_1,r_2$ respectively intersect at points $ A,B$,let $ d$ be an arbitary line passing through $ A$ and intersect $ C_1$ at $ C$ and $ C_2$ at $ D$,let $ AC = m$ and $ AD = n$ find the maximum value of $ m.n$ when $ d$ variates.[/b]

first of all note that $ A,B$ are fixed points so the value of arc $ AB$ in $ C_1$ and $ C_2$ is constant,so $ \angle DCB$ and $ \angle CDB$ are constants,now according to $ \triangle BCD$ we get that the third angle is constant too,i.e. $ \angle CBD$ is constant...

now let $ \angle CBD = \theta$ so $ \theta$ is constant,now let $ \angle ABC = x$ we have:

$ \frac {m}{\sin x} = 2r_1\rightarrow m = 2r_1\sin x$

$ \frac {n}{\sin (\theta - x)} = 2r_2\rightarrow n = 2r_2\sin (\theta - x)$

now multiply the last two equations,we get tat:

$ mn = \underbrace{4r_1r_2}_{\text{constant}}.\sin x.\sin (\theta - x)$

so $ mn$ will be maximum iff $ \sin x.\sin (\theta - x)$ is maximum...

but we know that:

$ \sin x.\sin (\theta - x) = \frac {\cos (2x - \theta) - \overbrace{\cos \theta}^{\text{constant}}}{2}$

so $ \sinh.\sin (\theta - x)$ is maximum iff $ \cos (2x - \theta)$ is maxumum but we know that $ \cos \alpha\leq 1$ hence $ \cos (2x - \theta)$ is maximum iff $ 2x - \theta = 0$ i.e. $ x = \frac {\theta}{2}$ i.e. $ \angle ABC=\angle ABD$...

now in order to draw the line $ d$ its sufficient to find points $ C,D$,now construct two lines in both side of $ B$ with the angle $ \frac {\theta}{2}$ where $ \theta = 180 - \angle C - \angle D = 180 - \frac {AB_1}{2} - \frac {AB_2}{2}$ to intersect $ C_1$ and $ C_2$ at $ C,D$...

[quote="toanIneq"]Oh, you have a good solution but,in problem $ I\in (C_1)\cap (C_2)$ so I need a new solution  :lol:[/quote]
what do you mean? :maybe:
note that $ C_1\cap C_2\equal{}\{A,B\}$

[quote="Amir.S"]easy problem.
if $ NN'$ pass through the ex-homothecy center of $ C_1,C_2$ there is a circle that is tangent to $ C_1,C_2$ at points $ N,N'$ hence $ IN.IN'$ is its maximum.[/quote]
exactly...
because then let $ P$ be the power of point $ I$ wrt this circle then we have $ P\equal{}IN.IN'$,and for any $ N_1,N'_1$ we would have $ IN_1.IN'_1<p<IN.IN'$

[quote="BaBaK Ghalebi"][color=darkred]Let $ C_1=C(O_1,r_1)$ and $ C_2=C(O_2,r_2)$ be two secant circles for which denote $ \{A,B\}=C_1\cap C_2$. A mobile line $ d$ which passes through the point $ A$ intersects again the circles $ C_1$ and $ C_2$ in the points $ C$, $ D$ respectively. Find the maximum value of the product $ AC\cdot AD\ .$[/color][/quote]
[color=darkblue][b]Proof.[/b] Denote $ \{\begin{array}{c}
m(\widehat {ABC})=u\\\
m(\widehat {ABD})=v\end{array}$. Observe that $ u+v=\phi$ (constant),

where $ \phi=m(\widehat {O_1BO_2})\ .$ Observe that $ \cos\phi=\frac {r_1^2+r^2_2-a^2}{2r_1r_2}$, where $ a=O_1O_2\ .$

Thus, $ AC\cdot AD=(2r_1\sin u)\cdot(2r_2\sin v)$. Therefore,  $ AC\cdot AD$- [b]Max.[/b] $ \Longleftrightarrow$

$ \sin u\sin v$- [b]Max.[/b]. $ \Longleftrightarrow$ $ [\cos(u-v)-\cos\phi]$- [b]Max.[/b] $ \Longleftrightarrow$ $ u=v=\frac {\phi}{2}\ .$

In conclusion, the maximum value of the product $ AC\cdot AD$ is

$ M=4r_1r_2\sin^2\frac {\phi}{2}=2r_1r_2(1-\cos\phi)$ $ \implies$ $ \boxed {\ M=a^2-(r_1-r_2)^2\ }\ .$[/color]

thank you for a your solution.But it isn't a solution I need. I'm sorry to say that
Note that: $ I\in (C_1)\cap (C_2)$ not  ${ (C_1)\cap (C_2)=\{A,B}$ (This is a problem and I need a diffirent solution)

Sorry, what means $ I\in (C_1)\cap (C_2)$ ?? $ I\in \{A,B\}$ ?! Really $ (C_1)$ and $ (C_2)$ are disks ?!
Then in this case you must denote $ D_1$ ana $ D_2$ ... I don't understand. What wish you ? Say more clearly, please !


function
linear algebra
matrix
linear algebra unsolved

Suppose $f: M_n(\mathbb F) \rightarrow M_m(\mathbb F)$ is a surjective function which for each $X,Y\in M_n(\mathbb F)$ we have $f(X)f(Y)=f(XY)$. Prove that $m\leq n$

$f(0)=0$ and $f(M^k)=f(M)^k$. So, if $f(N)=M \in M_m(\mathbb{F})$ is a nilpotent matrix of order $m$ (it exists because $f$ is surjective), then $N \in M_n(\mathbb{F})$ is also nilpotent of order $m$. This concludes the pb.

Easily you could prove that $\forall A,B\in M_n(\mathbb F)$ if $Rank (A)=Rank(B)$ then $Rank f(A)=Rank f(B)$. This proves everything

and if $m \leq n$, does such a function exist ?

An intersting one. :) I think it's correct. When I got sure my example is correct, I'll send it.


function
probability
Support
integration
limit
real analysis
real analysis unsolved

Anybody knows if it is possible for a function f, which is monotonously increasing in a neighbourhood of zero and goes to zero at zero that

$ \frac {f(ax)}{f(bx)}\to 0$
as $ x\to 0$, $ 0 < a < b < \infty$. Are there any examples?

For $ x > 0,$ let $ f(x) \equal{} e^{ \minus{} \frac1x}.$ Let $ f(0) \equal{} 0$ and extend this to negative $ x$ as an odd function. It does exactly what you asked for.

Now, can you explain why this would be impossible for a real-analytic function?

Because I forgot to mention that 1-f(x) must be a Laplace transform of a probability measure on the positive real line. Sry I formulated the question properly (first I wanted to exlude as many bothering details as possible, but I wasn't right):

Is there a distribution function F with positive support, with Laplace transform $ \varphi(z) \equal{} \int e^{ \minus{} zx}dF(x)$ and

$ \frac {1 \minus{} \varphi(az)}{1 \minus{} \varphi(bz)}\to 0\qquad z\to 0$

with $ 0 < a < b < \infty$

Clearly with finite expectation, or regularly (or even slowly) varying upper tail- distribution this doesn't happen, an I am interested if it can happen at all.

P.S. I posted this here, because this is a purely analitycal question despite its probabilistic context, but if moderators feel otherwise put it somewhere else.

Try $ f(0) \equal{} 0$, $ f(x) \equal{} x^n$ for $ x \in \left ( \frac {1} {n\plus{}1}, \frac {1} {n} \right ]$, $ n \equal{} 1,2,\ldots$.

[quote="Moosquibe"]P.S. I posted this here, because this is a purely analitycal question despite its probabilistic context, but if moderators feel otherwise put it somewhere else.[/quote]
The analysis forum is the correct place for this. The thread stays right here.

So you're asking this: if $ \phi$ is such a Laplace transform (a.k.a. moment generating function), can $ \phi(z)$ behave like $ 1\minus{}e^{\minus{}\frac1z}$ near zero?

Of course, since  for such a $ \phi,$ we'd have the $ \phi'(0)\equal{}0,\phi''(0)\equal{}0,\phi^{[k]}(0)\equal{}0$ for all $ k\ge 1.$ All moments equal to zero??

That can't happen; for anything other than a point mass at the origin, the second moment $ \lim_{t\to 0}\frac{\phi(t)\plus{}\phi(\minus{}t)\minus{}2}{t^2}$ is either positive and finite or equal to $ \plus{}\infty$. (It always makes sense, even when $ \phi$ takes infinite values)

If the first moment is finite and nonzero, we get $ \lim_{t\to 0}\frac{1\minus{}\phi(at)}{1\minus{}\phi(bt)}\equal{}\frac ab$. If the first moment is zero and the second moment is finite, we get $ \lim_{t\to 0}\frac{1\minus{}\phi(at)}{1\minus{}\phi(bt)}\equal{}\frac{a^2}{b^2}$. If the first moment is zero and the second is infinite, we have $ \frac{1\minus{}\phi(at)}{1\minus{}\phi(bt)}>\frac{a^2}{b^2}$ for small values of $ t$; the limit might not exist, but it's certainly not zero. If the first moment doesn't exist, we again might not have the limit exist, but we can certainly say it's not zero.

"If the first moment doesn't exist, we again might not have the limit exist, but we can certainly say it's not zero."

Now that is exactly what I'd like to know how to prove. As I wrote, the finite first moment and the regularly varying case, i.e.

$ 1 \minus{} F(x) \equal{} x^{ \minus{} \alpha}L(x)$
where L(x) is a slowly varying function at infinity and $ \alpha\in[0,1]$

is easy, but I'd like to prove that for the general case.

Umm... on second thought, if the Laplace transform exists at all in a neighborhood of zero, infinitely many moments exist and are finite.

What if you only insist on a one-sided interval ending at 0 rather than a neighborhood of 0?

What if you make it the Fourier transform rather than the Laplace transform?

The one-sided problem is a simple one. If $ f(a)\equal{}\int_0^\infty e^{\minus{}at}\,d\mu(t)$ $ (a>0)$ for some probability measure $ \mu$, then $ f$ is convex and, thereby, $ 1\minus{}f$ is concave whence $ 1\minus{}f(2a)\le 2(1\minus{}f(a))$.


Problems with the search function that really need to be solved:

- be able to search any LaTeX stuff:
it should thus _not_ see "+", "{", "\" or whatever as seperators; if some symbols, e.g. + and -, are necessary for some function, they should only be interpreted this way if spaces are on both sides of them.

- interpret quotation marks the right way:
try searching "Olicrosse Godemite", you will get a single result, but this result does not contain the exact string searched for; this is just artficial, but if you want to find the "three squares theorem"...

- the delay between two searches must vanish:
if you don't use the correct keywords, you often get too many or the wrong results. It's plainly wrong to believe that everything can be found with one till three tries (and I would call myself experienced with searching ;) ). But after some time, you really hate it, in fact I do much more often just tell the people to use the search function instead of doing myself; but I doubt that all that people have more luck.
If you really fear flooding, leave it there for people with less than 47 posts and on the forum for less than 1 month.

This isn't exactly a problem but I think you should include blog entries and posts in addition to threads when you're searching. They do carry some interesting content.  :D

I agree with ZetaX. The problem is that I have to wait 15 seconds for seachgin again.

[quote="ZetaX"]As it seems that Valentin is back again: up with this :D[/quote]I'll soon make the necessary modifications for the search functions so that it won't take that much time between two searches.

As for the + and - issue: it's not me, but MySQL developers that you have to talk to :) It's simply not possible to avoid it.

[quote="ZetaX"]But why did it work in the past¿[/quote]I'm not quite sure what you think worked in the past, but it was a totally different system of search like 2 years ago (one that got very very very slow and server demanding as the site growed). Basically the old system kept a table with all the words ever used in posts which was taking 40% of the table space alone.

I will try to think of a way to overcome this (though I see no realistic way, as those symbols are simply not indexed in MySQL, or in other search engines in general for that matter).

[quote="ZetaX"]Since Valentin once claimed that there was never the ability to search TeX code/formulas:
See http://www.mathlinks.ro/Forum/viewtopic.php?t=63482 and try to search now, you have no chance.

PS: now the search for "Godemite Olicrosse" already gives three results :D[/quote]

That'an example, there are more.

It's just... well, this site really needs the possibility to search the TeX code.  :( I've no clue about MySQL, but I would prefer slow searches to what they are now.

The problem is not the search is slow, the problem is that it slows down the site. If two people search for something like that at the same time, the site will shut down because the CPU cannot undertake that much stress (and also apache is kinna instable as a general thought).

And I really do not understand your point about Godemite Olicrosse :) Can you please explain it more clearly? Seems to be working fine for me. ..

[quote="ZetaX"]The Godemite Olicrosse was just to show that "" don't have any effect. But the link in that quote shows that a year ago, TeX-searching was possible.[/quote]Ah, they will have effect soon, it's a thing I haven't completely implemented, but will be present in the near future.


trigonometry
geometry
integration
complex numbers
real analysis

[hide="A hint"]Setting $ z\equal{}\cos 1^{\circ}\plus{}i\sin 1^{\circ}$. Now see at $ z\plus{}z^{2}\plus{}...\plus{}z^{44}$.[/hide]

I didn't think of using complex numbers, but how does $ z\plus{}z^{2}\plus{}\ldots\plus{}z^{44}$ equal the original fraction?  :)

Use that hint , we can find the values of $ \sin 1^{\circ}\plus{}\sin 2^{\circ}\plus{}...\plus{}\sin 44^{\circ}$ and $ \cos 1^{\circ}\plus{}\cos 2^{\circ}\plus{}...\plus{}\cos 44^{\circ}$ .  :wink:

[quote="aidan"]What is the exact value of $ \frac{\cos 1^{\mathrm{o}}\plus{}\cos 2^{\mathrm{o}}\plus{}\ldots\plus{}\cos 43^{\mathrm{o}}\plus{}\cos 44^{\mathrm{o}}}{\sin 1^{\mathrm{o}}\plus{}\sin 2^{\mathrm{o}}\plus{}\ldots\plus{}\sin 43^{\mathrm{o}}\plus{}\sin 44^{\mathrm{o}}}$?[/quote]

[hide="Another hint without using complex numbers"]Compute $ \cos(x)\plus{}cos(\frac{\pi}{4}\minus{}x)$ and  $ \sin(x)\plus{}sin(\frac{\pi}{4}\minus{}x)$[/hide]

We get $ z\plus{}z^{2}\plus{}\ldots\plus{}z^{44}\equal{}\cos 1^{\mathrm{o}}\plus{}\cos 2^{\mathrm{o}}\plus{}\ldots\plus{}\cos 44^{\mathrm{o}}\plus{}i\sin 1^{\mathrm{o}}\plus{}i\sin 2^{\mathrm{o}}\plus{}\ldots\plus{}i\sin 44^{\mathrm{o}}$ and i don't see how to apply it. Sorry I'm lousy at applying complex numbers =P

With the computation $ \cos\frac{\pi}{4}\plus{}\cos (\frac{\pi}{4}\minus{}x)$ and $ \sin\frac{\pi}{4}\plus{}\sin (\frac{\pi}{4}\minus{}x)$, we have $ \sin A\plus{}\sin B \equal{} 2\sin\frac{A\plus{}B}{2}\cos\frac{A\minus{}B}{2}$ and $ \cos A\plus{}\cos B \equal{} 2\cos\frac{A\plus{}B}{2}\cos\frac{A\minus{}B}{2}$.

Since we have $ A\plus{}B \equal{}\frac{\pi}{4}$ and $ A\minus{}B \equal{}\frac{\pi}{2}\minus{}x$ the fraction simplifies to $ \frac{(\frac{\pi}{2}\minus{}x)(22\cdot 2\cos 45^{\mathrm{o}})}{(\frac{\pi}{2}\minus{}x)(22\cdot 2\sin 45^{\mathrm{o}})}\equal{} 1$?

There has to be mistake somewhere =P

[quote="aidan"] With the computation $ \cos\frac{\pi}{4}\plus{}\cos (\frac{\pi}{4}\minus{}x)$ and $ \sin\frac{\pi}{4}\plus{}\sin (\frac{\pi}{4}\minus{}x)$, we have $ \sin A\plus{}\sin B \equal{} 2\sin\frac{A\plus{}B}{2}\cos\frac{A\minus{}B}{2}$ and $ \cos A\plus{}\cos B \equal{} 2\cos\frac{A\plus{}B}{2}\cos\frac{A\minus{}B}{2}$.[/quote]

Yes

[quote="aidan"] Since we have $ A\plus{}B \equal{}\frac{\pi}{4}$ and $ A\minus{}B \equal{}\frac{\pi}{2}\minus{}x$ [/quote]

No

[quote="aidan"] There has to be mistake somewhere =P[/quote]

Yes

[quote="aidan"]What is the exact value of $ \frac{\cos 1^{\mathrm{o}}\plus{}\cos 2^{\mathrm{o}}\plus{}\ldots\plus{}\cos 43^{\mathrm{o}}\plus{}\cos 44^{\mathrm{o}}}{\sin 1^{\mathrm{o}}\plus{}\sin 2^{\mathrm{o}}\plus{}\ldots\plus{}\sin 43^{\mathrm{o}}\plus{}\sin 44^{\mathrm{o}}}$?[/quote]

[hide="Solution without using complex numbers"]We have to compute $ A\equal{}\frac{\sum_{k\equal{}1}^{44}\cos(k\frac{\pi}{180})}{\sum_{k\equal{}1}^{44}\sin(k\frac{\pi}{180})}\equal{}\frac{N}{D}$

$ N\equal{}\sum_{k\equal{}1}^{44}\cos(k\frac{\pi}{180})$ $ \equal{}\sum_{k\equal{}1}^{22}(\cos(k\frac{\pi}{180})\plus{}\cos(\frac{\pi}{4}\minus{}k\frac{\pi}{180}))$ $ \equal{}\sum_{k\equal{}1}^{22}2\cos(\frac{\pi}{8})\cos(k\frac{\pi}{180}\minus{}\frac{\pi}{8})$ $ \equal{}2\cos(\frac{\pi}{8})\sum_{k\equal{}1}^{22}\cos(k\frac{\pi}{180}\minus{}\frac{\pi}{8})$
$ D\equal{}\sum_{k\equal{}1}^{44}\sin(k\frac{\pi}{180})$ $ \equal{}\sum_{k\equal{}1}^{22}(\sin(k\frac{\pi}{180})\plus{}\sin(\frac{\pi}{4}\minus{}k\frac{\pi}{180}))$ $ \equal{}\sum_{k\equal{}1}^{22}2\sin(\frac{\pi}{8})\cos(k\frac{\pi}{180}\minus{}\frac{\pi}{8})$ $ \equal{}2\sin(\frac{\pi}{8})\sum_{k\equal{}1}^{22}\cos(k\frac{\pi}{180}\minus{}\frac{\pi}{8})$

And so $ S\equal{}\frac{N}{D}\equal{}\cot(\frac{\pi}{8})\equal{}\sqrt{2}\plus{}1$[/hide]

[quote="aidan"]Alright  :P  But anyway i get that answer also by integrating the area beneath the graphs of $ y \equal{}\sin x$ and $ y \equal{}\cos x$ from $ 0^{\mathrm{o}}$ to $ 44^{\mathrm{o}}$. Coincidence?[/quote]

I don't understand.

What exactly have you computed ?

I did $ \int_{k\equal{}0}^{k\equal{}\frac{\pi}{4}}\sin x$ and divided it by $ \int_{k\equal{}0}^{k\equal{}\frac{\pi}{4}}\cos x$.

[quote="aidan"]I did $ \int_{k \equal{} 0}^{k \equal{}\frac{\pi}{4}}\sin x$ and divided it by $ \int_{k \equal{} 0}^{k \equal{}\frac{\pi}{4}}\cos x$.[/quote]

But it does not match the required expressions!

The required expression is clearly the Riemann sum of cosine over 1 to 44 divided by the Riemann sum of sin in the same domain. I wonder if it is possible to show that the error in them are the same.

That was said already...but anyway, we can easily show both results with the following lemma:

$ \frac{\cos x\plus{}\cos 45\minus{}x}{\sin x\plus{}\sin 45\minus{}x}\equal{}\cot22.5\equal{}\sqrt{2}\plus{}1$.

Well i had a hunch that what i calculated was the same as what the question asked =P. Thanks for the lemma =)


Let f(x) = x^3 + 17. Prove that for each natural number n bigger than 2, there is a natural number x for which f(x) is divisible by 3^n but not 3^(n+1)
maybe someone comes up with a nice solution ?

this problem was already answered, you can find it here:
http://www.mathlinks.ro/viewtopic.php?p=2005&highlight=#2005


inequalities
integration
trigonometry
function
advanced fields
advanced fields unsolved

Use the Cauchy Schwarz inequality in an appropriate space to show that $ \int_{\pi/2}^{\pi} \frac{\sin x}{x} d x \leq \frac{1}{2}$.

Hint: the appropriate space is (nice enough) functions on $ [\frac{\pi}{2},\pi]$, with inner product $ \langle u,v\rangle \equal{} \int_{\pi/2}^{\pi} u(x)v(x)\,dx$. A natural choice of functions to apply the inequality to will probably get you a stronger inequality.

Is $ L^2[\pi/2,\pi ]$, the set of all square integrable functions good enough? or shall i look elsewhere?
thanks

$ L^2$ is overkill - we don't need completeness here, and all of the functions we'd ever consider are Riemann integrable. We can make the space be $ C[\pi/2,\pi]$ with that inner product. Yes, that's a proper (dense, not closed) subspace of $ L^2,$ but I said that we don't need completeness.

Of course, you'd be correct if you said $ L^2.$

[quote="jmerry"]A natural choice of functions to apply the inequality to will probably get you a stronger inequality.[/quote]
Hmmm. The choice of $ f(x)\equal{}\sin x,g(x)\equal{}\frac1x,$ which is about as natural a choice as I can think of, leads to
\[ \int_{\frac{\pi}2}^{\pi}\frac{\sin x}x\,dx\le\left[\int_{\frac{\pi}2}^{\pi}\sin^2x\,dx\right]^{\frac12}\left[ \int_{\frac{\pi}2}^{\pi}\frac1{x^2}\,dx\right]^{\frac12}\equal{}\left(\frac{\pi}4\right)^{\frac12}\left(\frac1{\pi}\right)^{\frac12}\equal{}\frac12.\]


After this interesting problem from Bulgaria
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32602
------------------------------------------------------------------------------

jivko777 do you know if the following case was given in Bulgaria Maths Olympiad for
students ? Does the author Vladimir Babev write an article about the following ?

$x_0>0$

$x_{n+1}=\frac{x_n^b}{e^{x_n}-1}$ where $b \neq 2$ real

1/ Study the comportement of $x_n$ convergence or divergence ?

2/ Give an equivalent with two or more terms ?

For $b>2$ it converges rapidly (faster than linear) to zero.
For $b<2$ the only interesting question is whether the unique positive fixed point is stable. There is a constant $a$ approximately equal to .4338664 such that the fixed point is stable for $b>a$ and unstable for $b<a$. For $b<a$, the sequence eventually diverges, alternating between tiny and huge values.

In both cases, this is much easier than figuring out what happens at $b=2$.
There was a very good reason for the original problem to have $x^2$.

If $b>2$ $x_n$ tends to zero

$x_{n+1}=\frac{x_n^b}{e^{x_n}-1}$

(1) $x_{n+1}=x_n^{b-1}-\frac{1}{2}x_n^b +o(x_n^b)$

From (1) how get an equivalent of $x_n$ ?

[quote="Moubinool"]From (1) how get an equivalent of $x_n$ ?[/quote]
Don't even think about it. The dominant term is $a^{(b-1)^n}$, multiplied by some terms that are likely exponential in $n$, and $a$ depends on initial conditions.

As I said, it's faster than linear convergence.


Hey folks,
Here is what a friend asked on a community forum and I thought he will get his answer from some of our latex experts here.

He wrote: "Me and other friends are interesting in creating math and science forum. We find a free host and successfully create a forum for ( trail ) but we run into some obstacle when trying to post math and chemistry symbols .

PHP won't recognize math symbol like 2X^2. Now we have to put everything off until we install latex to the forum. Then again when we install the latex; we run into another problem. The server that suppose to host it is telling us that they won allow clients to upload their files with such extension.

The person I talk to did not even know what I was talking about. Another word they don't host latex even with pay services. Can anybody please help giving us info about server that have latex install. I am not a programmer but I know HTML, Front page, little C++ and little Java. I appreciate all your help"

There are 8 hosts that offer LaTeX listed or linked to at [url=http://sixthform.info/steve/wordpress/?p=63]LaTeX Web Hosting[/url] though you will need to judge their quality and prices for yourself.
[url=http://www.forkosh.dreamhost.com/source_mathtex.html]mathTeX[/url]'s public server can also be used without having LaTeX on the server though clearly the speed won't be as fast.

I have 50mb of space available to me from the University (which is down to about 10mb now!). I was just wondering, if I was to "install" LaTeX into my "P drive" would it be possible to get LaTeX working on a phbb forum?

[quote="stevem"]You need to be able to run programs from the P drive and I wonder if the University allows that? Also 10MB is nowhere near large enough for a LaTeX installation whch may require hundreds of MB. If the University won't let you have access to their LaTeX then [url=http://www.forkosh.dreamhost.com/source_mathtex.html]mathTeX[/url] is your answer. You already know that mathTeX works well and there are instructions (in the paragraph headed [i]mathTeX plugins...[/i]) on how to us it in phpBB.[/quote] I don't think we can run programs. (It has allowed me to create a homepage and put files online, but that is all.)

Thanks though. The university are just passing the ball around the court at the moment! We need the final OK so I just need to sit tight and wait... (Although I will see what I can make of mathTeX).


I certainly hope so!!!  Although, I haven't heard from the tournament coordinator Jai so as far as I know, we don't have any problems yet, which is a potential problem.  However, I imagine we will meet once school starts in September and work things out.

Adding on to this question: Could someone give me a link to the web page for JHMT? I tried finding it through Google search, and there was nothing. I know there used to be a website for JHMT, but it seems to be gone now.

Yeah, their old website http://www.tripleintegral.org/jhmt/ appears to have evaporated . . . that's not a good sign.  This has always been a tournament that totally reinvents itself every year, though, so they'll probably pull something together.

Yes, I recall not being able to access the site last year.  I'm not sure why this has happened, and why I have not been notified about problem writing.  When I get back in September, I will contact Jai and post what he said.

Okay; will a temporary site be set up?

And also, is there a site where we can download problem sets from past years?

The site would be the one you know: the tripleintegral site.  I'm not sure what is going on this year.  I'll be back in Baltimore next week, so I'll find out then.  I do have last year's problems in PDF form if you'd like to see those (I may have 2007 as a hard copy but I'd have to look).  Give me a week or so.

Bump.

So is JHMT going to be happening this year, or is it still up in the air? Just curious.  :lol:

I heard that JHMT is happening in the spring this year.

That is good news, although it is kind of odd that I never heard anything.

Any updates? All I've been able to find online is the searching for a webmaster.

I'm not sure who is currently running the tournament-the person who used to graduated.

I have heard from one of my coaches that Daniel Cho is the director, and that the tournament will be held in the spring.

could someone please post an old JHMT? That is, at least 3 years old. I would really appreciate that. :)


AMC
USA(J)MO
USAMO
AIME
AMC 10
AMC 10 A
AMC 10 B

A number of comments and threads have been critical of the selection process used by CAMC for this year's USAMO.

I am a coach and a parent.  But, the manner of this year's selection did not have an impact on any of my students, so I think my remarks are from a neutral perspective.

There is little doubt the CAMC is aware of the inequities caused by this year's unusual selection process.  On the other hand, rule 4 of the USAMO Selection Process, as given in the Teacher's Manual is pretty precise:

[quote="USAMO Teachers Manual (emphasis added)"]4.  The lowest AIME score among those 240 first selected will determine a floor value.  [i]The second selection of approximately 120 USAMO participants will be among students in the 10th grade and below who received an AIME score at least as high as the floor value.[/i]  [b]If there are more than 120 young students with a score above the floor value, then approximately 120 students will be selected from this group by using the USAMO index.[/b][/quote]
My hunch is that there were many more than 120 students in the second selection who received an AIME score of 8 or higher.  If that was the case, it would seem the CAMC had little choice but to follow their published rule, and resort to using the defined USAMO index.

AMC Director Dunbar has announced that the first selection of 240 students had an index of "217.0".  He has announced that the floor value was "8".  And he has announced that the second selection of 120 students had an index of "212.0".

Assuming they followed the second (bold) part of the rule to the letter, some of the (unexpected?) impacts are as follows:

1.  A hypothetical young student who scored 101.5 on the AMC-12 and 11 on the AIME would NOT have qualified under the [b]bold[/b] rule, but would have qualified under the [i]italicized[/i] rule.  (USAMO Index = 211.5.)

2.  A hypothetical young student who scored 146.5 on the AMC-12 and 7 on the AIME would NOT have qualified under either rule.  (AIME score below floor.)

3.  Under the bold rule, a hypothetical young student who scored 131 on the AMC-12 and 8 on the AIME would NOT have qualified, whereas a similar student who scored 132 on the AMC-10 and 8 on the AIME WOULD have qualified.  Under the italicized rule, both students would have qualified.

The members of the CAMC do not live under rocks.  As I suggested before, my hunch is that the peculiar distribution of this year's test scores probably caused the unexpected invocation of the second (bold) part of rule #4.

For future years, if someone were to suggest an alternate (but simple) selection process, that was fair under all possible score distributions, I bet CAMC would listen.  They have tinkered with the selection process in the past.  [Don't just fire out some random ideas.  Give some thought to a comprehensive set of rules.  I think you'll find it is very difficult to define a set of rules that is fair under all circumstances.]

For those students in grades 10 and below, who scored 8 or more points on the AIME, the bold part of rule 4 does appear to give equal weight to the AMC-10 and -12.  The letter of rule 4 may violate the spirit of item 7:  [Personally, I don't view Item 7 as a "rule", so I would remove it from the Selection Process.]

[quote="USAMO Teachers Manual (emphasis added)"]7.  The selection process is designed to favor students who take the more mathematically comprehensive AMC 12A and AMC 12B contests.[/quote]

It's also worth mentioning that AoPS funded the expansion of USAMO by about 50%.  [Thank you, AoPS!  I hope your funding the expansion didn't aggravate any inequities in this year's USAMO selection process.]  At the time the expansion was announced, it was stated that there was a low correlation between AIME scores and USAMO scores.  The purpose of the expansion was partially to get more students who might have undiscovered USAMO talents.  Individual inequities aside, [b]any[/b] selection criteria that includes 50% more students from near the top will certainly be an improvement to the overall pool of students available for USAMO, MOSP, and IMO.

Thanks for your thoughtful and carefully worded post, rcv.

The difficulties faced by the CAMC are considerable.  There simply is no easy way to make a selection that everyone will regard as fair.   In any given year, there are at least 10 different score paths students can take before the USAMO

 {AMC10A,AMC10B,AMC12A,AMC12B,USAMTS} X  {AIME I, AIME II}

The statistical difficulties in calibrating the equivalence of difficulty of the different exams is a daunting task.  There is a very short time window for compiling all the data, analyzing it, and making the selection.

I believe the CAMC is full of smart and thoughtful people who do the best they can under the circumstances.   In the one case that I know of where there was widespread belief that a very serious error had been made in the selection process (in 2002), that error was remedied by later inviting a large number of additional students to write the USAMO, at great budgetary cost (since those additional students were also invited to MOP.)

I am very grateful to the people who serve on the CAMC.   They are dedicated volunteers with a difficult and often thankless job.   As I said, I believe they are smart and thoughtful people who care very much about this superb program and work hard to do the best they can under difficult circumstances.   No matter what judgments they make, there will always be someone critical.

We do not have all the data the CAMC had.  I agree with rcv that it's possible that there were a very large number of students in 10th grade and below with scores of 8, too large to be easily accommodated.   I would guess that there simply was no "perfect" way to make that cut.   I believe the CAMC tried hard to do the best they could.

In the interests of accurately describing the process to future students,  I do believe that it would be good to edit the AMC webpage on USAMO selection to eliminate the text describing "Rule 7" and the subsequent advice to young students that taking the AMC12 will be to their advantage in qualifying for USAMO.

As a coach, I personally would advise future young USAMO-hopefuls to take the AMC12 on the A date and the AMC10 on the B date, to cover all possible bases as completely as possible.  (EDIT:  I'm sorry to say that I have not given this advice in the past.  I believed that the wording under "Rule 7" meant that students would never be at a disadvantage due to the fact their score on the possibly more difficult AMC12 might be lower than it would have been on the AMC10.   Actual past practice reinforced this belief.  But I agree with rcv that it's not really a "rule.")  I would also advise students to take the USAMTS, of course, but primarily because it's an excellent learning experience, as are all the AMC exams.

Thanks to everyone who works hard to make this program possible, especially all the volunteers on the CAMC!

EDIT: I would also like to add my thanks and appreciation to AoPS for their funding of the additional USAMO-takers this year!  I also appreciate this terrific website, the excellent articles and links the in Resource section, and the books and classes AoPS has created.

Another parent weighing in...

I do think the complaints about the younger student index are valid, but not iron clad.

Any method for choosing the students near the edge of eligibility are going to result in winners and loosers, with everyone second guessing decisions made long ago. I think they need to accept they were marginal. I stress to my kids that there is a lot of luck in academic competitions as well as talent and hard work, and you have to accept the bad luck along with the good luck. Hard work is the only part of the equation that is under their control.

If the AMC had made the whole tiebreaker rule clearer, then more kids would have taken the 10 over the 12. In the first selection process, there would have been less younger students picked. The overall result would have been less younger students among the 360 kids.  I think that firmly confirms that the 12 is a better idea for the stronger math students, and does favor them.

Taking both tests is the obvious answer, but again - good luck rears its ugly head. Not all kids are fortunate enough to be in a school that values the AMC and sees the advantages to offering it twice. Heck, most schools do not offer it at all. Our kids are very lucky to have the AMC and AoPS.

Thank you for this much-needed post.

You're right on in pointing out that we do not have the data that AMC has.  Thus we can only judge the apparent fairness of the selection; only if we have the data can we make any judgment about (a) whether it is actually fair, and (b) if a better method was available.  I have great respect for Prof. Dunbar and the other AMC staff, and thus I suspect that the answer to (b) is "no".  I won't speculate on the answer to (a), especially since it's irrelevant if (b) is "no".

When the prospect of too many people qualifying through the floor came up, I was hoping that it either wouldn't happen, or the AMC would tweak the selection procedure, as occurred in 2000, 2001, and 2002.  However, note that in 2001 and 2002, the procedure was changed, but it was changed to include more students.  Anyone who would have qualified under the system as originally expounded still qualified.  (As a side note, in 2000, the system was changed from what they said it was going to be, but the likely result is that anyone who would have qualified still qualified.  See the AMC website for the original and revised procedures that year.)  It is definitely a good idea, if not an obligation on AMC's part, to qualify all those who were going to qualify based on the published system.  This year, unfortunately, I wouldn't be surprised if it were not possible to add more qualifiers in a fair way, due to either the lack of a good way to add qualifiers, or the fact that any fair method of adding people would add far too many to be feasible.

The moral of my post:
1. AMC has more information than we do, and said information is crucial to making judgments about the selection process.
2. The people at AMC are not out to get you; they are good and fair people.
3. Therefore, we can assume that the selection procedure used was the best one available, given the circumstances.

On behalf of the CAMC, thanks for the kind words and the encouragement.

The post by rcv has it generaly correct, but let me clarify further.  The italicized rule is applied first, AND then the bold rule is applied second as necessary.  Note the "AND".  Both conditions apply.

This year, for the first time in my experience, we needed to additionally apply the bold rule.  For at least the last two years, the number of students, 10th grade and below, with an AIME above the floor was the number we intended to invite.  This year we had about 200 or more above the floor value.  Therefore, we had to additionally invoke the bold rule.

rcv sketches three scenarios:
1.  In scenario 1, an AMC 12 score of 101.5 and an AIME of 11 (index 221.5)would not have qualified, since the young student did not have a USAMO index greater-or-equal to 212.0.  Fortunately such an unfortunate and unusual combination did not occur.

2.  Scenario 2 did occur.  I regret that, but I am bound by the rules.

3. Scenario 3 did occur.  I regret that too, but I am bound by the rules.

 If I bend or break the rules, the system is corrupted.  That won't happen.

The problem mathematically boils down to this:  We have a 2-vector of information, AMC score, and AIME score.  We are trying to induce a linear order on this information, and make a dividing line in that linear ordering.  I don't see how to induce a linear order on a set of 2-vectors that does not introduce some "paradox" as in the Scenarios above.  And on any boundary or dividing line there will be people on each side, and very close.

Really, in contrast to what Sly Si suggests, we don't have additional information, although we see all the information at one time.  We try to run a transparent competition, with the rules published and known to all.  We are not hiding anything, nor do we have access to privileged information that would make it all clear.  We try to be fair, in fact, I would say we are fair, but the two facts above are bound to create some unhappiness.  I regret that, too.

Ultimately, I want people to enjoy mathematics, that's our real goal.

Steve Dunabr
 AMC Director

OK, look, I do not want this thread to devolve into a cross-examination of Dr. Dunbar.  A question beginning with "Is it true however,..." is pretty accusatory.  A much more respectful, neutral way to ask the question would be "Do you have any statistics on how many students who scored an 8 on the AIME took the AMC 10 versus the AMC 12"?

Even so, I doubt whether they have these stats on hand, and I really doubt that they're going to take the time to generate them right now.  I'm confident that the AMC will publish the stats from the AIME as soon as they are available, whatever subset of stats that they choose to publish.

The AMC10 and AMC12 were introduced in 2000.  Before that, there was only a single preliminary high school exam, the AHSME.

The CAMC was initially confronted with questions from teachers and students wondering which exam younger students should choose, at a time when there was a great deal of uncertainty as to what the two exams would look like.

Nobody had ever seen an AMC10 exam since they were brand new.

There had always been a handful of strong young students who had done very well on the AHSME (now redubbed the AMC12) and the AIME and qualified for the USAMO under the old regime.  Indeed two ninth graders (Reid Barton and Gabriel Carroll) had won gold medals at IMO in 1998.   An eighth grader (Po Ru Loh) had made USAMO Honorable mention and won the Clay prize for elegance on the USAMO in 1999.  Several exceptional seventh graders (Dani Kane and Reid Barton) had even made 30-person MOP (in 1998 and 1996, respectively.)

Clearly, given the track record of such students on the old AHSME, it would be silly to encourage ALL young students to take the AMC10.   For students like the ones I listed above (and others with strong backgrounds aspiring to follow in their footsteps), it would be silly to encourage them to take the AMC10.

Students like the ones I listed above had sufficient preparation to take the AMC12 long before 11th grade.   The AMC12 would be the better learning experience for them and such students should clearly be encouraged to take the AMC12.

But how to encourage stronger students to take the AMC12 when other students in their grades had the option of taking the easier AMC10 was a difficult challenge?

This was an especially difficult issue in 2000 and 2001, because there was no option to take A-date and B-date exams.  There was only one date for the AMC exams in those years, so students had a one-shot choice:  take the AMC10 OR the AMC12, not both.

I believe the CAMC had this problem in mind when they devised the language which still appears under "Rule 7" on the USAMO selection page.

The CAMC adhered to the "Rule 7" principle in 2000 and 2001.  They created a selection system in which a first round of students were selected by index total without regard to exam choice (but with lower cutoffs for younger students) AND then AFTER that round, they added a number of students with lower indices who had taken the AMC12.

The current USAMO archives do not spell out all these details completely, but I have paper copies and vivid memories as the parent of an eighth grader in 2000 who was very surprised and delighted to qualify for USAMO that year.  She took the AMC12 and was invited to write the USAMO with an index that would not have qualified if she gotten the same index based on the AMC10.  She had great good luck that year and even managed to qualify for MOP, which was the beginning of many incredible adventures.  (She is one of several examples I know that illustrate that students with marginal indices sometimes do surprisingly well on the USAMO.)

In 2001, the CAMC again followed a similar principle, selecting a first group of students based on index regardless of exam taken (with graduated cutoffs for different grades) and THEN an additional group of students who had taken the AMC12 and gotten lower indices.

Since 2002, students have had the option to take both exams because of the introduction of the A-date and the B-date, and CAMC policies have evolved with a growing body of experience and an increasingly complex set of paths to the USAMO.

I think that overall the CAMC does a magnificent job of providing a wonderful program that inspires and encourages so many students.

They are in the unenviable position of making difficult judgments.  We all know that there is no perfect system, so there will always be grounds for criticizing whatever system is in place.

I am deeply appreciative of the countless hours of volunteer effort put in by the members of the CAMC.

[quote="rcv"]Sophia points out there are at least 10 paths to the USAMO.  Meanwhile, the AMC Web Site has now posted 2006 AIME statistics.

In the vicinity of the floor value, it looks to me like the AIMEs were at least 2 questions different in difficulty, with the AIME-I being the more difficult test.  On the AIME-II, 5.58% of the participants scored 9 or more.  On the AIME-I, 4.71% of the participants scored 7 or more.

If my analysis is correct, this is essentially a 20-point advantage in index score to those who took the AIME-II.  I am quite surprised to see such a difference!  Is my analysis incorrect?[/quote]

My guess is that the indices indicated smarter students taking the AIME II.

[quote="pieterminate"][quote="rcv"]Sophia points out there are at least 10 paths to the USAMO.  Meanwhile, the AMC Web Site has now posted 2006 AIME statistics.

In the vicinity of the floor value, it looks to me like the AIMEs were at least 2 questions different in difficulty, with the AIME-I being the more difficult test.  On the AIME-II, 5.58% of the participants scored 9 or more.  On the AIME-I, 4.71% of the participants scored 7 or more.

If my analysis is correct, this is essentially a 20-point advantage in index score to those who took the AIME-II.  I am quite surprised to see such a difference!  Is my analysis incorrect?[/quote]

Yes, and offtopic, but for many people(such as me), our school only offers amc test and only one aime test, so the "10 paths" seems exaggerated.  There are really just two paths for many people: amc 10~~>aime, or amc 12 ~~> aime.[/quote]

I don't think she means that someone can make the USAMO 10 different ways, but that there are 10 different paths to the USAMO, and when deciding on the USAMO index, it's hard to make it exactly fair to the people who are in each of the 10 ways to qualify.

[quote="sophia"]Thanks for your thoughtful and carefully worded post, rcv.

The difficulties faced by the CAMC are considerable.  There simply is no easy way to make a selection that everyone will regard as fair.   In any given year, there are at least 10 different score paths students can take before the USAMO

 {AMC10A,AMC10B,AMC12A,AMC12B,USAMTS} X  {AIME I, AIME II}

[/quote]

You can qualify through the USAMTS?

[quote="DPatrick"][quote="Xantos C. Guin"]You can qualify through the USAMTS?[/quote]You can qualify for the AIME via the USAMTS.  I believe that the AMC treats it as an AMC 12 score of 100.[/quote]

That is rather low for a much harder competition... :|

Whatever.  I don't really care about all this; I think I made a 0 on the AIME.  Anyway, I just wanted to say thank you to Mr. Dunbar for all his hard work.  Maybe next year I'll get a positive AIME score, and have something to complain about!

Comments on rcv's analysis of the AIME I vs. AIME II data.

The population taking AIME II is likely to be on average stronger and more likely to qualify for USAMO.

Why?

Because taking AIME II means that the school has to go out of its way to order the special exam, to pay extra for the exam fee and for the overnight postage fee.  That's more likely to happen if the school thinks there are students with a significant likelihood of doing well enough to qualify for USAMO.

If a student is ill or on a field trip the day of AIME I but has little prospect of making USAMO, the math club advisor is less likely to be inclined to take the trouble and expense of setting up AIME II.

Indeed the AMC's published statistics show that the average AMC score of AIME I takers was 110.0 while the average AMC score of AIME II takers was 112.3.   The mean AIME I score was 2.71 while the mean AIME II score was 2.87.

These mean differences show a certain amount of consistency.  The AIME II pool was apparently, on average, a bit stronger than the AIME I pool.

Now rcv saw a much more dramatic diference at the right tail of the distribution, where USAMO qualifiers come from.   But that's what you would expect if you know statistics.  Small differences in the means of two populations can translate into very substantial differences in the proportions at the extremes.

[quote="sophia"]
Now rcv saw a much more dramatic diference at the right tail of the distribution, where USAMO qualifiers come from.   But that's what you would expect if you know statistics.  Small differences in the means of two populations can translate into very substantial differences in the proportions at the extremes.[/quote]

It would transform into a more substantial difference, but not to the extent of 2 points.  So a more plausible reason would be because the large magnet schools (such as TJ) took the aime II and that significantly bolstered the aime II distribution at the tail end since ~12 qualified from TJ alone.

The following graph shows a comparison of the 2005 AIME-I versus 2005 AIME-II.  Part of my surprise in this year's comparison (posted previously) is that last year's curves were almost indistinguishable near the floor score of 9.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=754[/img]

The following graph overlays the 2005 AIME comparison on top of the previously-posted 2006 AIME comparison.  [Both AIME-I curves are dark blue.  Both AIME-II curves are magenta.  The 2005 curves are solid.  The 2006 curves are dashed.]

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=756[/img]

In 2005, the AIME II takers did not differ nearly as much from the AIME I takers in their mean AMC scores.

Average AMC score of 2005 AIME II students was 110.4 vs. 109.9 for AIME I = mean difference of 0.5 points.

Average AMC score of 2006 AIME II students was 112.3 vs. 110.0 for AIME I = mean difference of 2.3 points.

So there was a far more substantial difference in the mean AMC scores of the two pools of students this year (which could indeed have been driven in part by factors like a large magnet school taking their exam on the AIME II date.)  That's going to have a huge effect on the right tail, and a much more muted (but still quite noticeable and statistically significant) effect on the mean.


calculus
integration
function
number theory
relatively prime
AMC

[quote="Calculus"]What does it mean if a number is a pairwise relatively prime positive integer?

Also, what is a monotonically increasing integral function?[/quote]

1.  [b]A set of numbers[/b] is pairwise relatively prime if, for [b]every[/b] $a, b$ in the set, $gcd(a, b) = 1$.

2.  A monotonically increasing function is a function that always increases (i.e. $f(x+1) > f(x)$).  An integral function is a function that only takes integer values.


A small crate of weight 5.0N is released from rest at a height 2.0m up a smoyj inclined plane. The crate slides down the plane and across a smooth 1.50m flooe to a rough plane where a spring is located.
The bottom of the spring is located at a height of 0.5m. The spring constant is 20N/m .
The coefficient of kinetic friction between the rough plane and crate is $\frac{1}{\sqrt{3}}$ and the coefficient of static friction is $\frac{1}{\sqrt{2}}$

(a) What is the max height the crate reaaches on the right hand inclined plane?
(b) Where does the crate permanently come to rest? How many times does the crate travel up to the rough plane?

Because it has a lengthy solution :) .

Anyway, as you have suggested, you can use energy approach to this problem. At the beginning, the crate has some potential energy. If there was no friction and no spring, it would get again to the height of 2.0 m. However, the crate loses some of its energy to the spring - this energy is exactly $\frac{1}{2}kx^2$ where [i]k[/i] is your spring constant and [i]x[/i] is the distance the spring is pushed. The crate gets this energy back as the spring gets to its original length. Then there's some energy lost to friction and this energy is exactly $Fl$ where [i]F[/i] is the friction force and [i]l[/i] the distance the crate has traveled on the rough surface. (You don't get this energy back and you lose it again as the crate goes down.)

$l=\frac{h}{\sin30^\circ}$

$F=f\times F_n$

where [i]h[/i] is the max height to which the crate can get, [i]f[/i] is the coefficient of kinetic friction and [i]F_n[/i] is the normal force.

$F_n=F_g\times\sin30^\circ$

$F_g=mg$

so you get:

$E_{p1}=E_{p2}+E_s+E_f$

$E_{p1}$ is the original potential energy, $E_{p2}$ is the final potential energy, $E_s$ is the energy of spring and $E_f$ is energy lost to friction:

$mgh_1=mgh_2+\frac{1}{2}kx^2+F_fl$

$x$ is the distance the crate travels beyond $\frac{\text{0.5 m}}{\sin30^\circ}=\text{1 m}$:

$mgh_1=mgh_2+\frac{1}{2}k(l-\text{1 m})^2+fmg\sin30^\circ l$

etc. - you do the rest. You can use the same approach for question (b). Just don't forget to disregard $E_s$ if $x$ is negative.


Find the value of $\frac{x}{y}$ if $\frac{3}{\sqrt{y}}-\frac{1}{\sqrt{x}}= \frac{2}{\sqrt{x}+\sqrt{y}}$

[hide]
$\frac{3}{\sqrt{y}}-\frac{1}{\sqrt{x}}= \frac{2}{\sqrt{x}+\sqrt{y}}$

$\frac{3\sqrt{x}-\sqrt{y}}{\sqrt{xy}}= \frac{2}{\sqrt{x}+\sqrt{y}}$

$(3\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) = 2\sqrt{xy}$

$3x+3\sqrt{xy}-\sqrt{xy}-y = 2\sqrt{xy}$

$3x = y$

therefore

$\frac{x}{y}= \frac{x}{3x}= \boxed{\frac{1}{3}}$
[/hide]

wow I guess doing these when you're tired isn't very good

that's a cool solution. I tried using the conjugate of the 2nd part, but it didn't work. I see where I stumped up.  :blush:

Yeah I actually tried rationalizing the denominators after I posted the solution and just ended up with a nasty equation that took far far too many steps to solve.

I wonder why having roots in denominators is a problem anyway?

I also tried to solve it imagining that the square roots weren't there.

I think roots in denominators just makes it harder to solve, as they can get ugly if not approached in the right way.

[quote="Glen Mann"]
I wonder why having roots in denominators is a problem anyway?[/quote]
I was wondering that too.
My algebra teacher called it "radical pants".  :D

the equation homogenous in x and y, so we can wlog y=1...then it is a normal problem...clear the numerator...solve the quadratic

Radicals in a denominator can make algebra slightly painful as well as the fact that it is a tradition left over from when you had to look up all square roots in tables; $\frac{\sqrt{3}+1}{2}$ is a lot easier to calculate than $\frac{\sqrt{2}}{\sqrt{6}-\sqrt{2}}$.


It depends on how experienced you are.

For example, MOEMS is too easy for some, but quite challenging for beginners.

Uh, "easy"? It seems to me that, for example, Classroom Math is easy for you, then, oh, I don't know, the College topics just might count as hard.  :roll:

AoPS is not hard. In fact, it isn't even a problem. What is hard is what people put on AoPS.

[quote="Cyrazeno"]not yet  :P[/quote]That's the spirit!

[quote="tjhance"][quote="pingala"]i mean what you think about AoPS.[/quote]

Do you mean the website or the books?[/quote]
It would make a lot more sense if it was the books.

Even the books vary in levels. The AoPS book 1 is much easier than book 2

[quote="pingala"]is AoPS too easy?[/quote]
MS forums: easy
HS forums: challenging
Olympiad & college: mostly very hard, but some of the problems are do-able

I put hard. I really think it is challenging, but not hard, but u didn't have that as an option...

still really hard... trying to push Intermediate Geo into my Anti-Intermediate-Geo head.

I think it's hard. But, I [b]AM[/b] in 5th grade. Some of the Classroom math topics make me confused, because all 6th grade stuff I learned was Pre-Algebra. I looked at my brother's AoPS books, and I gotta say, some of it looks pretty familiar. But that doesn't stop me from not knowing the questions  :oops:

Intros and AoPS book 1: perfect (sometimes slightly challenging, not hard)
MC, MOEMS, and other MS: easy
Intermediates, AoPS 2: hard

I can't remember what I put......(bad memory :oops: )

Aops vol. 1 = moderately easy
Aops Vol. 2 = quite challenging

[quote="Lawrence Wu"]y'all are calling International Contests& PRograms easy? dang geniuses...i voted very hard, im junk at that imo stuff[/quote]

Most likely, he's talking about the books. The site varies too much and there are too many unsolved problems. If one solves all those unsolved problems, he or she would be really pro.


Where I can find some information about these inequalities:

$ \sqrt[k]{\frac{a_1\plus{}a_2\plus{}...\plus{}a_n}{n}}\ge\frac{\sqrt[k]{a_1}\plus{}\sqrt[k]{a_2}\plus{}...\plus{}\sqrt[k]{a_n}}{n}$

AND

$ \sqrt[k]{a_1a_2...a_n}\ge\cfrac{k}{\frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}...\plus{}\frac{1}{a_n}}$ ?

 :)

[url]http://www.artofproblemsolving.com/Wiki/index.php/RMS-AM-GM-HM[/url]

The second inequality is the GM-HM,

and the first seems to be the power mean.

For the first one, we can do one of two things:

[b]1.[/b] Let $ b_i = \sqrt [k]{a_i}$, so that the inequality is equivalently:

$ \sqrt [k] {\frac {b_1 ^k + b_2 ^k + \ldots + b^n ^k}{n}} \geq \frac {b_1 + b_2 + \ldots + b_n}{n}$

which is true for integers $ k \geq 2$ and $ a_i \geq 0$ by the Power Mean Inequality

[b]2.[/b] Use Jensen's Inequality on the function $ f(x) = \sqrt [k]{x}$


algebra
polynomial
superior algebra
superior algebra unsolved

Find the degree and a basis for Q(sqrt3 + sqrt5) over Q(sqrt15)..

I don't know how to do this..

Thanks in advance..

you have to find the minimal polynomial of $ a \equal{} \sqrt{3} \plus{} \sqrt{5}$ over $ \mathbb{Q}(\sqrt{15})$. to do this, expand $ a^2$.


[quote="Moubinool"]$A,B\in M_n(R)$  suppose there exist $k\in R^{*}$ such that $A^2+B^2=kC$
with $C=AB-BA$

prove that $(C_n^1k^{n-1}-C_n^3k^{n-3}+C_n^{5}k^{n-5}-...)det(C)=0$[/quote]

Since $(A+iB)(A-iB)=A^2+B^2+i(BA-AB)=(k-i)C$ (1) and using the fact that $det(A+iB)det(A-iB)=|det(A+iB)|^2 \in \mathbb{R}$, thus from (1)  we can see that $Im((k-i)^n detC)=0$ which means that $(C_n^1k^{n-1}-C_n^3k^{n-3}+C_n^{5}k^{n-5}-...)det(C)=0$


inequalities
quadratics
algebra
polynomial
inequalities proposed

Given that $ a,b,c \geq 0$, and $ a\plus{}b\plus{}c \equal{} 3$, prove that

\[ a\plus{}ab\plus{}2abc < 5
\]

Please show full proof  :)

[hide="Remark"]I think, with the same conditions, following sharper inequality holds
\[ a\plus{}ab\plus{}2abc \leq \frac{9}{2}
\][/hide]

[quote="babyloverain"]$ 1 \plus{} b \plus{} 2bc \le \frac {5}{a}$
$ b(2c \plus{} 1) \plus{} 1 \le \frac {5}{a}$
$ \frac {(2b \plus{} 2c \plus{} 1)^2}{2} \le \frac {5}{a}$
$ \frac {(7 \minus{} 2a)^2}{2} \le \frac {5}{a}$
It is true because
$ 4a(7 \minus{} 2a)(7 \minus{} 2a) \le \frac {14^3}{27} < 40$[/quote]

are you sure?  :wink:

What about: $ \frac {14^3}{27} \equal{}101.62963 > 40$

[quote="shaaam"]Given that $ a,b,c \geq 0$, and $ a \plus{} b \plus{} c \equal{} 3$, prove that
\[ a \plus{} ab \plus{} 2abc < 5
\]
Please show full proof  :)

[hide="Remark"]I think, with the same conditions, following sharper inequality holds
\[ a \plus{} ab \plus{} 2abc \leq \frac {9}{2}
\]
[/hide][/quote]
You're right, shaam. We can show that $ a \plus{} ab \plus{} 2abc \le \frac {9}{2}$ as follow:
Replacing $ b \equal{} 3 \minus{} a \minus{} c,$ then we have to prove
$ a \plus{} a(3 \minus{} a \minus{} c) \plus{} 2ac(3 \minus{} a \minus{} c) \le \frac {9}{2},$
or equivalently,
$ f(a) \equal{} (2c \plus{} 1)a^2 \plus{} (2c^2 \minus{} 5c \minus{} 4)a \plus{} \frac {9}{2} \ge 0.$
We see that $ f(a)$ is a quadratic polynomial of $ a$ with the highest coefficient is positive. Moreover, its disciminant is
$ \Delta \equal{} (2c^2 \minus{} 5c \minus{} 4)^2 \minus{} 18(2c \plus{} 1) \equal{} (2c \minus{} 1)^2(c^2 \minus{} 4c \minus{} 2) \le 0,$ as $ 0 \le c \le 3.$
Therefore, $ f(a) \ge 0$ and our proof is completed. Equality holds if and only if $ a \equal{} \frac {3}{2},b \equal{} 1,c \equal{} \frac {1}{2}.$ :)

Dear Can, thank you for your proof  :)

Recently, I have found a proof based on AM-GM Inequality to this inequality. Here it is:
\[ a \plus{} ab \plus{} 2abc \leq \frac {9}{2}
\]

\[ ab(1 \plus{} 2c) \leq \frac {9}{2} \minus{} a
\]

\[ 2ab\left(\frac {1}{2} \plus{} c \right)\leq \frac {9}{2} \minus{} a
\]

\[ 2a\left(\frac {b \plus{} \frac {1}{2} \plus{} c}{2}\right)^2 \leq \frac {9}{2} \minus{} a
\]

\[ \frac {a}{2}\left(3 \minus{} a \plus{} \frac {1}{2}\right)^2\leq \frac {9}{2} \minus{} a
\]
which is equivalent of proving:
\[ a(7 \minus{} 2a)^2\leq 36 \minus{} 8a
\]
But
\[ 36 \minus{} 8a \minus{} a(7 \minus{} 2a)^2 \equal{} (4 \minus{} a)(2a \minus{} 3)^2\geq 0
\]
Since $ a\leq 3$, proof is complete.  :)

[quote="shaaam"]$ ab(1 \plus{} 2c) \leq \frac {9}{2} \minus{} a$

$ 2ab\left(\frac {1}{2} \plus{} c \right)\leq \frac {9}{2} \minus{} a$
$ 2a\left(\frac {b \plus{} \frac {1}{2} \plus{} c}{2}\right)^2 \leq \frac {9}{2} \minus{} a$

[/quote]
Are you sure about this step? :)

[quote="ith_power"][quote="geniusbliss"]i think it has to be given,
$ a,b,c$ > 0 ,otherwise we cant use AM-GM directly[/quote]

but by am-gm,
 we have,
$ 2ab\left(\frac {1}{2} \plus{} c \right) \le 2a\left(\frac {b \plus{} \frac {1}{2} \plus{} c}{2}\right)^2$
and we have,
$ 2ab\left(\frac {1}{2} \plus{} c \right)\leq \frac {9}{2} \minus{} a$
this two equations do not imply the third.[/quote]
No. But the third imply both of those, don't they ? ;)

[quote="geniusbliss"]well someone plz clear my doubt,
shaam has done it by to prove that
and he has succeded in proving that value greater than the LHS is
lesser than the RHS therefore he is done but one small thing
how canyou apply AM-GM for NONEGATIVE REALS
THEY ARE APLICABLE ONLY FOR POSTIVE REALS right?  :([/quote]
Hmm.. When you are in doubt what a theorem says, you should probably try to prove it !
The AM-GM inequality says $ \frac{x_1\plus{}x_2\plus{}...\plus{}x_n}{n} \ge \sqrt[n]{x_1...x_n}$ for any non-negative reals $ x_1,x_2,..,x_n$.
The case $ \frac{x_1\plus{}x_2}{2} \minus{} \sqrt{x_1x_2} \equal{} \frac{(\sqrt{x_1}\minus{}\sqrt{x_2})^2}{2} \ge 0$, proves the case $ n\equal{}2$.
Now we can use that:
$ \frac{x_1\plus{}x_2\plus{}x_3\plus{}x_4}{4} \ge \frac{\sqrt{x_1x_2}\plus{}\sqrt{x_3x_4}}{2} \ge \sqrt[4]{x_1x_2x_3x_4}$, and in this way you can prove it for all $ n \equal{} 2^m$. (And it is obvious here that it holds for non-negative reals)
To prove it for an abritary $ n$. Let $ 2^r \ge n$. And let $ X \equal{} \frac{x_1\plus{}x_2\plus{}...\plus{}x_n}{n}$.
Then we now that:
$ \frac{x_1\plus{}x_2\plus{}...\plus{}x_n\plus{}(2^r\minus{}n)X}{2^r} \ge \sqrt[2^r]{x_1x_2..x_nX^{(2^r\minus{}n)}}$
Notice that $ \frac{x_1\plus{}x_2\plus{}...\plus{}x_n\plus{}(2^r\minus{}n)X}{2^r} \equal{} X$, and hence we have:
$ X \ge \sqrt[2^r]{x_1x_2..x_nX^{(2^r\minus{}n)}} \equal{} \sqrt[2^r]{x_1..x_n}X^{(2^r\minus{}n)/2^r} \iff$
$ X^{2^r} \ge x_1..x_nX^{2^r\minus{}n} \iff$
$ X^n \ge x_1...x_n \iff$
$ X \ge \sqrt[n]{x_1..x_n} \iff$
$ \frac{x_1\plus{}x_2\plus{}...\plus{}x_n}{n} \ge \sqrt[n]{x_1x_2...x_n}$.
But we have only used that $ x_1,x_2,...,x_n \ge 0$, and thus it holds for abritary non-negative numbers !

well mathias_dk
i did prove the theorem on very similar lines but then
i think i was misguided by some sites which described the reals as
positive and i got used to it ,so i never really tried for nonegative
i never even noticed that i had never significantly used positive reals for the proof!!!
thanks mathias_dk

p.s. i am extremely sorry cuz the previous posts where just spam!!!!(rubbish)

The same ineq:
Given $ a,b,c \ge 0$ and a+b+c=4.
Prove that:
$ 2a\plus{}ab\plus{}abc \le 9$
With this ineq,we can use AM-GM.

[quote="secrets"]The same ineq:
Given $ a,b,c \ge 0$ and a+b+c=4.
Prove that:
$ 2a + ab + abc \le 9$
With this ineq,we can use AM-GM.[/quote]
$ 2a + ab + abc = a(2+b(c+1)) \le a(2 + (\frac{b+c+1}{2})^2) \le 9 \iff$
$ a(8+(b+c+1)^2) = a(8+(5-a)^2) = 33a - 10a^2 + a^3 \le 36 \iff$
$ -a^3 + 10a^2 - 33a + 36 = (4-a)(3-a)^2 \ge 0$, which is obviously true. Equality iff $ a=3,b=1,c=0$ :)

Let  $ a,b,c \geq 0$  and $ a+b+c = 3.$ Prove that
$$a+ab+kabc  \leq 4$$
Where $0<k<1.$ Equality holds when $a=2,b=1,c=0.$
$$a+ab+kabc  \leq \frac{(k+1)^2}{k}$$
Where $k\geq1.$ Equality holds when $a=\frac{k+1}{k},b=1,c=\frac{k-1}{k}.$
[url]https://artofproblemsolving.com/community/c6h1233736p23191118[/url]
[hide=k=1]Let  $ a,b,c \geq 0$  and $ a+b+c = 3.$ Prove that
$$a+ab+abc  \leq 4$$
[url]https://artofproblemsolving.com/community/c6h265079p1438609[/url]
[/hide]


Show that [1+(1/m)]^m>[1+(1/n)]^n where m>n are positive integers.

AM-GM for $1+n$ numbers, in that $n$ numbers equal to $1+\frac{1}{n}$ and $1$ numbers equal to $1$.


The 2008 MathPath Qualifying Quiz is now available on MathPath's website.

The quiz can be found [url=http://www.mathpath.org/Quiz/MP08test.pdf]here[/url].

Please do not discuss the problems on the quiz as each is suppose to do the quiz on their own.

If you would like to attend MathPath in 2008, I highly suggest you attempt the quiz, even if you personally feel you can't answer all of the questions.  Those who are responsible for grading and admitting campers look more for evidence of mathematical thinking and problem solving than, final, correct answers.

are we allow to ask for clarifications here?

Didn't you go last year, abacadaea? If you did, you don't need to do the application quiz again. Anyway, if you need clarifications, just e-mail Dr. Thomas directly.

Eh, sorry to revive but when can we discuss? Not until the actual camp?

Cause there is discussion on Mathcamp...


inequalities
induction
inequalities proposed

Proof that for all positive $ x,a_{1},...,a_{n}$ holds
$ \frac {1}{a_{1} \plus{} x} \plus{} \frac {a_{1}}{(a_{1} \plus{} x)(a_{2} \plus{} x)} \plus{} ... \plus{} \frac {a_{1} \cdot ... \cdot a_{n\minus{}1}}{(a_{1} \plus{} x) \cdot (a_{2} \plus{} x) \cdot ... \cdot (a_{n} \plus{} x)} \le \frac {1}{x}$

I don't see the pattern.

The first term is $ \frac1{a_1\plus{}x}$ and the second is $ \frac{a_1}{(a_1\plus{}x)(a_2\plus{}x)}$. But why is the last term $ \frac{a_1\ldots a_n}{(a_1\plus{}x)(a_2\plus{}x)\ldots(a_n\plus{}x)}$ ? Is it not $ \frac{a_1\ldots a_{n\minus{}1}}{(a_1\plus{}x)(a_2\plus{}x)\ldots(a_n\plus{}x)}$ ?

I'll prove it with induction on $ n$. Clearly it is true when $ n \equal{} 1$. Now assume that the inequality holds for some $ n$. Multiplying the inequality by $ a_1 \plus{} x$ it is sufficient to prove:

$ 1\plus{}\frac{a_{1}}{(a_{2}\plus{}x)}\plus{}...\plus{}\frac{a_{1}\cdot ...\cdot a_{n}}{(a_{2}\plus{}x)\cdot ...\cdot (a_{n\plus{}1}\plus{}x)}\le\frac{a_1}{x} \plus{} 1$ take the one from each side and divide by $ a_1$:

$ \frac{1}{a_{2}\plus{}x}\plus{}\frac{a_{2}}{(a_{2}\plus{}x)(a_{3}\plus{}x)}\plus{}...\plus{}\frac{a_{2}\cdot ...\cdot a_{n}}{(a_{2}\plus{}x)\cdot (a_{3}\plus{}x)\cdot ...\cdot (a_{n\plus{}1}\plus{}x)}\le\frac{1}{x}$ which is just the case for $ n$ numbers $ a_{2},\ldots,a_{n\plus{}1}$.


P is any point on altitude CD of $\Delta ABC$. AP and BP meet sides CB and CA at points Q and R, respectively. Prove that $\angle QDC \cong \angle RDC$.

Blanchet's Theorem

Above $C$ draw a parallel to $\overline{AB}$, next extend $\overline{AQ}$ and $\overline{BR}$ till meet the parallel at points $M$ and $N$ respectively. Finally, apply similar triangles and Ceva, and the proof is completed.

Really, you gotta prove that $\triangle DMN$ is isosceles.


Let $ (a_{n})$ be a sequence such that $ a_{n}$ $ \in\mathbb{Z}_2$ for all $ n\in \mathbb{N}$.
If $ a_{n}$=$ a_{n\minus{}1}$ +$ a_{n\minus{}2005}$ ,prove that this sequence becomes eventually periodic($ a_{1},a_{2},\dots..,a_{2005}$ are arbitrary elements of $ \mathbb{Z}_2$).

What happens when some string of 2005 values occurs twice?  Why must this always happen?

This problem can be understood in very simple terms without even referencing the notion of a finite field.  In those terms, what we could say is that the characteristic polynomial splits over a large enough field extension $ \mathbb{F}_{2^n}$, so the powers of its roots become eventually periodic.

[quote="timothi_math"]Let $ (a_{n})$ be a sequence such that $ a_{n}$ $ \in\mathbb{Z}_2$ for all $ n\in \mathbb{N}$.
If $ a_{n}$=$ a_{n \minus{} 1}$ +$ a_{n \minus{} 2005}$ ,prove that this sequence becomes eventually periodic($ a_{1},a_{2},\dots..,a_{2005}$ are arbitrary elements of $ \mathbb{Z}_2$).[/quote]

The essence of the matter is quite simple and general:

[b]HINT[/b]  The shift map is a permutation of a finite set
so its orbits are the [b]cycles [/b] of the permutation.

For much further discussion see my prior post
http://google.com/groups?selm=y8zoe5857bx.fsf%40nestle.csail.mit.edu


$ O$ is the center of circles containing $ \widehat{APB}$ and $ \widehat{CED}$. Both circular arcs have the same length. Find the ratio of $ OA$ to $ OC$. Express your answer as a common fraction.
[asy]size(150); defaultpen(linewidth(0.8)); import graph; import geometry_dev;
draw(Arc((0,0),5,90,360));
draw(Arc((0,0),15,0,90));
draw((0,15)--origin--(15,0));
markrightangle((0,5),origin,(5,0),10);
dot(Label("$O$",align=SW),origin); dot(Label("$A$",align=SW),(0,5)); dot(Label("$E$",align=NE),(7.5*sqrt(2),7.5*sqrt(2))); dot(Label("$P$",align=SW),(-5/sqrt(2),-5/sqrt(2)));
label("$C$",(0,15),NW); label("$B$",(5,0),SE); label("$D$",(15,0),SE);[/asy]

\begin{eqnarray*} 2\pi OA \times \frac {270}{360} & = & 2\pi OC \times \frac {90}{360} \\
OA \times \frac {3}{4} & = & OC \times \frac {1}{4} \\
\frac {OA}{OC} & = & \boxed{\frac{1}{3}} \end{eqnarray*}


calculus
integration
logarithms
calculus computations

Find the value of $a$ such that $\int_{0}^{a}\frac{1}{e^{x}+4e^{-x}+5}\ dx=\ln \sqrt[3]{2}.$

The integral is equivalent to $\int_{0}^{a}\frac{e^{x}\ dx}{e^{2x}+5e^{x}+4}$

Make the substitution $u=e^{x}\Rightarrow dx=\frac{du}{u}$ and the integral becomes

$\int_{1}^{e^{a}}\frac{du}{u^{2}+5u+4}= \frac{1}{3}\int_{1}^{e^{a}}\left(\frac{1}{u+1}-\frac{1}{u+4}\right) \ du = \frac{1}{3}\ln\left(\frac{5(e^{a}+1)}{2(e^{a}+4)}\right)= \frac{1}{3}\ln(2)$

And thus $a=\ln(11)$.


graph theory
combinatorics proposed
combinatorics

Let $K_{n+1}$ be complete graph with vertices $1,2,3,...,n+1$ let$i_1,\dots,i_{n+1}$ be permutation of $1,\dots,n+1$ .suppose we color the edge s joining vertex $k+1$ to vertices $\{\ 1,\dots k\}\$ with color k,independently color the edges joinig vertex $i_{k+1}$ to vertices $\{i_1,\dots,i_k\}$ with color $k'$,($k=1,\dots,n$).then $K_{n+1}$ has spanning tree $T$ such that no two edges of $t$ have the same color in the first coloring and no two edges of $T$ have the same color in the second coloring.

Basically, the problem asks for this:

Given two total orders on the set $\{1,2,\ldots,n\}$, prove that we can find a graph (which will automatically be a tree) with vertices $1,2,\ldots,n$, s.t. each vertex is connected to exactly one vertex smaller than it, in both orderings (except for the smallest vertices in the two orderings, of course).

It's easy to see that the following procedure gives the desired graph (assume the first order is the usual one, $1<2<\ldots<n$, and the second one is some order $<_1$):

In the $k$'th step ($k=\overline{1,n}$) connect $k+1$ to the smallest $t\in\{1,2,\ldots,k\}$ for which there is no $t'\ne t,\ t'\in\{1,2,\ldots,k\}$ s.t. $t>_1t',k+1$.