I want to learn stuff, know stuff, be able to understand what's going on in the world... but the whole formal education thing isn't all that appealing to me. I'd rather make a practical place for myself in the world.

Anyone got any ideas about a good approach to how much education I should get--financially, morally, and personally?

(High school right now, 15 yrs old, going to a rather unsophisticated state university next year. When I go to math competitions, I think about the people there, and when I solve math problems, I think about my method of thinking more than about the actual solution to the problem, just to give you a general idea of who I am.)

I'm kinda worn out from thinking about the future and my purpose in life right now. Just fishing for suggestions.

Oh, and do you think I'm necessarily more stupid than mathematics people, since I'm more of a humanities person? I took Calculus BC last year without difficulty... Made 800 verbal and 650 math on the SAT when I was 12 without any preparation, so I have some degree of intelligence, but I know a lot of people here are VERY intelligent--but are there things I understand that some of the mathematical achievers would have difficulty grasping?

What are the most important things to put mental energy into?

Put mental energy into whatever draws you personally. Do not listen to anyone who says "You must devote your time and energy to X because of Y." Choose what [i]you[/i] like.

Thanks for the reply! I'm wasting too many people's time posting on these forums, looking for an answer that I already know. I'm not mathematically inclined, and now I'm being forced to say goodbye to problem solving, which I've already more-or-less given up, or to take it up again--but I don't think I have enough interest to take it up again, and I don't want to admit that it's not part of my life anymore, so I've been on here posting random things, trying to find an answer to why I should hold on.

But it's hard to get in touch with what really matters to me...

I'm sure you aren't wasting anyone's time, we are all glad to help you out.

What if you did problem solving on the side while focusing on whatever it is you want to focus on (which sounds like Humanities unless I am mistaken)? A hobby is often refreshing, especially if you ever feel burned out with regards to whatever it is you are focusing on (which can happen even if you really enjoy it most of the time).

[quote="akkaren"]But it's hard to get in touch with what really matters to me[/quote]
In that case- why are you going to college early? Where's the advantage if you don't know what you want to do already?

[quote="akkaren"]do you think I'm necessarily more stupid than mathematics people?[/quote]
I've never really believed in 1-dimensional intelligence; there's a correlation between different fields of learning, but we don't have to express that by choosing a single "primary" axis of intelligence. This forum is full of people who are very focused on mathematics; they may also have excellent verbal skills, but that's secondary. Your strengths in humanities are real; we're not better than you at everything.

Well, I don't actually know what to focus on. I suppose I'm a very passionate, guardian-of-humanity sort of person.

What I really want? I want to contact and work with people of all ages, social stations, abilities, occupations and cultures, to promote the good of society while keeping the good of the individual in mind first. I'm not quite sure how to go about that, or how to shape it into a practical lifestyle.

My hobby is jumping from idea to idea... or, at least, that's what burns up all my freetime.

I suppose, with goals like that, it's not suprising I'm stumped.

@jmerry (you posted while I was writing this post) I'm going to college early because I don't have any classes left to take in high school, except for one completely superfluous language class, Spanish IV, and intro level electives, which would hardly put me in a good position for applying to and attending college the year after.

[quote="Elemennop"]You think about people while at tournaments HUH?[/quote]
I'm more concerned with wondering about all the people I see, who they are, what they think about, how they live, what they've been through, why they're there, etc, then about competing. It's... I'm a people-person, not an academic or achievement-oriented person. I'd rather meet new people and make friends than anything else. I like learning things from other people, acquiring abilities from people who focus on those areas, but I don't have any motivation myself, just what I pick up from the people around me.

[quote="akkaren"] I like learning things from other people, acquiring abilities from people who focus on those areas, but I don't have any motivation myself, just what I pick up from the people around me.[/quote]

Sounds like high school

Sounds like wishful thinking to me. *sociophobe* Yes, only an apathetic idiot teenager would post such desperately useless, self-contradictory, inconsistent posts as I do... At least they don't have me on beta blockers anymore...

[quote="akkaren"]
What I really want? I want to contact and work with people of all ages, social stations, abilities, occupations and cultures, to promote the good of society while keeping the good of the individual in mind first. I'm not quite sure how to go about that, or how to shape it into a practical lifestyle.
[/quote]

I think a good job match for you would be the pope.  Well, except maybe a better match for the pope would be keeping the good of the individual in soul first though I'm not well enough versed with the philosophical mindsets of the popes to comment about past popes...

On a more serious note, college is a good place to connect with diverse people.  I think it's kinda unfortunate that you're going to a state university (which one is it, btw?).  It's not that they aren't good places to learn, it's just that it doesn't seem like a state university is the best fit for you based on the very small amount of stuff I know about you.

Another thing that may be confusing you is that you want to promote goodness in society but do you have a clear picture of what you think is "good"?  This is a philosophical problem and there are many possible answers.

You can still do problem solving as a hobby - nothing wrong with that.

Actually, I'm only going to a state university next year because it has an early entrance program, and I didn't feel confident in graduating two years early and going to college normally at 16, so I won't necessarily be staying long. I can't stay in high school because I've exhausted the coursework.

It's not unusual for a teenager to be so chronically and chaotically depressed, but it's annoying to me. I've not seen enough of the world to know 100% what's good--but has anyone? I just want to be able to reach out and touch something positive, something meaningful--it makes the lows less terrible, when you know they're not all of your existence.

I've thought a lot about various political, economic, and personal factors in the "good" of society, but I'd like to know more about the economics especially, so that I know what to support. For me, though, many of the things that torment people in this world are simple, and many of the controversial topics are clearcut--I'm hardly unique in this, but it always helps for someone else to be out there persuading others.

Simply? I'd like to remove some of the confusion from the world. I know it's a little farfetched for me, since I allow myself to sink into emotional pits so easily, as demonstrated, but that's a natural result of loneliness, I think.

Consider the fact that pleasure, pain, joy, anguish, etc., human emotions, are  "felt" the same way by everyone. It is possible for two people to have the same feeling of inner turmoil and anxiety with completely different causes: perhaps one living well in the U.S. and another struggling to survive elsewhere. This is what links us all.

So before you try to help the world, help yourself: find the joy in your life, and ask "why." Then maybe you can understand what could bring joy elsewhere. Sure, some problems are clear cut, and you can devote plenty of service time to fixing them. But what makes it meaningful to you is exactly causing the joy in others that you have seen in yourself.

Study academics if this makes you happy. Understand fulfillment when it comes, then you can deliver it elsewhere.

[quote="curl"]
So before you try to help the world, help yourself: find the joy in your life, and ask "why." [/quote]
Everyday.

Awesome reply, by the way. Thanks.

Have you taken any philosohpy classes?

I reccomend it.

I can't quite answer you definitively without a bit more specifics (I love generalities, also...) but I'd say this: If you are looking for a way to improve things (globally and not...) consider a couple bits:

1. assuming you are an intelligent person based on the fact that yer graduating early, got good scores, and whatnot, you are probably most valuable for your mental, rather than physical aspects.

2. from an outside perspective, you are most likely to do best at something that genuinely interests you, so even if it may seem pointless to pursue seemingly inaplicable, ivory tower academia, it may be useful in ways you would not initially think of.  Anything which helps you develop yer potential as a thinker will be useful later-on.  Being 15, odds are you still have a lot of potential for mental growth.

....... pre-mature list ending.......

I guesse the main point I've got is that at this point, it's difficult to have a specific idea of how you can help "promote the good of society while keeping the good of the individual in mind first".  I'd say try to continue working on self-development/"discovery"..... I don't know if your going to get any specific carreer advise..... um...... ........ but try very hard not to lose track of your current goal, general though it may be; there's a lot of things which really wouldn't fit into it....

LUCK.

Don't be afraid to take your time.  Explore different areas that look interesting.  Don't expect yourself to know right now what you want to do with your life.  You've got a while to figure that out.

And have some fun--you won't be a kid forever.

So you seem to me to be a genuinely altruistic person who is worried because they aren't sure how their interests are going to be able to help other people, if I understand you right.

I wouldn't worry about it. In every profession, if you truly love what you are doing, you will find a way to do anything you set your mind to. If it is your goal to help people, and you are working at something you love, you will have no difficulty turning it to the good of humanity.

I'm not sure whether you are religious or not, but I like to think of it as, you should use what the Lord has provided you with, and he will then also provide you with a way to accomplish His work. Just do whatever you feel like doing and that will find its own purpose.

[quote="akkaren"]...It's not unusual for a teenager to be so chronically and chaotically depressed, but it's annoying to me. I've not seen enough of the world to know 100% what's good--but has anyone? I just want to be able to reach out and touch something positive, something meaningful--it makes the lows less terrible, when you know they're not all of your existence...[/quote]

From what you say, it sounds like a community service career would be good for you.  These careers can take many forms, and are very good for all types of people who care about others.  Also, have you considered a boarding school?  Not to get away from home, but for the many courses and community options they offer?

Good luck,
lyra

EDIT:I mean a highly academic boarding school like Exeter, Andover, etc.

Have you tried looking at going to Simon's Rock College of bard?  it's a college meant for people like you, who want to leave high school early and go to college.  I don't know too much about it, but you should check it out.

I think Exeter or Andover would be a better environment than Simon's Rock--but correspondingly harder to get into. Both schools have great financial aid for families who need it. (That said, my son decided to stay in town during his teenage years, except for summer programs.)

High school debate (one of my son's new activities) is all about solving world problems, and would be a good way to develop some new social experiences and knowledge of the world before going off to college.


Given a function $ f(x)$ satisfying $ f(x)\plus{}2f(1/(1\minus{}x))\equal{}x$, find $ f(2)$.

Thanks in advance!

Do not give up too soon. [hide="hint"]If you make $ x\equal{}2$, you get an equation relating $ f(2)$ and $ f(\minus{}1)$.
If you then try $ x\equal{}\minus{}1$ you get another equation relating $ f(\minus{}1)$ and $ f(1/2)$.
Do not give up there. Try making $ x\equal{}1/2$, and keep trying.
If you ever go full cycle and find yourself back to $ f(2)$, you'll have a system of equations.
Solving that system, you can find $ f(2)$, $ f(\minus{}1)$, and so on.
Feel free to call them $ a$, $ b$, $ c$ and so on $ (a\equal{}f(2)$, $ b\equal{}F(\minus{}1)$, and $ c\equal{}f(1/2))$ to save yourself some ink and confusion.[/hide]

[hide="Solution"]
$ (1) \ \ f(x) \plus{} 2f(\frac {1}{1 \minus{} x}) \equal{} x$

Substituting $ x$ with $ \frac {1}{1 \minus{} x}$ in (1) we get

$ (2) \ \ f(\frac {1}{1 \minus{} x}) \plus{} 2f(\frac {x \minus{} 1}{x}) \equal{} \frac {1}{1 \minus{} x}$

again substitute $ x$ with $ \frac {x \minus{} 1}{x}$ in (1) and we get

$ (3) \ \ f(\frac {x \minus{} 1}{x}) \plus{} 2f(x) \equal{} \frac {x \minus{} 1}{x}$

solving equations (1), (2) and (3) we get

$ f(x) \equal{} \frac {x^3 \plus{} 3x^2 \minus{} 6x \plus{} 4}{9(x^2 \minus{} x)} \implies \boxed{f(2) \equal{} \frac {2}{3}}$[/hide]

[hide="a slightly less algebraically intensive version"]
$ f(x)\plus{}2f\left(\frac{1}{1\minus{}x}\right) \equal{}x \Leftrightarrow f(x)\equal{}x\minus{}2f\left(\frac{1}{1\minus{}x}\right)$
$ f(2)\equal{}2\minus{}2f(\minus{}1)$
$ f(\minus{}1)\equal{}\minus{}1\minus{}2f\left(\frac{1}{2}\right)$
$ f\left(\frac{1}{2}\right)\equal{}\frac{1}{2}\minus{}2f(2)$

Substituting everything back in,

$ f(2)\equal{}2\minus{}2(\minus{}1\minus{}2\left(\frac{1}{2}\minus{}2f(2)\right)$
$ f(2)\equal{}2\plus{}2\plus{}4\left(\frac{1}{2}\minus{}2f(2)\right)$
$ f(2)\equal{}4\plus{}2\minus{}8f(2)$
$ 9f(2)\equal{}6$
$ f(2)\equal{}\boxed{\frac{2}{3}}$

[/hide]

Well, this was the math department problem of the month for the month of... I think it was December... (mabye November?) at my school.  I got a small prize last month for being the first to turn in a full and correct solution.  So I'm just wondering, where's this problem even from?  also, if you just substitute $ x \equal{} 2$, $ x \equal{} \minus{} 1$, and $ x \equal{} 1/2$ into the origional equation to get a system of three equations as KMST suggested, then substitute back in and solve for $ f(2)$, it's just as simple as laughinghead's solution, there isn't really a need to rearange the origional equation first.

Here's my solution:
$ f(2) \plus{} 2f( \minus{} 1) \equal{} 2$
$ f( \minus{} 1) \plus{} 2f(1/2) \equal{} \minus{} 1 \implies 2f( \minus{} 1) \equal{} \minus{} 4f(1/2) \minus{} 2$

So $ f(2) \minus{} 4f(1/2) \minus{} 2 \equal{} 2$

Also, $ f(1/2) \plus{} 2f(2) \equal{} 1/2$

Solving for $ f(2)$ gives $ 2/3$.


Let S equal the number of five-pointed stars on the front cover of this test. This number S is a factor of 2002 - N  , where N is a whole number. What is the smallest possible value of N?

A.6 B.8 C.10 D.12 E. 16

I think you make a mistake while posting the problem.  :|
From what you said, the answer should be the smallest possible :$ \boxed{A}$
Maybe the problem meant that $ \mathcal{S}$ was the number of points of the stars.

Maybe there is a test, and on the front cover, maybe a logo or something, there are a bunch of stars?   :huh:


calculus
derivative
calculus computations

You must construct a closed rectangular box wit volume 576 in.^3 and with it's bottom twice as long as it is wide. find the dimensions of the box that will minimize it's total surface area.

We recently discussed a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=121587]very similar problem[/url].

[b]its[/b]


integration
logarithms
limit
calculus
derivative
calculus computations

Let  $I(s) = \int_{0}^{1}\frac{x^{s}}{1+x}\; dx$. By simple calculation, we have

1. $I(s) = \frac{1}{s+1}-I(s+1)$

2. $\lim_{s\to\infty}I(s) = 0$

So we can write $I(s)$ as

$I(s) = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{s+n}$

and its derivative is

$I'(s) = \sum_{n=1}^{\infty}\frac{d}{ds}\frac{(-1)^{n-1}}{s+n}= \sum_{n=1}^{\infty}\frac{(-1)^{n}}{(s+n)^{2}}$.

Then the value of theintegral we want to calculate is given by $I'(0)$ and

$I'(0) = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{2}}=-\eta(2) =-\frac{\pi^{2}}{12}$.


topology
algebra
function
domain
advanced fields
advanced fields unsolved

Disclaimer: This question is probably really dumb to people who know anything about cohomology.

I am just starting to learn about the De Rham cohomology. I know that there are other cohomologies out there. Sometimes I hear people refer to "the cohomology of *insert something*." (This happened in our weekly topology seminar last week).

Question(s): If someone refers to the cohomology, is there one that they usually will be referring to? Does this typically depend on the context? Are they all related somehow? Are they all the same? (The last one probably is no). If they aren't all the same what is the point of having so many?

You don't have to answer all these...I'm mostly interested in the first question. The other ones just kind of followed as generalizations.

My understanding is that for "nice" spaces (smooth manifolds) the various cohomology groups are isomorphic, which is why one can talk about "the" cohomology. (For me, de Rham cohomology is the first I think of.)

Different cohomologies may have different domains of definition: for example, de Rham cohomology requires differentiable structure, while the Čech cohomology makes sense in arbitrary topological spaces. Also, some definitions may allow for easier computation of cohomology groups of particular spaces.

[quote="HilbertThm90"]Are they all related somehow? Are they all the same?[/quote]

I think this touches http://en.wikipedia.org/wiki/Motive_%28algebraic_geometry%29.


geometry
angle bisector
perpendicular bisector
geometry unsolved

Find all triangles $ABC$ for which $\angle ACB$ is acute and the interior angle bisector of $BC$ intersects the trisectors $(AX, (AY$ of the angle $\angle BAC$ in the points $N,P$ respectively, such that $AB=NP=2DM$, where $D$ is the foot of the altitude from $A$ on $BC$ and $M$ is the midpoint of the side $BC$.

What do you mean by "the interior angle bisector of BC"?

Dear ORL,
please, could you tell us what you mean by "the interior angle bisector of BC" because this problem seems quite interesting and I think that, if you mean the internal bisector of <ABC, the problem has no solution.
Kind regards, sincerely. JPE

Patience is a virtue.  :)

[hide="Pierre here you are."]In a triangle $ABC$ with an acute angle at $C$, let $H$ be the feet of the altitude from $A$, $M$ be the midpoint of $BC$, $x$ and $y$ be the trisectors of the angle $A,$where $\angle BAx = \angle xAy = \angle yAC$, and $N, P$ be the intersection points of the perpendicular bisector of side $BC$ with rays x and y, respectively. Find all triangles ABC with the property that $AB = NP = 2HM$.[/hide]

Many thanks for your answer!
As C is acute, we have MD = AB (sin A/sin C - 2cos B)/2 , thus
AB = 2 MD <=> B = 2C.
Now, if P' = AP ^ BC, as A/3 = 60° - C, we have <AP'C = 120° and DM = AP cos 60°;
hence AP = AB.
Thus AB = NP <=> APN isosceles; as <PAN = 60° - C and <PNA = C - 30°, it follows that
AB = NP <=> C = 45°
Solution : ABC is isoceles rectangular at B.


geometry
geometric transformation
reflection
trigonometry
perpendicular bisector
geometry unsolved

Let $ X,Y$ be the orthogonal projections of $ D$ on $ AB,AC$ and $ B'$ the reflection of $ B$ about $ AD.$ Then it follows that $ B'$ lies on the circle centered at $ D$ that goes through $ B,C,$ due to $ \angle BB'C \equal{} 90^{\circ} \plus{} \frac {_1}{^2}\angle A.$ Hence, $ DY$ is perpendicular bisector of $ B'C.$ WLOG assume that $ AC > AB$  $ \Longrightarrow$ $ B'Y \equal{} CY \equal{} \frac {_1}{^2}(AC \minus{} AB).$ Therefore

$ AX \equal{} AY \equal{} AB' \plus{} B'Y \equal{} AB \plus{} \frac {_1}{^2}(AC \minus{} AB) \equal{} \frac {_1}{^2}(AB \plus{} AC) \ \ (1)$

On the other hand, by Casey's theorem for  $ (A,B,C,\omega),$ all tangent to $(O),$ we have

$ AT \cdot BC \equal{} AC \cdot BM \plus{} AB \cdot CM \equal{} \frac {_1}{^2}BC(AB \plus{} AC)$

$ \Longrightarrow AP \equal{} AT \equal{} \frac {_1}{^2}(AB \plus{} AC) \ \ (2)$

From $ (1)$ and $ (2),$ we conclude that $ X$ and $ P$ are identical $ \Longrightarrow \angle APD \equal{} 90^{\circ}.$

Let $ M$ be midpoint of $ BC$ and $ D_1,D_2$ respectively be the projections of $ D$ onto $ AB$, $ AC$. Then by [i]Simpson line theorem[/i], we have $ D_1,M,D_2$ are collinear. Let $ X\equiv AD\cap D_1D_2$. In the other hand, since $ AD$ is the internal bisector wrt $ \angle BAC$ of $ \triangle ABC$. Therefore, $ AD\perp D_1D_2$ at $ X$, which implies $ \angle MXD \equal{} 90^{\circ}$. Thus $ X\in (W)$. Moreover, we have $ AD_1^2 \equal{} \overline {AX}.\overline {AD} \equal{} \mathcal {P}_{A/(W)} \equal{} AT^2$ $ \Longrightarrow$ $ AD_1 \equal{} AT$, which implies that $ D_1\equiv P$, which leads to the result of the problem. $ \square$

Let $ M$ be the midpoint of $ BC$.
$ \frac{AD}{CD}\equal{}\frac{AT}{CM} \Rightarrow \frac{AD}{AP}\equal{}\frac{CD}{CM}$ and $ \angle MCD\equal{}\angle DAB \Rightarrow \angle APD\equal{}\angle CMD\equal{}90^o$

[quote="shoki"]1-Let $ \triangle ABC$ be a triangle and $ (O)$ its circumcircle. $ D$ is the midpoint of arc $ BC$ which doesn't contain $ A$. We draw a circle $ W$ that is tangent internally to $ (O)$ at $ D$ and tangent to $ BC$.We draw the tangent $ AT$ from $ A$ to circle $ W$.$ P$ is taken on $ AB$ such that $ AP = AT$.$ P$ and $ T$ are at the same side wrt $ A$.PROVE $ \angle APD = 90^\circ$.[/quote]
Complete trig bashing!
Let $ DP' \perp AB$. Then it is enough to prove that $ AP'=AP=AT$.
We have $ AP'=AD \cos \frac B2$. Sine law in $ \triangle ABD$ gives,\begin{align*} &AD=2R \sin (B+A/2) \\ \Longrightarrow & AP'=R\cdot 2\sin (B+A/2) \cos \frac B2=R(\sin C+\sin B)=\frac{b+c}{2}\end{align*}
Let $ M$ be the midpoint of $ BC$. Now from power of point, $ AT^2=AW^2-MW^2$
From Cosine law in $ \triangle AMD$, \begin{align*}
& AW^2=AM^2+MW^2-2AM\cdot MW\cos \angle AMD
\\ \iff & AT^2=AW^2-MW^2=AM^2+2AM\cdot MW\cos \angle AMH'\end{align*} Here $ AH' \perp MO$. We can see that $ H'M=AH=AM \cos \angle AMH'$
We know that $ AM^2=\frac{2b^2+2c^2-a^2}{4}=\frac{b^2+c^2}{4}+\frac{b^2+c^2-a^2}{4}=\frac{b^2+c^2}{4}+\frac{bc \cos A}{2}$.
Also, $ AH=b \sin C$, and $ MD=\frac a2 \tan \frac A2$
So, \begin{align*} 2AM\cdot MW\cos \angle AMH'= & AH\cdot MD \\ = & \frac 12 ab \sin C \tan \frac A2 \\ = & \frac{bc}{2} \sin A \frac{\sin \frac A2}{\cos \frac A2} \\ = & \frac{bc}{2}\cdot 2 \sin^2 \frac A2 \end{align*} At last, \begin{align*} AP^2= & \frac{b^2+c^2}{4}+\frac{bc \cos A}{2}+\frac{bc}{2}\cdot 2 \sin^2 \frac A2 \\ = & \frac 14 \left (b^2+c^2+2bc (\cos A+2\sin^2 A/2)\right ) \\ = & \frac{(b+c)^2}{4}\end{align*}
So, $ AP=AP'=\frac{b+c}{2}$, and we are done!

[quote="livetolove212"]
$ \frac {AD}{CD} \equal{} \frac {AT}{CM}$[/quote]
Why?

Let $M$ be midpoint of $BC$ and $BC \cap AD = X$
$W$ is clearly tangent to $BC$ at $M$.
$AD \cap W = K$
$AT^2 = AK.AD = AP^2 \Longrightarrow PK \perp AD \Longrightarrow P,K,M$ are collinear
$\Longrightarrow \angle PBD = \angle \frac {A} {2}+ \angle C = \angle AXB = \angle DXM = \angle PMD$
So $P,B,M,C$ are concyclic and $ \angle BPD = 180 - \angle BMD = 90$

[hide=Solution][quote=Iran MO Round 3 2009 G1]1-Let $ \triangle ABC$ be a triangle and $ (O)$ its circumcircle. $ D$ is the midpoint of arc $ BC$ which doesn't contain $ A$. We draw a circle $ W$ that is tangent internally to $ (O)$ at $ D$ and tangent to $ BC$.We draw the tangent $ AT$ from $ A$ to circle $ W$.$ P$ is taken on $ AB$ such that $ AP \equal{} AT$.$ P$ and $ T$ are at the same side wrt $ A$.PROVE $ \angle APD \equal{} 90^\circ$.[/quote]
[b][color=#000]Solution:[/color][/b] Let $F$ be the foot of perpendicular from $D$ to $AB$. Let $M$ be midpoint of $BC$ and Let $FM$ $\cap$ $AC$ $=$ $E$. Let $G$ be midpoint of $EF$. Clearly, $G$ $\in$ $\overline{AD}$. Further, $\angle MGD$ $=$ $90^{\circ}$ $\implies$ $G$ $\in$ $\omega$. Also, $AF$ is tangent to $\odot (FGD)$ \begin{align*} AF^2=AG \cdot AD=AT^2 \implies AT=AF \end{align*}
Hence, $F \equiv P$ as desired $\qquad \blacksquare$
[/hide]


Define a sequence $ \left\{a_{n}\right\}$ of integers by $ a_{1} \equal{} c$ and $ a_{n \plus{} 1} \equal{} 3a_{n} \plus{} 1$ if $ a_{n}$ is odd and $ a_{n \plus{} 1} \equal{} \frac {a_{n}}{2}$ if $ a_{n}$ is even.

Prove that for all $ c$, there is always some integer $ n\geq 1$ such that $ a_{n} \equal{} 1$.

I probably read the problem statement wrong, but isn't this an open problem - [url=http://en.wikipedia.org/wiki/Collatz_Conjecture]the Collatz Conjecture[/url]?

Edit: Ah, ok, that makes sense.  :lol:

Uh, now that you posted that, I think so.

I got this from my precalc class a year ago, but we just had to plug in numbers and look at the pattern. I wondered how you would go about proving it. And since it has been a year and I haven't found a way, I decided to post it. But that answers my question (and makes me feel better about not solving it). Thanks for the help


Let $ a,b,c>0$ such that $ abc\equal{}1$.Show that:
$ \sqrt[4]{{\frac{{{a^2} \plus{} bc}}{{a\left( {b \plus{} c} \right)}}}} \plus{} \sqrt[4]{{\frac{{{b^2} \plus{} ca}}{{b\left( {c \plus{} a} \right)}}}} \plus{} \sqrt[4]{{\frac{{{c^2} \plus{} ab}}{{c\left( {a \plus{} b} \right)}}}} \ge \sqrt {12 \minus{} \frac{{3ab \plus{} bc \plus{} ca}}{{a \plus{} b \plus{} c}}}$

[quote="leedt26"]Let $ a,b,c > 0$ such that $ abc \equal{} 1$.Show that:
$ \sqrt [4]{{\frac {{{a^2} \plus{} bc}}{{a\left( {b \plus{} c} \right)}}}} \plus{} \sqrt [4]{{\frac {{{b^2} \plus{} ca}}{{b\left( {c \plus{} a} \right)}}}} \plus{} \sqrt [4]{{\frac {{{c^2} \plus{} ab}}{{c\left( {a \plus{} b} \right)}}}} \ge \sqrt {12 \minus{} \frac {{3(ab \plus{} bc \plus{} ca)}}{{a \plus{} b \plus{} c}}}$[/quote]
I am sr. I've edit it above.


inequalities
Vieta
inequalities proposed

x^2+y^2+z^2+xy+yz+zx >=3+ x + y + z,where x,y,z is positive and xyz=1  [/code]

hello, do you meant $ x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx\geq3\plus{}x\plus{}y\plus{}z$ where $ x,y,z>0$ and $ xyz\equal{}1$.
Sonnhard.

We multiply with 2 and we have
$ (x \plus{} y \plus{} z)^2 \plus{} (x \minus{} 1)^2 \plus{} (y \minus{} 1)^2 \plus{} (z \minus{} 1)^2\geq9$
But $ (x \plus{} y \plus{} z)^2\geq9$because $ xyz \equal{} 1$
And the equality holds when $ x\equal{}y\equal{}z\equal{}1$

ineq$ \Leftrightarrow p^2\minus{}p\minus{}3\minus{}q\geq 0$
that $ q\leq\frac{p^2}{3}\Rightarrow$ ineq$ \Leftrightarrow \frac{2}{3}p^2\minus{}p\minus{}3\geq 0\Leftrightarrow(p\minus{}3)(\frac{2}{3}p\plus{}1)\geq 0,Right$

$ x^3 \minus{} x^2 \minus{} x \plus{} 1 \equal{} (x \minus{} 1)^2(x \plus{} 1)\geq 0$ for $ x > 0\Longleftrightarrow x^2 \plus{} \frac {1}{x} \minus{} x\geq 1\Longleftrightarrow x^2 \plus{} yz \minus{} x\geq 1$

$ \therefore x^2 \plus{} yz \minus{} x \plus{} y^2 \plus{} zx \minus{} y \plus{} z^2 \plus{} xy \minus{} z\geq 3$
$ \Longleftrightarrow x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} yz \plus{} zx\geq 3 \plus{} x \plus{} y \plus{} z.$ Q.E.D.

dear karis
can u explain what is p and q

$ p \equal{} x \plus{} y \plus{} z,\ q \equal{} xy \plus{} yz \plus{} zx,\ r \equal{} xyz$ which are factor of Vieta's formula, or basic symmetric expression for 3 variables $ x,\ y,\ z$.


topology
function
calculus
derivative
advanced fields
advanced fields unsolved

Suppose that $f: M\to N$ is a $C^{\infty}$ map, $M$ and $N$ are compact connected $n$-manifolds and $rank(df)=n$. Show that $f$ is a covering map.

uhm..
ok, elementary topology tells us that a continuous map from a compact space to an hausdorff one is open (ok, in this case it was guaranteed also from the rank hypothesis): the image $\emptyset \neq f(M)\subset N$ is then open and compact, hence closed, so it's the whole $N$.
the rank hypothesis tells us that it's locally invertible (by rank and inverse function theorems), so it's a (topological) covering..
looking deeper into the thing, local inverses for $f$ are not just $C^{1}$, but even $C^\infty$ (there's some kind of formula, which should be well-known to exist, that gives further derivatives, and which requires just invertibility of the first derivative.. as in $n=1$ case..), so it's a $C^\infty$ covering.

am i missing anything?

The map $f: [-1,1] \to \mathbb{R}$ by $x \mapsto x^{2}$ is a continuous function from a compact space to a hausdorff space, but the image of the open set $(-1,1)$ is the not so open $[0,1)$, which means $f$ is not an open map. I think you meant to say that $f$ must be a closed map.

uh, right...
i usually use the compact-hausdorff lemma to verify that some function is a homeomorphism, so the two things are actually equivalent...
anyway, $f$ is continuous, closed, open (inverse function theorem) between compact connected manifolds, so it's surjecive. by rank and inverse function theorems, it's a local diffeomorphism.
and that should be enough...


calculus
integration
limit
real analysis
real analysis unsolved

Does anybody happen to know how to compute this limit :
$ \lim_{n\to\infty} (\int\limits_{0}^{1} x^n(1\minus{}x)^n dx)^{\frac{1}{n}}$

Hint:
Try to prove $ \lim_{n\to\infty} \sqrt [n]{\int\limits_{0}^{1} f(x)^n dx} \equal{} \max_{x\in[0,1]}(f(x))$
where $ f: [0,1] \to [0,\infty)$.

$ I_n\equal{}\int_0^1 x^n \, (1\minus{}x)^n\, dx\equal{}B(n\plus{}1,n\plus{}1)\equal{}\frac{n!\, n!}{(2n\plus{}1)!}\equal{}\frac{1}{2n\plus{}1}\, \frac{1}{{2n\choose n}}$

So $ I_n^{1/n}\equal{}\frac{1}{(2n\plus{}1)^{1/n}}\, \frac{1}{{2n\choose n}^{1/n}}$ . For large $ n \quad (2n\plus{}1)^{1/n}$ tends to  $ 1$.

And $ {2n\choose n}^{1/n}$ tends to $ 4$ because the central binomial coefficient $ {2n\choose n}$

has the same asymptotic behaviour as $ \frac{2^{2n}}{\sqrt{\pi n}}$ .

You can show this with Stirling's formula.

In general: Let $ f: [0,1]\to [0,\infty)$ be continu.

Let $ a_n : \equal{} \sqrt [n]{\int_0^1 f^n(x) \ dx}$ for all $ n \in \mathbb{N}$

Suppose that $ M$ is the  maximum of $ f$.
Let $ \epsilon > 0$ and $ A \equal{} [f > M \minus{} \epsilon]$, then, because of the continuity of $ f$, $ 0 < \lambda(A)\le 1$.
Then
$ \sqrt [n]{\lambda(A)}(M \minus{} \epsilon) \equal{} \sqrt [n]{\int_0^1 (M \minus{} \epsilon)^n1_A(x)\ dx} \le$
$ \le \sqrt [n]{\int_0^1f^n(x)\ dx} \le \sqrt [n]{\int_0^1M^n1_{[0,1]}(x)\ dx} \equal{} M$
$ M \minus{} \epsilon \le \liminf_{n\to\infty}a_n \le \limsup_{n\to\infty}a_n\le M$
This is true for every $ \epsilon > 0$, so $ \lim_{n\to\infty}a_n \equal{} M$

Right ! I recall this equation from my calculus classes, thanks alot PeterP.
Doe John~ Your solution is a bit complex but thanks anyway.


Take an angle between 2 rays $L_1$ and $L_2$ with vertex O. Take a point A inside the angle. Consider all triangles with vertex O , and two sides of which belong to $L_1$ and $L_2$, and the third side of which pass through the point A. Prove that the area of the triangle will be the smallest if the point A is a middle of the third side.

Let $R$ and $S$ be points on $L_{1}$ and $L_{2}$ respectively such that $RA=SA$. Let $A'$ be points on $L_{1}$ such that $AA'//L_{2}$. $R'$ and $S'$ are points on $L_{1}$ and $L_{2}$ respectively such that $R'S'$ passes through $A$.

If $R',O$ are on different sides of $R$, then $S',O$ will be on the same side of $S$. Since $\frac{R'A}{AS'}=\frac{R'A'}{A'O}>\frac{RA'}{A'O}=1$, so $R'A>AS'$. Hence \[[R'OS']=[ROS'A]+[R'AR]=[ROS'A]+\frac{1}{2} \cdot AR \cdot AR' \cdot sin \angle RAR' > [ROS'A]+\frac{1}{2} \cdot AS \cdot AS' \cdot sin \angle SAS' =[ROS'A]+[S'AS]=[ROS].\]
If $R',O$ are on the same side of $R$, then $S',O$ will be on different sides of $S$. Since $\frac{R'A}{AS'}=\frac{R'A'}{A'O} < \frac{RA'}{A'O}=1$, so $R'A<AS'$. Hence \[[R'OS']=[R'OSA]+[S'AS]=[R'OSA]+\frac{1}{2} \cdot AS \cdot AS' \cdot sin \angle SAS' > [R'OSA]+\frac{1}{2} \cdot AR \cdot AR' \cdot sin \angle RAR' =[R'OSA]+[R'AR]=[ROS].\]
Therefore $\triangle ROS$ is the triangle whose third side passing through $A$ with the smallest area.


inequalities
induction
quadratics
geometry
3D geometry
function
Gauss

Hello Everyone,

i'm trying to make a strong marathon for number theory .. which will be in Pre-Olympiad level

Please if you write any problem don't forget to indicate its number and if you write a solution please indicate for what problem also to prevent the confusion that happens in some marathons.

it will be preferred to write the source of the problem.
please , show detailed solutions , and please post some challenging Pre-Olympiad problems..
remember ,, different approaches are most welcome :)

now .. let's get our hands dirty :lol: :

let $ f(n)$ denote the sum of the digits of $ n$. Now let $ N \equal{} {4444^{4444}}$.
Find $ f\left( {f\left( {f\left( N \right)} \right)} \right)$.

$ 4444^{4444}<10000^{4444}\equal{}10^{17776}$

So $ 4444^{4444}$ has at most 17776 digits, which means that f(N) can not be greater than $ 9\cdot17776\equal{}159984$,
Here we see that $ f(f(N))$ can not be greater than $ 9\cdot{5}\equal{}45$,
$ f(f(f(N)))$ can not be greater than $ 3\plus{}9\equal{}12$,

also We know that $ {f(f(f(N)))}\equiv{f(f(N))}\equiv{f(N)}\equiv{N}(mod9)$,

$ 4444^{4444}\equiv{(\minus{}2)^{4444}}\equal{}2^{4444}\equal{}2^{4440}\cdot2^{4}\equal{}64^{740}\cdot16\equiv1\cdot7\equiv7(mod9)$

Since $ f(f(f(N)))\le{12}$, $ f(f(f(N)))\equal{}7$

New problem:
Prove that if $ p$ and $ p^2\plus{}8$ are prime, then $ p^3\plus{}8p\plus{}2$ is also prime.

[quote="Bachukas"]

New problem:
Prove that if $ p$ and $ p^2 \plus{} 8$ are prime, then $ p^3 \plus{} 8p \plus{} 2$ is also prime.[/quote]

if $ p > 3$ then we have $ p^2\equiv 1 (\mod3) \implies p^2 \plus{} 8\equiv 0 (\mod3)$ which means that $ 3 \mid p^2 \plus{} 8$ which is a contradition
Thus $ p \equal{} 3 \implies p^3 \plus{} 8p \plus{} 2 \equal{} 53$  :oops:
The case $ p \equal{} 2$ is trivial

[i][b]Problem: 3[/b][/i]: Find all positive integers $ n$ such that for all odd integers $ a$.if $ a^2\le n$ Then $ a\mid n$

my try for p.3:

clearly if such an $ n$ exists then it must satisfy the inequality: $ (2k \plus{} 1)^2 \le n \equal{} 1\cdot 3 \cdot 5 \cdot \cdot \cdot (2k \plus{} 1) \cdot m < (2(k \plus{} 1) \plus{} 1)^2$ for some integer $ m \ge 1$ . Now $ m < \frac {(2(k \plus{} 1) \plus{} 1)^2}{1\cdot 3 \cdot 5 \cdot \cdot \cdot (2k \plus{} 1)} < 1 \;\; \forall k \ge 3$ ( can be proven easily by induction ) which means that we need $ k \equal{} 0,1,2$ for $ m$ to exist and, consequently, for $ n$ to exist. Now study each case alone to get:

$ k \equal{} 0 \Longrightarrow m \equal{} 1,2,3,4,5,6,7,8 \Longrightarrow n \equal{} \lbrace 1,2,3,4,5,6,7,8 \rbrace$

$ k \equal{} 1 \Longrightarrow m \equal{} 3,4,5,6,7,8 \Longrightarrow n \equal{} \lbrace 9,12,15,18,21,24 \rbrace$

$ k \equal{} 2 \Longrightarrow m \equal{} 2,3 \Longrightarrow n \equal{} \lbrace 30,45 \rbrace$

are the only solutions. $ \Box$

$ \text{Problem} \;\; 4$:

Find all $ n \in \mathbb{Z}^ \plus{}$ such that $ 2^n \plus{} n \mid 8^n \plus{} n$.

[b]The solution of problem4[/b]
$ \begin{gathered} ({8^n} \plus{} n,{2^n} \plus{} n) \hfill \\
\equal{} ({8^n} \plus{} n \minus{} ({4^n})({2^n} \plus{} n),{2^n} \plus{} n) \hfill \\
\equal{} (n \minus{} ({4^n})n,{2^n} \plus{} n) \hfill \\
\equal{} (n \minus{} ({4^n})n \plus{} (n{2^n})({2^n} \plus{} n),{2^n} \plus{} n) \hfill \\
\equal{} (n \plus{} ({n^2})({2^n}),{2^n} \plus{} n) \hfill \\
\equal{} (n \plus{} ({n^2})({2^n}) \minus{} ({n^2})({2^n} \plus{} n),{2^n} \plus{} n) \hfill \\
\equal{} (n \minus{} {n^3},{2^n} \plus{} n) \hfill \\
\equal{} ({n^3} \minus{} n,{2^n} \plus{} n) \hfill \\
\end{gathered}$
so we need $ ({2^n} \plus{} n)|{n^3} \minus{} n$ and so we need $ {n^3} \minus{} n \geqslant {2^n} \plus{} n$
but the inequality has limitation, it turns when $ n \equal{} 10$ ..
now we have to use induction on $ {n^3} \minus{} n < {2^n} \plus{} n$ with base case $ n \equal{} 10$
assuming that it's true for $ n \equal{} k$ and trying it for $ n \equal{} k \plus{} 1$ ,, we have $ (k \plus{} 1)^3 < 2(2^k)$
so proving $ 3{k^2} \plus{} 4k < {2^k} \plus{} k$ is equivalent or $ 3{k^2} \plus{} 5k < {k^3}$
$ 3{k^2} \plus{} 5k < 3{k^2} \plus{} 10k \leqslant 3{k^2} \plus{} kk \equal{} 4{k^2} < 10{k^2} \leqslant k{k^2} \equal{} {k^3}$
since $ 10 \leqslant k$
trying $ n\equal{}1,2,...,9$ .. which works for $ ({2^n} \plus{} n)|{n^3} \minus{} n$ and $ {n^3} \minus{} n \geqslant {2^n} \plus{} n$
we find that it works only for $ n\equal{}1,2,4,6$

and we are Done  :lol:

[b]Problem 5[/b] : Find all nonegative integers n such that there are integers a and b with the property:
$ {n^2} \equal{} a \plus{} b$ and $ {n^3} \equal{} {a^2} \plus{} {b^2}$

[hide]
$ n^2 \equal{} a\plus{}b \Longrightarrow a\equal{}n^2 \minus{} b$ substitute this into $ n^3\equal{}a^2\plus{}b^2$ and rearrange as a quadratic in $ b$ to get: $ 2b^2 \minus{} 2n^2b \plus{}n^4\minus{}n^3\equal{}0$ and we need $ \Delta \equal{} 4(2n^3\minus{}n^4) \ge 0$ or $ n^3(2\minus{}n) \ge 0$ and since $ n \ge 0$ we get $ n\equal{}0,1,2$ and that produces $ b\equal{}0,1,2$ and $ a\equal{}0,0,2$ respectively. $ \Box$
[/hide]

$ \text{Problem} \; \; 6$ :

Find all pairs of positive integers $ (n,m)$ satisfying $ 3n^2\plus{}3n\plus{}7\equal{}m^3$.

Solution N:6
$ 3n^2 \plus{}3n \plus{}7 \equal{} 1[6]$ then $ m \equal{}1[6]$
then we get $ n^2\plus{}n \plus{}2 \equal{} 0[6]$ but this is not true ! so there exist no solution
Probelm 7:  $ a, b, c, d$ are integers such that:
$ a < b \le c < d$ , $ ad \equal{} bc$ and $ \sqrt d \minus{} \sqrt a \le 1$
show that $ a$ is a perfect square!

thanks [b]kirwa01[/b] for your response
another [b]Solution for Problem6 [/b]:

if $ \begin{gathered}
  {n^2} \plus{} n \equiv 0mod(3) \Leftrightarrow n \equiv 0,2mod(3) \hfill \\
  {n^2} \plus{} n \equiv 2mod(3) \Leftrightarrow n \equiv 1mod(3) \hfill \\
  \therefore {n^2} \plus{} n \equiv 0,2,3,6mod(9) \hfill \\
  \therefore 3{n^2} \plus{} 3n \equiv 0,6mod(9) \hfill \\
\end{gathered}$
Adding 7:
$ 3{n^2} \plus{} 3n \plus{} 7 \equiv 4,7mod(9)$
but all cubes are $ \equiv  \minus{} 1,0,1mod(9)$ ,, so no solution exists :)
thanks for the nice problem :)

[quote="BadrGhamdi"][quote="kirwa01"]Problem N:4
$ 2^n \plus{} n| 8^n \plus{} n$ implies $ 2^n \plus{} n| \minus{} 8^n \minus{} n \plus{} 4^n(2^n \plus{} n)$
the same argument implies $ 2^n \plus{} n | 2^n n^2 \plus{} n$
then $ 2^n \plus{} n| n^3 \minus{} n$
but we know for n>9 $ \frac {n}{ln(n)} > \frac {3}{ln(2)}$  which means $ 2^n > n^3$
then all solutions must be in $ {1,2,..,9}$ the solutions if i am not mistaken are: $ {1,2,4,6}$

Probelm 5:  $ a, b, c, d$ are integers such that:
$ a < b \le c < d$ , $ ad \equal{} bc$ and $ \sqrt d \minus{} \sqrt a \le 1$
show that $ a$ is a perfect square![/quote]
can you please edit your post , and remove your problem ? :) .. i think you were late .. but your elegant solution is most welcomed :)
thanks in advance mate[/quote]

done!

[quote="kirwa01"]Solution N:6
$ 3n^2 \plus{} 3n \plus{} 7 \equal{} 1[6]$ then $ m \equal{} 1[6]$
then we get $ n^2 \plus{} n \plus{} 2 \equal{} 0[6]$ but this is not true ! so there exist no solution
Probelm 7:  $ a, b, c, d$ are integers such that:
$ a < b \le c < d$ , $ ad \equal{} bc$ and $ \sqrt d \minus{} \sqrt a \le 1$
show that $ a$ is a perfect square![/quote]

solution:

[hide]for $ a\equal{}n^2$ we get $ d\equal{}(n\plus{}1)^2$, $ b\equal{}c\equal{}n(n\plus{}1)$ is a possible solution. Now suppose $ a\equal{}n^2\plus{}x$ where $ 1\le x \le 2n$. Suppose $ b\equal{}n^2\plus{}y$, $ c\equal{}n^2\plus{}z$, $ d\equal{}n^2\plus{}w$. Now from $ ad\equal{}cb$ we get $ n^4\plus{}(w\plus{}x)n^2 \plus{}wx\equal{}n^4\plus{}(y\plus{}z)n^2\plus{}yz$ and we get $ wx\equal{}yz$ and $ w\plus{}x\equal{}y\plus{}z$ from which we get $ w\equal{}y\plus{}z\minus{}x$ and substituting we get $ x^2\minus{}(y\plus{}z)x\plus{}yz\equal{}0$ or $ (x\minus{}y)(x\minus{}z)\equal{}0$ and then $ x\equal{}y$ or $ x\equal{}z$ but then $ a\equal{}b$ or $ a\equal{}c$. contradiction. [/hide]

If that's correct then:

$ \text{Problem} \; 8$ :
Let $ a_1,a_2,...,a_{n}$ be $ n$ integers satisfying $ \sum_{k\equal{}1}^{n} a_k \equal{} 2009$.  Find $ \lbrace a_1,a_2,...,a_n \rbrace$ and $ n$ that makes the product $ a_1 \cdot a_2 \cdot \cdot \cdot a_n$ obtain its maximum possible value.

By AM-GM inequality we have
$ (\frac {\displaystyle\sum_{k \equal{} 1}^{n}\ a_{k} }{n})^{n} \ge a_{1}a_{2}.......a_{n}$

$ (\frac {2009}{n})^{n} \ge a_{1}a_{2}...a_{n}$
For the maximum value
$ (\frac {2009}{n})^{n} \equal{} a_{1}a_{2}...a_{n}$
Since the R.H.S is an integer ,we have $ n \equal{} 1,7,49,41,2009$
For the maximum value of R.H.S ,It is easy to verify that $ n \equal{} 41$ yields the maximum value .
So we have $ a_{1}a_{2}........a_{n} \equal{} 49^{41} \equal{} 7^{82}$
 But the equality in AM-GM can hold good only when $ a_{1} \equal{} a_{2} \equal{} ......... \equal{} a_{n}$
Which implies $ a_{1} \equal{} a_{2} \equal{} ........ \equal{} 49$

[quote="srinath.r"]By AM-GM inequality we have
$ (\frac {\displaystyle\sum_{k \equal{} 1}^{n}\ a_{k} }{n})^{n} \ge a_{1}a_{2}.......a_{n}$

$ (\frac {2009}{n})^{n} \ge a_{1}a_{2}...a_{n}$
For the maximum value
$ (\frac {2009}{n})^{n} \equal{} a_{1}a_{2}...a_{n}$
Since the R.H.S is an integer ,we have $ n \equal{} 7,49,41$
For the maximum value of R.H.S ,I think $ n \equal{} 41$
So we have $ a_{1}a_{2}........a_{n} \equal{} 49^{41} \equal{} 7^{82}$
Now  I think it is easy for finding $ a_{1},a_{2}.........a_{n}$[/quote]
[color=red]please , show detailed solutions[/color]

$ \text{Problem 9}$
Let $ P$ be a prime number which can be expressed as a sum of two integer squares ,prove that there are no other integer squares ,whose sum is $ P$

The question means 'Whenever there are positive integers $k,m,n$ such that $nm=k^2+k+3$, there must exist such $x,y$' . There is absolutely no contradiction with $n=3,m=5,k=3$ since $4.5=20=3^2+11.1^2$. Try to understand what he means before you make baseless fake claims. It's from a highly reputed source (as far as I remember). I could not do it when I first saw it but I will try it again tomorrow.

What a nice problem, Shivang. I hope my solution is correct...

$mn=k^2+k+3 \implies 4mn=(2k+1)^2+11$

This is truly screaming 'Quadratic Fields' but it also should remind one of the most famous techniques used on Quadratic Forms - Fermat's Method of Infinite Descent.

Suppose that $(m,n,k)$ is a triple such that $\nexists a,b,c,d$ with $a,b,c,d$ all not of the same parity satisfying $4m=a^2+11b^2, 4n=c^2+11d^2,2(2k+1)=ac+11bd,|ad-bc|=2$ and the sum $m+n+2k+1$ is minimum over all other triples of its kind. WLOG, let $m \le n$.

[b]Case I:[/b] $2k+1 < 2m$. Then, $2m-1 \ge 2k+1 \implies m \ge k+1 \implies n \ge k+1 \implies 4mn \ge 4(k+1)^2$.
$\implies (2k+1)^2+11 \ge 4(k+1)^2 \implies 4k+12 \ge 8k+4 \implies k \le 2$. This leads to $2$ subcases.

[u]Subcase I:[/u] $k=1 \implies 4mn=20 \implies mn=5 \implies m=1,n=5$. But then, $4n=3^2+11.1^2$.

[u]Subcase II:[/u] $k=2 \implies mn=9 \implies m=3,n=3$ or $m=1,n=9$. But, in the first case $4m=4n=1^2+11.1^2$ and in the second, $4n=36=5^2+11.1^2$.

[b]Case II:[/b] $2k+1 > 2n \implies (2k+1)^2+11 > 4n^2+11 >4n^2 \ge 4mn$, a contradiction.

[b]Case III:[/b] $2m+1 \le 2k+1 \le 2n-1$. Since $4mn=(2k+1)^2+11 \implies (2k+1-2m)^2+11=4m(n-2k-1+m)$.

Since the LHS is positive and so is $4m$, $n-2k-1+m$ must also be positive.

But, $(2k+1-2m)+(m+n-2k-1)+m=n<m+n+2k+1$.

So, $\exists a,b,c,d$ such that $a^2+11b^2=4m, c^2+11d^2=4(n+m-2k-1), 2k+1-2m= (ac+11bd)/2$ and $|ad-bc|=2$, not all of which have same parity.

But then, $4n=4m+4(m+n-2k-1)+4(2k+1-2m)=(a+c)^2+11(b+d)^2$.

Therefore, $a,b$ have same parity and $c,d$ have same parity but these pairities are different. So, $a+c$ and $b+d$ are both odd, with $|a(b+d)-b(a+c)|=2$ and $2k+1=(a^2+11b^2+ac+11bd)/2=(a(a+c)+11b(b+d))/2=2k+1$, contradiction.

[quote="dibyo_99"]
But, $(2k+1-2m)+(m+n-2k-1)+m=n<m+n+2k+1$.
[/quote]
What exactly are you trying to say here?
Are you disposing of case 3 because you found another triple $k, m, n$ such that $ k^2 + k + 3 = mn $ which has a lower $ m+n+2k+1 $?

EDIT: Right I see $k-m, m, m+n-2k-1$ are the new $k, m, n$
the condition that $ k^2 + k + 3 = mn $ is satisfied because $(2k-2m+1)^2 + 11 = 4m(m+n-2k-1)$

[i]Prob. 429[/i]: Prove that given any prime $p$, there exists $x,y,z,t\in\mathbb{N}, 0<t<p$ so that $x^2+y^2+z^2-tp=0$.

[b]Solution #430:[/b] $x^2-14xy+y^2=-12 \implies (x-7y)^2=12(2y+1)(2y-1)$
$\implies (\frac{x-7y}{2})^2=3(2y-1)(2y+1)$.

Since $(2y-1,2y+1)=1$, we must have $2y-1=3m^2,2y+1=n^2$ or $2y-1=m^2,2y+1=3n^2$. First case is impossible since $n^2 \ncong 2 \pmod{3}$.

So, $m^2-3n^2=-2$ and it is easy to see that every solution of this Pell-like equation corresponds to a solution of the given condition.

[b]Solution #431:[/b](sdsert, please change the numbering) http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=522651

Problem 429 is from Korea 2000, problem 1:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2599280&sid=92188c8e77cef647a12d4cf0895a5a5a#p2599280

Problem #433: Let the sequence $ \{a_n\}_{n\ge1} $ be defined : $a_1=0$ and $ a_{n}=a_{\lfloor \frac{n}{2}\rfloor} + a_{\lfloor \frac{n}{3}\rfloor} +a_{\lfloor \frac{n}{4}\rfloor}+\cdots +  a_{\lfloor \frac{n}{n}\rfloor}+1$ . Prove that there are infinitely integer n satisfy: $ a_{n}\equiv a_{n-1}\pmod{2^{2013}}.$

P434
I am trying to figure out my friend's 7-digit phone number. I can ask questions about it, but my friend can must answer "yes" or "no". What is the least amount of questions required if my friend is allowed to (not required to) lie at most twice? Give a rigorous proof as to why your answer is the minimum

[hide="HINT"][hide="Are you sure you want a hint? You would learn more if you figure it out on your own"][hide="This is your final warning"]Here is your hint: BINARY[/hide][/hide][/hide]

[quote="dibyo_99"][b]Problem #432:[/b] Let $p$ be a Fermat Prime. Find the number of ordered pair of integers $(m,n)$ such that $0 \le m<n <p-1$ satisfying $2^m+3^m \equiv 2^n+3^n \pmod{p}$.[/quote]

Could you kindly read the thread a little before posting in it....

Problem #432 is unsolved.

Moderator  , please do something!! :D :D
The thread is going all mess

[size=85][color=#FF0000][Moderator: Thanks, dear back-seat moderator.][/color][/size]

Is this Fun Factory or something? I can't even imagine how many posts should be removed from this topic. 20, 30, or even more?

I think we have to set this rule in all of running marathons: Be Mature.

Topic locked.


[quote="James Potter"]Let be given triangle $ABC$ and a point $M$ inside it.Let $D,E,F$ be the orthogonal projection of M on the line $BC,CA,AB$ respectively.
 Prove that:$EF+FD+DE \le \frac{BC+CA+AB}{2}$[/quote]

Wrong. Moreover, for an [i]equilateral[/i] triangle ABC, we have $EF+FD+DE\geq\frac{BC+CA+AB}{2}$ for every point M, since the orthic triangle is the inscribed triangle with minimal perimeter.

  Darij


Each can of tennis balls has $ x$ balls.  Initially, Sam has 14 cans of tennis balls, plus 3 extras.  Jerri initially has 7 cans of tennis balls, plus 8 extras. If Jerri doubles the number of tennis balls she has, how many more balls will she have than Sam has?

Sam has $ 14x\plus{}3$ total tennis balls.  Jerri has $ 7x\plus{}8$ tennis balls.  When Jerri doubles the amount she has, she'll have a total of $ 2(7x\plus{}8)\equal{}14x\plus{}16$ tennis balls.  The difference between Jerri's new total and Sam's total is:
\[ 14x\plus{}16\minus{}(14x\plus{}3)\equal{}14x\plus{}16\minus{}14x\minus{}3\equal{}\boxed{13}\]

Or you could assume WLOG $ x\equal{}0.$ That way \[ 2(7\cdot0\plus{}8)\minus{}(14\cdot0\plus{}3)\equal{}16\minus{}3\equal{}\boxed{13}.\]


ratio
geometry
circumcircle
trapezoid
Last thread.
first post in ff

.. the math national olympiad in romania is taking place ... and I am part of the big event - this year it's going to be one-on the clock olympiad - as there isn't much time for all the tests to take place, and most important the paper evaluation is goiing to be made in a lot shorter time than usual

the subjects will be posted on the site ASAP - and as for the MathLinks Contest - the subjects will be posted (hopefully) monday in the morning ( +2 GMT Time )

I am once again sorry for not posting the solutions to the last two rounds, but I have just finished writing a book and preparing for this olympiad - more time will be available afterwards - and I plan to make the mathlinks site complete :D

but enough talking, good luck again and c'ya

Raicu Claudiu
Caraiani Ana
Negut Andrei
Stefanescu Andrei
Poradici Horia
Kreindler Gabriel (Kappa)

these 6 students were selected to represent Romania in the upcoming Balkan Math Olympiad  ... good luck to ya all

Can you send problems from national competition in a few days?
And can you send some informations for BMO?
I am from Yugoslavia, and I hope we'll see in Albania!

well - I'm not coming to Albania - but those 6 guys I wrote upthere are, so you'll probably meet them

the first selection tests from 2003 have been posted on the [url=http://www.mathlinks.go.ro]mathlinks[/url] web site - check them out there

and gluck to your endeavour to BMO !

after the contest the following results (for the Romanian Team) followed

Ana Caraiani            40p GOLD
Claudiu Raicu           40p GOLD
Andrei Negut            37p SILVER
Andrei Chites           35p SILVER
Andrei Stefanescu    33p SILVER
Horia Poradici           29p BRONZE

Congrats to all the participants!

PS Problem no.2 was proposed by me ...  :D

if i'm not mistaken u r tokin abt the balkaniad.i wud just like to know if  the official proof for the prob no 2 (it was that geometry problem abt collinearity,wasnt it?!!) was non-inversive or used inversion,i found only a inversive proof.
thanx
(congrats to all the winners)

Because both triangles DBE and DCF are right-angled triangles, it is sufficient to prove that they are similar to obtain D,E,F collinear. All we must prove is that DB/DC=BE/FC (1)

Now let's compute each ratio: for the first one, we'll use the fact that AD is tangent to the circumcircle of ABC. Therefore <DAB=<ACB thus the triangles DAB and DCA are similar, which gives us:
DB/DC=AB/AC=AD/BC => DB/DC = c2/b2  (2)

Let's compute now BE and EF. Let M be the midpoint of [BC]. Then in the right-angled trapezoid [EBMO] because <EOM=<A+<B it follows that <BEO = <C. Using the latter fact we can compute easily the length of
BE=(a/2)cotB+RcosA
and by the same arguments FC=RcosA+(a/2)cotC . Now we have to prove using (2) that:    (RcosA+(a/2)cotB)b2 = (RcosA+(a/2)cosC)c2   which is obvious after some computations.
Thus the three points are collinear.


superior algebra
superior algebra solved

$R$ is a ring. The product of any two non-zero ideals is not zero. Show that the product of any two non-zero left ideals is not zero.

Now, what doi you mean by "ideals"? Don't they include the left ideals? :) Or maybe when you say "ideal" you mean bilateral ideal (both left and right ideal)?

Oops, it seems that we meet different defintions again.

The ideal in my problem means ideals of two sides.

I hope that this time $R$ is an ideal of $R$.

Let $a,b$ be left ideals of $R$.
Suppose $ab=0$. Then let $\mathfrak a=a+aR$, $\mathfrak b=b+bR$.
$\mathfrak a$ and $\mathfrak b$ are ideals of $R$. So $\mathfrak a\mathfrak b\neq0$. But $\mathfrak a\mathfrak b=(a+aR)(b+bR)\subseteq ab+abR+aRb+aRbR$.
Since $b$ is a left ideal, $Rb\subseteq b$. So $\mathfrak a\mathfrak b\subseteq ab+abR+ab+abR=0$. This is a contradiction.

If this solution is correct then you should change the problem text from the problem "Nilpotent ideal", or that problem doesn't need the assumption L nilpotent.


Hey, I am working on AT problems and the following question comes up to me :

$ f : X\to Y$ and $ g : Y\to Z$ are not null-homotopic, is it true that $ g\circ f$ is not null -homotopic?

thanks

Nope.

Say $ f : S^1 \to S^1 \times S^1$ by $ x \mapsto (x,1)$. Then $ g : S^1 \times S^1 \to S^1$ by $ g(x,y) \equal{} y$. Then $ (g \circ f)(x) \equal{} 1$ which is a constant map, but neither $ f$ nor $ g$ is nullhomotopic.


geometry
geometric transformation
reflection
geometry proposed

Let $ CH$ be the altitude of triangle $ \triangle ABC$ with $ \angle ACB \equal{} 90$. The bisector of $ \angle BAC$ intersects $ CH$, $ CB$ at $ P$ and $ M$, respectively. The bisector of $ \angle ABC$ intersects $ CH$, $ CA$ at $ Q$,$ N$, respectively. Prove that the line passing through the midpoints of $ PM$ and $ QN$ is parallel to line $ AB$

Let $ D,E$ be reflection of $ C$ on $ AM,BN$ and $ X,Y$ be midpoint of $ PM,QN$. It is easy to see $ CP \equal{} CM$. So $ X,Y$ are midpoint of $ CD,CE$. Hence $ XY\parallel BC$.


What's your favorite fruit?

See Fav. fruit 1 and Fav. fruit 2 if your favorite fruit is not here. Don't choose the "other" option until you've done so. Please only vote in one of the polls.

Fav. fruit 1 is the main thread for discussion.

Sorry if your favorite fruit is bunched up in a category and you don't like the other fruits in that category.

*bump* for those who might not know about this and number 2...


conics parabola trigonometry calculus integration

delta v = at
delta v = 9.8*2
that gets you started

[quote="shobber"]Can that solved by analyzing a parabola?[/quote]

It could be.  Here, you solve the problem using the tools in math you learned.  You plan what tools you are going to need to solve it.

No, you're not.  What I refer to in the math problem is a scene from the anime [i]Ranma ½[/i] season 1 episode 4.  (Also on page 4-5, chapter 7 of the manga)  I was wondering how much force Akane could put into her kick.  It's been about a year since I've had Calculus and Physics, so I decided to seek help around the Internet.  Now I realize that force probably wasn't the correct measurement to use, but rather energy, since it's difficult to tell how fast Ranma went from the velocity of zero to his initial velocity in the parabola.  Energy seems more correct.  I found out websites where this kind of problem solving takes place and I found this and decided to toss it up--see what you guys came up with.

[quote="blahblahblah"][quote="beneficii"]What about this, in showing your work?

On Earth, Akane kicks Ranma, who weighs 180 pounds, up into the air at a 35º angle, with 2 seconds before Ranma reaches the maximum height. In her kick, how many pounds of force did Akane exert on Ranma?  Assume that the movement here is as for a particle; don't worry about complicating things like wind resistance, movement of the arms and legs, etc.

If you have any questions, just ask.[/quote]

Yes, my question is, 'are we doing your homework for you'?[/quote]

Right, I am asking the amount of energy or force that was imparted from the leg.  With force, now I don't think I gave you enough information.  How long did it take for Ranma to get from a velocity of zero to an initial velocity of 34.20 m/s in the parabola, from Akane's kick?  Remembering from the scene, it was about 1/10 of a second, probably even less.  Assuming 1/10 second, the acceleration of Ranma's body involved was 342.0 m/s^2.

1 lb (pound) = 4.448 N, so 180 lb = 180 (4.448 N) = 800.64 N = (81.6 kg) (9.81 m/s^2)

So the guy has a mass of about 80 kg.  I never knew his exact weight, so I'll use this approximation.  Because it has been a while since Calculus, I had forgotten how to use integrals to set up the parabola.  Tell me, it was

dv/dt = a

v = at + C (v0)

v = ds/dt

ds/dt = at + v0

s = ½at^2 + v0t + s0

Right?

The equation you seem to be using, ½mv^2, I'm not as familiar with.  That's the kinetic energy, Wnet = ½mv^2 - ½mv0^2.  Because the initial velocity was zero, that comes down to Wnet = ½mv^2.  So, using ½ (80 kg) (34.20 m/s)^2 = 46,786 J.  That's a lot of energy.  It took a tenth of a second to use up all this energy, so her power was 467,860 W.  To use the American system, that was about 627 horsepower.  A bit unrealistic, of course, but it's from the anime, so.

Anyway, Akane is very strong, as anyone watching the show can attest to.  In a couple episodes (nos. 2 and 3), she was able to singlehandedly fight off about 4 dozen men.

Oh, for force, with an acceleration of 342.0 m/s^2 (= 34.20 m/s / 0.1 s) and Ranma having a mass of 80kg, the amount of force used was 27,360 N.  1 N = 0.2248 lb, so 27,360 N = 27,360 (0.2248 lb) = 6150.5 lb.  That's more than 3 tons, quite a bit of force.

Thank you all for working with me.  I'm not sure if the purpose of the site is to work on things like this, but perhaps the people here got something out of it. ^_^

[quote="prof_daniel2000"]I don't quite quite get what you are asking, but I believe you are asking the amount of energy that is imparted from the leg to the 180 pound guy. I am not familiar with pounds (I only know metric calculation), so I will treat the guy as having 180kg of mass.

It takes 2s for velocity of the guy to reach 0. Since acceleration due to gravity is 9.81 initial upward velocity is 19.62m/s. (9.81 * 2)

The speed at which the person starts out can be calculated using trigonometry = 19.62 / (sin 35) = 34.20

The amount of energy the person has is therefore = 1/2 (180) (34.2)^2 = 105267 J

If I am wrong, please point it out to me, thanks[/quote]

I think this is more of a physics question that can be solved using the kinematics equations.


ratio
limit
calculus
calculus computations

Let $ a_n$ be a positive sequence and $ \frac{a_{n\plus{}1}}{a_n}\ge1\minus{}\frac1n$ for each $ n\ge1.$ Prove that $ \sum_{n\ge1}a_n$ diverges.

I was thinking to apply the ratio test to the series, but that'd tell me that $ \lim_{n\to\infty}\frac{a_{n\plus{}1}}{a_n}\ge1,$ and we have to cases here: if the limit equals 1, then the test gives nothing, but if the limit is greater than 1 then the series diverges.

How'd you approach this problem?

Claim: $ a_n\ge \frac{a_2}{n\minus{}1}$ for all $ n\ge 2.$ Try proving that by induction.

(Ignore $ a_1$ - you can't say anything about it but it's just one number. To make this a little cleaner, it would have been nicer to state the condition either as $ \frac{a_{n\plus{}1}}{a_n}\ge1\minus{}\frac1{n\plus{}1}$ or as $ \frac{a_n}{a_{n\minus{}1}}\ge1\minus{}\frac1n.$ )

Try to use one version of comparision test.

[b]Comparision test:[/b]
If $ \sum_{n\in\mathbb{N}} |b_n|$ diverges, and $ \left |\frac{a_{n\plus{}1}}{a_n}\right|\geqslant \left|\frac{b_{n\plus{}1}}{b_n}\right|$ for all sufficiently large $ n$ , then $ \sum_{n\in\mathbb{N}} a_n$  also fails to converge absolutely (though it could still be conditionally convergent, e.g. if the an  alternate in sign).

Next thing to consider:
The $ \sum_{n\in\mathbb{N}} \frac{1}{n}$ diverges (it should be know, and of course known to prove it).


[hide]well1/6 of 3=1/2

and 2/7 of 2 1/3 =2/3

you want the positive difference, so the answer is 2/3-1/2=1/6[/hide]

Here's how to do it[hide="My approach"]
well we take 1/6 of 3 and 2/7 of 2 and 1/3

lets not have them mixed

k,so (x is the positive difference)

$\frac{9}{3}*\frac{1}{6}=\frac{1}{2}$
$\frac{7}{3}*\frac{2}{7}=\frac{1}{3}$
$|\frac{1}{2}-\frac{1}{3}|=x$
$|\frac{3}{6}-\frac{2}{6}|=x$
$|\frac{1}{6}|=x$
$\frac{1}{6}=x$ $and$ $\frac{1}{6}=-x$

since we don't want a negative answer, we disregard the second choice
so....

$x=\frac{1}{6}$

[/hide]
-jorian

LMAO i was getting something different then it said.. i acxcidently thought 2 1/3 was 6/3..not 7/3 lmaoooooooo i am so brain dead sometimes.

it's ok. it happens  :)

especially when you are in a hurry. which i figured out last week

Don't worry, on a test in 6th grade i put $2*1=5$ in my work (dead serious!)
-jorian

omg we just did this problem!!!
:D 1/6

i feel stupid at MATHCOUNTS, too... first year...
  :oops:

It's easy.
1/3+1/6=1/2
1/3+2/6=2/3, since 2/6=1/3
[b]therefore[/b]
1/2+1/6=2/3

get it? :D

[size=75][color=darkred]Double post removed. Please avoid double posting in the future. Thanks.
-nebula42[/color][/size]

i never had the pleasure of doing mathcounts since I wasn't on math team. Anyways, our school didn't do mathcounts either, and most of the sschools in our area didn't care. I didn't hear of it this year.

Has anyone heard of/done Math Olympiad? Our middle school had a team and I never did it. It's sorta like MATHCOUNTS.

[quote="tinadoty"]What is the positive difference of 1/6 of 3 and 2/7 of 2 1/3
Express your answer as a common fraction.[/quote]
[hide]
Put the answer like this:

All they're asking is
$|\frac{1}{6}\times 3-\frac{2}{7}\times\frac{7}{3}|$.

$=|\frac{1}{2}-\frac{2}{3}|$

$=\frac{2}{3}-\frac{1}{2}$

$=\frac{4}{6}-\frac{3}{6}$

$\implies \boxed{\frac{1}{6}}$.[/hide]

[hide="solution"]The keyword "of" means multiplied by. so 1/6*3=1/2. 2/7*7/3=2/3. We subtract those:

2/3-1/2=1/6[/hide]

[quote="tinadoty"]What is the positive difference of 1/6 of 3 and 2/7 of 2 1/3
Express your answer as a common fraction.[/quote]

[hide]abs(1/2-2/3)=1/6[/hide]


first divisible by 2
second divisible by 3
third divisible by 5
fourth divisible by 7
fifth divisible by 11

[hide]788,789,790,791,792.[/hide]

I solved this by guessing and checking and by eliminating certain numbers to multiply by. Overall, a very crude way to solve this problem. i think you can do it with mods. Can anyone do it with mods?

-rts2007

[hide="General solution"]$ a\equiv 0\pmod 2$
$ a \plus{} 1\equiv 0\pmod 3$
$ a \plus{} 2\equiv 0\pmod 5$
$ a \plus{} 3\equiv 0\pmod 7$
$ a \plus{} 4\equiv 0\pmod {11}$

Let $ a \equal{} 2k$.
Then $ 2k \plus{} 1\equiv 0\implies 2k\equiv \minus{} 1\equiv 2\pmod 3$. Multiply both sides by 2 to get $ k\equiv 1\pmod 3\implies k \equal{} 3j \plus{} 1$.
Then $ 2(3j\plus{}1)\plus{}2\equiv j\equiv 1\pmod 5\implies j\equal{}5l\plus{}1\implies k\equal{}4\plus{}15l$
$ 2(4\plus{}15l)\plus{}3\equiv 4\plus{}2l\equiv 0\implies l\equiv 5\pmod 7\implies l \equal{} 7m\plus{}5\implies k\equal{}79\plus{}105m$.
$ 2(79\plus{}105m)\plus{}4\equiv 0\implies m\equiv 3\pmod {11}\implies m \equal{} 11n \plus{} 3$.

Then $ k\equal{}394\plus{}1155n$, so $ a\equal{}788\plus{}2310n$.

Taking $ n \equal{} 0$ gives an answer:
$ \boxed{(788,789,790,791,792)}$

But we also have a general solution:
$ \boxed{(788\plus{}2310n, 789\plus{}2310n, 790\plus{}2310n, 791\plus{}2310n, 792\plus{}2310n)}$.
[/hide]

EDIT:
Thank you [b]JBL[/b], after fixing that error I recognized a couple of other issues.
This solution should work, and the general solution should hold for all $ n\in\mathbb{Z}$.

this is close... first three terms look great

I probably did something similar to rts2007 and I got the same answer: 788, 789, 790, 791, 792

Very crude:
I listed out even numbers and saw that the the even numbers that are followed by multiples of 3 are in the form 6k + 2.
so a = 6k+2 and that satisfies the first two conditions

now 6k+4 is the third term, so 6k+4 must be divisbile by 5: 6k+4=0mod5 : k=-4 mod 5 = 1 mod 5 = 1 + 5r
now 6k+5 is the fourth term, so 6k+5 must be divisible by 7: 6k+5=0mod7 : -k=-5mod7 : k=5mod7 = 5 + 7s
now 6k+6 is the fifth term, so for this one I just said 6(k+1)/11 is an integer so k+1 is divisible by 11 so k = 10 mod 11 = 10 + 11t

now we are looking for a k such that:

k = 1 + 5r
k = 5 + 7s
k = 10 + 11t

for positive integers r,s,t.

The way to properly solve this is with mods and the Euclidean Algorithm. I do not know how to do this, so I used some good logic to quickly get to k = 131 is the first value that worked

so then you plug into 6k+2, 6k+3, etc...

how do you solve euclidean algorithm way?

[quote="grn_trtle"]Then $ 2k\plus{}2\equiv 0$
...
I messed up somewhere[/quote]


logarithms
integration
function
trigonometry
calculus
real analysis
real analysis unsolved

Show that $ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^{n\minus{}1}}{n} \equal{} \ln 2$

$ \int \frac{1}{1\minus{}x} \equal{}\int \sum_{k\equal{}0}^\infty x^k$

at x=-1

That also requires the use of Abel's Theorem - a power series that converges at a boundary point of its interval of convergence equals the limit of the function coming from the interior.

Also take a look [url=http://www.mathlinks.ro/viewtopic.php?t=165931]here[/url]

Here is a rather elementary proof (which, however, lacks motivation):

Define $ I_n = \int_{0}^{\frac {\pi}{4}} \tan^n x \, dx$. Then

$ I_n + I_{n + 2} = \int_{0}^{\frac {\pi}{4}} \tan^n x \sec^2 x \, dx = \frac {1}{n + 1}$

and clearly $ I_1 \geq I_2 \geq \cdots \geq I_n \geq 0$. Thus $ 2I_{n} \leq I_n + I_{n - 2} \leq \frac {1}{n - 1}$ for large $ n$ and hence $ I_n \to 0$.

From these observations, we have

\begin{eqnarray*} I_m & = & \frac {1}{m + 1} - I_{m + 2} \\
& = & \frac {1}{m + 1} - \frac {1}{m + 3} + I_{m + 4} \\
& = & \frac {1}{m + 1} - \frac {1}{m + 3} + \frac {1}{m + 5} - I_{m + 6} \\
& = & \qquad \qquad \vdots \\
& = & \sum_{k = 1}^{\infty} \frac {( - 1)^{k - 1}}{m + 2k - 1} \end{eqnarray*}

Then plugging $ m = 1$ with $ I_1 = \tfrac{1}{2} \ln 2$ gives the desired result. (Plugging $ m = 0$ yields another famous series identity.)

Another elementary argument:
First, we note an algebraic identity: $ 1\minus{}\frac12\plus{}\frac13\minus{}\frac14\plus{}\cdots\plus{}\frac1{2n}\equal{}\frac1{n\plus{}1}\plus{}\frac1{n\plus{}2}\plus{}\cdots\plus{}\frac1{2n}$. To verify this, rearrange the terms in groupings such as $ 1\minus{}\frac12\minus{}\frac14\minus{}\frac18\minus{}\cdots\minus{}\frac1{2^m}$ with sum $ \frac1{2^m}$ or $ \frac13\minus{}\frac16\minus{}\frac1{12}\minus{}\cdots\minus{}\frac1{3\cdot 2^r}\equal{}\frac1{3\cdot 2^r}$, starting t each odd number. Each element of $ \{n\plus{}1,n\plus{}2,\cdots,2n\}$ is a product of an odd number and a power of two in exactly one way.

Now, we want to estimate $ \frac1{n\plus{}1}\plus{}\frac1{n\plus{}2}\plus{}\cdots\plus{}\frac1{2n}$. This is a right-endpoint Riemann sum for $ \int_1^2 \frac1x\,dx$ with $ n$ equally spaced intervals. As $ n$ goes to $ \infty$, it must converge to the integral value $ \ln(2)$.

The other half of the partial sums of the original series differ from the above by $ \frac1{2n}$. The differences go to zero, so they converge to the same limit.


[quote="math92"]Find the probability of at least 2 girls in 7 births. Assume that male and female births are equally likely and that the births are independent events.

a) 0.063
b) 0.164
c) 0.773
d) 0.938[/quote]

[hide]$1-\frac{8}{2^{7}}$
$\boxed{D}$[/hide]

[quote="daermon"][hide]d[/hide][/quote]
Don't post solely the answers.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ p$ be a prime, and $ n$ a positive integer. Let $ X$ be a set with $ p^n$ elements, and let $ G$ be a finite group acting transitively on $ X$. Prove that every Sylow $ p$-subgroup of $ G$ acts transitively on $ X$.

Since the action is transitive, $\forall x \in X, O_{x}=X$, whence $|X|=|O_{x}|=(G: N_{x})=p^{n}$, where $N_{x}$ and $O_{x}$ stand for respectively the stabilizer and the orbit of $x$. If I assume that the $p$-Sylows of $G$ have order $p^{n}$, I can conclude immediately since this would mean that $N_{x}$ has no element of order $p$, and hence no $p$-Sylow has a non-trivial intersection with $N'_{x}$.

Maybe someone can do it in the general case where $n$ is not necessarily maximal ?

Edit: sorry I'm a little tired  :D  this is of course completely solved since $(S_{p}: N'_{x}) \geq p^{n}$ anyhow, and thus the equality stands and the $p$-Sylow $S_{p}$ acts transitively on $X$ ($N'_{x}$ being the stabilizer of $x$ in $S_{p}$).


three variable inequality
Inequality
Cauchy Inequality
inequalities
China

The left part is easy. The term is larger than

$ \frac{a}{\sqrt{a^2\plus{}b^2}}\plus{}\frac{c}{\sqrt{c^2\plus{}a^2}}$.

We can assume $ c\geq b$. then it is larger than

$ \frac{a}{\sqrt{a^2\plus{}c^2}}\plus{}\frac{c}{\sqrt{c^2\plus{}a^2}} > \frac{a}{a\plus{}c}\plus{}\frac{c}{a\plus{}c} \equal{} 1$.

Misha

The inequality is cyclic so we can assume that c is the largest of a,b, c. In case of b>c you only have to replace all a's by c's all b's by a's and all c's by b's in my proof.

Misha

Simpler solution for the left part:

$ S > \frac{a}{\sqrt{a^2\plus{}b^2\plus{}c^2}} \plus{} \frac{b}{\sqrt{a^2\plus{}b^2\plus{}c^2}} \plus{} \frac{c}{\sqrt{a^2\plus{}b^2\plus{}c^2}}$.

Here is another proof of the left side (remember Engel ;-) ):

We have

$\sqrt{a^{2}+b^{2}}<\sqrt{a^{2}+b^{2}+c^{2}+2bc+2ca+2ab}=a+b+c$,

and thus

$\frac{a}{\sqrt{a^{2}+b^{2}}}>\frac{a}{a+b+c}$.

Similarly,

$\frac{b}{\sqrt{b^{2}+c^{2}}}>\frac{b}{a+b+c}$

and

$\frac{c}{\sqrt{c^{2}+a^{2}}}>\frac{c}{a+b+c}$.

Thus,

$\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1$.

As for the right side, after the substitution $A=a^2$, $B=b^2$, $C=c^2$, it becomes

$\frac{\sqrt{A}}{\sqrt{A+B}}+\frac{\sqrt{B}}{\sqrt{B+C}}+\frac{\sqrt{C}}{\sqrt{C+A}}\leq \frac{3\sqrt{2}}{2}$,

or, equivalently,

$\sqrt{\frac{A}{A+B}}+\sqrt{\frac{B}{B+C}}+\sqrt{\frac{C}{C+A}}\leq \frac{3\sqrt{2}}{2}$,

what can be rewritten as

$\sqrt{\frac{2A}{A+B}}+\sqrt{\frac{2B}{B+C}}+\sqrt{\frac{2C}{C+A}}\leq 3$.

And this is exactly Vasc's bloody inequality 113 in

  Titu Andreescu, Vasile Cîrtoaje, Gabriel Dospinescu, Mircea Lascu, [i]Old and New Inequalities[/i], GIL, Zalau 2004.

Bloody since I have been trying to solve it for an hour until I realized that it's in that book with two solutions, each one page long.

  Darij

I've seen quite a number of inequalities now in the form of
\[
\sum_{i=1}^{n} \frac{1}{\sqrt{1+x_i}} \leq S
\]
where $x_1 x_2 \cdots x_n$ is a constant (often 1, as in this case). If there a general (and neat) way to tackle them?

Darij, could you please post one of the two solutions you mention because I don't have the book (or just mention which inequalities were used in the solution).

Misha

[quote="Myth"]Is this solution distinct from Vasc's ones?[/quote]

Not only distinct; it is also two times shorter than any of the solutions in the book, and much more elegant than any of them.

[quote="Myth"]P.S. Are there copyrights for solutions?[/quote]

I hope not, but in any case, when I will refer to your solution, I will call it the Myth solution!

  Darij

[quote="darij grinberg"][quote="Myth"]Is this solution distinct from Vasc's ones?[/quote]
Not only distinct; it is also two times shorter than any of the solutions in the book, and much more elegant than any of them.[/quote]
Jensen forever!

[quote="darij grinberg"][quote="Myth"]P.S. Are there copyrights for solutions?[/quote]
I hope not, but in any case, when I will refer to your solution, I will call it the Myth solution!
[/quote]
I hope all people will respect solution's authors as well as you respect...

[quote="Myth"][quote="darij grinberg"][quote="Myth"]Is this solution distinct from Vasc's ones?[/quote]
Not only distinct; it is also two times shorter than any of the solutions in the book, and much more elegant than any of them.[/quote]
Jensen forever![/quote]

More precisely, [i]weighted[/i] Jensen forever! I myself had learnt it, as a method, from Peter Scholze in Athens (the inequality itself was known to me before, but I didn't really know what do with it  :blush: ), and it was immediately clear to me that it should be much more powerful that plain Jensen and it should have a lot of interesting applications. But until now I have seen only very few ones, including your ingenious proof of Vasc's inequality. I guess that most applications of the weighted Jensen inequality remain undiscovered just because people too seldomly try to apply it. And don't forget the weighted Karamata inequality (I believe it's also called Fuchs inequality) - I can't believe that it has no use!

  Darij

A very elegant and neat solution. Congratulations, Myth.

[quote="darij grinberg"][/quote][quote="Myth"][quote="darij grinberg"][quote="Myth"]Is this solution distinct from Vasc's ones?[/quote]
Not only distinct; it is also two times shorter than any of the solutions in the book, and much more elegant than any of them.[/quote]
Jensen forever![/quote][quote="darij grinberg"]

More precisely, [i]weighted[/i] Jensen forever! I myself had learnt it, as a method, from Peter Scholze in Athens (the inequality itself was known to me before, but I didn't really know what do with it  :blush: ), and it was immediately clear to me that it should be much more powerful that plain Jensen and it should have a lot of interesting applications. But until now I have seen only very few ones, including your ingenious proof of Vasc's inequality. I guess that most applications of the weighted Jensen inequality remain undiscovered just because people too seldomly try to apply it. And don't forget the weighted Karamata inequality (I believe it's also called Fuchs inequality) - I can't believe that it has no use![/quote]
I very often use Jensen's inequality (at least, I try such approch in almost each inequalities).

What is the "weighted Karamata inequality "??

[quote=WolfusA][quote=andyciup]
Therefore it suffices to show that \[ 2(ab+bc+ca)\cdot\left(\frac{1}{(a+b)(b+c)}+\frac{1}{(b+c)(c+a)}+\frac{1}{(c+a)(a+b)}\right)\leq\frac{9}{2},\]
[/quote]

Isn't it some Iran MO inequality from 1990's?[/quote]

Nooo, that's a bit different, if you mean the REALLY known one.

By the way, that was almost this one, but instead of $\frac{1}{(a+b)(b+c)}$, it had $\frac{1}{(a+b)^2}$ and others.

[quote=Ji Chen][quote="Vasc"]If $ a,b,c$ are positive numbers, then

$ \sqrt {\frac {a}{4a + 5b}} + \sqrt {\frac {b}{4b + 5c}} + \sqrt {\frac {c}{4c + 5a}}\leq 1$.[/quote][hide=*]Notate ${{ \frac {a}{4a + 5b}} = \frac {u^2}{(1 + 2u)^2},\frac {b}{4b + 5c}} = \frac {v^2}{(1 + 2v)^2},\frac {c}{4c + 5a} = \frac {w^2}{(1 + 2w)^2},$ where $ u,v, w$ are nonnegative.

Hence, $ \{\begin{array}{l}(1 + 4u)a = 5u^2b \\
(1 + 4v)b = 5v^2c \\
(1 + 4w)c = 5w^2a \end{array}\Longrightarrow (1 + 4u)(1 + 4v)(1 + 4w) = 125u^2v^2w^2.$

We need to prove that $ \frac {u}{1 + 2u} + \frac {v}{1 + 2v} + \frac {w}{1 + 2w}\leq 1\Longleftrightarrow 1 + w + v + u\geq 4uvw$.

In fact, $ 2[13 + 4(u + v + w)](1 + w + v + u - 4uvw) - (1 + 4u)(1 + 4v)(1 + 4w) + 125u^2v^2w^2$

$ = 25 + 30(u + v + w) + 8(u^2 + v^2 + w^2) - 168uvw - 32uvw(u + v + w) + 125u^2v^2w^2$

$ \geq 0$ holds for all nonnegative numbers $ u,v,w.$ See : http://www.mathlinks.ro/viewtopic.php?t=229616[/hide]
[quote="shalex"]For all $ x,y,z\in\mathbb{R_{ + }}$ such that $ xyz = 1$ and $ 0\leqslant\lambda\leqslant\frac {5}{4}$ there is

$ \frac {1}{\sqrt {1 + \lambda x}} + \frac {1}{\sqrt {1 + \lambda y}} + \frac {1}{\sqrt {1 + \lambda z}}\leqslant\frac {3}{\sqrt {1 + \lambda}}.$

The solution is nearly the same as Arwen's to the original one in which $ \lambda = 1$.[/quote]$ \lambda > \frac {5}{4}\Longrightarrow \frac {1}{\sqrt {1 + \lambda x}} + \frac {1}{\sqrt {1 + \lambda y}} + \frac {1}{\sqrt {1 + \lambda z}} < 2.$

holds for all positive numbers $ x,y,z$ with $ xyz = 1.$

[hide=**][b]Remark[/b] $ \sqrt {\frac {a}{4a + 5b}} + \sqrt {\frac {b}{4b + 5c}} + \sqrt {\frac {c}{4c + 5a}}\leq 1$

$ \Longleftrightarrow \cfrac {1}{\sqrt {4 + \cfrac{5b}{a}} } + \cfrac {1}{\sqrt {4 + \cfrac{5c}{b}}} + \cfrac {1}{\sqrt {4 + \cfrac{5a}{c}}}\leq 1$

$ \stackrel{a=x^2y,b=y^2z,c=z^2x}{\rightleftharpoons}\cfrac {1}{\sqrt {4 + \cfrac{5yz}{x^2}} } + \cfrac {1}{\sqrt {4 + \cfrac{5zx}{y^2}}} + \cfrac {1}{\sqrt {4 + \cfrac{5xy}{z^2}}}\leq 1$

$ \Longleftrightarrow \frac {x}{\sqrt {4x^2 + 5yz}} + \frac {y}{\sqrt {4y^2 + 5zx}} + \frac {z}{\sqrt {4z^2 + 5xy}}\leq 1.$[/hide]

See also here : http://www.mathlinks.ro/Forum/viewtopic.php?t=106970[/quote]
Let $a,b,c$ be nonnegative reals.  [url= https://artofproblemsolving.com/community/c6h561272p3269586then ]Prove [/url]

$$\sqrt{\frac a {4a+5b}}+\sqrt{\frac b {4b+5c}}+\sqrt{\frac c {4c+5a}} \ge \frac 1 2$$
If $a, b, c>0.$  [url=https://artofproblemsolving.com/community/c6h580951p3432029]Prove that[/url]
\[\sqrt{\frac{a}{4b+5c}}+\sqrt{\frac{b}{4c+5a}}+\sqrt{\frac{c}{4a+5b}}\ge\frac{2}{3}+\frac{ab+bc+ca}{(a+b+c)^2}\]
If $a,b,c$ are non-negative numbers such that $ab+bc+ca>0.$ [url=https://artofproblemsolving.com/community/c6h1631229p10242194]Find the minimum value of[/url] $\sqrt{\frac{a}{4b+9c}}+\sqrt{\frac{b}{4c+9a}}+\sqrt{\frac{c }{4a+9b}} .$
[url=https://artofproblemsolving.com/community/c6h531354p3035279]here[/url]

Suppose that $ a$, $ b$, $ c$ are positive real numbers. Prove that
$$ \frac {a}{\sqrt {a^{2} \plus{} b^{2}}} \plus{} \frac {b}{\sqrt {b^{2} \plus{} c^{2}}} \plus{} \frac {c}{\sqrt {c^{2} \plus{} a^{2}}}+\frac{\sqrt{2}}{16}\left(\frac{b^4}{a^4}+\frac{c^4}{b^4}+\frac{a^4}{c^4}\right)\geq\frac {27\sqrt {2}}{16}.$$
[url=https://artofproblemsolving.com/community/c6h17451p119168]IMO2001[/url]
[url=https://artofproblemsolving.com/community/c6h1783082p11759649]2008 Jiangxi Higher Education Entrance Examination[/url]
High-School Mathematics(China Tianjin) No.8(2007)：

[quote=Myth]I have just proved this inequality in excellent way, in my opinion. :)

We can write
\[
\sqrt\frac{a}{a+b}+\sqrt\frac{b}{b+c}+\sqrt\frac{c}{c+a}=
\]
\[
=\frac{a+c}{2(a+b+c)}\cdot\sqrt\frac{4a(a+b+c)^2}{(a+b)(a+c)^2}+
\frac{b+a}{2(a+b+c)}\cdot\sqrt\frac{4b(a+b+c)^2}{(b+c)(b+a)^2}+
\frac{c+b}{2(a+b+c)}\cdot\sqrt\frac{4b(a+b+c)^2}{(c+a)(c+b)^2}=S.
\]
Now we apply Jensen's inequality for $\sqrt{x}$ and obtain:
\[
S\leq \sqrt{\frac{2a(a+b+c)}{(a+b)(a+c)}+\frac{2b(a+b+c)}{(b+c)(b+a)}+\frac{2c(a+b+c)}{(c+a)(c+b)}}=\sqrt{S'}.
\]
So we are to show that $S'\leq \frac{9}{2}$. You can easily check this inequality, because it is equivalent to $\sum\limits_{\textrm{cyclic}}a^2b\geq 6abc$.

Is this solution distinct from Vasc's ones?

P.S. Are there copyrights for solutions?
Each author merited to be mentioned if his solution is used somewhere.[/quote]

indeed  very nice!

Suppose that $ a$, $ b$, $ c$ , $ d$ are positive real numbers, prove that
\[ \frac {a}{\sqrt {a^{2} \plus{} b^{2}}} \plus{} \frac {b}{\sqrt {b^{2} \plus{} c^{2}}} \plus{} \frac {c}{\sqrt {c^{2} \plus{} d^{2}}}\plus{} \frac {d}{\sqrt {d^{2} \plus{} a^{2}}}\leq3.\]

[quote=Vasc]I think this is more difficult.

If $a,b,c$ are non-negative numbers, then
$\sqrt{\frac{a}{4a+5b}}+\sqrt{\frac{b}{4b+5c}}+\sqrt{\frac{c}{4c+5a}}\leq 1$.[/quote]

Let $\frac{a}{4a+5b}=\frac{p^2}{9}$, $\frac{b}{4b+5c}=\frac{q^2}{9}$ and $\frac{c}{4c+5a}=\frac{r^2}{9}$
Hence, we need to prove that $p+q+r\leq3$ and since $\frac{5b}{a}=(-4+\frac{9}{p^2})$, we obtain $\{p,q,r\}\in(0;\frac{3}{2})$
Let $p+q+r>3$, $p=kx$, $q=ky$ and $r=kz$, where $k>0$ and $x+y+z=3$.
Hence, $k>1$, $\{x,y,z\}\in\{0;\frac{3}{2}\}$ and $125=\prod (-4+\frac{9}{k^2x^2})<\prod (-4+\frac{9}{x^2})$, which is contradiction because we'll prove now that:
$125\geq \prod (-4+\frac{9}{x^2})$
Let $3u=x+y+z$, $3v^2=xy+yz+zx$ and $w^3=xyz$
Since, we need to prove $f(w^3)\geq0$, where $f(w^3)$ is increasing function.
Hence, $f(w^3)$ reaches minimum value for minimum value of $w^3$, which happens when two of variables are equal.
$x=y$
$27(x-1)^2(7x^2+14x+9)\geq0$

[quote=mecrazywong]Suppose that $ a$, $ b$, $ c$ are positive real numbers, prove that
\[ 1 < \frac {a}{\sqrt {a^{2} \plus{} b^{2}}} \plus{} \frac {b}{\sqrt {b^{2} \plus{} c^{2}}} \plus{} \frac {c}{\sqrt {c^{2} \plus{} a^{2}}}\leq\frac {3\sqrt {2}}{2}
\][/quote]

first is my problem

[quote=xzlbq]Let $x,y,z>0$,prove

\[2\,\sqrt {{\frac { \left( x+y+z \right)  \left( xy+xz+yz \right) }{
 \left( y+z \right)  \left( z+x \right)  \left( x+y \right) }}}\geq \sqrt
{{\frac {x}{x+y}}}+\sqrt {{\frac {y}{y+z}}}+\sqrt {{\frac {z}{z+x}}}\]
<=>

\[2\,\sqrt { \left( x+y+z \right)  \left( xy+xz+yz \right) }\geq \sqrt {x
 \left( y+z \right)  \left( z+x \right) }+\sqrt {y \left( z+x \right)
 \left( x+y \right) }+\sqrt {z \left( x+y \right)  \left( y+z \right)
}\][/quote]

Note this a=a:

\[4\,{\frac { \left( x+y+z \right)  \left( xy+xz+yz \right) }{ \left( x+
y \right)  \left( y+z \right)  \left( z+x \right) }}={\it \sum}
 \left( z+x \right) {\it \sum} \left( {\frac {x}{ \left( z+x \right)
 \left( x+y \right) }} \right) \]

By C-S, is easy

Suppose that $ a$, $ b$, $ c$ are positive real numbers, prove that
\[ \frac {a}{\sqrt {a^{2} \plus{} b^{2}}} \plus{} \frac {b}{\sqrt {b^{2} \plus{} c^{2}}} \plus{} \frac {c}{\sqrt {c^{2} \plus{} a^{2}}}\leq\frac {3\sqrt {2}}{2}
\]
$$\iff$$
Suppose that $ a$, $ b$, $ c$ are positive real numbers such that $(1-a)(1-b)(1-c)=abc.$ Prove that$$\sqrt{a}+\sqrt{b}+\sqrt{c}\leq \frac{3\sqrt{2}}{2}$$
[url=https://artofproblemsolving.com/community/q4h365871p2012455]here[/url]

[quote=sqing]Suppose that $ a$, $ b$, $ c$ are positive real numbers, [url=https://artofproblemsolving.com/community/c6h1429668p8053801]prove that[/url]
$$\frac{a}{\sqrt{b^2+c^2}}+\sqrt{2}\left(\frac{b}{a+c}+\frac{c}{a+b}\right) \ge \frac{3\sqrt{2}}{2}$$[/quote]
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{\sqrt{2(a^2+b^2)}} \ge \frac{3}{2}$$
Suppose that $ a$, $ b$, $ c$ are positive real numbers. Prove that
$$\frac{a}{b+c}+\frac{b}{c+a}+\sqrt{\frac{2c}{a+b}} \ge {2}$$
[url=https://artofproblemsolving.com/community/c6h1064117p10988486]here:[/url]
$$ \frac{a}{b+c}+\frac{b}{c+a}+\frac{4(c+\sqrt{ab})}{a+b}\geq 4$$

$$ \frac{a}{b+c}+\frac{b}{c+a}\ge \frac{a+b}{2}\left(\frac{1}{b+c}+\frac{1}{c+a}\right)$$


Wrong forum, there's a forum for calculus.

Calculus problems should go in the calculus forum  :wink:
Mods please move this.

[quote="HFL"]Solve:
\[ I = \int\limits_{0}^{\pi/4}xtan(x)dx\]
[/quote]

Sorry ! I confused


We want iTest/PUMaC/HMMT teams. Since Berman and Vitek are gone, I will have to organize these now *shudder*

people who MUST be on these teams (no buts or ifs :P. ok, not must; but should like, definitely go?) [other than me]
Calvin
Bryce
Thomas
Pom

people who SHOULD be on these teams
Gray
Allen
Christine
Michael G.

SIGN YOURSELF UP. You don't have to be on these lists, but these people get preference due to previous results. email me, or pm me, or post here, or something. also, emailing me would probably be better because I'll want your email address.

I demand - no, expect - more 1st places. gogogo.

Also, other possible contests - "purple comet"? and other stuff. also dang NC teams should try to get top 4 at duke this year  :|

I should probably be able to do iTest... the rest I'm not sure... also I have heard that Enloe is trying to organize and send a team to PUMaC...

[quote="blackbelt14253"][quote="not_trig"]
people who MUST be on these teams (no buts or ifs :P. ok, not must; but should like, definitely go?) [other than me]
Calvin
Bryce
Thomas
Pom
[/quote]

hmm that makes me feel better about my math capabilities
calvin, you have no choice. you must go. MUST MUST MUST.
also I heard that mgao was in china for some reason?

EDIT: yes i will most likely be able to go.[/quote]

uh no mgao is not going to china

wait so when will applications/confirmations/etc be posted/emailed/etc?

I think I'll email out a list of people who have responded. maybe by tonight, maybe by... Tuesday (vacation woo?)

HELLO DAMIEN I CANNOT BE ON THESE TEAMS  I THINK D:

I'll sign up (though I probably wont get to be on any of the teams unless everyone else gets sick :P)
My email is jasonyliang@yahoo.com
Im assuming you know who I am by my email, but if not, ask calvin.

[quote="not_trig"]darn arren no you may not

you should calm down

or you should move to NC

that would be pretty cool?[/quote]

Dude i seriously thought you were talking to me since thats what half of Cary Academy calls me now x.x

[quote="not_trig"]darn guys

we did pretty mediocrely[/quote]
Maybe next year we could have NC A and NC B teams, instead of having people like Calvin and Nick not be part of the main NC team.

[quote="not_trig"]darn guys

we did pretty mediocrely[/quote]

Hmm, actually, I thought NC did ok, particularly given the split nature of our people.

4 individual finalists (2 from Enloe), 6th power, 7th overall, 9th team round, and 3 people placed in the top 3 in a subject...

Also, power would have probably gone better if more people had been working on it earlier...


Let $a;b;c$ are positive real numbers.Prove that:
$\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\geq \frac{1}{2}$

i think it's false.just try $a=b=c=0.1$ You might have forgotten a condition.

no we have  $\sum_{cyc}\frac{ab}{a+b}\leq\frac{2(a+b+c)}{4}$
so the sign $\geq$ must be $\leq$.


Let $ a$ , $ b$ , $ c$ and $ d$ be real numbers from interval $ [1,2]$. Prove that:
$ \frac{a\plus{}b}{c\plus{}d} \plus{}\frac{c\plus{}d}{a\plus{}b} \minus{}\frac{a\plus{}c}{b\plus{}d} \leq \frac{3}{2}$

This is the same problem as this one:

[list]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=291925[/list]


Find the necessary and sufficient condition for which the number of the integer, that is not devisible by 4 or 6 among $ 3N$ integers: $ 1,\ 2,\ 3,\ \cdots ,\ 3N$, is greater than $ 2N$.

The number of numbers divisible by 4 or 6:
$ \lfloor\frac{3N}{4}\rfloor \plus{} \lfloor\frac{3N}{12}\rfloor$
So according to the sum,
$ N> \lfloor\frac{3N}{4}\rfloor \plus{} \lfloor\frac{N}{4}\rfloor$
This always holds true for N not being a multiple of 4 (they become equal when it is).
So the answers is N is not a multiple of 4.

[quote="grafitti123"]The number of numbers divisible by 4 or 6:
$ \lfloor\frac {3N}{4}\rfloor \plus{} \lfloor\frac {3N}{12}\rfloor$[/quote]

That's incorrect.

Sorry, ill give it one more try. I got N =4k+1. If it isnt then just ignore my solution.
[hide="Ignore solution if wrong"]
The number of numbers divisible by 4 is:
$ \lfloor\frac{3N}{4}\rfloor$
The number of numbers divisible by 6 but not by 4 is( i hope this is right):
$ \lfloor\frac{N}{4}\plus{}\frac{1}{2}\rfloor$
So,
 $ N>\lfloor\frac{3N}{4}\rfloor\plus{} \lfloor\frac{N}{4}\plus{}\frac{1}{2}\rfloor$
so N has to leave a decimal part of less than .50 (but not 0)
So for the condition to hold N is of the form,
N=4k+1
[/hide]


[quote="archimedes1"][url]http://www.artofproblemsolving.com/Resources/Papers/SatoNT.pdf[/url][/quote]

Thanks. I needed it too. :lol:

You can download some notes of Reid Barton here: http://web.mit.edu/rwbarton/Public/mop/

[quote="Ronald Widjojo"]@Johan : yang Naoki Sato download aja, ntar tahap I Pak Nanang pakai itu (katanya Rudi sih gitu..)[/quote]Thanks infonya. Udah download sih, tapi belom pernah baca. :D

[quote="soumik"]I need something on non-linear diophantine equations - I mean not only qsns, it must have some explanations as well.[/quote]

you should take Diophantine Equations, of Titu Andreescu.

[quote="enndb0x"]Hi guys,
Can anyone of you tell me where I can find an ebook(free download/pdf) about Euler's fuction,Chinese Remainder Theorem etc  ...(IMO Level)[/quote]
See here http://trungtuan.wordpress.com/2010/06/02/topic-97/ .


For $ a,b,c\geq 0$,prove that
\[ \frac{bc(c\plus{}a)^2}{(b\plus{}c)^2(a^2\plus{}bc)}\plus{}\frac{ca(a\plus{}b)^2}{(c\plus{}a)^2(b^2\plus{}ca)}\plus{}\frac{ab(b\plus{}c)^2}{(a\plus{}b)^2(c^2\plus{}ab)}\leq \frac{3}{2}\]
A quite nice one:lol:

I hope there could appear a nice solution in the new year....Could some can help me :(


Two dinners, each with a cost of $ x$ dollars, are purchased with a ``buy one, get the second one at half price'' coupon. An $ 8\%$ sales tax added to the bill and finally a $ 15\%$ tip, based on the regular cost of the two meals, is added. Find the total cost of the two dinners purchased with this coupon. Express your answer as a decimal in terms of $ x$.

Before any tax or tip, the dinners cost $ 1.5x$.

Then, adding the tax, it costs $ 1.5 \cdot 1.08x\equal{}1.62x$.

Adding the tip gives $ 1.15 \cdot 1.62x\equal{}1.863x$.

[quote="BOGTRO"]Before any tax or tip, the dinners cost $ 1.5x$.

Then, adding the tax, it costs $ 1.5 \cdot 1.08x \equal{} 1.62x$.

[/quote]

This is correct, but the tip is 15% of the normal price (or 15% of 2x):

0.15 \cdot 2x = 0.3x$ Totaling:
1.62x$ + 0.3x$ \equal{} 1.92x$


Find all positive integrer numbers $n$ such that $\frac{499-n}{2n+1}$ is a positive integrer number.

[hide]Divide it out to get
$-\frac{1}{2}+\frac{999}{2(2n+1)}$
so we want $\frac{999}{2n+1}$ to be an integer
So $2n+1$ has to be a divisor of $999$
$2n+1=1,3,9,\text{BLAHBLAHBLAH},999$
when it's equal to 1, n=0 so it's not positive and when it's equal to 999 the expression is 0. in other cases it should work.
[/hide]

Welp, n \in [1, 499) by inspection.  And for the 498 possible n, you can eliminate 332 possible n considering that if n = 166, (499 - n) / (2n + 1) = 1.  Thus, (499 - n) / (2n + 1) < 1 for all n > 166.  As far as finding all n \in [1, 166) such that (2n + 1) | (499 - n), i can't think of anything off the top of my head.  Sorry  :oops:

[quote="José"]Thanks.

Any complete solution?[/quote]
The post above, your post,  for all practocal purposes , is      [b]complete [/b]solution! (each factor of 999 would lead you to one solution - 6 in all) :!:

We want both n and the expression to turn out positive. But when the factor is 1 or 999 one of them becomes 0 and therefore is not a solution.


calculus
integration
function
ratio
geometry
perimeter
real analysis

Hello, I am a high school student that is currently taking Calculus II.  I am part of an engineering program, and our current project is to design and build a hovercraft.  My friend and I wanted to calculate how high the hovercraft we designed would hover.  However, we ran into some problems.

[color=red]For the impatient readers, here is the problem we need to solve, without any explanations:

[img]http://www.xdemi.com/equation.gif[/img]

... where a, b, and c are constants.[/color]

Here is a summary of how this function came to be...

The hovercraft has two main parts: the frame and the skirt.  The frame simply holds the leaf-blowers which will power the hovercraft, and the pilot that will drive the hovercraft.  It is of no significance in this equation.  The skirt is an inflatable bag with an open bottom (the ground acts as a barrier so that air does not escape and inflates the bag).  As air enters, the skirt inflates and lifts the frame.  When the pressure gets high enough, the entire hovercraft will lift off the ground as the air tries to escape through the open bottom.  This pressure is a function of time, p(t).

However, the higher the hovercraft lifts, the larger the gap between the hovercraft and the ground.  The larger the gap, the higher the rate of air escape.  The higher the rate of air escape, the faster the drop in pressure.  And thus, we return to the function for pressure.

[color=red]So the higher the pressure, the faster the hovercraft rises.  The faster the hovercraft rises, the faster the pressure falls.  Doing a million replacements and substitutions, the equation displayed above is the final result, or basic form, of the equation.[/color]

The true equation replaces constants a, b, and c with a plethera of combinations of over 10 other constants.

[b][color=red]I would look this up in textbooks or online, but I have no idea what this is even called.  The function is the integral of itself?  So if anyone could at least point me in what topic this covers, I could look it up myself.  Or better yet, someone could show me how this can be solved.[/color][/b]

Thank you very much.

-Andrew Cheong

Two things:

It's a huge abuse of notation to write

$p(t)=\int^t_0 f(p(t))dt$: write $p(t)=\int^t_0 f(p(x))dx$ instead or something similar.

Anyways, the method is obvious if you've solved differential equations before.

We get:

$dp/dt=abp-ac$, which is a separable differential equation if you write the RHS as $ab(p-c/b)$

...Sorry about the notation thing?

And we haven't really solved any differential equations.  We're on our last unit, infinite series.  I guess I'll do differential equations when I go to college.

Thank you.  I don't entirely understand your answer, or what "RHS" stands for.  But I understand how you made it separable.  I'm going to explore it further before I ask anything else.  Thanks again.

There's a common sense test here: the pressure is not going to go to infinity, no matter how you start the system. If $ab$ is not negative, you've created nonsense with either a mistake in your model or in your algebra.

If $ab$ is negative, then the pressure will tend rapidly towards its limit $\frac cb$, and you have a stable hovercraft.

By the way, RHS means "right-hand side".

Allow me to explain my reasoning.

The pressure ratio at time t is equal to the volume of air inside the chamber divided by the volume of the chamber (assuming the chamber is non-expandable).  This ratio multiplied by the atmospheric pressure is the absolute pressure inside the chamber.  The atmospheric pressure subtracted from the absolute pressure is the gauge pressure (or the pressure differential), which I called p(t).  The volume of the chamber and the atmospheric pressure are constants.  However, the volume of air inside the chamber is the integration of the rate of air-flow, which I called q(t).

The rate of air-flow, q(t), is basically the rate of air-flow into the chamber (a constant, ignoring the pump-curve of our leafblower system) minus the rate of air-flow out of the chamber.  The rate of airflow out of the chamber is YCg(t)sqrt(2gp(t)/r) where Y is the net expansion factor (a constant), C is the orifice flow coefficient (a constant), g(t) is the area of the gap that allows air to escape, g is the gravitational acceleration constant, p(t) is the aforementioned pressure differential, and r is the density of air (a constant).

Thus, I already have the first loop back to p(t).

I must also deal with g(t), the gap between the ground and the chamber that allows air to escape.  This area is equal to 2(x+y)h(t) where x is the hull width, y is the hull length, and h(t) is the height.  (Basically, perimeter times height is the "hover gap", or g(t)).

Now, we must deal with h(t).  Acceleration and velocity is zero at start so h(t) is simply 0.5*a(t)*t^2, where a(t) is the acceleration.

a(t) is F(t), the net force upward, divided by the mass of the hovercraft, m (a constant).

F(t) is the force up minus the force down.  The force up is the pressure differential, p(t), times the area on the hovercraft it acts upon, which is basically xy (hull width times hull length).  The force down is mg (the weight of the craft).

This is how this whole system loops back to another p(t).

Substituting everything I explained eventually got me one equation.  I thought it no use to include constants like Y, C, m, g, x, y, etc. while posting on this forum, so I combined all of the constants into simply a, b, and c.

It is very possible that I could have made a mistake during the substitution process, so I will do it again and see what I come up with.

Thank you for the clarification, jmerry.