MATHCOUNTS
ARML
geometry
AMC
AIME
videos
Functional Analysis

[quote="goldendomer"]I was thinking, nobody in my school goes or even knows this website. I think its great, but are we nerds? I mean I'm not. All I want is to do good at MathCounts. I'm popular and I have a lot of friends. I'm not saying anything mean so dont get mad at me. Express yourself and how your like.[/quote]

I'm popular and have a lot of friends too but I am a NERD. Just cuz your cool doesn't mean you aren't a nerd.  :rotfl:
Seriously, I am a nerd and some of my friends are too, but the majority of my friends aren't nerds. They are all smart though. I am obsessed with mathcounts and am lucky that my school has a high gifted population. I don't feel strange about being smarter than others. The strange thing is most of the smart people in my school are cool/popular.

Anyone at my school who's smart is a miniscule piece of dirt.(really...)I don't really mind
[size=200][color=green]I love my geekiness/nerdiness/dorkiness!!![/color][/size]

I'm homeschooled and I go to a homeschool coop. Among my friends at the coop, normal is considered bad.

To G&FF.

Speaking of homeschoolers: There are about 4 classes of homeschoolers: Religious homeschoolers, gifted homeschoolers, special needs homeschoolers, and others. Normal for a homeschoolers is exceptional for a government-schooler. (I count in the first 2 groups of homeschoolers; if lacking long attention spans/ constantly fidgety counts, I fit in 3 too.)

Billy

i say im a geek...and im proud of it...my "friend(s)" all think im a  dork too =p

I am not a nerd, not a maths nerd at least.
I like maths and do well in Maths, but that is all.

There are things in Physics nothing in Maths can beat!

All I want to be known as is as an intellectual, talking of strings as well as of petty bourgeois!

I am a nerd, because I spend 2-3 hours a day mathing my head off, doing two grades of math last year.

I've always felt that being in IB by choice gives you a couple nerd points.

At my school, being a nerd is the norm. It's rather nice, actually, being able to discuss Dagobah for an extended period of time without anyone rolling their eyes in superiority.

[quote="mysmartmouth"]Whose Dagobah?[/quote]
It is not a person. It is Yoda's planet.

I am a part time Math Nerd. I spend time normally socializing on the bus and in class. But I'm nerdy during math class, math team, when students need to know what to make on the final to get a certain grade,  while on this forum, and other times that it helps to be nerdy. I make my brain run 2x as fast when I need it, and let it stop when I don't need it.
BTW: Being in a Magnet school makes you feel alot less nerdy and generally more accepted.

I'm definitely a math nerd, though I border on "band geek."

One question you should ask your self is, is What is a nerd?

So....whats the answer to that question?

[quote="beta9"]Math people are way different than nerds.  I define a nerd as someone who spends way too much time on school and has no real friends.  In contrast to this, most math people get by in school doing little work, just riding on their natural intelligence, and while they might not be part of the in-crowd (drunken football-players/cheerleaders), they have friends.

Also, why do math people not shave ever.  I know this Indian kid in 8th grade who has a full moustache and it just looks rediculous.[/quote]

Completely correct. I couldn't agree more. With both points  :P  ;)

"He who refuses to take part in modern mediocrity will forever stand alone, an outcast. He is the bringer of change. Changes that instill fear in all who go with fashion's flow. Their power comes from others, not within. They are false!"
(Manowar)
Math people are not nerds. They are different, yes, but not nerds.

[quote="perfectnumber628"]but if we're not nerds, who is?[/quote]

thor's right. all people are different. nerds don't exist. that's one big human lie. humans don't like any kind of diffrences. so they'll make up something just to isolate the ones who are above or under them, everybody who isn't average.
but, if there weren't these differences, humans would live in caves, hunting elephants with wooden sticks. and i don't like to fight with creatures bigger (and heavier) then me. (i don't like fights anyway)

[quote="deimos"]nerds don't exist. [/quote]
uh...you know there's a whole bunch of people who posted on this topic claiming to be nerds?  the thing is, it depends how you define the word "nerd": some people think it's offensive, and some people are proud of it.

however, when in doubt, we can just call ourselves "math people".

[quote="perfectnumber628"][quote="deimos"]nerds don't exist. [/quote]

however, when in doubt, we can just call ourselves "math people".[/quote]

well, yes.

true, it depends how do u see it. i explained my point of view when the word "nerd " is offensive.
i can explain my point of view when the word "nerd" is something good, nice and a desirable describing word

in my school other students don't say to me that i'm nerd or something like that. they say just that i'm "a bit strange, confusing being"...

At our school, a nerd is anyone that does extracurricular academic activities. But if someone says I'm a nerd, I just say, "I'm smarter than you" and that shuts them up.  :rotfl:
The term can be good or bad, offensive and complimental, it just matters where your viewpoint is. I don't really mind much though.

I am happy for anybody who is a nerd, in fact it is my wish that all people find something to be nerdy about.  To me it indicates that you like something enough to really want to pursue it and that makes me happy!  So I guess according to my definition you can be a band nerd, a math nerd, a sports nerd, a social nerd, etc.  It's all the same.  Based on this, I am a BIG NERD!  It is so much fun flaunting it, it's fun seeing people's reactions, surprising them, challenging perceptions.

[quote="goldendomer"]I was thinking, nobody in my school goes or even knows this website. I think its great, but are we nerds? I mean I'm not. All I want is to do good at MathCounts. I'm popular and I have a lot of friends. I'm not saying anything mean so dont get mad at me. Express yourself and how your like.[/quote]

I think that some MathLinkers are nerds and others aren't.... and the most nerds are like undercovered, and you don't know them...... I won't put examples, but just think about this.......

This would depend on how you define nerd. On one hand,I enjoy many things one might define as "nerdy" (math, computer science, video games, etc.) However, if you were to define a nerd as one who spends all of their time studying, I would not be a nerd. And no, this is not because all of my classes are easy (I go to TJ), but because I simply lack the capacity to do so.

nerds are always studying... that's the definition

I thought that nerds were people that wore glasses. I don't believe that.
Nerd is a fake word!
Nerd is a fake word!
Nerd is a fake word!
Nerd is a fake word!
Nerd is a fake word!
Nerd is a fake word!
Nerd is a fake word!
Nerd is a fake word!

[quote="surge"]The term "nerd" was actually invented by [url=http://en.wikipedia.org/wiki/Nerd]Dr. Seuss[/url].[/quote]

Nice link. However, the meaning is just to people that study all the time....... just that

[quote="perfectnumber628"]as a girl nerd, i agree with hmmm- nerds aren't ugly.  and yay for nerds![/quote]

Thing is, mostly girl nerds don't consider guy nerds/geeks unattractive. Although, that may be a good thing. Personally, I've gone from being a nerd and geek, by the previous definitions, to merely a nerd. People have gone to avoiding me to giving me some respect for my intellectual abilities. Also, I don't know if the two are related, but my interest in math has shot down and my interest in computer science has shot up. This may be because I'm making good amounts of money programming at the moment.


I found this in a file on Gabriel Carroll's site (I can't remember the name of the site; it was something about a gastropod  :D ):

ABC is a triangle and I is its incenter. D, E, ,F are the points where the incircle touches BC, CA, AB, respectively. AI intersects EF at P. K is the intersection of DF and BP. L is the intersection of DE and CP. Prove that KL || EF.

This is easy.
KL || EF
iff DK/KF=DL/LE
iff [BDP]/[BFP]=[DCP]/[PEC]
iff [BDP]/[CDP]=[BFP]/[CEP]
iff BD/CD=[BFP]/[CEP] (1)
But [BFP]/[CEP]=BF*FE*sin(BFE)/CE*PE*sin(CEF)=BF/CE.
So (1) iff BD/CD=BF/CE
The last is obvious since BD=BF and CD=CE.

[ABC] is the area of ABC.
I would have expected for a harder problem from Gabriel Carroll.

[quote="amfulger"]I would have expected for a harder problem from Gabriel Carroll.[/quote]

Okay: I would have expected for a harder problem from Gabriel Carroll's site.

Then why don't you try to solve other problems created by him? You will see that he can create extraordinary difficult problems, too!

Please give a link to his site (or at least to his problems  :D ) here.

Indeed, this isn't one of his problems, it was composed by some friend of his. Not only can he compose extremely difficult problems, there are some problems there created by him which he hasn't solved yet! Those could make interesting subjects for some research! :D


superior algebra
superior algebra solved

1) If $\alpha\in Z[i]$ is irreducible, show that it divides a prime number (integer).

2) Find all irreducible elements in $Z[i]$.

It's well-known that the primes in $\mathbb Z[i]$ are the integer primes of the form $4k+3$ and the numbers of the form $a+bi,\ a,b\in\mathbb Z$, where $a^2+b^2$ is a prime of the form $4k+1$.

$\mathbb{Z}[i]$ is euclidian so any non-invertible irreducible is prime.

The primes in $\mathbb{Z}[i]$ are the primes $\equiv 3(\mod 4)$ and $x+iy$ with $x^2+y^2\equiv 1(\mod 4)$ and prime in $\mathbb{Z}$.

First of all, let's prove that those numbers are indeed prime:

Assume first that $p$ is a prime of the form $4k+3$. If $p=ab$ in $\mathbb Z[i]$, then the norm of $p$, which is $p^2$, is equal to $N(a)\cdot N(b)$, and, since we assume that $a,b$ are not invertible, it means that $N(a)=N(b)=p$, which is impossible, since the norm of $a$ is a sum of two squares, and cannot be of the form $4k+3$. As for the numbers of the form $a+bi$, where $a^2+b^2$ is a prime, those must be prime because they're indecomposable (their norms are integer primes).

Conversely, we have to show that all the primes in $\mathbb Z[i]$ have that form.

Every Gaussian prime divides some rational integer, which can be decomposed in integer prime factors, so our Gaussian prime $p$ must divide one of these factors (this proves your first question). Our primes are thus divisors of integer primes. The integer primes of the form $4k+3$ are Gaussian primes (as proven above). Assume now $p$ is an integer prime of the form $4k+1$. Then it divides some integer of the form $a^2+1$, but it obviously divides none of $a+i,a-i$, meaning that it is not a Gaussian prime. Let $p=(a+bi)(c+di)$, with non-invertible $a+bi,c+di$. The norm of $a+bi$ must then be $p$, so $a^2+b^2=p$, and we find the other kind of Gaussian primes.

P.S. I repeatedly used the fact that in $\mathbb Z[i]$ the primes are precisely the irreducible elements (which I also called indecomposable from time to time, because that's what it sounds in Romanian :)).


Can anyone tell me the best free web hosting service? Thanks!

depends... you need php and such or just html?

I go to http://www.freewebs.com its fairly simple..though you don't get many options..its free!

Well, I run my own server at home.

Advantages: It's great for low-volume web traffic and as an email server. You have absolute control over the machine and you can also use it as a file server. You can also use it as a LAN server. Stick a huge hard disk into it and use it to store files that are too big for the other computers on the network or you want to be shared. I put things like digital photos on the server. And you could share a printer, too.

Disadvantages: You have to have enough technical expertise to be able to set up the server though...and a spare machine lying around...and you have to pay for the electricity. Unfortunately, these disadvantages are big enough to stop many people from running their own server.


Solve in positive integers

1/a+1/b+1/c=1

[hide]WOLOG $ a\leq{b}\leq{c}$.  Note that $ a$ cannot equal $ 1$ because $ 1/b,1/c > 0$.  When $ a \equal{} 2$, solutions for $ (b,c)$ are $ (3,6)$ and $ (4,4)$; there are no others since $ 1/b$ must be at least $ 1/4$.  When $ a \equal{} 3$, solutions for $ (b,c)$ are $ (3,3)$; there are no others since $ 1/b$ must be at least $ 1/3$.  Since $ 1/a$ must be at least $ 1/3$, these are the only solutions to the equation in positive integers:

$ (a,b,c) \equal{} (2,3,6)$ or $ (2,4,4)$ or $ (3,3,3)$[/hide]

Extension:

Let $ n\in N,r\in Q^\plus{}$, does there exist a couple $ (n,r)$ such that there are infinite many sets of $ (a_1,a_2,...,a_n)$ in positive integers for which

$ \frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}\cdots \plus{}\frac{1}{a_n}\equal{}r$　$ ?$

To Bovy11's questions:

[hide]The short answer: no.  It will be shown by induction on $ n$.  It will be shown by contradiction that there exists no such pair $ (n,r)$ when $ n\equal{}2$.  Assume there exists a pair $ (2,r)$ such that there are an infinite number of sets $ (a_1, a_2)$ such that $ \frac{1}{a_1}\plus{}\frac{1}{a_2}\equal{}r$.  WOLOG $ a_1\leq{a_2}$, so $ 1\geq{\frac{1}{a_1}}\geq{\frac{r}{2}}$.  There are only a finite number of $ a_1$ for which this inequality holds true, and for each $ a_1$ there can be at most one set $ (a_1, a_2)$ such that $ \frac{1}{a_1}\plus{}\frac{1}{a_2}\equal{}r$.  Thus the number of $ (a_1, a_2)$ is finite, and by contradiction, there exists no such pair $ (2,r)$.

Now assume that there exists no such pair $ (k,r)$ for a fixed positive integer $ k$.  It will be shown that there exists no such pair $ (k\plus{}1, r)$ by contradiction.  Assume there exists such a pair $ (k\plus{}1, r)$ such that there are an infinite number of sets $ (a_1, a_2,\ldots, a_{k\plus{}1})$ such that $ \frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}\ldots\plus{}\frac{1}{a_{k\plus{}1}}\equal{}r$.  Again, WOLOG assume $ a_1\leq{a_2}\leq{a_3}\ldots\leq{a_{k\plus{}1}}$.  There should still be an infinite number of sets $ (a_1, a_2,\ldots,a_{k\plus{}1})$ given this limitation.

Note that since $ \frac{1}{a_1}\geq{\frac{1}{a_2}}\ldots\geq{\frac{1}{a_{k\plus{}1}}}$, it follows that $ \frac{1}{a_1}\geq{\frac{r}{k\plus{}1}}$.  Also, since $ \frac{1}{a_2}>0$, it is known that $ \frac{1}{a_1}<1$.  There are only a finite number of $ a_1$ such that $ 1>\frac{1}{a_1}\geq\frac{r}{k\plus{}1}$.  Let there be $ m$ of these $ a_1$, and let them be labeled $ a_{11},a_{12},a_{13},\ldots{a_{1m}}$.

For each of these $ a_{1i}$ there are some number of sets $ (a_2, a_3, \ldots, a_{k\plus{}1})$ such that $ \frac{1}{a_2}\plus{}\frac{1}{a_3}\plus{}\ldots\plus{}\frac{1}{a_{k\plus{}1}}\equal{}r\minus{}\frac{1}{a_{1i}}$.  For each $ a_{1i}$, let the number of such sets be denoted $ S_i$.  Note that since $ r$ and $ \frac{1}{a_{1i}}$ are rational, so is $ r\minus{}\frac{1}{a_{1i}}$.  Thus each $ S_i$ is finite by the induction hypothesis.  Since the total number of sets $ (a_1, a_2,\ldots, a_{k\plus{}1})$ is the sum of $ S_1, S_2, \ldots, S_m$, and all $ S_i$ are rational, so is the number of sets $ (a_1, a_2,\ldots, a_{k\plus{}1})$.  This contradicts the initial assumption that there were an infinite number of such sets.  Thus, by contradiction, there exists no pair $ (k\plus{}1, r)$.

By induction, there exists no such pair $ (n,r)$.[/hide]

Talk about confusing, poorly expressed, nested arguments with too many variables.  But it does prove the point?

Yes that's right,jrshoch. My solution is as same as yours,using induction.


If a car engine delivers constant power, does the car have an accelerating acceleration?  If so, then how is power related to acceleration?

Constant power means energy is changing at a constant rate. However, since the kinetic energy is $ \frac{1}{2}mv^2$, the velocity will be changing at a rate proportional to the square root of the change in kinetic energy. Ok, I don't know if that made sense, but anyways, accelerating is decreasing, I think.

i got what your saying, it helped me figure out how to do a problem : )

Well, Power is equal to work over a period of time.
so Power times an amount of time is equal to the amount of work.
Work is also the change in kinetic energy, which is proportional to velocity squared.  So if your engine does work, which the constant power increases, then velocity will increase.  Therefore, your acceleration for that period of time will be positive because acceleration is the change of velocity and delta x is positive.

I think that sounds right...

EDIT:  didn't see the next post, oh well.

yeah acceleration is always > 0, but acceleration is decreasing.
In other words if f(x) = acceleration, then f'(x) < 0 if the power delivered is constant.  (this is based on my understanding of pascal12's post)

if you guys are curious, here is the problem i was trying to solve.

A car engine delivers constant power.  Accelerating from rest at time t = 0, it reaches an acceleration of X at time t = Y.  What is the acceleration of the car at time t = 2Y.

It's pretty simple, now that I know what power is.

I solved this, but I used calculus to find the acceleration function.

Can someone post a solution that does not use calculus?

Well, Power at any instant is Force multiplied by velocity
P = F * v
   = m*a*v
So Power is constant means a*v is constant or acceleration is inversely proportional to velocity.
That means that the magnitude of acceleration decreases as velocity increases.
If you supply cinstant power to a body starting from rest, then it's acceleration goes on decreases from an initial infinte value at t=0 as it's velocity goes on increasing as time passes. Moreover as time tends to infinity acceleration tends to zero and velocity tends to a fixed value

I did'nt know that P = F * v
interesting

still, what's a solution without calc?

ill post mine (which kinda uses calc) I guess

k(t) = bt where k(t) is the amoutn of kinetic energy the car has at time t
b is just some random constant.  the kinetic energy is constantly increasing with time because power is constant
because K(t) = 1/2 * m * v^2(t),
v(t) = square root of (2k(t)/m) = square root of (2bt/m)
a(t) = v'(t) = square root of (2b/m) * 1/(2 * square root of t)
a(2t) = square root of (2b/m) * (1/2 * square root of (2t))

thus if a(t) = x, then a(2t) = x/square root of 2

[hide="solution without calculus"]
constant power means KE difference from time 0 to Y to Y to 2Y is same.
Therefore, you get that $ \frac{v_1^2}{y}\equal{}\frac{v_2^2\minus{}v_1^2}{y}$
solve for $ v_2$ in terms of $ v_1$ ($ \sqrt{2}v_1\equal{}v_2$)
P=Fv, so P=m*a*v, so constant power means that a and v are inversely proportional.
Therefore, if a at time Y is X, then a at time 2Y is...
P=mX$ v_1$
P=m$ a_2v_2$=Pm$ a_2\sqrt{2}v_1$
Divide to get $ 1\equal{}\frac{X}{a_2\sqrt{2}}$
Solve to get $ a_2\equal{}\frac{X}{\sqrt{2}}$[/hide]

Calculus looks like a good way to get around the problem without the formula, though.

I don't know where you got the first equation from
the v^2/y one

Oh, I set the power equal to each other, canceled out 1/2m and Y is time.

oh i see

I was confused by the "y's," but they just cancel out lol

okay, got it.

Is power defined as F * V?

Defined?  Not really.  It's better to think of power as the difference in work over a certain amount of time and difference in KE.

You can derive it from $ P \equal{} \frac {W}{\Delta t}$.  Try it.

[hide="hint"]
remember, the v in power is average power throughout the whole time so it's defined as $ \frac {v_f \plus{} v_i}{2}$
[/hide]

is work the change in kinetic energy?
If so, I got it...i think

I get that

P = F * 1/2 (final velocity + initial velocity)
which means that
P = F * avg velocity

not sure if avg velocity is right

edit: I looked at your hint
Nice, looks like I did it right haha (easy algebra is the best)

The amount of work done is the change in the [i]total[/i] energy of the system - this includes kinetic, spring potential, gravitational potential, magnetic potential, heat (which is kinetic but on a molecular scale), etc., types of energy.  In one dimensional problems like a block accelerating without friction, the only type of energy that changes is kinetic energy, so work done is equal to change in kinetic energy.  If some energy were lost to friction or was used to raise the object, not all of the work would be "put into" kinetic energy.

Hopefully that helps.


geometry
3D geometry
tetrahedron
symmetry

Whe discussing Tetrahedral Geometry in terms of Crystal Field Theory, MO Theory etc. We construct a cube indicating the Cartesian axes. We place the corners of the tetrahedron at the origin and three other corners not lying directly on the axes. Why is that?

If you link those four corners you will get a tetrahedron. With molecular models that should be more easy to see.

Yeah I know that. It's just that originally there is another way to pick four corners giving a tetrahedron. My tutor just told me to place the central atom at the origin hence no matter how I pick the corners, the symmetry will resolve any ambiguity.


Two Chinese mathematicians claimed that they have solved Poincare conjecture, which says all simply connected and closed 3-manifolds are homeomorphic to 3-sphere(I'm have very limited knowledge on topology, so someone correct me if I'm wrong).
[url]http://news.xinhuanet.com/english/2006-06/04/content_4644754.htm[/url]
I think they are still checking the proof, so we don't know whether it's solved or not.

their work is mostly based on Perelman's proof.


I sat here forever trying to figure this out. I can't make any assumptions about it, since it doesn't tell me which angles are equal and if there's any 90 degree angles, so I'm completely stumped.
It's #22.

[img]http://i47.tinypic.com/mrtsih.jpg[/img]

22. $ m\angle{BFA} \equal{} m\angle{DFE}$ since they are vertical angles.
$ 3r \plus{} 12 \equal{} \minus{} 8r \plus{} 210$
Solving this gives:
$ r \equal{} 18 \Rightarrow \angle{BFA} \equal{} 66$

Since $ \angle{AFE}$ is supplementary to $ \angle{BFA}$
$ \angle{AFE} \equal{} 180 \minus{} 66 \equal{} \boxed{114}$

Thanks for the response. That sounds good, but I was always told not to assume anything about the angles unless there were marks indicating what they are.
The angles look like vertical angles, but how can I know for sure if there's no markings to indicate that they are? They may be off by .01 degrees.
I guess on a simple question like this I can assume they're vertical. But I still think there should be some markings so I know for sure. Something may look equal to something else, but it may not necessarily be equal.

Thanks again.

I got another one I ran into at the end of the page. This one is also driving me nuts. I don't see a solution. I keep getting 10=8 which doesn't make sense. It's #50.

[img]http://i46.tinypic.com/2mo6l21.jpg[/img]

First, notice that $ PQ \plus{} QR \equal{} PR$.  This should be common sense; it says that a part of a segment plus the other part equals the whole thing.  We are given that $ PQ \equal{} 1 \minus{} x$, $ QR \equal{} 4x \plus{} 17$, and $ PR \equal{} \minus{} 3x$.
Substitute these values into our equation $ (PQ \plus{} QR \equal{} PR)$, to get $ (1 \minus{} x) \plus{} (4x \plus{} 17) \equal{} ( \minus{} 3x)$.  Now we just solve:
$ (1 \minus{} x) \plus{} (4x \plus{} 17) \equal{} ( \minus{} 3x)$.
$ 3x \plus{} 18 \equal{} \minus{} 3x$
$ 6x \equal{} \minus{} 18$
$ x \equal{} \minus{} 3$
$ QR \equal{} 4( \minus{} 3) \plus{} 17 \equal{} 5$

By the way, you do not have to prove two angles are vertical because you can assume a line drawn in a diagram is a line.

[quote]First, notice that . This should be common sense; it says that a part of a segment plus the other part equals the whole thing. We are given that , , and .
Substitute these values into our equation , to get . Now we just solve:
[/quote]
I know, but for some reason I kept messing up the addition and subtraction like an idiot.
[quote]By the way, you do not have to prove two angles are vertical because you can assume a line drawn in a diagram is a line.
[/quote]
Oh ok, so I don't have to worry about a line being crooked by .01 degrees? Ok thanks. I guess I took it too literally.

[quote="leroyjenkens"]Here's one that's giving me trouble:

I have to find the perimeter of the rectangle.

"The length of the rectangle is half the width. The area is 25 square meters."

I don't see how I can answer that with it having two unknowns. I'm sure it's simple, but I just don't see it.[/quote]

Let $ 2w$ be the width.  Because the length is half of the width, the length is simply $ w$.  The perimeter of the rectangle is then $ 6w$.  IF we multiply the length and width, we get $ 2w \times w \equal{} 2w^2 \equal{} 25$.  Solving for $ w$ yields $ \frac{5\sqrt2}{2}$.  Plugging this back into to perimeter equation, we get $ 6 \times \frac{5\sqrt2}{2} \equal{} \boxed{15\sqrt2}$.

We could have also used $ w$ for the width, but that would have created some fractions, and I'm just really lazy.  There really aren't two unknowns, I suppose you could have $ 2w \equal{} l$ and $ lw\equal{}25$ as two equations, but substituting $ l$ in the second equation would yield to same result.


[quote="littlefermat23"]Hi,

This problem came across in a study/problem session with some friends.  Does anybody have any thoughts on it?

Let A,B,C,D be four consecutive points on a quadrilateral that lie on a circle.  Prove:

|A-C||B-D|= |A-B||C-D| + | B-C||A-D|[/quote]
This is known as Ptolemy's theorem:
http://en.wikipedia.org/wiki/Ptolemy%27s_theorem


http://www.onemorelevel.com/games.php?game=284

My best:

L1: 2
L2: 3
L3: 3
L4: 4
L5: 5

Those are probably the optimal best scores anyway.

Would have been nice if there were a few more levels though.


a) Prove that 10201 is composite in any base greater than 2.

b) Prove that 10101 is composite in any base.


geometry
circumcircle
incenter
geometric transformation
reflection
power of a point

This is a particular case of the following:

[b]Theorem[/b]. Let $ ABC$ be a given triangle and let $ P$ be a point in its plane. Let $ A'B'C'$ be the cevian triangle of $ P$ with respect to $ ABC$ and denote by $ A_{1}$ the intersection of $ BC$ with the perpendicular in $ P$ to the line $ AP$. The points $ B_{1}$, $ C_{1}$ are defined similarly.
1) The points $ A_{1}$, $ B_{1}$, $ C_{1}$ are collinear.
2) The line $ \overline{A_{1}B_{1}C_{1}}$ is perpendicular to $ PP'$, where $ P'$ is the isogonal conjugate of $ P$ with respect to triangle $ A'B'C'$.

The line $ \overline{A_{1}B_{1}C_{1}}$ is known as the orthotransveral line of $ P$ with respect to $ ABC$. Many proofs for the collinearity of $ A_{1}$, $ B_{1}$, $ C_{1}$ are known. I prefer the following:

[i]Proof of Theorem 1)[/i] (by Jacques Hadamard; see Darij Grinberg, Problem: The ortho-intercepts line, available on [url=http://www.cip.ifi.lmu.de/~grinberg/]his website[/url]). Let $ \mathcal{K}$ be an arbitrary circle centered at $ P$; the polars of the points $ A$, $ B$, $ C$, $ A_{1}$, $ B_{1}$, $ C_{1}$ w.r.t. $ \mathcal{K}$ are called $ a$, $ b$, $ c$, $ a_{1}$, $ b_{1}$, $ c_{1}$. After the construction of polars, we have $ a \perp PA$, $ b \perp PB$, $ c \perp PC$, $ a_{1} \perp PA_{1}$, $ b_{1} \perp PB_{1}$ and $ c_{1} \perp PC_{1}$. Since the point $ A_{1}$ lies on BC, the polar $ a_{1}$ passes through the intersection $ b \cap c$. From $ a_{1} \perp PA_{1}$ and $ PA_{1} \perp PA$, we have $ a_{1} \| PA$; from $ a \perp PA$, thus we get $ a_{1} \perp a$. Hence, $ a_{1}$ is the line passing through the intersection $ b \cap c$ and orthogonal to $ a$. Analogously, $ b_{1}$ is the line passing through the intersection $ c \cap a$ and orthogonal to $ b$, and $ c_{1}$ is the line passing through the intersection $ a \cap b$ and orthogonal to $ c$. Therefore, $ a_{1}$, $ b_{1}$ and $ c_{1}$ are the altitudes of the triangle formed by the lines $ a$, $ b$ and $ c$; consequently, the lines $ a_{1}$, $ b_{1}$ and $ c_{1}$ concur. From this, we derive that the points $ A_{1}$, $ B_{1}$ and $ C_{1}$ are collinear.

Theorem 2) is due to treegoner. For the proof, see [url=http://reflections.awesomemath.org/2006_4/2006_4_solutions.pdf]Khoa Lu Nguyen, Solution to Problem O13, Mathematical Reflections, 4 (2006), 20-21[/url].

the generalization of the first was proposed also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=200084]here[/url]

Pohoatza, I am not quite sure how my problem can yield this. Also in your post, it should be $ P'$ is the isogonal conjugate of $ P$ wrt $ A'B'C'$. Actually, if you apply this result and my problem, then you will obtain a solution to your problem http://www.mathlinks.ro/viewtopic.php?t=216926 (which I think there is a typo $ P \equal{} I$)

One can generalize this as follows.

$ \textbf{Proposition}$: If $ P$ is a point, define $ A'$ to be the intersection of the tangent at $ P$ of the circumcircle $ BPC$ and $ BC$. Define $ B', C'$ analogously. Then $ A', B', C'$ are collinear and this line is perpendicular to $ OP$.

$ \textit{Proof}$: Obvious. It is the radical axis of the circumcircle and the point $ P$.

[quote="treegoner"]Pohoatza, I am not quite sure how my problem can yield this. Also in your post, it should be $ P'$ is the isogonal conjugate of $ P$ wrt $ A'B'C'$. Actually, if you apply this result and my problem, then you will obtain a solution to your problem http://www.mathlinks.ro/viewtopic.php?t=216926 (which I think there is a typo $ P \equal{} I$)[/quote]
I've edited both posts. Thanks.

Dear Mathlinkers,
after the nice approach, I have seen before, I present my synthetic proof.
Let IaIbIc the excentral triangle of ABC and Be the Bevan point of ABC (The center of the circumcircle of IaIbIc).
1. IaIbIc and ABC are perspective at I ; let (P) be the perspectrix.
2. IaIbIc and ABC are orthologic with pole I and Be
3. According to Sondat's theorem (see : http://perso.orange.fr/jl.ayme  article: Le théorème de Sondat  vol. 1) IBe is orthogonal to (P).
4. According a result from Nagel (article : Cinq théorème de Nagel vol. 3) I, Be and O are collinear.
5. The proposition in the Treegoner Message which can be proved with the Thalès's theorem, gives (P) // A'B'C'
and we are done
Sincerely
Jean-Louis

[quote="Fang-jh"][quote="Fang-jh"]Let $ O,I$ be the circumcenter, incenter of a scalene triangle $ ABC$, respectively. The line passing through $ I$, and perpendicular to $ AI$ intersects the side $ BC$ at $ X$, define $ Y,Z$ similarly. Prove that
(I): the threes points $ X,Y,Z$ are collinear.
(II): $ OI\perp XYZ$[/quote]
Denote by $ \Gamma$ the circumcircle of triangle $ ABC$, $ R$ the radius of $ \Gamma$. Let $ P$ be the second intersection of $ AI$ with $ \Gamma$, Let $ Q$ be the tangency point of $ A \minus{}$mixtilinear incircle of triangle $ ABC$ with $ \Gamma$.
Lemma I: $ IQ\perp PQ$
Lemma II: $ PQ$ passes through $ X$.
By Lemma I and Lemma II, we have $ XI^2 \equal{} XQ\cdot XP \equal{} XB\cdot XC \equal{} XO^2 \minus{} R^2$, thus $ XO^2 \minus{} XI^2 \equal{} R^2, (1)$. similarly,
$ YO^2 \minus{} YI^2 \equal{} R^2 ,(2), ZO^2 \minus{} ZI^2 \equal{} R^2 , (3)$
From $ (1),(2)$ and $ (3)$, we conclude the points $ X,Y,Z$ lie on the line that is perpendicular to $ OI$, qed
PS: the solution belongs to $ tomchen$[/quote]
In order to prove $ XI^2 \equal{} XB\cdot XC$, it is not necessary using Lemma I and  Lemma II,
we can show it by $ \bigtriangleup XIB\sim\bigtriangleup XCI$  directly,  the desired result follows once.

Nice problem: Clearly, $\overline{IX}$ is tangent to $\odot(BIC)$. Thus, $XI^2 = XB \cdot XC$ so $X$ lies on the radical axis $\gamma$ of the circumcircle and the degenerate circle at $I$. In a similar way, we can prove that $Y, Z \in \gamma$. Hence, $X, Y, Z$ are collinear and $\overline{XYZ} \perp \overline{OI}$.


Going on what happened this year and my preferences, I think Nathan will win written next year.

Nathan - because he was second place overall this year, and because we have the same name. :lol: :lol:

[quote="NoSoupForYou"]Nathan, though Neal Wu will probably win Countdown.[/quote]

No, Kevin will win!!! :D. At least I hope he does, since he's from CA :D

[quote="NoSoupForYou"]If you think about it, Kevin has a pretty good chance. After all, he'll probably place in top 12 and he is extremely fast during Countdown.

Top 12 + Fast = WIN!!![/quote]

Not as fast as me :D and I'm not as fast as Patricia, and Patricia is not as fast as Neal...

Edit: Actually, Patricia should've owned Neal

[quote="NoSoupForYou"]... and Neal isn't as fast as me, obviously.[/quote]

Yes, of course, Alex, how could I forget to mention that... :lol:

You're right Bob. How could you?

Oh wait, that was sarcasm. OK, that's it. NO SOUP FOR YOU!!!

[quote="kpavery"]too bad you'll be in 9th grade alex, otherwise you could [i]easily[/i] win[/quote]

Not if I'm there to stop him :)

[quote="NoSoupForYou"]probably[/quote]

I beat you this year, so I can beat you next year too :) logic :D I'm jk Alex

Hey, we tied! I just had a really bad sprint score, which knocked me down ten spots.

Me :D [hide]Just kidding maybe nathan[/hide]

So I'm guessing David is also homeschooled. (or was?)

This is my guess: based on what happened this year, Nathan will get written; Neal will get Countdown; Andrew will get Masters. However, a lot can change in a year. Only one other seventh grader has been National Champion (Ricky liu) and he got eighth in eighth grade.

Ryan Ko was first in 7th grade as well ;) . Plus he got in fourth in 8th grade.

I seriously have a shot at champion next year. I'm gonna probably study around an hour a day before Chapter, 90 minutes a day before States, and two hours a day before Nats.

I made 12 point worth of careless errors this year :blush: .

[quote="Phelpedo"]I find that broadcasting such statements is bad luck.[/quote]

Well, okay, but for the less superstitious among us...

I doubt that Neal Wu will win again. You just can't go into that room with the confidence that being a National champion gives you and expect to win. It's inhuman.

A new bracket this time, though- and I'm pretty sure he will at least make Countdown, so even if he is like 9th written again, he will have a good chance. If he gets too confident, though, he probably won't win because there will be a lot of other people out there aiming for him. But being champion already gives him quite an advantage, and, if he works just as hard as everyone else, he will win (probably).

It depends. In vocab, you just have to open up the dictionary and there is a nice word sitting on that page that would make perfect sense to use and you don't know it. Easy to quash that pride. In math, you open up a calc book and say "wow, I can't understand it, but it isn't going to be on the test, so I'm doing great! I'm gonna WIN! I'm unbeatable!". So I would say that Neal Wu will not win again.

Of course, that doesn't mean that it can't happen - I know of one very ameliorative property that if Neal has, he may well be unbeatable. He's just got to conquer any pride or cockyness that he might have. Hard to do.

Billy

[quote="solafidefarms"]It depends. In vocab, you just have to open up the dictionary and there is a nice word sitting on that page that would make perfect sense to use and you don't know it. Easy to quash that pride. In math, you open up a calc book and say "wow, I can't understand it, but it isn't going to be on the test, so I'm doing great! I'm gonna WIN! I'm unbeatable!". So I would say that Neal Wu will not win again.
[/quote]

What?


An elegant answer can be derived from the concept of proportion.

Like $ \frac{a}{b}$ = $ \frac {p}{q}$ $ \implies$ each = $ \frac {a \plus{} p}{b \plus{} q}$ or $ \frac {a \minus{} p}{b \minus{} q}$


$ \bullet$ An organization of $ 100$ people set up a telephone call system so that the initial contact person calls three people, each of whom calls three others,and so on, until all have been contacted.  What is the maximum number of people who do not need to make a call?

[hide]$ 67$[/hide]

$ \bullet$ From a two-digit number $ N$ we subtract the number with its digits reversed and find that the result is a positive perfect cube.  For many such $ N$ is this true?

[hide]$ 7$[/hide]

$ \bullet$ A set of $ 17$ points is chosen on the circumference of a circle.  Find the maximum number of segments which can be drawn connecting pairs of points such that each segment intersects all others inside, not on, the circle.

[hide]$ 8$[/hide]

Thanks in advance!

[hide="first problem"]First person calls 3 people. Then each of those three people call 3 people bringing it to a total of 1+3+9=13 people who are contacted. Then each of the 9 people call 3 more people each bringing the total to 40. Now, out of the 27 people who were just contacted, only 20 of them need to call three people because there are only 60 people who are still not contacted. The total of number of people who didnt call anyone is 60 plus the 7 people giving the desired answer of 67.[/hide]

[hide="2"]
Let $ N\equal{}ab$ $ (a\neq 0)$
$ 10a\plus{}b\minus{}10b\minus{}a\equal{}k^3$. $ k$ is positive integer
$ 9(a\minus{}b)\equal{}k^3$
clearly $ a\minus{}b\equal{}3$
so $ 7$ nos $ b\equal{}0,1..6$[/hide]


Let $ f$ be a mapping of the unit ball $ S\equal{}\{ x: \parallel{}x\parallel{} \leq 1 \}$ into $ \mathbb{R}$ such that $ f(\alpha x\plus{}\beta y)\equal{}\alpha f(x)\plus{}\beta f(y)$ whenever $ x,y$ and $ \alpha x\plus{}\beta y$ are in $ S$. Show that $ f$ may be extended to a linear functional on all of $ X$.

Define $ f(x) \equal{} \|x\|f(x/\|x\|)$. Show that this agrees with the the previous definition if $ \|x\|\leqslant1$. Then show that $ f$ defined thus is linear on $ X$.


calculus limit logarithms algebra polynomial function domain

evaluate ((1+1/2+1/3+...+1/n)^2)/n as n goes to infinity

The limit??? Isn't this a calculus question???

Though it may be solved with calculus, it is also solvable using non calculus methods

Youcant' separate parts of a limit if one is not defined!!!

The limit is zero.  We may bound $\sum_i 1/i$ by multiples of $\log x$.  But since $\log x$ goes to $\infty$ slower than any polynomial, we are done.

[quote="nat mc"]Well, since $n$ goes to $\infty$, wouldn't it be $0$?
Because $\lim_{n\to\infty}{\frac{\left(\sum_{a =1}^n{\frac1a}\right)^2}{n}}=\lim_{n\to\infty}{\left(\sum_{a =1}^n{\frac1a}\right)^2} \cdot \lim_{n\to\infty}{\frac1n}$, and $\frac1n$, goes to zero, while $\left(\sum_{a =1}^n{\frac1a}\right)^2$, limits at an irrational number when $n\to\infty$.  Is that correct?[/quote]

Hmmmmmmmmm.........

Yes, you are right in some sense that the series of the form 1/[i]n^2[/i] is convergent therefore has a limit. However, such a convergency as[i] n [/i]goes large i.e. infinity, has to fall within certain condition to which it is convergent, i.e its Domain of Definition that is, say an interval of some real numbers to which a limit can be found.

I agree with you that such a sequence of series has a limit.


Prove that, for any reals a and b,

[tex]2(a^3+b^3)^2\ge (a^4+b^4)(a+b)^2[/tex].

Why does everyone work backwards?  It's [i]so[/i] risky with inequalities!

Doing things backwards instead of fowards is something I do without thinking about it. :oops: I'll fix it.

Well, me too, of course.  But especially with inequalities, there are lots of things that you can do that aren't reversible steps, or aren't even valid steps at all.  A very important part of solving them by manipulation (rather than by reference to relatively obscure theorems) is to write it forwards so that you can check every single step carefully for that kind of mistake.

[quote="belenos"]An other solution

expanding

(a+4ab+b)(a-b)(a+b) >= 0

Gives wanted ineg[/quote]

Not sure how this implies the original inequality.

[b]nr1337[/b]'s proof looks good to me.

In fact, LHS - RHS = (a-b)^4(a+b)^2  :ge: 0.

I was able to solve it from the trivial inequality, as nr1337 did.  However, that was my second approach - using AM-GM on individual terms had little effect.  Can anyone tell me if I was just not being creative enough, or if AM-GM was just not a valid approach?

[quote="cats..."]where did you use it?  It doesn't say positive numbers, so that could be a problem.[/quote]

I expanded and compared terms.  I didn't notice the bit about positive numbers.

Hey nr, what made you think to start with (a-b)^4 and working backwards from there?

He didn't think to start with that. What he did (and then changed after comments), was start with the inequality, go through those steps backwards, and realise you get to (a-b)^4. Then, as has been talked about many times, when you write up the proof you of course start from (a-b)^4 and work until you get the solution, making sure everything still works on the way.

So why not just write the solution goign forwards from 2*whatever >= stuff? Instead of starting at the solution to the inequality (a-b)^4 >=0?

Well, the way he did it was to start with (a-b)^4 >= 0, which is true, and prove the inequality thats asked for.
But if you did it that way, you would assume the thing which is asked to be proved is true, and then shown that (a-b)^4 >= 0! Which is not what you wanted to do at all.

It is not a proof if you assume what you are trying to prove is true and end up at a true statement. You want to prove that this thing is true, starting at true statements.

Here is a pretty obvious example which "proves" that n>4 for every integer n.

If n>4 for every integer n, then n>4 when n=5. So 5>4. This is a true statement, so n>4 for every integer n.

Now you should be able to see why thats wrong, and thats the same idea as before.

Actually, you can see a brief conversation about exactly the point you are raising in the three proofs directly under nrl337's post with the solution.

[quote="perfect_radio"][quote="TripleM"]But if you did it that way, you would assume the thing which is asked to be proved is true, and then shown that (a-b)^4 >= 0! Which is not what you wanted to do at all.[/quote]

Well not really... If I'm not mistaken, we can use equivalences in this case, i.e. the given inequality is true iff blabla iff blabla iff blabla iff $(a-b)^4 \geq 0$. but the last statement is true, so the first one must be too, right? (looking at nr1337's soln , I didn't discovers any irreversible steps)[/quote]
If you read carefully, post #14 responds to post #13 only, explaining why proving $p\rightarrow q$ is not the same thing as proving $q\rightarrow p$. When TripleM said "that's wrong" he did not say nr1337's solution was wrong, he said the idea of proving the converse is wrong because you are proving another implication.


Ioana has three ropes whose lengths are 39 inches, 52 inches and 65 inches. She wants to cut the ropes into equal length pieces for magic tricks. No rope is to be wasted. What is the greatest number of inches possible in the length of each piece?

It should be just the greatest common factor, which is $ \fbox{13}$, unless the question means something else...


let $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}1$ prove that

 $ \sum\frac{1\plus{}a}{1\minus{}a}\geq\frac{2a}{b}\plus{}\frac{2b}{c}\plus{}\frac{2c}{a}$


ok so how many questions are going to be on this year's prelim exam? still 25?

and I'm going through fundamentals of physics by halliday, resnick, walker - is this good enough for the prelim mechanics exam (content and difficulty of content)......I'm able to do about 95 % of the end of chapter problems by myself without looking at the solutions....so do I have a good chance of getting to the second round (semifinals)

and....what were the previous cutoffs to advance from the first exam?

But they took out quarterfinals. (so we have to jump from F=ma to semis)

Any expectations?  estimations?

Also, any idea on how may people actually take the test?
I think last year it was around ~1000 for quarter finals and 500 or 400 to semi finals (USAPHO)

How many take the F=ma exam though?  that might help in estimates.

Just looking at the problems, it doesn't seem nearly as difficult as the AMC and a 20+ score is definitely posible.

Since they took out the quarter finals, is there gonna be 25 MC on the F=ma?

Wiki says that the cutoff score for 2008 F=ma was around 16, although I have no idea how they got that statistic.
And the 2008 homepage says that over 2000 people took the F=ma test.

ok, I e-mailed the people who run the competition:
the preliminary exam is still 25 q and 75 min
I'm still waiting for the response to how many people will be admitted to semifinals (less because of no quarterfinals? or the same?) and what the cutoff scores, for the f=ma exam, were for previous years in order to get to quarterfinals, and if the test will be harder since there is no quarterfinals

Wow, what a smart idea to email them! I seriously would have never thought of that.

The website says: "The top 400 to 500 students based on the combined score on the preliminary exam will be invited to take this exam." So it seems similar to previous years, although I have no idea what they mean by combined score.

ok its official from the reply to the e-mail i sent: 400-600 will advance to semifinals.  It will be 25 questions, 75 minutes, and harder than last year.

Ouch. Do you know if it's gonna be the same level of problems as the quarterfinals of previous years? Or just in between?

I don't know for sure...I'm guessing it will be in between...it's still mechanics only though

hey! does anybody know the estimated score of this year,2010? i think it will be ok as long as we don't discuss any of the problems. the problems this year is much harder than last year. Will the cutoff be lower?

the cutoff is probably similar to past years.
Although the problems were much harder, they are selecting less people.

I ACTUALLY THINK THE PEOPLE THEY SELECT WILL REMAIN THE SAME,TOP400_500. LAST YEAR THEY SELECTED 400 Quarterfinal .THIS YEAR THEY WILL SELECT 400 FOR SEMIFINAL. uc, same number,different name.

I thought you could understand what I am talking about. Sorry if I brought you any inconvenience. I know neither the cutoff of 2009 or of 2010. Whatever, we will see.

Well, they're not advancing less people even though an intermediate round has been removed...it's virtually the same: 400-600 (this is official: straight from AAPT via e-mail).  Considering the questions were harder, relative to past years, I'm guessing the cutoff will be a couple points lower (or maybe this is just me desperately hoping I made it :maybe:)

Why in the world are they disallowing discussion of the test for so long? Shouldn't teachers have turned it in by now?

When are we allowed to start discussing it?  Speculating on the cutoff is fine right?
---The teachers should've turned it in by January 30th: final allowed postmark date.

[quote="pascal12"]Firstly, learn to spell. Don't just use "uc" in the place of you see.
Secondly, don't type in all caps. It does not add anything to your post.
Thirdly, I think the cutoff will be similar to previous years. If lower, it will not be significant, maybe 1 point lower.[/quote]

don't be so mean... there is nothing wrong with spelling using shorthand and using caps...

anyway... i think the cutoff will be  around 13-13.25. just to be safe, i will not disclose my actual view of the test, but when i emailed AAPT, they said that it would be harder, so i think that the cutoff will be lower.

also, webassign says that we are not allowed to discuss the test until February  20th, but when i emailed the AAPT, they said that we will have the test back by February the 8th. so yeah...

what sucks is that the day that we are supposed to know our score is 1 day before AMC 10/12, so right now, i don't know whether to study for semifinals or AMC 10/12.

Ok, sorry, I shouldn't have been so mean, but I still stand behind my views. I'm not against the use of shorthand, just the use of spelling such as if this was a text.

Dang, 13 is pretty low, I would have thought maybe 14, high 13's.
And dang, the results get back in less than a week! I am freaking out!

I would study for AMC, because the semifinals are still in March. But do you think that the semifinals are also going to be more difficult than those of previous years?

i know for a fact that the cutoff was 13.75 and this test was clearly harder than 2008, so the cutoff will be at most 13.75 (i hope).

@pascal12, thank you for understanding :)

but aren't they cutting out quarterfinals this year?

Yes, but they're still taking the same amount of people for semifinals this year that they took for quarterfinals last year.

Here is the quotation from AAPT:
"The test was more difficult this year, but the location of the cutoff will depend on the scores of all the students who took the test.  It could be higher or lower."

Is he just being official or he actually mean it?

Who took the 2009 test? Is it harder than the test in 2008? It seems that since the cutoff of 2009 is higher, the problems should be easier. But what if it's harder and the cutoff is still higher? It could happen this year too. We will see!

I took the 2009 test. The MC might have been about as difficult as 2008's test, but the Quarterfinal round ended up being much easier.
Is the F=ma contest's cutoff set by the AAPT as a whole, or by each state division, if state divisions exist? I know that the US National Chemistry Olympiad is done this way.

Not sure...hopefully on a state basis - it would probably help my chances.

I'm guessing there is one cutoff for everyone though, considering not that many people do this olympiad...although do a lot of people, compared to the Physics Olympiad, do the Chemistry Olympiad?

I believe that Chemistry Olympiad is even less known than the Physics Olympiad.

Physics Olympiad doesn't do anything on a regional/state basis as far as I know, or at least, they don't say so on their website.

Does anybody know how they will inform us the result?

I'm guessing that aapt gives the result to the teachers.

last year, the scores came up on webassign... but that's only for people who took it online.

According to the aapt,
"February 8th is the deadline for the students answers to be received by the grader.  Semi-finalist will be posted approximately March 3rd."

[quote="cognos599"]see my latest post on the "2010 fnet = ma" topic. =). sorry about that.[/quote]


Theoretically, if we send enough electrons to the moon, we can pull the moon closer to Earth. Then, if we can find a way to DE-accelerate the moon and make it a safe touch-down, we will have a new world. And this is, maybe the realistic solution for our over populated, energy hunger near future.

When we pull the moon come closer and closer to the Earth, the moon will circle the Earth faster and faster. The attraction force between the two bodies will be equal to the centrifuge force.

Maybe we can arrange a controled touch-down in some way, then, we will have two balls that bite into each other, ocean will form around the bite ring and most sea bed will become dry land.

http://shoutthemoon.com

Any feedback?

Can someone give this idea some math view?

Off topic, but that site looks very unprofessional...

Well, just another lousy webmaster I am.

All I want is answers about the idea.

I think that Earth would look really scary having a shape like that :). And people living close to the new Moon won't be too happy about the direction of grav field in their houses, the Moon would make a huge shade during the day and won't shine during the night so it would be pretty dark, the atmosphere could do funky things near the Moon and the whole moving of oceans would probably screw the whole bioenvironment of Earth - just lots of things you can't even think of!

Are you calling me a trouble maker?  :D

Maybe when we get busy we won't fight people that much.

Assuming that you are taking this seriously and not just joking around, (well, you are serious enough to make a website out of it)

1) How exactly do electrons pull the moon to the Earth?(!)

2) If electrons can pull the moon to the earth, how do you "send" the electrons to the moon?

3) What is the point of dragging the moon closer to the Earth?

4) How does dragging the moon closer offer us a solution for our over populated energy hungry humans?

5) The moon is $7.349*10^{22} kg$.  A controlled "touchdown" is definitely out of the question.

6) Yes, the moon will circle faster and faster around the Earth, but according to you, the attraction force between the two bodies is equal to the centrifugal force.

6)a) There is no such thing as a centrifugal force - it is a centripetal force (centre-seeking force) - this so called centrifugal force is not a force at all... but merely a consequence of newton's first law.

6)b) The attraction force (provided by gravity) is indeed equal to the centripetal force.  If you do the maths using Newton's Law of Universal Gravitation, $F=\frac{GMm}{r^2}$, you'll realise that this is a VERY LARGE FORCE.

[b]That is, the moon will effectively become the largest meteorite ever to hit earth[/b]... Goodbye Earth.  Mind you, it will solve humanity's ever increasing need for energy... [u]seeing as there won't be a humanity anymore, and hence, no need for energy![/u]

Now what about the social, ethical and moral side?

7) so, where's the moon going to land?  On Iraq or America?

8)The prospect of using so much resources to complete the project will mean a VAST amount of resources diverted from other areas which would be more beneficial - say, in medical research

9) What if things go wrong?

10) I'm sure there are even religious implications for doing something like this...

11) The environmentalists would be fuming if they realised the moon was just about to wipe out thousands of entire species of plants and animals (not to mention, of course, humans).

There is probably more to say, but I'll leave it for someone else.

P.S. Interesting, bizzare (... and without offense) a stupid idea... but its always good to see a bit of scientific imagination out there.

Well, if I forget about all the problems with the "controlled touchdown" and just imagine the Moon sitting quietly in the middle of Pacific Ocean, I find it actually quite amusing to think about what would be Earth like. Just as Mr. hello1 said, the  problem with overcrowded Earth could be solved by populating now empty ocean beds since all the Earth's water would be somwhere near the Moon. Of course, living without water is somewhat problematic but anyway....

you know whats even better... this rather amusing idea will DEFINITELY solve the overpopulation problem! (at the expense of, say, 6 billion people ;) )

 :P

If we glue two balls together and spin them around mass certer, when speed up to a point, the centrifugal force will break them apart.

If we pull the moon into smaller orbit, the spin speed of the Earth will increase, when touch down, the moon's orbitting speed will equal to Earth's spin speed.

So there should be no impact.

Maybe we can pull the moon to just as close as we want, say, 500 miles away.

Okay.  I don't know why you're taking this seriously.

1)  You're talking about charging the moon.  Putting enough electrons on the moon, such that it'll attract the earth and go closer(keep in mind that the Earth is electrically neutral overall, so that'll be a lot of dipole action) would probably be, well, impossible.  What's to keep the excess electrons from being repelled from each other and flying off the moon?  How would we SEND the electrons over?  Do you have any idea how much charge it'd take to make the oribital radius of the moon that much smaller?

2)  How are you ever going to de-celerate the moon?  It's mass is 7.36 × 10^22 kilograms- you're going to need a lot of rockets (and you cant count on drag to slow it down) to do anything to that.

3) You made this web page, yes?

All right... theres too much to go for- I'll just pick one more

[quote="hello1"]
If we pull the moon into smaller orbit, the spin speed of the Earth will increase, when touch down, the moon's orbitting speed will equal to Earth's spin speed.

So there should be no impact. [/quote]

The two bodies are rotating about the system's center of mass, the barycenter.  Although they both have the same angular velocity, because the earth is 81 times as massive as the moon, it will also be 81 times close to the center of mass.  Therefor, the moon will be moving much, much faster than the earth.

Theres a lot more. gosh

hello1, you've still left a lot of questions unanswered.

p.s. there's no such thing as centrifugal force.

I am college droped out, this is just an idea that came out few days ago after few beers.

Hopefully, someone in this forum will answer all the questions in a more scientific way.  :D


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hallo guys, i was tryng to do this exercise from Herstein: find an example of group G, where there is a subgroup H and an element g of G so that $ gHg^{\minus{}1}$ is a proper subgroup of H. I've tought a way that bring me to a problem i want to propose you: Does there exist a group G that acts on a set E with this property: there are x,y in E, in the same orbit but the stabilizer of y is a proper subgroup of the stabilizer of x.? Clearly this suffices to have an example for the exercise because stabilzers of thing in the same orbit are coniugated.
p.s: 1)I've tryed some strange example, that didn't work, but on the other way i don't see a clear reason of impossibility, so i hope this question is a little bit interesting  :)

Let G be the semi-direct product of a normal subgroup N isomorphic to the additive group of rationals and complement group K isomorphic to the multiplicative group of nonzero integers, where an element of K acts as multiplication on N.

Let H be an infinite cyclic subgroup of N and g the element of K corresponding to 2.  Then gHg^-1 is an index 2 subgroup of H.

As matrices, G = { [ a, b ; 0, 1 ] : a,b in Q, a≠0 }, H = < [1,1;0,1] > = { [1,n;0,1] : n in Z }, g=[2,0;0,1], and gHg^-1 = < [1,2;0,1] > = { [1,2n;0,1] : n in Z }.

Letting G act on the cosets of H gives a permutation group with stabilizer H such that the point x=1H and y=gH are in the same orbit, but the stabilizer of x, H, properly contains the stabilizer of y, gHg^-1.


probability logarithms geometry perimeter function conics hyperbola

[color=#F8F8F8]=========================[/color][size=150]Problem Set 1[/size]
[color=#F8F8F8]===================[/color][size=150]12/03/05 through 12/11/05[/size]

1. The points of intersection of $xy=12$ and $x^2+y^2=36$ are joined in succession. Give the most detailed name of the resulting figure.

2. In a certain figure, triangle $ABC$ is inscribed in a circle $O$ with $AB=4$, $BC=8$, and $AC=9$. Segment $BY$ bisects arc $AC$ and meets the circle at $Y$. If $D$ is the point of intersection between $BY$ and side $AC$, find $DC$.

3. If $f(n)=\frac1 3 n(n+1)(n+2)$, what is $f(r)-f(r-1)$ in simplest form?

4. Mr. Williams is going for a jog. The probability that he will see a car in the next 20 minutes is $\frac{609}{625}$. If the probability of seeing a car at any time is uniform, what is the probability Mr. Williams will see a car in the next $5$ minutes?

5. Find the larger of two four digit numbers with product $4^8+6^8+9^8$.

6. In the diagram below, points $B$ and $C$ are centers of two tangent circles, points $E$ and $F$ are points of tangency, $BF=2$, and $BD=10$. Lines $AD$ and $AE$ intersect at $A$. Find $AE$.
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=426[/img]

7. If $a$ and $b$ are positive numbers for which $\log_9a=\log_{12}b=\log_{16}(a+b)$, find $\frac b a$.

8. On a circle $O$ with diameter $5$, points $A$, $B$, $C$, and $D$ lie on the perimeter of the circle, and point $X$ is on the diameter $AD$. Given that $BX=CX$ and $3\angle BAC= \angle BXC = 36^\circ$, express $AX$ in terms of trigonometric functions of acute angles.

9. An equation $x^4-16x^3+94x^2+Ax+B=0$ has two double roots. Find $2(A+B)$.

10. Find the last three digits of $3^{2006}$.

I think it is end now? Already Sunday evening... 4everwise posted the problems exactly the same time last week!

But I live in the California, so it's only lunch time.

That means people in Asia have until Monday Morning.  :lol:

[quote="eryaman"]are we allowed to use calculators?[/quote]You can use anything you want, except other people. :D Same as USAMTS.

As for solutions, you can just e-mail or pm me.  :)

I'l like to inform you all that Silverfalcon is now sponsoring this competition with me.  :)

From now on, we will each stick to designing specific categories of problems.

Me: Probability, Geometry
Silverfalcon: Algebra, Number Theory

We both work on anything else that may show up.

[hide="Problem 1 Solution"]Start by noticing that we are given a hyperbola and a circle. Because they have centers at the same point, we have symmetry at the intersection points, and our resulting figure is definately a parallelogram.

However, we see that the distance betweeen opposite intersection points is equal, so our parallelogram is a rectangle.
[/hide]
[hide="Problem 2 Solution"]Because segment $BY$ bisects the arc $AC$, we see that $BY$ is an angle bisector in triangle $ABC$. Thus, we apply the angle bisector theorem to get $DC=6$. [/hide]
[hide="Problem 3 Solution"]Plug $r$ and $r-1$ into the function. Take the equal terms and factor to get $r(r+1)$.
[/hide]
[hide="Problem 4 Solution"]The probability that Mr. Williams doesn't see a car in 20 minutes is $\frac{16}{625}$. Thus, the probability he doesn't see a car in 5 minutes is $\sqrt[4]{\frac{16}{625}}=\frac 2 5$.
Since the probability he doesn't see a car in 5 minutes is $\frac 2 5$, the probability he does is $1- \frac 2 5 = \frac 3 5$.[/hide]
[hide="Problem 5 Solution"]First, express everything in terms of 2 and 3 so that we get $2^{16}+2^83^8+3^{16}$. Then factor to obtain $(2^8+3^8)^2-(2^83^8)=(2^8+3^8)^2-(2^43^4)^2$.

Applying differences of squares, our two four digit numbers are 8113 and 5621, so the larger of these is 8113.
[/hide]
[hide="Problem 6 Solution"]Let the radius of circle $C$ be $R$ and the radius of circle $B$ be $r$, so that $10=BD=2R+r=2R+2\rightarrow R=4$. Because $R=2r$, we see that $BF$ is the midsegment of $\triangle ACE$. Apply Pythagoras to get $AE=\sqrt{AC^2-CE^2}=8\sqrt{2}$. [/hide]
[hide="Problem 7 Solution"]Let each logarithm equal x, so that $9^x=a$, $12^x=b$, and $16^x=a+b$. Thus, $16^x=12^x+9^x$.

Dividing by $9^x$, we get the quadratic $\left(\frac 4 3\right)^{2x}-\left(\frac 4 3\right)^x-1=0$ and solving it, our answer is $\frac 1 2 (1+\sqrt5)$, also known as the golden ratio.[/hide]
[hide="Problem 8 Solution"]First, we prove several triangles congruent. Draw the center $O$ and triangles OXB and OXC are congruent by SSS congruency. Next, we see that triangles BXD and CXD are congruent, by SAS congruency.
In right triangle ABD (inscribed with a side as diameter), we apply the law of sines to get $AB=5\sin 84$.
Next, in triangle AXB, we use the law of sines to get $\frac{AB}{\sin 162}=\frac{AX}{\sin 12}$. Rearranging, our answer is $AX= \frac{5 \sin12\sin84}{\sin162}= \frac{5 \sin12\sin84}{\sin18}$. [/hide]
[hide="Problem 9 Solution"]Apply your knowledge of the properties of coefficients in polynomials. If the doubled roots are $r$ and $s$, we know that $A=-(2r^2s+rs^2)=-2rs(r+s)$ and $B=r^2s^2$. We can then find $A$ and $B$ if we find $r+s$ and $rs$.

Looking at the coefficient of $x^3$, we see that $2(r+s)=16 \rightarrow (r+s)=8$. From the coefficient of $x^2$, we see that $r^2+s^2+4rs=94$. By squaring $r+s$, and then two systems of equations, we see that $rs=15$.
Now, $A=-240$ and $B=225$, so $2(A+B)=-30$.[/hide]
[hide="Problem 10 Solution"]We don't want to expand this, so there must be a better way. We see that $3^{2006}=9^{1003}=(10-1)^{2003}$ and that our problem is the same thing as finding the value of this $\pmod{1000}$.
Applying the binomial theorem and throwing away all but the last two terms, we get $-271$. However, the last three digits can't be negative, so the answer is $729$. [/hide]

Edited by mod

[hide="Faster Solution for Round 1 Number 10"]Using euler's generalizaiton of FLT, we have

$3^{2006} \equiv 3^6 \equiv 729 \mod 1000$[/hide]

[quote="Altheman"]for number 8 week 2, do you mean maximum number of triangles, because it varies on the arrangement of the points...[/quote]Yeah. Sorry for not being clear about that.

[color=#F8F8F8]=========================[/color][size=150]Problem Set 2[/size]
[color=#F8F8F8]===================[/color][size=150]12/14/05 through 12/21/05[/size]

1. Let $abc$ be three digit number such that $a$ is an even digit (0 doesn't count) and $abc$ is divisible by 45. Find the number of all such $abc$ three-digit numbers.

2. If $n$ is a randomly chosen 3-digit integer, what is the probability that both $n^2$ and $(n+100)^2$ have the same number of digits.

3. Define $f(p) = (p+1)^4$ for all positve integers. For $1 \leq p \leq 100$, find the number of values of $p$ that satisfy the inequality $f(p+1) - f(p) \geq 2006$.

4. Compute the number of sets of three distinct elements, namely $e_1,e_2,$ and $e_3$, from the set $\{3^1,3^2,3^3,...3^{2005},3^{2006} \}$ such that there exists a positive integer $d$ that satisfy $\frac{e_3}{e_2} = \frac{e_2}{e_1} = d$ given that $e_1 < e_2 < e_3$.

5. For all positive integers $a, b,$ and $n$, there is a function $g(a,b,n) = \frac{b}{a^n}$ where $n$ is the largest integer that makes $\frac{b}{a^n}$ an integer. Find the value of $\gamma$ for $g(2,2006!, \gamma)$.

6. If $\displaystyle \sum^{2006}_{n=1} \frac{1}{\sqrt[3] {n^2} + \sqrt[3] {n^2+n} + \sqrt[3] {n^2+2n+1}} = \sqrt[3] {a} - \sqrt[3] {b}$, find the value of $a+b$.

7. Suppose that 2006 points are selected on the circumference of a circle and all ${n \choose 2}$ chords connecting a pair of these points is drawn. Given that no three chords pass through the same point inside the circle, find the maximum number of triangles that are formed by portions of the chords inside the circle and do not have any of the $n$ points as vertices.

8. In triangle $ABC$, $\angle B$ is obtuse and $AB > BC$. An angle bisector of an exterior angle at A meets line $BC$ at $D$ and an angle bisector of an exterior angle at $B$ meets line $AC$ at $E$. If $AD=AB=BE$, find the measure of $\angle BAC$.

9. In the figure below, the two circles are the incircles of triangles $ADB$ and $ADC$. The incircles are tangent to $AD$ and each other at $G$. If the circles have radii of lengths $7$ and $11$, find the length of $DE$.
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=425[/img]

10. On AoPS, 4everwise and Silverfalcon built the Weekly Potpourri sequence called $W_p$. For $p \geq 1$, $\displaystyle W_p = \frac{p(p+1)}{2} \left[\sin \left(\frac{p(p+1) \pi}{2}\right) + \cos \left(\frac{p(p-1) \pi}{2} \right) \right]$. To celebrate the beginning of 2006, find the value of $\left| \sum \limits^{2006}_{p=3} W_p \right|$.

4everwise, did you get my pm???
basically it says that the polynomial one in round 1(number 7, i believe) has problems.  if we plug in $x=1$ in the equation, we get $79 + A + B = 0$, and therefore $2(A+B) = -158$.  but it contradicted your solution....

You can't assume $x = 1$ is one root!  ;)

nvm..I thought it holds true for all $x$...

Somebody requested that I post this, so I will post the top 5 scores and the score distribution. If you were unhappy about your score, don't worry because I won't post your username with your score.

The Top 5!  :D
1. Altheman
2T. Pkerichang
2T. Calc Rulz
3. Boy Soprano II
5. 1234567890

Score Distribution out of 20 points: (1 point is docked for small, but crucial errors)
20- 1
19- 2
18- 1
17- 1
16- 6
15- 0
14- 4
13- 0
12- 1
(none below 12)

You all did a very nice job. :thumbup:

And for number 1, I can't tell you the only logical possibilities. :D

Edited by the roaming moderator

You still didn't edit your post though! Click the little "edit" button at the bottom right hand corner. :)

Ahhh!!! Panic!!!!  :o

Actually, it's ok. I just edited it for him. :D

I think he thought that this is something like the problem solving marathon...

on number 5, does "circular segment" means, semi-circle arc, quarter-circle arc, or anything??? Because I think the area changes base on the arc...

And on number 6, the denominator of $f(y)$, is it $y^2 + 5y + 6$ or $y^2 + 5x + 6$???

Where are the solutions (partial solutions) for the last week potpourri?

Sorry...I'm new...
But at least I know how the contest works now  :)

What happened?? I tried deleting the poll that I posted, and it deleted the first two posts...  :?

That's not good... :dry:

[quote="4everwise"]Due to the lack of answers submitted for Round 3, I'm extending the due date for this round to January 6, 2006. :dry:  :roll:
[quote="varun"]I discovered this contest today. :P
Can I participate in it.

If yes, what I have to do. :roll:[/quote]Sure. You can participate whenever you want. Just turn in answers to the current round of problems. When there's a new round, you then submit answers to those problems. That simple. :)[/quote]

sorry...I have to finish my english video project so I didn't even get a chance to read the questions...btw, did you see my questions for the problems in the previous posts??

Reply:
5) It's tough for me to explain, but a circular segment is the area between a chord and the arc it intercepts. Did you catch that?  :?
6) They're all y terms

I'm sorry, but due to the fact that few people have been turning in answers, I'm temporarily stopping this competition. I realize that USAMTS is due, school is going on, and all that other stuff so everybody is really busy. :noo:

To those of you who enjoyed these problems, sorry. :(

I think I'll find some other way to contribute instead of running yet another competition. There are enough of those already. ;)


Can someone please explain to me why the product of the logarithms is equal to the sum of the logarithms?

I assume you mean $\log x+\log y=xy$. If so, that's a basic property of logarithms. It works because logarithms are related to exponents. If you say $\log x+\log y=\log xy$, and $\log xy=3$, then that means that $10^3=xy$. So, say $\log x=1$ and $\log y=2$. That means that $x=10$ and $y=100$ because $10^1=x$ and $10^2=y$, and, lo and behold, $xy=1000=10^3$. :)

Let $r=\log_b{x}$ and $s=\log_b{y}.$  Therefore $b^r=x$ and $b^s=y$.  Multiplying, we get \[b^r\cdot b^s=b^{r+s}=xy\Rightarrow r+s=\log_b{xy}=\log_b{x}+\log_b{y}.\]


algebra proposed
algebra
arithmetic sequence

We have n numbers in arithmetic sequence.We know that there exists 10 of then in geoemtric sequence.Prove that ,in this conditions, n must be at least 512.

First note that we may suppose that the first term of the arithmetical sequence is also the first of the geometrical one, if not we delete all the first terms of the aritmetical sequence which clearly not increases n...(and then the first term is not 0)
Moreover, wlog we may assume that the sequence is formed by positive numbers (if not replace each term a by -a), and that the sequence is increasing (if not just relabel the terms a_i in a'_(n-i) ).
Then, the sequence is a_i = a + i*r where i = 0,1,...,n-1 and r > 0.
And the geometrical one is b_i = a*q i  for i = 0,1,...,9, where q > 1 (since increasing)and  b_i = a + r*p_i with p_0 = 0 < p_1 < ... < p_9  \leq  n-1, are integers.

It is easy to see that :
(i) aq = a + r*p_1, and
(ii) a + r*p_(i+1) = q*a*q i  = q(a + r*p_i) for i = 1,...8.
Thus : p_(i+1) = p_1 + q*p_i, which leads to p_i = p_1*(q i  - 1)/(q-1).

Moreover q = (p_2 - p_1)/p_1 then q is rational.
Let q = x/y with gcd(x,y) = 1, x > y > 0.
From (ii), we deduce that p_9 = (p_1/y 8 )*(x 9  - y 9 )/(x-y).
Since gcd(y 8  ,  x 9  - y 9) = 1, and that p_9 is an integer, we deduce that y 8  divides p_1.

- If y = 1 : Then q is an integer and q > 1. So that q  \geq  2, from which we deduce that p_9 = p_1(1+q+...+q 8 )  \geq  1+2+...+2 8  = 2 9  - 1, hence n  \geq  2 9  = 512.

- If y  =  2 then x  \geq  3 and q  \geq  3/2. Moreover p_1 \geq 2 8.
The same reasoning as above leads to p_9  \geq  2 8 *(1 + 3/2 + ... + (3/2) 8 ) = 2 9 * ( (3/2) 9  - 1) > 2 9 , and we are done in that case.

- If y  \geq  3 then p_1  \geq  3 8  > 512, and we are done again.

Pierre.


"A purple cow is a confirming instance of the hypothesis that all crows are black."

What does it mean?

(Sorry if I'm posting this in the wrong forum; I don't know where to post anywhere else.)

probably either "cow" is a typo of "crow" or vice versa

See [url=http://www.geocities.com/CapitolHill/Lobby/3022/hempel.html]this page[/url] and [url=http://en.wikipedia.org/wiki/Raven_paradox#The_paradox]this one[/url].

go read AoPS volume 1 last chapter. inverses arent necessarily true.

Inverses aren't the issue here.  Read the Wikipedia article; the paradox is much more subtle.

Oh. Actually, technically in that case it is true, but it it's like a science experiment: the accepted true one is the one that has the most evidence (like in the first page: black crows are black, but all non-black and non-white non-crows prove both.)

Seeing a purple crow settles the hypothesis in the negative, but part of the argument being raised here is that a purple [b]cow[/b] can be thought of as evidence for the [i]truth[/i] of the hypothesis since it is an example of a non-black non-crow.

ah, I see. thanks -- this thought experiment is all about inductive logic rather than deductive.

well dang I am just no good with inductive logic

Wow. I read the wikipedia page for probably an hour yesterday, and didn't get all the way through it. Now I'm fascinated with this paradox. Does anyone know if this appears in any books or have a book about it that goes into a good amount of detail/history/theory/etc?

The thing is that a purple cow being evidence for the proposition that all ravens are white MAY be a confirming instance, and that's allowed. But the problem is that to prove it further, we will have to continue looking for non-white non-ravens and eventually we'll come upon a black object that just happens to be a raven, thus destroying the hypothesis that all ravens are white.


limit
vector
probability
invariant
topology
real analysis
advanced fields

Suppose that $X$ is a compact metric space and $T: X\rightarrow X$ is a continous function. Prove that $T$ has a returning point. It means there is a strictly increasing sequence $n_i$ such that $\lim_{k\rightarrow \infty} T^{n_k}(x_0)=x_0$ for some $x_0$.

I found this to be a pretty neat approach :).

Consider the locally convex vector space of real regular Borel measures on $X$, as the space of real functionals on $C(X)$, with the corresponding weak-$*$ topology. The set of probability measures on $X$ is a convex compact set, on which $T$ acts as a continuous map via $T\mu(A)=\mu(T^{-1}(A))$ for all Borel subsets $A$ of $X$. By standard fixed point theory, there is a probability measure $\mu$ on $X$ fixed by $T$, i.e. such that $T$ is measure-preserving with respect to $\mu$.

By the so-called Recurrence Theorem, for every Borel subset $A$ of $X$, the set $S_{A}$ of those $x\in A$ for which only finitely many of the points $T^{n}(x),\ n\ge 1$ belong to $A$ has measure zero.

Now let $(U_{n})_{n}$ be a countable basis of open sets for $X$. $S=\bigcup S_{U_{n}}$ then has measure zero, being a countable union of sets of measure zero, so every point $x$ of $X\setminus S$ has the property that whenever $x\in U$ ($U$ being an open subset of $X$), infinitely many of the points $T^{n}(x)$ belong to $U$. The conclusion follows, since any such $x$ is a returning point.

I remembered this while working on something similar. Here's a shorter proof:

Consider the collection of all closed subsets of $ X$ invariant under $ T$ (i.e. mapped into themselves by $ T$). By Zorn's Lemma, a member $ F$ of this collection which is minimal with respect to inclusion can be found. It follows from minimality that for every $ x\in F$, the orbit $ \{T^{n}x\ |\ n\ge 0\}$ of $ x$ is dense in $ F$, and hence every point of $ F$ is a returning point.


parameterization
number theory
least common multiple

1)Solve in $ \mathbb{R}$ this equation :

$ |[x \minus{} a]| \plus{} [|x \minus{} a|] \equal{} 1$ , where $ a$ is a real parameter .

2)Solve in $ \mathbb{R}$ the equation :

$ \left[\frac{3 \plus{} [2x]}{4}\right] \equal{} \{x\}$

3)If $ a\ge 1 \plus{} \sqrt {3}$ , show that :

$ \left[\frac {\left[\frac {na^2 \plus{} 2}{a}\right] \plus{} 2}{a}\right] \equal{} n\ ,\ (\forall) n \in \mathbb{N}$

4)Solve this equation :

$ \left[\frac{x}{2}\right] \plus{} \left[\frac{x}{3}\right] \plus{} \left[\frac{x}{4}\right] \equal{} x$ , with $ x\in \mathbb{R}$

2)
Since $ LHS$ is an integer $ \{ x\} \equal{}0 \quad x\in \mathbb{Z}$
Our equation becomes
$ \left[ \frac{3\plus{}2x}{4}\right]\equal{}0$

$ 0\le \frac{3\plus{}2x}{4} < 1$

$ x \in [ \minus{}\frac{3}{2} ,\frac{1}{2})$

1)$ |[x-a]|=1-[|x-a|] \ge 0 \Leftrightarrow [x-a]= \pm 1$ and $ [|x-a|]=0$ .......(1)

                                                             or $ [x-a]=0$ and $ [|x-a|]=1$........(2)

For (1), $ [|x-a|]=0 \Leftrightarrow -1 < x-a<1$, Combine with $ [x-a]= \pm 1$, we have that $ -1<x-a<0$ is the solution.
For (2), $ [x-a]=0 \Leftrightarrow 0 \le x-a <1$, which contradicts $ [|x-a|]=1$, so no solution in this case.

3)$ a \ge 1+\sqrt{3}$ so $ (a-(1+\sqrt{3})(a-(1-\sqrt{3})=a^2-2a-2 \ge 0 \Rightarrow \frac{2}{a} \le a-2$

$ {\left[\frac {na^2 + 2}{a}\right]={\left[na+\frac {2}{a}\right] \le {\left[na+a-2 \right]}}}$ $ \Rightarrow {\left[\frac {na^2 + 2}{a}\right]+2 \le {\left[na+a \right]}}$
$ \Rightarrow \frac {\left[\frac {na^2 + 2}{a}\right] + 2}{a} \le \frac{[na+a]}{a} \le n+1$ by multplying both sides by $ a$
Equality holds iff $ a=1+\sqrt{3}$ and $ na+a$ is an integer, which is impossible.

Also $ \frac {\left[\frac {na^2 + 2}{a}\right] + 2}{a} \ge n$
$ \Leftrightarrow \left[\frac {na^2 + 2}{a}\right] \ge an-2$, which is obviously true, done.

4)Let $ f(x)=\left[\frac{x}{2}\right] + \left[\frac{x}{3}\right] + \left[\frac{x}{4}\right]$ and $ a$ be the smallest positive integer such that $ x=12k+a$ for some non-negative integer k

The equation now becomes $ k=a-f(a)$ as $ 12$ is the $ lcm$ of $ 2,3,4$

For $ a=0,1,2,...,11$, $ a-f(a)=0,1,1,1,0,1,0,1,0,0,0,1$ respectively

This gives $ x=0,13,14,15,4,17,6,19,8,9,10,23$ as the solutions.

Micf , you have a mistake . You have a contradiction in your solution . I let you to find out alone .  :lol:

For 3) your solution is very nice stephencheng . :P

[hide="mistake of micf"] $ x$ is an integer as you had found before $ \{x\}\equal{}0$. [/hide]


calculus integration inequalities trigonometry function algebra domain

Prove the following inequalities:

(1) $ x\minus{}\sin x\leq \tan x\minus{}x\ \ \left(0\leq x<\frac{\pi}{2}\right)$

(2) $ \int_0^x \cos (\tan t\minus{}t)\ dt\leq \sin (\sin x)\plus{}\frac 12 \left(x\minus{}\frac{\sin 2x}{2}\right)\ \left(0\leq x\leq \frac{\pi}{3}\right)$

[quote="kunny"]Prove the following inequalities:

(1) $ x \minus{} \sin x\leq \tan x \minus{} x\ \ \left(0\leq x < \frac {\pi}{2}\right)$

(2) $ \int_0^x \cos (\tan t \minus{} t)\ dt\leq \sin (\sin x) \plus{} \frac 12 \left(x \minus{} \frac {\sin 2x}{2}\right)\ \left(0\leq x\leq \frac {\pi}{3}\right)$[/quote]
I don't know if my solution for #1 is right since I didn't use integrals...but
[hide="1?"]Take the function $ \tan x\minus{}x$ as $ g(x)$. Let $ x\minus{}\sin x\equal{}f(x)$. This inequality is the same as proving the minimum of $ g(x)$ is equal to the maximum of $ f(x)$. The minimum of $ g(x)$ is when $ \sec^{2} x\minus{}1\equal{}0$ and this only occurs when $ x\equal{}0$. Therefore the minimum of $ g(x)$ is $ 0$. The maximum of $ f(x)$ occurs when $ 1\minus{}\cos x\equal{}0$ or when $ x\equal{}0$ so the maximum of $ f(x)$ is $ 0$. [/hide]

1) Consider the function
$ f(x)\equal{}\tan x \plus{}\sin x \minus{}2x$ in $ [0,\pi/2)$
In the open interval $ (0,\pi,2)$, we have
$ f'(x)\equal{}\sec^2x \plus{}\cos x \minus{}2 \equal{} \dfrac{\cos x\minus{}\cos2x}{\cos^2x} \equal{}\dfrac{2}{\cos^2x}\sin\frac{x}{2}\sin\frac{3x}{2}>0$
Hence, $ f$ is strictly increasing in $ (0,\pi/2)$. As such $ f(x)>f(0)$ giving us the desired inequality. The equality obviously holds for x=0.

ckck: what you've said is wrong. The functions $ x\minus{}\sin x$ and $ \tan x\minus{}x$ are both positive and increasing on $ (0,\pi/2),$ and both have a minimum value at $ x\equal{}0.$ You turned one of them upside down, and in so doing didn't prove anything.

Taylor's Theorem can be used.

Let $ f(x)\equal{}\sin x$ and $ g(x)\equal{}\cos x.$ Both are at least five times differentiable in the domains we need.

$ f(x)\equal{}x\minus{}\frac{x^3}6\plus{}\frac{f^{[5]}(\xi)x^5}{5!}.$

$ g(x)\equal{}x\plus{}\frac{x^3}3\plus{}\frac{g^{[5]}(\xi)x^5}{5!}.$

In the case of the sine, we have $ f^{[5]}(x)\equal{}\cos x>0$ on $ (0,\pi/2)$ so we conclude that $ \sin x>x\minus{}\frac{x^3}6$ or $ x\minus{}\sin x<\frac{x^3}6$ on that interval.

In the case of the tangent: each derivative of $ \tan x$ is a linear combination with positive integer coefficients of terms of the form $ \sec^ax\tan^bx$ for various positive integer powers $ a$ and $ b.$ We conclude that $ g^{[5]}(x)>0$ for $ x\in(0,\pi/2).$ We conclude that $ \tan x>x\plus{}\frac{x^2}3$ or $ \tan x\minus{}x>\frac{x^2}3.$

Hence for $ 0<x<\frac{\pi}2,$ we have

$ x\minus{}\sin x<\frac{x^3}6<\frac{x^3}3<\tan x\minus{}x.$

(I assume kunny is going to say something about whether or not Taylor's theorem should be available as a tool to apply to this. I don't care.)

For 2) we have
$ \begin{array}{*{20}{c}}
   {0 \le x \minus{} \sin x \le \frac{\pi }{3} \minus{} \frac{{\sqrt 3 }}{2}}  \\
   {0 \le \tan x \minus{} x \le \sqrt 3  \minus{} \frac{\pi }{3}}  \\
\end{array}$
That means $ x \minus{} \sin x,\tan x \minus{} x \in \left[ {0,\frac{\pi }{2}} \right)$
since cos is strictly decreasin at this interval we have
$ x \minus{} \sin x \le \tan x \minus{} x \Rightarrow \cos \left( {x \minus{} \sin x} \right) \ge \cos \left( {\tan x \minus{} x} \right)$
if and only if $ \cos \left( {\tan x \minus{} x} \right) \le \cos x \cdot \cos \left( {\sin x} \right) \plus{} \sin x \cdot \sin \left( {\sin x} \right)$
we also know that $ \left| {\sin \left( {\sin x} \right)} \right| \le \left| {\sin x} \right|$ and for x belonging on [0,π/3] we have $ \sin \left( {\sin x} \right) \le \sin x$

So our inequality becomes
$ \cos \left( {\tan x \minus{} x} \right) \le \cos x \cdot \cos \left( {\sin x} \right) \plus{} \sin x \cdot \sin \left( {\sin x} \right) \le \cos x \cdot \cos \left( {\sin x} \right) \plus{} \sin x \cdot \sin x,x \in \left[ {0,\frac{\pi }{3}} \right]$

Integrating gives
$ \begin{array}{l}
 \int\limits_0^x {\cos \left( {\tan t \minus{} t} \right)dt}  \le \int\limits_0^x {\left[ {\cos t \cdot \cos \left( {\sin t} \right) \plus{} {{\sin }^2}t} \right]} dt \Leftrightarrow  \\
 \int\limits_0^x {\cos \left( {\tan t \minus{} t} \right)dt}  \le \sin \left( {\sin x} \right) \plus{} \int\limits_0^x {{{\sin }^2}tdt}  \Leftrightarrow  \\
 \int\limits_0^x {\cos \left( {\tan t \minus{} t} \right)dt}  \le \sin \left( {\sin x} \right) \plus{} \int\limits_0^x {\frac{{1 \minus{} \cos \left( {2t} \right)}}{2}dt}  \Leftrightarrow  \\
 \int\limits_0^x {\cos \left( {\tan t \minus{} t} \right)dt}  \le \sin \left( {\sin x} \right) \plus{} \frac{x}{2} \minus{} \frac{{\sin \left( {2t} \right)}}{4} \\
 \end{array}$


It isn't horrible, it is very nice like all your inequalities.
I find it very difficult to solve.
I hope you will post as soon as possible the proof.

Thank you very much, Harazi.

Ok, I'm very busy for a few weeks, but I will post a solution for this one.
   Write the condition a^9*c^4/b^6+b^9*a^4/c^6+c^9*b^4/a^6=1. Put
x=a^9*c^4/b^6 and so on. Then we must show that for x+y+z=1 we have x^3*y^2+y^3*z^2+z^3*x^2<108/3125. Let x=max{x,y,z}. Since
x^3*(y+z)^2=x^3*y^2+y^3*z^2+z^3*x^2+(x^3-y^3)*z^2+x^3yz+(x^3yz-x^2z^3)>=x^3y^2+y^3z^2+z^3x^2, it is enough to prove that
 x^3(y+z)^2<=108/3125. But 1=x/3+x/3+x/3+(y+z)/2+(y+z)/2 and apply AM-GM.


Find the three largest primes of $ 555,555,555,555.$ I just want to know how you got the answer from a step.

Thanks

Do you mean the three largest prime [i]divisors[/i]?

Clearly, it is divisible by 3 and 5. $ \frac{555555555555}{15}\equal{}37037037037$, hence 37 is also a divisor. $ \frac{37037037037}{37}\equal{}1001001001$. This is divisible by $ 1001\equal{}7\times11\times13$. $ \frac{1001001001}{1001}\equal{}1000001$, which is divisible by 101 and 9901. It is not difficult to check that 101 and 9901 are both primes. Therefore $ 555555555555\equal{}3\cdot5\cdot7\cdot11\cdot13\cdot37\cdot101\cdot9901$. Thus the three largest prime divisors are 37, 101, 9901


$\triangle{ABC}$. $D$, $E$ are on $BC$. $AD || EG$($G$ is on the extension of $BA$), $EG$ meets $AC$ at $F$. If $EF+EG=2AD$, prove $AD$ is the median of $\triangle{ABC}$.

[hide]
Let $M$ be the midpoint of $FG$. Then $EF+EG=2EM$, so we have $AD=EM$. Since $AD\parallel EM$, $ADEH$ is a parallelogram.

From here, we can play around with area ratios.
Suppose $\frac{MF}{EF}=\frac{a}{b}$.
Then we can let $\triangle AFM=\triangle AGM=a^2$ and $\triangle CEF=b^2$.

Also, we have $\triangle AEF=ab$

[hide="Reason"]$\frac{\triangle AFM}{\triangle AEF}= \frac{\triangle AEF}{\triangle CEF} =\frac{MF}{FE}=\frac{a}{b}$[/hide]
So $\triangle ADE=\triangle AFM+\triangle AEF=a^2+ab$
In addition, since $\frac{GA}{AB}=\frac{ED}{DB}=\frac{a}{a+b}$, we have $\triangle ABD=(a+b)^2$.

Therefore, $\triangle ACD=\triangle ADE+\triangle AEF+\triangle CEF= a^2+ab+ab+b^2=(a+b)^2=\triangle ABD$, and thus $BD=DC$.[/hide]


[quote="t0rajir0u"][hide="Big hint"] Let $x = y-2$. [/hide][/quote]
 :wallbash:

I was thinking too hard.


[quote="airelemental135"]Let f(x)=2x+3. Find values for a and b such that the equation f(ax+b)=x is true for all values of x.[/quote]

[hide]
f(ax+b)=2(ax+b)+3=2ax+2b+3
2ax+2b+3=x
2b=-3
b=-1.5
2ax=x
2a=1
a=0.5[/hide]

How did you get from 2ax+2b+3=x to 2b=-3?

[quote="airelemental135"]Can somebody explain how he did that?[/quote]
Equating the coefficients. This is way below Intermediate Topics discussion. Can a mod move this please?

[quote="deej21"]I don't think this is really intermediate topics... anyway airelemental, you can't expect a reply within 5 minutes.

For the value to be equal for all values of x, the constant (2b + 3) must be 0.[/quote]

In general

 if the polynomials p(x) and q(x) (both of the same degree say n) are such that

 p(x)  =  q(x) for all values of x

then the corresponding co-efficients of the like powers of x in p and q are the same.

This is because any polynomial eqn. of degree n has exactly n roots (The fundamental theorem of algebra).These roots can again be all distinct  or all equal or mixed.

The equation p(x) = q(x) , when transformed becomes

 p(x) - q(x) = 0

 => A(x) = 0 where A(x) is a polynomial of degree n with coefficients equal to  pk - qk for k = 0,1,2,...n-1

 Now , since A(x) = 0 can't have more than n distinct roots,
so if A(x) = 0 holds for all values of x(real or complex), the coefficients of A(x) have to equal zero

In other words,
  pk = qk for k = 0,1,2,...n-1

U can try proving this result for small values of n ,say for example 2 when the equation reduces to a quadratic.


geometry
circumcircle
analytic geometry
geometry unsolved

In the plane, let $ A$ be the fixed point and $ d$ be the fixed line ($ A$ doesn't belong to $ d$). $ M,N$ are two points varying on line $ d$ such that $ \large \angle MAN$ is constant. Prove that the circumcircle $ \large AMN$ is always tangent to a fixed circle.

Let $ I$ be the center of circle $ (AMN)$ and $ X$ be the feet of perpendicular line from $ I$ to $ d$. We have:
$ \frac {IA}{IX} \equal{} \frac {IM}{IX} \equal{} \frac {1}{|cos\widehat{MAN}|} \equal{} k > 1$
Then locus of $ I$ is a hypecbol with $ 2$ focus $ A$ and $ B$.
Then $ |IB \minus{} IA| \equal{} R > 0:$const
Last circle $ (AMN)$ tangent with circle center $ B$ radius $ R$.

Sorry because I can't give a synthentic solution for your problem!

PS: 1) If $ \widehat{MAN} \equal{} 90^0$ then this problem is incorrect!
      2) If you want to show exactly $ (B;R)$ you can use coordinate system!

Consider $ f$ is the inversion has $ A$ to pole. The inversive image of $ d$ is a circle $ (O)$ passes through $ A$. The inversive image of $ (AMN)$ is a line $ M'N'$. Note that $ M',N' \in (O)$ and $ \angle MAN \equal{} constant \Rightarrow M'N' \equal{} constant$. So $ M'N'$ is a chord has fixed length, it always tangent to a fixed circle which has $ O$ to center ...


Given a point P and two lines g and h. Construct a line u through the point P such that if E and F are the points of intersection of this line u with g and h, then the point P is the midpoint of the segment EF.

I want to see as many different solutions as possible. I myself have a very freaky one ;-)

  Darij

I don't discuss the case g//h.

If O=g.h let Q be a point on OP so that OQ=2OP. Trace from Q the pallalels to g and h and call R and S the point of intersection between those and g and h. Is easy to see tha the line RS solve the problem.

The problem is equivalent to the following.

Construct a triangle of which are given the median from a vertex and the two angles in which the median divide the angle a that vertex.

We can before, easily construct (using a locus of points from where a fixed segment is "seen" by a fixed angle) a similar triangle to that requested and from this scaling in the sense is necessary we have the solution.

anhoter way, if O=g.h, is to construct(*)  a forth line i so that {u,g,OP,l} is harmonic then u is the parallel to l by P.

one of possible mode to do that is to take a point A on g and trace a line that cuts OP and h in K and C. Then take a point B on h and call D=BK.g and H=DC.AB. l=OH il the fourth harmonic wrt g,OP,h.

[quote="grobber"]I don't know if it's been posted before, but we could construct the symmetric of $g$ wrt $P$, see where it intersects $h$, and voila! :)[/quote]

That's the same as sprmnt21's second solution:

[quote="sprmnt21"]Anhoter way is to trace from P a generic line p which cuts g in R.

If R' is the simmetric of R wrt P, trace from R' the parallel to g, which cut h in Q. The line PQ solve the problem.[/quote]

Thanks for the many trivial solutions. I was a complete idiot, I thought the problem was hard! So let me finally tell my ridiculous freaky construction:

Let the perpendicular from the point P to the line g meet the line g at A and the line h at A'. Let the perpendicular from the point P to the line h meet the line h at B and the line g at B'. If O is the midpoint of the segment A'B', then the perpendicular to the line OP at the point P is the required line u.

Why? Well, by the butterfly theorem applied to the circle with diameter A'B' (whose the center is O, and AB is a chord in this circle)...

I was a real idiot... :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:

  Darij


Describe all solutions to:

$ 2^a\plus{}3^b\equal{}2^c\plus{}3^d$, where $ a\neq c$, $ b\neq d$.

Let's assume $ a > c$
$ \Longrightarrow d > b$
Now,
$ 2^a \minus{} 2^c \equal{} 3^d \minus{} 3^b$
$ 2^c\left(2^{a \minus{} c} \minus{} 1\right) \equal{} 3^b\left(3^{d \minus{} b} \minus{} 1\right)$
If $ c\neq0$ and $ b\neq0$, as $ 2^c$ and  $ 3^b$are coprime,
$ 2^c \equal{} 3^{d \minus{} b} \minus{} 1$ and $ 3^b \equal{} 2^{a \minus{} c} \minus{} 1$
$ 2^c \plus{} 2^{a \minus{} c} \minus{} 1 \equal{} 3^{d \minus{} b} \minus{} 1 \plus{} 3^b$
$ 2^c \plus{} 2^{a \minus{} c} \equal{} 3^{d \minus{} b} \plus{} 3^b$,which is absurd.
So we know $ b \equal{} 0 \lor c \equal{} 0$(It can't be both because $ 2^a\neq3^d$).
If $ c \equal{} 0$,
$ 2^a \minus{} 1 \equal{} 3^d \minus{} 3^b$
$ odd \equal{} even$, absurd.
$ \Longrightarrow b \equal{} 0$
Hmm, I don't know what to do next. I'll post when I find something new.

[quote="Mithril"]$ 2^c\left(2^{a \minus{} c} \minus{} 1\right) \equal{} 3^b\left(3^{d \minus{} b} \minus{} 1\right)$
If $ c\neq0$ and $ b\neq0$, as $ 2^c$ and  $ 3^b$are coprime,
$ 2^c \equal{} 3^{d \minus{} b} \minus{} 1$ and $ 3^b \equal{} 2^{a \minus{} c} \minus{} 1$ [/quote]

Wrong  :(