I have heard and read a lot of things about time, but my question remains unanswered: [b]What is time?[/b] Can anyone '[b]define time[/b]'? I dont understand, Isnt time just a human convention for convenience? :?:

[quote="Aashay THE ROKR"]We say that time has passed when there is a change(even as slight as at the nuclear level)in the structure of the universe.[/quote]

This may be very stupid :oops: , but still:
How do you define "Change"?
What I thought was something like all other 'things' being in rest with respect to a particular frame.
But consider that the universe consisted only of two particles moving with same velocities in the same direction.Whichever particle, you take to be your reference frame, the whole universe would look static. So, according to your definition, time would never flow(pass) in that universe, unless, the frame is outside the universe, which sounds absurd...

Also, another doubt:It is said that the universe is fast expanding.How is the 'boundary' of the universe visualized? Is it just the farthest most objects?

Thanks Aashay.
I just saw in Feynman Lectures about time:

"Within the past few years, just such a technique was used to measure the
lifetime of the pð°-meson. By observing in a microscope the minute tracks left in
a photographic emulsion in which p°-mesons had been created one saw that a
p°-meson (known to be travelling at a certain speed nearly that of light) went a
distance of about 10-7 meter, on the average, before disintegrating. It lived for
only about 10~16 sec. It should be emphasized that we have here used a somewhat
different definition of "time" than before. So long as there are no inconsistencies
in our understanding, however, we feel fairly confident that our definitions
are sufficiently equivalent.

By extending our techniques—and if necessary our definitions—still further
we can infer the time duration of still faster physical events. We can speak of the
period of a nuclear vibration. We can speak of the lifetime of the newly discovered
strange resonances (particles) mentioned in Chapter 2. Their complete life occupies
a time span of only 10-24 second, approximately the time it would take light
(which moves at the fastest known speed) to cross the nucleus of hydrogen (the
smallest known object).
What about still smaller times? Does "time" exist on a still smaller scale?
Does it make any sense to speak of smaller times if we cannot measure—or
perhaps even think sensibly about—something which happens in a shorter time?
Perhaps not. These are some of the open questions which you will be asking and
perhaps answering in the next twenty or thirty years."

Of course time exists on much smaller scales.This in fact is the goal of 21st century.We want to probe time scale as small as 10^-43 seconds after the big bang.This is the main aim of the large hadron collider,to detect the 'Higgs Boson',a theoretical particle with very very small lifespan.If we are successful in detecting it,then we may know what happened 'immediately' after the big bang which is called the 'Planck Time'.We will be a step closer to 'The theory of everything'.

Time can be simply defined as

Time is God's way of keeping things from all happening at once.....

Its short and sweet :P

Einstein in one of his books wrote that in physics giving a way of measuring something is as good as a definition – and we certainly all know how to measure time using lenght and the constant velocity of light.
But in a philosophical view, i believe time is simply the change in the system.

Time is God's way of keeping things happening apart.........a short and sweet one!! :D

[quote="Aashay THE ROKR"]@arjunarul
It is not necessary that we should be able to percieve the 'change'.There is always something happening,e.g.lowering of temperature of universe,radiation coming,expansion,etc.It is not necessary that if a thing we cannot percieve does not actually happen.
About boundary condition,ofcourse the boundary is the farthest objects.Vacuum cannot have a boundary.See in the Astronomy forum also for more discussion.[/quote]

Even though it doesn't have a boundary, it can still expand.

The universe isn't expanding [i]into[/i] anywhere, it is just expanding. The distance between any two points in the universe increases with time, but there is no boundary.


function
calculus
derivative
calculus computations

Suppose u and v are functions of x that are differentiable at x=0 and that u(0)=5, u'(0)=-3, v(0)=-1, v'(0)=2.

Find the values of the following derivatives at x=0.

a)d/dx(uv)
b)d/dx(u/v)
c)d/dx(v/u)
d)d/dx(7v-2u)

Well for the first one

$(uv)' = u'v+uv' = (-3)(-1)+5(2) = 13$

Just use the quotient rule for the next 2.


I suppose it's assumed that both functions are invertible.
Taking f of both sides, we have $ f(g(x^5))\equal{}2x\plus{}7$. Plug in $ \sqrt[5]{a}$ for $ x$ to get $ f(g(a))\equal{}2\sqrt[5]{a}\plus{}7$, then plug in $ g^{\minus{}1}(b)$ for $ a$ to get $ f(b)\equal{}2\sqrt[5]{g^{\minus{}1}(b)}\plus{}7$, or choice a.

[quote="Humbleman"]If $ f^{ \minus{} 1}(2x \plus{} 7) \equal{} g(x^5)$, then .......

a)   $ f(x) \equal{} 2\sqrt [5]{g^{ \minus{} 1}(x)} \plus{} 7$
b)   $ f(x) \equal{} 2[g^{ \minus{} 1}(x)]^5 \plus{} 7$
c)   $ f(x) \equal{} 2\sqrt [5]{g^{ \minus{} 1}(x)} \minus{} 7$
d)   $ f(x) \equal{} 2[g^{ \minus{} 1}(x)]^5 \minus{} 7$

How did u conclude that?[/quote]
    Assuming that the function is indeed invertiable we take $ f$ of both sides so we know have $ f(f^{\minus{}1}(2x\plus{}7))\equal{}f(g(x^5))$ or $ 2x\plus{}7\equal{}f(g(x^5))$ let $ x$ be $ \sqrt[5]{x}$ so we get...
$ 2\sqrt[5]{x}\plus{}7\equal{}f(g(x))$ according to what we know about inverse functions,specifically $ g(y)\equal{}x$ and $ g^{\minus{}1}(x)\equal{}y$ we can plug in $ g^{\minus{}1}(x)$ to simplify to get $ f(x)$ so we have $ f(g(g^{\minus{}1}(x)))\equal{}\boxed {f(x)\equal{}2\sqrt[5]{g^{\minus{}1}(x)}\plus{}7}$ or A!


algebra
polynomial
induction
algebra solved

Find all polynomials $p(x)$ such that:
\[ (x+1) \cdot p(x-1)+(x-1) \cdot p(x+1)=2x \cdot p(x) \] for all real $x$.

I have a solution about this:
the first: $ (x+1) \cdot p(x-1)+(x-1) \cdot p(x+1)=2 \cdot x \cdot p(x) $ iff $ (x+1) \cdot (p(x)-p(x-1))=(x-1)\cdot(p(x+1)-p(x)) $
Let $q(x)=p(x)-p(x-1)$, then we have $(x+1)\cdot q(x)=(x-1)\cdot q(x+1)$.

I think we can proceed this way:

$x=0\Rightarrow p(-1)=p(1)$
$x=1\Rightarrow p(0)=p(1)$
$x=-1\Rightarrow p(0)=p(-1)$

Now, I'm stuck. :(

Here's hopefully a more detailed solution...

[color=blue]Problem: Find all polynomials $p(x)$ with real coefficients such that
\[ (x+1)p(x-1) + (x-1)p(x+1) = 2x p(x)  \]
for all real $x$.[/color]

Rewrite the condition as

$(x+1)(p(x) - p(x-1)) = (x-1)(p(x+1) - p(x))$
$\iff (x+1)q(x) = (x-1)q(x+1)$
where $q(x) = p(x) - p(x-1)$

Note that $q(-1) = q(1) = 0$ and if $q(r) = 0$ ($r\neq \pm 1$), then $q(r \pm 1) = 0$ so $q(x)$ would have infinitely many roots.  Thus $q(x) = C(x^2-1)$ for some constant $C$.  Note that $p(x)$ satisfies the original condition iff $Cp(x)$ does so we may WLOG say that $p(x)$ is monic.  This implies

$p(x) - p(x-1) = x^2-1$ for all real $x$ so $p(x)$ has degree at most $3$ (this isn't too hard to verify).  Let

$p(x) = ax^3+bx^2+cx+d$ and we have

$p(x) - p(x-1) = a(3x^2-3x+1) + b(2x-1) + c = x^2 - 1$ so

$a = 1/3, b = 1/2, c = -5/6, d\in \mathbb{R}$ (and all solutions scaled by $C$)

ehh i think u have a few minor slipups?

first... we dont know that $q(-1) = 0$, rather, we know that $q(1) = q(0) = 0$

$q(-1) = p(-1) - p(-2) = -p(-2)$ but we know nothing of $p(-2)$ (plus, checking your $q(x) = C(x^2-1)$ in the original $(x+1)q(x) = (x-1)q(x+1)$ shows that it is not correct)

hence, using an adjusted form of ur reasoning, $q(x) = Cx(x-1)$

checking this in $(x+1)q(x) = (x-1)q(x+1)$ we see that this holds

then, again using your reasoning since $p(x)$ has degree at most $3$ (this is from setting $p(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0$, expanding, and noting that the $x^{n-1}$ terms don't cross out), we know

$p(x) - p(x-1) = x(x-1) = x^2-x$

setting $p(x) = ax^3 + bx^2 + cx + d$, we get:

$p(x) - p(x-1) = a(3x^2 - 3x + 1) + b(2x-1) + c = x^2 - x$

equating coefficients, we get:

$a = \frac{1}{3}$, $b = 0$, and $c = - \frac{1}{3}$ (again, all scaled by $C$)

so:

$p(x) = \frac{1}{3}x^3 - \frac{1}{3}x + d$ and letting $x = 0$, we know that since $p(0) = 0$, then $d= 0$ and finally that:

$p(x) = \frac{1}{3}x^3 - \frac{1}{3}x$ is a solution (checking we see that it holds)

also, i dont know if we have to include the fact, from $(x+1)q(x) = (x-1)q(x+1)$, when $q(x) = 0$, in which case we know $p(x) = p(x-1) = k$ for some constant $k$

checking this, we see that when $p(x)$ is a constant our original expression holds as well

thus we have our two solutions for $p(x)$?

Yeah oops.

Well... $d$ can be anything so that includes both of your solutions, scale it, and voila

Clearly, if p(x) is a solution, then p(x)+a is also a solution where a is some constant. So we can assume that p(0) = 0. Plugging 1 and -1 into the given will yield that p(1)=p(-1) = 0 as well so let p(x) = x(x-1)(x+1)q(x) and we get (x-2)q(x-1)+(x+2)q(x+1)=2xq(x). once again if q(x) is a solution, then so is q(x)+a for some constant a so assume q(-2) = 0. clearly q(-3) is also 0 which implies that q(-4) is 0 and induction can be used to show that q(-n) = 0 for arbitrarily large n so q has an infinite number of 0s and is thus the 0 polynomial. However we adjusted q by a constant so we have q(x) = c for some constant c which tells us that $\fbox{p(x) = cx(x-1)(x+1)+d}$ where the constant d comes from the fact that we shifted around p(x) by a constant factor at the start.


calculus
integration
trigonometry
limit
real analysis
real analysis unsolved

Admitting that:
$ \int_{0}^{\plus{}\infty} \frac{sin(t)}{t}  \equal{} \int_{0}^{\plus{}\infty} \frac{sin^2(t)}{t^2} \equal{}\frac{\pi}{2}$

Prove that this integral is convergent and compute it
$ \int_{0}^{\plus{}\infty} (\int_{x}^{\plus{}\infty} \frac{sin(t)}{t})dx$

NB: you may solve it without using the admitted result

The first integral may alternatively be computed using contour integrals.  I'll do your second integral first, assuming the first one.

$ \int_{0}^{\infty}\frac{\sin^{2}{x}}{x^{2}}\,dx=\sin^{2}{x}\left(\frac{-1}{x^{2}}\right)\bigg|_{0}^{\infty}+\int_{0}^{\infty}\frac{2\sin{x}\cos{x}}{x}\,dx=\int_{0}^{\infty}\frac{\sin{2x}}{x}\,dx$.
Set $ u=2x$, $ du=2\,dx$, so that $ du/u=dx/x$.  This transforms the integral into $ \int_{0}^{\infty}\frac{\sin{u}}{u}\,du$, which we know to be $ \frac{\pi}{2}$.

Here's a sketch for the first integral:
1) Use integration by parts to show that
$ \int_{a}^{b}\frac{\sin{x}}{x}\,dx=\frac{\cos{a}}{a}-\frac{\cos{b}}{b}-\int_{a}^{b}\frac{\cos{x}}{x^{2}}\,dx$,
and conclude that $ \int_{0}^{\infty}(\sin{x})/x\,dx$ exists by taking limits.
2) Show that $ \int_{0}^{\pi}\frac{\sin(n+\frac{1}{2})t}{\sin\frac{t}{2}}\,dt=\pi$ (this comes from the $ \sum_{i=1}^{n}\sin(ix)$ formula).
3) Prove that $ \lim_{\lambda\rightarrow\pi}\int_{0}^{\pi}\sin(\lambda+\frac{1}{2})t\left(\frac{2}{t}-\frac{1}{\sin\frac{t}{2}}\right)\,dt=0$ (use the Riemann-Lebesgue lemma).
4) Use the substitution $ u=(\lambda+\frac{1}{2})t$ and part (2) to reach the conclusion.

I'll say at the outset that FOURRIER's primary problem is so far from absolute convergence that extreme care must be taken with almost any possible manipulation, such as interchanging orders of integration. In fact, I haven't yet totally convinced myself that it converges - I think it does, but it's delicate.

Yes, it converges. You definitely can't interchange the order right away, as the interchanged version diverges.

Start with integration by parts on the inner integral: $ \int_x^{\infty}\frac{\sin t}{t}\,dt\equal{}\left[\frac{\minus{}\cos t}{t}\right]_x^{\infty}\minus{}\int_x^{\infty}\frac{\minus{}\cos t}{\minus{}t^2}\,dt$

We have
$ I\equal{}\int_0^{\infty}\int_x^{\infty}\frac{\sin t}{t}\,dt$
$ I\equal{}\int_0^{\infty}\left(\frac{\cos x}{x}\minus{}\int_x^{\infty}\frac{\cos t}{t^2}\,dt\right)\,dx$

Near $ x\equal{}0$, $ \int_x^{\infty}\frac{\sin t}{t}$ is bounded, and so we don't have to worry about convergence there. To handle large $ x$, we integrate by parts again:
$ \int_x^{\infty}\frac{\cos t}{t^2}\,dt\equal{}\left[\frac{\sin t}{t^2}\right]_x^{\infty}\plus{}\int_x^{\infty}\frac{2\sin t}{t^3}\,dt$
$ I\equal{}\int_0^{\infty}\left(\frac{\cos x}{x}\plus{}\frac{\sin x}{x^2}\minus{}\int_x^{\infty}\frac{2\sin t}{t^3}\,dt\right)\,dx$

Since $ \int_a^{\infty}\int_x^{\infty}\frac1{t^3}\,dt\,dx\equal{}\int_a^{\infty}\frac1{2x^2}\,dx\equal{}\frac1{2a}<\infty$, the double integral portion converges absolutely (away from zero, at any rate). The integral of the other terms converges conditionally, and $ I$ exists. Now, we interchange:
$ I\equal{}\lim_{a\to 0^\plus{}}\int_a^{\infty}\left(\frac{\cos x}{x}\plus{}\frac{\sin x}{x^2}\minus{}\int_x^{\infty}\frac{2\sin t}{t^3}\,dt\right)\,dx$
$ I\equal{}\lim_{a\to 0^\plus{}}\left(\int_a^{\infty}\frac{\cos x}{x}\plus{}\frac{\sin x}{x^2}\,dx\minus{}\int_a^{\infty}\int_a^t\frac{2\sin t}{t^3}\,dx\,dt\right)$
$ I\equal{}\lim_{a\to 0^\plus{}}\left(\int_a^{\infty}\frac{\cos x}{x}\plus{}\frac{\sin x}{x^2}\,dx\minus{}\int_a^{\infty}\frac{2(t\minus{}a)\sin t}{t^3}\,dt\right)$
$ I\equal{}\lim_{a\to 0^\plus{}}\int_a^{\infty}\left(\frac{\cos x}{x}\plus{}\frac{\sin x}{x^2}\minus{}\frac{2(x\minus{}a)\sin x}{x^3}\right)\,dx$
$ I\equal{}\lim_{a\to 0^\plus{}}\int_a^{\infty}\frac{x^2\cos x\minus{}x\sin x\plus{}2a\sin x}{x^3}\,dx$

We would like to eliminate $ a$, so we estimate. $ \int_0^{\infty}\frac{\sin x\minus{}x}{x^3}\,dx$ converges absolutely, so $ \int_a^{\infty}\frac{\sin x}{x^3}\,dx\equal{}\int_a^{\infty}\frac{1}{x^2}\,dx\plus{}\int_a^{\infty}\frac{\sin x\minus{}x}{x^3}\,dx\equal{}\frac1a\plus{}O(1)$ and
$ I\equal{}\lim_{a\to 0^\plus{}}\left(2a\left(\frac1a\plus{}O(1)\right)\plus{}\int_a^{\infty}\frac{x^2\cos x\minus{}x\sin x}{x^3}\,dx\right)$
$ I\equal{}\lim_{a\to 0^\plus{}}\left(2\plus{}O(a)\plus{}\int_a^{\infty}\frac{x\cos x\minus{}\sin x}{x^2}\,dx\right)$
$ I\equal{}2\plus{}\int_0^{\infty}\frac{x\cos x\minus{}\sin x}{x^2}\,dx$
Now, $ \frac{d}{dx}\frac{\sin x}{x}\equal{}\frac{x\cos x\minus{}\sin x}{x^2}$, so $ \int_0^{\infty}\frac{x\cos x\minus{}\sin x}{x^2}\,dx\equal{}\left[\frac{\sin x}{x}\right]_0^{\infty}\equal{}\minus{}1$, and $ I\equal{}1$.

[quote="jmerry"]Now, we [u]interchange[/u]:
$ I \equal{} \lim_{a\to 0^ \plus{} }\int_a^{\infty}\left(\frac {\cos x}{x} \plus{} \frac {\sin x}{x^2} \minus{} \int_x^{\infty}\frac {2\sin t}{t^3}\,dt\right)\,dx$
$ I \equal{} \lim_{a\to 0^ \plus{} }\left(\int_a^{\infty}\frac {\cos x}{x} \plus{} \frac {\sin x}{x^2}\,dx \minus{} \int_a^{\infty}\int_a^t\frac {2\sin t}{t^3}\,dx\,dt\right)$[/quote]

Idon't understand this line because I still don't know the rules of interchanging infinite integrals, but your method proves the convergence, is there another way to the computation, or we must necessary use the interchange?

There are two standard sufficient conditions: we can interchange the order when either the integrand is positive or the integral converges absolutely. Keywords: Fubini's theorem and Tonelli's theorem. In this case, the piece I interchanged converges absolutely. We can separate $ \int_a^{\infty}\left(f(x) \plus{} \int_x^{\infty}g(x,t)\,dt\right)\,dx \equal{} \int_a^{\infty}f(x)\,dx \plus{} \int_x^{\infty}g(x,t)\,dt\,dx$ by simple convergence of the two terms, and then switch the order for the second term by absolute convergence.

Proving convergence isn't exactly prt of the computation, but it's necessary to justify the manipulations. Is there another way to compute it? Maybe. I didn't find one, but I didn't look very hard either. There is a general principle of problem solving that you almost always want to switch the order when a double sum or integral appears.


Sign up for tourney three. If you know you will not come on every day or every other day PLEASE DO NOT SIGN UP. Inactive people that sign up for tournaments are very annoying.

post your highest rating and also confirm that you have played 50 games.

I'm hoping to have 16 people signup, but that seems too much. If not enough people sign up it will be an eight person tourney.

The tourney will be very simple. It will be single elimination and will function just like the NCAA basketball tourney. 1vs 16, 2 vs 15 and so on.

All rounds and games will be best 2 out of 3 cds until the sem-finals. The semi finals will be 3 out of 5 and the final will be be 4 out of 7.

RESULTS:

1st Place. AIME15 :first:
2nd Place. isabella2296 :winner_second:
3rd Place. r15s11z55y89w21 tied with David2143 :winner_third:

Ok i join my highest is 1446
(although i'm currently failing)

If you're highest rating changes during signup you can inform me before I say signups are over.

Also cgyao, please confirm you have played 50 games.

1664

Confirmation: Games Played Leaderboard

2053

Confirmation: very bottom of games played leaderboard :D

1618 (It'll keep on changing)

played 100+ games

1549 is my highest rating.
I confirm the 50+ games played for me.

ok 5 people so far. I'll let signups go until monday morning or afternoon. Its there is less than 16 people but more than 8 I'll just let the first 8 signers be in the tourney.

/me joins.

highest rating: 1455

confirm: 50+ games

Highest rating> everyone who is not AIME15

Games played: Uhhhhhhhhhh....hmmmmm

Motivation for playing: Why not play in the tourney I was the catalyst of?

I'm joining!!!!
My highest rating is 1201. :blush:
Hope its fun!!!!!!!!!!!!!!!!!! :P
Also I played more than 200 games! :o

Eight people have signed up!. Hopefully we can get 16. Again, signups go on until monday. If we don;t reach 16 people, these 8 will be in the tournament, and the extras will be replacement.

If we don't reach 16( I seriously doubt we have a change at all), this would be the bracket. STILL SIGN UP.

#1 Aime15 vs. #8 Pink Panther

 #4 David2143 vs. #5 jxl28

#3 r15s11z55y89w21 vs. #6 algebra2000

#2 Mewto55555 vs. #7 cgyao15

You can begin playing your games if you want. remember they are 2 out of 3 cd. But if we do get 16 signups the game would be void because a new bracket would be made.gl.

CONTINUE SIGNING UP!

General Note: Sign up in the "FTW Tourney Sign-Up Thread" next time, please. Thanks. :)

But since we're already doing this, I sign up...

i sign up!

highest: 1605...  i've played way more than 50 games

#2 isabella2206 vs. #3 r15s11z55y89w21. Winner vs. #1 Aime15

Izzy vs r15. best 3 out of 5 cd

Winner vs. Aime. best 4 out of 7 cd.

What do you guys think about having a third place match?

It would be loser of Izzy and r15. vs david.

#2 isabella2206 vs. #1 Aime15

for the championship.

Finals are 4 out of 7 cd.

:first: :winner_first: AIME15

 :winner_second: isabella2296

 :winner_third: r15s11z55y89w21 and David2143.

We finally have a finished tournament. Good job to all the contestants.

Thanks for running this tournament.

Shouldn't there be a 3rd place match between r15 and David?

yah, but they didn't feel like doing it.

tourneys start when enough pplz have signed up
there is no exact time

if you wish to participate in a tourney, keep an eye on the sticky called "sign ups for tourney" or something like that
if a tourney moderator is looking for sign ups, he/she'll post there

[quote="r15s11z55y89w21"]"Read the stickies before posting on ANY forum."[/quote]

And if you had done so, you would have known the best thing to do is PM the offender.

And yes, you could quote me and say the same thing, but alas, I have not read the stickies  :P

...or maybe because there really are that many people that need guidance towards the sticky threads so as to PM everybody wouldn't be so much of a pain?

EDIT: I'd hate to be direct, but it is clearly obvious that
1) you are the one who is turning this into an argument
2) what I said at first was also said by a moderator

And when you continue this argument, you become the very thing you set out to destroy.

PS: jjx there is a reason threads are locked when tourney is done.


If one wants to prepare for the Putnam, but he never participated in a math contest before, should he buy the AoPs guides? or should he just study Putnam exams?

If you've never done contests before, there are two main things to focus on: problem-solving techniques and proof writing. There are many resources for the first; that doesn't have to be at Putnam level to help. The second simply requires practice: solve problems, write up solutions, and get someone to read those solutions if you can.

Many colleges that do the Putnam have a class or a student group that meets regularly during the fall to prepare for the Putnam. If that's available, do it.

Also, don't be too discouraged if you end up with fewer than 10 points out of 120. The Putnam is hard.

Some [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11946]FAQs on Preparing for College Math Study[/url] could be helpful for Putnam preparation, as they mention resources about proof writing.

Here is an free online class for putnam prep:
http://ocw.mit.edu/OcwWeb/Mathematics/18-S34Fall-2004/CourseHome/index.htm


On the last math league competition, there was a question that required one to decide wheter or not 0 should be counted as a perfect square.  The math league people say that 0 is a perfect square and refuse to give credit for those people who did not count 0 as a perfect square.

First, of all I have not found convincing evidence that 0 is commonly accepted and well-known to be a perfect square.  Due to this ambiguity, I feel they should have given both answers credit.  But they did not and quite frankly, I'm ticked off.  What do you think?

By the way, after looking it up I concede that 0 is a perfect square.  However, I still disagree with their decision.

I have always followed the convention that 0 is a perfect square.  My definition of "perfect square" is the square of an integer.

Having said that, MathCounts has on at least one question said that 0 is not a perfect square.  For example, MathCounts 1993-94 Handbook, Warmup 18, Question 3.  See
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11090]www.artofproblemsolving.com/Forum/viewtopic.php?t=11090[/url]
for a statement of the problem and some discussion.

Thanks for giving the link.  I had read those before and was trying to find them :)  They are most of the reason why I considered that 0 was not a perfect square while I was taking the math league test...

Unless I'm mistaken, there aren't any huge ramifications from considering 0 to be a perfect square or not.  It's not like choosing one definition over the other produces a completely different number system like accepting or not accepting Euclid's fifth produces different geometry systems.  For this reason, I don't see the definition as all that important.  That is, until we come to math competitions.  The question should have specified or worked around the 'perfect square' term adequately as to avoid ambiguity.

What I disagree with most is that they refuse to accept the answer that me and tons of other people got.  Our answers are fully justified.  Additionally, the point of the problem was not to test whether or not we agree with their definition of a perfect square (it was some little combinatorics/nt problem).  I was even sitting there trying to decide whether or not 0 was a perfect square.  I had answers figured out for both cases.  I knew it would come down to a matter of reading their minds.

I doubt there's any way I can change their decision at this point.  And it's not like math league is all [i]that[/i] important.  I just find it very, very irritating.

I strongly agree.  I don't know if it was the NYML you were taking, but I had that problem on the NYML (New York math league).  I marked that it was a perfect square, and s did some others.  Many marked that it was not.  The head of the math department, also my teacher, ended up making an appeal to accept both answers.  Apparently, he has a book which states that 0 is not a perfect square...so that is proof.  So then they said "Okay, we'll accept those who said 0 is not a perfect square".  Which, of course, is essentially the same problem as before.  My teacher then said not for that to be the case, so they just blew him off and went their own way.  Although, in our school, my teacher posted up scores accepting both kinds of answers, the actual competition, officially, gave some people a 6/6 on that round, and some 5/6.

I thought that a perfect square is "a number which can be expressed as the square of an integer".  But it could be "a number which can be expressed as the square of a natural number" , which makes all the difference.

According to [url=http://mathworld.wolfram.com/SquareNumber.html]MathWorld[/url], $0$ is regarded as a perfect square...

Yes... I think 0 is a perfect square, albeit a very trivial one.

"A square number, also called a perfect square, is a figurate number of the form $S_n=n^2$, where  is an integer."
Well, 0 [i]is[/i] an integer...

But I wouldn't blame someone for thinking that the first perfect square is 1... it's probably just their definition of perfect square.

$\sqrt{0}$ is an integer.  So $0$ is a perfect square.  :)

(There was a problem like this on this year's AMC, too, wasn't there?)

[quote="pieterminate"]Why would 0 NOT be an integer?[/quote]

The point is that it's not a natural number. That's what most of the debate is about (see above posts).

The definition seems to be a matter of convenience: sometimes there's no harm in generalizing and including 0 as a perfect square, and sometimes a problem doesn't work as intended (or becomes trivial) if 0 is included.  I remember the question you're referring to, and I think the Math League should have given credit for those who answered correctly with the assumption that zero is not a perfect square - if the problem is not made significantly easier by that assumption, then it is not clear what definition should be used.  Personally, I think that if one says "perfect square", then they are including zero, and that "positive" should be specified otherwise, but sometimes it's not necessary to specify if it's obvious that the problem intends for only positive perfect squares to be used.  In all other cases, though, it should be specified to avoid conclusion.

[quote="joml88"][quote="José"]Well, if you are in a contest and you have a problem with perfect squares, try the case when the perfect square is $0$ and then try the another cases, so you don't make mistakes, and we're all happy.........[/quote]

what???  You cannot experimentally determine if 0 is a perfect square or not...it depends on the definition of a perfect square which you cannot derive.  I think we've all pretty much agreed that 0 should be considered a perfect square.  I still think that math league should have given credit to both answers.[/quote]

Yes... but if you have a problem with perfect squares, 1º of all try the case when the perfect square is $0$


D,E,F is the midpoint of AB,BC,CA.I is the center of the inscribed circle of ABC.AI,BI,CI intersects the circle at U,V,W.PROVE:UE,VF,DW meets at one point.

I will change the names of points. You can see the figure below.
[b]Lemma:[/b] Given triangle $ ABC.X$ lies on $ BC$ and $ \vec{u}// \vec{AX}, \vec{u} \equal{} \alpha \vec{AB} \plus{} \beta \vec{AC}$ then $ \frac {MB}{MC} \equal{} \frac {\beta}{\alpha}$
This lemma is easy so I leave for the readers.
The tangency of $ (I)$which passes through $ X$ cut $ AB, AC$ at $ D, E$
We have $ \vec{XA'} \equal{} 1/2(\vec{DB} \plus{} \vec{EC}) \equal{} \frac {1}{2}.\frac {DB}{AB}.\vec{AB} \plus{} \frac {1}{2}.\frac {EC}{AC}.\vec{AC}$
$ \equal{} \frac {DB}{AB}.\vec{B'A'} \plus{} \frac {EC}{AC}.\vec{C'A'}$
Applying that lemma above we get $ \frac {A''C'}{A''B'} \equal{} \frac {EC}{AC}: \frac {DB}{AB} \equal{} \frac {c}{b}.\frac {EC}{DB}$
On the other side, $ \frac {DB}{EC} \equal{} \frac {S_{DIB}}{S_{EIC}} \equal{} \frac {DI.BI.sin \angle DIB}{EI.IC.sin \angle EIC} \equal{} \frac {IB}{IC}$
So $ \frac {A''C'}{A''B'} \equal{} \frac {c}{b}.\frac {IC}{IB}$
Similarly and applying Ceva's theorem for triangle $ A'B'C'$ we complete the proof.
See also:
http://www.mathlinks.ro/viewtopic.php?t=287975

Dear Mathlinkers,
two pure synthetic proofs of this known result originated from Clark Kimberling has been put on my website
http://perso.orange.fr/jl.ayme  vol. 5  The second mid-arc point…
Sincerely
Jean-Louis


geometry
trigonometry
trig identities
Law of Cosines

Take two triangles with sides $(a,a,b)$ and $(a,a,c)$.

My intuition (however 'counter' it may be) tells me that if $b$ is not equal to $c$ then the two triangles have different areas.

Are there triangles with sides $(a,a,b)$ and $(a,a,c)$, with $b$ not equal to $c$, where the triangles have the same area?

Yes.

[hide="Method 1"] If the angle between the two sides of length $a$ and $a$ is $\theta$, then the area of the triangle is

$\frac{1}{2}a^{2}\sin \theta$

As $\theta$ ranges from $0$ to $180^{\circ}$.  But of course,

$\sin \theta = \sin (180^{\circ}-\theta)$

For a given triangle $(a, a, b)$ with such angle $\theta$, we have (by Law of Cosines)

$b^{2}= 2a^{2}(1-\cos \theta)$

Now, I showed above that the area is the same if we use $180^{\circ}-\theta$ in the triangle $(a, a, c)$, so

$c^{2}= 2a^{2}(1+\cos \theta)$

So we have the formula

$c^{2}= 4a^{2}-b^{2}$ [/hide]


Can nyone help me regarding engineering admissions. I would like to know d details of d various colleges so that I can decide where to apply. I would aLso like to know to d rank cutoffs for different colleges. Also I would like to know d cost involved.

"this is spam"
hmm...
rvce cs rank 1-10
pesit cs 11-20
pesit ec -20 -40
rv ec -21-41

Hey thaks for suggesting  this great site which answers most of my questions.it is http://www.colkonnect.com
They have the entire database of all the colleges including the ranks which they took in last year.

this is a really useful site
http://www.webcampuz.com/

try the mock online COMEDk and CET counselling


Imagine a long [tex] L*L[/tex] rectangle, where [tex]L[/tex] is an integer. Clearly, one can pack this rectangle with [tex]L[/tex] circles of diameter 1, and no more. (By pack we mean that touching is OK, but overlapping is not) On the other hand, it's not immediately obvious that [tex]2L[/tex] circles is the maximum number possible for packing a [tex]2*L[/tex] rectangle. Investigate this, and generalize to [tex]m*L[/tex] rectangles.
__________________________________

Really a logical problems...... :D

It's very fun to examine yourself for a short time...

Very interesting problems...   WoW~!!
--------and.. this is in the book THE ART AND CRAFT OF PROBLEM SOLVING..
which Paul Zeitz who led the American Dreamteam wrote..

I happen to know that for $2\times L$ there are some $L$s for which we could pack more than $2L$ such circles, but I never even tried to show it. I saw it as a problem on some Internet page. I'll try to look for it.


A group of children forms a circle and each child starts with an even number of pieces of candy. Each child then gives half of what he has to his right-hand neighbour. After the transaction, if a child has an odd number of pieces, he or she will receive an extra piece from an external source. Show that, after a finite number of such steps, each child will have the same number of pieces of candy.

[i]Peking Mathematics Contest, 1962[/i]

this one is famous (at least in China) and this game is introduced by a professor in Peking University.

We may find that after an adjusting, the maximum value leave unchanged while the minimum value increases. Thus, we should proof that

(i) During the process, the number of candy in one's hand keeps between 2m(the max value) and 2n(the min value).

Suppose a children have 2k pieces of candy and his right-hand neighbour has 2h pieces. since n\leq k\leq m, n\leq h\leq m, so after the adjustment, this children has h+k pieces of candy and 2n \leq h+k\leq 2m.
if h+k is even, the conclusion has been proved; it not, he will be given an extra piece, but h+k<2m, hence, 2n<h+k+1\leq 2m, the conclusion holds too.

(ii) if a children has more than 2n pieces of candy, he will still have more than 2n pieces of candy after an adjustment.

Suppose he has 2k > 2n pieces of candy, and his right neighbour has 2h.
after adjusting, this child has h+k pieces of candy. it is obvious that h+k>n+n=2n. if k+h is odd, then he will be given another piece of candy, thus k+h+1>2n.

(iii) there exists a child who has fewest pieces of candy; and after an adjustment, he has more than 2n pieces of candy, i.e., he get two more pieces.

Of course, there exists such a child that his candy is fewest. suppose he has 2n pieces. and his right neighbour has 2h>2n. after the adjustment, this child has h+n pieces. if h+n is even, then h+n\geq 2(n+1)=2n+2; if it is odd, h+n\geq 2n+1, after an extra income, h+n+1\geq 2n+2. hence, the minimum value increases.

It is in primary school that I know this problem for the first time.
You know, math olympics in primary school in China is also prosperous.

so it is in romania, and i'll bet that it's the same in Russia too :)


lol coincidentally we had to give our cat a bath today.

[img]http://icanhascheezburger.files.wordpress.com/2009/08/funny-pictures-cat-stares-at-toy-toilet.jpg[/img]

ehhh
The message is too small. Please make the message longer before submitting.


Numerele reale pozitive $x$, $y$, $z$ verifică: $\frac{x^{2}}{y^{2}}+\frac{y^{2}}{z^{2}}+\frac{z^{2}}{x^{2}}=\alpha , \alpha\geq 3.$
Să se afle cea mai mare valoare pe care o poate lua expresia: $\frac{z-y}{x}+\frac{x-z}{y}+\frac{y-x}{z}.$

$\left( {\frac{x}{y} + \frac{y}{z} + \frac{z}{x} - 1} \right)^2 \ge 0 \Rightarrow$

$\Rightarrow \left( {\frac{x}{y} + \frac{y}{z} + \frac{z}{x}} \right)^2 \ge 2\left( {\frac{x}{y} + \frac{y}{z} + \frac{z}{x}} \right) - 1 \Rightarrow$

$\Rightarrow \left( {\frac{x}{y}} \right)^2 + \left( {\frac{y}{z}} \right)^2 + \left( {\frac{z}{x}} \right)^2 + 2\left( {\frac{x}{z} + \frac{y}{x} + \frac{z}{y}} \right) \ge 2\left( {\frac{x}{y} + \frac{y}{z} + \frac{z}{x}} \right) - 1 \Rightarrow$

$\Rightarrow \frac{{\alpha + 1}}{2} + \frac{x}{z} + \frac{y}{x} + \frac{z}{y} \ge \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \Rightarrow$

$\Rightarrow \frac{{z - y}}{x} + \frac{{x - z}}{y} + \frac{{y - x}}{z} \le \frac{{\alpha + 1}}{2}$

Intr-adevăr, nu am ţinut cont de cazul de egalitate si a rezultat o problemă eronată  :blush: .


quadratics
modular arithmetic
continued fraction

[quote="cdymdcool"]Suppose for a positive integer n both 2n+1 and 3n+1 are perfect squares.  Show that n is divisible by 40.[/quote]

Let's $ 2n\plus{}1\equal{}m^2$, $ 3n\plus{}1\equal{}k^2$
We have that $ m^2\plus{}k^2 \equiv 2 (\bmod 5)$ Therefore we have
$ m^2 \equiv k^2 \equiv 1 (\bmod 5)$
so $ 5|n\equal{}k^2\minus{}m^2$

It's easy to see that $ n$ - is even.
So $ k^2 \equiv 1 (\bmod 8)$
therefore $ 8|3n\equal{}k^2\minus{}1$
and $ 40|n$

Poincare, do you mean that n is even? If n is odd, then 2(2k+1)+1 is a square, or a square is 3 mod 4, contradiction.

A square is either 0,1, or 4 mod 5, so they would both have to be 1.

oh ok  :D The message is too small. Please make the message longer before submitting.

[quote="cdymdcool"]why is k^2equiv1 (\bmod9)? when n is even?[/quote]

$ n\equal{}2x$ and $ k^2\equal{}(3n\plus{}1)^2\equal{}(3\times2x\plus{}1)^2\equal{}8(4x^2\plus{}x)\plus{}4x(x\plus{}1)\plus{}1$. Thus, $ k^2\equiv 1\mod{8}$.

[quote="SaYaT"][quote="cdymdcool"]Suppose for a positive integer n both 2n+1 and 3n+1 are perfect squares.  Show that n is divisible by 40.[/quote]

$ m^2 \equiv k^2 \equiv 1 (\bmod 5)$[/quote]

How can you know?

little,,Dragon!: the quadratic residue in mod 5 are $ 0,1,4$.  The only way to add two of these in mod 5 and get 2 is for them both to be 1.

How do we know that these are the only quadratic residues?  Well note that
$ (5k)^2 \equal{} 25k^2$
$ (1\plus{}5k)^2 \equal{} 1\plus{}10k\plus{}25k^2$
$ (2\plus{}5k)^2 \equal{} 4\plus{}20k\plus{}25k^2$
$ (3\plus{}5k)^2 \equal{} 9 \plus{} 30k \plus{} 25k^2$
$ (4\plus{}5k)^2 \equal{} 16 \plus{} 40k \plus{} 25k^2$

Since all integers have one of the forms on the left hand sides of these equalities, and the right hand sides go $ 0,1,4,4,1$ in mod 5, these are the only quadratic residue in mod 5.

[quote="cdymdcool"]Suppose for a positive integer n both 2n+1 and 3n+1 are perfect squares.  Show that n is divisible by 40.[/quote]

{Note:  I do NOT prefer the following method for this problem.
It is shown for contrast.}

Doing it by use of the contrapositive:
----------------------------------------------

If n is not divisible by 40, then at least one of 2n + 1 and
3n + 1 must not be a perfect square.

For k a nonnegative integer, n = 40k + 1, 40k + 2,...,40k + 39

Let m be an integer.
Let N be a positive integer.

The set of integers that n belongs to can be partitioned into three
classes of:  40k + 2m + 1, 40k + 4m + 2, and 40k + 4m.

First class
-------------

2(40k + 2m + 1) + 1  and 3(40k + 2m + 1) + 1
80k + 4m + 2 + 1 . . and 120k + 6m + 3 + 1
80k + 4m + 3  . . . . . and 120k + 6m + 4
4N + 3 ** . . . . . . . . and 120k + 6m + 4

** Not a perfect square form
========================================================

Second class
-----------------

2(40k + 4m + 2) + 1 and 3(40k + 4m + 2) + 1
80k + 8m + 4 + 1. . . and 120k + 12m + 6 + 1
80k + 8m + 5. . . . . . and 120k + 12m + 7
80k + 8m + 5. . . . . . and 120k + 12m + 4 + 3
80k + 8m + 5. . . . . . and 4N + 3 (not a perfect square form)

========================================================

Third class
--------------

2(40k + 4m) + 1 and 3(40k + 4m) + 1
80k + 8m + 1 and 120k + 12m + 1

Each of these is four times an integer, then plus one, so neither can
be dismissed outright.

If you look at the sqrt(1 + 12m + 120k) to be an integer,
you get (1 + 6m), with a remainder of 120k - 36m^2,
that must equal 0.

If you look at the the sqrt(1 + 8m + 80k) to be an integer,
simultaneously with the other square root, you get (1 + 4m),
with a remainder of 80k - 16m^2, that must equal 0.

Solve these simultaneously:

120k - 36m^2 = 0
  80k - 16m^2 = 0
---------------------

It turns out that:
k = m = 0

Recall:  n = 40k + 4m for this class

So, n = 0 produces the only situation where 2n + 1 and 3n + 1
are perfect squares at the same time, when n is of the form
40k + 4m, BUT this doesn't count (not allowed), because k
has to be nonnegative.  And we're considering just the
positive integers n.

[quote]In Vworld's post surely k^2 should be (3n+1) and not (3n+1)^2?[/quote]

yes.  Anyway, the proof that $ k^2 \equiv 1 \pmod{8}$ should be that $ k^2 \equal{} 3n\plus{}1$, and $ n$ is even makes $ k^2$ odd, and hence $ k$ is odd.  In mod $ 8$, all squares of odd numbers are equivalent to $ 1$:

$ (2a\plus{}1)^2 \equal{} 4a^2\plus{}4a\plus{}1 \equal{} 4a(a\plus{}1) \plus{}1$.  Since either $ a$ or $ a\plus{}1$ is even, $ 4a(a\plus{}1) \equiv 0 \pmod{8}$ and thus $ (2a\plus{}1)^2 \equiv 1 \pmod{8}$ as claimed

Just curious... can we find all values of $ n$ for which $ 2n \plus{} 1$ and $ 3n \plus{} 1$ are perfect squares?  I'm going to try:

$ 2n \plus{} 1 \equal{} m^2$
$ 3n \plus{} 1 \equal{} k^2$

$ 3m^2 \minus{} 2k^2 \equal{} 1$
$ (3m)^2 \minus{} 6k^2 \equal{} 3$

looks like a pell equation.  Letting $ n \equal{} 1$ we have the smallest solution with $ m \equal{} 9$, $ k \equal{} 11$.  So it appears $ 3 \equal{} (27 \minus{} 11\sqrt {6})(27 \plus{} 11\sqrt {6})$, i.e. $ N(27 \minus{} 11\sqrt {6}) \equal{} 3$ (N is the Norm function).

Now we need to find units in $ Z[\sqrt {6}]$.  The continued fraction is $ \sqrt {6} \equal{} \{2;\overline{2,4}\}$  And taking different convergents $ p/q$ we find $ 5/2$ gives $ 5^2 \minus{} 6\cdot 2^2 \equal{} 1$, i.e. $ N(5 \minus{} 2\sqrt {6}) \equal{} 1$.  Thus all units (other than $ 1$) in $ Z[\sqrt {6}]$ are $ (5\pm2\sqrt {6})^a$

Now we have $ 3m \minus{} \sqrt {6}k \equal{} (5 \minus{} 2\sqrt {6})^a(27 \minus{} 11\sqrt {6})$ and $ 3m \plus{} \sqrt {6}k \equal{} (5 \plus{} 2\sqrt {6})^a(27 \plus{} 11\sqrt {6})$,

Wait!  we also have $ N(3\pm\sqrt {6}) \equal{} 3$, i.e. we should be using this instead of $ 27\pm11\sqrt {6}$ (actually, it probably doesn't matter, since the only difference is a factor of a unit)

$ 3m \minus{} \sqrt {6}k \equal{} (5 \minus{} 2\sqrt {6})^a(3 \minus{} \sqrt {6})$ and $ 3m \plus{} \sqrt {6}k \equal{} (5 \plus{} 2\sqrt {6})^a(3 \plus{} \sqrt {6})$, so $ m \equal{} \frac {(5 \minus{} 2\sqrt {6})^a(3 \minus{} \sqrt {6}) \plus{} (5 \plus{} 2\sqrt {6})^a(3 \plus{} \sqrt {6})}{6}$ and $ k \equal{} \frac {(5 \plus{} 2\sqrt {6})^a(3 \plus{} \sqrt {6}) \minus{} (5 \minus{} 2\sqrt {6})^a(3 \minus{} \sqrt {6})}{2 \sqrt {6}}$

starting with $ a \equal{} 0,1,2,3,4 \ldots$ these produce the sequences $ m \equal{} 1,9,89,881,8721 \ldots$ and $ k \equal{} 1, 11,109,1079, 10681 \ldots$ for which we have $ n \equal{} 0, 40, 3960, 388080, 38027920 \ldots$

In general, $ n\equal{}\frac {(5 \minus{} 2\sqrt {6})^a(3 \minus{} \sqrt {6}) \plus{} (5 \plus{} 2\sqrt {6})^a(3 \plus{} \sqrt {6}) \minus{}6}{12}$

I like the approach by arrange your tan, however I disagree with the last part: 80k^2 -16m^2 need not be zero. For example m = 1 and k = 2 give (4m+1)^2 +80k -16m^2 = 5^2 + 144 = 5^2 + 12^2 = 13^2 and so sqrt(80k + 8m +1) = 13 is an integer.

[quote="facis"][quote]In Vworld's post surely k^2 should be (3n+1) and not (3n+1)^2?[/quote]

yes.  Anyway, the proof that $ k^2 \equiv 1 \pmod{8}$ should be that $ k^2 \equal{} 3n \plus{} 1$, and $ n$ is even makes $ k^2$ odd, and hence $ k$ is odd.  In mod $ 8$, all squares of odd numbers are equivalent to $ 1$:

$ (2a \plus{} 1)^2 \equal{} 4a^2 \plus{} 4a \plus{} 1 \equal{} 4a(a \plus{} 1) \plus{} 1$.  Since either $ a$ or $ a \plus{} 1$ is even, $ 4a(a \plus{} 1) \equiv 0 \pmod{8}$ and thus $ (2a \plus{} 1)^2 \equiv 1 \pmod{8}$ as claimed[/quote]

Er... Facis [i]clearly[/i] missed the step where I applied the bijective map $ \Re^\plus{} \rightarrow \Re^\plus{} : k \mapsto \sqrt{k}$... you must later apply the inverse map $ \Re^\plus{} \rightarrow \Re^\plus{} : k \mapsto k^2$ whence we have:

$ k^4 \equiv 1 \pmod{8} \Rightarrow k^2 \equiv 1 \pmod{2} \Rightarrow k \equiv 1 \pmod{2} \Rightarrow k^2 \equiv 1 \pmod{8}$ (we have quadratic residues $ \{0,1,4\}$ modulo $ 8$.)  :D


This weekend MCrawford and I put together a lot of the comments we made in the USAMO Math Jam and created a very long article about how to write a solution.  You can find the beginning of the article [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php]here[/url].

In this article, we discuss many items solution writers should know.

A few words of warning, some of the problems we use as examples are extremely tough (it's hard to write a sample bad solution for an easy problem!), so you shouldn't be discouraged if you have trouble solving the problems.

The articles are only lightly proofread, as we wanted USAMO participants to have a crack at the article before the USAMO.  We recommend these participants not spend too much time trying to do the problems right now, but rather that they soak up the style lessons and come back for the problems later.

Please let us know if you have any comments on the article.

Whew, cool, that is one long article! Thanks.

Just a general question (to anyone):

Do you think that writing USAMO solutions with a pencil will result in lead smudges once the pieces of paper are jammed into an envelope and shipped to the AMC (sort of how notes in pencil get smudged after they've been in your notebook for a while)?

I doubt it; pencil doesn't smudge [i] that [/i] easily.

[quote="rrusczyk"]This weekend MCrawford and I put together a lot of the comments we made in the USAMO Math Jam and created a very long article about how to write a solution.  You can find the beginning of the article [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php]here[/url]. . . . The articles are only lightly proofread, as we wanted USAMO participants to have a crack at the article before the USAMO.[/quote]

Wow! Great stuff. The articles may not be up to your own high standards yet, but they are already very good, and great food for thought. I particularly like the contrasting examples used in most of the subarticles--I like learning from examples.

I agree with TA and fierytycoon.  The article was very helpful, and not only did I learn how to write a proof, I learned a few formulas along the way.

Yeah, I agree. The article is very good, but the best feature is the large number of examples. They really clarify what you're trying to say. And thanks for using my proof, and thanks even more for fixing my proof!

[quote="Fierytycoon"]Just a general question (to anyone):

Do you think that writing USAMO solutions with a pencil will result in lead smudges once the pieces of paper are jammed into an envelope and shipped to the AMC . . . ?[/quote]

I thought [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2003-4pub/04USAMO/html/04-04usamo-text.html]the rule[/url] is that USAMO solutions have to be [b]faxed[/b] back to AMC headquarters. Sad experience with my fax machine has taught me that only [b]BLACK PEN[/b] faxes very well, among handwritten marks on paper, so I would advise a math competitor to write out a USAMO solution in pen.

Tokenadult is right; solutions are faxed in (at least, I think they are - I know my math teacher spent a while trying to get mine faxed in).

I liked the articles a lot.  In particular, the business of calling 'a large thin-shelled vehicle for a young fowl that was created by a huge female bird' Humpty-Dumpty made me laugh.

[quote="billzhao"]It seems to me that many of the "how not to write a solution" are completely correct, and would be given 6 or 7 point if it were put on an olympiad.

In an olympiad, we shouldn't to be obessed with trying to polish our proofs to a state of perfection. There just isn't the time for it. As long as it's right, we should just leave it at that and move on. Having three 6's is always better than just one perfect solution. Besides, solution-writing comes from experience. None of us are Reid.

Nonetheless, if there wasn't the time-limit, then the ideas introduced in the article could be very helpful.[/quote]

Indeed, many of them would get 6's or 7's on an Olympiad, but Olympiads aren't the only reason people write mathematics.  In fact, they are only a very, very small subset of mathematical writing.  Furthermore, I disagree with noting that it might not be important - my senior year this was all that separated 4th from 12th on the USAMO.  The USAMO winners were the top 8 then.  It can be very important to be polished even on contests.  Besides, once you're experienced writing in this style, it's usually easier and faster to write a polished solution than to write a sloppy one.

I think it's unlikely that many of the ``how not to write'' solutions would score 6 points. They might get 7 -- if it's perfect, they can't take off points. But if you annoy the graders AND you make a mistake, I would imagine that your penalty won't just be one point. It's pretty easy to make a minor error that would give a perfect solution a 6, so if the writeup is bad, then you might lose a great deal of points because the grader will not be lenient. Then again, all this stuff I said might be nonsense; I'm not a USAMO grader, so I don't know how these things work.

I posted different solutions to two problems in the article. If you are interested, you may find them [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12528]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=86151#86151]here[/url].

[quote]Problem: From vertex A of triangle ABC, perpendiculars AM and AN are drawn to the bisectors of the exterior angles of the triangle at B and C. Prove that MN is equal to half the perimeter of ABC.[/quote]

[quote]Problem: Let I be the incenter of triangle ABC. Prove that

(IA)(IB)(IC) = 4Rr2,

where R is the circumradius of ABC and r is the inradius of ABC.

(Problem due to Sam Vandervelde of the Mandelbrot Competition. Note that the incenter of a triangle is the center of the circle inscribed in the triangle. The inradius is the radius of this circle. The circumradius is the radius of the circle that passes through the vertices of ABC.)
[/quote]

I'm trying to clean up the top of the forum.  Should I just de-announceify this, or should I move it to another forum, since it has more to do with the AoPS website than the USAMO?

[Edit] This was previously an announcement at the top of the AMC forum.  It has since been moved.

Moving it to a forum appropriate to site announcements makes sense to me. Of course there should be a pointer to the underlying article in my FAQ and in your FAQ about preparing for AMC contests.


I have a question in front of me, which says:
A positive real number p is given Consider all pairs of positive real numbers (x,y) that satisfy the equation xy(x+y)=p. Find all pairs for which the value of the expression x(cubed)+y(cubed) is the least possible.
[b]Help needed ASAP. :? [/b]

$ x^2-xy+y^2 \ge xy$ with equality when $ x=y$.

$ x^3+y^3=(x+y)(x^2 -xy + y^2) \ge (x+y)xy = p $, with equality when $x=y$.

So $x^3+y^3$ is always bigger that $p$ and is equal to $p$ when $x=y$

So $x^3+ y^3$ is smallest only when $x=y= (\frac p2)^{\frac 13}$.

See you on monday....

Thanks Steve   :lol: , that wos da only one i was stuck on, the rest took me less than an hour to solve. I might meet u sunday night when u arrive. C U
By the way i'm Galin


geometry
incenter
circumcircle
Euler
rotation
similar triangles

A famous one, yet i have never seen it posted here.

Prove the incenter, the circumcenter, and the orthocenter of a triangle all lie on the same line

i know this is on here somewhere, but i have to find it out. It might be on MathWorld, although I doubt it. I guess the proof is somewhat similar to the proof of Euler Line, which is the line that connects orthocenter, centroid, and circumcenter. I have read proof of that on Geometry Revisited, but neither do i remember nor do i understand.

Hey,it's not the incenter it is the centroid.yes,is true this represents Euler's line in a triangle.If you want a proof
I'm right here,but I'll post it tomorow because today I'm  too tired and I have to go to sleep. ;)

[quote="cezar lupu"]Hey,it's not the incenter it is the centroid[/quote]

yes. I have said that i was only assuming seamus's problem was [b]similar [/b]to Euler line, not [b]same[/b]

I don't think the problem is true. I think he meant centroid in the original problem. I'm not completely sure, though.

[quote="yetti"]Let A', B', C' be the midpoints of the sides BC, CA, AB. Since the triangle midlines are parallel to the triangle sides, $C'B' \parallel BC,\ A'C' \parallel CA,\ B'A' \parallel AB$, the triangle $\triangle ABC$ and its medial triangle $\triangle A'B'C'$ are similar. Since the lines AA', BB', CC' connecting their corresponding vertices (medians of the triangle $\triangle ABC$) concur at its centroid G, these 2 triangles are even centrally similar. Their similarity coefficient h is

$h = \frac{B'C'}{BC} = \frac{C'A'}{CA} = \frac{A'B'}{AB} = -\frac 1 2$

(Since the segments A'B' and AB, etc., are oppositely oriented, the similarity coefficient is negative). Therefore, all lines connecting the corresponding points P', P of the triangles $\triangle A'B'C', \triangle ABC$ pass through the centroid G (the center of similarity). In addition, $\frac{P'G}{PG} = h = -\frac 1 2$, which means that the correcponding points P, P' lie on the opposite sides of the similarity center G and $|P'G| = \frac{|PG|}{2}$. Let H, O be the orthocenter and circumcenter of the triangle $\triangle ABC$. The circumcenter O of the triangle $\triangle ABC$ is the intersection of the perpendicular bisectors of the sides BC, CA, AB. Consequently, it is the intersection of the altitudes of the medial triangle $\triangle A'B'C'$, i.e., its orthocenter H'. As a result, the points $O \equiv H',\ H$ are corresponding points of the centrally similar triangles $\triangle A'B'C' \sim \triangle ABC$, hence, the line $HO \equiv HH'$ passes through their similarity center G and  $|OG| \equiv |H'G| = \frac{|HG|}{2}$.[/quote]

Wasn't that the proof of Euler Line, not Seamusboyle's problem???? My question is, is there such thing as what Seamusboyle is asking??

From a couple of sketches, I'm pretty sure it's not true.

If his problem is true, then the centroid, incentre, othorcentre and circumcentre all lie on a line. It is not true for every triangle.

Have a look under "Theorems and Formulas "in the Olympiad Geometry section.

  A post  on this  matter was entered by me under heading "Relation between incentre and Euler Line".

The incentre lies on the Euler line if and only if the triangle is equilateral.

Exercise: prove that if any pair of these points (incentre circumcentre orthocentre or centroid) coincide then the triangle is equilateral.

[quote="Michael Niland"]The incentre lies on the Euler line if and only if the triangle is equilateral.[/quote]

This is not correct, the incenter I lies on the Euler line HO (or conversely, the orthocenter H lies on the diacentral line IO) only if the triangle is isosceles, not necessarily equilateral. Moreover, an equilateral triangle has no Euler line. Let the circumcircle (O) and the incircle (I) be fixed and "rotate" the triangle around these 2 fixed circles according to Poncelet's porism. The radius of the 9-point circle (N) is $\frac R 2$, where R is the circumradius, because the 9-point circle is the circumcircle of the medial triangle. The 9-point circle (N) is internally tangent to the incircle (I) (Feuerbach's theorem). So, the 9-point circle of a fixed radius $\frac R 2$ rolls around the fixed incircle (I) with radius r. This means that the locus of the 9-point circle centers is a circle concentric with the incircle and with radius $\frac R 2 - r$. Since the 9-point circle center N is the midpoint of the segment HO, the locus of orthocenters is a circle (M) centered on the diacentral line IO at the distance MO = 2 IO from the circumcenter O and with radius R - 2r. This circle obviously intersects the diacentral line IO at 2 points, the orthocenters of the triangles having the Euler line HO coincident with the diacentral line IO (i.e., the points H, I, O collinear) and these triangles are both isosceles.

Whoops yeah the problem was wrong... i got a cool other one though....

There was another well known theorem about collinearity with the incenter... Can anyone refresh my memory?

The incentre, the Nagel point and the centroid are collinear.

The line (known as the Nagel line) joining the incentre $I$ to the centroid $G$ is an analogue of the Euler line. The Nagel point $Na$ corresponds to the orthocentre $H$, the Spieker centre $Sp$ corresponds to the nine-point centre $N$ and the incentre corresponds to the circumcentre $O$.

So just as $OG : ON : OH = 2 : 3 : 6$, $IG : ISp : INa = 2 : 3 : 6$.


floor function
logarithms
induction
number theory proposed
number theory

if  n!=(p1^a1)(p2^a2).......(pm^am)

that  pi  is  the  i,st  prim(p1=2,p2=3,p3=5......)
and ai is  the power of pi in  n!

then  WHAT IS THE MAXIMUM OF (pi^ai)  i=1,2,3,....,m
for any n.

Hi

Dear Hadi !

I solve your problem !   :P

If you need to answer of your problem , Send mail to Me. :o

Janan.pouransari@Gmail.com :arrow:

Bye Bye!

We claim that the answer is $2^{\nu_2(n!)}$ for $n\neq 3$ (where $\nu_p$ denotes $p$-adic order), and $3$ for $n=3$.

For $n\leqslant 39$, we check the above claim directly. Therefore, we shall prove that $2^{\nu_2(n!)}>p^{\nu_p(n!)}$ for any integer $n\geqslant 40$ and prime $p\neq 2$.

Let $p=3$. Since $\nu_p(n!)=\sum_{i=1}^{\infty}\left\lfloor \frac n {p^i}\right\rfloor$, it is enough to prove that \[\sum_{i=1}^{\infty}\left\lfloor \frac n {2^i}\right\rfloor>\log_2 3\cdot\sum_{i=1}^{\infty}\left\lfloor \frac n {3^i}\right\rfloor.\quad(1)\]
Indeed, it would then follow $2^{\nu_2(n!)}=2^{\sum_{i=1}^{\infty}\left\lfloor \frac n {2^i}\right\rfloor}>2^{\log_2 3\cdot\sum_{i=1}^{\infty}\left\lfloor \frac n {3^i}\right\rfloor}=3^{\sum_{i=1}^{\infty}\left\lfloor \frac n {3^i}\right\rfloor}=3^{\nu_3(n!)}$. In order to prove $(1)$, let us first prove \[\left\lfloor \frac n 4\right\rfloor\geqslant\frac {15} 7\left\lfloor\frac n 9\right\rfloor\quad(2)\] and \[\left\lfloor \frac n 2\right\rfloor\geqslant\frac {22}{15}\left\lfloor\frac n 3\right\rfloor.\quad(3)\] In order to prove $(2)$, it is enough to prove $\frac n 4 -1\geqslant\frac{15n}{63}$, what is equivalent to $n\geqslant 84$, while for $40\leqslant n\leqslant 83$ we check $(2)$ directly. Similarly, in order to prove $(3)$ it is enough to prove $\frac n 2-1\geqslant\frac{22n}{45}$, what is equivalent to $n\geqslant 90$, while for $40\leqslant n\leqslant 89$ we check $(3)$ directly. Let us now prove \[\left\lfloor\frac n 2\right\rfloor+\left\lfloor\frac n 4\right\rfloor>\log_2 3\cdot\left(\left\lfloor\frac n 3\right\rfloor+\left\lfloor\frac n 9\right\rfloor\right),\quad(4)\] or, equivalently, $\left\lfloor\frac n 2\right\rfloor+\left\lfloor\frac n 4\right\rfloor-\log_2 3\cdot\left\lfloor\frac n 3\right\rfloor-\log_2 3\cdot\left\lfloor\frac n 9\right\rfloor>0$. Using $(2)$ and $(3)$, we see that it is enough to prove $\left(\frac{22}{15}-\log_2 3\right)\left\lfloor\frac n 3\right\rfloor+\left(\frac{15}7-\log_2 3\right)\left\lfloor\frac n 9\right\rfloor>0$, what holds iff $\frac{\frac{15}7-\log_2 3}{\log_2 3-\frac{22}{15}}\left\lfloor\frac n 9\right\rfloor>\left\lfloor\frac n 3\right\rfloor$ (the numerator is positive since $\log_2 3<\frac 8 5$, and the denominator is positive since $\log_2 3>\frac 3 2$). We thus have $\frac{\frac{15}7-\log_2 3}{\log_2 3-\frac{22}{15}}>\frac{\frac{15}7-\frac 8 5}{\frac 8 5-\frac{22}{15}}>\frac{\frac{19}{35}}{\frac 2{15}}=\frac{57}{14}>4$, and it is therefore enough to prove $4\left\lfloor\frac n 9\right\rfloor>\left\lfloor\frac n 3\right\rfloor$. We have $4\left\lfloor\frac n 9\right\rfloor-\left\lfloor\frac n 3\right\rfloor>4\left(\frac n 9-1\right)-\frac n 3=\frac{n-36}9\geqslant 0$ for $n\geqslant 36$. This proves $(4)$.

Before finishing the proof of $(1)$, we need two propositions.

[i]Proposition 1:[/i] For any $n,a,b\in\mathbb{N}$ it holds $\left\lfloor\frac{\left\lfloor\frac n a\right\rfloor}b\right\rfloor=\left\lfloor\frac n{ab}\right\rfloor$.

[i]Proof:[/i] Let $n=ka+r$ where $0\leqslant r<a$. We have $\left\lfloor\frac{\left\lfloor\frac n a\right\rfloor}b\right\rfloor=\left\lfloor\frac{\left\lfloor\frac {ka+r}a\right\rfloor}b\right\rfloor=\left\lfloor\frac{\left\lfloor k+\frac r a\right\rfloor}b\right\rfloor=\left\lfloor\frac k b\right\rfloor$. Further, $\frac n{ab}=\frac{ka+r}{ab}=\frac k b+\frac{\frac r a}b<\frac{k+1}b$, and therefore $\left\lfloor\frac n{ab}\right\rfloor=\left\lfloor\frac k b\right\rfloor$.

[i]Proposition 2:[/i] For any integers $u\geqslant 2$, $v\geqslant 3$ and $b\geqslant 3$ we have $\frac{\left\lfloor\frac u 2\right\rfloor}{\left\lfloor\frac v b\right\rfloor}\geqslant\frac u v$.

[i]Proof:[/i] It is enough to prove $\frac{\left\lfloor\frac u 2\right\rfloor}{\frac v b}\geqslant \frac u v$, what is equivalent to $b\left\lfloor\frac u 2\right\rfloor\geqslant u$. This is easily checked for $2\leqslant u\leqslant 5$, and for $u\geqslant 6$ stronger $3\left(\frac u 2-1\right)\geqslant u$ holds.

Now, note that from $(2)$ follows $\left\lfloor \frac n 4\right\rfloor\geqslant\log_2 3\cdot\left\lfloor\frac n 9\right\rfloor$, and by induction it follows $\left\lfloor \frac n {2^i}\right\rfloor\geqslant\log_2 3\cdot\left\lfloor\frac n {3^i}\right\rfloor$ for all $i\geqslant 2$, the inductive step being $\frac{\left\lfloor \frac n {2^i}\right\rfloor}{\left\lfloor\frac n {3^i}\right\rfloor}\geqslant\log_2 3\Rightarrow\frac{\left\lfloor \frac n {2^{i+1}}\right\rfloor}{\left\lfloor\frac n {3^{i+1}}\right\rfloor}=\frac{\Big\lfloor\frac{\left\lfloor\frac n{2^i}\right\rfloor}2\Big\rfloor}{\Big\lfloor\frac{\left\lfloor\frac n{3^i}\right\rfloor}3\Big\rfloor}\geqslant\frac{\left\lfloor \frac n {2^i}\right\rfloor}{\left\lfloor\frac n {3^i}\right\rfloor}\geqslant\log_2 3$. (Propositions 1 and 2 were used. Clearly, the observed quotient makes sense only when the denominator is positive, but otherwise the assertion is trivial.) Together with $(4)$, this finishes the proof of $(1)$.

Let now $p\geqslant 5$. We are going to show that  for any $n\geqslant 31$ it holds \[\left\lfloor \frac n {2^i}\right\rfloor>\log_2 p\cdot\left\lfloor \frac n {p^i}\right\rfloor\] what is enough to finish the proof. By the inductive argument similar to the one already seen, we see that it is enough to show \[\left\lfloor \frac n 2\right\rfloor>\log_2 p\cdot\left\lfloor \frac n p\right\rfloor\quad(5)\]
Denote $f(n,p)=\frac n 2-1-\log_2p\cdot\frac n p$. Clearly, it is enough to show that $f(n,p)>0$. Since $\log_2 5<\frac 7 3$, we have $f(31,5)=\frac{31}2-1-\log_2 5\cdot\frac{31} 5>\frac{31}2-1-\frac 7 3\cdot\frac{31} 5=\frac{465-30-434}{30}=\frac 1{30}>0$. Further, we have $\frac{\partial f}{\partial n}(n,p)=\frac 1 2-\log_2 p\cdot\frac 1 p>0$ (since for $p> 4$ we have $\log_2 p<\frac p 2$), and $\frac{\partial f}{\partial p}(n,p)=-\frac 1{p\ln 2}\cdot\frac n p+\frac{\ln p}{\ln 2}\cdot\frac n{p^2}=\frac {n(\ln p-1)}{p^2\ln 2}>0$ (since $\ln p>1$). Therefore, $f(n,p)>0$ whenever $n\geqslant 31$ and $p\geqslant 5$, what completes the proof.


function
percent
geometry
number theory
relatively prime

I just turned on the famous program "who wants to be a millionaire"  and the question for [size=150]16 000[/size] dollars was:
[b]Which of the following square numbers can be written as a sum of two smaller square numbers?[/b]
A. $16$      B. $25$      C. $36$         D. $49$
The person answering the question [b]could not solve it[/b] :o  so he asked the audience.
In the audience, [size=200]50%[/size] said the answer was A, and on the Internet voting thing, [size=200]45%[/size] said the answer was A. :rotfl:
The person answering the questions followed the advice of the audience and picked A and lost.
(the correct answer is obviously B: 25=16+9)
 :o
What are your thoughts on this?

:|  :|  :rotfl:

That's REALLY FUNNY!!!! If he called me, I would have picked 25, and proved it.

just for the record, the phi function returns the number of numbers less than that number relatively prime to it.

Wait, or is the number of numbers not relatively prime to it?

Uh, oh.... :maybe:

[quote="Treething"]don't a lot of people know that the 16th president is abe lincoln?

i don't even know what circle thing(45) is[/quote]
That's sorta my point...

[quote="Iversonfan2007"][quote="ragnarok23"]. I highly doubt a random person would know what $\phi(45)$ is, but they might know the 16th president.[/quote]

uh, dude, the chances of a random aopser knowing about the phi function is sorta low too[/quote]

Ok, I changed it...

Hm I thought you mean $\frac{1+\sqrt{5}}{2}*45$.

This is nothing to laugh at. An average man from any country is expected not to know that 5 is a pythagorean hypotenuse, though they can actually check this out easily (that is, they know suffiecient maths for the required calculations).  Its a matter of interest.

[quote="ragnarok23"]People on these kind of shows don't know enough math to save their life. They mostly know random knowledge in history and pop culture and whatnot. I highly doubt a random person would know how to factor $x^{2}+4x+4$, but they probably know the 16th president.[/quote]

Few Americans know the presidents by number.

Middle School Academic Bowl Competition

[quote]Coordinator: "While serving as the 32nd president of the United States, this person"

Me: *buzz*

Coordinator: *calls on me*

Me: "Franklin D. Roosevelt."

Coordinator: "That is correct."

Other Team: "You answered before the entire question was read. You must have cheated to know the answer."

Me: "There is only one 32nd president of the United States." [/quote]

Read [url=http://www.nationalgeographic.com/roper2006/findings.html]this[/url] for some sad facts.

I also found out that a higher percentage of Mexicans (18-24 yrs old) can locate the U.S. on the map than Americans.

[quote="Xantos C. Guin"]
I also found out that a higher percentage of Mexicans (18-24 yrs old) can locate the U.S. on the map than Americans.[/quote]

Sorry if I offend anyone, but I just can't help it....

"because they have to know or else they can't jump over the right border"

really, I have nothing against Mexicans...(just know a lot of jokes about it)

Well something I've realized is that the people on there ever since they switched off the old version 1.0 (people on the sidelines have to put things in order in order to get on the chair), they have been "selecting" excessively stupid people.  If this was an episode with Meredith as the host, I bet everyone on there was a complete idiot, due to a lack of filtration as they did on the old episodes.  But yes, americans are definitely dumb :(.

The most shocking thing is the fact that it's not just the person in the chair who got the answer wrong, it was about $70$ percent of the audience(in the room and voting online) that also got the wrong answer...

wooooooooooooow lame. 9+16=25. probably an intro to alg student

I can understand the audience not knowing, cause unless you remember the pythaorgean theorom from school, it'd take a while to figure out, and you don't really have that much time to answer.

BUT, seriously, I don't get how you can not answer that right; just try every single possibility, and count on your fingers if you have to.

yes, but not everybody uses math in their lives, while everybody's job involves memorizing lists of former presidents

[quote="Iversonfan2007"][quote="lingomaniac88"]Back on the subject:

I was watching [i]Super Millionaire[/i] the other day, where a contestant redeemed himself.  The $\text{\$5,000}$ question was this (I'm probably paraphrasing):
[b]To find the radius of a circle, what must you multiply the diameter by?[/b]
A. $\pi$
B. $2\pi$
C. $2$
D. $1/2$
The contestant, after 10 seconds of thinking, chose D.  He got it right.  I don't remember what happened after that.[/quote]

the answer is actually A...it's even worse that the gameshow itself thought D was the right answer..sheesh[/quote]
If a bunch of little middle schoolers start failing geometry tests, I'm blaming you.

[quote="BoesFX"]To ensure that is your final answer, shake hand first :D [It supposed to that way or what?]

[10 seconds thinking for me is normal anyway, you need to calm down first before making decision or you will make mistake][/quote]

dude what A is the right answer...

[quote="Iversonfan2007"][quote="poetryinmath"]diameter x 1/2 = radius

diameter x pi = circumference[/quote]

um....since when?[/quote]
Come on!  ...
[hide]:spam:[/hide]

[quote="Iversonfan2007"][quote="lingomaniac88"]Back on the subject:

I was watching [i]Super Millionaire[/i] the other day, where a contestant redeemed himself.  The $\text{\$5,000}$ question was this (I'm probably paraphrasing):
[b]To find the radius of a circle, what must you multiply the diameter by?[/b]
A. $\pi$
B. $2\pi$
C. $2$
D. $1/2$
The contestant, after 10 seconds of thinking, chose D.  He got it right.  I don't remember what happened after that.[/quote]

the answer is actually A...it's even worse that the gameshow itself thought D was the right answer..sheesh[/quote]

I really, really, really hope you're being sarcastic.


Which one is your favorite?

I really like #24, for one.

I like 24, then 5, then 1.
I think all of us agree that 24 is the coolest.

in a thread where people mention they like Paganini's #24, Paganinio(me) wants to say that he likes the game Klebiam's 24.
funny reply should not be deleted.

I like # 5 and # 16. I play them on the guitar.. Paganini is the master! :)

Yeah agreed, 24 is the coolest. I can even play the arrangement for piano by Liszt. It includes Variations as well.

i like 15, i try to play this on flute. but.... i give it up.


A sphere with mass m is connected with two springs with inconsiderable mass and [b]l[/b] length. The coefficients are $c1$ and $c2$. Study the lagrange of the system.
This is my solution but i'm not sure. Maybe energy isn't correct.
$\begin{array}{l}T = m\mathop x^\bullet \\ V = \frac{{x^{2}}}{2}(c_{1}+c_{2}) \\ L = T-V \\ \mathop x^{ \bullet \bullet }-\frac{{(c_{1}+c_{2})}}{m}x = 0 \\ \end{array}$
[url=http://img402.imageshack.us/my.php?image=untitledpm1.jpg][img]http://img402.imageshack.us/img402/7978/untitledpm1.th.jpg[/img][/url]

well I dont think thats right.
The dimension of $T$ is not even right. :wink:

Actually, the final result IS right.

He just did $\frac{\partial T}{\partial \dot x}$ by mistake instead of $T$, but then didn't differentiate a second time, so it worked out.


[quote="stergiu"]I have seen this problem in a collection for aplications of complex numbers to geometry. But any solution is welcome![/quote]

[b]Probably this figure will help a little.[/b]

[hide="possible hint, not sure"]Analytic? The equations wouldn't be too ugly I think.

Edit: Tried. I think it works out, but the equations are very, very ugly.[/hide]

[quote="worthawholebean"][hide="possible hint, not sure"]Analytic? The numbers wouldn't be too ugly I think.[/hide][/quote]

 Unfortunately I have not a solution. I'll try again some days later, because now i'm very busy.

2 different solutions are at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=91654]Semicircle; not a difficult problem[/url].


A student collects $ 355\text{ cm}^3$ of oxygen saturated with water vapor at $ 27^{\circ}C$. The mixture exerts a total pressure of $ 775\text{ mmHg}$. (At $ 27^{\circ}C$, the vapor pressure of $ H_2O$ is $ 26.7\text{ mmHg}$.)

What is the partial pressure of $ O_2$ in the sample?
How many grams of $ O_2$ does the sample contain?

I don't get it. As far as I can tell I don't have the necessary values to complete the problem. Help?

Well the law of partial pressures says that

the partial pressure of the oxygen + partial pressure of the water vapor = total pressure

and it gives you the total pressure and the partial pressure of the water vapor, so solve.

$ P_\mathrm{O_2}\equal{}P_\mathrm{total}\minus{}P_\mathrm{H_2O}$

This should be enough to figure out the partial pressure of $ \mathrm{O_2}$.

Of course, the legendary formula is useful:

$ PV\equal{}nRT$

You are given $ V$, and $ T$.
You have found $ P$.
You know the molar mass of $ \mathrm{O_2}$

No what I don't understand is if the given partial pressure for $ H_2O$ is just for one mole or if that accounts for all water present. Can someone clarify that?

I can just plug and chug anything...

$ \frac{PV}{RT}\equal{}n\equal{}\frac{(0.355)(775)}{(300.15)(0.0821)(760)}\equal{}0.0147$ moles? What of?

I just don't understand what I'm getting. I mean, I know that
$ P_T\equal{}\sum^2_{k\equal{}1}P_k\equal{}P_{O_2}\plus{}P_{H_2O}$
 but I don't know how to use it here.

[quote="igiul"]No what I don't understand is if the given partial pressure for $ H_2O$ is just for one mole or if that accounts for all water present. Can someone clarify that?[/quote]

The given water vapor pressure is, in fact, refering to the partial pressure of all the water vapor present.

Basically, if you have water in some container at a given temperature, the water will evaporate/condense until the partial pressure exerted by the water vapor is equal to the water vapor pressure (which is just dependent on temperature). You're given that the vapor pressure of water vapor at $ 26^{\circ}C$ is $ 26.7$ mmHG, so that would be the partial pressure of $ P_{H_2O}$

Almost. The molar mass of $ O_2$ is 32, not 16.


Are students invited to the MOSP if they're also to USAMO? So is MOSP a summer program meant to help prepare students for USAMO?

MOSP  is for students who score well on USAMO.

MOP is more often described as training for the IMO.

The USAMO is slightly unique as an olympiad in that the test writers deliberately avoid problems that are made easier by knowledge of advanced theorems (#3 and #6 are often exceptions). I believe the idea is that people with little olympiad preparation but extraordinary ability can score well on the USAMO and make it to MOP, where they will learn all the stuff they need to know for the IMO.

The top twelve scorers ("winners") of the USAMO go to Black MOP, I believe. Six of those are the USA IMO team.

EDIT: Maybe I should explain this color thing:

Black MOP: As stated above.

Blue MOP: Just people who score high enough.

Red MOP: For freshman who score high enough.

In 2006 the USAMO cutoff scores were 25 for black MOP, 18 for blue MOP, and 9 for red MOP.

The usual cut offs are around-
Red Mop=7-9 or so
Blue Mop=17-21 or so
Black Mop=23-28 or something.
So if you get 1.5 problems right, you'll most likely make Red MOP;3 problems should make Blue MOP, and 4 will make Black.

Also:

ONLY freshmen can make red MOP. That is, no 8th graders, no sophomores.
SENIORS cannot make blue MOP.

grgh forgot I gotta qualify via blue this year

guess I actually need to study

What is this "s" business? In the (paraphrased) words of Arnav

"This is associated with a suspicious amount of s-ness"

Hmm I also need to study too. Blue... Argh...

What's so hard about making blue? :P

nah not really, I need to study too. Last year's was pretty optimal layout for me, but I need to study in case the easy ones are geo this year.

[quote="1=2"]What if on the extremely rare occasion that an 8th grader gets a perfect on the USAMO? Does that guy go to MOP?[/quote]
8th graders can make blue MOP.

There's only one 8th grade MOPper this year: Evan O'Dorney

Proof: http://www.unl.edu/amc/a-activities/a6-mosp/a6-1-mosparchives/2007-ma/07mosppart.shtml

[quote="mathcrazed"]Meh... I think 1=2 is like some 7th grade genius XD.  Probably better than me at Mathcounts when I was in 7th grade -_-.  But I suck at MC so ya ~_-.
It's seriously impossible to make MOP in 8th grade unless you're like that kid over there o.O.  I need to start studying also XD.[/quote]
No, you beat him at nationals last year when you were both 7th graders (he's an 8th grader now).
[quote="MellowMelon"]All grade levels can make Blue or Black MOP, with the possible exception of 9th making Blue: I don't know what their policy is, but Hamster was put in Red last year. Zuming said that back before they even had Red MOP (so it was just blue/black, though they didn't have the colors then), there were 7th graders making MOP. These people of course went on to make the IMO later.[/quote]
I thought Reid Barton was the only one?

uhh...

people, stop talking about people making MOP... go make MOP.

OK?

(Also 7th graders making MOP... that was in 2002, when like 90% of USAMO qualifiers made MOP. Obviously the guy was really good... his last name is Lawrence, hint hint)

There have been 7th graders besides Reid.I don't think the "big MOP" of 2002 counts.

[quote="not_trig"]people, stop talking about people making MOP... go make MOP.[/quote]

Agreed.


abstract algebra
calculus
derivative
algebra
polynomial
number theory
relatively prime

I really need help on this question:

If W is the solution space of the differential equation f''''(x) + 2*f''(x) + f(x) = 0, show that
W = Ker((D + 1)^2) © Ker((D + 1)^2).

Note : © is a direct sum

It's wrong. Does that help?

This is based on interpreting the derivative as a linear operator $D$, so our equation is $D^4f+2D^2f+f=(D^4+2D^2+1)f=0$. Factoring that polynomial, it's
$(D^2+1)^2f=(D+i)^2(D-i)^2f=0$. Since $(D+i)^2$ and $(D-i)^2$ are relatively prime as polynomials, their kernels are linearly independent and $\text{Ker}(D+i)^2\oplus\text{Ker}(D-i)^2=\text{Ker}(D^2+1)^2$.
The above assumed our field was $\mathbb{C}$; over $\mathbb{R}$, you can't go beyond $(D^2+1)^2$ anyway.

There's hardly anything there except notation anyway.


calculus
integration
trigonometry
trig identities
calculus computations

Compute the indefinite integral:

$ \int \sec^{4}(x) dx$

So far I have:

$ \int \big((\tan^{2}(x) \plus{} 1)^{2}\big)dx$
$ \equal{} \int \big(\tan^{4}(x) \plus{} 2\tan^{2}(x) \plus{} 1\big)dx$
$ \equal{} \int \big(\tan^{4}(x) \plus{} 2(\sec^{2}(x) \minus{}1) \plus{} 1\big)dx$
$ \equal{} \int \big( \tan^{4}(x) \plus{} 2\sec^{2}(x) \minus{} 1 \big)dx$
$ \equal{} \int \big( \tan^{4}(x) \big) dx \plus{} 2\tan(x) \minus{}x \plus{} C$

But I might have slipped up perhaps, and I don't know what to do with $ \int \tan^{4}(x)dx$, since trig identities don't seem to help here :(

Edit: My futile attempt simplifying $ \int \tan^{4}(x) dx$

$ \equal{} \int \big( \tan^{2}(x)\tan^{2}(x)\big) dx$
$ \equal{} \int \big( \tan^{2}(x)(\sec^{2}(x)\minus{}1)\big) dx$
$ \equal{} \int \big( \tan^{2}(x)\sec^{2}(x) \minus{} \tan^{2}(x) \big) dx$

Which just leaves the $ \tan^{2}(x)\sec^{2}(x)$ integral to compute, since I already know how to deal with $ \int \tan^{2}(x) dx$.

From here I'm clueless, I know $ \int 2\sec^{2}(x)\tan(x) dx \equal{} \sec^{2}(x)$ but that doesn't seem to help here...

$ \int \frac{1}{\cos ^ 4 x}dx\equal{}\int \frac{1}{\cos ^ 2 x}\cdot \frac{1}{\cos ^ 2 x}dx$

$ \equal{}\int (1\plus{}\tan ^ 2 x)(\tan x)'dx\equal{}\tan x\plus{}\frac{1}{3}\tan ^ 3 x\plus{}C.$

Try $ \sec^4x\equal{}\sec^2x(\tan^2x\plus{}1)$ and the substitution $ u\equal{}\tan x.$


function
integration
inequalities
college contests

[quote="pigfly"]Let be given continous function $f(x)$ in $[0,1]$ and satisfying the condition:
\[ \int^1_x f(t) dt \ge \frac {1-x^2}{2},\forall x \in [0,1]\]
Prove that:
\[\int^1_0[f(x)]^2 dx \ge  \int^1_0 xf(x)dx.\][/quote]

Since $\int_0^1[f(x)-x]^2dx\geq0$ $\int^1_0[f(x)]^2 dx \geq 2\int_0^1xf(x)dx - \frac{1}{3}$. So it is sufficent to prove that
$\int_0^1xf(x)dx \geq \frac{1}{3}$. Let's do it: As follows from the condition $\int_0^1 (\int^1_x f(t) dt)dx \geq \int_0^1 \frac{1-x^2}{2}dx$, hence $\int_0^1 xf(x) dx \geq \frac{1}{3}$ which proves our problem.


Prove that if $ a,b,c>0$ and $ abc\equal{}1$ then:

$ ( 1\plus{}\frac {a^2\plus{}b^2\plus{}c^2} {a\plus{}b\plus{}c} ) ( \frac {a^2} {bc\plus{}1}\plus{}\frac {b^2} {ac\plus{}1}\plus{}\frac {c^2} {ab\plus{}1} ) \ge 3$

[quote="moldovan"]Prove that if $ a,b,c > 0$ and $ abc \equal{} 1$ then:

$ ( 1 \plus{} \frac {a^2 \plus{} b^2 \plus{} c^2} {a \plus{} b \plus{} c} ) ( \frac {a^2} {bc \plus{} 1} \plus{} \frac {b^2} {ac \plus{} 1} \plus{} \frac {c^2} {ab \plus{} 1} ) \ge 3$[/quote]
$ LHS \geq  2(\frac{(a\plus{}b\plus{}c)^2}{ab\plus{}ac\plus{}bc\plus{}3})  \geq 6(\frac{ab\plus{}ac\plus{}bc}{ab\plus{}ac\plus{}bc\plus{}3}) \geq 3$


Ok I gotta show that the height of the next figure is given by the formula:

[i]h = d[/i] sin [i]a[/i] sin [i]b[/i] / [i]sqrt[/i] (sin^2 [i]a[/i] - sin^2 [i]b) [/i]

http://img292.imageshack.us/img292/8504/number10kr4.png

Edit: Sorry its ok now >_>

Sure you uploaded it correctly? I see no picture here. :what?:

Ok i had the italics thing after the link >_> its ok now

There exist 2 quantities A, B such that

\begin{eqnarray} \tan a & = & \frac {h}{A} \\
\tan b & = & \frac {h}{B} \\
d^{2} + A^{2} & = & B^{2}. \end{eqnarray}

From the above system we obtain that

$ B^{2} = \frac {d^{2}\tan^{2}a}{\tan^{2}a - \tan^{2}b}.$

Therefore,

$ \begin{eqnarray*}
h &=& B\tan b\\ &=& \frac {d \tan a \tan b}{\sqrt{\tan^{2}a - \tan^{2}b}}\\&=&  \frac {d \sin a \sin b}{\sqrt{\sin^{2}a \cos^{2}b - \sin^{2}b\cos^{2}a}}.
\end{eqnarray*}.$

Why don't you try to refresh the browser?  :arrow:

Is

sin^2 a cos^2 b - sin^2 b cos^2 a

The same as:

sin^2 a - sin^2 b ??

And... if you have any webpage that gives any help on this it would be great =/

From equations (1) and (2) we obtain that

h= A tan a, h=B tan b.

Therefore,

\begin{eqnarray*} h^{2} & = & A^{2} \tan^{2} a \\
h^{2} & = & B^{2} \tan^{2} b. \end{eqnarray*}

Then, $ 0 = h^{2} - h^{2} = A^{2} \tan^{2} a - B^{2} \tan^{2} b.$ Now, since $ A^{2} = B^{2} - d^{2}$ we conclude that

$ 0 = (B^{2} - d^{2})\tan^{2} a - B^{2} \tan^{2} b.$

Things must be clearer now. In any case, do not hesitate to let me know.

Is everything else clear (as an unmuddied lake)?

Please check the PM I just sent you.  :D

[quote="mathwizarddude"]But I'm afraid that the above post still didn't show that sin^2 a cos^2 b - sin^2 b cos^2 a = sin^2 a - sin^2 b.[/quote]

They are not equivalent...  :D  I guess there is something wrong with the formula that [i]xcenario[/i] put forward in the first place.

Its from a practice we got, we started trigonometry this week and we got this.

Ima tell you on monday if its true the formula, Ill show my teacher all this, in case the formula its true, ill post again.

Thanks anyways =]

Well, we discussed the problem, and we could actually finish it, but not using tan.

Lets put that the hypotenuse of angle [i]a[/i] is x and the hypotenuse of angle [i]b[/i] is y.

So:

1) sin a = h / x  --> x = h / sin a

2) B^2 = A^2 + d^2

3) x^2 = A^2 + h^2  substituing 1 with 3 we get:

4) h^2 / sin^2 a = A^2 + h^2  --->  A^2 = h^2/sin^2 a - h^2

5) sin b = h / y  --> y = h / sin b

6) y^2 = h^2 + B^2

7) h^2 / sin^2 b = h^2 + B^2

8) h^2 / sin^2 b = h^2 + A^2 +d^2

h^2 / sin^2 b = h^2 + h^2 / sin^2 a - h^2 + d^2

(h^2 / sin^2 b) - (h^2 / sin^2 a) = d^2

h^2 ( sin^2 a - sin^2 b / sin^2 a  sin^2 b) = d^2

h^2 = d^2 sin^2 a sin^2 b / sin^2 a - sin^2 b

h = d sin a sin b / sqrt (sin^2 a - sin^2 b)

Sorry for all this, just copied it all. And im going to start learning how to use LaTeX today. Thanks anyways =] [/i]

The solution I'd posted is correct, too.

From

$ d^{2} = B^{2} - A^{2} = \left(\frac {h}{\tan b}\right)^{2} - \left(\frac {h}{\tan a}\right)^{2},$

we obtain that

$ h^{2} = \frac {d^{2}}{\left(\frac {1}{\tan b}\right)^{2} - \left(\frac {1}{\tan a}\right)^{2}}.$

Since

\begin{eqnarray*} \left(\frac {1}{\tan b}\right)^{2} - \left(\frac {1}{\tan a}\right)^{2} & = & \frac {\tan^{2} a - \tan^{2} b}{\tan^{2} a \tan^{2} b} \\
& = & \frac {(\tan^{2} a + 1) - (1 + \tan^{2} b)}{\tan^{2} a \tan^{2} b} \\
& = & \left(\frac {(\sec^{2} a ) - (\sec^{2} b)}{\sin^{2}a \sin^{2} b}\right) \cdot (\cos^{2} a \cos^{2} b) \\
& = & \frac {\cos^{2} b - \cos^{2} a}{\sin^{2}a \sin^{2} b} \\
& = & \frac {(\cos^{2} b - 1) + (1 - \cos^{2} a)}{\sin^{2}a \sin^{2} b} \\
& = & \frac { - \sin ^{2} b + \sin^{2} a}{\sin^{2}a \sin^{2} b} \end{eqnarray*}

the veracity of my claim follows at once.

[b][size=150]Moral:[/size][/b] Different solution-approaches might lead to different formulas. One thing they all ought to have in common is correctness, though.  :lol:


calculus
geometry
integration
linear algebra
analytic geometry

I could really use some advice on self studying of Calculus. We homeschool and right now are working out of a Saxon Calculus textbook. We found this site and are ordering the problem solving texts.  Should we use a text for AP Calculus instead?  Thanks.

Age 17

Mathematical Maturity  Worked through Algebra 11 and Advanced Math by self. Student has great theoretical comprehension but soemtimes less than stellar performance on math tests. Too many "dumb" mistakes which he sees as soon as he gets his test back. (Frustrating for both of us!)  In last part of Advanced Math finally started telling me he didn't know how to do a problem. He works answers to everything, no solution manuals--only answer keys.

He thinks Applied Mathematics is what he wants to study in college. I wonder if everyone who pursues this degree starts off as a math whiz..............

Okay, I don't know what Algebra 11 or Advanced Math means, I'm from Sweden. But I'm going trough Principles of Mathematical Analysis right now, which is a really great text on analysis. I would recommend it before calculus, and any motivated high school student won't find it that difficult. And there are no prerequisites but some mathematical maturity. Maybe some linear algebra, but only the basics.

That Principles of Mathematical Analysis book might be too advanced for a high school student to start out with. That's a book typically used by honors college students here in the US once they've already learned calculus. I think it's good to master the basics of calculus and to get a feel for it before you jump into Analysis.

A good Calculus book to use is Stein and Barcellos, [u]Calculus and Analytic Geometry, 5th ed.[/u] It also comes with a solutions manual. This is the book used by first and second semester calculus students at many Universities and is well suited for preparing for the AB or BC AP Exams. I would also recommend getting an AP Calculus problem and solution book.

Another good, slightly more advanced book, is Serge Lang's [u]A First Course in Calculus, 5th ed.[/u].

Here's a bibliography of Calculus books you can look at - http://www.ericweisstein.com/encyclopedias/books/Calculus.html

As for whether everyone who pursues a math degree starts off as a math wiz, the answer is certainly no. I've met many people who had no idea what they wanted to get their degree in before they went to college, who ended up getting their degree in math and doing very well at it. As long as you have a good foundation, you can learn a great deal of math very rapidly once you get to college.

(I think an Advanced Math course  is also known as a Precalculus in the US.)

Thank-you for the recommendations. We are checking them out. If I can actually get the correct solutions manual for the Sherman, Stein, Barcellos text it might be the way for us. Our "student" worked through a college text for Chemistry and Physics but with some online instruction. Can a student really get through this text for Calculus by self-studying and a solutions manual?

 Have you heard of a book by Paul Foerster, Calculus Concepts and Applications published by Key Curriculum Press?  I've noticed in homeschooling circles that publisher is getting more notice.

I learned calculus by starting with [u]Calculus the Easy Way[/u] (sorry but I forget the author now), which is a rather fun introduction to calculus. I then got Loomis's calculus text, which I thought was better than many others. I asked my father to help me with a few things that I wasn't able to figure out, but I mostly understood it by myself. I think that sequence is good. I have heard the Apostol calculus text is quite good, and I enjoyed the Stewart text as it included some problem-solving aspects of calculus that many other books ignore.

I agree with gauss.  I read through Principles of Mathematical Analysis [i]after[/i] my calc course and it probably would've been a bit confusing to start off with.   I used Apostol's Calculus Volume I (and for multi, Volume II) in class, which is an awesome book, but I probably wouldn't recommend it as the primary text for someone who's learning the material on his own for the first time.  (Integration before differentiation... eek!)

I didn't say it was an easy task to read POMA before any standard calculus text, but it is doable. One can also read POMA and usa a standard calculus text as a reference or something.


In these questions a ,b and c are integers lying between 1 and 9

1. The number 13ab45c is divisible by 792, find a,b and c.

2. The number 62ab427 is divisble by 99 find a and b.

[quote="Michael Niland"]2. The number 62ab427 is divisble by 99 find a and b.[/quote]

I'm too tired to do part a.  Anything that involves me pulling out a pencil right now is too much work.

[hide]
62ab427 has to be divisible by 9 and 11, which means 6+2+a+b+4+2+7=21+a+b is a multiple of 9.  It can be either 27 or 36, so that a+b=6 or 15.  62ab427 also has to be divisible by 11, which means 6-2+a-b+4-2+7=13+a-b has to be a multiple of 11 which means a-b=-2.  The system a+b=n and a-b=m will have integer solutions if and only if n and m are of the same parity (divisible by 2-ness).  Therefore, we get that a+b must be 6.  Since a+b=6 and a-b=-2, we get a=2, b=4.[/hide]

$1$ isn't any harder, because $8\vert 1000$.


limit
trigonometry
logarithms
calculus
calculus computations

evaluate
lim x->0   (10^x - 2^x - 5^x + 1)/(x + tanx)

sorry i dont know LATEX.. :(

$ \lim_{x \rightarrow 0} \frac {10^x \minus{} 2^x \minus{} 5^x \plus{} 1}{x \plus{} tanx}$
By L'hopital's rule
$ \equal{} \lim_{x \rightarrow 0} \frac {10^xln10 \minus{} 2^xln2 \minus{} 5^xln5}{1 \plus{} sec^2x}$
$ \equal{} \frac {ln10 \minus{} ln2 \minus{} ln5}{1 \plus{} 1}$
$ \equal{} \frac {0}{1 \plus{} 1}$
$ \equal{} 0$

You can even approximate $ \tan x \approx x$ near zero to avoid it's derivation.

Just use $ \lim_{x\rightarrow 0} \frac {\tan x}{x} \equal{} 1,\ \lim_{x\rightarrow 0} \frac {a^x \minus{} 1}{x} \equal{} \ln a\ (a > 0,\ a\neq 1)$.


Since: $2yz \leq z^2+y^2$, so $\frac{x^2}{y^2+z^2+yz} \geq \frac{2x^2}{3(y^2+z^2)}$

Hence: $\sum \frac{x^2}{y^2+yz+z^2} \geq \frac23 \cdot \sum \frac{x^2}{y^2+z^2} \geq \frac{2}{3} \cdot \frac{3}{2} =1$
(According to Nesbits.)

I did not use $xyz=1$...Did I make any mistakes?

I solved it almost by the same way
without using $xyz=1$, too.

I think that this is the most direct solution!!!

With Adreescu's inequality we find that LHS>1 I think

I don't think so.
But write down your solution..

If xyz=n, then if you replace x,y, and z with $x/\sqrt[3]{n}$,$y/\sqrt[3]{n}$, and $z/\sqrt[3]{n}$, then the inequaltiy reduces to the exact same thing, except with xyz=1.  I think that's why your solutions didn't make any use of xyz=1.

the ineq is homogenous, so there can be no alike constraint. ;) that's only for non-homogenous inequalities. You should know the xyz=1 is fake even before you start! :)

[quote="probability1.01"]This inequality looks like there has to be a geometric interpretation.  Can anyone find one?[/quote]

I drew out a triangle like this:

Have line with length z, rotate 120 degrees, draw line with length y, rotate another 120, draw line with length x. It should be a graph of 3 lines meeting at a single point. Connect the other end of these lines to form a triangle. The inequality is the length of these lines over the side they "point" at.

I remember someone posting about the Fermat point how its a point insdie the triangle such that the distances to the 3 vertices are minimum? not quite clear right now...


There is a convex lens of focal length $ f \equal{} 10cm$.A point isotropic source of light($ \lambda \equal{} 5000 A^{0})$ is kept at a distance of $ a \equal{} 15cm$ from it on one side of the lens.A screen is placed at a distance $ D \equal{} 60 cm$ from the lens on the other side.Now the lens is cut [b]along it's principle Axis[/b] into two equal halves and the halves are moved apart a distance $ d \equal{} 0.5 mm$apart.Find the number of Interference Bands obtained on the screen.

Is the source of light on the (original) principal axis?? Otherwise this sum will become pretty tough.

anyone trying this....it is really good :)
[hide="hint"]
the images act as two objects for YDSE[/hide]

considering the middle .5mm his covered with black paper
it is same as YDSE with
D=30 cm
d=1mm
wavelength =5000A
no of fringes is 4001

no that's wrong :)
a just little more thought...
[hide="hint"]The images [b]don't[/b] behave as isotropic sources[/hide]


If the number $ \sqrt[3]{\sqrt[3]{2}\minus{}1}$ can be written as $ \sqrt[3]a\plus{}\sqrt[3]b\plus{}\sqrt[3]c$ , then what is $ a\plus{}b\plus{}c$ ?

Assuming that $ a,b,c$ don't have perfect cube factors (other than $ 1$)...

Since $ (a\plus{}b\plus{}c)^3\equal{}a^3\plus{}b^3\plus{}c^3\plus{}3(a\plus{}b)(b\plus{}c)(c\plus{}a)$, we can cube both sides of $ \sqrt[3]{\sqrt[3]2\minus{}1}\equal{}\sqrt[3]a\plus{}\sqrt[3]b\plus{}\sqrt[3]c$ to get

$ \minus{}1\plus{}\sqrt[3]2\equal{}a\plus{}b\plus{}c\plus{}3(\sqrt[3]a\plus{}\sqrt[3]b)(\sqrt[3]b\plus{}\sqrt[3]c)(\sqrt[3]c\plus{}\sqrt[3]a)$.

Because the product of sum of radicals on the RHS must multiply to $ \frac{\sqrt[3]2}3$, we can write this as

$ \minus{}1\plus{}\sqrt[3]2\equal{}a\plus{}b\plus{}c\plus{}\sqrt[3]2\implies a\plus{}b\plus{}c\equal{}\boxed{\minus{}1}$.

Of course, this isn't completely rigorous, but...

I see! Thanks a lot math154 I appreciate all your help.


Let a_n be the well-known Fibonacci sequence. Express a_(2n) in terms of only a_(n-1), a_n and a_(n+1).

the general form is

F_{n+m} = F_{m-1} F_n + F_m F_{n+1}


Hey guys, a friend referred me to this site for some help...
I need some  help on how we would use the Ceva's Theorem today, I tried looking around the internet for what we would use this theorem for, but honetly I see no point to it..
Thanks for your guys' help.  :lol:

Well how would you go about proving that the medians of a triangle are concurrent? Can you find a way to do it with Ceva's theorem?

It's not very useful outside of Olympiad problems to be honest. It can prove useful on AIMEs and AMCs though.
As tonypr mentioned, it's a good way for doing certain concurrency problems. If you have concurrency, you can use it backwards to find sides or angles (with trig Ceva).

Also, mass points are very usefull and quick too for AIME's. They're actually Ceva's in disguise.