trigonometry
function
absolute value
geometric series
real analysis
real analysis unsolved

$ a>1$, $ s\in(1,2)$, prove that
\[ \sum_{n\equal{}0}^{\infty}a^{n(s\minus{}2)}\sin(a^nx)\]
is continuous and nowhere differentiable on $ R$.

what can we say about
\[ \sum_{n\equal{}0}^{\infty}a^{n(s\minus{}2)}|\sin(a^nx)|\]
with the same assumptions ?
is it continuous and nowhere differentiable on $ R$ too ??

I believe so because all the absolute value sign does is flip one part of the graph over an axis.  The actual graph itself is not changing, just its orientation.

[quote="Leonard"]$ a > 1$, $ s\in(1,2)$, prove that
\[ \sum_{n \equal{} 0}^{\infty}a^{n(s \minus{} 2)}\sin(a^nx)
\]
is continuous and nowhere differentiable on $ R$.
[/quote]
I can do the easy part, proving it is continous. Note $ \left| a^{n(s\minus{}2)} \sin (a^n x) \right| \leq a^{n(s\minus{}2)}$.
Now $ \sum_{n\equal{}0}^{\infty} a^{n(s\minus{}2)} < \infty$ because $ \minus{}1<s\minus{}2 <0$ so $ 0< a^{s\minus{}2} < 1$ so it converges by geometric series test. By Weierstrass theorem we have a uniform limit of continous functions, which must be continous. Thus, the function defined by that series is a continous function everywhere.

For the differentiable part, that is quitre similar to the Weierstrauss function, as well as many others continuous but not differentiable.  I can't rigorously prove it though.

For a function to be differentiable at a point $ b$, its oscillation (difference of max and min) on intervals $ [b\minus{}\epsilon,b\plus{}\epsilon]$ must be at most $ C\epsilon$, where $ C$ remains constant as $ \epsilon\to 0$. This is violated by the $ n$th term of the series, where $ n$ is such that $ a^{\minus{}1}\epsilon \le 2\pi a^{\minus{}n}\le \epsilon$. Indeed, the oscillation of this term over its period is $ a^{n(s\minus{}2)}\ge \left(\frac{\epsilon}{2\pi a}\right)^{2\minus{}s}$, which is much larger than $ \epsilon$ due to $ 2\minus{}s<1$.

To complete the proof, one must show that the bad behavior of the $ n$th term is not magically canceled by the others, but I'll leave that part as an exercise.  :)

When does one generally learn how to rigorously complete epsilon-delta proofs?  Would it be Analysis?

Yes, in Real Analysis. It's sometimes called advanced calculus.

Let's rename $ n$ from my previous post as $ n_0$. We expect two things to happen: (1) the contribution of terms with $ n > n_0$ is small; (2) the terms with $ n < n_0$ are changing relatively slowly, and therefore have small oscillation on the interval $ [b \minus{} \epsilon,b \plus{} \epsilon]$.
For (1), notice that $ \sum_{n \equal{} n_0 \plus{} 1}^{\infty}a^{n(s \minus{} 2)}|\sin(a^n x)|\le \sum_{n \equal{} n_0 \plus{} 1}^{\infty}a^{n(s \minus{} 2)} \equal{} \frac {a^{(n_0 \plus{} 1)(s \minus{} 2)}}{1 \minus{} a^{s \minus{} 2}}$.
For (2), observe that $ a^{n(s \minus{} 2)}\sin(a^nx)$ is a Lipschitz function with constant $ a^{n(s \minus{} 1)}$, which implies that its oscillation on $ [b \minus{} \epsilon,b \plus{} \epsilon]$ is at most $ 2\epsilon a^{n(s \minus{} 1)}$. Summing over $ n \equal{} 0,\dots,n_0 \minus{} 1$, we get at most $ 2\epsilon \frac {a^{n_0(s \minus{} 1)} \minus{} 1}{a^{s \minus{} 1} \minus{} 1}\le 4\pi a^{1 \minus{} n_0}\frac {a^{n_0(s \minus{} 1)} \minus{} 1}{a^{s \minus{} 1} \minus{} 1} \equal{} 4\pi a\frac {a^{n_0(s \minus{} 2)} \minus{} a^{ \minus{} n_0}}{a^{s \minus{} 1} \minus{} 1}$.

Recall that the oscillation of the term with $ n \equal{} n_0$ was $ a^{n_0(s \minus{} 2)}$, and we would like the sum of estimates in (1) and (2) to be less than this number. Alas, that does not happen. We should modify the approach: instead of relying on a single term $ n \equal{} n_0$ to create large oscillation, we should consider the finite sum over $ n_0 \minus{} k\le n\le n_0 \plus{} k$, where $ k$ is chosen so that the tail estimates in (1) and (2) work. However, this will require a more careful oscillation estimate for the finite sum.


trigonometry
geometry
circumcircle
trig identities
Law of Sines
angle bisector
geometry proposed

There was some trivial angle problem in pre-olympiad section, now I would like to expand it a little

Prove: for that $\triangle ABC$ from here  http://www.mathlinks.ro/Forum/viewtopic.php?t=47637 it stands \[c(a+b+c)=a(a+b)\] where $a=BC,b=AC,c=AB$

I don't know trigonometry yet, so this solution is sertanly correct but i don't understand it, and you don't have to write elementary solution, I am goingt to try to solve it because i din'd, i just found that problem and posted it BUT
i am interested in $\"double---angle---theorem\"$ what is that?...

Double Angled Theorem: In $\triangle{ABC}$, if $\angle{B}= 2\angle{C}$, then $b^2=c^2+ac$.

The proof is easy, just draw the angle bisector of $\angle{B}$. This is an extremely useful theorem so I gave him the nave Double Angle Theorem.

Have you found a geometrical way to prove it?

I wasn't solving it i was doing some number theory... i'll try to solve as soon as I disconnect  :)  :)

Here it is, elementary solution, maybe some things from here be can be shown lot
easier but i don't have formal solution, so i have to write my own

First, this is notation: $\angle A=40^o,\angle B=60^o,\angle C=80^o$ and sides
$a=BC,b=AC,c=AB$, so we have to prove $a(a+b+c)=b(b+c)$. Now let's start:

First construct point $C_1$ on line $BC$ such that $B-C-C_1$ and $CC_1=CA=b$ so
$\angle C_1AC=\angle AC_1C=40^o=\angle A$ and so $BA$ is tangent line to
circumcircle of triangle $ACC_1$ and from power of the point B we have $c^2=a(a+b)$
or as you like it to call "Double angle theorem", we should prove that
\[a(a+b+c)=b(b+c)\] now when we have $c^2=a(a+b)$ and get that into equation we
obtain \[a(a+b)+ac=b(b+c)\Leftrightarrow c^2+ac=b(b+c)\Leftrightarrow
c(a+c)=b(b+c)\Leftrightarrow \frac{b}c=\frac{a+c}{b+c}\]

So that is the beginning, now it comes tricky part:

Let $M$ be the point such that $B-A-M$ and $AM=b$ so angles $AMC$ and $MCA$ are both
$20^o$. $N$ is point such that $C-A-N$ and $AN=c$ so angles $ANB$ and $ABN$ are also
$20^o$,so $\angle NBC=60^o+20^o=80^o$ so $NB=NC=c+b$. So we see that $MC||NB$ and
$NB=b+c$, $AN=c$ and $AC=b$ so all we have to prove now is $MC=a+c$.

Let $X$ be the point on segment $MC$ such that $MX=a$ (that point must exist,
because $MC$ is greater then $b$ and $b>a$), so $\triangle XBC$ is isosceles and
$\angle XCB=100^o$ so angles $BXC,XBC=40^o$, and from that $\angle MBX=20^o=\angle
BMX$, now $MB=b+c=NB$ so we obtain $\triangle MBX\cong \triangle ANB$ and from that
$MX=c$ so \[MC=MX+XC=a+c\]

So that's it... it maybe looks long but really isn't...

Bye!!!


So just memorize the Unit Circle.
[img]http://upload.wikimedia.org/wikipedia/commons/c/c9/Unit_circle_angles.svg[/img]

[quote="sizor"]In what cases can $ sec(\theta) \equal{} 1$ or $ \minus{}1$? I thought that the measure of the hypotenuse of a right triangle was always greater than that of the base.

Wouldn't it have to be $ r \equal{} |x|$ in order for $ sec(\theta) \equal{} 1$ or $ \minus{}1$?

The math book that I'm teaching myself from says that $ sec(\theta)$ can be $ 1$ or $ \minus{}1$ but doesn't give examples.

Thanks.[/quote]

The same case where $ \cos{\theta}\equal{} 1$ or $ \minus{}1$; 0 or 180 degrees.

[quote="Anaxerzia"][quote="sizor"]In what cases can $ sec(\theta) \equal{} 1$ or $ \minus{}1$? I thought that the measure of the hypotenuse of a right triangle was always greater than that of the base.

Wouldn't it have to be $ r \equal{} |x|$ in order for $ sec(\theta) \equal{} 1$ or $ \minus{}1$?

The math book that I'm teaching myself from says that $ sec(\theta)$ can be $ 1$ or $ \minus{}1$ but doesn't give examples.

Thanks.[/quote]

The same case where $ \cos{\theta}\equal{} 1$ or $ \minus{}1$; 90 degrees.[/quote]

Hmm. I thought $ \cos(90) \equal{} 0$ because $ x \equal{} 0$. Also, I thought $ \sec(90) \equal{} undefined$.

[quote]You have to expand your definition of trigonometric functions from the right triangle to the unit circle.

The unit circle allows you to define angles that are not in $ 0 <\theta < 90$
You need these other angles in order to get values like 1 or -1

If you look at the graph of the secant, you will see that the range is $ x\geq 1$ or $ x\leq\minus{}1$
use the unit circle to think about trig rather than right angles and it will be easier

also, on a side note, you can backslash trig functions in latex. It makes them more easily readable
$ \cos\theta$ is better than $ cos\theta$[/quote]

Hmm. So, if $ \theta \equal{} 0$ or $ \theta \equal{} 180$ then $ |\sec\theta| \equal{} 1$, right?

Thanks for the latex tip, by the way.


Hey! I am minerva, which is nothing like my real name.
To celebrate the new decade, I started an AoPS account. So yes, I am a new member to the AoPS community and proud of it!

Artemis Fowl FTW?

Maybe this should be moved to Games & FF...

I do nothing for the new year, really.
Except maybe go to the mall/movies.
Lol, look at [url=http://online.wsj.com/article/SB123051100709638419.html]this.[/url]

Started doing olympiad problems, which I am happy I did.

Sushi! Baguettes with goat cheese! Pina Colada! Eggnog infused with brandy! Night at the Museum! Etcetera!

Well, didn't do too much in the morning........but then I went to Chicago in the evening to go to an Italian restaurant and see "The Pillowman" play.

go watch a movie, and of course sleep in :D

It's not a new decade, since the calendar began at year 1. 2011 is a new decade.

well, to some people, it is considered a decade. To other people they do not consider it a decade, so it is opinion differences, as I always say :P

The new decade cycles depending on the units digit (0-9). So this indeed a new decade.

I kill time, literally.

Sigh........This decade was one of the worst decades in American history(particularly 2005-2008)

ohmygoshness. 2009 WAS A FAIL YEAR. 2010 must be better or else..:P
Ya I'm threatening the year, isn't that weird"

Many birthdays were celebrated in 2009, although slightly less than in 2008.

[quote="jjfun1"][quote="Maybach"]2009 wasn't that eventful.[/quote]
Are you kidding?
Lots of things happened in 2009.[/quote]

Yes, but I kind of considered 2008 more interesting.  It was the advent of the recession, ushered in  political election, and things were more turbulent.

But yes, 2009 was eventful.

i think we should start the year with ice cream! ! ! who doesnt like ice cream i mean seriously? its so yummy to mah tummy!  :D

[size=84][color=red]Please don't revive threads. Thanks. [/color][/size]


linear algebra
matrix
search
trigonometry
geometry
calculus
derivative

I hope I've started the thread in the correct forum. :)
I've tried to search for similar threads first, though I didn't find something helpful.

So.. Let the following space-state system:

$ \dot{x}(t)\equal{}Ax(t),\: \: x(0)\equal{}x_0$

where $ A\equal{}\left[\begin{matrix}0&1\\\minus{}\omega^2&0\end{matrix}\right]$

1) Find a similarity transform which diagonalizes matrix A.
2) Using the diagonalization above, find the fundamental matrix $ \Phi(t), \: \Phi(0)\equal{}I$ and if $ x_0\equal{}(\minus{}1\;\;0)^T$ find $ x(t)$, for $ t>0$.

-------

1) Simple enough.

Eigenvalues:
$ |\lambda I\minus{}A|\equal{}0\\
\left|\begin{matrix}\lambda&\minus{}1\\ \omega^2&\lambda\end{matrix}\right|\equal{}0\\
\lambda^2\plus{}\omega^2\equal{}0\\
\lambda\equal{}\pm\mathbf{j}\omega$

Eigenvectors:
$ V_1\equal{}\left[\begin{matrix}1\\{j}\omega\end{matrix}\right]\equal{}V_2^*$

And: $ T\equal{}\left[\begin{matrix}1&1\\{j}\omega&\minus{}{j}\omega\end{matrix}\right]$

Where $ x\equal{}Tz$.

So $ \dot{x}(t)\equal{}Ax(t),\: \: x(0)\equal{}x_0$ becomes: $ \dot{z}(t)\equal{}\Delta z(t),\: \: z(0)\equal{}T^{\minus{}1}x_0$

Where: $ \Delta\equal{}\left[\begin{matrix}{j}\omega&0\\0&\minus{}{j}\omega\end{matrix}\right]$

2) If I had real and distinct eigenvalues, I would find $ \Phi(t)$ like that:

$ \Phi(t)\equal{}e^{\Delta t}\equal{}\left[\begin{matrix}e^{\lambda_1t}&0\\0&e^{\lambda_2t}\end{matrix}\right]$

However, matrix $ \Phi(t)$ should be real.

[hide]
In fact, if I'm correct, using the original system, I get:
$ \Phi_A(t)\equal{}\mathcal{L}^{\minus{}1}((sI\minus{}A)^{\minus{}1})\equal{}\left[\begin{matrix}\cos(\omega t)&\minus{}\frac{1}{\omega}\sin(\omega t)\\ \sin(\omega t)&\cos(\omega t)\end{matrix}\right]$

Where $ \mathcal{L}^{\minus{}1}$ denotes inverse Laplace transform. This "hidden area" is just a parenthesis.[/hide]

So, any ideas of how could I continue with that? Thanks in advance.  :)  :)  :)

Check the formula in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=177357]this topic[/url].

I think you have a small error in what you wrote. It should be

$ \Phi_A(t)\equal{}\left[\begin{matrix}\cos(\omega t)&\frac{1}{\omega}\sin(\omega t)\\
\minus{}\omega\sin(\omega t)&\cos(\omega t)\end{matrix}\right]$

The discrepancy isn't that hard to spot: the derivative of this at $ 0$ should be the original matrix.


topology
calculus
real analysis
real analysis unsolved

Suppose that D is dense in R. Conjecture that D is the intersection of open sets and prove your conjecture.

What if the conjecture is false?  :maybe:

Are these just general topological spaces?  If so, this counterexample is as simple as they come:

Let $ R\equal{}\{x,y\}$, with topology $ \tau\equal{}\{\emptyset, R\}$.  Then $ D\equal{}\{x\}$ is dense in $ R$ but is not the intersection of open sets.

Of course, this kind of example is impossible in a metric space, which is probably how the question was intended...

Considering that this is a "Calculus - Real Analysis" forum, I conjecture that R is in fact $ \mathbb R$. (I'll leave it to the original poster to prove or disprove this conjecture.)

Well, every subset $ S$ of $ \mathbb R$ is the intersection of some family of open sets. Let $ O_x \equal{} {\mathbb R}\setminus\{x\}$. Then
\[ S \equal{} \cap_{x\notin S} O_x.
\]
(Also every set $ S$ equals the intersection of all open sets that contain $ S$.)

In general, however, $ S$ can be only represented as the intersection of (at least) continuum open sets. It is [b]not true[/b] that every dense set is the intersection of [b]countably[/b] many open sets.


Two students are watching an action film in which a car drives down a ramp onto the back of a moving lorry. Both are moving at high speed, the car slightly faster than the lorry.

one student complains that this is impossible because the car would not be able to stop before hitting the cab of the lorry.

The car has mass 1250kg and is moving at a speed of 28.0m/s. The lorry has mass 3500 kg and a speed of 25.5m/s. The length of the flat back of the lorry allows a braking distance of 5.0m.
[b]
By considering both momentum and energy show that the stunt is possible, provided a minimum force of about 600N slows the car down. You should support your explanation with calculation

Treat the situation as one in which two objects join together[/b]

Help with this would be much appreciated. Thanks!

What means "bump bump bump bump"?

You already got it?

No, sorry I thought this was universal. Maybe it's an English thing?

It 'bumps' the post up the ranks so it doesn't get lost/ ignored at the bottom. Not that it worked! haha.

oh come on people you know you want to.

Tell you what- I'll send you a nice packet of English earl grey tea if you help me. It's to die for!

oh okay, booze it is! Come on clever people- help me!

Success in school depends in figuring out what the teacher thinks is the correct answer. If they give you enough hints (or multiple choice questions), it is easier.
[quote="JasonSullivan"]The car has mass 1250kg and is moving at a speed of 28.0m/s. The lorry has mass 3500 kg and a speed of 25.5m/s. ........
[b]
By considering both momentum and energy .........
Treat the situation as one in which two objects join together[/b][/quote]
Before the "joining" the momentum is
p=(1250kg)(28.0m/s)+(3500 kg)(25.5m/s)=124250kg*m/s
Knowing that momentum is conserved (no external forces) and calling the new speed of the "joined" vehicles v:
(4750 kg)v=124250kg*m/s
Solving for v,
v=26.157895m/s
(I would round up to 26.2m/s, but it would not work)
[quote="JasonSullivan"]The length of the flat back of the lorry allows a braking distance of 5.0m.
[b]
By considering both momentum and energy show that the stunt is possible, provided a minimum force of about 600N slows the car down.[/b][/quote]We consider only kinetic energy (no mention/data about any other energy that could be changing), and any change must be due to energy "lost due to the work of a non-conservative force (the friction/braking force).
Before the "joining":
KE=(1/2)(1250kg)(28.0m/s)^2+(1/2)(3500 kg)(25.5m/s)^2 =490000J+1137938 =1627938J
After the vehicles are "joined":
KE=(1/2)(4750kg)(26.157895m/s)^2 =1625059J
The energy lost is 1627938J-1625059J=2879J, which equals the work of the force slowing down the car.
Here they rounded up to 3000J and made that equal to the work of a force F acting over a distance of 5 meters
3000J = F(5m)
Solving for F,
F= 3000J/5m =600N

Thank you very much, I did much the same but didn't get the last part. What do you mean about 'success in school'? I didn't say anything about school?

[quote="ffao"]That seems to be the intended answer, but I don't get the last part either. If we consider just the car, we have

$ \Delta E_k \equal{} 625(28^2 \minus{} 26.157895^2) \equal{} 62352.83...$
$ F \approx \frac {62353}5 \equal{} 12470.6$

I don't see why that wouldn't be right (if you consider the lorry, you get about the same force), and so the simplified "proof" would be meaningless. Could you clarify?[/quote]I can only hope to clarify what I tried to do with my posted "answer." I agree that the simplified "proof" is not convincing (to me or you), but it sounds like it could be a convincing explanation to a "teacher" looking for a 600N answer to a problem seen as "simple".

The problem did not look simple to me. Maybe it will be when we find the right point of view. Can you help?

The car is pushing forward on the lorry as the lorry pushes backwards on the car, while they are both moving forward. The distance of 5 m is the displacement of the car with respect to the lorry, which is not moving at a constant velocity with respect to the ground. I thought about it, and assumed (just educated intuition) that I could use any frame of reference as long as it was an inertial one (one moving at a fixed velocity). The problem here is that the only point (other than one on the ground) that I can assume to be moving with a fixed velocity is the center of mass of the car+lorry system. From the moment the two vehicles are "joined" until they cease to move with respect to each other, the center of mass of the car+lorry system is supposedly moving at a fixed velocity (approx. 26.2 m/s with respect to the ground). If I fix my set of coordinate axes on that center of mass, I will have an inertial reference system. In that system, the lorry is moving backwards and the car is moving forward for a short while, until they are not moving with respect to each other any more. That makes the problem more complicated than my simple "solution." It was more complicated than I could handle without further thought, and my head already hurt. I could not think of a better explanation (one that actually convinced me), but thought that maybe someone else could (maybe even I, if I think about it long enough). It may not be a matter of knowledge or genius, but of staying with the problem long enough. Can you? Can I? Can we? If we get to a convincing answer we will have trully learned.

JasonSullivan and ffao,
I was just hoping that my posting would instigate a better explanation. Maybe we can join forces to generate one. The TV show "Myth Busters" had a segment where they did an experiment similar to the problem, but I do not think they will acre to provide us with a scientific explanation.
What I meant by "school" is that artificial situation where one person behaves as having all the right answers and capable of doling out wisdom for the people around/in front to accept unquestioningly. (It happens painfully often, though not always). The pompous windbag could be an instructor in an educational institution, or a textbook author. My religious framwork does not include blind faith in those people. I believe that we can all learn from each other and that is what I am trying to do here. I'm open to all ideas, and I would welcome any convincing arguments. Just convince me of anything and I'll be on your side of the argument.

[quote="ffao"]I have just noticed one thing -- the car doesn't travel 5 meters. Instead, it travels [i]5 meters more than the lorry[/i], which is a huge difference. I still don't think that the solution above is correct, but it explains the difference in forces I got for the lorry and the car (small difference, but noticeable) and should give us enough grounds to reach a correct answer.

EDIT: Say the lorry traveled $ x$ meters. So, the differences in the kinetic energies of the car and the lorry are:

$ |\Delta E_{k_c}| \equal{} 625(28^2 \minus{} 26.157895^2) \equal{} 62352.83...$
$ |\Delta E_{k_l}| \equal{} 1750(25.5^2 \minus{} 26.157895^2) \equal{} 59474.57...$

using that $ F \equal{} \frac {W}{d}$ we have the system

$ F \equal{} \frac {59474.57}{x}$
$ F \equal{} \frac {62352.83}{x \plus{} 5}$

$ F \approx 575.62$

Now, this is an answer I'm satisfied with ;)[/quote]

yes this makes sense.

[b]"What I meant by "school" is that artificial situation where one person behaves as having all the right answers and capable of doling out wisdom for the people around/in front to accept unquestioningly. (It happens painfully often, though not always). The pompous windbag could be an instructor in an educational institution, or a textbook author. My religious framwork does not include blind faith in those people. I believe that we can all learn from each other and that is what I am trying to do here. I'm open to all ideas, and I would welcome any convincing arguments. Just convince me of anything and I'll be on your side of the argument." [/b] A very refreshing point of view KMST. My old maths teachers acted like they knew everything, when I started asking uni level questions they didn't have time for me! :D I'm doing physics now for fun. I actually have an exam in the morning (it's 00.03 here in England) so off to bed now. Wish me luck!

[quote="ffao"]I have just noticed one thing -- the car doesn't travel 5 meters. Instead, it travels [i]5 meters more than the lorry[/i], which is a huge difference.
.........
Say the lorry traveled $ x$ meters. So, the differences in the kinetic energies of the car and the lorry are:

$ |\Delta E_{k_c}| \equal{} 625(28^2 \minus{} 26.157895^2) \equal{} 62352.83...$
$ |\Delta E_{k_l}| \equal{} 1750(25.5^2 \minus{} 26.157895^2) \equal{} 59474.57...$

using that $ F \equal{} \frac {W}{d}$ we have the system

$ F \equal{} \frac {59474.57}{x}$
$ F \equal{} \frac {62352.83}{x \plus{} 5}$

$ F \approx 575.62$

Now, this is an answer I'm satisfied with ;)[/quote]I am satisfied with that too. The calculation and the end result are the same, but now there is an explanation for why the distance covered by the lorry does not matter, and we just divide by 5 meters. From
$ F \equal{} \frac {|\Delta E_{k_{lorry}}|}{x}$
$ F \equal{} \frac {|\Delta E_{k_{car}}|}{x \plus{} 5}$
or
$ Fx \equal{} |\Delta E_{k_{lorry}}|$
$ F(x\plus{}5) \equal{} |\Delta E_{k_{car}}|$
we get
$ 5F \equal{} |\Delta E_{k_{lorry}}\minus{}\Delta E_{k_{car}}|$
or
$ 5F \equal{} |\Delta E_{k_{car\plus{}lorry}}|$
The $ x$ distance cancels out and
$ F \approx \frac {2879J}{5m}\approx 576N$
We can round to more or less digits in the calculations, and I guess the 600 N result comes from a calculator that did not hold that many digits in its memory.


The bottom one.

Draw AC.  Then we can solve for x + y in terms of a variable z (if the vertex point of angle x is X, z degrees is angle BXA).

It's basically a (self-intersecting) heptagon.

Heptagons have total angle of 900 degrees. Since five of the angles are 28, 26, 74, 90, and 90, and the other two are 360-x and 360-y, add them up.

You get 1028-x-y=900, so x+y=128

AIME15, A,B,C are not necessarily collinear.

Can't we assume it is?  I think we'd get the same answer either way...

It's not rigorous, and james4l's solution is still very quick.


modular arithmetic
number theory solved
number theory

Let $S=\{a^2+b^2,(a,b)\in\mathbb{N}^2\}$.
For $L\in \mathbb{N^*}$, can we find $a\in \mathbb{N}$ such that $[a,a+L]\cap S=\emptyset$ ?

Related interesting question :  [url=http://www.mathlinks.ro/Forum/topic-4675.html][u]here[/u][/url]

Evidently, the answer to your question is 'yes.' To see this, choose distinct primes $p_1,p_2,\ldots,p_{L}$ of the form $4k+3$. By the Chinese remainder theorem, we may choose an integer $x$ such that $x \equiv p_j - j \pmod{p_j^2}$ for $1 \leq j \leq L$. Then, for each $j \in \{1,\ldots,L\}$, the integer $x+j$ will be divisible by $p_j$ but not by $p_j^2$, and $p_j \equiv 3 \pmod{4}$, so the set $\{x+1,\ldots,x+L\}$ is disjoint from $S$.

--Vesselin

Thanks!
And can we find $a\in \mathbb{N}$ such that $[a,a+L]\subset S$ ?

Related interesting question :  [url=http://www.mathlinks.ro/Forum/topic-4675.html][u]here[/u][/url]

[quote="Collins"]
And can we find $a\in \mathbb{N}$ such that $[a,a+L]\subset S$ ?
[/quote]

No, because of every four consecutive integers one is of the form $4k+3$ and thus cannot be expressed as the sum of two squares.


[quote="i_like_pie"][quote="Quevvy"]Juan and Kevin working together can paint a room in 7 hours. Kevin and Tad working together can paint the same room in 8 hours, and Tad and Juan working together can paint the room in 7 hours. What is the number of hours that it will take Juan, Kevin and Tad working together to paint 23 identical rooms?[/quote]
[hide]$\frac1j+\frac1k=\frac17$

$\frac1k+\frac1t=\frac18$

$\frac1t+\frac1j=\frac17$

Add everything together to get $\frac2j+\frac2k+\frac2t=\frac{23}{56}$.

Now, divide both sides by $2$ to make the numerators on the LHS $1$ again.

$\frac1j+\frac1k+\frac1t=\frac{23}{112}$

It will take them $\boxed{112\text{ hours}}$ to paint $23$ rooms.[/hide][/quote]
Wow, that makes it a lot simpler....  :roll:


I can't seem to find asy 1.44 for mac...anyone know how to get it?

The same Asymptote sourceforge page has the updated version. Just follow the installation steps again. :)

Thanks!

By the way there's a new version of mactex.  It's over 1GB to download :(

[url=http://www.tug.org/mactex/newfeatures.html]The changelog[/url] says that the new version is updatable over the internet, so it might be a worthwhile upgrade if you can get it.


A bracelet is made by stringing together four beads.  Each bead is either red or green.  How many different color patterns are possible for the bracelet, where patterns are considered the same if turning one will produce the other, as shown here?
[asy]

size(8cm,8cm);

draw(Circle((0,0),1));

filldraw(Circle((1,0),0.2),white);
label("$G$",(1,0));

filldraw(Circle((0,1),0.2),white);
label("$R$",(0,1));

filldraw(Circle((-1,0),0.2),white);
label("$R$",(-1,0));

filldraw(Circle((0,-1),0.2),white);
label("$G$",(0,-1));

draw((1.5,0)--(2.5,0),Arrow);

draw(Circle((4,0),1));

filldraw(Circle((5,0),0.2),white);
label("$R$",(5,0));

filldraw(Circle((4,1),0.2),white);
label("$R$",(4,1));

filldraw(Circle((3,0),0.2),white);
label("$G$",(3,0));

filldraw(Circle((4,-1),0.2),white);
label("$G$",(4,-1));

[/asy]

[hide]
You can either put a red bead or a green bead on the bracelet, that means there are 2 choices for each spot. To prevent overcounting you must divide by 4 for each position on the bracelet and 2 for flips. This leads to
\[ \frac{2^4}{4\cdot2}\equal{}\boxed{2}\]
[/hide]

if you use burnside's lemma you will achieve the following result
where:

In the following, let G be a finite group that acts on a set X. For each g in G let Xg denote the set of elements in X that are fixed by g. Burnside's lemma asserts the following formula for the number of orbits, denoted |X/G|:[2]
[img]http://upload.wikimedia.org/math/8/f/6/8f6fa913d4752b5a3fccdc24b40e16d4.png[/img]
(taken from wikipedia)

X = 2^4
G = 4
which comes down to:
(1/4)(2^4 + 2 + 2^2 + 2)
(1/4)(24)
6

[quote="aznluster"]$(n-1)!$

$\boxed{6}$[/quote]

why is it (n-1)! ? If we had 2 beads, that wouldn't work: (2-1)!=1, but there are obviously two different necklaces- one with a green, one with a red

i would have approached it like this:
0 red- 1 possibility
1 red- 1 possibility
2 red- 2 possibilities
3 red- 1 possibility
4 red- 1 possibility
total: 6, the answer

If we had two, we would only have one possibility because they are the same. It is a circular bracelet.

This question is an example of counting with Symmetry- It is with any question with circular combinations. Because no flipping of the ring will be required, and the turning doesn't really matter, we can just do 4!(The ways)/2!^2(The repeats) For a total answer of 6


Given is the positive integer $n > 2$. Real numbers $\mid x_i \mid \leq 1$ ($i = 1, 2, ..., n$) satisfying $\mid \sum_{i=1}^{n}x_i \mid > 1$. Prove that there exists positive integer $k$ such that $\mid \sum_{i=1}^{k}x_i - \sum_{i=k+1}^{n}x_i \mid \leq 1$.

It seems like this can be done as follows: suppose WLOG that $\sum_{i=1}^n x_i > 0$. Then, let $S_k = \sum_{i=1}^k x_i - \sum_{i=k+1}^n x_i$; we have $S_n = \sum_{i=1}^n x_i > 1$ and $S_0 = -\sum_{i=1}^n x_i < -1$. Suppose that we never have $-1 \le S_k \le 1$. Then let $k_0$ be the maximal value of $k$ for which $S_{k_0} < 0$; we know $k_0$ exists as $k = 0$ certainly satisfies the given property; additionally, $k_0 \ne n$ as $k = n$ does NOT satisfy the property, so $S_{k_0 + 1}$ is defined. By the assumption, we have $S_{k_0} < -1$, and as $k_0$ was the maximal value, we must have $S_{k_0 + 1} \ge 0$ and thus by the assumption $S_{k_0 + 1} > 1$. Thus $2 < S_{k_0 + 1} - S_{k_0} =$ $\left(\sum_{i=1}^{k_0+1} x_i - \sum_{i=k_0+2}^n x_i\right) - \left(\sum_{i=1}^{k_0} x_i - \sum_{i=k_0 + 1}^n x_i\right) =$ $2x_{k_0+1} \le 2|x_{k_0+1}| \le 2$ which is sort of false. Thus there exists a $k$ with the desired property, i.e. that $|S_k| \le 1$.

wow :lol:  beautiful problem and solution!

:)

Define $$S_k = \left|\sum_{i=1}^k x_i - \sum_{i=k+1}^n x_i\right|.$$ In particular, denote $S_0 = -\sum x_i$ and $S_n = \sum x_i$. WLOG suppose that $S_n > 1$; then, we necessarily have $S_0 < -1$.

Now, assume for the sake of contradiction that $|S_k| > 1$ for all $k$. Then there must exist some $k, k+1$ such that $S_k < -1$ and $S_{k+1} > 1$. But then $$2|x_{k+1}| > 1-(-1) = 2 \iff |x_{k+1}| > 1,$$ which is a contradiction.


Find the value of : $ \frac{1}{5}\cdot \frac{3}{7} \cdot \frac{5}{9} \cdot \frac{7}{11}\cdot \ldots \cdot \frac{99}{103} \cdot \frac{101}{105}$

This product telescopes.  All of the denominators from 5-101 cancel so we're left with
$ 1\cdot 3\cdot\frac{1}{103}\cdot\frac{1}{105}$

Which, of course, equals $ \boxed{\frac{1}{3605}}$.


calculus
integration
number theory unsolved
number theory

Find all positive integer triples $ (x,y,z)$ such that
$ xyz\equal{}x\plus{}y\plus{}z$

We can even solve it over the integers. Wlog, suppose $ |x|\le |y|\le |z|$. Then $ |xyz| \equal{} |x\plus{}y\plus{}z|\le |x|\plus{}|y|\plus{}|z|\le 3|z|$. So $ |xy|\le 3$, viz $ |x|\le 1$, and we're left with a very few cases to check. If $ x \equal{} 0$, then $ z \equal{}\minus{}y$. If $ x \equal{}\pm 1$, then $ (y\mp 1)(z\mp 1) \equal{} 2$. But $ |(y\mp 1)(z\mp 1)|\ge (|y|\minus{}1)(|z|\minus{}1)\ge (|y|\minus{}1)^{2}$, thus $ |x| \equal{} 1\le |y|\le 2\le |z|$. When dealing with $ (y\minus{}1)(z\minus{}1) \equal{} 2$, this means $ y \equal{} 2$ and $ z \equal{} 3$, whereas $ (y\plus{}1)(z\plus{}1) \equal{} 2$ yields $ y \equal{}\minus{}2$ and $ z \equal{}\minus{}3$. Thus we conclude that all integral solution to the equation $ xyz \equal{} x\plus{}y\plus{}z$ are given (up to a permutation) by the triples $ (0,a,\minus{}a)$ and $ (\pm 1,\pm 2,\pm 3)$, for arbitrary $ a\in\mathbb{Z}$.


function
topology
calculus
calculus computations

a)How do I show that an isometry is continuous?
b) If I wanted to show that $ [0,1]$ is not isometric to $ [0,2]$, is it enough to assume both closed sets on the real number line can't possibly have a one-to-one mapping just because of the fact that $ [0,2]$ has more elements than $ [0,1]$?

a) What is the definition of continuous?
b) The sets $ [0, 1]$ and $ [0, 2]$ have the same number of elements, and it is easy to construct bijections between them (e.g., the function $ f(x) \equal{}2x$ is one).  So, yes, it would be enough to assume this, since with a false premise you can prove anything  :P

a) defn of continuity would be the $ \epsilon \minus{}\delta$ argument
b) so am I right about saying that $ [0,1]$ is not isometric to $ [0,2]$? I 'm confused by the reply to this question

I don't think it's right to say "definition of continuity would be the $ \varepsilon$-$ \delta$ argument;" $ \varepsilon$-$ \delta$ is used for a lot more than that, and that is simply one definition-there are several other equivalent definitions.

The definition of an isometry is a function that preserves distances; $ \rho(f(a),f(b))\equal{}d(a,b)$.  With regard to the standard metrics, $ [0,1]$ and $ [0,2]$ are not isometric.  However, you might be able to (I haven't tried, maybe one can't) construct a metric so that they are isometric.  The point I'm making is that the notion of distance depends on the metric.

[quote="mstrip"]a) defn of continuity would be the $ \epsilon \minus{} \delta$ argument[/quote]  Which means what?  You're going to have to write down the definition at some point in order to prove that something satisfies it ... so start now, and see where that gets you.
[quote="mstrip"]b) so am I right about saying that $ [0,1]$ is not isometric to $ [0,2]$? I 'm confused by the reply to this question[/quote] Yes, you are right, but your proposed explanation was wrong in every respect and had nothing to do with solving the problem.  Think about the conceptual meaning of the term isometric, think about the sets [0, 1] and [0, 2], and think about what feature of those two sets could possibly prevent them from being isometric.

JRav, if $ X$ is any metric space and $ Y$ is any set in bijection with $ X$, we can construct a metric on $ Y$ to make it isomorphic to $ X$ in the completely obvious way.

1) From the definition of an isometry, for a bijective map $ f: M \rightarrow N$ that preserves distance for all $ p,q \in M$, we have $ d_{N}(f_{p}, f_{q}) \equal{} d_{M}(p,q)$. For some $ \delta > 0$ and for some $ \epsilon > 0$, I decided to make $ d_{M}(p,q) < \delta$. From this point, I don't see how I can justify saying $ d_{N}(f_{p}, f_{q}) < \epsilon$ if I set $ \delta \equal{} \epsilon$. I could simply say that $ d_{M}(p,q) < \delta$ implies $ d_{N}(f_{p}, f_{q}) < \epsilon$ for the sake of putting something down for the professor, but for personal understanding, how can I justify that? Any comments would be appreciated.

2) I can simply invoke the fact that an isometry is distance preserving and since $ [0,2]$ is of length 2 and $ [0,1]$ is of length 1, one length is greater than the other such that a map between the two would not be distance preserving, right?

What is the $ \epsilon$-$ \delta$ definition of continuity?  If for each $ \epsilon>0$, there exists $ \delta>0$ such that $ d(x,y)<\delta$ implies $ \rho(f(x),f(y))<\epsilon$.  If $ d(x,y)\equal{}\rho(f(x),f(y))$ by design, then clearly whenever $ d(x,y)<\delta$, so is $ \rho(f(x),f(y))$ since they are the same quantity.  So choose $ \delta\equal{}\epsilon$, like you said.

You have to be careful with what you mean by "length."  If you are using the standard Euclidean metric, then yes, that is why they are not isometric.

Please read Kent's lovely post http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1629469#1629469 -- when you're working with formal statements, you actually have to take the formality seriously.  Once you're more experienced (by the end of your current course, even), you'll know how to translate effortlessly between certain formal and informal statements, but at the beginning it's crucial that you take the formal statements seriously: don't pull out bits and pieces of them, don't handwave them, and certainly make sure you know which quantifier goes where.

You still haven't written down a proper definition of continuity (though JRav, who's more impatient than I am ;), has gone ahead and done it for you) -- "for some $ \delta$ and some $ \varepsilon$" is not part of the definition, and the quantifiers are both the trickiest and the most important part!

For the second question, [i]I[/i] know how to properly formalize the notion that one of those sets has longer length than the other into a proof that they are not isometric, but I don't think [i]you[/i] do.  As a hint, you're both going to want to state precisely what an isometry is, and proceed by contradiction.  (Well, we can rephrase this argument as a direct argument, but for purposes of solving the problem it's much more natural to proceed by contradiction.  Then at the end we can decide if we want to phrase it directly or not.)


[quote="Geehoon"]I cannot vote:

counter ex3) I know what go is
counter ex2) I know its an intelligent game
counter ex1) I do not know how to play, so I couldnt find its joy yet.

Hence, I cannot vote.

QED.[/quote]
Option added.

[quote="math92"]what's go?[/quote]
This is an old Chinese Chess game and it is played by many people today.

when i play Go, I use the force... place the seed in the place which has the most influence (=

[quote="K81o7"]The rules of Go are fairly simple.  However, there are all sorts of complexities which arise in the game, and all sorts of strategies, making it a very good one.[/quote]
Yep...I learned how to play Go in just a few minutes, but when really start playing games with other people, I found that I often lost more than 30...

By the way, can someone please tell me link where they introduce go in English?

[url]http://en.wikipedia.org/wiki/Go_%28board_game%29#Basic_rules[/url]

For more detail: [url]http://en.wikipedia.org/wiki/Go_rules[/url]

[quote="math92"]Can someone give me a link of how to play go?[/quote]
You can also check out the American Go Association website http://www.usgo.org

yea, go is pretty fun.  haven't played a real game in awhile.  Anyone who wants to play on yahoo or something, pm me.  I'm pretty bored these days.   :P

The fun math blog [url=http://simpler-solutions.net/pmachinefree/thinkagain/thinkagain.php]think again![/url] had Go problems as their theme last week.  (Page down to Monday August 22 to see the start of the thread.)

[quote="zmli"]While computer programs are getting stronger in playing GO, it is still the only game that no a single computer program is able to beat a human master (9-Da Level - a world ranking of GO players).

Unlike chess where the computer is now faster enough to calculate all the moves to catch the mistake of a human master player, alghrithm alone cannot win a GO game.  There is a large element of intuition involved in playing GO and it is harder for the computer to "calculate" intuition.

To learn how to play GO, follow the following link:

[url=http://www.usgo.org/index.asp]American GO Association[/url][/quote]
yes it is the only one.
many AI programmers think beating human is one of the final goals of AI because even the strongest computer can't calculate more than 5 moves.human has huge advantages in thinking. if computers want to beat human it has to think like human beings. that's y it's really hard...
it is not about the speed of the computer but the INTELLIGENCE of it.

I think the game can be solved in the following way:
1. Let play the computer against himself;
2. Let the program do every time a different sequence of moves and different moves after every match;
3. Compute and archive using a database all possible moves;
4. When all possible legal moves are written the database can be used to choose the perfect move to win the game because all moves have been yet tested in advance.
Maze solved!  :P

That presupposes that you have enough computing time to allow the computer to find all of them and enough storage space to hold them.  If someone did, the winning strategy for Go would be known already.

Today computer are fast enough to compute and storage all moves in a database, i think that these games have been yet solved.
Go moves (like chess moves) are not endless.
May be who solved them prefer to keep it secret.
Imagine a government strategy department that solved the chess game, can you imagine the potential of this country?
Weapon are nothing without a good strategy.  :D

[quote="shobber"][quote="Zyloch"]Go is interesting in that it's one of the only board games that still cannot be played particularly well by a computer :)[/quote]
I think the computer is pretty strong...[/quote]

I think the computer is pretty weak :P

Amusing story:  I was playing against a relatively strong go program (Aya).  I decided to be over-aggressive and ended up winning by over $300$ points.  (An average game between two human opponents of close skill tends to be $\pm 3$ points, and the maximum possible number of points is $361$.  :) )

[quote="lightman"]Is it possible to [b]compute[/b] all the possible games and moves?  :huh:[/quote]

Go is played on a $19 \times 19$ board.  There are $361$ possible first moves, $360$ possible second moves, ...

Personally, I love Go, although I hardly ever play it anymore.  It reminds me of the Mandelbrot set - relatively simple rules giving rise to complexity.  Someday my father wants to discover a mathematical theory of Go.  :)

[quote="JBL"]Computers do not have enough speed or memory to play all Go or Chess sequences.  The number of possible chess games is far, far greater than anything current computing power can attack directly.[/quote]
I think is possible create a simple program that play against himself to compute all moves sequences, i will start with chess. It's not a matter of speed or memory, it's a matter of time only, maybe it will take months or years with a simple computer, but at the end all moves will be computed. It must be done with more than one computer with many databases that can collect long series and sequences of different moves, so speed will be increased.
Can you imagine the potential of this results? I cannot.
After chess, go can be soved too.
I remembre that 20 years ago there were CRAY computers (few seconds of their work can be rent for money), i cannot imagine what kind of features is possible to rent today.  :)

I think you're really failing to grasp just how large the numbers involved are.  There are more than 361! possible Go games.  That's roughly $1.4\cdot10^{769}$ games.  The best current supercomputers can do a teraflop or so -- that's $10^12$ calculations per second.  Suppose, for sake of simplicity, that each game is one calculation long -- that means I have $10^{757}$ seconds of computation on one of these computers.  But you're not talking about just one computer -- let's suppose that I could use all of the computers in the United States, and each of them were doing a teraflop.  There are more than 100,000,000 computers in the US -- let's just round that up to one billion, $10^9$.  Now, I only have $10^{748}$ seconds of computation.  The current age of the universe is an estimated 15 billion years, which is (rounding up) $10^{19}$ seconds.  Are you beginning to see the problem?  Let's say the computation time was approachable, though, and let's suppose that we could store the outcome of each game very densely.  There are an estimated $10^{87}$ particles in the universe -- let's call it $10^{100}$, just in case the computations have been low.  That means I have to store $10^{669}$ games per particle.

[quote="Zyloch"]
Go is played on a $19 \times 19$ board.  There are $361$ possible first moves, $360$ possible second moves, ...
[/quote]
The number of moves depend from board squares and from game rules, i think the moves number is less high than you can expect, a lot of possible moves are not useful to win the game, so if the program is teached to compute only good moves, the moves list will be less long than we could expect. Or not?
A friend of mine told me that the game can be broken in smaller parts to compute moves, he says that in GO there are certain patterns for which the solution is already known.
I'd like to start with chess, how could i break the game in smaller parts?
For example can I calculate all the first five moves in a reasonable time with a normal computer?
Can you estimate the number of the first five moves?
I think that computing the first ten moves could be enough to have the first golden keys of the game.  :D

[quote="lightman"]...if the program is teached to compute only good moves, the moves list will be less long than we could expect. Or not?[/quote]

Well, yes, if the computer made only good moves, we would have fewer "silly games" -- but tell me, how do you teach a computer to make good moves?  The only way is if you have some winning strategy, which was the whole point of this exercize in the first place.  The problems that you are thinking of as intuitive or easy ("just have it play all games," "only have it make good moves") are exactly the most difficult, computation-wise.

The endgame of Go often divides into many smaller games, played simultaneously.  However, this isn't going to save you nearly the needed amounts of computation time or memory that we're talking about -- for one thing, you still have to play most of the game to get to those points, and then those are the places where the number of moves is numerically quite small.  In other words, Go has been solved for subgames in which your scheme doesn't result in absurdly high numbers; that doesn't really do anything to the whole game, because it takes off only a relatively small portion of the whole.  (The first 30 moves, after all, can happen in about $10^{76}$ ways.  Playing on a 7x7 board gives at least $10^{63}$ possible games -- these are more approachable number, to be sure, but still vastly too large for our purposes.)

Chess is not a game which can be broken down into smaller parts in the same way -- one move can't ever be independent of another.  Also, unfortunately, counting moves in chess is complicated -- it depends very much on the current position.  All possible first ten moves is probably quite a large number -- after all, opening up the queen tends to make available a dozen or more moves just from that piece.  At each move, a player in general has at least 16 places to go early in the game.  This (probably low) estimate gives $10^{24}$ possible first ten moves (each player playing 10), a number which is, in fact, small enough to be attacked by computer.  However, because very few of these games end in a decision (Scholar's mate aside), you would then need a person to go through and eliminate all the "silly" games.  In addition to potentially being falliable, I don't think this process will really give you as much insight as one might hope into the game of chess.

[quote="JBL"]Well, yes, if the computer made only good moves, we would have fewer "silly games" -- but tell me, how do you teach a computer to make good moves?  The only way is if you have some winning strategy, which was the whole point of this exercize in the first place.[/quote]
I think it is possible, we have to try thinking like a computer.
For example can you compute in advance all possible first 3 chess moves first to loose one piece of the board?
A good chess player can teach to "think" to a computer too, i have not your skill in math but i can play and i know the the game very well, i'm studying this game from a lot of time, together we can solve the maze i think.

To figure out all the number of Go games is so high and impractical that my brother told me that it gets [b]classified as infinite[/b].  There is quite a few more than 361! because you can take stones, get into Ko's... I have seen games go up around 350 moves before!

You can find me as hoshattack on [url=http://www.pandanet.co.jp/English/]IGS[/url]  :lol:

computer go programs are relatively weak because as ppl said they "calculate" their moves.

in go you usually start at the corners. and if you pay attention you will notice a "pattern" with computer programs. like they always play the same move in similar situations. this is something that would be different if you were playing with a human being.

[quote="kstan013"]Reading the wikipedia article, it looks like go is played like othello. Is this true?[/quote]

The only similarity they share is probably the pieces and the board they use. Othello is reversi which is nothing like Go in terms of rules and strategy.

[quote="Zyloch"]I think the computer is pretty strong too. Actually, I'm so horrible at go I'm embarassed to be even posting in a thread about go lol. My parents always wanted me to learn and all I used the pieces to do was throw them when I was angry to vent. Of course, a few of them are chipped now, and I'm a bit unhappy that the go set my uncle sent me from China is cracked so I've learnt my lesson (pillows work fine :))

Anyhow, I've read that compared to games such as chess and (especially) checkers, go computer players are relatively weak. Of course, this is due to the amazing amount of possible legal moves on a go board since computer programs mainly use brute forcing techniques so far, but it's still an interesting topic in AI :)[/quote]

throw precious go pieces?  :o

[quote="JBL"]I think you're really failing to grasp just how large the numbers involved are.  There are more than 361! possible Go games.  That's roughly $ 1.4\cdot10^{769}$ games.  The best current supercomputers can do a teraflop or so -- that's $ 10^{1}2$ calculations per second.  Suppose, for sake of simplicity, that each game is one calculation long -- that means I have $ 10^{757}$ seconds of computation on one of these computers.  But you're not talking about just one computer -- let's suppose that I could use all of the computers in the United States, and each of them were doing a teraflop.  There are more than 100,000,000 computers in the US -- let's just round that up to one billion, $ 10^{9}$.  Now, I only have $ 10^{748}$ seconds of computation.  The current age of the universe is an estimated 15 billion years, which is (rounding up) $ 10^{19}$ seconds.  Are you beginning to see the problem?  Let's say the computation time was approachable, though, and let's suppose that we could store the outcome of each game very densely.  There are an estimated $ 10^{87}$ particles in the universe -- let's call it $ 10^{100}$, just in case the computations have been low.  That means I have to store $ 10^{669}$ games per particle.[/quote]
at first I thought OMG so we cannot solve GO, but actually we can delete a game which ends with a lost or a draw..but it remains still problematic some what cause we cannot save all the games that exist..although we may be able to generate the one we seek for quickly..

ME ME ME!!!

used to take lessons and do competitions when i waz a vee lil kiddie.

but now im mediocre. still love it though.

I read somewhere that there are more possible games of Go than there are atoms in the universe o.O

I like go.

BTW, does anyone here go on the KGS servers? ([url]http://gokgs.com[/url]/)


[quote="rem"]Algebra+completing the square[/quote]

We now, but can you put a full solution, with all the steps?


invariant
conics
geometry
geometric transformation
rotation

Prove that the discriminant of the general conic equation $ Ax^2 \plus{} Bxy \plus{} Cy^2 \plus{} Dx \plus{} Ey \plus{} F \equal{} 0$ is invariant under rotation of axes.

[hide="Hint"]Rotate the axes of the general conic equation by an angle of $ \alpha$, then get the coefficients of the transformed conic equation. Then, show that the discriminant of the transformed equation is equal to the discriminant of the original conic equation. ^^[/hide]

[hide="Hint again"]Transform $ Ax^2 \plus{} Bxy \plus{} Cy^2 \plus{}Dx \plus{} Ey \plus{} F \equal{} 0$, $ B \neq 0$ into $ \bar{A}x^2 \plus{} \bar{B}xy \plus{} \bar{C}y^2 \plus{} \bar{D}x \plus{} \bar{E}y \plus{} \bar{F} \equal{} 0$. You must show that $ B^2 \minus{} 4AC \equal{} \bar{B}^2 \minus{} 4\bar{A}\bar{C}$.[/hide]


Prove that every positive rational number can be written as the sum of distinct unit fractions (i.e. fractions with numerator 1).

For our purposes, consider $ 1/1$ a valid unit fraction.

We can use the greedy algorithm as follows:

Take a number $ r_0\equal{}\frac pq$ where $ p$ and $ q$ are relatively prime. If $ p\equal{}1$ then obviously we are done. If not, then find the largest unit fraction which is smaller than $ \frac pq$ and not already in $ \{a_n\}$. Let this unit fraction be $ \frac 1{a_1}$. Then let $ r_0\minus{} \frac 1{a_1}\equal{}r_1$, and repeat the process until $ r_n\equal{}0$. Then the sequence $ \{a_n\}$ is the desired set of unit fractions.

This algorithm is guaranteed to terminate because $ r_0 \minus{} \frac 1{a_1}\equal{}\frac{pa_1\minus{}q}{qa_1}$. Observe that since $ \frac 1{a_1}\leq \frac pq<\frac 1{a_1\minus{}1}$, $ a_1\geq \frac qp>a_1\minus{}1 \implies pa_1\geq q>pa_1\minus{}p$. Therefore, $ pa_1\minus{}q \geq 0 > (pa_1\minus{}q)\minus{}p$, which means that $ pa_1\minus{}q<p$. Thus at each stage, the numerator strictly decreases even before cancellation. Since the numerator is a nonnegative integer, it must reach zero.


Use the existence and uniqueness theorem to find all solutions of the equation $ x''\plus{}4x\equal{}0$

Is this what you're looking for? From the auxiliary equation $ r^2\plus{}4\equal{}0$ we get that $ r\equal{}\pm 2i$, so our solutions are $ x(t)\equal{}C_1e^{2it}\plus{}C_2e^{\minus{}2it}$.

I always understood it as, if you find a family of solutions, then you know it is unique, the way you get there doesn't matter.


hi, does anyone know about "MU Alpha Theta???"  :huh: does anyone have tips for comp. geo next year so i don't end up being like last place??? :blush:  :D

Doing practice problems and reading the solutions is always a big plus.

Just make sure that you examine all types of problems and ask about ones you dont understand how to solve or prove.

Practice practice practice is the only way that works for me :fool: ... dunno about others.

It really depends...get Intro to Geo, but if you can't use books well, then just do lots of problems...but don't waste too much time on the AoPS forums...search like geometry problems everywhere...for a start go to http://mathcounts.saab.org/mc.cgi ...dunno.

The most important thing is to practice, practice, practice. It's the only way to get better. If you don't know how to solve a problem, make sure you understand the solution. And don't be afraid to ask questions!

thanks u guys, it sounds pretty easy.....for u guys. cuz ur all like amazingly smart ppl, but ill practice!

We can only show you the door, you have to walk through it.
 :rotfl:

[quote="27r"]Yeah, I competed in geometry Mu Alpha Theta this year.
Mu Alpha Theta basically just gets kids interested in math. It doesn't go deep into things, unlike other tests. If you want to do good, you just need to cover all the concepts. And don't make stupid mistakes.[/quote]

What's your name? I did also.

[quote="FTI"]Is it true that the sum of two squares may not be factored?[/quote]

The Fundamental Theorem of Integers
and Pythagoras
(factoring the sum of two squares)
by Brian S. McMillan

Application of the Integer Theorem in the pro-engineered factoring for the sum of two squares, takes the definitive inverse for the difference of two squares to establish a generalized case.

Where a postulate may be made concerning the factors comprising the base of the third term yield, as an extension for the sum of two squares never equal to a (4k + 3) type integer.  This is then taken further, in that when the third term yield may be evenly rooted to a least factor of the (4k + 3) integer type, then the complete first term consisting of a (4k + 1) integer may never possess an even root.  The reverse would also be true.

The General Equation (specific case)

FOR:  Integer k

WHERE:     (4k + 1) + (2k)^2 = (2k + 1)^2

THEN:         (2b + 1) + b^2 = (b + 1)(b + 1)

The Specific Equation (specific case)

FOR:  Some odd integer n

THEN:        (4 * (n^2 - 1) /4 + 1) + ((n^2 - 1) /2)^2 = ((n^2 - 1) /2 + 1)((n^2 - 1) /2 + 1)

Numerical Examples

IF:                3^2 = 9

AND:           (3^2 - 1) /4 = 2

THEN:          3^2 = (4 * 2 + 1) = (4k + 1)

WHERE:     (4k + 1) + (2k)^2 = (2k + 1)^2

THEN:         (4 * 2 + 1) + (2 * 2)^2 = (2 * 2 + 1)^2

HENCE:       3^2 + 4^2 = 5^2 = a^2 + b^2 = c^2

IF:                17^2 = 289

AND:           (17^2 - 1) /4 = 72

THEN:          17^2 = (4 * 72 + 1) = (4k + 1)

WHERE:      (4k + 1) + (2k)^2 = (2k + 1)^2

THEN:          (4 * 72 + 1) + (2 * 72)^2 = (2 * 72 + 1)^2

HENCE:        17^2 + 144^2 = 145^2 = a^2 + b^2 = c^2

OR:               17^2 + 12^4 = 145^2

IF:                 103^2 = 10609

AND:            (103^2 - 1) /4 = 2652

THEN:           103^2 = (4 * 2652 + 1) = (4k + 1)

WHERE:       (4k + 1) + (2k)^2 = (2k + 1)^2

THEN:           (4 * 2652 + 1) + (2 * 2652)^2 = (2 * 2652 + 1)^2

HENCE:         103^2 + 5204^2 = 5305^2

Brian S. McMillan Copyright 1997-2008

http://www.pumpraser.com/fermatbrief.html

Yahoo Answers:

http://answers.yahoo.com/question/index?qid=20080421122828AAOkoYn


Hey people, can somebody post the solution of problem 3 which earned the only special award?

It was posted three times I think.

The solution was \[ \frac{x^5 - x^2}{x^5 + y^2 + z^2} - \frac{x^5 - x^2}{ x^3(x^2+y^2+z^2} = \frac{(x^3-1)^2 x^2(y^2+z^2)}{x^3(x^2+y^2+z^2)(x^5+y^2+z^2)} \ge 0  \]
Therefore
\[ \sum \frac{x^5-x^2}{x^5+y^2+z^2} \ge \sum\frac{x^5-x^2}{x^3(x^2+y^2+z^2)} = \frac{1}{x^2+y^2+z^2}\sum{(x^2 - \frac{1}{x})} \ge \newline \ge \frac{1}{x^2+y^2+z^2}\sum{(x^2 - yz)} \ge 0  \]

The substitution $(x^2 - \frac{1}{x})$ for $(x^2 - yz)$ it's because of $xyz \ge 1$ and $\sum{(x^2 - yz)} \ge 0$ it's because $\sum{(x^2 - yz)} = x^2 + y^2 + z^2 - xy - yz - xz$ result easy to show by factorizing that it is $\ge 0$


Let $a,b,c$ be the side-lenths of a triangle such that $a\le b\le c$. Prove that

\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge 3(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-\frac 1{2}). \]

[quote="Sunjee"][quote="hungkhtn"]I think this problem follows the inequality
\[3+\sum \frac{a+b}{c}\ge 6\sum \frac{a}{b+c}, \]
which can be solved by SOS method.[/quote]

please show me how...[/quote]
If do you mean "how follows" then:
if $0<a\leq b\leq c$ then $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq\frac{a}{c}+\frac{b}{a}+\frac{c}{b}.$ :wink:

[quote="Sunjee"][quote="hungkhtn"]I think this problem follows the inequality
\[3+\sum \frac{a+b}{c}\ge 6\sum \frac{a}{b+c}, \]
which can be solved by SOS method.[/quote]

please show me how...[/quote]
This is a problem in VietNam TST 2006. You can find it easily :P
Above inequality equivalent:
$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\ge 6( \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})$

[quote="zaizai-hoang"][quote="Sunjee"][quote="hungkhtn"]I think this problem follows the inequality
\[ 3 \plus{} \sum \frac {a \plus{} b}{c}\ge 6\sum \frac {a}{b \plus{} c},
\]
which can be solved by SOS method.[/quote]

please show me how...[/quote]
This is a problem in VietNam TST 2006. You can find it easily :P
Above inequality equivalent:
$ (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})\ge 6( \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b})$[/quote]
See at http://www.mathlinks.ro/Forum/viewtopic.php?p=487332#p487332


[quote="amirhtlusa"]what is the remeinder of $17^{76}-8^{76}+1-9^{76}$ by $72$?[/quote]

I've actually never learned how to do these...

Would it help lots if 1 was changed to $1^{76}$ ?

[hide]Since $17^2 = 289$ then $17^2 \equiv 1 \pmod {72}$ so $17^{76} \equiv 1 \pmod {72}$.

$8^3 \equiv 8 \pmod {72}$ and also $8 \equiv 8 \pmod {72}$ so we divide the exponent by three until we're left with $8^2$ which is congruent to $64 \pmod {72}$

$9^2 \equiv 9 \pmod {72}$ and $9 \equiv 9 \pmod {72}$. We keep dividing the exponent by two until we're left with just $9^1$ which is $9 \pmod 9$.

$1 - 64 + 1 - 9 = -71$ so the answer is $1$[/hide]

[quote="pkothari13"]try Euler's Totient Function...
[url]http://mathworld.wolfram.com/EulersTotientTheorem.html[/url][/quote]
that's right :D

[hide]This is the way I did it.

First, taking the expression mod 9 we get 1.  Then taking it mod 8 we get 1.  So the remainded when it is divided by 72 is a nonnegative integer less than 72 that is 1 greater than a multiple of 8 and 1 greater than a multiple of 9.  This narrows it down to 1.[/hide]


Let $n$ be a positive integer with $n>1$
Prove that

a) $\frac{1}{2}+\cdots+\frac{1}{n}$ is not an integer

b) $\frac{1}{3}+\cdots+\frac{1}{2n+1}$ is not an integer

$\forall n>1$ exist prime $\frac{n}{2}<p\le n$.

Let $S(n) = \frac{1}{2}+\cdot+\frac{1}{n}$
Let $p$ be the largest prime less than or equal to $n$,
if $2p\leq n$, then by Bertrand's postulate there is a prime $q$ such that $p<q<2p\leq n$. That contradicts the maximality of $p$, so we have $2p>n$ ---(1)
Now we let $S(n)-\frac{1}{p}=\frac{a}{b}$ where $(a,b)=1$ and $b|\frac{n!}{p}$
By (1) we have $(\frac{n!}{p},p)=1$, thus, $(b,p)=1$
$S(n)=\frac{1}{p}+\frac{a}{b}=\frac{b+ap}{bp}$ but $p$ doesn't divide $b+ap$, thus $S(n)$ is not an integer.

[hide="a"]
Let $2^{m}$ be largest power of 2 less than or equal to $n$. Then when writing the given expression over a common denominator, there will be $m$ factors of 2 in the denominator, but none in the numerator! (To see this, $\frac{1}{2^{m}}$ will contribute an odd number to the numerator, all others contribute even number). So the sum cannot be an integer, and this even shows that the sum is an odd integer divided by a multiple of $2^{m}$.
[/hide]
[hide="b"]
Very similar to a, but use powers of 3 instead
[/hide]

More generally,

$\frac1a+\frac1{a+1}+\ldots+\frac1{b-1}+\frac1b$

is never an integer for integers $a<b$ (consider the largest power of 2 dividing any of $a,\ldots,b$).

If $h_{n}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}(n>1)$.
Assume that $2^{a}\leq n<2^{a+1}$ and $P=\prod_{d\leq n,\text{d is odd}}d$. Now see at $2^{a-1}Ph_{n}$.

Above hint is from Nathanson's book.


In a theme park, you walk into a giant can-shaped room that can rotate.  When the angular speed of the "room" reaches a certain value, the bottom "floor" drops out.  However, you don't fall down!  Instead, you "stick" to the wall.  Explain why in terms of forces involes.

If the coefficieent of static friction betwee you and the room is $ 2.5 m$, what is the minimum rotational speed of the "can" so that you can "stick" to the wall of the bottomless room?

This phenomenon doesn't not make sense to me since if you stand at the very center of the can-shaped room and then the room reaches at that angular speed, which part of the wall are you going to stick to since the system is symmetric?

The coefficient of friction is a dimensionless constant between 0 and 1, so your value is wrong.

I think you're missing the point of the question.  If you 're standing at the wall, then there are a few forces at work.
1) Gravity pulling down
2) The normal force of the wall pushing back at you
3) The force of friction, pushing up against you
4) A centripetal force pushing out (in the frame of reference we're working with)

In the horizontal direction, our equation of motion is
$ 0\equal{}N\minus{}F_{c}\equal{}N\minus{}m\omega^{2}R\Rightarrow N\equal{}m\omega^{2}R$
In the vertical direction, it is
$ 0\equal{}mg\minus{}f\Rightarrow f\equal{}mg$
The person will "stick" when the force of friction exceeds the force of gravity.  That is,
$ f>mg\Rightarrow \mu N\equal{}\mu m\omega^{2}R>mg\Rightarrow\omega>\sqrt{\frac{g}{R\mu}}$
So once $ \omega$ exceeds this value, the person will "stick."


inequalities
trigonometry
inequalities unsolved

$a,b,c > 0 : ab+bc+ac =1$

Find minium:
a, $\sum_{cyclic}\sqrt{a+1}$

b, $\sum_{cyclic}\sqrt{a^3+2}$

I'm very poor in inequalities so probably there is a smarter solution ....

Setting $x = 1/a$, $y = 1/b$, $x = 1/c$,
We have $x+y+z=x \cdot y \cdot z$ so by usual trick we may assume that $x = \tan X$, $x = \tan Y$, $x = \tan Z$

Now $x \rightarrow \sqrt{1+1/x}$ has the good idea of being convex so we can apply Jensen.

Min is $\sqrt{3} \sqrt{3 + \sqrt{3}}$ (so many $3$ and $\sqrt$  :P)

Possibly the other uses the same trick ($\sqrt{2+1/x^3}$ is very probably convex) .


Four red candies and three green candies can be combined to
make many different flavors. Flavors are different if the
percent red is different, so 3 red / 0 green is the same flavor
as 2 red / 0 green; and likewise 4 red / 2 green is the same
flavor as 2 red / 1 green. If a flavor is to be made using some
or all of the seven candies, how many different flavors are
possible?

we can divide this into simple casework: work on the number of green candies
If there are x green candies , then there can be y red candies (that are different from previous combinations)
x=0;y=1
x=1;y=0,1,2,3,4
x=2;y=1,3; and
x=3;y=1,2,4

so we have a total of 1+5+2+3=11 flavors possible.


function
trigonometry
Olimpiada de matematicas

Calcular el máximo valor de la función:

$f(x)=(senx+cosx)(cosx-senx)^{3}$

I'm not sure but the maximum should be for $f\left(-\frac{\pi}{12}\right)$.

Como lo sabes?

$(senx+cosx)(cosx-senx) (cosx-senx)^{2}$
$= (cos^{2}x-sen^{2}x)(1-2senxcosx)$
$= cos(2x)(1-sen(2x))$
$= cos(2x)-\frac{1}{2}sen(4x)$

y entonces $f(\frac{-\pi}{12}) = cos(\frac{-\pi}{6})-\frac{1}{2}sen(\frac{-\pi}{4}) = \frac{3\sqrt3}{4}$

preguntan el maximo valor de la funcion..

sabemos que el coseno lo maximo que llega es 1 y el minimo es -1, y el seno lo mismo, por lo que lla funcion que vale $ cos(2x)\minus{}\frac{1}{2}sen(4x)$ puede a lo mas tener valor $ 1\minus{}\frac{1}{2}(\minus{}1)\equal{}\frac{3}{2}$, ese el valor maximo que alcanza la funcion.

para determinar para que $ x$ se alcanza, habria que resolver la ecuacion trigonometrica asociada, o bien, derivar la funcion y encontrar un punto critico, pero limitemonos a la pregunta

saludos

Y estas seguro que si existe $ x$ tal que $ \cos 2x \equal{} 1$ y $ \sin 4x \equal{}\minus{}1$? Si $ \cos 2x \equal{} 1$ entonces $ \sin 2x \equal{} 0$ y $ \sin 4x \equal{} 2\sin (2x)\cos (2x) \equal{} 0$, entonces la respuesta no es $ \frac{3}{2}$. Lo unico que queda garantizado con eso que la respuesta es un numero menor o igual que $ \frac{3}{2}$.

llegue al mismo maximo pero con otro punto
$f\left( x \right) = \sqrt 2 \sin \left( {\frac{\pi }{4} + x} \right)\left( {\sqrt 2 } \right)^3 \cos ^3 \left( {\frac{\pi }
{4} + x} \right) = 2\sin \left( {\frac{\pi }{2} + 2x} \right)\cos ^2 \left( {\frac{\pi }{4} + x} \right)$

$ = 2\cos 2x\cos ^2 \left( {\frac{\pi }{4} + x} \right) = 2\cos 2x\frac{1}{2}\left( {1 - \sin 2x} \right) = \cos 2x - \cos 2x\sin 2x = \cos 2x - \frac{1}{2}\sin 4x$

$f'\left( x \right) =  - 2\sin 2x - 2\cos 4x = 0 \Leftrightarrow \sin 2x + 1 - 2\sin ^2 2x = 0$
$2\sin ^2 2x - \sin 2x - 1 = 0 \Rightarrow \sin 2x = \frac{{1 \pm \sqrt {1 + 8} }}{4} = \frac{{1 \pm 3}}{4}$

$\sin 2x = 1 \Rightarrow x = \frac{1}{2}\left( {\left( { - 1} \right)^k \frac{\pi }{2} + k\pi } \right) \vee \sin 2x =  - \frac{1}{2} \Rightarrow x = \frac{1}{2}\left( {\left( { - 1} \right)^{k + 1} \frac{\pi }{6} + k\pi } \right)$

$f\left( x \right) = \cos \left( {\frac{\pi }{6}} \right) - \frac{1}{2}\sin 2\left( { - \frac{\pi }{6}} \right) = \frac{{\sqrt 3 }}{2} + \frac{1}{2}\frac{{\sqrt 3 }}{2} = \frac{{\sqrt 3 }}{2}\left( {1 + \frac{1}{2}} \right) = \frac{{3\sqrt 3 }}{4}$


Let  f from C^1[0, 1] and f(0) = 0, |f'(x)| <= K|f(x)| (K be given constant). Show that f(x) = 0 for all x from [0, 1].

Choice n positive integer such that n > K , so 1 - K/n > 0.
Suppose there is u such that f(u)=0 .
f is bounded because of f is continue. Note M = sup(u =< x =< u + 1/n){|f(x)|}. We can write :
|f(x)| = |f(u) + (x-a)*f'(t(x))|  =<  K(x-u)*|f(t(x))| =< (K/n)*M , where u=< t(x) =< x, for all x  \in [u, u + 1/n]
Therefore  0 =< M*(1-K/n) =< 0, so  M=0 , so f(x)=0 for all x \in [u, u + 1/n]
Now , because of f(0)=0,  it's trivial to achieve the proof by induction .


This lovely problem came from Michael Albert, former Leader of the New Zealand IMO team.  I don't know its original source.

Let $a_0,a_1,a_2,\ldots$ be a sequence of nonnegative integers such that every nonnegative integer can be expressed in exactly one way as $a_i+a_j+a_k$, where $i$, $j$, $k$ are not necessarily distinct.  Find $a_{2002}$ [or $a_{2005}$, if you like].

Edit:  I meant to say $a_i+2a_j+4a_k$.

[quote="Agrippina"]I don't know its original source.[/quote]

I think it is close to one from ISL 1998.

Pierre.

Apologies; I just realised I made a mistake (now corrected) in posting.  I've checked the reference you gave and I don't think it's a related problem.  I'll post the solution later when I have time.

[hide]Let $ f(x)\equal{}x^{a_0}\plus{}x^{a_1}\plus{}\cdots$. The problem statement is equivalent to $ f(x)f(x^2)f(x^4)\equal{}\frac{1}{1\minus{}x}$.
Plugging in $ x^2$ for $ x$, we obtain $ f(x^2)f(x^4)f(x^8)\equal{}\frac{1}{1\minus{}x^2}$. Dividing our first equation by this gives us $ \frac{f(x)}{f(x^8)}\equal{}1\plus{}x \iff f(x)\equal{}f(x^8)(1\plus{}x)$. Plugging in $ x^8$, we obtain $ f(x^8)\equal{}f(x^{64})(1\plus{}x^8)$, and plugging this back in gives us $ f(x)\equal{}f(x^64)(1\plus{}x)(1\plus{}x^8)$. Obviously, this pattern will continue, so $ f(x)\equal{}(1\plus{}x)(1\plus{}x^8)(1\plus{}x^{64})\cdots$. Then $ f(x)$ is the binary number $ x$ read in base 8. Too lazy for the rest.[/hide]


hi, please help me for the following problem.

If you reversed the digits of a two digit number and then multiplied the new number by 10 the result would be 198 more than the orginal numbe.  What was the orginal two digit number?

[hide="Hint"]Try writing the two digit number as $ 10a \plus{} b$, and setting up an equation.

There will be more than one possible value for $ a$.[/hide]

[hide="Solution"]$ 10(10b \plus{} a) \equal{} (10a \plus{} b) \plus{} 198$
$ 100b \plus{} 10a \equal{} 10a \plus{} b \plus{} 198$
$ 99b \equal{} 198$
$ b \equal{} 2$
$ a \in \{1, 2, 3, 4, 5, 6, 7, 8, 9 \}$

$ 12, 22, 32, 42, 52, 62, 72, 82, 92$ all satisfy the conditions.[/hide]

[hide="Solution"]Let the two digit number be $ \overline{ab} \equal{} 10a \plus{} b$. Reversed and multiplied by ten, you have $ \overline{ba0} \equal{} 100b \plus{} 10a$.
$ 100b \plus{} 10a \equal{} 10a \plus{} b \plus{} 198\implies 99b \equal{} 198\implies b \equal{} 2$
$ a$ can be any integer in the range $ [1,9]$.

So the solutions are $ \boxed{12, 22, 32, 42, 52, 62, 72, 82, 92}$[/hide]

Edit: Beaten. Oh well.


geometry
circumcircle
absolute value
national olympiad

[b]1)[/b] Find the number of the connected graphs with 6 vertices. (Vertices are considered to be different)

[b]2)[/b] Two different points $A$ and $B$ and a circle $w$ that passes through  $A$ and $B$ are given. $P$ is a variable point on $w$ (different from $A$ and $B$. $M$ is a point such that $MP$ is the bisector of the angle $<APB$ ($M$ lies outside of $w$) and $MP=AP+PB$. Find the geometrical locus of $M$.

[b]3)[/b] $a, b, c$ are positive reals such that their sum is $1$. Prove that
$\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ac+2b^{2}+2b}$ is more then or equal to $\frac{1}{ab+bc+ac}$.

[b]4)[/b] $ABC$ is an acute angled triangle, $B_{1}, C_{1}, A_{1}$ are points respectively on $[AC, [BA, [CB$ such that $ABC$ is similar to $A_{1}B_{1}C_{1}$. Prove that orthocenter of $A_{1}B_{1}C_{1}$ coincides with circumcenter of $ABC$.

[b]5)[/b] $n$ is satisfying the above conditions
i) $n$ is a positive odd integer.
ii) There are some odd integers such that their squares' sum is equal to $n^{4}$.
Find all $n$ numbers.

[b]6)[/b] We write $1$ or $-1$ on each unit square of a $2007x2007$ board. Find the number of writings such that for every square on the board the absolute value of the sum of numbers on the square is less then or equal to $1$.

Please post these problems in sub-box of Olympiad box (Combinatorics, Geometry,etc) , too. And post links here, can you? Thanks  :wink:


Let a circle C with center O and an exterior point A with AO=d. Find the points B,C,D such that the convex ABCD has the max area

would you read the problem agian ?
what is the relation of the $B,D $ and the cirlce and what is $C$, a circle or a point?

hmm,perhaps you want to say that $B,C',D \in C$ (i changed the point $C$ to $C'$) :)

You have right! C is the name of the circle and B,C',D the points of the circle. Sorry for the mistake iwait your answers.

Let E is the point that AC and BD intersect
Let R is radius of C
we khow that  the area of  ABC'D=(AC'*BD*SIN(<AEB))/2
max BD = 2*R
max AC' = R+AO = R+d
max SIN <AEB=1 so max <AEB=90
so E and O are same point
in otherword we must draw the line that is perpendicular to AO in O that intersects circle in B,D and we must extend
AO until intersect circle in C' :lol:
it is good and easy :!:

Hi silouan
can you understand my solution :?:

Yes Naser. Your solution is very good. I have an onother with trigonometry


What is the sum of all integer values of $ n$ such that $ \frac{20}{2n \minus{} 1}$ is an integer?

We realize that 2n-1 must evenly divide 20. What value do so?

Let's list the factors of 20:

1,2,4,5,10,20. But wait! What about negatives? Thos work too!

So our list is -20,-10,-5,-4,-2,-1,1,2,4,5,10,20

So if 2n-1=-20,-10,-5,-4,-2,-1,1,2,4,5,10,20

Therefore 2n=-19,-9,-4,-3,-1,0,2,3,5,6,11,21

We could find all value of n here, but let's be tricky! Let's rearrange those values!

2n=-19,21,-9,11,-4,6,-3,5,-1,3,0,2

Look! When we sum groups of those, we get 2! Let's do that!

2n=2,2,2,2,2,2

Add those up

2n=12

n=6.

Thus the sum of all possible values is $ \boxed{6}$

I seem to have mistaken an integer for positive integer.

"all integer values of $ n$" $ \iff$ "$ n \in \mathbb{Z}$" ???

If $ n \in \mathbb{Z}$,
$ n\equal{}\minus{}2, 0, 1, 3$.

Thus, $ \minus{}2\plus{}0\plus{}1\plus{}3\equal{}\boxed{2}$.

Oh whoops I forgot n had to be an integer. Therefore 2n-1 must be an odd factor of 20.

2n-1=-5,-1,1,5 meaning 2n=-4,0,2,6 so n=-2,0,1,3 as Kouichi said.


Gauss
analytic geometry
vector
conics
ratio
induction
geometry

[hide]
I think it should be C.
$v'=\frac{M}{M-m} v_0$, so not uniformly...[/hide]

but then what happens to the sand. Isn't the sand moving after it leaves? Doesn't the sand take some energy away as it leaves? Some things to consider...

I guess spoon is on the correct track. shobber, i used your method but the answer turned out to be wrong. The key part of the problem must be that there are no frictional effects. But i need a logical explanation.

[hide="hint"]Think of it this way. Say you have a spaceship traveling through space at a constant velocity and no forces are acting on the ship. Instantly, the ship breaks in half. What happens to the velocity of each piece?[/hide]

[quote="ak007"]The velocity of the sand will be the same even after it leaves the cart and thus, the velocity of the whole system will remain the same. Since the momentum of the system remains the same, the velocity of the cart will remain constant.[/quote]
 ;)  Frictionless...
I solved a problem similar to this one several months before, and in that problem, we considered the object falling onto the ground would stop at last.

shobber, even if the sand falls to the ground and stops, the car would still remain at a constant velocity (assuming there was no friction between the car and the ground). Once the sand and car are separate, thye no longer interact. The sand leaving the car takes energy with it, and it is this loss of energy that compensates for the velocity not increasing.

[quote="ak007"]A train car is coasting at a constant speed along a long horizontal rail where frictional effects are negligible. A hole develops in the bottom of the car and sand starts spilling out onto the ground below at a constant rate. While the sand is spilling out the speed of the train car will

(A) increase uniformly
(B) decrease uniformly
(C) increase non-uniformly
(D) remain constant[/quote]

[hide]

The answer is C.

$p = \text{const.}\\
\\
\dot{p}=\dot{m}v + m\dot{v} = 0\\
\\
\Rightarrow \dot{v} = - \frac{\dot{m}v}{m}\\
\\
$

Assuming a constant cross section of the hole then $\dot{m}=\rho A\dot{z}=-\rho Agt$

While $m = M_{\text{car}}+\rho A\cdot(z_0-\frac{1}{2}gt^2)$

$\therefore \boxed{ \dot{v} = \frac{\rho Avgt}{M_{\text{car}}+\rho A(z_0-\frac{1}{2}gt^2)}}$
[/hide]

for once i agree with Spoon. i think you have your frame of reference messed up Dr. No.

you need to decide whether or not the sand is in the system or not. it should be but you are losing some of it. when you say:

[quote="Dr. No"]
$\dot{p}=\dot{m}v + m\dot{v} = 0\\$
[/quote]

you are making a mistake here. when you say $\dot{m}v$, you are implying that mass is lost. no mass is lost. the sand that pours out is still part of the system.

$\dot{p}= m\dot{v} = 0$

$\therefore \dot{v} = 0$

[quote="Spoon"]but momentum isn't constant... the sand has no momentum?[/quote]

The momentum of the sand is its [i]mass x velocity[/i]. The sand has a horizontal component of velocity $v$ equal to that of the train car, and a vertical component of velocity $\dot{z}$ as it pours out from the hole.

[quote="skylarth"]for once i agree with Spoon. i think you have your frame of reference messed up Dr. No.

you need to decide whether or not the sand is in the system or not. it should be but you are losing some of it. when you say:

[quote="Dr. No"]
$\dot{p}=\dot{m}v + m\dot{v} = 0\\$
[/quote]

you are making a mistake here. when you say $\dot{m}v$, you are implying that mass is lost. no mass is lost. the sand that pours out is still part of the system.

[/quote]

The boundary of the train car itself [u]is[/u] the boundary of the system. Mass is being lost as the sand exits the hole. Momentum is constant because there are no external forces on the system.

In principle it is the same idea as rocket propulsion. In other words, [i]Sometimes [b] it is[/b] rocket science.[/i]

I'm not sure it makes sense to define the system to be just the railroad car. Then mass would be leaving the system, taking energy and momentum with it. How can you then (logically) apply conservation laws?
I have a qualm with the frictionless issue. Does that mean the sand would move at the same speed as the train even after hitting the ground? Eventually, the sand would pile up and block the hole, so it's as if the sand never left the train. Unless we think about the components of the force of weight that would cause the sand to spread out.

[quote="Dr. No"]Momentum is a vector quantity. Gravity, an external force, acts in the z-direction. The train travels in the x-direction. There are no external forces in the x-direction. Momentum, in the x-direction, is conserved.[/quote]

That's right, in fact NOTHING changes in the $\hat{x}$ direction, not the sand, not the car, nothing. A force in the $-\hat{y}$ direction has NO affect on the $\hat{x}$ component. Everything stays the same and the car continues to move at V, as well as the sand continues to move at V.

[quote="Spoon"]Assume three balls. Take the back one away. The center of mass was within the middle ball. The middle ball will still have no reaction to removing the back ball. If the momentum started as 3MV and you remove the back ball, the remaining two will have momentum 2MV and they will have been unaffected. In order for a force to be applied to the other balls, there has to be some interaction. We are assuming that all forces such as gravity are negligible, therefore it's not possible, no matter what the balls configuration to affect any other ball by removing any other ball. Consider balls traveling in line in the form of a circle, or a rough circle formed of say 5 balls. If you remove any one without touching the others, the others will continue to move as if nothing had happened. They are separate entities that do not interact. Assume thousands of balls (sand) all moving together in a random configuration with no friction between them. If you reach in and magically grab one without touching another, no changes to the remaining balls will occur. This is even "removing" the ball from the system. You lose the mass of the removed ball along with it's velocity. The overall momentum of the group drops by that amount and they continue to move at the same rate in the same direction.[/quote]
The center-of-mass may have been within the middle ball, but there is no constraint which requires it to be so. When you remove the back ball, you apply a force to the system accelerating the COM backward. Though as you said, there is no interaction between the balls; the forward pair continues on merrily, unaware of the violence to the system. In such a case the COM is an abstraction divorced from any material consequence.  Its congruence with the position of any real particle within the system is coincidental.

[quote="Spoon"]I disagree that the sand is leaving the system. Where does it go? Please in words, not math, where does it go? It's still moving forward with the car, it's merely moving vertically downward! Just as the ball.[/quote]
In attempt to rebut you here I found my error.

[quote="A Dictionary of Physics, ISBN 0-19-280030-2"][b]thrust[/b], The propelling force generated by an aircraft engine or rocket. It is usually calculated as the product of the rate of mass discharge and the velocity of exhaust gases relative to the vehicle.[/quote]

Mass is being discharged from the system (as I said I can prove that by the Reynolds Transport Theorem). However since the train is moving horizontally and the discharge is vertical, the velocity of the exhaust gas (sand) relative to the vehicle (train car) is zero, in the x-direction. As you pointed out, the sand and the train are moving at the same speed in the x-direction. Thus $\text{Thrust } =\frac{dm_{sand}}{dt} \cdot 0 = 0$

Hey Rogue Physicist, I seem to recall a Foxtrot comic where there was a problem involving a train moving from point A to point B in a certain amount of time.  A trivial time/distance problem.  On the answer sheet, Jason wrote...

"Assuming a frictionless track, assuming a completely uniform rail, assuming that this is over the surface of the earth rather than through space, assuming that the train travels on a straight line over the surface of the earth, assuming that the train did travel the "long way" around over the earth, assuming that the earth is a perfect sphere, assuming that the clocks are accurate, assuming that the clocks are synchronized, assuming that the clocks are accurate, assuming that the stations are in the same time zones, disregarding the effects of relativity, etc, etc, etc, blah, blah, blah...

55 mph."

This is paraphrased, but it went something like that.

This is a physics problem.  It uses idealized conditions.  I could follow up that question that my math teacher gave me on one test, where the wording was not exactly precise, with the intended interpretation and then an alternative interpretation that was much, much more complex but allowed for the fuzzy wording.  I'm pretty sure that my math teacher would have either thrown away the extra work or marked me down for it.  I actually did stuff like that before, but the teachers told me to stop being a smartass and do the problem the way it was meant.  The whole world doesn't use your standards, you know.

[quote]I'm pretty sure that my math teacher would have either thrown away the extra work or marked me down for it. I actually did stuff like that before, but the teachers told me to stop being a smartass [/quote]I know exactly what you mean.
Sometimes you have to compromise the truth and your own values to just get past the idiots.

My son, who was quite smart and took physics and sciences seriously was in high-school class once.   The teacher told the class that gold had the greatest conductivity.  My son knew this was completely wrong, and spoke up, correcting him.  Silver actually has a lower resistance.  The teacher was furious and argued violently about it.  My son thought he had made a mistake after all.  He came to me and I opened my Handbook of Physics 21st Ed. and indeed confirmed for both of us that silver has the lowest resistance, the greatest conductivity.

My son, still struggling with the physics and with observations of industrial standards asked more about it.  Why did they use gold contacts, not silver?  This has nothing to do with conductance, I explained.  Gold doesn't corrode in air (oxidize), so it is the best overall choice for pressure-contacts in electrical circuits, like cards and connectors.   Silver is the best conductor, but is susceptible to corrosion.

The teacher was wrong.  And a bigger smartass for both arguing about it, and punishing my son for being right, accusing him of being a smartass when he was just being accurate, and preventing the teacher from misleading 30 other people indefinitely.

The educational system is fundamentally more concerned with authority (teaching students to fear authority) than truth or accuracy.  It is a flaw with the system, not the proper way to do science or educate.

I supported my son, who was right, and was not a smartass.  The teacher was the biggest smartass for not double-checking his own facts.

Firstly, your language is inappropriate, and I hope a an admin drops by.

Secondly, that situation you mentioned is not the same as looking too far into an assigned problem. Lot's of people get simple facts wrong and argue blindly that they are correct. So what. That has nothing to do with the fact that it's arrogant and just wrong to take a simple problem and expand it into something it obviously isnt.

The simplest example is air resistance. In almost all cases we assume no air resistance. If you were a given a problem where it wasn't necessarily stated yet was implied and you did the problem with air resistance, you should be marked wrong, even if your answer is correct for your interpretation of the problem.

[quote="Dr. No"]Mass is being discharged from the system (as I said I can prove that by the Reynolds Transport Theorem). However since the train is moving horizontally and the discharge is vertical, the velocity of the exhaust gas (sand) relative to the vehicle (train car) is zero, in the x-direction. As you pointed out, the sand and the train are moving at the same speed in the x-direction. Thus $\text{Thrust } =\frac{dm_{sand}}{dt} \cdot 0 = 0$[/quote]
I do not agree that the sand is being discharged. There MUST be an interaction for a discharge. One cannot exert a force on the other if there is no interaction, it simply isn't possible in classical mechanics. The sand is being pulled away by gravity, it is not being pushed by the car. There is not upward force exerted on the car by the sand.

[quote="Spoon"]Firstly, your language is inappropriate, and I hope a an admin drops by.[/quote]I only quoted the same language as was used against me here.
Anyway 'smartass' is hardly a serious 'swear-word'.  It's a common word used by adults to characterize children to their face, and it would be both hypocritical and unrealistic to expect children not to know or be allowed to use the word.  What might make such a word inappropriate in a forum would be if you or I started calling each other names here in the midst of a dispute.  That has not happened that I know of, and I certainly have not called anyone any names myself.

[quote]... it's arrogant and just wrong to take a simple problem and expand it into something it obviously isnt.[/quote] People will always differ on how deep to analyze a problem.  This has nothing to do with 'arrogance'.  That is just your perception.

[quote]The simplest example is air resistance. In almost all cases we assume no air resistance. If you were a given a problem where it wasn't necessarily stated yet was implied and you did the problem with air resistance, you should be marked wrong, even if your answer is correct for your interpretation of the problem.[/quote]Wrong.
When do you try to take account of air resistance?  When it is significant.
When do you ignore air resistance? When it is not significant.

When do you take into account hot air?  You must always filter it from a response.

[quote="Spoon"]I do not agree that the sand is being discharged. There MUST be an interaction for a discharge. One cannot exert a force on the other if there is no interaction, it simply isn't possible in classical mechanics. The sand is being pulled away by gravity, it is not being pushed by the car. There is not upward force exerted on the car by the sand.[/quote]
Try the following [i]Gedanke Versuch[/i], put the train car loaded with sand on a balance scale. Load the opposite arm of the balance with an equivalent weight. Now cut a hole in the bottom of the train car and allow the sand to drain out. Does the scale remain balanced? If not, why?

[quote]Try the following Gedanke Versuch, put the train car loaded with sand on a balance scale. Load the opposite arm of the balance with an equivalent weight. Now cut a hole in the bottom of the train car and allow the sand to drain out. Does the scale remain balanced? If not, why?[/quote]I agree with you Dr. No.
The result of your thought experiment is not in dispute.
That is why I suggested the train rises when the sand is lost, changing velocity (direction).

WHAT'S UR POINT WITH TTHE BAANCE EXPERIMENT NO GUY.....
IF THE HOLE IS MADE ,THE SAND DRIPS ON TO THE BALANCE,SO THE WEIGHT IS SAME,OR AM I INTERPRETING THIS WHOE THING WRONG? :oops:  :blush:

WHY IS THIS QUESTION GETTIN SO ONG?ISNT THE ANS SIMPY CONSTANT VEL?

Stop using caps... online it is interpreted as screaming. We would rather you not scream. Thank you.

I believe that the force of the sand suddenly hitting the balance would cause it to drop on that side slightly, but then it would recover and oscillate. I also do not see how this has any relation to the problem. The fact still remains that when sand is drawn away from the car by a force that doesn't act on the car, the car experiences no force. The car and sand is still the same as my balls example.

I also don't understand how the center of mass plays into this problem at all, as I have clearly shown above that any configuration of any amount of balls will still obey this behavior.

[quote="Spoon"]I believe that the force of the sand suddenly hitting the balance would cause it to drop on that side slightly, but then it would recover and oscillate. I also do not see how this has any relation to the problem. The fact still remains that when sand is drawn away from the car by a force that doesn't act on the car, the car experiences no force. The car and sand is still the same as my balls example.

I also don't understand how the center of mass plays into this problem at all, as I have clearly shown above that any configuration of any amount of balls will still obey this behavior.[/quote]
First, the balance in my thought experiment doesn't have to be a pan balance. It is conceivable to suspend the train car from above so that there is nothing to catch the spilling sand.

Second, I am loathe to admit it but I worked the wrong side of the equation when dealing with the center-of-mass. The rate of change of total momentum is zero. With respect to the center-of-mass, no mass is lost. Since this is the case the velocity of the center-of-mass does not change.

Third, rather than working the momentum equation from the standpoint of the COM one may write it in a straight forward way as: $p_{\text{total}}=m_{\text{car}}v_{\text{car}}+m_{\text{sand}}v_{\text{sand}}$

Considering the right hand side one can immediately write: $v_{\text{sand}}=v_{\text{car}}+v_{\text{rel}}$, which is valid even if $v_{\text{rel}}=0$

This means one can rewrite the total momentum as: $p_{\text{total}}=(m_{\text{car}}+m_{\text{sand}})v_{\text{car}}+m_{\text{sand}}v_{\text{rel}}$

The portion of total momentum whose velocity is $v_{car}$ can be rewritten as $(m_{\text{car}}+m_{\text{sand}})v_{\text{car}}= p_{\text{car}}$ such that $p_{\text{total}}=p_{\text{car}}+m_{\text{sand}}v_{\text{rel}}$

Finally, if one differentiates this expression with respect to time one obtains:

$0=\dot{p}_{\text{car}}+\dot{m}_{\text{sand}}v_{\text{rel}}+m_{\text{sand}}\dot{v}_{\text{rel}}$

This result implies that there is a force on the car only if the mass of the sand changes with time and has some velocity relative to the train car. Clearly, the sand has no velocity relative to the train car in the x-direction; therefore there is no force on the train car in the x-direction and the car does not accelerate in the x-direction.

The mass of the sand does change with respect to time and does have velocity relative to the car in the z-direction (vertical direction) due to gravity, but this force is overwhelmed by the weight of the train car itself. The weight of the train car, the mass flow of the falling sand notwithstanding, is counterbalanced by a vertical reaction force of the train track; thus the train car does not accelerate in the z-direction despite the spilling sand.

Now that makes much more intuitive sense :)

So it seems that the problem has been settled.

So I'd just make a remark :

[quote="Rogue Physicist"]

[quote]The simplest example is air resistance. In almost all cases we assume no air resistance. If you were a given a problem where it wasn't necessarily stated yet was implied and you did the problem with air resistance, you should be marked wrong, even if your answer is correct for your interpretation of the problem.[/quote]Wrong.
When do you try to take account of air resistance?  When it is significant.
When do you ignore air resistance? When it is not significant.[/quote]

Do you really think that there is a significant difference between considering a flat Earth and a round Earth in this problem ? Do you really think the error made by considering the Earth to be flat is significant ?

DROP IT ANTONY,other wise  rogue physicist will start another pointless speeech and make me sick.
seems like spoon has alot of problem with my messages,first the space,now the screamer ;)

i shouldnt be doin this,but will u guys try the questions in my totalinternalreflectinglight post,cause i ahve a doubt in one or 2 of them
thanx


PLEASE HELP ME WITH THIS ONE!!!!!!!!
which expression is equivalent to:
8c-80c+200

a) (2c-10) (4c+20)
b) 8(c-5)(c+5)
c) 8(c+5)2squared
d)8(c-5)2squared

please tell me how to find this out!!!!!!!
thank u so much

this isn't the appropriate section...  :)

I think you want to say $8c^{2}-80c+200$
[hide="click here"]$8c^{2}-80c+200=8(c^{2}-10c+25)=8(c-5)^{2}$[/hide]

8(csquared-10c+25) and you reduce!!!! :oops:


Given positive reals $x, y, z \le \frac{1}{\sqrt[3]{2}}$, prove that

$\frac{1}{x^2+y^2}+\frac{1}{y^2+z^2}+\frac{1}{z^2+x^2} \ge x+y+z$.


quadratics
LaTeX
algebra
quadratic formula

[quote="nonie"]What positive integer is 56 less than it's square?[/quote]
[hide]$x=x^2-56$
$x^2-x-56=0$
$(x-8)(x+7)=0$
$7$ doesn't work because it asks for a positive integer, so $\boxed{x=8}$[/hide]

[hide]$8^2-8=56$

$8$[/hide]

[size=84][color=green]Hide your answers please![/color][/size]

[hide]We can write this as $x-x^2=56$, or $x-x^2-56=0.$ This factors as $(x-8)(x+7)$, and taking the positive root we get $x=8$[/hide]

Or alternatively, use your common sense...
$7$ can't be right, because it's not even bigger than $50$
$9$ can't be right, it's far too big...
so the only logical answer is $8$, so we just simply test it, $64-8 = 56$ :P

Or you could simply, take the squareroot of $56$ in your head approximately top of my head, little more than $7.48$... so it can only be $8$, as $9$ will be again too large.

:P

[hide] $x^2-56=x$
$x^2-x-56=0$, from where $x=8$ or $x=-7$, but whe are searching for natural numbers, so $\boxed {x=8}$[/hide]

[quote="Ixeus"]Or alternatively, use your common sense...[/quote]

I am not too good at using my common sense. Not that I don't have any ;)... I used algebra becuase I have mastered the Art of Algebra Solving.(AoAS)

[quote="nonie"]Yeah [hide]8[/hide] is obviously right.

Here's another one. What fraction is 2/9 less than it's square?[/quote]

x=x^2-2/9
9x=9x^2-2
9x^2-9x-2
(3x-2)(3x+1) um...i am not sure if I wrote this correctly but it is [hide="THE"][hide="Answer"]2/3[/hide][/hide] Algebra rules.

Not quite...
[hide]$\frac{2}{3}$ is $\frac{-2}{9}$ less than it's square.[/hide]
Maybe try solving it by representing $\frac{x}{y}$ as the fraction?

Actually he just factored wrong. You dont need to do the x/y thing.
[hide]$9x^2-9x-2=0$
and i used the quadratic formula so $x=\frac{3\pm17}{6}$.[/hide]

[hide="Answer"]x=x^2 - 56; thus (x+7)(x-8)=0.The answer is 8![/hide]
P.S. I'm new so I don't know a lot about how to put latex on messages.