How many times in a day, are the hour and minute hand of a clock on top of each other?

[hide]is the answer 22?  :maybe: [/hide]

[hide="depends on how you count it i guess"]
Depending on how you count it, I think there might be 23...
[/hide]

[quote="i_like_pie"][hide]I say 23. Once every our, except midnight, which is the first one of next day.
How did you get 22, Splash?[/hide][/quote]

what about noon?

[quote="SplashD"]what about noon?
[/quote]

[hide]Huh? There are 24 hours in a day, so the two hands are in the same spot 24 times. However, midnight is not counted since it's the beginning of the next day. 23. Nothing about noon. :)[/hide]

[quote="i_like_pie"][quote="SplashD"]what about noon?
[/quote]

Huh? There are 24 hours in a day, so the two hands are in the same spot 24 times. However, midnight is not counted since it's the beginning of the next day. 23. Nothing about noon. :)[/quote]
That makes sense. But after midnight they match in the morning of the next day. So what happens to the 24th meeting?

[quote="bpms"]That makes sense. But after midnight they match in the morning of the next day. So what happens to the 24th meeting?
[/quote]
I AM CONFUSED.

Edit:

[hide]I got it! There are not 24 meetings of the two hands per day. There are 24 hours, but only 23 meetings. Explanation: Midnight marks the end of the 24th hour of day 1, but it's meeting is NOT counted since it is the beginning of day 2.[/hide]

[quote="i_like_pie"][quote="bpms"]That makes sense. But after midnight they match in the morning of the next day. So what happens to the 24th meeting?
[/quote]
I AM CONFUSED.

Edit:

I got it! There are not 24 meetings of the two hands per day. There are 24 hours, but only 23 meetings. Explanation: Midnight marks the end of the 24th hour of day 1, but it's meeting is NOT counted since it is the beginning of day 2.[/quote]

your logic is wrong. by the way, you are using a 12 hour clock, right?

Of course 12 hours. :|
How am I wrong?

Here's a better explanation:
[hide]The meetings occur in the intervals between two hours. 23 per day.[/hide]

[quote="SplashD"]the minute and the hour hand overlap once every hour except 11:00-11:59 am and pm[/quote]
wait what? I'm confused...

[hide]SplashD was correct the first time. The answer is 22, because the hands don't meet in two of the 1-hour intervals (11:00 - 11:59) each day. $24-2=22.$[/hide]

[hide]wait i still think 23! the first midnight is 1...so the last midnight isnt included but isnt that 23?[/hide]

That [i]is[/i] easier! Except that the hour hand goes around twice. :P

[quote="nebula42"]There is actually a much simpler way:
[hide="Solution"]The minute hand goes around the clock $24$ times in a day. The hour hand goes around $1$ time. So they meet $24-1 = 23$ times. However, this counts midnight of the next day, so we need to subtract that out to get $23-1 = \boxed{22}$ times, the answer.[/hide][/quote]

why doesn't the hour hand go around two times?

[hide]Ah. Sorry.  :oops:  The hour hand actually goes around $2$ times, but the answer is still $22$, though, because we actually [i]aren't[/i] counting midnight twice.  :roll: (It's just $24-2 = 22$. But anyway, you still get the basic idea of the solution, which is more important.)
Heh, I just made two mistakes that "canceled" each other out (and still got the right answer).  :P [/hide]

[quote="mtms5467"]Extension: how would you find the times, in a 12-hour period, that the hands form a 0 degree angle? And list the times please[/quote]
[hide]It'd be way too much work to find the times. The answer is 11. The reason is that the hour hand circles the clock once and the minute hand does 12 times. 12-1=11.[/hide]

[hide]There are 22 meetings per day, or 11 meetings in every 12-hour period. The hands are in the same position every 1:05:27. Here is a list of times:
12:00:00
1:05:27
2:10:54
3:16:21
4:21:48
5:27:15
6:32:42
7:38:09
8:42:36
9:48:03
10:54:30
It's 3 seconds off because the time difference is actually $65.\overline{45}.$[/hide]


[color=red]Use Trigonometric Substitutions[/color]
1)
$ x,y,z$ be positive real numbers such that $ xy\plus{}yz\plus{}zx\equal{}xyz$
Prove that
$ (x\minus{}1)(y\minus{}1)(z\minus{}1) \leq 6\sqrt{3} \minus{} 10$

2)
Let $ a,b,c$ be real numbers.
Prove that
$ (a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1)\geq(ab\plus{}bc\plus{}ca\minus{}1)^2$

[quote="colorfuldreams"][color=red]Use Trigonometric Substitutions[/color]
1)
$ x,y,z$ be positive real numbers such that $ xy \plus{} yz \plus{} zx \equal{} xyz$
Prove that
$ (x \minus{} 1)(y \minus{} 1)(z \minus{} 1) \leq 6\sqrt {3} \minus{} 10$

2)
Let $ a,b,c$ be real numbers.
Prove that
$ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)\geq(ab \plus{} bc \plus{} ca \minus{} 1)^2$[/quote]

$ x,y,z$ be positive real numbers such that $ xy \plus{} yz \plus{} zx \equal{} xyz$
Prove that
$ (x \minus{} 1)(y \minus{} 1)(z \minus{} 1) \leq 6\sqrt {3} \minus{} 10$

try $ a \equal{} b \equal{} c \equal{} 3$

 :lol:

I think  $ (x \minus{} 1)(y \minus{} 1)(z \minus{} 1)\geq 8$

Let $ a,b,c$ be real numbers.
Prove that
$ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)\geq(ab \plus{} bc \plus{} ca \minus{} 1)^2$

$ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)\geq(ab \plus{} bc \plus{} ca \minus{} 1)^2 \Longleftrightarrow ( \minus{} a \minus{} b \minus{} c \plus{} abc)^2\geq 0$

 :lol:

For (2) see also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=82579]here[/url]  :)


Let $a,b,c,d$ be reals. If $a^3+b^3+c^3+d^3=0$ and $a+b+c+d=0$
prove that a pair of $a,b,c,d$ must sum to 0.

that if you consider the 6 possible pairs of $a,b,c,d$ at least one of them, say $a$ and $b$ must have $a+b=0$

using that $ a+b = -{c+d) }$
I get that , $ ab(a+b) + cd (c+d) = 0 $

Note that this can be done for any arbitrary 2 - combination of a,b,c,d .

Rushil, why does that imply that there is a pair whose sum is 0?

i got that $a^2-ab+b^2=c^2-cd+d^2$ must imply that there is such pair. but i bit stuck from here

You've effectively done the same thing as I have. And I haven't concluded the resullt. i'm stuch there as well !!!!

oh ok i see... i'll try to finish it now.

let's see if we can combine both of our results

i was thinking to try it.. but don't know if it will lead us..

write a polynomial with roots $a, b, c, d$, $\sum{a}=0$, and you can show $\sum{abc}=0$.  factor that thing in different ways,  and use the fact $\sum{a}=0$, like one factorization is $ab(c+d)+cd(c+d)=0$, so $(c+d)(ab-cd)=0$, and you can do this for all pairs.   Then for, all the pairs, suppose $(c+d) \neq0$, that implies $ab=cd$, and do that similarly to get $ab=ac=ad=bc=bd=cd$.  Suppose one of $a,b,c,d$ is zero, then the given statements are true.  WLOG, let $a$=0, then the given equations say that $b+c+d=0$, $b^3+c^3+d^3=0$.

but $b^3+c^3+d^3-3abc=(b+c+d)(junk)$, so $3abc=0$ and one of $a, b, c,$ must be zero, so its true.

now suppose none of $a, b, c, d$ are zero, then you have $a=b=c=d$, but then from the given equations $4a=0$, contradiction.

Great solution!!!!

I was thinking of bringing TOE in but couldn't make it!!!

I thought that wouldn't be helpful but yo've proved me wrong!!!!


Let $ X$ be a Hausdorff space such that $ X \equal{} A \cup B$ where $ A$ and $ B$ are each homeomorphic to a torus, and $ A \cap B \equal{} \{x_0\}$.

What is the structure of $ \pi_1(X, x_0)$?

How do I describe the structure of this topological space's fundamental group?  I know how to draw it, etc.  But, I don't know much about this space's fundamental group.

Is $ \pi_1(X, x_0)$ isomorphic to the free product of $ \mathbb{Z} \times \mathbb{Z}$ and $ \mathbb{Z} \times \mathbb{Z}$?

In general, Van Kampen's theorem will tell you that

\[ \pi_1 \left( \bigvee_{i\equal{}1}^n X_i  \right) \equal{} \pi_1(X_1) * \pi_1(X_2) * \dots * \pi_1 (X_n).\]


are there any good math site other than saab.org or mathcounts.org with some good preparation problems and such?
thanks. :D

AMC website

http://www.unl.edu/amc (I think)

I know a good site but it's gonna be real hard..

P.S. It's not for Mathcounts nor AIME

Some old competitions at http://goodlymaths.0catch.com/Contests/MATHCOUNTS/mathcounts_archive.htm

[quote="Silverfalcon"]Very difficult site which I just know it but don't use:

http://cage.rug.ac.be/~hvernaev/olympiad.htm

P.S. This site is for USAMO plus othre international olympiad.[/quote]

P.P.S. That link doesn't work.

That site worked fine to me though I don't use it...

http://cage.rug.ac.be/~hvernaev/olympiad.html

Forgot one letter, sorry...
That's the right site..


In the plane are given 2005 points, such that no three of them are on a line and every two of them are connected with a line. Prove that the points can be coloured in 2 colours, in a way that for evry two point with a same colour, the number of the lines which separate them is an even number.

I'm somewhat confused. What do you mean by 'lines that separate them'? if you mean lines between them, then it is not a counting problem and also is not necessarily true. If you mean AB-->BC=1, AB-->BC-->CD=2, then then it it can be done in any positive integer number of steps. Please clarify.

A line separates two points if and only if the points do not lie on the line and they are on diferrent halfplanes in relation to the line


This is not putnam problem,
but Berkeley problems interesting, I put the links here

http://math.berkeley.edu/~desouza/pb.html

hmm.
Most of them are not so difficult as Putnam's. After all, they are not contest problems. They are exam problems.


On the AMC 8 contest Billy answers 13 questions correctly, answers 7 questions incorrectly and doesn't answer the last 5. What is his score? (There is no penalty for answering incorrectly or leaving answers incomplete.)

If Billy got thirteen questions right and each correct answer is a point and there is no penalty for guessing, then Billy's score is ...

13 x 1 = 13.

The answer is 13.

 :D

Wasn't this on an AMC 8 problem? I forgot what year.


vector
real analysis
real analysis unsolved

Does there exist a sequence $\{x_{n},\,n\ge1\}$ of vectors from $l_{2}$ with unit norm which satisfy $(x_{n},x_{m})<-\ds\frac1{2004}$ for $n\ne m$, $n,m\in\mathbb{N}$?
(A. Bondarenko)

No. We have $0 \leq (x_{1}+\dots+x_{n},x_{1}+\dots+x_{n})=n+2\sum_{i \neq j}{(x_{i},x_{j})}<n-\frac{n(n-1)}{2004}$, which tends to $-\infty$ as $n \to+\infty$.  :maybe:


function
calculus
integration
real analysis
real analysis solved

Can anyone help me with this problem?

f:R->R is a continous function, such that (F(x+h)-F(x-h))/2h>=f(x) for any real x and any positive h, where F is a primitive of f (I wrote it like that because I didn't know how to write integral from x-h to x+h).

A) Show that on any closed interval the function reaches its maximum in one of the extremities of the interval (I managed to show this)
B) Show that the function is convex (this is where the problem is)

C'mon dudes, can someone pleeeease help me? :D

Let a<b. I want to pruve that f(a)+f(b)>=2f((a+b)/2)
Let g(x)=f(x)-(x-a)(f(b)-f(a))/(b-a)
Notice that g(a)=g(b)=f(a)
(F(x+h)-F(x-h))/2h>=f(x) implies that (G(x+h)-G(x-h))/2h>=g(x)
From A) applied to g we get g((a+b)/2)<=g(a) rezulting that
f(a)+f(b)>=2f((a+b)/2)


[quote="Laputa"]With three positive number $a, b, c$ satisfy $a+b+c \leq 1$.  Prove that:

$(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c}) \geq 64$[/quote]
Easy.
$\prod(1+\frac{1}{a})\ge(1+\frac{1}{\sqrt[3]{abc}})^3\ge 64$

I went wrong in writing this message...  :blush:  now it's below  :D

[quote="Nameless"]
$\prod(1+\frac{1}{a}) \ge (1+\frac{1}{\sqrt[3]{abc}})^3\ge 64$[/quote]

$(1+\frac{1}{\sqrt[3]{abc}})^3\ge 64$ is obvious, instead

$\prod(1+\frac{1}{a}) \ge (1+\frac{1}{\sqrt[3]{abc}})^3$ is to be explained  :P

He used Holder's inequality.I think  that you also could use Jensen's ineq.

we only need Am-GM and AM-HM

it is easy to find $ab+ac+bc\leq \frac{1}{3}$ and $abc\leq\frac{1}{3}$ by AM-GM
and $\sum \frac{1}{ab}\geq \frac{9}{\sum ab}\geq 27$ , $\sum\frac{1}{a}\geq \frac{9}{\sum a}\geq 9$ by AM-HM

So we can get

LHS=$1+\sum\frac{1}{a}+\sum\frac{1}{ab}+\frac{1}{abc}\geq 1+9+27+27=64$


1.Solve the equation:
$ [\frac {a^2}{b}] \plus{} [\frac {b^2}{a}] \equal{} [\frac {a^2 \plus{} b^2}{ab}] \plus{} ab$
($ a, b \in N$)

2.Solve

$ \frac {x \minus{} y \sqrt {x^2 \minus{} y^2}}{\sqrt {1 \minus{} x^2 \plus{} y^2}} \equal{} \frac {2 \sqrt {3}}{\sqrt {5}}$
$ \frac {y \minus{} x \sqrt {x^2 \minus{} y^2}}{\sqrt {1 \minus{} x^2 \plus{} y^2}} \equal{} \frac {\sqrt {2}}{\sqrt {5}}$

hello, let us solve the following equation system:
$ \frac {x \minus{} y \sqrt {x^2 \minus{} y^2}}{\sqrt {1 \minus{} x^2 \plus{} y^2}} \equal{} \frac {2 \sqrt {3}}{\sqrt {5}}$
$ \frac {y \minus{} x \sqrt {x^2 \minus{} y^2}}{\sqrt {1 \minus{} x^2 \plus{} y^2}} \equal{} \frac {\sqrt {2}}{\sqrt {5}}$
It is easy to show that
$ \frac{(x\minus{}y\sqrt{x^2\minus{}y^2})^2}{1\minus{}x^2\plus{}y^2}\minus{}\frac{(y\minus{}x\sqrt{x^2\minus{}y^2})^2}{1\minus{}x^2\plus{}y^2}\equal{}x^2\minus{}y^2\equal{}\frac{12}{5}\minus{}\frac{2}{5}\equal{}2$.
By inserting $ x^2\minus{}y^2\equal{}2$ in our system we get
$ \frac{x\minus{}y\sqrt{2}}{\sqrt{\minus{}1}}\equal{}\frac{2\sqrt{3}}{\sqrt{5}}$
$ \frac{y\minus{}x\sqrt{2}}{\sqrt{\minus{}1}}\equal{}\frac{2}{\sqrt{5}}$
Solving this linear system we have found the solutions
$ x\equal{}\minus{}\frac{2}{5}i\sqrt{5}(\sqrt{3}\plus{}1)$
$ y\equal{}\minus{}\frac{1}{5}\sqrt{10}i(2\sqrt{3}\plus{}1)$
Sonnhard.


[quote="AIME15"]Haha, fun game.  The only part I don't like is the lack of control over the character.

http://erhter.mybrute.com[/quote]

That is the beauty of it. Its so awesome. Try multiple characters. I am trying to think of an optimal character, and maybe an optimal strategy.

I found out that you can fight someone 2 levels lower and get the same experience.

Lol.

I just found a great starting character. He totally owns.

http://ishti12.mybrute.com/fight/497107249

[quote="EggyLv.999"]asdf12321.mybrute.com

fails a lot[/quote]

man that brute is pretty bad.
Definitely not the optimal character I am looking for.

I rule! Won all fights but one...

http://amintnt.mybrute.com

Who dares fighting me? :huh:

[quote="bbgun34"]http://ihavnothing2do.mybrute.com

it beat you, amintnt...  :P

EDIT:
actually,
http://ihavenothing2do.mybrute.com might be better...(it's a different one)[/quote]

your weapons are pretty bad. My char can own u.

[quote="Chompy"]whats best weapon?
My spear is pretty good...for me...[/quote]

the hammer is just sick.

check out this fight.

http://ishti12.mybrute.com/fight/501426581

[quote="bbgun34"]http://ihavnothing2do.mybrute.com

it beat you, amintnt...  :P

EDIT:
actually,
http://ihavenothing2do.mybrute.com might be better...(it's a different one)[/quote]
No sign of ur winning, it's just pupil written there.

http://iiiiiiiiiii32.mybrute.com/

What I don't like is the sheer randomness and near-total lack of control*

*Can't choose what stats, what weapons, what skills, and how to fight!?

[quote="xpmath"]I have a lot, and I've won 5 of 6 in almost all of them...this is the only one where I've won all 6.

[url]http://lulzwhat8.mybrute.com/cellule[/url][/quote]

the knife speed combo seems to be superior.

http://nameistakenbleh.mybrute.com

the only thing that sucks about this game is that you can't actually control your brute..
other than that though, it's pretty fun  :lol:

It's not fair at all! A level 8 brute comes to fight my level 2 brute! :huh:

Well... Wasn't so lucky today... Lost three fights!  :oops:

[quote="EggyLv.999"][quote="Fraenkel"]I found this game yesterday and have been addicted to it since, and created multiple fighters.
The cool thing about this game is that your skills and battles are automated and random. (everything is random)
Thus it is difficult to get an optimal strategy.
Share with me your optimal strategies, if you find one.

You can play the game at this link.

http://ishti10.mybrute.com[/quote]

my guy beat yours...[/quote]

my brute totally owned your guy

http://ishti10.mybrute.com/fight/507099059

MyBrutes is down? I can't access it :maybe:

hmm... me 2!!
something about a locational error and trying again later...

the servers go down for an hour a day. You may have been trying to access your brute during that time.

OMG!!! I GOT A WOLF!!!!!!! AND AT LEVEL 8 TOO!!!!!!!!!

http://ishti10.mybrute.com/cellule

there is a 0.3% chance to get one.

Seems fun!! I'm gonna try it out....hopefully I wont get addicted :maybe:

Gah, i can't control my charecter, Is this supposed to happen?

http://jessi-bluebil.mybrute.com

Man, i suck at this game (or at least my brute does).

Is there even ability points that you can change?

lol....anyone new click here and challenge me:
http://king1aops.mybrute.com

I don't get the whole power-ups thing. They don't seem to work.

[url]http://toonlinkrocks.mybrute.com[/url]

[quote="polkadotman1"]Your right.

That's probably because you don't control the fighting.[/quote]

your logic is backwards

http://jessi-bluebil.mybrute.com

how can i get more pupils?

You get more pupils by having people click your link. So you should send your link to all your friends.

[url]http://moomman.mybrute.com[/url]

My brute.


complex analysis
geometry
geometric transformation
rotation
inequalities
function
topology

Supposed that $f$ is bounded and holomorphic on the open unit disc $\mathbb{D}$, and for every $\epsilon > 0$, there exists $\delta$ s.t. $0 < 1-|z| < \delta \Rightarrow |f(z)| < \epsilon$ for every $z$ in the sector $\theta_{1}< z < \theta_{2}$ (i.e. $f$ converges uniformly to 0 as $|z| \to 1$ in the sector).

Prove that $f \equiv 0$ on $\mathbb{D}$.

Let $n$ be an integer number s.t. $n(\theta_{2}-\theta_{1}) \geq 2 \pi$. Define "rotations" $f_{k}(z)$ of $f(z)$ as follows:
\[f_{k}(z) = f(e^{2\pi i k/n}z),\]
where $0\leq k < n$. Let $g(x) = f_{0}(z) \cdot \dots \cdot f_{n-1}(z)$. Since $f(z)$ is bounded and $f(z) \to 0$ as $|z|\to 1$ in the sector,
\[g(z) \to 0 \text{ as }|z| \to 1\]
(here $z$ is not restricted to a particular sector). Using the Maximum Modulus Principle, we get that $|g(z)|$ is bounded by $\varepsilon$ for every $\varepsilon > 0$. Therefore, $g(z) \equiv 0$. So $f_{k}(z) \equiv 0$ (for some $k$), and $f(z) = f_{k}(e^{-2\pi i k/n}z) = 0$.

This step requires more justification:

$g \equiv 0 \Rightarrow f_{k}\equiv 0$ for some $k$ (I suppose by analytic continuation.)

Also, you probably want the strict inequality $n(\theta_{2}-\theta_{1}) > 2\pi$, since otherwise in the case of equality you're missing the boundary of the rotated sectors, right?

[quote]This step requires more justification:

$g \equiv 0 \Rightarrow f_{k}\equiv 0$ for some $k$ (I suppose by analytic continuation.)
[/quote]

If $f_{k}(z)\not\equiv 0$, then $f_{k}(z)$ has only a finite number of zeros in $\{z: |z| < 1/2\}$ (why?). So if all $f_{k}(z)$ are not equal to $0$, then $g(z)$ has a finite number of zeros in $\{z: |z| < 1/2\}$.

[quote]
Also, you probably want the strict inequality $n(\theta_{2}-\theta_{1}) > 2\pi$, since otherwise in the case of equality you're missing the boundary of the rotated sectors, right?[/quote]
Non-strict inequality suffices. Hint: if a continuous function is bounded by $\varepsilon$ in an open set, then it is also bounded by $\varepsilon$ in the closure of this set.


Find all continuous functions $ f : \mathbb{R} \to \mathbb{R}$ such that
\[f(x\plus{}y) \equal{} \frac{ f(x) \plus{} f(y) \minus{} f(x)f(y) }{1 \plus{} 2f(x)f(y) },\quad \forall x, y  \in \mathbb{R}.\]

[quote="nguyenvuthanhha"][i]Find all continuous  function $ f \ : \ \mathbb{R} \mapsto \mathbb{R}$ such that :

        $ f(x + y) \ = \ \frac { f(x) + f(y) - f(x)f(y) }{1 + 2f(x)f(y) } \ \forall \ x; y \ \in \ \mathbb{R}$[/i][/quote]

Let $ P(x,y)$ be the assertion $ f(x+y)=\frac{f(x)+f(y)-f(x)f(y)}{1+2f(x)f(y)}$
Let $ f(0)=a$

1) the only constant solutions are $ f(x)=c$ with $ c\in\{-1,0,\frac 12\}$
==============================================

Constant solutions $ f(x)=a$ imply $ a=\frac{2a-a^2}{1+2a^2}$ and so $ a\in\{-1, 0,\frac 12\}$ and it is easy to check that these 3 values are indeed solutions.
Q.E.D.

Consider from now that $ f(x)$ is not a constant function.

2) Non constant solutions are such that $ f(0)=0$ and $ f(x)\in(-1,\frac 12)$ $ \forall x$
=======================================================
$ P(x,0)$ $ \implies$ $ a(f(x)+1)(2f(x)-1)=0$
If $ a\ne 0$, this implies either $ f(x)$ constant and so $ f(0)=a=0$
Suppose now $ \exists u$ such that $ f(u)=-1$. $ P(x-u,u)$ $ \implies$ $ f(x)=-1$ and $ f(x)$ would be constant
Suppose now $ \exists u$ such that $ f(u)=\frac 12$. $ P(x-u,u)$ $ \implies$ $ f(x)=\frac 12$ and $ f(x)$ would be constant
So $ f(x)=0$ and continuity implies $ f(x)\in(-1,\frac 12)$
Q.E.D.

3) non constant solutions are injective
=========================

If $ f(x)$ is continuous but not monotonous, we can always find $ u,v$ with $ |v-u|>0$ as little as we want such that $ f(u)=f(v)$

But, if $ f(u)=f(v)$, with $ u\ne v$ then $ P(u,v-u)$ $ \implies$ $ f(u)=f(v)=\frac{f(u)+f(v-u)-f(u)f(v-u)}{1+2f(u)f(v-u)}$ and so $ f(v-u)(f(u)+1)(2f(u)-1)=0$ and so $ f(v-u)=0$ and then $ P(x,v-u)$ $ \implies$ $ f(x+v-u)=f(x)$ and $ f(x)$ is periodic (and one period is $ v-u$)

So $ f(x)$ is periodic and there exist periods as small as we want. Then, since $ f(x)$ is continuous, $ f(x)$ is constant, which is impossible (we supposed $ f(x)$ is not a constant function)

So $ f(x)$ is injective and strictly monotonous.
Q.E.D

4) non constant solutions are $ f(x)=\frac{1-e^{ax}}{2+e^{ax}}$
============================================
Let $ g(x)=\frac{1-e^{x}}{2+e^{x}}$. $ g(x)$ is a bijection from $ (-\infty,+\infty)\to(-1,\frac 12)$.

[hide="How did I think to this?"]
================  hidden part =====================

I supposed $ f(x)$ differentiable and I wrote $ f(x+y)-f(x)=-f(y)\frac{2f(x)^2+f(x)-1}{1+2f(x)f(y)}$

So $ \frac{f(x+y)-f(x)}y=-\frac{f(y)}y\frac{2f(x)^2+f(x)-1}{1+2f(x)f(y)}$

Setting $ y\to 0$, I got $ \frac{f'(x)}{{2f(x)^2+f(x)-1}}=c$ which is easy to solve with solution $ f(x)=\frac{1-e^{ax+b}}{2+e^{ax+b}}$. Then $ f(0)=0$ gives $ b=0$

================ end of hidden part ===================
[/hide]

Let then $ h(x)=g^{[-1]}(f(x))$ and $ h(x)$ is continuous.

It's [rather] easy to verify that $ g(x)$ is a solution of the required equation and so that $ g(x+y)=\frac{g(x)+g(y)-g(x)g
(y)}{1+2g(x)g(y)}$

Then $ g(h(x)+h(y))=\frac{g(h(x))+g(h(y))-g(h(x))g(h(y))}{1+2g(h(x))g(h(y))}$ $ =\frac{f(x)+f(y)-f(x)f(y)}{1+2f(x)f(y)}$ $ =f(x+y)=g(h(x+y))$

So $ g(h(x)+h(y))=g(h(x+y)$ and, since $ g(x)$ is injective, $ h(x+y)=h(x)+h(y)$ and so $ h(x)=ax$ (remember $ h(x)$ is continuous)
Q.E.D

5) synthesis of solutions
================

$ f(x)=-1$ $ \forall x$

$ f(x)=0$ $ \forall x$

$ f(x)=\frac 12$ $ \forall x$

$ f(x)=\frac{1-e^{ax}}{2+e^{ax}}$

[i]Here is the official solution

   Although it was written in Vietnamese

   I think you can understand it , Pco :lol: [/i]


Hello all,

        I am thinking about taking the SAT's in march. My weakest sections are CR and grammar. I was wondering if anyone had any strategies or tips (or even traps to look out for). Im just hoping that we can make a successive list of tips for anyone who is in a similar position as I am.

[quote="t0rajir0u"]Read.  [i]A lot.[/i]  This is a habit you should build from as early an age as possible, so it's not a cramming strategy.[/quote]

Agree. The best "prep" I can recommend is a rigorous English teacher (or a disciplined approach towards reading) in the couple of years before the SAT.

A list of common grammar traps and commonly confused words would help, for example, the ones in The Cooper Hill Stylebook.

If you want to improve your grammar you should use Michael Vince's books from Mecmillan.


inequalities
trigonometry
geometry
geometric transformation
reflection

$\triangle ABC\ ,\ AB\perp AC\ : \ D\in BC\ ,\ AD\perp BC\ ,\ h=AD\ \Longrightarrow\boxed {h+\max\ \{b,c\}\le \frac{3\sqrt 3}{4} a}\ .$

[hide]W.L.O.G. let $b \geqq c$, and $a=1$.
Then $h=sinBcosB$ $B \in [45,90)$ and we have
$h+b=sinBcosB+sinB=f(B)$
$f'(B)=(cosB)^2-(sinB)^2+cosB=2(cosB)^2+cosB-1=(cosB+1)(2cosB-1)$
so we see $f(b)$ has $\large minimum$ when $B=60$
but as $\cos B$ is $\large decreasing$ for $B \in [45,90)$
this means that this is the $\large maximum$ for B.
So $f(B) \leqq f(60)=\frac{3 \sqrt 3}{4}$
Hope to see a geometrical (syntetical solution)[/hide]

[color=darkred][b]The first (algebraical) method.[/b] Denote $AB=x$, $AC=y$, $x^2+y^2=a^2$ (constant) and shall find the maxim for $h+b=\frac{xy}{a}+y$. But $\frac{xy}{a}+y\ -\ \max$ $\Longleftrightarrow$ $y(x+a)\ -\ \max$ $\Longleftrightarrow$ $y^2(x+a)^2\ -\ \max$ $\Longleftrightarrow$ $(a^2-x^2)(x+a)^2\ -\ \max$ $\Longleftrightarrow$ $(x+a)^3(a-x)\ -\ \max$ $\Longleftrightarrow$ $\left(\frac{x+a}{3}\right)^3(a-x)\ -\ \max$. Observe that the sum $\frac{x+a}{3}+\frac{x+a}{3}+\frac{x+a}{3}+(a-x)=2a$ (constant). Thus, the product $\frac{x+a}{3}\cdot \frac{x+a}{3}\cdot \frac{x+a}{3}\cdot (a-x)$ is maximum iff $\frac{x+a}{3}=a-x=\frac{2a}{4}$, i.e. $c=x=\frac a2$ and in this case $b=y=\frac{a\sqrt 3}{2}$, $h=\frac{a\sqrt 3}{4}$, $h+b=\frac{3a\sqrt 3}{4}$.[/color]

[color=green][b]The second (trigonometrical) method.[/b] Suppose $a=1$ and denote $m(\widehat {ABC})=u$. Thus, $h+b=\sin u(1+\cos u)$ and $h+b$ is maximum iff $\sin^2u(1+\cos u)^2$ is maximum iff $(1-\cos u)\left(\frac{1+\cos u}{3}\right)^3$ is maximum iff $(1-\cos u)=\frac{1+\cos u}{3}=\frac 12$ iff $\cos u=\frac 12$ iff $u=60^{\circ}$ a.s.o.[/color]

[color=darkblue][b]The third (syntetical) method.[/b] Denote the reflection $D$ of the point $A$ w.r.t. the line $BC$. Observe that $a=2R$. In the triangle $ADC$ apply the well-known inequality $2p\le 3R\sqrt 3$, i.e. $2h+2b\le 3R\sqrt 3$ $\Longleftrightarrow$ $h+b\le \frac{3a\sqrt 3}{4}\ .$[/color]


just get intermediate algebra from AoPS  :lol:


The percent that $ M$ is greater than $ N$ is:

$ \textbf{(A)}\ \frac {100(M \minus{} N)}{M} \qquad\textbf{(B)}\ \frac {100(M \minus{} N)}{N} \qquad\textbf{(C)}\ \frac {M \minus{} N}{N} \qquad\textbf{(D)}\ \frac {M \minus{} N}{M}$
$ \textbf{(E)}\ \frac {100(M \plus{} N)}{N}$

[hide="Solution"]
Suppose we chose values for $ M$ and $ N$, such as $ M \equal{} 110$ and $ N \equal{} 100$. Then to figure out how much greater $ M$ is than $ N$, we find the difference and divide it by $ N$. Then we multiply by 100 to get the percent. Thus, the answer is $ \boxed{(B)\frac{100(M \minus{} N)}{N}}$.
[/hide]


Triangle ABC is acute and angle A=30 degrees.  The half circle with diameter BC intersects AB at D and AC at E.  Find the area of ADE over the area of BCED.

[hide]$S_{ADE}=\frac{1}{4}AD.AE$
$S_{ABC}=\frac{1}{4}AB.AC$
So $S_{BCED}= \frac{1}{4}(AB.AC-AD.AE)$
And $\frac{S_{ADE}}{S_{ABC}}= \frac{1}{\frac{AB.AC}{AD.AE}-1}$
Also $AD.AB=AE.AC$
Then $\frac{S_{ADE}}{S_{ABC}}= \frac{1}{\frac{AC^{2}}{AD^{2}}-1}$
Moreover, $AD = \frac{\sqrt{3}}{2}AC$
Hope I didn't make any mistakes.[/hide]


f(x)=[sin x]^{cos x}+[cos x]^{sin x}
    where [X] denotes the greatest integer less than or equals to X,
              {X}=X-[X].
   Can we find the range of f(x) on its domain? If so find it. :D

Not really right for basics, but definitely would be good for pre olymp or at least intermediate. It looks like a challenging and fun problem.

just break it up into intervals...and do a bit of case work


Plz try this out
There are given $ n$ figures in plane. Let $ S_{{i_1}...{i_n}}$ be the area of the intersection of
figures indexed by $ i_1, . . . , i_n$ and $ S$ be the area of the part of the plane covered by the given
figures; $ M_{k}$ be the sum of all the $ S_{{i_1},{i_2}....{i_k}}$ . Prove that:
a) $ S \equal{} M_1 \minus{} M_2 \plus{} M_3......( \minus{} 1)^{n \minus{} 1}$ and
b) $ S\geq M_1 \minus{} M_2 \plus{} M_3.......( \minus{} 1)^{m \minus{} 1}M_m$ for $ m$ even and $ S\leq M_1 \minus{} M_2 \plus{} M_3........( \minus{} 1)^{m \minus{} 1}M_m$ for odd $ m$.

thanks  :D

hey guys plz try it out

i am unable to do it  :(

or at-least give some ideas to solve

I think(just intuition) that combinatorics is involved(esp PEI)

help

I'm afraid I was unable to understand the question. What do the n S terms represent? By intersection, we mean the intersection of at least 2 figures? How come we have just n such intersecting areas for sure? Or is that an assumption we are making? And what about multiple overlappings and stuff? I've got terribly confused it seems.  :rotfl: Please elaborate the question and tell the full form of PEI.  :?:

well this was the ques as given

but what i can infer:
by $ S_{12}$ we mean intersection of fig 1 and 2,by $ S_{123}$ we mean intersection of fig 1,2,3 and so on

And srry for typo,it is PIE(principle of Inclusion and exclusion)

thanks


Compute the sum of all the roots of $(2x-3)(x-4)+(2x-3)(x-6)=0$

[hide]Coefficient of $x^{2}$ is $2+2=4$. Coefficient of $x$ is $-8-3-12-3=-26$. From viete, sum of roots is $13/2$.[/hide]


algebra
polynomial
inequalities
algebra unsolved

Let $ p(x)$ be a polynomial with degree $ n \geq 2$ and $ p(x) \in R[x]$ such that $ |p(x)| \leq 1$ for all $ x\in[\minus{}1,1]$.Find the maximum value of the leading cofficient of $ p(x)$

The leading cofficient max  is $ 2^{n \minus{} 1}$
Use the the Chebyshev Polynomial .
$ \text{cos}(nx) \equal{} P(cosx)$
Where $ P(x)$ is a polynomial with the leading cofficient is $ 2^{n \minus{} 1}$


The volume of a rectangular box increased by $ 66\frac23\%$ when the length increased by $ 33\frac13\%$ and the height remained the same. By what percent did the width increase?

uh...right. so here is the solution

[hide]Increasing the length by 33 1/3% is like multiplying it by 4/3. Increasing the volume by 66 2/3% is the same as multiplying the volume by 5/3. Letting w=the scale factor for the width(as a fraction), we get 4/3w=5/3 (we can ignore the height because it remains the same, so it won't have an affect on the volume increase). Solving gets w=5/4, so the width increases by $\boxed{25\%}$[/hide]


Let $S=\{p-q|p,q$is prime$\}$.Prove that $S=N_{0}$

You mean prove the set is countably infinite? I'm not sure about your notation...

Yes, such as it's written, it's false, since not all odd numbers (being of the form $p-2$ for $p>2$) are contained in $S$.


limit
induction
search
algebra unsolved
algebra

Find the value of $ \sqrt{1\plus{}2\sqrt{1\plus{}3\sqrt{1\plus{}4\sqrt{1\plus{}5\sqrt{\cdots}}}}}$

Use $ \sqrt{1\plus{}n(n\plus{}2)}\equal{}n\plus{}1$ and result 3.

[quote="Rust"]Use $ \sqrt {1 \plus{} n(n \plus{} 2)} \equal{} n \plus{} 1$ and result 3.[/quote]

That's not rigorous enough. Else, let any sequence $ g(2) \equal{} u$ and $ g(n \plus{} 1) \equal{} \frac {g(n)^2 \minus{} 1}{n }$ and we could write :

"Use $ \sqrt {1 \plus{} ng(n \plus{} 1)} \equal{} g(n)$ and result $ u$".

[hide="Solution"]
Let $ f_i (n) \equal{} \sqrt {1 \plus{} (i \plus{} 1)\sqrt {1 \plus{} (i \plus{} 2)\sqrt {1 \plus{} \cdots \plus{} n\sqrt {1}}}}$.  We shall show that $ \lim_{n\to \infty} f_1 (n) \equal{} 3$.

First assume that $ f_i (n) > 3$ for all $ n\ge x$ for some integer $ x$.  If $ n\ge x$, we will show that $ f_i (n) > i \plus{} 2$ for all integers $ i\ge 1$ by induction; the statement is clear for $ i \equal{} 1$.  Assume that the statement is true for all $ i\le k$.  Note that\[ f_{k \plus{} 1} \equal{} \frac {(f(k))^2 \minus{} 1}{k \plus{} 1} > \frac {k^2 \plus{} 4k \plus{} 3}{k \plus{} 1} \equal{} k \plus{} 3\]completing induction.  For large enough $ n$,\[ n \plus{} 1 < f_{n \minus{} 1} (n) \equal{} \sqrt {1 \plus{} n}\]which is not possible for positive $ n$, so $ f_i (n) \le 3$ for all $ n$.

Now, assume that $ f_1 (n) < 3 \minus{} 3\varepsilon$ for some $ \varepsilon > 0$ and all $ n$ (note that $ \varepsilon < 1$ because $ f_1 (3) \equal{} \sqrt {5} > 2 \equal{} 3 \minus{} 1$).  We will prove, by induction, that $ f_i (n) < (i \plus{} 2)(1 \minus{} \varepsilon)$.  The result is clear for $ i \equal{} 1$; assume that it is true for all $ i\le k$.  Note that\[ f_{k \plus{} 1} (n) \equal{} \frac {(f_k (n))^2 \minus{} 1}{k \plus{} 1} \equal{} \frac {k^2 \plus{} 4k \plus{} 3}{k \plus{} 1} \minus{} \frac {2(k \plus{} 2)^2\varepsilon}{k \plus{} 1} \plus{} \frac {(k \plus{} 2)^2 \varepsilon^2}{k \plus{} 1}\]\[ < k \plus{} 3 \minus{} \frac {(k \plus{} 2)^2\varepsilon}{k \plus{} 1} < (k \plus{} 3) \minus{} \varepsilon(k \plus{} 3)\]completing the induction.  Hence, for a fixed constant $ m$, whenever $ n \ge m$, $ f_m (n) \le m \plus{} 1$.   We will now show, by induction, that $ f_{m \plus{} k} (n) \le m \plus{} 1 \minus{} k$ for all $ k\ge 0$.  For $ k \equal{} 0$, this is clear.  Assume it is clear for all $ 0\le i\le k$.  We have that\[ f_{m \plus{} k \plus{} 1} (n) \equal{} \frac {(f_{m \plus{} k})^2 \minus{} 1}{m \plus{} k \plus{} 1} \le \frac {(m \plus{} 1 \minus{} k)^2 \minus{} 1}{m \plus{} k \plus{} 1} \le m \minus{} k\]for all $ k\ge 1$, so induction is complete.  However, setting $ i \equal{} m \plus{} 1$, we get that $ f_{2m \plus{} 1} (n) \le 0$, which is impossible.

We therefore have a contradiction, so $ f_1 (n) \le 3$ for all $ n$, but $ 3 \minus{} f_1 (n)$ can get arbitrarily small.  This, coupled with the fact that $ f_1 (n)$ is monotone implies that $ \lim_{n \to \infty} f_1 (n) \equal{} 3$.
[/hide]

[quote="The QuattoMaster 6000"]
Let $ f_i (n) \equal{} \sqrt {1 \plus{} (i \plus{} 1)\sqrt {1 \plus{} (i \plus{} 2)\sqrt {1 \plus{} \cdots \plus{} n\sqrt {1}}}}$.  We shall show that $ \lim_{n\to \infty} f_1 (n) \equal{} 3$.  [/quote]

According to me;, the problem is not well written. The "..." in the original problem are not enough to solve the problem since they may be understood in a lot of manner.
In fact the question is to write :

Let $ f(n) \equal{} \sqrt {1 \plus{} 2\sqrt {1 \plus{} 3\sqrt {1 \plus{} \cdots \plus{} nh(n)}}}$.  We must find $ \lim_{n\to \infty} f (n)$.
"The QuattoMaster 6000" found the limit $ 3$ when $ h(n)\equal{}1$  $ \forall n$
But we obviously can find some other $ h(n)$ for which the limit is not $ 3$, but may be $ 4$ for example.
So the OP's problem was not well defined.

See also [url]http://www.mathlinks.ro/viewtopic.php?search_id=1112658757&t=302868[/url]
and also other posts on this topic on the fora.

Classical problem by Ramanujan, which indeed needs care in establishing the convergence value (see, on the above quoted url, my solution; another classical solution is by N.J. Fine in some A.M.M.).


Functional Analysis
real analysis
real analysis unsolved

consider $ \mathbb{R}^2$ with norm $ \parallel{}_\infty$ and let $ Y\equal{}\{(y_1,y_2)\epsilon \mathbb{R}^2: y_1\plus{}y_2\equal{}0\}$
if $ g: Y\to R$ is defined by $ g(y_1,y_2)\equal{}y_2$ for $ (y_1,y_2)\epsilon Y$
then what can we say about the hahnbanach extension
is it unique?

The extension is not unique. We can write $ g(y_1,y_2)\equal{}\minus{}ty_1\plus{}(1\minus{}t)y_2$ for any $ t\in[0,1]$ and it will do as required on $ Y$ without increasing the norm.

Is every linear functional $ f: \mathbb R^2 \to \mathbb R$ satisfying $ f(\minus{}1,1)\equal{}1$ is hahn extension of $ g$ to$ \mathbb R^2$

No, not all of them, which is why I specified $ t\in[0,1]$ in my previous post. If you choose $ g$ with $ t$ outside that range, then the norm of $ g$ is greater than the norm of $ f.$ A Hahn-Banach extension must not increase the norm - that's part of the theorem.

THE functionals $ f_1,f_2: \mathbb R^2 \to \mathbb R$ given by $ f_1(x_1,x_2)\equal{} x_2$ and $ f_2(x_1,x_2)\equal{}\minus{}x_1$ are both hahn banach extentions of $ g$ to $ \mathbb R^2$

If you'll check my post #2, those are, respectively, the $ t\equal{}0$ and $ t\equal{}1$ cases of what I wrote. They're the ends of the interval of allowable values of $ t.$


Let $a,b$ two positive integers such that for any prime number $p$ the residue of $a$ modulo $p$ is at most  the residue of $b$ modulo $p$. Then $a=b$.


Nice, I got a 138 =\

I can't believe I thought 94 was a multiple of 4 on #24  :wallbash:

I also, sort of pushed for time on #25, thought 1,3,9 all worked for b and got D.

I got 144, made stupid mistake on #20, forgot to do 3600/6, should've noticed that 25C3<3600. :(

*bump*

Eh, I almost missed #20. Originally I said we choose a point in 25 ways, then another in 16 ways, finally another in 9 ways, divide by 3! for the 3! possible arrangements gives 600. Then I said wait, we just choose three of the 5 columns and three of the 5 rows, which gives 100. Hm, let's draw this out. Oh, three columns and three rows do not uniquely determine a set of 3 points duh! :)...

[quote="jamieding51"]Lots of D's on the answer key...[/quote]

What did you get on the B?

[quote="jamieding51"]Lots of D's on the answer key...[/quote]
lol yea.
I was thinking, "theres too many D's, this cant be right"
i got a 145.5, btw, ran out of time on last problem because i decided to check the first 24 instead of spending time on 25.

132 here.  I took the 10 with the sole purpose of getting a 150.  Oh well.

I wasted some time on some of the early questions (ceilings are walls, right? :P).  Then #17 took me about 10 minutes to bash with Law of Cosines/Sines.  When I finally got to the end, I was rushed and didn't read "at least one of each" in #24.

It was still a good test, though.  I liked some of the problems, especially #25, which unfortunately I did not solve until after the test.

solafidefarms, congratulations on your 150.

132 yea for some reason i completely blanked on a bunch of those problems.

[quote="13375P34K43V312"][quote="asdf4"]
i got a 145.5[/quote]

N00B

dude never do that, just fill in a random answer.[/quote]

What if say, a friend of mine did fill it in and got a 144? And top score in the school gets a plaque?
Well, i was thinking about that, but I had no idea.

lol i know people who just do 19 problems and make sure they're right and skip the right. it works pretty good for most people.

...but that still isn't enough to get them to qualify for AIME if they make a stupid mistake :wink:

I'm a tard cuz I got 126 :( . Just gotta get to the floor at AIME now, which is highly unlikely :(


I was going to request a mexican community, but I think it would make more sense to create a latin american comunity, seeing as there is no such thing. Thank you :)

Why don't you go to the spanish community :?:

[quote="manuel"]we already use the spanish comunity![/quote]

Yes, see here: [url=http://www.mathlinks.ro/Forum/index.php?f=255]Spanish Communities[/url]

thanks...  :)

ya me parecía extraño el no haber encontrado una comunidad latina acá


induction
modular arithmetic
number theory unsolved
number theory

Prove that there exists a succession $a_1, a_2, ... , a_k, ...$, where each $a_i$ is a digit ($a_i \in (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)$ ) and $a_0=6$, such that, for each positive integrer $n$, the number $x_n=a_0+10a_1+100a_2+...+10^{n-1}a_{n-1}$ verify that $x_n^2-x_n$ is divisible by $10^n$.

By induction:
If we have $x$ such that $x^2-x\equiv 0\pmod{10^n}$, then we need to find a digit $k$ with $(10^nk+x)^2-(10^nk+x)\equiv 0\pmod{10^{n+1}}$. This can be done, since the congruence reduces to $10^n(2xk-k+\frac{x^2-x}{10^n})\equiv 0\pmod{10^{n+1}}$, ie $k(2x-1)\equiv -\frac{x^2-x}{10^n}\pmod{10}$, and $2x-1\equiv 1\pmod{10}$.

[quote="Rust"]$x_n=10^n-(5^{2^{n-2}}-1),(mod \ 10^n),n>1$[/quote]

That doesn't seem to work for n=2, or any n in fact..


Sa se determine $n\in N$ care verifica relatia $R_n\equiv\frac{1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \cdot \cdot (2n)}=\frac{3}{2\sqrt{7n+2}}$
si sa se arate ca $R_n<\frac{3}{2\sqrt{7n+2}},\ (\forall )\ n>2.$

After simplification we have
$\frac{1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \cdot \cdot (2n)}=\frac{(2n)!}{4n!n!}$
then
$\frac{(2n)!}{n!n!}=\frac{6}{\sqrt{7n+2}}.$
we know that
$\frac{(2n)!}{n!n!}\in N$ for all $n\in N$
then $\rm \sqrt{7n+2}|6$
and we find that $n = 1$ or $n = 5$  :D

the answer is wrong. it is not 4.[/code]


calculus
integration
trigonometry
logarithms
symmetry
limit
complex analysis

I noticed that $\int \frac{1}{(x-a-bi)(x-a+bi)} \ dx = \frac{\tan^{-1}\left(\frac{x-a}{b}\right)}{b}$.
Can anyone prove this?

Also I was wondering how to integrate $\frac{1}{1+x^4}$ and other similar integrals (denominators that have imaginary zeros).

Thanks.

Here's a useful identity:

$\arctan(z)=\frac1{2i}\text{Log}\left(\frac{1+iz}{1-iz}\right).$

Here I am using the complex logarithm:

$\text{Log}(w)=\ln|w|+i\arg(w).$

The complex logarithm has "branches" - many acceptable values, differing from each other by integer multiples of $2\pi i$. Of course (and notice the division by $2i$ in the formula above), the arctangent also has branches, differing from each other by $\pi.$

You could prove the formula above by differentiating, taking for granted that the complex logarithm has the algebraic and differential properties you think it ought to have. Alternatively, for $|z|<1$ and for the "principal" branches of the logarithm and arctangent ($\arctan(0)=0,\,\text{Log}(1)=0$) we can expand both sides of this identity as power series.

Let's make the last part of Jimmy's question into a challenge:

Compute $\int_{-\infty}^{\infty}\frac1{x^4+4}\,dx.$

Two notes:

(1). I made it $x^4+4$ instead of $x^4+1$ just to make the factorization a little nicer - don't need to write all those $\sqrt{2}$ 's.

(2). If you know the residue theorem from complex analysis, back off and let someone else answer this.

I can see the symmetry $\int_{-\infty}^{\infty} \frac{1}{x^4+4} \ dx = 2\int_{0}^{\infty} \frac{1}{x^4+4} \ dx $, but that's about it.  I tried partial fractions but can't figure out how to integrate each term without using the identity I mentioned.

What prerequisites are necessary for Complex Analysis?
Also, can someone recommend a good textbook on Complex Analysis?
Thanks.

[quote="kunny"]The answer is $\frac{\sqrt{2}}{2}\pi$.[/quote]
That's the right answer, but to the wrong question :lol:
$\int_{-\infty}^\infty \frac{1}{1+x^4} \ dx = \frac{\pi\sqrt{2}}{2}$.
$\int_{-\infty}^\infty \frac{1}{4+x^4} \ dx = \frac{\pi}{4}$.

[quote="Kent Merryfield"]Methods, anyone? And Jimmy: why not use that identity?[/quote]
I haven't proven it yet, but I'll give it a shot.

Edit: I've proven it by differentiating.

Using partial fractions, we have
$I=\int \frac{1}{4+x^4} \ dx=\int \frac{x}{8(x^2+2x+2)} \ dx+\int \frac{1}{4(x^2+2x+2)} \ dx-\int \frac{x}{8(x^2-2x+2)} \ dx+\ \frac{1}{4(x^2-2x+2)} \ dx = I_1+I_2+I_3+I_4$

By the identity,
$I_2 = \frac{\tan^{-1}(x+1)}{4}$
and
$I_4=\frac{\tan^{-1}(x-1)}{4}$
The first and third integrals can be rewritten as
$I_1 = \frac{1}{8}\int \frac{x+1}{x^2+2x+2} - \frac{1}{x^2+2x+2} \ dx = \frac{\ln(x^2+2x+2)-2\tan^{-1}(x+1)}{16}$
$I_3 = \frac{1}{8}\int \frac{1-x}{x^2-2x+2} - \frac{1}{x^2-2x+2} \ dx = \frac{2\tan^{-1}(x-1)-\ln(x^2-2x+2)}{16}$

$I=I_1+I_2+I_3+I_4 = \frac{1}{16} \left(\ln\left(\frac{x^2+2x+2}{x^2-2x+2}\right)+2\tan^{-1}(x+1)+2\tan^{-1}(x-1)\right)$

Now we have to evaluate this at the given values, by symmetry $\int_{-\infty}^{\infty} \frac{1}{x^4+4} \ dx = 2\int_{0}^{\infty} \frac{1}{x^4+4} \ dx$.

So we just need the limit of $2I$ as $x \to \infty$.
$\frac{1}{8}\lim_{x\to\infty}  \left(\ln\left(\frac{x^2+2x+2}{x^2-2x+2}\right)+2\tan^{-1}(x+1)+2\tan^{-1}(x-1)\right)$.
The terms inside the $\ln$ go to $1$ and thus the $\ln$ disappears , so now we have
$\frac{1}{8}\lim_{x\to\infty} \left(2\tan^{-1}(x+1)+2\tan^{-1}(x-1)\right) = \frac{2\pi}{8} = \frac{\pi}{4}$.

Here is a tricky solution.
We want to find out $I = \int_0^\infty \frac{dx}{1+x^4}$.
Change variable $x=\frac{1}{y}$ then we have
$I = \int_0^\infty \frac{dx}{1+x^4} = \int_0^\infty \frac{y^2}{1+y^4}dy$
Hence
$2I = \int_0^\infty \frac{x^2+1}{1+x^4} dx = \int_0^\infty \frac{x^2+1}{1+x^4} dx = \int_0^\infty \frac{1+\frac{1}{x^2}}{\frac{1}{x^2}+x^2}dx$.

(Tricky here):
$\int_0^\infty \frac{1+\frac{1}{x^2}}{\frac{1}{x^2}+x^2}dx = \int_0^\infty \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+2}=\frac{1}{\sqrt 2}\left.\arctan\left(x-\frac{1}{x}\right)\right|_0^\frac{\pi}{2} = \frac{\pi}{\sqrt 2}$
Hence
$I = \frac{\pi}{2\sqrt 2}$

Now we want to calculate $J=\int_0^\infty \frac{dx}{x^2+4}$.
$J=\int_0^\infty \frac{y^2dy}{4y^4+1} =\buildrel {u=\sqrt 2y}\over= \frac{1}{2\sqrt 2}\int_0^\infty \frac{u^2du}{u^4+1} = \frac{I}{2\sqrt2} = \frac{\pi}{8}$

Now we use residues to compute $2I=\int_{-\infty}^\infty \frac{dx}{x^4+1}$.

$f(z)=\frac{1}{z^4}+1$ has two poles, $z_1=e^\frac{\pi i}{4}$ and $z_1=e^\frac{3\pi i}{4}$

$2I=2\pi i\left(\mbox{Res}(f(z),z1) + \mbox{Res}(f(z),z2)\right) = 2\pi i\left(\frac{1}{f'(z_1)} + \frac{1}{f'(z_2)}\right) = 2\pi i\frac{1}{4}\left(\frac{1}{z_1^3}+\frac{1}{z_2^3}\right) = \frac{\pi}{\sqrt 2}$


I was wondering if anyone had any ideas on how to grade USAMOs by one self. I have the (easier) USAMOs 1972-86 w/solutions, but am not sure how to grade my solutions. I know WOOT would help with that, but its way too expensive. So, could anyone give me a detailed general rubric from 0 to 7 on how I should give points to my solutions? I would really appreciate that.

Post your solutions here for starters.

Later you'll get an instinct overall for how to grade solutions, but at first you need a little experience.

I can't speak with authority, but here's generally how I think about it:

0- self explanatory; not a 1
1- Substantial progress towards a solution.  Perhaps 1 direction of an iff, or an upper but not lower bound, or a result based on an assumption that doesn't make the problem trivial.
2- Correct proof with a mistake (i.e., WLOG that's not really WLOG, forgetting a case, &c., that would probably be fixable)
2'- Strong 1.
3-5- Rare
6- Correct proof with a small hole (something somewhat non-trivial not proven)
7- Correct proof.

What inteluser said is pretty good, and accurate. However, it is still difficult to judge those things without having a better "mastery" of a solution which you may not always have just from having read it. It doesn't hurt to post here!

[quote="jb05"]Yeah, if you can judge your own paper, you could get 7s instead of 1, 2, or 6. You could think that something is good enough when the normal judges wouldn't.[/quote]Very good point. I would say (as I have graded papers from National, TST, MathLinks Contest and this year International Olympiad :D) that the scheme inteluser presented is not that accurate in fact. To point out a few mistakes: you get 6 usually if you make a mistake or miss something that [b]is[/b] trivial, not [b]non-trivial[/b]. Furthermore 2 points for missing out a case or something fixable is not enough. Usually 2p represents the idea of the problem. But there's much more than that ...  :roll:

[quote="Valentin Vornicu"]this year International Olympiad :D[/quote]

:clap2:  :clap:  :heli:

I received a 4 on problem number four on this year's USAMO(at least i'm assuming so, i'm very confident because the other problems i either didn't attempt or i got the exact answer).

If you take a look at number 4, or recall the problem, you'll remember having to evaluate the sum of the first n or n+/-1 squares, i can't remember which. anyway, in evaluating the sum i made an incredibly stupid mistake, saying that the sum of the first n squares is n^2 + n if n is even and n^2 if n is odd (this is the rule for the first n odd/even integers  :mad: )
anyway, using the result from that i accounted for overcounting and everything the problem required and intended. so i got a 4.

just thought that might be a helpful example.

My proof for $k_1+k_3=k_2+k_4$ was not clear and rigorous and that cost me 3-4 points on that problem.

I had a solid coordinate proof for $k_1 + k_3 = k_2 + k_4$, but made an incredibly stupid mistake that I forgot when $k_1 + k_3$ exceeds $n$ you lose solutions that I evaluated $1^2 + 2^2 + ... (2n+1)^2$ instead, that cost me 4-5 points. So I think any kind of fixable, non-trivial mistake will cost 5 points.

Also, make sure you actually answer the question asked...  :wallbash_red: :wallbash_red: :wallbash_red:


function
LaTeX
algebra
functional equation

First of all, thanks to all who are helping me become accustomed to functional equations.

For all real $ x$ values, solve the equation $ f(x \plus{} t) \equal{} f(x) \plus{} f(t) \plus{} 2\sqrt {f(x)}\sqrt {f(t)}$ for $ f(x)$.

[hide="my approach"]I chose to let $ t\rightarrow x$ in the first equation.  Thus, $ f(2x) \equal{} 4f(x)$.  Letting $ g(x) \equal{} \frac {f(ax)}{x}$, we have that for nonzero $ a$, $ g(x) \equal{} g(2x)$.  Thus, the function must be periodic.  How do I express that as an answer?  All periodic functions?

Another substitution can easily show that $ f(0) \equal{} 0$, so $ g(x)$ can apply for all other $ x$ values.[/hide]

[quote="azn6021023"]
[hide="my approach"]I chose to let $ t\rightarrow x$ in the first equation.  Thus, $ f(2x) \equal{} 4f(x)$.  Letting $ g(x) \equal{} \frac {f(ax)}{x}$, we have that for nonzero $ a$, $ g(x) \equal{} g(2x)$.  Thus, the function must be periodic.  How do I express that as an answer?  All periodic functions?

Another substitution can easily show that $ f(0) \equal{} 0$, so $ g(x)$ can apply for all other $ x$ values.[/hide][/quote]

Unfortunately, $ g(x) \equal{} g(2x)$ does not imply the function is periodic [a periodic function satisfies $ g(x) \equal{} g(x \plus{} T)$]. The problem here is that you have shown a condition is true which is (likely) weaker than the problem statement, but you have not shown that it must hold. Indeed, there are many functions that satisfy $ g(x) \equal{} g(2x)$, most of which will not correspond to solutions in the original functional equation.

[hide]
We have $ f(x) \ge 0$ for all $ x$, so rewrite the equation in the form of $ f(x \plus{} t) \equal{} \left[\sqrt {f(x)} \plus{} \sqrt {f(t)}\right]^2 \Longrightarrow$ $ \sqrt {f(x \plus{} t)} \equal{} \sqrt {f(x)} \plus{} \sqrt {f(t)}$. Then let $ g(x) : \equal{} \sqrt {f(x)}$, where $ g(x) \ge 0$, gives $ g(x \plus{} t) \equal{} g(x) \plus{} g(t)$, which is [url=http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation]Cauchy's functional equation[/url]. We get $ g(x) \equal{} Cx, f(x) \equal{} Cx^2$, where $ C \ge 0$, and we easily see this works in the original equation.

Actually there are more solutions to Cauchy's, and I'm not sure that they would not be an issue, the only restriction being that $ g(x) \ge 0$.
[/hide]

[quote="azjps"]$ g(x \plus{} t) \equal{} g(x) \plus{} g(t)$, which is [url=http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation]Cauchy's functional equation[/url]. ...

Actually there are more solutions to Cauchy's, and I'm not sure that they would not be an issue, the only restriction being that $ g(x) \ge 0$. [/quote]  You were good until the "..." and you noted yourself what the error was at that point.  One could use your function $ g$ to get the solution, but it's not really necessary.  Here are some suggested substitutions into the original functional equation:
[hide="Not very useful on its own"]$ (x, t) \mapsto (0, 0)$[/hide]
[hide="Very useful when combined with the previous one"]$ t \mapsto \minus{}x$[/hide]

[quote="azjps"]we easily see this works in the original equation. [/quote]

:oops: $ \sqrt {f(x)} \plus{} \sqrt {f( \minus{} x)} \equal{} 0 \Longrightarrow f(x) \equal{} 0$

I still don't know why $ \sqrt {f(x)} \plus{} \sqrt {f( \minus{} x)} \equal{} 0$ and you can have $ f(x)\equal{}0$. Can you explain for me.

I think that follows from the first substitution $ x\equal{}t\equal{}0$.

ARGH!  I thought I was getting the hang of these!  But thanks for all your help.

Ok, now I get it. Because of $ \sqrt {f(x)} \ge 0$, $ \sqrt {f(\minus{}x)} \ge 0$ but we $ \sqrt {f(x)} \plus{} \sqrt {f( \minus{} x)} \equal{} 0$. So $ \sqrt {f(x)} \equal{} 0$, $ \sqrt{f(\minus{}x)} \equal{} 0$

f(x + t) = f(x) + f(t) + sqrt(f(x)) + sqrt(f(t))

            = [ sqrt(f(x)) + sqrt(f(t))]^2

sqrt(f(x+t)) = sqrt(f(x)) + sqrt(f(t))

define g(x) = sqrt(f(x)). then

g(x+t) = g(x) + g(t).

g(x) = cx , where c is a constant, works.

so, f(x) = kx^2 works.

I'm really interested to hear about how $ (1 \minus{} 1)^2 \equal{} (1)^1 \plus{} ( \minus{} 1)^2 \plus{} \sqrt {(1)^1}\cdot\sqrt {( \minus{} 1)^2}$.   :P  :roll:  (Repeating a correct solution to a problem, I understand.  Repeating a solution that's [i]incorrect[/i] after the errors have been pointed out is a bit stranger.)  Unrelatedly, your math would be more legible if you used [[LaTeX]] -- see that link or http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690

[quote="azn6021023"]Thus, $ f(2x) \equal{} 4f(x)$.  Letting $ g(x) \equal{} \frac {f(ax)}{x}$, we have that for nonzero $ a$, $ g(x) \equal{} g(2x)$.  [/quote]
I'm not following this; $ g(x)\equal{}g(2x)\implies \frac{f(ax)}x\equal{}\frac{f(2ax)}{2x} \implies 2f(ax)\equal{}f(2ax)$ for nonzero $ x$, but how did you get the last equality?  Shouldn't we get $ 4f(ax)\equal{}f(2ax)$ instead?

Why is it that $ f(x) \equal{} Cx^2$ satisfies the original equation but we still have $ \sqrt{f(x)} \plus{} \sqrt{f(\minus{}x)} \equal{} 0$ implying that f(x) = 0 is the only solution? What's going on with this problem?

Assume that $ f(x) \equal{} Cx^2\implies Cx^2 \plus{} Ct^2 \plus{} 2Cxt \equal{} Cx^2 \plus{} Ct^2 \plus{} 2C\sqrt {x^2}\sqrt {t^2}$.

However, $ \sqrt {x^2}\sqrt {t^2}$ can equal $ xt$ or $ \minus{} xt$, so if one of $ x,t$ were positive and the other negative, we would have $ 2Cxt \equal{} \minus{} 2Cxt\implies C \equal{} \minus{} C$, which is clearly a contradiction for all $ C\neq0$, and we can only have $ f(x)\equal{}0$.


Stanford college MIT Harvard Putnam geometry

I recently got into Stanford, Early Action. I am obviously very happy about this, and love a lot of things about Stanford. However, I am still applying to Caltech, MIT, and Harvard, as I don't know what my top choice is yet.

Does anyone know much about the math program at Stanford? I know that their Putnam team doesn't rank top 5 very often recently (though I obviously would change that  :D ), but would it be a good place to go for undergraduate? And also, would it be a good place to come from when applying for graduate school?

In terms of location, I really don't think it's possible to beat Stanford. (Of course, I grew up in the area and went to Stanford for various reasons on many occasions when I was in high school, so I'm somewhat biased.) Stanford's math program is good though. In the past few years, Stanford's Putnam team has been in the top 5, although that was largely due to the presence of Paul Valiant, who has now graduated, I believe. The Stanford Putnam seminars are excellent, and the man who runs (Ravi Vakil) is a four-time Putnam fellow. If you go to Stanford, say hi to him from me.

As a former Berkeley student, I have to say this.

Putnam results, 2001:
...
4. UC Berkeley
5. Stanford

Putnam results, 2002:
...
4. UC Berkeley
5. Stanford

Ha.

I was Berkeley's best those two years, and second to the ringer (not on the team, still a high school student) in 2000.

Stanford and Berkeley were still both top 10 in 2003, although I don't remember the exact results.

Stanford certainly has a good enough undergraduate math department that a top student can go from there to any graduate school he pleases.

The schools you mention have very different undergraduate experiences, and somewhat different departmental cultures in their math departments. Stanford is by no means a bad choice, but you will have to decide which school fits you the best if you are admitted to more than one.

Once you know your admission results in the spring, please share your application stories here for next year's applicants.

Congratulations, and best wishes for finding the right fit for your undergraduate studies.

If you're at all an outdoorsy type, you'll pick Stanford over Harvard/MIT/New England if you visit there (I probably would have had I visited).

If you are very interested in doing research as an undergrad, however, then MIT/CalTech are probably a better bet than Harvard/Stanford.

As for getting into grad schools, these four are essentially equivalent.

Congrats, and good luck with your impending decision.

[quote="Magnara"][quote="ComplexZeta"]As for grad schools, if you know where you want to go for grad school, don't go there as an undergraduate. Most universities want their undergraduates to go somewhere else for grad school.[/quote]

Why is that?  I've never understood.[/quote]

The idea is that the graduate schools want to draw together people with a fresh perspective, to bring new ideas to the university. But don't worry about it, there is always another good grad school, even if you made the "mistake" of choosing an undergraduate program with a very desirable grad school.

The same principle does NOT apply to professional schools (law school, medical school, etc.) For admission to professional school, just go to the best undergraduate school you can, and be sure to get good grades while there.

[quote="tokenadult"]The same principle does NOT apply to professional schools (law school, medical school, etc.) For admission to professional school, just go to the best undergraduate school you can, and be sure to get good grades while there.[/quote]

Hmm...could anyone touch on the difference between "professional" and "grad" school? I believe business would be a professional school, no?

Professional schools teach you a profession.  Examples are medical schools, law schools, business schools.  They each have a special degree associated with them: MD, JD, MBA.  Graduate schools teach an academic subject: graduate school in math, in history, in chemistry.  They result in the same final degree, either a Masters or a PhD.  They can be considered training for the profession of academia.  There are, of course, other differences, but those are the major defining ones.

[quote="ComplexZeta"]As for grad schools, if you know where you want to go for grad school, don't go there as an undergraduate. Most universities want their undergraduates to go somewhere else for grad school.[/quote]
MTU actually offers a (large) incentive to a select group of its undergraduates to attend MTU for grad school.

[quote="confuted"]
MTU actually offers a (large) incentive to a select group of its undergraduates to attend MTU for grad school.[/quote]

What's MTU? Do you mean MIT?

MTU = Michigan Technical University, http://www.mtu.edu


Find all positive integers solutions to (a,b) to a^2+b^2=2005


function calculus derivative algebra domain limit trigonometry

hi to all, sorry for my english i'm an italian student and i need an help
here the question:

this is the function  : f(x,y)=(x-y)^2*sin(1/(x-y))

1.determinare il dominio.
2.dire se e dove e prolungabile con continuita
3. studiare lesistenza delle derivate parziali della funzione prolungata.
4.studiare la differenziabilita della funzione prolungata

1 determine the dominion.
2.say if and where is prolungable with continuity?
3.study the existence of the partial derivatives of the extended function.
4.study the differentiability  of the extended function

i hope u understand me and u help me!!

[quote="Kent Merryfield"]On #1, the English word is "domain", which is clearly all of $\mathbb{R}^2$ except for the set where $x=y$. For the rest of it, notice that we can write $f(x,y)$ as $g(x-y)$ where $g(t)=t^2\sin(1/t)$. The function $g$ (extended by continuity so that $g(0)=0$) is a classic example of a function which is differentiable everywhere but whose derivative is not continuous.

That makes clear (by the chain rule) that $f(x,y)$ has first-order partial derivatives everywhere. By itself, that wouldn't be enough to show differentiability, but the function is nevertheless differentiable everywhere. I'll leave the demonstration of that up to you, since the first step will be to recall the exact definition of "differentiable" for a function of several variables.

It's also clear that $f\notin C^1$.[/quote]

ok for the domain....but the function is derivable in all the domain.....it's also derivable in the point Y=X???? if i apply the definition of partial derivate i find that the limit for h->0 exists and is finited, so the function has partial derivates in all R2....it's correct?????

The function does have partial derivatives everywhere, but that's by itself not the same as differentiable. A function of several variables is differentiable at a point iff it can be approxmiated by a linear map to within an error with the right kind of behavior. For the specific case of a function $f:\mathbb{R}^2\to\mathbb{R}$, we can say that $f$ is differentiable at $(x,y)$ iff $\exists a,b$ such that $\lim_{\sqrt{h^2+k^2}\to 0} \frac{f(x+h,y+k)-f(x,y)-ah-bk}{\sqrt{h^2+k^2}}=0$.

Differentiability does imply the existence of partial derivatives, and the $a$ and $b$ above must be those partial derivatives. The converse is not true.

A standard example: let $g(x,y)=\frac{2xy}{x^2+y^2}$ for $(x,y)\ne (0,0)$ and $g(0,0)=0$. Then $g$ has partial derivatives everywhere, including $(0,0)$, but is not differentiable, and in fact not even continuous, at $(0,0)$. This example does show that the existence of partial derivatives does not imply differentiability - but it's not the same as the problem before us.

ok for all.....
but what is the solution for this limit???? :

lim (x,y)->(0,0)  [((x-y)^2)*sin(1/(x-y))]/[sqrt(x^2+y^2)]

if i solve this, i found the solution to the problem!!!!!!

Size is everthing here. Don't deal with the details of the sine, just estimate it as $|\sin(\cdot)|\le 1$. Note that if $\rho=\sqrt{x^2+y^2}$, then $|x-y|^2\le C\rho^2.$

i try to calculate it.....
the result is zero??????

thanks


If $ x^{ \minus{} 1} \minus{} 1$ is divided by $ x \minus{} 1$ the quotient is:

$ \textbf{(A)}\ 1 \qquad\textbf{(B)}\ \frac {1}{x \minus{} 1} \qquad\textbf{(C)}\ \frac { \minus{} 1}{x \minus{} 1} \qquad\textbf{(D)}\ \frac {1}{x} \qquad\textbf{(E)}\ \minus{} \frac {1}{x}$

[hide] $ \boxed{D}$

Turn $ x^{\minus{}1} \minus{} 1$ into $ \frac {1 \minus{} x}{x}$ and cancel.[/hide]

EDIT: See below.

[quote="happyme"][hide] $ \boxed{D}$

Turn $ x^1 \minus{} 1$ into $ \frac {1 \minus{} x}{x}$ and cancel.[/hide][/quote]
[hide]Forgot the negative, so E.[/hide]


limit
inequalities unsolved
inequalities

Find min:$A=\frac{22ab+12bc+31ca}{(a+2b+3c)^{2}}$

where $a;b;c$ are non-negative and a numbers $\neq 0$

Take $a=1$ and $b=c=\varepsilon\rightarrow0.$ Then
\[\lim_{\varepsilon\rightarrow0}\frac{12\varepsilon^{2}+53\varepsilon}{(1+5\varepsilon)^{2}}=0\]
so $A$ does not have a lower bound and $\inf A=0.$


calculus
integration
trigonometry
calculus computations

Use the result

$ \int_{0}^{a} f(x)\ dx \equal{} \int_{0}^{a} f(a\minus{}x)\ dx$

to evaluate

$ \int_{0}^{\pi} \frac{x}{2\plus{}\tan^{2}x}\ dx$.

hello, we get
$ \int_{0}^{\pi}\frac{x}{2\plus{}\tan(x)^2}\,dx\equal{}\frac{1}{4}(2\minus{}\sqrt{2})\pi^2$.
Sonnhard.


function
ratio
limit
logarithms
trigonometry
complex analysis

I have the following question:

In what region does $ \sum_{n \equal{} 1}^{\infty} \frac {sin(nz)}{2^n}$ represent an analytic function? What about $ \sum_{n \equal{} 1}^{\infty} \frac {sin(nz)}{n^2}$?

Would this just be the radius of convergence so for the first part we take the ratio test: $ \lim_{n \equal{} \infty} |\frac {sin((n \plus{} 1)z)}{sin(nz)} \frac {2^n}{2^{n \plus{} 1}}| \equal{} \lim_{n \equal{} \infty} |\frac {sin(nz)cos(z) \plus{} cos(nz)sin(z)}{2sin(nz)}| < 1$? How would I find the radius of convergence for z in this limit?

For the first, show that if $ |\Im (z)| < \log 2$ then $ |\sin (nz)| < r^n$ for some $ r < 2$ and sufficiently large $ n$, and conclude that the series converges in this case.  Similarly, show that the series does not converge if $ |\Im (z)|\geq \log 2$ (by showing that in this case $ \lim_{n\to \infty}\frac{\sin nz}{2^n}\neq 0$).

For the second, show that for all $ z\in \mathbb{C}\backslash \mathbb{R}$, $ \lim_{n\to \infty}\frac {\sin nz}{n^2} \neq 0$, so it is not analytic anywhere.

Hint: $ \sin (x \plus{} iy) \equal{} \cosh y \sin x \minus{} i\sinh y \cos x$, so the real and imaginary parts of $ \sin (x \plus{} iy)$ are bounded by $ \pm \cosh (y)\approx \pm \frac {e^{|y|}}{2}$.

[quote="JoeBlow"]

For the second, show that for all $ z\in \mathbb{C}\backslash \mathbb{R}$, $ \lim_{n\to \infty}\frac {\sin nz}{n^2} \neq 0$, so it is not analytic anywhere.

Hint: $ \sin (x \plus{} iy) \equal{} \cosh y \sin x \minus{} i\sinh y \cos x$, so the real and imaginary parts of $ \sin (x \plus{} iy)$ are bounded by $ \pm \cosh (y)\approx \pm \frac {e^{|y|}}{2}$.[/quote]

For this one if all $ z\in \mathbb{C}\backslash \mathbb{R}$ then $ z\equal{}yi$ and $ \lim_{n\to \infty}\frac {\sin nz}{n^2} \equal{} \lim_{n\to \infty}\frac {\sin nyi}{n^2} \equal{} \lim_{n\to \infty}\frac {i\sinh ny}{n^2}$
Then using L'Hospital's Rule we have:
$ \lim_{n\to \infty}\frac {in\cosh(ny)}{2n} \equal{} \lim_{n\to \infty}\frac{i\cosh(ny)}{2} \equal{} \infty$ so it is analytic nowhere. What about the case where $ z\notin \mathbb{C}\backslash \mathbb{R}$?

$ \mathbb{C}\backslash \mathbb{R} \equal{} \{x \plus{} iy: x,y\in \mathbb{R},y\neq 0\}$, it is not the set of imaginary numbers.  Just use $ \sin (x \plus{} iy) \equal{} \cosh y \sin x \minus{} i \sinh y \cos x$ to show it diverges.  It's really not difficult -- the numerator grows exponentially while the denominator is polynomial.


Let $f(x)=1+\sum_{k=1}^n(a_{k}sinkx+b_{k}coskx)$ and $(\forall x)f(x)>0.$
Prove that $(\forall x)f(x)<n+1.$

[quote="zhaobin"]And this one will be similar ;)[/quote]
Yes! :)

[quote="barra"]$f(x)<2$ for all $x$ right?[/quote]
why :!:


Hi, all

I am a little bit confused for the concept "limit point of a set " and the concept "limit of a sequence".
Can anyone show me the relationship between these two concepts? thanks

A previous [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=72476]discussion[/url] of this issue.

Hi there,

jmerry gave you a link to a thread I started a while ago. I was asking (albeit not direct question) if a [b]limit[/b] of a sequence attained by a convergence sequence is a [b]limit point[/b]. The answer is [i]no[/i] as I found out later from a lecture note from Internet. Kent suggested a different term for [b]limit point[/b], that was [b]accumulation point[/b], to avoid confusion. I now agree with him.

The [url=http://mathworld.wolfram.com/LimitPoint.html]definition of [b]accumulation point/limit point[/b][/url] is in term of set. This is not a problem since we can construct a sequence (convergence sequence, to be more specific) out of the set.

The similarity between [b]limit[/b] of a sequence and [b]accumulation point[/b] (in term of sequence) is that eventually the sequence converges to them (that is the distance from a point in the sequence to its limit eventually becomes zero: $distance(x_k, \bar{x}) \to 0$ as $k \to \infty$ where $\bar{x}$ is the limit). The difference is that, the concept of [b]accumulation point[/b] requires the existence of at least one point $x_k \neq \bar{x}$ in the sequence such that $distance(x_k, \bar{x}) < \epsilon$ for [i]every[/i] $\epsilon > 0$ no matter how small. However, [b]limit[/b] of sequence does not impose this rule. So, a convergence sequence converges to a [b]limit[/b]. But the [b]limit[/b], which is the limit of the sequence, is not necessarily an [b]accumulation point[/b].

Let me clarify what I have said: A sequence $\{5, 4, 3, 2, 1, 0,0 ,0,\ldots\}$ obtained from a finite set $G = \{0, 1, 2, 3,4,5,6,7,8 \}$ converges to a [b]limit[/b], which is $0$. But the value $0$ is not an [b]accumulation point[/b]. On the other hand, for an infinite set $H = [0, 8]$,  the sequence $\{1, 1/2, 1/3, 1/4, 1/5, \ldots\}$ converges to a [b]limit[/b] $0$. Also this value $0$ is an [b]accumulation point[/b]. You may ask a sequence $\{5, 4, 3, 2, 1, 0,0 ,0,\ldots\}$ can also be drawn from $H$, so what is the difference? To answer this, notice that, as one example,  the sequence can be amended to become $\{5, 4, 3, 2, 1, 1/2, 1/3, 1/4, 1/5, \ldots\}$. This is not possible for set $G$.


geometry
incenter
cyclic quadrilateral
national olympiad

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112308]Problem 1[/url]
For $n\in\mathbb{N}$, let $a_{n}$ denote the closest integer to $\sqrt{n}$.
Evaluate $\sum_{n=1}^\infty{\frac{1}{a_{n}^{3}}}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112313]Problem 2[/url]
Let $ABCD$ be a cyclic quadrilateral $\angle{BAD}< 90^\circ$ and $\angle BCA = \angle DCA$. Point $E$ is taken on segment $DA$ such that $BD=2DE$. The line through $E$ parallel to $CD$ intersects the diagonal $AC$ at $F$. Prove that $AC\cdot BD=2AB\cdot FC$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112311]Problem 3[/url]
Let $n$ blue lines, no two of which are parallel and no three concurrent, be drawn on a plane. An intersection of two blue lines is called a blue point. Through any two blue points that have not already been joined by a blue line, a red line is drawn. An intersection of two red lines is called a red point, and an intersection of red line and a blue line is called a purple point. What is the maximum possible number of purple points?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112309]Problem 4[/url]
Let $f: \mathbb{R}^{+}\rightarrow \mathbb{R}+$ be an increasing function. For each $u\in\mathbb{R}^{+}$, we denote $g(u)=\inf\{ f(t)+u/t \mid t>0\}$. Prove that:
$(a)$ If $x\leq g(xy)$, then $x\leq 2f(2y)$;
$(b)$ If $x\leq f(y)$, then $x\leq 2g(xy)$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112312]Problem 5[/url]
Find the set of all ordered pairs $(s,t)$ of positive integers such
that $t^{2}+1=s(s+t)$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112314]Problem 6[/url]
The incircle of triangle $ABC$ touches $BC$ at $D$ and $AC$ at $E$. Let $K$ be the point on $CB$ with $CK=3BD$, and $L$ be the point on $CA$ with $AE=CL$. Lines $AK$ and $BL$ meet at $P$. If $Q$ is the midpoint of $BC$, $I$ the incenter, and $G$ the centroid of $\triangle ABC$, show that:
$(a)$ $IQ$ and $AK$ are parallel,
$(b)$ the triangles $AIG$ and $QPG$ have equal area.


$ x_1 \equal{} 1$ and
$ x_{n\plus{}1} \equal{} \frac{x_n\plus{}5}{x_n\plus{}1}$ for all $ n \in \mathbb{N}$.
Find an explicite expression for $ x_n$ (in terms of the Fibonacci sequence).

$ x \equal{} \frac{x\plus{}5}{x\plus{}1} \iff x^2 \minus{} 5 \equal{} 0 \iff x \equal{} \pm \sqrt{5}$.
\[ x_{n\plus{}1} \minus{} \sqrt{5} \equal{} \frac{x_n \plus{} 5}{x_n \plus{} 1} \minus{} \sqrt{5} \equal{} \frac{(1 \minus{} \sqrt{5})x_n \plus{} (5 \minus{} \sqrt{5})}{x_n \plus{} 1} \equal{} (1 \minus{} \sqrt{5}) \frac{x_n \minus{} \sqrt{5}}{x_n \plus{} 1}\]
\[ x_{n\plus{}1} \plus{} \sqrt{5} \equal{} \frac{x_n \plus{} 5}{x_n \plus{} 1} \plus{} \sqrt{5} \equal{} \frac{(1 \plus{} \sqrt{5})x_n \plus{} (5 \plus{} \sqrt{5})}{x_n \plus{} 1} \equal{} (1 \plus{} \sqrt{5}) \frac{x_n \plus{} \sqrt{5}}{x_n \plus{} 1}\]
$ \therefore \ \frac{x_{n\plus{}1} \minus{} \sqrt{5}}{x_{n\plus{}1} \plus{} \sqrt{5}} \equal{} \frac{1 \minus{} \sqrt{5}}{1 \plus{} \sqrt{5}} \cdot \frac{x_n \minus{} \sqrt{5}}{x_n \plus{} \sqrt{5}}$

Let $ \frac{x_n \minus{} \sqrt{5}}{x_n \plus{} \sqrt{5}} \equal{} y_n$, $ y_{n\plus{}1} \equal{} \left( \frac{1 \minus{} \sqrt{5}}{1 \plus{} \sqrt{5}} \right) y_n$, and $ y_1 \equal{} \frac{x_1 \minus{} \sqrt{5}}{x_1 \plus{} \sqrt{5}} \equal{} \frac{1 \minus{} \sqrt{5}}{1 \plus{} \sqrt{5}}$
\[ \therefore \ y_n \equal{} \left( \frac{1 \minus{} \sqrt{5}}{1 \plus{} \sqrt{5}} \right)^n.\]

$ \frac{x_n \minus{} \sqrt{5}}{x_n \plus{} \sqrt{5}} \equal{} y_n \iff x_n \equal{} \minus{}\sqrt{5} \cdot \frac{y_n \plus{} 1}{y_n \minus{} 1}$
Thus
\[ x_n \equal{} \minus{}\sqrt{5} \cdot \frac{\left( \frac{1 \minus{} \sqrt{5}}{1 \plus{} \sqrt{5}} \right)^n \plus{} 1}{\left( \frac{1 \minus{} \sqrt{5}}{1 \plus{} \sqrt{5}} \right)^n \minus{} 1} \equal{} \minus{} \sqrt{5} \cdot \frac{(1 \minus{} \sqrt{5})^n \plus{} (1 \plus{} \sqrt{5})^n}{(1 \minus{} \sqrt{5})^n \minus{} (1 \plus{} \sqrt{5})^n}\]

Another solution
Let $ y_n\equal{}x_n\plus{}1$ ,$ y_n\equal{}\frac{w_{n\plus{}1}}{w_n}$ for all $ n$
then we can easyly find $ w_{n\plus{}1}\equal{}2w_n\plus{}4w_{n\minus{}1}$
And now,the end is really simple


function
Euler
limit
logarithms
real analysis
real analysis unsolved


Timothy arrives on an island populated by 3 knights who always tell the truth, 3 knaves who always lie, and 3 pages who tell the truth only half the time. Meeting 3 people at random, what is the probability that they all tell the truth when asked a question?

Please explain your answer.

[hide="hint"]Use some case work:  3 knights, 2 knights / 1 page, 1 knight / 2 pages, 3 pages.[/hide]

[hide="solution"]CASE 1:  3 knights
The probability of meeting 3 knights is $ \left(\frac13\right)^3 \equal{} \frac1{27}$, and all three will tell the truth.
CASE 2:  2 knights / 1 page
The probability of meeting these three is $ 3\cdot\left(\frac13\right)^3 \equal{} \frac19$, and the page will tell the truth with probability $ \frac12$, so the total probability is $ \frac19\cdot\frac12 \equal{} \frac1{18}$.
CASE 3:  1 knight / 2 pages
$ 3\cdot\left(\frac13\right)^3\cdot\frac14 \equal{} \frac1{36}$
CASE 4:  4 pages
$ \left(\frac13\right)^3\left(\frac12\right)^3 \equal{} \frac1{216}$
Adding these up, we have $ \frac {8\plus{} 12 \plus{} 6 \plus{} 1}{216} \equal{} \frac 18$.[/hide]

EDIT:  fixed some arithmetic

controversial solution:
[hide]
the probability that a randomly picked person will tell a truth is $ \frac{3}{9} \cdot 1\plus{}\frac{3}{9} \cdot \frac{1}{2} \plus{} \frac{3}{9} \cdot 0 \equal{} \frac{1}{2}$
now cube this and get $ \frac{1}{8}$
the reason it is controversial is that it does not address the fact that the problem asks for picking people without replacement
[/hide]

but yes, it is 1/8... b-flat's arithmetic on the last line is flawed :D


Fie $ABCD$ un patrat de latura 1. Prin $C$ se duc doua drepte care taie $(BD)$ in $M$ si $N$, iar $(AB)$ si $(AD)$ in $P$ si $Q$. Urmatoarele afirmatii sunt echivalente:
[color=blue]a)[/color] $m(\angle PCQ)=45^0$;
[color=blue]b)[/color] $P[APQ]=2$;
[color=blue]c)[/color] $PQ=BP+QD$;
[color=blue]d)[/color] $MN^2=BM^2+ND^2$;
[color=blue]e)[/color] $PQ$ este tangenta la arcul $BD$ al cercului de centru $C$ si raza $CB$;
[color=blue]f)[/color] $CT=1$, unde $T$ este punctul de tangeta de la e);
[color=blue]g)[/color] $Aria[CMN]=\frac{1}{2} Aria[CPQ]$.
[img]http://i19.photobucket.com/albums/b200/mariusdamian/secantala45.gif[/img]

Lasa ca sunt frumoase problemele astea.Poate ma apuc de ea maine dupa scoala.Arata interesant problema si cred ca din moment ce am demonstrat ca una din afirmatzii duce la rezolvarea celorlaltor afirmatzii,putem recurge la metoda reducerii prin absurd.

[quote="Marius Damian"]Fie $ABCD$ un patrat de latura 1. Prin $C$ se duc doua drepte care taie $BD$ in $M$ si $N$, iar $AB$ si $AD$ in $P$ si $Q$[/quote]
Nu intzeleg un lucru in enunt:Cele 2 drepte care trec prin $C$ nu taie dreapta $AD$ in 2 pcte distincte si dreapta $AB$ tot in 2 pctre distincte?Nu intzeleg de ce sunt doar 2 pcte de concurentza dintre cele 2 drepte care trec prin $C$ si cele 2 drepte:$AB$ si $AD$ ,si nu 4,cate ar trebui sa fie :2 pe $AB$ si 2 pe $AD$.

Am completat cu figura. Problema este, de fapt, o suma de probleme deosebit de frumoase.

Si o ultima intrebare inainte de a ma duce la culcare:subpunctele sunt in ordinea dificultatii,nu?Caci nu ash vrea sa ma chinui la $a)\longrightarrow b)$ daca $a)\longrightarrow e)$ este mai simpla(desi ma indoiesc) :)
Noapte buna!

Problema de fata a fost prezentata de dl. prof. C. Mortici la cursurile de vara de la Busteni din 2003.
Multumesc pe aceasta cale dlui. profesor care mi-a aratat ca geometria este cea mai frumoasa ramura a matematicii.
Prezint cativa pasi din demonstratia problemei.
$d)\Rightarrow a)$ (Baraj 2001 - a dat 3 solutii);
$a)\Rightarrow d)$;
$a)\Leftrightarrow c)$;
$a)\Leftrightarrow b)$;
$a)\Leftrightarrow g)$ (M976 - Kvant/1986);
$b)\Leftrightarrow c)$.

In primul rand, $AB,AD,BD$ se refera la linie sau la segment? Intreb asta pentru ca in primul caz nu prea ies toate echivalentele (cred).

In al doilea rând, am o idee pentru rezolvarea neanalitica/fara trigonometrie a unora dintre echivalente: Fie $\mathcal C$ cercul de centru $C$ si raza $1$. De asemenea, fie $T \neq D$ a.i. $QT$ e tangenta la $\mathcal C$. Apoi avem $\angle PCQ = 45^\circ \, \, \Longleftrightarrow \, \, \triangle BPC \equiv \triangle TPC \, \, \Longleftrightarrow \, \, PT$ e tangenta la cerc, care e echivalent cu $d \left( C, PQ \right) = 1$. Apoi folosim arii pentru echivalenta punctelor [i]a[/i], [i]b[/i], [i]c[/i], [i]e[/i], [i]f[/i].

Echivalenta punctelor [i]d[/i], [i]g[/i] cu [i]a[/i] iese cu trigonometrie (ca toata problema de altfel), dar inca n-am gasit o solutie cat de cat eleganta pentru aceasta.

[quote="perfect_radio"]In primul rand, $AB,AD,BD$ se refera la linie sau la segment? Intreb asta pentru ca in primul caz nu prea ies toate echivalentele (cred).[/quote]
Eu nu am studiat cazul in care intersectiile au loc pe prelungirile laturilor patratului. Am modificat enuntul; trebuia sa scriu segmente.


find the 10000'th digit (from the right) of the fraction  $\frac{50^{10000}-1}{50-1}$

[quote="m4467"](in other words find the 10000'th digit (from the right) of the fraction given above)[/quote]

that expression finds the first 10,000 digits starting from the right of the fraction, not just the 10,000th one. are you sure you mean $\mod 10^{10000}$?

you are correct.

sorry about that...
just find the 10000th Digit...


counting
distinguishability
geometry
rectangle
analytic geometry

An 8 by 8 checkerboard is placed with its corners at (0,0), (8,0), (0,8), and (8,8). Find the number of distinguishable non square rectangles, with corners at points with integer coordinates, that can be counted on the checkerboard.

Answer:

I am assuming that by distinguishable, that two rectangles are the same if one has the same dimensions as another.

This is a question which is looking at how many different ways you can pick 2 items (sides) from a set of 8 different items.

Hence: number of ways = $ \binom{8}{2}= 28$

Edit: Sorry! I accidently pressed the wrong button on my calculator (permutation button).

[quote="mdk"]An 8 by 8 checkerboard is placed with its corners at (0,0), (8,0), (0,8), and (8,8).[/quote]
Edited: For a moment, I thought that was a 9 by 9 checkerboard. Sorry.
[quote="mdk"]Find the number of distinguishable non square rectangles, with corners at points with integer coordinates, that can be counted on the checkerboard.[/quote]
Must we only count the rectangles with sides paralel to the checkerboard's? That would be easier.
[quote="sludgethrower"]I am assuming that by distinguishable, that two rectangles are the same if one has the same dimensions as another.[/quote]
I would have assumed that two rectangles are distinguishable if they have a different set of 4 corners (notice that the order is not important here).

I assumed that the rectangles formed are those that are parallel to the sides, as these are bounded by the grid on the checkerboard...

you know jli will delete your post don't you?

[quote="13375P34K43V312"]$ (\binom{9}{2})^{2}-\sum_{k=1}^{8}k^{2}$[/quote]

Thats not counting teh slanted rectangles and you are overcounting since they are distinguishable from eachother (im assuming distinguishable is same dimentions)

BUT if distguishable is set of 4 corners andwe dont count slanted ones, then thats the right answer but the question is so ambigious so i cant tell


calculus
integration
logarithms
calculus computations

may be it is an improper integral .when we keep t=(x)^2 we get the expression as $ 1/ln(t)$ which i thinkis an improper integral.

$ \int_0^2 \frac{xdx}{\ln x} \equal{} \int_{\minus{}1}^{1} \frac{(y\plus{}1)dy}{\ln(y\plus{}1)}$ -- is diverges. Because in $ y_0\equal{}0$ we have $ \frac{y\plus{}1}{\ln(y\plus{}1)}$ is equvivalent to $ \frac{1}{y}.$

[quote="anik_andrew"]$ \int_0^2 \frac {xdx}{\ln x} \equal{} \int_{ \minus{} 1}^{1} \frac {(y \plus{} 1)dy}{\ln(y \plus{} 1)}$ -- is diverges. Because in $ y_0 \equal{} 0$ we have $ \frac {y \plus{} 1}{\ln(y \plus{} 1)}$ is equvivalent to $ \frac {1}{y}.$[/quote]

 Wonderful, thank you


inequalities
logarithms
function
analytic geometry
graphing lines
slope

If $x$ and $y$ are positive numbers, denote ${\mathbf X }=\frac{\ln{(1+x)}}{x}\; ,\;{\mathbf Y}=\frac{\ln(1+y)}{y}\; .$
[b]Prove or disprove that  $\; \; \;{\mathbf X}{\mathbf Y }< 1\; .$  [/b]

Consider the function $f(x)=\ln(1+x)$, limiting ourselves to nonnegative $x$'s.

The slope of $f$ at any $x$ in the interval $[0, b]$, $b\in\mathbb{R}^{+}$, is $f'(x)=1/(1+x)$, which is no larger than 1 (Property 1). Now, since $f$ is continuous on $[0, b]$ and $f(x)\in[0, b]$, it has a fixed point in the interval $[0, b]$. By a uniqueness theorem for fixed points, $f$ has a unique fixed point due to Property 1. It is easy to see that the fixed point is 0.

As a result, the curve of $f(x)$ is always "lower" than the line $y=x$, hence $f(x)-x\leq0$. Thus, $f(x)/x=\ln(1+x)/x<1$ for all positive $x$.

Since $\ln(1+x)/x<1$ and $\ln(1+y)/y<1$ for positive $x, y$, their product is less than 1.


hi all,
What is the procedure for admission to doctoral programme in mathematics in ENS,Though i am now in FRANCE,i do'nt speak french and I am from India,I will be 23 next year,hopefully i am still eligible for admission.I tried their website but could'nt make out much from there  :( .
thanks in advance  :) .

Most of the people who use this forum are American -- you might have better luck asking around the French subforum.

Is the École Nationale Supérieure program in mathematics at a specific place? JBL's suggestion to ask in the French Forum is good, but I do want to make clear that questions about schools of higher education anywhere in the world are welcome here. Just as a tip, using a full title for a school in your subject line makes it easier for readers to know what your question is about. I thought for a minute to figure out that ENS must mean "École Nationale Supérieure."

Best wishes for your application. I welcome any answers that informed persons have in this thread.

I think the N is actually for Normale, not Nationale.

Good luck with your search!


I am holding a contest that will test your brain on puzzles and brainteasers.  If you miss a question in a round then you are no longer in the competition.  There will be as many rounds as needed to determine a winner.  If you want to sign up just make a post saying "i'm in".  Have fun. :)

since nobody comes to this forum we'll wait until about 6 or 7 people to sign up.

If  A is B and B is C and C is D and D is E then im in!

G-unit, I'm sorry that you can't participate since A doesn't equal E

I'm in,

just PM me to notify me

[size=200]ROUND 1[/size]
1) You have a barrel, filled to the top with water, which weighs 150 pounds. What can you add to the barrel in order to make it lighter?

2)Can you replace the question marks with maths symbols (+-/x) and make the equation correct:

    (9 ? 1) ? (7 ? 6) ? (8 ? 4) = 3

3)Using only two 2's and any combination of mathematical signs, symbols and functions can you make 5?

Here are the ROUND 1 answers. I did not know what to do withthe answers so I am sending it in the attached document.


trigonometry inequalities quadratics algebra

Solve $10(25^{cos \pi x}-4^{\cos \pi x})=7(5^{\cos \pi x}-2^{\cos \pi x})$ and find all solutions that satisfy the inequality $x^4-6x^2-1 \leq 0$

[hide]Making the substitution $u=\cos{\pi x}$ the equation becomes $10(5^{2u}-2^{2u})=7(5^u-2^u)$ which implies either $5^u=2^u$ or $5^u+2^u=\frac{7}{10}$.  In the first case $u=0$ and thus $x=\frac{k}{2}$ for $k$ odd.  In the second case notice $\frac{7}{10}=\frac{1}{5}+\frac{1}{2}$ so $u=-1$ which means $x$ is an odd integer.  By the quadratic equation $-\sqrt{3+\sqrt{10}}\leq x \leq \sqrt{3+\sqrt{10}}$ which yields the only valid values for $x$ to be $-\frac{3}{2},-1,-\frac{1}{2},\frac{1}{2},1,\frac{3}{2}$[/hide]


function
algebra
domain
LaTeX
combinatorics unsolved
combinatorics

Given a set A containing n positive integers in the range [1 .. m] and a set B containing m positive integers in the range [1 .. n], there must exist a subset of A and a subset of B, denoted by A' and B' respectively, such that the sum of the elements in A' equals the sum of those in B'.

Some one asked me to prove this claim. I am not absolutely sure it is true yet doing some practice made me conjecture it is.

I think INTERVALL is better than RANGE. The range (of a function $ f(x): A \to B$ ) is usually the set $ \{f(x) \mid x\in A\}$. This set is also called  domain of a function.

[quote]
Given a set $ A$ containing $ n$ positive integers in the intervall $ [1 \dots m ]$ and a set $ B$ containing $ m$ positive integers in the intervall $ [1 \dots n ]$. Prove that there must exist a subset of $ A$ and a subset of $ B$, denoted by $ A'$ and $ B'$ respectively, such that the sum of the elements in $ A'$ equals the sum of those in $ B'$.[/quote]

No, I mean, A has n elements each of which has the value between 1 and m.
Similarly, B has m elements each of which has the value between 1 and n.

So b_i must not be too large...

p.s. excuse me as I am a newbie here and I have not figure out yet how to format the math formulas.

pfanni is having difficulty with the fact that the word "range" has more than one meaning.  (S/he was suggesting the alternate word "interval," with the same meaning, though s/he misspelled it.)  For information about [[LaTeX]], see that link and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]this one[/url].  Finally, just to clarify, by "set" here actually you mean "multiset," with elements allowed to repeat, right?

Yes I agree, it should be multiset. Thx for all the replies.


trigonometry
geometry
algebra
polynomial
angle bisector

This is Theorem 1 in http://www.mathlinks.ro/Forum/viewtopic.php?t=6088 post #2. See also http://www.mathlinks.ro/Forum/viewtopic.php?t=17 .

  darij

had a projective proof while doing this so i am reviving the old thread :D
clearly it suffices to prove $ (FE,DT)\equal{}\minus{}1$
Let $ EF \cap BC\equal{}T$ then we know $ (AP,DR)\equal{}\minus{}1$ where $ R\equal{}AP \cap BC$ so projecting through $ B$ we get $ (FE,DT)\equal{}\minus{}1$

[quote][color=darkred][b][u]Lemma.[/u][/b] Let $ d \equal{} AB$ be a line and $ C\in (AB)$ , $ D\in d$ so that $ B\in (AD)$ . For a point $ P\not\in d$ denote

$ \begin{array}{c} m(\widehat {APC}) \equal{} \alpha \\
\ m(\widehat {CPB}) \equal{} \beta \\
\ m(\widehat {BPD}) \equal{} \gamma\end{array}$ . Then the points $ C$ , $ D$ [u]are "harmonical conjugate" w.r.t. the points[/u] $ A$ , $ B$

i.e. $ \frac {CA}{CB} \equal{} \frac {DA}{DB}$ if and only if $ \boxed {\ \cos\ (\ \alpha\ \plus{} \ \gamma\ )\ \equal{} \ \cos\ (\ \alpha\ \plus{} \ \beta\ )\ \cdot\ \cos\ (\ \beta\ \plus{} \ \gamma\ )\ }$ .[/color][/quote]
[color=darkblue][u][b]Proof.[/u][/b]

$ \frac {CA}{CB} \equal{} \frac {DA}{DB}\ \Longleftrightarrow\ \frac {PA}{PB}\cdot\frac {\sin \widehat {CPA}}{\sin\widehat {CPB}} \equal{}$ $ \frac {PA}{PB}\cdot\frac {\sin \widehat {DPA}}{\sin\widehat {DPB}}\ \Longleftrightarrow\ \frac {\sin \alpha}{\sin \beta} \equal{}$ $ \frac {\sin (\alpha \plus{} \beta \plus{} \gamma )}{\sin \gamma}\ \Longleftrightarrow$

$ \sin \alpha\cdot\sin \gamma \equal{} \sin \beta\cdot\sin (\alpha \plus{} \beta \plus{} \gamma )$ $ \Longleftrightarrow$ $ \cos (\alpha \minus{} \gamma ) \minus{} \cos (\alpha \plus{} \gamma ) \equal{} \cos (\alpha \plus{} \gamma ) \minus{} \cos (\alpha \plus{} 2\beta \plus{} \gamma )$ $ \Longleftrightarrow$

$ 2\cdot\cos (\alpha \plus{} \gamma ) \equal{} \cos (\alpha \minus{} \gamma ) \plus{} \cos (\alpha \plus{} 2\beta \plus{} \gamma )$ $ \Longleftrightarrow$ $ \boxed {\ \cos\ (\ \alpha\ \plus{} \ \gamma\ )\ \equal{} \ \cos\ (\ \alpha\ \plus{} \ \beta\ )\ \cdot\ \cos\ (\ \beta\ \plus{} \ \gamma\ )\ }\ \ (*)$ .

[b][u]Remark.[/u][/b]

Prove easily that if $ C$ , $ D$ are harmonical conjugate w.r.t. $ A$ , $ B$ i.e. the relation $ (*)$ is truly, then

$ PA\perp PC$ i.e. $ \alpha \plus{} \beta \equal{} 90^{\circ}\ \Longleftrightarrow$ the ray $ [PB$ is the bisector line of $ \widehat {CPD}$  i.e. $ \beta \equal{} \gamma$ .[/color]

[color=darkred][size=117][b][u]An easy problem.[/u][/b][/size]

$ \triangle\ ABC$ and $ \left\|\begin{array}{ccccccc}
E\in AC & , & BE\perp AC & ; & F\in AB & , & CF\perp AB\\\\
M\in (BC) & , & BM\equal{}2\cdot MC & ; & C\in (BN) & , & CN\equal{}BC\end{array}\right\|$ $ \Longrightarrow$ $ \frac {FM}{FN}\equal{}\frac {EM}{EN}$[/color]

Proof:  Let $ EF \cap BC \equal{} X$ , $ AP \cap BC \equal{} Y$, applying Ceva and Menelaus theorem in triangle $ ABC$, we get that : $ BXCY$ is a harmonic range.
Consider the pencil $ A$ {$ BXCY$} .  This implies $ FDEX$ is also a harmonic range.  So if two rays of a pencil are perpendicular, they are the internal and external bisectors of the angles formed between the other two rays of the pencil. Here consider the pencil $ Y$ {$ FDEX$}. As internal and external bisector of an angle are perpendicular, we are done.

Virgil Nicula sir had already made the remark which is actually a theorem that I used in the proof. And I think pardesi's proof is similar, if not probably the same, but I don't know projective geometry, so I have written this proof.

Quite a few INMO problems become trivial if you know some important fact. In fact, some of them are special cases of well known results. A lot of the polynomial problems especially become one liners, but are we allowed to just state (without proof) advanced theorems which are never used in official solutions?

Yes Agr_94_Math both proofs are wonderful and same.
I am eager to know more examples of $Inversion$ and $Projective$ $geometry$
in $Inmo$ and $Rmo$ possibly $IMO$ too
Will anybody tell me?

Let $G$,$H$,$K$, and $M$ be the feet of the perpendiculars from $F$,$A$,$D$, and $E$, respectively, onto $BC$. We want to show that $\angle FKD= \angle EKD$, or that $\angle FKG= \angle EKM$, or that $\frac{FG}{GK}=\frac{EM}{MK}$.

By similarity we have
$\frac{FG}{AH}=\frac{BF}{AB}$
$\frac{EM}{AH}=\frac{CE}{AC}$
$\frac{GK}{MK}=\frac{FD}{DE}$
By Menelaus we have
$\frac{AF}{AB} \cdot \frac{BP}{PE} \cdot \frac{DE}{FD}=1$
$\frac{CE}{AC} \cdot \frac{AF}{FB} \cdot \frac{BP}{PE}=1$

Combining all five of these, we have
$\frac{FG}{GK}=\frac{EM}{MK}$

Which is what we wanted to show.

[quote="vanu1996"]use miquel point lemma to these points $F,K,E$.[/quote]
How does this idea help ?
Will anyone please explain :)

Let $AP \cap BC \equiv N $
$EF \cap BC \equiv L $
First we see that $ AN, BE, CF$ are concurrent Cevians so by this famous theorem:
[b][u]"In a triangle $ABC$ consider three points $X,Y,Z$ on the sides $BC, CA$ ,respective $AB$ . If $X$ is the point of intersection of $YZ$ with the extended side $BC$, then the four-part $(BXCX)$ forms and harmonic division if. and only if the Cevians $AX,BY$ and $CZ$ are concurrent."[/b][/u]
 $ (BNCL) $ are harmonic. $(FDEL) $ is also harmonic due to pencil $A(BNCL) $.
Now by another famous theorem :
[b][u]"Let four points $A,B,C$ and $D$, in this order, lying on d.Then, if two of the following three propositions are true, then the third is also true:
$(1)$ The division $(ABCD)$ is harmonic.
$(2)$ $XB$ is the internal angle bisector of $ \angle AXC$.
$(3)$ $XB \perp XD$."[/b][/u]
We see that $ KD \perp KL$ and $(FDEL) $ are harmonic so $ \angle FKD = \angle EKD$ i.e $DK$ bisects $ \angle FKE $
Proved.  :)

P. S -  I copied the so called famous theorems ( :P ) from
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=50&t=161310&start=0

Just a special case of [url=http://www.artofproblemsolving.com/community/c6h2928p8860]China TST 2002[/url]

@above Why you think it is a special case?? I think both problems are same :maybe:
[quote=INMO 2003 P1]Let $P$ be an interior point of an acute-angled triangle $ABC$. The line $BP$ meets the line $AC$ at $E$, and the line $CP$ meets the line $AB$ at $F$. The lines $AP$ and $EF$ intersect each other at $D$. Let $K$ be the foot of the perpendicular from the point $D$ to the line $BC$. Show that the line $KD$ bisects the angle $\angle EKF$.[/quote]
[hide=Solution]$-1=(A,P;D,R)\overset{B}{=}(F,E;D,T)$ completing the proof.[/hide]
[hide=P.S]Just saw #3 has the same solution as this one :wallbash:[/hide]
@below Yep! Earth is younger than Jupiter :lol:

[quote=Fermat -Euler]Let $P$ be an interior point of an acute-angled triangle $ABC$. The line $BP$ meets the line $AC$ at $E$, and the line $CP$ meets the line $AB$ at $F$. The lines $AP$ and $EF$ intersect each other at $D$. Let $K$ be the foot of the perpendicular from the point $D$ to the line $BC$. Show that the line $KD$ bisects the angle $\angle EKF$.[/quote]

I have a history of tryst with this question. A friend gave this question to me in class 9 and I tried for a week and left it. 7-8 months ago also I tried it but gave up, this happened once more until I learnt projective geo 2 weeks back and today I remembered this.
Let $BC$ meet $FE$ at $X$.
From perspectivity at $A$, $(F,E;D,J)=-1$ and since angle $DKJ=90$, we are done.

Let $EF \cap BC = L$ And $AP \cap BC = T$
$$-1 = (B,C;L,T) \stackrel{A}{=} (E,F;L,D) \ \text{and EGMO Lemma 9.18 finishes} \ \blacksquare$$

@above Even I had got scared by seeing the complex diagram with so many lines (I am not a fan of multiple lines) but now destroyed it within 15 seconds  :rotfl: (Not flexing)
[s]No one observed that the 2 people above using projective did not specify the points properly, until now :rotfl:[/s]


1. There is a list of 7 numbers.  The average of the first four numbers is 5, and the average of the last four numbers is 8.  The average of all seven numbers is $ \frac{46}{7}$.  Find the common number to both sets of four numbers.

2. Joe randomly selects two distinct numbers from {1,2,3,4} and Susan randomly selects one number from {1,2,3,4,5,6,7,8}.  Find the probability that Susan's number is larger that the sum of the two numbers chosen by Joe.  Express your answer as a common fraction.

3. The area of rectangle ABCD is 72.  Let E be the midpoint of BC and F the midpoint of CD.  Find the area of the triangle AEF.

Thanks. :lol:  :D  :)

[quote="ajain"]1. There is a list of 7 numbers.  The average of the first four numbers is 5, and the average of the last four numbers is 8.  The average of all seven numbers is $ \frac {46}{7}$.  Find the common number to both sets of four numbers.

2. Joe randomly selects two distinct numbers from {1,2,3,4} and Susan randomly selects one number from {1,2,3,4,5,6,7,8}.  Find the probability that Susan's number is larger that the sum of the two numbers chosen by Joe.  Express your answer as a common fraction.

3. The area of rectangle ABCD is 72.  Let E be the midpoint of BC and F the midpoint of CD.  Find the area of the triangle AEF.

Thanks. :lol:  :D  :)[/quote]
[hide]1. Call the common number x. In the first set of numbers, the total sum is 20. So the sum of the other three numbers is 20-x. In the other set of numbers, the total sum is 32. So the sum of the other three numbers is 32-x. The sum of all the numbers is 46. So 20-x+32-x+x=46. Solving for x, 52-x=46. Common number=6.
3. Length=a, and width=b. ab=72. Area of triangle ABE is then $ \frac{ab}{4}$. Area of triangle EFC is $ \frac{ab}{8}$. Area of triangle ADF is $ \frac{ab}{4}$. Sum of these is $ \frac{5ab}{8}$. The area of triangle AEF is 1-(the sum of the other triangles). And since ab=72, we now have
Area of Triangle AEF=$ 72\minus{}\frac{5\cdot72}{8}\equal{}72\minus{}45\equal{}27$.[/hide]
Don't feel like doing that probability problem...never liked probability anyways :P

for number 1:
[hide] if the average of the first 4 numbers is 5, the sum is 20. is the average of the last 4 numbers is 8, the sum is 32. so the sum  of the 8 numbers (one is counted twice; that is what we are looking for.) is 52. the average of the 7 numbers is 46/7, so the susm is 46. 52-46=6. the number counted twice was 6. ans,[/hide]

for number 2 its not too hard to just list all the possibilities..

number 3:
[hide]if you draw a picture, clearly triangle aed=abcd-abe-efc-adf
working in general terms, letting abcd=1 you get aed=1-1/4-1/8-1/4=3/8
so the area is 3/8 of abcd
abcd is 72, so the area is 3*9=27 ans.[/hide]

Hm...no one is doing #2.  I think it's 9/64....hm... :idea:  :wink:

[hide="1"]
The sum of the first four numbers is 20 and the last four is 32. This is 52, but the sum is 46 so the common number is 6.[/hide]

[hide="2"]
Joe can choose two numbers 6 ways while Susan can choose 8 numbers. This means there's 48 possibilities.
If Susan gets an 8, there are 6 ways for Susan to have a larger number than Joe's sum.
If it's a 7, there are 5 ways (4+3 doesn't work).
6: 4 ways because 4,3 and 4,2 don't work.
5: 4,3 4,2 4,1 3,2 don't work so there are 2 ways.
4: Only one way: 1,2
3: No ways
2: No ways
1: No ways

So the probability is $ \frac{6\plus{}5\plus{}4\plus{}2\plus{}1}{48} \equal{} \frac{18}{48} \equal{} \frac{3}{8}$[/hide]

[hide="3"]
The side lengths are $ a$ and $ b$.
$ ab \equal{} 72$
There are 4 triangles in the figure: AEF and three other right triangles.
One's area is $ \frac{ab}{8}$ and the other areas are both $ \frac{ab}{4}$
The equation now is $ 72 \equal{} A \plus{} \frac{5ab}{8}$
We also know that $ ab \equal{} 72$ so we have $ 72 \equal{} A \plus{} 45$
$ A \equal{} 27$.[/hide]

[hide="#2"]If you make a table, you can easily see that Joe gets:

1 possibility for 2
2 possibilities for 3
3 possibilities for 4
4 possibilities for 5
3 possibilities for 6
2 possibilities for 7
1 possibilities for 8.

Our answer is
$ \dfrac{1}{8}*\dfrac{1}{16}\plus{}\dfrac{1}{8}*\dfrac{3}{16}\plus{}\dfrac{1}{8}*\dfrac{6}{16}\plus{}\dfrac{1}{8}*\dfrac{10}{16}\plus{}\dfrac{1}{8}*\dfrac{13}{16}\plus{}\dfrac{1}{8}*\dfrac{15}{16}$

So I got $ \dfrac{3}{8}$[/hide]

Thanks.  I messed up by having the denominator 4*4*8 instead of 6*8 :blush:


algorithm
combinatorics unsolved
combinatorics

Let $A_{1}A_{2}B_{1}B_{2}$ be a convex quadrilateral. At adjacent vertices $A_{1}$ and $A_{2}$ there are two Argentinian cities. At adjacent vertices $B_{1}$ and $B_{2}$ there are two Brazilian cities. There are $a$ Argentinian cities and $b$ Brazilian cities in the quadrilateral interior, no three of which collinear. Determine if it's possible, independently from the cities position, to build straight roads, each of which connects two Argentinian cities ou two Brazilian cities, such that:

$\bullet$ Two roads does not intersect in a point which is not a city;
$\bullet$ It's possible to reach any Argentinian city from any Argentinian city using the roads; and
$\bullet$ It's possible to reach any Brazilian city from any Brazilian city using the roads.

If it's always possible, construct an algorithm that builds a possible set of roads.

The answer can be found in the IOI solution booklet...


An easier factorization:

$x+3\sqrt{x}+2=(\sqrt{x}+1)(\sqrt{x}+2),x+9\sqrt{x}+18=(\sqrt{x}+3)(\sqrt{x}+6)$, which gives

\begin{eqnarray*}168x &=& (\sqrt{x}+1)(\sqrt{x}+2)(\sqrt{x}+3)(\sqrt{x}+6)\\ &=& (\sqrt{x}+1)(\sqrt{x}+6)(\sqrt{x}+2)(\sqrt{x}+3)\\ &=& (x+7\sqrt{x}+6)(x+5\sqrt{x}+6)\\ &=& (x+6\sqrt{x}+6)^{2}-(\sqrt{x})^{2}\end{eqnarray*}

That gives $(x+6\sqrt{x}+6)^{2}=169x\implies x+6\sqrt{x}+6=\pm 13\sqrt{x}$

The two equations are $x-7\sqrt{x}+6=0, x+19\sqrt{x}+6=0$, but only the first gives positive solutions for $\sqrt{x}$, and we find $x\in\{1,36\}$