A word problem
There is a man who wants to marry the princess. But the king doesn't want him to so he tells everyone that he will put 2 slips  in a hat, one says married and the other says not married. So whichever one he picks will happen. The man hears that the king will make both slips say not married. When the time arrives for him to pick, a large crowd appears to watch him choose. What can the man do to marry the princess?

[quote="moogra"]goes up and erases the not in the not marry idon't knwo[/quote]

People would see him do that

prove that both sheets say not married, and renounce the king! and then take the princess away to a foreign land!

[quote="goldendomer"]EASY!!! Take a slip out and then eat it and swallow it without looking. :rotfl: Then the king would get mad and say he cant marry his daughter. Tell him to look at the other slip. He would show everybody and it would say "not married". Then the man could say the "marry" slip is in his mouth. It took me a long time to get that. :D[/quote]

Then it's not easy...
Yep thats it. Nice. :) I was thinking of him tearing it up but that works too.

[quote="nutz_for2.718281828"][hide="simple solution"]He closes his eyes, picks up the hat, throws it as fair as he can, and shows the tickes to everyone. Since it wasn't a far play, the man will make two tickets, show them to everyone, and put them inside the hat. He will put "He will marry" on the other side of the "He will not marry" ticket, and secretly flip it over. He then draws it form the hat, and puts it not marry side face-down onto the table. Voila![/hide][/quote]

no ur wrong

Say to everyone that he will pick out the one he won't choose.

the guy should make a condition and announce to the crowd that he will do the opposite of whichever chit he picks up... :P

[hide]Tell the king he's an old bag that is very selfish to the whole kingdom to the crowd. Then, you have to take both slips out, and show them to the crowd to prove that the king is a selfish old bag![/hide]

I have a feeling it is from a Naseeruddin Hodja story coz i saw something similar in a book last month...not sure of the solution though.

[hide]ok...the man who wants to marry the princess has a pair of scissors up his sleeve, and while he's getting it out of his hat he carefully cuts off the "not" part. [/hide]


quadratics
number theory unsolved
number theory

Find the largest integer N such that [N/3] has three digits, all equal, and [N/3] = 1 + 2 + 3 + ... + n for some n

This one is coming from Spain. Guess which year.

Let $x = [\frac N 3] = a + 10a + 100a = 111a$ for some $a \in \{1, \cdots , 9 \}$.
Then $\frac {n(n+1)}2 = 111a = 3 \times 37 \times a$.
It follows that :
$n=0$ or $n = 36 \mod [37]$ and $n=0$ or $n=2  \mod [3]$. (1)

Moreover $n^2+n-222a=0.$
Solving this quadratic in $n$, we obtain $n = \frac {-1 + \sqrt {1+888a} } 2.$
Since $1 \leq a \leq 9$, we deduce that $14<n\leq 44$, so that $n=36$ from (1).
Thus $a=6$ and $x=666.$
Therefore $3 \times 666 \leq N < 3 \times 667$, and the desired maximum is $N=2000.$

Pierre.


geometry
3D geometry
floor function
number theory open
number theory

is it possible for
a) square
b) cube
of natural number (in decimal system) to start with arbitrarily long sequence of 1's ?

Why not?  Round the square root of $ 1111...$.

For the case of squares a number starting with $ n$ threes will start with $ n \minus{} 1$ ones.
For cubes you can use the cube of $ \lfloor{10^n \frac {\sqrt [3]{3}}{3}}\rfloor$

thanks very much.

what about generalisation (higher powers) ?


inequalities
integration
real analysis
real analysis unsolved

$f(x)$ is continuous and $\frac{f(x)}{x}$ is increasing.
Show that
$xf(x) \geq 2\int_0^x f(t)dt$ for $x\geq 0$.

It is easy to prove if $f$ is differentiable. But how to prove it without the assumption?

Just integrate the inequality $f(t)\le t\frac{f(x)}x$ with respect to $t$ from $0$ to $x$.


let $a,b,c$ be positive real numbers
find the great value of$ x,y$ such that $x,y>0 $and $x+y=2$ satisfied
$a/b+b/c+c/a>= 3x/2 +y(a/(b+c)+ b/(a+c)+c/(a+b))$

Did you mean that : find the largest value of the product of x and y ?

It is a general problem of the Japanese inequality. :) but I didn't know how to solve

oh iam sorry !!! i mean find the great value of y


Check out this site http://www.compadre.org/psrc/evals/olympiad.cfm

Can you post some problems of previous USAPhO??

maybe tell us the kinds of things that come up?? like topics, types of problems... etc.

And shobber, the link that Mna gave us have problems... go look it.

How does your teacher do that?  I imagine we have plenty of students here who would like to participate, but would have to tell their teachers to nominate them (and how to do it).

I believe there is an associated fee, but it is pretty much up to the teacher to nominate as many as want to take the test. Supposedly the nomination already places one as one of the top 1,000+ (since that is the number of individuals who take the first round exam). Effectively, the nomination supplants an AMC style test.

This is an excellent website for the physics olympiad: http://www.aapt.org/Contests/olympiad.cfm

Here is the link to the 2005 application:
http://www.aapt.org/olympiad2005/pdf/blk-whtmailer-Appl.pdf

Check out the team pages for different years also.

on that website, http://www.compadre.org/psrc/evals/olympiad.cfm, there are only the exams.  does anyone know if the solutions are up anywhere on the internet or if they're there and i don't see them?  i'd like to be able to check some of my answers.  thanks.

Ask you friends on the AOPS physics forums :)

MY friend John Schulman from great neck, NY is in Top 24, and he'll be taking TST thing soon.
I believe he is on AoPS also.

[quote="qweretyq"]But does anyone have tips on how to study for it?
Just practice, practice, practice the questions?
Has anyone here ever been a semifinalist?[/quote]
I do not live in USA thus i can never participate USAPhO, for as i know, if one wants to study Physics well, he/she must do Mountains of exercises...

[quote="qweretyq"]But does anyone have tips on how to study for it?
Just practice, practice, practice the questions?
Has anyone here ever been a semifinalist?[/quote]

I think Yi Sun is on this site somewhere.

arnav, arent you a semifinalist??? or havent you taken USAPhO before?


limit
trigonometry
integration
calculus
calculus computations

$ \lim_{n\to\infty} \sum_{i\equal{}1}^n \frac{i}{n^2}\sin\left(\frac{i^2\pi}{n^2}\right)$

$ \lim_{n\to\infty} \sum_{i \equal{} 1}^n \frac {i}{n^2}\sin\left(\frac {i^2\pi}{n^2}\right)$

$ \equal{} \lim_{n\to\infty}\frac {1}{n}\sum_{i \equal{} 1}^n \frac {i}{n}\sin \left(\frac {i}{n}\right)^2\pi$

$ \equal{} \int_0^1 x\sin (\pi x^2)\ dx$

$ \equal{} \frac {1}{2\pi}\left[ \minus{} \cos (\pi x^2)\right]_0^1$

$ \equal{} \frac {1}{\pi}$.

Dr Kent Merryfield would give another solution.

No, I would have said the same thing. I have nothing to add.


number theory
prime numbers
algebra open
algebra

Prove or disprove the following statement:
There exists infinitely many prime numbers $ p$ that $ \frac{p\minus{}1}{2}$ is also a prime number.  :wink:

That's an open problem. Primes $ p$ such that $ 2p\plus{}1$ is also prime are called Sophie-Germain primes and it is not known if there are infinitely many of them.

Hehehe, knew that'd be to easy (if that lemma worked it'd niecely solve one problem). Anyway thanks :wink:


prove that $gcd(x^{n}-1,x^{m}+1) \leq 2$ with both m and n are natural numbers and n is odd

oh , i'm sorry
x,m,n are three natural numbers and n is odd :D


function
algebra
domain
vector
real analysis
real analysis unsolved

Under the assumption of Zermelo's Axiom of choice, prove the existence of discontinous convex functions. And prove that these functions are unbounded in every interval.

This begs the question of how you're defining "convex function."

I would normally use this definition: $ \varphi$ is convex on an interval iff for all $ a<b$ in the domain of $ \varphi$ and for all $ \theta\in[0,1],$ then $ \varphi((1\minus{}\theta)a\plus{}\theta b)\le (1\minus{}\theta)\varphi(a)\plus{}\theta\varphi(b).$

Under that definition, a convex function is always continuous on the interior of its domain.

That you are even posing this question implies that you are using a different definition - perhaps this one, which I would call "midpoint convex":

$ \varphi$ is midpoint convex on an interval iff for all $ a<b,$ then $ \varphi\left(\frac{a\plus{}b}2\right)\le
\frac{\varphi(a)\plus{}\varphi(b)}2.$

Now, any function that is additive, meaning that $ \varphi(x\plus{}y)\equal{}\varphi(x)\plus{}\varphi(y),$ is midpoint convex. We can use Zorn's Lemma to find a Hamel basis for $ \mathbb{R}$ as a vector space over $ \mathbb{Q},$ then a cardinality argument to show that there are more additive functions than linear functions.

To finish the problem, you would need to prove this theorem: If $ \varphi$ is midpoint convex and there is any interval on which $ \varphi$ is bounded above, then $ \varphi$ is convex.

yes. I use your definition of midpoint convex as my ususal convex definition. From this point of view, just like you said we can find Hamel Basis for real numbers then using this the coeficient of the representation to form such function.
 :)


ratio
analytic geometry
geometry unsolved
geometry

Let $ABC$ be a triangle with vertices at lattice points. Suppose one of its sides in $\sqrt{n}$, where $n$ is square-free. Prove that $\frac{R}{r}$ is irraational . The symbols have usual meanings.

I came up with $\frac{R}{r}= \frac{2abc}{(b+c-a)(a+c-b)(a+b-c)}$ and I'm stuck with this. :(  Anyone?

we know that every triangle with lattice coordinates has integer area.So we have to prove $2abc\sum a$ irrational.Now if we prove that $\sum ab$ is irrational is over.


The display on a digital clock reads 6:38 . What will the clock display twenty-seven digit changes later?

7:01
1h,3-10m,23m

Is there a faster way than to just guess then count from there?
i.e. 6:38 -> 6:58 -> 7:00 -> 7:01


abstract algebra
function
group theory
superior algebra
superior algebra unsolved

These two problems are starred in Herstein, although they both seem quite easy. Could someone please comment on whether or not my proofs are correct?

12. If M is an irreducible R-module such that $ rm \neq 0$ for some $ r \in R$ and $ m \in M$, prove that any R-homomorphism T of M into M is either an isomorphism of M onto M or that $ T(m) \equal{} 0$ for every $ m \in M$.

14. If M is an irreducible R-module such that $ rm \neq 0$ for some $ r \in R$ and $ m \in M$, prove that E(M) is a division ring.

[hide="Note on format"] This is copied word-for-word from Herstein, except that I used modern left-side function notation (i.e. T(m) intead of mT). [/hide]

[hide="My Solutions"]
First we note that the kernel of any homomorphism is a submodule of M (pretty easy; see problem 4 of Herstein in that section). Also, the image of a homomorphism is a submodule of M, since the image must be a subgroup (under addition), and if T(m) is an element of the image, then for any $ r \in R$, we have rT(m)=T(rm) is also in the image.

For problem 12, since the kernel of T is a submodule, it must be either (0) or M. If the kernel is M, then we have that T(m)=0 for every $ m \in M$. If T(m) is not 0 for some $ m \in M$, then the kernel is (0), so it is an isomorphism, and since the image is a submodule not equal to (0), the image must be M, so the homomorphism is onto. (Comment: That seemed [i]quite[/i] easy, no?)

Now we use this to prove problem 14. It is easy to see that the identity mapping I on M is a homomorphism and IT=TI=T for any $ T \in E(M)$. Now, any nonzero $ T \in E(M)$ is, by problem 12, a bijection, so it has an inverse, $ T^{ \minus{} 1}$. For any $ a,b \in M$ and $ r \in R$, we have $ T^{ \minus{} 1} (a \plus{} b) \equal{} T^{ \minus{} 1}(T T^{ \minus{} 1} (a) \plus{} T T^{ \minus{} 1} (b)) \equal{} T^{ \minus{} 1} T (T^{ \minus{} 1} (a) \plus{} T^{ \minus{} 1} (b) ) \equal{} T^{ \minus{} 1} (a) \plus{} T^{ \minus{} 1} (b)$, and we have $ T^{ \minus{} 1} (r a) \equal{} T^{ \minus{} 1} r T T^{ \minus{} 1} (a) \equal{} T^{ \minus{} 1} T r T^{ \minus{} 1} (a) \equal{} r T^{ \minus{} 1} (a)$, so $ T^{ \minus{} 1} \in E(M)$. Thus T has an inverse in the ring, and therefore E(M) is a division ring.
[/hide]

Except for misspelling "injective homomorphism" as "isomorphism" I don't see any problems.


geometry
rectangle
trigonometry
geometric transformation

Merry Christmas April , your problem is very very nice (cute and easy) . Here is my solution

Let the circle with center the midpoint of AM intersects the LM at the point Q . Then $ \angle{AQM}\equal{}90$ and so we have to prove that A,Q, C are collinear . By the power point theorem we have

$ LC^2\equal{}LN\cdot LB\equal{}LQ\cdot LM$ so $ \frac{LC}{LQ}\equal{}\frac{LM}{LC}$ so the triangles LQC and LMC are similar so

$ \angle{LQC}\equal{}90$ and we are done  :D

[quote="April"]Let $ ABCD$ be a rectangle, and $ M$ an arbitrary point in the side $ BC$. Circles with diameters $ AM$, $ BC$ intersect at $ B$ and $ N$. $ BN$ intersects $ CD$ at $ L$, prove that $ ML$ is perpendicular to $ AC$.[/quote]

 Son, many thanks for your wishes ! Merry Cristmas to you , too, if you are a Cristian.

 Στους νέους φίλους του φορουμ θέλω να υπενθυμίσω ότι ο APRIL είναι  λάτρης της γεωμετρίας και πολύ καλός λύτης . Νομίζω ότι είναι στην τελευταία τάξη του Λυκείου και θέλει να σπουδάσει computer scienses !
[color=red][b]
 Son , we wish  you good luck at your University exams![/b][/color]

Babis

April thank you very much for writing in our subforum :)

I also wish to all [color=red]merry Christmas[/color]

And here is my solution  in your very nice problem (I hope it's correct because I am not so sure :blush: )

First of all, ABNM is inscribed in a circle so $ \angle{MAN}\equal{}{MBN}\equal{}90\minus{}BLC\equal{}LCN$ and so their sines are equal.
That means that:
$ \frac{NL}{NC}\equal{}\frac{NM}{AN}\Longleftrightarrow \frac{AN}{NC}\equal{}\frac{NM}{NL}$ and using also the fact that $ \angle{ANC}\equal{}MNL\equal{}90\plus{}MNC$ we can concluse that the triangles $ ANC, MNL$ are similar, so there is a spiral similarity which can take $ ANC$ into $ MNL$. The result follows by observing that $ NL\perp AN$ and $ MN\perp NC$ which means that the third pair must also be perpendicular :wink:

Thank you very much for your wishes and for your solutions, too (the bilot4's one is the same with mine, but you can make it better, I think, as below). I’m not a Christian, and Christmas is not a common holiday in my country, but its atmosphere has traveled into Vietnam, and I also felt somewhat this warm air. I have some Greek friends, they are so kind, and very, very good at mathematics, especially in geometry. I also see that this community is very active, very ebullient, so I want to post some problems and solutions there, as thanks to all of you. Wish you a monumental new year. Thank you.

We have $ \angle LCN \equal{} \angle CBN \equal{} \angle MAN$. It follows triangle $ NCL$ and $ NAM$ are directly similar, so that triangle $ NCA$ and $ NLM$ are directly similar, too. It means $ \angle NAC \equal{} \angle NML$, i.e. points $ A$, $ M$, $ E$, $ N$ are cyclic, the result follows.


[quote="amature fermat"]for any real number a,b,c,d,e
proove next inequality
130(a^2+b^2+c^2+d^2+e^2)+62(ac+bd+ce+da+eb)>=179(ab+bc+cd+de+ea).

sorry for my bad English...
and sorry for too easy problem....
waiting for short cut..... :blush:[/quote]

$ 2(a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2)\plus{}ac\plus{}bd\plus{}ce\plus{}da\plus{}eb\minus{}3(ab\plus{}bc\plus{}cd\plus{}de\plus{}ea)\equal{}\frac{1}{2}((a\plus{}c)^2\plus{}(b\plus{}d)^2\plus{}(c\plus{}e)^2\plus{}(d\plus{}a)^2\plus{}(e\plus{}b)^2)\plus{}\frac{3}{2}((a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}d)^2\plus{}(d\minus{}e)^2\plus{}(e\minus{}a)^2)\ge0$,

so $ 2(a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2)\plus{}ac\plus{}bd\plus{}ce\plus{}da\plus{}eb\ge 3(ab\plus{}bc\plus{}cd\plus{}de\plus{}ea)$ and $ 120(a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2)\plus{}60(ac\plus{}bd\plus{}ce\plus{}da\plus{}eb)\ge 180(ab\plus{}bc\plus{}cd\plus{}de\plus{}ea)$, so
$ 130(a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2)\plus{}62(ac\plus{}bd\plus{}ce\plus{}da\plus{}eb)\ge 179(ab\plus{}bc\plus{}cd\plus{}de\plus{}ea)$, because  $ 10(a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2)\plus{}2(ac\plus{}bd\plus{}ce\plus{}da\plus{}eb)\plus{}(ab\plus{}bc\plus{}cd\plus{}de\plus{}ea)\equal{}(a\plus{}c)^2\plus{}(b\plus{}d)^2\plus{}(c\plus{}e)^2\plus{}(e\plus{}a)^2\plus{}\frac{1}{2}((a\plus{}b)^2\plus{}(b\plus{}c)^2\plus{}(c\plus{}d)^2\plus{}(d\plus{}e)^2\plus{}(e\plus{}a)^2)\plus{}\frac{17}{2}(a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2)\ge 0$, done.

[quote="MTM"][quote="amature fermat"]for any real number a,b,c,d,e
proove next inequality
130(a^2+b^2+c^2+d^2+e^2)+62(ac+bd+ce+da+eb)>=179(ab+bc+cd+de+ea).

sorry for my bad English...
and sorry for too easy problem....
waiting for short cut..... :blush:[/quote]

$ 2(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2) \plus{} ac \plus{} bd \plus{} ce \plus{} da \plus{} eb \minus{} 3(ab \plus{} bc \plus{} cd \plus{} de \plus{} ea) \equal{} \frac {1}{2}((a \plus{} c)^2 \plus{} (b \plus{} d)^2 \plus{} (c \plus{} e)^2 \plus{} (d \plus{} a)^2 \plus{} (e \plus{} b)^2) \plus{} \frac {3}{2}((a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} d)^2 \plus{} (d \minus{} e)^2 \plus{} (e \minus{} a)^2)\ge0$,

so $ 2(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2) \plus{} ac \plus{} bd \plus{} ce \plus{} da \plus{} eb\ge 3(ab \plus{} bc \plus{} cd \plus{} de \plus{} ea)$ and $ 120(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2) \plus{} 60(ac \plus{} bd \plus{} ce \plus{} da \plus{} eb)\ge 180(ab \plus{} bc \plus{} cd \plus{} de \plus{} ea)$, so
$ 130(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2) \plus{} 62(ac \plus{} bd \plus{} ce \plus{} da \plus{} eb)\ge 179(ab \plus{} bc \plus{} cd \plus{} de \plus{} ea)$, because  $ 10(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2) \plus{} 2(ac \plus{} bd \plus{} ce \plus{} da \plus{} eb) \plus{} (ab \plus{} bc \plus{} cd \plus{} de \plus{} ea) \equal{} (a \plus{} c)^2 \plus{} (b \plus{} d)^2 \plus{} (c \plus{} e)^2 \plus{} (e \plus{} a)^2 \plus{} \frac {1}{2}((a \plus{} b)^2 \plus{} (b \plus{} c)^2 \plus{} (c \plus{} d)^2 \plus{} (d \plus{} e)^2 \plus{} (e \plus{} a)^2) \plus{} \frac {17}{2}(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2)\ge 0$, done.[/quote]

Well I made this problem from
cyclic sum of (a-5b+7c-13d+4e)^2


For $ a,b,c>0$, prove that

$ (bc\plus{}ca\plus{}ab)\left(\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}\plus{}\frac{1}{(a\plus{}b)^2}\right)\geq \frac{9}{4}\plus{}\frac{60(b\minus{}c)^2(c\minus{}a)^2(a\minus{}b)^2}{(a\plus{}b\plus{}c)^6}$.

[quote="Parkdoosung"]For $ a,b,c > 0$, prove that

$ (bc \plus{} ca \plus{} ab)\left(\frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \plus{} \frac {1}{(a \plus{} b)^2}\right)\geq \frac {9}{4} \plus{} \frac {60(b \minus{} c)^2(c \minus{} a)^2(a \minus{} b)^2}{(a \plus{} b \plus{} c)^6}$.[/quote]
It's not hard because we have
$ \sum_{cyc}\frac {ab}{(a \plus{} b)^2} \equal{} \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} 1/4 \minus{} \frac {(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{4(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2}$

We also have
 $ LHS\ge 9/4 \plus{}\frac{15(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2}{4(a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2}$ with $ a,b,c\ge 0$

It is also true that for $ a,b,c>0$,

$ (bc\plus{}ca\plus{}ab)\left(\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}\plus{}\frac{1}{(a\plus{}b)^2}\right)$
$ \geq\frac{9}{4}\plus{}\frac{(b\minus{}c)^2(c\minus{}a)^2(a\minus{}b)^2}{(b\plus{}c)^2(c\plus{}a)^2(a\plus{}b)^2}\left(\frac{15}{4}\plus{}\frac{(b\minus{}c)^2(c\minus{}a)^2(a\minus{}b)^2}{(bc\plus{}ca\plus{}ab)^3}\right)$.


Find all strictly increasing functions: $f: \mathbb{N}\rightarrow\mathbb{N}$ satify condition: $f\left(n+f(n)\right)=2f(n) \forall n\in\mathbb{N}$

Let $a_{0}=1$, $a_{n+1}=a_{n}+f(a_{n})$. Then $a_{0}<a_{1}<\dots$, and $f(a_{n+1})=2f(a_{n})$, so $f(a_{n+1})-a_{n+1}=f(a_{n})-a_{n}$. Hence $f(n)-n$ is the same for infinitely many $n$, so since the function is strictly increasing it must be $f(n)=n+k$ for some nonnegative integer $k$.

[quote]Hence $f(n)-n$ is the same for infinitely many $n$, so since the function is strictly increasing it must be $f(n)=n+k$[/quote]

Congratulation! [b]jkramar[/b]
But now i still haven't understanded this argument! Please help me buy explaining carefully[/b][/quote]

This is really not so hard.
$f$ strictly increasing $\Rightarrow f(n+1)\geq f(n)+1\Rightarrow f(n+1)-(n+1)\geq f(n)-n\Rightarrow f(n)-n$ is increasing.
But [b]jkramar[/b] has shown, that for all $n\in\mathbb{N}$, there exists $m\in\mathbb{N},m>n$ with $f(m)-m=f(1)-1$.
But $f(1)-1\leq f(n)-n\leq f(m)-m$, so $f(n)-n=f(1)-1\in\mathbb{N}_{0}$ constant for all $n\in\mathbb{N}$.

Now i really understand above solution. Thanks [b]jkramar[/b] and [b] olorin[/b] for show me


ok... its from FTW
if two random numbers $ a$ and $ b$
are chosen from the set $ (2,3,4,5,6)$ edit: WITHOUT replacement!!!
what is the probability that the fraction $ \frac {a}{b}$
is less than 1 and can be expressed as a terminating fraction.
If your solution is not fast, i will mark it as  :spam: .

[hide:too hard to the whole thing so here]any fraction with a denominator that has 3 or 5 as a factor is non-terminable [/hide]

If you threaten to mark our solutions as spam, nobody will post.

I find that very very snobby.

And if you have a question from FTW, review the problems after the game and click "Ask on Forum".  Also, this should be moved to FTW Problems Discussion.

sorry about my snobbish-ness.
i just wanted a quick solution.

[hide="Solution"]
The possible terminating fractions less than 1:

2/4
2/5
3/4
3/5
4/5

There are 5 such fractions, and $ \binom{5}{2}$ to select two numbers, thus our answer is $ \frac{1}{2}$.

[/hide]


A yogurt shop sells four flavors of yogurt and has six different
toppings. How many combinations of one flavor and two
different toppings are available?

4 different flavors.  6C2=$ \frac{6!}{4!2!}=15$ different combinations of 2 toppings.

$ 4*15=60$


geometry
parallelogram
geometry unsolved

Given triangle ABC, show that there is a triangle whose sides are equal in length and parallel to the medians of triangle ABC, and find its area in terms of [ABC].

I think that this is very simple:

Call D,E,F the midpoint of BC,CA,AB. Then parallel line from D to BE meet FE on G. Clearly $ \overline{FE}\equal{}\overline{EG}$ so AFCG is a parallelogram and so $ \overline{CF}\equal{}\overline{GA}$. Then $ \triangle ADG$ is the triangle that we was serching for.
Call $ P: GD \cap CA$, we have:

$ [ADG]\equal{}2[ADP]\equal{}\frac{3}{2} [ADC]\equal{}\frac{3}{4} [ABC]$


Let a, b be integers such that \frac{47}{151}<\frac{a}{b}<\frac{34}{109}, what is the minimum of b?
(A)58 (B)61 (C)77 (D)93

Please don't post the same problem in multiple forums.


find all $f:R\rightarrow R$such that:
$\forall x,y \in R$   $f(x+yf(x))=f(f(x))+xf(y)$

Let x=0, a=f(0) then f(ya)=f(a) for all real y.
If a is not 0, then f(x)=f(a) holds for all real x. It leads to f(a)=0, contradiction.
So a=0
Let y=0 then f(x)=f(f(x)) for all real x.
We have f(x+yf(x))=f(x)+xf(y)--------(*)
Let f(x) takes x, y=-1 then f(x)(f(-1)+1)=0 for all real x.
so f(x)=0 always holds or f(-1)=-1.
For the latter case, we get f(x-f(x))=f(x)-x.
If x=f(x) doesn't hold for some x=b. Let f(b)=c then f(b)=f(c)=c.
f(b-c)=c-b.Then let y=1 x=b-c we have f(b-c)=(b-c)f(1), f(1)=-1.
But x=y=1 into (*) this makes f(0)=-2 contradiction.
So f(x)=0 and f(x)=x are the solutions.


algebra polynomial trigonometry function

Find a polynomial $F(x)$ with leading coefficient 1 such that $F(\cos a)=\cos 7a$ for any angle a. What is the coefficient of the $x^{2}$ term of $F$?

[hide="Hint"]Since ${(\cos(a)+i\sin(a))}^{7}= \cos(7a)+i\sin(7a)$, I expanded the left hand side and equated the real parts of the equation to find that the polynomial was:
$x^{7}-21x^{5}(1-x^{2})+35x^{3}{(1-x^{2})}^{2}-7x{(1-x^{2})}^{3}$

I don't know what the coefficient of the $x^{2}$ term is, though.[/hide]

[quote="vishalarul"][hide="Hint"]Here's the polynomial:
$x^{7}-21x^{5}(1-x^{2})+35x^{3}{(1-x^{2})}^{2}-7x{(1-x^{2})}^{3}$

I don't know what the coefficient of the $x^{2}$ term is, though.[/hide][/quote]

can you post the solution as well?

$F(\cos a)=\cos 7a$
$F(-\cos a) =F(\cos (-a))=\cos 7(-a)=-\cos 7a$

So we have

$F(-x)=-F(x)$ (at least for $-1 < x < 1$)

i.e. $F$ is an odd function, so coefficient of $x^{2}$ is 0.


inequalities
function
integration
logarithms
limit
inequalities unsolved


hi all
my major is teaching mathematics but here in Turkey "maths" and "teaching maths" programs are not so different. I have taken a big sum of maths courses and actually I like it as a science not only as a topic I will teach.
I think to make a master in teaching mathematics but someone advised me to think the "applied mathematics". I don't know the universities and their policies about selecting graduate students for faculties.
What do you think? Do I have a chance?if yes what do you recommend me to do?
thanks...

Are you planning on applying to American or Turkish universities?  Because that would have a substantial effect on the answers to your questions.

Universities in U.S
actually I have the chance to be accepted by almost all the Turkish Universities...
but I want to be qualified, in Turkey I came to some degree and in U.S I can make the better...


...assuming you have perfect recommendation forms.

I'm in 8th grade.

I'm not sure if I can even get a single question on the admission test right... a friend suggested I go...  :|

Well, considering that it's already 5:00 CDT when you posted, I would say try for the C test. Seriously, there little time to write the solutions now (since I infer you haven't started yet, correct me if I'm wrong). I took about 4-5 hours, maybe more. (really, you might be able to do it in 2-3, but I tried to play it safe and rewrote almost all my answers)

[quote="korammath"]i believe that its not how many you get right, but the love of math you show when you do it, and how many you did. Apparently you would like to aim for 6-10.[/quote]
but not getting even 1 would be pretty bad so yeah definitely go do test c when its out

If you do poorly on the test does that mean you will probably be too stupid to learn anything at the camp anyway? Do the have a class for the stupid people like me?

Yeah, I'm doing the C test.


Let $n\geq 3$ be an integer. Show that for all the positive real numbers
$x_1,..., x_n$, with$x_1 \leq ... \leq x_n$, the following inequality holds :

$\frac{x_nx_1}{x_2}+\frac{x_1x_2}{x_3}+....+\frac{x_{n-1}x_n}{x_1}\geq x_1+x_2+....+x_n$

[quote]
using equality TCHEBYTCHEV
[/quote]

i think your solution is wrong because it is possible that :
$\frac{x_2}{x_3}\geq \frac{x_1}{x_2}$

I agree with toetoe, can anyone explain it?

[quote="mathmanman"][hide="Hint."]For any $0 \leq x \leq y$ and $0 < a \leq 1,$ we have :
$x+y \leq ax+\frac{y}{a}.$[/hide][/quote]
Yeah, I solved the problem in the same way, but I still want a nice solution :P


hi
prouve that : $ sin(\pi/(2n\plus{}1))>\equal{} 2/(2n\plus{}1)$$ (n \in N)$

[quote="The Zuton Force"][hide]Let $ x \equal{} \frac {\pi}{2n \plus{} 1} \in \left( 0 , \frac {\pi}{3} \right]$.

The equivalent inequality is $ \sin x \geq \frac {2}{\pi} x$. We can show it to be true over the interval $ x \in \left( 0, \frac {\pi}{2} \right)$.

Note that $ \frac {2}{\pi}x$ is the secant line to the graph of $ \sin x$ from $ (0,0)$ to $ \left( \frac {\pi}{2}, \; 1 \right)$.

Since $ \sin$ is concave over the interval, it must be higher than its secant line.[/hide][/quote]
can we prouve   $ \sin x \geq \frac {2}{\pi} x$ by using the classical way $ f(x)\equal{}sin(x)\minus{}(2x)/(\pi)$ ?????


Find the value of the first term in the geometric sequence: \[ \{a,b,c,32,64\}.\]

$ 32\times2 = 64$ so we know the common ration between consecutive terms. Now we can just figure out terms backwards. The set becomes $ \set{a, b, 16, 32, 64}$, then $ \set{a, 8, 16, 32, 64}$, then finally $ \set{\boxed{4}, 8, 16, 32, 64}$


Find positive number(s) $ n$ and $ a_1 ,a_2 ,....a_n$ such that $ a_1 \plus{} a_2 \plus{} a_3 ... \plus{} a_n \equal{} 1000$ and the product $ a_1 a_2 a_3 ....a_n$ is as large as possible.

[hide]Answer is independent of n ofcourse[/hide]

Yes , the question doesnt say that but the answer does.

Please "Click to reveal hidden content" in my question.

I think by A.m.-G.m. inequality we can prove that 1000/n > (a1...................)^(1/n)

So we should find the greatest n
such that 1000 > (n+1)^n+1 / n^(n)

 :arrow: It can be proved using limits that (1+1/n)^n as n :arrow: infinity goes to e.
So what should be done next . plz help
 :blush:

befuddlers has come close.

By am-gm obviously, we must consider the function, $ f(x) \equal{} (\frac {1000}{x})^x$. It takes a maximum at $ x \equal{} \frac {1000}{e} \approx 368$.

So, $ a \equal{} e$

If a in integers. then, $ a \equal{} \frac {1000}{250} \equal{} 4$


Okay, this is how it works.
The person who posts before you, you must choose an image to pwn it with. When you do that, you post it.
Attachment or Direct Link. W/e.
For example, if someone posts an orange, then the person after that posts a juicer, then the person after that posts a short circuit... etc.
You just keep on going and going.
[b]Teh Rulez (we must have them)[/b]
Absolutely [b]NO[/b] Chuck Norris. Just no. If you DO post Chuck, then the person after you will just ignore it and pwn the image before it.
No repeated images by the same user (like, don't keep posting a 'hammer' after every time you post)

If there's something else, just say.
Previous 'Image Pwning' games;
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=291960 - Mewto55555
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=249868 - 1=2

So like, I'll start.
[img]http://redthnapper.files.wordpress.com/2009/02/burrito.jpg[/img]
Burritos FTW.

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

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

[img]http://i41.photobucket.com/albums/e284/tattila/godzilla-facepalm-godzilla-facepalm.jpg[/img]
I don't know if it works, its blocked  :mad:

[img]http://www.edgeofexistence.org/images/dead_gorilla.jpg[/img]

How was the facepalm beaten by a gorilla?

[img]http://sharetv.org/images/my_little_pony_tales-show.jpg[/img]
I don't even have to explain why this pwns.

[img]http://www.abbeyarchery.com.au/mm5/graphics/Bear_Lights_Out_DarkHorse09_med.jpg[/img]

Let's go a horse-huntin'!

Once Again...

[color=red]Mod notes:
[quote="meewhee009"]No repeated images by the same user[/quote][/color]

Oh Yeah?
This is my favorite:
[img]http://www.sg.hu/kep/2006_07/0713saw3.jpg[/img]

I was hoping not to have to do this...but now I have to:
[img]http://www.smashbros.com/en_us/gamemode/modea/images/modea17/modea17_080410a-l.jpg[/img]

Imma gonna have to kill you now...
[img]http://www.acepolls.com/uploads/0046/1495/CHUCKY.jpg[/img]

I lost the game.

[img]http://members.gamedev.net/trapperzoid/images/projects/penguin/060712-penguin.png[/img]

Evil shall destroy you.
[img]http://www.exclaim.ca/images/up-Pinhead_1.jpg[/img]

[img]http://charlespaolino.files.wordpress.com/2009/08/abbott-and-costello.jpg[/img]

For those of you that don't understand, watch Abbott and Costello Meet Frankenstein.

[quote="Richard_Min"]Hopefully, you realize where I got this from.
[img]http://farm3.static.flickr.com/2452/3843993804_05ac37f9c7.jpg[/img][/quote]

hmm want me to give you a membership?

[img]http://the-adventurers-club.typepad.com/photos/uncategorized/michael_myers_halloween.jpg[/img]

MICHAEL MYERS!


Because the bonding isn't really diene.

The ring bond is fundementially different in many ways.  The electrons move very easily across all of the carbon orbitals in the ring structure, or beyond if it is set up correctly.

[quote="jrwkc"]I'll Doubt if anyone knows this...

How come Benzene does not follow the regular woodward rules for dienes... its calculated absorption is  283 where the expermintal calculation is 198 such a huge difference[/quote]

Google aromaticity.

A really good explanation can be found at http://www.chemguide.co.uk/basicorg/bonding/benzene1.html


Consider the unit square [tex]S=[0,1]\times[0,1][/tex]. Pick countable subsets  H and V of [0,1]. Does there exist [tex]E\subset S[/tex] so that for any [tex]h\not\in H, v\not\in V[/tex], [tex](v\times[0,1])\setminus E[/tex] and [tex]([0,1]\times h)\cap E[/tex] are each countable?

An even harder problem: Is Simon capable of posing a problem P such that a non-negligible fraction of AoPS members can understand P without doing at least an hour of research on other sites? ;) But it looks (deceptively?) simple, now that I know what it means.

I think the problem's written wrong, just because of the "is and." Is the "is" not supposed to be there, or is there supposed to be something after it?

Oops, you're right. I'll fix it. Thanks.

Awesome...a yes/no question.  Um....I say no.

So, a very basic first step:  we can find a bijection between any two countable sets.  This bijection can be extended to a bijection from [0, 1] to [0, 1] if both countable sets are subsets of [0, 1].  Thus, without loss of generality, we may assume H and V to be any particular countable sets we choose, such as the rationals, or the numbers 1/n for integral n, or whatever.

Contrary to Magnara, I'm going to guess "yes," since uncountable*uncountable = countable*uncountable, give or take.  I'm trying to construct an example.

We first note that the following simplified question (which is the case [tex]H=V=\emptyset[/tex]) is equivalent to Simon's question:

Consider the unit square [tex]S=[0,1]\times[0,1][/tex]. Does there exist [tex]E\subset S[/tex] so that for any [tex]h, v\in [0,1][/tex], [tex](v\times[0,1])\setminus E[/tex] and [tex]([0,1]\times h)\cap E[/tex] are each countable?

To see this, first note that one can assume WLOG in Simon's question that [tex]E\subset ([0,1]\setminus V)\times([0,1]\setminus H)[/tex] by replacing [tex]E[/tex] with its intersection with this set, as this only changes the cardinality of the intersection of [tex]E[/tex] with a "row" ([tex][0,1]\times h[/tex]) or "column" ([tex]v\times [0,1][/tex]) by a countable amount. It then remains only to take bijections from [tex][0,1]\setminus H[/tex] and [tex][0,1]\setminus V[/tex] to [tex][0,1][/tex].

The question now is slightly simpler, and we realize now that the problem is undecidable:

Theorem: The answer to Simon's question is yes iff the continuum hypothesis holds.

Proof: First, suppose that the continuum hypothesis holds. Then [tex]c = \aleph_1[/tex], which is the cardinality of the set [tex]\omega_1[/tex] of countable ordinals, so there exists a bijection [tex]f \colon\, [0,1] \to \omega_1[/tex]. Let [tex]E = \{(a,b)\in S | f(a) < f(b)\}[/tex]. It is easily verified that [tex]E[/tex] satisfies the necessary conditions in the simplified question.
Next, suppose that [tex]E[/tex] satisfies the conditions in the simplied version of Simon's question. Since [tex][0,1][/tex] is uncountable, it has a subset [tex]A[/tex] of cardinality [tex]\aleph_1[/tex]. Now, each column in the "rectangle" [tex][0,1]\times A[/tex] must contain at least one element of [tex]E[/tex] (because [tex]E[/tex] is only missing a countable number of elements in each column), so card[tex](([0,1]\times A)\cap E) \ge c[/tex]. However, each row in this rectangle contains only a countable number of elements of [tex]E[/tex], so card[tex](([0,1]\times A)\cap E) \le \aleph_0 \cdot \aleph_1 = \aleph_1[/tex]. Thus [tex]\aleph_1 \ge c[/tex], so in fact [tex]\aleph_1 = c[/tex], the continuum hypothesis.

Since the continuum hypothesis is undecidable, so is this question.

[quote="ComplexZeta"]Very good! I'm impressed.[/quote]

He IS an IMO gold medalist, after all.

I know that, but I still find it impressive.


trigonometry
integration
algebra
partial fractions
calculus
calculus computations

$\int{\frac{{x^{2}}}{1+x^{2}}dx}$

So apparently, we're supposed to use partial fractions and an inverse trig substitution. However, when my teacher explained this, I wasn't able to follow how he got the answer of $x-\tan^{-1}(x)$

I do know that $\int{\frac{{du}}{a^{2}+u^{2}}}$ = $\frac{1}{a}\arctan{\frac{u}{a}}+C$

So I was thinking that we let $a = 1$ and $u = x$, which means $du = dx$, but I'm not even sure if this is right or how to proceed. Can anyone show me the full solution so I can figure out how to do this?

Noting that $1 = \frac{1+x^{2}}{1+x^{2}}$ you can say
$\frac{x^{2}}{1+x^{2}}= 1-\frac{1}{1+x^{2}}$
and integrate each term separately. It looks like you can finish from here :)

Thanks so much for that hint - I understand how to do it from here!


NaCl + H2SO4 + MnO2 → Na2SO4 + MnSO4 + H2O + Cl2

In this reaction, the reducing agent is:

What's the best way to work with problems like these: where you are asked to find the substance that is oxidizing or the substance that is reducting?

Place oxidation numbers on each substance.
Balance charges and ions!
It is easier to look at the net ionic formula imo.
Also, split the reaction into two half reactions.
Yay nontrivial reaction rules!

2 Cl- + MnO2 + 4 H+ -> Mn2+ + 2 H2O + Cl2

2 Cl- -> Cl2 + 2 e-
MnO2 + 4H+ + 2 e- -> Mn2+ + 2 H2O

OIL - oxidation = a loss of electrons
RIG - reduction = gaining electrons
And, a redox reaction is a transfer of electrons

So, the half reaction in which electrons are gained is the reduction reaction, and the product being reduced is MnO2.

There is absolutely no need to split into the half reactions. Simply calculate and write the oxidation numbers: then a decrease (or a [b]reduction[/b]) in oxidation number corresponds to a reduction, and increase in it corresponds to an oxidation. Of course, when you are experienced enough you can do all this in your head in a matter of seconds. Looking for you reaction its obvious that the ions/atoms Na+, SO42-, H and O remain unchanged. Here, the chloride ion gets oxidized, so $ \ce{Cl \minus{} }$ is the reducing agent.

Some remarks:

[quote="Jdlkem"]And, a redox reaction is a transfer of electrons[/quote]

More correctly, a redox reaction is a reaction that [i][b]can be viewed[/b][/i] as a net transfer of electrons between atoms, which we can confortably keep track with the concept of oxidation numbers. However, most redox reactions are quite different from (Brønsted) acid-base reactions - where we have an actual transfer of protons -, since in them we don't actually have electrons "swimming around" in the solution from one atom into another.

[quote="Jdlkem"]and the product being reduced is MnO2.[/quote]

MnO2 is a reagent, not a product.

In the mean time, here's a redox reaction that I like to propose to my students:

write the half reactions corresponding to the net reaction

$ \ce{As2S3}(s) \plus{} \ce{NO3\minus{}}(aq) \longrightarrow \ce{H3AsO4}(aq) \plus{} \ce{S}(s) \plus{} \ce{NO}(g)$

Assuming that this is in acid,

4H+ +NO3-  +3e- ---> NO + 2H2O

As2S3 + 8H2O ---> 2H3AsO4 +3S+10H+ +10e-

Multiply top by 10, and bottom by 3, then add the reactions, cancel out repeats algebraically.

Your answer is not complete. Check again.

Pay attention to this:

[quote="Carcul"]Simply calculate and write the oxidation numbers: then a decrease (or a reduction) in oxidation number corresponds to a reduction, and increase in it corresponds to an oxidation.[/quote]

 :juggle:

Isn't Cl2 the reucing agent as it gets oxidised , Mn4+ gets reduced to Mn2+ . These alone are undergoing change. All others are spectator ions . Am I right?

The first one I posted and break up:

As2S3 + 8H2O ---> 2H3AsO4 +3S+10H+ +10e-

Into :

2As+3 +8H2O--> 2H3AsO4+10H+ +4e-

And

3S-2 --> 3S + 6e-

The reducing/oxidizing agent is always one of the reagents, never a product.

Assuming acidic solution:

Reduction:
$ 3\mathrm{e^\minus{}}\plus{}4\mathrm{H^\plus{}}\plus{}\mathrm{NO_3^\minus{}}\rightarrow\mathrm{NO}\plus{}2\mathrm{H_2O}$

Oxidation #1:
$ 8\mathrm{H_2O}\plus{}\mathrm{As_2S_3}\rightarrow2\mathrm{H_3AsO_4}\plus{}3\mathrm{S^{2\minus{}}}\plus{}10\mathrm{H^\plus{}}\plus{}4\mathrm{e^\minus{}}$

Oxidation #2:
$ \mathrm{S^{2\minus{}}}\rightarrow\mathrm{S}\plus{}2\mathrm{e^\minus{}}$

Depending on the pH, one reaction will preside first. If there is enough oxidizing agent, then the second reaction will run. Balance each of the oxidations with the reduction to get two redox reactions. Then add the two resulting redox reactions to get the entire balanced reaction.

Yes, Feynman_wannabe, you are right, but your last equation could be divided by 3.

Jdlkem, in your first oxidation it is supposed not to include the sulphur atoms.

That would be fine, but As2S3 is, as far as i know, insoluble in water (it could very well be soluble though...). So, it would make more sense to oxidize one ion at a time from the precipitate: either arsenic or sulfide would be oxidized from the precipitate first.

1) What does solubility have to do with redox reactions?

2) Just because to write half reactions you have to artificially separate ions, that doesn't mean they are separated in reality. Don't forget, half reactions are just a confortable way of keep track of what's going on, and not a mirror of what really does happen.


I've been told that research experience is one of the most important tools you can have when applying for graduate school. But how do you obtain this experience? I've heard that US citizens get the experience by attending to REU programs. But how do the other ones not accepted to these programs get their experience? How do international students get the experience if they don't have programs as REU in their countries?

When I hear the words "research experience" I think of students helping a professor researching in a field, but then I ask myself "What does an undergraduate have to offer a professor in his research field?".
I've also been told that the experience is obtained when doing the master's thesis, but there has got to be more to that.

Could anyone clear this up for me?
Thank you.

[quote="kahsi"]When I hear the words "research experience" I think of students helping a professor researching in a field, but then I ask myself "What does an undergraduate have to offer a professor in his research field?"[/quote]

Professors typically have far more questions to consider than they have time to solve.  Many professors are also willing to help students out by setting them up with a problem and consulting with them as research progresses.

Is that the most common way students get their experience? What other ways do exist?

Most big (read: well-funded) colleges offer research opportunities to their undergrads.  The scope of these opportunities differs from school to school, but if you're aggressive enough you can almost certainly sign on to a research program being conducted by a professor at your school.  JBL has it right when he explains how you could be useful - professors need to do more things (often tedious work) that they would rather not do themselves or give to their grad students.  While the work you do initially may not be particularly interesting, often you can easily spin that research off into a new line of questioning that leads to interesting results.  The professor or grad student you're working under will usually be pleased to help you explore these new options, and you may even eventually get a paper published off of it.

I'm sure there are other research opportunities for college students, mainly during the summer, but I haven't begun looking into those yet.  I do know that it's very common to get research experience with the college you're currently at.

You as the student need to pursue internships.  Don't wait for a professor to ask you or for some program to match you up with a professor.  You should know what kind of stuff you want to research.  So look for professors in that area, ask them if they could use an extra person in the lab.  You'll also want to write a resume as many professors will want to see this before they pick you up.  If you can't find any profs at your school, try looking at other schools.  If you're qualified, it's definitely possible they'll take you.

Also, I think it's harder for undergrads to get research opportunities in math areas.  The more applied the math is, the more likely the professor will be able to put you to work.  In more theoretical math, there's not all that much you can do. [edit: I defer to JBL (see next post) who has more experience about this kind of thing :D ]

So my main advice is to talk to your professors.  Even if they don't need you they might know someone who does or they can help you started on the path to finding a research position.

I agree with almost everything joml just said; the point about having a resume and sending it to professors when you first contact them is definitely key: it lets them know that you're somewhat serious about this, and gives them a quick chance to see that you actually know what you're doing.  However, I disagree with the following statement:

[quote="joml88"]The more applied the math is, the more likely the professor will be able to put you to work.  In more theoretical math, there's not all that much you can do.[/quote]

More applied work is typically associated with a lab and a number of ongoing projects, some of which could almost certainly use some extra hands and brains.  Theoretical work doesn't have a lab associated with it, but as a result it is cheaper and easier to oversee.  A couple of hour-long conversations with a professor (with probably a trip to the library in between) is often enough to narrow in on a problem, and another hour every other week or so is all that is necessary to review what you've done and offer suggestions and possible directions.  This is a really minimal drain on a professor's time, and many professors will probably be willing to help you.  (Although, many probably will not.  Don't worry too much about rejections -- often, they come with suggestions for other professors who might be available.)  Hitting up younger professors may be more profitable than older professors because they are less likely to have many graduate students, and they are generally less intimidating, but "anyone who will help" is good enough.  ;-)

I'm pleased with all the answers. Thank you!

I'm currently doing some local research with a professor about socioeconomics. Professors don't have a lot of time, and they don't have 9000 clones of themselves. They have a little bit of time and a lot of work to do.

Don't doubt the idea that professor will think nothing of you. They'd be glad to find a way to put you into some type of research project you could handle. Sometimes they'll train you, as I have been trained by various professors.

I think you need to start asking professors questions. Sometimes they'll direct you to local industries around the city or town. You can do volunteer work at times, while other times you need some recommendations.

It's like applying for a job: Keep asking who's hiring.

There is good advice here: have confidence and look for researchers who may need your help.


Plastic bags are recycled at the local grocery store. For every 5 bags that are recycled by customers, 3 new bags can be made. How many new bags can be made if 800 bags are returned?

Slightly ambiguous because you can make more plastic bags.

For every $ 5$ there are $ 3$ new bags, there are $ \frac{800}{5}$ such sets of $ 5$ bags.

$ \frac{800 \cdot 3}{5}\equal{}\boxed{480}$

[hide="hopefully this is the right answer"]5bags=3 new bags
800bags =X new bags

800/5==160*3=480

so the answer is 480 or $\boxed{480}$[/hide]

[b][i][u]please rate my post thanks ![/u][/i][/b]

please be fair though [= thanks!

 :welcomeani:  :welcomeani:  :welcomeani:  :welcomeani:


geometry
inradius
incenter
perimeter
ratio
trigonometry
area of a triangle

For an arbitrary triangle $ UVW$, let $ r\left(UVW\right)$ mean the inradius (i. e., the radius of the inscribed circle) of triangle $ UVW$.

Let $ X$, $ Y$, $ Z$ be three points on the sides $ BC$, $ CA$, $ AB$ of a triangle $ ABC$ such that $ r\left(AYZ\right) \equal{} r\left(BZX\right) \equal{} r\left(CXY\right)$. Denote $ r\left(AYZ\right) \equal{} r\left(BZX\right) \equal{} r\left(CXY\right) \equal{} u$. Prove that $ r\left(XYZ\right) \plus{} u \equal{} r\left(ABC\right)$.

I have a pretty simple solution, but it will look too long and complicated for my taste if I try to explain everything properly :). There are also lots of notations, but draw a picture and everything will become clear.

Denote by $A',B',C'$ the incenters of triangles $AYZ,BZX,CXY$.

The desired conclusion is equivalent to showing that $XYZ,A'B'C'$ have equal inradii. Now, since all the distances $(A',YZ),(B',ZX),(C',XY),(X,B'C'),(Y,C'A'),(Z,A'B')$ are equal, we can easily see that the fact that $A'B'C'$ and $XYZ$ have equal areas is equivalent to the fact that they have equal perimeters (I won't write this in detail; it requires defining something like six more points, and I'm not willing to do this :)). This means that if we show that the two triangles have equal perimeters, we'll be done, since the inradius is the ratio between the area and the semiperimeter.

Let $A_B,A_C$ be the points where the circles $(B'),(C')$ touch $BC$, and let $X_B,X_C$ be the points where $XZ,XY$ touch the circles $(B'),(C')$, and define $B_C,B_A,Y_C,Y_A,C_A,C_B,Z_A,Z_B$ similarly. We have $XX_B+XX_C=XA_B+XA_C,YY_C+YY_A=YB_C+YB_A$, and $ZZ_A+ZZ_B=ZC_A+ZC_B$, so the perimeter of $XYZ$, which equals $XX_B+XX_C+YY_C+YY_A+ZZ_A+ZZ_B$, is also equal to $A_BA_C+B_CB_A+C_AC_B$, which, in turn, equals the perimeter of $A'B'C'$ because $A_BA_C=B'C',B_CB_A=C'A',C_AC_B=A'B'$.

P.S.

Sorry it looks so ugly. I really liked the argument before having to actually define all these points, when all I had to do was stare at my dynamic sketch :).

This problem is exactly same as CMO 1989\4.

This problem is very nice !
  The key of this problem is to compute $ S(O_1YO_3XO_2Z) $ by two ways . After that , prove that : $ p(XYZ) = p(O_1O_2O_3) $ and $ S(XYZ) = S(O_1O_2O_3) $ , where $ O_1 $ is the incenter of triangle $ AYZ $  :)

Note that this problem was posted by Romano on Nov 24, 2004 on 4:11 pm in the Geometry Proposed&Own Section under the subject "inscribed circle" and subsequently lost during the hacker attack.

In fact, this is an extremely beautiful problem. I remember seeing it in one of the newer issues of the American Mathematical Monthly, together with a very short solution - my own solution is not so short, so I am wondering whether anybody has a better one.

Here is my solution:

For any triangle UVW, apart from the inradius r (UVW), we denote by $ s\left(UVW\right) \equal{} \frac {VW \plus{} WU \plus{} UV}{2}$ the semiperimeter of triangle UVW, and by $ \left|UVW\right|$ the non-directed area of triangle UVW. Since the area of a triangle equals inradius $ \cdot$ semiperimeter, we have $ \left|UVW\right| \equal{} r\left(UVW\right)\cdot s\left(UVW\right)$ for any triangle UVW.

Particularly, for the triangle ABC, we get $ \left|ABC\right| \equal{} r\left(ABC\right)\cdot s\left(ABC\right)$, while for the triangle AYZ, we obtain $ \left|AYZ\right| \equal{} r\left(AYZ\right)\cdot s\left(AYZ\right) \equal{} u\cdot s\left(AYZ\right)$ and similarly, for the triangles BZX and CXY, we get $ \left|BZX\right| \equal{} u\cdot s\left(BZX\right)$ and $ \left|CXY\right| \equal{} u\cdot s\left(CXY\right)$; finally, for the triangle XYZ, we get $ \left|XYZ\right| \equal{} r\left(XYZ\right)\cdot s\left(XYZ\right)$. But

$ s\left(AYZ\right) \plus{} s\left(BZX\right) \plus{} s\left(CXY\right) \equal{} \frac {AY \plus{} AZ \plus{} YZ}{2} \plus{} \frac {BZ \plus{} BX \plus{} ZX}{2} \plus{} \frac {CX \plus{} CY \plus{} XY}{2}$
$ \equal{} \frac {\left(BX \plus{} CX\right) \plus{} \left(CY \plus{} AY\right) \plus{} \left(AZ \plus{} BZ\right)}{2} \plus{} \frac {YZ \plus{} ZX \plus{} XY}{2}$
$ \equal{} \frac {BC \plus{} CA \plus{} AB}{2} \plus{} \frac {YZ \plus{} ZX \plus{} XY}{2} \equal{} s\left(ABC\right) \plus{} s\left(XYZ\right)$,

and thus

$ r\left(ABC\right)\cdot s\left(ABC\right) \equal{} \left|ABC\right| \equal{} \left|XYZ\right| \plus{} \left(\left|AYZ\right| \plus{} \left|BZX\right| \plus{} \left|CXY\right|\right)$
$ \equal{} r\left(XYZ\right)\cdot s\left(XYZ\right) \plus{} \left(u\cdot s\left(AYZ\right) \plus{} u\cdot s\left(BZX\right) \plus{} u\cdot s\left(CXY\right)\right)$
$ \equal{} r\left(XYZ\right)\cdot s\left(XYZ\right) \plus{} u\cdot\left(s\left(AYZ\right) \plus{} s\left(BZX\right) \plus{} s\left(CXY\right)\right)$
$ \equal{} r\left(XYZ\right)\cdot s\left(XYZ\right) \plus{} u\cdot\left(s\left(ABC\right) \plus{} s\left(XYZ\right)\right)$
$ \equal{} u\cdot s\left(ABC\right) \plus{} \left(r\left(XYZ\right) \plus{} u\right)\cdot s\left(XYZ\right)$.

On the other hand, by one of the half-angle formulas (actually, the $ \tan\frac {A}{2} \equal{} \frac {r}{s \minus{} a}$ one), in triangle ABC, we have $ \tan\frac {\measuredangle CAB}{2} \equal{} \frac {r\left(ABC\right)}{s\left(ABC\right) \minus{} BC}$, and in triangle AYZ, we have $ \tan\frac {\measuredangle YAZ}{2} \equal{} \frac {r\left(AYZ\right)}{s\left(AYZ\right) \minus{} YZ}$. Since < CAB = < YAZ, the left hand sides of these two equations are equal, and thus we get $ \frac {r\left(ABC\right)}{s\left(ABC\right) \minus{} BC} \equal{} \frac {r\left(AYZ\right)}{s\left(AYZ\right) \minus{} YZ}$. In other words, $ \frac {r\left(ABC\right)}{r\left(AYZ\right)} \equal{} \frac {s\left(ABC\right) \minus{} BC}{s\left(AYZ\right) \minus{} YZ}$. Since r (AYZ) = u, this becomes

$ \frac {r\left(ABC\right)}{u} \equal{} \frac {s\left(ABC\right) \minus{} BC}{s\left(AYZ\right) \minus{} YZ}$.

Similarly, we find

$ \frac {r\left(ABC\right)}{u} \equal{} \frac {s\left(ABC\right) \minus{} CA}{s\left(BZX\right) \minus{} ZX}$;
$ \frac {r\left(ABC\right)}{u} \equal{} \frac {s\left(ABC\right) \minus{} AB}{s\left(CXY\right) \minus{} XY}$.

Now, by a well-known rule from algebra, if t, $ u_1$, $ u_2$, $ u_3$, $ v_1$, $ v_2$, $ v_3$ are seven reals such that $ t \equal{} \frac {u_1}{v_1}$, $ t \equal{} \frac {u_2}{v_2}$ and $ t \equal{} \frac {u_3}{v_3}$, then $ t \equal{} \frac {u_1 \plus{} u_2 \plus{} u_3}{v_1 \plus{} v_2 \plus{} v_3}$. Thus,

$ \frac {r\left(ABC\right)}{u} \equal{} \frac {\left(s\left(ABC\right) \minus{} BC\right) \plus{} \left(s\left(ABC\right) \minus{} CA\right) \plus{} \left(s\left(ABC\right) \minus{} AB\right)}{\left(s\left(AYZ\right) \minus{} YZ\right) \plus{} \left(s\left(BZX\right) \minus{} ZX\right) \plus{} \left(s\left(CXY\right) \minus{} XY\right)}$.

But

$ \left(s\left(ABC\right) \minus{} BC\right) \plus{} \left(s\left(ABC\right) \minus{} CA\right) \plus{} \left(s\left(ABC\right) \minus{} AB\right)$
$ \equal{} 3s\left(ABC\right) \minus{} \left(BC \plus{} CA \plus{} AB\right) \equal{} 3s\left(ABC\right) \minus{} 2s\left(ABC\right) \equal{} s\left(ABC\right)$,

and

$ \left(s\left(AYZ\right) \minus{} YZ\right) \plus{} \left(s\left(BZX\right) \minus{} ZX\right) \plus{} \left(s\left(CXY\right) \minus{} XY\right)$
$ \equal{} \left(\frac {AY \plus{} AZ \plus{} YZ}{2} \minus{} YZ\right) \plus{} \left(\frac {BZ \plus{} BX \plus{} ZX}{2} \minus{} ZX\right) \plus{} \left(\frac {CX \plus{} CY \plus{} XY}{2} \minus{} XY\right)$
$ \equal{} \frac {\left(BX \plus{} CX\right) \plus{} \left(CY \plus{} AY\right) \plus{} \left(AZ \plus{} BZ\right)}{2} \minus{} \frac {YZ \plus{} ZX \plus{} XY}{2}$
$ \equal{} \frac {BC \plus{} CA \plus{} AB}{2} \minus{} \frac {YZ \plus{} ZX \plus{} XY}{2} \equal{} s\left(ABC\right) \minus{} s\left(XYZ\right)$,

so this becomes $ \frac {r\left(ABC\right)}{u} \equal{} \frac {s\left(ABC\right)}{s\left(ABC\right) \minus{} s\left(XYZ\right)}$, and thus $ r\left(ABC\right)\cdot\left(s\left(ABC\right) \minus{} s\left(XYZ\right)\right) \equal{} u\cdot s\left(ABC\right)$. In other words, $ r\left(ABC\right)\cdot s\left(ABC\right) \minus{} r\left(ABC\right)\cdot s\left(XYZ\right) \equal{} u\cdot s\left(ABC\right)$, thus $ r\left(ABC\right)\cdot s\left(ABC\right) \equal{} u\cdot s\left(ABC\right) \plus{} r\left(ABC\right)\cdot s\left(XYZ\right)$. Comparing this with $ r\left(ABC\right)\cdot s\left(ABC\right) \equal{} u\cdot s\left(ABC\right) \plus{} \left(r\left(XYZ\right) \plus{} u\right)\cdot s\left(XYZ\right)$, we obtain $ r\left(ABC\right)\cdot s\left(XYZ\right) \equal{} \left(r\left(XYZ\right) \plus{} u\right)\cdot s\left(XYZ\right)$, and thus r (ABC) = r (XYZ) + u. And the proof is complete.

Another consequence of the above is the following: Transform $ \frac {r\left(ABC\right)}{u} \equal{} \frac {s\left(ABC\right)}{s\left(ABC\right) \minus{} s\left(XYZ\right)}$ into $ \frac {u}{r\left(ABC\right)} \equal{} \frac {s\left(ABC\right) \minus{} s\left(XYZ\right)}{s\left(ABC\right)}$. Now, since r (ABC) = r (XYZ) + u, we have r (ABC) - u = r (XYZ); thus,

$ \frac {r\left(XYZ\right)}{r\left(ABC\right)} \equal{} \frac {r\left(ABC\right) \minus{} u}{r\left(ABC\right)} \equal{} 1 \minus{} \frac {u}{r\left(ABC\right)} \equal{} 1 \minus{} \frac {s\left(ABC\right) \minus{} s\left(XYZ\right)}{s\left(ABC\right)}$
$ \equal{} \frac {s\left(XYZ\right)}{s\left(ABC\right)}$.

Equivalently, $ \frac {r\left(XYZ\right)}{s\left(XYZ\right)} \equal{} \frac {r\left(ABC\right)}{s\left(ABC\right)}$.

  Darij


Have you done for n=2 yet,Nickolas? When n=2,we have:
a1+a2-2 \sqrt (a1*a2) \geq  \sqrt a1- \sqrt a2
<=>( \sqrt a1- \sqrt a2)( \sqrt a1- \sqrt a2-1) \geq 0.
Oh la la,is it allways right???? :?

Nobody  posts  here  April  Fool's  jokes  here .

This is not a joke, but also it's not Kvant problem. As far as I know, the right inequality is with ( \sqrt a_1- \sqrt a_n) 2  instead of  \sqrt a_1- \sqrt a_n. Anyway, I have already solved this problem. As far as I remember, Kuba posted this problem in the analysis section, but I don't remember exactly. Anyway, the idea is that it reduces immediately to AM-GM.

You  can also  proove  that
(Σai)/n - (Πai)^(1/n)  \geq m/2 , where  m=min{ ( \sqrt ai -  \sqrt aj) 2 }
Thanks  anyway!


Find all triples $ (x,y,z)$ of integers, solutions to the system of equations
$ xy\plus{}z\equal{}100$
$ x\plus{}yz\equal{}101$

[hide="hint"]For exaample:
(2)-(1):
$ (x\minus{}z)\minus{}y(x\minus{}z)\equal{}1$
$ (x\minus{}z)(1\minus{}y)\equal{}1$[/hide]

Using LideR_M's hint...

[hide="Solution:"]We have $ (x\minus{}z)(1\minus{}y)\equal{}1$.  Thus, we have two cases...

Case 1): $ x\minus{}z\equal{}1\minus{}y\equal{}1$.
This implies that $ y\equal{}0$.  By plugging this back into (1) and (2), we get $ x\equal{}101$ and $ z\equal{}100$ (and $ x\minus{}z\equal{}1$, as desired).

Case 2): $ x\minus{}z\equal{}1\minus{}y\equal{}\minus{}1$
Now, we have $ y\equal{}2$ and $ x\equal{}z\plus{}1$.  Plugging this into (2) gives:
$ z\plus{}1\plus{}2z\equal{}101$, so $ z\equal{}\frac{101}{3}$ is not an integer and we are done.

Only triple of integers: $ (x,y,z)\equal{}(101,0,100)$[/hide]

[quote="krsattack"]
Case 2): $ x\minus{}z \equal{} 1\minus{}y \equal{}\minus{}1$
Now, we have $ y \equal{} 2$ and $ x \equal{} z\plus{}1$.
[/quote]

No, $ x\equal{}z\minus{}1$ and you'll find another solution.

[quote="pco"][quote="krsattack"]
Case 2): $ x\minus{}z \equal{} 1\minus{}y \equal{}\minus{}1$
Now, we have $ y \equal{} 2$ and $ x \equal{} z\plus{}1$.
[/quote]

No, $ x \equal{} z\minus{}1$ and you'll find another solution.[/quote]
You are correct using $ x\equal{}z\minus{}1$ we get the second solution of the system which is $ (x,y,z)\equal{}(33,2,34)$
So the complete list of the solutions is:
$ (x,y,z)\equal{}(101,0,100)$and $ (x,y,z)\equal{}(33,2,34)$

Sorry... I accidentally took x-z=-1 and, well, somehow got x=z+1.  Dumb mistake.


trigonometry
logarithms
complex analysis

Solve the following equation:

$ \cos z \equal{} 3 \minus{} 2i$.

My book says that

$ \cos^{\minus{}1}z\equal{} \minus{}i \log[z\plus{}i(1\minus{}z^2)^{\frac{1}{2}}]$.  It also says that $ (1\minus{}z^2)^{\frac{1}{2}}$ is a double valued fuction.  How do I use this formula or is there a better way to do this? I don't see how to get the double value it is talking about.

The square root may be double valued but the logarithm is (in the same sense) infinite valued. Do you understand that every nonzero complex number has two square roots? And that, since the exponential is periodic, every complex number has infinitely many logarithms? (And, of course, the arccosine will also be infinite-valued.)


Find the remainder r, 1 ≤ r ≤ 13, when 2^1985 is divided by 13.

[hide]
$ 2^{5}\equiv 6 \mod 13$
$ 1985 \equiv 0 \mod 5$
$ \therefore 2^{5}\equiv 2^{1985}\equiv 6 \mod 13$
$ 6$ is the remainder.
[/hide]

[quote="AstroPhys"][hide="Quoted Solution"]
$ 2^{5}\equiv 6 \mod 13$
$ 1985 \equiv 0 \mod 5$
$ \therefore 2^{5}\equiv 2^{1985}\equiv 6 \mod 13$
[/hide][/quote]
[hide="What?"]
You need to show some sort of [u]periodicity[/u] of $ 2^{n}\bmod 13$ before you just go off and say that $ 5|1985 \implies 2^{5}\equiv 2^{1985}\bmod 13$. And there isn't any periodicity at all when only looking at periods of five; i.e. $ 2^{5}\equiv 6 \bmod 13$ whereas $ 2^{10}\equiv 10 \bmod 13$, etc. It's only a coincidence that you obtained the correct answer, as $ 1985 \equiv 5 \bmod 12$.[/hide]

[hide="Solution"]
By Fermat's Little Theorem, we have: \[ 2^{13-1}=2^{12}\equiv 1 \bmod 13 \implies 2^{1985}= 2^{12 \cdot 165+5}\equiv 2^{5}\bmod 13 \equiv \boxed{ 6 \bmod 13}\][/hide]


geometry
geometric transformation
reflection
geometry unsolved

ABCD is inscribed in a circle , circlr O1 and O2 are inscribed in triangle ACD ,BCD MN is tagent to O1,O2 prove that : MN//AB.

Dear Mathlinkers,
first aid : if it is true, this situation look like the parallel tangent theorem.
Sincerely
Jean-Louis

Let $ E$ be the midpoint of arc $ CD$.  Then $ E$ is the circumcentre of both triangles $ CO_1 D$ and $ CO_2 D$.  Hence, quadrilateral $ CO_2 O_1 D$ is cyclic.  Let $ O_1 O_2$ and $ CD$ meet at $ P$.

Then $ \angle PDO_1 \equal{} \angle D/2$, $ \angle PO_1 D \equal{} 180^\circ \minus{} \angle O_2 CD/2 \equal{} 180^\circ \minus{} \angle C/2$, and $ \angle O_1 PD \equal{} (\angle C \minus{} \angle D)/2$.  Since $ MN$ is the reflection of $ CD$ in $ O_1 O_2$, $ \angle DPN \equal{} \angle C \minus{} \angle D$.  It follows easily that $ AB \parallel MN$.

[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/f97b4597625cb00ff135c384fad55725.png[/img]


This should probably be moved to the Puzzles & Brainteasers Forum, although I don't really see how it's challenging anyway...

Here's a website with an interestig discussion about another version of this problem:
[url]http://mathcentral.uregina.ca/mp/previous2002/dec02sol.html[/url]

[quote="archimedes1"][quote="i_like_pie"] I don't really see how it's challenging anyway...[/quote]
Isn't it impossible?  I mean, try "There are _ 1s in this sentence."[/quote]
Well the one you said is impossible, but this one is doing all 9 simultaneously, which is possible.

6 1s
3 2s
2 3s
1 4
1 5
2 6s
1 7
1 8
1 9

Nice one, MellowMelon. I posted this here because i thought there might be a systematic way of setting up equations and solving it =P. Thanks for the website, i think it includes the number of 0s in the sentence, which wasn't in the one i posted. :P


In the right triangle shown the sum of the distances $ BM$ and $ MA$ is equal to the sum of the distances $ BC$ and $ CA$.  If $ MB \equal{} x$, $ CB \equal{} h$, and $ CA \equal{} d$, then $ x$ equals:
[asy]size(200); defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
draw((0,0)--(8,0)--(0,5)--cycle);
label("C",(0,0),SW);
label("A",(8,0),SE);
label("M",(0,5),N);
dot((0,3.5));
label("B",(0,3.5),W);
label("$x$",(0,4.25),W);
label("$h$",(0,1),W);
label("$d$",(4,0),S);[/asy]$ \textbf{(A)}\ \frac {hd}{2h \plus{} d} \qquad \textbf{(B)}\ d \minus{} h \qquad \textbf{(C)}\ \frac {1}{2}d \qquad \textbf{(D)}\ h \plus{} d \minus{} \sqrt {2d} \qquad \textbf{(E)}\ \sqrt {h^2 \plus{} d^2} \minus{} h$

[hide="Click for solution"]
$ x\plus{}\sqrt{d^2\plus{}(h\plus{}x)^2}\equal{}h\plus{}d$, $ \sqrt{d^2\plus{}(h\plus{}x)^2}\equal{}h\plus{}d\minus{}x$, $ d^2\plus{}h^2\plus{}2hx\plus{}x^2\equal{}h^2\plus{}d^2\plus{}x^2\plus{}2hd\minus{}2hx\minus{}2dx$, $ 2hx\plus{}dx\equal{}hd$, $ x\equal{}\frac{hd}{2h\plus{}d}$ or $ \boxed{\textbf{(A)}}$.
[/hide]


A baseball league has nine teams. During the season, each of the nine
teams plays exactly three games with each of the other teams.
What is the total number of games played?

Hm...this was kind of vague.

I managed to misunderstand this as "each team plays 3 other teams" instead of "plays 3 times against each other team"

If there was only one game, the answer would be $ \frac{8 \times 7}{2}$ or $ 28$ (or $ \binom{8}{2}$).  However, you must multiply this by $ 3$, to get $ \boxed{84}$.

The answer is 108.

First do ${8+7+6+5+4+3+2+1}$ For the amount of games for one round.

To get all 3, multiply by 3 getting $\boxed{108}$

[hide="Another Solution"]First, count how many games played if each team plays each other once, then multiply that by $3$.  There are $\dbinom{9}{2}=36$ ways for the teams to play each other once.  Multiply $36$ by $3$ since each team plays each other $3$ times, so the answer is $36\times3=\boxed{108}$.[/hide]

One way to do it is by counting pairs, or what hesa did, to get 108.
But, if you don't know what $ \dbinom{9}{2} $ is, it is the same as this logic.
You have 9 teams, so if you pick 1 team, you have 8 teams to verse with.
Now that you've gotten rid of 1 team, you have 8 more teams.
You have 8 teams now and 7 to pair up with, and so on...
You will eventually find that 8+7+6+...+2+1=36 (that is the equation you should get).
So 36*3 is 108.

Answer: $ \boxed{108} $

BTW, you should learn the $ \dbinom{x}{y} $  form and how to solve it if you don't know it, it can be very useful.*

[hide = Solution]First let's count how many ways to pair up the teams. This is just $\binom{9}{2}$ since we're choosing $2$ teams out of the $9$ total.
If we don't know choose notation, we could instead think: $9$ choices for the first team, times $8$ for the second. But then we are counting each pair twice: one where one team is chosen first and one where the other team is chosen first.
Either way we end up with $\frac{9 \cdot 8}{2} = 36$ pairs. However, each pair plays $3$ games together.
So, the total games played is $36 \cdot 3 = \boxed{108}$.[/hide]

[color=#f0f][size=200][hide="solution"]$\frac{9 \cdot 8}{2} \cdot 3 = \boxed{106}$[/hide][/size][/color]

[color=#f0f][size=200][hide="KewlStuffByMe's Solution (fixed)"]$\frac{9 \cdot 8}{2} \cdot 3 = \boxed{108}$

Fixed By ShockFish.
[/hide][/size][/color]

[hide=My solution][color=#9a00ff]Each of the 9 teams plays 3 games with each of the 8 other teams, or 3 • 8. Then we divide by two so that we do not overcount the games. 9 • 3 • 8 ÷ 2 =[hide=this][color=#9a00ff][b][size=200]108[/size][/b][/color][/hide][/color][/hide]


geometry
circumcircle
incenter
analytic geometry
trigonometry
conics
hyperbola

Let a regular triangle $ ABC$ with $ X$ move on $ BC$. Let $ I,J$ are incenter of triangle $ AXB$ and $ AXC$. Find locus of circumcenter of triangle $ AIJ$.

If K is the circumcenter of triangle AIJ, then KX and BC perpendicular might help...
It seems the locus belongs to a circle centered high above A.

Best regards,
sunken rock

The locus is not a circle. I'll give the polar coordinates.

First note that $ A\hat{B}X\equal{}60^\circ$ is inscribed in the circle with centre $ I$, so $ A\hat{I}X\equal{}120^\circ$, and this implies that $ AIXC$ is cyclic, that is, $ I$ is on the circuncircle of $ ACX$. Similarly $ J$ is on the circumcircle of $ ABX$. Now it's easy too see that $ AIJ$ is an equilateral triangle.

The caculation of the polar coordinates is somewhat boring. Let the triangle $ ABC$ have side $ 2$. Let $ B\hat{A}I\equal{}\theta$. Then $ AI\equal{}\frac{1}{cos{\theta}}$. Let $ K$ be the circuncenter of $ AIJ$. Then $ AK\equal{}\left(\frac{AI\sqrt{3}}{2}\right).\left(\frac{2}{3}\right)\equal{}\frac{1}{\sqrt{3}\cos\theta}$. We also have $ B\hat{A}K\equal{}\theta\plus{}30^\circ$, and so the polar coordinates are $ \left(\theta\plus{}30^\circ,\frac{1}{\sqrt{3}\cos\theta}\right)$, or, writing in a friendlier way, $ \left(\theta,\frac{1}{\sqrt{3}\cos(\theta\minus{}30^\circ)}\right)$. Don't know what figure is it  :D

Nice observations, however <JAI = 30 degs, there4 AIJ cannot be equilateral!

Best regards,
S. Fulger

Why is $ J\hat{A}I\equal{}30^\circ$?

A and I belong to the circumcircle of ACX, so JA=JI. A and J belong to the circumcircle of ABX, so IA=IJ, and then AIJ is equilateral. A good picture shows this clearly.

Logically, I is on the bisector of the angle <BAX, whilst J on the one of <CAX, hence
<JAI = <JAX + <XAI = (<BAX + <CAX)/2 = <A/2 = 30 degs

Best regards,
sunken rock

(See also the post #3 at https://artofproblemsolving.com/community/c6h272602

There I have proved that, if K was the subject circumcenter, then KX and BC are perpendicular and, if C' was the circumcenter of the triangle XJC, then KX = IC'.
But AIXC'C is cyclic and <AIC' = 30 degs, hence the circumradius of AIXC'C is equal to KX  ( 1 ).
AX is a chord of this circle and <ACX = 60 degs, that is, AX = KX. sqrt(3), or AX^2 = 3KX^2  ( 2 ).

Let AB = 2a, hence OA = a.sqrt(3), OX = x and KX = y.
The relation (2) becomes:
3a^2 + x^2 = 3y^2; after dividing by 3a^2, we get: y^2/a^2 - x^2/3a^2 = 1, that is, a hyperbola with foci on the A - altitude of the equilateral triangle ABC. (Actually, a 'segment' of the branch intersecting the interior of the triangle, bounded by the perpendiculars at B and C onto BC).

Best regards,
sunken rock

I think locus is a part of circle. In my picture, locus is chord $ MN$ of circle $ (O)$

[img]http://i227.photobucket.com/albums/dd72/TPFC/Locus.png[/img]

Some another result:
-) $ AX$ pass through centroid $ G$ of triangle $ IJK$ and $ AG\equal{}2GX$
-) Reflection point of $ K$ over $ IJ$ is on $ A$-altitude of triangle $ ABC$

Dear tdl,

Kindly, post your solution, or the one you know!

Very interesting properties, the 2nd one I noticed as well, and quite easy to prove, the 1st one, I shall study.

Best regards,
Stan Fulger

[img]http://i227.photobucket.com/albums/dd72/TPFC/899fd6ad.png[/img]
Sorry because with this problem I only can solve it by coordinate system. But with two another resut I give, I can solve it synthenticly.
Call $ L$ is midpoint of $ IJ$ and $ G$ is centroid of triangle $ IJK$, $ XL\cap JK \equal{} M$.
We have:
$ \angle{KMX} \equal{} \angle{MLJ} \plus{} 60^0 \equal{} \angle{ILX} \plus{} 60^0 \equal{} 180^0 \minus{} 2\angle{XIJ} \plus{} 60^0$
$ \equal{} 2(\angle{AIX} \minus{} \angle{XIJ}) \equal{} 2\angle{AIJ} \equal{} \angle{AKJ}$
Then $ AK\parallel{}XL$
And $ AK \equal{} JK \equal{} IJ \equal{} 2XL$
But $ KG \equal{} 2GL$ then $ AGX$ are collinear and $ AG \equal{} 2GX$!

As already shown before, KX || AD, D being the midpoint of BC; then K', the symmetrical of K w.r.t. IJ lies on AD (OIK'J is cyclic, hence OK' is the angle bisector of <BOC). It's easy to see that M, the midpoint of IJ is equally apart of AD and KX, hence the circle (M, MI) passes through D and X, and intersects AD and KX at X' and D' respectively, both different of D and X. It's obvious also that X'DXD' is a rectangle, DD' and XX' are diameters of (M, MI), hence we get the following equalities: AK = XX' = DD', hence AX'XK is a parallelogram, having the equal segments AK and XX' with the extremities sliding on the parallel lines AD and KX. KX'K'X is a parallelogram as well, hence AX' = KX = K'X', hence AK' = 2.KX, and AX and KK' intersect at G, the centroid of the triangle IJK.

However, tdl, your proof for this issue is far more elegant!

Best regards,
sunken rock


algebra
polynomial
modular arithmetic
linear algebra
linear algebra unsolved

Let $A_i$ be a finite family of complex square matrices that have no eigenvalues in common and assume that each $A_i$ is similar to a real matrix. Let also $f_i$ some real polynomials. Prove that there is a real polynomial $f$ such that $ f_i(A_i)=f(A_i)$.

If $\varphi_i$ are the pairwisely coprime characteristic polynomials of the $A_i$'s, we can find $f$ s.t. $f\equiv f_i\pmod{\varphi_i}$, and this is the $f$ we're looking for ($f$ can be taken to be real because the $\varphi_i$'s are all real).


calculus
integration
logarithms
real analysis
real analysis unsolved

Let $ E\equal{}\left\{(y,s)\in \mathbb R^n\times \mathbb R\;: \; s\le 0,\; \frac{1}{\left(4\pi(\minus{}s)\right)^{n/2}}\cdot\exp\left(\frac{|y|^2}{4s}\right)\geq 1\right\}$.
The set $ E$ is called a [i]heat ball[/i]. Prove that $ \iint\limits_{E}\frac{|y|^2}{s^2}dyds\equal{}4.$

clearly $ 0 < \minus{}s \leq \frac{1}{4\pi}$ and $ E \cap \{ s \equal{} s_0\} \equal{} \{ (s_0, y): \|y\| \leq f(|s_0|) \}$ where $ f: (0; \frac{1}{4 \pi}] \to [0;\infty)$ is easily computable. Of course $ f(\frac{1}{4 \pi}) \equal{} 0$ and $ (\frac{1}{4\pi |s|})^{\frac{n}{2}} \exp(\minus{}\frac{f(s)^2}{4 |s|}) \equal{} 1$. Hence, for any $ s$ and any $ 0< \lambda < 1$:
\[ 1 \equal{} (\frac{1}{4\pi |\lambda s|})^{\frac{n}{2}} \exp(\minus{}\frac{f( \lambda s)^2}{4 |\lambda s|}) \equal{}  \frac{1}{\lambda^{\frac{n}{2}}} \exp(\frac{f( s )^2}{4 |s|}\minus{}\frac{f( \lambda s)^2}{4 |\lambda s|})\]
so that if $ s_0\equal{}\frac{1}{4 \pi}$ and $ f(s_0\equal{}\frac{1}{4 \pi}) \equal{} 0$:
\[ 1 \equal{} \frac{1}{\lambda^{\frac{n}{2}}} \exp(\minus{}\frac{f( \lambda s_0)^2}{4 |\lambda s_0|})\]
which gives:
\[ f(\frac{ \lambda}{4 \pi}) \equal{} \frac{n}{2 \pi} \lambda |\ln(\lambda)|\]
The end is straightforward:
\[ I \equal{} \int_{t\equal{}0}^{ \frac{1}{4\pi}} \frac{1}{t^2}\int_{|y| \leq f(t)} \|y\|^2 \, dy \, dt \equal{} \omega_{n\minus{}1} \int_{t\equal{}0}^{ \frac{1}{4\pi}} \frac{f(t)^{n\plus{}2}}{(n\plus{}2) t^2} \, dt \equal{} \ldots\]

I also came to that point but I am not sure why the result is $ 4$, could you show it more details?  :)


Let positive integer  $ n$
Prove that : $ \gcd(1,2,...,n)\geq 2^{(n\minus{}1)}$

I am sorry
Let positive integer  $ n$
Prove that : $ \ Lcm(1,2,...,n)\geq 2^{(n\minus{}1)}$

Have a look here
http://www.mathlinks.ro/Forum/viewtopic.php?p=136686#136686

Pierre.


They had a problem like this at ARML

if you multiply it out, its a sum of powers of x, if i remember right.

haha lemem try:

(im right)

${x}^4 + {x}^3 + {x}^2 + x + 1$


advanced fields
advanced fields unsolved

Let a and B be subsets of a metric space where A is compact and B is closed. Show that A $ \cap$B is compact

$ A$ is compact so it's totally bounded and complete.  $ B$ is closed so it's complete.  What does that say about the intersection of $ A$ and $ B$?


show
         n C k=m^l has no sol. in naturals where k>=4 and <= n-4. :?:

finally i got it.the key is following sylvester lemma if n>=2k,then one of the nos. n,n-1,...,n-k+1 is multiple of prime >k. :D


integration
logarithms
function
calculus
calculus computations

Derive the formula

$\int_{0}^{\infty} \frac{e^{-bt}-e^{-at}}{t}dt= log(a/b)$

for all $a,b>0$

[quote="Disco Very"]Derive the formula

$ \int_{0}^{\infty} \frac {e^{ \minus{} bt} \minus{} e^{ \minus{} at}}{t}dt \equal{} log(a/b)$

for all $ a,b > 0$[/quote]
Here is the solution that I can see, there might be an easier way. Someone should probably verify.

[hide]$ \frac{e^{\minus{}bx}\minus{}e^{\minus{}ax}}{x}\equal{}\int_b^{a}e^{\minus{}yx}\text{ }dy$

So we have

$ \int_0^{\infty}\int_b^{a}e^{\minus{}yx}\text{ }dy\text{ }dx\equal{}\int_b^{a}\int_0^{\infty}e^{\minus{}yx}\text{ }dx\text{ }dy$

Now we see that

$ \int_0^{\infty}e^{\minus{}yx}\text{ }dx$

Converges iff $ y>0$

And further more supposing that this condition is met we have that

$ \int_0^{\infty}e^{\minus{}yx}\text{ }dx\equal{}\frac{\minus{}e^{\minus{}yx}}{y}\bigg|_0^{\infty}\equal{}0\minus{}\frac{1}{y}\equal{}\frac{1}{y}$

So we then have

$ \int_b^{a}\int_0^{\infty}e^{\minus{}yx}\text{ }dx\text{ }dy\equal{}\int_b^{a}\frac{dy}{y}\equal{}\ln(y)\bigg|_b^a\equal{}\ln(a)\minus{}\ln(b)\equal{}\ln\left(\frac{a}{b}\right)\quad\blacksquare$[/hide]

[hide=" Another solution"]Let's find the function f(t) for which

$ f(t) \equal{} \mathcal{L} ^{\minus{}1} \left(\ln \frac{s\plus{}a}{s\plus{}b}\right)$.

Now, if f(t) is a piecewise continuous function on $ [0,\infty)$ and is of exponential order, and if $ \mathcal{L} [f(t)] \equal{} F(s)$, then we can write

$ f(t) \equal{} \minus{} \frac{1}{t} \mathcal{L} ^{\minus{}1}\left[\frac{d}{ds} F(s)\right]$.

On the present case, $ F(s) \equal{} \ln \frac{s\plus{}a}{s\plus{}b}$, and so $ \frac{d}{ds} F(s) \equal{} \frac{1}{s\plus{}a} \minus{} \frac{1}{s\plus{}b}$. Therefore,

$ f(t) \equal{} \minus{}\frac{1}{t} \mathcal{L} ^{\minus{}1}\left(\frac{1}{s\plus{}a} \minus{} \frac{1}{s\plus{}b}\right) \equal{} \minus{}\frac{1}{t} (e^{\minus{}at} \minus{} e^{\minus{}bt}) \equal{} \frac{e^{\minus{}bt} \minus{} e^{\minus{}at}}{t}$.

But this means that

$ \mathcal{L} \left(\frac{e^{\minus{}bt} \minus{} e^{\minus{}at}}{t}\right) \equal{} \int_0^{\infty}\frac{e^{\minus{}bt} \minus{} e^{\minus{}at}}{t} \cdot e^{\minus{}st}\,dt \equal{} \ln \frac{s\plus{}a}{s\plus{}b}$.

If, on this last equality, we take the limit when $ s \to 0^\plus{}$, we obtain

$ \int_0^{\infty}\frac{e^{\minus{}bt} \minus{} e^{\minus{}at}}{t}\,dt \equal{} \ln \frac{a}{b}$.

Note that this is also a special case of Froullani's integral.[/hide]

[quote="Disco Very"]Derive the formula

$ \int_{0}^{\infty} \frac {e^{ \minus{} bt} \minus{} e^{ \minus{} at}}{t}dt \equal{} log(a/b)$

for all $ a,b > 0$[/quote]

http://mathworld.wolfram.com/FrullanisIntegral.html  :P


function
algebra
domain
real analysis
real analysis solved

Let $f:R ->R$ a continous,non-constante function and $F:R*R ->R$ such that
$f(x+y)=F(f(x),f(y))$.Show that f is strictly monotonous.

Very nice indeed.

It's easy to see that $f(a)=f(b)\iff b-a$ is a period of $f$. Assume we have two such numbers $a<b$. Now try to show (it’s not hard at all) that there are $x_n,y_n\in [0,b-a]$ s.t. $f(x_n)=f(y_n)$ and $y_n-x_n=\frac{b-a}n$ ($\forall n\ge 1$), and it’s a well known fact that if a continuous function has a sequence of periods which tends to $0$, then the function is constant, which, in our case, is a contradiction.

I was asked to prove the existence of $x_n,y_n$.

Let $k=b-a$. We have $f(0)=f(k)$. Consider the function $g:[0,k-\frac kn],\ g(x)=f(x+\frac kn)-f(x)$. We have $g(0)+g(\frac kn)+\ldots+g(k-\frac kn)=f(k)-f(0)=0$, so the continuous function $g$ changes sign on its domain, meaning that at some point it must be $0$. If $g(x_n)=0$, we have $f(x_n)=f(x_n+\frac kn)$ and we can take $y_n=x_n+\frac kn$.


inequalities
trigonometry
inequalities unsolved

let $ ABC$ a triangle and $ A$ , $ B$ , $ C$ the angles , prove that :

$ \frac {\cos(A) }{ (\sin(B) + \sin(C))^2} + \frac {\cos(B)}{ (\sin(C) + \sin(A))^2 } + \frac {\cos(C)}{ (\sin(A) + \sin(B))^2 } \le \frac {1}{2}$

${{ (\frac {\cos(A) }{\sin(B)\sin(C)})^2 + (\frac {\cos(B)}{\sin(C)\sin(A)}})^2 + (\frac {\cos(C)}{\sin(A)\sin(B)}})^2 \geq \frac {4}{3}$

$ \frac {\sqrt {\cot(B)\cot(C)}}{\sin(A)} + \frac {\sqrt {\cot(C)\cot(A)}}{\sin(B)} + \frac {\sqrt {\cot(A)\cot(B)}}{\sin(C)} \le 2$

1) Follows $ AM\geq GM$ we have

$ LHS\leq\sum {\frac {\cos A}{4\sin B\sin C}} \equal{} \frac {\sum {\sin A\cos A}}{4\sin A\sin B\sin C} \equal{} \frac {\sum{\sin2A}}{8\sin A\sin B\sin C} \equal{} \frac {4\sin A\sin B\sin C}{8\sin A\sin B\sin C} \equal{} \frac {1}{2}$

 2) By CBS we get $ LHS \geq\frac {1}{3}\left(\sum {\frac {\cos A}{\sin B\sin C}}\right)^2 \equal{} \frac {1}{3}\left(\frac {4\sin A\sin B\sin C}{2\sin A\sin B\sin C}\right)^2 \equal{} \frac {4}{3}$

 3) Similarly

 $ LHS\equal{}\frac{\sum {\sqrt{\cos B\cos C\sin B\sin C}}}{\sin A\sin B\sin C}\equal{}\frac{\sum {\sqrt{\sin 2B\sin 2C}}}{2\sin A\sin B\sin C}\leq\frac{\sum {\sin 2A}}{2\sin A\sin B\sin C}\equal{}2$


Prove that for all natural numbers $ n\ge 3$ there exist odd natural numbers $ x_n,y_n$ such that $ 7x_n^2 \plus{} y_n^2 \equal{} 2^n$

known pell-fermat application and it should be on number-theory or algebra section :S i dont think it has a simple solution


geometry
vector
analytic geometry
search
geometry unsolved

Several equal spherical planets are in outer space. On the surface of each planet is a set of points which is invisible from any of the remaining planets.
Prove that the sum of the areas of all these sets equals the surface area of one planet.

It's trivially true if we transpose the problem to the plane. Is it also true in higher dimensions?

here is some approach which i think yields sol . for two planet it's trivial.assume that $O_1,O_2,...,O_n$ are the centers of planet and R is their redius,it's sufficient to show that for any vector a such that |a|=R ,there exists exactly one point X on one of the planet , for instance i th planet , such that O_iX=a and X is invisible for others,
we can easily prove that such a point exists ,define  Ox as a horizon parallel to a the point which has the biggest coordinate is our goal.now u can prove that its unique,WLOG assume that O_iX=O_jY=a and X,Y are invisible for others and i <> j,but in the case of two planet if j th planet be invisible from X then ith planet isn't  invisible from Y,contradiction

Right. It seems that most IMO and ISL problems have already been solved here.
I'll search better before posting another one.


geometry
symmetry
3D geometry
IMO
reflection
IMO 1999
Convex hull

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

http://www.kalva.demon.co.uk/imo/isoln/isoln991.html

[b]Alternate solution :[/b]
Let us consider the convex hull $S'$ of the points $A_1, A_2, \ldots, A_n$ of $S$.
Since the convex hull of a convex polygon is a convex polygon, we have that $S'$ satisfies the conditions of the problem. Then, $S'$ is the set of the vertices of regular polygon.
Let's show that : $S = S'$. This means that $S'$ does not contain points inside $S$. Indeed, let us consider an inside point $A \in S$, $S'$ would not be invariant with regard to $s_{AA_i}$, where $AA_i = \min_{1 \le j \le n} AA_j$.

[quote="epitomy01"]I think the problem statement here is different from the real problem.
In the real problem, the points were in a plane, and not in space, so there was nothing about a tetrahedron or an octahedron.[/quote]
You're right, but the problem appeared as IMO 01 was adopted the main problem...i.e. space to plane.

Proof for the planar case, as asked in the competition. Denote by $\Omega$ the mass center of $S$; clearly $\Omega$ lies on each axis of symmetry. Since for $A,B \in S$ the perpendicular bisector of $AB$ is an axis of symmetry for $S$, it follows $\Omega A = \Omega B$, hence $S$ is made of concyclic points, lying on a circle of centre $\Omega$.

Consider any three consecutive (on the circle) points $A,B,C \in S$; since the perpendicular bisector of $AC$ is an axis of symmetry for $S$, it follows $B$ lies on it, hence $BA = BC$. Therefore $S$ is a regular polygon, and clearly all regular polygons fulfill the requirements.

[quote="orl"]A set $ S$ of points from the space will be called [b]completely symmetric[/b] if it has at least three elements and fulfills the condition that for every two distinct points $ A$ and $ B$ from $ S$, the perpendicular bisector plane of the segment $ AB$ is a plane of symmetry for $ S$. Prove that if a completely symmetric set is finite, then it consists of the vertices of either a regular polygon, or a regular tetrahedron or a regular octahedron.[/quote]

[hide="Solution"]
The question can be simplified in dealing with a 2D plane and a 3D plane. We will divide into two cases:
1) 2D plane
Consider 3 points A,B,C. Their perpendicular bisectors meet at a point O that is the center of the circumscribed circle. Reflect, say $A$ over the perpendicular bisector of BC to get another point $A'$ lying on the circle because the perpendicular bisector of  $AA'$ passes through $O.$
Essentially this means set $S$ is a set of points that lie on a circle. In an infinite set of points, we can continously add another point by reflecting it over a perpendicular bisector of 2 other points. However, set $S$ is finite, meaning that at some time, the "pattern" will repeat so that a point that was reflected over a perpendicular bisector is the same point as an already existing point. Because the pattern will stop at one point, we can take 3 consecutive points A,B,C in that order on the circle. Assume that AB>BC.  Say there exists a perpendicular bisector of any 2 other points that goes through B. Then the reflection of C must lie between the arc $AB$, but there exists no points, contradiction, meaning that AB=BC. Now say there doesn't exist a perpendicular bisector of any 2points that goes through B but there is a bisector that passes between arc AC, of which we can take the perp bisector of AC. Then B can be reflected to a point $B'$ on AC, which is a contradiction, so AB=BC. In either case, AB=BC, and so we can take any other 3 points, A,B,D (D on the opposite side of C) and arrive at the same conclusion.  lnductively, all the sides are equal and it is circumscriptible, so it is a regular polygon. $\blacksquare$
2) 3D plane
Take three points A,B,C and take a crosssection to form a circle around those 3 points. If a perp bisector of some point of the 3 reflects one of the 3 points to a point on the circle, then we have a regular polygon and it reverts to case 1. Take a point D such that, wlog, it is the reflection of C over the perp bisector of AB. argh... too wasted to finish this case, i'll finish it up when I get time, but I feel like you proceed the same way as case 1.[/hide]

I guess for the 3-d case it suffices to show that 1) convex hull of S = S is circumscribable in a sphere, and 2) every circle formed by taking 3 points of S is congruent. [this, combined with the 2-D case should be sufficient...]

Cute problem!

First, the global part. Let $O$ be the centroid of $S$. Since reflection across the perpendicular bisector of $AB$ fixes $S$, it must also fix $O$, so $AO=BO$ and $O$ is the circumcenter of $S$.

We now split the problem into two cases:

[b]Case 1:[/b] $S$ is coplanar.
[indent]
Let $\Omega$ be the circumcircle of $S$. Assume for the sake of contradiction that there are points $A,B,C,D\in S$ in that order around $\Omega$ with [list] [*] $A\ne D$, [*] $A$ and $B$ consecutive, [*] $C$ and $D$ consecutive, and [*] $\overarc{\ensuremath{ADB}}<\overarc{\ensuremath{CAD}}$. [/list]
[asy]
draw(unitcircle);
pair A=dir(162),B=dir(234),C=dir(354),D=dir(42);
pair Z=(5/4)*sqrt(A*D), Cp=A*D/C;
draw(-Z--Z,blue+dashed+(1/2)*bp);
draw(C--Cp,blue+(1/2)*bp,Arrow,Margins);
dot("$A$",A,A); dot("$B$",B,B); dot("$C$",C,C); dot("$D$",D,D);
dot("$C'$",Cp,Cp,blue);
[/asy]
Then, reflecting $C$ over the perpendicular bisector of $AD$ gives a point between $A$ and $B$ in $S$, a contradiction.

Therefore, the arcs between consecutive points in $S$ must all be congruent, so $S$ consists of the vertices of a regular polygon, as desired.
[right]$\square$[/right][/indent]

[b]Case 2:[/b] $S$ is not coplanar.
[indent]
Let $\Omega$ be the circumsphere of $S$. Take the convex polyhedron formed by $S$, i.e. the convex hull of $S$. Let $P\in S$ and let the neighbors of $P$ on the convex hull be $A_1,A_2,\ldots,A_n$ in that order for some $n\ge 3$.

By Case 1, the faces of the convex hull are regular polygons, so $$PA_1=PA_2=\ldots=PA_n.$$ Thus, $A_1,A_2,\ldots,A_n$ lie on the circle formed by the intersection of $\Omega$ and the sphere centered at $P$ containing $A_1,A_2,\ldots,A_n$. By Case 1, they then must form a regular polygon. Therefore, $$\angle A_1PA_2=\angle A_2PA_3=\ldots=\angle A_nPA_1,$$ so the faces containing these angles have the same number of sides and are congruent regular polygons. It follows that the convex hull of $S$ is a platonic solid.

The cube, dodecahedron, and icosahedron do not work because they are not preserved when reflecting across the perpendicular bisector of a space diagonal. Thus, $S$ consists of the vertices of either a regular tetrahedron or a regular octahedron, as desired.
[right]$\square$[/right][/indent]

Solved with [b]nukelauncher[/b] and [b]Th3Numb3rThr33[/b].

First, we verify the planar case: Let \(G\) be the centroid of the points in \(S\); then \(G\) lies on each perpendicular bisector \(AB\), so \(GA=GB\) for all \(A\), \(B\) in \(S\); i.e.\ the points in \(S\) lie on a circle centered at \(G\). Let the points of \(S\) lie on the circle in the order \(A_1\), \(A_2\), \ldots, \(A_n\). We can verify for each \(i\) by looking at the perpendicular bisector of \(\overline{A_{i-1}A_{i+1}}\) that \(A_iA_{i-1}=A_iA_{i+1}\), so the polygon is regular, as needed.

Then assume the points are not all planar. Let \(G\) be the centroid again, so we can verify analogously that the points in \(S\) lie on a sphere centered at \(G\). Consider the polyhedron formed by the points in \(S\).

Observe that any plane containing at least three points must form a regular polygon. We will verify the following claim:

[color=red][b]Claim:[/b][/color]      If we have a plane containing at least four points, and there is a point \(P\) not on the plane, then there is a point \(P\) on the other side of the plane as \(P\).

[i]Proof.[/i]      Let the plane contain regular polygon \(A_1A_2\cdots A_k\) with side length \(s\). If there is a point \(P\) ``above'' the plane, then we can select a point \(Q\) on the plane containing \(\triangle PA_1A_2\) yet still ``above'' the plane with \(QA_1=s\).

By \(k\ge4\), we have \(QA_1=A_1A_2<A_1A_3\), so the perpendicular bisector plane of \(\overline{QA_3}\) intersects segment \(A_1A_3\); that is, the reflection of \(A_1\) over the perpendicular bisector plane lies ``below'' the plane, and it must also be in \(S\). \(\blacksquare\)

It follows that every face of the polyhedron is an equilateral triangle, so the polyhedron is either a regular tetrahedron, regular octahedron, or regular icosahedron (by platonic solids). To rule out the icosahedron, take the space diagonal; the two pentagons do not pair up.

Solution from [i]Twitch Solves ISL[/i]:

Let $G$ be the centroid of $S$.

[b][color=red]Claim:[/color][/b]  	All points of $S$ lie on a sphere $\Gamma$ centered at $G$.
[i]Proof.[/i]  	Each perpendicular bisector plane passes through $G$. 	So if $A,B \in S$ it follows $GA = GB$. $\blacksquare$

[b][color=red]Claim:[/color][/b]  	Consider any plane passing through three or more points of $S$. 	The points of $S$ in the plane form a regular polygon.
[i]Proof.[/i]  	The cross section is a circle because we are intersecting 	a plane with sphere $\Gamma$. 	Now if $A$, $B$, $C$ are three adjacent points on this circle, 	by taking the perpendicular bisector we have $AB=BC$. $\blacksquare$
If the points of $S$ all lie in a plane, we are done. Otherwise, the points of $S$ determine a polyhedron $\Pi$ inscribed in $\Gamma$. All of the faces of $\Pi$ are evidently regular polygons, of the same side length $s$.

[b][color=red]Claim:[/color][/b]  	Every face of $\Pi$ is an equilateral triangle.
[i]Proof.[/i]  	Suppose on the contrary some face $A_1 A_2 \dots A_n$ 	has $n > 3$. 	Let $B$ be any vertex adjacent to $A_1$ in $\Pi$ 	other than $A_2$ or $A_n$. 	Consider the plane determined by $\triangle A_1 A_3 B$. 	This is supposed to be a regular polygon, 	but arc $A_1 A_3$ is longer than arc $A_1 B$, 	and by construction there are no points inside these arcs. 	This is a contradiction. $\blacksquare$
Hence, $\Pi$ has faces all congruent equilateral triangles. This implies it is a regular polyhedron --- either a regular tetrahedron, regular octahedron, or regular icosahedron. We can check the regular icosahedron fails by taking two antipodal points as our counterexample. This finishes the problem.

Let $G$ be the centroid. Note that since the entire figure is symmetric across each perpendicular bisector, the centroid must be on every perpendicular bisector. Thus, $S$ is conspheric.
\\$~$\\
Let $T$ be a subset of $S$ that lies on a circle. Let $A_1$, $A_2$, $A_3$, $A_4$, $A_5$ be adjacent points appearing in that order, taking cyclics if needed. By the perpendicular bisector of $A_3A_4$, $A_2A_3=A_4A_5$. By the perpendicular bisector of $A_2A_4$, $A_1A_2=A_4A_5$ which means $A_1A_2=A_2A_3$, and therefore the points on $T$ form a regular polygon.
\\$~$\\
Clearly any regular polygon works, so suppose there was some other point $P$. Then its neighbors form a regular polygon and therefore all the angles including $P$ of the convex hull are equal. Furthermore all the faces of this solid are regular polygons so we have ourselves a tetrahedron, cube, octahedron, dodecahedron and icosahedron and it is easy to check only the tetra and octa hedrons work.


Trivial for some, right?
My other problems were moved here so I assume this is where I should put these kind of problems.

From where did you conclude that the value is less than $ \frac{1}{\sqrt{101}}$ , I didn't get that

Well, I would personally place this in Intermediate Topics, but I suppose it could be argued that it works here.

As for being less than $ \frac{1}{\sqrt{101}}$, from the second line $ \left(\frac {1}{2}\cdot\frac {3}{4}\cdot\cdots\cdot\frac {99}{100}\right)^2<\frac {1}{2}\cdot\frac {2}{3}\cdot\cdots\cdot\frac {99}{100}\cdot \frac {100}{101}\equal{}\frac{1}{101}$, hence the result follows.

hello, in the book from P.P. Korowkin "Ungleichungen" we can find the following inequality
$ \frac{1}{2}\cdot \frac{3}{4} \cdot \frac{5}{6}\cdots\frac{2n\minus{}1}{2n}\le \frac{1}{\sqrt{3n\plus{}1}}$.
Sonnhard.

I presume Dr. Graubner means $ \frac{1}{\sqrt{2n\plus{}1}}$; this can be proved by extending the same method I used above.

Remark: This bound is surprisingly sharp; sharper than using AMGM and smoothing to $ \left(\frac{49}{50}\right)^{100}$.

I understand it now, thanks a lot temperal.

[hide="another solution"]
we know $ \frac{n\minus{}1}{n} < (\sqrt{\frac{n\minus{}1}{n}})(\sqrt{\frac{n}{n\plus{}1}})$       for n>1    (you can check it by cross multiplication)
for n=2 we have
$ \frac{1}{2}< (\sqrt{\frac{1}{2}})(\sqrt{\frac{2}{3}})$
for n=4 we have
$ \frac{3}{4}< (\sqrt{\frac{3}{4}})(\sqrt{\frac{4}{5}})$
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.
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for n=100 we have
$ \frac{99}{100}< (\sqrt{\frac{99}{100}})(\sqrt{\frac{100}{101}})$
multiplying the LHS and RHS of the above inequalities we have
$ (\frac{1}{2})(\frac{3}{4})...(\frac{99}{100}) < \sqrt{\frac{1}{101}}<\frac{1}{10}$            (as succesive numerators and denominators get cancelled exept the last one)[/hide]

hello, no i meant $ \frac{1}{\sqrt{3n\plus{}1}}$ and the given inequality can be proved by induction.
Sonnhard.

Let  $ P \equal{} \frac {1}{2}\cdot\frac {3}{4}..............\frac {99}{100}$ (1) then $ P < \frac {2}{3}\cdot\frac {4}{5}....\frac {100}{101}$ (2)

By muliplying then squareing (1) with (2) we get the desired inequality

[quote="Dr Sonnhard Graubner"]hello, no i meant $ \frac {1}{\sqrt {3n \plus{} 1}}$ and the given inequality can be proved by induction.
Sonnhard.[/quote]

Whoops; that is indeed a stronger statement, with equality when $ n\equal{}1$.

[quote="Dr Sonnhard Graubner"]hello, no i meant $ \frac {1}{\sqrt {3n \plus{} 1}}$ and the given inequality can be proved by induction.
Sonnhard.[/quote]
How was the inequality derived?


6. A point is randomly selected inside a rectangle with sides 2' and 3'. What is the probability that the point is more than 1' from every vertex of the rectangle? Use 3.14 as an approximation for $ \pi$ Express your answer as a decimal to the nearest hundredth.

[hide="Answer"]0.48[/hide]

[hide]
Well, the point cannot be 1 foot away from each vertex, so make each vertex the center of a quarter circle with radius one. Now, the point has to be somewhere in the rectangle but outside the quartercircles which there are four of. So the total possible area the point can be in is:

$ 6\minus{}\pi$

Now, since the total area is 6, and we have to (apparently) Find the probability to the nearest hundredth:

$ \frac{6\minus{}\pi}{6}\equal{}0.476401224$

or

$ \boxed {0.48}$
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