function
modular arithmetic
number theory unsolved
number theory

$ m,n$ are positive integers. If $ \phi (5^m\minus{}1)\equal{}5^n\minus{}1$, show that $ m$ and $ n$ aren't coprime.

[hide]Suppose $ m,n$ are coprime.  Then the only common factor that $ 5^m\minus{}1,5^n\minus{}1$ can have is $ 4$.  If they have any other common prime factors, $ p$, other than $ 2$ then $ \phi(p)$ divides $ n$ and $ m$, and thus $ n,m$ are not relatively prime.  If $ 2$ divides both $ \frac{5^m\minus{}1}{4},\frac{5^n\minus{}1}{4}$ then we have

$ 0\equiv5^{m\minus{}1}\plus{}5^{m\minus{}2}\plus{}\cdots\plus{}5\plus{}1\equiv1^{m\minus{}1}\plus{}\cdots\plus{}1\equiv m\pmod{2}$

Similarly we find $ n\equiv0\pmod{2}$ and thus $ m,n$ aren't coprime, contradiction.  Thus, $ 2^m\minus{}1,2^n\minus{}1$ only share $ 4$ as a common factor.

Now, let $ 5^m\minus{}1\equal{}2^{e_1}p_2^{e_2}\cdots p_n^{e_n}$.  Then we have

$ (5^m\minus{}1)\cdot\frac{(p_2\minus{}1)\cdots(p_n\minus{}1)}{2p_2\cdots p_n}\equal{}5^n\minus{}1$

If we let $ a\equal{}\frac{5^m\minus{}1}{4},b\equal{}\frac{5^n\minus{}1}{4}$ then we have

$ a\cdot\frac{(p_2\minus{}1)\cdots(p_n\minus{}1)}{2p_2\cdots p_n}\equal{}b$

Notice that $ a,b$ are relatively prime.  We prove that $ 2\not|a$.  Suppose it does.  Then let $ 2^k|a$ where $ k$ is maximized.  Notice that the 2 in the denominator of the fraction cancels.  If $ a$ has any odd prime factors $ p_i$, then $ p_i\minus{}1$ is even, and thus $ b$ is even, a contradiction.  If $ a$ has no odd prime factors, then it must equal $ 2$.  If it is a higher power of 2, then $ b$ would have to be even, a contradiction.  Thus, $ a\equal{}2\Rightarrow5^m\minus{}1\equal{}8$, which is impossible.  So, $ 2\not|a$.  Thus, $ e_1\equal{}2$.

If $ a\equal{}1$, then $ m\equal{}1\Rightarrow \phi(5^m\minus{}1)\equal{}2\Rightarrow5^n\minus{}1\equal{}2$, which is impossible.  Thus $ a$ has some odd prime factor $ p_i$ which is also a prime factor of $ 4a\equal{}5^m\minus{}1$.  Thus, $ 5^m\equiv1\pmod{p_i}$ and so $ p_i\minus{}1|m$.  Thus, since $ p_i\minus{}1$ is even, so is $ m$.  Let $ m\equal{}2t$.  Then $ 5^{2t}\minus{}1\equal{}(5^t\minus{}1)(5^t\plus{}1)$.  But, $ 4|5^t\minus{}1$ and $ 2|5^t\plus{}1$, thus $ 8|5^m\minus{}1$ so $ e_1>2$, a contradiction to the fact that $ e_1\equal{}2$.

Thus, $ m,n$ aren't coprime.

[/hide]

$ 5^m\equiv 1\pmod{p}$ doesn't require $ p\minus{}1\mid m$.

[quote="Umut Varolgunes"]$ 5^m\equiv 1\pmod{p}$ doesn't require $ p \minus{} 1\mid m$.[/quote]

As long as $ p$ is prime and $ 5\not\equiv 0,\pm1\pmod{p}$ I think it does.

ahh yes.  Then my solution is definitely wrong.  I'll try find a way to fix it, but if anyone else has a solution, please post it.


If $ x,y\in \mathbb{Z}$ , solve this equation :

$ x^6\plus{}x^5\plus{}4\equal{}y^2$

[quote="alex2008"]If $ x,y\in \mathbb{Z}$ , solve this equation :

$ x^6 \plus{} x^5 \plus{} 4 \equal{} y^2$[/quote]
I have found $ (x,y) \equal{} (0,2),(0, \minus{} 2)$

[quote="alex2008"][quote="Dimitris X"]
I have found $ (x,y) \equal{} (0,2),(0, \minus{} 2)$[/quote]

 :maybe:  That is good , but there are six other solutions[/quote]
Yes... :oops:
I gave an ugly solution....
$ x^5(x \plus{} 1) \equal{} (y \minus{} 2)(y \plus{} 2)$
So we have:
$ x \equal{} y \minus{} 2$ and $ x^5 \plus{} x^4 \equal{} y \plus{} 2$, $ x \equal{} y \plus{} 2$ and $ x^5 \plus{} x^4 \equal{} y \minus{} 2$ ,$ x^2 \equal{} y \minus{} 2$ and $ x^4 \plus{} x^3 \equal{} y \plus{} 2$ etc....I think this way works but it is really ugly.....There sould be a better solution....

My solution is this :

Our equation it is equivalent with :

$ 64x^6 \plus{} 64x^5 \plus{} 256 \equal{} 64y^2$

If $ |x|\ge 3\Rightarrow (8x^3 \plus{} 4x^2 \minus{} x)^2 < 64x^6 \plus{} 64x^5 \plus{} 256 < (8x^3 \plus{} 4x^2 \minus{} x \plus{} 1)^2$

$ \Rightarrow x\in \{ \minus{} 2, \minus{} 1,0,1,2\}$

So , $ (x,y)\in \{( \minus{} 2,\pm 6);( \minus{} 1,\pm 2);(0,\pm 2);(2,\pm 10)\}$


In fact,you can write it in this way :
$\displaystyle\lim_{x->0} \frac{(a^{1}_{1}a^{2}_{2}***a^{n}_{n})^{x}-1}{x}.$
Then the result is obvious,i.e.,$ln(a^{1}_{1}a^{2}_{2}***a^{n}_{n}).$


it strikes me that you use the vanishing homology of euclidean space a good deal in electromagnetism. This is perhaps most obvious when finding potentials, as you're directly using that closed differential forms implies exactness, i.e. vanishing of the de Rham cohomology. It also seems like you also use it implicitly in a lot of other contexts too, though. For example, when talking about the vector area (such as when defining magnetic dipole moment), you consider a loop and then integrate something over the surface that the loop bounds, but the existence of this surface, i.e. the fact that any $ S^1$ is a boundary, is again topological information. so how much does electromagnetism change when working over a topologically nontrivial space?

by the way, http://www.math.unc.edu/Faculty/jds/survey.pdf

I will freely admit I do not understand much of it

Is this a sequence of some conversation or it starts  a new topic? Seems that te beginning is missing...

R. Bott's "On some recent interactions between mathematics and physics" in the canadian mathematical bulletin is much more readable; I might have gotten through more than ten pages before dying.


limit
trigonometry
real analysis
real analysis unsolved

$a_k=\frac{\sin(k)}{|\sin(k)|}$

Compute

 $\lim_{n\rightarrow+\infty}\frac{\sum_{k=1}^{n}a_ka_{k+1}}{n}$

Some post was deleted between 21/11 to 3/12

I know. It was a joke!  :lol:  :lol:  :lol:

I'm very sorry to have kept you waiting for you and my delaying reply, Moderator [b]Moubinool[/b].

I will answer as follows.

First ,there exist integers $m\geqq0$ such that  $m\pi <n+1<(m+1)\pi$ $\ \cdots\ $(☆)   for any natural numbers $n$.

Therefore since there is the number of $m$ for $k$ which satisfies $\sin k\sin (k+1)<0\ (1\leqq k\leqq n)$,

let $S_n=\frac{1}{n}\sum_{k=1}^n a_k a_{k+1}$,we have $S_n=\frac{1}{n}\{(n-m)+(-1)\cdot m\}=1-2\frac{m}{n}$

From (☆), we obtain $\frac{1}{\pi}\left(1+\frac{1}{n}\right)-\frac{1}{n}<\frac{m}{n}<\frac{1}{\pi}\left(1+\frac{1}{n}\right)$ . By Archimedes' axiom $\lim_{n\to\infty}\frac{m}{n}=\frac{1}{\pi}$

Consequently the desired limit is,  as [b]Kent Merryfield [/b]obtained, $\lim_{n\to\infty}S_n=1-\frac{2}{\pi}$

kunny

hey, this is the first time i see some person who calls a formula question mark :D

Change your fonts, because it is asterisk.  :D  :D  :D


calculus
integration
geometry
geometric transformation
rotation
function
real analysis

Hello, i'm new around here. I was having trouble with a problem, i thought i could look for help on the net. Anyway here's the problem:

Calculate the volume of a solid obtained by the rotation around Ox of all points (x,y) in RxR where y >= x*x, y <= square root of x and y <= 1/(8x).

What I did was:
1/(8x) = x*x so x = 1/2
1/(8x) = square root of x so x = 1/4

And then i found the integrals which are:
+ Integral of square root of x times dx between 0 and 1/4
- Integral of x * x times dx between 0 and 1/4
+ Integral of 1/(8x) times dx between 1/4 and 1/2
- Integral of x * x times dx between 1/4 and 1/2

So this gives me the area, but i have no idea on how to calculate the volume.
Hope someone can help.

Tks...
And i'm sorry about my english, this is my first time writing math in english.

Let  $V_{1}$ be the volume obtained after rotation of $\sqrt{x}$  between the points $0$ and $\displaystyle{ \frac{1}{4}}$ (that is the point of intersection of $\sqrt{x}$ and  $\frac{1}{8x}$ )

$V_{2}$ be the volume obtained after rotation of $\frac{1}{8x}$ between the points $\displaystyle{\frac{1}{4}}$ and $\displaystyle{\frac{1}{2}}$ (the intersection of $\frac{1}{8x}$ and $x^2$ ) and $V_{2}$ be the volume

obtained after rotation of $x^2$ between the points $0$ and $\displaystyle{\frac{1}{2}}$ .

Then the volume you want is $\displaystyle{V=V_{1}+V_{2}-V_{3}}$

The volyme of revolution of the function $y=f(x)$ beetween the poins $x=a, \,\, x=b$ is given by $\int_{a}^{b} \pi y^{2}\, dx$

A quick calculation gives $V = \frac{ 29 \cdot \pi }{30 \cdot 32 }$ .
[b]
Check again everything ![/b] Your english are good, mine are not so i hope you understund what i write
and sorry if i am wrong.

 :)


inequalities
geometry
geometric transformation
reflection
inequalities proposed

Thank you, Can. My Yahoo Message is false these days, so I can not connect you. Your problem is very nice and the same as an old problem I sent you,

If $a,b,c$ are three length sides of an triangle then
\[4(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3) \ge \frac{a^{2}+b^{2}}{a^{2}+c^{2}}+\frac{a^{2}+c^{2}}{b^{2}+c^{2}}+\frac{b^{2}+c^{2}}{a^{2}+b^{2}}. \]

Very nice, 10maths. It is the same as my solution. Thanks for notice my mistake in typing. One of a nice application is the following inequality I sent to Math Reflection (not issued) before

If $a,b,c$ are positive and $k\ge max(ab,bc,ca)$ then
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{a^{2}+k}{b^{2}+k}+\frac{b^{2}+k}{c^{2}+k}+\frac{c^{2}+k}{a^{2}+k}.\]

When does the equality occur?

then when does the equality hold, 10 math?  :lol:

[quote="10maths_tp0609"]a=b=c
a=c,k=ab
b=c,k=ab
...[/quote]

More exactly, the equality holds for $a=b=c$ or $a=b\le c, k=ac$ (and permutations).

[quote="toanhocmuonmau"]If $a,b,c$ are three sides of a triangle, prove that
\[\frac{a^{2}+b^{2}}{a^{2}+c^{2}}+\frac{b^{2}+c^{2}}{b^{2}+a^{2}}+\frac{c^{2}+a^{2}}{c^{2}+b^{2}}\ge \frac{a+b}{a+c}+\frac{b+c}{b+a}+\frac{c+a}{c+b}\]
:)[/quote]
In fact, we can prove that the inequality still holds in case $\max\{a^2,\;b^2,\;c^2\} \le ab+bc+ca.$


For $ a,b,c\geq 0$ Prove that
$ \frac{1}{a\plus{}b}\plus{}\frac{1}{a\plus{}b}\plus{}\frac{1}{a\plus{}b}\ge\frac{9\sqrt{3}}{2\sqrt{a\plus{}b\plus{}c}(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})}$

I think that you mean:
$ \frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\geq\frac{9\sqrt{3}}{2\sqrt{a\plus{}b\plus{}c}(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})}$.

$ \frac{5}{2}\geq\frac{9.\sqrt{3}}{4.\sqrt{2}}\Leftrightarrow\sqrt{\frac{25}{4}}\geq\sqrt{\frac{243}{32}}\Leftrightarrow\frac{25}{4}\geq\frac{243}{32}\Leftrightarrow{6\frac{1}{4}}\geq{7\frac{19}{32}}$    $ {FALSE!}$

Oh yes mihai miculita ,so Harry Potter'problem is no true  :)

Sorry  :blush:
Problem true is Erken 's Post  :blush:


Is the answer

$ \frac{1}{a\minus{}1} \minus{} \frac{2(2^n \minus{} 1)}{a^{2^{n\plus{}1}} \minus{}1} \minus{} \frac{2}{a^{2^{n\plus{}1}} \minus{}1}$  ?

Multiply the numerator and denominator of the first term of the question by $ (a\minus{}1)$, then add it with the second term, then add the entire with third term ...
Then apply $ (a^{2^{n\plus{}1}}\minus{}1)\equal{}(a\minus{}1)(a^{2^{n\plus{}1}\minus{}1}\plus{}a^{2^{n\plus{}1}\minus{}2}\plus{}..\plus{}1)$

$ \frac {a^{2^{n} \plus{} 1} \minus{} a \plus{} 2^{n}(1 \minus{} a)}{(a^{{2^n} \plus{} 1} \minus{} 1)(a \minus{} 1)}$

i have posted only the solution because it will take many pages to post the whole solution

write all terms, take their lcm and the denomenator will be a gp
$ a^{2^n} \plus{} a^{2^n \minus{} 1} \plus{} ... \plus{} a \plus{} 1$
and the numerator will be an agp
$ a^{2^n \minus{} 1} \plus{} 2a^{2^n \minus{} 2} \plus{} 3a^{2^n \minus{} 3} \plus{} ... \plus{} (2^{n} \minus{} 1)a\plus{}2^n$
evaluate them separately and we'll get the above solution

note:
$ (a \plus{} 1)(a^2 \plus{} 1)...(a^n \plus{} 1) \equal{} a^{2^n} \plus{} a^{2^n \minus{} 1} \plus{} ... \plus{} a \plus{} 1$


The demand function for do-it-yourself replicas of Stonehenge is given by p=1000(3/\-q) where p is a price in dollars and q is a quantity of replicas in hundreds demanded at that price. What quantity will be demanded at the price of 12340.00?

I have no idea what to do with this problem

Once you get to there, fireemblem13, you can use $\log_a b = \frac{\ln b}{\ln a}$

[quote="DanK"]Or $\log_ab=\frac{\log a}{\log b}$...

That's what I did too, but I got the demand to be $\approx -2.3$...[/quote]

It's called the "change of base rule". Umm... $\log{\frac{1}{12.34}}$ must be negative and $\log{3}$ must be positive. So... :?

It's just the question itself that is flawed. In order for the price to be greater than 1000, $3^{-q} > 1$ which will only happen if q is negative.

Perhaps if it was $q$ instead of $-q$, the answer would be $\frac{\log{12.34}}{\log{3}}=2.287$ Multiplying this by $100$ as said in the original problem, we get $\boxed{229}$.

Yeah, I was thinking it might be something like that...


abstract algebra
group theory
superior algebra
superior algebra unsolved

[quote="ZetaX"]Since it's not transitive, you can't even define the quotient group.[/quote]

ok. But here we can, since there is also given that $M \lhd G$, right? Does the rest hold too then?

[quote="jerrific"]If $M,N \lhd G$, $M \lhd N$ and $G/N$ is cyclic, can I conclude that $G/M$ is abelian (cyclic)? and $G/M \leq G/N$ ? and what if $|M/N|=2$?
Since the property 'normal subgroup' is not transitive, I'm not sure at all.[/quote]

It is clear that $N/M \cong Z_{2}$ is abelian. Furthermore is $N/M$ a subgroup of $G/M$, that is contained in $Z(G/M)$. It is easy to check that if $\frac{G/M}{N/M}\cong G/N$ is cyclic, and this is the case, $G/M$ is abelian.

[quote="blue_se7en"]why the property "normal subgroup" is not transitive?Can some one give me one example  :([/quote]

http://planetmath.org/encyclopedia/NormalityOfSubgroupsIsNotTransitive.html


Given triangle $\triangle ABC$, let $f(A,B,C)=\sin A+\cos A\sin B+\cos A\cos B\sin C$. Find:

(a) the maximum of $f(A,B,C)$
(b) the minimum of $f(A,B,C)$

A guess:
[hide]Minimum: $\frac{7\sqrt{3}}{8}$?[/hide]

And why is that the minimum? I say your answer is wrong unless you present your proof :D

[hide]The min and max should be from the extremes (a $60^\circ-60^\circ-60^\circ$ triangle and a degenerate triangle).  We then find $f(60,60,60)=\sin 60+\cos 60\sin 60+\cos 60\cos 60\sin 60 = \frac{7\sqrt{3}}{8}$
We then test some arbitrary triangle which is not an extreme. We use a $30^\circ-60^\circ-90^\circ$ triangle for simplicity. We find the value of $f(30,60,90)$ to be $\frac{11 + \sqrt{3}}{4}$, which is greater than $\frac{7\sqrt{3}}{8}$. Thus $\frac{7\sqrt{3}}{8}$ is the minimun.

The angles of a degenerate triangle (line) confuse me so I didn't evaluate it.[/hide]

But, $f(90,60,30)=1$? Sorry but there are exceptions in your argument and you have work through carefully.

More importantly, it's not even clear what boundary values are when you have three variables which are dependent on each other.

I agree with what JBL is saying, so I'll allow degenerate triangles (usually this happens with min/max problems). So in this case, $A+B+C=180$ with $0\le A,B,C\le 180$.


For $ a,b,c \in \mathbb{R^ \plus{} }$ prove inequality:

$ \frac {a^2b \plus{} b^2c \plus{} c^2a}{3(a^3 \plus{} b^3 \plus{} c^3)} \plus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} ac \plus{} bc} \ge \frac {4}{3}$

I forgot to say before. Please don't write official solution. I'm looking for a more beautiful. If someone wants to see [b]ALL BSMC 2008. and 2009. PROBLEMS WITH SOLUTIONS[/b] you can find them [url=http://web.studenti.math.hr/~mornik/BSMC.htm][b]here[/b][/url].


calculus
parameterization
advanced fields
advanced fields unsolved

1 - This is easy algebra using that the parameters are constants.

2 - Assuming that this is meant after taking expectations (which coincides with taking conditional expectation w.r.t. the trivial $ \sigma$-algebra augmented by all null-sets) note that the limit stems from the first summand only due to the symbolic relations $ dW_t\cdot dW_t\equal{}dt$ and $ dW_t\cdot dt\equal{}0$. They are derived from the defining properties of the Brownian motion. The second term trivially tends to 0.

3 - Don't what the author is asking for here since the should be a bunch of assumptions to be put in place if, e.g., $ \sigma$ and $ r$ are (time-dependent) processes.


MATHCOUNTS
geometry
3D geometry
number theory
prime numbers
exterior angle

I was just reviewing the MATHCOUNTS Problem Series transcripts and found the part:

"aanjohri (20:48:02)
is there a trick to find out if a number is prime or not?

rrusczyk (20:48:27)
This is a big, rich area of mathematics - ask this on the message board and we will discuss it!"

I couldn't find this in the message board. "Is" there a trick?  :D  I got all of the dividing tricks down for 7, 13, 17, 19, and all those hard ones, but I haven't found a trick for immediate knowing if a number is prime or not...

BTW, "After being painted, a solid wooden cube whose edge is 4 cm is cut into 64 small one-centimer cubes. How many of these small cubes will have exactly three painted faces?" Two painted faces? One painted face? At least 2? At least 1?

I always logic and such on these problems. How do ya do it geometrically?

Oh yeah, I don't really get the stuff about arcs and such. Can u ever find the length (without the knowing of the circumference) or can u only know the degree? Yeah, my geom. really sux....

I don't really know if there is a way to find if a number is prime, but about your second question, Only corners have 3 painted faces so 8 cubes, only edges have 2 painted faces so 24, only centers on edges have 1 painted face so 24. For at least 2, you do amount with 2 + amount with 3, or 32, with at least 1, take amount with 1 + amount with 2 + amount with 3 or 56.

about arcs, can you give an example?

to find prime numbers you just kinda have to check divisibility. have them memorized up to 200 and you should be fine.

second question: 8 for 3 painted, 24 for 2 painted, 24 for one painted, and 8 for no painted.

third question: you use degrees and the radius (you need both) unleess you are given other lengths

edit: ihatepie beat me to it

If you want to find if a number is prime or not this method usually works:
- find its square root
- if it is not divisible by any of the prime numbers up to its square root it means the number is [u][b]most likely [/b][/u]to be prime.

Example: lets check if 113 is prime or not: its square root is close to 11. and 113 is not divisible by 11 and any prime that is less than 11. so it is a prime number.

[quote="al-kharizmi"]- find its square root
- if it is not divisible by any of the prime numbers up to its square root it means the number is [u][b]most likely [/b][/u]to be prime.[/quote]

Actually, if a number isn't divisible by any of the prime numbers up to its square root, then it's [b]guaranteed[/b] to be prime.  This is because a composite number must have a nontrivial factorization $ n \equal{} ab$ where $ a, b > 1$ and then either $ a$ or $ b$ must be less than or equal to $ \sqrt{n}$.

This is all you'll need to do in contest situations, but the divide-and-check method becomes very cumbersome for numbers with four or more digits because it gets harder to divide by large primes by hand.  There are [b]very[/b] sophisticated algorithmic tests for determining whether a number is prime, but they're beyond the scope of this forum.

Also any prime number $ p$ is in the form:

$ p\equal{}6n\pm1$

So thats one quick way to check if the number is [b]not[/b] prime (it doesn't guarantee if $ p$ is prime).

[quote="happyme"]Also any prime number $ p$ is in the form:

$ p \equal{} 6n\pm1$

So thats one quick way to check if the number is [b]not[/b] prime (it doesn't guarantee if $ p$ is prime).[/quote]

Any prime number greater than $ 3$.  This is equivalent to checking divisibility by $ 2$ and $ 3$, both of which are very easy anyway.

[quote="toadoncart"]Alright, thx. I think i got it. BTW, what's the exterior angle of a polygon? Like for an =lateral triangle, the exterior angle is 360-60=300, right? or is it 180-60?[/quote]

*cough**cough*, still not answered....(lol)  :lol:

[quote="uldivad9"]Happyme, I think you have something wrong with that. Are you trying to say that all numbers expressible that way are prime (25?) or that all numbers that are not expressible that way are composite (3?)?[/quote]

1. The sum of the exterior angles is always $ 360^\circ$, and if the polygon is regular, then each of the exterior angles has the same measure. In that case, you can divide.

2. If a number cannot be expressed as:

$ p \equal{} 6n\pm1$

then the number is NOT prime. However if the number can be expressed in that form, it doesn't guarentee that the number is prime. (If $ p > 3$). Sorry for any ambiguity.

exterior=180-interior
now how hard was that?
now plz join the im a PHAILURE club?

[quote="stevenmeow"]
now plz join the im a PHAILURE club?[/quote]

How?? Oh btw look at the hax comment I made on your blog... :lol:


trigonometry
trig identities
Law of Cosines
geometry
angle bisector
AMC

In $\triangle ABC$, $M$ is the midpoint of side $BC$, $AN$ bisects $\angle BAC$, $BN\perp AN$ and $\theta$ is the measure of $\angle BAC$.  If sides $AB$ and $AC$ have lengths $14$ and $19$, respectively, then length $MN$ equals

[asy]
size(230);
defaultpen(linewidth(0.7)+fontsize(10));
pair B=origin, A=14*dir(36), C=intersectionpoint(B--(9001,0), Circle(A,19)), M=midpoint(B--C), b=A+14*dir(A--C), N=foot(A, B, b);
draw(N--B--A--N--M--C--A^^B--M);
markscalefactor=0.1;
draw(rightanglemark(B,N,A));
pair point=N;
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$M$", M, S);
label("$N$", N, dir(30));
label("$19$", (A+C)/2, dir(A--C)*dir(90));
label("$14$", (A+B)/2, dir(A--B)*dir(270));
[/asy]

$\displaystyle \text{(A)} \ 2 \qquad \text{(B)} \ \frac{5}{2} \qquad \text{(C)} \ \frac{5}{2} - \sin \theta \qquad \text{(D)} \ \frac{5}{2} - \frac{1}{2} \sin \theta \qquad \text{(E)} \ \frac{5}{2} - \frac{1}{2} \sin \left(\frac{1}{2} \theta\right)$

[hide="Hint 1"]
Extend $BN$[/hide]
[hide="Hint 2"]
Let $P$ be the intersection of $\overleftrightarrow{BN}$ and $\overline{AC}$.
What kind of triangle is $\triangle APB$? ;) [/hide]

[hide="My solution"]Extend $BN$ and let its intersection with $AC$ be $P$.  Now we have $\Delta APB$, which is an isosceles triangle, with $PB = (2)(14)\sin (\angle A/2) = 28 \sin \alpha$, where $\alpha = \angle A/2$.

Now let's examine $\Delta PCB$.  Let $CM = MB = x$.   Also let $\angle CBP = \gamma$ and apply the Law of Cosines to get:

$5^2 = (28 \sin \alpha)^2 + (2x)^2 - (2)(28 \sin \alpha)(2x) \cos \gamma$

$\Rightarrow \cos \gamma = \frac{4x^2 + 784 \sin^2 \alpha - 25}{112 x \sin \alpha}$

Now let's examine $\Delta BNM$ and apply the Law of Cosines one more time:

$MN^2 = (14 \sin \alpha)^2 + x^2 - (2)(14 \sin \alpha)(x) \cos \gamma$

$= 196 \sin^2 \alpha + x^2 - (x^2 + 196 \sin^2 \alpha - 25/4) = \frac{25}{4}$

$\therefore MN = \frac{5}{2}$[/hide]

[hide]There is a lot of info in this problem.  Let's just try to take advantage of it.  We know that AN is an angle bisector, so we want to use the angle bisector theorem.  Either we extend AN or BN to do that.  It turns out that extending BN leads to a very quick solution.  As soon as we do that and hit AC at the point D we stare at the picture for a bit.  Hopefully we didn't draw a picture like the misleading one chess64 tried to give us.  It should be pretty clear that since angles ANB and angles AND are both right angles, that our triangle APB is in fact isosceles.  Thus, AD = 14 and DC = 5.  Now notice that BMN is similar to and half the size of BCD and so our length is MN = 2.5[/hide]

[quote="pdiao"][hide]Now notice that BMN is similar to and half the size of BCD [/hide][/quote]Hmm... that would have been really helpful to know.  I didn't think to look at the fact that they were similar - that would have simplified my solution enormously.  I do suck at geometry...  :blush:

[quote="frt"]Since $BN=ND$ and $BM=MC$, we have $MN\parallel CD$ and $MN=\frac12 CD$. It's called Midline theorem.[/quote]
Yes, I see how the proof worked - it just took me a moment to see the similarity (and Chess's diagram definitely didn't help  :P ).

[quote=djmathman]The problem statement needs to be adjusted slightly, because $\theta$ is actually defined in the original version:

[quote]
In $\triangle ABC$, $M$ is the midpoint of side $BC$, $AN$ bisects $\angle BAC$, $BN\perp AN$ and $\theta$ is the measure of $\angle BAC$.  If sides $AB$ and $AC$ have lengths $14$ and $19$, respectively, then length $MN$ equals
[/quote][/quote]

my guy you just bumped a 17 year old thread [hide]use trig[/hide]


Find the largest real $ T$ such that for each non-negative real numbers $ a,b,c,d,e$ such that $ a\plus{}b\equal{}c\plus{}d\plus{}e$: \[ \sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\plus{}e^{2}}\geq T(\sqrt a\plus{}\sqrt b\plus{}\sqrt c\plus{}\sqrt d\plus{}\sqrt e)^{2}\]

I think T = 1/ sqrt(125) . Is it correct?

I don't think so. My guess is [hide]$ \frac{\sqrt5}{(\sqrt{2}\plus{}\sqrt{3})^{2}}$.[/hide]

by putting $ a\equal{}b\equal{}3$ and $ c\equal{}d\equal{}e\equal{}2$ we find $ \frac{\sqrt{30}}{6(\sqrt{3}\plus{}\sqrt{2})^{2}}\geq T$
to prove that for such a $ T$ the inequality always holds notice that $ x^{2}\plus{}y^{2}\geq 2(\frac{x\plus{}y}{2})^{2}$ and $ x^{2}\plus{}y^{2}\plus{}z^{2}\geq 3(\frac{x\plus{}y\plus{}z}{3})^{2}$ which means that the worst case is when $ a\equal{}b$ and $ c\equal{}d\equal{}e$ for which we have equality  :)


Let us bash committees.

BASH! BASH! BASH!

Yay! We have successfully bashed committees.

...next?

[quote="suma_milli"]Let us form the committee for the resurrection of the Illinois forum.

:P

(Apparently no one checks the state forums for spam, proof of this is shown by the missouri forum. :D )[/quote]
well then maybe we should not make the IL forum spammy :P

True!  :lol:

Here's a random problem from the mathcounts book I have next to me:

A function $ (x,y)$ is defined by the following rules:
$ A(0,n) \equal{} n \plus{} 1$
$ A(m,0) \equal{} A(m\minus{}1, 1)$
$ A(m,n) \equal{} A(m\minus{}1, A(m,n\minus{}1))$
for m and n natural numbers.

A(2,3) = k, where k is a whole number.  Find the value of k.

Problem:  If we make the IL forum spammy, fewer people will come.  If fewer people will come, the forum will die.

Problem:  If we don't make the IL forum spammy, we won't think of enough stuff to post.  If we don't think of enough to say, the forum will die.

So either way, the forum will die.

I'm such an optimist. :rotfl:

That is quite a dilemma..

Maybe we should start an Illinois math marathon, b/c people from IL are cool enough to have their own math marathon. :P

Yup, that's it.  You should hope nobody who answers potd sees your post :P

Did u like my long solution? And the way I put in mr. greens as i went along?

Haha I'm so funny. :D  :D  :D (Usually that's just me online, but when u meet me in reality, I don't think that I seem that funny; i seem rather grave) :huh:

Guassy,  rather grave?  .___. *gasp*

When you meet me I'm sure I seem like a complete idiot!  :rotfl:  :P

Why do you think so?

I always imagined you look like  :D

Or how about this: wear green shirts or paint our faces green and then go meet at states and wave a Mr. Green flag? :D

That's a nice idea too.  Not sure if my teammates will like it since I think green is the color of a rival school :P

Lol, green is actually one of my school's colors :D

My school's colors are blue and white.  Like the school spirited arrow :arrow:

lol maybe the MO people will come and try to sabatoge us.  Then again..we're the only people that ever post in the IL forum.   :roll:

Because we're the only active Illinois AoPSers! :D

Congratulations everyone we have succeeded in the renewal of the IL forum! :D

Yes we did. Thanks to myself, webster, rose, kevin, and...MOST IMPORTANTLY:

[hide]Smillie/Smiley Monster :P  :D [/hide]

for the effort!

xD

 :rotfl:

wow, you're actually referring to me by my name. that's rather rare on these forums.
although i think going by a username alone makes me seem rather MYSTERIOUS and B.A.

...

rawr.

lololol I have a feeling that nearly all of the people who post here regularly know MY real name... :maybe: and perhaps we all know each other's... :huh:

[quote="Aryth"]hey when you go to ARML tryouts try to find me. i won't be able to find you because

1. i will probably forget
2. your facebook photos are not very helpful for identification purposes

so yeah, say hello. although, it could be very difficult to look for people because the classroom (singular) we all take the tryout in is rather crowded.[/quote]
well i might be able to find u
i've seen lots of pics of u

[quote="avec_une_h"]I probably can't find you, but I'll wear my AoPS shirt.[/quote]
Good call. :D  :D  :D

just randomly yell out "KHU" when you think you see me. if the person in question turns around in a few circles, it's probably me  :wink:

[quote="suma_milli"]Well since gauss already knows what I look like she can find me ( :| ) and then we can find aryth and we'll pretty much be able to tell who avec is unless there are multiple people wearing AOPS shirts.[/quote]
i doubt anyone would find me :|  :huh: b/c i epicly failed at usamts.
only 12 on round 3. :mad: 3/3/2/1/3.


There will be some admins of the ML/AoPs site at the IMO 2006: [url=http://imo2006.dmfa.si/sub_Coordinators.html]Valentin,[/url] [url=http://imo2006.dmfa.si/participants/CAN.html]Naoki,[/url] and [url=http://imo2006.dmfa.si/participants/USA.html]Richard.[/url]

[b]Richard:[/b]

[img]http://imo2006.dmfa.si/participants/Image757.jpg[/img]

[b]Naoki:[/b]

[img]http://imo2006.dmfa.si/participants/Image213.jpg[/img]

[b]Valentin:[/b]

[img]http://imo2006.dmfa.si/participants/ThumbImage905.jpg[/img]

Yeah, blame basketball and ageing for that :D

[quote="joml88"]Wow, Valentin looks much different than in the picture on this page: http://www.artofproblemsolving.com/AboutUs/AoPS_A_Company.php[/quote]

Indeed! In fact, I was a bit tired and did not realize it was he who took that picture of me at last year's hurricane party until after he'd left.  :blush:


A machine gives out five pennies for each nickel inserted into it. The machine also gives out five nickels for each penny. Can you, starting out with one penny, use the machine in such a way as to end up with an equal number of nickels and pennies?

When you put one coin into the machine and gain five, the parity of the amount of coins you have remains the same. You start out with an odd amount of coins, so you will always have an odd amount. You can only have an equal number of both coins if you have an even amount of total coins, so it is impossible.


prove if x,y,z is a real number then
  \sum x 2 - \sum xy \geq 3/4(x-y) 2  :D

It can be written (a-c)^2+(b-c)^2>=(a-b)^2/2 and it's trivial.


[quote="Yarka"]Hi,
How can put on a part of a latex document (a few  lines) more margin on the left?

Thanks.[/quote]

Try this site:

[url]http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Layout#Layout_of_the_Page[/url]

You set the parameters (described on the above link) to set margins, amount of paragraph indent, etc.

Those links deal with margins for the whole document rather than for a few lines.

A very simple way is to use \parbox. Suppose you want your text to be 6cm wide on the right (so a large margin on the left of that text). Put [code]\hfill \parbox{6cm}{Put the text you want on the right with a large margin here}[/code]


This was once this very unusual man named John. What made him unusual was that whenever he was given an instruction, he would always obey it *exactly*. No matter what it was or what it said, he would carry out that instruction. One day he was sitting in the park when he noticed a sign. It was not an unusual sign - you could easily find it in any park. On seeing this sign, John went up to someone and said, very angrily, "There is absolutely no way that you are putting me in that bin!".

The puzzle: what was on the sign?

[and if you think there are various answers, I will accept anything that fits these conditions. You may be surprised.]

the sign told him to say those exact words in that exact manner. cause um...sum ppl put it there as a joke?

[quote="TripleM"]It was not an unusual sign - you could easily find it in any park.[/quote]

(When I say easily find it, it could of course have slightly different wording, but thats not important. The important point was that the message on the sign was not unusual for a park.)

Too hard? Maybe I'll have to start giving out hints..

"Keep off the grass"? (If the bin was the only way to do so...)

:) I think every single person that I've seen try this puzzle has fallen into the same trap (including myself). That answer can't work. Read the puzzle statement more carefully.

Does it have anything to do with "john" being slang for a bathroom?

hm....is it a picture of sumone throwing this paper into a bin?

A sign saying "put the porta-potty in the recycling bin" ?

In my world, you see that all the time :P .

Nah, I just made up the name John. Its irrelevant. However, still none of your answers so far would cause John to act that way..

Oh!

A sign saying "Throw me away in this bin with an arrow pointing to a trash can"
A small sign right below shows a picture of a piece of garbage.

Please place the trash in the bin....

Is John trash?  :D

Still doesn't work. I didn't give you much information in the puzzle, but the most important part makes that answer not work, like the others :) hehehe.. don't worry, as I said, I kept making the same mistake as well.

Is the "someone" relevant to the solution?  Is it a particular someone whose identity isn't disclosed, or is it some arbitrary stranger?

Nope, its just an arbitrary person. Theres no tricks in the wording - everything you need to know is in that paragraph.

MithsApprentice wrote:TripleM wrote:OK, I think its time to reveal the answer. In spoiler.
[hide]
Refuse to be placed in the bin.

(of course, 'refuse' being another word for rubbish.)
[/hide]

Yea that's definitely a New zealand thing...lol...i dunno of anyone in the US that uses refuse for garbage...

This is nothing to do with New Zealand.. its not used commonly here either (this puzzle was made up by an American and solved by Americans ) but you'll definitely find it in your dictionary. See - LynnelleYe knew it

[quote="TripleM"]This is nothing to do with New Zealand.. its not used commonly here either (this puzzle was made up by an American and solved by Americans :P) but you'll definitely find it in your dictionary. See - LynnelleYe knew it :P[/quote]

Well no, I mean I've heard of it and I knew what it meant...it's just...so...uncommon...

And ur right it probly is in the dictionary...but oh well...

[quote="Fiachra"]I agree with the above. It doesn't even seem obvious after you know it (which most answers to this type of problem do).
Most Irish people would know and perhaps use the word "refuse" in that sense, if only very infrequently.[/quote]

As was discussed later in this thread (not here anymore) its basically the 'official' word for rubbish - rubbish would be used in everyday language, where 'refuse' is used when talking a lot more formally. I'm very surprised that someone would not be able to understand the answer.

The "slogan" on the garbage trucks around here is "We don't refuse refuse."

[quote="RobinHood3000"]In Taiwan (or at least in Keelung), the garbage trucks play "Fr Elise."  Weird; you'd think the smell would be fair warning enough...[/quote]

 :lol:
At least that name contains the letters of 'refuse' ;) How about.. "I L Refuse"..

[quote="RobinHood3000"][quote="Treething"]That would be funny if they played the same song as the ice cream truck :P .[/quote]

Jeez...a blind man with no sense of smell would be left standing on the street going, "Hey...this ice cream tastes like toilet paper..."[/quote]

Why are blind people even on the streets?

If you don't have a sense of smell, can you actually taste anything anyway? Or was that what you meant :P

No...the implication was that it was ACTUAL toilet paper...

And as for blind people in the streets, the blind need to shop too...


I need to see a solution for this one:
Find all solutions of

$x^{\log_{10}x}=\frac{x^{3}}{100}$

[hide]
$x^{\log_{a}x}=a$
Following basic log rules.
QED.[/hide]

Taking logarithm with base 10 of both sides, you will get the quadratic equation with respect to $\log_{10}x$.

[quote="CircleSquared"]
$x^{\log_{a}x}=a$
Following basic log rules.
QED.[/quote]

dude wtf are you high?  Thats not a log rule

I'm sorry, but I do not know how to use latex. Anyways, here is my solution:
[hide]
take log base x of both sides and get
logx=logx(x^3)/100
simplifying the right side, you get
logx=3-logx100
move the log to the left side of the equation
2logx10 + log x = 3
set a=log x10
2a + 1/a =3
solve for a and get a = 1/2 or 1
so logx10 = 1/2 or 1
solving for x, you get x= 10 or 100[/hide]

[quote="CircleSquared"]:huh:  Sorry bout that usaha (no I'm not high, that's illegal)
I thought $x^{\log_{a}x}=a$ but maybe I'm tired.
Will someone correct me please?[/quote]
No, you're getting that mixed up with $a^{\log_{a}x}=x$

:wallbash_red:
Ok, that explains it, thanks diophantient.
Please ignore my solution.

usaha was joking..but anyways
a solution

take $\log_{10}$ to both sides and you get

$(\log_{10}{x})^{2}=\log_{10}{x^{3}}-\log_{10}{100}$ so now you call $\log_{10}{x}=y$ so its a quadratic equation and

$y^{2}-3y+2=0$, so $y=\log_{10}{x}=$$1$ and $2$, so $x=10$ and $100$

note I skipped a few steps involving knowing basic log properties such as $\log_{a}{b^{x}}=x\log_{a}{b}$

[quote="Whatthedeuce"]I'm sorry, but I do not know how to use latex. [/quote]

that's perfectly alright... as long as you start learning now  :D. I think that you can study, practice and use it well under half an hour.

heh, im also one of those who dont use latex very well, i've been trying to get around to learning it, but im sort of a procrastinator in that aspect, but back to the topic:

im a beginner with logarithms, and basically just know what they are, so im not sure what you guys meant with "take log 10 to both sides" ?. But here's my method without that:
[hide]
Let a=log x , then we get:
10^a= x, so
we have 10^ (a^a) = 10 ^ (3a-2), so by taking away the base of 10's we get a ^ a = 3a - 2 , so a = 1 and 2 (this part is sort of messy, and unreliable, so im sure you had better solutions) , so 10^a = 10 and 100.
That was pretty messy without latex, i should probably learn fast...
Could someone explain how to take log of both sides, that would probably get rid of the messy part of my solution
Thanks [/hide]

I'm not sure why taking the log of both sides makes it simpler either, but it seems really important to know and I would like to know it...could someone show me? How exactly, does taking log base 10 of x simplify the left side of the equation?

Sorry, I phrased it wrongly. I get why it simplifies the left side, but I don't get how it actually ends up as (log x)^2.

EDIT: But I think I figured out how it simplified the left side...Basically, by taking the log of both sides, you end up with (log x) to the (log x) power, and using one of the properties, you end up with (log x) times (log x), which is (log x)^2, right?

Takign the $\log$ of both sides helps make the $x$ in the LHS no longer in the exponent.

thanks, i kind of get it now, assuming the right side is supposed to be log sub 10 x.


linear algebra
matrix
search
linear algebra unsolved

Prove the following theorem about stochastic matrices:

Let $ A\equal{}[a_{ij}]$ be a stochastic matrix, that is $ a_{ij}>0$, $ \sum_{j\equal{}1}^{n}{a_{ij}} \equal{} 1$, then:

1) A has got an unitary eigenvalue and the dimension of the associated eigenspace is 1
2) Let $ A^n \equal{} [a_{ij}^{(n)}]$ and let $ u\equal{}(u_1, ..., u_n)$ be a normalized eigenvector with eigenvalue 1, then: $ \displaystyle lim_{n \to \infty}{a_{ij}^{(n)}} \equal{} u_i$

1) Clearly 1 is an eigenvalue since $ Av\equal{}v$,where $ v\equal{}(1,1, \ldots ,1)^T$.
Since $ A$ is primitive, by Perron-Frobenius theorem, the dimension of the associated eigenspace is 1.

2) See http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=846370849&t=43670


number theory
least common multiple
prime factorization

what is the LCM of 1,2,3,4,5,6,7,8,9, and 10
hint: the lcm of 10, 9 and 8 are multiples of 1, 2, 3, 4, 5, 6, 8, 9, 10 :starwars:

[hide]Write out the prime factorization of each term.  The highest power of 2 is 3.  The highest power of 3 is 2.  All other primes only have 1.  That makes the lcm $2^3\times3^2\times5\times7=2520$.[/hide]

[hide]
2, 4, and 6 are multiples of 8.
3 and 6 is a multiple of 9.
2*5=10
So, the answer is
$8\cdot9\cdot5\cdot7=2520$.[/hide]


floor function
logarithms
number theory open
number theory

Define a sequence $a_{n}=a+(n-1)d$ where $(a,d)=1$ $,a,d,n\in\mathbb{N}$ and $a_{i}>0$ for all $1\leq i\leq n$ .

among all the terms in $a_{n}$ , we define $p_{i}$ ,$1\leq i\leq n$ as the largest prime divisor of $a_{i}$ such that they form a set
$S=\{p_{1},p_{2}\cdots ,p_{n}\}$ .Suppose there exist a $p_{k}$ such that $p_{k}=\text{max}\{p_{1},p_{2}\cdots ,p_{n}\}$ . Prove or Disprove that $p_{k}$ is unique (or prove that if $p_{k}=p_{j}$ $\implies k=j$).If it is false , give a counter-example .

Let me make a first step to the proof.

Assume the contrary, i.e., that there is a sequence $a_{1},\dots,a_{l}$ with two elements divisible by the maximum prime $p=p_{m}$ (I will use indexes to denote primes in their natural order: $p_{1}=2,\ p_{2}=3,$ etc.). Without loss of generality, we assume that $p$ divides $a_{1}$ and $a_{l}.$

Since $(a,d)=1$ we have:
$\gcd(a_{i},a_{j}) = \gcd(a_{i},a_{i}+(j-i)d) = \gcd(a_{i}, j-i)$ for any two indexes $1\leq i<j\leq l$;
and, in particular,
(*) $\gcd(a_{i},a_{j})$ divides $j-i.$

Since $p$ divides both $a_{1}$ and $a_{l},$ it must divide $l-1$ as well (according to the property (*)). We can focus on the case when $l=p+1$ (since it is the minimum possible length).

So, in the sequence $a_{1}, \dots, a_{p+1},$ the elements $a_{1}$ and $a_{p+1}$ are divisible by $p$ and all prime factors of intermediate elements are smaller than $p$ (i.e., belong to the set $\{ p_{1}, \dots, p_{m-1}\}$).

The property (*) implies that there are at most $1+\lfloor\frac{p}{2}\rfloor$ elements among $a_{1}, \dots, a_{p+1}$ that are divisible by $p_{1}=2$; there are at most $1+\lfloor\frac{p}{2^{2}}\rfloor$ elements divisible by $2^{2}$; ... there are at most $1+\lfloor\frac{p}{2^{k}}\rfloor$ elements divisible by $2^{k}$ (where $k$ is chosen such that $2^{k}\leq p<2^{k+1}$, i.e., $k=\lfloor\log_{2}p\rfloor$). And there is at most one element that is divisible by $2^{k+1},$ we denote such element $b_{1}.$
The total maximum power of $2$ dividing all elements $a_{1},\dots,a_{p+1}$ (except $b_{1}$ if it exists) is bounded by
\[(1+\lfloor\frac{p}{2}\rfloor)+\dots+(1+\lfloor\frac{p}{2^{k}}\rfloor) = k+\lfloor\frac{p}{2}\rfloor+\dots+\lfloor\frac{p}{2^{k}}\rfloor = k+\sum_{i=1}^{\infty}\lfloor\frac{p}{2^{i}}\rfloor.\]

Similarly we can consider $p_{2}=3$ and define element $b_{2}$ (if it exists) which is divisible by $3^{1+\lfloor\log_{3}p\rfloor}.$ The rest elements together are divisible by at most the following power of $3$:
\[\lfloor\log_{3}p\rfloor+\sum_{i=1}^{\infty}\lfloor\frac{p}{3^{i}}\rfloor.\]

etc.

Therefore, the product of all elements except $b_{1}, \dots, b_{m}$ (if they exist, so the total number in the product is $p+1-m$) is bounded by
\[\prod_{j=1}^{m}p_{j}^{\lfloor\log_{p_{j}}p\rfloor+\sum_{i=1}^{\infty}\lfloor\frac{p}{p_{j}^{i}}\rfloor}= p!\cdot \prod_{j=1}^{m}p_{j}^{\lfloor\log_{p_{j}}p\rfloor}\leq p!\cdot p^{m}.\]
From this we can conclude that $a_{1}\leq (p!\cdot p^{m})^{\frac{1}{p+1-m}}.$


A regular polygon with exactly 20 diagonals is inscribed in a circle. The area of the polygon is 144 :sqrt: 2. Find the area of the circle.

[hide="Hint"]The shape is an octagon![/hide]

this a very easy geo problem
to find the number of sides of the polygon, you know that:
$\frac{n(n-3)}{2}=20 \rightarrow n=8$
so you have a octogon.
so if you inscribe this octogon in an circle you will have eight isoceles triangles that have sides $r,r,a$ where $a$ is the lenght of the side of a octogon, you can easily see that the angles in these triangles is equal to $45,67.5,67.5$, now based on the area of the octogon, you can see that the area of each triangle is : $18\sqrt{2}$
and you also know that the area of the isoceles triangle is :
$\frac{1}{2} r^2 \sin(45) \rightarrow r=3\sqrt{8}$ so the area of the circle is:
$\boxed{72\pi}$


We have a prime p such that r, the remainder when p is divided by 210 (0<r<210), is a composite number which can be written as the sum of two squares. Find all possible values of r.

[hide]

umm... gcd(r,210)=1

and any prime divisor of r of the form 4k+3 has an even exponent in r?

too much work for me to list each and single r

did i include too much/not enough?

[/hide]


The teacher said to divide up into groups of 3 to 5 people. My group has 5 people including me but another group only has 2 people. Since we're the only group of 5, the teacher says we have to send a person to the other group, but do it randomly. So the teacher thinks of a random number from 1 to 5 and writes it down. Whoever guesses the number has to go to the group of 2. Supposes I can choose what position I want to be to guess, (I can choose to be the 1st, 2nd, 3rd, 4th or 5th person to guess a number) What is the probability that I will get to stay in my group when I guess in each of those positions? If I want to stay in my group, which position would I want to be? Assume no one can guess the same number somebody else before them already guessed.

:? Even though there's a lot of words, it isn't that hard.  :huh:

Here's my solution:
[hide]1st: $\frac{1}{5}$
2nd: $(\frac{4}{5})(\frac{1}{4})=\frac{1}{5}$
3rd: $(\frac{4}{5})(\frac{3}{4})(\frac{1}{3})=\frac{1}{5}$
4th: $(\frac{4}{5})(\frac{3}{4})(\frac{2}{3})(\frac{1}{2})=\frac{1}{5}$
5th: $(\frac{4}{5})(\frac{3}{4})(\frac{2}{3})(\frac{1}{2})(\frac{1}{1})=\frac{1}{5}$

Therefore, you have an equal probability of being chosen no matter what position you're in.[/hide]

I was going to solve it but now there is no point since you answered your question.

[quote="b-flat"][quote="SplashD"][quote="pgpatel"]WHy'd you answer your own question?[/quote]

'cause nobody answered it.[/quote]

Next time, wait more than one day for people to respond.  Give it 4 or 5 days, or if it goes off the front page of problems.  Then post your solution so you can give other people a chance to solve it.[/quote]

yes.  People could have probably already solved it but they didn't feel like posting it as well.  Others probably haven't seen it yet.


A hollow piece of cylindrical pipe has an outside radius of 5.6 inches and an inside radius of 5.3 inches. The pipe is 6 feet long. To the nearest tenth, how many square inches are in the total surface area of the pipe.

Any number can be written as the sum of distinct powers of two. I.e. 73= 2^6+2^3+2^0.
Write 1001 as the sum of distinct powers of two.

If the product of 9*HATBOX=4*BOXHAT, find the 6 digit numbers of BOXHAT and HATBOX

1001= 512 +256+128+64+32+8+1

2^9+2^8+2^7+2^6+2^5+2^3+2^0.

The answer to the first problem is pi * outer diameter * height =  pi * 11.2 * 72 which equals, to the nearest tenth,  2533.4 square inches!

Good job SnowHawk! I think you are correct!

Good JOB!!! But I was wondering how did you get 72?


Using some position in lectures of Hojoo Lee on his his website!  :wink:


calculus
topology
advanced fields
advanced fields unsolved

Exactly how dow e prove the generalized stokes theorem and why is it true(intuitively speaking?).

Intuitively, it's the Fundamental Theorem of Calculus.

Have you looked in Spivak's [i]Calculus on Manifolds[/i]?


So twice in the last week I have bought a Pepsi with an advertisement for a chance to win free music downloads from iTunes.  The odds of winning a free song are given as 1 in 3.  I won both times.

What is the probability I will win free songs with both of the next two sodas I buy??(theoretically of course :) )

Hmm, that really wasn't the question I meant to ask(way too simple).  I meant to ask something a littel less straight forward like :

What is the probability that I would win on at least 2 of the next three?

[hide="HINT"]

The ways you can win on at least 2 of the next 3 are:

WWW
WWL
WLW
LWW

(with W being a win and L being a loss)
[/hide]
[hide="BIG HINT"]
The prob. of the first one happening is $\left(\frac{1}{3}\right)^{3}=\frac{1}{27}$

The prob. of each of the next three are $\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)=\frac{2}{27}$
[/hide]
[hide="ANSWER"]
Adding we get $\frac{1}{27}+3\left(\frac{2}{27}\right)=\frac{7}{27}$.
[\hide][/hide]

[hide]the probability that you win on exactly 2 of the three: 2/9 (2/3*1/3*1/3*3)
probability that you win on exactly three out of the three: 1/27
so probability that you win at least two free songs is: 1/27 + 2/9 = 7/27... which is just over a 22% chance[/hide].... not looking too good.


calculus
derivative
function
integration
calculus computations

Use the Fundamental THeorem of Calculus to find the derivative of the function:

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/edba28b7dc20a91778af22008d65ae0d.gif[/img]

im not sure where to go, i rewrote it like this first:

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/4e7b4e12c738cba52a5778f14e440800.gif[/img]

and from here i went to:

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/125c2830099ac227dffd7b9e4281af2b.gif[/img]

but im at a loss of...what to do?

any ideas?

thanks alot :)

You're trying to find an expression for the function you want to differentiate (and doing it wrong). Don't: you already have what you need. What does the FTC say about the derivative of an integral?

hmm a bit confused as to where to go then, i know that the FTC integration and differention are inverse procedures. so do i need to differentiate the integral first? that sounds like it would be odd to do..
o.o

Hmm random ideas-

1) Factor $u^{2}-1$ but i don't think it helps =)
2) Cancel out some terms
3) use $\text{u}$ and $\text{du}$ to substitute but i dno't think it works for problems like these?

edit- nvm don't think canceling out will work...but experiment with the $\text{u}$ and $\text{du}$?

i don't remember but use the Power Rule to solve for a derivative and then you can solve your integral

We're only dealing with the Second Fundamental Theorem of Calculus here, for differentiating definite integrals.

[hide="hint"]It has to do with the limits of integration.[/hide]

well I think i could do it this way....

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/ec003c31a1d6ba7ef55c403efb6c289a.gif[/img]

could i not? but i still cant integrate it yet either...

If $F(t)=\int_{0}^{t}\frac{u^{2}-1}{u^{2}+1}\,du$, what's $F'(t)$? Can you write the integral you wish to differentiate in terms of $F$?


[quote="Silverfalcon"]Solve simultaneously:

$(r+1)(s+1) = 24\\
(s+1)(t+1) = 30\\
(t+1)(r+1) = 20$

Find the value of $r,s,t$.[/quote]

If you are asking for the anser, just unhide, otherwise, continue trying by yourself

[hide]
Note that the numbers $(r+1)$, $(s+1)$ and $(t+1)$ are either all positive or all negative.

Now, for simplicity we are going to rename $(r+1)=x$, $(s+1)=y$ and $(t+1)=z$, this renames the equations in $xy = 24,\ yz=30,\ zx=20$, therefore, in this new set of variables, the solutions are symetric, that is, if $(x_0,y_0,z_0)$ is a solution, then $(-x_0,-y_0,-z_0)$ is also a solution, therefore, we are going to assume that the numbers are positive.

Multipliying all the equations we have that $(xyz)^2=14400$, hence $xyz=120$ and therefore $z=\frac{xyz}{xy}=\frac{120}{24}=5$, and this implies that $y=6$ and $x=4$.

Finally, all the solutions are $(\pm4,\pm6,\pm5)$, and then, the solutions in the set of variables $r$, $s$ and $t$ are $(3, 5, 4)$ and $(-5,-7,-6)$.
[/hide]

Enjoy math!  :D

[quote="Silverfalcon"]Solve simultaneously:

$(r+1)(s+1) = 24\\
(s+1)(t+1) = 30\\
(t+1)(r+1) = 20$

Find the value of $r,s,t$.[/quote]

[hide]
R=3 OR -5
S=5 OR -7
T=4 OR -6
[/hide]

[quote="djimenez"][quote="DarkKnight"]
[hide]
R=3
S=5
T=4[/hide][/quote]
DarkKnight,
[hide]Nobody said that the numbers had to be positive![/hide]Best,[/quote]

Yeah. I realized that and fixed it. :P

[hide]
multiply the three equations together to get:
(r+1)^2*(s+1)^2*(t+1)^2=14400, take the square root of both sides
(r+1)(s+1)(t+1)=120
and since we know (r+1)(s+1)=24, we know t+1 must equal 120/24 or 5
do that with the others to get:
[b]r=3
t=4
s=5[/b]

edit: oh, they can be negative too, didnt realize that till aftter seeing djiminez's post :blush:  :P
[/hide]


quadratics algebra polynomial calculus integration geometry Support

[hide="Problem #9"]
[quote]Consider the function $f(x)=x^4-(2k+4)x^2+(k-2)^2$, where $k$ is an integer. Suppose that $f(x)$ can be written in the form $f(x)=P(x)Q(x)$, where $P(x)$ and $Q(x)$ are quadratic polynomials with integer coefficients. If $k\le2005$, then the largest possible root of $P(x)$ can be written in the form $a+\sqrt{b}$, where $a$, $b$, and $c$ are integers, and $c$ is square-free. Compute $a+b$.[/quote][/hide]
[hide="Solution"]Seeing so many squares, we might try to complete the square and then use the difference of two squares. Indeed, this can be done in several ways:[list]$f(x)=(x^2-(k-2))^2-(2k+4)x^2+2(k-2)x^2$
$f(x)=(x^2+(k-2))^2-(2k+4)x^2-2(k-2)x^2$
$f(x)=(x^2-(k+2))^2+(k-2)^2-(k+2)^2$[/list]
[size=134][b]Case 1:[/b][/size] $f(x)=(x^2-(k-2))^2-(2k+4)x^2+2(k-2)x^2$
$f(x)=(x^2-(k-2))^2-8x^2$
$f(x)=(x^2-(k-2)-2\sqrt{2}x)(x^2-(k-2)+2\sqrt{2}x)$
However, these quadratics will never have integer coefficients so we do not need to consider any further.

[size=134][b]Case 2:[/b][/size] $f(x)=(x^2+(k-2))^2-(2k+4)x^2-2(k-2)x^2$
$f(x)=(x^2+(k-2))^2-4kx^2$
$f(x)=(x^2+(k-2)-2\sqrt{k}x)(x^2+(k-2)+2\sqrt{k}x)$
Hence, $k$ must be a perfect square. From the quadratic formula, we see that the roots of the first quadratic are \[ r=\frac{2\sqrt{k}\pm\sqrt{8}}{2},  \] and the roots of the second quadratic are \[ r=\frac{-2\sqrt{k}\pm\sqrt{8}}{2}.  \] Hence, the largest possible root of these two quadratics occurs with the first polynomial and the largest value of $k$. Thus, we choose $k=44^2=1936$, and find $r=\frac{2\cdot44+\sqrt{8}}{2}=44+\sqrt{2}$.

[size=134][b]Case 3:[/b][/size] $f(x)=(x^2-(k+2))^2+(k-2)^2-(k+2)^2$
$f(x)=(x^2-(k+2))^2-8k$
$f(x)=(x^2-(k+2)-2\sqrt{2k})(x^2-(k+2)+2\sqrt{2k})$
Since $b=0$ in the quadratic formula, the largest root occurs with the most negative value of $c$, which comes from the first quadratic. Hence, \[ r=\frac{\sqrt{4\cdot(k+2+2\sqrt{2k}})}{2}=\sqrt{k+2+2\sqrt{2k}}.  \] Since $k\le2005$ and all coefficients must be integral, the largest possible value is $k=2\cdot31^2=1922$, which corresponds to $r=\sqrt{2048}$.

Since $\sqrt{2048}<45.3$, the largest possible root of $P(x)$ is $44+\sqrt{2}$, and the desired answer is $\boxed{46}$.[/hide]

[hide="Problem #5"][quote]A pomegranate is a spherical fruit filled with seeds and support material. The volume taken up by the support material is directly proportional to the surface area of the fruit. A pomegranate of radius $2$ inches contains $72$ seeds, which fill up $90\%$ of its volume. Based on this information, how many seeds should a pomegranate of radius $r$ inches contain?[/quote][/hide]
[hide="Solution"]The pomegranate of radius $2$ inches has surface area $4\pi(2)^2=16\pi$. The support material takes up $10\%$ of the volume, or $(\frac{1}{10})(\frac{4}{3})\pi(2)^3=\frac{32\pi}{30}$. This means the seeds take up volume $(\frac{9}{10})(\frac{4}{3})\pi(2)^3=\frac{288\pi}{30}$

The pomegranate of radius $4$ inches has surface area $4\pi(4)^2=64\pi$. Since this pomegranate's surface area is four times the surface area of other one, its support material takes up four times as much volume, $4\cdot\frac{32\pi}{30}=\frac{64\pi}{15}$. The volume of this pomegranate is $(\frac{4}{3})\pi(4)^3=\frac{256\pi}{3}$. Hence, the seeds take up volume $\frac{256\pi}{3}-\frac{64\pi}{15}=\frac{1216\pi}{15}$. Since $72$ seeds take up volume $\frac{288\pi}{30}$, then let $x$ seeds take up volume $\frac{1216\pi}{15}$: \[ \frac{x}{(\frac{1216\pi}{15})}=\frac{72}{(\frac{288\pi}{30})},  \] \[ x=608.  \] Hence, a pomegranate of radius $4$ inches contains $\boxed{608}$ seeds.[/hide]

I'll take any solutions that have already been posted here for the Solution Guide, as well, though it would be best if you emailed them to me as well (or did both).

If two people post the same solution to the same problem, whichever one I get first will be the one recognized.

i'll do 1 :P
[hide="problem_1"]
[quote]
a rectangular prism with integral sides has volume 101. what is the surface area of the prism?[/quote]
[/hide]

[hide="solution1"]
let $a,b,c$ be the sides of the prism. since the volume is $abc$ and since $101$ is prime, the only way we can have integral sides is to have $a,b,c$ equal to $1,1,101$.
the surface area is:  $2(ab+bc+ac)=2(203)=406$
[/hide]


Source: 2000 AMC10 #23

When the mean, median, and mode of the list 10, 2, 5, 2, 4, 2, x are arranged in increasing order, they form a non-constant arithmetic progression.  What is the sum of all possible real values of x?

(A) 3
(B) 6
(C) 9
(D) 17
(E) 20

OK, [hide]so the mode is 2 no matter what since even if x was one of the numbers mentioned, it would only appear twice. The median, however, will depend on x. If x:le:2, median is 2, and if x:ge:4, median is 4. If 2<x<4, then median is x. In this case A (3) is the only one in between 2 and 4, and it works (mode, median, and mean is 2,3, and 4, respectively)

If the median was 2, then it would be constant since the mean also equaled 2, and if 4 x have to be 31 in order to satisfy the condition. B through E is both over 4 and not equal to 31, so the answer is A, 3.[/hide]

The fire is dying...

Your terminology was all wrong.
Also, I followed you up until the last 3 lines.  Could you edit them for clarity of english?

Mean = average = sum / number of elements
Mode = most frequently appearing element
Median = the element which, when everything is arranged in increacing order, is in the middle.

Sad, sad mistake...although I doubt that'll be a problem on a multiple choice, but oh well. Just to show that skipping grades and doing math in languages other than your natural screws up your thinking.
Example: Cuntos hay dos ms tres?

*sigh* can't believe how stupid I am...did I at least get the answer correctly?

I still don't follow your last paragraph -- partly I think some of the same mean-median-mode problem, partly you just need to explain more.  Like where did 31 come from?

Hmmm

I got [hide]C) 9

So, you have 2, 2, 2, 3, 4, 10, and x

Mode is obviously 2.

The median is either 2, 4, or x.

Average is (25+x)/7

So all those have to be greater than or equal to stuff...solve, and ya get 9

[/hide]

Why are full solutions so out of vogue?  Write down what you did!

Ok, ya want the right answer? Full solution?

The answer asks for all real solutions. Thus the answer is infinity.

Wait, with C:
mode=2
median=4
mean=(25+9)/7=4 6/7
2, 4, and 4 6/7 is not an arithmetic progression I think...

i dont think either of your answers are right.  you're partially right tare.

Tare: the question asks for the [i]sum[/i] of all solutions -- SouthPaw was giving 9 as the sum, not as a solution itself.  He claimed solutions were 3 and 6.  You can check for yourself whether both work.  There's one valid solution that no one has found, however.

I still don't understand why 6 could be a legitimate answer...I got the sequence 2, 4, and 4 3/7

Tare -- You are extremely close... I think your "31" number is wrong, though.

(and you are correct about 6 not working -- although it looked in your previous answer that you had shown that 9 didn't work, not 6)

Doy! A calculation error...subtracted 25 from 42 and somehow got 31 , but that's not surprising if you see the test I got back today: 93, 7 points off for misspelling "triangle"

Anyways, [hide]17[/hide] also works, so the answer is [hide]E, 20 (3+17)[/hide]

This is funnier than the time I hit myself with the ball I hit in baseball , thanks for all your help everyone (and not laughing  )


function
integration
real analysis
real analysis theorems

If a function $f: [a,b] \rightarrow R$ is bounded and has at most finitely many discontinuities on [a, b], prove that $f \in R[a, b]$.
Here is how I proceeded:
The interval can be partitioned so as to get
$[a, b] = [a, c_{0}] \cup (c_{0}, c_{1}] \cup (c_{1}, c_{2}] \cup ... \cup (c_{n}, b]$.
Since $f$ is bounded on the entire interval [a, b], it is bounded on each subinterval. Thus there exists a positive integer M  and n nonnegative integer such that
$|f_{n}| \leq M$, which implies $-M \leq{f_{n}}\leq M.$
Let $\epsilon = 3M(b-a) > 0.$ Then $\int^{b}_{a}{(M-(-M))}= 2M(b-a ) < \epsilon.$ Thus by Squeeze Theorem, $f_{n}$ is integrable, where $(n = 0, 1, 2, 3, ...)$, but the sum of integrable functions is integrable and since $\int^{b}_{a}{f}= \int^{c_{0}}_{a}{f_{0}}+...+\int^{b}_{c_{n}}{f_{n+1}}$, $f \in R[a, b]$.

Can anyone tell me if or not the preceding argument is valid, please?

No, your choice of epsilon does not make sense. Epsilon should be arbitrary.

You are right. May be I should have used $\delta$ which depends on epsilon?
Does anyone have any suggestion or hint on how to fix the problem?

It's not going to come this easy. First of all, the discontinuities may be at the endpoints, so your splitting does not work. But modifying slightly what you did it is enough to show that if $f: (a,b) \to \mathbb{R}$ is continuous and bounded then it's integrable. What you should do is to take a compact subinterval $[a+\epsilon, b-\epsilon]$. On this interval you can use uniform continuity of $f$ to create upper and lower darboux sums with value arbitrarily close to each other. Then we add to these sums two pieces for the intervals $(a,a+\epsilon)$ and $(b-\epsilon,b)$ where you just use the constants from the boundedness. If you do this appropriately you can let $\epsilon \to 0$ to get upper and lower darboux sums on the interval $(a,b)$ arbitrarily close to each other in value.

Can't you use the Riemann criterion and the fact that f is bounded? Should be no sweat since you are dealing with finitely many discontinuities.


calculus
integration
number theory unsolved
number theory

Given a positive integer $n$ and an odd prime $p$.Let $a,b$ be integers s.t.$a^{2}+b^{2}=p^{2n}$.Denote $S=\{(x,y)|(x-\frac{a}{2})^{2}+(y-\frac{b}{2})^{2}=\frac{p^{2n}}{4},x,y\in %Error. "mathbox" is a bad command.
{Z}\}$.Show that there is a subset $T$ of $S$ with $|T|\geq \frac{|S|}{2}$ such that for any elements $u=(x_{1},y_{1}),v=(x_{2},y_{2}) \in T$,we have the property that $|u-v|=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$ is an integer.


superior algebra
superior algebra unsolved

For which $ n \geq 2$ does $ a^n \equal{} a$ for every $ a \in R$, where $ R$ is a ring, imply that $ R$ is commutative.

All of them, see http://www.mathlinks.ro/viewtopic.php?t=129148 .


Is there a way to prove that

$\lim_{x\rightarrow\infty}\left(\frac{F_{x+1}}{F_{x}}\right) = \frac{1+\sqrt{5}}{2}$?

When $F$ represents the Fibonnaci series?

This is more of a curiosity thing than a question.

[hide="to do it rigorously"]
Let $\phi=\frac{1+\sqrt{5}}{2},\ \bar{\phi}=\frac{1-\sqrt{5}}{2}$ so that $x^{2}-x-1$ has roots $\phi$ and $\bar{\phi}$, and $d_{n}=\left| \phi-\frac{F_{n+1}}{F_{n}}\right|$ so we need to show that $d_{n}$ approaches 0.

Now we have
\[\frac{1}{\phi F_{n}-F_{n+1}}=\frac{\bar{\phi}F_{n}-F_{n+1}}{(\phi F_{n}-F_{n+1})(\bar{\phi}F_{n}-F_{n+1})}\]
Since $\phi$ and $\bar{\phi}$ have sum 1 and product -1, the denominator of the above fraction is equal to $F_{n+1}^{2}-F_{n}^{2}-F_{n}F_{n+1}$. But it's easy to prove by induction that this expression equals $(-1)^{n}$ (if $F_{0}=0$) because
\[(F_{n}+F_{n-1})^{2}-F_{n}^{2}-F_{n}(F_{n}+F_{n-1})=-(F_{n}^{2}-F_{n-1}^{2}-F_{n}F_{n-1})\]
Therefore, we have
\[d_{n}=\frac{|\phi F_{n}-F_{n+1}|}{F_{n}}=\frac{1}{F_{n}(F_{n+1}-\bar{\phi}F_{n})}\]
(Note that $-\bar{\phi}$ is positive.) Now it's clear that the absolute value of the difference tends to 0 because Fibonacci numbers get arbitrarily large.
[/hide]

we could just use binet's formula

[expects a negative post rating...although there is nothing else to say...]

[quote="buzzer11"]whats binet's formula?[/quote]
The closed form for the Fibonaccis. Using the notation in my solution,
\[F_{n}=\frac{\phi^{n}-\bar{\phi}^{n}}{\sqrt{5}}\]


topology
complex analysis
real analysis
advanced fields
advanced fields theorems

I'm a high school student and am beginning a self-study of topology. This question is sure to be standard and I'm sorry if this is not the right place!!!

Anyways, I want to prove that the open set in the Eucildean space i.e.

Given a point [b]p[/b] of [b]X[/b] and a non-negative real
number r, the open ball $B_X (p; r) $ in X of radius r about [b]p [/b]i s defined to be
the subset of X given by
${ B_X (p; r) \equiv { x \in X : |x - p}| < r} $

, is a topologcial space. Also, why is the closed ball in Euclidean space NOT a topological space.

Basically I am asking for a proof of the fact that the definiton of open sets which comes from topogical spaces( i.e. if $(X, T) $ is a topolgical space, then the elements of $T$ are called open sets) is the same as that which comes intuitively (in real spaces). Also, why is the definiton of topology $T$ relate to "open " sets as we know them only?? Why can't closed sets be included. Also, why is it that we consider a finite collection to form the intersection in the definition of topology??? Why don't closed sets form topological spaces??

Can anybody give recomendations of sites or web pages which cover introductory topology at a very basic level!!!

What do you want to prove, that the open sets (in the traditional sense) in $\mathbb{R}$ form a topological space?

This is not hard.  $\mathbb{R}$ is open, as is $\emptyset$.  Now you just need to show that finite intersections of open sets are open.  Say that some point $x$ lies in the intersection of finitely many open sets $S_i$.  Then for each $i$, there is an open ball $B(x,r_i)$ such that $B(x,r_i)\subset S_i$, and it is clear that $B(x,\min_i r_i)\subset \cap_i S_i$.  So finite intersections are open.  Finally, for unions, if $x$ lies in the union of some open sets, then $x\in S_i\subset\cup_i S_i$, where $i$ may run over any set of indices we like.

For closed sets, we would like to show that some union of closed balls is not closed.  This is NOT true if the union runs over a finite number of balls (prove it!).  But over countable unions, this is not a problem - we have

$\cup^{\infty}_{i=1} [1/i,1]=(0,1]$

which is not closed.

Finally, a topological space is quite an abstract concept.  Open sets in a general topological space may be quite different than the open sets in $\mathbb{R}$.  This is because $\mathbb{R}$ has many intrinsic properties that make working with it 'nice' - it is, for example, a locally compact Hausdorff space - and they don't generalize to other spaces.

A final note is that topological spaces do not necessarily have metrics.  Thus things like distance, which we use without thought in $\mathbb{R}$, do not necessarily have analogues in topological spaces (although metric spaces are topological spaces).

This seems to have a lot of general knowledge on topology.

http://en.wikipedia.org/wiki/Topological_space

So the best definiton of open sets is the one which comes from the definiton of a topolgy $T$.. Thanks for your proof. Just a thing.. I still haven't understood clearly why closed sets(in a general case) don't make up a topolgy. Can you elaborate a bit more??

Also, I was asking for some excellent introductory lecture notes and e-books. Do you have any in mind???

I would also like to ask why we take finite intersections and arbitrary no, of unions in the definiton of topology???

I'm not clear regarding infinte sets , infinte intersection. unions and such arbitrary operations on sets. Can you suggest some resource on the net(lecture notes etc.) from where I can clarify my concept???

Because underneath all the generalizations, we would like our definitions to conform with nice, simple examples that we know already (e.g the real line as a metric space)

I don't know any such notes - the topology I know is derived from Rudin's Real and Complex Analysis and the topology portion of Folland's Real Analysis: Modern Techniques and their Applications.

But as always, MIT's opencourseware is a good place to start.

Can you please elaborate. I don't fully understand your statement. Also is my Understanding(as I have posted in my post #2) of open sets etc. correct. ...

I checked OCW (always first) but the topology course material does'nt contain the starting chapters (basic topology!!!) Any other suggestions???

Try this online book
http://uob-community.ballarat.edu.au/~smorris/topbookchaps1-8.pdf

At least the typesetting is nice :)
If you are not happy with that one, you can try another one here:
http://www.econphd.net/roessler/notes.htm

Thanks a lot!!! Can you answer my doubts!!!

[quote]Basically I am asking for a proof of the fact that the definiton of open sets which comes from topogical spaces( i.e. if $(X, T) $ is a topolgical space, then the elements of $T$ are called open sets) is the same as that which comes intuitively (in real spaces). Also, why is the definiton of topology $T$ relate to "open " sets as we know them only?? Why can't closed sets be included. Also, why is it that we consider a finite collection to form the intersection in the definition of topology??? Why don't closed sets form topological spaces??[/quote]

I don't know if this can answer your question.
1) Yes, the definition of topology (in some sense) is the abstraction of the behaviour of open intervals in $\mathbb{R}$. Why we only take finite intersection? because if we take infinite intersection of open intervals the result might be not longer an interval. Example:
\[
\bigcap_{i=1}^\infty \left(-\frac{1}{n},\frac{1}{n}\right)=\{0\}
\]
(since if $-1/n<x<1/n$ for every $n$ implies $x=0$)

2) As pointed out by blahblahblah for closed intervals, if we take arbitrary unions of closed intervals the result migh be not longer a closed interval.

[quote="Rushil"]So , I want to confirm waht I have understood now.

1) The definiton of a topolgy is such that the things are as general as possible while at the same time, the definition confirms and corroborats out intuitive understanding of the real number system.

2) The precise definiton of an open set is the one which comes from the definiton of topology , i.e. an open set is the one which is an element of a topology $T$ of a set $X$.

3) The definiton regarding open and closed sets(i.e. union\intersection etc.) is again to keep things ok

I hope my final understanding is ok!!![/quote]

Thats OK! :)


$ p\in\mathbb{Q}$, $ w\in(1,\infty)\setminus\mathbb{Q}$.

prove that $ \{\minus{}kw\}\plus{}\{pkw\}<1$ for some $ k\in\mathbb{N}$.


There is a set $ \{a_1,a_2,a_3,...,a_{201}\}$, where $ a_1,a_2,a_3,...,a_{201}$ - different positive integers less than 300.

Prove that there is always such terms $ a_i$ and $ a_j$, such that $ \frac {a_i}{a_j} \equal{} 3^k$, $ k\ge1$.

Thank you  :)

My logic is tragical  :oops:

[quote="azjps"][hide]
Follows like [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=240840]this[/url], for example;

Divide the $ 201$ $ a_i$s through by their largest power of $ 3$; then they each become one of the $ 200$ numbers $ < 300$ not divisible by $ 3$; by Pigeonhole, at least two are the same. These give us the desired $ a_i,a_j$.
[/hide][/quote]

Although your analysis is so good , this may not be always  true.For example ,

From $ 1$ to $ 299$ ,there are $ 299$ different postive integers, if we delete $ \{3,9,12,15,...297\}$ from these $ 299$ integers ,then the rest $ 299 \minus{} 98 \equal{} 201$ integers to form $ \{a_i\} ,i \equal{} 1,2,3,..,201$.
How can we find such $ a_i ,a_j$ that $ \frac {a_i }{a_j } \equal{} 3^k, k\ge 1$ ?

Note: The total numbers of $ \{3,9,12,15,...297\} \equal{} \frac {297}{3} \minus{} 1 \equal{} 98$

Edited.
 :P  :blush:

How about $ (2,6)$?  :wink:

Sorry, let me re-phrase.

Consider the parallel set $ \{b_1, \cdots, b_{201}\}$ where each $ b_{i} \equal{} a_i/3^{e_i}$, where $ e_i$ is the greatest power of $ 3$ dividing into $ a_i$. It follows that $ 3 \nmid b_i$ and $ 0 < b_i < 300$; there are $ 299 \minus{} \left(\frac{297}{3}\right) \equal{} 200$ such numbers. So we must have $ b_i \equal{} b_j$ for some $ i \neq j$, but then $ \frac{a_j}{a_i} \equal{} 3^{e_j \minus{} e_i}$ as desired.

[quote="azjps"]How about $ (2,6)$?  :wink:

Sorry, let me re-phrase.

Consider the parallel set $ \{b_1, \cdots, b_{201}\}$ where each $ b_{i} \equal{} a_i/3^{e_i}$, where $ e_i$ is the greatest power of $ 3$ dividing into $ a_i$. It follows that $ 3 \nmid b_i$ and $ 0 < b_i < 300$; there are $ 299 \minus{} \left(\frac {297}{3}\right) \equal{} 200$ such numbers. So we must have $ b_i \equal{} b_j$ for some $ i \neq j$, but then $ \frac {a_j}{a_i} \equal{} 3^{e_j \minus{} e_i}$ as desired.[/quote]

Clearly got. Thanks.:thumbup:

 :trampoline:


geometry
trapezoid
angle bisector
congruent triangles
geometry solved

ABCD is a trapezoid.  AB is parallel to CD.  If DA = AB + CD, and AE is the angle bisector of angle A,  Show that BE = EC.

[color=red][[b]Moderator edit:[/b]

[b]1.[/b] The passage "AE is the angle bisector of angle A" should be understood as: "Let the angle bisector of the angle DAB meet the line BC at E".

[b]2.[/b] The problem was cross-posted with http://www.mathlinks.ro/Forum/viewtopic.php?t=29127 .][/color]

Take $T\in AD$ s.t. $AT=AB$. We then get $DT=DC$. $AE$ is a bisector and thus an altitude in the isosceles triangle $ABT$, meaning that it passes through the midpoint of $BT$ and is perpendicular to $BT$. A quick angle chase shows that $TC\perp BT$, and this means that $AE\|TC$, and, since $AE$ passes through th midpoint of $BT$, it also passes through the midpoint of $BC$.

With the general reference of A, B, C and D in the diagram, produce AE to meet at DC(ray) at M where AB//DM
Given angle(DAC) = angle(BAM) [angle bisector]
...and angle(BAM) = angle(AMC) [AB//CM] (I)
Therefore angle(DAC) = angle(AMC/D)
Therefore DA = MD [equal sides in isos. DMA]

DA = AB + CD [condition]
Therefore MD = AB + CD

Since MD = CD + MC [M, C, D are col.]
Therefore AB = MC (II)

In triangles AEB, MEC:
angle(BAM) = angle(AMC) [by (I)]
angle(AEB = angle(MEC) [equal vert. opp. angles]
AB = MC [by (II)]
Therefore triangles are congruent [AAS]

Hence BE = ME [equal corresponding sides in congruent triangles]


"Nature laughs at the difficulties of integration" was really all I could see in the Quote box here without major straining to pick out "Pierre-Simon de Lapiac" on the bottom.  I am using a PC w/Windows ME on a Firefox (about to be 1.0.4 from -1) browser viewing the AoPS through a 1024 by 768 screen.  (That was said in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=237761#p237761]How to use the new board[/url].)  OK?  Can you make this better--bigger enough so I can see or even a different font type with thinner lines...

thanks - -

Huh?  :huh: Can explain perhaps in more detail what your problem is exactly? Maybe add a screenshot? Thanks!

I think he's trying to read some stuff, such as the little quotes at the bottom (AoPS interface).  The problem is that his browser somehow makes it very small.  I think it is a problem of his browser though--you probably can't do much about it.

Ah ok then :) I can't help you more than by telling you to use IE 6.0 (latest packs), Firefox 1.0.4 or Opera 8 :) These browsers seem to have no problem rendering correctly this site :P :)

[quote="fedwinri"]It's displaying the quotes; it just seems that the names at the bottom are always a little scrunched (particularly 'r's and 'l's and 'i's).  Frankly, the letters run into eachother much of the time on my screen.  Since this may be just an old monitor issue and it displays at least slightly better on other screens, I think I'll let it go.  Btw, I do think my browser has almost nothing to do with the rendering as it is not affected by the Text Size setting.

It would also be really nice to be able to grab a nice quote with a click or at least by selecting it (in one frame?) al at once.

Thanks[/quote]No really I don't understand what's going on with your computer. And what do you mean by grabbing a quote?!

[quote="fedwinri"]It would also be really nice to be able to grab a nice quote with a click or at least by selecting it (in one frame?) al at once.[/quote]I can do this in my browser (Opera 8) simply by mousing over the text, right-clicking, and choosing "Select All".  Although I do have to do this separately for the quote text and the author.

[quote="MCrawford"]I think he means copy the quote, but that's not easy because the output is not text output (it's an image of text).[/quote]It [b]is[/b] actual text (though embedded in a Flash movie), and can be selected as such (for example, double-clicking on a word selects that word).

[quote="DPatrick"][quote="sonar"][quote="SnowStorm"]Did you try making it very large using Ctrl++ already?  That ought to at least temporarily allow you to read it.[/quote]

That will not help this problem.  Ctrl++ will have the browser make the text larger, but the problem here is a flash movie, so hitting Ctrl++ will not effect the quotes.[/quote]It does on my browser (Opera 8).[/quote]

Ah, well, it works neither on Firefox 1.0.4 nor IE 6.0.2900 (ctrl + scroll wheel).

Sorry for all the hassle everyone.  It's just that I 1: would like to be able to select the quote as one string (rather than seperately as DPatrick) and to 2: see at least the author's name more clearly.  I'm on a different computer--in a completely different part of the State--and the names are still scrunched together.  This only matters becuase some often times, I do not know the person (by name) to which the quotation is ascribed.

(btw, I haven't been able to get on in a while; should be more after a week or so.)

Also, I am currently using IE 6.0, rather than Firefox, so it may be just my personal preference in reading things easily.

Eddy

If you can get the source code of the Flash somehow, you win :P .


trigonometry
function
logarithms
integration
superior algebra
superior algebra unsolved

Unsolved Problems? I think this one is trivial...

Since, if x = tan A and y = tan B, then ${x \star y = \frac{x+y}{1+xy} = \frac{\tan A + \tan B}{1 + \tan A \tan B} = \tan \left(A+B\right)}$.

Thus, $x^{\left[n\right]} = x \star x \ldots \star x = \tan \left(n \arctan x\right)$.

  Darij

nearly right, but $\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$, not with +.
just find some function such that $f(x+y)=\frac{f(x)+f(y)}{1+f(xy)}$. but it's much too late to construct it for me.

[quote]just find some function such that $ f(x+y)=\frac{f(x)+f(y)}{1+f(xy)} $[/quote]

The hyperbolic tangent.

[quote="Lagrangia"]$H = (-1,1) \subset R$ and $\displaystyle x \star y = \frac{x+y}{1+xy}.$
Compute $x^{[n]} = x \star x \ldots \star x.$ (n times)
cheers! :D :D[/quote]

Hi Lagrangia :)
Sorry to ask this but, what is the purpose of $H$ in this problem? Maybe it is $x,y\in H$ :D

Continuing the discussion by Kent Merryfield, the hyperbolic tangent possesses that property, i.e. that $\tanh(x+y)=\frac{\tanh x+\tanh y}{1+\tanh x\tanh y}$.
Let $x=\tanh\alpha$. Then $\alpha=\tanh^{-1}x$ and so
$\displaystyle x^{[n]} = x \star x \cdots \star x=\tanh(n\alpha)=\tanh\left(\frac{n}{2}\ln\left|\frac{1+x}{1-x}\right|\right)=\frac{\left|\displaystyle \frac{1+x}{1-x}\right|^{\frac{n}{2}}-\left|\displaystyle \frac{1+x}{1-x}\right|^{-\frac{n}{2}}}{\left|\displaystyle \frac{1+x}{1-x}\right|^{\frac{n}{2}}+\left|\displaystyle \frac{1+x}{1-x}\right|^{-\frac{n}{2}}}$.
Assuming $x\in H$, we have finally $\displaystyle x^{[n]} = \frac{(1+x)^n-(1-x)^n}{(1+x)^n+(1-x)^n}$.
:cool:  :blush:  ;)

Yes, you're right, I did my calculations too quickly.

  Darij

[quote="3X.lich"] Assuming $x\in H$,  $x^{[n]} = \frac{(1+x)^n-(1-x)^n}{(1+x)^n+(1-x)^n}$.
[/quote]

Let $f_n(x)= x^{[n]}$

Study the simply and uniform convergence of this sequence function on $H$

[quote]Study the simply and uniform convergence of this sequence function on $ H $[/quote]
It converges pointwise to the signum function: 1 for x positive, -1 for x negative, zero for x zero. Since this is a sequence of continuous functions converging to a discontinuous limit, the convergence is not uniform.

The addition formula for the hyperbolic tangent is also the formula for the addition of velocities in special relativity.

$f_n$ converge simply to $f$ the signum function

$f(x)=1, if \0<x<1$
$f(0)=0$
$f(x)=-1, if \-1<x<0$

Call $\overline{f}$ the function obtained by 2-periodicity
What is $S(\overline{f})$ the Fourier serie of $\overline{f}$
Write Parseval equality

[quote]Call $ \overline{f} $ the function obtained by 2-periodicity
What is $ S(\overline{f}) $ the Fourier serie of $ \overline{f} $
Write Parseval equality[/quote]
That doesn't seem to have much to do with the rest of the problem, but we might as well have an answer so the the thread can be closed.

In real form, the Fourier series is $\frac4{\pi}\sum_{k=0}^{\infty} \frac{\sin((2k+1)\pi x)}{2k+1}$

Parseval's identity (assuming that we have a function of average value zero) says that
$\frac1P\int_{-P}^P\overline{f}(x)^2dx=\sum_{n=1}^{\infty}(a_n^2+b_n^2).$

In our case, $2=\left(\frac4{\pi}\right)^2\sum_{k=0}^{\infty}\frac1{(2k+1)^2}.$

From this, $1+\frac1{3^2}+\frac1{5^2}+\frac1{7^2}+\cdots = \frac{\pi^2}8$, from which we can derive $\sum_{n=1}^{\infty}\frac1{n^2}=\frac{\pi^2}6.$

This is all in many different textbooks.


Find all positive integer solutions to
\[ \frac{1}{a}\plus{}\frac{a}{b}\plus{}\frac{1}{ab}\equal{}1\]

[hide]Rearrange to get
$ a^{2}\plus{}1\equal{}ab\minus{}b\Leftrightarrow a^{2}\minus{}2a\plus{}1\equal{}ab\minus{}2a\minus{}b\Leftrightarrow (a\minus{}1)^{2}\plus{}2\equal{}(a\minus{}1)(b\minus{}2)$, or
$ (a\minus{}1)(b\minus{}a\minus{}1)\equal{}2$. Since $ a,b\in\mathbb{N}$, we have $ (a,b)\equal{}(2,5)$ or $ (3,5)$.[/hide]


calculus
integration
inequalities
trigonometry
real analysis
real analysis solved

Prove the following inequality.

\[\int^{1}_0 \cos \left(\frac{\pi}{2}x^{\frac{1}{1000}}\right)dx>\frac{\pi}{2004}\]

Perfom substitution $x=t^{1000}$ and integrate by parts some times to obtain $\int >\frac{\pi}{2002}-\frac{\pi^3}{2^3\cdot 1001\cdot 1002}\cdot\frac{1}{5}>\frac{\pi}{2004}$.


Determine all positive integers $ (x,n)$ such that $ 1\plus{}x\plus{}x^{2}\plus{}...\plus{}x^{n}\equal{}40$.

[hide="Solution"]Note that, for $ x\neq 1$ we have

$ \frac {x^{n \plus{} 1} \minus{} 1}{x \minus{} 1} \equal{} 40$

$ x(40 \minus{} x^n) \equal{} 39$.

Case 1. $ x \equal{} 1$, obviously, $ n \equal{} 39$ (not from the derived equalities, but form the original one).

Case 2. $ x \equal{} 39$ gives $ 40 \minus{} 39^n \equal{} 1$, i.e. $ n \equal{} 1$.

Case 3. $ x \equal{} 3$ gives $ 40 \minus{} 3^n \equal{} 13$ i.e. $ 3^n \equal{} 27$, ergo $ n \equal{} 3$.

Case 4. $ x \equal{} 13$ gives $ 40 \minus{} 13^n \equal{} 3$ i.e. $ 13^n \equal{} 37$ which yields no natural $ n$.

To sum up, solutions are

$ (1,39),(39,1),(3,3)$.[/hide]


Let $ABC$ be a triangle with $BC=a$ $AC=b$ $AB=c$. For each line $\Delta$ we denote $d_{A}, d_{B}, d_{C}$ the distances from $A,B, C$ to $\Delta$ and we consider the expresion $E(\Delta)=ad_{A}^{2}+bd_{B}^{2}+cd_{C}^{2}$. Prove that if $E(\Delta)$ is minimum, then $\Delta$ passes through the incenter of $\Delta ABC$.

[hide]Prove that if $\Delta$ doesn't pass through $I$, and $\Delta'$ passes through $I$, and $\Delta\parallel \Delta'$ then $E(\Delta)>E(\Delta')$[/hide]


Let's say that we "concatenate" two numbers when we write one beside of the other one; e.g. when we concatenate 200 and 6, we get 2006, and when we concatenate 6 and 200 we get 6200. Find two six-digit numbers such that the number obtained by concatenating them is divisible by their product.

Let x and y be two six-digit such that the number obtained by concatenating them is divisible by their product, i.e.

$(1) \; xy \, | \, 10^6 x + y.$

This implies that x|y, i.e. y = xz for a natural number z. Hence

$(2) \; xz \,|\, 10^6 \: +\: z$

according to (1). Now $z \,|\, 10^6$ and z = y/x < 10 since the integers x and y have the same number of digits. So $z \in \{1,2,4,5,8\}.$ Futhermore $xz \: =\: y \: < \: 10^6 \: < \: 10^6 \: +\: z,$ which combined with (2) gives

$(3) \; nx \;=\; \frac{10^6}{z} \: +\: 1$

where n is an integer in the intervall [2,10]. Obviously $\frac{10^6}{z}$ is an even integer, so nx is an odd integer by (3). Consequently $n \, \geq \, 3$, which again implies that $nx \: \geq \: 3\cdot10^5 \: +\: 1$. Therefore $z \: \leq\: \frac{10}{3}$ by (3), which leaves us with two possible choices for z:

$\bullet \; z = 1 \; \Rightarrow \; nx = 1000001$, which is impossible since 1000001 don't have odd divisors < 10.

$\bullet \; z = 2 \; \Rightarrow \; nx = 500001$. The only odd divisor of 500001 which is < 10, is 3. So the only possible solution is n = 3, giving

[b]x[/b] = 500001/3 = [b]166667[/b] and [b]y[/b] = xz = 2x = 2*166667 = [b]333334[/b].


inequalities
number theory unsolved
number theory

Find all positive integer solutions of the following equation

x^3-y^3=xy+61

[hide="hint"]
use inequality and try to reduce the maximum value that $ \max \{x,y\}$ can take write the equation in this form :
$ (x\minus{}y)(x^2\plus{}y^2\plus{}xy)\equal{}61\plus{}xy$ or in this form $ x^3\plus{}(\minus{}y)^3\plus{}(\minus{}1)^3\minus{}3(x)(\minus{}y)(\minus{}1)\equal{}60\minus{}2xy$
[/hide]

Since $ x>y$,we let $ x\equal{}y\plus{}z$,where $ z>0$.The problem is much easier then.The equation has the degree of $ 2$.


[quote="felipesa"]If xE $[0,2\pi]$ , the numbers of solutions $sen^3x-senx+1=cos^2x$ is equal ?[/quote]

$cos^2 x=1-sin^2 x$ and we can see that $sin^3 x +sin^2 x -sin x=0$ which gives us $2$ solution $sin x =0$ and $sin x =\frac{-1+\sqrt{5}}{2}$

Right idea, I believe, but since it asks for the number of solutions for $x$ and not $\sin{x}$ you have to consider the specifed interval.
[hide]Sketching a graph of $y=\sin{x}$ from $0$ to $2\pi$ it is clear that there are 5 solutions.[/hide]


For $a,b,c$ positive real numbers such that $a+b+c=6$ prove that

$\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a} \geq 2 $

Isn't this trivial??

From Cauchy,

$\sum_{cyclic} \frac{a}{1+b} = \sum_{cyclic} \frac{6a}{a+7b+c}$

$\geq \frac{6(a+b+c)^2}{a(a+7b+c)+b(b+7c+a)+c(c+7a+b)} \geq 2$,

since the last inequality is equivalent to $a^2+b^2+c^2 \geq ab+bc+ca$.


Sorry to bother you guys again..

There's something called "creativity essay" in the application.. Out of curiosity, how many of you (who got accepted to MIT) wrote this completely optional essay?

huh.  I completely overlooked that...  :blush:
well, lucky me, I guess.

who said it had to be an essay? you can send pictures, diagrams, equations, whatever...

Just that it's quite difficult without the formatting to give equations unless you are sending them by post or anything.

I was wondering...for text without formatting, is it a convention to write
a_(n-1) for $a_{n-1}$?

If not, what do people normally do?

[quote="ak007"]
I was wondering...for text without formatting, is it a convention to write
a_(n-1) for $a_{n-1}$?

If not, what do people normally do?[/quote]

Yeah, that's usually how people write subscripts.  Although, when I'm talking to someone else who know $\LaTeX$ I'll often times just write it as $\LaTeX$ code.  So, for example, I would write a_{n-1}.

So, why dont you guys post what you sent along in the creative essay... the deadline's past now so no danger of plagiarism!!!

Anyways, I discussed a problem on Special Relativity that I thought about (look around mathlinks - there's some cool discussion on it here!) and a famous painting that I like ( 'Relativity' by M.C. Escher!) I correlated them and discussed my problem!

Do you think its good!! Remember, i'm an international!!!
Opinions please!! Thanks!

Also, love to hear about your submissions!!

That's really a nice idea!
  You've got a good chance alright, there's no doubt about that!


hello, do you mean $ \sinh(x)\equal{}\minus{}\sinh(\minus{}x)$?
Sonnhard.

hello, let $ f(x)\equal{}\frac{e^x\minus{}e^{\minus{}x}}{2}$, it follows $ f(\minus{}x)\equal{}\frac{e^{\minus{}x}\minus{}e^x}{2}\equal{}\minus{}\frac{e^x\minus{}e^{\minus{}x}}{2}\equal{}\minus{}f(x)$.
Sonnhard.

hello, why? which step is not clear?
Sonnhard.

well what you've said is true but it doesn't seem to answer the question. You will, no doubt, point something very obvious out to me and I will feel silly, but there we go :)

hello, we say that $ f(x)\equal{}\sinh(x)$ is a odd function, because it is $ \sinh(\minus{}x)\equal{}\minus{}\sinh(x)$ for all real $ x$.
Sonnhard.

$ \sinh(x) \equal{} \frac {e^x \minus{} e^{ \minus{} x}}{2}$ by definition.
So $ \minus{}\sinh( \minus{} x) \equal{} \minus{} \frac {e^{ \minus{} x} \minus{} e^x}{2} \equal{} \frac {e^x \minus{} e^{ \minus{} x}}{2} \equal{} \sinh(x)$.


function
Euler
number theory
totient function
number theory unsolved

[quote="tringo"][quote="nguyenvuthanhha"]Prove that  for all positive real numbers $ a , b$ such that : $ a > b$

then there exists a positive integer $ n$ such that :

  $ b < \frac {\varphi(n \plus{} 2)}{\varphi(n) } < a$[/quote]

Could you define the function, please?[/quote]
In number theory $ \varphi(n)$ is normally defined as [url=http://en.wikipedia.org/wiki/Euler%27s_totient_function]Eulers phi function[/url]

very nice problem but hard indeed. i've been working quite a lot on this one some time ago and failed to solve it. i also think it has been posted before but without a solution.

its not hard to proove that $ \frac{\phi (n\plus{}2)}{\phi (n)}$ can get arbitrary small or arbitrary large. for example how to get arbitrary large... take $ n\plus{}2$ to be a prime number congruent to $ 2$ by modulo $ \prod_{i\equal{}2}^{k} p_i$ where $ p_i$ denotes the i-th prime number. such a number exists by dirichlet's theorem. if we send $ k$ to infinity $ \frac{\phi (n\plus{}2)}{\phi (n)}$ gets arbitrary large.

very simmilar for arbitrary small. i however faild to proove from this that kvotient can get arbitrary close to any real number

[i] I have asked some excellent Number theory lovers i known about this problem but they told me that they cannot solve it

    Darij , Peter ,Can you help me ? :( [/i]

Well, I won't say anything new to the topic, but we've been struggling with a weaker one ($ a,b\in \mathbb{N}$) on a polish math forum for a few months, and we came up with nothing besides what Jure wrote. What is more I gave that problem to a really great mathematician and he failed to solve it (he was trying for a few hours).
nguyenvuthanhha, you say it's from Italian olympiad? It seems impossible, the problem looks a lot too hard even for the IMO... But that's only my opinion :wink: I'd love to hear what ZeTaX thinks about it... Maybe it is a well-known problem?

very nice problem...I wanna see the proof.

[i]I am not sure about the source but i guess that  this problem is right

  It has been appeared in Pen  and Komal ( without solution :)  )[/i]

   You can see here :

 http://www.komal.hu/verseny/feladat.cgi?a=honap&h=200701&t=mat&l=en

[i]Give up !

  I am really far from convinced that Italy - a country not famous for Mathematics is author of this problem[/i]

well looking at the statistics of that problem tells you that 2people did solve it so... either it is solvable in some unexpected but straightforward way or they got something wrong because it seems very hard

Some trivial remark: it is enough to consider rational $ a$ and $ b$ (not sure that it simplifies the problem...).
As Jure said, a similar problem, with $ n\plus{}1$ instead of $ n\plus{}2$ appears on PEN - problem $ J 9$. The source of the problem is mentioned, and I checked it, and the problem is mentioned very briefly, as a result by Schinzel and Sierpinski.
So I think that either the $ n\plus{}2$ makes it a lot easier, or it is very hard. In either case, we shall keep looking for a solution. It is a very original problem and that what makes it interesting.

Well, Schinzel and Sierpiński are very famous in Poland (and not only), they were known as a pair of faboulous number theory problem solvers with numerous publications etc.
I can hardly believe there exists an elementary way to approach that problem.
On my forum one of the user's (xiikzodz ) tried to consider a convex framing and pawn the problem in a very non-elementary way... (of course it was a problem based upon natural numbers)
http://matematyka.pl/77192.htm
If anyone shoulod need it, I could translate fragments of ideas :wink:

[i]Two months have passed . I am looking for a solution[/i] :lol:

Sir , Nam Dũng , where are you ? :maybe:

You could prove this beatiful result by using a cool theorem deu to Schinzel: For any $ b_1,b_2,\ldots,b_m$ nonegative number then exist a sequence $ \{a_k\}$ so that $ \frac {\phi(a_k \plus{} i)}{\phi(a_k \plus{} i \minus{} 1)}\to b_i$  when $ k\to\infty$ for all $ i \equal{} 1,...,m$.

Indeed, take $ n \equal{} a(a \plus{} 3)$ then $ n \plus{} 2 \equal{} (a \plus{} 1)(a \plus{} 2)$ where $ 3\nmid a$. So $ \frac {\phi(n \plus{} 2)}{\phi(n)} \equal{} \frac {\phi(a \plus{} 1)}{\phi(a)}\times\frac {\phi(a \plus{} 2)}{\phi(a \plus{} 3)}$. Take $ h\geq 7$ to large enough so that such $ a_k \plus{} i$ is not divisible by $ 3$ exist and $ i \plus{} 4 < m$. The result follows.

I suppose that there exist a more simple solution.

Let $ n \equal{} 4m$, then must be
(1) $ 2b < \frac {\phi(2m \plus{} 1)}{\phi(2m)} < 2a$.
We can chose $ 2m \plus{} 1 \equal{} (\prod_{2 < p < c}p)^k$. Then $ \frac {\phi(2m \plus{} 1)}{2m \plus{} 1} \equal{} \prod_{1 < p < c}(1 \minus{} \frac 1p) \equal{} t(c)$ - independent from k, $ \frac {\phi(2m)}{2m} \equal{} \frac 12 \prod_{q|d^k \minus{} 1}, d \equal{} d(c) \equal{} \frac 12 \prod_{p < c}p$.
We can chose c and k, suth that satisfyed (1).


geometry
geometric transformation
reflection
analytic geometry
number theory proposed
number theory

Four frogs are positioned at four points on a straight line such that the distance between any two neighbouring points is 1 unit length. Suppose the every frog can jump to its corresponding point of reflection, by taking any one of the other 3 frogs as the reference point. Prove that, there is no such case that the distance between any two neighbouring points, where the frogs stay, are all equal to 2008 unit length.

if I got it correctly,the problem seems to be too simple.
As it traditionally happens,we assume the contrary,so after some "jump" we get the described situation, and let's make all jumps in a reverse order, but as it is easy to observe that starting from that position,any distances between frogs must be divisible by $ 2008$,contradiction.

Another solution - use invariants. Colour the points on x (with integer coordinates) black, white, black, white... When we start 2 frogs are on white and 2 on black points. We see that after every jump the color of the frog point does not change, but if all distances are equal to 2008 all four points ale coloured the same - contradiction.

My solution much uglier than above two :)
Assume contrary.Denote points in line as A,B,C,D in turn.Choose any point  in our line as origin of coordinates.Denote coordinates of A,B,C,D after n-th "jumping" as $ a_n$,$ b_n$,$ c_n$,$ d_n$ respectively.denote $ l_n$ as sum of distances between A and B,B and C,C and D after n-th "jumping".$ l_0\equal{}3$.Notice that $ l_n\equiv b_n\minus{}a_n\plus{}c_n\minus{}b_n\plus{}d_n\minus{}c_n \equiv d_n\minus{}a_n\mod 2$.and it is easy to understand that $ d_n\minus{}a_n\equiv d_{n\plus{}1}\minus{}a_{n\plus{}1}\mod2$,so as we can see $ l_n \equiv l_0\mod2$.We get contradiction.