number theory
greatest common divisor
geometry
circumcircle
rotation
function
incenter

[u][b]Day1[/b][/u]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=139760url]Problem1[/url]
m$\in$N.Let gcd$(m,6)=1$. $S_{m}$ is the set of numbers which is smaller than m and coprime to m.And $\sum_{{n}\in{S}_{m}}\frac{1}{i}=\frac{A}{B}$ , ${A,B}\in{N}$. Prove that $m^{2}$ divide $A$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=139771url]Problem2[/url]
$S=\{1,2,3...n\}$ and ${A}_{1},{A}_{2},...{A}_{3}$ are subsets of $S$ and has $k$ elements.${B}_{1},{B}_{2},...,{B}_{m}$ are subsets of $S$ and has $l$ elements. And let ${k}+{l}\le{n}$. If $%Error. "forral" is a bad command.
{i}\not={j}:{A}_{i}\cap{B}_{i}=0,{A}_{i}\cap{B}_{j}\not=0$ prove that ${m}\le{C}^{l}_{k+l}$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=139777/url]Problem3[/url]
$P,Q$ is midpoint of $AC, BD$ diagonals of convex quadrilateral $ABCD$, $PQ$ intersects with $AB, CD$ at $N,M$ respectively.Prove that circumcenters of triangle $NAP, NBQ, MQD, MPC$are concurrent.
[u][b]Day2[/b][/u]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=139774url]Problem1[/url]
${A}_{n}$ is number of $a_{1}, a_{2},..,a_{n}$ permutate of $1,2,3..n$ such that $\mid{a}_{k}-{k}\mid=0 or 1 or 2$.  When $n\ge6$ prove that ${A}_{n}={2A}_{n-1}+2{A}_{n-3}-{A}_{n-5}$.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=139779/url]Problem2[/url]
Incircle of triangle $ABC$ tangents $AC$ at $D$. Prove that the circle which tangents rays $[BD), [DC)$ and circumcircle of $ABC$ equal to excircle of $ABC$ of $\angle{B}$.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=139765/url]Problem3[/url]
$1,2,..,40$ are located on circle such that for any sequential numbers $a,b,c$ , ${41}\mid{b}^{2}-{a}{c}$. How many distinct locating which is possible?( Rotate and symmetrical locating is same)

[b]10th grade[/b]
[i][b]Day1[/b][/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=139911url]Problem1[/url]
Find all function such that ${f}:{Q}\rightarrow{R}$ and for $\forall{x,y}\in{Q}$ ,$f(xy)=f(x)f(y)$ and${f(x+y)}\le{max(f(x),f(y))}$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=139914url]Problem2[/url]
Given a sequence$\{{x}_{n}\}$ such that ${x}_{1}=2,{x}_{2}=4,{x}_{n+1}={x}_{n}^{2}-{x}_{n-1}({n}\ge{2})$. Is there infinitely many members which is multiple of $2004$ ?
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31693]Problem3[/url]
A quadrilateral is cyclic and has incircle. Prove that intersection of diagonals, circumcenter , incenter are collinear.
[i][b]Day2[/b][/i]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=139925url]Problem1[/url]
Let $a,b,c$ are positive nmbers. Prove that
 $\frac{ab}{3a+b}+\frac{bc}{b+2c}+\frac{ac}{c+2a}\le\frac{2a+20b+27c}{49}$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=139920url]Problem2[/url]
On a plane given triangle $ABC$ and point $O$ .The lines which perpendicular  with $OA, OB, OC$ through $O$ intersect with $BC, AC, AB$ at ${A}_{1},{B}_{1},{C}_{1},$ respectively. Prove that ${A}_{1},{B}_{!},{C}_{1}$ are collinear
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=139926url]Problem3[/url]
Let $p>2$ be prime. Note that ${a}^{-1}$ is $x$ such that  $ax\equiv1(mod p)$. Prove that $\frac{2^{p-1}-1}{p}\equiv1^{-1}+3^{-1}+5^{-1}+...+{(p-2)}^{-1}(modp)$


find all $(x,y)$ such that $\ x|2^{y-1}+1, y|2^{x-1}+1$ where

a) $x,y$ are all primes
b) $x,y$ integers

( the post has been edited)

hallo,
i think a) is easy: (x, y) = (2, 2) .

I somehow think that it shall be $x|2^{y-1}-1$ and $y|2^{x-1}-1$or something similar... (the one stated above is trivial).

i have edited it,  :oops:    :wallbash_red:


A base-10 three-digit number $n$ is selected at random. Which of the following is closest to the probability that the base-9 representation and the base-11 representation of $n$ are both three digit numerals?

$\left(A\right) 0.3$
$\left(B\right) 0.4$
$\left(C\right) 0.5$
$\left(D\right) 0.6$
$\left(E\right) 0.7$

[hide="Solution"] This might look somewhat daunting, but the easy thing is to calculate ranges.  (All ranges are inclusive.)

In base $10$, three-digit numbers range from $100$ to $999$.

In base $9$, three-digit numbers range from $81$ to $728$.

In base $11$, three-digit numbers range from $121$ to $1330$.

A number in the first range is chosen randomly.  What is the probability that it also falls into the second and third ranges?

The intersection of those ranges occurs from $121$ to $728$.  This range has length $608$, so the probability we desire is

$\frac{608}{900}\approx 0.7$

Making our answer $\boxed{ E }$. [/hide]


1.  A sealed envelope contains a card with a single digit on it. 3 of the following statements are true, and the other is false.

I. The digit is 1.
II. The digit isn't 2.
III. The digit is 3
IV. The digit is not 4.

Which of the following must necessarily be correct?

(A) I is true     (B) I is false     (C) II is true      (D) III is true     (E) IV is false.

2. What is the max. number of points of intersection of the graph of thwo different fourth-degree polynomials, y=p(x) & y=q(x), each with leading coefficient 1?   (A) 1  (B) 2  (C) 3  (D) 4  (E) 8

Thanks!

[hide]

b) this isn't rigorous, but it's a perfectly reasonable method for the ahsme

just draw some pictures.  you realize that it has to be at least four because the 4th degree polynomial p(x) can have up to 4 roots (intersect the line y=0x^4+0x^3+0x^2+0x+0  4  times.)

now realize that if q(x) intersected p(x) even once more, the answer would have to be d.  so you add a little curve do your y=0x^4+0x^3+0x^2+0x+0 and voila!  you can make it intersect a 5th time!

a) being "necessarily correct" implies that there CANNOT be a valid case when the opposite of the statment is true.  so we go through, looking at the opposites of the given answer choices.   [/hide]

Are you sure, because x-values of all intersections must be a real root to p(x)-q(x)=0, which can have at most 4 real roots.

hmmm... maybe you're right.

ack!  LEADING COEFFICIENT 1

i suppose this is where both of us go wrong?

[hide]

p(x)=q(x)

p(x)-q(x)=0

due to both p and q having the same leading coefficient, the result is a third degree polynomial r(x)=0

so 3 intersection points?

[/hide]

TRICKY TRICKY TRICKY EVIL person who wrote this problem :)

1. number 1 is very easy

[hide]3 things need to be true, out the four.

That must mean one of them can't be true.

Two of the statements are "The digit is 1" and "The digit is 3"

Since theyboth can't be right, one has to be wrong.  Since we can only have one statement to be false, II and IV have to be true.

Since answer (C) II is true is correct, and there can only be one correct answer, the answer is c.[/hide]


Find all functions from R to R such that
$f(xy)+f(x+y)=f(x)f(y)+f(x)+f(y)$  for all $x,y\in R$

This problem is not easy, this is solution of Phan Duc Chinh.
We will assume that $f\not\equiv 0$.
[hide="Sketch"]
Step 1: $f(0)=0$
Step 2: $f(x+1)=f(1)(f(x)+1)$ and $f(-1)=-1$.
Step 3: $f(1)=1$ and $f(-x)=-f(x)$
Step 4: $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$
Step 5: $f$ is increasing.[/hide]


[quote="kunny"]The answer is $(m.n)=(-3,0),(-1,-1),(0,-3),(2,5),(3,3),(5,2)$.

kunny[/quote]

How do we know this? Did you find a solution or just use guess and check? (which is perfectly fine)

O.K.Transformationing $mn-m-n=3$ into $m(n-1)-n=3$,.......Let's try!

kunny

[quote=Fawxx]/bump $~~~~~$[/quote]

Why tho

[quote=Fawxx]/bump $~~~~~$[/quote]

Don't bump a 16 year old thread just for entertainment.
That's including truth or dare.

[quote=dineshs]Why did you bump a 16-year old thread? Do you have a question about the problem? If not, please don't do this again.[/quote]

It was part of a truth or dare thread.

Wait I just realized this was upvoted by etvat. Coincidence?

[quote=Fawxx]LOL it was a dare why u so mad  :o  :P[/quote]

There's something called a rule which states
*Avoid bumping old posts.
And people dislike it when rules are broken.

wait this was posted like 16 years ago and it only has 13 (14) posts?!

[quote=rrusczyk]Find all integers m and n which satisfy mn = m + n + 3.[/quote]

We rewrite this equation as $mn-m-n=3$. Adding $1$ to both sides and factoring gives $(m-1)(n-1)=4$. Now, just consider the factor pairs of $4$. The solutions are $\boxed{(-3,0), (0,-3), (-1,-1), (2,5), (5,2), (3,3)}$.

Wait did SFFT come out after this post? If so, rr has legendary prediction skills ;)

[quote=OlympusHero][quote=mathfanatic]I think the most efficient method is to [hide]rearrange to get mn - m - n + 1 = 4, or (m-1)(n-1) = 4, so (m-1,n-1) = (-4, -1), (-2, -2), (-1, -4), (1, 4), (2, 2), (4, 1), so (m, n) = (-3, 0), (-1, -1), (0, -3), (2, 5), (3, 3), (5, 2).[/hide][/quote]

Also, this post uses SFFT.[/quote]

oops didn't see that lol

samrocksnature,

I don't think sfft was invented 16 years ago
when did Simon invent it?

mathfanatic uses the method, but he didn't say it was sfft, so it's not :P

MPPEL

[quote=samrocksnature][quote=williamxiao][quote=Fawxx]LOL it was a dare why u so mad  :o  :P[/quote]

so? that's not a good reason to bump a 16 year old post. If you don't have a question about the problem, then don't bump it[/quote]

nice 500th post :dry:[/quote]

bruv

[quote=samrocksnature]WOAHWOAHWOAH did nobody seriously ever consider SFFT until now???[/quote]

to be fair the algebra book wasn't published until 2007 I believe
but yeah it can be solved so much quicker using it, didn't even think of it until franzliszt did it

[quote=samrocksnature]MPPEL?

No like not invented but that guy Simon[/quote]
take the first letter of every word from my username

I meant maybe simon invented his factoring trick after 2004
cuz only aops ppl call it that

I'm pretty sure that Simon was still a student in 2004 so SFFT didn't exist, or at least it wasn't called Simon's Favorite Factoring Trick.

anyways why the bump ;-;

Simon might have come up with the name and parts of the technique but the basis of the technique is actually factoring by grouping

[quote=samrocksnature][quote=mathfanatic]I think the most efficient method is to [hide]rearrange to get mn - m - n + 1 = 4, or (m-1)(n-1) = 4, so (m-1,n-1) = (-4, -1), (-2, -2), (-1, -4), (1, 4), (2, 2), (4, 1), so (m, n) = (-3, 0), (-1, -1), (0, -3), (2, 5), (3, 3), (5, 2).[/hide][/quote]

So Simon just took credit for something he didn't make :dry:

[quote=Fawxx]sry[/quote]

FTFY[/quote]

well how are you going to find who made this
it's a technique
it's probably been a thing for hundreds of years
you want to track it, lol?

simon was the one who formalized it and stuff, so i think he gets the credit
there are tons of other things that are named after people who weren't actually first (e.g. cauchy-shwarz)

[quote name="peace09" url="/community/p19287623"]
[quote=nicarmt][quote name="Fawxx" url="/community/p19285741"]
/bump $~~~~~$
[/quote]

why did you bump?

[size=150][color=#FF8800][font=Comic Sans MS]-nicarmt [img]https://cdn.artofproblemsolving.com/attachments/b/8/f5af6316833f6816b20455babdec5605fec062.png[/img][/font][/color][/size][/quote]

were you not the one who bumped as well? >15 years or an hour... no difference really

reporting this and @above
[/quote]

i never bumped a 15 year old topic :huh:

also why did you report my post

[quote=hashtagmath][quote=franzliszt]Whatever, I am gonna solve rrusczyk's problem :coolspeak:

Use SFFT to get $(m-1)(n-1)=4$ where you can then guess and check to find that $(m, n)=(-3, 0);(-1, -1);(0, -3);(2, 5);(3, 3);(5, 2)$.[/quote]

@fransliszt you have violated the third commandment of the AoPS Middle School Math Official Forum Rules
[quote=AoPS Middle School Math Official Forum Rule #3]
Hide solutions!
[/quote][/quote]

Lots of people do that. They sometimes forget. Just remind them. Dont go like [b][u][i][size=200][color=#f00]tHiS iS vErY bAd FoLlOw RuLeS oMg[/color][/size][/i][/u][/b] (not trying to be insulting, more on the humor side) and go like "hey u remember hide ur solutions :)"

[quote=rrusczyk]Find all integers m and n which satisfy mn = m + n + 3.[/quote]

[hide=S]Use sfft to get (m-1)(n-1)=4 (3,3);(5,2);(2,5)

[quote=peace09][quote=nicarmt][quote name="Fawxx" url="/community/p19285741"]
/bump $~~~~~$
[/quote]

why did you bump?

[size=150][color=#FF8800][font=Comic Sans MS]-nicarmt [img]https://cdn.artofproblemsolving.com/attachments/b/8/f5af6316833f6816b20455babdec5605fec062.png[/img][/font][/color][/size][/quote]

were you not the one who bumped as well? >15 years or an hour... no difference really

reporting this and @above[/quote]

no he was not

funny how so many old threads are being bumped nowadays!

SFFT was already known on AoPS...

It was introduced [url=https://artofproblemsolving.com/community/c4h8215p43468]here.[/url]


Does anyone here know how many games you must play in order to earn the 5th game badge? I am seeing users with as many as 6000 games but they still have 4 badges (earned after 350 games) Is this a glitch or is the 5th badge not released yet?

go [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=258276]here[/url]

there is most likely no 5th games badge


[quote="ifai"]How to prove that $L(x)=\log_x (x+1),\ x>1$ is a decreasing function without calculus?[/quote]
Are you joking?
I thjnk this mention is useful.
\begin{eqnarray*} L(x)&=&\log_{x} (x+1) \\ x^{L(x)} &=& x+1 \end{eqnarray*}
Taking derrivative, then the question is easily shown.

[quote="ifai"]How to prove that $L(x)=\log_x (x+1),\ x>1$ is a decreasing function without calculus?[/quote]

You can change the base (I don't know exactly how it is called in English) of the logarithm and get:

$L(x)=\frac{\log x+1}{\log x}$ which is obviously decreasing.

[quote="Lovasz"][quote="ifai"][quote="eLeM"]$L(x)=\frac{\log x+1}{\log x}$[/quote]
 :?:[/quote]
Yes,
\[ \begin{eqnarray*} L(x)&=&\frac{\log x+1}{\log x} \\ &=& 1+\frac{1}{\log x}  \]
Easily, $f(x)=\log x$ is increasing function if $x>1$ (property of loga function).[/quote]
$\log(x+1)=\log x+1$??????????

[quote="eLeM"][quote="ifai"][quote="eLeM"]$L(x)=\frac{\log x+1}{\log x}$[/quote]
 :?:[/quote]

Sorry. Should be

$L(x)=\frac{\log (x+1)}{\log x}$[/quote]
But why it is obviously decreasing?

[quote="ifai"][/quote][quote="eLeM"][quote="ifai"][quote="eLeM"]$L(x)=\frac{\log x+1}{\log x}$[/quote]
 :?:[/quote]

Sorry. Should be

$L(x)=\frac{\log (x+1)}{\log x}$[/quote][quote="ifai"]
But why it is obviously decreasing?[/quote]

In this case logarithms doesn't change anything (I doubt :) )so it is the same as $f(x)=\frac{x+1}{x}$ which is decreasing.

[quote="eLeM"][/quote][quote="ifai"][/quote][quote="eLeM"][quote="ifai"][quote="eLeM"]$L(x)=\frac{\log x+1}{\log x}$[/quote]
 :?:[/quote]

Sorry. Should be

$L(x)=\frac{\log (x+1)}{\log x}$[/quote][quote="ifai"]
But why it is obviously decreasing?[/quote][quote="eLeM"]

In this case logarithms doesn't change anything (I doubt :) )so it is the same as $f(x)=\frac{x+1}{x}$ which is decreasing.[/quote]

Exactly... they do not change anything I think.......

[quote="eLeM"]
In this case logarithms doesn't change anything (I doubt :) )so it is the same as $f(x)=\frac{x+1}{x}$ which is decreasing.[/quote]
In which case logarithms changes something?

[quote="ifai"][quote="JBL"]$\log(x+1) - \log x = \log (1 + \frac1x)$ is decreasing, and $\log x$ is increasing, so $\frac{\log (x+1) - \log x}{\log x} = \frac{\log (x+1)}{\log x} - 1$ is decreasing, so $\frac{\log (x+1)}{\log x}$ is decreasing.[/quote]
$-x$ is decreasing, and $x^2$ is increasing when $x>0$.
But $\frac{-x}{x^2}=-\frac 1 x$ is NOT decreasing.[/quote]

Yes, you are right, I also need that they are both positive.  But they are, so it works.

The same argument with exponentials instead of logarithms: let $\log_x(x + 1) = 1 + \epsilon(x)$, where $\epsilon(x)$ is some positive function of $x$.  Then $x^{1 +\epsilon(x)} = x + 1$ so $x^{\epsilon(x)} = 1 + \frac1x$.  The RHS is decreasing in $x$.  If $\epsilon(x)$ were ever non-decreasing, the left-hand side would be increasing in $x$ (since for positive exponent $a$, $1 < x < y \Rightarrow x^a < y^a$), a contradiction.

[quote="ifai"]How to prove that $L(x)=\log_x (x+1),\ x>1$ is a decreasing function without calculus?[/quote]

take f(x), f(x^2)

$\frac{\log x+1}{\log x}>\frac{\log x^2+1}{2\log x}$

using x>1, we can cancel the logs, giving

$2\log{x+1}>\log{x^2+1}$
$x^2+2x+1>x^2+1$
$x>0$ which is true

now i am not sure if this can be made rigirous, but it still works, since we have that $x^2>x$, so we can use this on any x>1, and make the sequence $f(x)$, $f(x^2)$, $f(x^4)$, $f(x^8)$, etc that will be decreasing...and that should be enough as the function is continous for $x>1$

[quote="ifai"]How to prove that $L(x)=\log_x (x+1),\ x>1$ is a decreasing function without calculus?[/quote]

$a_n = \frac{ \log (n + 1)}{ \log n}$ is a decreasing sequence if

$b_n = \log n$ grows more slowly than an exponential function, which is obviously true.  :)

[quote="ifai"][quote="JBL"]
Yes, you are right, I also need that they are both positive.  But they are, so it works.
[/quote]
You mean if both $f(x)$ and $g(x)$ are decreasing and positive when $x\in I$, then $f(x)g(x)$ is also a decreasing function when $x\in I$?[/quote]

Yes.


PROBLEM (Lokman GÖKÇE): In $ABPE$ convex quadrilateral, $m(PBA)=63^{o}, m(BAE)=108^{o}, AB=AE=\sqrt 2.PB$.
Find the measure of $\angle PEA$.


Find four distinct positive integers, $a$, $b$, $c$, and $d$, such that each of the four sums $a+b+c$, $a+b+d$,$a+c+d$, and $b+c+d$ is the square of an integer. Show that infinitely many quadruples $(a,b,c,d)$ with this property can be created.

We can just solve a group of equations:

a+b+c=p^2, b+c+d=q^2, c+d+a=r^2, d+a+b=s^2

then we will have: a + b + c + d = (p^2 + q^2 + r^2 + s^2) / 3

let 3 be the divisor of p, q, r, s, then then sum be an integer, that is, a, b, c, d are integers.

because p, q, r, s are infinite, (a, b, c, d) is infinite

i think you should at least give an example

this is a more complete proof(using the notations from libra_gold's post):
let us say that a<b<c<d.then p<q<r<s.we saw that( p^2+q^2+r^2+s^2)/3.let us believe that p,q,r,s are all divisible by 3.seeing 3 i thought that this could happen:q=p+3,r=p+6,s=p+9.let's try it.
then d-c=9+6p,c-b=27+6p,b-a=45+6p <=>b=a+45+6p
c=a+72+12p
d=a+81+18p.then p^2=3a+18p+117 delta=792+12a
then p=(18+sqrt(792+12a))/2<=>p=9+sqrt(198+3a) all we must do now is finding a such that 198+3a=k^2
let k=3j then 66+a=3j^2 but this is simple.we just put any j and find a,b,c,d.that's all.the easiest exemple is 9,198,369,522 :P

Let $n \ge 8$ be a positive integer.

Choose
$a = 3 n^2 - 18 n - 39 > 0$
$b = 3 n^2 + 6$
$c = 3 n^2 + 18 n + 33$
$d = 3 n^2 + 36 n + 42$

Then you have
$a + b + c = (3n+0)^2$
$a + b + d = (3n+3)^2$
$a + c + d = (3n+6)^2$
$b + c + d = (3n+9)^2$

There are infinite pairwise distinct a, b, c , d, since the polynomials are not equal (pairwise!) for $n \ge 8$.


Yo, I think I'll have a go at this one. Though I must say that mold in the lab has been getting to me recently so tell me if this makes sense or not...

[hide]1) Factor(B - 1 + 1/C)  =  B(1 - 1/B + 1/(BC))  = B(1 + A - 1/B)

Which means, (A - 1 + 1/B)(B - 1 + 1/C) = B(A^2 - (1 - 1/B)^2) <= B*A^2 so squaring the product of all three <= B*A^2 * C*B^2 * A*C^2 = 1

If we had a negative left side then how would we multiply the inequalities? I couldn't answer that part.

But I dealt with the case (A - 1 + 1/B), (B - 1 + 1/C), (C - 1 + 1/A) are not all positive.

If one of the three terms is negative, then the other two must be positive.

For example, if A - 1 + 1/B < 0, then A < 1, so C - 1 + 1/A > 0, and B > 1, so B - 1 + 1/C > 0. But if one term is negative and two are positive, then their product is negative and hence less than 1. [/hide]

[color=cyan]Without doing a thorough check of everything, I'll say that if you're right, that's really nice.  If the left side is negative, then you're done because negative < 1.  I think that's what your question was, no?[/color]

Yeah, i think we've got it solved..  and check out the new font size feature, it rox!


factorial
geometry
3D geometry
AMC
AIME
number theory
prime factorization

The third contest is way over now, i think... so feel free to post your solutions to those problems.

Could someone post solutions for the last 2 factorial ones...

I wish more people would pitch in, but for now, I'll post 12.

12. What is the number of perfect square and cube factors of 15!?

The prime factorization is

2^11 * 3^6 * 5^3 * 7^2 * 11 * 13

The number of perfect square and cube factors of 15! is

# ^2 + # ^3 - # ^6

since 6th powers are both square and cubes

You can get the answer by using the basic #of factors forumula

the exponents plus one, then multiplied together.

only this time, they can only be square or cube or 6th powers

there are 1 and 2, 2^2, 2^4, 2^6, 2^8, 2^10 and
3^2, 3^4, 3^6... so on...

so its 6 * 4 * 2 * 2 = 96

in a similar manner, # ^3 is 4*3*2 = 24 and # ^6 is 2*2 = 4

so the answer is

96 + 24 - 4 = 116

there was an aime problem like this, sorta

I think I saw that problem on 2003 AMC 12A # 23 or something, only that problem was somewhat easier than this one because it asks for square factors.

By the way, can someone please show me how to do #13. I've had problem with it.

I didn't create this problem, Repmax sent it to me, then 13 was came up by me misunderstanding his question when he asked me to verify the answer... so i have no clue...

As for 13.

You know the forumula for the sum of factors?

so its just

the sum of square factos + sum of cube factors - sum of ^6 factors

Oh, that was cool. I did not see the formula for some reason.

Umm..still lost, can u post a full solution...

I'm going to wait for someone to post it, if they don't in a day or two, i'll post it... meanwhile, any solutions to the other problems?


geometry
circumcircle
geometric transformation
reflection
cyclic quadrilateral
geometry proposed

For the triangle $ABC$ denote: the circumcircle $C(O,R)$; the incircle $C(I,r)$; the exincircle $C(I_a,r_a)$ w.r.t. the vertex $A$; the projection $D$ of the vertex $A$ on the sideline $BC$. Prove that $\boxed {\ D\in OI\Longleftrightarrow R=r_a\ }\ .$

Let $X$ be the tangency point of the ex-circle with $BC$.
It can be proved that $XI$ passes through the midpoint of the altitude from $A$.
Therefore, if $Y$ is the reflection of $I_a$ in $X$ then $Y\in{OI}$.
Consequently, $R=r_a$.

$\overline {\underline {\left|\mathrm {Preliminary\ study.}\right| }}$ Denote $F\in AI\cap DO$, the circumcircle $w=C(O,R)$ and $E\in (AI\cap w$. But $AD\parallel OE$ $\Longleftrightarrow$ $\boxed {\frac{FA}{FE}=\frac{h_a}{R}}\ .$ From the [u]Ptolemeu's theorem[/u] applied in the cyclic quadrilateral $ACEB$ and with the wel-known relation $EB=EC=EI$ we obtain $a\cdot AE=(b+c)\cdot IE$, i.e. $\boxed {\frac{AE}{b+c}=\frac{IA}{b+c-a}}\ .$

$\overline {\underline {\left|\mathrm {Demonstration.}\right| }}$ $R=r_a$ $\Longleftrightarrow$ $\frac{FA}{FE}=\frac{h_a}{r_a}$ $\Longleftrightarrow \frac{FA}{FE}=\frac{b+c-a}{a}\Longleftrightarrow$
$\boxed {\frac{AE}{b+c}=\frac{FA}{b+c-a}}\Longleftrightarrow FA=IA\Longleftrightarrow F\equiv I\ .$

here is my solution(?):

  Vertex A is the external center of similitude of ex-circle and incircle.

  The external center X for incircle and circumcircle is on OI line and

  such that line AX is parallel to OOa <---- OI goes thru base of altitude.

  Thus ex-circle and circumcircle have equal radii.

     This idea was communicated to me in PM by M. D'Alambert himself,

     but rather vaguely.

  T.Y.

  M.T.


Hi,

Next year I will be attending the University of Texas at Austin, majoring in computer engineering. However, I am very seriously considering double majoring in some area of math.

My question is, does anyone know how good UT's math department is? And also, is it recommended that I double major?

Paul Graham has [url=http://paulgraham.com/hs.html]good advice for high school seniors[/url] (including why to major in math).

Have fun at UT.

Thanks tokenadult. That's a terrific article!


5 - 2 abs(x + 1) < -1  solve for x and give the answer in interval notation

[hide="answer"]
$ 5\minus{}2|x\plus{}1|<\minus{}1\Longrightarrow\minus{}2|x\plus{}1|<\minus{}6\Longrightarrow |x\plus{}1|>3$
now we have to cases to solve for:
$ x\plus{}1>3\Longrightarrow x>2$
and
$ x\plus{}1<\minus{}3\Longrightarrow x<\minus{}4$

so the answer in interval notation is
$ (\minus{}\infty,\minus{}4)\cap (2,\infty)$[/hide]

I think you meant "union" $ \cup$ not "intersect", since if two intervals intersect the answer can be simplified to 1 interval. Otherwise, your solution seems correct  :D

oops.. :blush: . that's right, union. the intersection of those two inequalities would be the null set.. my mistake, it's $ \cup$


A uniform disc has mass M and radius R. A circular hole of radius r is made in the disc with its centre at a distance x from the centre of the disc. Find the moment of inertia of the remaining portion of the disc about an axis passing through the centre of the hole and perpendicular to the plane of the disc.

It's just superposition principle and parallel axis theorem.


[quote="FOURRIER"]I need a link where I can find belarussian olympiad.

Thanks.[/quote]

See here.

http://www.imomath.com/index.php?options=oth|Blr|Ne|Ne&ttn=Belarus&p=0


geometry
trigonometry
MATHCOUNTS
ratio
trig identities
Law of Cosines
angle bisector

Triangle ABC is inscribed in a circle with center O.  $ m\angle AOB \equal{} 120^\circ, BC \equal{} 3, AB \equal{} 3\sqrt {3}$.  Construct the bisector of $ \angle C$ and let it intersect the circle at point $ M$. Find the length of $ MC$.

Im pretty sure that it is a 30 - 60 - 90 triangle, , so AC is 6. This would make c 60 degrees... I think that makes MC 6 as well, b/c i think the intersection of all the angle bisectors is the center of the circle, making MC a diameter....... correct me if I am wrong( probably am)

Intersection of angle bisectors is the center of the inscribed circle.

Ummm...here's how I did it...
Use law of cosines to find really messy expressions for AM and MB in terms of MC and another messy expression for AB.
Use Ptolemy's and solve the (I think it's quartic) equation to get 0 and $ 3\plus{}\sqrt{3}$.
So is the answer $ 3\plus{}\sqrt{3}$?

Nope.  It's a nice solution.

[hide="Solution"]Since $ \angle AOB\equal{}120^\circ,\angle ACB\equal{}60^circ$ by the inscribed angle theorem.

Obviously, $ \angle CAB\equal{}30^\circ$ because of the law of sines.  Thus we have a $ 30\minus{}60\minus{}90$ right triangle.  This implies that $ AC$ is a diameter of the circle.  We also know that it must be $ 6$ due to the $ 30\minus{}60\minus{}90$ triangle.

Let the intersection of the angle bisector of $ \angle C$ with side $ AB$ be $ D$.  We then have that $ CD*DM\equal{}AD*DB$ by power of a point.  However, we also know that $ \bigtriangleup ADC$ is isosceles with $ AD\equal{}DC$ because $ \angle DAC\equal{}\angle ACD$.

Since $ \angle DCB\equal{}30^\circ$, we have another $ 30\minus{}60\minus{}90$ right triangle, namely $ \bigtriangleup DBC$.  Since $ BC\equal{}3,DB\equal{}\sqrt{3}$ and $ DC\equal{}2\sqrt{3}\longrightarrow AD\equal{}2\sqrt{3}$ from our isosceles triangle.

Since $ AD\equal{}DC$ and $ CD*DM\equal{}AD*DB$, we have that $ DM\equal{}BD\longrightarrow DM\equal{}\sqrt{3}\longrightarrow MC\equal{}3\sqrt{3}$[/hide]

Extra: You can actually find the length in general for an arbitrary triangle. It's not a good formula to remember though.  Also, note that $ MA\equal{}MB$ for any triangle in general.  This is relatively easy to prove by using the inscribed angle theorem or by the property of cyclic quadrilaterals.

[hide="Proof/Solution"]Let $ AB\equal{}c,AC\equal{}b,BC\equal{}a$.  The angle bisector of $ A$ intersects $ BC$ at $ D$ and the circle at $ M$.  First, we have that $ \angle CAM\equal{}\angle CBM$ because it is a cyclic quadrilateral.  Similarly, $ \angle BCM\equal{}\angle BAM$.  However, we know that $ AM$ is an angle bisector which implies that $ \angle CBM\equal{}\angle BCM$.  Thus, we have that $ MC\equal{}MB$. This completes the first part of the proof.

To find the length of $ AM$, we consider Ptolemy's theorem, which states that the product of the diagonals is greater than or equal to the sum of the products of the opposite sides (equality holds only in a cyclic quadrilateral). We thus have that $ a*AM\equal{}b*BM\plus{}c*CM$
Let $ BM\equal{}CM\equal{}x$.

Then we have that $ AM\equal{}\frac{x(b\plus{}c)}{a}$
We then wish to find $ x$.

Let $ \angle BAM\equal{}\angle CAM\equal{}\angle BCM\equal{}\angle CBM\equal{}\alpha$.
Then by laws of cosines on  triangle $ BCM, a^2\plus{}x^2\minus{}2xa\cos\alpha\equal{}x^2\longrightarrow x\equal{}\frac{a}{2\cos\alpha}$
To find $ \cos\alpha$, we take $ \bigtriangleup ABC$ and apply laws of cosines again (there may be a better way, but I'm doing this off the top of my head).

$ b^2\plus{}c^2\minus{}2bc\cos2\alpha\equal{}a^2\longrightarrow \cos2\alpha\equal{}\frac{b^2\plus{}c^2\minus{}a^2}{2bc}$
Since $ \cos2\alpha\equal{}2\cos^2\alpha\minus{}1, 2\cos^2\alpha\equal{}\frac{(b\plus{}c\minus{}a)(b\plus{}c\plus{}a)}{2bc}$ (using difference of squares to simplify)

$ \cos\alpha\equal{}\pm\frac{\sqrt{(b\plus{}c\minus{}a)(b\plus{}c\plus{}a)}}{2\sqrt{bc}}$.  Since $ \alpha<90^\circ$ because it is half of one of the angles of a triangle, we take the positive value.  We can now plug this into our equation for $ x$, etc etc... There is a better way to find it, but you can do it yourself =)[/hide]

does this really belong in the mathcounts forum?

The question is probably around state 25-30 level. I think. But the extra proof is way above.

i dont see a way to do it without trig, so unless there is, it wont appear in mathcounts

any problem that requires the use of the law of cosines probably won't be on mathcounts, but tis still a good problem, and nice proof.

Err... It didn't use law of cosines... that's the extra part. The law of sines was simply to explain, but it's obvious that since it has a ratio of a 30-60-90 right triangle, it MUST satisfy that. There is no need for trig to solve it.

Oh...
woops...
I read it as $ AC \equal{} 3\sqrt{3}$...
No wonder it was so messed up...


Which pair of the following expressions are never equal for any natural number $ x$: $ x, x^2, 2^x, x^x$?

For each pair, there is some way for them to be equal. For example, $ 2^x$ and $ x^2$ can be equal if $ x$ is 2. We find that the only exception is $ \boxed{x \text{ and } 2^x}$.

EDIT: Yeah, I meant 2. Typoed.  :oops:

[quote]For example, $ 2^x$ and $ x^2$ can be equal if $ x$ is $ 1$.[/quote]

umm...
don't u mean when $ x$ equals $ 2$?

because $ 2^1$ does not equal $ 1^2$

In FTW, the way to express this would be:

$ x$ $ 2^x$

Actually, it would be $ x, 2^x$.

EDIT: Probably both.

[quote="isabella2296"]Actually, it would be $ x, 2^x$.[/quote]

Really?, cuz it accepts $ x$ $ 2^x$.  And its easier to type.


search
combinatorics unsolved
combinatorics

Let A be any set of n residues mod $ n^2$. Show that there is a set B of n residues mod $ n^2$ such that at least half of the residues $ mod n^2$ can be written as a + b with a in A and b in B.


[hide]From the information given, we can deduce that $ x_1,x_2,x_3$ are solutions to $ x^3\plus{}ax^2\plus{}bx\equal{}t_1$ where $ t_1\equal{}x_1x_2x_3$ and $ y_1,y_2,y_3$ are solutions to $ x^3\plus{}ax^2\plus{}bx\equal{}t_2$ where $ t_2\equal{}y_1y_2y_3$.

Consider the graph of $ x^3\plus{}ax^2\plus{}bx$.  We are looking for its intersections with the line $ y\equal{}t_1$ and $ y\equal{}t_2$.  Since we are assuming $ x_i,y_i\in\mathbb{R}$ we know that the largest root will be after the relative minimum.  Notice that in cubics, the section after the relative minimum is increasing.  Thus, $ x_1\ge y_1\Rightarrow t_1\ge t_2$.  A knowledge of cubics also reveals that the least root, if all roots are real, is before the relative maximum.  Since the section before the relative maximum is increasing, we have $ t_1\ge t_2\Rightarrow x_3\ge y_3$, and we're done.[/hide]


Can anyone illustrate the fact that the function $e^{-x^{2}}$ does not have an elementary antiderivative?
I've read that the proof is difficult, could anyone give a sketch of its proof?

Take a look at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107861]this topic[/url]. It's three topics under yours :D There are some useful links there.


Given the three numbers $x, y=x^x, z = x^{(x^x)}$ with $.9 < x < 1.0$.  Arranged in order of increasing magnitude, they are:
\[ \text{(A)}\ x, z, y\ \ \ \ \text{(B)}\ x, y, z\ \ \ \ \text{(C)}\ y, x, z\ \ \ \ \text{(D)}\ y, z, x\ \ \ \ \text{(E)}\ z, x, y \]

[hide]The answer is $A$, but I'm not sure how to do it without a calculator.[/hide]

im not sure if i know what magnitude really is but..
[hide]
when and exponent is less than one, and the base is greater than one, the result will be less than the original base right (i think), but since the base is less than one (.9<x<1) the result will be larger than the original. Thus:
if y=x^x and x is less than one, then y is greater than x, and z=x^(x^x) (however, if you have a base less than one and a larger exponent (x^x is larger than x) the result will be less).
thus y is greater than z as well as x, but z is greater than x
so the answer is [b]A[/b][/hide]

How do I delete a post on my own, I wanna try it.  :blush:

This is how I approached it, though I think I may be wrong.

 $\displaystyle y=x^x$
Since  $\displaystyle x$ is less than $\displaystyle 1$, then anything to the power of $\displaystyle x$ will be less than it used to be.

so  $\displaystyle x > y$

Now that we know that, we need to find the relationship between  $\displaystyle y$ and  $\displaystyle z$

Assume  $\displaystyle y>z$

Then  $\displaystyle x^x>x^{(x^x)} \Rightarrow \sqrt[x] {x^x}>\sqrt[x] {x^{(x^x)}} \Rightarrow x>x^{x^{x -1}}$

Now here, in my opinion, is the key part of the question.

Since $\displaystyle x<1$ the statement simplifies to $\displaystyle x>x^{\frac1{x^{1-x}}}$

Since $\displaystyle 0<1-x<1$, then $\displaystyle x^1>x^{\frac 1{1-x}} \Rightarrow 1>\frac 1{1-x} \Rightarrow 1-x>1 \Rightarrow 0>x>.9$ ($\displaystyle x>.9$ is given). Contradiction.

So $\displaystyle z>y$
$\displaystyle x>y$ (from before)

So now we need the relationship between $\displaystyle x$ and $\displaystyle z$

Since $\displaystyle 1>x>y$, then $\displaystyle x^1>z=x^y$.

So $\displaystyle x>z>y$

Is this $D$?

[quote]you are forgetting that a number less than one to a number less than one power will be greater[/quote]

What?
 :?  ;)  :(

[hide="hint"]$\ln{x}$ is a strictly increasing function.[/hide]
[hide="solution"]First $0<x<1\Longrightarrow\ln{x}<\ln{1}=0$.

Next $x<1\Longrightarrow{x}\ln{x}>\ln{x}\Longrightarrow\ln{x^x}>\ln{x}\Longrightarrow\mathbf{x^x>x}\Longrightarrow{x}^x\ln{x}<x\ln{x}\Longrightarrow\ln{x^{(x^x)}}<\ln{x^x}\Longrightarrow\mathbf{x^{(x^x)}<x^x}.$

Finally $x>0\Longrightarrow{x}\ln{x}<0\Longrightarrow{x}\ln{x}<\ln{1}\Longrightarrow\ln{x^x}<\ln{1}\Longrightarrow{x}^x<1\Longrightarrow{x}^x\ln{x}>\ln{x}\Longrightarrow\ln{x^{(x^x)}}>\ln{x}\Longrightarrow\mathbf{x^{(x^x)}>x}.$

Thus, by the three boldfaced inequalities, $x^x>x^{(x^x)}>x,$ or $y>z>x,$ or (A).[/hide]

Well I feel like an idiot now.
 :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:

Does the range $\displaystyle x>0.9$ have to do with anything? Can someone with a calculator test $\displaystyle 0.95^{0.95}$ or $\displaystyle (\frac{19}{20})^{\frac{19}{20}}$ to see if it's greater than $\displaystyle 0.95$ or $\displaystyle \frac{19}{20}$

[quote="hiitsme"]According to you................... ($ \displaystyle x^x>x$)
Test: $ \displaystyle \frac 12^{\frac 12} > \frac 12 \Rightarrow \sqrt {\frac 12}> \frac 12$. Which is untrue
 Counterexample................[/quote]$\sqrt{\frac{1}{2}}>\frac{1}{2}$ is not untrue: $2>\sqrt{2}\Longrightarrow\frac{1}{\sqrt{2}}>\frac{1}{2}\Longrightarrow\sqrt{\frac{1}{2}}>\frac{1}{2}$.

[quote="hiitsme"]How can $ \displaystyle x^x > x$?
$ \displaystyle x$ is less than 1, anything positive to the power of $ \displaystyle x$ should be less.[/quote]Not when "anything" is also less than 1.

[quote="hiitsme"]Does the range $\displaystyle x>0.9$ have to do with anything? Can someone with a calculator test $\displaystyle 0.95^{0.95}$ or $\displaystyle (\frac{19}{20})^{\frac{19}{20}}$ to see if it's greater than $\displaystyle 0.95$ or $\displaystyle \frac{19}{20}$[/quote]

Not really. As long as $0<x<1$, the answer is A. Anyways, 0.95 is within the range 0.9<x<1.0

[quote="hiitsme"]Does the range $\displaystyle x>0.9$ have to do with anything?[/quote]The restriction $.9<x<1\Longrightarrow0<x<1,$ and then everything I said follows, since my solution is based on this assumption (technically the assumption that $\ln{x}$ is strictly increasing also).

I've been toying around with this problem, and this is what I came up with.

Let $x\uparrow_{(n)}x$ be the $n\text{th}$ term in the sequence $\{x,x^x,x^{x^x},\ldots\}$, where $x$ is the $0\text{th}$ term (I saw the $\uparrow$ notation on mathworld but I'm too lazy to check if I'm using it right.. doesn't matter anyway). The sequence can be defined as \[a_0=x; a_{n+1}=x^{a_n}, n\ge0,\]so that $a_n=x\uparrow_{(n)}x$. Assume $0<x<1$ (and thus $\log{x}<0$) for everything that follows.

First it can be proved by induction pretty easily (I hope :D) that $a_{n+1}>a_n$ for even $n$ and $a_{n+1}<a_n$ for odd $n$, and that $a_{n+2}>a_n$ for even $n$ and $a_{n+2}<a_n$ for odd $n$, using the solution I offered in my previous post. I was going to write up the proofs but it would take way too long. Maybe later..

So we have for even $n$: $x\uparrow_{(n+1)}x>x\uparrow_{(n+2)}x>x\uparrow_{(n)}x$, and for odd $n$: $x\uparrow_{(n)}x>x\uparrow_{(n+2)}x>x\uparrow_{(n+1)}x.$

So we have (sorry I'm not explaining anything in depth..) \[x^x>x^{x^x}>x\]\[x^x>x^{x^{x^x}}>x^{x^x}>x\]\[x^x>x^{x^{x^x}}>x^{x^{x^{x^x}}}>x^{x^x}>x\]\[x^x>x^{x^{x^x}}>x^{x^{x^{x^{x^x}}}}>x^{x^{x^{x^x}}}>x^{x^x}>x,\]etc. So apparently if you define $x\uparrow_{(\infty)}x=\lim_{n\to\infty}x\uparrow_{(n)}x$, then it [edit: CONverges, not diverges :blush:] for all $0<x<1$. For real $x$ between $0$ and $1$ \[x^{x^{x^{x^{x^{x^\ldots}}}}}\]is actually a real number. Incredible.


Another one at [url=http://hallpass.com/media/antbuster.html]ant busters[/url]. This one is very complex

ehhh, got 9692, could've done quite a bit better if i utilized better placement and better towers

It's so easy. I'm at level 21, and no ants stole any cake since what? Level 11? And I only have 5 towers!!!

The ants are almost indestructible at level 52. I can go to that level only because the ants are stupid.

Try getting to 56. That's when the real fun begins!!!

I get to level 61 with tons of triple cannon.
Score: 10096

Ice cannon + triple canoon
Level 64
Score: 10816

Quick+ice+triple is even better
Score: 11909

What's the best way to place towers? The ants always slip around or between my towers.

level 65
score:  11972
not too hard :D
man the ants are really dumb, a few times one got really close to the thing with plenty of life left then turned back to follow another
just like little sheep :P

lets see... i used 3 missile launchers, 1 triple, 2 lasers, and 2 electrics

Missle launcher is really crazy :D

score: 13606
Level 69

2 x ice, 1 x spray, n x missile
score: 15918
level 75 (the ants have 5000 something energy)


calculus integration logarithms trigonometry function geometry symmetry

To do that, split the original square into two triangles.

$\int_{0}^{1}\int_{0}^{1}\frac{dx\,dy}{\sqrt{1+x^{2}+y^{2}}}=2\int_{0}^{1}\int_{0}^{x}\frac{1}{\sqrt{1+x^{2}+y^{2}}}\,dy\,dx$

$=2\int_{0}^{\frac{\pi}4}\int_{0}^{\sec\theta}\frac{r}{\sqrt{1+r^{2}}}\,dr\,d\theta= 2\int_{0}^{\frac{\pi}4}\left(\sqrt{1+\sec^{2}\theta}-1\right)\,d\theta$

So the remaining challenge is to compute $\int_{0}^{\frac{\pi}4}\sqrt{1+\sec^{2}\theta}\,d\theta.$

[quote="Kent Merryfield"]So the remaining challenge is to compute $\int_{0}^{\frac{\pi}4}\sqrt{1+\sec^{2}\theta}\,d\theta.$[/quote]

And we can do this using the substitution $\sin \theta = \sqrt{2}\sin \phi$. Then

$\int_{0}^{\frac{\pi}{4}}\sqrt{1+\sec^{2}\theta}\, d\theta = \int_{0}^{\frac{\pi}{6}}( 1+\sec 2\phi ) \, d\phi$.

The remaining computation is easy, but this gives

$-\int_{0}^{1}\int_{0}^{1}\frac{1}{\sqrt{1+x^{2}+y^{2}}}\;dx\;dy\;=\; \frac{\pi}{6}-\log (2+\sqrt{3})$,

I got this answer yesterday, but I hesitated to post this result because of this. It is exactly one fourth of the result of misan's.

To start with, let's get rid of that negative sign out front and just talk about

$\int_{0}^{1}\int_{0}^{1}\frac{1}{\sqrt{1+x^{2}+y^{2}}}\,dx\,dy$

misan claims this is $-\frac{2\pi}3-4\ln\left(2-\sqrt{3}\right)\approx 3.17$

But we are integrating a function which takes value in $[0,1]$ over a square of area $1.$ The answer must lie in the interval $[0,1].$ Hence, we know for sure that misan's answer is incorrect.

I get that $\int_{0}^{\frac{\pi}4}\sqrt{1+\sec^{2}\theta}\,d\theta = \frac{\pi}6+\frac12\ln\left(2+\sqrt{3}\right).$

(I got that by following sos440's method.)

Then the original integral would be

$2\left(\frac{\pi}6+\frac12\ln\left(2+\sqrt{3}\right)\right)-\frac{\pi}2=-\frac{\pi}6+\ln\left(2+\sqrt{3}\right)\approx 0.793$

Conclusion: sos440 is correct.

[quote="Kent Merryfield"]To do that, split the original square into two triangles.

$ \int_{0}^{1}\int_{0}^{1}\frac {dx\,dy}{\sqrt {1 \plus{} x^{2} \plus{} y^{2}}} \equal{} 2\int_{0}^{1}\int_{0}^{x}\frac {1}{\sqrt {1 \plus{} x^{2} \plus{} y^{2}}}\,dy\,dx$

$ \equal{} 2\int_{0}^{\frac {\pi}4}\int_{0}^{\sec\theta}\frac {r}{\sqrt {1 \plus{} r^{2}}}\,dr\,d\theta \equal{} 2\int_{0}^{\frac {\pi}4}\left(\sqrt {1 \plus{} \sec^{2}\theta} \minus{} 1\right)\,d\theta$[/quote]
Two questions:

1) Why when reversing the integration order, the integration limits of the inner integral of the RHS are from 0 to $ x\,?$

2) How did you find the new integration limits when you applied polar transformation?

1: Notice the $ 2$ out in front. Notice also my words at the top. We take the original square $ [0,1]\times [0,1]$ and split it in half along the line $ y\equal{}x.$ I am claiming a symmetry - that the integral on each triangle is the same. (This comes from the radial symmetry of the integrand.) Those limits of integration saying that we are integrating on the triangle bounded below by $ y\equal{}0,$ on the right by $ x\equal{}1,$ and on the upper left by $ y\equal{}x.$

2. Take those boundary lines and convert them to polar coordinates. $ y\equal{}0$ become $ \theta\equal{}0.$ $ y\equal{}x$ become $ \theta\equal{}\frac{\pi}4.$ And $ x\equal{}1$ becomes $ r\cos\theta\equal{}1$ or $ r\equal{}\sec\theta.$

Having a sketch of the region in question helps.

[quote="Kent Merryfield"]1: Notice the $ 2$ out in front. Notice also my words at the top. We take the original square $ [0,1]\times [0,1]$ and split it in half along the line $ y \equal{} x.$ I am claiming a symmetry - that the integral on each triangle is the same. (This comes from the radial symmetry of the integrand.) Those limits of integration saying that we are integrating on the triangle bounded below by $ y \equal{} 0,$ on the right by $ x \equal{} 1,$ and on the upper left by $ y \equal{} x.$[/quote]
Could you show me how'd you work that writing the integrals?

[quote="Kent Merryfield"]2. Take those boundary lines and convert them to polar coordinates. $ y \equal{} 0$ become $ \theta \equal{} 0.$ $ y \equal{} x$ become $ \theta \equal{} \frac {\pi}4.$ And $ x \equal{} 1$ becomes $ r\cos\theta \equal{} 1$ or $ r \equal{} \sec\theta.$[/quote]
This definitely made a lot of sense to me. So for the inner limits, you convert them to the polar coordinates $ x\equal{}r\cos\theta$ & $ y\equal{}r\sin\theta,$ but for the outer limits you only took the upper one, why didn't you choose the lower one?

[quote="sos440"][quote="Kent Merryfield"]So the remaining challenge is to compute $ \int_{0}^{\frac {\pi}4}\sqrt {1 \plus{} \sec^{2}\theta}\,d\theta.$[/quote]

And we can do this using the substitution $ \sin \theta \equal{} \sqrt {2}\sin \phi$. Then

$ \int_{0}^{\frac {\pi}{4}}\sqrt {1 \plus{} \sec^{2}\theta}\, d\theta \equal{} \int_{0}^{\frac {\pi}{6}}( 1 \plus{} \sec 2\phi ) \, d\phi$.

The remaining computation is easy, but this gives

$ \minus{} \int_{0}^{1}\int_{0}^{1}\frac {1}{\sqrt {1 \plus{} x^{2} \plus{} y^{2}}}\;dx\;dy\; \equal{} \; \frac {\pi}{6} \minus{} \log (2 \plus{} \sqrt {3})$,

I got this answer yesterday, but I hesitated to post this result because of this. It is exactly one fourth of the result of misan's.[/quote]

without a doubt, your answer is correct !!  :coolspeak:

indeed,
$ \int_{0}^{1}\int_{0}^{1}\frac {1}{\sqrt {1 \plus{} x^{2} \plus{} y^{2}}}\;dx\;dy\; \equal{} \; \log (2 \plus{} \sqrt {3}) \minus{} \frac {\pi}{6}$

[quote="Kent Merryfield"]$ \int_{0}^{1}\int_{0}^{1}\frac {1}{\sqrt {1 \plus{} x^{2} \plus{} y^{2}}}\,dx\,dy$

misan claims this is $ \minus{} \frac {2\pi}3 \minus{} 4\ln\left(2 \minus{} \sqrt {3}\right)\approx 3.17$

But we are integrating a function which takes value in $ [0,1]$ over a square of area $ 1.$ The answer must lie in the interval $ [0,1].$ Hence, we know for sure that misan's answer is incorrect.
[/quote]

point well taken.  :)

Professor Kent:

Now I understood the questions that I had posted before.

Thanks a lot for your explanation.

P.S.: may be that's why you don't wanted to post anymore, so that I realize of the situation.


geometry
parallelogram
geometry unsolved

you should not post problems into the "number theory " section .

[moderator: agreed -> moved]

[quote="primoz2"]Let ABCD be a konvex qadrilateral. Let K and L be the midpoints of BC and DA. LEt M be a point on the paralel to AB throuh L, such that AB=LM. And let N be a point on the paralel to DC through L , such that LN=DC.M and N are both on the side of AD, where also BC is. Prove, that M,N and L are kolinear.[/quote]
I believe you mean that $ M$, $ K$, and $ N$ are collinear?
[hide="Solution"]
$ AB\equal{}LM, AB\parallel ML$, which implies that $ ABLM$ is a parallelogram.  Thus, $ AM\equal{}BL$ and $ AM\parallel BL\implies AM\parallel BC$.  Also, $ CLND$ is parallelogram.  Thus, $ DN\equal{}LC$ and $ DN\parallel LC\implies DN\parallel BC$.  Now, $ DN\equal{}LC\equal{}BL\equal{}AM\implies DN\equal{}AM$ and $ DN\parallel BC\parallel AM\implies AM\parallel DN$.  So, $ AMDN$ is a parallelogram, so $ MN$ passes through the midpoint of $ AD$, which is $ K$.  Thus, $ M$, $ K$, and $ N$ are collinear.  [/hide]


Este problema AM 161, pag. 256, culegere admitere Poltehnica din Timisoara (infernala carte):

"f este o functie derivabila pe R astfel incat limita cand x tinde la plus infinit din f(x) este egala cu a, numar real.
Se mai stie ca exista (atat!)  si limita din xf'(x) tot cand x tinde la infinit. Sa se calculeze limita din xf'(x) cand x tinde la infinit".

Pentru a diferit de zero am facut-o, dar nu-mi iese pentru a = 0. Multumesc.

Tinand cont de ipoteza si de regula lui L'Hospital, avem
\[ a^2 \plus{} a \plus{} 1 \equal{} \lim_{x\to\infty}\frac {x(f(x) \plus{} a^2 \plus{} 1)}{x} \equal{} \lim_{x\to\infty}(f(x) \plus{} a^2 \plus{} 1 \plus{} xf'(x)) \equal{} a^2 \plus{} a \plus{} 1 \plus{} \lim_{x\to\infty}xf'(x)
\]
deci $ \lim_{x\to\infty}xf'(x) \equal{} 0.$

[quote="jeanvaljean"]Dar nu e bine, va rog, pentru ca L'Hospital se poate aplica numai daca suntem in cazul zero pe zero sau infinit pe infinit (neaparat asta si raportul derivatelor sa aibe limita).

La noi insa: f(x) tinde la a real la numarator si lnx tinde la infinit cand x tinde la infinit. ![/quote]

Am corectat. Acum e O.K.

[quote="jeanvaljean"]Ideea cu a^2 +a+1 care nu se anuleaza, asta era. Va multumesc si va doresc un An Nou Fericit![/quote]
Multumesc! Si tie, un an plin de realizari!
LA MULTI ANI!


trigonometry
geometry
circumcircle
geometric transformation
reflection
cyclic quadrilateral
geometry unsolved

Give quadrilateral $ ABCD$ with $ \angle CAD \equal{} 45^o$, $ \angle ACD \equal{} 30^o$, $ \angle BAC \equal{}\angle BCA \equal{} 15^o$, find the value of $ \angle DBC$.

[quote="Moonmathpi496"]Give quadrilateral $ ABCD$ with $ \angle CAD \equal{} 45^o$, $ \angle ACD \equal{} 30^o$, $ \angle BAC \equal{} \angle BCA \equal{} 15^o$, find the value of $ \angle DBC$.[/quote]
By trigonometry we can show that $ AD\equal{}AB$, hence $ \triangle{ABD}$ is equilateral. Hence $ BD\equal{}BC$. Since $ \angle{BCD}\equal{}45^o$, so $ \angle{DBC}\equal{}90^o$.

I haven't found a synthetic solution yet......

Let the triangle $ ABC$ with  $ BAC \equal{} BCA \equal{} 15$.
 Let the circle of center $ B$ and radius $ BC \equal{} BA$.
Let $ D'$ on this circle so that $ D'BC \equal{} 90$.
 Then $ D'AC \equal{} 45$ . So the triangle $ ABD'$ is equilateral so $ ABD'\equal{}60$. Then angle $ ACD'\equal{}30$.
 By UNIQUENESS of construction, we have that $ D' \equal{} D$
So angle $ DBC \equal{} 90$ :wink:

Let $ P$ be the orthogonal projection of $ C$ onto $ AD$ and $ X\equal{}PB \cap CD$.

We have that $ \angle PCA\equal{}\frac{\pi}{2}\minus{}\angle CAP\equal{}\frac{\pi}{2}\minus{}\frac{\pi}{4}\equal{}\frac{\pi}{4}$, so $ PA\equal{}PC$, and $ BAPC$ is a kite.
$ \angle PCB\equal{} \angle PAB \equal{} \frac{\pi}{3}$ (since $ BAPC$ is a kite), so $ \angle PCD\equal{}\frac{\pi}{3}\minus{}\frac{\pi}{4}\equal{}\frac{\pi}{12}$
$ \angle CDP\equal{}\frac{\pi}{2}\minus{}\frac{\pi}{12}\equal{}\frac{5\pi}{12}$
$ \angle PXD\equal{}\pi\minus{}\angle XDP \minus{}\angle DPX\equal{}\pi\minus{}\frac{5\pi}{12}\minus{}\frac{\pi}{4}\equal{}\frac{\pi}{3}$
$ \angle DXB\equal{}\pi\minus{}\angle PXD\equal{}\pi\minus{}\frac{\pi}{3}\equal{}\frac{2\pi}{3}$
Since $ \angle DXB \plus{} \angle BAD\equal{}\frac{2\pi}{3}\plus{}\frac{\pi}{3}\equal{}\pi$, $ ADXB$ is a cyclic quadrilateral, so:
$ \angle XBD\equal{}\angle XAD\equal{} \angle XAP\equal{} \angle XCP\equal{} \frac{\pi}{12}$

$ \angle  CBP\equal{} \pi\minus{} \angle  BPC \minus{}\angle  BCP\equal{} \pi\minus{}\frac{\pi}{4}\minus{}\frac{\pi}{3}\equal{}\frac{5\pi}{12}$

Finally, $ \angle CBD\equal{} \angle CBP\plus{}\angle XBD\equal{}\frac{5\pi}{12}\plus{}\frac{\pi}{12}\equal{}\frac{\pi}{2}$.

See that B is the circumcenter of (ACD), hence $ \angle DBC \equal{} 2 . m(\angle DAC) \equal{} \pi/2$.

Best regards,
sunken rock

Reflect C over the line AB to form point F. Then since $ \angle FBC\equal{}60$, $ \triangle FBC$ is equilateral. Also, since $ \angle AFC\equal{}75$ and $ \angle ADC\equal{}105$, ADCF is cyclic. However, since $ AB\equal{}BC\equal{}BF$, B is the circumcenter of $ \triangle ACF$, so it is the circumcenter of AFCD. So $ BD\equal{}BA\equal{}BF\equal{}BC$, so $ \angle DBC\equal{}90$.


limit
calculus
integration
logarithms
calculus computations

Evaluate

\[\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{n+k\sqrt{-1}}\]

We can convert this sum into a definate integral:

Let $dx = \frac{i}{n}$ and let $x=\frac{n+ki}{n}$ (I also pulled out a factor of $-i$)

We then have $\lim_{n\to\infty} \frac{n+i}{n} = 1$ and $\lim_{n\to\infty} \frac{n+ni}{n} = 1+i \Rightarrow \frac{1+i-1}{n}=\frac{i}{n}$, so our choice of $x$ and $dx$ are ok.

Thus $\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{n+k\sqrt{-1}} = -i\int_{1}^{1+i} \frac{1}{x} \ dx = \frac{\pi}{4}-\frac{\ln 2}{2}i$


hello
u can easily prove it by induction.Try it urself .
Thank u.

no 2^n+1 will not divide this one

hence ur assumption is wrong

try something else
??????????

hello
I didn't saw the second question.
$ 2^{n\plus{}1}$ doesn't divide
Thank u.


linear algebra
matrix
abstract algebra
superior algebra
superior algebra unsolved

A cute result:

Prove that a finite simple group has no irreducible complex representations of degree $2$.

$G$ is not abelian.

Let $\rho: G \to \mathrm{GL}_{2}(\mathbb{C})$ be an irreducible representation. It is faithful since $G$ is simple. If $\det\rho \equiv 1$ every $\rho(g)$ is similar to a diagonal matrix $\mathrm{diag}(\zeta_{n}, \zeta_{n}^{-1})$, $n = |G|$. $G$ has an element of order $2$ by Feit-Thompson, so there is an $g \in G$ with $\chi(g) = 0$, so ...?

So $\ker\det \neq G$ and also $\ker\det \neq 1$ since then $G$ would be abelian.

Suppose for a contradiction that $G \leq \textrm{GL}_{2}(\mathbb{C})$ is a finite non-abelian simple group.  By Feit-Thompson, there is some $g\in G$ of order 2, which must be either $-I$, or else similar to $\textrm{diag}(1,-1)$.  In the former case $1<Z(G)<G$, and in the latter, $1 < \ker\det < G$.


I believe it would be:
\[ \sum_{k=1}^nk^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} \]

Masoud Zargar

also one thing that works is that the sum of x^k can be put into a closed form with a degree of k+1, so you can just plug in points and solve for the coefficients, but if you wanted to prove it then (since knowing that the degree will be k+1 is not convincing), you can use induction to assure yourself that it works

[quote="sirorange"]Can you find a formula, in terms of x, for 1^4+2^4. . . .x^4? If so, what is it?[/quote]
Please use SEARCH feature of this website, this has been asked many times in AoPS/ML

Here's a general formula for the sum of any powers.
\[ S=1^x+2^x+...+n^x=\sum_{i=1}^xS(x,i)\left(\begin{array}{rr}n+1\\i+1\end{array}\right)i! \]

Masoud Zargar

[quote="tµtµ"]Have a look there : http://mathworld.wolfram.com/PowerSum.html[/quote]
Oh, thanks.  I didn't see that in there, however, thanks.  The mathematics is very interesting. :)

Masoud Zargar

[quote="sirorange"]Don't get me wrong, I know how to do this, I was just posting this as an interesting problem.[/quote]

Oh, ok, sorry, it's not always easy to separate "request for help" from "challenges".


We say the set $ \{1,2,\ldots,3k\}$ has property $ D$ if it can be partitioned into disjoint triples so that in each of them a number equals the sum of the other two.

(a) Prove that $ \{1,2,\ldots,3324\}$ has property $ D$.

(b) Prove that $ \{1,2,\ldots,3309\}$ hasn't property $ D$.

(a) suppose that if $ D_{n}\equal{}\{1,2,...,n\}$ has property $ D$, then so have $ D_{4n}$ and $ D_{4n\plus{}3}$ :

suppose $ D_{n}$ has property $ D$.
$ D_{4n}$ can be partitioned into the following sets (in which the last integer is the sum of the two first :
$ \{1,2n\plus{}n,2n\plus{}n\plus{}1\}$
$ \{3,2n\plus{}n\minus{}1,2n\plus{}n\plus{}2\}$
$ \{5,2n\plus{}n\minus{}2,2n\plus{}n\plus{}3\}$
...
$ \{2n\minus{}1,2n\plus{}1,4n\}$
and eventually the even integers : $ \{2,4,6,...,2n\}\equal{}2D_{n}$
if $ D_{n}$ has property $ D$, $ 2D_{n}$ also has property $ D$ and hence $ D_{4n}$ has property $ D$

$ D_{4n\plus{}3}$ can be partitioned in a similar way :
$ \{1,2n\plus{}n\plus{}2,2n\plus{}n\plus{}3\}$
$ \{3,2n\plus{}n\plus{}1,2n\plus{}n\plus{}4\}$
$ \{5,2n\plus{}n,2n\plus{}n\plus{}5\}$
...
$ \{2n\plus{}1,2n\plus{}2,4n\plus{}3\}$
and the even integers : $ \{2,4,6,...,2n\}\equal{}2D_{n}$

so we have proved that if $ D_{n}$ has property $ D$, so have $ D_{4n}$ and $ D_{4n\plus{}3}$
now it is sufficient to check that :
- $ D_{3}$ has property $ D$
- 3x4=12
- 12x4+3=51
- 51x4+3=207
- 207x4+3=831
- 831x4=3324
From that we conclude $ D_{3324}$ has property $ D$.

(b) It is clear that $ D_{n}$ cannot have property $ D$ if the sum of its elements is odd : suppose $ D_{n}$ has property $ D$ ; by construction, the sum of the elements of each triple in the partition is even so the same must be true of $ D_{n}$. So $ D_{3309}$ cannot have property $ D$ because the sum of its elements is an odd integer.

I am stuck in a similar problem. It is as follows:
Find all values of n for which one can divide the set $X=\{1,2, . . . , 3n\}$ into n triples such that in each of the triplets the sum of some two of them equals the third.

Can anyone help me?


Assume that a,b,c are positive real numbers satisfying the condition:[tex]a^2+b^2+c^2=3[/tex].Prove that:[tex]\frac{a^2}{ab^4+2}+\frac{b^2}{bc^4+2}+\frac{c^2}{ca^4+2}\geq\ 1[/tex]

$\sum \frac{a}{b^2+3}\geq\frac{3}{4}$ with $a+b+c=3$
You can solve it with the same method

Can you post that solution ? My solution is so long .

[tex]\frac{2a^2}{ab^4+2}=a^2-\frac{a^3b^4}{ab^4+2}
\geq\ a^2-\frac{a^3b^4}{3 \sqrt [3]{ab^4}}[/tex]
And the well-known inequality is:if [tex]a+b+c=3[/tex] then [tex](ab)^m+(bc)^m+(ca)^m\leq\ 3[/tex],[tex]m=4/3[/tex].The method which is used in this proof is similar to the method in Vasc's proof for the inequality:If [tex] a+b+c=3[/tex] then:[tex] \frac{a}{ab+1}+\frac{b}{bc+1}+\frac{c}{ca+1}\geq\ \frac{3}{2}[/tex]

[quote="Beat"]Can someone explain me how to finish this problem..?
$\frac{2a^2}{ab^4+2}=a^2-\frac{a^3b^4}{ab^4+2} \geq\ a^2-\frac{a^3b^4}{3 \sqrt [3]{ab^4}}$
and then what..  :huh:[/quote]
$\frac{2a^2}{ab^4+2}=a^2-\frac{a^3b^4}{ab^4+2}=a^2-\frac{a^3b^4}{ab^4+1+1}\geq\ a^2-\frac{a^3b^4}{3 \sqrt [3]{ab^4}}$

[quote="spdf"]$\frac{2a^2}{ab^4+2}=a^2-\frac{a^3b^4}{ab^4+2} \geq\ a^2-\frac{a^3b^4}{3 \sqrt [3]{ab^4}}$
And the well-known inequality is:if $a+b+c=3$ then $(ab)^m+(bc)^m+(ca)^m\leq\ 3$,$m=4/3$.The method which is used in this proof is similar to the method in Vasc's proof for the inequality:If $a+b+c=3$ then:$\frac{a}{ab+1}+\frac{b}{bc+1}+\frac{c}{ca+1}\geq\ \frac{3}{2}$[/quote]

can you show me how to prove the last inequality:
$a+b+c=3$ then $(ab)^m+(bc)^m+(ca)^m\leq\ 3$,[tex]m=4/3
I can't find the method By Vasc.
thanks :)

$\sum_{cyc}\frac{2a^2}{ab^4+2}=\sum_{cyc}a^2-\frac{a^3b^4}{ab^4+2}=\sum_{cyc}a^2-\frac{a^3b^4}{ab^4+1+1}\geq \sum_{cyc}a^2-\frac{a^3b^4}{3 \sqrt [3]{ab^4}}\geq [..]\geq1$
What is in the [..] please post full solutions .

Yes.The following inequal can be also solved with this way:
Let a,b,c be non-negative s.t:a+b+c=3.Prove that:
$\sum\frac{a}{2+b^2}\geq 1$
But what about this inequal for the condition:1/a+1/b+1/c=3


probability
symmetry
pifinity is ze best

What is the probability that a two-digit number selected at random has a tens digit less than its units digit? Express your answer as a common fraction.

Solution 1: Proceed by cases, and find the pattern.

Tens Digit is 1, then ones can be 2-9, 8 numbers.
Tens Digit is 2, then ones can be 3-9, 7 numbers
...
...
Tens Digit is 9, then ones can be 0 numbers.

This yields $ \frac {8 \cdot 9}{2} \equal{} 36$ numbers, and since there are $ 90$ two-digit numbers, we get a probability of $ \frac {36}{90} \equal{} \frac {2}{5}$.

Solution 2: There are 90 total two-digit numbers. We suppose we have 100, by counting 00 through 99. 10 of these have ones and tens digit the same, so subtract these for 90. Now, by symmetry, half of the remaining numbers have ones digit larger than tens. This gives 45. However, we cannot count 01 through 09, or 9 numbers, yielding $ 45 \minus{} 9\equal{} 36$, and thus we get $ \frac{2}{5}$.


geometry
LaTeX
geometric sequence
geometric series

Circle A has radius 1. A circle is tangent to Circle A with one of it's edges. This circle has a diameter equal to half that of Circle A. another circle is tangent to the circle that is tangent to circle A. This circle has diameter equal to half that of the circle tangent to circle A. An Infinite line of circles are tangent to one another, each with a diameter equal to half of the circle before it.

a) What is the area of the 25th circle in this sequence?

b) What is the area of the 50th circle in this sequence?

c) What is the sum of the areas of all the circles? (rounded to the nearest whole number)

here's a link to a diagram for help
http://img363.imageshack.us/my.php?image=circle3gs.png

[quote="B4k4ka"]Circle A has radius 1. A circle is tangent to Circle A with one of it's edges. This circle has a diameter equal to half that of Circle A. another circle is tangent to the circle that is tangent to circle A. This circle has diameter equal to half that of the circle tangent to circle A. An Infinite line of circles are tangent to one another, each with a diameter equal to half of the circle before it.

a) What is the area of the 25th circle in this sequence?

b) What is the area of the 50th circle in this sequence?

c) What if the total area of all the circles? (rounded to the nearest whole number)[/quote]

[hide="a"]$\left(\frac{1}{2}\right)^{24} *\pi=\frac{\pi}{2^{24}}$[/hide]

[hide="b"]$\left(\frac{1}{2}\right)^{49} *\pi=\frac{\pi}{2^{49}}$[/hide]

[hide="c"]The area of the circles are a geometric sequence: $\pi$, $\frac{\pi}{4}$, $\frac{\pi}{16}$, $\frac{\pi}{64}$ Using the formula for infinite geometric series, the area of all the circles is $\frac{\pi}{3}$ [/hide]

a and b are really close, but wrong

c- i said round to the nearest whole number

HINT: If circle A has radius 1, area = 1squared pi=1 pi

if the diameter = half of that, the radius is also half of that... so the radius is 1/2... and 1/2 squared is 1/4...

Hmm... isn't the radius of the "i"th circle (1/2)^i?
well.. $\sum_{i=0}^{\inf} (1/2)^i = 2$
so the area is pi * 4
(for c)

hmm.. i think it may be that i worded the problem wrong... or something...

Circle A's area = 1 pi
second circle is 1/4 of that
and then the third circle is 1/4 of the previous, so it is 1/16

you see how that works?

so the Xth circle's area is..

$1/4^{(X-1)}$

I hope that helps...

here's something that might help

http://img363.imageshack.us/my.php?image=circle3gs.png

[quote="236factorial"][hide]so a is $\frac{\pi}{4^{24}}$ and b is $\frac{\pi}{4^{49}}$? [/hide][/quote]

yup! That's correct!

Now can you try C?

[hide="EDIT-scroll down please"]$x=\frac{\pi}{4}+\frac{\pi}{4^2}+\frac{\pi}{4^3}+....$

$4x=\pi+\frac{\pi}{4}+\frac{\pi}{4^2}+...$

$4x-\pi=x$

$3x=\pi$

$x=\frac{\pi}{3}\approx1$

Thank you for the latex practice  :D [/hide]

[quote="236factorial"][hide="C"]$x=\frac{\pi}{4}+\frac{\pi}{4^2}+\frac{\pi}{4^3}+....$

$4x=\pi+\frac{\pi}{4}+\frac{\pi}{4^2}+...$

$4x-\pi=x$

$3x=\pi$

$x=\frac{\pi}{3}\approx1$

Thank you for the latex practice  :D [/hide][/quote]
[hide]
I dont think thats the correct answer...

heres what i had in mind:

Circle A = 1pi
the next circle = 1/4 pi
and then 1/16 pi and so on...

try to picture a pie with area 2 pi. circle takes up half of that. and then add on the other circles' area. You get 1.25/2 then 1.3125/2 then 1.328125. so it will always be less than 1.5 even if you add infinite circles. So it would be about 1.4 pi which would round to about 4. So you were pretty close.

[/hide]

[quote="236factorial"]What's wrong with my solution?  :([/quote]
you forgot to add the first circle. which has an area of pi

[quote="236factorial"]I CANNOT BELIEVE I WAS SO STUPID!!! sorry for the shout.
So the answer is $\frac{\pi}{3}+\pi=\frac{4\pi}{3}\approx4$[/quote]yup thats right :)

[hide="SOLUTION"]
if we consider the areas of the circles then...
$\pi,\frac{\pi}{4},\frac{\pi}{16}....$
and so by this geometric series the answer to the $1st$ one is...
$\boxed{\frac{\pi}{4^{24}}}$
and the 2nd one is....
$\boxed{\frac{\pi}{4^{49}}}$
and the third...
$\boxed{\frac{4\pi}{3}}$
[/hide]


Let $ \Delta$ be the Cantor set and $ x,y\in\Delta$.  Let $ f: \Delta\rightarrow [0,1]$ be the Cantor function defined by
$ f\left(\sum_{n\equal{}1}^{\infty}\frac{2b_{n}}{3^{n}}\right)\equal{}\sum_{n\equal{}1}^{\infty}\frac{b_{n}}{2^{n}}$
where each $ b_{n}\equal{}0$ or $ 1$.  If $ f(x)\equal{}f(y)$, show that $ x$ has two distinct binary decimal expansions.

I know that this means that $ x$ and $ y$ are two endpoints of a removed middle third, and such points always have two binary decimal representations, but I'm not sure how to prove that.

I believe a proof is supplied here: http://www.mathematics.jhu.edu/415/Scan001.PDF in problem 26.  Unfortunately, it is cut off so I can't quite read it.  Could someone please either just show me how to prove it outright, or patch up the proof in the PDF for me?  Thank you.

It's a typo: it should say, "show that $ f(x)$ has two distinct binary representations."  By the way, "binary decimal" is really awful wording -- "decimal" means "base 10," not "written out as digits."

Thanks.  Should it be that, or that $ x$ has two ternary representations?, since I think that could be true as well.


conics
parabola
quadratics
special factorizations
algebra
quadratic formula

Use the discriminant http://en.wikipedia.org/wiki/Discriminant

In the quadratic formula, the part sqrroot(b^2-4ac) must be negative for no solution. Hence,

9k^2-36k < 0

k < 4

Any value of k less than 4 will yield you no solution.

For one solution, k=4, so the discriminant is 0 and there is only one solution.

In other words, this equation must be a square.

Answer is k=4,0.

[quote="undertaker"]Greetings, I'm new to the forum and here comes my first question that I need some help with. In my textbook I found the following

$y=9x^{2}+3kx+k$
a) Find the value of $k$ so that the parabola only has 1 solution.

I know that the vertex must be at the X axis but how do I find them. I know one solution is $0$, It says tehres two though...
I tried factoring by completing the square and came to the following
$9(x+\frac{1}{3}k)^{2}+k=0$
I have two unknowns. How do I find k?[/quote]

1. You need two unrelated equations with 2 or less unknowns to find each unknown.
2. There are infinitely many solutions to that equation, unless there is a specified y in mind.

I'm going to assume that you meant "$0=9x^{2}+3kx+k$ has exactly one solution":

[hide]This is of the form $ax^{2}+bx+c=0$, where $a=9$, $b=3k$, and $c=k$.

If we want only one solution, the discriminant, $b^{2}-4ac$, should be 0.

Therefore $(3k)^{2}-4(9)(k)=0$.
$9k^{2}-36k=0$
$9k^{2}=36k$
$k^{2}=4k$
So k=0 or 4[/hide]

[hide]I used the quadratic formula, which is any general formula in the form of $ax^{2}+bx+c$. Given from the problem, $a=9$, $b=3k$, and $c=k$.

The discriminant, as someone else mentioned, is used in a quadratic. In this case, the discriminant, $b^{2}-4ac=0$. After that, I hope you understand my following steps:
$\\ (3k)^{2}-4(9)(k)=0\\ 9k^{2}-36k=0\\ 9k^{2}=36k\\ k^{2}=4k\\ \therefore k=\{0, 4\}$

You mentioned that you already knew that 0 was a solution, so the other one is $\boxed{4}$.[/hide]


group theory
abstract algebra
superior algebra
superior algebra solved

I need to find all of the subgroups of $\mathbb{Z}_{6}$.  I am assuming that each must contain a $1$ or a $5$ (the generators).  Is there a quick way to do this (or for that matter, $\mathbb{Z}_{n}$ in general) or is it really just guess-and-check using the properties of a subgroup?

When the generators are in the subgroup, then it's no proper subgroup anymore but the whole group itself.
The subgroups of $\mathbb{Z}/n\mathbb{Z}$ correspond bijectively to the divisors $d$ of $n$.
To see this, note that all multiples of $d$ (still a divisor of $n$) build a subgroup (generated by $d$), and all those subgroups are different.
Conversey, one can take the smallest positive representant of the elements of a subgroup and see that it is indeed a divisor of $n$ (more generaly: the element $t$ generates the same subgroup as $\gcd(n,t)$).

OK.  So if I understand correctly, by Lagrange's Theorem the order of each subgroup will be one of $2,3,6$.  For each of these numbers, there will only be one subgroup.  Therefore all of the subgroups of $\mathbb{Z}/6\mathbb{Z}$ are:

$H_{1}=\left\{0\right\}$

$H_{2}=\left\{0,3\right\}$

$H_{3}=\left\{0,2,4\right\}$

$H_{4}=\left\{0,1,2,3,4,5\right\}$

Is this correct or did I misinterpret?  Thanks again for the help.

Edit:  Changed subgroups.

Those are not subgroups (except the last)- any subgroup which contains 1 contains everything. Look at what ZetaX said- the multiples of $d$ form a subgroup for each divisor $d$.

Also, the identity element is a subgroup- that corresponds to $d=6$ in this case..

I edited the groups in my previous post using the divisors $1$, $2$, $3$, and $6$ as generators.  I hope I got it right this time :)


Let $S= \{ 1,2,3,4,5,6,7 \}$. Divide this set in 2 subsets $A$ and $B$. Show that, at least one between $A$ and $B$ contains at least a pair of number whose difference is in the same set. (for example, if $A=\{ 1,2 \}$ that's good, because $2-1=1 \in A$)

[hide]Proof by contradiction

We will try to create 2 subsets that doesn't have a pair of numbers whose difference is in the same set.

Without loss of generality, let 1 be in set A.
Now, no two consecutive numbers can be in set A.

Because of this, 2 must be in set B in order to not have a pair of numbers whose difference is in the same set.

Now, 4 cannot be in set B, 4-2=2.
Four is in set A.

Now, 3 canot be in set A, 4-3=1
Three is in set B.

Now, 5 cannot be in set B, 5-3=2
But 5 cannot be in set A, 5-4=1

So it is impossible (I think even with only the first 5 numbers, I hope) to create two subsets that do not have a pair of numbers whose difference is in the same set.[/hide][/hide]

try to generalise
[hide] we can also check all the possibilities for 1...5 and see it ain't work, and because of that also 1..7 cannot work (also 1...n,n>5)[/hide]

This is a famous problem. We'll cal a [b]set without differences [/b]a set of natural numbers $A$ such that if $a$ and $b$  are in $A$ and $b>a$ then $a-b$ is not in $A$.
Given the integers $1,2,....n$ divide them in $k$ subsets such that any subset is a  [b]set without differences[/b].
Given $k$ the number of subsets find the bigest $n(k)$ when it is possible two split $1,2,.....n$ in $k$ subsets.
As far as i know this is solved for $k=2$ then $n(2)=4$ as u gyus correctly found :D , try to solve it for $k=3$ (i can post a solution) and by computer for $k=4$.
So if u can make a good computer program maybe u can solve this for $k=5$ (when i was your age computers were two slow, and now I'm too dumb :oops: ). Good luck :wink:

[quote="ehehheehee"]But how would you know that all the possibilities for 1 through 5 don't work unless you prove it first?

I am too lazy to write down about 32 possibilities and check each one if I am not sure if it gets me any where (what about 4,3,2,1,6,7?)[/quote]
it's ain't difficult to check-about 1 minute!


Find the number of odd terms in the $n$th row of Pascal's Triangle.

[hide] I got $2^{m}$, where $m$ is the number of ones in the base $2$ representation of $n$. [/hide]

It was posted very many times, for example
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=102805[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=107668[/url]


A rectangle with vertices at (2, 5), (7, 5), (7, 8) and (2, 8) is revolved around the line y = 5 to obtain a cylinder. What is the volume of the cylinder? Express your answer in terms of π

The radius of the cylinder is 8-5=3, and the height is 7-2=5. Thus, the volume is $ 3^2\times 5\pi\equal{}\boxed{45\pi}$


arithmetic sequence
number theory
relatively prime
number theory proposed

Let $ n\in N^*$. Prove that exist an arithmetic sequence have $ n$ addend , all addend be composite and pairwise relatively prime

I suppose that you ask to prove that, for each integer $ n>0$, there exists an arithmetical sequence of length $ n$ formed by composite positive integers which are pairwise coprime.

Let $ n>3$ be a positive integer (the cases $ n\equal{}1,2,3$ are easy). Let $ 2b$ be the product of all primes not greater than $ n\minus{}1$.
Let $ p_1,p_2, \cdots, p_{2n}$ be pairwise distinct primes, each of them greater than $ n\minus{}1$.
From the Chinese remainders theorem, there exists a positive integer $ a$ such that
$ a\equal{}1\mod[2b]$ and $ a\equal{}\minus{}2kb\mod[p_{2k\plus{}1}p_{2k}]$ for $ k\equal{}0,1,\cdots, n\minus{}1$.
Now, let $ u_k \equal{} a\plus{}2kb$.
The positive integers $ u_0, \cdots, u_{n\minus{}1}$ form an arithmetical sequence, none of them being prime since divisible by at least two different primes.
Assume that there exist $ 0\leq k<m\leq n\minus{}1$ such that $ u_k$ and $ u_m$ are not coprime. Let $ p$ be a common prime divisor. Then $ p$ divides $ u_m\minus{}u_k \equal{} 2b(m\minus{}k)$.
- If $ p$ divides $ 2b$ then, since $ p$ divides $ u_k$ it has to divide $ a$ too. But $ a$ and $ 2b$ are coprime from the first condition. A contradiction.
- If $ p$ does not divide then it has to divide $ m\minus{}k$, with $ 0<m\minus{}k\leq n\minus{}1$. Thus $ p\leq n\minus{}1$ and $ p$ divides $ 2b$, a contradiction.
Therefore the $ u_k$'s are pairwise coprime and we are done.

Pierre.


Euler
trigonometry
function
geometry
geometric transformation
rotation
algebra theorems

I don't know how we have this equality:
$ e^{\frac{i\pi}{2}}\equal{}i$
Please help.May be you can give me the proof or hint.Thank you very much.

It directly follows by the well-known Euler formula $ e^{ix} \equal{} \cos x \plus{} i\sin x$, which can be proved using power series (or the definition of sine and cosine) or integration.  See [url=http://mathworld.wolfram.com/EulerFormula.html]this[/url] for a brief discussion of this formula.

Here is a nice, short proof that uses differential equations:

Observe that $ f(z) \equal{} i sin(iz) \plus{} cos(iz)$ satisfies
$ f(z) \plus{} f'(z) \equal{} 0$ as well as the initial value condition $ f(0) \equal{} 1$
It can be proven that any function defined over C and that satifies both of these conditions must be equal to $ e^{ \minus{} z}$

Hence $ e^{ \minus{} z} \equal{} f(z) \equal{} i sin(iz) \plus{} cos(iz)$ Replace $ x \equal{} iz$ and obtain Euler's equality.

See my comments about rotation matrices [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=216343]here[/url].


geometry
rectangle
function
floor function
algebra

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=274251]Puerto Rico PRMO Team Selection Test 2009[/url]

The entries on an $ n$ × $ n$ board are colored black and white like it is usually done in a chessboard, and the upper left hand corner is black.

We color the entries on the chess board black according to the following rule:
In each step we choose an arbitrary $ 2$×$ 3$ or $ 3$× $ 2$ rectangle that still contains $ 3$ white entries, and we color these three entries black.

For which values of $ n$ can the whole board be colored black in a finite number of steps?

[hide="Big Hint"]Suppose that you can solve it on $ k$ steps, then you have colered black $ 3k$ squares and the total amount of white squares is $ \frac{n \times n}{2}$[/hide]

[hide="Solution"]$ 3|\frac{n \times n}{2} \iff 6|n$. Example for $ n\equal{}6$, by putting $ 3 \times 2$ non overlapping blocks of $ 2 \times 3$[/hide]

$ n$ can also be in the form of $ 6k\plus{}1$ and $ 6k\plus{}5$, because when you apply the floor function to $ \dfrac{(6k\plus{}1)^2}{2}$ and $ \dfrac{(6k\plus{}5)^2}{2}$ you get a multiple of $ 3$.

By the way the total amount of white squares is $ \lfloor \dfrac{n^2}{2}\rfloor$


modular arithmetic
number theory proposed
number theory

Inspired by http://www.mathlinks.ro/Forum/viewtopic.php?p=135064
I was thinking about following problem:
prove that for every integer a, there are infinitely many positive
integers n, so that $a^2 + 2^n$ is not a prime.

Clearly (if $a>0$), there is some $k$ for which $a^2+2^k$ is not a power of $2$. Let $p$ be an odd prime which divides $a^2+2^k$. Then $p|a^2+2^{k+t(p-1)},\ \forall t\in\mathbb N$, and we are done.

Yes, it's pretty clear that if a is odd, then a^2 + 2^n is not a power of 2 (although i can't see why this is important) and i can't see in your solution, why a^2 + 2^k can't be prime. Maybe you could explain a bit more.

I want $2^{p-1}\equiv 1\pmod p$, so $p$ should be an odd prime. I took $p$ to be some divisor of $a^2+2^k$, so it's important for me that $a^2+2^k$ [b]has[/b] odd prime divisors, i.e. it is not a power of $2$.

I don't care whether $a^2+2^k$ is prime or not. Just take an odd prime $p$ which divides it (it can be the whole number, that is not important). Then $p$ also divides $a^2+2^{k+t(p-1)}$, and when I make $t$ large, that number is divisible by some fixed prime $p$, but it's larger than that prime, so for all $t>0$, $a^2+2^{k+t(p-1)}$ is not a prime (disregard $t=0$; in this case $a^2+2^k$ could indeed be a prime).

Is it clear now?


[quote="mathsrox"]If 2s=a+b+c , then show tthat :
[2bc+(b^2+c^2+a^2)] / [2bc-(b^2+c^2+a^2)] = [s(s-a)] / [(s-b)(s-c)]

If 2s=a+b+c , then show that:
1/(s-a) +1/(s-b) +1/(s-c) +1/s= abc/s(s-a)(s-b)(s-c)[/quote]
[hide="hint"]geometry[/hide]

[hide="Are you sure?"]
The first one can be rearranged to $-\frac{a^{2}+(b+c)^{2}}{a^{2}+(b-c)^{2}}$ which, by the Trivial Inequality, will always be negative.  However, the right side is not neccessarily negative, which is a contradiction, so the statement is not neccesarily true.[/hide]

In fact, you can see that both of those are false simply by setting $a=b=c=1$.

Where did you get these problems?


Let $ a,b,c$ be positive integers such that
\[ c(ac \plus{} 1)^2 \equal{} (5c \plus{} 2b)(2c \plus{} b)\]
Prove that $ c$ is an odd perfect square.

$ \gcd(b,c)\equal{}d$
we replace $ c$ by $ dc$ and $ b$ by $ bd$ and we get
$ c^2(adc\plus{}1)^2\equal{}(5c\plus{}2b)(2c\plus{}b) \implies c  | 2b^2 \implies c |2 \implies c\equal{}1,2$
let's see what happen if $ c\equal{}2$
we'll have $ 2(2ad\plus{}1)^2\equal{}(b\plus{}5)(b\plus{}4) \implies x\equal{}b\plus{}4 \implies \frac{x(x\plus{}1)}{2}\equal{}(2ad\plus{}1)^2$
now if we put $ d\equal{}17$ and $ a\equal{}1$ and $ x\equal{}49$ we get that $ 49.50/2\equal{}35^2$
so one solution is $ d\equal{}17,c\equal{}2,a\equal{}1,b\equal{}45$ but we obviously don't have $ 34\equal{}dc\equal{}(2m\minus{}1)^2$
which proves that your problem is wrong unfortunately,
plz tell me if i'm wrong :blush:

From working out, I think the correct problem would be :
$ a,b,c$ are positive integers.
$ c(ac\plus{}1)^2 \equal{} (5c\plus{}2b)(2c\plus{}b)$. Prove that $ c$ is an odd perfect square.

Sorry Agr_94_Math... :blush:  its edited....
@Shoki - Please don't assume $ c\equal{}dc$ it makes the solution all confusing.....


Ross Mathematics Program
calculus
integration
calculus computations

This is a worked example from Ross Prob 7th ed section 7.2 2a. The problem is that I need to find E[X-Y]. X is uniformly distributed on interval length L, as is Y. X is independent of Y, so I want to find the expected distance between X and Y.

The full integral to be evaluated is...

$ E(|X-Y|) = \frac{1}{L^{2}}\int_{0}^{L}\int_{0}^{L}|x-y| dydx$

So first step is...

$ \int_{0}^{L}|x-y| dy = \int_{0}^{x}(x-y) dy+\int_{x}^{L}(y-x) dy$

So my question is - why is this integral split into 2? When we take integrals of absolute values, we must ensure that all values are positive. So we expect the integrand (x - y) when x > y and (y - x) when x < y. The issue is, x and y are uniformly distributed and independent of each other, so then how would x - y depend on the position of x?

$ y$ is uniformly distributed independently of $ x$, so the distribution of $ y$ doesn't depend on the value of $ x$.  But $ x-y$ [i]has an $ x$ in it[/i], so of course it will depend on the value of $ x$.

$ \int_{0}^{L}|x-y| dy = \int_{0}^{x}|x-y| dy+\int_{x}^{L}|x-y| dy = \int_{0}^{x}(x-y) dy+\int_{x}^{L}(y-x) dy$

Oh yeah - holy crap - I was somewhat careless here. (since we have a dy here, y ranges from 0 to x so x - y > 0 for y less than x).


real analysis
topology
irrational number
real analysis unsolved

A set X in a metric space is perfect if it is closed and every point of X is a limit point of X. Does every perfect set in the metric space of real numbers contain a rational number?

I'm stumped, but intuitively I think the answer is yes...

With some measure theory:

Consider translates of the Cantor set $\mathcal{C}$, which you know is perfect by the discussion in that chapter of Rudin.  In particular, the Cantor set has Lebesgue measure zero, and so if the union of translates $x_{\alpha}+\mathcal{C}$ is to contain $[0,1]$, then $\{x_{\alpha}\}$ must be an uncountable set (by countable additivity).  If each $x_{\alpha}+\mathcal{C}$ contains a rational, then there would be uncountably many rationals in $[-1,2]$, an absurdity.  So there exists an $x$ such that $x+\mathcal{C}$ has the desired properties.

Without measure theory:

Set $X$ to be the set of numbers in $[0,1]$ which contain only $4$ and $7$ in their decimal expansions.  $X$ is perfect (prove this, this is Q17 of the same chapter).  Put $x=0.1010010001000\ldots$.  Then $x+X$ is perfect and contains no rational.

A different approach (but also based on a measure argument):

First we prove that a closed subset of a metric space with a countable basis can be written as the union of a perfect set and a countable set. This is simple: we just eliminate the isolated points. Every isolated point is the only isolated point in a basis element (from a countable basis), so there are at most countably many. What is left is clearly a perfect set. In particular, we have shown that every uncountable closed subset of a metric space with a countable basis has an uncountable perfect subset.

Now, for our purposes, it suffices to exhibit an uncountable real closed set composed only of irrationals. Let $(q_n)_n$ be an enumeration of the rationals, and take out an open interval of length $\frac 1{2^n}$ around $q_n$ for each $n$. We're left with a closed set of postive measure, which must thus be uncountable, and which contains no rational points.

Thanks guys. Nice solutions.

Grobber:
[quote]Every isolated point is the only isolated point in a basis element (from a countable basis)[/quote]

This isn't obvious to me...What is the definition of a countable basis?

Also, does every perfect set of real numbers contain an irrational number? I can't seem to find a counterexample to this.

A countable basis is just that: a basis of open sets of the topological space which is countable.

As for your other question, the answer is "yes": every non-empty perfect real set must contain an irrational number. In fact, try to prove that every non-empty perfect subset of $\mathbb R$ has cardinality $c$ (i.e. the same cardinality as the real line). It's a nice exercise.

Rudin contains the proof that every nonempty perfect set is uncountable.  If there were a perfect set $X$ in $\mathbb{R}$ with no irrational, then we would have $X\subseteq \mathbb{Q}$, an absurdity.

By the way, some books use the phrase countable base in place of countable basis, if that rings a bell.


The sequence $ x_n$ is said to be of bounded variation if there is $ C>0$ such that for every $ n\in\mathbb{N}$:
\[ |x_2\minus{}x_1|\plus{}|x_3\minus{}x_2|\plus{}\cdots \plus{}|x_n\minus{}x_{n\minus{}1}|\leq C.\] Prove that $ x_n$ is a Cauchy sequence.

Denote $ |x_2\minus{}x_1|\plus{}|x_3\minus{}x_2|\plus{}\cdots \plus{}|x_n\minus{}x_{n\minus{}1}|\equal{}s_{n\minus{}1}$, $ n\in \mathbb{N}\setminus\{1\}$.

We see that $ \{s_{n}\}^{\infty}_{n\equal{}1}$ is a bounded monotonic increasing sequence. Therefore it is convergent and Cauchy.

So for each $ \epsilon$ we can find $ n_{0}$, such that $ |s_{m}\minus{}s_{n}|<\epsilon$ for $ m, n \geq n_{0}$.

But $ |s_m \minus{} s_n|\equal{}|x_{m\plus{}1}\minus{}x_m|\plus{}\ldots\plus{}|x_{n\plus{}2}\minus{}x_{n\plus{}1}|>|x_{m\plus{}1}\minus{}x_{n\plus{}1}|$.

Hence for $ m,n > n_{0}$, $ |x_{m}\minus{}x_{n}|<\epsilon$. So also $ \{x_{n}\}^{\infty}_{n\equal{}1}$ is Cauchy.


Hello friends,

                       This turing machine problem is really getting on my nerves. Please help me out:-

(1) Design a turing machine (deterministic) to calculate[b] n![/b]  (n-factorial).

(2) Design a turing machine (deterministic again) to calculate ceiling func of [b]lg(n) [/b]
      where lg(n) means log(n) calculated in base-2...

Thank you for your co-operation.

-Goodbye

To part a): Do you know how to multiply two nubers? It is not too hard, you can figure it out. Well - first I'd try to create a sequence of numbers $n$, $n-1$, $n-2$, ..., $1$. This is quite simple, you just copy the previous number and then substract one. After that, you multiply the last two numbers ($2\times1$) and erase them, leaving only their product. Then again you multiply the last two numbers ($3\times2$) and leave only their product, and so on... you end when you reach the beginning of the tape with no numbers left to multiply. If you're not sure how to implement any part of this just ask, meanwhile I'll be thinking about lg(n) :) ...


modular arithmetic
combinatorics proposed
combinatorics

The positive integer $n$ is given and for all positive integers $k$, $1\leq k\leq n$, denote by $a_{kn}$ the number of all ordered sequences $(i_1,i_2,\ldots,i_k)$ of positive integers which verify the following two conditions:

a) $1\leq i_1<i_2< \cdots i_k \leq n$;

b) $i_{r+1}-i_r \equiv 1 \pmod 2$, for all $r \in\{1,2,\ldots,k-1\}$.

Compute the number $a(n) = \sum\limits_{k=1}^n a_{kn}$.

[i]Ioan Tomescu[/i]

In other words, we are asked to find the number of increasing sequences of elements in $[n]$ which are alternatively even and odd. Now, I just guessed the answer, and this makes things a lot easier :).

For each $n$, let $o_n$ be the number of such sequences which begin with an odd number, and $e_n$ the number of sequences which begin with an even number (so the value we're trying to find is $a(n)=o_n+e_n$). We have $o_n=(o_{n-1}+1)+(o_{n-3}+1)+\ldots$ (we define $o_0=0$, and the sum goes on for as long as the indices are non-negative), because if your sequence begins with $1$, then you can either leave it like this, or extend it in $o_{n-1}$ ways, if it begins with $3$, you can leave it like this or extend it in $o_{n-3}$ ways, and so on. It is also obvious that $e_n=o_{n-1}$.

We can observe now that $x_n=F_{n+2}-1$ has the same initial values as $o_n$ and satisfies the same recurrence, where $F_0=0,F_1=1,F_n=F_{n-1}+F_{n-2},\ \forall n\ge 2$ is the Fibonacci sequence. This means that we have $o_n=F_{n+2}-1$, and we thus have $a(n)=o_n+e_n=F_{n+3}-2$.


Hi, need some help. How often does Easter fall on (Sunday) March 26? Maybe too easy for you, but too tough for me - help.

Hanoi, Vietnam

Wait a second. You join the day after easter, to ask why easter was the day before? Anyways March 26 is a Saturday. The 27, the day you joined was easter. Maybe you were looking at an old calendar. No matter that. I don't see how this is number theory.


Given $ a, b, c \geq\ 0$ and $ a\plus{}b\plus{}c\equal{}2$. Prove that:  $ \sqrt[3]{\frac{a\plus{}b}{a^2\plus{}kab\plus{}b^2}}\plus{}\sqrt[3]{\frac{b\plus{}c}{b^2\plus{}kbc\plus{}c^2}}\plus{}\sqrt[3]{\frac{c\plus{}a}{c^2\plus{}kca\plus{}a^2}} \geq\ 2\plus{}\frac{1}{\sqrt[9]{2}}$
With $ k \equal{} 2\sqrt[3]{2}\minus{}2$. Hope that you'll like it, my dear friend  :)

thank you very much for you nice gift my friend :lol:  ,  I'm really suprised ,that's very kind  of u  , I will try it after sunset because I'm fasting , and I can't  think without eating  :D

Sorry for being late I had some issues with internet  :blush:

I prove your ineq by the classical ugly way  :( , indeet , let $ x\equal{}a\plus{}\frac{c}{2} ,y\equal{} b\plus{} \frac{c}{2}$ then it is fairly easy to prove that :

$ \frac{a\plus{}b}{a^2\plus{}kab\plus{}b^2}  \geq  \frac{x\plus{}y}{x^2\plus{}y^2\plus{}kxy}$

$ \frac{a\plus{}c}{a^2\plus{}kac\plus{}c^2}  \geq  \frac{1}{x}$

$ \frac{b\plus{}c}{b^2\plus{}kbc\plus{}c^2}  \geq \frac{1}{y}$,    the ineq is reduced to 2 variable inequality with the constraint $ x\plus{}y\equal{}2$ it can be easy solved by AM-GM  :wink: , now let's discuss the general problem , wa can easly prove that :(with the same conditions)

$ \sum{ \sqrt[3]{\frac{a\plus{}b}{a^2\plus{}b^2\plus{}kab}}} \geq K$ when K is achieved in the case $ a\equal{}b\equal{}c$  for $ k\geq 2$

$ \sum{ \sqrt[3]{\frac{a\plus{}b}{a^2\plus{}b^2\plus{}kab}}}  \geq K'$ when K' is achieved in the case $ a\equal{}b ,c\equal{}0$  for $ k\leq 1$

for $ 1\leq k \leq 2$ the ineq :$ \sum{ \sqrt[3]{\frac{a\plus{}b}{a^2\plus{}b^2\plus{}kab}}} \geq min(K,K')$ is not true ( I can't remember the counter exemple  :D ) , how ever I think that I managed to prove the following:
with the same condition , if $ k\leq \frac{4}{3}$ and $ max(a,b,c) \geq  min(a,b,c)\frac{k}{4\minus{}3k}$  then $ \sum{ \sqrt[3]{\frac{a\plus{}b}{a^2\plus{}b^2\plus{}kab}}}  \geq K'  (*)$  ,I think that   $ (*)$ is true for $ k\equal{}2\sqrt[3]{2}\minus{}1$ but I couldn't prove it  :blush:  maybe I will try later :wink:  , dear nguoivn , can your am-gm tricks help us with this situation? :) ,
I believe that is interesting to determine the biggest k in $ (1,2)$ so that  $ (*)$ is true , may be Ji chen or Vasc will help me ??!! :maybe:


War von Euch auch jemand in einem der Standtorte dabei??

Also unsere Schule hatte 2 Teams in Heppenheim. Eigentlich waren wir mit dreien angemeldet haben aber 2 draus gemacht da 3 leute von team 3 net gekommen sind

Naja Ergebnis war dass im Gruppenwettbewerb usere Schule einen Doppelsieg eingefahren hat 1. Platz Gruppe von Friedrich Feuerstein (und mir) 3 Leute 2. Platz Gruppe von Yasin Zähringer 5 Leute
und im Einzelwettbewerb so ähnlich 1.Preis Friedrich 2.Preis Yasin und der 5.te ging auch noch an einen Helmhöltzer dessen Namen ich aba nicht kenn.

Weiß von Euch jemand wo die Aufgaben, Lösungen, Statistiken usw. im Netz veröffentlicht werden? Weil mich würd mal die Lösung von dieser Aufgabe mit dem Dreieck (E3) intressiern die hat nähmlich bei unserer Schule Niemand komplett gelöst.

Idril

ich war in karlsruhe dabei... gruppe des mgg horb, bestehend aus roland riester, anja randecker, lisa wolf, michael matiu und mir, und wir hatten den 2. platz, obwohl wir so ziemlich bei jeder zweiten aufgabe es geschafft haben nen fehler reinzubauen... im einzelwettbewerb hatte ich auch nen 2. platz, mit einem abgezogenen punkt wegen nem rechenfehler (bei der E3; ich glaub, ich hatte $44^2+4\cdot 3\cdot 64$ falsch ausgerechnet)...

gratuliere euch, insbesondere friedrich, ihr habt ja ordentlich gearbeitet... bei uns wars ein totales chaos, wir waren beim speedwettbewerb voll gestresst und haben lauter dumme fehler gemacht...

was die E3 angeht, ich hab sie in der klausur völlig hässlich durchgeholzt und mich dabei entsprechend verrechnet, aber es gibt eine halbwegs gescheite lösung: http://www.mathlinks.ro/Forum/viewtopic.php?p=184503 . ob das ganze irgendwann mal offiziell im netz veröffentlicht wird weiß ich nicht...

  dg

dari, dari dari, man man man.... es wird ja immer schlimmmer.. jetzt machst du auch noch bei DG fehler :P
also ich hab nen 2. beim Einzelwettbewerb in Konstanz, wegen dem gleichen Fehler wie Dari (also nicht unbedingt genau gleicher, aber trotzdemrechenfehler bei der E3) , und das Team bestehend  aus mir (G1, G2 ohne aufschreiben, G3) und 3 weiteren aus meiner schule (aufschreiben G2, G4, aber muss ich sagen, schön aufgeschrieben :lol: , und die 4. voll korrekt.. weil ich hab mich bei der 1 allem Anschein nach verrechnet) :D , wir haben den 1. beim Gruppenwettbewerb, und damit endlich die 3jährige vorherrschaft der cantonschule Romanshorn abgeschafft..*yuhoooo* (hab mir des vor 2 jahren als ziel gesetzt , und jetzt hats endlich geklappt^^*freu*)

Ehm....
 ich habe gerade mal mir die Statistiken angesehen

leider sind in der Statistik wohl nur alle standorte in Hessen gewertet worden d.h. weder Karlsruhe noch Konstanz sind dabei.

In der Gesammtwertung liegen unserer beiden Schulteams auf Platz 3 eigentlich müsste Friedrichs Gruppe auf platz 2 sein, leider wurde der Hürdenwettbewerb nicht gewertet da in Darmstadt zu dem Zeitpunkt Feueralarm war. Und vorallem hat die Schule auf platz 1 angeblich 36 Punkte im Gruppenwettbewerb wo doch eigentlich bekannt ist 4*8=32 nicht 36. nunja würde die schule deshalb noch ausscheiden wären wir platz 1 *g*

aber was mich ein wenig überascht hat

Feuerstein Friedrich Helmholtz-Gymnasium Heidelberg 26  Platz16  (Nur Heppenheim Platz 1)

und dann gibt es da noch so sachen wie

Zähringer Yasin Helmholtz-Gymnasium Heidelberg 25 Platz23 (Nur Heppenheim Platz 2)
Rusanov Radoslav Helmholtz-Gymnasium Heidelberg 22 Platz54 (Nur Heppenheim Platz 5)
Zuber Marcus Helmholtz-Gymnasium Heidelberg 13 Platz212 (Nur Heppenheim Platz 15)
Walter Sebastian Helmholtz-Gymnasium Heidelberg 9 Platz379  (Nur Heppenheim Platz 34)
[b]Bachmann Stefanie Helmholtz-Gymnasium Heidelberg 7 Platz550 (Nur Heppenheim Platz 60)[/b]  *g*
Ylli Klevis Helmholtz-Gymnasium Heidelberg 6 Platz624 (Nur Heppenheim Platz 69)
Stengele Oliver Helmholtz-Gymnasium Heidelberg 1 Platz993 (Nur Heppenheim Platz 111)

ich hätte ja prinzipiell eher ein Ergebniss mit 0 1 oder 2 Punkten erwartet.  d.h. mit meinen 7 Punkten lag ich ja schon etwas über der Hälfte Teilnehmer insgesammt waren 1222


vector
topology
Functional Analysis
real analysis
real analysis solved

Prove that every metric space can be imbedded isometrically into a Banach space as a linearly independent set.

We just build an appropriate space. Consider the vector space $V$ of finite formal linear combinations of the points in our metric space $X$ with real coefficients. We want to give $V$ a norm that induces the metric space on its basis.
Choose $p\in X$, and let $\|p\|=1$. For any $u=\sum a_ix_i$ with $x_i\in X$ and $\sum a_i=0$, let $\|u\|=\inf \{\sum c_{ij}d(x_i,x_j): c_{ij}\ge0\text{ and }\sum_jc_{ij}-\sum_jc_{ji}=a_i\}$
For everything else, $\|ap+u\|=|a|+\|u\|$ if $u$ is as above.

This is a norm on $V$, and the $x_i$ are linearly independent. To make a Banach space, take the completion of $V$.


geometry
inequalities
function
search
graph theory
number theory
national olympiad

[url=http://www.mathlinks.ro/Forum/download.php?id=1186]PDF Version[/url]
The problems are located at

Problem 1
http://www.mathlinks.ro/Forum/viewtopic.php?t=21496

Problem 2
http://www.mathlinks.ro/Forum/viewtopic.php?t=21497

Problem 3
http://www.mathlinks.ro/Forum/viewtopic.php?t=21498

Problem 4
http://www.mathlinks.ro/Forum/viewtopic.php?t=21499

Problem 5
http://www.mathlinks.ro/Forum/viewtopic.php?t=21500

Problem 6
http://www.mathlinks.ro/Forum/viewtopic.php?t=21502

Problem 7
http://www.mathlinks.ro/Forum/viewtopic.php?t=21504

Problem 8
http://www.mathlinks.ro/Forum/viewtopic.php?t=21505

(maybe someone could compile a .pdf file with these?)

Wow. Amazing. All problems by Pierre. Congrats.  :love:

Pierre, Darij did the first step and talked about the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21011]German way[/url] up to the IMO. Now it is your turn...  :)

No, I'm not the proposer of the problems. I've just taken them from my archives  :P

It is really hard for me to explain how the IMO selection will be organised in France because I do not know the answer, and I'm not sure that even Claude Deschamps, our Team leader who organizes it, does know.

For the first time, we have organised a pre-TST to select the people who will be allowed to participate to the training. But this will only start in January 2005.

The 8 problems were to be solved in 4 hours. They were quite easy, absolutely not at an IMO level. But, it's the beginning. Let's hopewe readh IMO before July...

Toumaf, which is sometimes on ML, has got the only perfect score (56).
Igor, which is sometimes on ML too and with a bronze medal at IMO 2004, did a very bad score (21, a real bad surprise) and has been very near to be ejected  :D . Hopefully, we took 21 people and he ranked 19.

We hope that our contestants will read all the courses (Plane Geometry, Inequalities, Graph theory, Functional equations...) which have been written for IMO training. The first part of the Number Theory course is over and will be available soon on http://www.animath.fr

Pierre.

[quote="pbornsztein"]Igor, which is sometimes on ML too and with a bronze medal at IMO 2004, did a very bad score (21, a real bad surprise) and has been very near to be ejected  :D . Hopefully, we took 21 people and he ranked 19.[/quote]
Seems like almost the story of grobber, but this one ended happily (for the student).

Valentin Vornicu wrote:
[quote](maybe someone could compile a .pdf file with these?)[/quote]

I have done this work just now !

Thanks, Pierre, for the nice problems - although, at least speaking about the geometry problems, I wouldn't call them so difficult  ;)  Anyway better than the German TST problems (most of those whom I know, including myself, solved only 3 of the 6 problems completely, but many others made lots of partial scores - hell knows who will succeed to the IMO training). Anyway, how much time did the participants have the for the problems?

  Darij

4 hours. I think that was the main difficulty of this pre-TST.
In another hand, only one perfect score, only two other contestants marked at least 40 points (each problem is 7 points worth), and only 10 others maked at least 30 points... :(
Thus, this allows us to select people more easily...

To be honest, our goal is that french contestants can solve IMO problems 1,2,4,5. It would be a nice achievement.

Pierre.

the singapore team could be in a similar plight... sadly...  :(

and of course 1,2,4,5 would be nice.  partial scores then for qn 3 and 6 would put one in good stead for a gold medal!

kenneth

[quote="pbornsztein"]
Igor, which is sometimes on ML too and with a bronze medal at IMO 2004, did a very bad score (21, a real bad surprise) and has been very near to be ejected  :D . Hopefully, we took 21 people and he ranked 19.
[/quote]

Yes, hopefully ... And I got 23, not 21  :blush:

And Pierre, you didn't post problem 9 :)

[quote="kueh"]hmm i did 7 of the problems in 35 minutes (misread question 6...)[/quote]

 :D  You are too bad with Igor.

Pierre.

[quote="Valentin Vornicu"][quote="kueh"]hmm i did 7 of the problems in 35 minutes (misread question 6...)[/quote]were you at the IMO last year? :D[/quote]

yes, but for singapore :blush: one of the defectors... displaced by circumstance !  ;) this year will be my last year - hoping for a gold!

[quote="kueh"]actually they are quite instructive problems in a sense? hmm i did 7 of the problems in 35 minutes (misread question 6...)[/quote]

Actually, I misread question 6 the first time I did it too... :blush: I read it as maximum instead of minimum (which actually makes the question just an insignificant little bit harder! ;) )

[quote]yes, but for singapore :blush: one of the defectors... displaced by circumstance !  ;) this year will be my last year - hoping for a gold![/quote]

*sniff* :( Valentin, if you remember the two people who were carrying our teddy bear around everywhere, Andre was the other person who did most of the carrying :D .

[quote="Valiowk"]Somebody who represents one country in a particular year and another country the next year.[/quote]Hmm, we are seeing many of these persons lately :D :D I wonder why ...  :P

Actually, I doubt that the trend is any more prevalent now than it used to be in the past.  I think the reason why we are reading about this more frequently is because there is a tendency (especially on the part of the former teammates :oops: ) to blow the matter up.

[quote="kueh"]yeah i love the teddy bear (= btw, is there still the function that will allow one to see if anyone replied to any of your posts?[/quote]This is getting really off-topic, so I'll soon split this :) Anyway what you are lookingfor is this: http://www.mathlinks.ro/Forum/search.php?search_id=egosearch

[quote="Valiowk"]Actually, I misread question 6 the first time I did it too... :blush: I read it as maximum instead of minimum (which actually makes the question just an insignificant little bit harder! ;) )[/quote]

Actually, its still doable i think, if you colour it in four colours and just start arguing each case.