function
calculus
derivative
real analysis
real analysis unsolved

Is the Dirichlet function f:(0,1) -> R

\[f(x) = \left\{ \begin{array}{l}0{\rm }\\ \frac{1}{q}{\rm }\\ \end{array}\right. \]

where at 0, x does not belong to Q
where at 1/q, x belongs to Q and x is expressed in lowest terms as p/q

differentiable at some point
\[x_{o}\in (0,1) \]

If so, what does the proof look like?

if it were differentiable at $x_{0}$, is it possible that $x_{0}\in \mathbb{Q}$ ? And what would be the derivative ? From here you should be able to conclude..


Which digit should replace $ a$ in the units place so that $ 98\;675\;42a$ is divisible by 11?

For a number to be divisible by $ 11$, the sum of every other digit, starting from the one's digit, subtracted from the sum of every other digit, starting from the ten's digit, must be divisible by $ 11$.  The first part of our requirement is $ 9 \plus{} 6 \plus{} 5 \plus{} 2 \equal{} 22$.  The second part is $ 8 \plus{} 7 \plus{} 4 \plus{} a \equal{} 19 \plus{} a$.  Because it would be impossible for the difference to be greater than $ 6$ (if $ a\equal{}9$, then $ 28\minus{}22\equal{}6$), the difference must be $ 0$.  So we have $ 22 \minus{} 19 \minus{} a \equal{} 0\implies a \equal{} \boxed{3}$.  Sorry if my explanation is a little hard to follow.


let $ {a.b.c} \in \aleph$ solve the system

$ a^3\equal{}b^3\plus{}c^3\plus{}3abc$
$ a^2\equal{}2(b\plus{}c)$
 please i need more that one solution

hello, i have
$ a\equal{}2,b\equal{}2\minus{}c,c\equal{}c$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, i have
$ a \equal{} 2,b \equal{} 2 \minus{} c,c \equal{} c$.
Sonnhard.[/quote]
please can you post the hall solution ,

First is $ a^3\equal{}(b\plus{}c)^3\plus{}3(a\minus{}b\minus{}c)dc$ or $ (a\minus{}b\minus{}c)[a^2\plus{}a(b\plus{}c)\plus{}(b\plus{}c)^2\minus{}3bc]\equal{}0$.
If $ a\equal{}b\plus{}c$, then
1. $ a\equal{}2,b\equal{}2\minus{}c,c\equal{}c$ and
2. $ a\equal{}0, b\equal{}\minus{}c,c\equal{}c$.
Else $ a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}ac\plus{}bc\equal{}0$. But it equvalent to $ (a\plus{}b)^2\plus{}(a\plus{}c)^2\plus{}(b\plus{}c)^2\equal{}0$ or $ a\equal{}b\equal{}c\equal{}0$. These solution belong 2.

[quote="Pablo09"]Hi, this is my solution:
 rewrite :
$ a^3 \minus{} b^3 \minus{} c^3 \minus{} 3abc \equal{} 0$, then it is equivalent to
$ a^3 \minus{} b^3 \minus{} c^3 \minus{} 3abc \equal{} (a \minus{} (b \plus{} c) ((a \plus{} b)^2 \plus{} (a \plus{} c)^2 \plus{} (b \minus{} c)^2) \equal{} 0$ so it must be $ a \equal{} b \plus{} c$ , ->

$ a^2 \equal{} 2(b \plus{} c) \equal{} 2a$
$ a \equal{} 2$ -> $ b \plus{} c \equal{} 2$ - > $ b \equal{} c \equal{} 1$

The solution is (a,b,c)=(2,1,1)[/quote]

hi $ b \plus{} c \equal{} 2$ doesn't mean that $ b \equal{} c \equal{} 1$ :(  and i don't thinks that

$ a^3 \minus{} b^3 \minus{} c^3 \minus{} 3abc \equal{} (a \minus{} (b \plus{} c) ((a \plus{} b)^2 \plus{} (a \plus{} c)^2 \plus{} (b \minus{} c)^2) \equal{} 0$  is true (see rust what he's got !!)

[quote="Rust"]First is $ a^3 \equal{} (b \plus{} c)^3 \plus{} 3(a \minus{} b \minus{} c)dc$ or $ (a \minus{} b \minus{} c)[a^2 \plus{} a(b \plus{} c) \plus{} (b \plus{} c)^2 \minus{} 3bc] \equal{} 0$.
If $ a \equal{} b \plus{} c$, then
1. $ a \equal{} 2,b \equal{} 2 \minus{} c,c \equal{} c$ and
2. $ a \equal{} 0, b \equal{} \minus{} c,c \equal{} c$.
Else $ a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} ac \plus{} bc \equal{} 0$. But it equvalent to $ (a \plus{} b)^2 \plus{} (a \plus{} c)^2 \plus{} (b \plus{} c)^2 \equal{} 0$ or $ a \equal{} b \equal{} c \equal{} 0$. These solution belong 2.[/quote]
 hi rust

if $ a$different to $ b \plus{} c$ so  $ [a^2 \plus{} a(b \plus{} c) \plus{} (b \plus{} c)^2 \minus{} 3bc] \equal{} 0$.

that mean $ a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} ac$ [size=200]$ \minus{}$[/size]$ bc \equal{} 0$  !!

we can say that $ b^2 \plus{} c^2 > \equal{} bc$ so  $ a \equal{} b \equal{} c \equal{} 0$

i'm still waiting for other solutions  :P

hello, by substituting $ c\equal{}\frac{a^2}{2}\minus{}b$ in $ a^3\equal{}b^3\plus{}c^3\plus{}3abc$ and factoring we get:
$ \minus{}\frac{1}{8}a(a\minus{}2)(a^4\plus{}2a^3\minus{}6a^2b\plus{}4a^2\plus{}12b^2)\equal{}0$ and hence we have:
$ a\equal{}0$ or $ a\equal{}2$ or $ b\equal{}\frac{1}{2}\left(\frac{1}{2}a\plus{}\frac{2}{3}\sqrt{3}i\plus{}\frac{1}{6}\sqrt{3}ia\right)a$ or $ b\equal{}\frac{1}{2}\left(\frac{1}{2}a\minus{}\frac{2}{3}\sqrt{3}i\minus{}\frac{1}{6}\sqrt{3}ia\right)a$
Sonnhard.

Ibn Rochde wrote:
[quote]$ a^3 \minus{} b^3 \minus{} c^3 \minus{} 3abc \equal{} (a \minus{} (b \plus{} c) ((a \plus{} b)^2 \plus{} (a \plus{} c)^2 \plus{} (b \minus{} c)^2) \equal{} 0$ is true (see rust what he's got !!) [/quote]

oppss i lost $ 1/2$ ! it should be
$ 0 \equal{} a^3 \minus{} b^3 \minus{} c^3 \minus{} 3abc \equal{} 1/2 (a \minus{} (b \plus{} c)) ((a \plus{} b)^2 \plus{} (a \plus{} c)^2 \plus{} (b \minus{} c)^2)$ sorry!

Ibn Rochde wrote :
[quote]hi $ b \plus{} c \equal{} 2$ doesn't mean that $ b \equal{} c \equal{} 1$[/quote]

Mate it means. beacause $ a,b,c \in N$, the equation $ b \plus{} c \equal{} 2$ has only one soultion b=c=1 in natural :)

edit: of course i assume 0 is not a natural number. Greetings

[quote="Ibn Rochde"]let $ {a.b.c} \in \aleph$ solve the system

$ a^3 \equal{} b^3 \plus{} c^3 \plus{} 3abc$
$ a^2 \equal{} 2(b \plus{} c)$
 please i need more that one solution[/quote]

1)   $ a>b$ and $ a>c$ ==> $ 2a>b\plus{}c$ so $ 4a>a^2$ that mean $ 4>a$

$ a\in {0.2}$ $ S\equal{}{(0.0.0)(2.11)(2.2.0)(2.0.2)}$

2) $ a^3\equal{}b^3\plus{}c^3\plus{}3abc$ ==> $ 2a^3\equal{}a^2(b^2\plus{}c^2\minus{}bc)\plus{}3abc$

$ 2a^2\minus{}a(b\plus{}c)^2\equal{}3bc(2\minus{}a^)$ ==> $ a^2(8\minus{}a^3)\minus{}12bc(2\minus{}a)\equal{}0$

$ a\equal{}2$ or $ a^2(a^2\plus{}2a\plus{}4)\equal{}12bc$  and $ a^2>3bc$ tha mean $ a^2\plus{}2a\plus{}4>\equal{}4$ ==>$ a\equal{}0$

and we conclude


limit
trigonometry
logarithms
calculus
calculus computations

This is our problems.

$ \lim_{x\to 0}\frac {\sqrt {1 \plus{} x^3} \minus{} 1}{1 \minus{} \cos \sqrt {x \minus{} \sin x}}$

$ \lim_{x\to 0}\frac {\sqrt {1 \plus{} \tan x} \minus{} \sqrt {1 \plus{} \sin x}}{x\ln(1 \plus{} x) \minus{} x^2}$

I would multiply numerator and denominator by $ \sqrt{\cdots}\plus{}\sqrt{\cdots}$, and then use Taylor series if necessary.

[quote="kg2100"]$ \lim_{x\to 0}\frac {\sqrt {1 \plus{} x^3} \minus{} 1}{1 \minus{} \cos \sqrt {x \minus{} \sin x}}$[/quote]

$ \lim_{x\to 0}\frac {\sqrt {1 \plus{} x^3} \minus{} 1}{1 \minus{} \cos \sqrt {x \minus{} \sin x}} \equal{} \lim_{x\to 0}\frac {1}{\frac {1 \minus{} \cos \sqrt {x \minus{} \sin x}}{(\sqrt {x \minus{} \sin x})^2}}\cdot \frac {1}{\frac {x \minus{} \sin x}{x^3}}\cdot \frac {1}{\sqrt {1 \plus{} x^3} \plus{} 1}$

$ \equal{} \frac {1}{\frac {1}{2}}\cdot \frac {1}{\frac {1}{6}}\cdot \frac {1}{1 \plus{} 1} \equal{} \boxed{6}$

[quote="kg2100"]$ \lim_{x\to 0}\frac {\sqrt {1 \plus{} \tan x} \minus{} \sqrt {1 \plus{} \sin x}}{x\ln(1 \plus{} x) \minus{} x^2}$[/quote]

$ \lim_{x\to 0}\frac {\sqrt {1 \plus{} \tan x} \minus{} \sqrt {1 \plus{} \sin x}}{x\ln(1 \plus{} x) \minus{} x^2}$

$ \equal{}\lim_{x\to 0}\frac{\tan x}{x}\cdot \frac{\frac{1\minus{}\cos x}{x^2}}{\frac{\ln (1\plus{}x)\minus{}x}{x^2}}\cdot \frac{1}{\sqrt{1\plus{}\tan x}\plus{}\sqrt{1\plus{}\in x}}$
$ \equal{}1\cdot \frac{\frac{1}{2}}{\minus{}\frac{1}{2}}\cdot \frac{1}{1\plus{}1}$

$ \equal{}\boxed{\minus{}\frac{1}{2}}$


geometry
circumcircle
cyclic quadrilateral
similar triangles

Hey, i came across this problem and i cant get it. I gave it to the whole maths staff and they struggled. Here it is

$ ABCD$ is a quadrilateral inscribed in a circle. $ BA$ and $ CD$ when produced meet at $ P$.
$ O$ is the centre of the circle $ PAC$. Prove that $ BD$ is perpendicular to $ OP$. (see image
here http://img241.imageshack.us/my.php?image=slide1ww4.jpg)

[quote="Bictor717"][hide="Hint"]Let $ \overline{PQ}$ be a diameter of circle $ O$ (i.e., let $ Q$ be diametrically opposite from $ P$ in circle $ O$). Draw some helpful lines and chase some angles.[/hide]
[hide="Solution minus details"]Let $ X$ be the intersection of $ \overline{BD}$ and $ \overline{OP}$.
$ m\angle PQA \equal{} m\angle PCA \equal{} m\angle DCA \equal{} m\angle DBA \equal{} m\angle XBP$. Then $ \triangle PQA\sim\triangle PBX$. Since $ \angle PAQ$ is right, so is $ \angle PXB$.[/hide]
Questions?[/quote]

Whats $ m$?

[quote="Bictor717"]$ m\angle A$ is the measure of angle $ A$.
$ m\angle A \equal{} m\angle B$ is equivalent to saying $ \angle A\cong\angle B$.[/quote]

So isnt it the same as saying $ \angle A \equal{} \angle B$?

[hide]Let the intersection of $ BD$ and $ PO$ be $ H$. Let $ PO$ intersect circle $ O$ again at $ Q$. By cyclic quadrilateral $ ABCD$, we have $ \angle BDC \equal{} \angle BAC$. By cyclic quadrilateral $ ACQP$, we have $ \angle BAC \equal{} 180^\circ \minus{} \angle CAP \equal{} \angle CQP$. Thus $ \angle CQP \equal{} \angle BDC$, so $ \angle QHD \plus{} \angle QCP \equal{} 180^\circ$. Since $ PQ$ is diameter of circle $ O$, $ \angle QCP \equal{} 90^\circ$. Therefore, $ \angle QHD \equal{} 90^\circ$.
[/hide]

Put Q on PA so $ OQ \perp PA$, and R is the intersection of PO and DB. $ \angle POQ \equal{} \frac {1}{2} \cdot \angle POA \equal{} \angle PCA \equal{} \angle DBA$, hence ORBQ is cyclic, so since $ OQB \equal{} 90$, $ ORB \equal{} 90$.

I thought to do this because I wanted to use the fact that in a circle, the perp from the center to a chord bisects it (OQ bisects PA) and the fact that the inscribed angle is half the central angle, OPA.

[quote="kurt.math"]Bictor717, could you please tell me what thoughts you had to get to the end solution please.

See, what i thought was that, because the question mention a cyclic quadrilateral that it would revolve around that, so i began to find angles in terms of y's and x's to no avail.

What made you think, "hey, why dont we make the diameter and connect Q to A and A to C and find similar triangles"?[/quote]Calculating x's and y's doesn't seem very productive. Ultimately, you'd have to draw some perpendiculars or diameters involving O to make use of the given fact that O is the circumcenter of PAC, but that would boil down to a proof that could be done without calculating x's and y's.

I guess it's from experience. Similar triangles are always nice, so I also began by looking at the angles of ABCD and searching for similar triangles. However, there were no right angles, so it wouldn't be easy to get the relevant similarity. Furthermore, I hadn't done anything with O at this point, so drawing the diameter to get a right angle seemed natural. It's also a bit of working backwards, thinking about what would be needed to give you the desired result.


I don't understand the binet-cauchy theorem. Could someone give me an example, say for A (2X3), B (3X2)...

[quote="Slizzel"] http://planetmath.org/encyclopedia/CauchyBinetFormula.html   ;)[/quote]

Maybe this will help you.

yes, i saw the formula and i've tried to apply it, but it doesn't work  :blush: ...can u pls give me an example for a particular case?

There are too many minors to be written.
How to prove this : Suppose $C=AB$ then
$c_{ij}=\sum_{k=1}^n a_{ik}b_{jk}$. Every minor of order $k$ of $C$ is of the form obtainde by the intersection of the lines $i_1,i_2,...,i_k$ and columns $j_1,j_2,...,j_k$:
$\xi=\left( \begin{array} {cccc} {\sum_{h=1}^n a_{i_1h}b_{hj_1}}&{\sum_{h=1}^n a_{i_1h}b_{hj_2}}&{...}&{\sum_{h=1}^n a_{i_1h}b_{hj_k}}\\{\sum_{h=1}^n a_{i_2h}b_{hj_1}}&{\sum_{h=1}^n a_{i_2h}b_{hj_2}}&{...}&{\sum_{h=1}^n a_{i_2h}b_{hj_k}}\\{...}&{...}&{...}&{...}\\{\sum_{h=1}^n a_{i_kh}b_{kj_1}}&{\sum_{h=1}^n a_{i_kh}b_{hj_2}}&{...}&{\sum_{h=1}^n a_{i_kh}b_{hj_k}} \end{array} \right )$.
Now because every element of $\xi$ is the sum of $n$ terms , $\xi$ can be decomposed  as sum of $n^k$ minors of the form:

$\alpha=\left ( \begin{array}{cccc} {a_{i_1 h_1}b_{h_1j_1}}&{a_{i_1 h_2}b_{h_2j_2}}&{...}&{a_{i_1 h_k}b_{h_kj_k}}\\{a_{i_2 h_1}b_{h_1j_1}}&{a_{i_2 h_2}b_{h_2j_2}}&{...}&{a_{i_2 h_k}b_{h_kj_k}}\\{...}&{...}&{...}&{...}\\{a_{i_k h_1}b_{h_1j_1}}&{a_{i_k h_2}b_{h_2j_2}}&{...}&{a_{i_k h_k}b_{h_kj_k}} \end{array} \right )= \prod_{z=1}^k b_{h_zj_z} D$
where $D$ is the matrix $D=(a_{i_zh_v})_{z,b=\overline{1,n}}$(I mean just factorize the $\alpha$ matrix).

So if you were to right the minor of order $n$ ,the determinant, it would be the sum of $n^n$ determinants.

Hope it`s clear now.

This is also a nice way to prove Sylvester`s inequality:$rank(AB)\leq \min(rank(A),rank(B))$


floor function
inequalities
least common multiple
number theory unsolved
number theory

Each of the positive integers $a_1,...,a_n$ is less than 1951. The least common multiple of any two of these is greater than 1951. Show that :
$\frac{1}{a_1}+...+\frac{1}{a_n}<2$

Davron

The title "Number" you used here and in http://www.mathlinks.ro/Forum/viewtopic.php?t=86839 says nothing and really nothing to the people and is very hard to refind (especially one has to open the thread to even get a clue what is happening).
Please change both!

Well, in fact it is still true if we replace 1951 with $N$. ;)

For $1 \leq i \leq n$, we denote :
\[ A_i = \{p \in \{1, 2, \ldots, N\} \mid a_i | p\}.  \]
From the problem we know that $A_i \cap A_j = \emptyset$ for all distincts $i, j \in \{1, 2, \ldots, n\}$. So :
\[ \left|\bigcup_{i=1}^n A_i \right| = \sum_{i=1}^n |A_i|,  \]
but since $\bigcup_{i=1}^n A_i \subseteq \{1, 2, \ldots, N\}$, we have :
\[ \sum_{i=1}^n |A_i| \leq N.  \]
But, $|A_i| = \left \lfloor \frac N{a_i} \right \rfloor$, $i \in \{1, 2, \ldots, n\}$, hence :
\[ N > N \left[\sum_{i=1}^n \frac 1{a_i} \right] - n,  \]
that is :
\[ \sum_{i=1}^n \frac 1{a_i} < 1 + \frac nN \leq 2.  \]

And, here's a little challenge left to the reader :) :
 :idea: This bound is not optimal. Find a better (or the best if you can) bound.  :idea:

Could you please give me the page on which this problem can be found?  I would like to observe two different solutions. Thanks in advance.

Masoud Zargar

"Post subject: Engel
Source: a book"

or

"Arithmetic"

is not really better than before.
It says nothing about the problem (except that it's from Engel's book, but that complete belongs to the source).
And also change the others (e.g. "Numbers 3"), too, and use better ones in future!
(see http://www.mathlinks.ro/Forum/viewtopic.php?t=21455 for the problem described for the inequalities subforum)


Let $x;y;z$ be non-negative real numbers and satisfyinf $x^{2}+y^{2}+z^{2}=1$.Find max:
$A=x+y+z-3xyz$

hello, by the method of Lagrange multipliers i have got that the searched maximum is $ \sqrt{2}$.
Sonnhard.


LaTeX
quadratics
algebra
difference of squares
special factorizations

Solve in $\mathbb{C}$ the equation: $(x-a)^4+(x-b)^4=(a-b)^4$.

[hide="My Solution"]

Ok, first divide through by $(a-b)^4$ to get two fractions on the left whose sum is one:

$\Big(\frac{x-a}{a-b}\Big)^4+\Big(\frac{x-b}{a-b}\Big)^4=1$
Or in other words:
$\Big(\frac{2x-a-b}{a-b}\Big)^4=1$
Now we set $2x-a-b=(a-b)(j)$ where $j=1,-1,i,-i$

So now solve each of this for $x$ and get $x=a$, $x=b$, $x= \Big(\frac{a(1+i)+b(1-i) }{ 2}\Big)$ or $x= \Big(\frac{b(1+i)+a(1-i)}{2}\Big)$[/hide]  This Latex looks ok to everyone right?  it is my first post with latex :)

oh yeah sorry stupid mistake.. . :blush: i'm going to go try this one again.

[quote="Hexaditidom"]Remember fermats theorem... isn't there no a, b, and c such that a^4+b^4=c^4 ?[/quote]

those were integers

here are complex numbers

Hint: There are two obvious solutions.  It might be profitable to divide these out and see what we're left with.  Alternately, there's a difference of squares that you might want to play around with.

Is this the answer by any chance?

[hide="My Feeble Attempt"]
$x=a$, $x=b$, or $x=\frac{a+b\pm(a-b)i\sqrt{7}}{2}$
[/hide] :?

This way works, but it was kind of a pain.
[hide="My Way"]

Expand everything, and gather on the left (some things cancel, and you can divide by 2, etc.)
$x^4-2x^3(a+b)+3x^2(a^2+b^2)-2x(a^3+b^3)+ab(2a^2-3ab+2b^2)$
that is a mess.

Ok looking at the first equation it is pretty obvious $x=b$ and $x=a$ work so this has to have a factor of $x^2-x(a+b)+ab$

Then we know the other factor has to be $x^2-x(a+b)+(2a^2-3ab+2b^2)$ so then just solve with the quadratic formula.  I got this factorization by checking the $x^3$ coefficient  that would result.
[/hide]

mkkool64, you're solution is quite correct.
i posted this problem because it's nice and it has three solutions at least (none of them mine)

another one would be noticing that we can rewrite it $(a-x)^4+(x-b)^4=(a-b)^4$ and then solving it in $u=a-x$, $v=x-b$, :).

[quote="blahblahblah"]No,

$a^4+b^4\not=(a+b)^4$

(well, it would be true if $4$ were prime and we were working $\bmod 4$, but clearly none on these situations hold)[/quote]

Isn't this called 'Freshman's Dream?'

 :D

hey, surely we've all written it like that once or twice in our lives... :blush:


integration
calculus
induction
inequalities solved
inequalities

Let a[i] be n reals.
Then
 \sum (a[m]*a[n])/(m+n)  >=0
Where the sum is taken with all the choices of the indices m & n.

All choices of m and n such what??? Because it is untrue for example if m = 1 and n = 1,2 for some choices of a_1 and a_2....

Pierre.

I think that the statement is the following:
  Let a_1,...,a_n reals. Then  \sum (i,j=1,n) a_i*a_j/(i+j)  \geq 0.
  The solution is simple, since it comes from the fact that  \int (0,1) ( a_1x^(1-1/2)+a_2*x^(2-1/2)+...+a_n*x^(n-1/2))^2>=0.

Yeah, Sorry for the mistake.
This is indeed a solution,but what about a 10th grade student that doesn't know calculus? (at all)
It is quite hard from that point of view :), but a solution exists.

I saw once a solution based on induction, but I wouldn't dare to prove it like this. I just like short solutions.


calculus integration quadratics conics hyperbola algebra

How about rubbin our hands at some simple questions-

1)solve-
   x^2+(y-z)^2=a^2
   y^2+(z-x)^2=b^2
   z^2+(x-y)^2=c^2

2)One morning is starts to snow at a constant rate. Later, at 6:00am, a snow plow sets out to clear a straight street. The plow can remove a fixed volume of snow per unit time, in other words its speed it inversely proportional to the depth of the snow. If the plow covered twice as much distance in the first hour as the second hour, what time did it start snowing?

NO replies for 2 days?????c'mon geniuses,these questions arent tooo tough. :)

The 1st one isnt but the 2nd one is hard!
Did you get it???? :P

I think the 2nd one will involve a differential eqn. and probably should be in Calculus forum. Again,I might be wrong!!!

this problem is from mathproblems.info and the solution contains some differentation.... I think because "d" variable was coming in the solution by the author.

In order to get the proper solution, you have to solve a differential eqn. Although , by making some silly assumptions(which I hate) you might be able to understand. This is because, the problem has a format similar to a ' first order reaction' in Physics. I think this is more appropriate for Physics if you don't know integration etc...

[quote="riddler"][quote="shadysaysurspammed"]Yeahhh the second one is sorta weird...But i know the solution but didnt get it...i think it needs some calculus n stuff....

How about the first one? :)[/quote]
not coming, I get....
$z^2+\frac{{(a^2-b^2)}^2}{{(2z)}^2}=c^2$
But then I cannot solve for $z$[/quote]
I haven't read the problem at all. I don't know what a, b and c represent either. I am just solving your equation.
$z^2+\frac{(a^2-b^2)^2}{(2z)^2}=c^2$
$4z^4+(a^2-b^2)^2=4c^2z^2$
$4(z^2)^2-4c^2(z^2)+(a^2-b^2)^2=0$
Consider this as a quadratic equation in $z^2$
$z^2=\frac{4c^2\pm \sqrt{(4c^2)^2-4(4)(a^2-b^2)^2}}{8}=\frac{c^2\pm \sqrt{c^4-(a^2-b^2)^2}}{2}$
Therefore $z=\pm \sqrt{\frac{c^2\pm \sqrt{c^4-(a^2-b^2)^2}}{2}}$

why didnt I think of that  :blush:  :blush:  :blush:  :blush:  is it right???? nil?

for part b, the speed of the plow is described by a hyperbola, so you have to find the integral of the hyperbola.  Calculus.  Or there's some otehr way.

[b]Do not post Calculus problems in the Intermediate Forum![/b]

Dude joml,I'm reeeeeeeaaaaaallllyyyyyyy sorry for that...I myself didnt know thats calculus,so i wanted to know the solution.I dont even know calculus....so thats why :lol:

It's ok.  I'm going to leave this one hear but I wanted to make sure that everyone knows that Calculus is not appropriate for this forum.

Thanks.

[quote="shadysaysurspammed"]Here's the solution to the 2nd one....

Let the depth of snow at time t to be t units. The speed of the plow at time t will be 1/t. Define t=0 as the time it started snowing and t=x the time the plow started.
The distance covered in the first hour is the integral from x to x+1 of 1/t dt. The antiderivative of 1/t is ln(t) so the total distance covered in the first hour is ln((x+1)/x).
By the same reasoning the distance covered in the second hour in ln((x+2)/(x+1)).
Using the fact that it the plow traveled twice as far in the first hour as the second: ln((x+1)/x) = ln((x+2)/(x+1))2
Exp both sides and you have (x+1)/x = ((x+2)/(x+1))2.
Solving for x you get x=(51/2-1)/2, which is the number of hours that elapsed between the time it started snowing and the snow plow left.

COuld someone writ the full sol of the first one...cause my ans is differnt from riddler's :lol:[/quote]
But this is directly from [url]http://mathproblems.info/[/url] :P

SO what riddler,even i know that,Didnt i say that even i dont know calculus and since i didnt know that the solution had calculus,I posted it to confirm that it neede calculus.....DO the first one with full solution.

I think I have done that! the solution is up above!


Is there a proof of Pick's Theorem other than the proof [url=http://en.wikipedia.org/wiki/Pick%27s_Theorem]here?[/url] If there is, I have a feeling that it involves tesselating the Cartesian Coordinate grid with unit squares centered at the lattice points.

Pick's Theorem can be proven as a consequence of [url=http://en.wikipedia.org/wiki/Minkowski%27s_theorem]Minkowski's theorem[/url]: the proof is given in a past issue of AMM.  [url=http://www.cs.drexel.edu/~crorres/Archimedes/Stomachion/Pick.html]Another issue[/url], if I've read the title right, gives a proof based on Euler's formula $ V \minus{} E \plus{} F \equal{} 2$; the $ \minus{}1$ in Pick's theorem is really a consequence of the [url=http://en.wikipedia.org/wiki/Euler_characteristic]Euler characteristic[/url] of the plane.


So, who is going to watch the new Harry Potter movie? It just released today. I think that I'm going to get tickets online and then go. It really takes a long time to get tickets in the actual building...

C'mon! Watch the new Harry Potter movie!

If you do, tell me about it!

[color=white]Please do![/color]

Eh...I'll probably go eventually.  I'm in no rush.  Probably during thanksgiving or winter break. :D

Edit: my $2^{10}$ post!

probably during thanksgiving break cuz theres no rush

Yea, so I watched the whole movie.  Like the whole thing.  Like the credits too.  Like there were ushers in the theater cleaning up while I was still watching the movie.  And I found out that "No dragons were harmed in the making of this movie."

Good I thought that their might have been dragons harmed

lol... did ane one see it at midnite on the day it came out

heh. I guess Harry Potter is really popular...

i watched it with my skool friends (including b4b4ka)

I am going to see it next week. Hope it worths.

i gonna watch it tomorrow and i think it is a good movie. :D

It was a freaking ripoff. They cut out the World Cup!.

I saw it on Thanksgiving.  If you read the book, you will know that a lot of very important things were left out.  It needed to be made into 2 movies instead of 1.  They left out:  Quiditch world cup, Harry giving 1000 galleons to Fred and George (which lets them open the joke shop in the 5th book), Capturing Rita Skeeter, Harry getting caught looking at the Maurderers map, and more not so important things.  Still, it was a good movie.

[quote="111222333444555"]Well, I saw it yesterday. It was good, but the book is much better than the film.[/quote]
i didn't see it but i can guess what you mean and i think you are right becasue in the movie something has cuted,and in the book there is more explain about everything and it has more things.there is something else,the book is written by J.K.ROLING but the director of movie is someone else,so they have a little difference. :D

i saw it on Saturday. It was good. It had nice special effects.

[quote=Altamira140]hello. i think i wasn't even 1 year old when you guys were talking about this. now, harry potter has 8 movies and i've watched them all. you were talkin' bout' the 4th one. om! that was 10 years ago!!!  :P[/quote]

PLEASE STOP REVIVING!!!!!!!!!!!

[quote=apolloheo]lol, aops active 10 years ago[/quote]

Yeap, i was 1 when the movie was released. This is like time travel!!!!! LOL

Harry potter is the best!!!!! I heard they are making a 8th and/or 9th book!!!!

[quote=Altamira140]hello. i think i wasn't even 1 year old when you guys were talking about this. now, harry potter has 8 movies and i've watched them all. you were talkin' bout' the 4th one. om! that was 10 years ago!!!  :P[/quote]

Guys please no necropost (spelling?)

[quote=Altamira140]hello. i think i wasn't even 1 year old when you guys were talking about this. now, harry potter has 8 movies and i've watched them all. you were talkin' bout' the 4th one. om! that was 10 years ago!!!  :P[/quote]

I was basically born at that time

[quote=Altamira140]I FEEL LIKE TIME TRAVEL!!!!!![/quote]

Yeah.  We can see what aops users where like ten years ago


[quote="muks"]Here is a problem posed by me:( Warning: This may be a very simple one as I have just started learning problem-solving). :)  :D

$ ABC$ is a right angled triangle, right angled at $ A$.Bisectors of $ \angle B$ and $ \angle C$ intersect at $ O$.Perpendiculars $ OP$ and $ OL$ are dropped on $ BC$ and $ AC$ respectively.Bisectors of $ \angle POC$ and $ \angle LOC$ meet $ BC$ and $ AC$ at $ G$ and $ M$ respectively,.Prove that $ OM \equal{} OG.$

Your comments and solutions are welcome.[/quote]

$ \triangle POC \cong \triangle LOC  \Rightarrow \triangle POG \cong \triangle LOM$, done.

Actually $ O$ can be any point on bisector of $ \angle B$ inside the triangle, not necessary the intersection.

Another way of looking at it:

[hide]O is the incenter of the triangle, and the points dropped from perpendiculars on the sides of the triangle are where the incircle is tangent. It's elementary that the triangles LOC and POC are congruent, and thus corresponding parts of congruent triangles are congruent. This result should hold even if ABC isn't a right triangle.[/hide]

[quote]Actually $ O$ can be any point on bisector of  inside the triangle, not necessary the intersection.[/quote]
Is that right?  It seems to me that if $ O$ was anywhere else in the triangle, $ O$ would not be the incenter, and the triangles would not be congruent.
But I agree with Lulze that any triangle would work


Can you write the equation more clearly?

[quote="ashegh"]Let N be a positive integer. Define the sequence a0, a1, a2, ... by a0 = 0, an+1 = N(an + 1) + (N + 1)an + 2( N

(N+1)an(an+1) )1/2. Show that all terms are positive integers[/quote]

$\frac{(N+1)a_n(a_n+1)}{2}$


function
algebra
polynomial
trigonometry
real analysis
real analysis unsolved

Find all sequences $a_{0}, a_{1},\ldots, a_{n}$ of real numbers such that $a_{n}\neq 0$, for which the following statement is true:

  If $f: \mathbb{R}\to\mathbb{R}$ is an $n$ times differentiable function
  and $x_{0}<x_{1}<\ldots <x_{n}$ are real numbers such that
   $f(x_{0})=f(x_{1})=\ldots =f(x_{n})=0$ then there is $h\in (x_{0}, x_{n})$ for which \[a_{0}f(h)+a_{1}f'(h)+\ldots+a_{n}f^{(n)}(h)=0.\]

I guess the answer is: the sequence such that:
$C(1+kx)^{n}=a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n}$
$C,k$ is any number.

I guess the answer is:

[hide="answer"]The sequence is such that the polynomial $P(x)=a_{0}+a_{1}x+\ldots+a_{n}x^{n}$ has only real roots.[/hide]

[hide="Proof."]
Let $b+ci$ be a complex root of $P$ and  $a_{0}=\ldots=a_{k-1}=0, a_{k}\ne0.$ Then $f(x)=\alpha x^{k}+e^{bx}\cos{cx}$ satisfy $a_{0}f(x)+a_{1}f'(x)+\ldots+a_{n}f^{(n)}(x)\equiv \alpha\cdot a_{k}k!\ne0$ and $f(x)$ has at least $n+1$ zeroes for sufficiently small $|\alpha|.$

Let $P(x)=(x-c_{1})\ldots(x-c_{n}).$ By Rolle's theorem $f_{c}(x)=e^{cx}(f(x)e^{-cx})'=f'(x)-cf(x)$ has a zero between consequtive zeroes of $f.$ It remains to notice that
$a_{0}f(x)+a_{1}f'(x)+\ldots+a_{n}f^{(n)}(x)=a_{n}\cdot(\ldots(f_{c_{1}})_{c_{2}}\ldots)_{c_{n}}$ and apply induction.  :) [/hide]

what do you think is this kind of answer really what people should expect to see in the solution? should the answer be given in some closed form because like this it looks like we only show to which problem it is equivalent?
how hard is to determine whether the mentioned polynomial has all roots real? what do we know about the set of such polynomials?


trigonometry algebra polynomial calculus

[quote="LordoftheMorons"]$\frac{sin x}{cos x} = sin x + .5$
$\frac{sin x}{cos x} = sin x + sin 30$
$sin x = (sin x)(cos x) + (sin 30)(cos x)$

And from there I don't know >_>[/quote]

Are you sure abt that?

[quote="Sporally"]How do i solve this equation?

$\tan x=\sin x+0.5$

It doesn't seem that difficult with the exception that you have $x$ two times and i don't know how to merge it down to one $x$.

Thx in advance for your help.[/quote]

$\frac{\sin x}{\cos x} = \sin x + 0.5$
$\sin x = \cos x ( \sin x + 0.5 )$
$\sin^2 x = \cos^2 x ( \sin^2 x + \sin x + 0.25 )$
$\sin^2 x = (1 - \sin^2 x)( \sin^2 x + \sin x + 0.25)$
$\sin^2 x = \sin^2 x + \sin x + 0.25 - \sin^4 x - \sin^3 x - 0.25\sin^2 x$
$4\sin^4 x + 4\sin^3 x + \sin^2 x - 4\sin x - 1 = 0$

And that's a quartic polynomial in $\sin x$. You can find the roots for that, then find the values of $x$.

I try using the substitution

$\sin x=\frac{2t}{1+t^2}$ , $\tan x=\frac{2t}{1-t^2}$ where $t=\tan\frac{x}{2}$

and come out with $t^4+8t-1=0$ and I dont think we can solve this equation other than calculus .....

Great. And i can see it makes sence. However, then i got to here where i want to transform it into fractions:

$t=\sin x$

$t^4+t^3+0.25t^2-t-0.25$

I tried finding numbers that could fit into this equation and end up with 0 when putting it into $t$'s spot, but as far as i know that isn't that easy. I found that we can find the solution through this equation:

$\frac{u}{(1/4)^4}=t$

$\frac{u}{(1/256)}=t$

$u$ is an even number that we can put into the equation.

Do you have a better solution to this problem than experimentate with this and see which number hits $x_0=0$?


trigonometry algebra polynomial quadratics function limit calculus

I'm guessing there must be easier way than simply squaring and squaring because eventually, you get $(2-x^4)^2 = 2+x$ which isn't something very fun to do.

[quote="arqady"][quote="Silverfalcon"]I'm guessing there must be easier way than simply squaring and squaring... [/quote]
You right! ;)  :)[/quote]

Here is some more progress I made.

[hide="Progress"]
Inspection on above equation shows that $x = -1$ is easily a root of the equation. Now, under first square, we have $\sqrt{2+x}$ which means that $\sqrt{2+x} \geq 0$ and then it meant $2+x \geq 0 \rightarrow x \geq -2$. Also, note that if we set $x = 2$, on the second square root, we get $\sqrt{2- \sqrt{2+2}} = \sqrt{2 - 2} = 0$. This tells that if $x$ is bigger than 2, the result on the second square root will be negative. So, $x \leq 2$.

So, I finally found the interval $-2 \leq x \leq 2$. [/hide]

I don't know how to go from here though...

[quote="Silverfalcon"]
Inspection on above equation shows that $x = -1$ is easily a root of the equation.[/quote]
No. If $x$ is root of the given equation then $x>0$, and equation $(2-x^4)^2=2+x$ is wrong.
[quote="Silverfalcon"]I don't know how to go from here though...[/quote]
$0<x<2$ is half of solution! :)

Oh crap.

I forgot the $x$ in the RHS. Eh...

isn't this equal to $\cos{\frac{2\Pi}{7}}$?

sorry I don't know the actual solution; i am just curious because I think i've seen this expression before and i was wondering how to solve it last time too.

[quote="agolsme"]isn't this equal to $\cos{\frac{2\Pi}{7}}$?
[/quote]
No. :juggle:

[quote="arqady"]Solve the equation: $\sqrt{2+\sqrt{2-\sqrt{2+x}}}=x.$[/quote]
Unsolved in 2005. ;)

Here is an ugly ugly brute force method where I've squared and solved polynomials
[hide]
Squaring over and over again gives the equation $x^8 - 8x^6 + 20x^4 - 16x^2 - x + 2 = 0$. By inspection -1 and 2 are roots to this equation. However, they are false roots to the original equation. Division by $(x-2)(x+1)$ now gives $x^6 + x^5 - 5x^4 - 3x^3 + 7x^2 + x - 1 = 0$.

Proposition of a factorization and comparing coefficients gives that this is equivalent to $(x^3 - 3x + 1)(x^3 + x^2 - 2x - 1) = 0$.

Now, $x^3 - 3x + 1 = 0$ is quite easy to solve! Let $x = z + \frac{1}{z}$. Plugging this in and multiplying by $z^3$ yields a quadratic equation $(z^3)^2 + z^3 + 1 = 0$ with solutions $z^3 = -\frac12 \pm i\frac{\sqrt{3}}{2} = e^{\pm\frac23 \pi i}$.

Solving the binomial equations and plugging back in x gets us three solutions:
$x_1 = 2\cos\left(\frac29 \pi\right)$
$x_2 = 2\cos\left(\frac89 \pi\right)$
$x_3 = 2\cos\left(\frac{14}{9} \pi\right)$

$x_2$ can immediately be disqualified since it's negative. $x_3$ can also be shown to be wrong by estimating the surd expression downwards and showing that it is larger than itself. This does require a numerical approximate first ($0.34 < x_3 < 0.35$), so without a calculator this is really troublesome.

$x_1$ is the correct answer, but I don't know how to show that. I think it's impossible to find an exact value for it, so... Well, you could solve the other cubic too and see that its solutions are false. By considering strictly increasing/decreasing functions it's obvious that there is one and only one solution to the original equation, and so it would have to be $x_1$. But solving the other cubic too is too much work.

edit: See Raluca's post for how this solution can be verified. Use the trigonometric identities that she uses.
[/hide]

[quote="Kalle"]Here is an ugly ugly brute force method where I've squared and solved polynomials
[/quote]
OK, Kalle. But I found very nice solution for the equation. :)

Why don't you show us? No one has come up with anything better than the ugly method I just showed (and didn't complete because the last verification calculations would take too long).

[quote="Silverfalcon"]So, I finally found the interval $-2\leq x\leq 2$.
I don't know how to go from here though... [/quote]
Well, you can write $x=2\cos t$ for some $t\in [0,\pi]$.

[hide="Solution"]
For any $\alpha\in [0,\pi]$, we have:
(1)  $\sqrt{2+2\cos\alpha}=2\sqrt{\frac{1+\cos\alpha}{2}}=2\cos\frac{\alpha}{2}$
and
(2)  $\sqrt{2-2\cos\alpha}=2\sqrt{\frac{1-cos\alpha}{2}}=2\sin\frac{\alpha}{2}$
Now put $x=2\cos t,\ t\in [0,\pi]$.
By (1) we get first
$\sqrt{2+x}=2\cos\frac t{2}$,
then by (2) we get
$\sqrt{2-\sqrt{2+x}}=\sqrt{2-2\cos \frac t{2}}=2\sin \frac t{4}$
and by (1) again
$\sqrt{2+\sqrt{2-\sqrt{2+x}}}=\sqrt{2+2\sin \frac t{4}}=\sqrt{2+2\cos(\frac\pi{2}-\frac t{4})}=2\cos(\frac\pi{4}-\frac t{8})$

The equation becomes
$cos(\frac\pi{4}-\frac t{8})=\cos t$,
hence,taking into account that $t\in [0,\pi]$, we obtain
$t=\frac{2\pi}{9}$
and finally $x=2\cos\frac{2\pi}{9}$. :)

[/hide]

Cool. That is also a nice way to verify my solution :P. I probably wouldn't have thought of such a substitution without first seeing my answer, so for me they are connected.

Well, I suppose I wouldn't have been able to solve the equation $x^3-x+1=0$... (I wonder how can you call it "quite simple" :roll:   ).

Well, it isn't hard. Try doing the method yourself. It's very simple. You get two solutions for the quadratic, but you only need to consider one solution, since $x = z + z^{-1}$ is symmetric so that the choice of sign in the solution is irrelevant.

Basically, when you have an equation of the form $x^3 + ax = b$ you set $x = z - \frac{a}{3z}$, substitute it in and multiply the equation by $z^3$. Now solve this quadratic equation and only take one solution and write it on polar form. Then solve this binomial equation $z^3 = re^{i\theta}$ and substitue the solutions back to get them in x. If the coefficients aren't "nice", you will get angles that can only be expressed by inverse trigonometric functions, but they still are the solutions.

A very similar method can be used for (some) fourth degree equations.

Very well, Raluca!
Try it:
Solve the next equation $8x(2x^2-1)(8x^4-8x^2+1)=1.$ ;)

Well, I think the trick here is putting $x = \cos\theta$, and without doing it I *think* it will turn into something of the form $A\cos(\theta)\cos(2\theta)\cos(4\theta)$, but I don't want to wade through all the details.

Ok, I had to finish it. Note that I'm just assuming that all solutions are such that $|x| \le 1$. I don't know how to show that before hand.

More specifically the equation turns into $8\cos(\theta)\cos(2\theta)\cos(4\theta) = 1$. Multiplying both sides by $\sin\theta$ and using the double angle formula we get that $\sin(8\theta) = \sin(\theta), \sin(\theta) \ne 0$ with the last condition because we've introduced that solution on our own.

Now then, we get that either $\theta = \frac{2\pi}{7}n$ for integer n, or $\theta = \frac{\pi}{9} + \frac{2\pi}{9}n$ for integer n.

Due to symmetric properties of $\cos$, we can state all solutions from the first equation a $x = \cos\left(\frac{2\pi}{7}n\right), n=1,2,3$ and from the second we get $x = \cos\left(\frac{\pi}{9} + \frac{2\pi}{9}n \right), n=-1,1,2,3$

There are no more solutions than these seven solutions, which also shows that the substitution works.

Cool, Kalle!
Try it:  $\sqrt{8-\sqrt{8+\sqrt{8-x}}}=x$ ;)

This post is much like that of Raluca, but there is a enough different that it is worth posting (P.S. I will do both problems.)

If we infinitely substitute $x$ back into the radical, which we are allowed to do, we get $x = \sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\cdots}}}}}}$ where the signs alternate between two plus signs and a negative.

We have 3 different, but related, radicals to find our answer:
$(++-) = L = \sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\cdots}}}}$

$(+-+) = N = \sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+\cdots}}}} = x$

$(-++) = M = \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}}$

We easily have three equations from these 3 radicals:

$L^2 - 2 = N$

$N^2 - 2 = M$

$2 - M^2 = L$

We wish to find $N$ and if we just substitute twice we get our nasty 8th degree polynomial that Kalle worked with, but we won't go that way, we will go back to the trig, as Raluca did.
First, we note two things:
$\text{1)} 8\cos^4 \theta - 8\cos^2 \theta + 1 = 2(4\cos^4 \theta - 4\cos^2 \theta + 1) - 1 = 2(2\cos^2 \theta - 1)^2 - 1$

$= 2(\cos 2\theta)^2 - 1 = \cos 2(2\theta) = \cos 4\theta$
The second thing is this, we have an abundance of 2's in our equations, so we wonder, what if I make a substitution, say $x = N = 2k$. Doing so, we find:

$\text{2)} M = N^2 - 2 = 4k^2 - 2 = 2(2k^2 - 1)$

and that is when we recognize our trig substitution. $2k^2 - 1$ immediately reminds us of the double-angle identity for $\cos$ and makes us want to say $k = \cos A$, so we do.

Doing so, we get:

$x = N = 2k = 2\cos A$

$M = N^2 - 2 = 2(2\cos^2 A - 1) = 2\cos 2A$

$N = L^2 - 2 = (2 - M^2)^2 - 2 = M^4 - 4M^2 + 2 = (2\cos 2A)^4 - 4(2\cos 2A)^2 + 2$

$= 16\cos^4 2A - 16\cos^2 2A + 2 = 2(8\cos^4 2A - 8\cos^2 2A + 1)$

Now, we use what the identity we found in 1) for $\theta = 2A$ to get $N = 2\cos 4(2A) = 2\cos 8A$.

So, we now have $2\cos A = N = 2\cos 8A$.

It's now easy and we are just left to solve $\cos A = \cos 8A$.

Edit: I have finished my researching.

Also, I have solved the 2nd question: $x = 2\sqrt{3} \sin\frac{\pi}{9} + 1$ but I don't wish to post my solution just yet, because I think there are still some holes(except checking it with a calculator, which works.)

And, since this post is getting so long, I will start a new one for my "generalized method".

[quote="Pork_Chop8"]
If we infinitely substitute $x$ back into the radical, which we are allowed to do[/quote]
Is it obvious that such continued radicals always converge? While it is quite obvious that the solution to the original equation will be the limit of the continued radical if it exists, I'm not totally sure why it must exist.

I just thought that was something to consider since you were talking about generalized results.

[quote="Kalle"]
Is it obvious that such continued radicals always converge?[/quote]
No! This isn't obvious. I agree with you. But for the next one "radicals  converge"!
Find a volue of the expression: $\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-...}}}}.$
Id est, if $a_{n+2}=\sqrt{2+\sqrt{2-a_n}}, a_1=\sqrt2,a_2=\sqrt{2+\sqrt2},$ then
1) prove that $\exists\lim_{n\to\infty}a_n.$
2) Find this is limit. :)

My Method :)   :blush:   :(

First we introduce a string of signs as representing each infinite nested radical. Assuming the signs change in a pattern that repeats in finite blocks, we can say that the desired nested radical is:
$(s_1 s_2 s_3 \ldots s_{n-1} s_n) = A_1 = \sqrt{2+ s_1\sqrt{2+ s_2\sqrt{2+ s_3\sqrt{2 + \cdots}}}}$
where for all $i$, $s_i$ is either 1 or -1(the sign), which automatically tells us that $s_i = (s_i)^{-1}$. Now our block of $n$ signs must cycle through $n$ different starting points -- yielding $n$ possibly different infinite nested radicals -- until it finally returns to $A_{n+1} = A_1$. So, for all $i$ we say:
$(s_i s_{i+1} s_{i+2} \ldots s_{i-2} s_{i-1}) = A_i = \sqrt{2+ s_i\sqrt{2+ s_{i+1}\sqrt{2+ s_{i+2}\sqrt{2 + \cdots}}}}$
where all indices cycle mod n.
So, from this, we get:
$s_i(A_i^2 - 2) = \frac{A_i^2 - 2}{s_i} = \frac{2+ s_i\sqrt{2+ s_{i+1}\sqrt{2+ s_{i+2}\sqrt{2 + \cdots}}}-2}{s_i}$
$= \sqrt{2+ s_{i+1}\sqrt{2+ s_{i+2}\sqrt{2 + \cdots}}} = A_{i+1}$
of cource, for all $i$.
Now, we have $A_i = \sqrt{2+ s_i\sqrt{2+ s_{i+1}\sqrt{2+\cdots}}} \leq \sqrt{2+ \sqrt{2+ \sqrt{2+ \cdots}}} = 2$ since negative signs only take away from the value under each of the radicals after them (those closer to the "surface".) So, for all $i$ we have $0 \leq A_i \leq 2$. Now, we can say $A_1 = 2 \cos a$ for $\frac{-\pi}{2} \leq a \leq \frac{\pi}{2}$. Using our identity, we find $A_2 = s_1(A_1^2-2) = 2s_1(2\cos^2 a - 1) = 2s_1\cos 2a$ and now we assert that for all $i \geq 2$, $A_i = 2s_{i-1} \cos (2^{i-1}a)$ which can be easily proven by induction. We have already completed our first case and now we can complete our inductive step, assuming our argument is true, we have:
$A_{i+1} = s_i(A_i^2 - 2) = s_i((2s_{i-1} \cos (2^{i-1}a))^2 - 2) = s_i(4(s_{i-1})^2 \cos^2 (2^{i-1}a) - 2)$
$= 2(s_i)(1)(2 \cos^2 (2^{i-1}a) - 1) = 2s_i(\cos 2(2^{i-1}a)) = 2s_i \cos (2^i a)$
and our proof is completed. By this second identity and the fact that the indices cycle mod n, we get:
$2 \cos a = A_1 = A_{n+1} = 2s_n \cos (2^n a)$

So, we either have $\text{1)} \cos a = \cos (2^n a)$ or $\text{2)} -\cos a = \cos (2^n a)$.
We can use four equations which we won't prove here to "finish" the problem:
$\cos a = \cos(2\pi k + a)$
$\cos a = \cos(2\pi k - a)$
$-\cos a = \cos((2k+1)\pi + a)$
$-\cos a = \cos((2k+1)\pi - a)$
where $k$ is an integer.

For 1):
$\cos(2\pi k + a) = \cos a = \cos (2^n a) \Rightarrow a = \frac{2\pi k}{2^n - 1}$
$\cos(2\pi k - a) = \cos a = \cos (2^n a) \Rightarrow a = \frac{2\pi k}{2^n + 1}$

For 2):
$cos((2k+1)\pi + a) = -\cos a = \cos (2^n a) \Rightarrow a = \frac{(2k+1)\pi}{2^n - 1}$
$cos((2k+1)\pi - a) = -\cos a = \cos (2^n a) \Rightarrow a = \frac{(2k+1)\pi}{2^n + 1}$

Beyond this I'm not sure we can simplify any further other than on a case-by-case basis eliminating the angles that don't work. If anyone can find a way to find the [b]exact[/b] angle $a$ that gives $A_1 = 2\cos a$ please submit it. Thank you.

To prove it converges, look at the sequence $a_0 = \sqrt{2}$, $a_1 = \sqrt{2+s_1\sqrt{2}}$, $a_2 = \sqrt{2+s_1\sqrt{2+s_2\sqrt{2}}}$, $\cdots$, $a_n = \sqrt{2+s_1\sqrt{2+s_2\sqrt{2 + \cdots + s_n\sqrt{2}}}}$ and $a_{n+i} = \sqrt{2+s_1\sqrt{2+s_2\sqrt{2 + \cdots + (s_n)(a_i)}}}$.

P.S. I was saddened to find even with the problem with 8's I could not use or change my method to fit it.  Then I tried $\sqrt[3]{2+\sqrt{2+\sqrt[3]{2+\sqrt{2+\cdots}}}}$ and again was devastated to find that even though I could smash it down to $\cos^2{b} = 2\cos^3 4b$ this did not have nice solutions and I was forced to only deal with the number 2 and the square root.
I think this may be due to the fact that 2 is the only solution to $k = \sqrt{k + \sqrt{k + \sqrt{k + \sqrt{k + \cdots}}}}$ but don't quote me on that.

P.S.S. I know this is kind of weak, but, without proof, I believe I have cracked the recursion that will find us the angle for a given string of signs. Say, the function $R(s_1,s_2,s_3,\ldots,s_{n-1},s_n) = \sqrt{2 + s_1\sqrt{2 + s_2\sqrt{2 + s_3\sqrt{2 + \cdots + s_n\sqrt{2 + s_1\sqrt{2 + \cdots}}}}}}$ is the infinite nested radical with given the given sign-string, as we have been talking about. Now, we define a quantity and a sequence. Say $P_k = \displaystyle\prod_{i=1}^{k} s_i$ and for a sequence ${r_i}$ we have $r_1 = \frac{1-s_1}{2}$ and $r_{i+1} = 2r_i + P_i\frac{1 - s_{i+1}}{2}$ for all $i$ after 1. With this, I am nearly 100% sure we can say: $R(s_1,s_2,s_3,\ldots,s_{n-1},s_n) = 2\cos\left(\frac{r_n \pi}{2^n - P_n}\right)$. In other words, we have $a = \frac{r_n \pi}{2^n - P_n}$. By inspecting our sequence quickly, we see that all $r_i$ must be integers, which will be important for 1 last step. Going backwards on the recursion, we can find the radical based on the given angle. If we know the angle, we know $r_n$ and $P_n$ (the demoninator of the angle must always be in the from $2^k + 1$ or $2^k - 1$ for integer $k$ since $P_n$ is always plus or minus 1, and some given angles will have to be put in equalivent forms, e.g. $\frac{\pi}{13} = \frac{5\pi}{65} = \frac{5\pi}{2^6 + 1}$, so that the demonitor stays in form.) Also, we know if $r_n$ is even, which is case 1) above, $s_n = 1$ and from case 2) if $r_n$ is odd, $s_n = -1$, or, equivalently, $s_n = (-1)^{r_n}$. So, given these we have:
$r_{i} = 2r_{i-1} + P_{i-1}\frac{1 - s_{i}}{2}$
$\Rightarrow r_{i-1} = \frac{2r_i - P_{i-1}(1 - s_{i})}{4}$
Since we know each $r_i$ must be integral, and we know 3 of the 4 variables needed to find it, we can use it to solve for $s_{i-1}$ and also this "decursion" can help use find each succeeding sign all the way down to $s_1$. I don't think I'll be able to prove this but I am like 94.1% sure it holds. :) If anyone has anything more to add -- like a [size=200][b]PROOF[/b][/size] -- please submit, submit submit.
THE END (I hope)

[quote="Pork_Chop8"]
P.S. I was saddened to find even with the problem with 8's I could not use or change my method to fit it. [/quote]
Trigo again, but harder. ;)

[quote="arqady"]Try it:  $\sqrt{8-\sqrt{8+\sqrt{8-x}}}=x$ ;)[/quote]
I find the answer to be $x = 2\sqrt{3}\sin\frac{\pi}{9} + 1 = 2\sqrt{3}\sin\frac{8\pi}{9} + 1 = 2\sqrt{3}\cos\frac{7\pi}{18} + 1$

I think there is a way to prove this by saying:
$f(\theta) = 2\sqrt{3}\sin\theta + 1$

But there is an identity I am searching for and can't find that "should" go something like:
$8 - f(\theta)^2 = f(m\theta + n)$

My problem is I can't find an $m$ and $n$ that make this an identity.

This springs from:
$\sqrt{8-\sqrt{8+\sqrt{8-x_2}}}=x_1$
$\sqrt{8-z}=x_1$
$\sqrt{8+y}=z$
$\sqrt{8-x_2}=y$
$\Rightarrow z = 8 - x_1^2$
$\Rightarrow -y = 8 - z^2$
$\Rightarrow x_2 = 8 - y^2$

Where $x_1 = f(\frac{\pi}{9})$, $z = f(\frac{2\pi}{9})$, $-y = f(\frac{14\pi}{9})$, and $x_2 = f(\frac{8\pi}{9})$

Or possibly, we could have $g(\theta) = 2\sqrt{3}\cos\theta + 1$ along with $8 - g(\theta)^2 = g(p\theta + q)$ and $x_1 = g(\frac{7\pi}{18})$, $z = g(\frac{5\pi}{18})$, $-y = g(\frac{17\pi}{18})$, and $x_2 = g(\frac{11\pi}{18})$

These are not the only angles that produce these signs and cosines, so using different angles may produce the identity we seek. I feel this will find me the answer, but I don't have a lot of free time, so if anyone could help me reach the finish line, I would be much obliged.

When can we use trigo to solve an equation? And how?

Thanks in advance!

If the problem is formulated in such a way that you can show $|x| \le a$ for some real $a$ then you may consider a substitution of the form $x = a \cos \theta$ or $x = a \sin \theta$.

But in this case, the flag that signals a trigonometric substitution is the extensive use of radicals -  you can use a trigonometric substitution anytime it makes the problem simpler  :D

More equation with nice real root:
$\sqrt{4+\sqrt{4+\sqrt{4-x}}}=x.$


[quote="xpmath"]Dude, I don't know if you've noticed, but there is a whole contest part on the top.
And they have solutions.
[url]http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2006[/url]That's the 2006 AIMEs.[/quote]
dude, i don't know if u noticed but that problem doesn't have a solution. and maybe the solution given wasn't clear enough. either way, what im trying to say is, abacadae...this problem too hard for me.  :P
i can start it though.

[quote="abacadaea"]it doesnt have solutions[/quote]

If you click on xpmath's link and click on the link labeled as "12", you can get the solution at:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=464986[/url]

(see natmc's post)

Better yet, just search "2006 AIME 12" instead of posting a new topic.


[quote="nr1337"]Why does the AoPS website now have a lot of the MathLinks boards?[/quote]
The appear to have merged. I don't like it.

gah! holy crud! this'll take some time to get used to

also, is the text auto-centered now? it seems that there's usually a space before where the text box starts at the top.

[quote="orl"]I am registered with the same user name at AOPS and ML. But only my AOPS login still works. The ML login seems to be disabled.[/quote] Do you have the same numbers of posts? Are they the same? If no, then we have a problem. If yes, then there is no problem :) Just go to the profile menu and change your password, email.

[quote="Valentin Vornicu"][quote="nr1337"]And for some reason, some of our avatars aren't working anymore.[/quote] We are sorry for that, but it seems that some of the avatar files were lost in the merging process. Those of you missing avatar files, you can just upload them back to the server.[/quote]
I get these errors when trying to upload my avatar:

Warning: copy(./images/avatars/145330611540ec06b4572f1.gif): failed to open stream: Permission denied in /var/www/Forum/includes/usercp_avatar.php on line 227

Warning: Cannot modify header information - headers already sent by (output started at /var/www/Forum/includes/usercp_avatar.php:227) in /var/www/Forum/includes/page_header.php on line 526

Warning: Cannot modify header information - headers already sent by (output started at /var/www/Forum/includes/usercp_avatar.php:227) in /var/www/Forum/includes/page_header.php on line 528

Warning: Cannot modify header information - headers already sent by (output started at /var/www/Forum/includes/usercp_avatar.php:227) in /var/www/Forum/includes/page_header.php on line 529

Should we start a thread somewhere for bugs and suggestions? (it'd be great if you could remove some of the ancient stickies cluttering up the top of some boards)

Can you give that avatar thing another try?

Hey, AoPS took on the MathLink's ranking thing (i.e. the stars and stuff) instead of the AoPS question marks and different theorem things.  I like the AoPS ranking better. :-P  Oh well.

Wow I got startled.I haven't come to aops in 2 days and a come back and I had to look for  F&G that was strange.  I'm not quite sure what I think of this right now.

[quote="SingTheSorrow"]Hey, AoPS took on the MathLink's ranking thing (i.e. the stars and stuff) instead of the AoPS question marks and different theorem things.  I like the AoPS ranking better. :-P  [/quote]

I'm with you. I hope the new combined forces management brings back the AoPS rank designations, which are more mathematical and spark more discussion of math in the forums.

I guess I'm OK with this merging thing, but it would be nice if we had the old AoPS ranks.

Ditto about the AoPS ranks.

I'm generally all right with the merger - though as one who just joined AoPS last week, I must say it was quite disorienting when I got on the website Wednesday morning. Hopefully the organizational problems can be fixed with time, and I'm sure the administration is working on it. My biggest concern would be that the addition of the Mathlinks boards heavily tilts this forum towards Olympiad-level problems. I hope that AoPS will strongly work to provide problems and help for beginning and intermediate problem solvers so that we won't get scared off by all the problems which are way over our heads.

I like the merging. It's better to be harder than to be easy or we won't improve.

[quote="joml88"]Yes, I too am a little worried about this turning into a predominantly olympiad level board.  Just as long as we try to keep the Getting Started and Intermediate level boards at where they were before we will be fine.  I think that it might take a little bit of time for the Mathlinks people to figure out the difficulty level of these two lower levels but they should get the idea after not too long.  Also, moderators and frequent users of these forums should and I am sure they will do their best to keep the difficulty at the right level.  We'll eventually get to the right place.[/quote]

We and the mods will actively be monitoring those forums to move inappropriately advanced topics into higher-level forums.

[quote="pascal012"]My biggest concern would be that the addition of the Mathlinks boards heavily tilts this forum towards Olympiad-level problems. I hope that AoPS will strongly work to provide problems and help for beginning and intermediate problem solvers so that we won't get scared off by all the problems which are way over our heads.[/quote] I think that it doesn't "heavily tilt" the forum towards Olympiad-level problems, but gives people who developed from AoPS a chance to learn more in the olympiad direction, practice their skills etc., or just watching, without having to re-register on another forum, without having to change their look of the forum, etc.

The same happens for beginners on MathLinks -> not all Mathlinkers are IMO-ers :).

We (Richard and me) hope that you guys will come to appreciate this merge at it real value in the future.

About rasing levels in some of old AoPS forums, I can tell you that should be of no concern for you. As for old MathLinks users complaining that the old MathLinks forums will be downgraded again there should be no concern. There will be still moderated, and whenever a topic will not be in the right place it will be moved. :)

I hope we can all have a nice time on the new board.

But it's so weird becuase now we can enter from two different websites, and they're both different.

You can always pick one you like and stick with that.

I must say I second (er, fifth? I wasn't counting!) going back to the mathematical ranks with the question marks. I think they are more creative. Other than that- change is change, we must make it a change for the better ourselves, although it's a little overwhelming to come back and there are 112 new posts instead of the usual 30-40.

[quote="pascal012"]My biggest concern would be that the addition of the Mathlinks boards heavily tilts this forum towards Olympiad-level problems. I hope that AoPS will strongly work to provide problems and help for beginning and intermediate problem solvers so that we won't get scared off by all the problems which are way over our heads.[/quote]

Taking care of that is part of my job description as a moderator of Getting Started Forum. You can be sure that the site owners are very concerned about helping absolute beginners, so I feel very supported in the cause of posting EASY problems as challenge problems on Getting Started and moving too-difficult problems up to higher levels.

[quote=eriedaberrie]BOI. Just BOI.

 :clap2:[/quote]

You should see my other 14 year bumps.

[quote=Euclidean345]That's a lot of scrolling...[/quote]

I use tags.

"RT-2PM2 Topol-M Silo-based intercontinental ballistic missile hits streak"


function
algebra
polynomial
quadratics
number theory
prime factorization
algebra unsolved

find all zeros of:

[img]http://img85.imageshack.us/img85/3876/2dbd6b06fb186653f870b32qw5.gif[/img]

i need to find all zeros... not simply factor

Why are you attempting to solve polynomials of degree 7 if you cannot solve it in factorised form with nothing higher than quadratic?   Surely if you know what a factor is, you can solve??

lol, obviously $x=2,3$ are two of the zeros.

The other two zeros are $\alpha=\frac{-4 \pm \sqrt{16-52}}{2}$
These are complex, and are $-2 \pm 3i$.

RIP, may I ask how you factored that? It's very impressive if you could do it by hand...

Most probably he started from the prime factorization of the free term - $624=2^{4}\cdot 3\cdot 13$ - with some help of long division.

Or perhaps he differentiated a few times and discovered that f(2) = f'(2) = f''(2) = f'''(2) = 0.

I just noticed that $f(2)=0$, divided $f(x)$ by $x-2$, $f(x)=(x-2)g(x)$, noticed that g(2)=0, etc.


No matter how much I try, I can't do it. Can anyone write their own easy instructions so i have a hope of getting it.

I've never tried it, but it is surely "the sign of a true nerd".

I can only pass it from the thumb to the ring finger and back again via the index and middle fingers. I wish I could do more.

I was determined to learn how to spin pencils in fifth grade. I tried it in school and near hit one of my friends in the face!

my friends instructions fir the sonic:
do a basic pen twirl and then push down your middle finger
of course i cant do it. but you should probably be able to do it

[quote="mz94"]well, all i can say is keep practicing. it took me a year to get the hang of one trick (Sonic). but then its come easier like i can do stuff really complex and really fast. i do it when im bored. like i did it for 2 hours straight at one point. and i practice in math class :P[/quote]

Heh, I knew you'd comment on this thread, mz xD

[youtube]GdVVt_jXJvM[/youtube]
yeah, i still can't get it either.


X plays the following game.
He generates a sequence where he initially puts down (deliberately) the values 1 and 4 as the 1st 2 terms of the sequence.
The rules of the game are :
1. Whenever you have a perfect square which is deliberately put down by you as the last term of a sequence, then the next term of the sequence is got by adding it and the previous term of the sequence.
2. If the last term of the sequence is a non perfect square
    Case 1 : not divisible by 3
    Put deliberately the next perfect square of the sequence (next to the last perfect square in the      sequence) and follow step1.
    Case 2 : divisible by 3
    Subtract 5 to get the next term of the sequence
3. If the last term of the sequence is a perfect square got by 2 (i.e. not deliberately added) then add to the sequence the next perfect square and follow step 1.
For Example the sequence proceeds as :
1,4
1,4,5
1,4,5,9
1,4,5,9,14
1,4,5,9,14,16
1,4,5,9,14,16,30
1,4,5,9,14,16,30,25
1,4,5,9,14,16,30,25,36
1,4,5,9,14,16,30,25,36,61
1,4,5,9,14,16,30,25,36,61,49
1,4,5,9,14,16,30,25,36,61,49,110
1,4,5,9,14,16,30,25,36,61,49,110,64
1,4,5,9,14,16,30,25,36,61,49,110,64,174
1,4,5,9,14,16,30,25,36,61,49,110,64,174,169
...............................................................

The question is : While playing such a game, how many perfect squares does X miss till 1,00,00,000. :P  :D

Well no one with the answer yet ?????  :(
Hint
[hide]
Follow the pattern by generating the sequence.
For example you will find that $ 9^2 , 10^2 , 11^2 \& 12^2$ are missing from the sequence first.
Try to generate a pattern for no. of nos missing at a go each time and hence try to deduce the pattern
[/hide]


Find all functions $ f$ such that
It is a function of $ R$ to $ R$.
For all $ x$:
$ f(x) \plus{} x f(x\minus{}1)\equal{}x^2\minus{}1$

Note that the function is fully determined once it is defined on a segment of length 1.

Well, let $ g$ be any function from $ [ \minus{} 1,0]$ to $ \mathbb{R}$ such that $ g(0) \equal{} \minus{} 1$. Then extends the function to the entire $ \mathbb{R}$ to get $ f$:

$ f(x) \equal{} g(x) \forall x \in [ \minus{} 1,0]$

$ f(x \plus{} 1) \equal{} (x \plus{} 1)^2 \minus{} 1 \minus{} (x \plus{} 1)f(x)$

$ f(x \minus{} 1) \equal{} \frac {x^2 \minus{} 1 \minus{} f(x)}{x} \forall x \not \equal{} 0$.

Remark: extending the function to positive infinity yields no problem while extends backward require $ x \not \equal{} 0$. So $ \minus{} 1$ must be included in the definition of $ g$.


[quote="rd5493"][quote="funcia"]The product of four consecutive integers is 17160 what is their sum?[/quote]

is it
[hide]46[/hide][/quote]
dude yeah.

semi-brute force, u know all greater than 10, and at least one multiple of five and one of eleven and only one mult. of three

You can find the prime factorization of  17160, and group it into consecutive numbers.

brute-forcing is quicker. =)
prime factorizing a 5-digit number is no joke.

[quote="funcia"]The product of four consecutive integers is 17160 what is their sum?[/quote]
[hide="solve via factoring"]
Factoring shouldn't be hard.  At the National level, halving and doubling should be (almost) instant operations in your head.
Four consecutive numbers means at least 3 multiples of 2.  (Why?)
Four consecutive numbers means at least one multiple of 3.  (Why?)
Ending with a zero means at least one multiple of 5.
To factor:  Divide by 10 (to 1716).  Halve (to 858).  Halve (to 429).  Divide by 3 (to 143).
143 is still way too big to be one of the consecutive integers.  So it must be the product of at least two factors.
We divided out $120$.  We were left with $143=11\times 13$.  So $17160=10\times 12\times 11\times 13,$  The sum is $\boxed{46}$.
[/hide]
[hide="solve via roots"]
I'm thinking the average of the four numbers will be near the fourth root of 17160.  (Why?)  So, we want $x^{4}\approx 17160.$
$10^{4}= 10000$
$11^{4}= 14641.$  Compare $11^{n}$ with Pascal's triangle.
$12^{4}> 19000.$  Too hard to calculate, exactly, but I know $12^{4}$ will be at least 4641 more than $11^{4}.$  (Why?)

Therefore $11<x<12.$

Without actually solving the problem, the numbers are probably $10, 11, 12, 13$.  Their sum is $\boxed{46}$.  Check your work if time permits.[/hide]

i just looked at the number and knew that it had to be divisible by a power of 10.  $17610/10=1716$.  Then, i figured out that its divisible by 11.  $1716/11=156$.  From here its easy to see that $12*13=156$.  So the numbers are $10+1+12+13=46$.

EDIT: @rcv $12^{4}=144^{2}=20736$ i think, but i did that in my head, so dont trust me...


modular arithmetic
quadratics
LaTeX
algorithm
algebra
polynomial
induction

hello guys we lack math in the threads here

except for Aryth's random inequalities which are either unfortunately easy or wrong :wink:

Here is a random problem that I found. It reminds me of an old USAMO problem... (I need to solve it though >.< it  is combo)
(Label the problem numbers in big bold!)

[size=150][b]#1.[/b][/size]
Each grid point of a cartesian plane is colored with one of three colors, whereby all three colors are used. Show that one can always find a right-angled triangle, whose three vertices have pairwise different colors.

hello I have solved it

[hide]
Consider triangle $ ABC$ with $ A$ red, $ B$ green, $ C$ blue. Assume for contradiction that such a triangle does not exist. Let $ D$ be the foot of the altitude from $ A$. Then $ D$ must be red, because otherwise either $ ABD$ or $ ACD$ is a 3-colored right triangle. Now, let $ E$ be the intersection of the perpendiculars to $ AB$, $ BC$ at $ A,C$ respectively. $ E$ must be green, because otherwise, either $ ECB$ or $ EAB$ is a 3-colored right triangle. But then, $ ECD$ is a 3-colored right triangle - contradiction.
[/hide]

any other solutions? (don't read this if you haven't solved it unless you've spent like an hour on it... it's pretty easy)

Sorry if this is obvious, but how do we know $ E$ is a grid point?

oh man I think maybe I should've read the problem

but grid points should be easy too

hmm

blah probably pigeonhole... let's see

[hide]
ok consider two points of different color on the same vertical or horizontal line (WLOG horizontal). They must exist because otherwise the whole lattice is one color, contradiction.

Assume they are blue and green (and the third color is red)

Assume the blue point is $ (b,k)$ and the green is $ (g,k)$; WLOG $ g>b$.

So no points on $ x\equal{}b$ or $ x\equal{}g$ are red.

Case 1. All points on $ x\equal{}b$ are blue, and all on $ x\equal{}g$ are green.

Let the red point $ R$ be at $ (r,0)$. (Adjust the x-axis so that $ R$ is on it.)

Subcase 1. $ r>g>b$ or $ g>b>r$; they are both the same, pretty much. Just reflect one over the line $ x\equal{}(b\plus{}g)/2$ to get the other.

So WLOG $ r>g>b$. Pick point $ G'\equal{}(g,(g\minus{}b)(r\minus{}g))$ and point $ B'\equal{}(b,(g\minus{}b)(r\minus{}g)\minus{}1)$. Then $ RG' \perp G'B'$ and we're done.

Subcase 2. $ g>r>b$.

Let point $ B'\equal{}(b,(g\minus{}r)(r\minus{}b))$ and $ G'\equal{}(g,(g\minus{}r)(r\minus{}b)\minus{}1)$. Then $ RG' \perp RB'$ and we're done.

Case 2. There is a green point on $ x\equal{}b$.

Again, shift the plane so that $ R\equal{}(r,0)$. But then no matter what color $ (b,0)$ is (as it must be blue or green, from above), we have a tricolor right triangle.

[/hide]

define grid point? Is it like the integer places on the cartesion plane? Or is it isometric? Or whatever.

[/hide]

isometric?

idk, I think it's lattice point

that's what it usually means

hello guys

more problems

DO THE IMO PROBLEMS

in particular I have reduced problem 2 to

[hide]
\[ \sum{x^2} \plus{} 9 \plus{} \sum{xy} \minus{} 6\sum{x} \ge 0\]

where $ x,y,z \in \mathbb{R}$, $ x,y,z \neq 1$, $ xyz\equal{}1$.

yay for bashing... but now I am like darn SOS fails because I suck at putting stuff as sum of squares :(
[/hide]

ok instead
[size=150]
#2
[/size]
A "strip" is the region between two parallel lines. Let the length of a strip be the distance between the two parallel lines. A unit circle is covered with some strips. Prove that the sum of the lengths of the strips is at least 2.

(btw #1 was from Russia 04)

EDIT: oh man I also finally solved IMO #2 part a

[hide]That inequality in my earlier post is $ (x \plus{} y \plus{} z \minus{} 3)^2 \ge 0$  :blush:

note that x,y,z are the reciprocals of the variables in the problem[/hide]

Here is my solution to the lattice points/right triangle problem (I didn't read your solution before writing this)

[hide]
I was stuck on this problem for awhile, considering the possibility in which all columns are monocolored.  It seemed impossible, and then I was like,  :idea:  :!:  :!:  :!:   :)

You don't need the legs of the right triangle to have slope 0 and undefined- they just need to be perpendicular to each other- any negative reciprocal pair of slopes.

Here's my casework-bash:

Let the three colors be blue, red, and green.
[u]Case I: All the columns of lattice points are monocolored.[/u]
In this case, select a point on one of the columns.  WLOG, it is blue.  Name this point B.  We then draw the lines through B with slope 1 and -1.  Since all colors occur and all columns are monocolored, these lines will eventually pass through a red column and a green column.  Select a red point and a green point in this manner, naming them R and G, respectively.  They can be chosen so as to not make the three points collinear since the pair of lines will intersect every column of lattice points (except B's) twice.  Since -1 is the negative reciprocal of 1, we have angle RBG = 90 degrees, so triangle RBG is right and contains all three columns, Case I is solved.

[u]Case II: There is at least one multicolored column of lattice points.[/u]
In this case, select two points of different color on this column.  WLOG, they are blue and red.  We then draw the horizontal lines through each of these points.  If either of these lines passes through a green point, the problem is solved.  Clearly, there must exist at least one green point somewhere.

[u]Case IIa: There is a green point not located in this column, and no green point located in this column.[/u]
Then, draw the horizontal line through this green point, G.  It will intersect the chosen multicolored column in a point that is red or blue.  WLOG, it is blue.  Name this point B.  Since the chosen multicolored column has no green points, it must have a red point in it somewhere.  Choose this red point, R, as the third vertex of the right triangle.

[u]Case IIb: There is a green point located in this column, and it is the only green point.[/u]
Then, all other columns are made of red and blue points.  If there is another multicolored column (necessarily just red and blue), proceed as in Case IIa.  If this multicolored column with the green point in it is the only multicolored column, all the other columns are all red or all blue.  Select the green point, and proceed as in Case I with the intersection of the perprendicular lines being the green point.

[u]Case IIc: There is a green point in this column, but it is not the only one.[/u]
Name this green point G and draw the horizontal line through it.  Since this column is not monocolored, it must have at least one point of another color- WLOG, it's blue- name this point B.  Draw the horizontal line through B.  [i]If either of these two lines intersects a red point, we are done.[/i]  Otherwise, assume these two lines are made of only blue and green points.  Since there must be at least one red point, select it and name it R.  Draw the vertical line through it- it will intersect the two previously mentioned lines in two places.  These intersection points are necessarily blue or green.  If the intersection with the line through B is green, the problem is solved, and it is also solved if the intersection with the line through G is blue.  Otherwise, assume that the intersection of the vertical line through R with the horizontal line through B is blue, and that the intersection with the horizontal line through G is green.  Then, if the line through B contains a green point, the problem is solved.  If the line through G contains a blue point the problem is solved.  Otherwise, assume that the line through B is all blue and that the line through G is all green (see italicized text, this was a while back).  Then, draw the lines with slope 1 and -1 through R.  They will intersect the two parallel lines of all green and all blue points in a two green points and two blue points (one blue, one green on each side).  Then, choose a blue or green point from one of the two intersection points on the left.  Choose the third point of the right triangle of the third color from one of the two intersection points on the right.  Then, we are done.  [b]Quod Erat Casework-Bashed.[/b]
[/hide]

Ok I'm assuming no overlaps

[hide]
(all of)The lines must be parallel. To completely cover the circle, the diameter perpendicular to the lines=the minimum distance (lines may be outside circle). Yay

[/hide]

Ok that was bad.

Any solutions? I was given a hint like...

[hide]
start from the center, and whenever you enter a new region (where regions are defined by the lines), you travel perpendicularly to one of the sides of the strips you haven't been in before. You try to make sure that you never make turns that are > 90 degrees. You try to reach the edge of the circle. Of course, you could do this in at least 2 ways depending on which direction you start out in (take 1 strip containing and look at the paths in both directions.) So the distance you go is at least 2 by triangle inequality (or something) and, as you are somehow avoiding going over the same strip twice, the distance is <= the total length of all strips.

Note this is a hint/solution sketch thingy, and is not rigorous... at all.
[/hide]

Hello. No one seems to like doing math. Hence, I will give another problem.

USA TST #4 08

Show that for all integers $ n$, $ n^7\plus{}7$ is not a square.

Unfortunately, I suck at bounding arguments. Also the odd power kind of sucks...

[hide="3"]Given $ n^7 \plus{} 7 \equal{} k^2$, we see $ n^7 \plus{} 128 \equal{} k^2 \plus{} 121$. Note that $ n\equiv 1 \pmod{4}$, and n+2 divides $ k^2 \plus{} 121$, so since this is a sum of two squares and $ n \plus{} 2\equiv 3\pmod{4}$, $ \frac {k^2 \plus{} 121}{n \plus{} 2}$ must not be relatively prime to $ n \plus{} 2$ (for then a 3 mod 4 prime would come up an odd number of times in a sum of two squares). So we want ($ n^6 \minus{} 2n^5 \plus{} 4n^4 \minus{} 8n^3 \plus{} 16n^2 \minus{} 32n \plus{} 64$, $ n \plus{} 2$)$ \neq 1$. But $ n^6 \minus{} 2n^5 \plus{} 4n^4 \minus{} 8n^3 \plus{} 16n^2 \minus{} 32n \plus{} 64 \equal{} ($blah$ )(n \plus{} 2) \plus{} 448$. Hence n+2 is divisible by 2 or 7. Since $ n\equiv 1\pmod{4}$, we conclude that $ n\equiv \minus{} 2\pmod{7}$. However, then $ n^7 \plus{} 7\equiv 5\pmod{7}$ is not a quadratic residue. Contradiction.[/hide]

Okay, this is only for people who know basic propositional logic, but it's still a decent problem.

Let $ P$ be a nonempty set of propositions, and let $ p \in P$.  Given the four standard binary operators (conjunction, disjunction, implication, and the biconditional), but not the unary operator (negation), show that we cannot construct some formula $ N(P)$ which has a truth table equivalent to that of $ \sim p$.

Furthermore, show that given negation and exactly one binary operator other than the biconditional, we can construct all of the remaining three operators.

[hide]If you mean proposition in the sense of formal variable, this is easy. Make all the propositions true. Then your formula can't help but be true. Contradiction.

For the second one, let's show and->or, or->if, if->and, if->iff. Then we will be done. In the interest of avoiding latex, I will use ~, &, |, >, = for not, and, or, implies.
(a|b)=(~((~a)&(~b)))
(a>b)=((~a)|b)
(a&b)=(~(a>(~b)))
(a=b)=((~((a>b)>(~(b>a)))))[/hide]

Correct as usual, and yes, I do mean in the sense of formal variable.

A problem that I have almost no idea how to even start on, but was wondering about: Given propositions $ p_i\; (1 \le i \le n)$, it's obvious that the truth table for any formula over the propositions $ p_i$ has exactly $ 2^n$ rows.  Let $ f(k)$ be the minimum number of binary operators such that, for any subset $ S$ of the rows of the truth table over the $ p_i$ containing exactly $ k$ rows, there exists an formula $ E$ containing at most $ f(k)$ binary operators (and as many negations as necessary) which is true exactly for the rows in $ S$ but for no other rows.  Compute $ f(k)$ in terms of $ k$ and $ n$.

hmm guys we need more porblems.

let's start with a fairly easy geo (may require knowledge though):

Let $ \Omega$ be the circumcircle of $ ABC$. Let $ \Omega_A$ be internally tangent to $ \Omega$ at $ D$, and tangent to $ AB, AC$ at $ E, F$ respectively. Show that $ I$ is the midpoint of $ EF$.

generalization (one of many)

Let $ \omega_a$ and $ \omega'_a$ be two circles that are both internally tangent to $ \Omega$ at $ D$; the former is tangent to $ AB$ at $ E$, and the latter is tangent to $ AC$ at $ F$. Show that $ I$ lies on $ EF$.

Darn somehow before you posted this problem here I couldn't do it, then after you posted it, I was like whoa what this is not that bad.
Since I'm not from NC I'll just write almost all of the solution:
[hide]
Consider any chord AB of a circle O and draw the point bisecting the shorter arc AB, and name it X. Now pick any point on the longer arc AB and name it K. Let the intersection of XK and AB be E. Let O_1 be the point on OK such that O_1E is perpendicular to AB. We see that XOK~EO_1K, and so the unique circle O_1 tangent to both O and AB is tangent at K and E, respectively.
[/hide]

Man you guys need some serious assistance from California in getting math in your forum :O
PROBLEM:
Find all sets of integers such that for any 2 members of the set, $ i,j$, $ \frac{i\plus{}j}{gcd(i,j)}$ is also in the set.
(stolen from a certain contest that stole it from a certain other contest)

[quote="1=2"]You guys have a twisted sense of what's AIME/USAMO level.

Besides no negative numbers are in that sequence. And 0 is most definitely NOT in there.

[hide="5"]We need to prove that every POSITIVE rational number $ \frac {p}{q}$ with $ GCD(p,q) \equal{} 1$ can be written as
\[ \frac {1}{\frac {1}{\frac {1}{\cdots \plus{} x_1} \plus{} x_2} \plus{} x_3} \plus{} x_4
\]
Where $ x_i$ are non-negative integers.
\[ \frac {1}{\frac {1}{\cdots \plus{} x_1} \plus{} x_2} \plus{} x_3 \equal{} \frac {q}{p \minus{} qx_4}
\]
And the process repeats. This is essentially the Euclidean Algorithm, and given $ GCD(p,q) \equal{} 1$, the numerator of the fraction will become 1 and we will have the reciprocal of an integer on our hands, and the reciprocal of all integers is in the sequence, hence all rational numbers are in the sequence.[/hide]

Yeah it's not that easy to understand.
Since this is Olympiad level marathon, I shall post an Olympiad level question.

Solve the system of equations given that $ a_k\in R$ for $ 1\leq k\leq n$ and $ k,n\in N$:

$ a_1 \equal{} \frac {2a_2}{1 \minus{} a_2^2}$

$ a_2 \equal{} \frac {2a_3}{1 \minus{} a_3^2}$
...
$ a_k \equal{} \frac {2a_{k \plus{} 1}}{1 \minus{} a_{k \plus{} 1}^2}$
...
$ a_n \equal{} \frac {2a_1}{1 \minus{} a_1^2}$

I made this easy enough for you guys to get in a fairly short amount of time, unlike some other problems.[/quote]
Dude you have a fairly twisted sense of how beastly NC people are...

tangent double-angle bash fail w00t well fail

Dude this should be easy:

Let f be a polynomial, and let g=f(f(.....f))), where f appears k times. Prove that f(x)=x has a solution iff g(x)=x has a solution.

Mabbitt you fail...

Only if is trivial. If is harder.

/me looks up the topic
uuummm.... you've been doing really hard problems.

[quote="PI-Dimension"][quote="dnkywin"][quote="1=2"]You guys have a twisted sense of what's AIME/USAMO level.

Besides no negative numbers are in that sequence. And 0 is most definitely NOT in there.

[hide="5"]We need to prove that every POSITIVE rational number $ \frac {p}{q}$ with $ GCD(p,q) \equal{} 1$ can be written as
\[ \frac {1}{\frac {1}{\frac {1}{\cdots \plus{} x_1} \plus{} x_2} \plus{} x_3} \plus{} x_4\]
Where $ x_i$ are non-negative integers.
\[ \frac {1}{\frac {1}{\cdots \plus{} x_1} \plus{} x_2} \plus{} x_3 \equal{} \frac {q}{p \minus{} qx_4}\]
And the process repeats. This is essentially the Euclidean Algorithm, and given $ GCD(p,q) \equal{} 1$, the numerator of the fraction will become 1 and we will have the reciprocal of an integer on our hands, and the reciprocal of all integers is in the sequence, hence all rational numbers are in the sequence.[/hide]

Yeah it's not that easy to understand.
Since this is Olympiad level marathon, I shall post an Olympiad level question.

Solve the system of equations given that $ a_k\in R$ for $ 1\leq k\leq n$ and $ k,n\in N$:

$ a_1 \equal{} \frac {2a_2}{1 \minus{} a_2^2}$

$ a_2 \equal{} \frac {2a_3}{1 \minus{} a_3^2}$
...
$ a_k \equal{} \frac {2a_{k \plus{} 1}}{1 \minus{} a_{k \plus{} 1}^2}$
...
$ a_n \equal{} \frac {2a_1}{1 \minus{} a_1^2}$

I made this easy enough for you guys to get in a fairly short amount of time, unlike some other problems.[/quote]
Dude you have a fairly twisted sense of how beastly NC people are...

tangent double-angle bash fail w00t well fail

Dude this should be easy:

Let f be a polynomial, and let g=f(f(.....f))), where f appears k times. Prove that f(x)=x has a solution iff g(x)=x has a solution.

Mabbitt you fail...[/quote]

Only If: Let g(a)=0. Then f(f(...f(a))...) with k-1 'f's is a solution to f(x).

If: each time you do another iteration of f() the graph moves away from the x-axis (unless its odd, where it's trivial)?[/quote]

false? can you say $ (x^2\minus{}1)^2\minus{}1$ is further than $ x^2\minus{}1$? rofl

you should actually think more before posting. though your first part is ok.

also wait what

this is not f(x) = 0, it is f(x) = x

what?

[quote="dnkywin"] Dude this should be easy:

Let f be a polynomial, and let g=f(f(.....f))), where f appears k times. Prove that f(x)=x has a solution iff g(x)=x has a solution.

Mabbitt you fail...[/quote]

[hide="solution?"]If $ f(x)\neq x$ for all $ x$, then since $ f$ is continuous, $ f(x) > x$ for all $ x$ or $ f(x) < x$ for all $ x$. Then $ g(x) > x$ for all $ x$ or $ g(x) < x$ for all $ x$.

Hopefullly I'm not smoking... [/hide]

[quote="pythag011"][quote="PI-Dimension"]I'm bored... and i can't figure out the one before so here's another one

[size=200][b]#5 [/b][/size][size=67][b](i think)[/size][/b]

Uhm

Let $ a_1 \equal{} 1$, $ a_{2n} \equal{} a_n \plus{} 1$, and $ a_{2n \plus{} 1} \equal{} \frac1{a_{2n}}$. Prove that all rational numbers are in that sequence.[/quote]

Dude this is olympiad level.

Also the if part can be fixed fairly easily? Just contradiction + wlog f(0) > 0.

Wait, actually the problem above is clearly false even if it's positive integers......

Lemma: 2>=a_2n>1 and 1 >= a_2n+1 > 0 by induction[/quote]

you have the same last name as me

also, it's positive rational numbers, which would make it correct (you can also prove they are unique if you feel like it)

[hide="PI-Dimension problem"]Notice $ a_{m\cdot 2^k} \equal{} a_m \plus{} k$, so it suffices to show the sequence contains all rationals $ \frac {q}{p}$ for $ q < p$. Use strong induction on $ p$. For $ p \equal{} 1$ this is true. Suppose for $ 1,2,...,p \minus{} 1$ this is true. Thus $ \frac {p \minus{} q}{q} \equal{} a_n$ for some $ n$, by induction and the fact $ a_{m\cdot 2^k} \equal{} a_m \plus{} k$. Then $ a_{2n \plus{} 1} \equal{} \frac {q}{p}$. QED
[/hide]

I guess I'll post a problem:
(ISL 2000 A3) Find all pairs of functions $ f, g : \mathbb R \to \mathbb R$ such that $ f \left( x \plus{} g(y) \right) \equal{} x \cdot f(y) \minus{} y \cdot f(x) \plus{} g(x)$ for all real $ x, y$.

[quote="serialk11r"]Man you guys need some serious assistance from California in getting math in your forum :O
PROBLEM:
Find all sets of integers such that for any 2 members of the set, $ i,j$, $ \frac {i \plus{} j}{gcd(i,j)}$ is also in the set.
(stolen from a certain contest that stole it from a certain other contest)[/quote]

For this solution, we take the greatest common divisor of any two nonzero integers to always be positive.  We further adopt the conventions $ \gcd{(0,k)} \equal{} |k|$ for all nonzero integers $ k$ and $ \gcd{(0,0)} \equal{} 1$, to avoid the degenerate expression $ 0/0$.

[hide="solution"]
Suppose that $ S$ is such a set.  We have of course the solution $ S \equal{} \emptyset$.  If $ S$ has exactly one element $ i$, then we must have $ \frac{2i}{\gcd{(i,i)}} \equal{} i$.  We have $ i \equal{} 0$ as a solution; otherwise we must have $ 2 \equal{} \gcd{(i,i)}$.  But by our convention $ \gcd{(i,i)} \equal{} |i|$ for $ i \neq 0$, hence we have the solutions $ i \equal{} \pm 2$.  Thus we also have the sets $ S \equal{} \{\minus{}2\}, S \equal{} \{0\}, S \equal{} \{2\}$ as solutions.

For convenience, denote $ f(i,j) \equal{} \frac{i\plus{}j}{\gcd{(i,j)}}$.  Further write $ \mathbb{Z}^\plus{}(k) \equal{} \{z \in \mathbb{Z} | z \ge k\}$ and $ \mathbb{Z}^\minus{}(k) \equal{} \{z \in \mathbb{Z} | z \le k\}$.

Now suppose that $ S$ contains two elements $ i, j$ with $ i \neq j$.  We have three cases.  Case 1 is that one of $ i,j$ is equal to zero.  Case 2 is that $ i,j$ are both nonzero and have the same sign. Case 3 is that $ i, j$ have opposite signs.

Case 1: Without loss of generality let $ j \equal{} 0$.  In this case we have $ f(i,0) \equal{} \frac{i}{|i|} \in S$.  We divide into two subcases; either $ i > 0$ or $ i < 0$.

Subcase 1: $ i > 0$.  Then $ 1 \in S$.  It immediately follows that $ S$ contains all non-negative integers.  If $ S$ contains some negative integer $ m$, then we have $ f(m,1) \equal{} m\plus{}1$ in $ S$.  Hence $ \minus{}1 \in S$, so all negative integers are in $ S$.  Thus we have either $ S \equal{} \mathbb{Z}$ or $ S \equal{} \mathbb{Z}^\plus{}(0)$.

Subcase 2: $ i < 0$.  In this case $ \minus{}1 \in S$ and we proceed as above to obtain the solutions $ S \equal{} \mathbb{Z}$ and $ S \equal{} \mathbb{Z}^\minus{}(0)$.

Case 2: Without loss of generality suppose that $ i, j > 0$, and that $ i < j$.  Then we have $ f(i,i) \equal{} 2 \in S$.  We note that if $ i \equal{} 1$, then all positive integers are in $ S$.  If $ i > 1$, then there must be some $ m \in S$ with $ m$ the least positive integer in $ S$ such that $ m > 2$.  We claim that $ m \equal{} 3$.  Suppose for a contradiction that this is not the case.  If $ m$ is odd, then we have $ m\plus{}2, m\plus{}4, \cdots, m\plus{}2m \in S$.  Hence we have $ f(m,3m) \equal{} 4 \in S$.  Then $ f(2,4) \equal{} 3 \in S$.  Thus $ m \equal{} 3$.  Now if $ m$ is even, then we have $ f(m,2) \equal{} \frac{m\plus{}2}{2} \in S$.  As $ m > 3$, we have $ \frac{m\plus{}2}{2} > 2$.  But we must have $ \frac{m\plus{}2}{2} \ge m$, hence $ m \le 2$.  Contradiction.  It follows that $ m \equal{} 3$.  Hence all odd numbers at least three are in $ S$.  But then for any $ k \ge 2$, we have $ 2k\minus{}1, 2k\plus{}1 \in S$ and $ \gcd{(2k\minus{}1,2k\plus{}1)} \equal{} \gcd{(2,2k\minus{}1)} \equal{} 1$.  Hence $ f(2k\minus{}1,2k\plus{}1) \equal{} 4k \in S$.  Similarly, we have $ 2k\minus{}1, 2k\plus{}3 \in S$ and $ \gcd{(2k\minus{}1,2k\plus{}3)} \equal{} \gcd{(4,2k\minus{}1)} \equal{} 1$.  Hence $ f(2k\minus{}1,2k\plus{}3) \equal{} 4k\plus{}2 \in S$.  Hence $ S$ contains all even numbers greater than or equal to 8.  Thus $ S$ contains all positive integers greater than or equal to 2, except possibly 6.  But we have $ f(2,10) \equal{} 6$.  Hence $ S$ contains all positive integers greater than or equal to 2.

We see that if $ S$ contains any negative integer $ m \le \minus{}2$, then $ f(m,\minus{}m) \equal{} 0 \in S$.  Hence $ f(m,0) \equal{} \minus{}1 \in S$ and $ f(\minus{}m,0) \equal{} 1\in S$.  Hence $ S$ is the set of all integers.  Similarly, if $ \minus{}1 \in S$, then we have $ f(2,\minus{}1) \equal{} 1 \in S$.  Hence $ f(\minus{}1,1) \in S$, so $ S$ is the set of all integers.

Now suppose that $ S$ contains no negative integers.  Then we need only analyze which of $ \{0,1\}$ are in $ S$.  We see that if $ 0 \in S$, then we have $ f(2,0) \equal{} 1 \in S$.  However, if $ 1 \in S$, then it is not necessarily the case that $ 0 \in S$.  Finally, it is possible for neither of 0,1 to be in $ S$.

Note that if $ i,j < 0$, then we have the same solutions, except that all signs are reversed.  Hence we have the solutions $ S \equal{} \mathbb{Z}, \mathbb{Z}^\plus{}(0), \mathbb{Z}^\plus{}(1), \mathbb{Z}^\plus{}(2), \mathbb{Z}^\minus{}(0), \mathbb{Z}^\minus{}(\minus{}1), \mathbb{Z}^\minus{}(\minus{}2)$.

Case 3: Without loss of generality, suppose that $ i < 0 < j$.  Then we have $ |f(i,j)| \le |i\plus{}j| < |i|,|j|$, and $ f(i,j) \in S$.  Then either $ f(i,j) \equal{} 0$, or one of the two sets $ \{f(i,j),i\}, \{f(i,j),j\}$ contains two elements with opposite signs.  In the latter case, we repeat our above step on this pair, obtaining either 0 or yet another pair with opposite signs and decreasing minimal absolute value.  This process must terminate, giving $ 0 \in S$.  Hence $ f(i,0) \equal{} \minus{}1 \in S$ and $ f(j,0) \equal{} 1 \in S$.  It follows that $ S \equal{} \mathbb{Z}$.

After all of our analysis, we find exactly eleven such sets.  Namely, we may have $ S \equal{} \emptyset, \{\minus{}2\}, \{0\}, \{2\}, \mathbb{Z}^\plus{}(0), \mathbb{Z}^\plus{}(1), \mathbb{Z}^\plus{}(2), \mathbb{Z}^\minus{}(0), \mathbb{Z}^\minus{}(\minus{}1), \mathbb{Z}^\minus{}(\minus{}2), \mathbb{Z}$.

[/hide]

hmm sup guys

[b]#7[/b]?

100 ambassadors sit around a round table; there are 4 ambassadors from each of 25 countries present. Show that there exists a partitioning of the 100 ambassadors into 4 groups so that: a) no neighbors at the table are in the same group, and b) no countrymen are in the same group.

also 7b (supposed to be easier, I think?):
Now there are only 2 ambassadors from each of 50 countries. Show that there exists a partition of the 100 ambassadors into 2 groups so that: a) no three consecutive people at the table are in the same group, and b) no countrymen are in the same group.

oops timwu has a problem; I'll try to do it...

hmm hello guys

the IMO problems don't seem too bad (1, 2, 4 anyways)

do them (unless you're Calvin and have already done them)


cot(x+y) = (cotx coty - 1)/(cot x + cot y)

can someone help me solve this, I have a start but can't quite get it.

make tg(x+y)=(tgx+tgy)/(1-tgxtgy)

you know cotgx=1/tgx
so you have:
1/(tg(x+y))=  ( 1/(tgx*tgy)  -1 )/( 1/tgx+ 1/tgy)=

( (1-tgx*tgy)/tgx*tgy )/( (tgy +tgx)/(tgx*tgy))=

(1-tgx*tgy)/(tgx+tgy)    <=>tg(x+y)=(tgx+tgy)/(1-tgx*tgy)

what is a true


I was just doing past problems of the day cause I am a huge nerd, and I didn't get how you would graph #4, so if anybody could help me there. The problem is:

Find the area in the plane contained by the graph of $ |x\plus{}y|\plus{}|x\minus{}y|\le4$.

[hide]The figure enclosed is a square with vertices at $ (\pm2,0),\ (0,\pm2)$. So it’s a square with sides of length 4; hence its area is 16.

The way to sketch the graph is as follows. First, draw the lines $ x \plus{} y \equal{} 0$ and $ x \minus{} y \equal{} 0$. These lines divide the Cartesian plane into four regions: a top region (N), a right region (E), a bottom region (S), and a left region (W). In each region, $ |x \plus{} y|$ and $ |x \minus{} y|$ are separately positive or negative. For example:

In Region N, both $ |x \plus{} y|$ and $ |x \minus{} y|$ are positive, so $ |x \plus{} y| \equal{} x \plus{} y$ and $ |x \minus{} y| \equal{} x \minus{} y$. Hence, in this region, $ |x \plus{} y| \plus{} |x \minus{} y|\leq4$ $ \Rightarrow$ $ x \plus{} y \plus{} x \minus{} y\leq4$ $ \Rightarrow$ $ x\leq2$.

In Region E, $ |x \plus{} y|$ is positive but $ |x \minus{} y|$ is negative, so $ |x \plus{} y| \equal{} x \plus{} y$ and $ |x \minus{} y| \equal{} \minus{} (x \minus{} y)$. Hence, in this region, $ |x \plus{} y| \plus{} |x \minus{} y|\leq4$ $ \Rightarrow$ $ x \plus{} y \minus{} (x \minus{} y)\leq4$ $ \Rightarrow$ $ y\leq2$.

Now do the same thing for the other two regions. You should find that in Region S, the graph is $ x\geq \minus{} 2$ and in Region W, $ y\geq \minus{} 2$. Putting everything together, we have the figure being defined by $ \minus{} 2\leq x\leq2$, $ \minus{} 2\leq y\leq2$.[/hide]

Sorry but how would we use that to graph it?

Yea you would have to graph it and then restrict the x and y values to get a region. It would just be the lines x=2, x=-2, y=2, y=-2.

[hide="hint for another approach"] Square both sides[/hide]


[quote="Victory.US"]Let $ a,b,c > 0$ such that:$ a \plus{} b \plus{} c \equal{} 1$. Show that

$ \frac {{{a^2} \plus{} {b^2}c}}{{b \plus{} c}} \plus{} \frac {{{b^2} \plus{} {c^2}a}}{{c \plus{} a}} \plus{} \frac {{{c^2} \plus{} {a^2}b}}{{a \plus{} b}} \ge \frac {2}{3}$[/quote]

substitution  $ a\equal{}\frac{x}{x\plus{}y\plus{}z},b\equal{}\frac{y}{x\plus{}y\plus{}z},c\equal{}\frac{z}{x\plus{}y\plus{}z},x,y,z>0$

the inequality is equivalent to
$ \minus{}8x^2z^2y\minus{}8xy^2z^2\minus{}8x^2y^2z\plus{}4y^4z\minus{}3y^3x^2\plus{}4x^4z\minus{}3x^3z^2\plus{}4x^4y\plus{}3x^5\plus{}4z^4x\plus{}4z^4y\minus{}3z^3y^2\plus{}4y^4x\plus{}3y^5\plus{}3z^5\geq 0$

assum $ x\equal{}max{(x,y,z)}$

$ \minus{}8x^2z^2y\minus{}8xy^2z^2\minus{}8x^2y^2z\plus{}4y^4z\minus{}3y^3x^2\plus{}4x^4z\minus{}3x^3z^2\plus{}4x^4y\plus{}3x^5\plus{}4z^4x\plus{}4z^4y\minus{}3z^3y^2\plus{}4y^4x\plus{}3y^5\plus{}3z^5$

$ \equal{}(3x^3\plus{}(7z\plus{}7y)x^2\plus{}(4z^2\plus{}11yz\plus{}7y^2)x\plus{}3y^2z\plus{}4y^3\plus{}4z^3)(x\minus{}y)(x\minus{}z)\plus{}(8(y\plus{}z)^2x\plus{}3(y^2\plus{}yz\plus{}z^2)(y\plus{}z))(y\minus{}z)^2\geq 0$

so.........

since $ a \plus{} b \plus{} c \equal{} 1$,$ a,b,c\le1$,
By AM-GM we have $ \frac {a}{2} \ge \sqrt {a(1 \minus{} a)}$
or $ \frac {a^2}{4} \ge a(1 \minus{} a)$
which re-arranges to -
$ \frac {a^2}{1 \minus{} a} \ge 4a$
or ,
$ \frac {a^2}{b \plus{} c} \ge 4a$
and if we do similarly for the others then we get,
$ LHS \ge \sum_{cyc}\frac {b^2c}{b \plus{} c} \plus{} 4(a \plus{} b \plus{} c) \equal{} \sum_{cyc}\frac {b^2c}{b \plus{} c} \plus{} 4 > \frac {2}{3}$!!! :o

[quote="Victory.US"]Let $ a,b,c > 0$ such that:$ a \plus{} b \plus{} c \equal{} 1$. Show that

$ \frac {{{a^2} \plus{} {b^2}c}}{{b \plus{} c}} \plus{} \frac {{{b^2} \plus{} {c^2}a}}{{c \plus{} a}} \plus{} \frac {{{c^2} \plus{} {a^2}b}}{{a \plus{} b}} \ge \frac {2}{3}$[/quote]
[hide="My Solution"]
Rewrite the inequality as \[ \sum \frac{a^2}{b\plus{}c}\plus{}\sum\frac{b^2c}{b\plus{}c}\geq \frac 23\]
[b]Claim[/b]
It is very well - known that $ \sum \frac{a^2\plus{}bc}{b\plus{}c}\geq a\plus{}b\plus{}c \forall a,b,c>0$
[b]Proof[/b]
$ \sum \frac {a^2}{b\plus{}c}\plus{}\sum \frac{bc}{b\plus{}c}\geq \sum \frac{b^2}{b\plus{}c}\plus{}\sum \frac{bc}{b\plus{}c}$
$ \equal{}a\plus{}b\plus{}c$ (from rearrangement inequality). Hence our claim is proved.

Using this, $ \sum \frac {a^2}{b\plus{}c}\geq 1\minus{}\sum \frac{bc}{b\plus{}c}$ (from our claim and $ a\plus{}b\plus{}c\equal{}1$).
So the inequality we are required to prove equivalents:
\[ 1\minus{}\sum \frac{bc}{b\plus{}c}(b\minus{}1)\geq \frac 23 \implies \sum \frac {bc(c\plus{}a)}{b\plus{}c}\leq \frac 13\]
Observe that from the trivial inequality, $ 4bc\leq (b\plus{}c)^2$. Hence we are only required to prove that
\[ \frac{1}{4}\left[\sum ((b\plus{}c)(c\plus{}a))\right] \leq \frac 13\]
But we see that $ \frac{1}{4}\left[\sum ((b\plus{}c)(c\plus{}a))\right]\equal{}\frac 14 \sum [(1\minus{}a)(1\minus{}b)] \equal{}\frac 14 (3\minus{}2(a\plus{}b\plus{}c)\plus{}ab\plus{}bc\plus{}ca)$
$ \equal{}\frac 14 (1\plus{}ab\plus{}bc\plus{}ca)\leq \frac 14 \left(1\plus{}\frac 13\right)\equal{}\frac 13$
Since, from the trivial inequality we  have $ (ab\plus{}bc\plus{}ca)\leq \frac 13 (a\plus{}b\plus{}c)^2$

Hence we are done. :)

I hope my proof is correct :)
[/hide]

[quote="Victory.US"]Let $ a,b,c > 0$ such that:$ a \plus{} b \plus{} c \equal{} 1$. Show that

$ \frac {{{a^2} \plus{} {b^2}c}}{{b \plus{} c}} \plus{} \frac {{{b^2} \plus{} {c^2}a}}{{c \plus{} a}} \plus{} \frac {{{c^2} \plus{} {a^2}b}}{{a \plus{} b}} \ge \frac {2}{3}$[/quote]

$ \sum_{cyc}\left(\frac {4a^2}{b \plus{} c} \plus{} 4a \plus{} \frac {4b^2c}{b \plus{} c}\right) \equal{} \sum_{cyc}\left(\frac {4a(a \plus{} b \plus{} c)}{b \plus{} c} \plus{} 4bc \minus{} \frac {4bc^2}{b \plus{} c}\right)\ge \sum_{cyc}\left(\frac {4a}{b \plus{} c} \plus{} 4bc \minus{} (b \plus{} c)c\right) \equal{} \sum_{cyc}\left(\frac {4a}{b \plus{} c} \plus{} 3bc \minus{} c^2\right)$

and hence it's enough to show that :

$ \sum_{cyc}\frac {4a}{b \plus{} c} \plus{} 18(ab \plus{} bc \plus{} ca) \ge \frac {23}{3} \plus{} 13(ab \plus{} bc \plus{} ca)$ since $ \sum(3bc \minus{} c^2) \equal{} 5(ab \plus{} bc \plus{} ca) \minus{} 1$
or we have
$ \sum_{cyc}\left(\frac {4a}{b \plus{} c} \plus{} 9a(b \plus{} c)\right)\ge \sum 12a \equal{} 12 \equal{} \frac {23}{3} \plus{} \frac {13}{3} \ge \frac {23}{3} \plus{} 13(ab \plus{} bc \plus{} ca)$

[quote="spanferkel"][quote="geniusbliss"]since $ a \plus{} b \plus{} c \equal{} 1$,$ a,b,c\le1$,
By AM-GM we have $ \frac {a}{2} \ge \sqrt {a(1 \minus{} a)}$

 :o[/quote]

here. By AM-GM we have $ \frac {1}{2} \ge \sqrt {a(1 \minus{} a)}$  :lol:[/quote]
 :oops: that was embarrasing.. sry abt tht post i cant even delete it


algorithm
coding theory
search
combinatorics

A magician asks you to pick any five cards from a standard deck(52 cards, with 13 denominations 2,3,...J,Q,K,A, and four suits). You do so, and then show them to the assistant, who places one of the five cards back in the deck and puts the remaining four cards into a pile. The magician is blindfolded, so he does not witness any of this, then he takes off his blindfold, takes the pile of four cards, reads the four cards that the assistant has arranged, and is able to find the fifth card in the deck(even if you shuffle the deck after the assistant puts the card into the deck). Assum that neither the magician nor the assistant have supernatural powers, and that the deck of cards is not marked. How is the trick done? If you are the assistant, you pick which of the five cards to be put back into the deck?

Are the cards symetrical through a $180^{\circ}$ rotation?

Because if they are not symmetrical, the assistant could use a rotation of the $n^{th}$ card to show him what colour it is,and afterwards indicate the number of the card by ordering the suits and using a reverse binary search technique.

[quote="i_like_pie"]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=144943

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=145230

No point in posting the same thing 3 times.[/quote]

Furthermore, it has appeared on the forum before.  Do a search.

Anyone has a solution to this question? Sorry for the 13 years bump :blush:

Nice 13yr bump lol.

[hide=Solution(correct?)]I think I can exploit a loophole in this problem. (that is, it is not mentioned whether we have to keep the cards in the pile face up or face down)

Number the cards from to $1$ to $52$ from Ace of Spades to King of Clubs. Then $5$ cards are chosen and $1$ is put back in the deck, and $4$ are put in a pile. Now, do the following :

$a)$ Number the cards in the pile from $49$ to $52$ and maintain their order as in the original numbering

$b)$ Then, number the other cards from $1$ to $48$, by shifting and maintaining positions as in original numbering.

Now, define a 'Permutation Number' $P_i$ of the $4$ cards. There are $4!=24$ ways to permute, so we'll get $24$ such $P_i's$ as following:

$P_1=\{49,50,51,52\}$

$P_2=\{49,50,52,51\}$

$\vdots$

$P_4=\{52,51,50,49\}$

Now, the assistant does the following:

$i)$ If the $5th$ card (which the magician has to identify) has a number from $1$ to $24$, the assistant arranges the pile with all cards face up, and place the cards (from top) according to the permutation number. So, the magician can identify that the hidden card has the number $i$, by looking at $P_i$ and choose that card.

$ii)$ If the $5th$ card has a number from $25$ to $48$, the assistant arranges the pile with all cards face down, and place the cards (from top) according to the permutation number. So, the magician can identify that the hidden card has the number $(i+24)$, by looking at $(P_i+24)$ and choose that card.

Thus, the trick is complete.

[/hide]

I don't understand the last line of the question (and I think it isn't reqd.)


calculus
integration
trigonometry
logarithms
calculus computations

Evaluate

\[\int_{\frac{\pi}{2}}^{\frac{2\pi}{3}} \frac{1}{\sin x \sqrt{1-\cos x}}dx\]

Thank you for your reply, Jimmy.

Your answer is correct.

kunny