Ok, the kalva archive is now dead (wayack machine), I know it will probably start working again but I think I remember seeing a grey miit website that was a copy of kalva, I've been searching forever and can't find it, does anyone know where it is?

Apparently, you now need to allow scripts from archive.org when browsing the Internet Archive (in case you are using NoScript or something similar). But it works for me with scripts allowed.

  darij

http://www.mat.itu.edu.tr/gungor/IMO/www.kalva.demon.co.uk/ is all I could find

and wayback machine doesnt work for me either =o

Yes, that was the website I was talking about, thanks a lot.
EDIT: actually, I think there may be another grey site like that with more olympiads, not sure.


modular arithmetic
algebra
polynomial
greatest common divisor
number theory solved
number theory

1) Let S be a set of primes such that a,b are in S (a and b need not be distince) implies ab+4 is an element of S. Show that S must be empty.

2) Given an integer n, show that an integer can always be found which contains only the digits 0 and 1 and which is divisible by n.

3) Show that if n divides a single Fibonacci number, then it will divide infinitely many Fibonacci numbers.

Look at the first one $\pmod 3$. You'll eventually find numbers which are divisible by $3$, so they're not primes.

For the second one, we may add as many zeros at the end of the number we are looking for, so we may very well consider $(n,10)=1$. If $n=3^at,\ 3\not |\ t$, there is a $k$ s.t. $10^k=1\pmod t,\ 10^k=1\pmod 3^{a+2}$, which means that $\displaystyle n|\frac {10^k-1}9$, which is written only with $1$s. As I have already said, if $(n,10)\ne 1$, then add the apropriate number of $0$s at the end of $\displaystyle\frac {10^k-1}9$.

As for the third one, use the well-known relation $(F_m,F_n)=F_{(m,n)}$, where $(x,y)$ is the greatest common divisor of $x,y$. In particular, if $m|n$, then $F_m|F_n$, and this easily solves the problem. A quick way to prove this is to consider the fact that $\displaystyle F_n=\frac{a^n-b^n}{a-b}$, where $a>b$ are the roots of the polynomial $x^2-x-1$.


algebra
polynomial
number theory unsolved
number theory

Determine all polynomials with integer coefficients such that for every positive integer $ n$, the number $ f(n)$ divides $ 2^n\minus{}1$


The equation $ x^{4}-16x^{3}+94x^{2}+px+q = 0$ has two double roots. Find $ p+q$

[hide="Solution"]
Let the roots be A and B.
We know that we can write a polynomial out by the product of it's roots, which are
$ (x-a)(x-a)(x-b)(x-b)$
Expanding this out, we get
$ x^{4}-2bx^{3}-2ax^{3}+b^{2}x^{2}+4abx^{2}+a^{2}x^{2}-2ab^{2}x-2a^{2}b x+a^{2}b^{2}$

$ x^{4}-(2b+2a) x^{3}+(b^{2}+4ab+a^{2}) x^{2}-(2ab^{2}+2a^{2}b) x+a^{2}b^{2}$

So we have $ (2b+2a) = 16$, and $ (b^{2}+4ab+a^{2}) = 94$
The first implies that $ a+b = 8$, the second can be factored as $ (a+b)^{2}+2ab = 94$ which leads to $ 64+2ab = 94$, implying
$ ab = 15$
and we already know
$ a+b = 8$

$ P =-(2ab^{2}+2a^{2}b) =-2ab(a+b) =-30 * 8 =-240$
$ Q = a^{2}b^{2}= (ab)^{2}= 15^{2}= 225$
$ P+Q =-15?$
[/hide]

[hide="My Solution"]Since the equation has two double roots, then it can be factored as
$ (x^{2}+ax+b)^{2}$, because each root of the quadratic will be squared.  Expanding, we have
$ x^{4}+2ax^{3}+(2b+a^{2})x^{2}+2abx+b^{2}$.  Clearly, $ a=-8$, and $ b=15$.  Thus, $ p=2ab=-240$ and $ q=b^{2}=225$, so $ p+q=\boxed{-15}$.[/hide]


function
linear algebra
matrix
calculus
derivative
advanced fields

$\psi(y)=\frac1{|\det G|}\varphi(G^{-1}y).$

A density function is a derivative; this is just the chain rule.

Hello

[quote="jmerry"]$\psi(y)=\frac1{\det G}\varphi(G^{-1}y).$

A density function is a derivative; this is just the chain rule.[/quote]

Thank you jmerry. I reallize that like two or three hours after I posted the problem. I was, I think, a little tired to think. But the whole problem that I was solving I believe that is done. Two or three weeks to finish the experimental evidency and to write carefully the paper, and I hope I will  send my first article for review to a journal by middle or end of march  :D .

Best,

[quote="vuhung"][quote="djimenez"]This is just a weird idea that I just had trying to solve a problem.

Suppose that you have $X$ a random variable $\mathbb{R}^n$ whose density function is $\varphi$. Let $G\in\mathcal{M}_{n\times n}$ an invertible matrix, and define the random variable $Y$ as $Y=GX$, whose density function is $\psi$. I need to express $\psi$ in terms of $\phi$ and $G$ somehow. Any help would be appreciated.

Best regards,[/quote]
Are you working in Markov chains?[/quote]

Not exactly. I am working in a Analytic Coding Problem, but actually I already could solve it.

Best,


If $ \tan{\theta} > 1$ and   $ \sum_{n \equal{} 1}^{\infty}{\frac {1}{\tan^{n}\theta}} \equal{} \sin{\theta} \plus{} \cos{\theta}$, find $ \theta$ in the interval $ {0}\le{\theta}\le{\pi}$

$ \tan \theta > 1\ (0\leq \theta \leq \pi) \Longleftrightarrow \frac {\pi}{4} < \theta < \frac {\pi}{2}\Longrightarrow0 < \frac {1}{\tan \theta } < 1$, thus the infinite geometric series converges and has the sum:

$ \sum_{n \equal{} 1}^\infty \frac{1}{\tan ^ n \theta} \equal{} \frac {\frac{1}{\tan \theta}}{1 \minus{} \frac {1}{\tan \theta}} \equal{} \frac {\cos \theta}{\sin \theta \minus{} \cos \theta}$.

$ \therefore \sum_{n \equal{} 1}^\infty \frac {1}{\tan ^ n \theta} \equal{} \sin \theta \plus{} \cos \theta$

$ \Longleftrightarrow \cos \theta \equal{} \minus{} \cos 2\theta$

$ \Longleftrightarrow \cos \theta \equal{} 1\minus{}2\cos ^ 2 \theta$

$ \Longleftrightarrow 2\cos ^ 2 \theta \plus{} \cos \theta \minus{} 1 \equal{} 0$

$ \Longleftrightarrow (\cos \theta \plus{} 1)(2\cos \theta \minus{} 1) \equal{} 0$

$ \Longleftrightarrow \cos \theta \equal{}  \frac {1}{2},\ \minus{}1$

Since $ \frac {\pi}{4}\leq \theta < \frac {\pi}{2}$, yielding $ \theta \equal{}\frac{\pi}{3}$.

I'm sorry, the problem was incorrect as stated. The limits of summation should be $ 1$ and $ \infty$, not $ 0$ and $ \infty$; Fixed.


function
group theory
abstract algebra
set theory
superior algebra
superior algebra unsolved

[quote="Gilad.R"]the problem:
[code]let A,B groups. let f:A->B , g:B->A functions.
it is known that f◦g is reversable.

does it necceseraly mean that g◦f is reversable?
if so, what is it's reverse function?[/code]
this is from an introduction to set theory course.
what can I conclude from knowing that [size=150]f◦g[/size] is reversable?

thanks.[/quote]

Are $ A$ and $ B$ truly supposed to be groups ? And, in this case, aren't $ f$ and $ g$ supposed to be homomorphisms ?
In any case, if $ f \circ g$ is reversible, $ f$ is surjective and $ g$ is injective, but $ g \circ f$ is not necessarily reversible.
Take for $ A$ a group of order 2 and for $ B$ the subgroup of $ A$ with only the identity element. For $ f$, take the only function from $ A$into $ B$, i.e. the trivial homomorphism. For $ g$, take the only homomrphism from $ B$ to $ A$, i.e. the canonical injection.
Then $ f$ and $ g$ are homomorphisms, $ f \circ g$ is reversible (it is the identity automorphism of $ B$) and $ g \circ f$ has no inverse function (a fortiori no inverse homomrphism), since it is not injective (it applies both elements of $ A$ on the unique identity element of $ A$).

Friendly,
P.


For your first one,
$\sum_{cyc}\sqrt{x^{2}+xy+y^{2}}\leq \sum_{cyc}(x+y) =2$.


induction
algebra
polynomial
linear algebra

assume that $m,n $ are positive integer and we have $m$ ,n*n matrices $B_1,\dots,B_m$. we know that $\sum_{i=1}^{m}B_i$ is non singular prove that there exists $S\subseteq \{\ 1,...,m\}\ $ such that $|S|\leq n$ ,and $\sum _{i\in S} B_i$ is nonsingular.

I see it is in the proposed problems section, so you can delight us with your solution, samn. If someone had found something about this magnificient problem, he would have posted it. Could you please post a solution?

i remember that i tried but i fail ,i know the official solution but let others try it then i send official solution :)

beautiful problem! actually, it suffices to take $m=n+1$ (the rest can be done by induction; and when going down up from $m$ to $m+1$, you can as well take $n=m$ by putting some extra 1's on the diagonal (0's elsewhere) of our matrices). assume the statement is wrong. considering $f(t_1,...,t_{n+1})=\det(\sum t_iB_i)$, we then see that we got a degree $n$ polynomial in $n+1$ variables which vanishes for $t_i\in \{0, 1\}$ except for $t_1=...=t_{n+1}=1$. we have to show that it vanishes there at well. let $a_{i_0,...,i_k}$ be the sum of the coefficients of the monomials in $f$ which exactly include the variables $t_{i_0},...,t_{i_k}$. a straightforward induction on $k$ then shows that $a_{i_0,...,i_k}$ is 0. but since $f$ has only degree $n$, $f(1,...,1)$ includes only these sums and therefore is 0.

peter

[Edit: little typo corrected]

Great!! I think of using polynomials too but cannot make it!

welldone  :D  it  is official solution too :)


Let u be differentiable function of x. use the fact that absolute value of u = the square root of u^2. to prove that
d/dx[absolute value of x]  = u' u/absolute value of u. and u cannot equal zero.
(some 1 feel free to rewrite)

Rewritten to make it easier to read:

Let u be a differentiable function of x.  Use the fact that
[tex]

\displaystyle|u| = \sqrt{u^2}
[/tex]

to prove that

[tex]
\displaystyle\frac{d|x|}{dx} = u^{\prime} \cdot \frac{u}{|u|}.
[/tex]

Chris, where did you get these problems?

y = (x^2)^(1/2)
u = x^2

dy/du = 1/(2sqrt(u))
du/dx = 2x

But sqrt(u) = 1/2y and so dy/dx = x/y and y = abs(x).


The angles of a convex n-gon $ A_1A_2...A_n$ form an arithmetic sequence.  One of the angles is 144 degrees.  Compute the largest possible value for $ n$.

the highest angle is bounded by 180 (as an upper bound but the highest angle cannot BE 180) and the lowest angle is bounded by 144 (as un upper bound but the lowest angle can be 144)
so the sum of the largest and smallest angles or $ A_1\plus{}A_n$ is 324 so the average is at most 162
so the interior angle of a regular n-gon must be less than 162 (note it cannot equal 162)
so we have the following inequality
$ \frac{(n\minus{}2)180}{n} < 162 \implies \frac{n\minus{}2}{n} < .9 \implies 1\minus{}\frac{2}{n} <.9 \implies \frac{2}{n}<.1 \implies n<20$
so n is 19

Oh thanks, but quick question

To find the $ d$ in the arithmetic sequence, do I just take $ a_1\equal{}144$ and $ a_n\equal{}180$ and solve for $ d$?

Why doesn't and number greater than 19 work?  Does it have something to do with too many numbers in the arithmetic sequence?

[quote="Ketvoman"]To find the $ d$ in the arithmetic sequence, do I just take $ a_1 \equal{} 144$ and $ a_n \equal{} 180$ and solve for $ d$? [/quote]  Sure, but in this problem I don't think it works since $ a_n$ cannot be a straight angle.

[quote="Ketvoman"]Why doesn't and number greater than 19 work?  Does it have something to do with too many numbers in the arithmetic sequence?[/quote]
Note that the measures of the angles are in arithmetic progression, and so we can just pair up and average the largest and smallest (and 2nd largest and smallest and so on) to get a regular n-gon.  The sum of the interior angles of an n-gon is always fixed, which equals to (n-2)*180 degrees.


Let $n\ge 2$ and $f: [0;a]\rightarrow R(a>0)$
$f(x)\le 0\forall x\in [0;a]$
$\forall x_1,...,x_n\in [0;1]\longrightarrow \sum_{i=1}^nf(x_i)\ge f(\sum_{i=1}^nx_i)$
Prove that $f(x)\ge \frac{2f(a).x}{a}\forall x\in [0;a]$


limit
real analysis
real analysis unsolved

Be $ x_{n}$ the solution of $ \arctan x+x=\frac1n$. Compute the following limits:
$ \lim_{n\to \infty}x_{n}$ and $ \lim_{n\to \infty}nx_{n}$

Definitely that $ x_{n}>0$. Hence

$ 0<x_{n}<x_{n}+\arctan x_{n}=1/n$.

Thus $ x_{n}\rightarrow 0$.

Now

$ nx_{n}\cdot \frac{\arctan x_{n}}{x_{n}}+nx_{n}=1$.

Passing to the limit in the preceding equality we get that

$ L+L=1$ or $ L=\lim nx_{n}=1/2$.


modular arithmetic
number theory unsolved
number theory

The periodic decimal digits of fractions $ \frac{1}{7}$, $ \frac{1}{17}$, $ \frac{1}{27}$, $ \frac{1}{37}...$, always end in 7

Why?

Let $ n\equal{}10a\plus{}7$.

$ 10^{\phi(n)}\equiv1\pmod{n}$. So if we take $ k\equal{}\phi(n)$, then $ 999\ldots9$ (with $ k$ 9's) is divisible by $ n$. So $ \frac1n$ can be written as $ \frac{m}{999\ldots9}$ for some integer $ m<999\ldots9$. But $ m$ is exactly the period of decimal digits of $ \frac1n$.

$ m(10a\plus{}7)\equal{}999\ldots9$. $ 7\times1,7\times2,7\times3,7\times4,7\times5,7\times6,7\times8,7\times9,7\times0$ does not end with 9. So $ m$ must end with 7.

Q.E.D.


The first three terms of a geometric progression are $\sqrt{2}$, $\sqrt[3]{2}$, $\sqrt[6]{2}$.  The fourth term is:

(A) 1
(B) $\sqrt[7]{2}$
(C) $\sqrt[8]{2}$
(D) $\sqrt[9]{2}$
(E) $\sqrt[10]{2}$

[hide]To find the common ratio in a geometric progression, divide any term by the preceding term. In this case, you would divide  by . To make it easier, write them as 2^(1/2) and 2^(1/3). Now the common ratio is

Multiply numerator and denominator by 2^(1/2)

Divide numerator and denominator by 2, making the ratio

Check by dividing the third term by the second term. It works. So the next term will be

Which is 1. Answer: A.[/hide]


solve equation $ log_{3}(1\plus{}2x)\equal{}3^x\minus{}x\minus{}1$

Hello,

we have: $ log_3(1\plus{}2x)\equal{} 3^x \minus{} x \minus{}1$

Using $ log_3(1\plus{}2x)\equal{}log(1\plus{}2x)/log3$ we get

$ log(1\plus{}2x)\equal{}3^x log3 \minus{} xlog3 \minus{} log3$

$ log(6x\plus{}3)\equal{}(3^x \minus{} x)log3$

From there: $ 3^{3^{x}\minus{}x}\equal{}6x\plus{}3$

Obvious solutions are $ x\equal{}0$ and $ x\equal{}1$,
tho I cannot prove it  :blush: .


Find all continuous functions $f:R^+->R^+$ satisfying:
$(1):f(2x)=2f(x);\forall x \in R^+$;
$(2):f(f(x)^2)=xf(x); \forall x \in R^+$;
$(3): f(x) \in N^*; \forall x \in N^*$;(Where $N^*$ be set of positive integer numbers).

First of all, we can easily show that [tex]f(2^kx)=2^kf(x)[/tex], for all [tex]k \in Z[/tex] and for all [tex]x \in R^+[/tex] using [tex](1)[/tex] condition. Therefore, [tex]\lim_{x\rightarrow \infty}{f(x)}=+\infty[/tex] and [tex]\lim_{x\rightarrow 0}{f(x)}=0[/tex]. Furthermore, [tex]f[/tex] is injective, since if [tex]f(x_1)=f(x_2)=a>0[/tex], then [tex]x_ia=x_if(x_i)=f(f(x_i)^2)=f(a^2), i=1,2[/tex], so [tex]x_1a=x_2a[/tex] or [tex]x_1=x_2[/tex]. Hence, [tex]f[/tex] is strictly increasing and surjective. Let [tex]f(y)=1[/tex], so putting [tex]x=y[/tex] in [tex](2)[/tex] we get [tex]f(1)=y[/tex]. If [tex]1 < y \Rightarrow f(1)=f(y) \Rightarrow y < 1[/tex]. Contradiction. Similarly [tex]y < 1 < y[/tex] contradiction takes place. Thus, [tex]f(1)=1[/tex]. Then [tex]f(2)=2, \@ f(4)=4[/tex], and using [tex](3)[/tex] we arrive at [tex]f(3)=3[/tex]. Then [tex]f(6)=6[/tex] and [tex]f(5)=5[/tex], and so on...(simple induction). So, [tex]f(\frac{p}{2^k})=\frac{p}{2^k}[/tex], where [tex]p[/tex] is odd positive integer. Since [tex]f[/tex] is continuous, [tex]f(c)=\lim_{x\rightarrow c}f(x)=c[/tex] using sequence of form [tex]\frac{p}{2^k}[/tex]. It is left to check that [tex]f(x)=x[/tex] fulfils the conditions.


In the convex  quadrilateral $ ABCD$,the points $ G_1,G_2,G_3,G_4$ are center of gravity for next triangles $ BCD,ACD,ABD$ and $ ABC$.Prove that lines  $ AG_1,BG_2,CG_3,DG_4$ have a common point.

\bump. I am not reviving this thread! I was just trying to solve this problem and wanted to know what ideas other people had.

This can be solved with coordinate bash.
Let $A (0,0$, $B (a,0)$, $C (b,c)$, and $D (d, e)$. Note that the center of gravity is just the centroid. To compute the centroid, we find the average of the coordinates. You should be able to solve with this.


linear algebra
matrix
linear algebra unsolved

Let $A$ be an $n$ by $n$ matrix over a field of characteristic $0$ such that $\mathrm{trace}(A^i) = 0$ for all natural numbers $i$. (Here, $0$ does not count as a natural number). Show that $A$ is nilpotent.

If $\lambda_{1},\dots,\lambda_{n}$ are eigenvalues of $A$, then $\sum_{i=1}^{n}\lambda_{i}^{k}=0$ for each $k$. So $\lambda_{1}=\dots=\lambda_{n}=0$. By Cayley Hamilton $A^{n}=0$

I'm not sure that is correct over fields of nonzero caracteristic!!

Omid's solution works no longer, since in Newton's formulas we have some $k$ multiplied by $S_{k}$, so if $k$ is the caracteristic of the field, we have troubles. Think about the identity matrix with entries in the field with two elements!

Over $\mathbb{Z}_{p},$ the $p\times p$ identity matrix has all of its powers equal to itself (it is the identity matrix, after all), and its trace is zero. (Which is what harazi just said, slightly generalized.)

Oh, I supposed the field to be $\mathbb R$ (as the who's post is not correct for fields with characteristic zero). [color=red][[b]Moderator edit:[/b] The problem has been edited now to explicitly require the field to have characteristic zero.][/color] For field $\mathbb F$ with charasteristic $p \neq 0$ my solution is not correct. Although when $n<p$ my solution works and we can prove that $\lambda_{1}=\lambda_{2}=\dots=\lambda_{n}=0$, when $n>p$ for the matrix \[A=\begin{pmatrix}I_{p}&0\\ 0&0\end{pmatrix}\] we know that $trace(A^{k})=0$ for each $k\in\mathbb N$, but $A^{i}=A\neq0$ and $A$ is not nilpotent.


In francophon system , 1rst  preparatory year for the High engineering schoolS , MPSI = Maths physics and engineer science (SI) .


Find the  of the complex number:
\[ \left( {\frac{{2 \plus{} i\sqrt 5 }}
{{2 \minus{} i\sqrt 5 }}} \right)^{10}  \plus{} \left( {\frac{{2 \minus{} i\sqrt 5 }}
{{2 \plus{} i\sqrt 5 }}} \right)^{10}
\]

Let $ \theta\equal{}\arctan\frac{\sqrt{5}}{2}$.

We find our expression equal to:
$ |e^{20i\theta}\plus{}e^{\minus{}20i\theta}|\equal{}|e^{\minus{}20i\theta}\parallel{}e^{40i\theta}\plus{}1|\equal{}|e^{40i\theta}\plus{}1|$

I don't think there's a good way to express this though (without trig functions, etc, in the expression)

[quote="Xaenir"]Find the  of the complex number:
\[ \left( {\frac {{2 \plus{} i\sqrt 5 }} {{2 \minus{} i\sqrt 5 }}} \right)^{10} \plus{} \left( {\frac {{2 \minus{} i\sqrt 5 }} {{2 \plus{} i\sqrt 5 }}} \right)^{10}
\]
[/quote]

Can this Q be solved geometrically? I haven't attempted (yet) but just noticed that they are conjugate expressions and addition of vectors on an Argand diagram?


[hide]The total is 738, he paid 123 at purchase, meaning he has to pay 615 in the next 12 months.

That's 51.25 per month.  [/hide]

[quote="aznmonkey1992"]at the store, bob bought 2 video game systems for 299 dollars each, and 6 controllers for 140 dollars ( the controllers' price added up to 140 dollars) if he paid 1/6 of the total cost at purchase and the balance in 12 equal monthly payments, how much did he pay every month?[/quote]
[hide]
2x299+140 = 598 +140 = 738
1/6x738 = 123
738 - 123 = 615

615/12 = 51.25
[/hide]

[quote="aznmonkey1992"]dude, the total is 738[/quote]

Duh duh duh duh :wallbash:
How did I miss that though? Typo that screwed the rest up? :furious:


2009 Chapter Target Round #7:

There is a total of 70 squares of three sizes whose vertices are points on this rectangular 3*n grid of points. What is the value of n? (sorry this is the best diagram I could draw using the computer  :| )

O   O   O   ...   O   O
O   O   O   ...   O   O
O   O   O   ...   O   O

Consider how many new squares there are every time a new vertical row of dots is added.

[quote="Thunder365"][hide="Solution"] Jbatterson's advice is the best.
in a 3x1 grid, there are 0 squares.
in a 3x2 grid, there are 2 squares-2 1x1 squares
in a 3x3 grid there are 6 squares-4 1x1 squares, 1 2x2 square, 1 sqrt(2)xsqrt(2) square
in a 3x4 grid there are 10 squares-6 1x1, 2 2x2, 2 sqrt(2)xsqrt(2)
3x5: 14

now we see a pattern. Starting with the 3x2 grid, each one is 4 more than the previous. That gives the formula:$ y \equal{} 4x \minus{} 6$ when $ x > 1$ . Since $ y \equal{} 70$, $ 70 \equal{} 4x \minus{} 6$, $ x \equal{} 19$[/hide][/quote]
Curious, how did you write up that formula $y \equal{} 4x \minus{} 6$?

[quote="Thunder365"][b]in a 3x2 grid, there are 2 squares[/b]-2 1x1 squares
[b]in a 3x3 grid there are 6 squares[/b]-4 1x1 squares, 1 2x2 square, 1 sqrt(2)xsqrt(2) square
[b]in a 3x4 grid there are 10 squares[/b]-6 1x1, 2 2x2, 2 sqrt(2)xsqrt(2)

now we see a pattern. Starting with the 3x2 grid, each one is 4 more than the previous. That gives the formula:$ y \equal{} 4x \minus{} 6$ when $ x > 1$ .[/quote]
By checking small cases, he made sure it was linear. The common difference is $4$, making that the coefficient of $x$. Then we add the constant term $6$ to get the correct values.

He also probably setted it up as a input and output table. You could figure out patterns and eventually come across a formula that is useful.


Let the triangle $ ADB_1$ s.t. $ m(\angle DAB_1)\ne 90^\circ$.On the sides of this triangle externally are constructed the squares $ ABCD$
and $ AB_1C_1D_1$ with centers $ O_1$ and $ O_2$, respectively.Prove that the circumcircles of the triangles $ BAB_1$, $ DAD_1$ and
$ O_1AO_2$ share a common point, that differs from $ A$.

Dear Mathlinkers,

1. X the second point of intersection of BAB1 and DAD1
2. X, D and B are collinear and also for X, B1 and D1
3. O1, O2 being the midpoints of DB, B1D1, according the midcircle theorem, we are done.

Sincerely
Jean-Louis


Find all functions $ f : \mathbb{R^{\plus{}}} \ \to \ \mathbb{R^{\plus{}}}$ satisfies two conditions :

1 / If $ 0 < x < y$ then $ 0< f(y) \ \leq \ f(x)$

 2 / $ f(xy) f \left( \frac{f(y)}{x} \right) \ \equal{} \ 1 \ \forall \ x;y \ \in \ \mathbb{R^{\plus{}}}$

[hide="edit"] there was some nonsense here, i had misread the problem. [/hide]

[quote="nguyenvuthanhha"]Find all functions $ f : \mathbb{R^{ \plus{} }} \ \to \ \mathbb{R^{ \plus{} }}$ satisfies two conditions :

1 / If $ 0 < x < y$ then $ 0 < f(y) \ \leq \ f(x)$

 2 / $ f(xy) f \left( \frac {f(y)}{x} \right) \ \equal{} \ 1 \ \forall \ x;y \ \in \ \mathbb{R^{ \plus{} }}$[/quote]

Let $ P(x,y)$ be the assertion $ f(xy)f(\frac{f(y)}x)\equal{}1$

$ P(\sqrt{f(1)},1)$ $ \implies$ $ f(\sqrt{f(1)})\equal{}1$ and then $ P(\frac 1{\sqrt{f(1)}},\sqrt{f(1)})$ $ \implies$ $ f(1)\equal{}1$

Now two possibilities :

1) $ \exists u\neq 1$ such that $ f(u)\equal{}1$. Then :
$ P(u,1)$ $ \implies$ $ f(\frac 1u)\equal{}1$
$ P(u,u)$ $ \implies$ $ f(u^2)\equal{}1$
And so $ f(u^{2^n})\equal{}f(u^{\minus{}2^n})\equal{}1$ and so $ f(x)\equal{}1$ $ \forall x$ since $ f(x)$ is non increasing.
And it is easy to check that this indeed is a solution.

2) $ f(u)\equal{}1$ $ \iff$ $ u\equal{}1$. Then :
$ P(\sqrt{\frac{f(x)}x},x)$ $ \implies$ $ f(\sqrt{xf(x)})\equal{}1$ $ \implies$ $ \sqrt{xf(x)}\equal{}1$ $ \implies$ $ f(x)\equal{}\frac 1x$
And it is easy to check that this indeed is a solution.

Hence the two solutions :
$ f(x)\equal{}1$ $ \forall x$

$ f(x)\equal{}\frac 1x$ $ \forall x$


Hi!!!

I have some figures that I have already created in Microsoft Word and Microsoft Excel. My questions are:-

1. Is there a way to convert the Microsoft Word figure into an eps figure and insert it into a latex document?

2. Similarly, is there a way to convert a Microsoft Excel figure into an eps figure to insert into a latex document?

3. If the method to do either one of the above is difficult, is there another way to use an existing application to create a an eps figure?

Thanks to anybody that can relieve me of this problem!!!

Why do you want to create an eps file? You can include a jpeg file in your source, and you can create those with paint.

Thanks ComplexZeta!!!

I'm still quite new to latex so I'm still a little blur about whatever that's going on. Thanks for that tip really. I originally thought that we have to convert images to eps images because that's what latex reads in. What about for graphs made in excel? Is there a way to convert it into jpeg maybe or something like that and insert it into a latex document?

Thanks a bunch!!

What I sometimes do to make jpegs is to take screenshots and then copy+paste those into paint. I forgot to mention that unscaled jpegs don't look very good in $\text{\TeX}$ files, so you might want to try making the dimensions of the image twice as large as what you want and then using a scalebox.

Actually I use .png files which scale correctly in .pdf files created with pdftex. In the end it's your choice :)

Hi,
   The way I convert Excel charts (and this is kind of the long way around, but the only way I know how) is to use Powerpoint.
SSN.

I'm not sure if this is of any use to you, but I use an application called wmf2eps for generating eps files from wmf/emf files.

Simply select the image/text from word,excel (practically any microsoft application) and copy. When in wmf2eps paste the image. You can then edit the bounding box and save as eps

There is a link to it along with some other useful tools I found at
http://www.donaldoyle.com/Support/latex.php

Hope this is of help.
Peterpan

To Convert figures from Microsoft Word into eps:
1. print to .prn file using generic postscript printer, select current page and print to file, to file fig.prn
2. change file fig.prn to fig.ps
3. open fig.ps by a text editor (such as textpad or notepad), search for setpagedevice
delete the block containing "setpagedevice", beginning by "featurebegin" and ending by "featurecleanup", save file, close text editor.
4. open fig.ps by gsview, use file--export---ps to eps to create file fig.eps
5. insert fig.eps to latex.

For excel chart, follow the same procedure

Hope it helps

vnns

To Convert figures from Microsoft Word into eps:
1. print to .prn file using generic postscript printer, select current page and print to file, to file fig.prn
2. change file fig.prn to fig.ps
3. open fig.ps by a text editor (such as textpad or notepad), search for setpagedevice
delete the block containing "setpagedevice", beginning by "featurebegin" and ending by "featurecleanup", save file, close text editor.
4. open fig.ps by gsview, use file->PS to EPS to create file fig.eps
5. insert fig.eps to latex.

For excel chart, follow the same procedure

Hope it helps

vnns


Hey all,
I made this post in the AMC forum, but I figured I would post it here as well.

There is a practice contest going on right now (through sunday) on my website http://matharena.net/

If you'd like to participate, go there and sign up for an account!

Thanks
Snowstorm

No offense but please do not double post.

And who made this as announcement? I'm putting this as regular thread now.

How does this classify as double posting?

You posted the Practice AMC thread TWICE in AMC forum. And then, you posted this thread in Getting Started forum. It is called "double posting" if you post same thread (or thread with same topic) in two or more forums.


function
logarithms
Euler
geometry
geometric transformation
reflection
real analysis

is it possible to get a [b]closed[/b] sum for

$ 1\plus{}\sum_{n\equal{}1}^\infty \;\dfrac{1}{\textbf p_n^2}\;\equal{}\;1\plus{}\dfrac{1}{2^2}\plus{}\dfrac{1}{3^2}\plus{}\dfrac{1}{5^2}\plus{}\dfrac{1}{7^2}\plus{}\dfrac{1}{11^2}\plus{}\dfrac{1}{13^2}
\plus{}\dfrac{1}{17^2}\plus{}\dfrac{1}{19^2}\plus{}\dfrac{1}{23^2}\plus{}\dfrac{1}{29^2}\plus{}\ldots$

where $ \textbf p_n$ is the $ n^{th}$ prime number ?  :!:  :|

It seems to me that if we had a closed form for primes, then this problem would be made easier but of course, isn't that a goal of the Riemann-Zeta function anyway?  I have no idea how to go about this.

The best I can offer is; since,
$ \log (\zeta(s)) \equal{} \sum_{n \equal{} 1}^{\plus{}\infty} \frac{\zeta_{p}(sn)}{n}$

from the Euler product, we can Mobius invert to get;

$ \zeta_{p}(s) \equal{} \sum_{n \equal{} 0}^{\plus{}\infty} \frac{\mu(n)}{n} \log(\zeta(sn))$

with $ s\equal{}2$ you have the series you want on the left hand side(this being the zeta function of primes). While this isn't a closed form it might help!

Actually on reflection I feel that this can't be expressed in closed form. Can't show it yet...

http://mathworld.wolfram.com/PrimeZetaFunction.html

the table given at that link puts your sum at approximately 1.452 (to 3 decimal places) but its just one of those constants that haven't yet been expressed in terms of better defined constants.

Compare this though with,
http://mathworld.wolfram.com/PrimeSums.html

equations (32) to (34)

Errata- http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1118903#1118903
that is in post no.3 of this thread I have the bottom limit wrong after inversion- it should be $ n \equal{} 1$ to $ \infty$,

i.e-
$ \zeta_{p}(s) \equal{} \sum_{n \equal{} 1}^{\plus{}\infty} \frac{\mu(n)}{n} \log(\zeta(sn))$

Sorry about that! :oops:


I'm afraid I don't understand your question.  Are you an enrolled student looking for information on how the course works?  If so, you should have received an email with course information.  (If you have not received such an email, make sure your email address is correct in your profile - the email will be re-sent on Tuesday.)

[quote="1/(ln x)"]I enrolled into the AIME A class today, I've only got the order confirmation email so far. When did you guys send the email with the course info?[/quote]You should have received the info by now.  Please let us know if you haven't.


[hide="Solution"]

Part A)
For this we will use modulo $ 19$,
\begin{eqnarray*} 5^{2n + 1}2^{n + 2} + 3^{n + 2}2^{2n + 1} & \equiv & (5\cdot 25^n)(4\cdot 2^n) + (9\cdot 3^n)(2\cdot 4^n)\pmod{19} \\
& \equiv & (5\cdot 6^n)(4\cdot 2^n) + (9\cdot 3^n)(2\cdot 4^n)\pmod{19} \\
& \equiv & (20\cdot 4^n\cdot 3^n) + (9\cdot 3^n)(2\cdot 4^n)\pmod{19} \\
& \equiv & (4^n\cdot 3^n) + (18\cdot 3^n\cdot 4^n)\pmod{19} \\
& \equiv & (4^n\cdot 3^n) - (3^n\cdot 4^n)\pmod{19} \\
& \equiv & 0\pmod{19} \end{eqnarray*}
Part B)
For this we will use modulo $ 31$,
\begin{eqnarray*} 6^{2n + 1} + 5^{n + 2} & \equiv & (6\cdot 36^n) + (25\cdot 5^n)\pmod{31} \\
& \equiv & (6\cdot 5^n) + (25\cdot 5^n)\pmod{31} \\
& \equiv & (6\cdot 5^n) - (6\cdot 5^n)\pmod{31} \\
& \equiv & 0\pmod{31} \end{eqnarray*}
[/hide]

You can solve them using induction too .

a-)

For $ n\equal{}1$ ,we have :

$ 5^{2n\plus{}1} 2^{n\plus{}2} \plus{}3^{n\plus{}2} 2^{2n\plus{}1} \equal{}5^3 2^3 \plus{}3^3 2^3\equal{}10^3 \plus{}6^3 \equal{}38\cdot 32$

Suppose it is true for $ n\equal{}k$  ,and we need to prove it's true for $ n\equal{}k\plus{}1$

$ 5^{2(k\plus{}1)\plus{}1} 2^{(k\plus{}1)\plus{}2} \plus{}3^{(k\plus{}1)\plus{}2} 2^{2(k\plus{}1)\plus{}1}\equal{}5^{2k\plus{}1} 2^{k\plus{}2} \cdot 50 \plus{}3^{k\plus{}2} 2^{2k\plus{}1} \cdot 12 \equal{}50(5^{2k\plus{}1}2^{k\plus{}2} \plus{}3^{k\plus{}2}2^{2k\plus{}1}) \minus{}38\cdot 3^{k\plus{}2} 2^{2k\plus{}1}$

Since by hypothesis of the induction $ 5^{2k\plus{}1}2^{k\plus{}2} \plus{}3^{k\plus{}2}2^{2k\plus{}1}$ is dividable by $ 38$ , then we're done

You can solve b-) using induction as an exercise


Hey i have attached the problem
please i need its solution urgently

recently i just finished a college algebra class and we did a lot of matrices/determinants in it.
first of all, i have to ask u, do u know the general way to find the inverse of a 3x3 matrix. u have to write an augmented matrix with that matrix (let's call it A) on the left and the corresponding identity matrix (all 1's on the principal diagonal and zeros everywhere else) on the right side. now u have to use row operations to transform A into its identity matrix and watever the right side becomes after the same row operations is the inverse.
try this, cause right now i myself am too lazy to do all that work. :D


geometry
circumcircle
number theory
national olympiad

i guess this is not exactly a national but its a test to gain access to nationals so i´ll post it here is the first day the second is about to be held in about an hour and a half so wish me luck!! because i had a very rough day yesterday (day one) hoping to get a perfect day today  :D :

1. - Prove that all primes except 2 and 3 belong to the sequence of numbers defined by
$a_n$= $sqrt(24n+1)$

2. - Does there exist a solution to ($sqrt(2005)$- $sqrt(2004)$= $sqrt(N)$ - $sqrt(N-1)$

3. - ABC  is isoceles AB=AC. let M be the midpoint of BC. let O  on the line AM  such that  OB  is perpendicular to AB  Q  a point  on  BC  and  E  and  F  on AB  and AC  such that E,Q,F  are collinear. Prove that OQ  is perpendicular to EF  if  and only if EQ=QF

oops!!  :blush:  i forgot for problem number 2 the left hand side should be elevated at the 2006 exponent. also some comments on the problems:

as you can see problem number one is trivial and pretty well known so i got 7 points on that
problem two is easy but has a trick to it one that didn ocurr to me in the test so ill have 0 to 1 on that problem .
problem 3 is easy but somehow i couldnt complete it  :mad:  anyway ill get 3 to 4 points on that one , as you can see i didnt have a good day yesterday and my score was vey below my expectations (18 to 21) nevertheless i think my bad day was because i couldnt get geometry and algebra is a very waek topic for me in the olympiad bit im hoping there is a number theory one today and that i could solve the geometry today too!!  :P  the combi one i probably wont get even with manuel' :D s useful tutoring  :P

here is the second day. im very happy because i got a perfect day today!! anyway here are the problems:

Define $a_n$ to be the sum of all numbers who have the same residue and cocient when divided by n. Prove  that :

a) $a_n$ + $a_(n+1)$ = $3(1^2+2^2+3^2...n^2)$

b) Find all n such that $a_n$ / $a_(n+1)$ is an integer.

Let ABC  be an acutangle triangle and let $L$ be an arbitrary line in the interior of angleBAC . Suppose tha $L$  cuts the perpendicular bisectors of  AB  and AC  at P  and  Q respectively. Let O  be the circumcenter of ABC. Prove that the circumferences of ABC , BPO and  CQO  have a common point.

Find  all sequences of 2005 positive integers such that the sum of the fourth potence of any 4 of them is divisible by their product.

Number one is easy, number 2 is also easy if you know how to construct the common point wich i found. number is also pretty easy i guess its a standard diophantine equation.

marko, no combinatorics  :mad:
your state needs to open its eyes and see where the beautiful problems are  :lol:

MANUEL:  i can garantee some combi next tst!!! :D

i hope you are prepared !!  :D   ( if not, blame WOOT)


A stream of water from a faucet decreases in diameter as it falls. Derive an equation for the diameter of the stream as a function of the distance y below the faucet, given that the water has speed v_0 when it leaves the faucet, whose diameter is D.

I think this requires Bernoulli's equation and the continuity principle.

$Let A_{0}= \frac{\pi D^{2}}{4}$, the area when it has speed $v_{0}$.

Then, when the stream has velocity $v$, we have $vA = v_{0}A_{0}$.

Thus, cancelling the $\frac{pi}{4}$ from both sides, we have $vd^{2}= v_{0}D^{2}$.

Therefore, $d = D %Error. "srqat" is a bad command.
{\frac{v_{0}}{v}}$.

But we know that from conservation of energy, $v = \sqrt{v_{0}^{2}+2gy}$ after it falls a distance $y$.

Thus, $d = D \sqrt{\frac{v_{0}}{\sqrt{v_{0}^{2}+2gy}}}$.


Find all real solutions to $ x^3 \minus{} 3x^2 \minus{} 8x \plus{} 40 \minus{} 8\sqrt[4]{4x \plus{} 4} \equal{} 0$

[quote="DuongChinh_K41SPHN"]Use AM-GM
very easy[/quote]
$ \sqrt[4]{4x \plus{} 4}\le \frac {4x \plus{} 52}{4} \equal{} x \plus{} 13$
then $ x^3 \minus{} 3x^2 \minus{} 8x \plus{} 40\le x \plus{} 13$
$ (x \minus{} 3)^2(x \plus{} 3)\le 0$
$ x \equal{} 3$

$ x^3 - 3x^2 - 8x + 40 = 8\sqrt [4]{4x + 4}$ so $ (x^3 - 3x^2 - 8x + 40)^4 = 16384\((x + 1)$ when we expand and groupize we see it's possible to apply ededi silsile boyukdi hendasi silsile qaydasi


For positive real number a,b,c satisfy: $ \frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c}  \geq \frac{27}{8}$
Prove that:   $ \sqrt{abc}  \leq \frac{a\plus{}b\plus{}c}{3}$

[hide="Solution"]$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\ge \frac{27}{8}\implies 8ab\plus{}8bc\plus{}8ca\ge 27abc$.

We wish to prove that $ \frac{a\plus{}b\plus{}c}{3}\ge \sqrt{abc}\implies 3(a\plus{}b\plus{}c)^2\ge 27abc$.

However, $ 3(a\plus{}b\plus{}c)^2\ge 8ab\plus{}8bc\plus{}8ca$, and we are done.[/hide]

I believe the equality does not occur.

If you expand the last line of my solution, we will have
$ 3a^2\plus{}3b^2\plus{}3c^2\plus{}6ab\plus{}6bc\plus{}6ca\ge 8ab\plus{}8bc\plus{}8ca$$ \implies 3a^2\plus{}3b^2\plus{}3c^2\ge 2ab\plus{}2bc\plus{}2ca$

However, since $ a, b, c$ are positive reals, $ 3a^2\plus{}3b^2\plus{}3c^2>2a^2\plus{}2b^2\plus{}2c^2\ge 2ab\plus{}2bc\plus{}2ca$ $ \implies 3a^2\plus{}3b^2\plus{}3c^2>2ab\plus{}2bc\plus{}2ca$.

Hence, the equality does not occur.

Please correct me if I am wrong.

Equality does not occur because the inequality is not tight.

FantasyLover, doesn't your proof hold even when $ \displaystyle\frac {1}{a} \plus{} \displaystyle\frac {1}{b} \plus{} \displaystyle\frac {1}{c} \ge \displaystyle\frac {9}{4}$?

[quote="darkdieuguerre"]FantasyLover, doesn't your proof hold even when $ \displaystyle\frac {1}{a} \plus{} \displaystyle\frac {1}{b} \plus{} \displaystyle\frac {1}{c} \ge \displaystyle\frac {9}{4}$?[/quote]
Yeah, it does. As you said, this inequality is weak.


This is a rather simple problem, I think.

I need a formula that can express the number of integers between, say, n^2 and (n+1)^2.

It wasn't obvious what a formula or a pattern would be (I worked on the problem for about 5 minutes), so instead I want to see if any of you have any insight into this and perhaps can come up with more elegant formula than I can, which I am sure a lot of you can.

Thanks in advance

$ (n \plus{} 1)^2 \equal{} n^2 \plus{} 2n \plus{} 1$; there are therefore $ \boxed{2n}$ integers strictly between $ n^2$ and $ (n \plus{} 1)^2$.

Another way to see this is to note that

$ 1^2 \equal{} 1$
$ 2^2 \equal{} 1 \plus{} 3$
$ 3^2 \equal{} 1 \plus{} 3 \plus{} 5$
$ 4^2 \equal{} 1 \plus{} 3 \plus{} 5 \plus{} 7$
$ 5^2 \equal{} 1 \plus{} 3 \plus{} 5 \plus{} 7 \plus{} 9$
...
$ k^2 \equal{} 1 \plus{} 3 \plus{} 5 \plus{} 7 \plus{} ... \plus{} (2k \minus{} 1)$

Did you try making a table of values?  The linear progression should've popped out after three or four values.

Also, more generally is the formula:

(n+k)2 = n2 + 2nk +k2 ==>there are k2 + 2nk - 1 integers strictly between the square of n and the square of (n+k). :)

I don't understand this problem :
There are $ m\minus{}n\plus{}1$ integer interval $ [n,m]$.

[quote="TTsphn"]I don't understand this problem :
There are $ m \minus{} n \plus{} 1$ integer interval $ [n,m]$.[/quote]

What can't you understand?

He doesn't understand why it was posted because it seems obvious to him.

$ x^2 \minus{} y^2 \equal{} (x \plus{} y)(x \minus{} y)$ so there are $ (x \plus{} y)(x \minus{} y) \minus{} 1$ integers in the interval.


calculus
integration
trigonometry
real analysis
real analysis solved

Hi :)

I'm new here so I hope I'm posting at the right place :lol:

[tex]\int_{0}^{\frac{\pi}{8}}(2cos^22x)dx[/tex]

I'm having problem with getting the integrale.

Any advice? I would preciate if you could show the steps instead of just giving me the answer :)

We have Calculus - Analysis section for this.
You need to use the identity $2\cos ^22x=\cos 4x+1$, so your integral becomes
\[\int_0^{\pi/8}(\cos 4x+1)dx=\frac{1}{4}\sin 4x+x\big|_0^{\pi/8}=\frac{1}{4}+\frac{\pi}{8}.\]

Thanx for the reply Myth :)

How did you get [tex]\displaystyle 2\cos ^22x=\cos 4x+1[/tex] so fast? How did you calculate it?

Since $\cos 2t=2\cos^2 t-1$, set $t=2x$.

[quote="kahsi"]Ahhh...so [tex]cos^2x = cos2x[/tex]? I didn't know that.[/quote]
???

So that's not correct.

Then I have no idé how you calculated it :blush:

[tex]2cos^22x = 2(cos2x)^2 = 2(2cos^2t - 1)^2[/tex] Then what?
I'm not that good at math, just trying to learn more :(

use the formula: $\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$

then $\cos(2t)=\cos^2(t) - \sin^2(t)$ but $\sin^2(t) = 1 - \cos^2(t)$ so plug in to get the

result.

Using $\cos 2t=2\cos^2 t-1$, we obtain $\cos 2\cdot 2x=2\cos^22x-1$, i.e. $\cos 4x=2\cos^2 2x-1$.


$ \mbox{Q.1} \  \mbox{If} \ x^{15}\minus{}x^{13}\plus{}x^{11}\minus{}x^9\plus{}x^7\minus{}x^5\plus{}x^3\minus{}x\equal{}7 \ ,  \mbox{show that } \ x^{16} > 15$
$ \mbox{Q.2} \  \mbox{If} \ a_1\equal{}1 \ \mbox{and for} \  n>1 , a_n\equal{}a_{n\minus{}1} \plus{} \frac{1}{a_{n\minus{}1}} , \mbox{then show that } a_{75} \in (12,15)$

1. The expression can be written as $ x\frac {x^{16} \minus{} 1}{x^2 \plus{} 1} \equal{} 7 \Rightarrow x^{16} \equal{} \frac {7x^2 \plus{} x \minus{} 7}{x}$. Now it is easy to show that this must be more than 0, as our progress is equivalent to the statement $ (x \minus{} 1)^2 > 0$, which is true for all x != 1, but if x=1, the expression is NOT equal to 7, so we are done.

How'd you get to $ \left(x\minus{}1\right)^2>0$? I got $ x^2\minus{}2x\minus{}1>0$.

[quote="123456789"]How'd you get to $ \left(x \minus{} 1\right)^2 > 0$? I got $ x^2 \minus{} 2x \minus{} 1 > 0$.[/quote]

Hmm... if you got $ x^2 \minus{} 2x \minus{} 1 > 0$, then $ x^2 \minus{} 2x \plus{} 1$ must be greater than 0 as well!

He meant the other way around; he got down to $ x^2 \minus{} 2x \minus{} 1 > 0$ which is not necessarily true.  That would be because of a typo in Jarle10's post, where it should say $ x^{16} \equal{} \frac {7x^2 \plus{} x \plus{} 7}{x} > 15$.

[hide="2"]
$ a_i$ is a strictly increasing sequence. Bound the sequence by $ b_n \equal{} b_{n \minus{} 1} \plus{} \frac {1}{\lfloor b_{n \minus{} 1} \plus{} 1 \rfloor} \le a_n \le c_n \equal{} c_{n \minus{} 1} \plus{} \frac {1}{\lfloor c_{n \minus{} 1} \rfloor}$, with $ a_1 \equal{} c_1 \equal{} 1$ and $ b_{18} \equal{} 6 < a_{18}$; then we get something along the lines of $ b_{ \minus{} 3 \plus{} (n \plus{} 1)n/2} \equal{} n$ and $ c_{1 \plus{} (n \minus{} 1)n/2} \equal{} n$ [yeah im tired]. Thus $ a_{75} > b_{75 \equal{} 12 \cdot 13/2 \minus{} 3} \equal{} 12$ and $ a_{75} < a_{92 \equal{} 13\cdot 14/2 \plus{} 1} < c_{92} \equal{} 14$.

Edit: the lower bound does seem pretty strong, since $ a_{75} \approx 12.32422351$ according to Excel.
[/hide]

Thanks for those soutions...anyone has a different idea about $ \mbox{Q.2}$?


geometry
incenter
geometric transformation
reflection
circumcircle
perpendicular bisector

In $\triangle ABC$, let $I$ be the incenter, let $B'$ be the midpoint of arc $AC$ which does not contain $B$, and $C'$ be the midpoint of arc $AB$ which does not contain $C$. Show that $B'C'$ is the perpendicular bisector of $AI$.

[asy]size(210); pair A,B,C,M,N,I,D,E; A=(10,50); B=origin; C=(70,0); I=incenter(A,B,C); draw(A--B--C--cycle); draw(circumcircle(A,B,C)); M=point(circumcircle(A,B,C),arctime(circumcircle(A,B,C),0.4*arclength(A--C))); N=point(circumcircle(A,B,C),arctime(circumcircle(A,B,C),2.3*arclength(B--A))); draw(A--I); draw(M--N); D=intersectionpoint((A--B),(M--N)); E=intersectionpoint((A--C),(M--N)); draw(A--M); draw(A--N); dot(A); dot(B); dot(C); dot(D); dot(E); dot(I); dot(M); dot(N); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,SE); label("E",E,S); label("M",M,NE); label("N",N,NW); label("I",I,SW);[/asy]

let $B'=M$ and $C'=N$. also let $AB\cap MN=D$ and $AC\cap MN=E$. since $\angle AMN=\angle BAN$ and $\angle ANM=\angle CAM$, we get $\angle ADE=\angle AED$. also, since $AI$ is the bisector of $\angle DAE$ of isosceles triangle $ADE$, $AI\perp DE$. now it's not difficult to see that $\triangle AMN$ is the reflection of $\triangle IMN$ on $MN$. hence the conclusion.

if $AI \cap B'C'=O$we want to prove that  $AO=OI$ and $\angle AOB'=\angle B'OI=\angle AOC'=\angle IOC'$
WE know that :$\angle B'AB=\frac{\angle C}{2}=\angle B'BA=\angle AC'B'$AND $\angle AB'C'=\angle ACC'=\angle C'AC=\angle IB'O=\frac{\angle B}{2}$ then in triangle $ABC$ we have $\angle OAB'+\angle OB'A=90 \Longrightarrow \angle AOB'=90$
so triangle  $\bigtriangleup AB'O=\bigtriangleup OB'I \Longrightarrow OA=OI$
$\angle AIB'=\frac{arc A'C+arc AB'}{2}=\frac{\angle A+\angle C}{2}$.

We have: $\widehat{B'IA}=\widehat{ICA}+\widehat{IAC}=\frac{A+C}{2}$
On the otherhand, $\widehat{B'AI}=\widehat{B'AB}+\widehat{BA}=\frac{A+C}{2}$
Thus $B'A=B'I.$
Similaly, we also have: $C'A=C'I,$ so $AI\perp B'C'$

i was a fool not to notice the following: it's well known that $BI$ meets the circumcircle of $\triangle ABC$ at the midpoint of arc $AC$ not containing $B$. thus $\angle AB'C'=\angle IB'C'$ and $\angle AC'B'=\angle IC'B'$. thus $\triangle AB'C'$ is the reflection of $\triangle IB'C'$ on $B'C'$ thus $B'C'$ is the perpendicular bisector of $AI$.


There was this question on my trig book that ask me to find the a and d for the function f (x) = a cos x + d by giving the picture of a certain graph.  But I dont' know how to figure out the a and d.  I know how to find the interval like for cosine, it's:

bx - c = 0 and bx - c = 2pi

That's one question.

Second is:

Use a graphing utility to graph h, and use the graph to decide whether h is even, odd, or neither.

a --> h (x) = cos^2 x
b --> h (x) = sin^2 x

Well, I got a is even because I remember reading cosine and cosecant functions are even while rest of them are odd.  But this book had b as even too!  Then what's this mean?  I'm struggle to understand this.

Help! :maybe:

A function is even if f(-x)=f(x)
A function is odd if f(-x)=-f(x)

Cosine and secant are the only even trig functions( out of sin, cos, tan, cot, csc, sec) the others are odd.

So we want to know if sin^2(x) and cos^2(x) are even or odd.

Let f(x)=sin(x)
We know that (in this case) f(x) is odd.  We want to know if (f(x))^2 is even or odd.  So what we do is look at (f(-x))^2

Since f(x) is odd we can replace f(-x) with -f(x).  Doing this we find that (-f(x))^2=(f(x))^2.   Therefore (f(x))^2 is even and since f(x)=sin(x), sin^2(x) is even.

We can make a similar argument for cos.

y = a cos x + d.

Start with y = cos x. You know what that looks like.
When we multiply by a, instead of y going from -1 to 1, it goes from -a to a.
Then when we add d, it shifts everything up by d.

See if you can work it out from there..

Also, silverfalcon, if you look at the graph of the function, you might be able to tell if the graph is even or odd. Even graphs are symetrical about the y axis, and odd graphs are symetrical about the origin...


Παιδιά έχει κανείς υπ΄όψη το θέμα αυτό ?

ΘΕΜΑ.

Αν  ABC είναι ένα ισόπλευρο τρίγωνο και  Μ ένα σημείο που βρίσκεται στον περιγεγραμμένο κύκλο, τότε το άθροισμα

$ MA^4\plus{}MB^4\plus{}MC^4$,....είναι σταθερό.

Πολύ γνωστό θέμα...
Την πρώτη φορά που το είχα λύσει, είχα χρησιμοποιήσει και την ταυτότητα του de Moivre, αυτή που χρησιμοποιούσε και το 4ο θέμα του Ευκλείδη.

ΜΠΡΑΒΟ ΡΕ ΝΙΚΟ ...   :)  και ΄μένα για De Moivre μου έκανε ... τώρα που έδωσες και βοήθεια θα το λύσω πάλι ...

Είχα κάνει μια λύση με αναλυτική γεωμετρία που τελικά βγήκε , αλλά άσε ... χάλασε ο κόσμος .

[quote="Nick Rapanos"]Πολύ γνωστό θέμα...
Την πρώτη φορά που το είχα λύσει, είχα χρησιμοποιήσει και την [u]ταυτότητα του de Moivre[/u], αυτή που χρησιμοποιούσε και το 4ο θέμα του Ευκλείδη.[/quote]

Ωραίο hint Νίκο :wink:

Κατ' αρχάς είναι γνωστό (αποδεικνύεται εύκολα π.χ. με Πτολεμαίο) ότι (έστω ότι το $ M$ βρίσκεται ανάμεσα σε $ B,C$ ) $ MA\equal{}MB\plus{}MC$

Σύμφωνα με αυτό το hint του Νίκου και την ταυτότητα de Moivre έχουμε:

$ 2(MA^2MB^2\plus{}MB^2MC^2\plus{}MC^2MA^2)\minus{}(MA^4\plus{}MB^4\plus{}MC^4)\equal{}0$
αφού μία απ' τις παρενθέσεις στο δεξί μέλος της ταυτότητας θα είναι η $ MB\plus{}MC\minus{}MA\equal{}0$.

Αρα:
$ MA^4\plus{}MB^4\plus{}MC^4\equal{}2(MA^2MB^2\plus{}MB^2MC^2\plus{}MC^2MA^2)\iff$

$ 2(MA^4\plus{}MB^4\plus{}MC^4)\equal{}(MA^2\plus{}MB^2\plus{}MC^2)^2\equal{} (3R^2\plus{}a^2)^2$ όπου R η ακτίνα του περιγεγραμμένου και a η πλευρά του τριγώνου. Η τελευταία σχέση ισχύει από τη σχέση Leibniz αφού το περίκεντρο ταυτίζεται με οτ βαρύκεντρο στο ισόπλευρο.

Οπότε:
$ MA^4\plus{}MB^4\plus{}MC^4\equal{}\frac{(3R^2\plus{}a^2)^2}{2}\equal{}c$ δηλαδή σταθερό :)

Εχω βρει και μια λύση χωρίς De Moivre που κινείται βέβαια στην ίδια λογική αλλά είναι αρκετά πιο χρονοβόρα.


geometry
3D geometry
complementary counting

A cube is painted red and then cut into 1000 congruent cubes. How many of these cubes are painted red on at least two faces?

Cube is
$ \text{Face}: 6$
$ \text{Edge}: 12$
$ \text{Vertex}: 8$

Thus, $ 6 \times 10^2 \minus{} 12 \times 10 \plus{} 8 \times 1 \equal{} \boxed{488}$.

However, the cubes in the $8\times8$ square on each face do not have at least two faces painted red, so we must subtract that out to get $488-6\times8^2=488-6\times64=488-384=\boxed{104}$.

Another method is complementary counting (count the inside cube of dimensions $8\times8\times8$ then add in the $8\times8$ square on each face, then subtract from $1000$).


Hi all,

I'll be posting teams saturday!!!

Mr C  :)

I have posted the ARML teams for NYC on the NYC math team site!!!  The first practice is saturday May 10 at Stuyvesant at 10:30 AM.  See you then!

Mr C :lol:

I made a mistake on one of the teams.   :o   The updated list is on http://www.nycmathteam.com  My apologies to the student affected.  Also, for the alternates, space on a bus is not guaranteed, more will be known as ARML approaches.

Mr. C


If you guys made it on your own, how does you manage to do it?
is there something turns the eps file into png?
can you show me some of your code? I want to make a math site too and with asy, it would be way easier. thx.

I've done it as well. It's a pretty simple modification of the latexrender code. Basically you simply want to change the references to latex to references to asy and change the appropriate file formats. Imagemagick (convert) should be able to convert eps to png just like it can convert dvi to png.

If AoPS did it some other way, I'd also be interested in what you did.

[quote="Temperal"]I'm sure this is exactly what they did, based on this bug: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=175015 which shouldn't happen if they did it otherwise.[/quote]
Huh?  What? That's just because they replace math tags with dollars.

well I'm just wondering... I heard ImageMagick is a huge program and it's quite CPU intensive.
for the Latex, dvipng would do a nice work
but Asymptote cannot output dvi, thus dvipng can not be used, so is ImageMagick the only alternative?


[quote="ragnarok23"]Pratik is shooting laser beams at Patrick who is at point (5,6). Pratik is at point (3,2). Because of a barrier, Pratik cannot shoot Patrick directly, but can shoot at a point on the x-axis such that it will bounce off and hit Patrick. what is this point?[/quote]
A third problem I would call "nice"
[hide]So we need to find the point (x,0) such that when a line is drawn through this and the other two points the two lines are perpendicular.
$\frac{2-0}{3-x}=\frac{x-5}{6-0}$
$-x^{2}+8x-15=12$
$x^{2}-8x+3=0$
$x=4+\sqrt{13}$
Uh oh, a radical... I think I am wrong. Can someone else point out why in the very likely case I am?[/hide]
PS Nice title :D.

[quote="bpms"][quote="ragnarok23"]Pratik is shooting laser beams at Patrick who is at point (5,6). Pratik is at point (3,2). Because of a barrier, Pratik cannot shoot Patrick directly, but can shoot at a point on the x-axis such that it will bounce off and hit Patrick. what is this point?[/quote]
A third problem I would call "nice"
[hide]So we need to find the point (x,0) such that when a line is drawn through this and the other two points the two lines are perpendicular.
$\frac{2-0}{3-x}=\frac{x-5}{6-0}$
$-x^{2}+8x-15=12$
$x^{2}-8x+3=0$
$x=4+\sqrt{13}$
Uh oh, a radical... I think I am wrong. Can someone else point out why in the very likely case I am?[/hide]
[b]PS Nice title [/b]:D.[/quote]
I stole it from Daesun.  :P
Try reflecting Patrick's point over the x-axis, and form a line using Pratik's (me) and Patrick's point.

[quote="ragnarok23"][quote="bpms"][quote="ragnarok23"]Pratik is shooting laser beams at Patrick who is at point (5,6). Pratik is at point (3,2). Because of a barrier, Pratik cannot shoot Patrick directly, but can shoot at a point on the x-axis such that it will bounce off and hit Patrick. what is this point?[/quote]
A third problem I would call "nice"
[hide]So we need to find the point (x,0) such that when a line is drawn through this and the other two points the two lines are perpendicular.
$\frac{2-0}{3-x}=\frac{x-5}{6-0}$
$-x^{2}+8x-15=12$
$x^{2}-8x+3=0$
$x=4+\sqrt{13}$
Uh oh, a radical... I think I am wrong. Can someone else point out why in the very likely case I am?[/hide]
[b]PS Nice title [/b]:D.[/quote]
I stole it from Daesun.  :P[/quote]
I know, I remember that. That is the reason I was calling it nice, you were able to use a somewhat inside joke and use it in proper context :P. Am I right?


trigonometry
logarithms
inequalities proposed
inequalities
109

Let $ \theta$ be a number with $ 0<\theta<\frac{\pi}{4}$.

Show that　$ (\cos\theta)^\cos\theta$ $ >(\sin\theta)^\sin\theta$ .

[quote="Bovy11"]Let $ \theta$ be a number with $ 0 < \theta < \frac {\pi}{4}$. Show that

$ (\cos\theta)^{\cos\theta} > (\sin\theta)^{\sin\theta}.$ [/quote]$ \Longleftrightarrow f(\theta) \equal{} \cot{\theta}\ln{\cos\theta} \minus{} \ln{\sin\theta}\geq0.$

$ \Longleftarrow f\left(\frac {\pi}{4}\right) \equal{} 0$ and

$ f'(\theta) \equal{} \minus{} \csc^2{\theta}\ln{\cos\theta} \minus{} 1 \minus{} \cot{\theta} < 0.$

$ \Longleftrightarrow g(\theta) \equal{} \ln{\cos\theta} \plus{} \sin^2{\theta} \plus{} \sin{\theta}\cos{\theta} > 0.$

$ \Longleftarrow g(0) \equal{} 0$ and

$ g'(\theta) \equal{} (1 \minus{} \tan{\theta})(1 \plus{} \sin{2\theta}) > 0,$

which is obviously true for $ 0 < \theta < \frac {\pi}{4}.$
[quote="Cai Ni"]$ x\ln{x}$……[/quote]$ \cos{\theta}\ln{\cos{\theta}} > \frac {\sqrt {2}\ln{2}}{ \minus{} \pi}\theta > \sin{\theta}\ln{\sin{\theta}}$

holds for $ 0 < \theta < \frac {\pi}{4}.$


We have to evaluate $\sum_{3\leqslant i\leqslant 1000}\sqrt{1+\frac{1}{(i-1)^{2}}+\frac{1}{i^{2}}}$. It's easy to see that $1+\frac{1}{(i-1)^{2}}+\frac{1}{i^{2}}= \left(\frac{i^{2}-i+1}{i(i-1)}\right)^{2}$ (just take the common denominator), so the sum we have to compute is $\sum_{3\leqslant i\leqslant 1000}\frac{i^{2}-i+1}{i(i-1)}= \sum_{3\leqslant i\leqslant 1000}\frac{i(i-1)}{i(i-1)}+\sum_{3\leqslant i\leqslant 1000}\frac{1}{i(i-1)}= 998+\sum_{3\leqslant i\leqslant 1000}\frac{1}{i(i-1)}$. To evaluate the last sum, see [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=76327[/url]. The final answer is $997.499$.


How many instructors are there for Mathcounts Math Jams?
Because I have competition this weekend and need a looooooot practices and I've neeeeever been MathJams and next one is coming on April.. So, if there are many instructors or good community memebers, can anyone do quick MathJams during this week before Friday?

there are three. but i don't think that is possible. but anyway you should first read the old mathjams first.


geometry
circumcircle
geometric transformation
rotation
geometry unsolved

[color=darkblue]Let $ ABC$ is a triangle and let $ AB>AC$. Let $ M$ is a change point and $ M$ lie on side $ AC$. Let $ N$ is a change point and $ N$ lie on the line pass two point $ A$, $ B$ and $ N$ lie out region of $ \triangle ABC$.

Know that $ BN \equal{}CM$. Prove that circle $ (AMN)$ always pass a fixed point other point $ A$.[/color]

Let $P$ be the 2nd intersection of $(O)$ and $\odot(AMN).$ Note that $ \angle MPN \equal{} \angle BPC \equal{} 180^{\circ} \minus{} \angle A$ $ \Longrightarrow$ $ \angle BPN \equal{} \angle CPM$ and since $ ABPC$ is cyclic $ \Longrightarrow$ $ \angle PMC \equal{} \angle NBP,$  which implies that $ \triangle PBN$ and $ \triangle PCM$ are congruent, because $CM \equal{} BN.$ As a result, $ PB \equal{} PC$ $\Longrightarrow$ $P$ is the midpoint of the arc $BC$ of circumcircle $(O),$ obviously fixed.

It is a well known property:

[i] if M, N lie on (AC), (AB) so that CM = BN, then the triangle MNP is isosceles, P being the midpoint of the arc BC containing A of (ABC), with the angle <MPN = <BAC, while when M and N lie as in subject problem, then the triangle MQN is isosceles, Q being the midpoint of the arc BC not containing A of (ABC). [/i]

Proof: If P is the midpoint of arc BAC, then BP = PC and <ABP = <ACP = |(<B - <C)|/2 and, with CM = BN we get PM = PN and
<BPN = <CPM, which gives <MPN = <A.

For 2nd part (our problem), just to change the solution, see that a (180 degs - <A) rotation around Q will map C to B and M to N, so ANQM is cyclic.

Best regards,
sunken rock


A particle of mass $ m$ is projected vertically upwards with speed $ ums^{-1}$ and returns to its starting point with speed $ Vms^{-1}$. The resistance to its motion in Newtwons has magnitude $ \frac{mv}{10}$ when its speed is $ vms^{-1}$. The acceleration due to gravity is$ 10ms^{-}2$.

 (i)For the ascent of the particle, show that its acceleration $ ams^{-2}$ is given by $ a=-10-\frac{v}{10}$. Hence show that the distance travelled $ x$ metres is given by $ x=10(U-v)-1000ln(\frac{100+U}{100+v})$

(ii) For the descent, show that its acceleration$ ams^{-2}$ is given by $ a=10-\frac{v}{10}$
Hence show that the distance fallen ( from the point of maximum height) $ x$ metres is given by $ x=-10v+1000ln(\frac{100}{100-v})$

(iii) Find the greatest height reached and show that $ U+V=100ln(\frac{100+U}{100-V})$

Question wasn't that long.

i) The particle is ascending. Resistance opposes the motion and gravity acts donwards so $ a =-10-\frac{v}{10}= \frac{-100-v}{10}$ since $ a= v. \frac{dv}{dx}$ we have$ -\frac{10v}{100+v}.dv =dx$.

Integrate both sides: $ \int^{x}_{0}.dx = x = \int^{U}_{v}(10-\frac{1000}{100+v}).dv = 10v-1000 \ln{|100+v|}]^{U}_{v}$

So: $ x= 10(U-v)-1000 \ln{|\frac{100+U}{100+v}|}$.

ii) Same reasoning and we get $ a= 10-\frac{v}{10}= \frac{100-v}{10}= v. \frac{dv}{dx}$.

So, $ x= \int^{0}_{v}(10-\frac{1000}{100-v}).dv=-10v+1000 \ln{|\frac{100}{100-v}|}$.

iii) Looking back at the equation from i), we set $ v=0$ and the greatest height would be:

$ x=10U-1000 \ln{|\frac{100+U}{100}|}$ now take this value and plug it into equation from ii)

So, $ 10U-1000 \ln{|\frac{100+U}{100}|}=-10V+1000 \ln{| \frac{100}{100-V}|}$ $ \Rightarrow$ $ U-100 \ln{|\frac{100+U}{100}|}=-V+100 \ln{|\frac{100}{100-V}|}$

Therefore: $ U+V = 100 \ln{|\frac{100+U}{100-V}|}$.

I meant reading wise.

Nice solutions... Did you do 4 unit /


inequalities
calculus
derivative
inequalities unsolved

Let $ x,y \in [0,1]$, $ r,s \in (0,1]$ and $ t\equal{}rs/(1\plus{}r\plus{}s)$. Prove that
\[ 2(1\minus{}t)\minus{}(1\minus{}r)(1\minus{}s) \geq (1\plus{}r)(1\minus{}s)x \plus{} (1\minus{}r)(1\plus{}s)y\minus{}\left[2(1\plus{}t)\minus{}(1\plus{}r)(1\plus{}s)\right]xy\]

After substituting the formula for $ t$ into the inequality, simplifying it, and using the change of variables $ X \equal{} 1\minus{}x$, $ Y \equal{} 1\minus{}y$, $ R\equal{}1\plus{}r$ and $ S\equal{}1\plus{}s$, in such a way that $ X,Y \in [0,2]$ and $ R,S \in [1,2]$, we get to the following (much) simpler inequality:

\[ 2R(R\minus{}1)X\plus{}2S(S\minus{}1)Y\minus{}RS(R\plus{}S\minus{}3)XY\geq 0\]

This should not be difficult to prove... Anyone?

I will consider the inequality in its original form. It is fairly easy to check that the right
hand side is increasing both in $ x$ and $ y$. For instance differentiation with respect to $ x$
gives something that is linear in $ y$ and to check positivity for all $ y$ it suffices to check
positivity for $ y\equal{}0$ and $ y\equal{}1$.

It remains to show the inequality for $ x\equal{}y\equal{}1$. But an easy simplification shows that
in this case equality holds for all $ s,t$.


THe way prime numbers are taught, it only includes positive numbers because we are taught prime numbers before we are taught negative numbers (and the same is true with history).

Now... Are there are negative prime numbers? The definition of a prime number is any number divisible only by itself and 1. So...

How about -1?

-1 is only divisible by itself and 1. So would -1 be prime?

But then of course, -2 isn't prime. Becuase it's divisible by -2, 2, -1, and 1. And then any number below that isn't prime.

Thanks!

Look here [url]http://mathworld.wolfram.com/PrimeNumber.html[/url]

Basically, the number must be positive.


Do this mentally: $41^2$. Explain your thinking  :)

Hehe, cool, I hadn't noticed this fact. :)

Follow up: what is $95^2$ without paper? It's harder.

true, but learn this trick:
[hide]when squaring a number ending in 5, take everything but the last digit, add 1, square it, and take away the number you squared. Finally, add a 25 back on.

So $95^2$:
9+1=10, 10^2=100, 100-10=90, so the answer is 9025.  :) [/hide]

That's neat.  :) You learn something new everyday I guess.

[hide]This isn't too bad.
Formula for the square of the sum of 2 numbers is:
(a+b)^2 = a^2 + 2ab + b^2

Make 95 = 90 + 5.

(90 + 5)^2 = 90^2 + 2(90)(5) + 5^2

= 8100 + 900 + 25

= [b]9025. There's your answer.[/b][/hide]

[quote="236factorial"]What is $195^2$?

This is fun  :D[/quote]
[hide]Using your trick for multiples of 5
$195^2= 19+1=20^2=400-20=380$put 25 on the end$=38025$[/hide]

Yay!!! Correct!!! Hope you won't do $(190+5)(190+5)$ this time  :D

[quote="236factorial"][hide="mine is very similar"]40*40 is 1600. Now do 41*40, which is 1640. Now add another 41 to make it 41*41, and it's 1681. [/hide]
Fact: 1681 is the only 4-digit perfect square in which the first two and last two digits are also perfect squares.[/quote]

This kind of thing was in my siggy.  :D

[quote="236factorial"]what's siggy?  :?[/quote]
i think it means signature

Yes, it does mean signature, and, I particularly like 236!'s one  :P

[quote="aznmonkey1992"]236, your formula is strange and wierd, but it works :huh:[/quote]
The same formula is in the MATHCOUNTS toolbox see it here
http://www.mathcounts.org/webarticles/articlefiles/105-toolbox.pdf

Alternatively, you can square the first digit and add the original digit to it.

[hide="proof for two digits"]Let the number be 10x+5.
$(10x+5)^2=100x^2+100x+25$
The reason there are 100's instead of 10's is because we have to make an extra space for that "2" in 25 we add on later.
Now the trick is easily seen.

You may prove it for three-digit integers, but it is much harder than the basics forum. [/hide]


algorithm
algebra
polynomial
sum of roots

Can anyone explain (in simple terms) how Newton Sums work? I read the section in Art of Problem Solving on this, but I don't understand it at all.

Thanks

It's what you get when you expand (r_1+r_2+r_3+...+r_n)^k and then factor and use Vite's formula. I haven't memorized it, so I can't give you the formula off the top of my head. What you're trying to do is find r_1^k+r_2^k+...+r_n^k. There's a recursive algorithm you can use to get S_1 (sum of roots), S_2 (sum of squares), etc.

im not positive, but i think this is it... someone can tell me if im wrong later

if the polynomial is a_0^n+a_(1)^(n-1)+...a_n, with roots r_1, r_2, ...r_3, and s_n denotes r_1^n + r_2^n +r_n^n, then

a_0*s_1 + a_1 = 0

a_0 * s_2 + a_1*s_1 + 2*a_2 = 0

a_0*s_3 + a_1*s_2 + a_2*s_1 + 3*a_3 = 0

and etc.


Solve for z in the triangle shown here
http://img.villagephotos.com/p/2006-12/1234083/6sb12.gif

any ideas?

This isn't a calculus problem.

Do you know how to find the sines, cosines, and tangents of angles (that aren't a right angle) in a right triangle?

This problem can be easily done by writing the cosine of the $39^{\circ}$ angle and rearranging to solve for $z$:

$\cos{39^{\circ}}= \frac{23}{z}$
$z = \frac{23}{\cos{39^{\circ}}}$


Find all functions $f: R \to R$,where $R$ denotes the field of real numbers,
satisfying
$f(x^2+xy+f(y))=[f(x)]^2+xf(y)+y$ for all [b]real[/b] numbers $x,y$.

Ok,here is my solution of this problem:
Put $x=0$ we get $f(f(y))=y+f(0)^2$ for all $y \in R$ (*)
Put $f(0)=a$.When $x=0,y=0$ we have $f(a)=a^2$
when $x=a,y=0$ we have $f(a^2+a)=a^4+a^2$
when $y=a,x=0$ we have $f(a^2)=a^2+a$.So that $f(f(a^2))=2a^2=f(a^2+a)=a^4+a^2 \Longleftrightarrow a^4=a^2 \Longleftrightarrow a=0 \vee a=1 \vee a=-1$
The case $a=1$ doesn't satisfy because $f(0)=1 \Longrightarrow f(1)=1$.But from (*) $f$ is injective so $0=1$ (Contradition)
Similar to the case $a=-1$,it doesn't satisfy
so we have only case $f(0)=0$.From here we have $f(f(x))=x$ for all $x \in R$
Put $y=0$ we have $f(x^2)=f(x)^2$ for all $x \in R$
Put $x=-y$ we have $f(f(y))=f(-y)^2-yf(y)+y=y \Longrightarrow f(-y)^2-yf(y)=0 \Longrightarrow f(y)^2=yf(y)$ (because $f(x^2)=f(x)^2$ for all $x \in R$) $\Longrightarrow f(y)=y$ for all $y \in R\{0\}$
So that $f(x)=x$ for all $x \in R$ :)  :)


function
geometry
geometric transformation
abstract algebra
algebra proposed
algebra

Find all continuous functions $ f : \mathbb{R}^\plus{} \rightarrow \mathbb{R}^\plus{}$ such that for all $ x,y,z \in \mathbb{R}^\plus{}$,

$ f(x\plus{}y) \neq y \plus{} f(z).$

($ \mathbb{R}^\plus{}$ is all positive real numbers)

[hide="Probably not Rigorous"]
We show that only constant functions $ f$ satisfy the condition.

Assume for the sake of contradiction that $ f(a)\equal{}f(b)\plus{}k$, for $ k>0$.  If $ a>k$, then we can let $ x\equal{}a\minus{}k$, $ y\equal{}k$, and $ z\equal{}b$ and we have a contradiction of the condition.  If $ a\leq k$, then by the Intermediate Value Theorem, for some $ c$ between $ a$ and $ b$ we have $ f(a)\equal{}f(c)\plus{}(a\minus{}j)$, for some $ 0<j<a$.  Then we can let $ x\equal{}j$, $ y\equal{}a\minus{}j$, and $ z\equal{}c$ for a contradiction.
[/hide]

Well, there is a problem with functions that are not differentiable (there are so many of them...).
Maybe some idea with approximating by smooth functions will be ok?

I don't follow.  The Intermediate Value Theorem requires only continuity.

Ahh, ok. I probably didn't think about argumentation, just found 'intemediate value theorem', which to me means Lagrange's theorem and for $ e^{\pi i}\equal{}\minus{}1$ it means Darboux property ;) I just confused it with 'mean value theorem', because I'm not native English speaker and in my language translation of 'intermediate value theorem' and 'mean value theorem' would be the same...
Sorry for confusion.


logarithms
trigonometry
induction
function
number theory
greatest common divisor
geometry

[color=red][Admin edit] For solutions and other discussions of this contest please check http://www.mathlinks.ro/Forum/viewtopic.php?t=40038[/color]

ONLY in the first week you will send PM's to me with your solutions. So here it goes:
Calculators are  [b][color=red]NOT[/color][/b] allowed!!

[b][size=134][color=blue]First week[/color][/size][/b]

[i][size=117][b]Day 1(algebra)[/b][/size][/i]

Solve the following equations:

(a) $x^{\log_{2}{(x-4)}}+(x-4)^{\log_{2}{x}}=2x^2$ [i](2 points)[/i]

(b) $5^{x}+5^{2x}+...+5^{nx}+(4\cdot 5^{n})^{x}=4^{x}$ [i](3 points)[/i]

Should we send in solutions as we finish problems?

[quote="Slizzel"]Yes. You will justify your answers as well, otherwise you will be un-scored :yup:[/quote]

Of course :)  I finished the problems and sent them to you.  When will we know our scores?

Which forum can we discuss the problems from previous days? Here?

I think we solve for x with n as a constant.

[i][size=117][b]Day 2(Complex numbers)[/b][/size][/i]

(a) Compute the absolute value of $z+z^{2}+z^{3}+...+z^{19}$ if $z=\cos a+i\sin a$ with $a\in (0,\pi)$ [i](2 points)[/i]
(b) Find all $z\in\mathbb{C}$ such that $(\cos{\alpha}+z\sin{\alpha})^{n}=\cos{n\alpha}+z\sin{n\alpha} \ $ where $n\in\mathbb{Z}^{*}$ and $\alpha\in\mathbb{R}$  [i](3 points)[/i]

EDIT: i've changed (1) because it might be wrong.

[quote="JesusFreak197"]I don't understand the second problem... what are we solving for?[/quote]

We just find all integers n and real numbers x that make the equation true.

I'm pretty sure #1 isn't true.  Take $z_1 = 1 + 2i$, $z_2 = 1/3 - 6i$, and $z_3 = \frac{5 \sqrt{14}}{3}$.  The sum of their squares is zero, but $z = \frac{999 + 432i}{8125 \sqrt{14}}$.

EDIT:  A simpler example: $z_1 = z_2 = 1$, $ z_3 = \sqrt{2} i$.  The sum of their squares is obviously zero, but $z = -\frac{3 i}{\sqrt{2}}$.

I had just decided as much, having just multiplied out the quantity I stated in my pm to him. This is almost always the case it appears, with just a few values of x, y, z, satisfying this.

Can we discuss yesterday's problems now?

Yes you can't discuss problem from the previous day always :). For #1 of Day 2 the statment is corect :)

[quote="Slizzel"] #1 of Day 2 the statment is corect :)[/quote]

But I gave 2 counterexamples!

Hmm...the ones for yesterday didn't seem so hard...but anyhow, i'll ease it up some more.  People will see the solutions on saturday, and probably see that the one for day 3 is a lot easier than they thought.
And for those people who don't know, the Law of Sines states that $\dfrac{a}{\sin{A}}=\dfrac{b}{\sin{B}}=\dfrac{c}{\sin{C}}=2R$ in a triangle $ABC$ with $a$ being the length of $BC$, $b$ that of $AC$, $c$ that of $AB$, and $R$ being the circumradius.  Although the $2R$ part isn't really significant for this problem...
I'll try to pick something at lower AIME level, then?

They seemed pretty easy to [i]me[/i]. I just didn't have enough time...so much homework.... :(

I'm not posting problems for today, have no time :(
Anyways, still waiting to see some more solutions for days 2 and 3.
I'll try to post some weekend problems.

[color=blue][size=150]Weekend 14 Problems[/size][/color]

A) We have a unit cube, $ABCDEFGH$ such that $A$ is opposite $H$.  We inscribe a regular tetrahedron in the cube, $AXYZ$, such that $X$, $Y$ and $Z$ are equidistant from $H$, and each is on a face of the cube.  Find the length of an edge of this tetrahedron. (3 pts)

B) There exists $ABCD$, a unit square, in the plane.  We inscribe a circle in it, such that it touches $AB$, $BC$, $CD$ and $DA$ at $E$, $F$, $G$ and $H$, respectively.  The point $I$ exists on this circle such that the area of $EHI$ is maximized.  $EI$ and $EH$ are extended to hit line $CD$ at $J$ and $K$, respectively.  Find the area of quadrilateral $HIJK$. (2 pts)

This contest will be continued as soon as the following poll deciced that there is enough interest in having such a contest ...

 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=54459]www.artofproblemsolving.com/Forum/viewtopic.php?t=54459[/url]

peeta

Hi
and it started again. The Intermediate Daily Contest has new problems. to all future moderators, dont forget that the problem need ato be appropriate for the Imtermediate Level. the rules can be found [url=http://http://www.artofproblemsolving.com/Forum/viewtopic.php?p=331006#p331006]here[/url] and [url=http://http://www.artofproblemsolving.com/Forum/viewtopic.php?p=335463#p335463]here[/url], but dont forget .. the winner of week $n$ will be moderator of week $n+2$ (or needs a substitute), so u have one week to find a substitute or prepare to give nice (!) problems to keep the contest attractive ...
well i ve said enough, here are theproblems for day 1:

[size=150][color=blue]Day 1 week 15[/color][/size]

a) (2 pts)

find all real solutions for:
\[ \begin{array}{rcl}(x+y)^2+3(x+y)&=&4\\ \frac1x+\frac1y &=& -\frac16\end{array} \]

b) (3 pts)

prove the existence of a real solution for:
\[ \sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x \]

please pm the solutions to me, peeta

solve the problem for the day and pm them to peeta.
that simple :)

hey guys ...
i unfortuanatly dint recieve [b]any[/b] solutions ...
i dont think the questions were that hard, if u think otherwise please let me know (pm me or post here) i will handle it as before u have time to send in solutions until today 12:00 (noon).
[b]if in this week there will have been less than 6 participants, the contest will probably not revive until winter or summer break![/b]

here the problems for today:

[size=150][color=blue]Day 2 week 15[/color][/size]

a) (2 pts)

Show that if $a,b,c$ are positive reals with $a+b+c=1$, then
\[ \left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2+\left(c+\frac1c\right)^2\ge \frac{100}3 \]

b) (3 pts)

If $a$ and $b$ are nonzero real numbers, prove that at least one of the following inequalities holds:
\[ \left|\frac{a+\sqrt{a^2+2b^2}}{2b}\right| <1,\qquad \left|\frac{a-\sqrt{a^2+2b^2}}{2b}\right|<1 \]

[size=150][color=blue]Day 3 week 15[/color][/size]

a) (2 pts)

Imagen you have a bar of chocolate. You want to break it into $n\times m$ pieces. How many times do you have to break the bar and then the pieces at least. Prove your answer.

b) (3 pts)

There are two teams of different people. The sum of their members is 20. Some of the people have personal contacts to members of the other team. Is it possible that 5 people have personal contacts to 4 members of the other team, 7 have contacts to 2, and 6 people dont have any contacts to the other team.

Here is the last problem i will post until winter for this Contest if there wont be any responses ...

(4 pts)

Let $\alpha, \beta, \gamma$ be the angles of a triangle. Prove that:
\[ \cos \alpha +\cos \beta + \cos \gamma \le \frac 32 \]

(1 pt)

Find the equality.

This is the solution for the question for 4 points(& 1 point)

Consider L.H.S,
  cosA + cosB + cosC
       = 2cos[(A+B)/2]cos[(A-B)/2] + cosC   ------ [1]

   Now A,B,C are the angles of a triangle,
        A+B=180-C
   Therefore eqn. 1 becomes
          2sin(C/2)cos[(A-B)/2] +cosC <= 2sin(C/2) + cosC ------[2]
     [This is because cos[(A-B)/2] <= 1 ]
           Now consider R.H.S of 2,
       cosC = 1 - 2sin^2(C/2)
            Therefore R.H.S = 2sin(C/2) + 1- 2sin^2(C/2)
    Completing the square,

         -2[sin^(C/2) - 2*(1/2)*sin(C/2) + (1/4)] +1+(1/2)

        =  -2[sin(C/2)-(1/2)]^2 + 3/2
     The min. value of a square is 0.
           Therefore R.H.S of eqn. 2 <= 3/2
      Hence proved.
      The equality holds when cos(A-B)/2 = 1
         i.e. when A=B


[img]6904[/img]

Point $T$ is to be placed at random within the quadrilateral $ABCD$, shown above. If $\triangle ABY$ is equilateral, $\overline{XY}\perp\overline{AD}$, $\overline{AD}\parallel\overline{BC}$, $\overline{BX}\cong\overline{XY}$, and $Y$ is the midpoint of $\overline{AD}$, what is the probability that the point $T$ will be placed within $\triangle BXY$. Prove your answer.

[hide]It can easily be proven that $\triangle BCY$ and $\triangle CDY$ are both equilateral and are congruent to eachother and to $\triangle ABC$. The probability of point $T$ being in $\triangle BXY$ is $\boxed{\frac16}$ because $[BXY]=\frac{[ABCD]}6$.[/hide]

That trapezoid isn't possible.
[hide] Since ${XY}\perp{AD},{XY}\perp{BC}$, because AD is parrellel to BC.  Therefore, since BX=XY, we have that $\angle XBY=\frac{180-90}{2}=45$.  Therefore, because AD is parrellel to BC, Angle BYA is also equal to 45 degrees.  That can't be true because Triangle BYA is equilateral and all angles of an equilateral triangle is 60 degrees.  Therefore, $45=\angle BAC=60$, but $45\ne60$.  Therefore, the trapezoid is not possible.  [/hide]

[quote="anirudh"]I thought so. It was some practice SAT we took and it was on there (I copied it word for word since we received the book) and I couldn't get it there or now.[/quote]

um, since when does sat have you prove stuff... :|

I added the Prove part because I didn't want conjectures and Guess/Check method etc.

The SAT dude says guess and check works :mad:

[quote="mathgeniuse^ln(x)"]Sats Diagrams are not drawn to scale, obviously.

So just do it normally and see what it would be.[/quote]
It doesn't matter if they aren't drawn to scale because you give them values that are either given or can be found.

The answer is 1/6 though but I cannot assume somethign that was given as right as wrong


Suppose you could observe the configurations of a Turing machine as it
executes. Suppose that in each move you observe, the tape head moves
right. Can you tell if the machine is in an infinite loop? How long might
you have to watch in order to draw this conclusion?

I don't think it is possible to decide this.... well, perhaps it depends on weather the input is finite or not. If a turing mashine input must be finite, then you'd have to wait until the head gets past the furthest point to the right it has ever reached before and then wait until it gets into the same state twice... and from that point on it is in an infinite loop.

Isn't this equivalant to the halting problem (which is undecidable)? What do you mean it move right? Do you mean it always moves right for every move?


How do you make tables in $ \text{\LaTeX}$? Can you please give me a few examples? Thanks in advance!  :)

There are lots of examples of tables on the net (including AoPSWiki's [[LaTeX:Layout#Tables]]) but a good place to start is [url=http://www.andy-roberts.net/misc/latex/latextutorial4.html]Getting to Grips with Latex - Tables[/url]. Try out the ideas in [i]Very basic table[/i] for your own tables before progressing on to the more advanced stuff later on that site.
There's lots more out there including a [url=http://en.wikibooks.org/wiki/LaTeX/Tables]Wikibook[/url] and a stunning [url=http://www.tug.org.in/tutorial/chap06-scr.pdf]presentation[/url].

Hi,

I don't like the [i]Getting to Grips[/i] text concerning tables much. Though it covers the basics it uses vertical lines extensively. I don't like vertical lines when reading rows from left to right, they just distract, that's not a good start.
The also mentioned Wikibook is based on it but adds something: have a look at the section [url=http://en.wikibooks.org/wiki/LaTeX/Tables#Professional_tables]Professional Tables[/url] near the end, it will give some hints regarding the design.

There's a german presentation I like, perhaps have a look at it, just to view at the layout of some tables: [url=http://userpage.fu-berlin.de/~latex/Materialien/tabsatz.pdf]tabsatz.pdf[/url]. Even if it's a foreign language one could get an impression of possible layouts and perhaps even understand the LaTeX source code behind.

Here's a newer article, introducing typesetting tables: [url=http://www.packtpub.com/article/creating-tables-latex]Creating Tables with LaTeX[/url]. And if you would like to do something fancy: [url=http://texblog.net/latex-archive/graphics/tikz-table/]Fancy tables with TikZ[/url].

Stefan

Oh! So we can make a table use Latex. Thank you much much. :)  :)  :)

$ \begin{tabular}{|l\parallel{}c|c|c|c|c| |c|}\hline & Blah & Blah & Blah & Blah & Testing & Total \\
\hline This & 1 & 5 & 5 & 2 & 4 & 3 \\
\hline Is & 0 & 2 & 0 & 3 & 3 & 4 \\
\hline Just & 0 & 5 & 2 & 7 & 6 & 9 \\
\hline A & 6 & 2 & 3 & 4 & 6 & 5 \\
\hline Test & 1 & 3 & 5 & 7 & 9 & 3 \\
\hline \end{tabular}$

The |c| means one collum and the | |c| means a double line.
The hline goes above the row after it.


MATHCOUNTS
Euler
number theory
modular arithmetic

if 3^1993=100k+n, where n<100 is a nonnegative integer, what is n

sigh, it is a team round, but bah humbug

Make a list of the final two digits of the powers of 3.  After 20 numbers the pattern starts to repeat.  Divide 1993 by 20: remainder is 13.  13th term of the sequence is 23, so n = 23.

[quote="WarpedKlown1335"]Am I allowed to do this?[/quote]
I believe naught since you're >8th...(nor qualify for it) Oh well, if it's a hard problem I think it's OK.

Hey everyone, I am a coach, but really enjoy this forum.  I read in this thread that some one asked if they were allowed to answer.  Do you not want people above 8th grade posting solutions??  I'll hide the solution on this one and if you don't want me posting solutions I won't.

[hide]I use modular arithmetic for this one.

3 :equiv: 3(mod 100)

3 ^5 :equiv: 243(mod 100)

3 ^5  :equiv: 43( mod 100)

3^10  :equiv: 1849(mod 100)

3^10 :equiv: 49(mod 100)

3^20 :equiv: 49^2 :equiv: 2401 :equiv: 01(mod 100)

3^20 :equiv: 1(mod 100)

3^1980 :equiv: 1(mod 100)

3^1980 x 3^10  :equiv: 1 x 49(mod 100)

3^1990 :equiv: 49(mod 100)

3^1990 x 3^3:equiv: 49x27(mod 100)

3^1993 :equiv: 1323 :equiv: 23(mod 100)

so 3^1993 =100n + 23

[hide][/hide][/hide]

yeah, I basically did the same thing and was weary about people putting the calculator solution.

always look for suave modular manipulations for a question like that

This isn't really especially appropriate for MathCounts or useful for this problem, but you can show that 340 :equiv: 1(mod 100) since  :phi: (100)=40, and then we just use Euler's Theorem.

heh, I think that was in a book I was reading in school today.  But I would not have had the chance to use that method when I did the problem some days back.  Three cheers for Euler phi!


MATHCOUNTS
geometry
geometric transformation
reflection
calculus
AMC
AIME

I have seen recently many people that are in 9th grade but whose problem solving skills are beyond their grade level. But the issue is people often post very hard problems in forums like Getting Started and MATHCOUNTS, even though they should not be there. They think the problems are easy but when they are matched with the forum's level of problems, they are unusually harder. I think there are many confusions as to what is considered 'easy' and 'hard'. These two words, which seem to describe a lot, are very vague. It depends on what you think, not the appropriate level for the forum. When it says 'Getting Started', they tend to think it means 'easy', as opposed to the actual meaning of it.

My purpose here is: I want to see you guys discuss and reflect on issues in this forum that I think many people need to explore. What does it mean by problem level? Does it have to do with grade level? Does age matters in problem solving? Why are you posting this problem in a particular forum in which you think is the appropriate level?

It is very hard to know whether a question is appropriate for the forum, especially if it's like a borderline question.

Sometimes, it's not just the skills that are hard; it's the depth that makes it hard. that's why there isn't really a grade level thing for a problem. A problem that uses elementary skills can be on the high school forums (but the reverse isn't always true).

I don't think it's ever a bad thing to post easier questions on a harder forum. If you just can't decide between intermediate and getting started, post it on intermediate. There's actually a great range of overlap between the forums.

But, it's nearly never a good thing to post harder problems on an easier forum, so when in doubt, move up.

My problem solving skills aren't even near highschool  :roll: , that's why I'm only moderator in the basics  :blush:

[quote="236factorial"]My problem solving skills aren't even near highschool  :roll: , that's why I'm only moderator in the basics  :blush:[/quote]
Have you considered the possibility of moving up yourself, instead of staying forever in Middle School? From your older posts, the level I have seen are very similar to mine. I am going to be 9th grade too. And yet, I have left Middle School because I think it's time to move. A lot of people are comfortable where they are, unmoving, or perhaps lazy/procrastinate. And believe it or not, I know some people in here say they are taking Calculus classes and they are younger than me! I'm just going to take Precalculus this upcoming year. My point here in reply is two-fold: 1) Problem solving has its own level of difficulty; 2) Classes cannot determine the problem solving level.

I am posting a lot more in the high school forums now (the getting started forum that is, intermediate forum is off). I still stick around in the classroom math when I'm bored (I can easily be a mod there).

I think my level is post-MATHCOUNTS/pre-getting started level, as long as the problem level in the getting started forum doesn't elevate, like they have been doing recently  ;) .

[quote="h78reg"]is getting started level like AOPS volume 1, intermediate is volume 2, and pre-olympiad is art and craft of problem solving? because volume 1 seems easier than getting started..............[/quote]

That's because the level was elevated.

[quote="krustyteklown"]236, that's wierd. I remember some Mathcounts being pretty tough near the end. Were you a "Mathlete" who aspired to do every problem, or simply a competitor?[/quote]

I know, I think I'm right on early national level, but still have some trouble on states. that's still a pretty late level for Mathcounts, given that 200 out of 500000 ever make it to nationals.
I'm a competitor, hoping to make it to states next year  :P

i wish my school offered any math competitions :(

I'm on intermediate level right now. Most Getting started problems I can do fairly easily.

What's annoying is that a lot of people who are good at math stay in the classroom/basics forums

[hide="this might offend you"][hide="you might not want to read it"][hide="are you sure"][b]math92[/b]
and

236!

well 236! is not doing it as much anymore, so that's good :) [/hide][/hide][/hide]

What is ----## doing in the classroom forums all the time?? my fifth grade sister goes in there, and some of those problems are definitely not classroom level. And ----## goes and solves them????? he should definitely get out of those forums now! :roll: those problems should be left for 6th graders (and less) who are just getting started with problem solving. same with basics! this is really annoying.

Chess64, do you think I am ready for intermidiate?

Anyway......

Its not a matter of real age, it's a matter of mental age.

[quote="236factorial"]I know, I think I'm right on early national level, but still have some trouble on states. that's still a pretty late level for Mathcounts, given that 200 out of 500000 ever make it to nationals.
I'm a competitor, hoping to make it to states next year  :P[/quote]

You still got time then :D

I made it to state, I placed, don't really remember what place tho, in cali north, in 7th grade. I slacked off and made lotsa mistakes at chapter in 8th, so didn't even advance. I'm going into 10th right now.
Just keep participating and having math in your head, I think that's what my past experience has taught me. :D

Actually, I've always thought that the classroom forum should include just homework type problems.  So you should be able to have 8th grade math in the classroom level, as long as it's something youy would get for homework, and still have 6th grade math in the basics, since it'll make you think a little more.

i'm guessing intermediate problems are similar to AIME, but, what about pre-olympiad? they seem very hard, but, not like USAMO problems either.

[quote="chess64"]What's annoying is that a lot of people who are good at math stay in the classroom/basics forums
[/quote]

Well, I'm a mod in the Basics, so I just look over that forum a lot.
I just post in the marathons to keep it alive.

Besides, what's annoying about being bad  :P at math and staying in the low level forums? Isn't that what you're supposed to do  :D

[quote="chess64"]----## is not bad at math. :roll:[/quote]

I mean me  :blush:

chess64 loves it when I do that.  :P

[quote="PenguinIntegral"]Chess64, do you think I am ready for intermidiate?
[/quote]

PenguinIntegral, look on the intermediate forums a little and see how those problems are for you.

[quote="PenguinIntegral"]Its not a matter of real age, it's a matter of mental age.[/quote]

Well said.  If you are mentally advanced, you should be able to go up a level, or many levels.  But if you are just fine where you are, you should be allowed to stay.  Calendar age or grade level should not have any relevance in this forum.

[quote="chess64"]

[quote="PenguinIntegral"]Chess64, do you think I am ready for intermidiate?
[/quote]

PenguinIntegral, look on the intermediate forums a little and see how those problems are for you.[/quote]
Those problems scare me. :huh:
But really, what is the average AoPS age for each of the fourms?
Im hoping being able to do 5/6 of the problems in "The basics" and age 12.5 is good, but, this is AoPS. For all I know some of you could be training hamsters that are smarter than I. :ninja:

[quote="h78reg"]well, i think the average age for intermediate is probably 14-15 and pre-olympiad is 15-16.olympiad is probably 15-18 and college playground is most likely 16 and above.[/quote]
Thank you :lol:

How about getting started? 13-14? Mathcounts? 12-13? Basics? 11-12, Classroom? Maybe 11 and under (although I still post  :blush: )

[quote="h78reg"]i guess those numbers are fairly accurate. keep in mind, if you want to be an IMO gold medalist, you probably want to be on the olympiad or pre-olympiad section probably by age 14-15 ;)[/quote]
That should be no problem for me, Im already thinking of moving out of GS.

I like it in middle school how they divide it into 4 steps, while in high school, there are only really 3 steps to cover a much broader topic range. thus the big jump. If there was a forum between getting started and intermediate, what would it be called? Is it that necessary (more mods though  :P )?

[quote]I like it in middle school how they divide it into 4 steps, while in high school, there are only really 3 steps to cover a much broader topic range. thus the big jump. If there was a forum between getting started and intermediate, what would it be called? Is it that necessary (more mods though  )?
[/quote]

well, yes, it is necessary. as you've mentioned in your blog, you can solve 50% of GS problems, but only 2% of intermediate problems. shouldn't the last percentage be equal to 15-20%?as for what the forum would be called, i don't know. maybe the forums should be renamed after tests. GS would be AMC 10. the new one i was talking about could be AMC 12-USAMTS Rounds 1 and 2. intermediate would be AIME-USAMTS Rounds 3 and 4. pre-olympiad could be re-named pre-USAMO or something. however, that wouldn't be very friendly for international users.

Mentioning this is easy, but do you know how much work that will be for the admins?

[quote]Called the one between GS and intermidiate"Experienced Beginner"
Like GS but with logs and trig and more advanced things that GS, but still pre Intermidiate.[/quote]

i agree. that would be a great idea. basically, harder than AOPS volume 1, but not as hard as volume 2.

I posted a topic on this in the Questions/Suggestions forum just to let everyone know. Let's let the admins make the final decision.

[quote="h78reg"]there should really be something more intermediary between getting started and intermediate. GS seems like the standard high school curriculum, and maybe some easier AMC 12 material. intermediate is already AIME level problems and proofs. it seems difficult to make the transition.[/quote]

Getting Started is meant for most AMC level problems.  Most AMC level problems [b]are curriculum[/b] [i]with just a touch of problem solving[/i].

The hardest AMC level problems and nearly all AIME problems belong in Intermediate.

If there are proofs that are harder than ordinary AIME problems in Intermediate, the moderators should move them.

I don't see the need for a forum in-between.  I just see the need for forum members to help others remember the appropriate level of difficulty of the problems.

Do we need more moderators?

I have always thought that these issues contribute to excessively difficult problems being posted in each forum:

1) newbies to AoPS who are experienced problem-solvers (or persons educated overseas) not realizing how "easy" is easy enough for the most beginning levels,

and

2) old hands on AoPS getting better at problem-solving (which is the point here, right? :)  ) forgetting that as their own level gets better, they should post in higher-level forums.

To deal with both issues, moderators have to be AGGRESSIVE in "promoting" problems, otherwise everyone thinks that the correct level of each forum is the level of most problems posted to it.

I got a lot of moderator burn-out by trying to keep Getting Started sufficiently easy, back when it was the lowest-level forum. Now that there are even MORE low-level forums, we still have the issue of each forum, at each level, containing too many problems that are too hard.

We DEFINITELY don't need any more forum levels. Most users can't keep the levels straight already. We may need a few more moderators, especially at the lowest levels. And we probably need more encouragement of moderators to resolve all doubts in favor of "promoting" problems, as I still see the main defect being problems that are too hard for each forum, NOT problems that are too easy.

I might rejoin the fray as a lower-level moderator if I were encouraged to do so by management, and assured that I could be aggressive in promoting problems.

tokenadult, we would be happy to have you rejoin the ranks of the moderators in the Middle School forums.

Pick your forums and I'll add you.


search
geometry
geometric transformation
reflection
rotation
calculus

Draw a spinner with 6 letters A, B, C, D, E, and F. Each letter will represent 6 different whole numbers.
The Challenge:
Find values for the 6 sectors so that the number in a sector, or the total of the numbers in a set of adjacent sectors gives all of the whole numbers from 1 to 28. The only operation you may use with adjacent sectors is addition.
Note: Adjacent sectors share a common radius.

This was a supposedly a fun problem for us.  :roll:

I hope someone will get the answer (and possibly have seen this question before). I was wondering if you can solve it algebraically. Reply if you can explain. Thanks!

EDIT: I added an attachment; it is the diagram of the spinner.

i could be wrong, but by computer search i dont think its possible

Whoa, that's crazy. I can't believe our Grade 9 math teacher butchered us with this difficult problem.

On another note, [b]can someone recommend some tutoring schools in the GTA area?[/b] I really want to start going to Saturday classes because I'm not very productive when learning math at home. Reply soon? Thanks.

Try:
1,3,2,7,8,10

I don't know what Alex is on... there are infinitely many solutions to this if you allow negatives as well.
Another: -1,2,3,8,9,6

Program to find them: (PASCAL)

program spinner;
uses wincrt;

var
   min,max:integer;
   v1,v2,v3,v4,v5,v6:integer;
   sums:array[1..31] of integer;
   i,j:integer;
   good:boolean;
   tgood:boolean;

begin
     min:=1;
     max:=15;
     for v1:=min to max do
     for v2:=v1 to max do
     for v3:=v1 to max do
     for v4:=v1 to max do
     for v5:=v1 to max do
     for v6:=v1 to max do
     begin
          sums[1]:=v1;
          sums[2]:=v2;
          sums[3]:=v3;
          sums[4]:=v4;
          sums[5]:=v5;
          sums[6]:=v6;
          sums[7]:=v1+v2;
          sums[8]:=v2+v3;
          sums[9]:=v3+v4;
          sums[10]:=v4+v5;
          sums[11]:=v5+v6;
          sums[12]:=v6+v1;
          sums[13]:=v1+v2+v3;
          sums[14]:=v2+v3+v4;
          sums[15]:=v3+v4+v5;
          sums[16]:=v4+v5+v6;
          sums[17]:=v5+v6+v1;
          sums[18]:=v6+v1+v2;
          sums[19]:=v1+v2+v3+v4;
          sums[20]:=v2+v3+v4+v5;
          sums[21]:=v3+v4+v5+v6;
          sums[22]:=v4+v5+v6+v1;
          sums[23]:=v5+v6+v1+v2;
          sums[24]:=v6+v1+v2+v3;
          sums[25]:=v1+v2+v3+v4+v5;
          sums[26]:=v2+v3+v4+v5+v6;
          sums[27]:=v3+v4+v5+v6+v1;
          sums[28]:=v4+v5+v6+v1+v2;
          sums[29]:=v5+v6+v1+v2+v3;
          sums[30]:=v6+v1+v2+v3+v4;
          sums[31]:=v6+v1+v2+v3+v4+v5;
          good:=true;
          i:=0;
          repeat
               i:=i+1;
               tgood:=false;
               j:=0;
               repeat
                     j:=j+1;
                     if (sums[j]=i) then
                        tgood:=true;
               until ((j=31) or (tgood=true));
               if tgood=false then good:=false;
          until ((i=28) or (good=false));
          if good=true then
          begin
             write('FOUND IT: ');
             write(v1,' ');
             write(v2,' ');
             write(v3,' ');
             write(v4,' ');
             write(v5,' ');
             writeln(v6,' ');
          end;
     end;
end.

yes, my bad, i didnt count the possibility that the numbers could be out of order for some reason

[quote="Elyot"]I don't know what Alex is on... there are infinitely many solutions to this if you allow negatives as well.
Another: -1,2,3,8,9,6
[/quote]
[quote="d.lam.86"]Draw a spinner with 6 letters A, B, C, D, E, and F. Each letter will represent 6 different whole numbers. [/quote]

It says "whole numbers", so negative numbers are not allowed.

Wow, cool. Our teacher said that there are a lot of solutions. I was wondering, what method did you use? Is it mainly trial and error and PASCAL (what's that?  :oops: )? I've heard of it at Programming Club before but I don't really know exactly what it is.

I know Math League is looked down upon on this forum.  :roll:  Did anyone do it by any chance? Any thoughts? I found it was a lot easier than the practice material my teacher gave me.

[quote="Elyot"]I guess so.

In that case there are 6 unique solutions:

1 2 5 4 6 13
1 2 7 4 12 5
1 3 2 7 8 10
1 3 6 2 5 14
1 3 11 5 2 6
1 7 3 2 4 14

each of these has a reflection and 6 rotations, so there are a total of 2*6*6 solutions... or 72 solutions.[/quote]
Does reflection refer to, for example: 13 6 4 5 2 1, in the first solution? When you mean rotation, does that mean that I can shift the numbers around into six different sets? I'm not quite sure how that works; the sum would be different if I shifted them around.  :huh: I might be wrong after all...  :blush:

reflection means that 1 2 3 4 5 6 becomes 6 5 4 3 2 1.
rotation means that 1 2 3 4 5 6 becomes 2 3 4 5 6 1, etc.
The sum does not change.
You can change a couple lines of the program to either show all solutions or only the unique ones.

[quote="d.lam.86"] [b]can someone recommend some tutoring schools in the GTA area?[/b] I really want to start going to Saturday classes because I'm not very productive when learning math at home. Reply soon? Thanks.[/quote]

Check out 'don valley academy', just google and you'll find there number. The math faculty is really bad at LHS ( just ask Alex ), I so far have enjoyed both group or single lessons for some of the G12 cources. Correct me if I'm wrong, but I think all of the math teachers are at least Grad stundents in math/sciences with math. At least mine was.

Wierd look on my face, when i find the guy teaching me Calculus and Descreate Was going for his phd in Math at UFT, untill NASA offered him big $\$$ for his work on flight computers. Only reason he is teaching kids like me: he had to quit down in florida, becuase his family was sick back here in toronto =(.


[quote="math92"]divide (1+4)+(1+8)+(1+12)+(1+16) by 4.  what is the remainder?[/quote]

[size=150]R=0

A good mental math warm-up for students.

Pencil and paper (optional)[/size]


Prove that ($ a,b,c,x,y,z\in\mathbb{R^\plus{}}$): \[ x^2\left(\frac{b\plus{}c}a\right)\plus{}y^2\left(\frac{a\plus{}c}b\right)\plus{}z^2\left(\frac{a\plus{}b}c\right)\ge 2(xy\plus{}xz\plus{}yz)\]

[quote="dogcat"]Prove that ($ a,b,c,x,y,z\in\mathbb{R^ \plus{} }$):
\[ x^2\left(\frac {b \plus{} c}a\right) \plus{} y^2\left(\frac {a \plus{} c}b\right) \plus{} z^2\left(\frac {a \plus{} b}c\right)\ge 2(xy \plus{} xz \plus{} yz)
\]
[/quote]

$ x^2\left(\frac {b \plus{} c}a\right) \plus{} y^2\left(\frac {a \plus{} c}b\right) \plus{} z^2\left(\frac {a \plus{} b}c\right)\equal{}[x^2(\frac ba)\plus{}y^2(\frac ab)]\plus{}[x^2(\frac ca)\plus{}z^2(\frac ac)]\plus{}[y^2(\frac cb)\plus{}z^2(\frac bc)]\geq2(xy\plus{}yz\plus{}zx)$ by AM-GM.


See http://www.mathlinks.ro/Forum/viewtopic.php?t=17549 beginning at post #7.

  darij