geometry
3D geometry
analytic geometry
graphing lines
slope
geometric transformation
rotation

find the critical time period of revolution in terms of relevant variables that must be achieved to keep a person stuck to the side of the wheel

for those that aren't aware, this thing is a cylinder which you go inside, and put your back to the wall. It them spins round and after a certain angular frequency has been achieved the floor falls and you stay stuck to the wall. great fun, athough I strained my neck trying to look at my feet for too long!

[hide]A body is stuck to the surface (the inside wall) of a vertically spinning cylinder.
There are three forces acting.
The weight of the body acting vertically downwards, the friction force from the surface of the cylinder's inner wall, acting vertically upwards and the centripetal force from the wall of the cylinder directed radially inwards and which provides the normal force, of the cylinder against the body, used to calculate the friction.

$F_c = \frac{mv^2}{R} = m\omega ^2 R$

$F_r = \mu F_c = mg$ ($F_r$ is the friction, and $F_c$ is the normal force)

$\mu m \omega ^2 R = mg$
$\omega ^2 = \frac{g}{\mu R}$
$\omega = \sqrt{\frac{g}{\mu R}}$
$\frac{2\pi}{T} = \sqrt{\frac{g}{\mu R}}$
$T = 2\pi \cdot \sqrt{\frac{\mu R}{g}}$
=============[/hide]

neat.

what about if the cylinder is replaced with a cone, how will the angle of banking affect it?

Slightly more complicated :)

And there are two situations to consider.

1) The body slides down - friction acts up the slope
2) The body slides up - friction acts down the slope

I've made up a couple of sketches to facilitate the explanation.

In Fig 1, I've simply drawn a body lying on a sloped surface (the inside surface of the cone) and the forces acting. There is the weight of the body acting down, mg. The normal reaction of the cone surface, Fn. And in this case I'm assuming the speed of rotation is slow enough that the body will tend to slide downwards and so friction will act upwards.

In Fig 2, I've resolved the forces acting on the body into horizontal and vertical components.
We have Fn.sinα acting radially inwards and Fr.cosα acting radially outwards. The difference between these two forces is what provides the centripetal force. i.e. Fc = Fn.sinα - Fr.cosα.

[u]Force Ananlysis[/u]
[u]Vertical forces[/u]
$F_n\cos\alpha + F_r\sin\alpha = mg ---------------------(1)$

[u]Horizontal forces[/u]
$F_n\sin\alpha - F_r\cos\alpha = F_c = \frac{mv^2}{R} = m\omega ^2R ---------------(2)$

[u]Friction force[/u]
$F_r = \mu F_n$

Substituting for $F_r = \mu F_n$ in (1) and (2),

$F_n(\cos\alpha + \mu\sin\alpha) = mg$
$F_n(\sin\alpha - \mu\cos\alpha) = m\omega ^2R$

Dividing 1st eqn by 2nd,

$\frac{(\cos\alpha + \mu\sin\alpha)}{(\sin\alpha - \mu\cos\alpha)} = \frac{g}{\omega ^2R}$
$\omega ^2 = \frac{g}{R}\cdot\frac{(\sin\alpha - \mu\cos\alpha)}{(\cos\alpha + \mu\sin\alpha)}$
$\frac{2\pi}{T} = \sqrt{\frac{g}{R}}\cdot\sqrt{\frac{\sin\alpha - \mu\cos\alpha}{\cos\alpha + \mu\sin\alpha}}$
$T = 2\pi\sqrt{\frac{R}{g}}\cdot\sqrt{\frac{\cos\alpha + \mu\sin\alpha}{\sin\alpha - \mu\cos\alpha}}$
=========================

[u]When friction acts down[/u]
$F_n\cos\alpha - F_r\sin\alpha = mg$
$F_n\sin\alpha + F_r\cos\alpha = F_c = \frac{mv^2}{R} = m\omega ^2R$

You can see that there is simply a change in two of the signs compared to the eqns for the first case. This results in a final expression for the time period as,

$T = 2\pi\sqrt{\frac{R}{g}}\cdot\sqrt{\frac{\cos\alpha - \mu\sin\alpha}{\sin\alpha + \mu\cos\alpha}}$
=========================

A little further analysis will show that in order to provide real solutions for T, the discriminant needs to be non-zero. Giving,

$\cos\alpha - \mu\sin\alpha > 0$
$\cos\alpha > \mu\sin\alpha$
$\tan\alpha < \frac{1}{\mu}$

This gives a limiting value of α=90 deg (cone = a straight-walled cylinder) and would need a frictionless surface for the inner wall of the cone.
This tends to confirm what would be expected in practice.


Putnam
linear algebra
matrix
geometry
rectangle
algorithm
graph theory

In all unit squares of a $2000*2000$ board there is a number $1$ or $-1$, if the sum of all numbers in the board is non negative, prove that there exist $1000$ columns and $1000$ rows in the board such that the sum of the numbers in the $1000^{2}$ unit squares of the intersections is at least $1000$.

First, let me give you a way to find the solution. Back when I was on the Putnam team at Michigan State University, I learned Herzog's rule: look at seven easy cases and try to find useful properties in them (Herzog was a former coach of that team). In this problem, the best way to find easy cases is to generalize the problem to a board of variable size, and then examine small, manageable cases.

In all unit squares of a $2n\times 2n$ board there is a number $+1$ or $-1$. If the sum of all numbers in the board is non-negative, prove that there exist $n$ columns and $n$ rows in the board such that the sum of the numbers in the $n^{2}$ unit squares of the intersections is at least $n$.

The $2\times 2$ case is boringly trivial, so let's look at the $4\times 4$ case. We need a reasonably balanced matrix for an example.
\[\begin{matrix}+1&+1&-1&-1\cr+1&-1&+1&-1\cr-1&+1&+1&+1\cr+1&-1&-1&-1\cr \end{matrix}\]
We want 4 entries arranged in a rectangle whose sum is 2 or more. That's easy, the upper left hand square works: it has three $+1$'s and one $-1$. Or the intersection of the first two columns and the top row plus any other row works too. That pair of $+1$'s in the top row is what pushes the sum up to 2.

Of course, a pair of $+1$'s in a row is not always enough. The  pair of $+1$'s in the second and third entries of the third row are matched by a pair of $-1$'s in the second and third entries of the fourth row, so that square has a sum of 0. But pairs of $-1$'s can't be overwhelmingly common, because the sum of all the entries is non-negative.

Let's move up to a $6\times 6$ example.
\[\begin{matrix}+1&+1&+1&-1&-1&-1\cr+1&+1&-1&+1&-1&-1\cr+1&-1&-1&+1&+1&-1\cr+1&-1&-1&+1&+1&-1\cr-1&+1&-1&+1&-1&+1\cr-1&-1&-1&+1&+1&+1\cr \end{matrix}\]
Here we want the sum to be 3 or more. Once you find a good set of rows and columns, swap a $+1$ and $-1$ so that it doesn't work anymore, and try again. See what properties you find.

At this point, the right approach looks like a greedy algorithm rather than graph theory (warning: the greedy algorithm does not count as a proof, so you have construct a separate proof along with it). I'll post a solution next week, unless someone else beats me to it, or you ask for it sooner.

Erin Schram, mathematician

This is not a difficult problem.
We denote the matrix by $a$
$b_{i}=\sum_{j=1}^{2n}a_{i,j}$,we can see that $b_{i}$ is even.
$b_{1}\geq b_{2}\geq ...\geq b_{2n}$
if $b_{n}\geq 2$ then $\sum_{j=1}^{n}b_{j}\geq 2n$  (1)
the problem is solved already.(Think about why)
if $b_{n}\leq-2$ then $\sum_{j=n+1}^{2n}b_{j}\leq-2n$
then $\sum_{j=1}^{n}b_{j}\geq 2n$
same as (1)
So we can only consider the situation when $b_{n}=0$
We can suppose $a_{n,1}=a_{n,2}=...=a_{n,n}=1,a_{n,n+1}=a_{n,n+2}=...=a_{n,2n}=-1$
We denote $c_{i}=\sum_{j=1}^{n}a_{i,j}$
We can rearrange the matrix so that $n=c_{1}\geq c_{2}\geq ...\geq c_{2n}$
if $c_{n}\geq 0$ then we have one $n*n$ submatrix satisfied the condition.
if $\sum_{j=1}^{n}c_{j}<n$ then
$c_{n+1}\leq-2,\sum_{j=n+1}^{2n}c_{j}\leq-2n$
$\sum_{j=1}^{2n}c_{j}<-n$
We denote that $d_{i}=b_{i}-c_{i}$
then $\sum_{j=1}^{2n}d_{i}>n$ (2)
We can rearrange the matrix so that $d_{1}\geq d_{2}\geq ...\geq d_{2n}=-n$
if $\sum_{j=1}^{n}d_{j}<n$ then $d_{n+1}\leq 0$
then $\sum_{j=n+1}^{2n}d_{j}<0$
$\sum_{j=1}^{2n}d_{j}<n$
contradict with (2)
here we done

P.S. (To upstairs)
  Beat you dude

[quote="Bourne"]Beat you dude[/quote]
Yes, you did.

Bourne's solution matches my first solution, except I had the steps in different order (I had started with a row that was at least half $+1$'s). The basic idea is to try to construct a solution in a greedy way, and if the construction fails, the opposite part of the matrix must have a sum greater or equal to $n$ to make up for all the $-1$'s in the failed construction.

However, I later found a second solution that does not split into cases like that.

We will select the $n$ columns and $n$ rows by crossing out rows and columns one by one in $n$ two-part steps. We start step $k$ with a $(2n-k+1)\times(2n-k+1)$ matrix. First, we cross out the row with the lowest sum, leaving a $(2n-k)\times(2n-k+1)$ matrix. Second, we cross out the column with the lowest sum, leaving a $(2n-k)\times(2n-k)$ matrix.

When we cross out a row or column, if its entries have a negative sum, then the sum of the entries in the remaining matrix increases. If its entries have a positive sum, then all the other rows or columns have positive sum too and the sum of the entries in the remaining matrix is at least $2n-k$. If its entries have a zero sum, the sum in the remaining matrix does not change. However, we cross out a row of length $2n-k+1$ and a column of length $2n-k$. One of them has odd length, so its sum is odd and cannot be zero.

So after each step, either the sum of the entries in the remaining matrix has increased, or the sum is greater than or equal to $2n-k$. Thus, after step $k$ for $1\leq k\leq n$ the sum of the entries in the remaining matrix is at least $k$. Therefore, after step $n$, where we have $n$ rows and $n$ columns left, the sum of the intersection of the $n$ rows and $n$ columns is at least $n$.

Erin Schram


group theory
abstract algebra
superior algebra
superior algebra unsolved

Aim: shows |Aut(S6) : Inn (S6)|<= 2

a) Let K be the conjugacy class of transpositions in S6 and let K' be the conjugacy class of any element of order 2 in S6 that is not a transposition. Prove that |K|=/=|K'| unless K' is the conjugacy class of products of three disjoint transpositions. Deduce that Aut(S6) has a subgroup of index at most 2 which sends transpositions to transpositions.

b) Prove that |Aut(S6) : Inn(S6)|<= 2.

For part a) ..I can only do the [b]|K|=/= |K'| unless K' is class of products of three disjoint transpositions[/b] part. I've no idea on how to do the deduce part....can some one help me??
Also, I've no clue on b)

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=16615]here[/url] for more information; this is also linked in Xevarion's thread, which you should have looked at before posting this.

um...... :maybe:
so any element from the subgroup should send K to K
or K to K'?????

But I don't get the "which send transposition to transposition" part...

I mean...the whole structure of this problem still seems so blurry to me.....


calculus
derivative
function
trigonometry
search
real analysis
real analysis unsolved

Prove that if f is of clase $ C^3$ and $2\cdot\pi$ periodic then the function $ f'+f'''$ has at least 3 zeroes on $[0,2\cdot\pi]$ . Nice problem!

[quote]$f$ is $2\pi$ periodic and $\mathcal{C}^3$ so :
\[\lim_{h\rightarrow 0, h>0} \frac{f(h)-f(0)}{h}=\lim_{h\rightarrow 0, h>0} \frac{f(2\pi+h)-f(2\Pi)}{h}\Rightarrow f'(0)=f'(2\pi)\]
\[\lim_{h\rightarrow 0, h>0} \frac{f(h)-f(0)}{h}=\lim_{h\rightarrow 0, h<0} \frac{f(h)-f(0)}{|h|}=\lim_{h\rightarrow 0, h<0} -\frac{f(2\pi+h)-f(0)}{h}\Rightarrow f'(0)=-f'(2\pi)\]
So $f'(0)=f'(2\pi)=0$. By same reasons we show that $f''(0)=f''(2\pi)=f'''(0)=f'''(2\pi)=0$
So $(f'+f''')(0)=(f'+f''')(2\pi)$ and by theorem of finites "acroissements" (sorry I don't know the english word) it exists $x_0, 0<x_0<2\pi$ such $(f'+f''')(x_0)=0$[/quote]
False

I didn't read it thoroughly, but if you got $f'(0)=0$ there must be an error. Take $f(x)=\sin x$, for example. It satisfies every condition, and $f'(0)=\cos 0=1$.

Oups that's only $f'(0)=f'(2\pi)$
The second line is false, it's a big mistake sorry. I search good solution ...

It's easy to show there is two zeros but three ... I don't see how.


a uniform disc of radius $ R$ is placed over another uniform disc of radius $ 2R$ which has the same thickness and density as the former.Find the center of mass of the system

$ M \equal{} \rho \pi 4R^2 h \\
 \\
m \equal{} \rho \pi R^2 h \\
 \\
x_{_{C.M.}} \equal{} \frac {M(0) \plus{} m(h)}{M \plus{} m} \equal{} \frac {h}{5}$

sorry rajiv but at the  back of the book the answer is given as $ \frac{R}{5}$.

Is there a diagram in the book? Is mine correct? Maybe the setup is different or the book might say $ R \equal{} h$.

the diagram is not given in the  book. its given that the peripheries of the discs touch each other.
but your diagram satisfies this . ithink.

No. My diagram is wrong.

[url=http://www.thefreedictionary.com/periphery]Click to reveal hidden content[/url]

Nevertheless, the method to find location of C.M remains same. You could use what I did above as a guide and get the answer. In case of 3D setup, you have to find $ x_{_{C.M.}}$, $ y_{_{C.M.}}$, and $ z_{_{C.M.}}$ to get the co-ordinates of the center of mass.


calculus
derivative
integration
function
rate problems
calculus computations

b) To find inflection points, recall that you need to find where all of the second derivatives are zero.  So we need $ g''$.  But this is easy by FTC:
We know $ g(x)\equal{}\int_{2}^{x}f(t)\,dt\Rightarrow g'\equal{}f(x)\Rightarrow g''\equal{}f'(x)$.
So you need where $ f'(x)\equal{}0$.  This is easy just by looking at the picture (you can do this).

c) Rate of change is given by the derivative.  Average rate of change is given by the MVT for integrals:
$ f(c)\equal{}\frac{1}{b\minus{}a}\int_{a}^{b}f(x)\,dx$
Now the derivative of $ g$ is just $ f(x)$ from above, so evaluating the integral is easy since we have the graph (you can do this).

d) You can do this.

[quote]c) Rate of change is given by the derivative. Average rate of change is given by the MVT for integrals: $f(c)=\frac{1}{b-a}\int_{a}^{b}f(x)\,dx$

Now the derivative of  g is just  f(x) from above, so evaluating the integral is easy since we have the graph (you can do this). [/quote]

in order to find the equation for the graph of f(x), would you substitute, u=x from the g(x) equation and solve then using the average rate of change? so that it looks like : (1/6) integral from 2 to 8 f(u) ?

In this case, the explicit equation of $ f$ is irrelelvant.  The graph of $ f$ is a series of straight lines, which means you can evaluate the integral by using triangles.  That is, just look at the picture.

[quote="cacti22"]I get a) as 0 because g'(x)=f(x) and f(1)=0.

but i don't understand b c d and how to find what they are asking for (inflection points, avg rate of change) for g when we are only given the graph of f.

:([/quote]
Would you post the ans. to the questions for me to check?

 :lol:  :o

[quote="cacti22"]I get a) as 0 because g'(x)=f(x) and f(1)=0.

but i don't understand b c d and how to find what they are asking for (inflection points, avg rate of change) for g when we are only given the graph of f.

:([/quote]

[hide="Solution to $ a)$"]
We have the formula:
$ \frac {d}{dx} \left( \int^{h(x)}_{g(x)} f(t)dt \right) \equal{} foh(x) \cdot h'(x) \minus{} fog(x) \cdot g'(x)$

Using this, we get:
$ g'(x) \equal{} f(x)(1) \minus{} f( \minus{} 2)(0) \equal{} f(x)$

Hence, $ \boxed {g'(1) \equal{} f(1) \equal{} 0}$
[/hide]

[quote="I"]
[hide="Solution to $ a)$"]
We have the formula:
$ \frac {d}{dx} \left( \int^{h(x)}_{g(x)} f(t)dt \right) \equal{} foh(x) \cdot h'(x) \minus{} fog(x) \cdot g'(x)$

Using this, we get:
$ g'(x) \equal{} f(x)(1) \minus{} f( \minus{} 2)(0) \equal{} f(x)$

Hence, $ g'(1) \equal{} f(1) \equal{} 0$
[/hide][/quote]
[hide="Solution to $ b)$"]
Since the graph of $ f$ consists of straight lines, the graph of $ g$ which is the value of the integral of $ f$ will consists of curves. Hence we need to find points at which $ g$ changes nature from [b]convexity to concavity[/b] or vice versa. And that is when $ g''(x) \equal{} 0$ and since $ g''(x) \equal{} f'(x)$ the points at which $ f'(x) \equal{} 0$ are the points of inflection for $ g$ which means no points of inflection.  :D
[/hide]

Can someone explain what is meant by [b]Average Rate of Change of $ g$ ?[/b]
 :mad:

The average rate of change of some function $ g$ on an interval $ [a,b],\ b > a$ is given by $ \frac {\Delta y}{\Delta x} \equal{} \frac {g(b) \minus{} g(a)}{b \minus{} a}$.

I this case, you need to handle it for integrals.


LaTeX
inequalities
algebra
polynomial
inequalities unsolved

I will help you latex!
Let $ a,b,c\ge 0$ and $ a^2\plus{}b^2\plus{}c^2\equal{}1$. Prove that:
$ \frac{a}{\sqrt {1\plus{}bc}}\plus{}\frac{b}{\sqrt{1\plus{}ac}}\plus{}\frac{c}{\sqrt{1\plus{}ab}}\le \frac{3}{2}$.

please help me . Why don't anybody solve it ?

[quote="thanhnam2902"]I will help you latex!
Let $ a,b,c\ge 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. Prove that:
$ \frac {a}{\sqrt {1 \plus{} bc}} \plus{} \frac {b}{\sqrt {1 \plus{} ac}} \plus{} \frac {c}{\sqrt {1 \plus{} ab}}\le \frac {3}{2}$.[/quote]
[url=http://www.mathlinks.ro/viewtopic.php?t=117906]See my solution here[/url] :)

[quote="thanhnam2902"]I will help you latex!
Let $ a,b,c\ge 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. Prove that:
$ \frac {a}{\sqrt {1 \plus{} bc}} \plus{} \frac {b}{\sqrt {1 \plus{} ac}} \plus{} \frac {c}{\sqrt {1 \plus{} ab}}\le \frac {3}{2}$.[/quote]
that's my solution
use Cauchy-Schwarz
$ (\sum {\frac{a}{{\sqrt {1 \plus{} bc} }}} )^2  \le (\sum a )(\frac{a}{{1 \plus{} bc}})$
but $ \sum {\frac{a}{{1 \plus{} bc}}}  \minus{} \sum a  \equal{}  \minus{} abc\sum {\frac{1}{{1 \plus{} bc}}}  \le  \minus{} abc\frac{9}{{\sum {ab}  \plus{} 3}} \le  \minus{} \frac{9}{4}abc$
Setting p=a+b+c,q=ab+bc+ca,r=abc.
q(q-9/4r)\le  9/4
$ \Leftrightarrow 4\left( {q \minus{} 1} \right)^2  \ge 0$
 :)


Find Max of  $ S(x,y) \equal{} inf(x, \frac {1}{y}, y \plus{} \frac {1}{x})$  , $ x,y > 0$.

[hide]Substitute $ w \equal{} \frac {1}{y}$.  Now the problem is even more nice:  we want to find $ M \equal{} \text{Max Min} \{ x, w, x^{ \minus{} 1} \plus{} w^{ \minus{} 1} \}$.  Assume WLOG $ w \le x$.  Now, if $ w$ is less than $ 1$, then clearly $ M \equal{} w < 1$.  But we can do better.

The other case is that $ w$ and $ x$ are greater than or equal to $ 1$.  Since $ x \ge w$, $ M$ can only possibly be $ w$ or $ 1/w \plus{} 1/x$.

If $ M \equal{} w$, then
$ 1/w \plus{} 1/x \ge w \iff w^2x \le x \plus{} w \le 2x \iff w \le \sqrt {2}$.  The maximum $ M$ in this case, then, is when $ w \equal{} x \equal{} \sqrt {2}$.

Otherwise, if $ M \equal{} 1/w \plus{} 1/x$, then
$ w \ge 1/x \plus{} 1/w \iff w^2x \ge w \plus{} x \ge 2w \iff wx \ge 2$.  The maximum $ M$ in this case also occurs when $ w \equal{} x \equal{} \sqrt {2}$.

Thus, the maximum of $ S(x,y)$ occurs at $ S( \sqrt{2} , \frac{1}{\sqrt{2}} ) \equal{} \sqrt{2}$.[/hide]

I think there's some problem in your proof in this equivalence here:
[quote="archimedes1"]$ w^2x \le x \plus{} w \le 2x \iff w \le \sqrt {2}$[/quote]

Anyway, here was my proof:
[hide]
As above, Let $ t \equal{} \frac {1}{x}$ then we are to find Max Of $ Inf(x,t,\frac {1}{x} \plus{} \frac {1}{t})$
assume $ x\le t$ then $ S(x,y) \equal{} Inf(x,\frac {1}{x} \plus{} \frac {1}{t})$
but $ S(x,y) \le \sqrt {x( \frac {1}{x} \plus{} \frac {1}{t})} \equal{} \sqrt {1 \plus{} \frac {x}{t}} \le \sqrt {2}$ and equality holds iff $ x \equal{} t$ and $ x \equal{} \frac {1}{x} \plus{} \frac {1}{t}$, ie: $ x \equal{} t \equal{} \sqrt {2}$

[/hide]

[quote="archimedes1"][quote="FOURRIER"]I think there's some problem in your proof in this equivalence here:
[quote="archimedes1"]$ w^2x \le x \plus{} w \le 2x \iff w \le \sqrt {2}$[/quote]
[/quote]
We assumed $ w \le x$ and we know that $ w,x$ are positive.  So where was the error?[/quote]

in the equivalence,

if $ w \le \sqrt {2}$ then prove $ w^2x \le x \plus{} w \le 2x$

It's only one implication, and the other is false.


The number $2000 = 2^4 \cdot 5^3$ is a product of $7$ prime factors. Let $x$ be the smallest integer greater than $2000$ with this property (also a product of $7$ prime factors) and $y$ be the largest integer smaller than $2000$ with this property. What is the value of $x - y$?

(A) $100$
(B) $64$
(C) $280$
(D) $203$
(E) $96$

Source: Australian Math Competition


Να βρεθεί η κλειστή θήκη του συνόλου $ A \equal{} \{(x,y,z)\in R^3|1 < x^2 \plus{} \frac {y^2}{9} \plus{} \frac {z^2}{4} < 4\}$
Εντάξει τώρα λογικά είναι το $ B \equal{} \{(x,y,z)\in R^3|1 \leq x^2 \plus{} \frac {y^2}{9} \plus{} \frac {z^2}{4}\leq4\}$ αλλά πώς το δείχνουμε;Με την τομή της ανοιχτής μπάλας με το σύνολο κτλ δεν μπόρεσα να βγάλω άκρη...μπορεί να βοηθήσει κανείς;

Θες να δείξεις 4 πράγματα

1) Αν $ x^2 \plus{} y^2/9 \plus{} z^2/4 > 4$, τότε υπάρχει ανοιχτή μπάλα γύρω από το $ (x,y,z)$ που δεν τέμνει το Α.
2) Αν $ x^2 \plus{} y^2/9 \plus{} z^2/4 < 1$, τότε υπάρχει ανοιχτή μπάλα γύρω από το $ (x,y,z)$ που δεν τέμνει το Α.
3) Αν $ x^2 \plus{} y^2/9 \plus{} z^2/4 \equal{} 4$, τότε όλες οι ανοιχτές μπάλες γύρω από το $ (x,y,z)$ τέμνουν το Α.
4) Αν $ x^2 \plus{} y^2/9 \plus{} z^2/4 \equal{} 4$, τότε όλες οι ανοιχτές μπάλες γύρω από το $ (x,y,z)$ τέμνουν το Α.

Υπάρχει κάποιο συγκεκριμένο από αυτά που σε δυσκολεύει;

Έτσι που το λες τα δύο πρώτα δεν τα έκανα γιατί έδειξα ότι το $ B$ είναι κλειστό και μετά βγαίνει νομίζω πιο εύκολα
Τα 3,4 δεν μπορώ να δείξω. :(

Πάρε $ (x,y,z)$ με $ x^2 \plus{} y^2/9 \plus{} z^2/4 \equal{} 1$ και μπάλα $ B$ ακτίνας $ 2\varepsilon > 0$ γύρω από το $ (x,y,z)$. Μπορείς να υποθέσεις ότι $ \varepsilon \leq |x|$. Αν $ x \geq 0$, δείξε ότι το $ (x \plus{} \varepsilon,y,z) \in A \cap B$. Αν $ x \leq 0$, δείξε ότι το $ (x \minus{} \varepsilon,y,z) \in A \cap B$.

Ευχαριστώ πολύ για τη βοήθεια :) Θα το τσεκάρω τώρα :lol:

Νομίζω στη λύση σου Demetres πρέπει να πάρουμε και $ \varepsilon \leq \frac{1}{2}$...εμένα τουλάχιστον δε μου βγαίνει αλλιώς :maybe: Και αν $ x\equal{}0$ δεν δουλεύει...αλλά τότε πάμε ανάλογα για $ y$ κτλ;
Να ρωτήσω και κάτι άλλο πάνω σε αυτή την άσκηση...το σύνολο $ A$ είναι κυρτό;Αν όχι ποια είναι η κυρτή θήκη του;
Αυτό πώς βγαίνει; :maybe:

Νομίζω πως $ \varepsilon \leq |x|$ πάλι δουλεύει. (Έχουμε $ |x| \leq 1$.)  Φυσικά αν σε βολεύει να υποθέσεις $ \varepsilon \leq 1/2$ για να το αποδείξεις, δεν υπάρχει κανένα πρόβλημα  :)  Αν το $ x\equal{}0$ έχεις δίκιο πως δεν δουλεύει. Δεν το είχα προσέξει. Όπως είπες όμως πάμε ανάλογα.

Για την κυρτότητα σκέψου το γεωμετρικά. Πια νομίζεις πως είναι η κυρτή θήκη του;

Δεν έχουμε $ |x| \leq 1$ αφού π.χ. με $ x\equal{}\minus{}\frac{5}{4},y\equal{}z\equal{}0$ ισχύει η ανισότητα.Έχουμε $ |x|<2$ και γι'αυτό δε βγαίνει αν δεν πάρουμε και $ \varepsilon\leq\frac{1}{2}$.
Για την κυρτή θήκη πώς να το σκεφτώ γεωμετρικά; :huh: Δεν μπορώ καν να φανταστώ πώς θα είναι το σχήμα εδώ :blush:

[quote="giannis18"]Δεν έχουμε $ |x| \leq 1$ αφού π.χ. με $ x \equal{} \minus{} \frac {5}{4},y \equal{} z \equal{} 0$ ισχύει η ανισότητα.Έχουμε $ |x| < 2$ και γι'αυτό δε βγαίνει αν δεν πάρουμε και $ \varepsilon\leq\frac {1}{2}$.
Για την κυρτή θήκη πώς να το σκεφτώ γεωμετρικά; :huh: Δεν μπορώ καν να φανταστώ πώς θα είναι το σχήμα εδώ :blush:[/quote]

Πιο πάνω μιλούσα για την περίπτωση $ x^2 \plus{} y^2/9 \plus{} z^2/4 \equal{} 1$. Το σημείο (-5/4,0,0) ανήκει στο Α, άρα και στην κλειστή θήκη του Α. Αυτό που σκεφτόμουνα είναι:

Αν $ x^2 \plus{} y^2/9 \plus{} z^2/4 \equal{} 1$, τότε $ (x \plus{} \varepsilon)^2 \plus{} y^2/9 \plus{} z^2/4 \leq (2x)^2 \plus{} y^2/9 \plus{} z^2/4 \equal{} 1 \plus{} 3x^2 \leq 4$. (Μάλλον $ \varepsilon < |x|$ ήθελα  :lol: )

Για την κυρτή θήκη. Τα $ \{(x,y,z): x^2 \plus{} y^2 \plus{} z^2 \leq 1\}$ και $ \{(x,y,z): x^2 \plus{} y^2 \plus{} z^2 < 4\}$ είναι μπάλες. Το $ \{(x,y,z): 1 < x^2 \plus{} y^2 \plus{} z^2 < 4\}$. Είναι μια μεγάλη μπάλα με μια μικρή ομόκεντρη μπάλα αφαιρεμένη από μέσα. Το $ \{(x,y,z): 1 < x^2 \plus{} y^2/9 \plus{} z^2/4 < 4\}$ είναι ελλιψοειδές με ένα μικρότερο ελλιψοειδές αφαιρεμένο από μέσα. (Σαν ένα αυγό με αφαιρεμένο τον κρόκο   :)  ) Άρα η κυρτή θήκη πρέπει να είναι το $ B \equal{} \{(x,y,z):  x^2 \plus{} y^2/9 \plus{} z^2/4 < 4\}$. Για να δείξουμε ότι όντως αυτή είναι η κυρτή θήκη πρέπει να δείξουμε

1) Το Β είναι κυρτό.
2) Κάθε $ (x,y,z)$ με  $ x^2 \plus{} y^2/9 \plus{} z^2/4 \leq 1$ ανήκει στη κυρτή θήκη του A.

[quote="giannis18"]Καταρχήν καλή χρονιά σε όλους και ευχαριστώ πολύ Δημήτρη για τη βοήθεια. :)
[quote="Demetres"]
2) Κάθε $ (x,y,z)$ με  $ x^2 \plus{} y^2/9 \plus{} z^2/4 \leq 1$ ανήκει στη κυρτή θήκη του A.[/quote]
Για αυτό υπάρχει κάποιο θεώρημα που απλοποιεί τα πράγματα; :maybe: Εκείνο που σκέφτηκα(δεν ξερω όμως κατά πόσο είναι σωστό) είναι:Έστω $ (x,y,z)$ με  $ x^2 \plus{} y^2/9 \plus{} z^2/4 \leq 1$ τότε θεώρησα $ l_1,l_2$ ώστε $ (l_1x)^2 \plus{} (l_1y)^2/9 \plus{} (l_1z)^2/4 \equal{} 2$ και $ (l_2)^2 \plus{} (l_2y)^2/9 \plus{} (l_2z)^2/4 \equal{} \frac {3}{2}$ και μετά πήρα έναν κυρτό συνδυασμό των $ (l_1x,l_1y,l_1z)\in B$ και $ (l_2x,l_2y,l_2z)\in B$ που να δίνει το $ (x,y,z)$.Μήπως υπάρχει και άλλη λύση πιο απλή; :maybe:[/quote]

Κάτι τέτοιο θα έκανα και εγώ. Για να δείξουμε ότι ένα σημείο ανήκει στην κυρτή θήκη του Β, παίρνουμε δύο σημεία του Β και δείχνουμε ότι αυτό το σημείο είναι κυρτός συνδυασμός των δύο σημείων του Β. Το μόνο που πρέπει να προσέξεις είναι να πάρεις το ένα από τα $ \ell_1, \ell_2$ θετικό και το άλλο αρνητικό.


topology
advanced fields
advanced fields unsolved

Some terminology:

Let $ \Gamma$ be a group acting on a Hausdorff space.

The action is cocompact if there is a compact set $ K\in X$ such that
                    $ \Gamma K=\bigcup_\gamma \gamma K=X$

The action is proper if each $ x\in X$ has a neighborhood U such that $ \gamma U \cap U=\emptyset$ for all but finitely many $ \gamma$

And a problem:

Let $ \Gamma$ be a group acting properly and cocompactly by isometries on a connected metric space X.
Using the properness of the action, show that there is $ r>0$ such that the set $ U=\{x: d(x,K)<r\}$ has the property that there are only finitely many $ \gamma$ for which $ \gamma U\cap U=\emptyset$.

[url=http://en.wikipedia.org/wiki/Lebesgue%27s_number_lemma]Lebesgue's number lemma[/url] should help.

Lebesgue's number lemma is for compact metric spaces...


Euler
geometry
circumcircle
geometry unsolved

Let $ (O)$ be a circle. Point A lie on circle and point M lie inside this circle .A arbitrary chord pass M meet $ (O)$ at $ B$and $ C$.
   prove that the Euler circle of triangle ABC tangent to fix circle .

We denote as $ N,$ the nine point center of $ \bigtriangleup ABC$ and as $ K,$ the midpoint of the segment $ AM.$

The line segment $ AN,$ intersects the midperpendicular of the side segment $ BC$ of $ \bigtriangleup ABC,$ at one point so be it, $ P.$

Because of $ NO \equal{} NH,$ where $ O,$ is the circumcenter and $ H,$ is the orthocenter of $ \bigtriangleup ABC,$ and because of $ OP\parallel AH,$ we have that $ AN \equal{} NP$ and $ OP \equal{} AH.$

But as well known $ OMa \equal{} AEa \equal{} EaH,$ where $ Ea,$ $ Ma,$ the midpoints of the segments $ AH,$ $ BC,$ respectively.

So, we conclude that $ OMa \equal{} MaP$ $ \Longrightarrow$ $ MO \equal{} MP$ $ ,(1)$

We denote as $ L,$ the intersection point of the nine point circle $ (N),$ of $ \bigtriangleup ABC,$ from the line segment $ KN.$

From $ KN\parallel MP$ $ ($ because of $ AK \equal{} KM$ and $ AN \equal{} NP$ $ ),$ we have that $ KN \equal{}\frac{MP}{2}\equal{}\frac{MO}{2}$ $ ,(2)$

Hence, we conclude that $ KL \equal{}\frac{MO\plus{}R}{2}\equal{} constant$ $ ,(3)$ $ ($ because of $ NL \equal{}\frac{R}{2},$ where $ R,$ is the circuradius of $ \bigtriangleup ABC$ $ ).$

From $ (3)$ we conclude that the nine point circle $ (N)$ of $ \bigtriangleup ABC,$ tangents internally to the fixed circle $ (K),$ centered at midpoint $ K$ of $ AM,$ with radius $ R' \equal{} KL \equal{}\frac{MO\plus{}R}{2}$ and the proof is completed.

Kostas Vittas.

Let $ M_a$ be the midpoint of BC. Since $ BC \equiv MM_a \perp OM_a,$ locus of $ M_a$ is a circle (P) with diameter OM and radius $ PO \equal{} \frac{OM}{2}.$ Let G, N be the centroid and 9-point circle center of the $ \triangle ABC.$ Since $ AG \equal{} \frac{2}{3} AM_a,$ locus of G is a circle (Q) centrally similar to (P) with similarity center A, coefficient $ \frac{2}{3},$ and with radius $ QG \equal{} \frac{2}{3}PO.$ Thus $ AQ \equal{} \frac{2}{3}AP.$ Since in addition, P is the midpoint of OM, Q is centroid of the $ \triangle AOM.$ Since OGN is the Euler line of the $ \triangle ABC$ and $ ON \equal{} \frac{3}{2}OG,$ locus of N is a circle (K) centrally similar to (Q) with similarity center O, coefficient $ \frac{3}{2},$ and with radius $ KN \equal{} \frac{3}{2}QG \equal{} PO.$ Thus $ OK \equal{} \frac {3}{2} OQ.$ Since Q is centroid of the $ \triangle AOM,$ K is the midpoint of AM. Since the locus of N is (K) with radius $ \frac{OM}{2}$ and since the 9-point circle (N) has fixed radius $ \frac{R}{2},$ it is tangent to a circle centered at K with radius $ \frac{OM \plus{} R}{2}.$


While trying (In vain) to prove Goldbach's Conjecture (That all even numbers greater than 4 can be written as the sum of two primes and that all odd numbers can be written as three primes), I found that if someone can prove the following conjecture, then they could prove the Goldbach's Conjecture. Nobody has actually proven either of these conjectures, so congratulations to anyone who proves both.

I propose a challenge:

Prove that any number greater than or equal to 4 can be written as the average of two primes.

After proving this, prove Goldbach's Conjecture (In general, not just for odds or just for evens).

Two things.

1. Wrong section, post it somewhere else
2. I doubt that we would stand much of a chance at solving Goldbach's conjecture where many others have failed.

[quote="Ihatepie"]post it somewhere else[/quote]  No: moderators will take care of sending it off to a more appropriate section of the forum.  In particular, I'm sending it to Number Theory, and then the moderators there can decide if they want to send it somewhere else  :)

The moderators (well, only me now...) decide three things:

a) Exactly that statement has already been posted.
b) It's trivially equivalent to Goldbach. Any reasonable attempt should do much more than a one step equivalence.
c) It may stay here. But eventually, it will probably get deleted.


Let $ a,b,c>0$ and $ abc\equal{}1$. Prove that

$ (a^2\plus{}b^2\plus{}c^2)^3\geq3(a^3b\plus{}b^3c\plus{}c^3a)(a\plus{}b\plus{}c)$

-------------------------
 :blush: Panagiote Ligouras

[quote="Ligouras"]Let $ a,b,c > 0$ and $ abc \equal{} 1$. Prove that

$ (a^2 \plus{} b^2 \plus{} c^2)^3\geq3(a^3b \plus{} b^3c \plus{} c^3a)(a \plus{} b \plus{} c)$

-------------------------
 :blush: Panagiote Ligouras[/quote]

We have:
$ (a^2\plus{}b^2\plus{}c^2)^2 \ge 3(a^3b\plus{}b^3c\plus{}c^3a)$
$ a^2\plus{}b^2\plus{}c^2 \ge \frac{(a\plus{}b\plus{}c)^2}{3} \ge a\plus{}b\plus{}c$


calculus
integration
trigonometry
calculus computations

find $\int\frac{\sqrt{1-k^{2}sin^{2}x}}{sin^{4}xcosx}dx$ where $k$ is real.
10 stars :lol:

[quote="sos440"][hide="Partial Answer"]
$\int\frac{\sqrt{1-k^{2}\sin^{2}x}}{\sin^{4}x \cos x}\, dx$  $=$ $-\frac{\left(1-k^{2}\sin^{2}x\right)^{3/2}}{3\sin^{3}x}-\frac{\sqrt{1-k^{2}\sin^{2}x}}{\sin x}+\sqrt{1-k^{2}}\tanh^{-1}\left(\frac{\sqrt{1-k^{2}}\sin x}{\sqrt{1-k^{2}\sin^{2}x}}\right)$$+C$

where $|k|\leq 1$
[/hide]

[hide="Idea"]
use the substitution $k \sin\theta = \sin\phi$[/hide][/quote]
Everybody can find answer by computer and say Idea is $ksinx=siny$ ( because in integral we have $\sqrt{1-k^{2}sin^{2}x}$ obviusly find it.
It's not new.  :wink:


induction
number theory unsolved
number theory

Let [tex]f ( x ) = x^3 + 17[/tex]. Prove that for each natural number [tex]n \geqslant 2[/tex], there is a natural number [tex]x[/tex] for which [tex]f ( x )[/tex] is divisible by [tex]3^n[/tex] but not [tex]3^{n + 1}[/tex].

My solution is:

We use induction on n
for $n=2$ it is clear that $1^3+17$ is not divisible by 27.
Now suppose that there exist an integer t such that $3^n\mid t^3+17$ and $3^{n+1}$doesn't divide $t^3+17$.
it mean that $t^3+17=k(3^n)$ where $k$ isn't divisible by $3$.
Now get
\[(r\times3^{n-1}+t)^3+17=r^33^{3n-3}+r^23^{2n-1}t+r\times3^nt^2+t^3+17\]
\[\equiv3^n(rt^2+k)\mbox{      mod3^{n+2}}\]
and now it is clear we can choose r in such a way that $rt^2+k\equiv 3\mbox{      mod9}$
because $t^2\equiv1\mbox{     mod3}$.
Now Prove is complete. :D
_____________
Good luck. :lol:


calculus
inequalities proposed
inequalities

[quote="nguoivn"]Given $ a, b, c \geq\ 0$ satisfy $ ab \plus{} bc \plus{} ca \equal{} 3.$
Find Minimum: $ P \equal{} \frac {1}{\sqrt []{a^2 \minus{} ab \plus{} b^2}} \plus{} \frac {1}{\sqrt []{b^2 \minus{} bc \plus{} c^2}} \plus{} \frac {1}{\sqrt []{c^2 \minus{} ca \plus{} a^2}}$
When does equality occur ?
Please don't use calculator because it exists a njce proof by Am-Gm  :)[/quote]
Assume $ c \equal{} \min\{a,b,c\}$, then
\[ P \ge \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{\sqrt {a^2 \minus{} ab \plus{} b^2}}
\]
Now, we will find the minimum of
\[ P\equal{}\sqrt{ab} \left( \frac{1}{a}\plus{}\frac{1}{b} \plus{}\frac{1}{\sqrt{a^2\minus{}ab\plus{}b^2}}\right)\]
And get the result (we can do it easy if we set $ t\equal{}\sqrt{\frac{a}{b}} \plus{}\sqrt{\frac{b}{a}} \ge 2$) :)

But you still have to calculus f'(t).
I meant with the final ineq, I can solved by Am-Gm only  :)


Nice, by AM-GM.

We have also:
$a \geq 0 , b \geq 0 , a+b=2$ $\Longrightarrow$ $\sqrt{8a^{3}+ab}+\sqrt{8b^{3}+ab}\ge 2\sqrt{16-7ab}\ .$


advanced fields
advanced fields unsolved

Let $\Omega\subset\mathbb R^{n}$ be open. Are the following statements equivalent for a sequence $(u_{n})\subset L_{\mathrm{loc}}^{1}(\Omega)$:

1. $u_{n}=o(1)$ in $L_{\mathrm{loc}}^{1}(\Omega)$;
2. $u_{n}=o(1)$ in $\mathcal D'(\Omega)$?

Here $D'(\Omega)$ is the space of distributions.

[b]P.S.[/b] This month I'm able to use my computer quite few days so don't bother about absence of my immediate reaction to your answers.

They're certainly not equivalent.

Consider $u_{n}(x)=n\cos(n^{2}x).$ (This is a one-dimensional example.)

Plainly, $u_{n}$ does not tend to zero in $L_{\text{loc}}^{1},$ but $u_{n}$ does tend to zero in the sense of distributions.


\[ \begin{array}{c} {x^{2} \plus{} y^{2} \plus{} z^{2} \equal{} 21} \\
{x \plus{} y \plus{} z \plus{} xyz \equal{} \minus{} 3} \\
{x^{2} yz \plus{} y^{2} xz \plus{} z^{2} xy \equal{} \minus{} 40} \end{array}
\]
The number of real triples $ \left(x,y,z\right)$ satisfying above system is

$\textbf{(A)}\ 0  \qquad\textbf{(B)}\ 3  \qquad\textbf{(C)}\ 6  \qquad\textbf{(D)}\ 12  \qquad\textbf{(E)}\ \text{None}$

[hide]Let $ s_1$, $ s_2$, and $ s_3$ be the first, second, and third elementary polynomials of $ x$, $ y$, and $ z$, respectively. Then the equations are equivalent to

$ s_1^2 \minus{} 2 s_2 \equal{} 21$
$ s_1 \plus{} s_3 \equal{} \minus{}3$
$ s_1 s_3 \equal{} \minus{}40$

From the last two equations (solving a quadratic via substitution), we can deduce that $ (s_1, s_3) \equal{} (5, \minus{}8), (\minus{}8, 5)$ are two possible solutions for $ s_1$ and $ s_3$.

If $ s_1 \equal{} 5$, then $ s_2 \equal{} 2$. If $ s_1 \equal{} \minus{}8$, $ s_2 \equal{} \frac{43}{2}$. In the first case, we have that $ x, y, z$ are the roots of $ a^3 \minus{} 5a^2 \plus{} 2a \plus{} 8 \equal{} 0$, which does indeed have 3 real solutions (-1, 2, and 4, to be precise.)

In the second case, $ x, y, z$ are the roots of $ a^3 \plus{} 8a^2 \minus{} \frac{43}{2}a \minus{} 5 \equal{} 0$. This also has 3 solutions (we can verify by checking the locations of the extrema.)

Therefore, there are $ 2*3! \equal{} 12$ solutions, so the answer is $ D$.
[/hide]

Hope I didn't do anything wrong. :D

hello, i have found
$ x\equal{}4,y\equal{}2,z\equal{}\minus{}1$
$ x\equal{}4,y\equal{}\minus{}1,z\equal{}0$
$ x\equal{}2,y\equal{}4,z\equal{}\minus{}\frac{1}{3}$
$ x\equal{}2,y\equal{}\minus{}1,z\equal{}\minus{}2$
$ x\equal{}\minus{}1,y\equal{}4,z\equal{}2$
$ x\equal{}\minus{}1,y\equal{}2,z\equal{}5$
Sonnhard.

Just by checking those through the first equation, you can tell that none of those work except for the first and fifth ones.

@ Zhero: your first case is correct but your second case has an error [quote]$ a^3 \plus{} 8a^2 \minus{} \frac{43}{2}a \minus{} 5 \equal{} 0$[/quote] should be $ a^3 \plus{} 8a^2 \plus{} \frac{43}{2}a \minus{} 5 \equal{} 0$. This polynomial has one real root and two complex ones, leading to the answer of 6 or $ C$.

$$\begin{array}{rcccl}
x+y+z = -8 &\Rightarrow& xy+yz+yz = \dfrac{43}2 &,& xyz = 5 \\
x+y+z = 5 &\Rightarrow& xy+yz+yz = 2 &,& xyz=-8
\end{array}$$
By Vieta Theorem, $x, y, z$ are the roots of the equations either
$$ a^3 - 5a^2 + 2a+ 8 = 0 $$ or
$$ a^3 + 8a^2 + \dfrac {43}{2}x - 5 = 0 $$ or

For the first case:
Since $ a^3 - 5a^2 + 2a+ 8 = (a+1)(a-2)(a-4)=0$, there are $3!=6$ solutions for $(x,y,z)$.

For the second case:
$f(a)=a^3 + 8a^2 + \dfrac {43}{2}x - 5$.
Since $f(0)<0$ anf $f(1)>0$, there is a real root on the interval $(0,1)$.
$f'(a) = 3a^2 + 16a + \dfrac{43}2=0 \Rightarrow \Delta = 16^2 - 4\cdot3\cdot \dfrac {43}2 = -2 < 0  $. So the graph does not have a tangent with slope $0$. This concludes the other roots are complex.

There is another solution for the second case that does not require calculus.

Since $xyz = 8 > 0$ ve $x+y+z = -5 < 0$, exactly two of $x,y,z$ are negative.
Assume $x \leq y < 0 < z$.

By $AM \geq GM$, $x^2 + y^2 \geq 2xy$.

Add $x^2 + y^2$ to both sides, $2(x^2 + y^2) \geq (x+y)^2$

Since $x^2 + y^2 + z^2 = 21$, we get $21 > x^2 + y^2$.

$42 > 2(x^2 + y^2) \geq (x+y)^2 = (-8-z)^2 = (8+z)^2 > 64$. Contradiction.

So for the second case, ($x+y+z = -8$ and $xyz = 5$) there is no real solution.


From Moldova Mathematical Olympiad.

For a,b,c strictly positive numbers prove that

ab/(c(c+a)) + bc/(a(a+b)) + ca/(b(b+c)) \geq a/(c+a)+b/(a+b)+c/(b+c)

Doing the calculation we obtain

a^2b^4 + b^2c^4 +c^2a^4 +a^3b^3 + b^3c^3 + c^3a^3  \geq

a^3bc^2 + a^2b^3c + ab^2c^3 + 3a^2b^2c^2

But by AM-GM we obtain

a^3b^3 + b^3c^3 +c^3a^3  \geq 3a^2b^2c^2

and

2a^3b^3 + c^3b^3  \geq 3a^2b^3c

2b^3c^3 + c^3a^3  \geq 3ab^2c^3

2a^3c^3 + a^3b^3  \geq 3a^3bc^2

so our inequality is correct.


geometry
trigonometry
area of a triangle
counting
combinatorics
IMO Longlist
IMO Shortlist

Consider all segments dividing the area of a triangle $ABC$ in two equal parts. Find the length of the shortest segment among them, if the side lengths $a,$ $b,$ $c$ of triangle $ABC$ are given. How many of these shortest segments exist ?

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

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=336213#336213]www.mathlinks.ro/Forum/viewtopic.php?p=336213#336213[/url]

if $a>b>c$, there are two segments $PQ,P'Q$, with equal lenghts:

$P,P' \in a$
$Q,Q' \in b$

$CP=CQ'$
$CP'=CQ$

so, the segments $PQ,P'Q'$ are symmetric through the bisector of $\angle C$

For $D\in \{A,B,C\}$ I note $DP=x$, $DQ=y$ and $\frac 12 xy\sin D=\frac 12 S$, i.e. $xy=k$ (constant). But $PQ^2=x^2+y^2-2xy\cos D$. Therefore, $PQ$ is minimum $\Longleftrightarrow x^2+y^2$ is minimum ($xy$ - constant) $\Longleftrightarrow$ $x=y=\sqrt{\frac{S}{\sin D}}$ a.s.o.

Nice and quick proof!

Why should we check three cases? We don't want to find a minimum segment for every angle, but for the entrire triangle.
I think that the smallest segment corresponds to the smallest angle of the triangle

YES ! Prove easily that

$\min PQ=\min_{D\in \{A,B,C\}} \sqrt{2S\tan \frac D2}=\sqrt{2S\tan \frac{\min \{A,B,C\}}{2}}.$

[u]Remark.[/u]

In my opinion, although this problem is nicely, the its level is ** (from ******).


How do you show that the dot product in n dimensions is equal to $x_1x_2 + y_1y_2 + z_1z_2 ... + n_1n_2$?

um,,,, isn't that they way its defined????

you can also use the defenition of the dot product to show that it must be true.

also, one interesting about this is that u can use this dot product to prove C-S inequality very easily :)

More generally, in $\mathbb{C}^n$, if $x=(x_1,x_2,\ldots,x_n)$ and $y=(y_1,y_2,\ldots,y_n)$, then
\[ \langle x,y\rangle = \sum^n_{i=1} x_i\overline{y_i}  \]
This is a definition, not something to be proved.

How come AoPS vol 2 proves this for 2-dimensional dot product?


Solve the inequality: $\left| f(x) \right| + \left| g(x) \right| \leq c$.

What exactly did you mean by solve? Find c in terms of f(x) and g(x)?


LaTeX
quadratics
algebra
quadratic formula

I am sad so I walk off a building of height h. I fall the last h/n in t seconds. What is the height of the building? Have fun!!

Um.. h?

Do you mean in terms of n and t or something?

Yes duhh... h/n is the fraction of h and t is the time it took for me to fall the last h/n

[hide]
At distance $ \frac{h(n-1)}n$, we have $ t_1=\sqrt{\frac{2h(n-1)}{ng}}$.
At distance $ h$, we have $ t_2=\sqrt{\frac{2h}{g}}$.

So $ t = t_2-t_1=\sqrt{\frac{2h}{g}}-\sqrt{\frac{2h(n-1)}{ng}}=\sqrt{\frac{2h}g}\left(1-\sqrt{\frac{n-1}{n}}\right)$.

Let $ k=\sqrt{\frac{n-1}{n}}$.

So now we solve for $ h$. We have:
\[ \sqrt{\frac{2h}{g}}=\frac t{1-k} \\
\frac{2h}{g}=\frac{t^2}{\left(1-k\right)^2} \\
h=\frac{gt^2}{2\left(1-k\right)^2}\]

So in closed form, $ \boxed{h=\frac{gt^2}{2\left(1-\sqrt{\frac{n-1}{n}}\right)^2}}$.
[/hide]

I wish I could have the time to learn latex, but your solution works. I did mine a little differently..

Ok let T be the time for it to fall the first h-h/x
I got h-h/x=4.9T^2 and h=4.9(T+t)^2
This resulted in T^2(x)/(x-1)=(T+t)^2
Solving for T using the quadratic formula I get
t(x-1)+t*sqrt(x-1)(x)
Factoring and and t (Because T+t is the total time) I get
h=4.9{t[x+sqrt(x(x-1))]}^2

Thanks Yongyi and sorry for making you look at this crap. Our answers are probably the same after you rationalize the denominator. Oh yea in your opinion how hard do you think that problem was? I was doing many problems in the form of this and decided to derive a formula.

Hm I'd say medium level. There are some pretty beastly physics problems out there.


algebra
polynomial
trigonometry
Vieta
complex numbers

[hide]$a_0=1$
$a_1=2$
$a_2=1$
$a_3=-1$
$a_4=-2$
$a_5=-1$
$a_6=1$
$a_7=2$

and then you are back to where you started. so just find the remanider when n is divided by 6, and use that.[/hide]

How would you solve it using characteristic equation?

what's a "characteristic equation" and why would you need it for this problem?

Well, the characteristic equation for this problem is $r^2-r+1=0$.
[url=http://en.wikipedia.org/wiki/Recurrence_relation]en.wikipedia.org/wiki/Recurrence_relation[/url]
I need it because I need an example how to use the characteristic equation for $\Delta < 0$

Actually, you don't have to deal with a linear recurrence relation differently if the roots of its characteristic polynomial are nonreal.  You will want, I'm sure, to simplify the final formula to remove the nonreal numbers, but you could leave the nonreal numbers in the formula, if you wanted to, and it would still work.  In this case, after simplication you get
$a_n=\cos\frac{n\pi}{3}+\sqrt{3}\sin\frac{n\pi}{3}$,
which we see, as perfectnumber628 wrote, repeats every 6th iteration.

just memorize the using of a characteristic equation, but there is an elegant proof that i found using newton's sums

here is an outline solution, generalize for other cases

let $P(x)=x^2+a_1x+a_0$, and $P(m)=P(n)=0$
define $s_k=m^k+n^k$
easily, we find that $s_1=-a_1$, and $s_2=a_1^2-2a_0$ through newton's sums or vieta's
now, the in the newton's sum, the n at the end drops out, and we get
$s_n+a_1*s_{n-1}+a_0*s_{n-2}=0$ for $n>2$
This is a recursion about $s_k$. What if we want to find $s_k$ explicitly?
We defined $s_k$ above, so we have the explicit form of the recursion!!!
Now, we can reverse this process given the recursion, and find the explict form!

[quote="Frozen"]So, back to the characteristic equation. $r^2-r+1=0$ has solutions $\epsilon$ and $\bar{\epsilon}$,so $a_n=A\epsilon^n+B\bar\epsilon^n$. Because $a_1=A\epsilon+B\bar{\epsilon}\in R$ we have the the imaginary part of $a_1$ is $0$, so $A=B$. We get $a_0=2A$ and $a_1=2Acos\frac{\pi}{3}\Rightarrow\frac{a_0}{a_1}=2$,which is false. So, where is my mistake?[/quote]

because $A,B$ is not real number . They are complex :)

in fact if the characteristic equation yields roots $p+qi$ then we can write the general solution as

$a_n=A(p+qi)^n+B(p-qi)^n\\=AR^n(\cos n\theta+i\sin n\theta)+BR^n(\cos n\theta-i\sin n\theta)\\=R^n(A+B)\cos n\theta+R^n(A-B)i\sin n\theta\\=R^n(C\cos n\theta+D\sin n\theta)$

where $R=\sqrt{p^2+q^2}$ , $C=A+B$ , $D=(A-B)i$

Lol, I was so stupid I thought $A,B\in\mathbb{R}$. Thx a lot :)


algebra
polynomial
trigonometry
geometric series

these are two similar(yet don't seem to be) problems I found in an olympiad book

express these two indefinite products finitely:

$(1+x)(1+x^{2})(1+x^{4})\cdot\cdot\cdot(1+x^{2^{n}})$

$sin{(x)}sin{(2x)}sin{(4x)}\cdot\cdot\cdot sin{(2^{n}x)}$

[hide="Hint for 1"] Multiply by $1-x$. [/hide]

[hide=" #1 (Method 1--t0rs hint) "]
When you multiply the expression by $(1-x)$ it will become $(1-x^{2})(1+x^{2})(1+x^{4})\ldots$, and similarly will continue to contract until it finally becomes $(1-x^{ 2^{n+1}})$. Then we need to divide by the $1-x$ we initially multiplied in, so our resulting expression becomes $\frac{1-x^{ 2^{n+1}}}{1-x}$ [/hide]

[hide=" #1 (Method 2)"]
Look at multiplying the initial two or three terms together:
$(1+x)(1+x^{2})(1+x^{4})\ldots=(1+x+x^{2}+x^{3})(1+x^{4})\ldots$
$=(1+x+x^{2}+x^{3}+\ldots+x^{7})\ldots$

This trend will continue as we multiply further since each polynomial will be of degree one less than the next term, and as we look at the highest-degree term of the resulting polynomial we see that it is one less than the next power of two (from the last term being multiplied). Summing up this geometric series we get the expression equal to $\frac{x^{2^{n+1}}-1}{x-1}$.[/hide]

(My solutions appear on my browser even though I've put in the "hide" coding-- I'm sorry if they show up unhidden.)

[quote="junggi"]You can't have a ' inside the hide tag thing
So t0r's should be written as t0rs[/quote]

Ok, its fixed now. Thanks!  :)

[quote="13375P34K43V312"]I want to see a solution to #2, t0rajir0u's hint doesn't seem to be helping me.[/quote]

I'm not sure myself...  I get it to the form $\frac{\sin^{2}(2x) \sin(4x)\ldots}{\cos(x)}$, and then i don't want to expand out the sines with larger angles because of the $\cos(2^{m}x)$ terms getting ugly.

On the other hand, if it were $\cos(x) \cos(2x) \ldots \cos(2^{n}x)$ then something similar would work perfectly to contract the expression.

Oops.  Haha.  I assumed it was the $\cos$ problem, which is more common.   :oops:

The cos problem is just combining into double-angle expressions, isn't it?

o my, I am really sorry, it was cos all along...my apologies

can anyone find a solution for the sine problem though? it looks very interesting.


analytic geometry
geometry
parallelogram
algorithm

Hi,
I tell you my problem:
I have a regular figure of 4 vertex, I need normalize all of the positions inside of this figure(I want to turn all his coordinates into a number between 0 and 1).
I have this regular figure of 4 vertexs,I know these 4 vertexs, these are the coordinates of figure:

vertex one: x=100 y=300----normalized x=0,y=0
vertex two: x=500 y=400----normalized x=1,y=0
vertex three: x=120 y=100----normalized x=0,y=1
vertex four:x=520 y=200---normalized x=1,y=1

For example I want to know the position normalized of the coordinates x=200 y=200, I need to know the value of x normalitzed and y normalized. I am trying to make one mathematical formula but I have problems.
can anybody help me please?

Since that figure is not a parallelogram, there's no natural way to make things work.
Are you sure those are the right coordinates?

I am Sorry,
I edited my topic now the coordinates of figure are right, can you help me?
thanks for the reply.

You want an affine transformation:
$x'=ax+by+c$ and $y'=dx+ey+f$, where $(x,y)$ is the original point and $(x',y')$ is the new point. Fitting the corners to this model gives you eight linear equations in the six unknowns $a,b,c,d,e,f$; you can ignore one corner to make that six equations. There are routine algorithms for solving linear systems.

It's actually easier than what I said above; the variables separate so we have two 3 by 3 linear systems.

Thanks for the reply I am testing your equations and I think that is perfect. For example in the coordinates x=200, y=200 I calculate the result with your formula and the results are x'=0.21945 y'=0.39024, I think that this is right, if you see some error please say me it.

Thanks!!!


analytic geometry
geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

i don't know if this is the correct place for this question, but:

in spherical geometry, how can one find the distance between two points $ (r, \theta_{1}, \phi_{1})$, and $ (r, \theta_{2}, \phi_{2})$?

In spherical coordinates themselves, I don't know.  However, if you convert your coordinates into the usual Cartesian system, then you can just use the distance formula and it works out easily.

There is an arc length formula for spherical coordinates. If you can parameterize a curve from $ (r, \theta_1, \phi_1)$ to $ (r, \theta_2, \phi_2)$ along the great arc running through these two points, then you have the distance.

I'm not sure if there is a way to do this arbitrarily, though...

I would think that trying to parametrize this arbitrary curve might be difficult.  I would just convert into a coordinate system that you know how to work with, like cartesian.

I guess it depends on the interpretation of the question. I interpreted it as two points are on a sphere of radius $ r$. What is the distance between these points along this sphere.

If it is really just: given two points in $ \mathbb{R}^3$ (this time specified by their spherical coordinates), find the straight line distance, then do it like JRav says.

I'm not sure the context of the question. If this is a calc III type class then do it the straight line way. If this is a differential geometry class, then it is probably trying to get you to use the spherical geodesic, so it must be done the hard way.

To find the distance along the sphere, the points need to have the same $ rho$ and $ \varphi$ or $ \theta$ components, or else it gets a lot more difficult.


The measure of one angle of an isosceles triangle is twice that
of another. What is the positive difference between the greatest
possible degree measure and the least possible degree measure of the
remaining angle?

The only two solutions for this is if the non-equivalent angle is $ 90^\circ$ or $ 36^\circ$. The positive difference is 54.

i don't know how to explain, but i know for a fact that the answer is $ 27$

This is very elementary:  Call the first two angles discussed A and 2A.  To be isocoles, the last angle is either A or 2A.

Case 1:  The third angle is 2A
Adding the angles up, 5A=180 degrees, so A is 36, meaning that the third angle 2A is 72.

Case 2: The third angle is A
Adding the angles up, 4A=180 gives us A=45.  So, the third angle is 45.

The difference 72-45=27, your answer.

[color=#FF0000][size=89]Do not quote the entire post immediately above yours.

Do not post messages without content.

Do not revive topics without sufficient reason. [/size][/color]


A baseball diamond is a square with sides of length 90 feet. A player is one-third of the way from second base to third base. If this player throws the ball to first, how many feet will the ball travel (horizontally)? Round your answer to the nearest foot.

I'm a little confused what they mean by "horizontally"...the most probable interpretation yields $ \sqrt{30^2\plus{}90^2}\equal{}30\sqrt{10}\approx\boxed{95}$ (Pythagorean Theorem).

Horizontally basically means 2-D. Think of it as a diagram you would sketch on paper. It simply means to not think about the altitude of the baseball.


Why did you post it to the Olympiad Section?
I will move it to Intermediate.

I got that it factorised as:

$ \displaystyle (\frac {x^2}{\sqrt 2}+1){(\sqrt 2 x^2-1)}^2=0$


function
calculus
derivative
real analysis
real analysis solved

Let f:R->R be a function s.t. it transforms any bounded and closed interval into a bounded and closed interval and any boudned and opened interval into a bounded and opened interval.Prove that f is continuous.

prove that f has Darboux and it is injective!

try not to post pictures of the problems please. Use regular text instead.
Put pictures only if they are strictly necesary.

[quote="Lagrangia"]... f has Darboux...[/quote]

What say exactely Darboux theorem
thx Lagrangia

Darboux Theorem states that the derivative of a function has the Intermediate Value Property, but I think Lagrangia was talking about DP (Darboux's Propery = Intermediate Value Property - that's how we call it in Romania).

Darboux property.

Let E be an interval. It is said that the function f:E-->R has the property of Daboux on E, if for any a<b in E and any real number alpha situated between f(a) and f(b), there exists a least one x_alpha from ]a,b[ so that f(x_alpha)=alpha.

cheers!

I think that it is quit obvious that f has Darboux property.
For the injectivity suppose we have f(x)=f(y)
Let f([x,y])=[a,b]
Then f((x,y))=(a,b), which means that (f(x)=a and f(y)=b) or (f(x)=b and f(y)=a), but a<>b, contradiction so f injective. :)  :)


How many positive integers less than 1,000 have at most two different digits?

Good luck. :thumbup:

$280$ ( with 0 ) , $279$ without the 0.
.I'll give the explanation later.

here is how i did it. there are $648$ three digit numbers in a form of $abc$, so there is $352$ three digit numbers that satisfy what we want, since it has said that there can be a maximum of two different digits, all of the two digits would satisfy what we want because the two digit numbers are either in a form of $ab$ or $aa$, also all of the one digit numbers would satisfy what we want, so the total would be:
$352+99=451$

I said there were three different ways to arrange the three-digit numbers

$aaa$

$abb \cdot 3$

$aab \cdot 3$

So I did $\frac {9!}{2!7!} = 36 \cdot 6 = 216$

$216 + 9 = 225$

All the two-digit numbers worked so

$225 + 99 = 324$

Then counting the 0 cases

$aa0$

$a0a$

$a00$

So $9 \cdot 3 = 27$

$324 + 27 = 351$

I must have messed up before ...

[quote="tarquin"]I said there were three different ways to arrange the three-digit numbers

$aaa$

$abb \cdot 3$

$aab \cdot 3$

So I did $\frac {9!}{2!7!} = 36 \cdot 6 = 216$

$216 + 9 = 225$

 ...[/quote]
why did you add $9$?

In order to count the $aaa$ case I added 9.

:wallbash_red:
right :)
i donno what to say...guess just have to wait..

Here's my explanation -
First , the single digit integers and the 2 digit integers all fall under this category.That makes up 100 integers , ( if you include the 0 . If you dont want to , the answer is 279 )
Anyway ,
In the three digit integers in the form $abc$ ,
$ a $ has 9 possibilities.
$b$ has 10 possibilities.
$c$ has 2 possibilities ( It's either $a$ or $b$)
So the total number of three digit integers are $ 9 \times 9 \times 2 = 180 $

The total number of positive integers is nothing but , well ,
$180 + 100 = 280 $
(or excluding the 0 )$ 180 + 99 = 279$.
Since you've specified positive integers , I think 279 will be better because 0 is neither positive nor negetive....

I think there's no other way then case-by-case.

When you do $9 \cdot 10 \cdot 2$, what if the second digit is the same as the first? Then you can have more possibilities for the third.

If tarquin's answer is correct, there is a much easier way to do the problem.   :D

[hide="Hint"]Think complements.  (I think that's what they call it, where P(b)=1-P(a))[/hide]

That's exactly what amirhtlusa did just that he probably forgot how many three-digit numbers there were and subtracted from 1000 instead of 900 and thus getting exactly 100 more than the correct answer :)

ahhhhhhhhhhh :wallbash:
yes, the answer is $351$ because there are only $900$ three digit numbers...(i subtracted out of 1000 first time)
so there are $648$ three digit numbers that would not fit in our category, so there are $252$ three digit numbers plus the $99$ equals $351$

:oops: Now I get the answer.....As far as my method is concerned , it's worked out as - $9 \times 9 \times 2$ + $9 \times 1 \times 10$ + $99$.....

And 351 is the right answer. Good job!

Amirhtlusa did it right in the first explanation, just added an extra 100, but nice method. Always count the opposite!


I downloaded the packet on Inequalities from the AMC website in Latex, but when I click on it, it is not opening. I'm pretty sure that this isn't the right forum for this, so mods, feel free to move this. Does anyone have any idea how I could open it?

From the site, it looks like one can either download it in LaTeX or Postscript formats.  If you clicked on the LaTeX option, then you'll probably need to get some sort of LaTeX program (look [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]here[/url]).  I'm not really familiar with Postscript, but I think you'll need some sort of specialized program for that too.

If you posted the link I might be able to help you and even pdf the file for you.

well, let's see:
http://www.unl.edu/amc/a-activities/a4-for-students/s-index.html

You can find the link on this page :)

Well I can send you the file if you wish? I'm assuming you have Adobe Acrobat Reader.

EDIT: I have already emailed you the Inequality Packet. My email is dbpopov@gmail.com just so you don't think it's spam.

Yup, and I'd absolutely love it if you sent it to me. :D

Thank you sooooo much :lol: ! Seriously, I really appreciate you taking the time to do this. Mods, feel absolutely free to delete this topic!

But you should get LaTeX anyways.  Situations like this, where the pdf file is not given, are very common.

I recommend Ghostscript followed by Winedt. That is what I always use and it works perfectly (can go from tex>>latex>>dvi>>ps>>pdf etc.)

I did get Latex, but tried to open the packet, and it didn't work.

This is because Latex and tex are different although I opened the file and had no trouble converting it to latex?! ;)

While we're at it, can somebody help me solve this page? (Attached)

From the cauchy schwarz inequality
we have
$\(x+y+z)(\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z})\geq\(\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1})^2$

however $\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}=1$
and we are complete!
 :)  :D  :)


1. A very large reservoir contains a water layer (d = 1 g/mL) with a height of 20 cm, and a layer of oil (d = 0.6 g/mL) with 50 cm height. In the bottom of the reservoir there is a small hole opened to the atmosphere. Compute the drainage velocity through the hole. (Neglect the viscosity of the fluids)

(Source: Biophysics last exam, Audiology course, 2004/2005)

2. The picture bellow shows a system composed by two cylindrical tubes with radii 2 cm and 1 cm, and lenghts 2 m and 4 m, respectively, linked together. In the left of the first tube there is a faucet, and in the exit of the right tube the water drains to the atmosphere.

(a) Find the hydrodinamic resistance of this system.
(b) Knowing that the water leaves the second tube with velocity 10 m/s, compute the manometric pressure immediatly at the exit of the faucet. What is the volume of water that leaves the faucet per second?

Data: d(water) = 1g/mL; $ \eta(water) \equal{} 1\times 10^{\minus{}3}\, Pa\,s$.

3. A sound source generates a complex periodic sound that propagates on air. The amplitude spectrum of that sound is shown bellow (horizontal axis: frequency; vertical axis: acustic pressure amplitude).

(a) Find the fundamental frequency, period, and the harmonics that compose the sound generated by the source, and write the time expression for the pressure, p(t).

(b) Compute the effective acustic pressure generated by the source and find the corresponding level of sound intensity ($ Z_{c}(air) \equal{} 442\, kg\, m^{\minus{}2}\, s^{\minus{}1}$, $ I_{o}\equal{} 10^{\minus{}12}\, W/m^{2}$).


algebra
polynomial
Vieta
calculus
integration
quadratics
irrational number

The polynomial $x^3 - 2004x^2 + mx + n$ has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of $n$ are possible? \[ \text {(A) } 250,000\qquad \text {(B) } 250,250\qquad \text {(C) } 250,500\qquad \text {(D) } 250,750\qquad \text {(E) } 251,000 \]

[hide]

Interesting problem...

Let $a,b,c$ be the roots. We have $2004 = a+b+c$ where $a = b+c$ so $a = 1002$. Then $m = 1002^2+bc$ and $n = -1002bc$ from Vieta's Formulas. So basically $bc$ is integral but $b$ and $c$ are not and $b+c = 1002$.

So let's think. If $b,c$ are rational and not integral, their product will have a denominator bigger than $1$ and won't be an integer. So that's out. $b,c$ must be irrational.

Thinking again, we have $b = 501+\sqrt{k} > 0$, $c = 501-\sqrt{k} > 0$ for some nonsquare $k$. That's the only way the radicals will cancel each other out and always sum to $1002$.

Then $n = -1002bc = -1002(501^2-k)$ so we find number of possible values of $501^2-k$ such that $k$ is not a square. So $k$ ranges from $1$ to $501^2-1$ and we take out the squares from $1$ to $500$.

We get $k$ can take $501^2-501 = 501(500) = 250500$ values.[/hide]

Nice problem!

Here is my solution:

[hide]Let $a$ be the zero of the polynomial that is an integer. Since the coefficient of $x^3$ is $1$, there can be no other rational zeros, which means that the other two zeros have to be $\frac{a}{2} \pm r$ for some irrational number $r$. The polynomial will then be:

$x^3-2ax^2+\left(\frac{5}{4}a^2-r^2\right)x-a\left(\frac{1}{4}a^2-r^2\right)$

Therefore $a = 1002$ and the polynomial is:

$x^3-2004x^2+(5(501)^2 - r^2)x-1002((501)^2-r^2)$.

All coefficients are integers iff $r^2$ is an integer, and the zeros are positive and distinct iff $1 \le r^2 \le 501^2-1 = 251000$. Because $r$ can't be an integer, there are $251000 - 500 = 250500$ possible values of $n$.Therefore, the answer is $\boxed{C}$.

[/hide]

[hide]pretty clear that this breaks down into a quadratic and linear factor, both with integer coefficients,

and in the quadratic, if $a+\sqrt{b}$ is a root, then $a-\sqrt{b}$ is a root (this guarantees integer coefficients), with $a^2>b$ (to be positive) and later I show a,b are integers
using the coefficients, $2a+y=2004$ since they are equal, then y=1002, a=501, $y(a^2+b)=n$ looking at that, b must be an integer, but not a square of an integer, otherwise there would be more than 1 integer root,
we have $251,001>b>0$, and b is not a square, since b<501^2, there are 500 squares that are in that range

this gives us $251,000-500=250,500 \implies \boxed{C}$[/hide]

[hide="Answer"]We have that the roots are $r$, $s$, and $r+s$. We know that $2r+2s=2004\Rightarrow r+s=1002$. $n=rs(r+s)=1002rs$, which must be an integer, and $m=rs+r(r+s)+s(r+s)=rs+1002^2$, so we must simply have $rs$ is an integer with $r+s=1002$. We need both $rs$ and $r+s$ integral without either an integer. With some thought, we see that this is impossible through fractions, so we must have square roots. All numbers will be of the form $501\pm \sqrt{k}$, and they must all be positive. $501^2=251001$, so we have $251000$ possible values for $k$, except that we cannot have $k$ a perfect square, as that makes both $r$ and $s$ integers. Obviously, this occurs for $500$ values, so our answer is $251000-500=250500\Rightarrow \boxed{C}$.[/hide]

[quote="JesusFreak197"]With some thought, we see that this is impossible through fractions, so we must have square roots.[/quote]

Ok explain your thought process.  It's not exactly a trivial idea.

Basically, if two fractions, both in simplest form, sum to an integer, then they must have the same denominator. However, when multiplied together they can never create an integer, as you will simply have the denominator squared on the bottom. This will never be canceled out if the two original fractions were in lowest terms, because otherwise those fractions would have canceled out their denominators before being multiplied together.

It seems fairly simple to me. *shrug*

Another version of that:
Suppose $r,s$ are rational and both $rs$ and $r+s$ are integers. Then $x^2-(r+s)x+rs$ is a polynomial with rational zeros $r$ and $s$. The leading coefficient is 1, so by the Rational Root Theorem, $r$ and $s$ are integers.
If the roots aren't integers, they must be irrational by this argument.


Please prove for any Pythagorean triples, twoo of the numbers will have a sum or diff divisible by 7
(My way is to consider all combinations)

[hide]Note that all quadratic residues $ \mod 7$ are $ 1,2$ and $ 4$. Hence if $ a^2\plus{}b^2\equal{}c^2$, one of $ a$ and $ b$ must be divisible by $ 7$. WLOG $ 7\mid b$. Then $ 7\mid (c\plus{}a)(c\minus{}a)$, implying either $ 7\mid c\plus{}a$ or $ 7\mid c\minus{}a$.[/hide]

[hide]If $ x^2\equiv y^2 \pmod 7$, then $ x^2\minus{}y^2\equal{}(x\plus{}y)(x\minus{}y)\equiv 0 \pmod 7$. Since 7 is prime, 7 divides either the sum or difference (possibly both). All that's left is to consider a triple whose squares are three different quadratic residues, which is impossible.[/hide]

[quote="worthawholebean"][quote="nayel"][hide]Note that all quadratic residues $ \mod 7$ are $ 1,2$ and $ 4$. Hence if $ a^2 \plus{} b^2 \equal{} c^2$, one of $ a$ and $ b$ must be divisible by $ 7$. WLOG $ 7\mid b$. Then $ 7\mid (c \plus{} a)(c \minus{} a)$, implying either $ 7\mid c \plus{} a$ or $ 7\mid c \minus{} a$.[/hide][/quote]What if $ a^2\equiv b^2\equiv2\pmod7$ and $ c^2\equiv4\pmod7$?[/quote]

Yes, sorry I missed those completely. :oops:

So three more cases are possible: $ (a,b,c)\equiv (1,1,2),(2,2,4)$ or $ (4,4,1)\pmod 7$. In each case, $ 7\mid a^2 \minus{} b^2$ and using the same argument we are done.

However, Bictor's solution above is really nice so we don't need such casework anymore. :)


Show, using the binomial expansion, that $ (1\plus{}\sqrt{2})^{5} < 99$. Show also that $ \sqrt{2} > 1.4$. Deduce that $ 2^{\sqrt{2}} > 1 \plus{} \sqrt{2}$.

Hello,

first we will prove that $ \sqrt {2} > 1.4$. Squaring that we get that $ 2 > 1.96$ which is true and we 'll prove that $ 1.5 > \sqrt {2}$, which is also trivial when we square it.

Now $ (1 \plus{} \sqrt {2})^{5} < 99$. --> $ (1 \plus{} \sqrt {2})^5 < (1 \plus{} 1.5)^5 \equal{} 97.65625 < 99$

$ 2^{\sqrt {2}} > 1 \plus{} \sqrt {2}$ is trivial by Bernoulli's inequality...Rewrite number $ 2$ from left side of inequality in form $ (1\plus{}1)$ :)

Thanks. I think the question wants us to deduce the final statement using both of the other two.

Thanks RobRoobiks. Got it: once I realised that you meant to write $ 2^{1.4} > 99^{\frac{1}{5}}$ as your first line.

So the previous parts now give

$ 2^{\sqrt{2}} > 2^{1.4} > 99^{\frac{1}{5}} > 1 \plus{} \sqrt{2}$.


Fie m,n numere neturale nenule, (m,n)=1 si 4|m. Sa se demonstreze ca daca [code]\frac{m}{n}>\sqrt{2}[/code] atunci avem chiar inegalitatea mai "tare":  [code]\frac{m}{n}-\frac{7}{mn}>\sqrt{2}[/code]

[quote="St. Elena"][color=darkblue] Fie $ m,n$ [b]numere naturale nenule[/b] astfel incat $ (m,n) \equal{} 1$ si $ 4|m.$ Sa se demonstreze ca daca $ \frac {m}{n} > \sqrt {2}$ atunci avem inegalitatea mai "tare":  $ \frac {m}{n} > \sqrt {2} \plus{} \frac {7}{mn}.$[/color][/quote]

[b]Solutie.[/b] Notam $ m \equal{} 4k,$ unde $ k\geq 1.$ Avem: $ \frac {m}{n} > \sqrt {2}\Longleftrightarrow 8k^2 > n^2.$ Cum resturile patratice modulo $ 8$ sunt $ 0,1$ si $ 4$ ne ramane sa analizam cazurile $ 8k^2 \equal{} n^2 \plus{} 4$ si $ 8k^2 \equal{} n^2 \plus{} 7.$

[b]1)[/b] $ 8k^2 \equal{} n^2 \plus{} 4\Longrightarrow 2|n.$ Dar $ (4k,n) \equal{} 1$ ceea ce ne indica ca ecuatia nu are solutii care sa respecte conditiile enuntului.

[b]2)[/b] Ecuatia $ 8k^2 \equal{} n^2 \plus{} 7$ are solutia principala $ (1,1).$ De aici deducem ca $ 8k^2\geq n^2 \plus{} 7.$

Acum: $ 8k^2\geq n^2 \plus{} 7\Longleftrightarrow \frac {16k^2}{n^2}\geq 2 \plus{} \frac {14}{n^2}\Longleftrightarrow \frac {m^2}{n^2} \minus{} \frac {14}{n^2} \minus{} 2\geq 0\Longleftrightarrow$ $ [(\frac {m}{n})^2 \minus{} 2\cdot\frac {m}{n}\cdot\frac {7}{mn} \plus{} (\frac {7}{mn})^2] \minus{} \frac {49}{m^2 n^2} \plus{} \frac {14}{n^2} \minus{} \frac {14}{n^2} \minus{} 2\geq 0\Longleftrightarrow$

$ \Longleftrightarrow (\frac {m}{n} \minus{} \frac {7}{mn})^2\geq 2 \plus{} \frac {49}{m^2 n^2} > 2\Longrightarrow \frac {m}{n} \minus{} \frac {7}{mn} > \sqrt {2}.$

Doua probleme asemanatoare(ceva mai grele) gasiti la urmatoarele link-uri:

$ \bullet$ [url=http://www.mathlinks.ro/viewtopic.php?p=1348687#1348687/]Stelele Matematicii 2008 problema 5[/url]

$ \bullet$ [url=http://www.mathlinks.ro/viewtopic.php?search_id=1314593109&t=242201/]Belarus 2000[/url]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Why is $S_{6} = (\mathbb{Z}/2\mathbb{Z}) \cdot (\mathbb{Z}/3\mathbb{Z})$?
Also, why is $A_{5} = A_{4} \cdot (\mathbb{Z}/5\mathbb{Z})$?

I remember seeing them somewhere. Perhaps I saw wrongly.
In any case, I am looking for two examples.
(i) M and N are nilpotent subgroups of G but [i]not both[/i] normal in G such that MN is not nilpotent.
(ii) M and N are solvable subgroups of G but [i]both not[/i] normal in G such that MN is not solvable.

Right I didn't. Sorry if my notation caused any confusion.


A bored student walks down a hall that contains a row of closed lockers, numbered 1 to 1024.  He opens the locker numbered 1, and then alternates between skipping and opening each closed locker thereafter.  When he reaches the end of the hall, the student turns around and starts back.  He opens the first closed locker he encounters, and then alternates between skipping and opening each closed locker thereafter. The student continues wandering back and forth in this manner until every locker is open.  What is the number of the last locker he opens?

I might have misunderstood the question because my answer seems a little too simple, but here goes.
[hide="Answer?"]The first time down the hall, the student opens all of the odd numbered lockers. Because there are an even number of lockers, the first one he opens while returning is even, and he will end that trip with an even locker, 2. Since he opened all of the odd lockers the first time, and all of the even lockers the second time, they're all open, and so he ended with the last locker opened on his trip back, number 2.[/hide]

Hokkage, it means that he on the way back, he only takes into account closed lockers, not the ones he opened already. Read my solution (if it's at all understandable :P).

My solution is a little long, but I'm kind of going through the whole thing to keep track of where I am. :P

[hide="Answer"]He opens $512$ lockers to begin with. On the way back, he opens $256$, then $128$, etc., until he opens the last one. On his first pass, he opens every other locker (all odds) beginning with $1$ and ending with $1023$. On the second, he opens every fourth locker (all evens) beginning with $1002$ and ending with $4$. On the third, he opens every eighth locker (all evens) beginning with $2$ and ending with $1018$. On the fourth, he opens every sixteenth locker (all evens) beginning with $1022$ and ending with $14$. On the fifth, he opens every thirty-second locker (all evens) beginning with $6$ and ending with $998$. On the sixth, he opens every sixty-fourth locker (all evens), beginning with $1014$ and ending with $54$. On the seventh, he opens every $128^{th}$ locker, beginning with $22$ and ending with $918$. On the eighth, he opens every $256^{th}$ locker, beginning with $982$ and ending with $214$. On the ninth, he opens every $512^{th}$ locker, those being $86$ and $598$. Finally, on the tenth, he opens every $1024^{th}$ locker, that being $\boxed{854}$.[/hide]

Yes, I definitely misread the problem  :oops:
Nice solution by the way.

Intuitive mod. solution I guess.

[hide]First, he leaves all the $2\pmod{2}$ numbered lockers closed, then on his return, since there are an even number of terms, the first term is unaffected, so he leaves $2\pmod{4}$ numbered lockers closed.

Then there's a gap because he closes the first locker, and so you are left with from $2\pmod{4}$, the next which is $2\pmod{4}$ numbered as the first closed locker. This is $6$. So now we have all lockers $6\pmod{8}$ closed going up, and the $6\pmod{16}$ lockers coming back.

There's another gap. Since the $6$ is opened, so the next $6\pmod{16}$ locker is $22$. So going up, the $22\pmod{32}$ numbered are closed, and coming down, $22\pmod{64}$ are closed.

Then a gap. The next $22\pmod{64}$ is $86$. So that one is closed. So going up $86\pmod{128}$ are closed, and coming back $86\pmod{256}$ are closed.

Then going up $342\pmod{512}$ are closed, and coming back $342\pmod{1024}$ are closed. And so the last closed one is $342$. Pretty tedious, but accurate.[/hide]

Ohh, I got off at one point. I see my mistake. I'm a little lost on how to fix it, though...

[hide]
Let's write each term in base 2:

1, 10, 11, 100, ..., 111111101, 111111110, 1111111111, 10000000000

After going through the lockers once, the closed lockers will be:

10, 100, 110, 1000, ..., 1111111010, 1111111100, 1111111110, 10000000000

Notice that they all end in 0. After going through twice, the closed lockers will be those that end in 10. After three times, the closed lockers will end in 110. 4 times leaves those that end in 0110 closed. 5 times leaves those that end in 10110 closed. 6 times leaves those that end in 010110 closed. 7 times leaves those that end in 1010110 closed, 8 times leaves those that end in 01010110, 9 times leaves those that end in 101010110, and the last one to be opened is $0101010110_2 = \boxed {342_{10}}$.
[/hide]

[quote="chess64"][hide]
Let's write each term in base 2:

1, 10, 11, 100, ..., 111111101, 111111110, 1111111111, 10000000000

After going through the lockers once, the closed lockers will be:

10, 100, 110, 1000, ..., 1111111010, 1111111100, 1111111110, 10000000000

Notice that they all end in 0. After going through twice, the closed lockers will be those that end in 10. After three times, the closed lockers will end in 110. 4 times leaves those that end in 0110 closed. 5 times leaves those that end in 10110 closed. 6 times leaves those that end in 010110 closed. 7 times leaves those that end in 1010110 closed, 8 times leaves those that end in 01010110, 9 times leaves those that end in 101010110, and the last one to be opened is $ 0101010110_2 \equal{} \boxed {342_{10}}$.
[/hide][/quote]

a great method

We will look at the direction of his 'passes' through the lockers, and the type of lockers that remain closed.

He first moves forwards (F) and leaves lockers $0 \pmod{2}$.

He moves backwards (B) and leaves lockers $2 \pmod{4}$.

This pattern continues and in general we can consider his $n$th pass as an ordered pair $(a_n, 2^n)$ where he leaves all lockers congruent to $a_n \pmod{2^n}$.

If $n$ is odd, he moves forwards and $a_n = a_{n-1}+2^{n-1}$. If $n$ is even, he moves backwards and $a_n = a_{n-1}$. Now we can compute the ordered pairs.

(0,2)
(2,4)
(6,8)
(6,16)
(22,32)
(22,64)
(86,128)
(86,256)
(342,512)
(342,1024)

Therefore, he closes locker $342$ last.

Can someone please explain the second solution here: [url]http://www.artofproblemsolving.com/Wiki/index.php/1996_AIME_Problems/Problem_9[/url].
I don't quite understand how they arrived at the equation and how $L_{n-1} $ would correspond to $ L_n $ under the equation.

@carelessmistakes, if you aren't very familiar with recursion, then that solution will be hard to understand and you won't get much out of it. I recommend checking out any of the solutions above-I for one enjoyed chess's solution.

p1:

First time: Opens odd numbered lockers; even numbered lockers remain.
Second time: Opens lockers numbered $0 \pmod 4$. $2 \pmod 4$ numbered lockers remain.
Third time: Opens lockers numbered $2 \pmod 8$. $6 \pmod 8$ numbered lockers remain.
Fourth time: Opens lockers numbered $14 \pmod {16}$. $6 \pmod {16}$ numbered lockers remain.
Fifth time: Opens lockers numbered $6 \pmod {32}$. $22 \pmod {32}$ numbered lockers remain.
Sixth time: Opens lockers numbered $54 \pmod {64}$. $22 \pmod {64}$ numbered lockers remain.
Seventh time: Opens lockers numbered $22 \pmod {128}$. $86 \pmod {128}$ numbered lockers remain.
Eighth time: Opens lockers numbered $214 \pmod {256}$. $86 \pmod {256}$ numbered lockers remain.
Ninth time: Opens lockers numbered $86 \pmod {512}$. $342 \pmod {512}$ numbered lockers remain.
Tenth time: Opens lockers numbered $854 \pmod {1024}$. $\boxed{342} \pmod {1024}$ numbered lockers remain.

This is a good problem, [quote=orangesyrup]what happened if we have 1000 lockers?[/quote]

use the same method

[quote=Altheman][hide="solution"]define $f(n)$ as the last locker to touch with $2^{2n}$ lockers
say you have $2^{2(n+1)}$ lockers, after going back and forth, , and under the numbering system, there will only be $2 \mod 4$ left, thus there will be $2^{2n}$ lockers left, under a new numbering system, there we know that locker $f(n)$ will be last, and under our old system, that is $4(f(n)-1)+2=4f(n)-2=f(n+1)$
$f(1)=2$, $f(2)=6$, $f(3)=22$, $f(4)=86$, $f(5)=\boxed{342}$[/hide]

Apply Your Knowledge: suppose we have $2^{2n}$ lockers, what locker will you end at? (give a closed form in terms of $n$)[/quote]
a closed form for $f(n)$ is [hide=closed form]$f(n)=\frac{4^n+2}{3}$[/hide], which can be easily proved by induction


How many four-digit even numbers can be formed if no digit is repeated?

The textbook has the answer, but the solution is not explained, so I don't really know how to do it at all.

Thanks to any help

[hide]Condition 1: last digit is 0

1st number: 9 (1-9)
2nd number: 8 (1-9, except digit 1)
3rd number : 7 (1-9, except digit 1)

so 9*8*7

Condition 2: last digit is 2, 4, 6, 8

1st number 8 (1-9, except last digit)
2nd number 8 (0-9, except digit 1 and 4)
3rd number 7 (0-9, except 1, 2, 4)
4th number 4 (2, 4, 6, 8)

so 8*8*7*4+9*7*8=1792+504=2296?[/hide]

[quote="Math Geek"]Spellitout, what method did you use to get your answer.  I got the solution using Walk Around the River's method of casework, but it seems like yours looks much easier.   :lol:[/quote]

Yeah id like to know as well.

yeah, it looks interesting, how'd you do it ?? ?? ?? ?? ?? ??


I think something is wrong here -- the degree of the left side is $ 3$, and of the right side is $ 9$... By the way, this looks a lot like Holder.

This inequality doesn't hold
I think u meant in the left side $ a_i^3 ; b_i^3 ; c_i^3$

If this is the case then let's prove!
By cauchy we have:
$ (a_1^4 \plus{}a_2^4\plus{}.....\plus{}a_n^4)(b_1^4 \plus{}b_2^4\plus{}...\plus{}b_n^4)(c_1^4\plus{}c_2^4\plus{}...\plus{}c_n^4)(d_1^4\plus{}d_2^4\plus{}...\plus{}d_n^4)>\equal{}(a_1^2b_1^2 \plus{} a_2^2b_2^2 \plus{}...\plus{}a_n^2b_n^2)^2 (c_1^2d_1^2 \plus{} c_2^2d_2^2 \plus{}...\plus{}c_n^2d_n^2)^2>\equal{}(a_1b_1c_1d_1 \plus{} a_2b_2c_2d_2 \plus{}...\plus{} a_nb_nc_nd_n)^4$

Now we just need to adjust the variables by setting $ a_i \equal{} \sqrt[4]{x_i^3} ; b_i\equal{}\sqrt[4]{y_i^3} ; c_i \equal{} \sqrt[4]{z_i^3}$ and $ d_i \equal{} \sqrt[4]{x_iy_iz_i}$ and substituing we get the result

[quote="primoz2"]Prove:    (a_1 +a_2+.....+a_n)(b_1 +b_2+...+b_n)(c_1+c_2+_...c_n)>=(a_1*b_1*c_1+....+a_n*b_n*c_n)^3[/quote]
I think you are wrong
the inequality you post, I think that is Holder inequality but not right
The Holder inequality:
 (a1^3+a2^3+.....+an^3)(b1^3+b2^3+.....+bn^3)(c1^3+c2^3+.....+cn^3)>=
 (a1*b1*c1+a2*b2*c2+.....+an*bn*cn)^3

[/b][/code]


function
linear algebra
matrix
algebra proposed
algebra

1)$ x_{1}$
$ x_{n\plus{}1}\equal{}\frac{x_{n}^{3}\plus{}3x_{n}}{3x_{n}^{2}\plus{}1}$
Find thegeneral cacus of this sequence.
2)$ x_{1}$
$ x_{n\plus{}1}\equal{}\frac{2x_{n}}{x_{n}^{2}\plus{}1}$
Find the general cacus of this sequence
3)$ x_{1}\in R^{\plus{}}$
$ a > 0$
Find the general cacus of following sequence
$ x_{n\plus{}1}\equal{}\frac{1}{2}(\frac{a}{x_{n}}\plus{}x_{n})$

There are easy execizes.
1)$ x_{n}\equal{} th(3^{n\minus{}1}arcth(x_{1}))$,
2)$ x_{n}\equal{} th(2^{n\minus{}1}arcth(x_{1}))$,
3)$ x_{n}\equal{}\sqrt a\ \ th(2^{n\minus{}1}arcth(\frac{x_{1}}{\sqrt a})).$

Why do you find out it.I don't think you induction.

Other problem
$ x_{1}\in R$
$ x_{n\plus{}1}\equal{}\frac{ax_{n}\plus{}b}{cx_{n}\plus{}d}$
Find the general cacus of  this sequence .
A proof is let
$ x_{n}\equal{}\frac{p_{n}}{q_{n}}$
Don't use this solution.

Obviosly
\[ x_{n}\equal{}\begin{pmatrix}a & b\\ c & d\end{pmatrix}^{n\minus{}1}\begin{pmatrix}x_{1}\\ 1\end{pmatrix}.\]


The original price of an item was $ \$50$. The store deducted $ 20\%$, and then deducted an additional $ 20\%$ off the reduced price. How many dollars more would a consumer save if the store had simply reduced the original price by $ 40\%$?


probability rotation function algebra polynomial AMC

From CofC Math Meet 2003, Level 3

13. Nestor flips a fair coin eight times.  What is the probability that the resulting sequence of heads and tails looks the same when viewed from either the beginning or from the end?  (For example, HTHTTHTH is one such sequence.)

(A) 1/32
(B) 1/16
(C) 3/16
(D) 3/64
(E) 3/32

15. Find the number of real solutions to the equation x^5 + x^4 + x^3 + x^2 + x + 1 = 0.

(A) 5
(B) 4
(C) 3
(D) 2
(E) 1

17. If the graph of the polar equation r = 1 + 3 sin:theta: is rotated :pi:/2 radians counterclockwise, find an equation for the resulting graph.

(A) r = 1 + 3 sin(:theta: + :pi:/2)
(B) r = 1 + 3 cos(:pi:/2 - :theta:)
(C) r = 1 - 3 sin(:theta: - :pi:/2)
(D) r = 1 - 3 cos:theta:
(E) r = 1 + 3 cos:theta:

21. The function (x+1)/(x^3 + 1) can be written as the sum of an even function and an odd function.  Find the even function.

(A) (x^4 - 1)/(x^3 + 1)
(B) (x^4 + 1)/(x^6 + 1)
(C) (x^4 - 1)/(x^6 + 1)
(D) (x^4 + 1)/(x^6 - 1)
(E) (x^4 - 1)/(x^6 - 1)

Please include your solutions.

[hide]13. First four don't matter, last four can only have one arrangement, based on the first four. (1/2)^4=1/16 B

15. By Descarte's Rule of Signs, there are 0 positive roots, and 5, 3, or 1 negative roots. Since it's a fifth degree polynomial, it can have at most 5 roots. Use the rational roots test and synthetic division to factor it to (x+1)(x^4 + x^2 + 1) The second term clearly has no roots, so the only root is -1, E.

17. Take a nice easy reference point, like the one at :theta: = 0. That's (r, :theta: ) = (1, :theta: ). The new position will obviously be (1,:pi:/2) after it's rotated. So B is obviously wrong. Now look at (4, :pi:/2), which will become (4,:pi:). A and E are obviously wrong. Then look at (1, :pi:), which becomes (1, 3:pi:/2). C works, D does not. C.

21.[/hide]

mathfanatic wrote:21. The function (x+1)/(x^3 + 1) can be written as the sum of an even function and an odd function. Find the even function.

(A) (x^4 - 1)/(x^3 + 1)
(B) (x^4 + 1)/(x^6 + 1)
(C) (x^4 - 1)/(x^6 + 1)
(D) (x^4 + 1)/(x^6 - 1)
(E) (x^4 - 1)/(x^6 - 1)

Spoiler:[hide]Obviously an even function will have and even degree, f(-x) will not equal f(x). So the bottom must be x^6-1, which factors to (x^3-1)(x^3+1). the sum of the two functions will yield (x+1)(x^3-1) on top, which expands to x^4 - 1 + x^3-x. then x^4 - 1 is the top of the even function , so the answer is (E) (x^4 - 1)/(x^6 - 1) [/hide]

mathfanatic wrote:13. Nestor flips a fair coin eight times.  What is the probability that the resulting sequence of heads and tails looks the same when viewed from either the beginning or from the end?

[hide]

1/16 ?

Such sequence is uniquely defined by the 4 first flips. So there are 2^4 of them.

There are 2^8 sequence.

So probability = 2^4/2^8=1/2^4 = 1/16[/hide]

There appears to be a problem with number 17, namely [hide]the correct answer, C, happens to equal the function D.[/hide]

Haha, yes, I was just...um, testing you.  Yeah.  I made that mistake on purpose to see who would catch it first.  </jk>

Joel is right.  C should have been "r = 1 - 3 sin(:theta: - :pi:/2)

15: [hide] as in a question posted in intermediate, you just multiply both sides by x-1, and the result: x^6-1=0 clearly only has  :pm: 1 as real solutions, but we ignore 1, and see -1 is the only real root.[/hide]


Let $a,b$ be positive real numbers such that $0<a<b,a+b=1$.
Compare the following value and four equations.

\[\frac{1}{2}\ ,a\ ,b\ ,2ab\ ,a^2+b^2\]

In order from largest to smallest:

[hide]

$b, a^2+b^2, \frac {1}{2}, 2ab, a$

Why?

Since a<b, a and b can't be equal to 1/2. Let's say that a = 1/4 and b = 3/4. Then:

1/2=1/2
a=1/4
b=3/4
2ab=3/8
a^2+b^2 =5/8

Thus, rearranging them, we'll have answer.

$b, a^2+b^2, \frac {1}{2}, 2ab, a$[/hide]

Thank you for your reply, siverfalcon.

Please justify your solition by showing the proof of inequality. ;)

kunny

In order from largest to smallest:

[hide]$b, a^2+b^2, \frac{1}{2}, 2ab, a$[/hide]

Proof:

[hide]First off, $0<a<b<1$.  $a$ and $b$ can't be both smaller than $\frac{1}{2}$, and can't both be larger than $\frac{1}{2}$, so $a$ must be smaller than $\frac{1}{2}$, and $b$ must be larger than $\frac{1}{2}$ from the inequality given above.

Also, $\frac{1}{2}-a=b-\frac{1}{2}$, so let $a=\frac{1}{2}-k$, and $b=\frac{1}{2}+k$, where $k$ is a positive real number and $k<1$.

replacing the list (in the order that kunny put it), we would have

$\frac{1}{2}, \frac{1}{2}-k, \frac{1}{2}+k, \frac{1}{2}-k^2, \frac{1}{2}+k^2$.

Since $k<1$, $k^2<k$, so we can order the list as follows:

$\frac{1}{2}+k,\frac{1}{2}+k^2, \frac{1}{2},  \frac{1}{2}-k^2, \frac{1}{2}-k$

which would be
$b, a^2+b^2, \frac{1}{2}, 2ab, a.$[/hide]

You are right,dhuang614.Thank you!

kunny


Howdy neighbor. This is the game where you start a sentence saying I am so [b] insert word [/b] that ....(bla bla bla). Obviously, I'll start by giving an example.

I am so [b] weird [/b] that I twist my nipples every time I think there's an ant in the room.

I am so eccentric I spent an hour talking to a tree.

I am so stupid that I am wasting my time posting on this topic.

I am so [b]skinny[/b] that I died when I got hit by a piece of thread.

I m so [b]normal[/b] that i m so normal. :ninja:

That post was so [b]original[/b] that it imitated every conformity accepted by our uniform society.

I am so [b][i][u]bored[/u][/i][/b] that i decided to post here, and died of boredom...

I am so [b]pissed off[/b] that I bought cow meat and swallowed it hoping to die from mad cow diseases. I then got so [b]ecstatic[/b] that I punched a hole in my wall and spat in it. A faerie came out and licked my toe 298 times.

I am so [b]annoyed[/b] at how Prime Factorization continues to use the words that are forbidden on these forums.

[size=25]I am so [/size][size=200][b][i][u][color=olive]FRUSTRATED[/color][/u][/i][/b][/size][size=25]for no apparent reason.[/size]

I am so [b]HAPPY[/b] because I am!!! :huh:

i'm so normal that i pick my nose when i think about michael vick.

i am so surprised that michael vick is getting sued 60some billion dollars for illegal dogfights

i am so confounded that miyomiyo does not understand that even though this is an online community of math geeks. (no offence) people will get excited...

I am to $ \text^{\text{cool}}$ to be with you. :P

I am so[hide][hide][hide][b]bored[/b][/hide][/hide][/hide] that I just posted this.

Whee, added more hide tags because I got [color=blue][b]bored[/b][/color] again.

I am so [size=59][color=white]bored[/color][/size] that i actually decided to open all of CheeseIsGood's hide tags

[size=9]I am so[/size][size=200]SATED[/size][size=9]right now because I am at my computer doing what I want.[/size]


calculus
integration
geometry
trigonometry
derivative
calculus computations

Let $a>1.$ Find the area $S(a)$ of the part surrounded by the curve $y=\frac{a^{4}}{\sqrt{(a^{2}-x^{2})^{3}}}\ (0\leq x\leq 1),\ x$ axis , $y$ axis and the line $x=1,$ then  when $a$ varies in the range of $a>1,$ then find the extremal value of $S(a).$

I think that [hide]after the substitution $x = a \sin t$ we get $a^{2}\int_{0}^{\sin^{-1}\frac1a}\frac{dt}{\cos^{2}t}$ which is $a^{2}\tan\sin^{-1}\frac1a = \frac{a^{2}}{\sqrt{a^{2}-1}}$.  This simplifies to $(a^{2}-1)^{\frac12}+(a^{2}-1)^{-\frac12}$, a form in which I can take the derivative, and I find that it reaches a minimum of 2 when $a = \sqrt 2$.[/hide]


Let $x\in R$, $n\in N$ and $\cos{nx}, \cos{(n+1)x} \in Q$.
Does it follow that $\cos x \in Q$?

I'm not sure but I don't think it is correct.

Perhaps with stronger conditions it's true, but, as it, it's false : take $x= \pi / 6, \ n = 2$


What is the remainder when the sum of the first five primes is
divided by the sixth prime?

The sum of the 1st five primes are 2+3+5+7+11=28.
The 6th prime is 13.
The remainder when 28/13 is 2.


سلام
آیا می توان 100 عدد کوچکتر یا مساوی 25000 انتخاب کرد به طوری که جمع دو بدو ی آن ها متمایز باشد؟
میشه ببینید راه حل من درسته یا نه؟
راه حل من:
اگر اعداد انتخابی
a1,a2,...,a100
باشند چون جمع هیچ دو تایی نباید برابر باشد پس تفاضل هیچ دوتایی هم نباید یکی باشد و
پس اگر ما اعداد را به صورت زیر انتخاب کنیم چون فاصله ی بین هردو متمایز است پس جمع هر دوتایی نیز متمایز است:
a1=1
a2=a1+1=2
a3=a2+2=4
a4=a3+3=7
a5=a4+4=11
.
.
.
a100=a99+99
اگر طرفین همه تساوی ها را جمع کنیم و مساوی ها را از طرفین حذف کنیم به دست می آید:
a100=1+(1+2+3+4+5+6+7+...+99)=2476
و چون حتی بزرگترین این 100 عدد از 25000 کمتره پس میشه این اعداد را انتخاب کردم
اما چیزی که من رو نسبت به راه حلم مشکوک کرده اینه که من عددها را کوچکتر از 2500 آوردم اما صورت سوال گفته کمتز از 25000
پس یعنی  میشه که سوال این همه آسون باشه
میشه اگه اشکالی داره راه حلم بگید

My weblog:
http://www.mathlinks.ro/weblog.php?w=1774

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ممنون


Let a,b,c  and k be possitive real numbers such that $ a\plus{}b\plus{}c\equal{}3$. Prove the Inequality:
$ \frac{1}{k\minus{}ab}\plus{}\frac{1}{k\minus{}bc}\plus{}\frac{1}{k\minus{}ca} \leq \frac{3}{k\minus{}1}$

[quote="nguoivn"]It's only trues with $ k \ge 6$. $ k \equal{} 6$ is the best constant  :)
It's trues by Schur's ineq[/quote]

it only prove  $ k \equal{} 6$

substitution $ a \equal{} \frac {3x}{x \plus{} y \plus{} z},b \equal{} \frac {3y}{x \plus{} y \plus{} z},c \equal{} \frac {3y}{x \plus{} y \plus{} z}$

$ 216zx^5 \plus{} 216xz^5 \minus{} 432z^3y^3 \minus{} 108z^2y^4 \plus{} 216zy^5 \minus{} 108z^2x^4 \minus{} 432z^3x^3$ $ \minus{} 108y^4x^2 \minus{} 432y^3x^3 \plus{} 216yz^5 \plus{}$

$ 216y^5x \plus{} 216yx^5 \minus{} 108y^2z^4 \minus{} 108x^2z^4 \minus{} 108y^2x^4$$ \plus{} 108x^6 \plus{} 108y^6 \plus{} 108z^6 \plus{} 135y^2zx^3 \plus{} 405yzx^4\plus{}$

$ 135yz^3x^2 \minus{} 1701y^2z^2x^2 \plus{}$ $ 135yz^2x^3 \plus{} 405y^4zx \plus{} 135y^3z^2x$ $ \plus{} 405yz^4x \plus{} 135y^2z^3x \plus{} 135y^3zx^2\geq 0$

$ \Longleftrightarrow 108\sum(x^4(x \minus{} y)( \minus{} z \plus{} x)) \plus{} 324\sum(x^3(y \plus{} z)(x \minus{} y)( \minus{} z \plus{} x)) \plus{}$

$ 216(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \plus{} 1377xyz\sum(x(x \minus{} y)( \minus{} z \plus{} x)) \plus{}$ $ 756xyz\sum((y \plus{} z)(x \minus{} y)( \minus{} z \plus{} x))\geq 0$

 :lol:

[quote="AngelHolder"]Let a,b,c  and k be possitive real numbers such that $ a + b + c = 3$. Prove the Inequality:
$ \frac {1}{k - ab} + \frac {1}{k - bc} + \frac {1}{k - ca} \leq \frac {3}{k - 1}$[/quote]

[b]we have Maclaurin's ine...[/b]

$ \sqrt {d_2}\le d_1\rightarrow \sqrt {\frac { ab + bc + ca}{3}} \le \frac {a + b + c}{3}$

$ f(x): = \frac {1}{6 - x}\rightarrow f"(x): = - \frac {2}{(6 - x)^3} < 0$

$ \right \sum_{cyclic} \frac {1}{6 - ab}\le \frac {3}{6 - \frac {\sum_{cyclic}ab}{3}}\le \frac {3}{5}$

[quote="khanhsy"][quote="nguoivn"]  It's only trues with $ k \ge 6$. $ k \equal{} 6$ is the best constant  :)
It's trues by Schur's ineq[/quote]

I think : It's only true with $ k \ge \frac {9}{4}$ I can not true :D[/quote]

$ k\equal{}3$

try $ a\equal{}\frac{3}{11},b\equal{}\frac{15}{11},c\equal{}\frac{15}{11}$


USAMTS
geometry
geometric transformation
inequalities
email

Are books from Gil Publishing written in English? (The website is in a different langauge, but it is listed under the "Resources" section of AoPS)

Hurm.  There used to be an English site.  Some of the books they produce are in English.  Valentin can probably tell you more (Gil published his first book, which is in Romanian).

I believe one of the USAMTS award books was published by GIL (not the one I got, so maybe someone can confirm this?) In any case, one of their books is by Andreescu and Andrica.

mathnovice, if you need a translation of a short passage, I can help you.

There are currently 2 books that I am aware of published in English by Gil Publishing House. Both of them are signed Titu Andreescu and Dorin Andrica. One of them is called something like 360 problems, and the other one is about Diophantine Equation. A third book is now being prepared, and will be announced on the forum (it's made by members of this site).

What is about the book " Old and new inequalities " (English version) ? Valentin , can you please post the ISBN of this book (English version) ?

Oh yes, that one too  :blush:  I forgot about it when I wrote this post. The ISBN is 973-9417-35-3.

GIL Publishing House published in english:
1. T. Andreescu, D. Andrica - An Introduction to Diophantine Equations [ ISBN 973-9238-88-2 ]
2. T. Andreescu, D. Andrica - 360 Problems for Mathematical Constests [ ISBN 973-9417-12-4 ]
3. T. Andreescu, V. Cartoaje, G. Dospinescu, M. Lascu - Old and New Inequalities [ ISBN 973-9417-35-3 ]

We are now working at the english version of the website.
More details will be available at http://www.gil.ro in the next couple of weeks. Also our e-mail address is gil1993@zalau.astral.ro.

is there anyone who know how to buy books from GIL?

i sent an email to GIL 4 times, and he didn't reply me.

how am i supposed to order those books???

is there any other store which sell books from GIL?

Here is a list published in English by GIL Publishing House:
1. T. Andreescu, D. Andrica - An Introduction to Diophantine Equations [ ISBN 973-9238-88-2 ]
2. T. Andreescu, D. Andrica - 360 Problems for Mathematical Contests [ ISBN 973-9417-12-4 ]
3. T. Andreescu, V. Cartoaje, G. Dospinescu, M. Lascu - Old and New Inequalities [ ISBN 973-9417-35-3 ]
4. Negut A. - Problems for the Mathematical Olympiads [ ISBN 973-9417-52-3 ]
5. Cirtoaje V. - Algebraic Inequalities. Old and New Methods [ ISBN 973-9417-66-3 ]

You can find more details on http://www.gil.ro and by clicking the button English you will see our English website. For orders e-mail at sales@gil.ro.

Regards!

Thank you for your english version!

How do Americans order books from you? Should we email your sales people?

sales@gil.ro is a mail where anyone from around the world can send e-mail with order of books. Depending on the days that the person needs the books, it is chosen if we send the books using DHL services (which are expensive) or romanian postal services (which are about half of DHL prices).

Order the books at sales@gil.ro and you will recieve them. The only problem is with the shipping costs and now there are some issues with the romanian postal office so DHL is the only way to deliver the books.

GIL Publishing House, as it is announced on the English version of the website will be present also at the IMO so who comes in Slovenia at the IMO can buy books without paying the shipping, saving some money this way.

Regards!


I have my school competition tomorrow and I really need some helpful websites with problems to study!!!!!! :maybe:  :(  :o

Previous competition papers....school round or any other rounds would help too
-sprint round
-target round
-team round

Advance thank you :P

If you want to practice for school rounds, you might be better off practicing chapter competitions...

Get them from Minnesota Mathcounts. (google it)

Your school actually goes out of its way and competes on Saturday? Wow my school is pathetic.

thanks for the Minnesota site it helped but is the chapter one's supposed to be more difficult than the school rounds :?:

this may answer ur question:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244691

Actually, one could make an argument about how school is actually supposed to be at the same level, or perhaps even harder than chapter, as some schools make their students take it and use it to find the team. I would say that the school and chapter are more or less the same level, or so it's appeared to me in the last two years that I've taken them both.

in our city the school competition takes place in a high school and all the school's in our city compete there. So it's like 5 middle school competing, but it's still considered as the school round.  :|

Does that actually use the official format? I mean, do you actually have a real countdown with the top scorers? I would guess not, because it would take quite a while, as it has more people, and it's unofficial, but I would love to do that. It would let you test yourself against the other people that you would meet at chapter and it wouldn't really count.

hey xpmath.  Yeah, we do have a countdown round.  There was about 60 students there total and the top ten go to the countdown. I didn't make it in the countdown though... :(   I'm in seventh grade...there's always another year :maybe:


Yesterday, 28 students took a test. The arithmetic mean of those 28 scores was 72 points. Two students who were absent yesterday took the test this morning, and the arithmetic mean of all 30 test scores is 73 points. If the difference of the two scores from this morning is 22 points, what is the lower score from this morning?

If the arithmetic mean of those 28 scores was 72 points, then the sum of all the scores is $ 28 \times 72 \equal{} 2016$. Similarly, including the two absent students, the total sum of all the scores is $ 30 \times 73 \equal{} 2190$. That means the sum of the two students' scores is $ 2190 \minus{} 2016 \equal{} 174$.

Let the scores be $ a$ and $ b$. We have the equations \[ a \plus{} b \equal{} 174 \] and \[ a \minus{} b \equal{} 22 \] Adding the two equations together gives us \[ 2a \equal{} 196 \implies a \equal{} 98 \] Therefore, the lower score was $ 98 \minus{} 22 \equal{} \boxed{76}$.


A-> n-(2-carboxyethyl)-phtalimide
B-> phtalic acid
C-> glycine
d-> it isn’t mentioned in the question but assuming  ‘one which is red in alkaline medium and colourless in acidic medium’ is D, it wud be phenolphthalein.