Call a positive integer $n$  [b]practical[/b] if every positive integer less than or equal to $n$ can be written as the sum of distinct divisors of $n$.

For example, the divisors of 6 are 1, 2, 3, and 6. Since
\[ \centerline{1={\bf 1}, ~~ 2={\bf 2}, ~~ 3={\bf 3}, ~~ 4={\bf 1}+{\bf 3}, ~~ 5={\bf 2}+ {\bf 3}, ~~ 6={\bf 6},} \]
we see that 6 is practical.

Prove that the product of two practical numbers is also practical.

It is obviosly. Let m and n are practikal, and k=ln+r,r<n,l<m. Then l is summ of divisors m, therefore ln is sum divisors mn not less n, r - is sum divisors n (or mn) less than n. All divisors are distinct.

see that all powers of 2 are practical...
any power of a practical no. is also practical...
Prove these two statements without using the fact that the product of two practical no.s is also practical... :roll:

$ 2$ is practical number because $ 1\equal{}1$ and $ 2\equal{}2$.
If $ n$ is practical number then $ 2n$ is also a practrical number.
(Positive integer less than $ n$ can be expressed by sum of positive divisor less than $ n$ thus positive integer less than $ 2n$ can be a sum of distinct divisors of $ n$. And $ 2n$ is a positive divisor of $ 2n$ itself.)

Let $ n$ is a practical number.
$ n \equal{} 2^m \cdot k$
If $ k_1$ is the smallest prime factor of $ n$ then it is less than $ 2^m\plus{}1$. (because $ k_1$ is larger than $ 2^m\plus{}1 \minus{}1$, $ 2^m\plus{}1$ cann't be a sum of distinct positive divisors of $ n$)
So positive integer less than or equal $ k_1$ can be a sum of positive divisors of $ 2^m$.
And we can easily show that the other prime factors and multiple of these can be a sum of divisors of  $ n$.

If $ n, m$ are practical numbers. We can adapt above steps to every odd divisors of $ n, m$ and multiple of power of $ 2$. And we are done!


calculus
integration
function
trigonometry
calculus computations

We know if an improper integral is absolutely convergent, then it converges. But how about the inverse?
Consider the functions like $ f(x) \equal{} \frac1x\sin x$.

That example pretty much answers the question.

$ \int_0^{\infty}\frac{\sin x}x\,dx$ converges but does not converge absolutely.

Its lack of absolute convergence was a topic [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184483]just today.[/url]

Its convergence as an improper integral has been a topic here a half dozen times at least. (My plan: integrate by parts, integrating the $ \sin x$ and differentiating the $ \frac1x.$)

Another question about improper integrals. By the Direct Comparison Test, we know:

[i]Suppose $ f(x)$ is continuous on $ [a,\infty), a>0$ and $ f(x)\geq0$. If there exist constants $ M>0, p>1$ such that $ f(x)\leq\frac M{x^p}$ for $ x\geq a$, then $ \int_a^{\infty}f(x)\,dx$ converges; if there is a constant $ N>0$ s.t. $ f(x)\geq\frac Nx$ for $ x\geq a$, then $ \int_a^{\infty}f(x)\,dx$ diverges.[/i]

Can we rewrite it as following and why?

[i]Suppose $ f(x)$ is continuous on $ [a,\infty), a>0$ and $ f(x)\geq0$. Then
$ \int_a^{\infty}f(x)\,dx$ converges $ \Longleftrightarrow$ $ \exists p>1, M>0$ such that $ f(x)\leq\frac M{x^p}$ for $ x\geq a$.[/i]

There is no way you're going to get an "if and only if" here. It's possible to construct absolutely convergent improper integrals in which the function doesn't even go to zero.
An example: Let $ f(x)$ be the piecewise linear function with vertices at $ (n\minus{}\frac1{n^3},0),(n,n),(n\plus{}\frac1{n^3},0)$ for each $ n\ge 2$ (and $ f(x)\equal{}0$ for all smaller $ x$). The integral $ \int_0^\infty f(x)\,dx$ converges, with value $ \sum_{n\equal{}2}^{\infty}\frac1{n^2}$.


function
absolute value
algebra unsolved
algebra

Suppose $a$ and $b$ are two positive real numbers such that the roots of the cubic equation $x^3 - ax + b = 0$ are all real. If $\alpha$ is a root of this cubic with minimal absolute value, prove that \[ \dfrac{b}{a} < \alpha < \dfrac{3b}{2a}.  \]

Well sorry to revive this old topic.
But I do not understand what the question tells in $ \alpha$ being the root with minimum absolute value. Could anyone please explain?
Its very urgent,

Thanks in advance.

All roots being real needs the discriminant to be nonnegative.
http://en.wikipedia.org/wiki/Cubic_function

Hence, we have $ 4a^3\ge27b^2$.

It is easy to check there is one root in each of the following three intervals:
$ ( \minus{} \frac {3b}{2a}, \minus{} \frac ba)$, $ (\frac ba,\frac {3b}{2a})$, and $ (\frac {3b}{2a},\infty)$.
Note the two positive roots coincide at $ \frac {3b}{2a}$ when $ 4a^3 \equal{} 27b^2$.
If you look at the absolute value of these three roots, you will see the root has the smallest absolute value has
So $ \frac ba < \alpha < \frac {3b}{2a}$.

[hide=MUCH, MUCH better solution that I'm genuinely proud of!]

Because of
\[\alpha = \frac{\alpha^3 + b}{a}\]
it is sufficient to show that \(0 < \alpha^3 < b/2\)

Firstly, we demonstrate the fact that \(\alpha\), the root closest to the origin, is necessarily positive.
To do this, we assume (for the sake of contradiction), that \(\alpha\) is negative. Note that \(\alpha\not = 0\) because that would force \(b=0\), which contradicts the hypothesis.
Also, suppose that \(\beta\) and \(\gamma\) are the other two roots.
Then, we must have
\[\beta + \gamma = -\alpha > 0\]
and also
\[\beta\gamma = \frac{-b}{\alpha} > 0\].
As \(\beta + \gamma\) and \(\beta\gamma\) are both positive, \(\beta\) and \(\gamma\) must both be positive. But then, it happens that
\[\lvert \alpha \rvert = \lvert \beta + \gamma \rvert > \lvert \beta \rvert\],
which contradicts the minimality of \(\alpha\). Therefore, it is necessary for \(\alpha\) to be positive.

Now, we show that \(\alpha^3 < b/2\).

Let \(f(x) = x^3 - ax + b\), so that \(f'(x) = 3x^2 - a\). Now, one can see that
\begin{align*}
\alpha^3 &< \frac{b}{2} = \frac{a\alpha - \alpha^3}{2} \\
\Leftrightarrow \alpha^2 &< \frac{a}{3} \\
\Leftrightarrow f'(a) = 3\alpha^2 - a &< 0
\end{align*}

Thus, it suffices to show that \(f'(\alpha) < 0\). By definition, \(f'(\alpha) < 0\) means that the function \(f\) is decreasing at \(x=\alpha\), and it is enough to show the same.

Consider the graph of \(y=f(x)\). Because \(f(0) = b > 0\), and there is no root in the interval \((0,\alpha)\) (because of the minimality of \(\alpha\)), the function switches signs from positive to negative at \(x=\alpha\). That is to say that the function is decreasing at \(x=\alpha\), which finishes the demonstration.
[/hide]


trigonometry
complex numbers
geometry unsolved
geometry

[b] Let's take the triangle ABC whith AD,BE and CF medians.If we know that m(<GAB)+m(<GBC)+m(<GCA)=90, prove that the triangle has two equal sides, where {G}=AD intersectated with BE and CF.[/b] :)

see http://www.mathlinks.ro/Forum/viewtopic.php?t=65289 .

  dg

Sorin, you can use the following properties:

$1.\blacktriangleright\boxed {\ \triangle ABC\Longrightarrow\tan A=\frac{4S}{b^2+c^2-a^2}\ }\ .$

$2.\blacktriangleright \boxed {\ \{x,y,z\}\subset \left[0,\frac{\pi}{2}\right)\ ,\ x+y+z=\frac{\pi}{2}\Longrightarrow \tan x\cdot \tan y+\tan y\cdot\tan z+\tan z\cdot\tan x=1\ }\ .$

[b]See the following message ![/b]

Sorin, you can use the following properties:

$1.\blacktriangleright\boxed {\ \triangle ABC\Longrightarrow\tan A=\frac{4S}{b^2+c^2-a^2}\ }\ .$

[u]Proof.[/u] $2S=bc\sin A$ and $b^2+c^2-a^2=2bc\cos A\Longrightarrow (1)\ .$

$2.\blacktriangleright \{x,y,z\}\subset \left[0,\frac{\pi}{2}\right)\ ,\ x+y+z=\frac{\pi}{2}\Longrightarrow$
$\boxed {\ \tan x\cdot \tan y+\tan y\cdot\tan z+\tan z\cdot\tan x=1\ }\ .$

[u]Proof.[/u] $x+y+z= \frac{\pi}{2}$ $\Longrightarrow$ $x+y=\frac{\pi}{2}-z$ $\Longrightarrow$ $\tan (x+y)\cdot\tan z =1$ $\Longrightarrow$

$\tan z\cdot (\tan x+\tan y)=1-\tan x\cdot \tan y\Longrightarrow (2)\ .$

[b][u]Proof of the proposed problem.[/u][/b]

$\tan \widehat {GAB}=\frac{4[GAB]}{c^2+\frac 49m^2_a-\frac 49m^2_b}=\frac{12S}{9c^2-4\left(m^2_a-m^2_b\right)}=\frac{4S}{3c^2+b^2-a^2}$ a.s.o.

Therefore, $\sum m(\widehat {GAB})=\frac{\pi}{2}$ $\Longleftrightarrow$ $\sum \tan \widehat {GAB}\cdot \tan \widehat {GBC}=1$ $\Longleftrightarrow$

$\sum \frac{16S^2}{(3c^2+b^2-a^2)(3a^2+c^2-b^2)}=1$ $\Longleftrightarrow$

$16S^2\cdot \sum (3a^2+c^2-b^2)=\prod (3a^2+c^2-b^2)$ $\Longleftrightarrow$

$\left(2\sum a^2b^2-\sum a^4\right)\cdot 3\sum a^2=\prod (3a^2+c^2-b^2)\ .$

We must obtain $(a^2-b^2)(b^2-c^2)(c^2-a^2)\boxed {\ldots \bigoplus\ldots}=0\ .$

Indeed, with the substitutions $x: =a^2\ ,\ y: =b^2\ ,\ z: =c^2$ we obtain :

$3\sum x\cdot \left(2\sum xy-\sum x^2\right)=\prod (3x+z-y)\Longleftrightarrow$

$8(y-z)x^2-8(y^2-z^2)x+8yz(y-z)=0\Longleftrightarrow$

$(y-z)(x-y)(x-z)=0\Longleftrightarrow x=y\ \vee\ y=z\ \vee\ z=x\Longleftrightarrow$

$\boxed {\ a=b\ \vee\ b=c\ \vee\ c=a\ }$ q.e.d.
\[ \mathrm {I\ wish\ you\ every\ success\ !}  \]

[quote="Virgil Nicula"]Sorin, you can use the following properties:

$1.\blacktriangleright\boxed {\ \triangle ABC\Longrightarrow\tan A=\frac{4S}{b^2+c^2-a^2}\ }\ .$

[u]Proof.[/u] $2S=bc\sin A$ and $b^2+c^2-a^2=2bc\cos A\Longrightarrow (1)\ .$

$2.\blacktriangleright \{x,y,z\}\subset \left[0,\frac{\pi}{2}\right)\ ,\ x+y+z=\frac{\pi}{2}\Longrightarrow$
$\boxed {\ \tan x\cdot \tan y+\tan y\cdot\tan z+\tan z\cdot\tan x=1\ }\ .$

[u]Proof.[/u] $x+y+z= \frac{\pi}{2}$ $\Longrightarrow$ $x+y=\frac{\pi}{2}-z$ $\Longrightarrow$ $\tan (x+y)\cdot\tan z =1$ $\Longrightarrow$

$\tan z\cdot (\tan x+\tan y)=1-\tan x\cdot \tan y\Longrightarrow (2)\ .$

[b]Proof of the proposed problem.[/b]

$\tan \widehat {GAB}=\frac{4[GAB]}{c^2+\frac 49m^2_a-\frac 49m^2_b}=\frac{12S}{9c^2-4\left(m^2_a-m^2_b\right)}=\frac{4S}{3c^2+b^2-a^2}$ a.s.o.

Therefore, $\sum m(\widehat {GAB})=\frac{\pi}{2}$ $\Longleftrightarrow$ $\sum \tan \widehat {GAB}\cdot \tan \widehat {GBC}=1$ $\Longleftrightarrow$

$\sum \frac{16S^2}{(3c^2+b^2-a^2)(3a^2+c^2-b^2)}=1$ $\Longleftrightarrow$

$16S^2\cdot \sum (3a^2+c^2-b^2)=\prod (3a^2+c^2-b^2)$ $\Longleftrightarrow$

$\left(2\sum a^2b^2-\sum a^4\right)\cdot 3\sum a^2=\prod (3a^2+c^2-b^2)$ $\Longleftrightarrow$

$(a-b)(b-c)(c-a)\boxed {\ldots \bigoplus\ldots}=0$ $\Longleftrightarrow a=b\ \ \vee\ \ b=c\ \ \vee\ \ c=a\ .$

\[ \mathrm {I\ wish\ you\ every\ success\ !}  \][/quote]

    [b] Thank you, Mr Nicula.I know those properties, but my teacher said the problem is quite easy whith complex numbers.I know something about  complex numbers, but I didn't know to solve it.Anywhay, it's an ingenious solution. :) I must be ready, because I'll go to the contest <[u][i]Ion Ciolac[/i][/u]> at Craiova and I've heard that it's a difficult contest.[/b]

From Darij I discovered that this problem belongs to [b]Toshio Seimiya, Japan[/b],
it appeared in [b]Crux Mathematicorum[/b] (#2427,1999) and it was solved
in the same journal by  the [b]Nikolaos Dergiades, Greece[/b].

I shall reproduce here [u]the nice and very short Dergiades' proof[/u]
which is (rather) similarly with the my above solution.

[quote][b]Denote:[/b] the intersection $M\in AG\cap BC\ ;$ $\alpha =m(\widehat {GAB})\ ,$ $\beta=m(\widehat {GBC})\ ,$ $\gamma =m(\widehat {GCA})\ ;$

$x=\cot A\ ,$ $y=\cot B\ ,$ $z=\cot C\ .$ But $A+B+C=\pi$ $\Longrightarrow$ ${\boxed {xy+yz+zx=1}\ \ (1)\ .}$ Thus,

$2=\frac{BC}{BM}=\frac{c}{m_a}\cdot \frac{\sin A}{\sin \widehat {GAB}}=\frac{\sin (B+\alpha)}{\sin C}\cdot \frac{\sin A}{\sin \alpha }$  $\Longrightarrow$ $\frac{\sin (B+\alpha)}{\sin \alpha}=\frac{2\sin C}{\sin A}$ $\Longleftrightarrow$

$\frac{\sin (B+\alpha)}{\sin \alpha\sin B}=\frac{2\sin (A+B)}{\sin A\sin B}$ $\Longrightarrow$ $\cot B+\cot \alpha =2(\cot A+\cot B)$ $\Longrightarrow$

$\boxed {\ \cot \alpha =2x+y\ }\ \ (2)\ \ \mathrm {a.s.o.\ }$ Therefore, $\alpha +\beta +\gamma =\frac{\pi}{2}$ $\Longleftrightarrow$

$\cot\alpha +\cot \beta +\cot \gamma =\cot \alpha\cdot\cot\beta\cdot\cot\gamma$ and the relations $(1)$ and $(2)$ $\Longleftrightarrow$

$3(x+y+z)=(2x+y)(2y+z)(2z+x)$ $\Longleftrightarrow$

$3(x+y+z)(xy+yz+zx)=(2x+y)(2y+z)(2z+x)\Longleftrightarrow$

$(x-y)(y-z)(z-x)=0\mathrm {\ q.e.d.}$[/quote]


In general position $n$ points are given in the space. Some pairs of these points are connected by line segments so that the number of segments equals $\dsp \left[
\frac{n^2}{4} \right],$ and a triangle exists. Prove that any point from which the maximal number of segments starts, is a vertex of a triangle.

A cute one :

Let's forget the geometrical meaning, and consider the problem in graph-theoretic approach.
Let $d$ be the maximal degree, and $A$ be one of the point with degree $d$.

Note that
$ {[} \frac {n^2} {4} {]}$ = number of edges = $\frac 1 2 ($ sum of the degrees $) \leq \frac {nd} 2$, so that $d \geq \frac 2 n [ \frac {n^2} 4 ]$. (1)

Now suppose, for a contradiction, that there is no triangle with vertex $A$.
Let $A_1, \cdots , A_d$ be the points joined by $A$ with an edge. The remaining points are $B_1, \cdots , B_{n-1-d}$.
Since there is no triangle with vertex $A$ there is no $i,j$ such that $A_i$ and $A_j$ are joined by an edge. Thus the degree of any of the $A_i$'s is at most $n-d$.
The degree of any of the $B_j$ is at most $d$ from the maximality of $d$.
Thus :
$ {[} \frac {n^2} {4} {]}= \frac 1 2 ($ sum of the degrees $) \leq \frac 1 2 (d + d(n-d) + d(n-1-d) ) = d(n-d)$. (2)

- If $n = 2k$ is even :
From (1) and (2), we have $d \geq k$ and $k^2 \leq d(2k-d) = k^2 -(d-k)^2 \leq k^2$.
It follows that $d=k$ and that in the above reasoning, all the majorations are equalities. Then, each of the $A_i$ has degree $n-d=k$, which forces each of the $A_i$ to be joined by an edge to any other point which is not an $A_j$. Thus, since each of the $B_i$ has degree $d$ no two of them are joined by an edge. Therefore, there is no triangle at all, contradicting the statement of the problem.

- If $n = 2k+1$ is odd :
From (1) and (2), we have $d \geq \frac {2k(k+1)} {2k+1} = k + \frac k {2k+1} > k$ thus $d \geq k+1$, and $k(k+1) \leq d(2k+1-d) = (k+ \frac 1 2)^2 -(d-\frac {2k+1} 2)^2$.
It follows that $d=k+1$ and that in the above reasoning, all the majorations are again equalities. Then, each of the $A_i$ has degree $n-d=k$, which forces each of the $A_i$ to be joined by an edge to any other point which is not an $A_j$. Thus, since each of the $B_i$ has degree $d$ no two of them are joined by an edge, leading to the same contradiction as above.

Thus, there is a triangle with vertex $A$, and we are done.

Pierre.


geometry
rectangle
trigonometry
AMC
AIME
trig identities
Law of Cosines

[hide]As it's "shown in the figure to the right" :D  :roll:  :D, all is happened in a triangle rectangle with the hypotenuse $1$, leg $r$ (where $r$ is the
radius of each disk) and opposite angle (to $r$) of  15°.By sinus we find \[ r=\sin{15}=\frac{\sqrt{6}-\sqrt{2}}{4} \]
BTW we have $12r^2=a-b\sqrt{c}$ and by little calculation we find \[ a-b\sqrt{c}=6-3\sqrt{3} \]
weach means \[ a+b+c=12 \]. :)
Sorry about  "°" wich miss to "15" ! [/hide]

not quite...in your solution, the disks will overlap

[quote="Farenhajt"]I don't think they will. Basically, the centers of the disks are the vertices of the inscribed regular dodecagon, and the radii are the half-side of the dodecagon, if I understand the setup correctly. That's what mstoenescu calculated.[/quote]
Thanks Farenhajt for the complement. I was thinking at the inscribed regular dodecagon but I haven't precised it !
I insist, [b]we must see the (future ?) figure at the right [/b] like chess64 told ... :oops:

Yes, I also agree with mstoenescu. The only thing that's annoying me right now is that my answer is different from Kalva's :?

Here's how I did it:

After drawing our figure, we can connect each successive center of each successive circle and create a regular dodecagon (each side of the dodecagon is composed of the two radii of successive circle, so the length is equal to diameter of each of the smaller circles). Call the side of the dodecagon $2r$. Create an iscoceles triangle with the radii of the bigger circle as the two equal sides and the base as one of the sides (so the triangle has sides $2r, 1$, and $1$). The opposite angle of $2r$ is $30^\circ$, so apply law of cosines:

$(2r)^2=1^2+1^2-2(1)(1)(\cos{30^\circ}) \Longrightarrow (2r)^2=2-\sqrt{3} \Longrightarrow r^2=\frac{2-\sqrt{3}}{4}$.

There are twelve circles, so the area of all twelve is $12(\frac{2-\sqrt{3}}{4})(\pi)=6-3\sqrt{3}\pi$.
The answer is thus $6+3+3=\boxed{12}$

EDIT: Hah, guess kalva was right. My picture was wrong (Kalva's pic says the triangle can't have sides 1,1, 2r since the dodecagon vertices are outside the big circle)

I'll post the right solution later

[hide="Solution"]
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=952[/img]

Consider the above diagram.  We know that

\begin{eqnarray*} r &=& \tan 15^\circ\\ &=& \tan(45-30)^\circ\\ &=& \frac{\tan 45^\circ -\tan 30^\circ}{1+\tan 45^\circ\tan 30^\circ}\\ &=& \frac{1-\frac{\sqrt{3}}3}{1+\frac{\sqrt{3}}3}\\ &=& 2-\sqrt{3} \end{eqnarray*}

Thus, the total area is \[ 12\pi(2-\sqrt{3})^2=\pi(84-48\sqrt{3})\Rightarrow a+b+c = \boxed{135.}  \][/hide]

yeah, it would be nice if someone with GS can make the picture for the problem itself :)

[quote="chess64"]Twelve congruent disks are placed on a circle $C$ of radius 1 in such a way that the twelve disks cover $C$, no two of the disks overlap, and so that each of the twelve disks is tangent to its two neighbors. The resulting arrangement of disks is shown in the figure below.  The sum of the areas of the twelve disks can be written in the from $\pi(a-b\sqrt{c})$, where $a,b,c$ are positive integers and $c$ is not divisible by the square of any prime. Find $a+b+c$.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=951[/img][/quote]

Wow, where did this picture come from  :?:   :D

Not so astute observation, shobhit :P

Could somebody please post the original diagram to this problem? I'm planning to compile all the problems from the AIME 1991, but am missing this diagram.

fyi, the image was lost when AoPS switched servers.

Thanks.  :)

I don't have THE original diagram, but it's no problem to draw twelve disks around the circle.

[quote="4everwise"]Hmm, would somebody be so kind as to post a diagram for this problem? Thanks.[/quote]
Hope this one is good enough.

Oh, sorry 4everwise. I forgot :(

But I made it now :lol:

[hide="metapost code"]
[code]
verbatimtex
\documentclass[12pt]{amsart}
\usepackage{mflogo}
\begin{document}
etex
\input boxes
beginfig(-1)

pair A,B,C,D,E,F,G,H,I,J,K,L;
A:=(1cm,0);
B:=(0.866025cm,0.5cm);
C:=(0.5cm,0.866025cm);
D:=(0,1cm);
E:=(-0.5cm,0.866025cm);
F:=(-0.866025cm,0.5cm);
G:=(-1cm,0);
H:=(-0.866025cm,-0.5cm);
I:=(-0.5cm,-0.866025cm);
J:=(0,-1cm);
K:=(0.5cm,-0.866025cm);
L:=(0.866025cm,-0.5cm);

k:=2; %scaling constant
A:=k*A;
B:=k*B;
C:=k*C;
D:=k*D;
E:=k*E;
F:=k*F;
G:=k*G;
H:=k*H;
I:=k*I;
J:=k*J;
K:=k*K;
L:=k*L;

draw A withpen pencircle scaled 2bp;
draw B withpen pencircle scaled 2bp;
draw C withpen pencircle scaled 2bp;
draw D withpen pencircle scaled 2bp;
draw E withpen pencircle scaled 2bp;
draw F withpen pencircle scaled 2bp;
draw G withpen pencircle scaled 2bp;
draw H withpen pencircle scaled 2bp;
draw I withpen pencircle scaled 2bp;
draw J withpen pencircle scaled 2bp;
draw K withpen pencircle scaled 2bp;
draw L withpen pencircle scaled 2bp;

%% the diameter should be 2*k

draw fullcircle scaled 4cm;

%% the diameter should be 0.267949192*k

draw fullcircle scaled 1.07179677cm shifted A;
draw fullcircle scaled 1.07179677cm shifted B;
draw fullcircle scaled 1.07179677cm shifted C;
draw fullcircle scaled 1.07179677cm shifted D;
draw fullcircle scaled 1.07179677cm shifted E;
draw fullcircle scaled 1.07179677cm shifted F;
draw fullcircle scaled 1.07179677cm shifted G;
draw fullcircle scaled 1.07179677cm shifted H;
draw fullcircle scaled 1.07179677cm shifted I;
draw fullcircle scaled 1.07179677cm shifted J;
draw fullcircle scaled 1.07179677cm shifted K;
draw fullcircle scaled 1.07179677cm shifted L;

pair O; O:=(0,0);
draw O--A--B--O;

endfig
verbatimtex
\end{document}
etex
end
[/code]
[/hide]

[url=http://mathideas.org/public/aime_1998_11.pdf]pdf[/url]

how does this get to be a problem #11???

the only observation to notice is the tangent points of the smaller circles lie on the larger circles, using good ol' trig, $r=\tan 15$, then the area is $12\pi\tan^{2}15$... $\tan15=2-\sqrt{3}$...$(84-48\sqrt{3})\pi$, so the answer is $84+48+3=135$

Thanks guys.  :)

Chess, I see you're pretty good with MetaPost.  :D


[quote="Arvind_sn"]How do you do those problems where it says something like

Jim and Bob are running in different directions around a circular track that is (insert number here) feet long. They meet after (insert number here) and (insert number here) seconds. What are their speeds?

I have heard that you can only find the difference in speeds, but plz tell me anything you know, especially specific problems of this sort and their solutions.[/u][/quote]
[hide]The key to solving such problems is to use relative speeds. For example, if Jim is running at $ x$ mph and Bob is running at $ y$ mph along the same direction, where $ x > y$, Jim's relative speed with respect to Bob is $ x\minus{}y$ mph. In other words, you can consider Jim running at $ x\minus{}y$ mph while Bob is standing still for all your computations. For example, if Bob is 5 m ahead of Jim, the time it takes for Jim to catch up with Bob is $ \frac{5}{x\minus{}y}$ hrs.[/hide]

But is that still for different directions?


vector
analytic geometry
graphing lines
slope

The point on the circle $ {(x\minus{}5)}^2 \plus{} {(y\minus{}4)}^2 \equal{}4$, which is closest to the circle $ {(x\minus{}1)}^2 \plus{}{(y\minus{}1)}^2 \equal{}1$
is?

[hide="hint"]

If we draw a line between the centers of the circles, is the intersection of that line with the second circle the closest point to the first circle?
[/hide]

Yes it is  :o  I am getting better at seeing these things..slowly.

Yeah, I see that too, but is it possible to prove it, though?

(Using the given idea, just construct a vector starts at (5,4) with magnitude 2 and slope 3/4; the end point is the answer.)

Let A be the center of a circle, B be a point on the circle, C be a point on another circle that does not overlap with the first, and D be the center of the circle that C is on.

We know that AB=R and CD=r, the fixed radii of the circles. We wish to minimize BC.

This is the same as minimizing BC+R+r = AB+BC+CD. But obviously AB+BC+CD is minimized when A, B, C, and D are collinear, at which point AB+BC+CD = AD.

Thus, the minimum value of BC occurs when B and C are on the line AD, in which instance BC = AD-R-r.


Ahh! Munch's "The Scream" has been stolen! This is my avatar!

http://www.foxnews.com/story/0,2933,129678,00.html

Yeah I heard it too, it's incredible! :o

What are they up to doing with it? You needn't be an expert to recognize that one, so who the world would want to buy that? :?

Perhaps they're gonna blackmail the museum or anything, but that painting has no value for them :? (unless they found a rich idiot that wants to hang it in his collection :D)

it was stolen before... maybe the theives will be caught.

the security in that museum isn't so good. i don't think there were even alarms.

cold entre, do you know what you're talking about? They came in with automated guns... I don't think an alarm will stop those :D

you should really read a little bit before you post so suddenly.
The museum in question is a bit notorious now for having shoddy security. There were no alarms attached to this priceless piece of art, just some wire that held it in place. ANd to think its the 2nd! time that the Scream has been stolen.

MagnusRex, I think you should delete your current avatar and replace it with an empty square until the painting is recovered  ;)


geometry
geometric transformation
reflection
inequalities proposed
inequalities

For $ a,b,c$positive reals

$ \frac{8abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\plus{}\frac{(a\plus{}b\plus{}c)}{\sqrt[3]{abc}} \geq 4$

Hi,

Because of the fact that

$ \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \frac {(a \plus{} b \plus{} c)}{\sqrt [3]{abc}} \equal{} \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \sum \frac {a \plus{} b}{2\sqrt [3]{abc}}$

the result follows from AM-GM.

[quote="manlio"]For $ a,b,c$positive reals

$ \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \frac {(a \plus{} b \plus{} c)}{\sqrt [3]{abc}} \geq 4$[/quote]

By AM-GM we have :

$ LHS\equal{}\frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \frac {(a \plus{} b \plus{} c)}{3\sqrt [3]{abc}} \plus{}\frac {(a \plus{} b \plus{} c)}{3\sqrt [3]{abc}}\plus{}\frac {(a \plus{} b \plus{} c)}{3\sqrt [3]{abc}}$

$ \geq 4\sqrt[4]{\frac{8(a\plus{}b\plus{}c)^3}{27(a\plus{}b)(b\plus{}c)(c\plus{}a)}} \geq 4$

Done !

:)


let $a_1<a_2<a_3....$ be a sequence of natural numbers. and define $s_n=a_1+a_2...+a_n$. It is known that for $n>2005$, $a_n|s_{n-1}$. prove $a_n=s_{n-1}$ for larger enough $n$.

Please always add the year and if possible the round.  :)

If we put $k_n=\frac{s_{n-1}}{a_n}$, from the fact that $a_n$ is strictly increasing, we get $\frac{S_n}{S_{n-1}}>\frac{k_{n+1}}{k_n}$. We have $S_n=a_n+S_{n-1}=\frac{S_{n-1}}{k_n}+S_{n-1}$, so the LHS is $\frac 1{k_n}+1$, which must then be $>\frac{k_{n+1}}{k_n}$, so $k_n+1>k_{n+1}$, meaning that the sequence $k_n$ is decreasing. Assuming it's not eventually $1$, it must be stationary and assume a constant value $k>1$ for $n$ large enough. This means $k|S_{n-1},a_n$ (because $k|S_{n-1},S_n=S_{n-1}+a_n$), and $k=\frac{S_{n-1}}{a_n}$ for large enough $n$'s. We quickly reach a contradiction (we get something like $k^t|S_{n_0},\ \forall t$ and for some constant $n_0$).

It seems a but murky :). Is it Ok?


trigonometry
calculus
calculus computations

Find $\inf_{0<\alpha<\pi}\{\sup_{n\in\mathbb Z}|\sin n\alpha |\}$

Which one? Well, $\inf_{n\in\mathbb Z}\{\sup_{0<\alpha<\pi}|\sin n\alpha |\}=0$ is kind of silly. I'll make a wild guess: $\inf_{0<\alpha<\pi}\{\sup_{n\in\mathbb Z}|\sin n\alpha |\}=\frac{\sqrt{3}}{2}$.

Sorry that, the problem is the second one.

Well if $\alpha$ is an irrational multiple of $\pi$, $n\alpha$ will get arbitrarily close to $\frac{\pi}{2}$ so $\sin{n\alpha}$ will get arbitrarily close to $1$.

If $\alpha$ is a rational multiple of $\pi$, $\alpha = \frac{p}{q}\pi$ so $n\alpha$ will cycle through $0, \frac{1}{q}\pi, \frac{2}{q}\pi, \ldots, \frac{2q-1}{q}\pi$. Clearly $q$ is odd or $\frac{\pi}{2}$ will appear in the sequence. Also $q > 1$ because $q = 1$ gives $\alpha = 0, \pi$ which are not allowed. Lastly we claim $q = 3$ because for $q = 2k+1 > 3$ we take $\frac{k}{2k+1}\pi$, which is closer to $\frac{\pi}{2}$ than $\frac{\pi}{3}$ is. Hence $q = 3 \Rightarrow \inf_{0 < \alpha < \pi}\{\sup_{n \in \mathbb{Z}}|\sin{n\alpha}|\}= \frac{\sqrt{3}}{2}$.

Right. The solution I had in mind: suppose $\alpha, \beta \in (0,\pi/2]$ are such that $\sin n\alpha\le \sin\beta$ for all $n\in\mathbb Z$. Letting $n=1$, find $\alpha\le\beta$. Since the sequence $\{n\alpha\}$ jumps over the interval $(\beta,\pi-\beta)$, it follows that $\alpha\ge\pi-2\beta$. Hence $\pi-2\beta\le\beta$, which implies $\sin\beta\ge \frac{\sqrt{3}}{2}$.


x+1/x= 1+sqrt5/2
So what is x^2000 + x^-2000?
I think the answer is two but i am not that sure why it is.

I would like to clarify something.

Does 1+sqrt5/2 mean $ 1\plus{}\frac{\sqrt{5}}{2}$ or $ \frac{1\plus{}\sqrt{5}}{2}?$

\[ x \plus{} \frac{1}{x} \equal{} \frac{1\plus{}\sqrt{5}}{2} \approx 1.something\]

if [b]x[/b] was real, then \[ x \plus{} \frac{1}{x} \geq 2\] .... {follows from A.M >= G.M}

so, here x is a complex no.

i figured this ... if  |x| = 1, then x + 1/x = 2 * Real(x)

or

[img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/0/7/a/07a3d048099f12b0549477145446edf0fbe9c7c7.gif[/img]

we know \[ cos 36^o \equal{} \frac{1 \plus{} \sqrt{5}}{4}\]

this gives \[ \theta \equal{} 36^o\]

De-Moivre's Theorem finally gives u

[img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/9/d/a/9da06424f318c7c0071c588462a0ae005f968dc4.gif[/img]

I wanna know the simpler method  :ninja:

I wonder if there is a simpler way. But sorry to ask. where did u get the cos 36 from

oh well, i just remember trigonometric ratios for some common angles

( 18, 36, 75, 22.5, 15, 72 degrees )

apart from 30, 45, 60, 90 degrees ...

they are useful at times  :trampoline:

I don't think there's a simpler way. Next time, varun, use $ \text{\LaTeX}$. It's not that hard to incorporate $ \text{\LaTeX}$, simply surround your text with dollar signs and know the basic commands in $ \text{\LaTeX}$, like \frac{A}{B} for A/B and \cdot for the multiplication bullet.

[hide="Extension"]
The sequence $ x^n + \frac 1{x^n}$ when $ x + \frac 1x = \phi$ is interesting. Consider the table below.
\begin{align*} x + \frac 1x & = \frac {1 + \sqrt 5}2 = \phi \\
x^2 + \frac 1{x^2} & = \frac { - 1 + \sqrt 5}2 = \phi^{ - 1} \\
x^3 + \frac 1{x^3} & = \frac {1 - \sqrt 5}2 = - \phi^{ - 1} \\
x^4 + \frac 1{x^4} & = \frac { - 1 - \sqrt 5}2 = - \phi \\
x^5 + \frac 1{x^5} & = - 2 \\
x^6 + \frac 1{x^6} & = \frac { - 1 - \sqrt 5}2 = - \phi \\
x^7 + \frac 1{x^7} & = \frac {1 - \sqrt 5}2 = - \phi^{ - 1} \\
x^8 + \frac 1{x^8} & = \frac { - 1 + \sqrt 5}2 = \phi^{ - 1} \\
x^9 + \frac 1{x^9} & = \frac {1 + \sqrt 5}2 = \phi \\
x^{10} + \frac 1{x^{10}} & = 2 \end{align*}
You can see for yourself that $ x^{11} + \frac 1{x^{11}} = x + \frac 1x$, and in general $ x^n + \frac 1{x^n} = x^{n + 10} + \frac 1{x^{n + 10}}$.

This result is a direct consequence of the values of $ 2\cos 36n^\circ$ for $ n\geq 1$.
[/hide]


geometry
geometric inequality
circumcircle
perimeter
Triangle
IMO Shortlist

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]  :)

This is one problem I had in one TST in 2000. Solution:

Consider the following lemma: let $ABC$ be a triangle and let $M$ be a point on the side $AC$ such that $\angle BMC \geq 90^\circ$. Then the following inequality takes place:

\[ BM+MC < BA + BC \]

[hide="The proof is obvious"]
If $\angle BAC \leq 90^\circ$ let us consider the point $A'$ on the segment $AM$ such that $BA' \perp AC$.
We have $BM \leq A'B+ A'M $ and thus $BM + MC \leq A'B+A'C < BA' +BC $, because $ BC$ is the largest side the triangle $A'BC$.
But $BA' \leq BA$, so we are done.

If $\angle BAC > 90^\circ$ we have $BM \leq AB+ AM $ and thus $BM + MC \leq AB+AC < BA +BC $, because $ BC$ is the largest side the triangle $ABC$.
[/hide]

Back to the problem, we notice that at least two of the angles  $\angle AMB$, $\angle BMC$, $\angle CMA$ are larger than $90^\circ$. Indeed, if  two of them are less than $90^\circ$, since their sum is $360^\circ$, one of them would be larger than $180^\circ$ which is an obvious contradiction.

So, let us suppose WLOG that $\angle AMB \geq 90^\circ$ and $\angle AMC \geq 90^\circ$. Also consider $M' = AM \cap BC$. We apply the lemma for the triangles $ABM'$ and $ACM'$ and we obtain
\[ AM + MB < AB + BM' \eqno (1) \]
\[ AM + MC < AC + CM' \eqno (2) \]
From (1) and (2), by addition, we obtain
\[ AM+BM+ CM + AM  < AB + BC + CA \] which solves our problem.

Also there is the official solution, much complicated which I will post later.

That's a beautiful solution! Just a little remark about your proof of the lemma:

[quote="Valentin Vornicu"]Consider the following lemma: let $ABC$ be a triangle and let $M$ be a point on the side $AC$ such that $\angle BMC \geq 90^\circ$. Then the following inequality takes place:

\[ BM+MC < BA + BC \]

[hide="The proof is obvious"]
If $\angle BAC \leq 90^\circ$ let us consider the point $A'$ on the segment $AM$ such that $BA' \perp AC$.
We have $BM \leq A'B+ A'M $ and thus $BM + MC \leq A'B+A'C < BA' +BC $, because $ BC$ is the largest side the triangle $A'BC$.
But $BA' \leq BA$, so we are done.

If $\angle BAC > 90^\circ$ we have $BM \leq AB+ AM $ and thus $BM + MC \leq AB+AC < BA +BC $, because $ BC$ is the largest side the triangle $ABC$.
[/hide][/quote]

In this proof, you needn't distinguish between the cases $\angle BAC \leq 90^\circ$ and $\angle BAC > 90^\circ$. In both cases, your argument works:

[hide="The proof is obvious"]Since $\angle BMC \geq 90^\circ$, let us consider the point $A'$ on the ray $MA$ such that $BA' \perp AC$.
We have $BM \leq A'B+ A'M $ and thus $BM + MC \leq A'B+A'C < BA' +BC $, because $ BC$ is the largest side the triangle $A'BC$.
But $BA' \leq BA$, so we are done.[/hide]

  Darij

[quote="Valentin Vornicu"]Also there is the official solution, much complicated which I will post later.[/quote]

I don't know what you had mind but these solutions are from the IMO1999 booklet I typed once. I always pasted them on ML as solutions, usually by grobber, were given.  :)

[b]Lemma[/b]
If $M$ is an interior point of the convex quadrilateral $ABCD$ then $MA+MB<AC+CD+DB$.

[b]Proof of the Lemma[/b]
The ray $AM$ intersects the quadrilateral in $N$; suppose, for
instance, that $N \in [CD]$. Then
\[MA + MB < MA + MN + NB \leq AN + NC + CB \leq AD + DN + NC + CB = AD + DC + CB.\]

[b]Offical Solution:[/b]
The median triangle $ D E F $ divides triangle $ ABC $ into four regions. Each region is covered by at least two of the convex quadrilaterals $ABDE, BCEF, CAFD$. If, for instance, $M$ belongs to  $[ABDE]$ and $[BCEF]$ then $MA + MB < BD + DE + EA$ and $MB + MC < CE + EF + FB$. By adding these two inequalities we get $MB+(MA+MB+MC) \leq AB + BC + CA$, which implies the required conclusion.

[b]Another Offical Solution:[/b]

Let $O$ be the circumcenter of $\triangle ABC$.

- [i]Case 1:[/i]

$O \in [ABC]$ . Suppose $M \in [ADOE]$. Then $MA = min\{MA,MB,MC\}$ and $MA +MB \leq NA + NB, MA+MC \leq NA + NC$. Since $NA + NC \leq AD + DC$. (for $O([ANC]$; otherwise $NA + NC < BD + DC$) it is enough to prove that (with the usual notations). Since $m_{c} < (a+b)/2$, it is enough $m_{c} + (c/2) + 2R < a+b+c$ to prove that $4R < a+b+c$ (in an nonobtuse-angled triangle). This reduces to $\sin A+\sin B+\sin C >2$, or

\[ \sin \frac{A}{2}\cos \frac{A}{2}+ \cos \frac{A}{2} \cos
\frac{B-C}{2} \frac{}{}>1.
\]

For fixed A the minimum value of the left member is obtained when $B-C$ has maximum value, that is $B= \pi/2, C+A =\pi/2$ and the minimum is $(1/2)(\sin A + 1 + \cos A)$ therefore it is greater than $(1/2)(1+1)=1$

- [i]Case 2:[/i]

$ABC$ is obtuse-angled. Suppose for instance that $m(\widehat{B})>90^{\circ}$. Let $P$ and $Q$ be the intersections of the perpendicular bisectors of the sides $[AB]$ and $[BC]$ with $AC$.

[i]Case 2.1:[/i]

For $M \in [ADP], min\{MA,MB,MC\}=MA$ and, as above, $MA + MB \leq PA + PB = 2 \cdot AP, MA + MC \leq m_{c} + (c/2) < (a + b + c)/2$. It is enough to prove that $4 \cdot AP < a+b+c$, which is implied by $4AP < 2b < a+b+c$. A similar argument works in the case $M \in [CEQ]$.

[i]Case 2.2:[/i]

For $M \in [BEQPD]$, let $BM  \cap AC = N$. Then $min\{MA,MB,MC\}= MB$ and $MB \cdot MC \leq NB + NC, MB+MA \leq NB + NA, NB \leq max\{PB,QB\}$, therefore it is enough to prove that $2 \cdot PB < AB + BC$. Since $PB < BD + DP < BD + PP'$, the needed relation is obvious.

We can also prove that
\[ 2(\min \{MA,MB,MC\} \plus{} \max \{MA,MB,MC\})\le AB \plus{} BC \plus{} CA
\]
Equality holds only if the triangle is degenerate (but not iff).

Proof:

[b]Lemma:[/b]

Given two lengths $ a\ge b$, and a point $ M$, let $ XYZ$ be the triangle (possibly degenerate) with smallest perimeter such that $ MX \equal{} a$ and $ MY \equal{} MZ \equal{} b$ and $ M$ lies in the interior of $ XYZ$ or on it's sides
(If there are more than one then let one of them $ XYZ$)

We shall prove that the perimeter of $ XYZ\ge 2(a \plus{} b)$

Denote the circle with radius $ b$ and center $ M$ to be $ (C)$
Denote the ellipse with foci $ X,Y$ passing through $ Z$ to be $ (E)$

(If the ellipse is degenerate then $ X,Y,Z$ collinear and easy to see that $ 2(a \plus{} b)\le XY \plus{} YZ \plus{} ZX$, so we assume the contrary)
$ (C)$ and $ (E)$ has (at least) one common point $ B$. If they ($ (C)$ and $ (E)$) are not tangent to each other then there also exists a point $ Z'$ which lies on the circumference of $ (C)$ and in the interior of $ (E)$
Hence $ YZ \plus{} XZ > YZ' \plus{} XZ'$

Then The triangle $ XYZ'$ satisfies $ MX \equal{} a$ and $ MY \equal{} MZ' \equal{} b$ but the perimeter of $ XYZ'$ is less than that of $ XYZ$. Contradiction.

Hence $ (C)$ and $ (E)$ are tangent to each other. By the well known
reflective property of ellipse $ \angle XZM \equal{} \angle YZM$. By symmetry $ \angle XYM \equal{} \angle ZYM$

Hence $ M$ is the incenter of $ XYZ$. Also $ MY \equal{} MZ$ implies $ XYZ$ is isosceles.

Let $ XM$ extended meet $ YZ$ at $ T$. $ XT$ is also median, altitude of triangle $ XYZ$.

Let $ YT \equal{} ZT \equal{} x$ and $ \angle MYT \equal{} \angle MZT \equal{} \cdots \equal{} \theta$
We can obtain that
\[ YZ \equal{} 2x, MT \equal{} x\tan \theta , XT \equal{} x\tan 2\theta , a \equal{} XM \equal{} x\tan 2\theta \minus{} x\tan \theta
\]

\[ b \equal{} MY \equal{} MZ \equal{} x\sec \theta , XY \equal{} XZ \equal{} x\sec 2\theta
\]
So:
\[ 2(a \plus{} b)\le XY \plus{} XZ \plus{} YZ
\]

\[ \iff 2[(x\tan 2\theta \minus{} x\tan \theta ) \plus{} (x\sec \theta )]\le (x\tan \theta ) \plus{} (x\tan \theta ) \plus{} 2x
\]

\[ \iff 1 \plus{} \tan \theta \plus{} \sec 2\theta \ge \tan 2\theta \plus{} \sec \theta
\]

\[ \iff 1 \plus{} \frac {1}{\tan 2\theta \plus{} \sec 2\theta }\ge \frac {1}{\tan \theta \plus{} \sec \theta }
\]
which is true since $ \theta \in (0,\frac {\pi}{4})$.

[b]End Lemma[/b]

Now let $ ABC$ any arbitrary triangle. (unlike $ XYZ$).
WLOG $ MA \equal{} \max \{MA,MB,MC\}$ and $ MB \equal{} \min \{MA,MB,MC\}$

Cut $ MC'$ equal to $ MB$ from $ MC$. Perimeter of $ ABC\ge$ that of $ ABC'$
From lemma with $ a \equal{} MA$ and $ b \equal{} MB \equal{} MC'$, we have perimeter of $ ABC'\ge 2(MA \plus{} MB)$
Hence perimeter of $ ABC\ge 2(MA \plus{} MB)$

[b]QED[/b]

We will show that for all $M$ not in the exterior of $\triangle ABC$, $\min(2MA + MB + MC, MA + 2MB + MC, MA + MB + 2MC\) < AB + BC + CA$.For any $X$ (not necessarily in the interior of $\triangle ABC$), let $f_1(X) = 2AX + BX + CX$, $f_2(X) = AX + 2BX + CX$, $f_3(X) = AX + BX + 2CX$, and $f(X) = \min(f_1(X), f_2(X), f_3(X))$. We wish to show that $f(X) < AB + BC + CA$. We will first prove three lemmas:

[b]Lemma 1[/b]: If $O$ is the circumcenter of acute $\triangle ABC$, $f(O) < AB + BC + CA$.
[i]Proof:[/i] Let $R$ be the circumradius of $\triangle ABC$. $f(O) = 4R$, so we simply wish to show that $\frac{AB}{2R} + \frac{BC}{2R} + \frac{CA}{2R} > 2$. Dividing by $2R$ and using the law of sines, we find that the inequality is equivalent to $\sin A + \sin B + \sin C > 2$.  Because $\frac{d^2 \sin x}{dx^2} = - \sin x \leq 0$ for $x \in [0, \frac{\pi}{2}]$, $\sin$ is concave over $[0, \frac{\pi}{2}]$.

Without loss of generality, let $A \geq B \geq C$. It is clear that $(\frac{A+B+C}{2}, \frac{A+B+C}{2}, 0)$ majorizes $(A, B, C)$, so by Karamata's inequality $\sin A + \sin B + \sin C > \sin \frac{\pi}{2} + \sin \frac{\pi}{2} + \sin 0 = 2$.

[b]Lemma 2[/b]: Let $S$ be an ellipse with foci $C_1$ and $C_2$ and major axis with endpoints $A_1$, $A_2$, and let $P$ be any point on the major axis of $S$. Then for any point $X$ on the ellipse, $PX \leq \max(PA_1, PA_2)$.
[i]Proof:[/i] Without loss of generality, let $C_1 = (-c, 0)$, $C_2 = (c, 0)$, $A_1 = (a, 0)$,  $A_2 = (-a, 0)$, with $a,c > 0$, $P = (p, 0)$, and let $b = \sqrt{a^2 - c^2}$. It is sufficient to show that the distance from $P$ to $X = (x,  b \sqrt{1 - \frac{x^2}{a^2}})$, where $-1 \leq x \leq 1$, is maximized when $x = 1$ or when $x = -1$. For $x \in [-1, 1]$, let $f(x) = PX^2 = (x - p)^2 + b^2 \left(1 - \frac{x^2}{a^2}\right)$. $f(x)$ is a quadratic function with leading coefficient $1 - \frac{b^2}{a^2} > 0$, so $f$ is maximized when $x$ is at an endpoint of $f$'s domain, i.e., when $x = 1$ or $x = -1$.

[b]Lemma 3[/b]: If $P$ is on the perimeter of $\triangle ABC$, then $f(P) < AB + BC + CA$.
[i]Proof:[/i] Without loss of generality, let $P$ be on side $BC$. Let $E$ be the ellipse with foci $BC$ passing through $A$. Let $E$ intersect line $BC$ at $A_1$ and $A_2$, with $B$ between $A_1$ and $C$ and $C$ between $B$ and $A_2$. By lemma 2, $PA < \max(PA_1, PA_2)$.  We may suppose without loss of generality that $PA < PA_2$. Suppose for the sake of contradiction that $f(P) \geq AB + BC + CA$. Then $f_2(P) = 2PB + PC + AP = BC + PB + AP \geq AB + BC + CA$, whence $PB + AP \geq AB + BC$. Because $AP < AP' = PC + CA_2 = PC + \frac{AB + AC - BC}{2}$, we find that $PB + PC + \frac{AB + AC - BC}{2} > AB + BC$, so $AB + AC + BC > 2AB + 2BC$, so $BC > AB + AC$, which contradicts the triangle inequality.

---------------------------------------------------------------------------------------------------------------------

Let $O$ be the circumcenter of $\triangle ABC$. If $M = O$, then $O$ must be in the interior of $\triangle ABC$, whence $\triangle ABC$ must be acute. By lemma 1, $f(O) < AB + BC + CA$, so we may suppose that $M \neq O$. Without loss of generality, let $O$ be the origin. Let $u$ be the vector from $O$ to $M$, and for any nonnegative real $t$ let $\alpha(t) = tu$. (The image of $\alpha$ would be the ray $\overrightarrow{OM}$.) Let  $g_i(t) = f_i(\alpha(t))$ for $i = 1,2,3$ and $g(t) = \min(g_1(t), g_2(t), g_3(t))$. We may suppose without loss of generality that $g_1(1) = g(1)$. We claim that $g_1(t) = g(t)$ for all nonnegative $t$. If $g_1(t) \neq g(t)$ for some $t$, then $g_j(t) < g_1(t)$ for some $j$, which we may suppose without loss of generality equals 2. Because $g_2(1) \geq g_1(1)$, the intermediate value theorem implies the existence of some $c \in [\min(t, 1), \max(t, 1)]$ such that $g_1(c) = g_2(c)$. Noting that $t \neq 0$ (since $g_1(0) = g_2(0)$ while $g_1(t) \neq g_2(t)$), we find that $c \geq \min(t, 1) > 0$.

Letting $X = \alpha(c)$, we find that $2AX + BX + CX = f_1(X) = f_2(X) = AX + 2BX + CX$, so $AX = BX$. It follows that $X$ must be equidistant from $A$ and $B$, so $X$ must lie on the perpendicular bisector of $AB$. Because $X$ also lies on the line through $\alpha(t)$ and $O$, we find that either $X = O$ or $\alpha(t)$ lies on the perpendicular bisector of $AB$. The former case is impossible since $c \neq 0$. In the latter case, we find that every point on the image of $\alpha$ is equidistant from $A$ and $B$, so $g_1(t) = g_2(t) < g_1(t)$, which is impossible.

If $O$ lies inside or on the perimeter of $\triangle ABC$, let ray $\overrightarrow{OM}$ intersect a side of the triangle at $P$, and let $P = \alpha(p)$. Because $g = g_1$ is a convex function over $[0,p]$, the maximum value of $g$ is attained at either $0$ or $p$. Since $M$ is between $O$ and $P$, $0 \leq 1 \leq p$, so $f(M) = g(1) \leq \max(g(0), g(p)) = \max(f(O), f(P))$. By lemmas 1 and 3, $f(O), f(P) < AB + BC + CA$, so our proof in this case is complete.

If $O$ lies outside of $\triangle ABC$, let $\overrightarrow{OM}$ intersect two sides of the triangle at $Q$ and $R$. Let $Q = \alpha(q)$ and $R = \alpha(r)$, and suppose without loss of generality that $q < r$. Because $g = g_1$ is a convex function over $[q,r]$, the maximum value of $g$ is attained at either $q$ or $r$. Since $M$ is between $Q$ and $R$, $q \leq 1 \leq r$, so $f(M) = g(1) \leq \max(g(q), g(r)) = \max(f(Q), f(R))$. By lemma 3, $f(Q), f(R) < AB + BC + CA$, so our proof is complete.

Let $D$ be the midpoint of $AB$.  Then $M$ lies within either $\triangle CDA$ or $\triangle BDA$.  If $M$ lies inside $\triangle BDA$, then $AM +BM < AD+BD < (AB+BC)/2 + BD = s$, where $s$ is the semi-perimeter of $\triangle ABC$.  Similarly $\min{(AM+MC,BM+MC)}<s$.  The desired result follows.

I finally found a solution after several hours; this problem was much harder than it looked at first :o

WLOG let $AX=\text{min}\{AX, BX, CX\}$ and WLOG $AB\le AC$. Let $\Gamma$ be the circle with center $A$ passing through $X$, and let $\mathcal{E}$ be the ellipse with foci $B,C$ passing through $X$. Let $D=\Gamma \cap AB$ and $E=\Gamma \cap AC$, and finally let $B'\in AC$ s.t $AB=AB'$. We see that $$BD+CD = B'E+CD < B'C+B'D+B'E = B'D+CE=BE+CE$$ so if $E$ is in the interior of $\mathcal{E}$, then $D$ is also. However, since $X\in \mathcal{E}$, $D, E$ cannot both be in the interior of $\mathcal{E}$, so $E$ must be outside of $\mathcal{E}$. Thus, $$2AX+BX+CX < 2AX+BE+CE=2AE+BE+CE=AE+AC+BE=AD+AC+B'D < AD+AC+CD < AB+AC+BC$$ done. (All inequalities in this solution are applications of triangle inequality, or properties of ellipses.)

@Below you are right. The minimality of AX should prevent this though, let me try to fix my solution.

[quote=bobthesmartypants]$B'C+B'D+B'E = B'D+CE$[/quote]

Wrong here: You have to assume that $B'$ is between $C$ and $E$, which may not true.

[quote=orl]Let ABC be a triangle and $M$ be an interior point. Prove that

\[ \min\{MA,MB,MC\}+MA+MB+MC<AB+AC+BC.\][/quote]
Let $M$ be a point inside triangle $ABC$. [url=https://artofproblemsolving.com/community/c6h1663379p10564930]Prove that[/url]
$$\frac{a+b+c}{2} < MA + MB + MC < a+b+c.$$
Let $ABC$ be a triangle and $M$ be a point in $\triangle ABC.$ Denote $a=BC,b=CA,c=AB,$ [url=https://artofproblemsolving.com/community/c6h406364p2269163]show that[/url]
$$a\cdot MA+b\cdot MB+c\cdot MC\le 2\max\{bc,ca,ab\}.$$
[url=https://artofproblemsolving.com/community/c6h296677p1607293]here[/url]

Let's use barycentric coordinates with $A = (1:0:0), B = (0:1:0), C = (0:0:1)$ and $P = (x:y:z).$ Notice that $x, y, z > 0$, and WLOG let $x$ be the largest of $x, y, z.$ Also assume WLOG that $AB \le AC.$ We claim that $2 \cdot PA + PB + PC < AB + BC + CA$, which would suffice.

Let $P_1, P_2$ lie on $AB, AC$ respectively so that $P_1P_2 || BC.$ Let $Q_1, Q_2$ be points so that $BP_1PQ_1, CP_2PQ_2$ are parallelograms.

Observe that $BP_1 + P_1P > BP, CP_2 + P_2P > CP$ by the Triangle Inequality. These give that $BP_1 + CP_2 + BQ_1 + CQ_2 > BP + CP$, and so we just have to show that $AP_1 + AP_2 + x \cdot BC \ge 2 \cdot PA.$

[b] Claim. [/b] $P$ is closer to $P_2$ than to $P_1$.

[b] Proof. [/b] If not, then we have that $AP \le \max (AP_1, AM)$, where $M$ is the midpoint of $P_1P_2.$ It's easy to see that $AP_2 + AP_1 \ge 2 AP_1$ and we get from Triangle Inequality that $AP_1 + AP_2 \ge 2 AM.$

$\blacksquare$

If $x \ge \frac12$, then $AP_1 + AP_2 + Q_1Q_2 \ge AP_1 + AP_2 + P_1P_2 = (AP_1+P_1P) + (AP_2+P_2P) > 2 \cdot AP.$ Otherwise, suppose that $x < \frac12$, and let $Q$ be the point on segment $P_1P_2$ so that $P_1QQ_2Q_1.$

[b] Lemma. [/b] $P$ is on segment $QP_2$, and is closer to $P_2$ than $Q$.

[b] Proof. [/b] If $P$ is not on segment $QP_2$, then we have from the claim that $AM < AP < AQ < AP_2$, and so $Q_1Q_2 + AP_1 + AP_2 = (AP_1+P_1Q) + AP_2 > AQ + AP_2 > 2 \cdot AP.$ Now, if $P$ is on segment $QP_2$ but is closer to $Q$, letting $N$ be the midpoint of $QP_2$, we have that $Q_1Q_2 + AP_1 + AP_2 = (AP_1+P_1Q) + AP_2 > AQ + AP_2 > 2 \cdot AN \ge 2 \cdot AP$. Hence we must have that the lemma is true.

$\blacksquare$

As a result of the lemma, we know that $\frac{P_2Q}{P_1P_2} \le \frac{\frac12 - x}{1-x}$. Since $x < \frac12$, we have that $\frac{\frac12 - x}{1-x} < 1- 2x$, which implies that $\frac{[\triangle AQP_2]}{[\triangle AP_1P_2]} < 1-2x$. However, this means that $y = \frac{[\triangle AQC]}{[\triangle ABC]} < 1 - 2x$. This is absurd, because $z = 1 - x - y \ge 1 - x - x = 1-2x$.

$\square$

Let $AM$ intersect $BC$ at $N.$ $AM+BM<AN+BN.$ Suppose WLOG that $N$ is closer to $B$ than $C$ then if $D$ is midpoint of $BC$ then $AN+BN<AD+BD.$ By doubling the median $AD$ and applying triangle inequality we get $2AD<AB+AC$ so we get $AM+BM<s.$ Now, extend $CM$ to intersect $AB$ and apply the same lemma. We get $AM+CM$ or $BM+CM<s.$ This implies result.

Without loss of generality, let $AM \leq BM \leq CM$.
Construct $\triangle DEF$, the medial triangle of $\triangle ABC$ as shown below:

[asy]
size(7cm);
pair A = dir(110), B = dir(210), C = dir(330);
draw(A--B--C--cycle);
pair D = (A+B)/2, E = (B+C)/2, F = (A+C)/2;
draw(D--E--F--cycle);

label("$A$", A, A);
label("$B$", B, B);
label("$C$", C, C);
label("$D$", D, dir(135));
label("$E$", E, dir(-90));
label("$F$", F, dir(45));
pair M = (0,0);
dot(M);
draw(A--M--B);
draw(C--M);
label("$M$", M, dir(30)*1.5);
[/asy]

If $M$ lies in trapezoid $AFEB$, we have

$$AM + MB < AF + FE + EB = \frac{AB + BC + CA}{2}. (1) $$

If $M$ lies in trapezoid $BDFC$, we have

$$BM + MC < BD + DF + FC = \frac{AB + BC + CA}{2}. (2) $$

If $M$ lies in trapezoid $ADEC$, we have

$$AM + MC < AD + DE + EC = \frac{AB + BC + CA}{2}. (3)$$

Since $M$ lies in at least 2 of these regions, it is guaranteed that at least one of (2) or (3) is true.

If (2) is true, then $$2AM + BM + CM \leq 2BM + 2CM < AB+ BC + CA.$$ If (3) is true, then $$2AM + BM + CM \leq 2AM + 2CM < AB+BC+CA.$$ Either way, we get that $\text{min}\{AM,BM,CM\} + AM + BM + CM < AB + BC + CA$.

Construction: Draw a equalateral triangle such that $M,$ is the circumcenter.
Proof: Let the side length be $2.$ (since all equalateral triangles are congruent we can assume side length) Then, we have $\min\{MA,MB,MC\}+MA+MB+MC=\frac{8}{\sqrt{3}},$ and $AB+AC+BC=6.$ Now, note that $\frac{8}{\sqrt{3}}<6,$ and we are done.

note that i am assuming this gives the minimum value of $AB+AC+BC-(\min\{MA,MB,MC\}+MA+MB+MC).$

Add in the medial triangle $DEF$. Of course suppose that WLOG $AM$ is the minimum.

It follows that $M$ lies on the parallelogram $AFED$. Also remark that $AEDB$ and $AFDC$ are convex, so thus $AM + BM < \max(AD + BD, BE + AE)$. Likewise, $AM + CM < \max(AD + CD, CF + AF)$.
Let's verify these four cases manually.

$AD + BD + AD + CD < AB + BC + CA$ iff $2AD < AB + AC$, which is a well known inequality.
$AD + BD + CF + AF < AB + BC + CA$ iff $AD + CF < AC + BC/2 + BA/2$, where we are given that $AD + BD > BE + AE$ and $AD + CD < CF + AF$.
From the latter we wish to demonstrate
\[ 2CF + AF - CD < AC + BC/2 + BA/2 \]
which rearranges to $2CF < AC + BC$. This is true.
Similarly the other 'mixed' case can be handled in the exact same way.
Lastly,
$BE + AE + CF + AF < AB + BC + CA$ iff $BE + CF < BC + AB/2 + AC/2$. However this is due to the two inequalities $2BE < AB + BC$ and $2CF < AC + BC$. Done.


function
floor function
integration
calculus
parameterization
real analysis
real analysis unsolved

${(0<a,b<1)\land (a<b)}$

${x(n)=a 10^{n}-\left\lfloor a 10^{n}\right\rfloor }$
${y(n)=b 10^{n}-\left\lfloor b 10^{n}\right\rfloor }$

Evaluate

${\int_{a}^{b}\sqrt{x'(t)^{2}+y'(t)^{2}}\, dt}$

I know that $x,y$ are caotic functions for some values of $a,b$ and non-continuous in some points etc.
But we can assume that $x,y$ are differentiable for some values etc.

Can we evaluate this integral?!

I Have a solution but i don't think this is rigth.

If someone knows if we can evalute this integral and how can I proceed

Thank

I don't see a proper definition of $x(t)$ and $y(t).$ This problem is badly posed.

$n$ usually denotes an integer-valued variable. If we assume that the $n$ are the same as the $t$ and look between a pair of jumps, this is merely an unusual parametrization of a line segment. If we ignore the jumps, we can simply splice together these line segments end to end, and that eliminates the $\lfloor a\cdot 10^{t}\rfloor$. This is not a very interesting question.


calculus integration algebra polynomial AMC USA(J)MO USAMO

Let $a,b,c$ denote three distinct integers, and let $P$ denote a polynomial having all integral coefficients.
Prove that it is impossible that $P(a)=b,(P(b)=c,$ and $P(c)=a$

Making an attempt
[hide]By substitution then

$P(P(P(a))) = a$

$P(P(P(b))) = b$

$P(P(P(c))) = c$

If we substitute further then we have an infinite number of $P$'s and at least one of the the above must diverge.[/hide]

This is an old USAMO problem, I think a number 1.

[hide="Solution"]
Assume there is a solution a, b, c.  So P(x) = (x - a)Q(x) + b = (x - b)R(x) + c = (x - c)S(x) + a for 3 integer polynomials Q(x), R(x), S(x).

Ok now plug in b for x and get
(b - a)q(b) + b = P(b) = c, so (b - a) divides (P(b) - b)=(c - b).

Repeat with c and a to get
(c - b) divides (a - c), and
(a - c) divides (b - a).

But (b - a) divides (c - b) divides (a - c) implies that (b - a) divides (a - c). So (b - a) and (a - c) have to divide each other. Hence (b - a) = ±(a - c).

2 cases:
If b - a = -(a - c), then b = c, a contradiction becasue they have to be distinct.

If b - a = a - c, then b - c = (b - a) + (a - c) = 2(a - c).  But we knew (c - b) divides (a - c)  so we cant have b - c=2(a - c) unless both are zero, which is impossible bacause of the disctinct condition, so that's a contradiction.  Thus, no solutions.  I learned this off of Kalva a while back, and there might be other ways.[/hide]


I need help with the complete solution of this induction proof, the base case is done

$ 2\cdot \sum_{k=1}^{n}k!  >  n\cdot \sum_{k=1}^{n-1}k!$

Let $ f(n)$ be the value of the LHS for a particular $ n$, and $ g(n)$ be the value of the RHS for a particular $ n$. You've already shown $ f(1) > g(1)$. Now we want to show $ f(n\plus{}1) > g(n\plus{}1)$ given $ f(n) > g(n)$.
\[ f(n\plus{}1) \equal{} f(n) \plus{} 2(n!) > f(n) \plus{} n! > g(n) \plus{} n! \equal{} g(n\plus{}1)\]
where the inductive hypothesis is used in the second inequality. We're done.

Your final equality does not hold:\begin{align*} g(n + 1) - g(n) & = (n + 1)\sum_{i = 1}^{n} i! - n\sum_{i = 1}^{n - 1} i! \\
& = (n + 1)! + \sum_{i = 1}^{n - 1} i! \\
& \neq n! \end{align*}

Ah, what the heck. That's not the first time I've made that mistake. Not to mention all of my $ n!$s should have been $ (n\plus{}1)!$s. Let's try this again.

Define $ S(n) \equal{} \sum_{i\equal{}0}^{n} i!$. So the problem is equivalent to showing $ 2 S(n) > n S(n\minus{}1)$. Suppose it's true for $ n$, then we want to show $ 2 S(n\plus{}1) > (n\plus{}1) S(n)$.

Note that $ S(n\plus{}1) \minus{} S(n) \equal{} (n\plus{}1)!$. Rewrite the inequality as
\[ 2 S(n) \plus{} (n\plus{}1)! > n S(n\minus{}1) \plus{} S(n\minus{}1)\]
$ 2 S(n) > n S(n\minus{}1)$ by the inductive hypothesis, so if $ (n\plus{}1)! > S(n\minus{}1)$, we're done. Let's show that $ (n\plus{}2)! > S(n)$ by another induction. $ n \equal{} 1$ gives $ 6 > 1$, so base case is done. If it's true for $ n$, then add $ (n\plus{}2) \cdot (n\plus{}2)!$ to the LHS and $ (n\plus{}1)!$ to the RHS. This doesn't change the sign of the inequality and results in
\[ (n\plus{}2) \plus{} (n\plus{}2) \cdot (n\plus{}2)! > S(n) \plus{} (n\plus{}1)!\]
\[ (n\plus{}3)! > S(n\plus{}1)\]
So we're done.

[hide]$ 2\sum_{k\equal{}1}^n k! > 2n! > n(1! \plus{} \cdots \plus{} (n \minus{} 1)!)$?

$ \implies 2(n \minus{} 1)! > 1! \plus{} \cdots \plus{} (n \minus{} 1)!$
$ \implies (n \minus{} 1)! > 1! \plus{} \cdots \plus{} (n \minus{} 2)!$

$ \implies (n \minus{} 2)!(n \minus{} 1) > 1! \plus{} \cdots \plus{} (n \minus{} 2)!$, since RHS has n-2 terms, all $ \le (n \minus{} 2)!$.[/hide]


crancen?-sintetic nu  mi-am dat seama,dar vectorial iese destul de simplu :)

[b][color=red]Am o solutie pentru aceasta problema:[/color][/b]
-----------------------------------------------------------------------------------------------
[color=blue]Daca
[tex]PB = c,PC = b,BC = a,BM = MC = \frac{a}{2}[/tex]
[tex]m\left( {\widehat{PBK}} \right) = m\left( {\widehat{PCL}} \right) = \alpha [/tex]
[tex]PK \bot BK,PL \bot CL[/tex]
Demonstrati:
[tex]MK=ML[/tex]
[/color]
[img]http://img.photobucket.com/albums/v218/hudrea/matholympicman01.jpg[/img]
-----------------------------------------------------------------------------------------------
[u][b]Solutie:[/b][/u]
[tex]MK^2  = ML^2  \Leftrightarrow [/tex]

[tex] \Leftrightarrow \frac{1}{4}BC^2  + BK^2  - 2 \cdot \frac{1}{2} \cdot BC \cdot BK \cdot \cos \left( {\alpha  + B} \right) = [/tex]
[tex] = \frac{1}{4}BC^2  + CL^2  - 2 \cdot \frac{1}{2} \cdot BC \cdot CL \cdot \cos \left( {\alpha  + C} \right) \Leftrightarrow [/tex]

[tex] \Leftrightarrow BK^2  - BC \cdot BK \cdot \cos \left( {\alpha  + B} \right) =  CL^2  - BC \cdot CL \cdot \cos \left( {\alpha  + C} \right) \Leftrightarrow [/tex]

[tex] \Leftrightarrow c^2  \cdot \cos ^2 \alpha  - ac \cdot \cos \alpha  \cdot \cos \left( {\alpha  + B} \right) = [/tex]
[tex] = b^2  \cdot \cos ^2 \alpha  - ab \cdot \cos \alpha  \cdot \cos \left( {\alpha  + C} \right) \Leftrightarrow [/tex]

[tex] \Leftrightarrow c^2  \cdot \cos \alpha  - ac \cdot \cos \left( {\alpha  + B} \right)  = b^2  \cdot \cos \alpha  - ab \cdot \cos \left( {\alpha  + C} \right) \Leftrightarrow [/tex]

[tex] \Leftrightarrow \sin ^2 C \cdot \cos \alpha  - \sin A \cdot \sin C \cdot \cos \left( {\alpha  + B} \right) = [/tex]
[tex] = \sin ^2 B \cdot \cos \alpha  - \sin A \cdot \sin B \cdot \cos \left( {\alpha  + C} \right) \Leftrightarrow [/tex]

[tex] \Leftrightarrow \sin ^2 C \cdot \cos \alpha  - \sin A \cdot \sin C \cdot \cos B \cdot \cos \alpha  =[/tex]
[tex]= \sin ^2 B \cdot \cos \alpha  - \sin A \cdot \sin B \cdot \cos C \cdot \cos \alpha  \Leftrightarrow [/tex]

[tex] \Leftrightarrow \sin ^2 C - \sin \left( {B + C} \right) \cdot \sin C \cdot \cos B = \sin ^2 B - \sin \left( {B + C} \right) \cdot \sin B \cdot \cos C \Leftrightarrow [/tex]

[tex] \Leftrightarrow \sin ^2 C - \sin ^2 C \cdot \cos ^2 B  = \sin ^2 B - \sin ^2 B \cdot \cos ^2 C \Leftrightarrow [/tex]

[tex] \Leftrightarrow \sin ^2 C \cdot \left( {1 - \cos ^2 B} \right) = \sin ^2 B \cdot \left( {1 - \cos ^2 C} \right)[/tex]
evident adevarata.

Demonstrat fiind enuntul de mai sus, problema se rezolva usor. [ a.) inclusiv b.) ]

[b]Solutie vectoriala:[/b] (pt. [i](a)[/i])

Avem $\overrightarrow{MN}=\overrightarrow{ON}-\overrightarrow{OM}=\frac{\overrightarrow{OB} + \overrightarrow{OC} - \overrightarrow{OA} - \overrightarrow{OD}}{2}=\frac{\overrightarrow{AB}+\overrightarrow{DC}}{2}$.

Stim ca $\overrightarrow{OE} \cdot \overrightarrow{AB} = 0$ si $\overrightarrow{OF} \cdot \overrightarrow{CD}=0$, deci

\begin{eqnarray*} \overrightarrow{MN} \cdot \overrightarrow{EF} &=& \frac12 \left( \overrightarrow{AB} + \overrightarrow{DC} \right) \cdot \overrightarrow{EF} \\ \ &=& \frac12 \left( \overrightarrow{AB} \cdot \overrightarrow{EF} + \overrightarrow{DC} \cdot \overrightarrow{EF} \right) \\ \ &=& \frac12 \left( \overrightarrow{AB} \cdot \overrightarrow{EO} + \overrightarrow{AB} \cdot \overrightarrow{OF} + \overrightarrow{DC} \cdot \overrightarrow{EO} + \overrightarrow{DC} \cdot \overrightarrow{OF} \right) \\ \ &=& \frac12 \left( \overrightarrow{AB} \cdot \overrightarrow{OF} + \overrightarrow{DC} \cdot \overrightarrow{EO} \right) \\ \ &=& \frac12 \left( AB \cdot OF \cdot \cos \measuredangle \left( \overrightarrow{AB}, \overrightarrow{OF} \right) + DC \cdot EO \cdot \cos \measuredangle \left( \overrightarrow{DC}, \overrightarrow{EO} \right) \right) .\end{eqnarray*}

Putem demonstra foarte usor ca $\triangle AEO \sim \triangle DFO$, deci $\frac{EO}{OF}=\frac{AE}{DF}$. In mod asemanator, demonstram ca $\frac{EO}{OF}=\frac{EB}{FC}$. Asadar, $\frac{EO}{OF}=\frac{AE}{DF}=\frac{EB}{FC}=\frac{AE+EB}{DF+FC}=\frac{AB}{DC}$, ceea ce implica $EO \cdot DC = AB \cdot OF$.

\begin{eqnarray*} \overrightarrow{MN} \cdot \overrightarrow{EF} &=& \frac12 \left( AB \cdot OF \cdot \cos \measuredangle \left( \overrightarrow{AB}, \overrightarrow{OF} \right) + DC \cdot EO \cdot \cos \measuredangle \left( \overrightarrow{DC}, \overrightarrow{EO} \right) \right) \\ \ &=& \frac12 \cdot AB \cdot OF \cdot \left( \cos \measuredangle \left( \overrightarrow{AB}, \overrightarrow{OF} \right) + \cos \measuredangle \left( \overrightarrow{DC}, \overrightarrow{EO} \right) \right) .\end{eqnarray*}

Fie $\alpha = \measuredangle \left( \overrightarrow{AB}, \overrightarrow{DC} \right)$. Atunci avem $\measuredangle \left( \overrightarrow{AB}, \overrightarrow{OF} \right) = \frac{\pi}{2} + \alpha$ si $\measuredangle \left( \overrightarrow{DC}, \overrightarrow{EO} \right) = \frac{\pi}{2} - \alpha$, adica cosinusurile au acelasi modul, dar semne diferite. In concluzie, $\overrightarrow{MN} \cdot \overrightarrow{EF} = 0$, i.e. $MN \perp EF$.

[b]Lema.[/b] Daca, in conditiile problemei, $S \in AB$ si $T \in DC$ a.i. $\frac{\overline{SA}}{\overline{SB}}=\frac{\overline{TD}}{\overline{TC}}$, atunci $L$, mijlocul segmentului $ST$, se afla pe $MN$.

[i]Demonstratie:[/i] Fie $k \in \mathbb{R}$, $k=\frac{\overline{SA}}{\overline{SB}}=\frac{\overline{TD}}{\overline{TC}}$. Avem $\overrightarrow{OS}=\frac{\overrightarrow{OA}-k\overrightarrow{OB}}{1-k}$ si $\overrightarrow{OT}=\frac{\overrightarrow{OD}-k\overrightarrow{OC}}{1-k}$. Notand cu $s=\frac{1}{1-k}$, $\overrightarrow{OS}=s \overrightarrow{OA}+(1-s)\overrightarrow{OB}$ si $\overrightarrow{OT}=s \overrightarrow{OD}+(1-s)\overrightarrow{OC}$.

\begin{eqnarray*} \overrightarrow{OL} &=& \frac12 \cdot \left( \overrightarrow{OS} + \overrightarrow{OT} \right) \\ \ &=& \frac12 \cdot \left( s \left( \overrightarrow{OA} + \overrightarrow{OD} \right) + (1-s) \left( \overrightarrow{OB} + \overrightarrow{OC} \right) \right) \\ \ &=& s \overrightarrow{OM} + (1-s) \overrightarrow{ON} \end{eqnarray*}

Ultima relatie inseamna ca $L \in MN$. $\square$

Folosind asemanarile $\triangle AEO \sim \triangle DFO$ si $\triangle BEO \sim \triangle CFO$, observam ca $\frac{\overline{EA}}{\overline{EB}}=\frac{\overline{FD}}{\overline{FC}}$. Aplicand [b]Lema[/b], descoperim ca mijlocul lui $EF$ se afla pe $MN$. Dar din moment ce $MN \perp EF$, $MN$ este mediatoarea lui $EF$.

Fie L si K mijloacele segmentelor OB, respectiv OC.
[img]http://i19.photobucket.com/albums/b200/mariusdamian/4.png[/img]
Cum triunghiul EOB este dreptunghic, iar KN este linie mijlocie in triunghiul COB, avem:
\[EL=\frac{1}{2}OB=KN\] si analog \[LN=FK.\]
Congruenta unghiurilor ELN si FKN este evidenta daca tinem cont de conditia de inscriptibilitate a patrulaterului ABCD si astfel este asigurata congruenta triunghiurilor ELN si NKF.
Urmeaza EN=FN si analog EM=FM de unde obtinem imediat ca MN este mediatoarea segmentului EF.

It is a good problem and the given solutions  are very good , as well.

  This problem is a different form from an old problem taken from the British Olympiad( between 1970 and 1980 , if I remember well. If BA and CD inersect at P , then .... The official solution is like the last one)

Indeed, the problem is not quite new; it was given, e. g., in the BWM (Bundeswettbewerb Mathematik = German mathematical contest) in 2003 (2nd round, problem 3), and an equivalent version was in the IMO Shortlist 1982 (see http://www.mathlinks.ro/Forum/viewtopic.php?t=30926 and http://www.mathlinks.ro/Forum/viewtopic.php?t=16761 ).

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=39107 .

  Darij

Construim paralelogramele $ADCR$ si $ARPB$ ca in figura.
[img]http://i19.photobucket.com/albums/b200/mariusdamian/1.png[/img]
Pentru a demonstra ca $MN\perp{EF}$ este suficient sa probam asemanarea triunghiurilor $OEF$ si $RPA$.
Avem: $\triangle{OAB}\sim\triangle{ODC}$, de unde $\frac{OE}{OF}=\frac{AB}{DC}$ si cum $AB=RP$, $DC=AR$, obtinem $\frac{OE}{OF}=\frac{RP}{AR}$.
Congruenta unghiurilor $EOF$ si $ARP$ rezulta usor din conditia de inscriptibilitate a patrulaterului ABCD si am asigurat astfel $\triangle{OEF}\sim\triangle{RAP}$.
De aici deducem imediat ca $AP\perp{EF}$ adica $MN\perp{EF}$.

Felicitari, Domnule Damian ! Ati oferit un exemplu de solutie compatibil cu cele discutate prin mes. priv. Frumusetea solutiei evidentiaza si mai mult frumusetea crancena a problemei. Pur si simplu mi-ati spulberat orice tentativa de a gasi si eu o solutie pentru aceasta problema. Trebuie sa remarc ca aveti si talent (cunostinte) de grafica pe calculator.

intr-adevar. solutia lui este una incredibil de simpla pt. o problema ce parea destul de greu de rezolvat fara trigonometrie ori vectori

Asa cum a afirmat Darij, problema propusa este cunoscuta si am gasit-o in fisele mele de geometrie (dupa indelungi cautari), insa ca o consecinta imediata a urmatoarei probleme a carei demonstratie nu este metrico-trigonometrica (ce cuvant nesuferit !):

[u]Enunt.[/u] Pe laturile $AB,\ AC$ ale triunghiului $ABC$ se construiesc in afara acestuia triunghiurile dreptunghice $ABP\ (PA\perp PB),\ ACN\ (NA\perp NC)$ respectiv astfel incat $m(\widehat {BAP})=m(\widehat {CAN})=x$. Atunci, pentru mijlocul $M$ al laturii $BC$ exista relatiile  $MN=MP,\ m(\widehat {NMP})=180^{\circ}-2x$.

[u]Demonstratie.[/u] Notam mijloacele $U,\ V$ ale laturilor $AB,\ AC$ respectiv. Astfel,
$UP=UB=VM,\ UM=AV=VN\Longrightarrow UP=VM,\ UM=VN;$
$m(\widehat {MUP})=A+2\cdot m(\widehat {BAP}),\ m(\widehat {MVN})=$ $A+2\cdot m(\widehat {CAN})\Longrightarrow \widehat {MUP}\equiv \widehat {MVN}.$
Asadar, $\triangle UPM\equiv \triangle VMN$ de unde rezulta $MN=MP,\ \widehat {PMU}\equiv \widehat {MNV},$
$\widehat {MPU}\equiv \widehat {NMV},\ m(\widehat {NMP})=A+m(\widehat {PMU})+m(\widehat {VMN})=$
$A+180^{\circ}-2x-A=180^{\circ}-2x.$
In concluzie, $MN=MP,\ m(\widehat {NMP})=180^{\circ}-2x$.

[u]Observatie.[/u] Revenind la problema initiala (in notatiile proprii acesteia) se obtine suplimentar ca
$m(\widehat {EMF})=2\cdot m(\widehat {BAC}),\ m(\widehat {ENF})=2\cdot m(\widehat {ACD}),$
$m(\widehat {EMF})+m(\widehat {ENF})=360^{\circ}-2\cdot m(\widehat {AOB})$ si faptul ca patrulaterul $MENF$
este inscriptubil daca si numai daca patrulaterul $ABCD$ este ortodiagonal.

[quote="Marius Damian"]Patrulaterul ABCD este inscriptibil iar diagonalele sale se taie in O. Daca M, N sunt mijloacele laturilor AD, respectiv BC, iar E, F sunt picioarele perpendicularelor din O pe AB, respectiv CD,

a) să se demonstreze că MN $\bot$ EF.
b) Este MN mediatoarea segmentului EF?[/quote]
    Da, este o problema frumoasa, dar nu cred ca e asa de grea.Totusi, mai lasa-ma sa ma mai gandesc la ea si iti dau o alta rezolvare. :)


A piece of cheese is located at $ (12,10)$ in a coordinate plane. A mouse is at $ (4, \minus{} 2)$ and is running up the line $ y \equal{} \minus{} 5x \plus{} 18.$ At the point $ (a,b)$ the mouse starts getting farther from the cheese rather than closer to it. What is $ a \plus{} b?$

$ \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 14 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 22$

[hide]The shortest distance bewteen $y=-5x+18$ and the cheese is the line that goes through the cheese and is perpendicular to $y=-5x+18$.

The equation of this line is $y=\frac15x+b$. Plugging in $(12,10)$, we find $b=\frac{38}5$.

Solving the system, we get $x=2$ and $y=8$.

$x+y=10\Rightarrow\boxed{\text{B}}$[/hide]


function
algebra
domain
algebra unsolved

Prove that there exist infinitely many functions $ f$ such that the domain is the set of natural numbers and
\[f(nf(k)\plus{}kf(n))\equal{}f(n^2\plus{}k^2)f(n\plus{}k\minus{}1)\]
for any two natural numbers $ k$ and $n$.

[quote="discredit"]Prove that there exist infinitely many functions $ f$ such that the domain is the set of natural numbers and

$ f(nf(k) \plus{} kf(n)) \equal{} f(n^2 \plus{} k^2)f(n \plus{} k \minus{} 1)$

for any natural numbers $ k,n$.[/quote]
[hide="My solution"]
Let $ f_a(x)$, for any $ a\in\mathbb N$ defined as :
$ f_a(x)\equal{}1$ for all odd $ x$
$ f_a(x)\equal{}2a$ for all even $ x$

$ f_a(x)$ is a solution ($ nf(k)\plus{}kf(n)$ is always even, so LHS is $ 2a$ and $ n^2\plus{}k^2$ and $ n\plus{}k\minus{}1$ are either even - odd, either odd - even, so RHS is always $ 2a$ too).

And there are infinitely many different $ f_a$, hence the result.

[/hide]


logarithms
number theory open
number theory

Hello, I want to know the answer to this problem.

Imagine all the numbers that have form $ 2^a 3^b$ with $ a,b$ integers and $ b \geq 0$.
Prove or disprove: we can approximate any real number arbitrarily close by choosing $ a$ and $ b$ accordingly.

Ex: to approximate 5, we can choose $ \frac{9}{2}$ which is quite close, but $ \frac{81}{16}$ is even closer. I have not found anything that's closer.

$ \frac {3^{45}}{2^{69}}$ and a better one
$ \frac {3^{351}}{2^{554}}$ thats the limit with the excel...

...@excel

Show that it suffices to show that we can approximate any positive real number arbitrary good by numbers of type $ a \log 2 \minus{} b \log 3$. For this, see http://www.mathlinks.ro/viewtopic.php?p=375120 or http://www.mathlinks.ro/viewtopic.php?p=171749 .


function
algebra
domain
algebra proposed

Find all functions $f, g: \mathbb R \to \mathbb R$ such that
\[f(x\plus{}g(x))\equal{}g(x)\]
holds for all real $x$.

Was going to ask for the domain and codomain, but in fact the answer is completely unchanged as long as it's something with addition and subtraction.

Let $ h(x) \equal{} g(x) \plus{} x$, so the equation becomes $ f(h(x)) \equal{} h(x) \minus{} x$. If $ h(a) \equal{} h(b)$, then $ f(h(a)) \equal{} f(h(b))$, so $ h(a) \minus{} a \equal{} h(b) \minus{} b$ and $ a \equal{} b$. Thus $ h$ is injective; we show all of these work. For each $ x$ in the domain of $ h$, define $ f(h(x))$ to be $ h(x) \minus{} x$; this is well-defined since $ h$ is injective. This defines $ f$ on the range of $ h$; on anything outside the range define $ f$ arbitrarily. This gets the entire solution set, which happens to be quite large.

[quote="mathematikos"]HI!
find all functions $ (f,g)$ such that:
$ f(x \plus{} g(x)) \equal{} g(x)$
thanx.[/quote]

Let $ h(x)\equal{}x\plus{}g(x)$

$ h(x)\equal{}h(y)$ $ \implies$ $ f(h(x))\equal{}f(g(y))$ $ \implies$ $ g(x)\equal{}g(y)$ $ \implies$ $ h(x)\minus{}g(x)\equal{}h(y)\minus{}g(y)$ $ \implies$ $ x\equal{}y$ $ \implies$ $ h(x)$ is injective.

Let then $ A\equal{}h(\mathbb R)$. $ h(x)$ is then a bijection from $ \mathbb R \to A$ and so $ \exists$ inverse function $ h^{[\minus{}1]}$ from $ A\to\mathbb R$

And we have $ g(x)\equal{}h(x)\minus{}x$ and $ f(x)\equal{}g(h^{[\minus{}1]}(x))$ $ \forall x\in A$

[b][u]And this is obviously the general solution [/u][/b]:
Take any injective function $ h(x)$ from $ \mathbb R\to\mathbb R$
Let $ A\equal{}h(\mathbb R)$ and the inverse function $ h^{[\minus{}1]}$ from $ A\to\mathbb R$

$ \forall x\in \mathbb R$ : $ g(x)\equal{}h(x)\minus{}x$
$ \forall x\in A$ : $ f(x)\equal{}g(h^{[\minus{}1]}(x))$
$ \forall x\notin A$ : $ f(x)\equal{}$any value.

The fact that this is a solution is obvious.
The fact that all solutions are of this form is a consequence of the beginning of this post.

Hope this will help you  :wink:


Prove or disprove: If x and y are real numbers with $y \ge 0$ and $y(y+1) \le (x+1)^2$, then $y(y-1) \le x^2$

Solving for x, $x = -1 \pm \sqrt{y^2 + y}$.  So $x \ge -1 + \sqrt{y^2 + y}$ or $x \le -1 - \sqrt{y^2 + y}$

===
First case:  $x \ge -1 + \sqrt{y^2 + y}$

Then we have $x^2 \ge (-1 + \sqrt{y^2 + y})^2 = y^2 + y + 1 - 2\sqrt{y^2 + y}$.

But $y^2 + y + 1 - 2\sqrt{y^2 + y} \ge y^2 - y \Leftrightarrow 2y + 1 \ge 2\sqrt{y^2 + y} = \sqrt{(2y + 1)^2 - 1}$.

===
Second case:  $x \le -1 - \sqrt{y^2 + y}$

Obviously, $x \le 1$.  So x is in $[-1, -inf]$ for sure.  Thus, to minimize $x^2$, we should choose the largest x possible.

Now $x^2 \ge (-1 - \sqrt{y^2 + y})^2 = y^2 + y + 1 + 2\sqrt{y^2 + y}$.

But $ y^2 + y + 1 + 2\sqrt{y^2 + y} \ge y^2 - y \Leftrightarrow 2y + 1 + 2\sqrt{y^2 + y} \ge 0$.

==========

and we are done.

Say that $y(y-1)\geq x^2$.  Adding that to the original inequality gives $2y\leq 2x+1$.  If $x\geq 0$, then this gives

$\left(y-\frac 12\right)^2\leq x^2$

but $y(y-1)\geq x^2$ can be written as

$\left(y-\frac 12\right)^2\geq x^2+\frac 14$

so we have a contradiction.

If instead $x<0$ then we must have $x\geq -\frac 12$

From $y(y-1)\geq x^2$, it is obvious that $y\geq 1$.  But then $y(y+1)\geq 2$, whereas $(x+1)^2\leq \frac {1}{4}$.


quadratics
trigonometry
vector
function
LaTeX
complex numbers

1. If $ z \in \mathbb{C}^{*}$ and $ \left| z^{3}+\frac{1}{z^{3}}\right| \leq 2$, then prove that $ \left| z+\frac{1}{z}\right| \leq 2$.

Find all $ z \in \mathbb{C}$
2. Such that $ \left| z \right| = 1$ and $ \left| z^{2}+\overline{z}^{2}\right| = 1$.

3. Such that $ 4z^{2}+8 \left| z \right|^{2}= 8$.
[hide="Wrong Approach"]
?
Quadratic to $ z=-\overline{z}\pm \sqrt{\overline{z}^{2}+2}$ , $ 3a^{2}-2abi+b^{2}-2=0$.
[/hide]

4. Such that $ z^{3}=\overline{z}$.
[hide="This method does not get anywhere it seems"]
$ (a+bi)^{3}= a-bi$.
[/hide]

[quote="AstroPhys"]Find all $ z \in \mathbb{C}$ 4. Such that $ z^{3}=\overline{z}$.[/quote][hide]EDIT:Let $ z=a+bi$, where $ a,b\in \mathbb{R}$. $ z\overline{z}=|z|^{2}\implies z^{4}=|z|^{2}$
Since $ |z|\geq 0$, $ z^{2}=\pm |z|$.
If $ a^{2}-b^{2}+2abi=a^{2}+b^{2}\implies abi=b^{2}$, but $ a$ and $ b$ are both reals, so $ a=b=0$ or $ b=0,a=\pm 1\implies z=\pm 1$.
If $ a^{2}-b^{2}+2abi=-a^{2}-b^{2}\implies-abi=a^{2}$, so $ a=0,b=\pm 1 \implies z=\pm i$.
so $ z=0,1,i,-1,-i$
(sorry, there was a mistake in my first solution for those who saw it.. :blush: )
[/hide]

Two views on #2, a hint for #3, and a polar approach to #4:

[hide="2."]
Trigonometrically, since $ |z|=1$, just let $ z=\cos{\theta}+i\sin{\theta}$ and a little algebra will quickly yield the answer.

Another approach (it won't be any faster, but no time spent visualizing complex numbers is time wasted) would be to let $ z=e^{i\theta}$.  Then we want $ |e^{2i\theta}+e^{-2i\theta}| = 1$.  A sketch of $ e^{2i\theta}+e^{-2i\theta}$ (two oppositely angled vectors of equal magnitude) shows that $ e^{2i\theta}+e^{-2i\theta}\in \mathbb{R}$ and since it has magnitude 1, $ e^{2i\theta}+e^{-2i\theta}= \pm1$.
Now its hard to explain without drawing, but since $ |e^{i(2\theta)}| = |e^{i(-2\theta)}| = |\pm1| = 1$, a sketch of $ e^{2i\theta}+e^{-2i\theta}= \pm1$ would consist of 4 equilateral triangles, one in each quadrant.  From there it can be solved.  For example, in the first quadrant we see that $ e^{i(2\theta)}= e^{i(\frac{\pi}{3}+2k\pi)}$, so $ \theta = \frac{\pi}{6}$ or $ \frac{7\pi}{6}$.

A combination approach for a fast answer if you are familiar with hyperbolic trig functions would be: $ |e^{2i\theta}+e^{-2i\theta}| = |2\cosh{2i\theta| = |2\cos{2\theta}| = 1}$, so $ \cos{2\theta}= \pm\frac{1}{2}$.[/hide]

[hide="3."]First step: divide by 4  :P
So $ z^{2}= 2-2|z|^{2}$.  But $ 2-2|z|^{2}\in \mathbb{R}$, so $ z^{2}$ must be real.  What conditions does that force on $ z$?  After figuring that out, break it down into two cases and the rest is basic algebra.[/hide]

[hide="4."]As usual, let $ |z|=r$ and $ \arg{z}=\theta$.  Then $ z^{3}= \bar{z}\Rightarrow r^{3}e^{i(3\theta)}= re^{i(-\theta)}$, so $ r^{3}= r$ and $ 3\theta =-\theta+2k\pi$

$ r^{3}= r$ means $ r=0$ or $ r=1$ ($ r=-1$ is the same as $ r=1$ but in the opposite direction).  Since only one number (zero) has magnitude $ 0$, we only need to consider the angle when $ r=1$.
$ 3\theta =-\theta+2k\pi \Rightarrow \theta = \frac{\pi}{2}k$, and $ e^{\frac{\pi}{2}ki}= i^{k}$.

Thus $ z \in \{0,1,i,-1,-i\}$.[/hide]

[quote="minsoens"][hide]If $ a^{2}-b^{2}+2abi=a^{2}+b^{2}\implies abi=b^{2}$, but $ a$ and $ b$ are both reals, so $ a=b=0$ or $ b=0,a=\pm 1\implies z=\pm 1$.
If $ a^{2}-b^{2}+2abi=-a^{2}-b^{2}\implies-abi=a^{2}$, so $ a=0,b=\pm 1 \implies z=\pm i$.
so $ z=0,1,i,-1,-i$[/hide]
[/quote]

[hide="Question"]
If $ b=0$ then $ ai=0 \Rightarrow a=0 \not = \pm 1$ , how does $ a= \pm 1$?  What about the $ b=\pm 1$ in the second case?  It is ovbiously the right answer though.
[/hide]

[quote="gotztahbeazn"]for #3 would it just be...
[hide]
$ z^{2}+2|z|^{2}= 2$

let $ z = a+bi$, hence:

$ z^{2}+2|z|^{2}= a^{2}-b^{2}+2abi+2(a^{2}+b^{2}) = 3a^{2}+b^{2}+2abi = 2$

comparing real to imaginary parts implies that $ 2ab = 0$ and $ 3a^{2}+b^{2}= 2$

hence, we have two cases where either $ a = 0$ or $ b=0$ and so:

$ z = \pm i \sqrt{2}$ or $ z = \pm \sqrt{\frac{2}{3}}$

[/hide][/quote]

Correct.  [hide="What I was getting at:"]
$ z^{2}= r^{2}e^{i(2\theta)}\in \mathbb{R}\Rightarrow 2\theta = k\pi \Rightarrow \theta = \frac{\pi}{2}k$, so $ z$ is either real or pure imaginary.  Then $ z=x$ or $ z=xi$ where $ x$ is real.
For the first case, $ x^{2}+2x^{2}= 2$ so $ z=x=\pm\frac{\sqrt6}{3}$.
For the second case, $ -x^{2}+2x^{2}=2$ so $ z=xi=\pm\sqrt{2}i$.[/hide]

2. We have that $ z+\overline{z}=2\Re(z)$...so $ \Re(z^{2})=\frac{1}{2}$. Then we just go line $ z=a+bi$; $ a,b\in R$ and solve the quadratic.

(NB : How to hide a solution, if I may know that ???)
Number one :  :P  :)
Notice that : |z^3 + 1/z^3|=|z+1/z||z^2+1/z^2-1|,

[quote="AstroPhys"][quote="minsoens"][hide]If $ a^{2}-b^{2}+2abi=a^{2}+b^{2}\implies abi=b^{2}$, but $ a$ and $ b$ are both reals, so $ a=b=0$ or $ b=0,a=\pm 1\implies z=\pm 1$.
If $ a^{2}-b^{2}+2abi=-a^{2}-b^{2}\implies-abi=a^{2}$, so $ a=0,b=\pm 1 \implies z=\pm i$.
so $ z=0,1,i,-1,-i$[/hide]
[/quote]If $ b=0$ then $ ai=0 \Rightarrow a=0 \not = \pm 1$ , how does $ a= \pm 1$?  What about the $ b=\pm 1$ in the second case?  It is ovbiously the right answer though.[/quote]I think I forgot to show that $ |z|=1$ but from $ z^{3}=\overline{z}$ and using the substitution $ z=re^{i\theta}$ you can show $ |r|=1$, where r is real. $ abi=b^{2}\implies b(b-ai)=0$ so if $ b=0$, $ a$ can be anything and still satisfy the equation. But then $ |z|=1$, so you must have $ a=\pm 1$. and similarly for $ -abi=a^{2}$.

@Qruni: you can learn a lot about Latex and BBCode by clicking "Quote" by someone's post and reading the code. To hide, surround your text with [code][hide]text[/hide][/code]

1.
 \[\begin{gathered}
  \left| {z^3  + \frac{1}
{{z^3 }}} \right| \leqslant 2\mathop {}\limits^{}  \Leftrightarrow \mathop {}\limits^{} \left| {z + \frac{1}
{z}} \right|\left| {\left( {z + \frac{1}
{z}} \right)^2  - 3} \right| \leqslant 2 \hfill \\
  but \hfill \\
  \left| {\left( {z + \frac{1}
{z}} \right)^2  - 3} \right| \geqslant \left| {\left| {z + \frac{1}
{z}} \right|^2  - 3} \right| \hfill \\
  so \hfill \\
  \left| {z + \frac{1}
{z}} \right|\left| {\left( {z + \frac{1}
{z}} \right)^2  - 3} \right| \leqslant 2\mathop {}\limits^{}  \Rightarrow \mathop {}\limits^{} \left( {\left| {z + \frac{1}
{z}} \right|} \right)\left| {\left| {z + \frac{1}
{z}} \right|^2  - 3} \right| \leqslant 2\mathop {}\limits^{}  \Rightarrow  \hfill \\
   \hfill \\
  \left| {\left| {z + \frac{1}
{z}} \right|^3  - 3\left| {z + \frac{1}
{z}} \right|^2 } \right| \leqslant 2\mathop {}\limits^{}  \Rightarrow  \hfill \\
   \hfill \\
  \left( {\left| {z + \frac{1}
{z}} \right| + 1} \right)\left( {\left| {\left| {z + \frac{1}
{z}} \right|^2  - \left| {z + \frac{1}
{z}} \right|} \right| - 2} \right) \leqslant 0\mathop {}\limits^{}  \Rightarrow  \hfill \\
   \hfill \\
  \left| {z + \frac{1}
{z}} \right|^2  - \left| {z + \frac{1}
{z}} \right| - 2 \leqslant 0\mathop {}\limits^{}  \Rightarrow \mathop {}\limits^{} \left| {z + \frac{1}
{z}} \right| \leqslant 2 \hfill \\
\end{gathered}
\]

4. we have that if $z=r(cos\theta+isin\theta)$ , then $r=0 ,or,1$
if $r=0$ , then $z=0$. if $r=1$ , $cos3\theta=cos\theta$ and $sin3\theta=-sin\theta$ and the rest of the job is trivial. :D

[quote="10000th User"][quote="Diehard"]4. Since $|z|=1$, multiplying by $z$ gives $z^{4}=1$.[/quote]
How do you know that $|z|=1$ for problem #4? Did you read carefully before posting?[/quote]

It's pretty obvious...

$z^3=\overline{z}\implies |z^3|=|\overline{z}|=|z|\implies |z^2|=1\implies |z|=1$.

However, the one flaw is that we divided by $|z|$, so we have to consider $|z|=0$ as well.

How is this is even a debate? Obviously Diehard was wrong because that doesn't yield $ z = 0 $. He should have qualified his post at least. Then mcdonalds106_7 seems to be disagreeing with 1000th user by saying "It's pretty obvious" and then actually proving 10000th user right... The flaw you mention is exactly what 1000th user was referring to. It really doesn't need to be argued over.

[color=darkred][b][u]Extension[/u].[/b] Prove that $(\forall )n\in\mathbb N^*$ and $ (\forall ) z\in\mathbb C^*$ there is the implication $\left| z^{n}+\frac{1}{z^{n}}\right| \leq 2\implies \left|z+\frac{1}{z}\right| \leq 2$ .[/color]

See http://www.artofproblemsolving.com/blog/43631 and http://www.artofproblemsolving.com/blog/32174


How do I prevent TeXnic Center from resizing images? I tried[code]\includegraphics[scale=1.00]{imagename.png}[/code]and[code]\includegraphics[width=1.546cm]{imagename.png}[/code]None of them worked (1.546 cm is the width of the image).

\includegraphics{imagename.png} shouldn't resize at all, but if it does could you give a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]complete minimal example[/url] with a sample image that does resize? Maybe something else is affecting it.


Let P(x) be a polynomial with integer coefficients. Suppose that there are integers a, b, and c such that P(a)=0, P(b)=1, and P(c)=2. Prove that there is at most one integer n such that P(n)=4.

It too late to solve the whole thing, but i came across the interesting fact that $ P(a)\equal{}0, P(b)\equal{}1, P(c)\equal{}2$ implys $ a,b,c$ are in arithmetical order!


analytic geometry
calculus
integration
percent
LaTeX
SFFT
special factorizations

14. Find the sum of the x-cordinates of all possible positive integral solutions to $ \frac1{x} \plus{} \frac1y\equal{}\frac17$

26. After a price reduction of $ x$% an item has its price increased to its original value. What was the percent of increase? Express your answer as a common fraction in terms of $ x$.

[hide="Answers"]14. 78

26. $ \frac {100x}{100\minus{}x}$ [/hide]

number 26 is pairly easy (no offense)
the price is (100-x)/100 of the original price after the initial discount
the mark-up must make the item 100/(100-x) or 10000/(100-x) % of the discount but this is not the answer.
the actual percent raise is (10000/(100-x))-100 or (10000/(100-x))-100(100-x)/(100-x)
appying the distributive property and simplifying, you get 100x/(100-x)

[hide="hint for #14"]try factoring[/hide]

[hide="#14"]
$ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{1}{7}$
$ \frac{x\plus{}y}{xy}\equal{}\frac{1}{7}$
$ xy\equal{}7x\plus{}7y$
$ xy\minus{}7x\minus{}7y\equal{}0$

Here you can apply Simon's Favorite Factoring Trick:

$ xy\minus{}7x\minus{}7y\plus{}49\equal{}49$
$ (x\minus{}7)(y\minus{}7)\equal{}49$

49 has factors 1, 7 and 49 so x would be 8, 14 and 56.

$ 8\plus{}14\plus{}56\equal{}\boxed{78}$
[/hide]

I assume my solution for #26 is ok and correct? otherwise it'd be corrected probably...

[quote="infinity4ever"][hide="#14"]
$ \frac {1}{x} \plus{} \frac {1}{y} \equal{} \frac {1}{7}$
$ \frac {x \plus{} y}{xy} \equal{} \frac {1}{7}$
$ xy \equal{} 7x \plus{} 7y$
$ xy \minus{} 7x \minus{} 7y \equal{} 0$

Here you can apply Simon's Favorite Factoring Trick:

$ xy \minus{} 7x \minus{} 7y \plus{} 49 \equal{} 49$
$ (x \minus{} 7)(y \minus{} 7) \equal{} 49$

49 has factors 1, 7 and 49 so x would be 8, 14 and 56.

$ 8 \plus{} 14 \plus{} 56 \equal{} \boxed{78}$
[/hide][/quote]
Simon's Favorite Trick.. I can't think that far.. hehe..

@Stevenmeow: Your solution makes perfect sense. It was hard for me, at least, to visualize how to solve this.

Thanks to both of you.

[quote="abacadaea"][quote="stevenmeow"]
the actual percent raise is (10000/(100-x))-100 or (10000/(100-x))-100(100-x)/(100-x)
appying the distributive property and simplifying, you get 100x/(100-x)[/quote]
where did you get this last part. why cant you just leave it at 100/(100-x)? and where did that extra x come from. i would guess there is an easier way to do that last part.[/quote]
My guess is that he put (10000/(100-x))-100[b](100-x)/(100-x) [/b] into make the variables equal, like when you do it with the radical and rationalizing.

...and ended up with a different answer...

People are so lazy these days. Use the LaTeX resource on the left ~_-.

If you reduce an item by $ x%
$, you are multiplying it by $ (100-x)%=\frac{100-x}{100}
$. Let's assume the original price was $ 1$ dollar for the sake of simplicity.

To get back to the original price, we need to add $ 1-\frac{100-x}{100}$ dollars back. This is $ \frac{x}{100}$

To find the percent of increase with respect to the new price ($ \frac{100-x}{100}$), we take $ \frac{\frac{x}{100}}{\frac{100-x}{100}}*100=\frac{100x}{100-x}$
We multiplied by 100 for the percentage.

sorry for not using LaTeX i am not really familiar with it  :blush:  :blush:   :blush: ,but i think bomboyage had the right idea about what the fractions meant.

[quote="mathcrazed"]I wasn't specifically referring to you Mr. Shen. xD. Oh and nice location:[hide][hide="omg"][hide="ok last one"] Houston, Texas. [/hide][/hide][/hide][/quote]

dude stop picking on the 7th grader XD


In $ \triangle ABC$, D is on AC s.t. $ AD\equal{}BC$, $ BD\equal{}DC$, $ \angle DBC\equal{}2x$, $ \angle BAD\equal{}3x$. Find $ x$

[hide="Trigonometric solution"]Denote $ a: \equal{}BC\equal{}AD, p: \equal{}CD\equal{}BD$

From $ \triangle ABC$: $ {a\plus{}p\over\sin 5x}\equal{}{a\over\sin 3x}$

From $ \triangle BCD$: $ a\equal{}2p\cos 2x$

$ {2p\cos 2x\plus{}p\over\sin 5x}\equal{}{2p\cos 2x\over\sin 3x}$

$ 2\sin 3x\cos 2x\plus{}\sin 3x\equal{}2\sin 5x\cos 2x$

$ \sin 5x\plus{}\sin x\plus{}\sin 3x\equal{}\sin 7x\plus{}\sin 3x$

$ \sin x\equal{}\sin 7x\minus{}\sin 5x$

$ \sin x\equal{}2\sin x\cos 6x$

$ \cos 6x\equal{}{1\over 2}$

$ x\equal{}10^\circ$[/hide]

Of course, all of these problems can be bashed with the law of sines. Here is another trig solution, possibly with an idea towards a synthetic solution?

[hide]Extend BD to F and BC to G s.t. DF=AD and CG=CD. Then $ \triangle DCG$ is isoceles ($ \angle CDG \equal{} \angle CGD \equal{} x$), ABCF is an isoceles trapezoid ($ \angle DAF \equal{} \angle DFA \equal{} 2x$, $ \angle BAC \equal{} \angle BFC \equal{} 3x$), and $ \triangle FBD$ is isoceles ($ \angle BFG \equal{} \angle BGF$). We now can figure out all angles in quadrilateral DCGF in terms of x, and thus our dependency on the sides being equal is removed and we have a pure angle problem. By a known lemma for such quadrilaterals, we have
$ sin x sin 5x cos 2x sin 3x \equal{} sin3x sin 7x sin x cos 4x$
$ sin 5x cos 2x \equal{} sin 7x cos 4x$
$ sin 7x \plus{} sin 3x \equal{} sin 11x \plus{} sin 3x$
$ sin 11x \equal{} sin7x$
$ x \equal{} 10$.[/hide]

Perhaps this can be done in a simpler way, but this is what I came up with.

See the attached diagram. Notation: $ a: = BC = AD, p: = CD = BD$

Since $ \triangle BDC$ is isosceles, we have $ \angle DBC = 2x$ and $ \angle BDA = 4x$.

Let $ M$ be the point such that $ \angle MAD = \angle MDA = 2x$. Since $ \triangle MDA\cong\triangle DBC$ (by ASA), we have $ MA = MD = p$.

Let $ N$ be the point on $ AD$ such that $ ND=p$.

Since $ \triangle DBN$ is isosceles, we have $ \angle NBD = 90^\circ - 2x$

Since $ \triangle DBM$ is isosceles, we have $ \angle MBD = 90^\circ - x\implies\angle MBN = \angle MBD - \angle NBD = x$

Extend $ AB$ over $ B$ to the ray $ As$ and construct the point $ T$ such that $ BT\parallel AC$ and $ BT = p$. Since $ \angle sBC = 5x$ (external angle) and $ \angle sBT = \angle sAC = 3x$, we have $ \angle TBC = 2x$. Therefore $ \triangle TBC\cong\triangle MAD$ (by SAS), hence $ TC = p$ and $ \angle TCB = 2x$

Let $ CAUV$ be the trapezoid congruent to $ ACTB$ - ie. its symmetric image with $ A$ and $ C$ reflecting to each other. Since $ \angle VCA = \angle BAC = 3x$, it follows that $ CV$ bisects $ \angle TCB$, and so we have $ \angle TVC = 3x$ (external angle in $ \triangle BVC$). On the other hand, $ \angle UAC = \angle TCA = 4x\implies\angle UAB = x$.

By the construction, $ \triangle TVC\cong\triangle UBA$, hence $ UA = p$. But that in turn yields $ \triangle UAB\cong\triangle MAB$ (by SAS), hence $ \angle ABM = 3x$

Now from the angles summing up to the $ \angle ABs$, we get

$ 3x + x + 90^\circ - 2x + 2x + 2x + 3x = 180^\circ\iff 9x + 90^\circ = 180^\circ\iff x = 10^\circ$


The equation MATH = COU + NTS can be made true if each of the letters M, A, T, H, C, O, U, N, and S is replaced by a different digit and the equation is seen as the sum of two three-digit integers resulting in a four-digit integer. If C = 2, a solution can be found in which one of the three-digit integers is a multiple of 23. In that case, what four-digit integer does MATH represent?

We need a way to do this in the team round's alloted time while still leaving time for other problems and a method other than brute force and preferably not casework with trial and error.

[hide="I now have hostile relations with this problem"]I wasn't listening when the smart guy that helps out with my team was explaining this so I am going try to do this myself.

$c=2$ given
$m=1$ 1998 is the largest number made by adding up two three digits numbers
$n=8 or 7$ n needs to be big enough to add up two 1000+ but not big enough to use a digit in the hundred's digit that is already used.
$a=0$ one and two are already taken, 3 is unreachable
$o=9$ o+1+t=modular 10 t
COU is not a multiple of 23     299 is not possible
$m=7$ we carried to get t and a is o
look at possibilities for multiples of 23 between 700-799 inclusive
713 nope
736 possible
759 nope
782 nope

NTS-736
numbers used are 0,1,2,3,6,9
look at possibilities
294+736 nope
295+736 nope
298+736=1034 possible

$\displaystyle{\boxed{1034}}$[/hide]

[hide="another problem of the same kind"]

..MATH
[u]+EXAM[/u]
..OVER

(use digits 1-9 for A, E, H, M, O, R, T, X)

once determining the value for each letter, find the numeric value of

..REMOVE
+MARTHA             (as in Martha Stewart but I didn't have the right letters)[/hide]

fun fun fun

There are so many alphametics that we can see on Math Contests.

But I like this one from ARML

LARRY
[u]+ GIL[/u]
YUCCA

Wasn't there a problem like this on the 2nd or 3rd to last 2006 chapters?
I remember I wasted alot of time on it.

Yeah expect it was A LOT easier. That's why it's on a Chapters Sprint and not a Nationals Team.

[quote="Ignite168"]Yeah ex[b]p[/b]e[b]c[/b]t it was A LOT easier. That's why it's on a Chapters Sprint and not a Nationals Team.[/quote]

ex[b]c[/b]e[b]p[/b]t


algebra
function
domain
real analysis
real analysis unsolved

Prove that there exists a bijection that domain $ [0,1]$ and range into $ (0,1)$.

$ 0\mapsto\frac{1}{2}\mapsto\frac{1}{3}\mapsto\cdots$, $ 1\mapsto\frac{2}{3}\mapsto\frac{3}{4}\mapsto\cdots$ and $ x \mapsto x$ for all other $ x \in [0,1]$.

Let's denote your bijection $ f$. Then, what is $ f\left(\frac{1}{2}\mapsto\frac{1}{3}\mapsto\frac{1}{4}\cdots\right)$ and $ f\left(\frac{2}{3}\mapsto\frac{3}{4}\mapsto\frac{4}{5}\cdots\right)$? Please, tell me more precisely.

Obviously you misunderstood my notation. $ 0\mapsto\frac {1}{2}\mapsto\frac {1}{3}\mapsto\cdots$ means that $ f(0)\equal{}\frac{1}{2}, f \left (\frac{1}{2} \right) \equal{} \frac{1}{3}$ and so on. You can read about Hilbert's paradox of the Grand Hotel, it's roughly the same idea.


Calculate:
$A = \left(\begin{array}{ccccc}{cc...c}a_1&x&x&...&xx&a_2&x&...&xx&x&a_3&...&x\\cdotsx&x&x&...&a_n\end{array}\right)$
I'm beginner in Matrix Algebra so can you help me?

Sorry for that what are we to calculate, determinant or the invers of $A$? :?

[hide="Answer"]$ det(A) \equal{} \left[a_n \plus{} x(a_n \minus{} x) \sum_{i \equal{} 1}^{n\minus{}1} \frac{1}{a_i \minus{} x}\right] \cdot \prod_{i \equal{} 1}^{n\minus{}1}(a_i \minus{} x)$.

To see this, add to each column the symmetric of the last column, and then add to the last line each of the other n-1 lines multiplied by $ \minus{} \frac{x \minus{} a_n}{a_i \minus{} x}$, i = 1, 2, ..., n-1.[/hide]


In honour of Saint Valentine's Day $n$ loving couples of
chess-players held tournaments of boys and girls, i.e. each two boys
and each two girls played a single game. It is known that no game
ended in draw and each couple of lovers won $n-1$ games. Prove that
the number of triples of boys $(A,B,C)$ such that $A$ beat $B,$ $B$
beat $C$ and $C$ beat $A$ is equal to the number of triples of girls
$(D,E,F),$ such that $D$ beat $E,$ $E$ beat $F$ and $F$ beat $D.$

We say that a set of three people $A, B, C$ with $A$ beating $B$, $B$ beating $C$ and $C$ beating $A$ is an unusual set. Note that for any set of 3 people $S= \{ A, B, C \}$, either $S$ is an unusual set or $A, B, C$ can be labeled $x, y, z$ in some order so that $x$ beats $y, z$ and $y$ beats $z$ - this will be an usual set.

Let a hook be the event "some person $A$ beats two people $B, C$". Let $A$ be the center of such a hook.
Now count the number of unusual sets, among the boys. The crucial point is that an unusual set contains no hooks, while any other set contains exactly one hook. So the number of hooks is equal to the number of usual sets. If $A$ has won $a$ games, there are $\binom{a}{2}$ hooks with $A$ as its center. There are $\sum \binom{a}{2}$ hooks in total, where the sum is taken over the degrees of all players. To prove that the numbers of unusual sets among the boys and girls are equal, it is enough to prove that the numbers of usual sets are equal. Since $\sum a= \binom{n}{2}$ in both groups, this reduces to proving $\sum g_i^2=\sum b_i^2$, where the LHS is taken over the girls and the RHS over the boys. The RHS is actually $\sum (n-g_i-1)^2$ since each couple wins $n-1$ games.

Now, $(n-g_i-1)^2=n^2-2ng_i-2n+g_i^2+2g_i+1$. Since $\sum g_i=\binom{n}{2}$, the overall sum of this becomes $n^3-n^2(n-1)-2n^2+ \sum g_i^2+n(n-1)+n$. This is just $\sum g_i^2$, after cancelling.


$ x^{5}\equal{}243\equal{}243cis0 \\\\ x \equal{} 3cis\left ( \frac{2k\pi}{5} \right ), \,\, k\epsilon \jmath \\\\ For \,\, simplicity, \,\, take \,\, k\equal{}0,\pm 1,\pm 2 \\\\ \therefore The \,\, roots \,\, are \,\, x\equal{}3, \,\, 3cis\left ( \frac{2\pi}{5} \right ), \,\, 3cis\left ( \minus{}\frac{2\pi}{5} \right ), \,\, 3cis\left ( \frac{4\pi}{5} \right ), \,\, 3cis\left ( \minus{}\frac{4\pi}{5} \right )$

We have $ \left(\frac x3\right)^5\equal{}1$. If we let $ u\equal{}\frac x3$, we have that $ u\equal{}e^{2\pi i k/5}$ (the fifth roots of unity), so $ \boxed{x\equal{}3e^{2\pi ik/5},\ k\in\mathbb{Z}}$ (this cycles, giving 5 unique roots for $ k\equal{}0,\ldots,4$)

ohh... i see how its done now :)
thanks :)


Given that $ a,b,c \geq 0$, Prove that,
\[ (a \plus{} b \plus{} c)^3(ab \plus{} bc \plus{} ca) \leq (a \plus{} 2b)(b \plus{} 2c)(c \plus{} 2a)(a^2 \plus{} b^2 \plus{} c^2)\]
My proof is by expanding and using Am-Gm. Can you find a nice proof?  :)

Dear shaaam, long expanding is a method of proving inequality and no matter how ugly it looks. It represents an aspect of problem solving. I have an article about this and if you wish to see along, I will privately send it to you since it contains some materials that theirs due dates are still far away.

[quote="shaaam"]Given that $ a,b,c \geq 0$, Prove that,
\[ (a \plus{} b \plus{} c)^3(ab \plus{} bc \plus{} ca) \geq (a \plus{} 2b)(b \plus{} 2c)(c \plus{} 2a)(a^2 \plus{} b^2 \plus{} c^2)\]
My proof is by expanding and using Am-Gm. Can you find a nice proof?  :)[/quote]
The correct statement should be

\[ (a \plus{} b \plus{} c)^3(ab \plus{} bc \plus{} ca) \leq (a \plus{} 2b)(b \plus{} 2c)(c \plus{} 2a)(a^2 \plus{} b^2 \plus{} c^2)\]

oops,  :blush:

I edited it. Thanks, Guest. Problem is true. Please try to prove it now