Prove that one can place a circle of radius $S/P$ inside a convex quadrilateral with perimeter $P$ and area $S$.

let the quad be two triangles stuck together with sidelengths (a,b,p), (c,d,p) and areas x, y.

Suppose the triangle with lengths (a,b,p) admits a bigger inradius.  We use the $sr = \Delta$ formula.

We have (inradius of (a,b,p) ) = $\frac{2x}{a+b+p} \ge \frac{x}{a+b+p} + \frac{y}{c+d+p}$

where we just need to show

$\frac{x}{a+b+p} + \frac{y}{c+d+p} \ge \frac{x+y}{a+b+c+d}$

which we do by brute force.  We compare the coefficents of x from both sides, since the y terms are the same.

$x(c+d+p)(a+b+c+d) \ge x(a+b+p)(c+d+p)$

$\Leftrightarrow a+b+c+d \ge a+b+p$

$\Leftrightarrow c+d \ge p$  which is true by triangle inequality.

Thus, we can fit a circle with radius S/P in the triangle of sides (a,b,p).

Here is a continuation of this problem:

A convex polygon of area S(1) and perimeter P(1) lies inside a convex polygon area S(2) and perimeter P(1)

Prove 2.S(1)/P(1) >S(2)/P(2)

On each side, draw a rectangle of area $S/P$ (inwards) on that side. The sum of the areas of these rectangles is $S$. Obviously there is some overlap between these rectangles, so there must be some part of the quadrilateral which is not covered by any of the rectangles. Take the center of the circle with radius $S/P$ in this part and we're done.


function
ceiling function
algorithm
induction
logarithms
floor function
algebra unsolved

This problem was written on a blackboard at my universities physics department. Its driving me insane...Does anyone have any idea of how to solve it

Try to write n as a binary number and see how many "one"'s you can find in that expression.

[quote="bishnu"]F(0)=0
F(2n)=F(n)
F(2n+1)=F(n)+1
Find F(n)[/quote]

$F(1)=1$ so $F(2n+1)=F(2n)+F(1)$
So let's write some values and note that $F(2^n)=1$ and $F(2^n-1)$ is the maximum value in $[1;2^n-1]$;so $\forall$ $x \in \mathbb{N}$ $F(x)=F(\left\lceil{\frac{x}{2}\right\rceil)}$. Notice that it as a sort of geometric-logaritmic period and it has a sort of proportionality with the residues of n mod $2^i$, for each i in N such that $2^{i+1}>n$ We can notice that finding f(kx) basing upon f(x), with k big as we want, it 's the algorithm for the change of base from 10 to 2. Pratically F(n) is the sum of the digit of $n_2$

oH,i'm late, i just see crux solution, very nice :) Could you explain?

[quote="bishnu"]F(0)=0
F(2n)=F(n)
F(2n+1)=F(n)+1
Find F(n)[/quote]

Note that $2n$ and $n$ have the same number of $1$s in binary expansion, and $2n+1$ has one more $1$, and further that $0$ has no $1$s :)

In fact, when you write n as a binary number you have $n=\overline{n_1n_2...n_k}$ for some $n_i\in\{0,1\}$ and a natural number $k$. Given $n$ and with this notation we have that $2n=\overline{n_1n_2...n_k0}$ and $2n+1=\overline{n_1n_2...n_k1}$. So you can easily check that the function that gives the number of "1" in the binary expression of n is a good function. Actually it's the only one because $f(n)$ is defined by induction for all $n$. Your solution says that $f(n)$ is equal to the sum of digits of $n_2=\overline{n_1n_2...n_k}$ is obviously true, but it's difficult to generalize the problem with this solution. Now you can try to solve this problem :

Let $f: \mathbb{N}\rightarrow\mathbb{N}$ and $g: \mathbb{N}\rightarrow\mathbb{N}$ be two functions such that:
\[ f(0)=0,\\ f(2n)=f(n),\\ f(2n+1)=f(n)+1,  \] and
\[ g(0)=1,\\ g(2n)=g(n)+1,\\ g(2n+1)=g(n)  \]for all natural number $n$.
Find $(f+g)(n)$.

[hide]Answer : $\left\lceil \log_{2}{n} \right\rceil$[/hide]

Indeed - $g(n)$ gives the number of $0$s in the base-2 expansion, so $f(n) + g(n) = d_2(n)$ where $d_k(n)$ denotes the number of digits in the base $k$ expansion of $n$, which can be written $\lfloor \log_k n \rfloor$.

Well, listen that: obviously the F(x)-F(x-1) \in \{0;1\}, so  let 2 the couple (2;3) and i (i;i+1). So let's take the interval [0;2^n]. between each power of 2 and the other one we'll have the couple i in this way. Between 2^j and 2^{j-1} we'll have as many couples i as the binomial of the i-th coefficent in (1+1)^j says. in each quadruple of couples we have first i then i+1 with molteplicity 2 and then i+2, continue in this way until i is exausted, then continue the chart with i+1, i+2 with molteplicity 2, i+3...
so the logarithm in base 2 and the correspondant residue we have the function, i'lldo the graph later...i'm dead tired :D


[quote="bolpoin"]What is the maximum value of $ n$ such that $ 2n\plus{}1 | 100\plus{}n^{2}$[/quote]

$ \frac{n^{2}\plus{}100}{2n\plus{}1}\equal{}\frac{1}{4}(2n\minus{}1\plus{}\frac{401}{2n\plus{}1})$

401 is prime, so $ n \equal{} 1,200$. $ 200>1$, and plugging $ n \equal{} 200$ back in, it is indeed a solution.

[quote="Lazarus"][quote="bolpoin"]What is the maximum value of $ n$ such that $ 2n\plus{}1 | 100\plus{}n^{2}$[/quote]

$ \frac{n^{2}\plus{}100}{2n\plus{}1}\equal{}\frac{1}{4}(2n\minus{}1\plus{}\frac{401}{2n\plus{}1})$

401 is prime, so $ n \equal{} 1$.[/quote]

But $ n\equal{}200$ fit .... .


[quote="Astronum"]Prove that $1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^{n}}< 2$.[/quote]
[hide] $S=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^{n}}$
$2S=2+1+\frac{1}{2}+...+\frac{1}{2^{n-1}}$
Subtract: $S=2-\frac{1}{2^{n}}$
Since $\frac{1}{2^{n}}$ is always positive, $S=2-\frac{1}{2^{n}}<2$.

I think the formula solution is boring :P , but...
$S=\frac{1(1-\left(\frac{1}{2}\right)^{n+1})}{1-\frac{1}{2}}=\frac{1-\frac{1}{2^{n+1}}}{\frac{1}{2}}=2-\frac{1}{2^{n}}$

[hide="Rigourously"] $\frac{1}{2^{n}}>0$
Multiply by $(2^{n})^{2}$ which by the trivial inequality is $\ge0$
$2^{n}>0$
This must be true because there is no solution to $2^{n}=-a$, where $a$ is a nonnegative number, since that would require the logarithm of a nonpositive number.
(Anyone have a more concrete proof??) [/hide][/hide]

[hide="solution"]If n is a finite number, it is less than 2. If n is not a finite number, it is equal to 2.[/hide]

[hide]This is a finite geometric series.  If it were infinite, the sum of all the terms would be $\frac{1}{1-\frac{1}{2}}$ or $2$.  However, since there are only a finite number of POSITIVE terms (since for any positive integer $n$, $\frac{1}{2^{n}}>0$), we subtract a positive number (although it may be very small), making the sum less than 2.
QED[/hide]

[hide]if the sequence goes on infinitely, it equals two.  so, the equation is always smaller than two if it terminates :D [/hide]


i have a few math problems that i'm having trouble with, can anyone help me solve them
1. Rewrite the expression in terms of the first power of the cosine
sin^2xcos^4x
answer: 1/32(2+cos2x-2cos4x-cos6x)
2.Find the exact zeros of the function in the interval 0,2pie
g(x)=sin6x+sin2x
3. Verify the identity algebraically
sec2θ=(sec^2θ)/(2-sec^2θ)
4. Verify the identity algebraically
sin4β=4sinβcosβ(1-2sin^2β)

[hide="HINTS"]
Use the Pythagoreans $\sin^{2}x+\cos^{2}x = 1$, and the others that can be gotten by dividing by $\sin^{2}x$ or $\cos^{2}x$.

Use the sum of angles formulas $\cos \left( \alpha \pm \beta \right) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta$
$\sin \left( \alpha \pm \beta \right) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta$
$\tan \left( \alpha \pm \beta \right) = \frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}$

Other strategies, factor or multiply out.
[/hide]

Sure, if I interpreted them correctly.

$1)$ Rewrite in terms of first powers of cosine: $\sin^{2}x\cos^{4}x$

Answer: $\frac{1}{32}(2+\cos(2x)-2\cos(4x)-\cos(6x))$

$2)$ In the interval $[0, 2\pi]$, find the exact zeroes of
$g(x) = \sin(6x)+\sin(2x)$

$3)$ Verify the identity: $\sec(2\theta) = \frac{\sec^{2}\theta}{2-\sec^{2}\theta}$

$4)$ Verify the identity: $\sin(4\beta) = 4\sin\beta\cos\beta(1-2\sin^{2}\beta)$


conics ellipse geometry quadratics trigonometry ARML algebra

This problem is in the AoPS book volume II, i am quite confused.

A circle has the same center as an ellipse and passes through the foci $F_1$ and $F_2$ of the ellipse. The two curves intersect in $2$ points. Let $P$ be any point of intersection. If the major axis of the ellipse has length 15 and the area of triangle $PF_1F_2$ is 26, compute the distance between the foci.

[hide]When it says the two curves intersect, does it mean that the circle is tangent to the ellipse and the circle is inscrbed in the ellipse so that the whole graph looks like an eye?[/hide]

Denoting the focal points $E, F$ (rather that $F_1, F_2$). The triangle $\triangle PEF$ is right, because the angle $\angle P$ spans an arc on the diameter $EF$. Its area is $\triangle = \frac 1 2\ PE \cdot PF = \frac{cd}{2} = 26$. The ellipse is a locus of points with a constant sum of distances $c + d = PE + PF$ from 2 fixed points $E, F$, the ellipse focal points, equal to the ellipse major axis. Thus $PE + PF = c + d = 15$, These are 2 equations for 2 unknowns $c, d$:

$\frac{cd}{2} = 26,\ \ c + d = 15$

$c + \frac{52}{c} = 15,\ \ c^2 - 15c + 52 = 0$

${c = \frac{15 \pm \sqrt{225 - 208}}{2} = \frac{15 \pm \sqrt{17}}{2}}$

The same quadratic equation can be obtained for $d$, hence, $c$ is one root and $d$ the other. Using Pythagorean theorem for the right triangle $\triangle PEF$,

$EF = \sqrt{PE^2 + PF^2} = \sqrt{c^2 + d^2} =$

$= \frac 1 2 \sqrt{15^2 - 30 \sqrt{17} + 17 + 15^2 + 30 \sqrt{17} + 17} = \frac 1 2 \sqrt{484} = 11$

But doesn't the area of the triangle $F_1F_2P$ equal to $r^2$? Since the point of intersection is right above the circle?

[quote="yetti"]When I first looked at the problem, I wondered what are the tow points? (Hey, here is a geometry term I am not familiar with.) Under the given conditions, how could the circle passing through the focal points $F_1, F_2$ be tangent to the ellipse? Then the right angle triangle $\triangle PF_1F_2$ would be isosceles, $PF_1 + PF_2 = 2r \sqrt 2 = 15$ would be equal to the ellipse major axis and the triangle area would be $\frac 1 2 PF_1 \cdot PF_2  = \frac 1 2 \left(r \sqrt 2\right)^2 = r^2 = 26$. From the 1st equation, $r = \frac{15 \sqrt 2}{4} \doteq 5.3033$ and from the 2nd equation $r = \sqrt{26} \doteq 5.0990$. This is clearly impossible.[/quote]

This exactly what i meant. This problem is from AoPS volume II. I don't know whether we made mistakes in our reasoning.  :?

The ellipse and the circle passing through the focal points [b]intersect[/b] at four points. If they were tangent, the problem would state that they [b]touch[/b] at two points. But if the ellipse and the circle touched, the problem would be over-constrained and consequently, it would not have a solution.

These problems are usually rigged to yield a nice result. When four intersection points are accepted, the distance between the focal points comes out as an integer, which is another indicator. The author of the problem either considered just the upper half of the plane, or made a mistake to say that the [b]two[/b] curves intersect at [b]two[/b] points (an easy mistake to make), or the problem contains a misprint ([b]2[/b] instead or [b]4[/b]).

I am still REALLY confused. I drew two possible graphes of this problem. Which one is it?  :?

[quote="yetti"][quote="henryyangrui"]A circle has the same center as an ellipse ...[/quote][/quote]
Did you use geometer sketchpad?
I don't how to draw an ellipse using a GSP.  :?
[hide]
[quote]The two curves intersect in $2$ points.[/quote] Your graph has 4 intersections.[/hide]

[quote="henryyangrui"]
Did you use geometer sketchpad?
I don't how to draw an ellipse using a GSP.  :?[/quote]
Yes, I did.

[hide]
[quote="henryyangrui"][quote="henryyangrui"]The two curves intersect in $2$ points.[/quote] Your graph has 4 intersections.[/quote]
I give up. [/hide]

Omg, i am sure there is no typo. Can someone please help us?  :(

OK, use the picture that looks like an eye. I think that's right. We know $PF_1+PF_2 = 15$. But since the tangent point is directly between the foci, $PF_1=PF_2=\frac{15}{2}$. Then $[PF_1F_2] = \frac{1}{2}(PF_1)(PF_2)\sin{\theta} = \frac{225}{8}\sin{\theta} = 26$, where $\theta$ is the angle between $PF_1$ and $PF_2$. So $\sin{\theta} = \frac{208}{225}$... from here it gets very ugly, which is uncharacteristic of an ARML problem, especially if it's in the Individual Round. But you can solve for $\cos{\theta}$ or $\sin{\frac{\theta}{2}}$ to get $F_1F_2$ I think.

EDIT: but wait... since $P$ is on the circle, $\angle F_1PF_2 = 90^{\circ}$... I think there is something wrong with the problem...

Yetti is right, there was a typo in the book, the two curves intersect in $4$ points.

If it's in AoPS Vol II, why is in the Olympiad Geometry section of the forum?


logarithms
function
limit
algebra
functional equation

Prove or disprove (finding all other families of solutions) that the only solutions $f$ to the functional equation: $2 f(x) = f(x^{2}) \ \ \forall x >0$ are $f(x)=k$ and $f(x)=\log_{a}(x)$ for some real $k$ or $a$.

[i](Followups: Same question, kinda:

[b](1)[/b] $\ \ t \cdot f(x) = f( x^{t})$ for a specific, real $t$.
[b](2)[/b] $\ \ t \cdot f(x) = f( x^{t})$ for all real $t$ at the same time.) [/i]

For the original problem, we can let $S$ be the set of positive real numbers of the form $2^{2^{n}}$ where $n$ is any integer, and then the function
\[f(x)=\left\{ \begin{array}{rl}\log_{2}x,&x\not\in S\\ 0,&x\in S \end{array}\right.\]
is a counterexample. Many other counterexamples to this problem and Follow-up 1 hold.

[quote="scorpius119"]For the original problem, we can let $S$ be the set of positive real numbers of the form $2^{2^{n}}$ where $n$ is any integer, and then the function
\[f(x)=\left\{ \begin{array}{rl}\log_{2}x,&x\not\in S\\ 0,&x\in S \end{array}\right. \]
is a counterexample. Many other counterexamples to this problem and Follow-up 1 hold.[/quote]

Alright.  :)  What if the problem statement were changed to "Prove or disprove (finding all other families of solutions) that the only [i]continuous[/i] solutions $f$ to the functional equation..."?

[hide="for the continuous case..."]
(b) is the most general version. If $t=1$, then [i]anything continuous[/i] works, and if $t=-1$, then the condition becomes $f(1/x)=-f(x)$, which is attained by anything of the form
\[f(x)=g(x)-g(1/x)\]
where $g$ is a continuous function. (These are the only cases that work. If $f(1/x)=-f(x)$, then let $g(x)=\frac{f(x)}{2}$ and $f(x)=g(x)-g(1/x)$.)

Other than these exceptions, $f(x)$ must equal $C\log x$ for some constant $C$. To prove this, let $h$ be the continuous function defined by
\[h(x)=\left\{ \begin{array}{rl}\frac{f(x)}{\log x},&x\neq 1\\ \lim_{h\rightarrow 1}\frac{f(h)}{\log h},&x=1\\ \end{array}\right.\]
The functional equation becomes $h(x)=h(x^{t})$. In particular, for any positive real number $x$, we have
\[h(x)=h(x^{t})=h(x^{t^{2}})=\cdots =h(x^{t^{n}})=\cdots\]
if $|t|<1$ (if $|t|>1$ use powers of $1/t$ instead of $t$).

Since the sequence $x^{t^{n}}$ (or $x^{1/t^{n}}$ if $|t|>1$) converges to 1 for all positive real numbers $r$, we obtain $h(x)=h(1)$ and $f(x)=h(1)\log x$ for all positive $x$.
[/hide]


calculus
Stanford
college
ARML
function
modular arithmetic
Putnam

I am thinking about starting a seminar in advanced mathematical analysis over the summer. We will begin with [i]Fourier Analysis: An Introduction[/i] by Elias Stein and Rami Shakarchi. Not much experience in advanced mathematics will be needed: if you have a knowledge of calculus at the advanced high school level and skim through Charles Pugh's [i]Real Mathematical Analysis[/i], you should be able to understand. If all goes well, it will meet at Stanford, five days a week (Monday through Friday) from 9AM to 4PM with a lunch break. As a result, we should have plenty of time to go through a large amount of material very rigorously. Post here if you're interested.

If I have time (i.e. it falls during a week when I'm home), then it sounds interesting. I'm really disappointed that we normally don't have any math events at Stanford, and I have to go to Berkely for everything, which I don't do because it's too far away. I guess there's ARML practices, but I have orchestra rehearsal right at that time and they're verrrryyy strict about attendance.

What is Fourier Analysis anyway?

I can probably arrange it so that it starts after MOP ends. That's probably better for me anyway since it gives me more time to prepare it and work out the logistics. At any rate, when will you be home? That information would be useful for me to know.

In Fourier analysis, we write functions as an infinite sum of trigonometric functions, just like in Taylor series we write functions as power series. There is a lot of fascinating theory relating to Fourier analysis, and hopefully we will be able to get to the proof of Dirichlet's Theorem (if $a$ and $n$ are relatively prime, then there are infinitely many primes congruent to $a\pmod n$).

Incidentally, Stanford has Putnam seminars in the fall once a week in the evenings, starting around 5 or 6. High school students are allowed to attend. (I attended in my junior and senior years of high school.) Ravi Vakil (a 4-time Putnam fellow and 3-time IMO gold medalist, if I recall correctly) runs these practices.

Which orchestra are you in?

My complete summer plans aren't quite firmed up right now, but it's looking like the week right after MOSP (the week of June 3) would be the best time for me. I'm assuming the seminars would only last one week. Otherwise, I might be going on a trip to China in the two weeks following that, so that week is probably my only free one. Hopefully I won't be too braindead from MOSP!

Fourier analysis sounds interesting. I'm in BC Calculus right now, and it's quite easy, so I think I should be prepared for it. Do you recommend that I "skim" that book that you mentioned?

Who else are you expecting at the seminars? Other Stanford students? It looks like nobody else has seen this post, so I might be your only recruit from AOPS!

I'm interested in Stanford's Putnam lectures, so if you had any more information (or updates later in the year), that would be great!

I'm co-principal flutist of the California Youth Symphony. Have you heard of them?

It would probably be best if we talked by email after this, seeing that there isn't much point on having more conversation on a public message board. My email is fatansn@aol.com.

i think i might do it if i can somehow coordinate it with the stanford summer program (the summer school). As far as content what kind of appproach will we be taking, mostly theoretical or applied or a mix?

It ended up not happening last year, but I'd be happy to consider it again for next year. It would be purely theoretical and would probably focus on analysis. (I thought of Fourier analysis for last year mainly because I wanted an excuse to learn more about it myself, but I'm certainly up for other options.)

im actually going to be a freshmen there next year, so if i know more people i figure the easier the transition will be.

No, I'm not actually at Stanford. But if you would encourage Stanford students to participate in the seminar, then you would get to know more Stanford students, and there would be a greater chance of it actually happening (and they might be able to find us a room). So we would all win in that case.


...yes, we are talking about football ( soccer to non-Europeans  :lol: )


Any one know what is HM?

I heard that AM >= GM >= HM.. But what is HM?  Can anyone include an example?

The harmonic mean $H_n$ of $n$ numbers $a_1,a_2,\ldots,a_n$, where $a_i\not=0$ for each $i$, is defined to be
\[ H_n=n\left(\sum^n_{i=1} \frac{1}{a_i}\right)^{-1}  \]
For example, when $n=3$, we get
\[ H_3=\frac{3}{\frac 1a+\frac 1b+\frac 1c}  \]

Even more generally, the mean of order $p$ is

$m_p(a_1,\dots,a_n)=\left(\frac1n\sum_{k=1}^na_k^p\right)^{\frac1p}$

This doesn't make sense for $p=0,$ but we can find (if we take the appropriate limits) that

$\lim_{p\to0}m_p(a_1,\dots,a_n)=\left(\prod_{k=1}^na_k\right)^{\frac1n}$

  -- in other words, the geometric mean. It then makes sense to define $m_0$ to be the geometric mean. It also makes sense (more limits) to define $m_{\infty}$ to be the maximum of these numbers and to define $m_{-\infty}$ to be the minimum of those numbers.

On this scale, the $p=1$ case is the arithmetic mean and the $p=-1$ case is the harmonic mean.

One of the theorems you can prove: for fixed positive numbers $a_1,\dots a_n,$ $m_p$ is an increasing function of $p,$ strictly increasing unless all the $a_k$'s are equal. That includes the observation that the harmonic mean ($p=-1$) is less than the geometric mean ($p=0$) is in turn less than the arithmetic mean ($p=1$).

Here's an exercise. Draw a trapezoid (WLOG, an isosceles trapezoid) with bases $a$ and $b.$
The length of any segment across the trapezoid parallel to the bases is some kind of mean of $a$ and $b.$ Identify which mean each of the following is:

(1) The segement that bisects the sides of the trapezoid.

(2) The segment that divides the trapezoid into two smaller trapezoids that are similar to each other.

(3) The segment that bisects the area.

(4) The segment that passes throught the point of intersection of the diagonals of the original trapezoid.

[quote="Kent Merryfield"]  -- in other words, the [b]harmonic[/b] mean. It then makes sense to define $m_0$ to be the [b]harmonic[/b] mean.[/quote]

Don't believe everything he says :D The bold words should read "geometric".

D'oh! Typed too fast, proofread badly. Fixed now.

[quote="Kent Merryfield"]
Here's an exercise. Draw a trapezoid (WLOG, an isosceles trapezoid) with bases $a$ and $b.$
The length of any segment across the trapezoid parallel to the bases is some kind of mean of $a$ and $b.$ Identify which mean each of the following is:

(1) The segement that bisects the sides of the trapezoid.

(2) The segment that divides the trapezoid into two smaller trapezoids that are similar to each other.

(3) The segment that bisects the area.

(4) The segment that passes throught the point of intersection of the diagonals of the original trapezoid.[/quote]

To check if your answers are right :) : http://www.mathlinks.ro/Forum/viewtopic.php?p=433528#p433528


\includegraphics*[viewport=100 200 500 650]{myimage}

What coordinates do those numbers represent? Is the bottom left corner (100, 200) and the upper right (500, 650)?

Edit: Never mind, I understand now

Edit: I just changed the coordinates from 100 0 460 250 to
120 0 480 250, and my image moved to the left. I guess I don't understand.

viewport gives a new bounding box whose origin is relative to the bounding box of the image. Imagine a frame round the picture which only shows that picture inside the frame. So if the viewport moves right the picture appears to move left.


hi,
Find all series $ (x_{n})_{n \geq 1}$ with $ x_{1}\equal{}\sqrt{3}$ and
\[ x_{n} \minus{} \frac{1}{x_{n}} \equal{} 2\frac{x_{n\minus{}1} \plus{} 1}{x_{n\minus{}1} \minus{} 1},\]
$ \forall n \in \mathbb{N}, n \geq 2$. :P


show that z^a = exp(a log(z)) is analitic function on complex with the exclusion of the non-positive axis real

why isn't z^a=e^alnz true for all values of z except 0?

You do have to worry about multiple logs but if you pick a suitable defintion such as -pi<imaginary part=<pi


A regular polyhedron with [i]f [/i]faces, [i]v[/i] vertices and [i] e[/i] edges is made by cutting a regular tetrahedron with an edge length of one unit from each of the four corners of a tetrahedron with an edge length of three units. Compute [i]f[/i] +[i]v[/i]+[i]e[/i]

Basically, the cuts add 4 faces, triple the number of vertices, and add 12 edges.  A tetrahedron has 4 faces, 6 edges, and 4 vertices to start with.  So the resulting polyhedron has 8 faces, 12 vertices, and 18 edges.  8+12+18=38.


function
topology
calculus
calculus computations

[b]Problem:[/b] Show that any mapping of a discrete metric space, i.e. a space with metric given by
\[ d(x,y)=\begin{cases}1 \quad &\text{if}\quad x\neq y \\ 0\quad &\text{if} \quad x=y \end{cases}, \]
into any other metric space is continuous.

[b]My work:[/b] Ok, so I consider functions $f: X\to Y$ where the metric on $Y$ is $d'$. Then I use the result which states that a mapping $f$ is continuous at $x_0$ iff $x_n\to x_0$ implies that $f(x_n)\to f(x_0)$.

So to show continuity at the point $x\in X$, let $(x_n)$ be a sequence such that $x_n\to x$. But in the discrete metric this means that $x_n=x$ for all $n\ge N\in \mathbb{N}^+$, say. Then
\[ d'(f(x),f(x_n))=d'(f(x),f(x))=0 \quad \text{for }n\ge N, \]
which show us that $f(x_n)\to f(x)$ as $n\to \infty$.

Am I correct?

[quote="fredbel6"]I like the fact that you always write what you have found on your own already, but here you are looking too far

consider $x_0$ and $f(x_0)$
if you get a $\delta$, just choose $\epsilon=1$

if $x\in B(x_0,1)$ , then $x=x_0$ and thus $f(x)=f(x_0)$[/quote]

First of, thank you very much for the feedback!

You wrote ''if you get a $\delta$, just choose $\varepsilon =1$''; but shouldn't it be the other way around (following the general use of $\varepsilon$ and $\delta$), i.e. that I'm given an $\varepsilon >0$ and am to show the function values are within this distance if only the values are within $\delta$ from each other. Of course I see what you mean, and yes, I did look somewhat too far, but it wasn't much work anyway. So to have $d(f(x),f(y))<\varepsilon$ we need only choose $\delta <1$, for then $d(f(x),f(y))=0$.

Thanks.

yes usually it is the other way around
you give a $\epsilon$ as distance from the value

my mistake :roll:


Find the positive constant $ c_{0}$ such that the series $ \sum_{n=0}^{\infty}\frac{n!}{(cn)^{n}}$ converges for $ c > c_{0}$ and diverges for $ 0 < c < c_{0}$

This question is question #9 located here:

http://web.mit.edu/hmmt/www/datafiles/problems/2004/pcalc04.pdf

The answer is located here:

http://web.mit.edu/hmmt/www/datafiles/solutions/2004/scalc04.pdf


greatest common divisor
geometry
3D geometry

What's the problem 4/80 ??  :ninja:
What is talking about ?

It could be referring to a problem that is in a textbook, like chapter 4, problem 80.

[quote="Raindiamond51526"]Can anyone tell me how to figure out the problem 4/80?? I don't know how to figure it out. Help would be greatly appreciated.

Stacey[/quote]

I'm confused do you mean like what is 4 divided by 80. (that's 1/20 which is .05?)

you find that $2^2$ is the GCF of 4 and 80, so you divide both sides by that, and you get $\frac{1}{20}=0.05$

[hide]If you're talking about this:

$\frac{4}{80}=\frac{1}{20}=0,05$[/hide]

AoPS book?

chapter 4, problem 80:
If x varies as the cube of y, and y varies as the fifth root of z, then x varies as the nth power of z. What is n?

I think I understand.
[hide]15
x = y^3
y = z^5
x = z^n
x = z^5^3
x = z^15[/hide]

[quote="pohoatza"]lol Jose...I think he meant something else :)[/quote]

Well, if he did not specify, I could not know..

[quote="Raindiamond51526"]Can anyone tell me how to figure out the problem 4/80?? I don't know how to figure it out. Help would be greatly appreciated.

Stacey[/quote]

dear, dont u want to come and stop, confusing us? ;)

Well, the [probable] question has been posted. Anyone else want to try the problem instead of posting useless messages?

i hope u dont mean me ;) .

becasue he himself should told us the original problem, not the probable problem.


probability
percent
blocked by some stupid people

A bag contains 12 red marbles and 6 blue marbles. Two marbles are selected at random and without replacement. What is the probability that one marble is red and one is blue? Express your answer as a common fraction.

We have two cases:

[b]Case 1:[/b] The first marble is red.

[b]Case 2:[/b] The first marble is blue.

Let's look at Case 1 first. If the first marble is red, then there is $ \frac{3}{4}$ chance of picking it, and then a $ \frac{6}{17}$ chance of picking the second blue one. We multiply these two fractions together to get $ \frac{9}{34}$.

Now look at Case 2. If the first marble is blue, then there is a $ \frac{1}{3}$ chance of picking it, and then a $ \frac{12}{17}$ chance of picking the second red one. Multiplied, this is $ \frac{4}{17}$.

We add these two fractions together.

$ \frac{9}{34} \plus{} \frac{4}{17} \equal{} \boxed{\frac{1}{2}}$.

Someone please correct me if I'm wrong.

[color=blue]* AIME15 agrees with isabella2296 AND ernie[/color]

[color=blue]* AIME15 wasted a post[/color]

[color=blue]* isabella2296 asks that AIME15 stop spamming, please.[/color]

how did you get 6/17 in the first sentence?
sorry my brain is turned off =[

Since we already removed one marble and didn't replace it, there's only 17 marbles left instead of 18.

[quote="myyellowducky82"]also why did you multiply instead of adding?[/quote]

I am not really sure how to say this.

[hide="The Long Way"]You multiply if you want both scenarios to happen.

For example, if you flip a coin two times, the chances of getting a head on the first flip is $ \frac {1}{2}$. (unless the coin is weighted, but it isn't.)  The chances of getting a head on the second flip is also, obviously $ \frac {1}{2}$.  However, if your coin lands on a tail on the first flip, there is zero percent chance of flipping two heads in two flips.  Thus, the probablilty should be $ \frac {1}{2}$ of the second probablility.  To find what the new probability is, you would have to multiply the two fractions together.[/hide]

[hide="The Short Way"]What are the chances of getting a head on a flip?  [hide="The Answer"]$ \frac {1}{2}$, unless you are dealing with weighted coins.[/hide]

There are 4 outcomes out of two flips.

HH
HT
TH
TT

This clearly shows that getting 2 heads in 2 flips has $ \frac {1}{4}$ chance.

So, if you added the fractions, what would the supposed probablilty be?  If you added you would get one, which is clearly not equal to $ \frac {1}{4}$.

But if you multiplied, you would get $ \frac {1}{4}$.  Thus, you would have to multiply.[/hide]

Thank ernie, not me, my explanation sucked.  :roll:

Nice one, though ernie!

But ernie your "long way" isn't THAT long...and izzy i like your explanation too!

@ernie, I think your long reason and short reason are just about the same length.

@ernie: i think your long explanation makes more sense =]

You're welcome! :thumbup:

And really... I should edit the names...

Unless you have anymore questions, please let this topic die. Thank you.  :)

[quote="myyellowducky82"]what names....[/quote]

"The Long Way"

"The Short Way"

[quote="isabella2296"]If we have a probability of A, and a probability of B, and we're looking for the probability of A and B, we multiply them.

(Did that make sense?  :huh:  It's explained better in Intro to C and P.)[/quote]
the probabilities are fractions right?

Yes.

Unless you have something like 1 or 0.

It's just not a fraction. You can see it works:

For example, say we were finding the probability that if we reached into a bag of only blue marbles, we would draw a blue marble and then a red marble. The probability of drawing a blue marble is 1, and the probability of drawing a red marble is 0. To find the probability of drawing both, we multiply $ 1 \times 0 \equal{} 0$, so there would be a 0 chance of drawing a red marble and a blue marble.


There is a car odometer with 6 digits.
1) to start, the last four numbers of the odometer are palindromic ( like race car)
2) 1 mile later, the last 5 numbers are palindromic.
3) 1 mile after that, the middle 4 numbers are palindromic.
4) 1 mile after that, all 6 numbers are palindromic.
So... what is the starting number, and what is the ending number?

[hide="hint"]
The last number has to be 8 or 9 (why?)
[/hide]
[hide="solution"]
198888, 198891

OR

199999, 200002
[/hide]

The second given solution definitely works.

19[b]9999[/b]
2[b]00000[/b]
2[b]0000[/b]1
[b]200002[/b]

[quote="Jean-Louis"]How do you arrive to the solution? is there an algebraic method?[/quote]

No, not that I know of. I used the fact that the last digit has to be 8 or 9 and intuition to find the answers. There might be more that I missed, though.


Find all reals x,y,z that satisfay
1/xy=x/y+1
1/yz=y/z+1
1/xz=z/x+1
I found that the condition can be rewroten as follows
x(x+y)=1
y(y+z)=1
z(z+x)=1
and i think we have to make a substitution , put for exemple a=x+y, b=y+z, c=z+x
and deduce
does anybody have an idea?

hm... it is known to me - this problem appeared on Czech-Poland Olympiad and was discussed already on the forum - try to search for it

[edit] I overlooked one thing - I meant this topic http://www.mathlinks.ro/Forum/viewtopic.php?p=174072&highlight=#174072 but the problem is a bit different

:) thank you very much, there is almost no diffrence


By how long, I mean from when you first started studying for Mathcounts specifically.

[quote="nat mc"]Yeah since beginning/middle of 6th grade.[/quote]

Me too.

I've been studying since the beginning of 5th grade (I was in a 6th grade math class) and I haven't been to Nats yet. So 2.5 years.

I've been studying since 6th grade, but I only studied seriously after I got 3rd in state this year (that was when I found out about AoPS).

This is the first year I have ever done Mathcounts.

I moved from California to Nebraska April of 2004, too late for Mathcounts. In California, at the school I was going to ( in Davis ), either they didnt offer Mathcounts or I didn't hear about it.

It was here that I learned about it, and for a first year, 49th at Nationals isn't too bad. :P

Hey Dnas, I scanned your website and noticed that you're missing a rather well-known proof. I'm too lazy to submit it, so I'll just post it here (Courtesy of Alex Vasilyev of MathPath) :

Postulates: time=money
                   money is the root of all evil

1. Girls take time and money
    Girls=time*money
2. Time=money
    By substitution, Girls=money^2
3. As the famous saying goes, "money is the root of all evil"
    money=sqrt(evil)
4. Girls=sqrt(evil)^2

Therefore, girls=evil

For all the girls on this site, I did not write/endorse this proof. Otherwise...

[quote="Phelpedo"]Your avatar is...never mind.

Is this also your last year?[/quote]

Yeah, the Avvie is awesome.

I'm in 8th grade, so yes.

No more than 18 months; my first year was 7th grade. I wish I had known about it in 6th grade.

My first year was in 7th grade, but I've really only studied for it since the beginning of 8th grade, 6 mos-1 year.


Uhh ya, I need to know how to space twice so I can begin writing my equations in another section of my page, I know that \\ gets me one space, but how do you do two or more?

-thanks in advance

Use \linebreak

blah blah first line\\
\linebreak
hey this is the next line

...great suggestion... LOL  :P

Use
[code]\renewcommand{\baselinestretch}{2.0}[/code]
before
[code]\begin{document}[/code]

I do not remember where I found it.


I'm trying to figure out what I'm going to take for math next year.  I can either take stats or take classes (multivariable, linear algebra, ...)at a local college.  From what I've read here, it doesn't really sound like stats is something I want to take, so I'm leaning towards the college courses.  But my concern is that the college class will probablly not be that great and I will most likely end up having to take the classes again once I am done with high school.  Is it a bad idea to go with the classes at the college then?  I figure that it's better that I at least get exposed to these materials, then if I have to take them again I'll have a better understanding the 2nd time through.

Any advice as to what I should do?

Would you be taking them at a state school, a private school, or a community college?

private, but for math it is not much (if any) better than a normal state school.

You're going to have to take stuff again, no matter what.  It'll be easier and more fun if you've already done it once.  I've already re-taken 2 classes I took and will be re-taking another one next year.  Let me tell you, doing differential equations theoretically is [i]so[/i] much easier when you've seen them done in the "Well, if we examine this equation carefully, we note that A*cos(x) looks like a possible solution.  And hey, it works!  And look, so does B*sin(x)!  So all the solutions look like a*cos(x) + b*sin(x)."  way.

I agree with JBL. If you end up having to take the class again, it may be a class with the same name but it won't be the same material. I wouldn't worry about that.

Don't you get college credit for college classes, i.e. aren't credits transferable from one place to another?

It depends on the college. My college accepts all sorts of things (AP scores, classes at other colleges, probably some other stuff), but some places don't. I have a friend who graduated from high school a year early and went to a state college for two years. Now he is trying to transfer to Berkeley, but Berkeley wants him to repeat many of the classes he took at the state college.

Don't take stats (unless for some reason it interests you a lot).  There aren't many (if any) good textbooks, and it's sort of just plugging in stuff and such.  It's a good subject, but you should probably take other courses first to understand the math part.  Although it can be very useful in engineering and other stuff, if you enjoy it, it won't be in the same way as other math.

Then again, this is just my opinion, so...

thanks for advice.  I think i'll take the college classes.

Yea, if it is the same quality as a state school they will most likely transfer.  You have to get an A/B in the class though I think.


inequalities
function
algorithm
inequalities open

Recently I've read that Chebyshev can be generalized for n sequences and I'm sure that the maximum of the function \[f(a,b,\ldots,z)=a_{1}b_{1}\cdots z_{1}+\cdots+a_{n}b_{n}\cdots z_{n}\] is obtained when all the sequences are increasing (or decreasing).

However, a problem occurs when we try to find the minima. Ordinary rearrangement tells us that the products of the sequences should make as much monotonic sequences as possible when the minima occurs (this [b]should,[/b] but may not be true). Can we find a short algorithm which will tell us the order of the elements in the sequences such that the above function obtains its minima?

bodan, you named the sequences $a,b, \cdots ,z$. So, are there $n$ or $26$ sequences?

if it is mentioned that all numbers in the sequence are positive, we can apply rearrangement inequality to n sequences


integration
LaTeX
trigonometry
calculus
calculus computations

Plz evaluate this:
[hide]
$ \int \frac{d x}{cos x \plus{} \sqrt{cos 2 x}}$[/hide]

ThX in advance.

I'd try substituting $ u^2\equal{}cos(2x)$

HINT
[hide]
Multiply and divide by its conjugate
[/hide]

You guys have stopped positng any replies, I still didn't get how to solve this! I thought maybe I'd bumb with my piece of work, so here it is (hope it helps):
[hide]$ \int \frac{dx}{cos x \plus{} \sqrt{cos 2 x}}
\equal{} \int \frac{cos x \minus{} \sqrt{cos 2 x}}{(cos x \plus{} \sqrt{cos 2 x})(cos x \minus{} \sqrt{cos 2 x})}dx
\equal{}\int \frac{cos x \minus{} \sqrt{cos 2 x}}{cos^2 x \minus{} cos 2 x} dx
\equal{}\int \frac{cos x \minus{} \sqrt{cos 2 x}}{cos^2x \minus{} (2cos^2x\minus{}1)} dx
\equal{}\int \frac{cos x \minus{} \sqrt{cos 2 x}}{1 \minus{} cos^2 x} dx
\equal{}\int \frac{cos x \minus{} \sqrt{cos 2 x}}{sin^2 x \plus{} cos^2x \minus{} cos^2 x} dx
\equal{}\int \frac{cos x \minus{} \sqrt{cos 2 x}}{sin^2x} dx
\equal{}\int \frac{cos x}{sin x}dx \minus{} \int \frac{\sqrt{cos2x}}{sin^2x}dx$
$ \equal{}\int cot x cosec x dx$ $ \minus{} \int \frac{\sqrt{cos 2 x}}{sin^2x} dx
\equal{} \minus{} cosec x \minus{} \int \frac{\sqrt{cos2x}}{sin^2x}dx
\equal{} \minus{}cosec x \minus{} I$

To evaluate$ I \equal{} \int\frac{\sqrt{cos2x}}{sin^2x}dx
\equal{}\int cosec^2 x \sqrt{cos2x} dx
\equal{}\int cosec^2x \sqrt{\frac{1\plus{}tan^2x}{1\minus{}tan^2x}}dx$

$ Let, cot x \equal{} t$
$ cosec^2x dx \equal{} dt$

Therefore, $ I \equal{} \int \sqrt{\frac{\frac{1 \plus{} tan^2x}{tan^2x}}{\frac{1 \minus{} tan^2x}{tan^2x}}}cosec^2x  dx
\equal{}\int\sqrt{\frac{cot^2x \plus{} 1}{cot^2x \minus{} 1}} cosec^2 x dx
\equal{}\int\sqrt{\frac{t^2 \plus{} 1}{t^2 \minus{} 1}} dt$[/hide]
I'm stuck there!
Please Help!  :(

A $ \text{\LaTeX}$ note: \sin and the like look better than without the backslashes.

I'm pretty sure this is an [url=http://en.wikipedia.org/wiki/Elliptic_integral]elliptic integral[/url], and can't be done with elementary functions.

Ty jmerry for Latex tip.

I guess this question involves elliptic integral, so probably faulty question.

I have posted another question, hope you guys can figure that out.


Consider a solution x(t) to x'=x^A with x(0) = 1, where A is a real number.  Show that x(t) -> infinity for an A.  Which values of A will x(t) go to infinity in finite time?

Hint: x(t) goes to infinity in finite time if there is an ρ>0 with lim t->ρ (x(t)) = infinity.

I can't quite tell but is this a differential equation?  Is it $ x'\equal{}x^{a}$?  Then use separation of variables to get
$ \frac{x^{1\minus{}a}}{1\minus{}a}\equal{}t\plus{}c\Rightarrow x^{1\minus{}a}\equal{}t(1\minus{}a)\plus{}C$,
where $ C$ is just a constant.  Then
$ x(t)\equal{}\sqrt[1\minus{}a]{t(1\minus{}a)\plus{}C}$

so then how do you show that it approachs infinity?


Let  $a,b\geq 0$ . Find the best constant  $M$ such that

$ab(a^{n-2}+b^{n-2})\leq M(a+b)^n$  where n is natural $n>2$ .

**I solved it for n=6

By the best constant M, do you mean s.t. $Mab(a^{n-2}+b^{n-2}) \le (a+b)^n$?
I'm tempted to say $2^{n-1}$, saying that AM-GM overrides Cauchy, but I know that's not the case...

I think it is more difficult. For example for n=6 the best constnt is
$M=\frac{5\sqrt{10}-14}{27}$  ;)

[quote="gdeskos"]I    think  that  he  want  to  write  that

       $ab(a^{n-2}+b^{n-2})\leq M(a+b)^n$[/quote]

Yes. That is the problem .Thank you for the correction.I edited now.


Given $ a,b,c \geq 0$ such that $ abc\equal{}1$.Prove that
$ (a\plus{}b\plus{}c)(a^2b^2\plus{}b^2c^2\plus{}c^2a^2) \geq \frac{3(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)}{ab\plus{}bc\plus{}ca}\plus{}2(a^2\plus{}b^2\plus{}c^2)$

[quote="dduclam"]Given $ a,b,c \geq 0$ such that $ abc \equal{} 1$.Prove that
$ (a \plus{} b \plus{} c)(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \geq \frac {3(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}{ab \plus{} bc \plus{} ca} \plus{} 2(a^2 \plus{} b^2 \plus{} c^2)$[/quote]
It's equivalent to $ \sum_{cyc}(a^4b^3\plus{}a^4c^3\minus{}a^4b^2c\minus{}a^4c^2b)\geq0,$ which Muirhead.

[quote="arqady"][quote="dduclam"]Given $ a,b,c \geq 0$ such that $ abc \equal{} 1$.Prove that
$ (a \plus{} b \plus{} c)(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \geq \frac {3(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}{ab \plus{} bc \plus{} ca} \plus{} 2(a^2 \plus{} b^2 \plus{} c^2)$[/quote]
It's equivalent to $ \sum_{cyc}(a^4b^3 \plus{} a^4c^3 \minus{} a^4b^2c \minus{} a^4c^2b)\geq0,$ which Muirhead.[/quote]
 too complex! :D
My solution using S.O.S method :)


Hello,

[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php[/url]

this webpage has a very interesting overview of all symbols,etc.. in LaTeX.  I have used this webpage for quite some time now and I've always enjoyed it.

But since very recently, it is loading extremely slow, severely disrupting my computer (it also disturbs the use of other programmes).  Closing the window (which isn't easy either) is the only solution.
I don't seem to have any problems with this forum though.

Is this normal?
Thanks!

It's working fine for me, and I'm on a (relatively) old computer...it's at least 5 years old.  But if you don't have high-speed internet that might cause problems with all the images...

[quote="Valentin Vornicu"]Loads fine for me in both Firefox 2 and IE 7. Takes a while though to load all the symbols, and maybe that's why your computer gets grumpy ...  :maybe:[/quote]
Hmm, I just did a test : the page loads almost immediately, I can see all the tables and all the text in it right away, but not all symbols.  It took about a minute, but now I can see all the symbols.
Weird, last time it was more trouble than that, my computer was severely disturbed (it was bought in 2002).
Well, if it stays like this, that's fine by me, even if I don't understand why it's better now. :P


$ m$ and $ n$ are natural numbers with $ 1 \leq m < n$.
If the last three digits of $ 1978^m$ are equal respectively to the last three digits of $ 1978^n$ , find $ m$ and $ n$ such that $ m\plus{}n$ has its least value.

[i]Thank You [/i]

 :)

1978 IMO, #1:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61017[/url]


algebra
polynomial
function
linear algebra
linear algebra unsolved

Hey,

Could anyone explain why:

p(f) is a linear function,

(p*q)(f) = p(f) o q(f)

(p(f) o f)(v) = (f o p(f))(v)

where p and q are arbitrary polynomials, f an endomorphism, v a vector.

Take a monomial $ m(x) \equal{} ax^n$. Then $ m(f) \equal{} af^n$ should be interpreted as $ af^n(v) \equal{} af(f(\cdots f(v)\cdots ))$, clearly linear. Since polynomials are sums of monomial, this answers your questions 1. and 2. As for question 3. there you have polynomials $ p$ and $ \textrm{id}$ (since $ \textrm{id}(x) \equal{} x$).


geometry
3D geometry
tetrahedron
geometry open

Four points (J, K, L, M) are chosen on each face of an arbitrary tetrahedron ABCD. A line is drawn from each of these points to the opposite point of the tetrahedron (the point on face ABC would have a line drawn to D).The original points are chosen such that these four lines all intersect at a point H. Now lets say four new lines are drawn such that each pass through a point (J, K, L, or M) and are perpendicular to the face this point resided on. Prove that these four new lines each intersect at a point I.

you sure you're not missing some information? take, for example, a point $ K$ on $ ABD$ and a point $ J$ on $ BCD$ so that $ CK$ meets $ AJ$ at $ H$. $ A, C, J, K$ thus form a plane, and thus $ AK$ and $ CJ$ meet $ BD$ at the same point, which we'll call $ P$. for the perpendiculars at $ K$ and $ J$ to the respective faces to meet, the projections of $ K$ and $ J$ on $ BD$ have to coincide. but (except for the case where $ CP\perp BD$, which we can arrange not to happen), you can fix $ K$ and move $ J$ along $ CP$ so that the projection of $ J$ on $ BD$ changes position, giving a configuration where the conclusion no longer holds.

You're right. I had a theory that this would work because it made intuitive sense to me. I tried several simple examples and it held true, but I couldn't prove it. That's why I posted it in the "open questions" forum. I guess I should have asked whether it could be proven or disproven.


Arqady, this is the [b]pre-olympiad[/b] forum, intended for students on the level of regional / subnational mathematical competitions. The problem was posted here as an exercise for students of this level. You, being clearly an experienced problem-solver (at least on inequalities!) should not post solutions here at all, since you are taking beginners their practice. (Of course, there are exceptions: If you feel that a problem posted in the pre-olympiad section is actually of olympiad level, then you can post a solution; or if you have some interesting comments on the problem to widen the horizon).

Please don't take this personally; almost every of us has done a similar mistake in past (I, for instance: http://www.mathlinks.ro/Forum/viewtopic.php?t=4348 ).

  Darij

Using arqady's hint:

LS $\ge \displaystyle \frac{ \Bigg(\sum \sqrt{ \frac{x_i}{x_{i+1}(x_i+x_{i+1}+x_{i+2}) } (x_i + x_{i+1} + x_{i+2}) } \Bigg)^2 }{ \sum x_i + x_{i+1} + x_{i+2} }$ by cauchy

where it remains to prove $\sum \sqrt{x_i / x_{i+1}} \ge n$ which is AMGM

Would it be possible to do some telescoping on the sum? I tried but I don't seem to find anything...

And is there anybody that has a solution using good old Chebychev? I've been searching, but didn't find a way out because of those $x_{n+1}$ and $x_{n+2}$ that aren't defined.

Ok for $x_n$ it is $\frac{x_n}{x_1 ( x_n + x_1 + x_2)}$

Oh and for Singlar's solution it is :

LE $\ge \displaystyle \frac{ \Bigg(\sum \sqrt{ \frac{x_i}{x_{i+1}(x_i+x_{i+1}+x_{i+2}) } (x_i + x_{i+1} + x_{i+2}) } \Bigg)^2 }{ 3\sum x_i + x_{i+1} + x_{i+2} }$

and it is the same solution i have


Prove that if $x>y>0$ then we have:

$ \frac{(x-y)^2}{8x} < \frac{x+y}{2} - \sqrt{xy} < \frac{(x-y)^2}{8y}$.

I remember this has been posted before...

yep!!  I  posted  it....Swedish  mo , 1985....


Does exists $ f:N\rightarrow N $ such that  f(f(n))=2n?
 :lol:

Yes such functions does exist. For example :
Let $f(1) = 3$, $f(3) = 2$, $f(2) = 6$ and so on $f(2^p)=3\times2^p$ and $f(3\times2^p) = 2^{p+1}$.
Let $m_0 = 1$.
Let $m_1$ be the least integer not appearing in the sequence above (here $m_1=5$ but we don't care). Define $f(m_1\times2^p)=m_1\times3\times2^p$ and $f(m_1\times3\times2^p) = m_1\times2^{p+1}$.
Now let $m_2$ be the least integer not appearing in the sequences above.
Define $f(m_2\times2^p)=m_2\times3\times2^p$ and $f(m_2\times3\times2^p) = m_2\times2^{p+1}$.

And so on....

I claim that
- At each step, if $m_i$ is defined then $m_i$ is odd.
Clearly, if $m_i = 2a$ is even, then from the minimality of $m_i$, it follows that $a$ has appeared before in the construction of some sequence. In that case, we have constructed $f(f(a)) = 2a = m_i$ later in the sequence, which contradicts that $m_i$ has not been considered before.
- No two sequences have a common term :
Suppose that for some $i<j$ we have
a) $m_i\times2^p=m_j\times2^q$. Since $m_i$ and $m_j$ are odd, it follows that $p=q$ and $m_i=m_j$. A contradiction.
b) $m_i\times2^p=m_j\times3\times2^q$. Note that from $i<j$ we deduce that $m_i < m_j$. But using parity again, we deduce that $m_i = 3m_j$. A contradiction.
c) $m_i\times3\times2^p=m_j\times2^q$. Using parity again, we deduce that $3m_i=m_j$, thus $m_j=f(m_i)$ has been considered in the step $i$ which contradicts that ^$m_j$ has not been considered before step $j$ in the construction.
d) $m_i\times3\times2^p=m_j\times3\times2^q$. Using parity again, we deduce that $m_i=m_j$, a contradiction.
Which proves the claim.
It follows that the function $f$ is well-defined.

- And, at last, it is obvious that, for every positive integer $n$, the number $f(n)$ is defined in the procedure above (from the minimality condition).

And we are done (if $N$ denotes ${0,1,....}$ then add $f(0)=0$)

Pierre.

[quote="grobber"]I'm pretty sure this was discussed before on the forum. I didn't follow everything Pierre wrote, but here's an example:

$f(4k+1)=4k+3$, $f(4k+3)=2(4k+1)$ and $f(2n)=2f(n)$ for all $n\in N$.

It looks like it works, but I might make some mistake somewhere (sorry if it turns out to be the same as in Pierre's post :D).[/quote]

If we change 2 by any natural number k then your answers is impossible

I'm not really sure what that means :): is the example wrong?

it is right, grobber, but one can't generalize your solution to give a solution for the equation $f(f(n))=kn$.
and i don't think pierre's proof can be generalized(i didn't study it in detail, so i may be wrong), so i conjecture that there are no such functions for k>2(it may be complete nonsense what i'm conjecturing ;) )...

Peter

There are, of course, the trivial cases when $k=a^2$, so $f(n)=an$ works.

I think we can find functions for which $f(f(n))=3n$, but there has to be something wrong (as always :D)..

We must obviously have $f(3n)=3f(n)$, so it's enough to define the function on numbers of the form $3k+1,3k+2$ (just like in the problem with $f(f(n))=2n$). We take the sequences $1\to 2\to 3\to 6\to 9\to 18\to \ldots,\ 4\to 5\to 12\to 15\to 36\to 45\to \ldots$ etc. where, in general, a sequence has the form $3k+1\to 3k+2\to 3(3k+1)\to 3(3k+2)\to 9(3k+1)\to 9(3k+2)\to\ldots$. If we show that these sequences are disjoint, then we can define our function ($f(3k+1)=3k+2,f(3k+2)=3(3k+1)\ldots$).

The sequences are obviously disjoint, because otherwise there would be $k_1\ne k_2$ s.t. $3^a(3k_1+\alpha)=3^b(3k_2+\beta)$, where $\alpha,\beta\in \{1,2\}$. This can't be true because it would mean $a=b\Rightarrow 3k_1+\alpha=3k_2+\beta\Rightarrow \alpha=\beta\Rightarrow k_1=k_2$, a contradiction.

If you look at it carefully, you'll see that it's basicly the same as what I did for the initial problem, so I guess that method can be generalized after all :).


Given a finite string $ S$ of symbols $ X$ and $ O$, we write $ *(S)$ for the number of $ X's$ in $ S$ minus the number of $ Os$.(for example $ *(XOOXOOX)=-1$ )We call a string $ S$ balanced if every substring $ T$ of(consecutive symbols) $ S$ has the property that $ -2\leq *T\leq 2$. Find the number of balanced strings of length $ n$.

Already appeared and solved here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153795


I am new at this.
How do you access the classroom during the times for the class?
How do you access the classroom during the times for open discussion?
Which links(buttons) do you click?
Where will the "Join" button appear?

Ten minutes before class time, log in to the forum, then click on Classroom on the sidebar.  If you have never been to an AoPS class, attend the first class 25-30 minutes early.

It's just like the Math Jams you've attended.  Just click on the "Classroom" button on the left sidepanel.


I just started relearning C++.  I have a Dev C++ compiler and when ever i type in a program and try to compile it, it comes up with errors that i don't know how to fix.  Here is my program source code:

#include <iostream>

int main()
{
cout << "Hello\n";
     return 0;
}

thats it...the thing says "expected unqualified-id before '{' token" and it also says "expected `,' or `;' before '{' token"
HOW DO I FIX THIS!?!?!?
please help. :)

You need the namespace to use the library. Add this line after the library declarations:

[code]using namespace std;[/code]

But when I compile the same code on my compiler (also Dev-C++) I don't get the same errors, so maybe that's not what's wrong.

i'll try that...thanks :)
EDIT:
this is what i wrote in:

#include <iostream>
using namespace std
;int main();
{
    cout<<"hello";
    return 0;
}

and i still get the same errors!!!!
EDIT#2:
i made another program and it compiled...now what do i do?
Do i press the run button?  Everytime i do something black just flashes.

You really shouldn't have a semicolon after the "int main()".  Here's how I would code it:

[code]#include<iostream>

using namespace std;

int main()
{
   cout<<"Hello\n";
   return 0;
}
[/code]

When you say that something black flashes, do you mean a black window pops up but then disappears?  If so you might need to add some sort of input to the program so that the window stays open until you hit the enter key to close the window.  In that case, I would try something like this (notice that I haven't tested this on my own compiler but I think it should work; I'm sure others will correct me if I'm wrong ;))

[code]#include<iostream>

using namespace std;

int main()
{
   cout<<"Hello\n";
   cin.get();   // Must hit a key on the keyboard for the program to terminate
   return 0;
}
[/code]

The cin.get() will get a character from cin, as its name implies, keeping the window open until you tell it to close.

wow!!! thanks!!!! that made it work!!!!! :D

No problem.  You can add that line to the end of just about any program if you feel it's necessary!

how do i fix the error that says "expected unqualified-id before '{' token" and it also says "expected `,' or `;' before '{' token"

Hm... I think that the problem is that your [b]#include <iostream>[/b] and [b]int main[/b] are on the same line. This could happen if you downloaded the code from somewhere and the "somewhere" used different end-of-line characters than windows uses - this way your editor [i]could[/i] display [b]#include <iostream>[/b] and [b]int main[/b] on different lines but to your compiler they would seem to be on the same line. Which is bad. Or it could be caused by something completely different. In any case, rewrite the whole code and try to compile again.

I wrote it again and i get the same errors! :(

#include <iostream>

using namespace std;

int main(){
cout << "Hello\n";
return 0;
}

works just fine with gcc on my machine. Maybe your compiler's broken?


[hide="well after all that work the inequality is wrong!!! and I think it is just a typo...consider $(a,b,c)=(.001,.001,.00001)$ when $f(a,b,c)=-.00001$"]
Let $a'=a,b'=b,c=c$ for notation then let $a'-b'=a,b'-c'=b$.
So $a,b\geq 0, c>0$. We only consider the strict inequality cases for now, those are easy to consider seperately afterwards.
We must show \[a(2c+2b+a)\left(\frac{1}{c}+\frac{1}{c+b}\right)-\frac{b(2c+b)}{a+b+c}\geq 3a-b \]
Indeed it is equivalent to \[2x^{3}y+3x^{2}y^{2}+2xy^{3}+x^{3}z+6x^{2}yz+5xy^{2}z-y^{3}z+3x^{2}z^{2}+5xyz^{2}-y^{2}z^{2}+2xz^{3}\geq  0\]
We let $y=1$ ($\not=y$) since its homogenous \[2x^{3}+3x^{2}+2x+x^{3}z+6x^{2}z+5xz-z+3x^{2}z^{2}+5xz^{2}-z^{2}+2xz^{3}\geq  0\]
Or \[f(z)=(2x)z^{3}+(3x^{2}+5x-1)z^{2}+(x^{3}+6x^{2}+5x-1)z+(2x^{3}+3x^{2}+2x)\geq 0\]
[/hide]


trigonometry
function
limit
geometry proposed
geometry

Find the real number $ x$ so that $ 2\sin\left( \frac{\sin x\plus{}x}{2}\right)\equal{}3x\minus{}\sin x$

hello, your solution is $ x\equal{}0$.
Sonnhard.

Here is my solution:
[hide]I'll prove that the unique solution is $ x\equal{}0$.Suppose that $ x$ isn't $ 0$.
We can easily see that we can  suppose WLOG that $ x>0$and $ x<\frac{\pi}{2}$.
We take the function $ f$ with $ x\rightarrow\frac{\sin x\plus{}x}{2}$ and then the equation is equivalent to
$ f(f(x))\equal{}x$ ;$ f$is increasing and we have that $ \ f(x)\equal{}x$.
Thus, $ \sin x\equal{}x$.But $ \lim_{x\rightarrow 0}\frac{\sin x}{x}\equal{}1$ .So, we only have the solution $ x\equal{}0$.[/hide]


Evaluate
$\tan^2 A+\tan^2 3A+\tan^2 5A + ...+\tan^2 89A$ where $A = 1$ degree.

i actually had this on my first mock aime...it is pretty difficult, the connection to polynomials is not obvious, refer here for help on a similar problem:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=tangents+sum&t=63539[/url]


i need an online sheet music database which provides violin music in pdf or gif form, is free, and requires no membership. preferably advanced/intermediate. any suggestions?
 :huh:

Though this site doesn't have that many pieces, it'll do.

I'll try to find more.

[url=http://www.8notes.com/violin/]I think you'll like this[/url]

i've visited that website before - many of the pieces are under member restriction and those are the difficult pieces i am looking for  :(
thanks anyway hwenterprise

[url=http://www.sheetmusicarchive.net]Sheet Music Archive, no membership required.[/url]

[url=http://www.music-scores.com]Membership required but is cheap.... (i'm a member)[/url]

Also you could type into Google Free sheet music, inculding your genre and stuff like that.

thanks  :D
the first link was exactly what i was looking for

Good! But I did find out, when I printed out the Back two-part invention no. 08 it was all on one page and really hard to read. I forgot that they also do not list slur marks, staccato, etc. But that was also on piano, it may be different for violin and such.

You can give these a try. I know there's some place I'm forgetting where I found a nice Paganini's 24th Caprice, which is just plain awesome, even though I can't come near to playing it, but its location slips my mind.  Good luck.

http://www.theviolinsite.com/violin_music.html

http://icking-music-archive.org/

http://www.classiccat.net/


Let $a,b,c$ be nonnegative real numbers satisfying $a^{2}+b^{2}+c^{2}=1$. Prove that
\[\underset{cyc}{\sum}\sqrt{\frac{a}{a^{3}+8bc}}\ge \frac{3\sqrt{3}}{\sqrt{1+8\sqrt{3}}}\]
:)

[quote="mathboy_cntHDVN"]$\widehat{BAC}=120^{0}$[/quote]
What do you mean?  :?:

i smell here imo 2001 :wink:  i will poste my solution

Could you post the details of your solution,fermat3?I don't think it can be solved by the way to solve that inequality


Also, I am afraid why this info was not sent:

(based on last year's experience for those [b]driving to campus[/b])

First, go to Waterview Apartment Office (as said in Altheman's post) and sign in.

Put your luggage in your rooms and wait and wait and wait.......

Then we will meet at student union around 4 or 5.

This may be helpful

Is there lunch provided for those arriving at the airport at half-past 11?  Should we pack something?

[quote="Altheman"]Hey, here is a map of the campus where Awesomemath is taking place. http://www.utdallas.edu/maps/ It might be nice to familiarize yourself with the layout of the campus (even though you won't have trouble getting around if you don't).

I'm pretty sure that we are going to be in phase 8 dorms (like we were last year). One thing that kinda helped me get around was noticing that there is a parking lot south of the dorms and then you go east to the classrooms and to eat. I'm not sure what buildings that stuff is going to be in.

One thing of note is that there are soccer fields and a track (its just pavement) south. I also remember playing frisbee there sometimes or in the field just south of parking lot I.[/quote]

You navigate via compass directions?

I just go "that way" or "this way"  :rotfl:

I'm a visual learner so when I'm going places that are new, it helps me to envision myself on a map.

So its more like okay I'll go to the right side of the map now. haha

Hmm... Maybe it's this way?  (Goes this way, and gets lost)

Then it's that way?  (Goes that way and gets lost, finally gets rescued by Altheman after 3 hours)

I'm good at getting lost.  :P

Anyone going to answer me about lunch?

Uhh I don't remember. I think that the staff were eating something at least...It doesn't hurt to just like pack a sandwich or something anyway.


How do you get equations and stuff into a post? Like the quad formula, fractions, etc.

Ima noob =D

BTW, if I were to take an AoPS class this summer to get better at math counts and amc 8, which one should i take? or which text to buy?

This might help-

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5261

By searching "LaTeX" on the AoPSWiki, you can also get a guide on codes and other useful information.

Sweet! Thanks

Can be locked I guess, 2nd question is optional...whatever

[quote="Urc"]BTW, if I were to take an AoPS class this summer to get better at math counts and amc 8, which one should i take? or which text to buy?[/quote]I would suggest taking a look at the INTRO classes (and AMC 10 Class).

Anyway check this page: http://www.artofproblemsolving.com/Classes/course_recommendations.php#g6up

and this one too: http://www.artofproblemsolving.com/Classes/course_recommendations.php#contest1
[quote]For students just getting started with MATHCOUNTS or the AMC 8, we recommend our Introduction series of classes. Review the recommendation information above to determine which of these classes best fits you. If you have some experience with MATHCOUNTS and are preparing for the State or National MATHCOUNTS competition, you are probably ready for our MATHCOUNTS Problem Series. The MATHCOUNTS Problem Series will also help students prepare for the AMC 8.[/quote]

Thanks soo much!

Lock pls! This site rocks!


Find the number of integers x which is an element of {-1997,....., 1997} for which

1997 | x^2 +(x+1)^2

[hide="Solution"]We need $ x^2+(x+1)^2=2x^2+2x+1\equiv{0} \pmod{1997}$ $ \implies{(2x+1)^2\equiv{-1} \pmod{1997}}$. $ 1997$ is a prime of the form $ 4k+1$ so this at least four solutions in the given interval. There cannot be more than four because $ 1997$ and $ 2$ are coprime so as $ x$ cycles through the residues $ \pmod{1997}$ so does $ 2x+1$.[/hide]


A stats reference I have quotes the variance in two different ways (one to correct the bias, one without).  One of them is

$\sigma^2 = \frac{\Sigma (x- \mu)^2}{n}$

They say this is equivalent to $\sigma^2 = \frac{\Sigma x^2}{n}-\mu^2$.

Can someone prove why? I am assuming is it just algebra, but for some reason, it's just not clicking for me.


[hide]$\frac{5}{32}$
Factor, group and add.[/hide]

Yeah, that's what i got...but you can group them into pairs, and then for every pair, the average of the outer two numbers gives the inner number, so for, for instance, $\frac{1}{8*11}+\frac{1}{11*14}$, you'd get
$\frac{8+14}{8*11*14}=\frac{2*11}{8*11*14}=\frac{1}{56}$

the correct answer, as K81o7 said, is $\frac{5}{32}$

I did it by telescoping.  :D  Yep, $\frac{5}{32}$ is correct.  :lol:

[quote="xxreddevilzxx"]telescoping does it, I think. It is just that the fractions dont come out to be 1 over something, but a little more work will get this simplified by telescoping.[/quote]
What is telescoping? Cancels out?

[quote="nr1337"][quote="xxreddevilzxx"]Telescoping is a method we use to cancel a lot of terms out, leaving us with first and last term. A great example would be

$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20}$

where

$\left(1-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)$

$1 - \frac{1}{5} = \frac{24}{25}$

From this example, we saw starting with the first set of fraction, the last one matched with the first one of the next set to cancel out. This matching, or the folding is why we call it telescoping, just like the telescopes fold in and out.[/quote]

1-1/5 = 4/5 :)[/quote]

 :rotfl:  But, thanx for telling me what telescoping is  :)

[quote="nr1337"][quote="xxreddevilzxx"]Telescoping is a method we use to cancel a lot of terms out, leaving us with first and last term. A great example would be

$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20}$

where

$\left(1-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)$

$1 - \frac{1}{5} = \frac{24}{25}$

From this example, we saw starting with the first set of fraction, the last one matched with the first one of the next set to cancel out. This matching, or the folding is why we call it telescoping, just like the telescopes fold in and out.[/quote]

1-1/5 = 4/5 :)[/quote]

oops. sorry about that. I was in a hurry.

Thanks!
I have learned about it in China and it seems to involve lots of skills.


inequalities
quadratics
LaTeX
algebra
quadratic formula

You seem to have taken the square root of every term individually, which doesn't work at all.

I have no idea what you're doing in the second part.

Are you sure the question is correct?  It would be much simpler as

$4y^{2}-4xy+x^{2}+2 = 0$

The question IS correct. And there must be a solution since my teacher has the answer sheet.

the expression is equivalent to: $(2xy-1)^{2}+x^{2}+1=0$

Mainly, like your teacher said, complete the square

$(2xy-1)^{2}+x^{2}+1 = 0$
By the trivial inequality, the squares must result in a positive number if they're real...therefore x and y are complex.

falconwing64 Could you further explain your answer. I looked up trivial inequalities on the AOPS wiki but it didnt explain much. Also Is it possible to factor this and actualy comple the square because in the answer sheet there was a quadratic formula looking answer.

trivial inequality says that
$x^{2}\geq0$
for any real x.

so in $(2xy-1)^{2}+x^{2}+1$, $(2xy-1)^{2}$ and $x^{2}$ can't be negative.
since you can't add up to 0 from nonnegatives and a positive, the equation has no solutions

[hide="completing"]$(2xy-1)^{2}+x^{2}+1=0\implies(2xy-1)^{2}+x^{2}=-1$

We know that for every real numbers n and m we have $m^{2}+n^{2}\geq0$

Knowing it , the first equation is imposible.[/hide]

[quote="Jeremic"]Hmm i got the wierdest question in one of my hmwk sheets... says
Show that the equation $4x^{2}y^{2}-4xy+x^{2}+2=0$ has no real solutions $(x,y)$. Hint: Complete the square.

I had an attempt but his is way too complicated.

Heres what i tried

$4x^{2}y^{2}+x^{2}+2=4xy$
$2xy+x+\sqrt{2}=2\sqrt{xy}$

from this i got

$\sqrt{xy}=x+\sqrt{2}$
$x^{2}-xy+2=0$

This got me no where... any hints?[/quote]
Here is something that I think should be pointed out.
$\sqrt{a+b}\neq \sqrt{a}+\sqrt{b}$.
What is the latex for does not equal to?
EDIT: I found the latex and edited my post.


Show that any two domino tilings of a rectangular board are connected by local flips, each happening in a 2 by 2 square.

Could you explain a bit more what you mean by "connected by local flips, each happening in a 2 by 2 square"? Maybe give an example? I don't understand what you mean.

This should be moved to the Combinatorics forum.

Could anyone explain this a little better please?

A "local flip" means taking two adjacent dominoes that form a 2 by 2 square and rotating them 90 degrees (flipping them from both vertical to both horizontal or vice-versa).  The problem asks you to show that by doing this repeatedly, you can get from any tiling of the board with dominoes to any other.

The answer is yes, but the proof requires some machinery.  One of the main tools used to analyze tilings is something called a "height function."  Here are some references:

[url]http://www.claymath.org/library/senior_scholars/stanley_ardila_tilings.pdf[/url] (page 13)
[url]http://arxiv.org/PS_cache/math/pdf/9801/9801111v1.pdf[/url]
W. P. Thurston, Conway’s tiling groups, Amer. Math. Monthly, 97, 8, 757-773 (1990).

bump after a decade... could anyone offer a detailed proof?

I think I have an intuition for a proof... but not sure how to write it formally...

It's exactly the same as this problem: [url]https://codeforces.com/problemset/problem/778/D[/url]

Here's my proof with no machinery (because I don't know of any machinery)  :

We'll use the term "switch" instead of "local flip"
First orient the board s.t. it has an even number of columns and let the number of columns and rows of the board be $2m,n$ respectively.
Call a tile "sleeping" or "standing" if it's horizontal or vertical respectively.
Let $d(a,b)$ denote the geometric distance between two standing tiles $a,b$ in the same two rows.
Let intersection of two objects mean the cells common to them.

Note : When I say that a line passes through a tile, I mean the line passes through the middle of the tile dividing the tile into two cells
-----
[b]Main Idea[/b] :

Let $T$ denote the tiling where all tiles are sleeping. (This exists as the number of columns is even)

We'll show that for any other tiling $T_0$ we can go to $T$ by a series of moves which always increases the number of sleeping tiles (via a new operation)

And this automatically implies that every two tilings are connected by switches
-----
Call a group of $2k+2$ tiles with $2k$ sleeping tiles and $2$ standing tiles a $k$-block where $k\geq 0$ if they look like this :
[img]https://i.ibb.co/gyWDndj/Tiling-connectedness.png[/img]
Call a $k$-switch of a $k$-block an operation formed of our normal switches which looks like this :
[img]https://i.ibb.co/Mg0Wwm0/Image-2-1.png[/img]
Note that every $k$-switch increases the number of sleeping tiles exactly by $2\quad\dots(1)$
-----
Let the given tiling be referred to as $T_0$
Now, the algorithm to go to $T$ looks like this :
[code]
while (A standing tile exists) :
	1. Find a k-block
	2. Perform a k-switch on it
[/code]
Now, if we can always do (1.) provided $\exists$ a standing tile, then by $(1)$ observe that as we increase the number of sleeping tiles every iteration, after some moves, we will reach a tiling with no standing tiles i.e. the $T$ tiling.

So, we require the following claim :
-----
[b]Claim[/b] : $\exists$ a standing tile $\implies$ $\exists$ a $k$-block

[b]Proof[/b] :

Let's first label the columns $1,2,\dots,2m$
Color the standing tiles red or blue according to if they are in an even column or an odd column.

We'll use the following lemma :

$\mathrm{Lemma\;1}$ : Every horizontal line passes through with the same number of blue tiles as red tiles

[hide = Hint]
Chessboard coloring
[/hide]

Now, call a pair of standing tiles "matching" if they are of different colors and belong to the same $2$ rows.
Observe that $2$ standing tiles $a,b$ in the same $2$ rows are matching iff $2|d(a,b)\quad\dots(2)$

We are given that a standing tile $t$ exists. WLG let it be colored blue
Consider the line which passes through $t$.
By Lemma 1, it passes through a red tile $t'$
$\therefore$ We have a matching pair $t,t'$

Now, consider the matching pair $t_1,t_2$ with the least distance say $2d$ (it's even cuz the tiles are of different colors).

Consider the $2\times 2d$ rectangle as shown below and let $s$ be its horizontal mid-segment.
Let $s$ divide $R$ into $2$ ($1\times 2d$) rectangles call the upper one as $U$ and the lower one as $D$
[img]https://i.ibb.co/Zd70YT0/Image-2.png[/img]
We'll show that the rectangle $R$ made up of only sleeping tiles.

$\mathrm{Lemma\;2}$ : $s$ does not pass through any tile
[hide = Proof]
If possible let $s$ pass through a tile $f$
$\therefore$ As $t_1,t_2$ are of different colors, one of $f,t_1$ and $f,t_2$ must be a matching pair and observe that this matching pair will have a smaller distance than $d(t_1,t_2)$, a contradiction.
[/hide]

$\therefore$ By $\mathrm{Lemma\;2}$, every standing tile intersects at most one of $U,R\quad\dots(3)$

$\mathrm{Lemma\;3}$ : $U$ is covered perfectly by sleeping tiles
[hide = Proof]
[img]https://i.ibb.co/VHzRYtd/Image-4.png[/img]

Let the sleeping tiles intersecting $U$ be $a_1,\dots,a_h$
Observe that as the two cells to the left and right of $U$ are covered by $t_1,t_2$, we have that each $a_i$ lies completely inside $U\quad\dots(4)$
And let the standing tiles intersecting $U$ be $b_1,\dots,b_k$ in that order.
Observe that by $\mathrm{Lemma\;2}$ we have that $b_i\cap U = \mathrm{Lower\;cell\;of\;b_i}\quad\dots(5)$
$\therefore$ Each of $b_1,\dots,b_k$ lies in the same two rows $\quad\dots(6)$

Now, for the sake of contradiction let $k\geq 1\quad\dots(7)$

Now, $(4),(5)$ and counting $\implies 2d = 2h +k\implies 2|r$
$\therefore$ By $(7), k\geq 2$

So, consider $b_1,b_2$
Observe that the range of cells between the lower cells of $b_1,b_2$ is covered perfectly by sleeping tiles.
$\therefore 2|d(b_1,b_2)$
$\therefore$ By $(2)$; $b_1,b_2$ is a matching pair with distance $\leq 2n-2<2n= d(t_1,t_2)$ a contradiction.
[/hide]

Similarly, we have that $L$ is also covered perfectly by sleeping tiles
$\therefore$ $R$ is perfectly covered by sleeping tiles
$\therefore$ The rectangle formed by considering the cells of $t_1,t_2$ as corners is a $k$-block. $\blacksquare$
-----
So, by this claim, the algorithm works fine and we are done!


By Cauchy Schwarz inequality, we have
\[\sum\frac{a^{2}}{b+c}\ge \frac{(a^{2}+b^{2}+c^{2})^{2}}{\sum a^{2}(b+c)}\]
Also by Cauchy Schwarz inequality
\[\sum a^{2}(b+c) \le \sqrt{\sum a^{2}\cdot \sum a^{2}(b+c)^{2}}\le \sqrt{\sum a^{2}\cdot \sum 2a^{2}(b^{2}+c^{2})}=2\sqrt{\sum a^{2}\cdot \sum a^{2}b^{2}}\le \frac{2}{\sqrt{3}}(a^{2}+b^{2}+c^{2})^{3/2}\]
Adding them up, we have the result. :)

[quote="Svejk"]If $a,b,c>0$ , $\sum \frac{a^{2}}{b+c}\geq \frac{\sqrt{3}}{2}\cdot \sqrt{a^{2}+b^{2}+c^{2}}\geq \frac{a+b+c}{2}$[/quote]

WLOG $a\ge b\ge c$. then from chebyshev\[\sum\frac{a^{2}}{b+c}\ge \frac{3(a^{2}+b^{2}+c^{2})}{2(a+b+c)}\]
so we need to show that \[\sqrt{3(a^{2}+b^{2}+c^{2})}\ge a+b+c\] which is trivial.
the right side also comes from the last inequality. :D


Let $ a,b,c\geq 0$ such that $ a^2\plus{}b^2\plus{}c^2\equal{}3$. Prove that
$ a^2b^2\plus{}b^2c^2\plus{}c^2a^2\plus{}ab^3\plus{}bc^3\plus{}ca^3\leq 2(a\plus{}b\plus{}c)$


Find an orthogonal $ Q$ so that $ Q^{\minus{}1}AQ \equal{} B$ if $ A$ is the 3 by 3 matrix of all $ 1$s and B is the 3 by 3 of all 0 entries except for the lower right corner, which has entry 3.

I started by finding that the eigenvalues of A were 3, 0, and 0, corresponding to matrix B with the eigenvalues on the diagonal. However, I found the eigenvector corresponding to $ \lambda \equal{} 3$, but I cannot find 2 orthogonal eigenvectors corresponding to $ \lambda \equal{} 0$. Can someone tell me how to find them, so I can put them as the columns of $ Q$?

Thanks

Can you find any two independent eigenvectors in that 2-dimensional eigenspace? Because if you can do that, you can always apply the Gram-Schmidt process to those.


Try add (+ x) and (- x) on the numerator.

There's a typo: In the numerator there should be a minus sign between the terms (otherwise the limit is just $\pm\infty$).

[hide="Solution of the correct version"]Add $3$ to and subtract $3$ from the numerator, and rearrange using $a-b={a^{2}-b^{2}\over a+b}, a-b={a^{3}-b^{3}\over a^{2}+ab+b^{2}}$:

$\begin{aligned}N &=\sqrt{x+2}-3-(\sqrt[3]{x+20}-3)=\frac{x+2-9}{\sqrt{x+2}+3}-\frac{x+20-27}{\sqrt[3]{(x+20)^{2}}+3\sqrt[3]{x+20}+9}\\ &=(x-7)\left(\frac{1}{\sqrt{x+2}+3}-\frac{1}{\sqrt[3]{(x+20)^{2}}+3\sqrt[3]{x+20}+9}\right)\end{aligned}$

Rearrange the denominator using $a-b={a^{4}-b^{4}\over a^{3}+a^{2}b+ab^{2}+b^{3}}$:

$D=\frac{x+9-16}{\sqrt[4]{(x+9)^{3}}+2\sqrt[4]{(x+9)^{2}}+4\sqrt[4]{x+9}+8}=\frac{x-7}{\sqrt[4]{(x+9)^{3}}+2\sqrt[4]{(x+9)^{2}}+4\sqrt[4]{x+9}+8}$

Therefore

\begin{eqnarray*}{N\over D}&=& \left(\sqrt[4]{(x+9)^{3}}+2\sqrt[4]{(x+9)^{2}}+4\sqrt[4]{x+9}+8\right)\left(\frac{1}{\sqrt{x+2}+3}-\frac{1}{\sqrt[3]{(x+20)^{2}}+3\sqrt[3]{x+20}+9}\right)\\ &\stackrel{x\to 7}{\longrightarrow}& 32\left({1\over 6}-{1\over 27}\right)\\ &=&{112\over 27}\end{eqnarray*}[/hide]

Of course, but the point of tricky limits is usually to avoid L'Hospitalization (if at all possible), since the limits in any case must be taught prior to the derivatives.

Sure, I was aware of that when posting. But it's just an alternative. Nice non-de-l'Hopital work, by the way  :)

As a side note: It's interesting that both "L'Hospital" and "L'Hôpital" (Guillaume de) are considered to be correct spellings... which seems to be a misconception?

This is quite off-topic, but: I've always heard that the "accent circonflexe" (^) was used where a [i]s[/i] disappeared near a vowel, when we compare French and Latin. Like

'fenestra' (Latin for window) -->  'fenêtre'
'asinus' (Latin for donkey)  -->  'âne'
'insula' (Latin for island)  -->  'île'

So that would explain the two spellings 'de l'Hôpital' and 'de l'Hospital'.


function
calculus
derivative
inequalities
algebra
domain
real analysis

Hi.
Where can i find(on the net) the proof of the well-known result that  every convex function has right and left derivatives and the inequalities related to this??
Thanks!

It's not that difficult a result. First, you need to show this: If $ f$ is the convex function in question, and $ x$ is a point in the interior of the domain of $ f,$ then $ g(h)=\frac{f(x+h)-f(x)}h$ is an increasing function on a punctured neighborhood of zero. The rest follows from the fact that monotone bounded sequences converge.


Find the value of w if

2x + y + z + w = 1
x + 3y + z + w = 2
x + y + 4z + w = 3
x + y + z + 5w = 25

The solution involves multiplying the equations by12, 6, 4, and 3 and then summing them all. I don't understand how someone would come up with those seemingly random numbers for multiplying through to yield 37(x+y+z+w)=111.

well its not random

say we want to have them all of the form k(x+y+z+w) = c
(something where we do have an equal coefficient of x, y, z and w)

then we want to multiply by a, b, c, and d on each inequality.

so  equating the coefficients

2a + b + c + d  =  a + 3b + c + d = a + b + 4c + d = a + b + c + 5d.

or:

a = 2b = 3c = 4d

but we see d < c < b < a.

3c = 4d => 3 | d = > d = 3k for some k.

so let k = 1

and we get

a = 12
b = 6
c = 4
d = 3

(dont these numbers look familiar)

I think I would have solved it like this.  Let s = x + y + z + w.  Then the four equations become x + s = 1, 2y + s = 2, 3z + s = 3, and 4w + s = 25.  Solving in terms of s, we get x = 1 - s, y = 1 - s/2, z = 1 - s/3, and w = (25 - s)/4.  Adding those four new equations gives s on the left side, so we get s = (1 - s) + (1 - s/2) + (1 - s/3) + (25 - s)/4.  Now we can solve that one equation for s.

This solution is less elegant than the official solution, but at least it is easy to come up with.

Without actually doing it, it looks like a prime candidate for Gaussian Elimination.


function
inequalities
vector
trigonometry
analytic geometry
quadratics

Let $ABC$ a triangle be given  with $BC = a$, $CA = b$, $AB = c$ ($a \neq b \neq c \neq a$). In  plane ($ABC$) take the points $A'$, $B'$, $C'$ such that:

[b]I.[/b] The pairs of points $A$ and $A'$, $B$ and $B'$, $C$ and $C'$ either all lie in one side either all lie in different sides under the lines $BC$, $CA$, $AB$ respectively;

[b]II.[/b] Triangles $A'BC$, $B'CA$, $C'AB$ are similar isosceles triangles.

Find the value of angle $A'BC$ as function of $a, b, c$ such that lengths $AA', BB', CC'$ are not sides of an triangle. (The word "triangle" must be understood in its ordinary meaning: its vertices are not collinear.)

Since it is not too likely that JHC will join this forum in the next time, let me write what he would probably have written on this problem (alas, not in his wonderful style):

My solution will require some knowledge about barycentric coordinates. Particularly, I will use some results from [1] (don't worry, only the first two pages of the article will be needed); hence, I rewrite the problem with the terminology from [1]:

[color=blue][b]Problem.[/b] Let ABC be a triangle. Assume that the triangle ABC is directed clockwisely (i. e. the angle < CAB is mathematically positive). Let $\phi$ be an angle (we use directed angles modulo 180°), and let $A^\phi$, $B^\phi$, $C^\phi$ be the apices of the isosceles triangles $BA^\phi C$, $CB^\phi A$, $AC^\phi B$ erected on the sides BC, CA, AB of triangle ABC, having the bases BC, CA, AB and the base angles

$\measuredangle CBA^\phi=\measuredangle A^\phi CB=\measuredangle ACB^\phi=\measuredangle B^\phi AC=\measuredangle BAC^\phi=\measuredangle C^\phi BA=\phi$.

Find all angles $\phi$ such that the segments $AA^\phi$, $BB^\phi$, $CC^\phi$ are not the sidelengths of a non-degenerate triangle.[/color]

[i]Solution.[/i] First we prove a lemma:

[color=blue][b]Lemma 1.[/b] Let ABC be a triangle, and let X, Y, Z be points in its plane such that the triangles BXC, CYA, AZB are all directly similar. Then, the segments AX, BY, CZ always are the sidelengths of a (possibly degenerate) triangle; this triangle degenerates if and only if the lines AX, BY, CZ are all parallel.[/color]

[i]Proof of Lemma 1.[/i] There are some synthetic proofs, but I am too lazy to explain them, so I will use complex numbers:

We work in the Gaussian plane. Let a, b, c, x, y, z be the affices of the points A, B, C, X, Y, Z, respectively. Since the triangles BXC, CYA, AZB are all directly similar, we have $\frac{x-c}{b-c}=\frac{y-a}{c-a}=\frac{z-b}{a-b}$. Denote $\frac{x-c}{b-c}=\frac{y-a}{c-a}=\frac{z-b}{a-b}=k$. Then, $\frac{x-c}{b-c}=k$ becomes $x-c=k\left(b-c\right)$, what quickly transforms into $x=kb+\left(1-k\right)c$; similarly, we get $y=kc+\left(1-k\right)a$ and $z=ka+\left(1-k\right)b$. Hence,

$\left(a-x\right)+\left(b-y\right)+\left(c-z\right)$
$=\left(a-\left(kb+\left(1-k\right)c\right)\right)+\left(b-\left(kc+\left(1-k\right)a\right)\right)+\left(c-\left(ka+\left(1-k\right)b\right)\right)=0$.

Hence, $\left(a-x\right)+\left(b-y\right)=z-c$. Thus, by the triangle inequality, $\left|a-x\right|+\left|b-y\right|\geq\left|\left(a-x\right)+\left(b-y\right)\right|=\left|z-c\right|$. In other words, $AX+BY\geq CZ$. Similarly, $BY+CZ\geq AX$ and $CZ+AX\geq BY$. Thus, the segments AX, BY, CZ are the sidelengths of a (possibly degenerate) triangle. This triangle degenerates if and only if at least one of the triangle inequalities $AX+BY\geq CZ$, $BY+CZ\geq AX$ and $CZ+AX\geq BY$ becomes an equality. But this happens if and only if the lines AX, BY, CZ are all parallel. In fact, for instance, the triangle inequality $AX+BY\geq CZ$ is equivalent to $\left|a-x\right|+\left|b-y\right|\geq\left|\left(a-x\right)+\left(b-y\right)\right|$ (as we saw above), and thus, if it becomes an equality, the vectors a - x and b - y must be collinear, what, of course, forces the vector c - z to be also collinear with both of them (since $\left(a-x\right)+\left(b-y\right)+\left(c-z\right)=0$), so that the vectors a - x, b - y and c - z are all collinear, i. e. we have AX || BY || CZ; conversely, if we have AX || BY || CZ, then the vectors a - x, b - y and c - z are all collinear, and if we WLOG assume that the vectors a - x and b - y have the same orientation (in fact, among three collinear vectors, one can always find two with the same orientation), then the triangle inequality $\left|a-x\right|+\left|b-y\right|\geq\left|\left(a-x\right)+\left(b-y\right)\right|$ becomes an equality, and thus the triangle inequality $AX+BY\geq CZ$ becomes an equality. Thus, Lemma 1 is proven.

Now, let's come to the solution of the problem. The triangles $BA^\phi C$, $CB^\phi A$, $AC^\phi B$ are isosceles triangles with the same (directed) base angles; thus, they are all directly similar, so that Lemma 1 applies. As a result, we see that the segments $AA^\phi$, $BB^\phi$, $CC^\phi$ always are the sidelengths of a (possibly degenerate) triangle, and that this triangle degenerates if and only if the lines $AA^\phi$, $BB^\phi$, $CC^\phi$ are all parallel.

So we are searching for the angles $\phi$ which make the lines $AA^\phi$, $BB^\phi$, $CC^\phi$ parallel. Actually, this can be done by direct trig bash, but there is also a nice way using the notations from [1]:

According to [1], §2, the lines $AA^\phi$, $BB^\phi$, $CC^\phi$ concur at the so-called [i]Kiepert perspector $K\left(\phi\right)$[/i] of triangle ABC. Hence, these lines are parallel if and only if this Kiepert perspector $K\left(\phi\right)$ lies on the line at infinity. Since the homogeneous barycentric coordinates of the point $K\left(\phi\right)$ are $K\left(\phi\right)=\left(\frac{1}{S_A+S_\phi}:\frac{1}{S_B+S_\phi}:\frac{1}{S_C+S_\phi}\right)$, this point $K\left(\phi\right)$ lies on the line at infinity if and only if $\frac{1}{S_A+S_\phi}+\frac{1}{S_B+S_\phi}+\frac{1}{S_C+S_\phi}=0$. This is equivalent to $\left(S_BS_C+S_CS_A+S_AS_B\right)+2\left(S_A+S_B+S_C\right)S_\phi+3S_\phi^2=0$. Using the formulas $S_BS_C+S_CS_A+S_AS_B=S^2$ and $S_A+S_B+S_C=S_\omega$, this rewrites as $S^2+2S_\omega S_\phi+3S_\phi^2=0$. This is a quadratic equation for $S_\phi$, and the two solutions are

$S_\phi=\frac{-S_\omega\pm\sqrt{S_\omega^2-3S^2}}{3}$.

Since $S_\theta=S\cdot\cot\theta$ for any angle $\theta$, this yields $\cot\phi=\frac{-\cot\omega\pm\sqrt{\cot^2\omega-3}}{3}$. Using $\cot\omega=\frac{a^2+b^2+c^2}{4\Delta}$, where $\Delta$ is the area of triangle ABC, and applying Heron's formula for $\Delta$, sooner or later we arrive at

$\cot\phi=\frac{-\left(a^2+b^2+c^2\right)\pm 2\sqrt{a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2}}{12\Delta}$.

This is the required value of $\phi$. Do you call this a good olympiad problem?

[color=red][b]References[/b][/color]

[b][1][/b] [url=http://forumgeom.fau.edu/FG2001volume1/FG200118index.html]Floor van Lamoen and Paul Yiu, [i]The Kiepert Pencil of Kiepert Hyperbolas[/i], Forum Geometricorum 1 (2001) p. 125-132[/url].

  Darij


Find all positive $x$ and $y$ such that

$\frac{x^{2}+y^{2}}{xy}$ is an integer.

also find all positive $x$ and $y$ such that
$x+y+\sqrt{x^{2}+y^{2}}=\frac{xy}{2}$

o yes...sorry i forgot to mention that.  :D

[hide="1"]$x^{2}+y^{2}= kxy$ with $k \in \mathbb{N}$
$x^{2}-kxy+y^{2}= 0 \Rightarrow x_{1/2}= \frac{ky \pm \sqrt{k^{2}y^{2}-4y^{2}}}{2}$.
Hence $k^{2}-4 = m^{2}\Rightarrow (k-m)(k+m) = 4$, that gives $k = 2$.
So $(a, a)$ is solution, with $a \in \mathbb{N}$.[/hide]

[hide="Hint for 2"]Pythagorean triples[/hide]

can we do this reasoning for 1?

[hide]

$xy\mid x^{2}+y^{2}\implies x^{2}+y^{2}\equiv 0 \mod x$

Thus, $y=kx$ for some $k$ integer

So the expression becomes $\frac{x^{2}+k^{2}x^{2}}{kx^{2}}=\frac{k^{2}+1}{k}$

but we have that $\gcd (k,k^{2}+1)=1$, so necessary $k=1$. Therefore, the only solutions are $x=y$ $\forall x \in \mathbb{N}$[/hide]

[quote="hydro"]

$x^{2}+y^{2}\equiv 0 \mod x$

Thus, $y=kx$ for some $k$ integer

[/quote]
Try $x=9$ and $y=6.$ ;)


He does not play fair. During the IMO 2006 Slovenia he does on and immediatelly off his pen almost every five second. That really bothers other people around, me too.

There was also some czech imo guy next to him, who asked two times some help to tell him to stop it.

IMO is here to give to everyone a chance. And it is not fair when someone intentionally disturbs other people during contest in the test room.

Shame on him. Who knows how did he win the gold with 42 points in Mexico..

This is a serius acussation... do you have any other people that can verify what you say??? Maybe I'm talking useless words, but he can be out the the competition

Yes I agree, setout first of all, who are you, I mean from what country are you and did you participate in IMO 2005?

And making that noise with a pen, does not have much to do with obtaining perfect score.

They will not throw him out of the competition for just this I think.

That can or cannot be a serious offense, depending on malicious intent.  However, IMO is a much higher level that local contests, and I'm not sure of precise rules.  Also, an internet forum is not the best place to bring up accusations, however true they may be.  I suggest talking to your IMO coach, or someone in charge of the contest.  They can take the neccessary steps to do whatever needs to be done.

[quote="orl"][quote="fredbel6"]Yes I agree, setout first of all, who are you, I mean from what country are you and did you participate in IMO 2005?
[/quote]

At least he posted from the Slovenian IMO site.[/quote]

I am not important. Ask the czech guy, he is cze6.

The second day of the contest he even bring with a special pen for Peter.., with it it was no possible to disturb other people.

Orl, what do you mean posted from the slovenian site?

Setout, your english is a little unclear : do you mean cze6 himself brought a special pen on the second day and gave it to Peter so he would use that pen instead and not mak[b]e[/b] that noise anymore?

Well, it's not because we know his name we can reach him.

[quote="ShoeFactory"][quote="fredbel6"]Setout, your english is a little unclear : do you mean cze6 himself brought a special pen on the second day and gave it to Peter so he would use that pen instead and not making that noise anymore?[/quote]

That is definitely what setout meant.[/quote]

Yeah.

Hey Setout, how did you do on the contest? Hope you did well.

And distracting other conestants during the most important high school math competition is just really mean and not nice.

Arne did you ever meet him in person?

I must say, being a nervous freak at times, i have been doing that too, I simply didn't know.  However, as soon as I noticed frustration in a girl in front of me I quit and even took another pen, without any hard feelings le[b]f[/b]t.  Maybe it is an accident, :maybe:

[quote="rem"]Hey Setout, how did you do on the contest? Hope you did well.[/quote]

Well, I did not manage to solve problem number 3, because of him. I have had one hour to it and I cannot fully concentrate. I do not say that without him I could give a solution but sure I can say that with him near me it was really harder to concentrate.

I don't know how often I have been annoying people by playing around with my pen during contests, usually I only note it myself if someone asks me to stop. If one has such a habit, it is quite difficult to avoid it for a longer time.

Interpreting that as an attempt to cheat is, IMHO, a bit... far-fetched, isn't it?

  darij

I have seen Peter Scholze in Slovenija.I have heard what people are talking about him and his treating others.I can only say that everything what I have heard about him and anything what I have seen that he is doing was so great and positive that I can say with 100 percent security that he didnt have any bad intents and that accusing him that he is cheater is very ridicoulus.

yeah, i know, but you remember... in BH... you did't have anything below your paper... and you wrote really fast and hard! but i can concentrate... it's my opinion that during the exams there should be hammerfall or any similar music!

HEUL DOCH!!!

Naphthalin

Well, my opinion is that it would have been more easily to told Peter Scholze: Please, what you do really disturbes me, cvan you stop please? Thanks. And he would have stopped. That's the hole thing. And, as Darij said, he had more important things to do in the contest that disturbing, he's right

[quote="José"]Well, my opinion is that it would have been more easily to told Peter Scholze: Please, what you do really disturbes me, cvan you stop please? Thanks.[/quote]

The Czech guy did twice, there came the person he told it, she was standing next to Peter about five minutes and controlled him. After she left he started again in few minutes.

It was told to him directly: "Please stop doing this sound with your pencil.". He heard it. Twice. And he ignored it.

According to my opinion that is not fair.

That is all.

I created this topic because of I would like to protect other future contestant from such bad behavior of other people.

[quote="setout"][quote="José"]Well, my opinion is that it would have been more easily to told Peter Scholze: Please, what you do really disturbes me, cvan you stop please? Thanks.[/quote]

The Czech guy did twice[/quote]

No need to talk about it.

I am really, really sorry for this!

Actually, I would have taken another pen if I had any and I actually tried to stop clicking with the pen - but somehow I started doing it again without noticing it in any way - I was too concentrated thinking about problem 3.

I fully understand that it can prevent you from any clear thought - Naphthalin told about the situation at the german TSTs (but I for myself don't have a real problem with that; however, I can understand people having problems with it). And also, I fully understand your anger at me - in the end, those 7 points prevent you from getting a better medal and I know the feeling of missing a medal by only few points(in Athens, I missed gold by one point). However, I want to stress that there was really NO BAD INTENT by no way!!! Actually, I took the other pen in the second exam; if I had had bad intents at all, I would surely have continued to click my own pen.

Again, SORRY for my behaviour, next year I will only take pens to IMO that one can't click

Peter

[quote="Peter Scholze"]
...next year I will only take pens to IMO that one can't click

Peter[/quote]

Assuming you make IMO... :rotfl: Just kidding.  Anyway, that should clear everything up nicely.

[quote="setout"][b]Moderators[/b] please lock this topics. Its task was done.[/quote]

The very next posting shows why the moderators should NOT lock topics so often.   The discussion suddenly reached resolution within hours of the request to lock.  Locking of threads tends to occur exactly at heated or "critical" points in the discussion, but those are precisely the times when interesting developments are likeliest.

Ok. Now the discussion is indeed end. So I think this time it is ok to lock it.

I would not even come to idea that Peter does this intentionally.
And yes, HammerFall would be nice :lol:

@all that want it to be locked:
Why should any topic be locked that just reaches end of discussion¿ (I agree in locking of flame wars or already posted stuff etc.)
Give any reason!
For the counterpart: it's always possible that someone wants to add something.

PS: this offtopic discussions about locking or not get much to frequently in the last days...

[quote="ZetaX"]
@all that want it to be locked:
Why should any topic be locked that just reaches end of discussion¿ (I agree in locking of flame wars or already posted stuff etc.)
Give any reason!
For the counterpart: it's always possible that someone wants to add something.
[/quote]

Isn't the US supposed to be the country of free speech.  :wink: Of course for the sake of economically advantageous issues the US sometimes mixes up peace with imposing their thoughts, values upon other countries that are not nuclear powers yet, called democracy, fighting terrorisms, latent cultural invasion aka McDonalds or whatever.
And yeah for German pen-terrorism! We impose our sounds upon everybody. But US music is first already. Too late. But seriously, the clicking pen might have been annoying but I guess you should not exaggerate its disturbing influence.  :)
But I think telling people to lock some arbitrary topic is yet another smart approach to boost one's post count.  :D

Everything I wanted was to tell other people (I mean future IMO contestants) is that such a behavior exists. No matter if intentionally or not. I told it. That is why I just wanted to lock it. That is all.

[quote="orl"]But seriously, the clicking pen might have been annoying but I guess you should not exaggerate its disturbing influence.  :)[/quote]

As it was already said in this topic, it can completely destroy your concentration.

[quote="Xantos C. Guin"][quote="Elemennop"][quote="Naphthalin"] it's my opinion that during the exams there should be hammerfall or any similar music!
[/quote]

This would be all kinds of awesome; they should seriously do this.[/quote]

Or allow cd player's / iPods after making sure nothing on them is useful. Then again, this idea is highly impractical.[/quote]

Once in one Olympiad, they asked us whether we prefered that kind of music. Only 1 student (a friend of mine) said he doesn't want, so they didn't put the music.

Using cd's players or iPod in a competition like IMO, for me, is useless. I mean, you can disturb your partner, because there's some sound you can hear, and you just go, take the exam, and then, like IMO, you have more than a week free. So, just do the exam quiet, and then listen as much music as you want. Just to avoid things that the one that start this topic, and to try to have a good memory of a tounament and not a bad one, like setout.