Και ναι !!!  Η πρώτη μου άσκηση γραμμένη σε Latex  :)   για αυτό θα το γιαρτάσω με μια ανισότητα

( Δύσκολη κατά τη γνώμη μου )   :juggle:

Αν  a,b,c > 0 να αποδείξετε ότι $ (ab\plus{}bc\plus{}ca)(a^5\plus{}b^5\plus{}c^5\plus{}12) \ge 45abc$

[quote="vangelismak"] ( Δύσκολη κατά τη γνώμη μου )[/quote]

Αυτό είναι πάντα σχετικό . από Αμ-Γμ $ ab\plus{}bc\plus{}ca\geq 3\sqrt[3]{a^2b^2c^2}$

kai $ a^5\plus{}b^5\plus{}c^5\plus{}12\equal{}a^5\plus{}b^5\plus{}c^5\plus{}1\plus{}1...\plus{}1\geq 15\sqrt[15]{a^5b^5c^5}$

Πολλαπλασιάζοντας ... το ζητούμενο :wink:

Ναί πράγματι σχετικό είναι  :)  Πως καταφέρνεις και ξεφτυλίζεις όλες τις ανισώσεις ρε παιδί μου  :P  Τέλος πάντων η δική μου ιδέα ήταν άλλη ( Πιο σύνθετη ... ) Αν θές σου τη στλενω ...

Ευχαριστώ για τη κομψή και πολύ σύντομη λύση σου :)


Let ABC is a triangle. AD is the bisector of angle BAC .B',D',C' is on AB, AD, AC. Prove that:
B',D',C' is collinear equiv    $ (\frac{1}{{AB}} \plus{} \frac{1}{{AC}})\frac{{AD}}{{AD'}} \equal{} \frac{1}{{AB'}} \plus{} \frac{1}{{AC'}}$

$ \vec{AD}\equal{}\frac{AD}{AD'}\vec{AD'}$

$ \vec{AB}\equal{}\frac{AB}{AB'}\vec{AB'}$

$ \vec{AC}\equal{}\frac{AC}{AC'}\vec{AC'}$

but $ \vec{AD}\equal{}\frac{AB}{AB\plus{}AC}\vec{AC}\plus{}\frac{AC}{AB\plus{}AC}\vec{AB}$ equivalent to

\[ \frac{AD}{AD'}\vec{AD'}\equal{}\frac{AC}{AB\plus{}AC}\frac{AB}{AB'}\vec{AB'}\plus{}\frac{AB}{AB\plus{}AC}\frac{AC}{AC'}\vec{AC'}\]

So $ B',C',D'$ are colinear iff

\[ \frac{AD}{AD'}\equal{}\frac{AC}{AB\plus{}AC}\frac{AB}{AB'}\plus{}\frac{AB}{AB\plus{}AC}\frac{AC}{AC'}\equal{}\dfrac{\frac{1}{AB'}\plus{}\frac{1}{AC'}}{\frac{1}{AB}\plus{}\frac{1}{AC}}\]


Express f(f(x)) as a simplified fraction:

$f(x)=(1-x)^{-1}$

[quote="SM4RT"]wouldnt

$\left(\frac{1-x-1}{1-x}\right)^{-1}=\left(\frac{-x}{1-x}\right)^{-1}=\left(\frac{x}{x-1}\right)^{-1}=\boxed{\frac{x-1}{x}}$[/quote]
Yes, sorry I'll edit...


conics
parabola
inequalities
geometry
rectangle
number theory
prime factorization

1. The symbol $ *$ represents a sequence of mathematical operations. If $ 12*6 \equal{} 35$, $ 4*2\equal{}3$, $ 11*14\equal{}8$, $ 5*7\equal{}3$ and $ 10*7\equal{}8$, what is the value of $ 8*4$?

2. If the graph of $ f(x)\equal{}2x^2\plus{}bx\minus{}3$ is symmetric about the line $ x\equal{}3$, what is the value of $ b$?

3. Point A is located at (5, 5) on the Cartesian plane. Point B is also in the plane and has integer coordinates. If $ 0 < AB \leq 4$, at how many points could B be located?

[hide="2"]If the parabola is symmetric about $ x\equal{}3$, then $ 2(2^2)\plus{}2b\minus{}3\equal{}2(4^2)\plus{}4b\minus{}3$, which simplifies to $ 2b \plus{} 5 \equal{} 4b \plus{} 29$.  Solving for b, you get $ b\equal{}\minus{}12$[/hide]

[hide="1"]Prime factorize each number: $ 35\equal{}5*7, 3\equal{}3*1, 8\equal{}4*2$

Notice that the arithmetic mean of the factors is equal to the difference of the numbers on the other side.

EX: $ \frac{5\plus{}7}{2}\equal{}12\minus{}6$

This holds true for all the equations, and we can then see that $ a*b\equal{}(a\minus{}b\plus{}1)(a\minus{}b\minus{}1)$

Thus, $ 8*4\equal{}5*3\equal{}15$[/hide]

For #2, another way is:

[hide]The minimum/maximum formula for a parabola is $ \frac{\minus{}b}{2a}$ so we need to get that to equal 3, and so we get:
$ \frac{\minus{}b}{2a}\equal{}3$

$ a\equal{}2$

$ \frac{\minus{}b}{4}\equal{}3$

$ \minus{}b\equal{}12$

$ b\equal{}\minus{}12$[/hide]

mathcrazed-what made you take the prime factorization of the numbers (what made you think of taking the numbers's prime factors)?

@mathcrazed - As $ *$ is already defined in the problem, perhaps you should maybe use $ \cdot$ or $ \times$.  I got confused for a moment.

For number one, you would also notice that each number is one off from a perfect square.
And you notice that that perfect square is formed by the difference of the two numbers squared.
So...
$ (8 \minus{} 4)^2 \minus{} 1 \equal{} 15$
I got that problem on my homework.
 :|
And for number three, do you mean the length of AB?

I think for number three I got 41...
But maybe my drawing skills just suck.
 :|

[quote="alanchou"][hide="3"]
Draw a circle of radius 4 with center at A.  Translate the circle so the center is at (0,0).  The equation of this circle is $ x^2 \plus{} y^2 \equal{} 16$, and just find all integer solutions to $ x^2 \plus{} y^2\leq 16$.  I got $ \boxed{48}$, although I might have counted incorrectly
[/hide]

@Peijin (not using your username...) - $ \overline{AB}$ is the segment AB.  $ AB$ is the length of the segment AB.  At least, that's how I learned it...[/quote]

That method works fine, thats how I did it.

[quote="ZhangPeijin"]
For number one, you would also notice that each number is one off from a perfect square.
And you notice that that perfect square is formed by the difference of the two numbers squared.
So...
(8 - 4)^2 - 1 = 15
I got that problem on my homework.
Neutral
And for number three, do you mean the length of AB?[/quote]

Most of these are from the harder Warm-Ups/Workouts from 2007-2008 and past handbooks.

[hide="3"]if we select a random point A on the plane, we can draw a right triangle using that point, the origin O, and a point on the x or y axis. obviouly, OA is the hypotenuse. we know that the smallest posible value of the hypotenuse of a right triangle with integer sides is 5.(3,4,5) since AB<4, we can only pick points on the x and y axis. we can just do the same thing and take everything relative to (5,5) i get 17 [/hide]

whoops nevermind. AB doesnt have to be an integer

[quote="alanchou"][hide="3"]
Draw a circle of radius 4 with center at A.  Translate the circle so the center is at (0,0).  The equation of this circle is $ x^2 \plus{} y^2 \equal{} 16$, and just find all integer solutions to $ x^2 \plus{} y^2\leq 16$.  I got $ \boxed{48}$, although I might have counted incorrectly
[/hide]

@Peijin (not using your username...) - $ \overline{AB}$ is the segment AB.  $ AB$ is the length of the segment AB.  At least, that's how I learned it...[/quote]

@alanchou: Did you actually count all of them? I think you can create an 45-45-90 triangle with hypothenuse of 4, but I think that includes all the possible values...

[quote="alanchou"]The inequality, now that I think of it, was kinda useless.  If you look at the circle, just find the number of lattice points contained by one quarter of the circle and multiple by 4.  Translating the circle is also pretty much useless, but for me, it's easier to count lattice points.  Don't know why.

@DK-Can you clear that up a little?[/quote]

Well, what I had in mind was this: I was just wondering if you could use the method of quadriture or try to make a square or rectangle in one quadrant and multiply it by 4.

To get the points, I think you can find the $ m$ by $ n$ (I forgot the mathmatically term for it..) and do $ (m\plus{}1)\times(n\plus{}1)$

[quote="mathcrazed"]Err... For #3: Since (5,5)  can be translated to (0,0)...circle of radius 4 has a total of 7^2 lattice points... subtract the origin, and voila: 48! XD. I think that there is a formula for a circle of radius r.  $ (2r \minus{} 1)^2$ from what I've observed. of course, that's only two cases XD.   Another way to count is to take 8*8 and subtract the points outside the circle.[/quote]

try 2 for that formula and its 9*9 (not 8*8) I think

[quote="mathcrazed"]Err... For #3: Since (5,5)  can be translated to (0,0)...circle of radius 4 has a total of 7^2 lattice points... subtract the origin, and voila: 48! XD. I think that there is a formula for a circle of radius r.  $ (2r \minus{} 1)^2$ from what I've observed. of course, that's only two cases XD.   Another way to count is to take 8*8 and subtract the points outside the circle.[/quote]
How did you get $ 7^2$ as the lattice points? Sorry, I am kind of slow..

I believe it's a general pattern that he observed.  Probably not proved though


Its not hard because:
$xy \ge yz \ge yz$ and $\frac{x}{z}\ge\frac{y}{x}$  and  $\frac{x}{z}\ge\frac{z}{y}$
1)if $\frac{z}{y}\ge \frac{y}{x}$ then we have two equal ordered lines and everything OK;
2)else we 'put in order':
$xy*\frac{x}{z}+yz*\frac{y}{x}\ge xy*\frac{y}{x}+yz*\frac{x}{z}$
then  $yz*\frac{x}{z}+xz*\frac{z}{y}\ge yz*\frac{z}{y}+xz*\frac{x}{z}$
and everything OK;

[b]Approach by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=10431]Gabriel(fr):[/url][/b]

Since $ x \geq y \geq z \geq 0$ we have that,

$ \frac {x^2y}{z} \plus{} \frac {y^2z}{x} \plus{} \frac {z^2x}{y} \equal{} \frac {x^3y^2 \plus{} y^3z^2 \plus{} z^3x^2}{xyz} \geq \frac {(x^3 \plus{} y^3 \plus{} z^3)(x^2 \plus{} y^2 \plus{} z^2)}{3(xyz)} \geq$
$ \frac {3xyz(x^2 \plus{} y^2 \plus{} z^2)}{3(xyz)} \equal{} (x^2 \plus{} y^2 \plus{} z^2)$ by Chebyshev's inequality

[b]Approach by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=12562]ductrung:[/url][/b]

First, note that
$ \sum_{cycl}\frac {ab(a \minus{} b)}{c} \equal{} \frac {(a \minus{} b)(a \minus{} c)(b \minus{} c)(ab \plus{} bc \plus{} ca)}{abc} \ge 0$
Hence
$ \sum_{cycl}\frac {a^2b}{c} \ge \sum_{cycl}\frac {ab^2}{c}$
and so
$ 2\sum_{cycl}\frac {a^2b}{c}\ge \sum_{cycl}\frac {ab(a \plus{} b)}{c}$
It remains to show that
$ \sum_{cycl}\frac {ab(a \plus{} b)}{c} \ge 2(a^2 \plus{} b^2 \plus{} c^2)$
or equivalently
$ \sum_{cycl}a^2b^2(a \plus{} b) \ge 2abc(a^2 \plus{} b^2 \plus{} c^2)$
But
$ \sum_{cycl}a^2b^2(a \plus{} b) \minus{} 2abc(a^2 \plus{} b^2 \plus{} c^2) \equal{} \sum_{cycl}c^3(a \minus{} b)^2 \ge 0$

[quote="chien than"]Prove that:
$ \frac {x^{2}y}{z} \plus{} \frac {y^{2}z}{x} \plus{} \frac {z^{2}x}{y}\geq x^{2} \plus{} y^{2} \plus{} z^{2}$
where $ x;y;z$ are real numbers saisfying $ x \geq y \geq z \geq 0$[/quote]
With $ x,y,z$ are length ò 3 side of a triangle .it becomes Mongolia TST 2006 :lol:

[quote=Nguyenhuyen_AG][quote=chien than]Prove that:
$ \frac {x^{2}y}{z} \plus{} \frac {y^{2}z}{x} \plus{} \frac {z^{2}x}{y}\geq x^{2} \plus{} y^{2} \plus{} z^{2}$
where $ x;y;z$ are real numbers saisfying $ x \geq y \geq z \geq 0$[/quote]
([url='https://artofproblemsolving.com/community/u53637']Toan_VN_LC[/url]) Stronger
\[\frac {x^{2}y}{z} \plus{} \frac {y^{2}z}{x} \plus{} \frac {z^{2}x}{y}\geqslant x^{2} \plus{} y^{2} \plus{} z^{2}+\frac{(x-y)^2(y-z)^2(z-x)^2}{xyz(x+y+z)}.\][/quote]

more nice stronger

[quote=chien than]Prove that:
$ \frac {x^{2}y}{z} \plus{} \frac {y^{2}z}{x} \plus{} \frac {z^{2}x}{y}\geq x^{2} \plus{} y^{2} \plus{} z^{2}$
where $ x;y;z$ are real numbers saisfying $ x \geq y \geq z \geq 0$[/quote]

Rearrangement inequality.

[quote=adityaguharoy]

Rearrangement inequality.[/quote]
Are you sure?

Yes you got me. I was a little sloppy (or sleepy) while making the post. Usually such problems can be framed into a form where Rearrangement inequality solves everything. But maybe putting it in that form is a little tricky, or maybe not possible in apparent terms.

arqady sir, do you think uvw helps here ?


For the following problems, assume $P(x)$ to be a polynomial.

True or false:
If $P(x) \in \mathbb{Z}$ for all $x \in \mathbb{Z}$, $P(x)$ has integer coefficients.

True or false:
If $P(x)$ has integer coefficients.  If $P(x) \in \mathbb{Z},$ then $x \in \mathbb{Z}$.

True or false:
If $P(x) \in \mathbb{Z}$ and $P(x)$ has integer coefficients, then $P(x^{2}) \in \mathbb{Z}$.

True or false:
If $P(x) \in \mathbb{Z}$ and $P(x)$ has integer coefficients, then $x \in \mathbb{Z}$.

I wouldn't call the topic "polynomial proofs"  :wink:
[hide="because..."]
they're all false!

(1) $P(x)=\frac{x(x-1)}{2}$ (if $x$ integer, either $x$ or $x-1$ is even, so its product is)

(2-4) $x=2+\sqrt{3}$ and $P(x)=x^{2}-4x+1$ (for this $x$, $P(x)=0$ and $P(x^{2})=70+40\sqrt{3}$)
[/hide]

Try proving:

$P(x)$ is integer for all integers $x$ if and only if $P$ is a linear combination of binomial coefficients. That is, we have
\[P(x)=a_{0}+a_{1}x+a_{2}\cdot \frac{x(x-1)}{2}+a_{3}\cdot \frac{x(x-1)(x-2)}{6}+\cdots+a_{n}\cdot \frac{x(x-1)(x-2)\cdots (x-n+1)}{n!}\]
where the $a_{k}$'s are integers.

Sorry - I meant to put this in HSB.  Erm, is there any way to move the first post only to HSB without the responses?

[quote="scorpius119"]$P(x)$ is integer for all integers $x$ if and only if $P$ is a linear combination of binomial coefficients.
\[P(x)=a_{0}+a_{1}x+a_{2}\cdot \frac{x(x-1)}{2}+a_{3}\cdot \frac{x(x-1)(x-2)}{6}+\cdots+a_{n}\cdot \frac{x(x-1)(x-2)\cdots (x-n+1)}{n!}\]
where the $a_{k}$'s are integers.[/quote]

What a fantastic result!

[hide="Solution"] We take as given the result that $P(x)$ can be written in that form for $a_{i}$ real coefficients.  That is, we have only to prove that $\forall x \in \mathbb{Z}: P(x) \in \mathbb{Z}\Leftrightarrow a_{i}\in \mathbb{Z}, i = 0, 1, 2... n$.

The right-to-left direction is easy, since $a_{k}{x \choose k}$ is an integer for integer $x$, hence a sum of such terms is an integer for integer $x$.

The other direction isn't hard either; substituting $x = 0, 1, 2, ... n$ in turn we have inductively that $a_{0}, a_{1}, a_{2}, ... a_{n}$ are integer.  In particular, substituting $x = k$ we have

$P(k) = a_{0}+a_{1}k+...+a_{k}\frac{k(k-1)(k-2)...(k-(k-1))}{k!}+0+... = a_{0}+a_{1}k+...+a_{k}\Leftrightarrow$
$a_{k}= P(k)-a_{0}-a_{1}k-...$

Which is an integer by strong induction.  QED.  [/hide]


geometry
geometric transformation
reflection

I propose the honourable administrator of this forum to start a new topic dealing with everyday life magics with mathematics. We can call it MATHE-MAGIC!
It will reflect the thinking ability and efficiency of the users. It will also be very enjoyable afterall! Thank u!
 :)

can give instances of mathe magic, i quite didnt understand what u were talking about.


group theory
abstract algebra
superior algebra
superior algebra solved

Let $G$ be a group and let $S=\left\{xyx^{-1}y^{-1}|x,y \in G \right\}$. Let $N$ be subgroup $\left\langle S \right\rangle$ generated by $S$, which is the commutator subgroup of $G$. Show that $N \triangleleft G$ and $G/N$ is Abelian

Denote $a^{-1}b^{-1}ab$ by $[a,b]$.

It is trivial that $N$ is a group. The identity element is $[e,e]$ and the inverse of $[a,b]$ is $[b,a]$. Now we prove
that it is normal, i.e., $xhx^{-1}\in N$ where $h\in N, x\in G$. Suppose $h=[a_1,b_1][a_2,b_2]\cdots[a_n,b_n]$, then $xhx^{-1}=[xa_1x^{-1},xb_1x^{-1}][xa_2x^{-1},xb_2x^{-1}]\cdots[xa_nx^{-1},xb_nx^{-1}]\in
N$. Therefore $N$ is normal.

For the second part,  it is sufficient to show that $(ab)N=(ba)N$. It is equivalent to $b^{-1}a^{-1}ba\in N$, i.e., $[b,a]\in N$, which holds obviously.


geometry
geometric transformation
reflection
function
number theory
totient function
relatively prime

In the figure, P is a point on the circular mirror. A light ray is emitted at P making an angle theta with the normal OP. the light ray is reflected every time it reaches the mirror making hte same angle with the normal as the incident path. Given 0<theta<90, find the number of possible values of theta such that the light ray will return to P at least once within 30 reflections.

Official answer = 138.
But i can never get such a large answer ...  :(

[quote="siuhochung"]Official answer = 138.
But i can never get such a larger answer ...  :([/quote]

Hmmm.  I get an answer of 138, if we allow up to 29 reflections.  I get an answer of 153, if we allow up to 30 reflections.  [Perhaps I have an "off-by-one" error, or perhaps there is a mistake in the translation of the problem.]

My point is that if you are getting less than 138, you need to look harder.  What method are you using?  Are you asking for assistance, or simply posting an interesting problem?

i am asking for help... can you show me your solution? thanks.

I have been considering the 180-2$\theta$ angle at the center. However, my answer is far less than 138..

[quote="siuhochung"]i am asking for help... can you show me your solution? thanks.[/quote]
[hide]Categorize the solutions by number of reflections.

With 1 reflection, the only solution is if the beam of light follows a diameter of the circle, and reflects back along the same path.  $\theta=0$.  However, the problem asks for $0<\theta<90^{\rm o}$, so we do not count this solution.

With 2 reflections, the only solutions are the light path that forms the equilateral triangle inscribed in the circle.  One could follow the sides of the equilateral triangle in either the clockwise or the counterclockwise direction.  I believe the condition, $0<\theta<90^{\rm o}$ is intended to count only one of the two directions.  So we count 1 of these 2 solutions.

With 3 reflections, we could follow a diameter of the circle across and back and across and back.  However, we already considered this path.  We could also follow the light path that forms a square inscribed in the circle.  Again, clockwise or counterclockwise.  We count 1 of these 2 new solutions.

With 4 reflections, the solutions include the light path that forms a regular pentagon inscribed in the circle.  We could also follow the light path that forms a regular 5-pointed star inscribed in the circle.  Each figure has two paths.  We count 2 of these 4 solutions.

For $n$ reflections, we consider light paths that contain $n+1$ segments.  It turns out for $n+1$ segments, the total number of paths that first return to point $P$ after $n$ reflections is $\phi(n+1)$, where $\phi$ is Euler's totient function.  In each case ($n>1$), we count half of the total, due to the constraint on $\theta$.

So the formula seems to be:
$\frac{1}{2}\sum_{n=2}^{30}\Big(\phi(n+1)\Big)=$

$\frac{1}{2}(2+2+4+2+6+\cdots+28+8+30)=\frac{306}{2}=153$[/hide]

[quote="siuhochung"]can u show me the proof that the no. of paths = phi (n)? or u r only hving a guess?[/quote]
(This may be getting a bit beyond "Intermediate"...)

[hide="On a silver platter..."]  The figure constructed by your light path reflecting inside a circular mirror is a closed regular $n{\rm {-sided}}$ figure, whose sides are permitted to intersect each other.  [Regular means the sides are of equal length and turning angles at each vertex are a constant, $\alpha$].  The sum of the turning angles at each of $n$ vertices must be a multiple of $360^{\rm o}$.  So we have $n\alpha=360^{\rm o}k$.  Since $0<\alpha<360^{\rm o}$, we have $0<k<n$.  Thus, there are $n-1$ distinct positive integer values for $k$.  For each distinct value of $k$, we have a distinct turning angle, $\alpha_{n,k}=360^{\rm o}\frac{n}{k}$, and thus a distinct figure.

Now, if $k$ and $n$ share any common factor, $f$, ($f>1$), we can rewrite the last equation as $\alpha_{n,k}=360^{\rm o}\frac{n}{k}=360^{\rm o}\frac{fn'}{fk'}=360^{\rm o}\frac{n'}{k'}=\alpha_{n',k'}$.

Thus, we will retrace a figure drawn with $n'<n$ segments.

The totient function, $\phi(n)$, tells you how many positive integers, $k$, less than $n$ are relatively prime to $n$.

BTW, the original test question is essentially asking you to count half the number of terms in the [url=http://mathworld.wolfram.com/FareySequence.html]Farey sequence[/url], $F_{31}$, excluding $\frac{0}{1},~\frac{1}{2},~\frac{1}{1}$.[/hide]

i know what's the problem with my solution now, thanks~!


Sambata , 17 Martie , s-a desfasurat la Oradea Concursul Interjudetean de Matematica si Informatica "Grigore Moisil" , editia XII .
         O problema interesanta la Clasa a IX-a , suna cam asa :

    Fie $n$ un numar natural nenul . Sa se determine toate numerele reale $x$ cu proprietatea {$x$}$\le${$nx$} unde {$a$} reprezinta partea fractionara a lui $a$ .

[hide="@turcas_c"]stii unde pot gasi subiecte, rezultate?[/hide]

[hide="@mychrom"] Nu stiu sa fie publicate inca pe web , de aceea o sa pun eu aici Subiectele si Rezultatele . Inca de Bareme nu am facut rost . [/hide]

[quote="turcas_c"]Fie $n$ un numar natural nenul . Sa se determine toate numerele reale $x$ cu proprietatea {$x$}$\le${$nx$} unde {$a$} reprezinta partea fractionara a lui $a$ .[/quote]

miroase a hermite, dupa ce scriem in functie de partea intreaga etc, obtinem ${\{x }\le \{nx\}\Longleftrightarrow \sum{|x+\frac{i}{n}|}\leq (n-1)x$, cu $i \in \overline{1,n-1}$, acuma minoram suma aia cu $|t| > t-1$, si cam atata.

PS: am notat cu $|t|$ partea intreaga din $t$, caci e destul de nasol de scris in latex semnu.

asta am "saltat-o" eu in concurs  :blush: . Vazusem recent identitatea lui Hermite , cand rasfoiam o carte ... nu i-am acordat prea multa atentie ... :)) si bineinteles ca "i-am dat eu o alta forma identitatii" ... fiindca numa' nu reuseam sa mi-o amintesc  :blush:


AMC
USA(J)MO
USAJMO
AMC 12
AMC 10
USAMO
Support

I have been reflecting further on this since my posts yesterday.

There is an "elephant in the room" that everyone is ignoring.

Overall, there is a disturbing trend over the past 15 years.  The number of students taking any AMC high school contest has fallen at the same time that the high school student age population has increased.

Here is the MAA's mission statement for the AMC:  "The American Mathematics Competitions (AMC) effort is dedicated to strengthening the mathematical capabilities of our nation's youth by identifying, recognizing, and rewarding excellence in mathematics through a series of national contests."

This is a much broader mission that just picking the optimal IMO team, and that mission properly encourages the AMC to cast a much wider net.

The number of students who have the opportunity to take the AMC contests is a very small fraction of the target population, and, even worse, that percentage has been decreasing over time.  Fifteen years ago, almost 2% of American high school age students took the AMC high school contest (then called AHSME.)  In 2009, about 1% of American high school age students took any AMC high school contest.

 In 2009, the AMC reported receiving a total of 220,953 high school contests total (AMC10A/12A/10B/12B combined.)    Well over 20,000 of those contests were taken by international students (China and Taiwan alone account for over 21,000 contests!)  It's great that international student can benefit from the contests, but a comparison with 1996 figures means we should consider only US and Canadian students, the original target population, since those were the only students who took the contest in significant numbers in the mid 90s.  The US and Canadian combined population age 14 to 17 is around 18 million (roughly 16 million US and 2 million Canada).   The number of contests taken by US and Canadian students in 2009 is fewer than 200,000.     Taking into account multiplicity (the fact the many students take a test on both dates), the number of individual US and Canadian students who actually took any 2009 high school AMC test is surely below 180,000, or under 1% of the current high-school age population.

In 1996, the AMC reported receiving a total of 275,887 AHSME contest forms.  There was no multiplicity back then (no A/B or 10/12), and the number of international test-takers was also very small back then, so 274K is probably a reasonable estimate of the number of US and Canadian students who took the AHSME that year.  The mid-90s is when the bottom of the "baby bust" trough hit, so the US and Canadian combined population ages 14 to 17 was lower then, roughly 14 million.  So in 1996, the number of individual US and Canadian students who took the AHSME was closer to 2% of the high school age population.  (274K/14 million = 1.96%)

If you look at AMC statistics for the total number of contests received over the last 15 years, it's hard to escape the conclusion that there has been a steady decrease in the total number of students taking the contest, even as the population of high school age students has increased steadily during the same time.

Total answer forms received all high school tests:

1996 275,887 (*)
1997 271,091
1998 257,998
1999 249,677
2000 281,507 (**)
2001 261,430 (***)
2002 240,595 (****)
2003 244,938
2004 240,305
2005 230 855
2006 227,559
2007 227,373
2008 225,320
2009 220,953

Notes:
*Prior to 2000, only a single contest, the AHSME, was offered.
** Starting in 2000, AHSME became AMC12 and the AMC10 was offered for the first time. Schools registering for the AMC12 were offered a free trial package of 10 AMC10 contests that year only.
*** 2001 was the year that the US hosted the IMO in Washington DC, with a good deal of attendant publicity.
**** 2002 was the first year of offering A and B dates (also the year of centralized USAMO and megaMOP)

Looking at the figures above, we can see an overall trend of declining numbers of students taking any  contest, despite the growth of options from 1 AHSME (prior to 2000) to 2 AMC10/12 (in 2000 and 2001) to 4 AMC10A/12A/10B/12B (from 2002 on), and despite the great growth of international students taking the contest since 2000.

We are currently in an economic crunch in which schools need to very carefully watch their budgets.   I personally know teachers who are not sure they will have jobs next year.  In almost every state, tax revenues are way down (due to decreased income and sales) and property owners are very concerned about their tax bills.  This is an economic climate in which organizations like the AMC need to make sure that they deliver as much value as possible to school districts.

Ultimately, we should all realize that the financial support that makes the AMC possible comes from a wide variety of school budgets all over the country, and it is important for the AMC to deliver a product that serves the overall mission of the MAA/AMC.

I think that the USAJMO is a great step in a good direction, which can greatly broaden the appeal and usefulness of the AMC to a much larger population, a population that includes many students who have yet to discover AoPS.

In the light of these reflections, I now believe that I can well understand and appreciate the MAA/AMC's rationale for a 270/230 split.

(Edited: to correct for consistent use of total US/Canada population denominators)

[quote]by identifying, recognizing, and rewarding excellence in mathematics[/quote]

The key word being excellence. Just because someone took the test doesnt mean they are good at math.(or atleast, that applies to my school)

Thank you for your comments.

Here at the AMC office and at the MAA Headquarters in Washington, we constantly think about "the elephant in the room".  We are making every effort possible in order to get the word out and get more students and schools involved.  As examples, a couple of years ago we started the "College and University administration sites program" to make the B-date more available to more students.  That program is slowly expanding but still needs more sites.   Making the documentary "Hard Problems" with George Czicsery took an enormous amount of time and energy and it still consumes some time almost three years later, but we hope that viewing the AMC program in a video format makes it more well-known.  The E.M. Sliffe Trust enables us to reward teachers with high-achieving programs and give them an award as well as the students they coach.

What really makes the AMC program work though is a dedicated teacher who encourages a whole class or even a whole school to participate, often for years.  We need more of them and we need students who provide them with the positive cooperation and positive feedback that they get so seldom.

Thanks, and we welcome any ideas you have for getting more schools to participate.

Steve Dunbar
MAA Director, American Mathematics Competitions

If I had any ideas, I'd be happy to tell you - that's if I had any ideas, which I don't.

I was a 10th grader in 1969, in Oklahoma. It was a pretty strong high school in a community that valued education, and in which a significant number of parents (including mine) were engineers or scientists - but it wasn't fundamentally all that unusual a place. At some point in the school year, I was just told that on such-and-such a day I was supposed to take some test, which turned out to be the AHSME. It was school business, and it was just something that was done. This was before there was any follow-on - no AIME, no USAMO, no U.S. participation in the IMO. But hey, when I got the highest score in my school as a 10th grader I thought it was a pretty big deal.

That's what's missing in what I see around me - the idea that "you will take the AMC" being ordinary school business at any more than a handful of places. Steve Dunbar mentions the program to offer it at college sites. That's not an answer, not in numbers that mean anything at all. I've done it for two years, right smack dab in the middle of one of the largest metropolitan areas in the country, with probably five million people living within 30 minutes drive - and in two years I've gotten maybe 6 or 8 people total who were new to the AMC exams. Most of the others - and not even all that many of them - were students from certain elite schools seeking a second chance. We do have our own "event" mathematics competition and we manage to get 8 or 10 schools to come (out of 30 or so total) that have no connection to any other math competition, including the AMC. I've sent messages to the teachers at those schools who serve as our connections for our contest, explaining what the AMC is and suggesting that they send some of their students over to take the AMC with us (at an after-school hour). So far those messages have netted me no students at all - in fact, no replies. Maybe those teachers did relay my message to a few students - but it was something those students would have had to have taken some initiative to follow through on, not something presented to them as ordinary school business.

The higher participation in earlier years that sophia showed in her post (and I think that if she'd gone further back in time, she'd have found even larger numbers) came about because of schools presenting it to students as it was presented to me - as a matter of ordinary school business, as something you're expected to do. (Perhaps only expected to do if you're in certain classes, or making certain grades in certain classes, but that's still a large population.)

The AMC's participation model pushes substantial costs onto the participating schools, and depends on schools being willing to bear those costs. I'm not talking about the fee charged for the exam materials (although I imagine there are enough bureaucratic obstacles to getting official reimbursement that more than a few teachers just pay for them out of their own pockets, and them maybe turn around and collect a buck or two from the students) - no, that fee is not much of an obstacle. The problem is the need to dismiss students from classes, to free a teacher up from classes to proctor, to find a room - all of that is money in another form. And it's not going to show up on the school's state mandated and highly publicized testing, and it's not going to show up on the school's SAT averages. The AIME may be a larger burden than the AMC - fewer students, but it is three hours. And the USAMO is a burden that some schools may be unwilling to bear at all, and some others are likely to offer rather substandard accommodations to the students taking the USAMO.

(A very modest suggestion for something to approach some donor for support: a fund explicitly targeted to support schools administering the USAMO and USAJMO: if you agree to meet certain conditions about testing environment, then you get some money to hire a proctor or a substitute teacher to free up a teacher to proctor, and maybe even some money to rent some space somewhere.)

[i]Hard Problems[/i] is a start, but we need a lot more publicity than that. How can we really get recognition? Let's say a freshman student tells some random collection of parents and fellow freshman that he got a 780 Math SAT and a 109.5 AMC12. How many people out there (besides me) are going to more impressed with that AMC12 score? It just doesn't have the widespread appreciation. I'm jealous of the attention that Academic Decathlon gets, and the National Spelling Bee - how come they get the press?

I'm actually afraid that the trend is towards the AMC becoming not a school activity at all, except at a tiny handful of schools. In some ways this site (AoPS) is part of the trend - school is not really where you go to learn this stuff. And those who found this site will be highly connected and know what to do about being able to take the AMC and other contests, while most students in most schools will never know such a thing exists. There are many schools where that AMC is only offered because some long-veteran teacher has always done it and insists on it. Three years after that teacher retires, not only will the school no longer participate, but no one will remember that the school ever participated. Perhaps there's enough life in a non-school model - but I have trouble seeing how that can be large enough or inclusive enough.

I should stop now - I'm depressing myself.

I think part of this may be an artifact of the shift in the tests themselves. Their topics aren't taught in schools anymore because the topics have shifted to become more difficult. If you look at AHSMEs from 1960 or so, you'll see that they are essentially math achievement tests - very little problem solving. So the question is: what is the mission of the AMC? Is its goal to interest a wider population or to create a good, interesting competition for the elite students? If it's the former, then Dr. Merryfield's trend is troubling. If the latter, then it's only natural, and I think the direction of the tests indicates this is the case.

In terms of media attention, USAMO is just less exciting than Academic Decathlon or Spelling Bee, which can be televised and which can be easily understood by anyone, really. A layperson cannot understand the difference between a USAMO-level problem and an AIME-level problem; they are equally inscrutable. Because of this, I don't think we can ever have a well-covered USAMO, aside from the occasional article such as the NYT one from three or four months ago.

I think the trend towards standardized testing has led to some level of test fatigue among students. Even though I left the public school system after 6th grade, the tests have started to pick up again as I approach the college admissions process. For those who stayed it's much worse; at least in my state, the statewide test, the CMT, has been expanded from every other year to every year. Even before that, we did practice CMTs in the off years. The constant test preparation and then test-taking end up taking huge amounts of time. In elementary school we prepared off and on for months. The tests are all essentially the same; they just increase in difficulty a little each year. I connect this all by saying that students are less enthusiastic in general about extra tests and therefore the AMC. I don't have a typical enough perspective to be able to say this is a factor with any degree of certainty, but this is what my analysis suggests. I don't think there has been a significant enough change in the motivation of teachers to meaningfully impact the numbers.

[quote]The AMC's participation model pushes substantial costs onto the participating schools, and depends on schools being willing to bear those costs. I'm not talking about the fee charged for the exam materials (although I imagine there are enough bureaucratic obstacles to getting official reimbursement that more than a few teachers just pay for them out of their own pockets, and them maybe turn around and collect a buck or two from the students) - no, that fee is not much of an obstacle. The problem is the need to dismiss students from classes, to free a teacher up from classes to proctor, to find a room - all of that is money in another form. And it's not going to show up on the school's state mandated and highly publicized testing, and it's not going to show up on the school's SAT averages. The AIME may be a larger burden than the AMC - fewer students, but it is three hours. And the USAMO is a burden that some schools may be unwilling to bear at all, and some others are likely to offer rather substandard accommodations to the students taking the USAMO. [/quote]

Our local school district has given the AMC contests for many years, but factors such as those cited by Kent have definitely constrained the number of students who can take it.  There just aren't enough rooms available and proctors available to offer it to everyone who I believe would benefit from it.  And students who would have to miss classes other than math to take it sometimes find their teachers aren't very happy about it.

And,. wholelottabean, many of my friends are teachers who can relate to testing fatigue too.  The first two weeks of May in a school that offers a lot of APs are very disruptive for teachers--different students are missing from your class every day because they take different APs.  So having students miss a lot of time the week before APs in order to take the USAMO compounds the disruption.  And then after all that testing, everyone is wiped out and exhausted, but there are still five weeks of school left (with another disruption for Regents Exam week in June.)

The Sliffe Awards are very nice--I was surprised and touched to receive one, but I think some of those awards should be set aside to recognize teachers who encourage students from family backgrounds who don't have all the advantages that so many of my students have.

Instead of just recognizing teachers whose top 3 students do very well, how about setting aside some of the awards to recognize teachers who have encouraged large numbers of students facing difficult circumstances to try the exams and to improve their scores over time?

[quote="sophia"]And,. wholelottabean, many of my friends are teachers who can relate to testing fatigue too.  The first two weeks of May in a school that offers a lot of APs are very disruptive for teachers--different students are missing from your class every day because they take different APs.  So having students miss a lot of time the week before APs in order to take the USAMO compounds the disruption.  And then after all that testing, everyone is wiped out and exhausted, but there are still five weeks of school left (with another disruption for Regents Exam week in June.)[/quote]I don't think the USAMO is really an issue - the students taking the USAMO are first of all few and far between and second of all are the type of students who can miss two days of class without issue.

[quote]I don't think the USAMO is really an issue - the students taking the USAMO are first of all few and far between and second of all are the type of students who can miss two days of class without issue.[/quote]

But sparing teachers' time to proctor during those two days of the USAMO before the AP weeks hit IS an issue.  (I have heard of a student in another state who was forced to take the USAMO along with the students in the in-school detention room (!!!) because there was no teacher available to proctor, and also no free room that could be made available for the entire time of the contest, and students in the detention room were, in theory, supposed to be working silently.)

And I do know students who were burdened by teachers with expectations of making up classwork missed during the USAMO.  I remember recommending to a student that he be excused from attending regular classes during the mornings of the USAMO so he could sleep in a little bit.  The bus in our area picks students up before 7 a.m., and it seemed fair to me that he should be able to miss the morning classes in order to be fresh at 12:30 for a 4.5 hour exam two days in a row.  He declined the offer, saying that he would just be overwhelmed with classwork to make up afterwards if he missed his morning classes as well as the afternoon classes for the USAMO.

Fair enough; I don't spend enough time thinking from the perspective of teachers and administration.

One idea would be to publicize the AMC/AIME/USAMO scores as an actual test of math ability and market it to colleges. Certainly you can make the argument that these tests differentiate people a lot more than the SATs. For example, an 800 on the SAT math is nice but it lots of people get that score. But getting a 10 on the AIME represents a lot more in terms of thinking ability.

I know that a few colleges look at the AMC test scores (like MIT) but if you could get a lot more colleges to recognize these test scores, then people would suddenly be a lot more interested in taking the tests. (They might actually learn something too.)

I am not sure if playing off the fact that people play "the college game" would be contrary to the mission of the MAA or something like that.

But wouldn't it be great if you could get people to prepare for the AMC exams like they do for the SATs :)

I'll be honest, and say that I certainly don't want to go there ("publicize scores as an actual test of math ability and market it to colleges") and I think I can say confidently that the MAA leadership doesn't want to go there either.  One reason is that to do so would "bureaucrat-ize" what is now a good mathematical experience and kill it.

To create a test that is an "actual test of ability" and to "market it", one has to demonstrate (at least) that it has  validity and reliability.  Validity means that a test measures what it claims to measure and reliability means that it measures it the same way each time, with the same results.  The AMC contests are created annually from scratch by a group of mathematicians and math educators proposing what they believe are interesting problems.  In fact, as mathematicians we have good reason to believe that the AMC contests are valid and measure math ability, but I would hate to have to prove it to somebody.  Likewise, I think our reliability is fair, but the annual debates about which A-date or B-date contest is harder may cause you to doubt it.  I have annual contact with the College Board people and hear a report about their efforts to make the SAT and the AP exams both valid and reliable.  They spend vast amounts of money and spend years in testing to create tests that are valid and reliable and fair.

I'd hate to see somebody say the AMC exams prevented them from getting into College X because we didn't take the time to make sure the test was valid or reliable or fair.

On the other hand, as a competition to measure your mathematical ingenuity against the interesting problems we create I  like the AMC  contests.  I hope that others do too, and spread the word, and learn some math and have some fun in getting challenged.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="Altheman"]But wouldn't it be great if you could get people to prepare for the AMC exams like they do for the SATs :)[/quote]Against the 0 hours I prepared for the SAT, I have to say my preparation for the AMC/AIME/USAMO is pretty big...


find minimal of expression $P= \frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}+\frac{(a+b+c)^{3}}{abc}$ where $a,b,c >0$

By EV-theorem or SMV-theorem we can conclude that the expression is only attained the minimum value if and only if $a=b$. But maybe we need more caculation.

[quote="April"][quote="zhaobin"]
$\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}+\frac{(a+b+c)^{3}}{abc}-10$

[/quote]I think it must
$\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}+\frac{(a+b+c)^{3}}{abc}-28$[/quote]
 :blush: Sorry,you are right.Thank you


geometry
trapezoid
trigonometry
geometry proposed

Let $ABCD$ be a isosceles trapezoid ($AD\parallel BC$), $AB=BC$, inscribed in the circle $C=C(O,1)$. I note the projections $S_b$, $S_c$, $S_d$ of the center $O$ on the lines $AB$, $AC$, $AD$ respectively.
Prove that $OS_b+OS_c+OS_d$ $\ge$ $1$.

[quote="Virgil Nicula"]Let $ABCD$ be a isosceles trapezoid ($AD\parallel BC$), $AB=BC$, inscribed in the circle $C=C(O,R)$. I note the projections $S_b$, $S_c$, $S_d$ of the center $O$ on the lines $AB$, $AC$, $AD$ respectively.
Prove that $OS_b+OS_c+OS_d$ $\ge$ $1$.[/quote]

The enunciation is not true.  Just take a circle with a very very small radius,

but a little bit longer than  epsilon, or it will be hard to inscribe a trapezoid in it

using compass and straight edge.

  Thank you.

   M.T.

Thanks, [i]Armpist[/i], you are true. I forgot $C(O,1)$, i.e. $R=1$. I corrected above.

[u]A solution with complex numbers.[/u]

$A(1),\ B(z),\ C(z^2),\ D(z^3)\in C(O,1)\Longrightarrow$

$OS_b=\frac 12 |1+z|,\ OS_c=\frac 12 |1+z^2|,\ OS_d=\frac 12 |1+z^3|.$

$2=|2z|=|(1+z)+z(1+z^2)-(1+z^3)|\le |1+z|+|1+z^2|+|1+z^3|$

$\Longrightarrow OS_b+OS_c+OS_d\ge 1.$

[u]Remark.[/u] I note $2x=\arg z$. Thus,

$x\in \left[ 0,\frac{\pi}{3}\right]\Longrightarrow \cos x+|\cos 2x|+|\cos 3x|\ge 1.$


geometry
3D geometry
superior algebra
superior algebra unsolved

x^3 + Bx^2 + Cx + D = 0

Hanssens Theorem

if C = (B^2)/3
then
x = ((B^3)/27 - D)^(1/3) - (B/3)

Does this mean that if C can be written as (B^2)/3, there is only one solution... and can all third degree equtaions with only one soloution be written like this:

x^3 + Bx^2 + ((B^2)/3)x + D = 0

?

The cube root has three values, in the complex plane.

If you're assuming the coefficients to be real and asking about real roots, then this condition implies that there is only one real root- but the converse is clearly false. A cubic with only one real root can be solved by taking a cube root (look up Cardano's formula), but other algebraic manipulation is needed, including a square root.

A random example: $x^3+x+34=0$. This has one real root, and it's not close to your form.


Let $ a_1$ be a natural number not divisible by $ 5$. The sequence $ a_1, a_2, a_3, . . .$ is defined by $ a_{n\plus{}1} \equal{} a_n \plus{} b_n$, where $ b_n$ is the last digit of $ a_n$. Prove that the sequence contains infinitely many powers of two.

[quote="outback"]Let $ a_1$ be a natural number not divisible by $ 5$. The sequence $ a_1, a_2, a_3, . . .$ is defined by $ a_{n \plus{} 1} \equal{} a_n \plus{} b_n$, where $ b_n$ is the last digit of $ a_n$. Prove that the sequence contains infinitely many powers of two.[/quote]
[hide="Solution"]
The second number of the sequence must be even since it is congruent to $ 2a_1\bmod 10$.  Notice that if $ a_2$ ends with a $ 2$, then we add $ 2$, so $ a_3$ ends with a $ 4$, so we add $ 4$, so $ a_4$ ends with an $ 8$, so we add $ 8$, and then $ a_5$ ends with a $ 6$, so we add $ 6$, and then $ a_6$ ends with a two again.  Thus, $ a_6 \minus{} a_2 \equal{} 20$.  In fact, we can do the same if $ a_2$ ends with a $ 4$, $ 6$, or $ 8$.  Analogously, $ a_{k \plus{} 1} \equal{} a_{k \plus{} 5} \plus{} 20$ for a positive integer $ k$.  Now, if $ a_{k \plus{} 1}\equiv 2^n\bmod 20$ for some $ n$, then we can continously add $ 20$ and get a number in the sequence that is a power of $ 2$.  In fact, we can do this infinitely times to get infinitely many powers of $ 2$ since $ 2^n\equiv 2^{n \plus{} 4}\bmod 20$ when $ n\ge 2$.  Thus, if $ a_{k \plus{} 1}$ is congruent to $ 4$, $ 8$, $ 12$, or $ 16\bmod 20$, then we can continuously add twenty.  This means that we are done if $ a_{k \plus{} 1}\equiv 0\bmod 4$.  If not, then if $ a_{k \plus{} 1}$ must be congruent to $ 2$, $ 6$, $ 14$, or $ 18\bmod 20$.  If it is congruent to $ 2\bmod 20$, then it ends with $ 2$, so we add $ 2$, so $ a_{k \plus{} 2}\equiv 4\bmod 20$, meaning we can repeat the process described above with $ a_{k \plus{} 2}$ rather than $ a_{k \plus{} 1}$.  Similarly, if $ a_{k \plus{} 1}\equiv 6\bmod 20$, then $ a_{k \plus{} 2}\equiv 12\bmod 20$, so we can do the process with $ a_{k \plus{} 2}$.  Now, if $ a_{k \plus{} 1}\equiv 14\bmod 20$, then $ a_{k \plus{} 2}\equiv 18\bmod 20$, so $ a_{k \plus{} 3}\equiv 6\bmod 20$, so $ a_{k \plus{} 4}\equiv 12\bmod 20$ and we can do the process with $ a_{k \plus{} 4}$.  Finally, if $ a_{k \plus{} 1}\equiv 18\bmod 20$, then $ a_{k \plus{} 3}\equiv 12\bmod 20$.  All the cases are finished, and for each case, there is a way to get infinitely many powers of two, so we are done. [/hide]


[quote="Potla"]What is the restriction on $ k$?
$ k\ge0$, I think......  :maybe:[/quote]
I'm sorry,I forgot,the restriction on $ k$ is $ k\ge0$

[quote="vo thanh van"][quote="vo thanh van"]Let $ a,b,c,k\ge 0$ such that $ a \plus{} b \plus{} c \equal{} 3$.Prove that $ \sum\frac {\sqrt {a^2 \plus{} kb}}{a^2 \plus{} kab \plus{} b^2}\ge \frac {3\sqrt {k \plus{} 1}}{k \plus{} 2}$[/quote]
How about this problem?[/quote]
It 's wrong,Try $ k\equal{}c \equal{}0,a\equal{}b\equal{}\frac{3}{2}$ :wink:


calculus
integration
geometry
function
real analysis
counting
distinguishability

[b] 1. [/b] Find the area, mass, and center of mass of a thin plate bounded by $ x^2\minus{}y^2 \equal{}1$ and $ x \equal{} 4$ with density $ \rho(x,y) \equal{} x$.

Computing these, should be OK. Just wondering about the limits of integration. Would it be $ x \equal{} 0$ to $ x \equal{} 4$ and $ y \equal{} 0$ to $ y \equal{} \sqrt{x^2\minus{}1}$?

[b] 2. [/b] Let $ f(x,y) \equal{} \begin{cases} 1, \ \ \ \ \ \text{if} \ x  \ \text{is irrational} \\ 4y^{3}, \ \ \text{if not} \end{cases}$

Show that $ \int_{0}^{1} \left(\int_{0}^{1} f(x,y) \ dy \right) \ dx \equal{} 1$, but that $ \int_{0}^{1} \left(\int_{0}^{1} f(x,y) \ dx \right) \ dy$ does not exist.

Then this implies that $ \int_{0}^{1} f(x,y) \ dy \equal{} 1$ or $ \int f(x,y) \ dy \equal{} y$. So we only integrate it when $ f(x,y) \equal{} 1$? And for the second case, the function may be similar to $ \int_{0}^{1} \frac{1}{x^2} \ dx$ which does not exist?

The only problem is, when we are integrating from $ 0$ to $ 1$, we integrate over both rationals and irrationals. So maybe the second case deals with this, and that is why it doesn't exist. Whereas, in the first case, we keep $ x$ constant, and integrate over the rationals between $ 0$ and $ 1$?

[quote="ragingbullrocky"]
Show that $ \int_{0}^{1} \left(\int_{0}^{1} f(x,y) \ dy \right) \ dx \equal{} 1$, but that $ \int_{0}^{1} \left(\int_{0}^{1} f(x,y) \ dx \right) \ dy$ does not exist.
[/quote]
this isn't true
it may not be continuous, but you can still easily take the integral. It is equal to 1

I suppose if depending on how you define integrals, you could say that the integral of a discontinuous function doesn't exist, but that is a bit weird. It's like saying something doesn't exist just because you don't feel like finding it.

$ \int f(x,y) \ dy \equal{} y$?
what is up with that? you can't have y outside the integral because it is the variable you are integrating with respect to

The second integral is not [url=http://en.wikipedia.org/wiki/Riemann_integral]Riemann integrable[/url] (the integrand can be written in terms of the characteristic function of the rationals), but it is likely [url=http://en.wikipedia.org/wiki/Lebesgue_integration]Lebesgue integrable[/url].

In the Lebesgue sense, it's indistinguishable from the constant function $ 1$. In the Riemann sense, it's discontinuous everywhere (except the line $ y^3\equal{}\frac14$), and that makes it hopeless.


algebra
polynomial
algorithm
calculus
integration
algebra theorems

It's true that every polynomial with coefficients equal only to $ \minus{} 1,0,1$ when will be present in products of irreducible polynomials then all this polynomials have too coefficients equal to $ \minus{} 1,0,1$ ?? Does exist method (algorithm) how to present polynomial with integral coefficient as product of irreducible polynomials??

False.  $x^{105}-1$ has a factor that violates the claim (specifically, $\Phi_{105}(x)$).  See http://en.wikipedia.org/wiki/Polynomial_factorization .


Look here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=%5C%5Cbc&t=35839
Post 11.

Thank you, and sorry about this useless post.

Hello arqady what do you think can we show this inequality by using Jenssens inequality ?  ;)
Lets try another ways Mathematicians ...

Davron

[quote="Davron"]Hello arqady what do you think can we show this inequality by using Jenssens inequality ?  ;)
Davron[/quote]
For you, Davron.   :lol:
$f(x)=\frac{1}{\sqrt x}$ is convex for $x>0.$ Hence,
$\sqrt x+\sqrt y+\sqrt z=3\cdot(\frac{x}{3\sqrt x}+\frac{y}{3\sqrt y}+\frac{z}{3\sqrt z})\geq\frac{3}{\sqrt{\frac{x^2+y^2+z^2}{3}}}.$
And  $\frac{3}{\sqrt{\frac{x^2+y^2+z^2}{3}}}\geq xy+xz+yz\Leftrightarrow27\geq(x^2+y^2+z^2)(xy+xz+yz)^2\Leftrightarrow$
$\Leftrightarrow(x+y+z)^6\geq27(x^2+y^2+z^2)(xy+xz+yz)^2.$ This is true.
 I found a nice proof  for the last inequality. ;)

[quote="Davron"]This problem is from Russian MO i dont remember which year but its from there

If you want i can put here a nicer solution ... :)

Davron[/quote]

Ιt was  Russian mo, 2002, grade 10.

[quote="arqady"]$\Leftrightarrow(x+y+z)^6\geq27(x^2+y^2+z^2)(xy+xz+yz)^2.$ This is true.
 I found a nice proof  for the last inequality. ;)[/quote]Please, may you share it with us?

[quote="spanferkel"]Please, may you share it with us?[/quote]
Certainly, only we shall do it here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78236

[quote=scytale]$\sqrt{x} + \sqrt{y} + \sqrt{z} \ge {xy} + {yz} + {zx}$ , where ${x + y + z} = 3$

I didn't managed to solve it, maybe someone from here could help me. Sorry if the idea of this ineq. was already in used in other problems posted on this forum...[/quote]
[url=http://www.artofproblemsolving.com/community/c6h76814p441058]RMO 2004, Grade 7, Problem 2:[/url]
The sidelengths of a triangle are $a,b,c$.

(a) Prove that there is a triangle which has the sidelengths $\sqrt a,\sqrt b,\sqrt c$.

(b) Prove that $\displaystyle \sqrt{ab}+\sqrt{bc}+\sqrt{ca} \leq a+b+c < 2 \sqrt{ab} + 2 \sqrt{bc} + 2 \sqrt{ca}$.

[quote=scytale]$\sqrt{x} + \sqrt{y} + \sqrt{z} \ge {xy} + {yz} + {zx}$ , where ${x + y + z} = 3$
[/quote]
Let $a,b,c\geq0 , a+b+c=3$ . [url=https://artofproblemsolving.com/community/c6h1064117p8392299]Prove that[/url]
$$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}+21\geq 8(ab+bc+ca).$$

[quote=iliyash][quote=scytale]$\sqrt{x} + \sqrt{y} + \sqrt{z} \ge {xy} + {yz} + {zx}$ , where ${x + y + z} = 3$

I didn't managed to solve it, maybe someone from here could help me. Sorry if the idea of this ineq. was already in used in other problems posted on this forum...[/quote]

$a,b,c\geq0 , a+b+c=3$ . Prove that:
$$3\ge\sqrt{a}+\sqrt{b}+\sqrt{c}\ge ab+bc+ca$$[/quote]
$$\sqrt{a}+\sqrt{b}+\sqrt{c}\le\sqrt{3(a+b+c)}$$

[quote=iliyash][quote=scytale]$\sqrt{x} + \sqrt{y} + \sqrt{z} \ge {xy} + {yz} + {zx}$ , where ${x + y + z} = 3$

I didn't managed to solve it, maybe someone from here could help me. Sorry if the idea of this ineq. was already in used in other problems posted on this forum...[/quote]

$a,b,c\geq0 , a+b+c=3$ . Prove that:
$$3\ge\sqrt{a}+\sqrt{b}+\sqrt{c}\ge ab+bc+ca$$[/quote]

The left inequality is just C-S. For the right see here:
https://artofproblemsolving.com/community/c6h497575

[quote=scytale]$\sqrt{x} + \sqrt{y} + \sqrt{z} \ge {xy} + {yz} + {zx}$ , where ${x + y + z} = 3$

I didn't managed to solve it, maybe someone from here could help me. Sorry if the idea of this ineq. was already in used in other problems posted on this forum...[/quote]

The problem is equivalent to that:
$x+y+z \ge {x^2y^2 +y^2z^2 + z^2x^2}$ , where ${x^2 + y^2 + z^2} = 3 (x,y,z\ge{0})$
By computer,we have
$$LHS-RHS=\frac{1}{2}\sum_{cyc}{x(x+2)(x-1)^2}\ge{0}$$


calculus
integration
logarithms
trigonometry
function
calculus computations

Calculate the following indefinite integrals.

[1] $\int x(x^2+3)^2 dx$

[2] $\int \ln (x+2) dx$

[3] $\int x\cos x dx$

[4] $\int \frac{dx}{(x+2)^2}dx$

[5] $\int \frac{x-1}{x^2-2x+3}dx$

Hello! diophantus.

You are right!  As For [5], since $x^2-2x+3>0$, it would be good that $\frac{1}{2}\ln (x^2-2x+3)+C$

Yeah.  I didn't determine the function was entirely positive.


probability
advanced fields
advanced fields unsolved

define formally a probabilty space with N-independent random variables taking values 1 and -1 ,and only these with equal probabilities

So... what's the problem?

It doesn't matter what the formal definition of a probability space is; only what can be done with it is meaningful.


Let $ H$ be the orthocenter of a triangle $ ABC$ with $ \widehat{B}\leq 90$ and $ \widehat{C}\leq 90$, which is inscribed in a circle of center $ K$ and radius $ R\equal{}1$. If $ S$ is the point where the lines $ HK$ and $ BC$ intersect, and $ SK(SK\minus{}SH)\equal{}1$, find the area of $ ABHCA$.  :)

Please give a detailed complex number solution :wink:

Double post. See [url]http://www.mathlinks.ro/viewtopic.php?t=197399[/url].


modular arithmetic
number theory unsolved
number theory

[quote="scarface"]If $ x \plus{} 19y \equiv (\bmod 23)$ and $ x \plus{} y < 69$ , what is number of  $ (x,y)$ positive integer pairs? (answer: 100)[/quote]

What's the meaning of $ x \plus{} 19y \equiv (\bmod 23)$  ?

$ x \plus{} 19y \equiv x \minus{} 4y \equiv 0 \mod{23}$

$ x \minus{} 4y \equal{} 23k \equal{} \{0,23,46\}$

I dunno if you can do something with that. :|

[quote="scarface"]If $ x \plus{} 19y \equiv 0(\bmod 23)$ and $ x \plus{} y < 69$ , what is number of  $ (x,y)$ positive integer pairs? (answer: 100)[/quote]

The equation is $ x\equal{}4y\pmod {23}$ and so we want to solve

$ y>0$
$ 4y\plus{}23k>0$ (value of $ x$)
$ 5y\plus{}23k<69$ (value of $ x\plus{}y$)

And so $ \frac{69\minus{}5y}{23}> k >\frac{\minus{}4y}{23}$

For $ y\equal{}1$, we get $ 2\geq k \geq 0$ and so $ 3$ solutions
For $ y\equal{}2$, we get $ 2\geq k \geq 0$ and so $ 3$ solutions
For $ y\equal{}3$, we get $ 2\geq k \geq 0$ and so $ 3$ solutions
For $ y\equal{}4$, we get $ 2\geq k \geq 0$ and so $ 3$ solutions
For $ y\equal{}5$, we get $ 1\geq k \geq 0$ and so $ 2$ solutions
For $ y\equal{}6$, we get $ 1\geq k \geq \minus{}1$ and so $ 3$ solutions
For $ y\equal{}7$, we get $ 1\geq k \geq \minus{}1$ and so $ 3$ solutions
For $ y\equal{}8$, we get $ 1\geq k \geq \minus{}1$ and so $ 3$ solutions
For $ y\equal{}9$, we get $ 1\geq k \geq \minus{}1$ and so $ 3$ solutions
For $ y\equal{}10$, we get $ 0\geq k \geq \minus{}1$ and so $ 2$ solutions
For $ y\equal{}11$, we get $ 0\geq k \geq \minus{}1$ and so $ 2$ solutions
For $ y\equal{}12$, we get $ 0\geq k \geq \minus{}2$ and so $ 3$ solutions
For $ y\equal{}13$, we get $ 0\geq k \geq \minus{}2$ and so $ 3$ solutions
For $ y\equal{}14$, we get $ \minus{}1\geq k \geq \minus{}2$ and so $ 2$ solutions
For $ y\equal{}15$, we get $ \minus{}1\geq k \geq \minus{}2$ and so $ 2$ solutions
For $ y\equal{}16$, we get $ \minus{}1\geq k \geq \minus{}2$ and so $ 2$ solutions
For $ y\equal{}17$, we get $ \minus{}1\geq k \geq \minus{}2$ and so $ 2$ solutions
For $ y\equal{}18$, we get $ \minus{}1\geq k \geq \minus{}3$ and so $ 3$ solutions
For $ y\equal{}19$, we get $ \minus{}2\geq k \geq \minus{}3$ and so $ 2$ solutions
For $ y\equal{}20$, we get $ \minus{}2\geq k \geq \minus{}3$ and so $ 2$ solutions
For $ y\equal{}21$, we get $ \minus{}2\geq k \geq \minus{}3$ and so $ 2$ solutions
For $ y\equal{}22$, we get $ \minus{}2\geq k \geq \minus{}3$ and so $ 2$ solutions
For $ y\equal{}23$, we get $ \minus{}3\geq k \geq \minus{}3$ and so $ 1$ solution
For $ y\equal{}24$, we get $ \minus{}3\geq k \geq \minus{}4$ and so $ 2$ solutions
For $ y\equal{}25$, we get $ \minus{}3\geq k \geq \minus{}4$ and so $ 2$ solutions
For $ y\equal{}26$, we get $ \minus{}3\geq k \geq \minus{}4$ and so $ 2$ solutions
For $ y\equal{}27$, we get $ \minus{}3\geq k \geq \minus{}4$ and so $ 2$ solutions
For $ y\equal{}28$, we get $ \minus{}4\geq k \geq \minus{}4$ and so $ 1$ solution
For $ y\equal{}29$, we get $ \minus{}4\geq k \geq \minus{}5$ and so $ 2$ solutions
For $ y\equal{}30$, we get $ \minus{}4\geq k \geq \minus{}5$ and so $ 2$ solutions
For $ y\equal{}31$, we get $ \minus{}4\geq k \geq \minus{}5$ and so $ 2$ solutions
For $ y\equal{}32$, we get $ \minus{}4\geq k \geq \minus{}5$ and so $ 2$ solutions
For $ y\equal{}33$, we get $ \minus{}5\geq k \geq \minus{}5$ and so $ 1$ solution
For $ y\equal{}34$, we get $ \minus{}5\geq k \geq \minus{}5$ and so $ 1$ solution
For $ y\equal{}35$, we get $ \minus{}5\geq k \geq \minus{}6$ and so $ 2$ solutions
For $ y\equal{}36$, we get $ \minus{}5\geq k \geq \minus{}6$ and so $ 2$ solutions
For $ y\equal{}37$, we get $ \minus{}6\geq k \geq \minus{}6$ and so $ 1$ solution
For $ y\equal{}38$, we get $ \minus{}6\geq k \geq \minus{}6$ and so $ 1$ solution
For $ y\equal{}39$, we get $ \minus{}6\geq k \geq \minus{}6$ and so $ 1$ solution
For $ y\equal{}40$, we get $ \minus{}6\geq k \geq \minus{}6$ and so $ 1$ solution
For $ y\equal{}41$, we get $ \minus{}6\geq k \geq \minus{}7$ and so $ 2$ solutions
For $ y\equal{}42$, we get $ \minus{}7\geq k \geq \minus{}7$ and so $ 1$ solution
For $ y\equal{}43$, we get $ \minus{}7\geq k \geq \minus{}7$ and so $ 1$ solution
For $ y\equal{}44$, we get $ \minus{}7\geq k \geq \minus{}7$ and so $ 1$ solution
For $ y\equal{}45$, we get $ \minus{}7\geq k \geq \minus{}7$ and so $ 1$ solution
For $ y\equal{}46$, we get $ \minus{}8\geq k \geq \minus{}7$ and so  no solution
For $ y\equal{}47$, we get $ \minus{}8\geq k \geq \minus{}8$ and so $ 1$ solution
For $ y\equal{}48$, we get $ \minus{}8\geq k \geq \minus{}8$ and so $ 1$ solution
For $ y\equal{}49$, we get $ \minus{}8\geq k \geq \minus{}8$ and so $ 1$ solution
For $ y\equal{}50$, we get $ \minus{}8\geq k \geq \minus{}8$ and so $ 1$ solution
For $ y\equal{}51$, we get $ \minus{}9\geq k \geq \minus{}8$ and so  no solution
For $ y\equal{}52$, we get $ \minus{}9\geq k \geq \minus{}9$ and so $ 1$ solution
For $ y\equal{}53$, we get $ \minus{}9\geq k \geq \minus{}9$ and so $ 1$ solution
For $ y\equal{}54$, we get $ \minus{}9\geq k \geq \minus{}9$ and so $ 1$ solution
For $ y\equal{}55$, we get $ \minus{}9\geq k \geq \minus{}9$ and so $ 1$ solution
For $ y\equal{}56$, we get $ \minus{}10\geq k \geq \minus{}9$ and so  no solution
For $ y\equal{}57$, we get $ \minus{}10\geq k \geq \minus{}9$ and so  no solution
For $ y\equal{}58$, we get $ \minus{}10\geq k \geq \minus{}10$ and so $ 1$ solution
For $ y\equal{}59$, we get $ \minus{}10\geq k \geq \minus{}10$ and so $ 1$ solution
For $ y\equal{}60$, we get $ \minus{}11\geq k \geq \minus{}10$ and so  no solution
For $ y\equal{}61$, we get $ \minus{}11\geq k \geq \minus{}10$ and so  no solution
For $ y\equal{}62$, we get $ \minus{}11\geq k \geq \minus{}10$ and so  no solution
For $ y\equal{}63$, we get $ \minus{}11\geq k \geq \minus{}10$ and so  no solution
For $ y\equal{}64$, we get $ \minus{}11\geq k \geq \minus{}11$ and so $ 1$ solution
For $ y\equal{}65$, we get $ \minus{}12\geq k \geq \minus{}11$ and so  no solution
For $ y\equal{}66$, we get $ \minus{}12\geq k \geq \minus{}11$ and so  no solution
For $ y\equal{}67$, we get $ \minus{}12\geq k \geq \minus{}11$ and so  no solution
For $ y\equal{}68$, we get $ \minus{}12\geq k \geq \minus{}11$ and so  no solution

And so $ 100$ solutions.

What an interesting problem !

Your conditions are $ x \equal{} 4y\mod 23$ and $ x \plus{} y < 69,x > 0,y > 0$.
If $ y \equal{} 0\mod 23$, then $ x \equal{} y \equal{} 23$ - 1 solutions.
Consider solutions $ y \equal{} 1,2,..,11, x \equal{} 4y\mod 23$. For all solution (x,y) we had exactly 9 solution:
$ (x,y),(23 \minus{} x,23 \minus{} y),(x \plus{} 23,y),(46 \minus{} x,23 \minus{} y),(x,y \plus{} 23),(23 \minus{} x,46 \minus{} y$ and
$ (x \plus{} 23,y \plus{} 23),(x \plus{} 46,y),(x,y \plus{} 46)$ if $ x \plus{} y < 23$
$ (46 \minus{} x,46 \minus{} y),(69 \minus{} x,23 \minus{} y),(23 \minus{} x,69 \minus{} y)$ -else.
Therefore we had $ 1 \plus{} 11*9 \equal{} 100$ solutions.
By these way we get, that number of solutions
$ ax\plus{}by\equal{}0\mod p, p\not |ab$ p - prime, $ x\plus{}y<kp,x>0,y>0$ is $ k\minus{}2\plus{}k^2\frac{p\minus{}1}{2}$.

[quote="Rust"] ... Therefore we had $ 1 \plus{} 11*9 \equal{} 100$ solutions.
By these way we get, that number of solutions
$ ax \plus{} by \equal{} 0\mod p, p\not |ab$ p - prime, $ x \plus{} y < kp,x > 0,y > 0$ is $ k \minus{} 2 \plus{} k^2\frac {p \minus{} 1}{2}$.[/quote]

Hello Rust!

I did understand nothing to your demo (as frequently, since you often give us very compact texts :) ) but what I'm sure about your formula is :

1) that it magically gives the good result for the required problem
2) that it does not always give the good result :

examples :
$ a\equal{}1$, $ b\equal{}2$, $ p\equal{}3$, $ k\equal{}1$ and your formula gives $ 0$ solutions instead of $ \plus{}1$ solution : $ x\equal{}y\equal{}1$
$ a\equal{}1$, $ b\equal{}1$, $ p\equal{}3$, $ k\equal{}2$ and your formula gives $ 4$ solutions instead of $ \plus{}2$ solutions : $ x\equal{}1,y\equal{}2$ and $ x\equal{}2,y\equal{}1$
$ a\equal{}1$, $ b\equal{}4$, $ p\equal{}5$, $ k\equal{}1$ and your formula gives $ 1$ solution instead of $ \plus{}2$ solutions : $ x\equal{}y\equal{}1$ and $ x\equal{}y\equal{}2$

Calculate number of solutions solutions (x,y), suth that $ p|x,p|y$. It give solutions (p,p),...(p,(k-2)p),(2p,p),...(2p,(k-3)p),..
$ \frac {(k \minus{} 2)(k \minus{} 1)}{2}$. (I have mistake, when write k-2).
Calculate solutions $ (x \equal{} \minus{} bt\mod p,y \equal{} bt\mod p$), suth that $ (0 < x < p,0 < y\le (p \minus{} 1)/2)$.
There are $ \frac {p \minus{} 1}{2}$ solutions.
For eath solution we have solutions $ (x \plus{} mp,y \plus{} lp),(cp \plus{} p \minus{} x,dp \plus{} p \minus{} y)$, were $ 0\le m,l, m \plus{} l\le k \minus{} 1$, $ 0\le c,d,c \plus{} d\le k \minus{} 2$,
 if $ x \plus{} y < p$ and $ 0\le m,l,m \plus{} l\le k \minus{} 2,0\le c,d,c \plus{} d\le k \minus{} 1$ else.
There are $ \frac {k(k \plus{} 1)}{2} \plus{} \frac {k(k \minus{} 1)}{2} \equal{} k^2$ solutions with eath solution $ (x,y)$, were $ k\ge 2$.
Therefore number of solutions is $ N \equal{} \frac {(k \minus{} 1)(k \minus{} 2)}{2} \plus{} k^2\frac {p \minus{} 1}{2}. (k\ge 2, p\ge 3)$

[quote="Rust"] ... Therefore number of solutions is $ N \equal{} \frac {(k \minus{} 1)(k \minus{} 2)}{2} \plus{} k^2\frac {p \minus{} 1}{2}. (k\ge 2, p\ge 3)$[/quote]

Thanks for the correction.

It still does not work for $ a\equal{}1, b\equal{}1, p\equal{}3, k\equal{}2$ : your formula still gives $ 4$ solutions and I'm sure, as I said in my previous post, that there are only $ 2$ : $ x\equal{}1, y\equal{}2$ and $ x\equal{}2, y\equal{}1$

 :oops:

You are right. When $ a \equal{} b$ these formula don't work, because $ p|ax \plus{} by\to p|x \plus{} y$ and I calculate solutions, suth that $ p\not |y,p\not |x \plus{} y$. I use, that if $ (x,y)$ solution $ 0 < x < p,0 < y < p$, then or $ x \plus{} y < p$ or $ p < x \plus{} y < 2p$ (except case $ x \plus{} y \equal{} p$).

[quote="Rust"]You are right. When $ a = b$ these formula don't work, because $ p|ax + by\to p|x + y$ and I calculate solutions, suth that $ p\not |y,p\not |x + y$. I use, that if $ (x,y)$ solution $ 0 < x < p,0 < y < p$, then or $ x + y < p$ or $ p < x + y < 2p$ (except case $ x + y = p$).[/quote]

Ok, :), and that's fine for me now, thanks. And I understood your idea (though I do not agree each line). I hope you'll allow me to rewrite your demo in a little bit different manner (but, sure, it's yours) :

1) Get rid of pathologic cases:
1.1) $ a=b=0\pmod p$
Then all couple $ (x,y)$ is a solution, and, for a fixed $ x$, we have $ y\in[1,kp-1-x]$ and so $ N=\sum_{i=1}^{kp-2}(kp-1-i)$ $ =\frac{(kp-1)(kp-2)}{2}$

1.2) $ a=0\pmod p$ and $ b\neq 0 \pmod p$ (or the inverse situation $ a\neq 0\pmod p$ and $ b= 0 \pmod p$
In this case, $ by=0\pmod p$ and so $ y=up$ and any $ x\in[1,kp-up-1]$, and so $ N=\sum_{u=1}^{k-1}(kp-up-1)$ $ =(k-1)(\frac{kp}{2}-1)$

1.3) $ ab\neq 0\pmod p$ but $ a=b\pmod p$
Then the equation becomes ${ x+y=0\pmod p}$ and so $ x+y=up$ for any $ u\in[1,k-1]$
For a given $ u$, we have exacty $ up-1$ ($ (1,up-1),(2,up-2)\to(up-1,1)$) such couples and so $ N=\sum_{u=1}^{k-1}(up-1)$ $ =(k-1)(\frac{kp}{2}-1)$

2) Consider now $ ab\neq 0\pmod p$ and $ a\neq b\pmod p$

2.1) So, let us count first the number $ N_1$ of solutions where $ x=0\pmod p$
$ x=0\pmod p$ $ \iff$ $ y=0\pmod p$
These are all couples $ (up,vp)$ where $ u>0$ and $ v>0$ and $ u+v<k$
For each $ u\in[1,k-1]$, we have exactly $ k-1-u$ possible such $ v$ and so $ N_1=\sum_{u=1}^{k-1}(k-1-u)$ $ =\frac{(k-1)(k-2)}{2}$

2.2) Let us now count the number $ N_2$ of solutions where $ x\neq 0\pmod p$
$ x\neq 0\pmod p$ $ \iff$ $ y\neq 0\pmod p$

To each solution $ (x,y)$, we can associates it's "base" solution $ (x_1,y_1)$ where $ x_1\in[1,p-1]$ and $ x_1=x\pmod p$ and, same, $ y_1\in[1,p-1]$ and $ y_1=y\pmod p$

The number of "base" solutions is exactly $ p-1$ since, to each $ x_1\in[1,p-1]$, we can associate a unique ${ y_1=-\frac{a}{b}x_1\pmod p}$ (this is meaningfull since $ p$ is prime and $ b\neq 0\pmod p$).

No "base" solution $ (x_1,y_1)$ is such that $ x_1+y_1=p$, else we would have $ a=b$

If a "base" solution $ (x_1,y_1)$ exists and is such that $ x_1+y_1<p$, then another different "base" solution $ (x_2,y_2)$ =$ (p-x_1,p-y_1)$ exists and is such $ x_2+y_2=2p-(x_1+y_1)>p$

So we have the same number (so $ \frac{p-1}{2}$) of "base" solutions where $ x_1+y_1<p$ and "base" solutions where $ x_1+y_1>p$

2.2.1) Let us now count the number $ N_{2.1}$ of solutions where $ x\neq 0\pmod p$ and "base" solution is such that $ x_1+y_1<p$
From the solution $ (x_1,y_1)$ and $ x_1+y_1<p$, we can obtain all solutions $ (x_1+up,y_1+vp)$ with $ u\geq 0$ and $ v\geq 0$ and $ u+v\leq k-1$.
So, for each $ u\in[0,k-1]$, we have exactly $ k-u$ possible $ v$ ($ 0, 1, ..., k-1-u$) and so $ \sum_{u=0}^{k-1}(k-u)$ $ =\frac{k(k+1)}{2}$ solutions whose "base" solution is $ (x_1,y_1)$

So $ N_{2.1}=\frac{k(k+1)}{2}\frac{p-1}{2}$

2.2.1) Let us now count the number $ N_{2.2}$ of solutions where $ x\neq 0\pmod p$ and "base" solution is such that $ x_1+y_1>p$
From the solution $ (x_1,y_1)$ and $ p<x_1+y_1<2p$, we can obtain all solutions $ (x_1+up,y_1+vp)$ with $ u\geq 0$ and $ v\geq 0$ and $ u+v\leq k-2$.
So, if $ k\geq 2$, for each $ u\in[0,k-2]$, we have exactly $ k-u-1$ possible $ v$ ($ 0, 1, ..., k-2-u$) and so $ \sum_{u=0}^{k-2}(k-u-1)$ $ =\frac{k(k-1)}{2}$ solutions whose "base" solution is $ (x_1,y_1)$

If $ k=1$, we have no such cases and the above value is still ok kor $ k=1$

So $ N_{2.2}=\frac{k(k-1)}{2}\frac{p-1}{2}$

And so $ N_2=N_{2.1}+N_{2.2}=\frac{k(k+1)}{2}\frac{p-1}{2} +$ $ \frac{k(k-1)}{2}\frac{p-1}{2}$ $ =k^2\frac{p-1}{2}$

And so $ N=N_1+N_2$ $ =\frac{(k-1)(k-2)}{2} + k^2\frac{p-1}{2}$

[u][b]And so the final result given by Rust[/b][/u] (given $ p$ prime and $ k>1$) :

If $ a=b=0\pmod p$ then $ N=\frac{(kp-1)(kp-2)}{2}$

If $ ab=0\pmod p$ and $ a\neq b \pmod p$ then $ N=(k-1)(\frac{kp}{2}-1)$

If $ ab\neq 0\pmod p$ and $ a=b\pmod p$ then $ N=(k-1)(\frac{kp}{2}-1)$

If $ ab\neq 0\pmod p$ and $ a\neq b\pmod p$ then $ N=\frac{(k-1)(k-2)}{2} + k^2\frac{p-1}{2}$

Hope this is clear enough,

And congrats, Rust, for this computation, much more clever than my rude own ugly calculus.

It is very clear, and you consider all cases.


modular arithmetic
number theory proposed
number theory

Let an integer a>1. Prove the sequence $ u_n\equal{}a^{n\plus{}1}\plus{}a^n\minus{}1$ contains infinite number of pairs $ (u_i,u_j)$ with their greatest comon divisor is 1.

Let $ m \equal{} (k \plus{} 1)d\,,n \equal{} kd$ for some positive integers $ k$ and $ d > 1$. Let $ (u_m,u_n) \equal{} g$. Then $ m \minus{} n \equal{} d$ and $ g\mid u_m \minus{} a^du_n \equal{} a^{m \plus{} 1} \plus{} a^m \minus{} 1 \minus{} a^d(a^{n \plus{} 1} \plus{} a^n \minus{} 1) \equal{} a^d \minus{} 1\,(*)$. But since $ d\mid n$ we have $ g\mid a^d \minus{} 1\mid a^n \minus{} 1$. Thus $ a^n\equiv 1\pmod g$ and $ a^{n \plus{} 1}\equiv a\pmod g$. Therefore $ u_n \equal{} a^{n \plus{} 1} \plus{} a^n \minus{} 1\equiv a\pmod g$. Since $ g\mid u_n$ we get $ g\mid a$ and thus $ a^d\equiv 0\pmod g$. Combining this with $ (*)$ we conclude that $ 1\equiv 0\pmod g$ and thus $ g \equal{} 1$. Therefore $ (u_m,u_n) \equal{} 1$ for all such $ m\,,n$.


[quote="chibi_tama_love"]Dudes, this is worth 10% of my grades. If I fail this thing, then I go down one full letter grade. I want to make sure I've got the right answers. I'm not going to tell you the answers I've got, though, because I don't want your answers to be biased towards mine.

1) Nine men earned in a certain time $1334.40. If one recieved$20.00, others $15.00, and the rest$5.40 for an hour's work, how long did they work?

2) The good ship Dingledore struck an unmarked rock as it neared the coast and began to leak. The crew wanted to take life boats, but the skipper decided to run for the shore. He calculated that he was 40 miles from shore, and that the ship was admitting 3&3/4 tons of water into her hold every 12 minutes. But if she admits 60 tons she will sink. Her pumps throw out 12 tons of water in an hour, and, if she makes for the shore at 3&3/8 miles an hour, how far from the shore will she sink?

Please, please, help.[/quote]
[hide="Number One"]$b+c=8$

$\$20.00x+\$15.00bx+\$5.40cx=\$1334.40$

$\$20.00+\$15.00b+\$5.40(8-b)=\frac{\$1334.40}{x}$

$\$20.00+\$15.00b+\$43.20-\$5.40b=\frac{\$1334.40}{x}$

$\$63.20+\$9.60b=\frac{\$1334.40}{x}$

Three variables, only 2 equations. I think you have to specify something else.[/hide]
[hide="Number Two"]The ship takes on $\frac{15}{4}$ tons of water per 12 minutes, or $\frac{5}{16}$ tons of water per minute. The pump gets rid of 12 tons per hour, or $\frac{1}{5}$ tons per minute. Combining that information, the ship takes on $\frac{9}{80}$ tons of water per minute. This means the ship has $\frac{80}{9}$ hours before it sinks. The speed of the ship is $\frac{27}{8}$ mph (although usually measured in knots). The ship will travel exactly 30 miles before it sinks, which is $\boxed{10\ \text{miles}}$ away from shore.[/hide]

Okay then, I got the second one right, at least. In the first one, I think that they all earned that amount of money in an hour. Like, one recieved $20 in an hour, some got $15 in an hour, and the rest got $5.40 in an hour. That's the way I worked it out.

[quote="chibi_tama_love"]Okay then, I got the second one right, at least. In the first one, I think that they all earned that amount of money in an hour. Like, one recieved $\$20$ in an hour, some got $\$15$ in an hour, and the rest got $\$5.40$ in an hour. That's the way I worked it out.[/quote]
In that case, just divide.

[hide="Number One"]$66.72,\ 88.96,\ 247.\overline{1}$[/hide]

The first time I read problem number 1 I understood that they worked an integer number of hours (I've never been paid for $0.11$ hours), so it was a diophantic equation.
[hide="Number 1"]$1334.40=20.00h+15.00nh+5.40(8-n)h$
$1668=(79+12n)h$.
We have $7h\equiv 0\bmod 12$ so $h=12k$,
$139=(79+12n)k$
and $139$ is prime so $k=1$ and $79+12n=139$
$\boxed{h=12,\ n=5}$.[/hide]


trigonometry
Euler
limit
geometric sequence
real analysis
real analysis unsolved

Prove that
\[\sum_{k=1}^{n}(-1)^{k}\sin(2kx)=\cos \left (\frac{n\pi}{2}+(n+1)x\right )\sec x\sin \left (\frac{n\pi}{2}+nx\right ).\]

I know that identity but how to evaluate
\[\sum_{k=1}^{n}\cos k\pi\sin 2kx?\]

$\sin \alpha \cos \beta =\frac{1}{2}\{\sin (\alpha+\beta )+\sin (\alpha-\beta )\}$.

$\sum_{k=1}^{n}\cos k\pi\sin 2kx=\frac{1}{2}\sum_{k=1}^{n}(\sin (k\pi+2kx)+\sin (2kx-k\pi))$

$=\frac{1}{2}\sum_{k=1}^{n}\sin (k\pi+2kx)+\frac{1}{2}\sum_{k=1}^{n}\sin (2kx-k\pi)$.

I don't know how to evaluate those sums.

Then Euler said :

$e^{ix}=\cos{x}+i\sin{x}$
I mean your sum you can rewrite  as
$\sum\sin(k(...))=Im\sum e^{ki(...)}$
last geometric progression :wink:

You can check that :
$Im \frac{e^{ia}}{e^{ib}-1}=\frac{\sin a}{2\cos{b}-2}$
It's not hard problem because it's include only some calculations and nothing original Idea.
It's looks something like this  :
for example:
prove that
(more calculations)
$\lim_{x\to 0}\frac{(\tan(\sin{x})-\sin(tan{x}))}{x^{7}}=\frac{1}{30}$

or ... find limit
$\lim_{x\to 0}\frac{(\tan(\sin{x})-\sin(tan{x}))-\frac{x^{7}}{30}}{x^{8}}=?$

(don't do this  :what?: )

[quote="yagaron"]extremal, can we do that?

${L=\lim_{x\to 0}\, \frac{-\frac{x^{7}}{30}-\sin (\tan (x))+\tan (\sin (x))}{x^{8}}}$

Making a substituition  $x \to-x$ and since the limit has both directions equal

${L=\lim_{x\to 0}\, \frac{\frac{x^{7}}{30}+\sin (\tan (x))-\tan (\sin (x))}{x^{8}}=-L}$

so we can conclude that $L=0$

we just need to proof that the limit is equal in both directions, dont we?[/quote]
yes  :lol:
there must be divide it on $x^{9}$


A line that contains the point (6,3) is perpendicular to the line whose equation is $ 3x\plus{}2y \equal{} 6$. What is the sum of the coordinates of the $ y$-intercept of this line?

We find that the slope of the perpendicular line is equal to $ \frac {2}{3}$. Using point slope form, we have that $ \frac {y \minus{} 3}{x \minus{} 6} \equal{} \frac {2}{3}$. Rearranging, we have $ 2x \minus{} 3y \minus{} 3 \equal{} 0$. The $ y$-intercept occurs when $ x \equal{} 0$. Plugging this in, we find that $ y \equal{} \minus{} 1$. $ 0 \minus{} 1 \equal{} \boxed{ \minus{} 1}$.


http://www.pathguy.com/chess/ChessVar.htm

Try four player chess!

isnt 4 player chess bug house or something?

Yeah, it is.
http://www.pathguy.com/chess/Bughouse.htm
it's really fun, except the program is pretty bad -- the computer always tries to check even if it needs to sacrifice the piece lol.

also, bughouse chess should be played with a clock because otherwise if it's mate in one on one boeard the guy on the other board can just stall and there would not be an end.

We play this every day at school. :)


Columbia
rotation
trigonometry
geometry solved
geometry

Let $ P$ be a point in the interior of the equilateral triangle $ \triangle{}ABC$ such that $ PA \equal{} 5$, $ PB \equal{} 7$, $ PC \equal{} 8$. Find the length of the side of the triangle $ ABC$.

We first rotate point $B,C,P$ $60^o$ anticlockwise from point $A$ so that $B\to C$ , $C\to C'$ , $P\to P'$ .

Since $AP=AP'=5$ and $\angle PAP'=60^o$ , so $\triangle PAP'$ is equilateral . Hence $PP'=5$

In triangle $PCP'$ , using cosine rules to get $\angle P'PC=60^o$ , hence the angle $\angle APC=120^o$

So

$AC^2=AP^2+AC^2-2AP\cdot AC\cos 120^o=25+64-2(5)(8)\frac{1}{2}=129$

$AC=\sqrt{129}$

See also
http://www.mathlinks.ro/Forum/viewtopic.php?t=77511
http://www.mathlinks.ro/Forum/viewtopic.php?t=67254

  Darij


[quote="mthiyaga"]The units digit of a six-digit number is removed, leaving a five-digit number. The removed units digit is then placed at the far left of the five-digit number, making  a new six-digit number. If the new number is $\frac{1}{3}$ of the original number, what is the sum of the digits of the original number?[/quote]

[hide]$100000a+b=(\frac{1}{3})(10b+a)$
$299999a=7b$

EDIT: continued on next post.[/hide]

[quote="SplashD"][quote="mthiyaga"]The units digit of a six-digit number is removed, leaving a five-digit number. The removed units digit is then placed at the far left of the five-digit number, making  a new six-digit number. If the new number is $\frac{1}{3}$ of the original number, what is the sum of the digits of the original number?[/quote]

[hide]$100000a+b=(\frac{1}{3})(10b+a)$
$299999a=7b$
$2+5(9)+7=\boxed{54}$[/hide][/quote]
Wait... First, that makes 7 digits, and second, can you explain your reasoning?

[hide]a could be 2. then b is 85714. so there's another number, 857142. the sum of the digits is still 27 because it has all the same digits but in a new order.[/hide]


function
limit
calculus
calculus computations

1)
let $ f$ a periodic function , such that $ \lim_{x \to\plus{} \infty } f(x)\equal{}l$ and $ D_f \equal{}\mathbb R$

prove that ( $ \forall x \in \mathbb R$ ) $ f(x) \equal{} l$

2)
let $ f$ a periodic and unconstante  function and $ D_f \equal{} \mathbb R$

prove that $ f$ doesn't  exept a limit when x tends  to $ \plus{} \infty$

1) let $ T>0$ be a period of $ f$, and $ x\in\mathbb R$.
Then $ f(x)\equal{}\lim_{n\to \infty}f(x\plus{}nT)\equal{}l$.

2) if $ f$ has a limit when $ x$ tends to infinity,
then according to part 1), $ f$ must be constant,
which is a contradiction.

Your avatar does not fool us. The avatar has a darker background than the stars you have earned, so we can notice.


limit
logarithms
number theory unsolved
number theory

Let $ T=\{p_{1},p_{2},\ldots\}$ be an infinite set of primes for which $ \sum_{j=1}^\infty 1/p_{j}$ converges.  Define $ T(x)=|\{p\le x: p\in T\}|.$  Prove that $ \lim \inf_{x\rightarrow\infty}T(x)/\pi(x) =0$.  Also, give an example such that $ \lim \sup_{x\rightarrow \infty}T(x)/\pi(x)=1$.

A related topic:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=141781

In the link given above, Fedja wrote:
Note that $ \sum_{a\in A}a^{-1}=\sum_{n\ge 1}\frac 1{n(n+1)}|A\cap\{1,\ldots,n\}|$.  Could someone explain this?  Thanks to everyone for the help.

If $ \lim \inf_{x\rightarrow\infty}T(x)/\pi(x) \not=0$, then exist A>0,c>0, suth that $ T(x)>c\pi(x), x>A$.  Therefore
$ \sum_{x>p\in T}\frac{1}{p}>cln(ln(x))-b.$
For example $ p\in T$ if and only if $ A_{2n-1}<p<A_{2n}$,were $ A_{0}=1,A_{n}=A_{n-1}b_{n}$. Then $ \sum_{A_{2n}>p\in T}\frac{1}{p}=\sum_{k\le n}ln(1+\frac{lnb_{2n}}{ln(A_{2n-1})})+O(1)<C$ if $ b_{2n}<A_{2n-1}^{1/n^{2}}$.
Let $ b_{2n-1}=2^{n^{2}}, b_{2n}=2^{n/4}$. It give $ \lim \sup_{x\rightarrow \infty}T(x)/\pi(x)=1.$

[quote="Rust"] $ \sum_{x>p\in T}\frac{1}{p}>cln(ln(x))-b.$[/quote]

Rust, I don't see why this is true.  We are talking about sets of primes.  Notice that $ \sum_{x>p\in T}\frac{1}{p}\neq \sum_{p\le cx}\frac{1}{p}$.  Nor, can we necessarily say either is bigger or smaller.  But, if we could then I do not believe that it is trivial as you stated.

Starting as you did, if $ \lim \inf_{x\rightarrow\infty}T(x)/\pi(x) \not=0$, then there exist $ C>0$ and $ x_{0}$ such that $ T(x)>C\pi(x)$ for all $ x>x_{0}$.  Now, I think we want to take a sequence $ p_{j}$ in $ T$ such that $ p_{j}>(4/C)p_{j-1}$.  Define $ N_{j}=T(p_{j})-T(p_{j-1})$.  It follows from a version of PNT, $ x/\log x < \pi(x) <(1.5)x\log x$ for large $ x$, that $ N_{j}> C'>0$ for large $ j$.  Altogether,
\[ \sum_{p\in T}\frac{1}{p}\ge \sum_{j=1}^\infty~ \sum_{p_{j-1}<p \le p_{j}}\frac{1}{p}.\]
As the inner sum on the RHS, for large $ j$, is $ >C'>0$, the series on the RHS diverges contradicting that the LHS converges.

I think, that is obviosly. Let $ k=[\frac{1}{c}]+1$, then
$ \sum_{x>p\in T}\frac{1}{p}>\sum_{j=1}^{(\pi (x)-pi(A))/k}\frac{1}{p_{\pi(A)+kj}}>\frac{1}{k}ln(ln(x)-b$.

[quote="baz"]$ \sum_{a\in A}a^{-1}=\sum_{n\ge 1}\frac{1}{n(n+1)}|A\cap\{1,\ldots,n\}|$.  Could someone explain this?[/quote]
Each element $ a\in A$ contributes $ a^{-1}$ to the left side and $ \sum_{n\ge a}\frac 1{n(n+1)}$ to the right side. Also, $ \sum_{n\ge a}\frac{1}{n(n+1)}=\sum_{n\ge a}\frac{1}{n}-\frac{1}{n+1}=\frac{1}{a}$.


Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=257229[/url]


Let $ \Delta ABC$. $ AH$ is a perpendicular of $ \Delta ABC$, $ K$ is on $ BC$ such that $ A,K,O$(circumcentre) are in a straight line. $ M,N$ are projections of $ H$ on $ AB,AC$ and $ P,Q$ are projection of $ K$ on $ AB,AC$.
1/Prove that $ M,N,P,Q$ is on a circle.
2/Prove that $ MN,PQ,AH$ are concurrent.

He probably means that $ AH$ is an altitude of $ \triangle ABC$ and that $ H$ is the foot of that altitude.


function
vector
probability
inequalities
linear algebra
matrix
induction

We are given the finite sets $ X$, $ A_1$, $ A_2$, $ \dots$, $ A_{n \minus{} 1}$ and the functions $ f_i: \ X\rightarrow A_i$. A vector $ (x_1,x_2,\dots,x_n)\in X^n$ is called [i]nice[/i], if $ f_i(x_i) \equal{} f_i(x_{i \plus{} 1})$, for each $ i \equal{} 1,2,\dots,n \minus{} 1$. Prove that the number of nice vectors is at least
\[ \frac {|X|^n}{\prod\limits_{i \equal{} 1}^{n \minus{} 1} |A_i|}.
\]

First of all, there are $ |X|^n$ ways to choose $ n$ values (not necessarily distinct) from set $ X$.
Since $ f_i: X\rightarrow\ A_i$, so we can choose the value of $ f_i(x_i)$ for $ |A_i|$ ways for each $ i\equal{}1,2,...,n\minus{}1$. Similarly, we can choose the value for $ f_i(x_{i\plus{}1})$ by $ |A_i|$ ways for each $ i\equal{}1,2,...,n\minus{}1$.
So there are $ |A_i|^2$ possibilities of choosing values for $ f_i(x_i)$ and $ f_i(x_{i\plus{}1})$ for each $ i\equal{}1,2,...,n\minus{}1$.
But since $ f_i(x_i)\equal{}f_i(x_{i\plus{}1})$ for each $ i\equal{}1,2,...,n\minus{}1$, we need only $ |A_i|$ ways of choosing the values for $ f_i(x_i)$ and $ f_i(x_{i\plus{}1})$ for each $ i\equal{}1,2,...,n\minus{}1$.
So the number of nice vectors are at least $ |X|^n(\frac{|A_1|}{|A_1|^2})(\frac{|A_2|}{|A_2|^2})...(\frac{|A_{n\minus{}1}|}{|A_{n\minus{}1}|^2})\equal{}\frac {|X|^n}{\prod\limits_{i \equal{} 1}^{n \minus{} 1} |A_i|}.$

We use a probabilistic method, so $ P(e(x): x\in S)$ denotes the probability that an event $ e$ dependant on some variable $ x$ happens as $ x$ varies [i]randomly with uniform probability[/i] (which I'll call 'uniformly') on a set $ S$ (though if dependencies are obvious we omit them). As is conventional, $ P(e|d)$ denotes the probability that $ e$ happens given $ d$ happens. We aim to show that
\[ P(\textbf{x}\text{ is nice}: \textbf{x} \in X^n ) \geq \frac {1}{\prod_{i \equal{} 1}^{n \minus{} 1} |A_i|}
\]
First, we fiddle our functions a little. We replace $ A_i$ with $ f_i(X)$ to force our functions to be surjective. We can then define a new set of functions $ g_i : A_i \rightarrow \mathbb{N}$ and sets $ G_i$
\[ G_i(a) \equal{} \{x \in X | f_i(x) \equal{} a\}, g_i(a) \equal{} |G_i(a)|
\]
We remark that every $ x\in X$ must be contained in precisely one G-set, so $ \sum_a g_i(a) \equal{} |X|$, a fact we shall need later.

Lemma: $ f_i(x_i) \equal{} f_i(x_{i \plus{} 1}) : \textbf{x} \in X^n$ and $ f_j(x_j) \equal{} f_j(x_{j \plus{} 1}) : \textbf{x} \in X^n$ are independent events for $ i \not \equal{} j$.

If $ |i \minus{} j| > 1$ the events are literally independent in that the variables each depend on are selected entirely independently of one another. It follows that we need only consider the case $ j \equal{} i \plus{} 1$.

We consider the probability of the event $ e_1: f_i(x_i) \equal{} f_i(x_{i \plus{} 1})$.
Suppose we know that event $ e_2: f_{i \plus{} 1}(x_{i \plus{} 1}) \equal{} f_{i \plus{} 1}(x_{i \plus{} 2})$ happens (and it does with positive probability). Consider the $ P(x_{i \plus{} 1} \equal{} y| e_2 : x_{i \plus{} 2} \in X)$ for some fixed $ y \in X$. If the probability is constant for all $ y$, we are done, because then $ x_{i \plus{} 1}$ continues to vary uniformly and so $ P(e_1) \equal{} P(e_1|e_2)$, because the only common value $ e_1$ and $ e_2$ depend on is $ x_{i \plus{} 1}$, which would vary uniformly if we did not know the outcome of $ e_2$.

Now, $ P(x_{i \plus{} 1} \equal{} y| e_2: x_{i \plus{} 2} \in X) \equal{} \frac {P(f_{i \plus{} 1}(y) \equal{} f_{i \plus{} 1}(x_{i \plus{} 2}): x_{i \plus{} 2} \in X)}{g_{i \plus{} 1}(f_{i \plus{} 1}(y))}$ (because if $ e_2$ is satisfied then this must certainly happen, but the probability it happens with $ x_{i \plus{} 1} \equal{} y$ is clearly only $ 1/g_{i \plus{} 1}(f_{i \plus{} 1}(y))$ times the probability it happens overall, because $ x_{i \plus{} 1}$ is chosen randomly uniformly and is just as likely to be any of the $ g_{i \plus{} 1}(f_{i \plus{} 1}(y)) \minus{} 1$ alternatives).
And this becomes $ \frac {P(y \in G_{i \plus{} 1}(f_{i \plus{} 1}(x_{i \plus{} 2})): x_{i \plus{} 2} \in X)}{g_{i \plus{} 1}(f_{i \plus{} 1}(y))} \equal{} \frac {g_{i \plus{} 1}(f_{i \plus{} 1}(y))}{|X|g_{i \plus{} 1}(f_{i \plus{} 1}(y))} \equal{} \frac {1}{|X|}$ because as $ x_{i \plus{} 2}$ varies uniformly it lands in the same G-set as $ y$ precisely $ g_{i \plus{} 1}(f_{i \plus{} 1}(y))$ times by definition of $ g$.

So we happily see that the probability $ x_{i \plus{} 1}$ still varies uniformly over $ X$ even if we know $ e_2$ happens, so the events are independent and the lemma is proven. :)

We next turn to directly estimating the probabilities $ P(f_i(x_i) \equal{} f_i(x_{i \plus{} 1}) : x_i, x_{i \plus{} 1} \in X)$, now knowing that we can simply take their product to give the probability that such an event holds for all $ i$. This will prove to be much less fiddly than the lemma.

$ P(f_i(x_i) \equal{} f_i(x_{i \plus{} 1})) \equal{} P(\exists a\in A_i \text{ s.t. } x_i, x_{i \plus{} 1} \in G_i(a))$.
Now if we fix $ a \in A_i$, as $ x\in X$ varies over all of $ X$, it is found in $ G_i(a)$ precisely $ g_i(a)$ times. As $ x_i, x_{i \plus{} 1}$ vary, they must be found in the same $ G_i(a)$ precisely $ (g_i(a))^2$ times. We therefore find the above probability evaluating as shown, which we then estimate using Chebyshev's (non probabilistic) inequality.
\[ P(f_i(x_i) \equal{} f_i(x_{i \plus{} 1})) \equal{} \frac {\sum_{a \in A_i}(g_i(a))^2}{|X|^2} \geq \frac {(\sum_{a\in A_i} g_i(a))^2}{|A_i\parallel{}X|^2} \equal{} \frac {|X|^2}{|A_i\parallel{}X|^2} \equal{} \frac {1}{|A_i|}
\]
We seek the probability $ P(\textbf{x}\text{ is nice}: \textbf{x} \in X^n )$, that $ f_i(x_i) \equal{} f_i(x_{i \plus{} 1})$ for all $ i$. However, since these events are independent by the lemma, we may simply take their product to give the final probability, thus:
\[ P(\textbf{x}\text{ is nice}: \textbf{x} \in X^n ) \equal{} \prod_{i \equal{} 1}^{n \minus{} 1} P(f_i(x_i) \equal{} f_i(x_{i \plus{} 1})) \geq \prod_{i \equal{} 1}^{n \minus{} 1} \frac {1}{|A_i|}
\]
This is the result required, and multiplying by the $ |X|^n$ different possibilities for randomly selected vectors, gives the result as stated in the problem.

Draft.
Let's analyze the case $ n \equal{} 2$.
Let $ A_1 \equal{} \{b_1,\ldots,b_t\}$ for some $ t$ and let $ |X| \equal{} k$. Denote by $ a_1,\ldots,a_t$ the number of $ x\in X$ for which $ f_1(x) \equal{} b_1,\ldots,b_t$ respectively. Clearly $ \sum_{i \equal{} 1}^t a_i \equal{} k$. It is straightforward to see that the number of nice vectors is $ \sum_{i \equal{} 1}^t a_i^2$, hence the inequality to prove becomes $ \sum_{i \equal{} 1}^ta_i^2\ge\frac {\left(\sum_{i \equal{} 1}^ta_i\right)^2}t \equal{} \frac {k^2}t$, and it follows directly from Cauchy.

Let a nice short vector of length $ m$ be a vector $ (x_1,\ldots,x_m)\in X^n$ which fulfills the statement $ f_i(x_i) \equal{} f_i(x_{i \plus{} 1})$ for all $ i \equal{} 1,2,\ldots,m \minus{} 1$, where $ 1\le m\le n \minus{} 1$. Let $ X \equal{} \{y_1,\ldots,y_k\}$. Assume that there are $ t_1,\ldots,t_k$ short nice vectors of length $ m < n \minus{} 1$ which end in $ y_1,\ldots,y_k$ respectively. Also for some $ y\in X$, let $ t_y$ denote the number of short nice vectors of length $ m$ ending in $ y$. Assume $ |A_m| \equal{} s$ and let $ X \equal{} X_1\cup X_2\cup\ldots\cup X_s$, where $ X_i\cap X_j \equal{} \emptyset$ if $ i\neq j$ be a partition of $ X$ satisfying: $ a,b\in X$ and $ f_m(a) \equal{} f_m(b)$ [i]if and only if[/i] $ a,b\in X_i$ for some $ i\in\{1,2,\ldots,s\}$. Then a short nice vector of length $ m$ ending in $ x\in X$ is the Root of $ |X_i|$ short nice vectors of length $ m \plus{} 1$, where $ x\in X_i$. Denote by $ b_i \equal{} |X_i|$.
Let
$ S_m \equal{} \begin{bmatrix}t_1 & t_2 & \ldots & t_k \\
											t_1 & t_2 & \ldots & t_k \\
											\ldots \\
											t_1 & t_2 & \ldots & t_k
			 \end{bmatrix}$.
Let $ U_m \equal{} (u_{ij})$, be a $ k\times k$ matrix with entries $ u_{ij} \equal{} 1$ if vectors $ x_i$ and $ x_j$ belong to the same $ X_l$, for some $ 1\le l\le s$ and $ u_{ij} \equal{} 0$ otherwise. In words, $ u_{ij} \equal{} 1$ if a short nice vector ending in $ x_i$ can be extended to a $ m \plus{} 1$ vector by adding $ x_j$ at the end.
It is easy to see that $ S_mU_m$ is a matrix $ S_{m \plus{} 1}$ of the same form as $ S_m$: $ S_{m \plus{} 1} \equal{} \begin{bmatrix}T_1 & T_2 & \ldots & T_k \\
T_1 & T_2 & \ldots & T_k \\
\ldots \\
T_1 & T_2 & \ldots & T_k\end{bmatrix}$. Moreover by the arguments above, we easily see that the number of nice vectors of length $ m \plus{} 1$ is $ \mathrm{tr}(S_{m\plus{}1})\equal{}T_1 \plus{} \ldots \plus{} T_k$. This is because $ T_i\equal{}\sum_{j\equal{}1}^k t_ju_{ij}$ and $ u_{ij}\equal{}1$ if and only if a short nice vector of length $ m$ ending in $ y_i$ can be added at the end the vector $ y_j$.
Hence $ S_{m \plus{} 1} \equal{} S_mU_m$, for each $ m \equal{} 1,2,\ldots,n \minus{} 1$. We easily see that $ S_1 \equal{} \begin{bmatrix}1 & 1 & \ldots & 1 \\
1 & 1 & \ldots & 1 \\
\ldots \\
1 & 1 & \ldots & 1\end{bmatrix}$ (meaning that there is exactly one short nice vector of length 1 ending in $ y_i$).
Hence $ S_n \equal{} S_1U_1\ldots U_{n \minus{} 1}$ where we take account of the fact that the multiplication of matrices is associative. All matrices involved in the product are square matrices of dimenstion $ k \equal{} |X|$.

Note now that the number of nice vectors is $ \mathrm{tr}(S_n)\equal{}g_1\plus{}g_2\plus{}\ldots\plus{}g_k$, where $ S_n\equal{}\begin{bmatrix}g_1 & g_2 & \ldots & g_k \\
g_1 & g_2 & \ldots & g_k \\
\ldots \\
g_1 & g_2 & \ldots & g_k\end{bmatrix}$

Now, $ \mathrm{tr}$ is a commutative operator, hence $ \mathrm{tr}(S_n)\equal{}\mathrm{tr}(S_1U_1\ldots U_{n\minus{}1})\equal{}\mathrm{tr}(S_1\cdot U_{\sigma(1)}U_{\sigma(2)}\ldots U_{\sigma(n\minus{}1)})$, where $ \sigma$ is any permutation of $ \{1,2,\ldots,n\minus{}1\}$. Our task is to show that $ \mathrm{tr}(S_1U_1U_2\ldots U_{n\minus{}1})\ge\displaystyle\frac{k^n}{|A_1|\cdot\ldots\cdot |A_{n\minus{}1}|}$.
now some induction with a suitable arrangement of the $ U_i$'s should work

Let $ X \equal{}${$ y_{i}$ for $ i \equal{} 1$ to $ |X|$}
$ f_{j}: X \to A_{j}$
let $ A_{j} \equal{}$ {$ a_{j,i}$  for $ i \equal{} 1$ to $ |A_{j}|$} for $ j \equal{} 1$ to $ n \minus{} 1$
good $ m$ vector implies a good vector with $ m$ dimensions
we define $ e_{j,i} \equal{}$ no of elements from X that satisfy $ f_{j}(y_{j}) \equal{} a_{j,i}$
we define an anologous function
$ g_{j}:$ good $ j \minus{} 1$ vectors $ \to A_{j}$
such that $ g_{j}(x_{1},x_{2},...,x_{j \minus{} 2},x_{j \minus{} 1}) \equal{} f_{j}(x_{j \minus{} 1})$
let $ m_{j,i} \equal{}$ number of elements from "good $ j \minus{} 1$ vectors" that satisfy $ g_{i}(w_{j}) \equal{} a_{j,i}$ for $ w_{j} \in$ good $ j \minus{} 1$ vectors for all $ j$
_________________________________________________________________________________
   for $ n \equal{} 2$
now the number of vectors $ (x_{1},x_{2}) \in X^{2}$
now we count the number of vectors for which $ f_{1}(x_{1}) \equal{} f_{1}(x_{2}) \equal{} a_{1,i}$
this is equal to $ e_{1,i}^{2}$ therefore the number of vectors for which $ f_{1}(x_{1}) \equal{} f_{1}(x_{2})$ is $ \sum_{i \equal{} 1}^{|A_{1}|}e_{1,i}^{2}$
by $ Tchbycheff's$   $ inequality$ this is greater than $ \frac {( \sum_{i \equal{} 1}^{|A_{1}|}e_{1,i})^{2}}{|A_{1}|}$
as $ \sum_{i \equal{} 1}^{|A_{1}|}e_{1,i}$ is the number elements in the domain $ \equal{} |X|$
hence number of vectors is $ \ge \frac {|X|^{2}}{|A_{1}|}$
therefore conjencture is true for $ n \equal{} 2$
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assume that the conjencture is true for $ n \equal{} k \minus{} 1$
i.e. number of vectors {$ x_{1},...,x_{k}$} $ \in |X|^{k}$ satisfying $ f_{i}(x_{i}) \equal{} f_{i}(x_{i \plus{} 1})$ for $ i \equal{} 1$ to $ k \minus{} 1$ is atleast $ \frac {|X|^{k}}{\prod_{i \equal{} 1}^{k \minus{} 1}|A_{1}|}$
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to prove the conjencture is true for $ n \equal{} k$

$ f_{k}: |X| \to A_{k}$
now we count the number of good $ k$ vectors $ ($ good $ k \minus{} 1$ vector$ ,x_{k})$ for which
$ g_{k}($good $ k \minus{} 1$ vector$ ) \equal{} f_{k}(x_{k}) \equal{} a_{k,i}$ for all $ i \equal{} 1$ to  $ |A_{k}|$
is $ m_{k,i} \cdot e_{k,i}$
therefore the number of good $ k$ vectors is $ \sum_{i \equal{} 1}^{|A_{k}|}m_{k,i} \cdot e_{k,i}$
observe that $ e_{k,i}$ and $ m_{k,i}$ are similarly ordered sequences
therefore $ \sum_{i \equal{} 1}^{|A_{k}|}m_{k,i} \cdot e_{k,i} \ge \frac {(\sum_{i \equal{} 1}^{|A_{k}|}m_{k,i}) (\sum_{i \equal{} 1}^{|A_{k}|}e_{k,i})}{|A_{k}|}$
by $ Tchbycheff's$  $ inequality$
now observe that $ (\sum_{i \equal{} 1}^{|A_{k}|}m_{k,i})$ is the number of good $ k \minus{} 1$ vectors and is $ \ge$ $ \frac {|X|^{k}}{\prod_{i \equal{} 1}^{k \minus{} 1}|A_{1}|}$ by our assumption
also $ (\sum_{i \equal{} 1}^{|A_{k}|}e_{k,i}) \equal{}$ number of elements in $ X \equal{} |X|$
hence $ \sum_{i \equal{} 1}^{|A_{k}|}m_{k,i} \cdot e_{k,i}$ $ \ge$ $ \frac {(\sum_{i \equal{} 1}^{|A_{k}|}m_{k,i}) (\sum_{i \equal{} 1}^{|A_{k}|}e_{k,i})}{|A_{k}|}$ $ \ge$ $ \frac {|X| \cdot |X|^{k}}{|A_{k}| \cdot \prod_{i \equal{} 1}^{k \minus{} 1}|A_{1}|} \equal{} \frac {|X|^{k \plus{} 1}}{\prod_{i \equal{} 1}^{k}|A_{1}|}$
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 :D   :lol:  :)
hence by induction number of good vectors is atleast $ \frac {|X|^{n}}{\prod_{i \equal{} 1}^{n \minus{} 1}|A_{1}|}$ for all integers $ n \ge 2$

I noticed that mathlinks contest round 1 was started today (4/27). When is the limit of sending? I think I'm too late.. :(
I tried some specific case.
When $ n=2$, Let $ \#f_{1}[X] = k$. we can assume that $ f_{1}[X]=\{1,2,\cdots , k\}$. there are $ a_{i}$ elements in $ X$ whose image is $ i$.
Then there are $ a_{1}^{2}+\cdots +a_{k}^{2}$ ( $ \ge \frac{(a_{1}+\cdots +a_{k})^{2}}{k} \ge \frac{|X|^{2}}{|A_{1}|}$ )  nice vectors....
So case $ n=2$ is proved.


LaTeX
inequalities unsolved
inequalities

Let $n \geq 2$. Let $a_1 , a_2 , a_3 , \ldots a_n$ be $n$ real numbers all less than $1$ and such that $|a_k - a_{k+1} | < 1$ for $1 \leq k \leq n-1$. Show that \[ \dfrac{a_1}{a_2} + \dfrac{a_2}{a_3} + \dfrac{a_3}{a_4} + \ldots + \dfrac{a_{n-1}}{a_n} + \dfrac{a_n}{a_1} < 2 n - 1 .  \]

That's wrong ! Take $x_2=...=x_n > 0$ and $x_1>0$ close to $0$

 :cool: What's the correct statement ? :cool:

After a bit of googling, it seems that the correct statement is:

[quote="Rushil"]Let $ n \geq 2$. Let $ a_1 , a_2 , a_3 , \ldots a_n$ be $ n$ real numbers all [b]greater[/b] than $ 1$ and such that $ |a_k \minus{} a_{k \plus{} 1} | < 1$ for $ 1 \leq k \leq n \minus{} 1$. Show that
\[ \dfrac{a_1}{a_2} \plus{} \dfrac{a_2}{a_3} \plus{} \dfrac{a_3}{a_4} \plus{} \ldots \plus{} \dfrac{a_{n \minus{} 1}}{a_n} \plus{} \dfrac{a_n}{a_1} < 2 n \minus{} 1 .
\]
[/quote]

a2/a1+a3/a2+....+an/a1<2n-1
subtracting n on both sides
a2/a1-1+a3/a2-1+....+an/a1-1<2n-1-n
a2-a1/a1+a3-a2/a2+....+an-a1/a1<n-1
put an-a1=(an-an-1)+(an-1-an-2)+....+(a2-a1)
a2-a1/a1+a3-a2/a2+....+(an-an-1)/a1+(an-1-an-2)/a1+....+(a2-a1)/a1<n-1
now we have these n-1 terms
(a2-a1/a1+a1-a2/a1)+(a3-a2/a2+a2-a3/a2)+.....+(an-an-1/an-1+an-1-an/a1)<n-1
it is easy to prove that these terms are each <1

P.S please tell me if there is anything wrong in my solution and if anybody understands it, put it in good latex form etc.

Just fix the typos. The proof is correct.

Without the typos it's
$n-1>\sum_{i=1}^{n-1}\left(\frac{1}{a_{i+1}}-\frac{1}{a_{i+2}}\right) (a_1-a_{i+1})$
And why is that correct?

Just use induction and see how the problem spoils

In the main inductive step
Make 2 cases,
Ak >A k+1 and the reverse