inequalities
combinatorics proposed
combinatorics

On the chessboard, $ 32$ black pawns and $ 32$ white pawns are arranged. In every move, a pawn can capture another pawn of the opposite color, moving diagonally to an adjacent square where the captured one stands. White pawns move only in upper-left or upper-right directions, while black ones can move in down-left or in down-right directions only; the captured pawn is removed from the board. A pawn cannot move without capturing an opposite pawn. Find the least possible number of pawns which can stay on the chessboard.

[quote="mszew"][hide="Hint I (Inequality)"]Color the chess board black and white as the usual one, notice that each pawn can eat only one pawn from each file (no moving backwards) then the minimum is 8 from the different rows. [/hide]
[hide="Hint II (Example)"]Put the white pawns in the bottom 4 white cells and the black pawns in the other 28 white cells, all those white pawns can eat those 28 black pawns. Similarly with the black cells. [/hide][/quote]
I think this answer is fasle. The answer must be 2.

Yes you are right, the answer is $ 2$. The coloring of the table with black & white only proves that the number is $ \ge2$, because one pawn must remain on a black and one must remain on a white square. And we have this example (0 - Black, 1 - White):

[color=blue]0[/color][color=black]0[/color][color=blue]0[/color][color=black]0[/color][color=blue]0[/color][color=black]0[/color][color=blue]0[/color][color=black]0[/color]
[color=red]0[/color][color=blue]1[/color][color=black]1[/color][color=blue]1[/color][color=black]1[/color][color=blue]1[/color][color=black]1[/color][color=green]0[/color]
[color=blue]1[/color][color=red]1[/color][color=blue]1[/color][color=black]1[/color][color=blue]1[/color][color=black]1[/color][color=green]1[/color][color=black]1[/color]
[color=red]0[/color][color=blue]0[/color][color=red]1[/color][color=blue]1[/color][color=black]1[/color][color=green]1[/color][color=black]0[/color][color=green]0[/color]
[color=blue]1[/color][color=red]1[/color][color=blue]0[/color][color=red]0[/color][color=green]0[/color][color=black]0[/color][color=green]1[/color][color=black]1[/color]
[color=red]0[/color][color=blue]0[/color][color=red]0[/color][color=green]0[/color][color=red]0[/color][color=green]0[/color][color=black]0[/color][color=green]0[/color]
[color=blue]1[/color][color=red]0[/color][color=green]0[/color][color=red]0[/color][color=green]0[/color][color=red]0[/color][color=green]0[/color][color=black]1[/color]
[color=red]1[/color][color=green]1[/color][color=red]1[/color][color=green]1[/color][color=red]1[/color][color=green]1[/color][color=red]1[/color][color=green]1[/color]

Take one color (red for example, so now we work just with the red pawns). First we capture with the 1-s at the bottom (without the corner) the 0-s until the diagonal. Then we capture with the 0-s at left the 1-s until the diagonal. Then we capture with the 1 in the corner the 0-s until the middle. We do the same with the other colors (blue, green, black).  So we will have(* - Empty):

********
********
********
***[color=blue]0[/color][color=black]0[/color]***
***[color=red]1[/color][color=green]1[/color]***
********
********
********

Two more moves and we have $ 2$ pawns :) .


1)When we solve the equation: $8x^3-4x^2-4x+1=0$

$a)cos\frac{\pi}{14}$ $b)sin\frac{3 \pi}{7}$ $c)sin\frac{2 \pi}{7}$ $d)cos\frac{5\pi}{14}$ $e)sin\frac{\pi}{14}$

2)Given:

$sin\frac{x}{2}+sin\frac{y}{2}+sin\frac{z}{2}=0$
$cos\frac{x}{2}+cos\frac{y}{2}+cos\frac{z}{2}=0$

then we have:

$a)\frac{x}{2}+\frac{y}{2}=z$ $b)\frac{x}{2}-\frac{y}{2}=3z$ $c)\frac{x}{4}+\frac{y}{4}=z$ $d)x+y=z$ $e)x-y=z$

3)Reduce the following expression:

$S=sec11^\circ sec19^\circ - 2 cot71^\circ$

$a)2cot11^\circ$ $b)2cot79^\circ$ $c)\frac{1}{2} tan11^\circ$ $d)tan19^\circ$ $e)\frac{1}{2} tan19^\circ$

4)In what kind of triangle ABC we have?:

$cosBcos\frac{A}{2}=sin\frac{C}{2}cos\frac{B}{2}$

$a)equilateral$ $b)isosceles$ $c)rectangle$ $d)A or B$ $e)acute$

hello, to 1.a) the solution is $ x\equal{}\sin\left(\frac{\pi}{14}\right)$.
Sonnhard.

If you are looking for a solution to #1 follow this link and just read pages 6,7, and 8 of the pdf.
http://nickjames8806.googlepages.com/LaTeX1.pdf


function
complex analysis
complex analysis unsolved

show that every function f which is analytic in |z|<=1 and satisfies |f|=1 on |z|=1
is a rational function.

thx

The proof on that other thead is a little hard to find, so summarizing here and making the proof more explicit:

We assume that by "analytic for $|z|\le 1$" you actually mean "analytic for $|z|<1,$ continuous for $|z|\le1.$" The zeros of an analytic function must be isolated, so the set of zeros of $f$ has no limit point in the interior of the disk, and since $|f(z)|=1$ for $|z|=1,$ the set of zeros also has no limit point on the boundary. The Bolzano-Weierstrass theorem says that any bounded infinite set must have a limit point; since the zero set of $f$ has no limit point, it must be a finite set.

Let $\{\alpha_1,\alpha_2,\dots,\alpha_m\}$ be a listing of the zeros of $f$ in the interior of the disk. Include in the list as many copies of each root as is the multiplicity of that root. Define

$B(z)=\prod_{k=1}^{m}\frac{z-\alpha}{1-\overline{\alpha}z}$

Note that $|B(z)|=1$ for $|z|=1.$

Now consider the function $g(z)=\frac{f(z)}{B(z)}.$

$g$ is analytic for $|z|<1,$ continuous for $|z|=1,$ has the property that $|g(z)|=1$ for $|z|=1$ and has no zeros on the closed disk. Apply the maximum modulus principle to both $g$ and $1/g$ and we conclude that $|g(z)|=1$ for all $|z|\le1.$ But an analytic function of constant modulus is constant.

So we conclude that $f(z)=\lambda B(z)$ for some $|\lambda|=1.$


abstract algebra
algorithm
Ring Theory
superior algebra
superior algebra unsolved

Show that $ D[x](x\minus{}a)$ is left maximal ideal of  ring $ D[x]$ , $ D$ is a division ring (I need this to say that $ D[x]/D[x](x\minus{}a)$ is simple $ D[x]$-module

What do you get when you take the quotient?

You don't take the quotient, because it's not a two-sided ideal.

Whoops.  Yes, it's not quite that simple.

But, it seems that it is still possible to perform the division algorithm in $ D[x]$, writing $ p(x)\equal{}q(x)(x\minus{}a)\plus{}b$ for any $ p$.  Hence any subring of $ D[x]$ that contains $ D[x](x\minus{}a)$ and some additional element, must contain a unit.


[quote="silouan"]Μια όμορφη άσκηση που βρήκα επίσης είναι η παρακάτω
Δίνονται άπειρα σημεία στο επίπεδο έτσι ώστε ανά δύο να απέχουν ακέραια απόσταση . Είναι σωστό ότι αυτά τα σημεία θα πρέπει να βρίσκονται σε ευθεία ?

Hope to enjoy it  :)[/quote]

Sosto...einai to Erdos-Anning Theorem. koitakste edo
http://www.ams.org/bull/1945-51-08/S0002-9904-1945-08407-9/S0002-9904-1945-08407-9.pdf
h opoudipote allou sto web..sigoura tha exoun tin apodeiksi.

Εξαιρετικά όμορφη και εξαιρετικά δύσκολη άσκηση. Αξίζει να την σκεφτείτε αρκετά προτού δείτε την λύση των Erdos και Anning.

Επαναφέρω την άσκηση διότι η λύση που γνώριζα δεν είναι αυτή των Erdos και Anning. Δίνω μια υπόδειξη για το πως μπορεί να λυθεί διαφορετικά (νομίζω πως και αυτή η απόδειξη είναι του Erdos)

Έστω 3 από αυτά τα σημεία Α,Β,Γ δεν βρίσκονται σε ευθεία. Πάρτε ένα άλλο σημείο Χ του συνόλου. Τι μπορείτε να πείτε για το ΧΑ - ΧΒ;

Η λύση είναι παρόμοια με του Δημήτρη


Find the moment of inertia about the line x=b of a lamina in the form of a segment of a parabola $ y62\equal{}4ax$ cut off by the line x=b.

He most likely meant $ y^2\equal{}4ax$ (note that ^ is Shift+6, and it is very probable that he just simply didn't hold Shift long enough when he pressed the 6.)

You can divide the lamina in infinitesimal thin strips with width $ dy,$ length $ l \equal{} b \minus{} x \equal{} b \minus{} {{y^2}\over{4a}},$ area (instead of mass) $ dA \equal{} l dy,$ moment of inertia $ {{l^2}\over 3}\;dA$ and sum for $ \minus{}2 \sqrt{ab} \leq y \leq 2 \sqrt{ab}$.


limit
logarithms
calculus
calculus computations

hi!!

evaluate this limit without using hospital's rule.

$ \lim_{x\to0}(\frac{\ln(x\plus{}1)\minus{}x}{x^{2}})$

thanx.

For $ x \neq \minus{}1$ and $ |x| \leq 1$, we have $ \ln (1 \plus{} x) \approx x \minus{} \frac{x^2}{2}$.


find all natural numbers a,b,c such that $ ab\plus{}bc\plus{}ca\equal{}a\plus{}b\plus{}c\plus{}1$

[hide="Solution"]Let $ x\equal{}abc$ so that
\[ (a\minus{}1)(b\minus{}1)(c\minus{}1)\equal{}x\minus{}(ab\plus{}bc\plus{}ca\minus{}a\minus{}b\minus{}c)\minus{}1\equal{}x\minus{}2\]
and
\[ 2\equal{}abc\minus{}(a\minus{}1)(b\minus{}1)(c\minus{}1)>(b\minus{}1)(c\minus{}1).\]
If $ b\equal{}1$ or $ c\equal{}1$, then $ x\minus{}2\equal{}0\implies x\equal{}abc\equal{}2$, which yields the permutations of $ \boxed{(2,1,1)}$ as solutions.
If $ b,c\ge2$, then if $ b\ge3$ or $ c\ge3$ we have a contradiction, so $ b\equal{}c\equal{}2\implies x\equal{}4a,\quad x\minus{}2\equal{}a\minus{}1\implies 2\equal{}3a\plus{}1\implies a\equal{}\frac13$, which is also a contradiction.[/hide]

[hide]

$ ab\plus{}bc\plus{}ca\equal{}a\plus{}b\plus{}c\plus{}1\ \Longleftrightarrow\ a(b\plus{}c)\plus{}bc\equal{}a\plus{}1\plus{}b\plus{}c$

$ \Longleftrightarrow\ (b\plus{}c)(a\minus{}1)\equal{}a\plus{}1\minus{}bc\ \Longleftrightarrow\ (b\plus{}c)(a\minus{}1)\equal{}a\minus{}1\plus{}2\minus{}bc$

$ \Longleftrightarrow\ (a\minus{}1)(b\plus{}c\minus{}1)\equal{}2\minus{}bc\ (1)$. Because $ 2\minus{}bc\in\mathbb{N}\ \implies bc\in\{0,1,2\}$

$ 1)$ $ bc\equal{}0$. Case $ b\equal{}c\equal{}0$ is false, so let $ b\equal{}0$.

Then, from $ (1)$ we obtain that $ (a\minus{}1)(c\minus{}1)\equal{}2$, which gives $ a\equal{}2, c\equal{}3$ or $ a\equal{}3,c\equal{}2$ .

$ 2)$ $ bc\equal{}1\ \implies b\equal{}c\equal{}1$ and $ a\equal{}2$ .

$ 3)$ $ bc\equal{}2\ \implies b\equal{}1,c\equal{}2$ or $ b\equal{}2,c\equal{}1$ and according to $ (1)$ we have $ a\equal{}1$ .

Therefore the solutions of the equation are $ (1,1,2);(0,2,3)$ and their permutations.

[/hide]

If we are taking natural numbers to include zero, then
[quote="Mateescu Constantin"]$ \Longleftrightarrow\ (a \minus{} 1)(b \plus{} c \minus{} 1) \equal{} 2 \minus{} bc\ (1)$. Because $ 2 \minus{} bc\in\mathbb{N}\ \implies bc\in\{0,1,2\}$[/quote]
is flawed because the possibility that $ a \equal{} 0$ has not been ruled out. Of course, it's easy to fix.

[quote="aadil"]$ 1,1,2$ is the only solution in natural numbers[/quote]

$ (1,1,2)$
$ (1,2,1)$
$ (2,1,1)$

These were equivalently referred to already in an earlier post in this thread.

aadil, you should use the phrase "positive integers" and not, unfortunately
the ambiguous phrase "natural numbers," because of its stance of the
inclusion of zero.  I always have it as not including zero, and it is
taught that way in Mathematics for Elementary Teachers courses,
for example whenever I come across them.

[quote]aadil, you should use the phrase "positive integers" and not, unfortunately
the ambiguous phrase "natural numbers," because of its stance of the
inclusion of zero. I always have it as not including zero, and it is
taught that way in Mathematics for Elementary Teachers courses,
for example whenever I come across them[/quote]
0 is definitely not natural.thats where the concept of whole numbers comes into existence

That convention is not universal.  When I was in elementary school (and maybe high school), we used $ 0 \not \in {\bf N}$, but computer scientists and combinatorialists (I'm the latter) use $ 0 \in \bf N$.  The symbol ($ \bf N$ or $ \mathbb{N}$) and the name "natural numbers" are both ambiguous, and it would be best to use unambiguous notations like $ {\bf Z}_{ > 0}$, $ {\bf Z}_{\geq 0}$, "positive integers," "nonnegative integers," etc.  The phrase "whole numbers" is not used beyond secondary school, as far as I can tell.  (There may also be differences between countries or continents about these notations; I'm speaking as an American.)

Is there any $ k$ such that there are no solutions to $ ab \plus{} bc \plus{} ca \equal{} a \plus{} b \plus{} c \plus{} k$?

[quote="JBL"]Is there any $ k$ such that there are no solutions to $ ab \plus{} bc \plus{} ca \equal{} a \plus{} b \plus{} c \plus{} k$?[/quote]

Take $ a\equal{}b\equal{}1, c\equal{}k\plus{}1$  :lol:

Okay, good.  There are some values of $ k$ for which this is the only solution in positive integers (like $ k \equal{} 1$, for example).  Can we classify these values of $ k$?  (I don't know the answer to this question.)

I know nothing exept trivial cases doubly (at least) presentable $ k$, like

$ k\equiv 0, 2 \pmod 3$
$ k\equiv 1, 3 \pmod 4$
..........
$ k\equiv i(m \plus{} 1 \minus{} i) \minus{} 1 \pmod m, \ \mbox{where} \ 1\leqslant i \leqslant \frac {m \plus{} 1}{2}$

for sufficiently large $ k$.


calculus
integration
algebra
difference of squares
special factorizations

[hide="hint"]Think binary.  Every positive integer can be expressed as the sum of distinct positive integral powers of two in exactly one way.  Therefore, we can turn the product into a geometric series.  Try expanding smaller similar products (replace the base $2$ with $x$ to see what I mean, e.g. $1+x$).[/hide]

multiple both sides by $2^{2^0}-1=1$, then then paired with the first term, $(2^{2^0}-1)*(2^{2^0}+1)=2^{2^1}-1$, repeat (using difference of squares) to get $2^{2^{51}}-1$

Nice "catalyst problem"!
Here's another one from (USSR Olympiad prob. book):
$\cos(x)\cos(2x)\cos(4x)...\cos(2^nx)$.

[quote="frt"][quote="kimby_102"]Here's another one from (USSR Olympiad prob. book):
$\cos(x)\cos(2x)\cos(4x)...\cos(2^nx)$.[/quote]
[hide]
$\cos(x)\cos(2x)\cos(4x)...\cos(2^nx)$

$=\frac{\sin(x)}{\sin(x)}\cos(x)\cos(2x)\cos(4x)...\cos(2^nx)$

$=\frac{1}{2\sin(x)}\sin(2x)\cos(2x)\cos(4x)...\cos(2^nx)$

$=\frac{1}{4\sin(x)}\sin(4x)\cos(4x)...\cos(2^nx)$

$=\frac{1}{2^n\sin(x)}\sin(2^nx)\cos(2^nx)$

$=\frac{\sin(2^{n+1}x)}{2^{n+1}\sin(x)}$

Can we simplify this further?
[/hide][/quote]
That is the correct answer! :)


Real numbers $x,y,z$ are not all equal, and
$x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}=k$

Find all possible $k$.

$y={1\over k-x}, z=k-{1\over x}$. Then

$y+{1\over z}={1\over k-x}+{x\over kx-1}=k$, which after rearranging yields

$(k^{2}-1)(x^{2}-kx+1)=0$

For $k=1$ we get $x+{1\over 1-x}=1\iff x=0\lor x=2$, but $x=0$ is extraneous because of $z=1-{1\over x}$. Hence for $k=1$ we find the triplet $(2,-1,{1\over 2})$.

Similarly, for $k=-1$ we find $(-2,1,-{1\over 2})$.

If $x^{2}-kx+1=0$, then $x=\frac{k+\sqrt{k^{2}-4}}{2}\lor x=\frac{k-\sqrt{k^{2}-4}}{2}$. In the first case we get $y={1\over k-x}=\frac{2}{k-\sqrt{k^{2}-4}}=\frac{k+\sqrt{k^{2}-4}}{2}=x$, and $z=k-{1\over x}=k-\frac{k-\sqrt{k^{2}-4}}{2}=\frac{k+\sqrt{k^{2}-4}}{2}=x$. Hence that possibility doesn't yield a solution, as $x,y,z$ can't be all equal. Similarly if $x=\frac{k-\sqrt{k^{2}-4}}{2}$.

Therefore $k=\pm 1$


Each vertex of an mxn grid is colored blue ,green or red in such way that all boundary vertices are red.We say that a unit square of the grid is properly colord if

(a) all the 3 color occur at the vertices of the square, and

(b)one side of the square has the endpoints of the same color.

prove that the number of properly colored squares is even.

[hide="an idea"]Let the diagram below be 9 vertexes of any 2*2 sub squares

ABC
DEF
GHI

I think this problem can be proved by showing whatever colors these vertexes have, the change of E to any other color will not change the parity of number of properly colored squares(within the 2*2 sub squares) Because any coloring of the m*n board can be attained from repeatedly changing one color of points not on the boundary.(start from all vertexes are red)

I fail to go further since there are still too much combinations.[/hide]

There is'nt any idea or solution? please help.

try giving weights to the segments
[hide = solution] for every edge with both endpoint colours same we weight it as $ 0$
else we weight it as $ 1$
now only good squares have an odd weight all the others have an even weight
and total weight of all squares= 2 (sum of weights of all edges)
(while concluding the last fact we use the fact that all the border vertices are coloured red)[/hide]

[quote="gauravpatil"]
now only good squares have an odd weight all the others have an even weight
[/quote]

How did you discover this beautiful relation! :idea:

beautiful discovery has only one answer weird experimentation :)


what is lim $2^n \cdot$arcsin$\frac{a}{2^n}$?

It's limit when $n$ goes to infinity is $a$ itself. To proof it, you can for example use the power series of $arcsin(x)$ or the well known theorem $\lim_{x \to 0} \frac{sin(x)}{x} =1$

that's too easy I think.
$lim{2^narcsin(\frac{a}{2^n})}=lim(a\frac{arcsin(x)}{x})$ when $x\rightarrow 0$ where $x=\frac{a}{2^n}$
which equals $a$  :?


function
limit
real analysis
real analysis unsolved

$ \phi: [0;1] \to \mathbb{R}^\plus{}$ a continuous increasing function satisfying:
1: $ \phi(0)\equal{}0$
2: $ \lim_{x \to 0^\plus{}} \frac{\phi(x)}{x}\equal{}0$

Define the operator $ T: \mathcal C([0;1]) \to \mathcal C([0;1])$ by:
$ Tf(x) \equal{} \inf_{y \in [0;1]} f(y) \plus{} \phi(|x\minus{}y|)$
Can one conclude that $ T^{k} f$ converge (pointwise, uniformly,..) towards $ \min_{x \in [0;1]} f(x)$.
(Rq: the case $ \phi(x)\equal{}x^2$ is quite easy)

Since $ T^kf(x)\equal{}\inf_{y_1,\dots,y_k}f(y_k)\plus{}\varphi(|y_k\minus{}y_{k\minus{}1}|)\plus{}\dots\plus{}\varphi(|x\minus{}y_1|)$, the conclusion is next to obvious: just build a chain from $ x$ to the point of minimum using small steps in the same direction and use the fact that $ \varphi(x)\equal{}o(x)$ as $ x\to 0\plus{}$ to see that all those $ \varphi$'s sum to almost $ 0$.


Let $ x_1\equal{}x_2\equal{}1$ and $ x_{n\plus{}2}\equal{}2x_{n\plus{}1}\plus{}8x_n\minus{}1$, $ n\geq 1$. Prove that $ x_n$ is a perfect square for any $ n\in N^*$

What does $ N^*$ denote?  I would assume $ N$ are the natural numbers but what does the * represent?

If your $ N$ includes 0 then your $ N^*$ doesn't; if your $ N$ doesn't, then your $ N^*$ does.  (Life lesson: use unambiguous notations, or define the ambiguous notations you use.)

[hide="Outline of Solution"]Let $ y_n \equal{} x_n \minus{} \frac {1}{9}$ to create a homogeneous linear recurrence of order two. Solve this for $ y_n$ and add back the $ \frac {1}{9}$ to get $ x_n \equal{} \frac {1 \plus{} 2 ( \minus{} 2)^n \plus{} 4^n}{9} \equal{} \left( \frac {2^n \minus{} ( \minus{} 1)^n}{3} \right )^2$

Clearly each $ x_i$ is an integer, so each must also be a square of an integer.[/hide]


Suppose that $ X$ is a normal random variable with mean 5. If $ P (X > 9) \equal{} .2$, approximately with is $ Var(X)$? I am given that $ E[X] \equal{} \mu$ and that $ Var(X) \equal{} \sigma^{2}$ and a table of values, for nonnegative $ x$, for the area $ \phi (x)$ under the standard normal curve to the left of $ X$. I also know that the Z-score is given by, or relates to the normal random variable by :

$ Z \equal{} \frac{ X \minus{}\mu}{\sigma}$
The frustrating thing is that I can't seem to find what what $ \sigma$ is because I can't relate it to the information I know. Any help?

From a table, $ P(Z>z)\equal{}.2$ for $ z\approxeq 0.846.$

So we simply need to have $ \frac{9\minus{}5}{\sigma}\approxeq 0.846.$

$ \sigma\approxeq\frac4{0.846}\approxeq 4.73.$

$ \sigma^2\approxeq \left(\frac4{0.846}\right)^2\approxeq 22.4$

what do the rows and columns of the table stand for, if the values within are the area under $ \phi (x)$

The rows and columns are both just values of $ z,$ locations. The table I was using was one I created myself for a course about 15 years ago, just so I wouldn't always have to be making Xerox copies from some textbook. I wrote this table in $ P(Z>z)$ form, and have appended it to this post.

The rows are for tenths in the values of $ z$ and the columns are in hundredths. I this case, if we look on the $ 0.8$ row we see $ .20045$ in the $ .04$ column and $ .19766$ in the $ .05$ column. That means $ P(Z>0.84)\approxeq .20045$ and $ P(Z>0.85)\approxeq .19766.$ The value $ z\equal{}0.846$ that I used above represents an interpolation of those values.


logarithms
number theory proposed
number theory

Let $In$ be the number of $p$ which satisfied to :

 - for every integrer $n$    $50^{n}<7^{p}<50^{n+1}$

Prove that $In=2$ or $In = 3$ .
Prove that there is infintely many integrers $n$ such that $In = 3$ and determine the smallest possible value of $n$ such that $In = 3$.

Excuse me if my english is bad, I am french......

Yes, it is very easy problem: $n<pa<n+1,a=\ln_{50}7$,because $\frac{1}{3}<a<frac 12$.


advanced fields
advanced fields unsolved

How much is really known about the mandlebrot? Are there any particular books or fields that are devoted to the study of its form? A computer can generate the fractal, certainly, but are there analytical theories that would allow a person to compute where a certain structure might be located, or, given a location, what kind of structure might be located there? I would be curious to know if any important and unexpected numbers, constants, or curves show up in its proportions.

I don't know that much about fractals (that part of my analysis course starts after the fall recess ;) but I know Mandelbrot has written a few books on fractals himself. You might want to check the nearest university library or something.

I would be not at all surprised if there were positive answers to some of your questions.


Two cars can be assigned to 3 parking spaces in 6 different ways.  In how many different ways can 4 cars be assigned to 6 different spaces?

[hide]We have 4 cars and 6 spaces to place them in.  In this problem, the order of the placement of cars matters, so we can solve this using permutations.
$(_6P_4)=360$

Therefore there are $\boxed{360}$ ways to place the cars in the spaces.[/hide]


modular arithmetic
number theory unsolved
number theory

[quote="dondigo"]Prove, that for each prime number $p\geq5$ we'll find integer numbers $x,y,k$ satisfying conditions:
$0<2k<p$ and $pk+3=x^2+y^2$[/quote]

Try as following :  $x,y=1,...,(p-1)/2$ Then $x^2$ and $3-y^2$ each attains $p$ remains modulo $p$, so there is $x,y$ which $x^2\equiv 3-y^2\pmod{p}$. (fact no such $x^2$ congruent modulo $p$). So $x^2+y^2-3=kp$. Now
$x^2+y^2-3\leq (p-1)^2/2-4<p^2/4$ so $k<p/2$. Cause $p\geq 5$ so $k>0$.


I use a TI-83 Plus.  It used to be able to plot many functions at once, even inequalities and stuff.  But now the symbol next to the y values that you normally high light to change the graph to dotted lines, inequalities, etc is a triangle and, after pressing enter, changes into symbols like look like fastforward, play, and pause.

How can I get it back to normal?  Do any of you have this problem?

Are you using the Inequality application or just the regular graphing?

Try clearing RAM (2nd+[MEM]+7+1+2) or deleting and reinstalling the Inequalz app. if worst comes to worst, clear all memory (2nd+[MEM]+->+->+1+2)

Temperal--Thanks!  It's all back to normal now.  I can graph equations and inequalities.  My world is complete.

7h3.D3m0n.117--I am calculator illiterate so I don't know what you mean. It wouldn't let me graph inequalities.  Fine now, though.  So was it malfunctioning because its memory was overloaded or something?  How did it get like that?

Yeah, with Temperal's solution it makes sense. You probably ran out of free RAM space to create a new file for graphing inqualities. Every once in a while, check your memory, using (2nd [MEM] 2 1). If you're under, say 1000, clear your RAM. It's best before you do this if you archive all your programs so you don't lose anything important.

[quote="7h3.D3m0n.117"]If you're under, say 1000, clear your RAM. It's best before you do this if you archive all your programs so you don't lose anything important.[/quote]

To archive a program, 2nd+[MEM]+5+PRGM+(program you want to archive)

To unarchive, 2nd+[MEM]+6+PRGM+(program you want to unarchive)

To check your memory, 2nd+[MEM]+2 and it'll say "RAM FREE (number)" at the top.


search
combinatorics proposed
combinatorics

There are $2N$ people at a party. Each knows at least $N$ others. Prove that one can always choose four people and place them at a round table so that ecah person knows both neighbours.

Pierre.

If $A$ knows $B$ does $B$ necessarily know $A$?

So we must show that there's always a $4$-cycle? :? Take two people which do not know each other. They must have at least two common acquaintances...

The problems I sent here are among the 99 problems appearing in the 'part 4' file you sent to me. Maybe the little paragaph (in russian) in the beggining would explain why they are given without solutions.

Pierre.

[quote="Myth"][quote="orl"]I think it is time to make up a topic for SPbMO problems and give links to the problems and Myth could be so nice to check for more information as year, round, grade and problem number in his book by Fomin. Thanks.  :)[/quote]
 :P  :P  :P  :P[/quote]

Why are you lauging ?  :alien: You always complain of reposting and that you have to search for JBMO problems and that you are not supposed to have knowledge where they are on the forum. That is why I want to organize all that stuff a little bit. Especially if there are a source where we have lots of problems as ISL/ILL, Kvant, JBMO, LMO etc. But you guys like to have everything in a haphazard fashion...  :?

[quote="Myth"]It is "a bit" difficult to search for problem within 250 pages...[/quote]

It is enough to look for the problems and not too look through the solutions when browsing through the book...  ;)  But luckily we have Pierre who can work with Russian sources quicker...  :P

[quote="pbornsztein"]????
I can do what? :?

Pierre.[/quote]

I meant you quickly discovered that the problems were those ones listed at the end of the book which are regarded as more challenging. There you can find the year in order to adopt the sources of your postings. Myth complained that he has to browse through 250 pages. But obviously a non-native Russian speakers as Pierre is faster...  ;)


The first number in a sequence is 15.  Each number after the first is three less than twice the previous.

What is the 10th number in the sequence?

Extension:  What is the nth number in the sequence?

[hide]Let $ a_i$ be the $ i$-th number in the sequence. Then $ a_1 \equal{} 15$ and $ a_{i \plus{} 1} \equal{} 2a_i \minus{} 3$, for $ i\geq1$.

So, $ a_2 \equal{} 2a_i \minus{} 3 \equal{} 2\times15 \minus{} 3$,
$ a_3 \equal{} 2a_2 \minus{} 3 \equal{} 2\times(2\times15 \minus{} 3) \minus{} 3 \equal{} 2^2\times15 \minus{} 2\times3 \minus{} 3 \equal{} 2^2\times15 \minus{} (1 \plus{} 2)\times3$
$ a_4 \equal{} 2a_3 \minus{} 3 \equal{} 2\times(2^2\times15 \minus{} (1 \plus{} 2)\times3) \minus{} 3 \equal{} 2^3\times15 \minus{} (2\times(1 \plus{} 2) \plus{} 1)\times3$

and now it's not hard to see that

$ a_n \equal{} 2^{n \minus{} 1}\times15 \minus{} (2^{n \minus{} 1} \minus{} 1)\times3$[/hide]

[color=darkred][size=75]Added hide tags.
-nebula42[/size][/color]

[quote="RJ Redd"]The formula is An=a1+d(n-1)
So on this problem the 10th term would be 15 + -3 (10-1)
=15 + -27
=-12

So for the nth term it would be 15 + -3 (n-1)
Hope this helps :lol:[/quote]

rj are you sure?  it's not an arithmetic sequence.

$ f(x)\equal{}6*2^{x}\plus{}3$
Base Case: x=1, f(x)=15. Good.

Inductive Step: $ f(x\plus{}1)\equal{}6*2^{x\plus{}1}\plus{}3\equal{}6*2^{x\plus{}1}\plus{}6\minus{}3\equal{}(6*2^{x}\plus{}3)*2\minus{}3\equal{}2*f(x)\minus{}3$ .

Which is what we wanted, so we're done.

10th number: f(10)=6*1024+3=[b]6099[/b]


Does anyone know where to get the [i]Instructor's Manual[/i] for Paul G. Hewitt's Conceptual Physics, tenth edition? An electronic version would be the best.[/i]

With most textbooks, you can't get the instructor's manual unless you provide proof that you are an instructor/teacher of the course.

I have just found out that the companion book, Practising Physics, which I have, has the answers to the exercises in the textbooks, which I primarily want.


Obviously I'm not normal because I don't understand how to convert from units N to units M. Could you explain how many N come out of one M for various kinds of reactions?

My book says that "We must know for what purpose the solution is being used, because it is the concentration of the reactive species with which we are concerned", but that is too erudite for me to understand...

Let's say we're talking about 0.5 Molarity solution of Sulfuric acid.

The formula for sulfuric acid is H2SO4, and for every mole of the acid, there is 2 moles of H+.

Let's say we're solving a neutralization reaction,and we're only interested in the H+'s from the sulfuric acid.  Therefore, we can convert the molarity of the solution to normality:

0.5 M * 2 mole's H+'s/ 1 mole H2SO4 = 1 N solution

Thus, the normality of the solution would be 1 N.

I think normality is most commonly used in titration reactions; I remember using it in chemistry last year for the titration lab.

That's only half of the story. The same solution of sulfuric acid can be used for BaSO4 precipitation and it will be only 0.5N for this reaction, that's why the book states that you have to know what the reagent will be used for.

Permanganate solution can be reduced in redox reaction to manganate, manganese dioxide or Mn(2+) - in each case normality of the solution will be different.

The responses you already received are correct, as far as they go.

If we stay with the example of H2SO4, lets consider something else.  The first H+ will come off the sulfuric acid easier than the second H+, so if we picked the correct weak base, we will only get one H+ from the sulfuric acid and the Normalicy will be half the Molarity.

Further, and not normally covered in high school chem, is the fact that the real world didn't read the text books.  Things tend to go a bit different from theory at times.  For example, some materials at high concentractions react as if the chemicals were a bit less pure than they react at low concentractions. particularly if one is calculating pH of a solution.

The obvious question of how high concentrations, is dependent on the chemical and greatly dependent on the degree of percision one is measuring.  Something that shows up in pH when measuring to five significant digits is irrelevant many times if one is measuring to three or two significant digits.  So in my pH example, is the concern the measured pH, or how much sodium hydroxide it will neutralize.  In the second case, you don't care the initial pH, just the reaction, so this effect is irrelevent.

Probably too much for this question, but I thought it might be useful later in the real world for someone reading.

[quote="wizard2378"]If we stay with the example of H2SO4, lets consider something else.  The first H+ will come off the sulfuric acid easier than the second H+, so if we picked the correct weak base, we will only get one H+ from the sulfuric acid and the Normalicy will be half the Molarity.[/quote]

Give an example. IMHO there is no such base.

The normality will never be half the molarity, but if we choose a base weak enough so that the second hydrogen is favored on the $\text{HSO}_4^-$ anion instead of accepted onto the base... $\text{HSO}_4^-$ has a $\text{p}K_\text{a}=2.0$, so dissolving $\text{H}_2\text{SO}_4$ in excess molten sodium methoxide ($\text{NaCH}_3\text{O}$, $\text{p}K_\text{b}=-1.5$) will show a very strong first dissociation step:

$\text{H}_2\text{SO}_4+\text{CH}_3\text{O}^-\leftrightharpoons\text{HSO}_4^-+\text{CH}_3\text{OH}$

But $\text{CH}_3\text{OH}$ has a $\text{p}K_\text{b}=16.4$, so the dissociation

$\text{HSO}_4^-+\text{CH}_3\text{OH}\leftrightharpoons\text{SO}_4^{2-}+\text{CH}_3\stackrel{+}{\text{O}}\text{H}_2$

(eww $\LaTeX$) will not be favored.

Does that work, or did I misunderstand normality?

EDIT: This doesn't work.

[quote="inteluser"]The normality will never be half the molarity[/quote]

What is normality of 1M silver solution used to precipitate chromate? Isn't it 0.5N?

[quote]in excess molten sodium methoxide[/quote]

In a way you've got me - I was referring to water solutions, but in general in other solvents one will be able to separate both protons and to get the two-step titration curve of sulphuric acid.


integration
calculus
real analysis
real analysis solved

the original problem asks us to prove that

for any a>0 it holds that

\int {1 to +\infty} f(a/x+x/a) (ln x) / x dx = ln a \int {1 to +\infty} f(a/x+x/a) 1/x dx

if the integrals in both sides are convergent.

===============================================

Take a = 2 and f(x)=1/x

\int {1 to +\infty} f(a/x+x/a) (ln x) / x dx = \int {1 to +\infty} 2 ln x / (4+x^2) dx

ln a \int {1 to +\infty} f(a/x+x/a) (ln x) / x dx = ln 2\int {1 to +\infty} 2 / (4+x^2) dx

Obviously they are not equal.

The original problem is wrong.

How to correct it?

the problem will be correct if we change the interval of two integrals to [0,\infty).
-- and it's easy to prove.. just change variable : t = ln (x/a)


Asymptote
function
abstract algebra
geometry

[asy]size(200);

path line (pair a, pair b) {return (a--b) ;}
path zline(real a, real b) {return line((a,b),(a,0)) ;}

real a=1, b=6, n=5, c= (b-a)/n;

real f (real x) {return sin(x)+0.5*(x) + 1  ;}

draw(graph(f,a,b),blue);

for (int i=0; i<5; ++i) {
for (int j=0; j<6; ++j) {
for (int k=1; k<6; ++k) {

real a=1, m= (6-1)/5;
draw( (a+ m*i, f(a+m*i))--( a+ m*(i+1), f(a + m*(i+1) ) ), red ) ;
draw(zline( a+m*j, f(a+m*j) ),dashed) ;
label("\scriptsize{$x_{"+string(k)+"}$}",(a+m*k,0), S);
dot ( (a+m*j,f(a+m*j)) );

}}}

draw ( (-1,0)--(8,0),Arrows);
draw( (0,-1)--(0,8),Arrows);

label("\scriptsize{$0$}",(0,0),SW);
label("\scriptsize{$x$}",(8,0),S);
label("\scriptsize{$y$}",(0,8),W);
label("\scriptsize{$y=f(x)$}",(4,4));
label("\scriptsize{$a$}",(a,0),S);[/asy]

I don't understand some of the things that happened to my graph. Why are the $ x_{i}$ labels bolded? And why do the red line segments look "fuzzy"?

[code]import math;
import olympiad;
size(250);

path line (pair a, pair b) {return (a--b) ;}
path zline(real a, real b) {return line((a,b),(a,0)) ;}

real a=1, b=6, n=5, c= (b-a)/n;

real f (real x) {return sin(x)+0.5*(x) + 1  ;}

draw(graph(f,a,b),blue);

for (int i=0; i<5; ++i) {
for (int j=0; j<6; ++j) {
for (int k=1; k<6; ++k) {

real a=1, m= (6-1)/5;
draw( (a+ m*i, f(a+m*i))--( a+ m*(i+1), f(a + m*(i+1) ) ), red ) ;
draw(zline( a+m*j, f(a+m*j) ),dashed) ;
label("\scriptsize{$x_{"+string(k)+"}$}",(a+m*k,0), S);
dot ( (a+m*j,f(a+m*j)) );

}}}

draw ( (-1,0)--(8,0),Arrows);
draw( (0,-1)--(0,8),Arrows);

label("\scriptsize{$0$}",(0,0),SW);
label("\scriptsize{$x$}",(8,0),S);
label("\scriptsize{$y$}",(0,8),W);
label("\scriptsize{$y=f(x)$}",(4,4));
label("\scriptsize{$a$}",(a,0),S);[/code]

[asy]import math;
import olympiad;
size(500);

path line (pair a, pair b) {return (a--b) ;}
path zline(real a, real b) {return line((a,b),(a,0)) ;}

real a=1, b=6, n=5, c= (b-a)/n, m= (6-1)/5;

real f (real x) {return sin(x)+0.5*(x) + 1  ;}

draw(graph(f,a,b),blue);

for (int i=0; i<5; ++i) {
for (int j=0; j<6; ++j) {

draw( (a+ m*i, f(a+m*i))--( a+ m*(i+1), f(a + m*(i+1) ) ), red ) ;
draw(zline( a+m*j, f(a+m*j) ),dashed) ;
dot ( (a+m*j,f(a+m*j)) );

}}

for (int k=1; k<6; ++k) {
label("\scriptsize{$x_{"+string(k)+"}$}",(a+m*k,0), S);
}

draw ( (-1,0)--(8,0),Arrows);
draw( (0,-1)--(0,6),Arrows);

label("\scriptsize{$0$}",(0,0),SW);
label("\scriptsize{$x$}",(8,0),S);
label("\scriptsize{$y$}",(0,6),W);
label("\scriptsize{$y=f(x)$}",(4,4));
label("\scriptsize{$a$}",(a,0),S);[/asy]

It was written normally when I wrote following code independently.
[code]for (int k=1; k<6; ++k) {
label("\scriptsize{$x_{"+string(k)+"}$}",(a+m*k,0), S);
}[/code]
And I think that we look it fuzzy because blue and red are near.

Because you draw them more than once in that triple-for loop (you overlap them) and because the graphic render isn't precise, it looks like it's bolded and fuzzy. In real, it's not. Do the labeling in single loop.

[code]import graph;
size(100);

xaxis(-10,10,Arrows);
yaxis(-10,10,Arrows);

real f (real x) {return x^2 ;}
path Z (real x) {return (x,f(x)) ;}

pair a= Z(4);

dot(a);[/code]

When asymptote compiles, I get the error "cannot cast path to pair". Why can't I use my new function Z when creating pairs?

Don't you define Z as a path? Then you try to tell it to be a pair.

Thanks.

I created a package called eric.asy, which simply contains

[code]
import olympiad;
pair z (real x) {return (x,0) ;} [/code]

I test it out and it doesn't work =(

[code]import eric;
size(100);

xaxis(-10,10,Arrows);
yaxis(-10,10,Arrows);

real f( real x) {return x^2 ;}

draw(graph(f,-2,2));[/code]

I keep getting "cannot load module 'eric' ".

Just put something like this (be careful about path, use .. and / instead of \):[code]import "eric.asy" as eric;[/code]

No one for the last question?

I have a new one which concerns with creating a particular function in asymptote. Particularly, I'm looking for a rule $ f$ such that:

$ f(1) \equal{} 0.2, f(2)\equal{} 0.6, f(3)\equal{} 0.75, f(4) \equal{} 0.8,$

$ f(5) \equal{}0.5, f(6) \equal{} 0.3, f(7) \equal{}0.5, f(8) \equal{} 0.2, f(9) \equal{} 0.85$

I'm guessing I'd need something like

[code]

real f (real x) {return x such that [insert all the conditions mentioned above] ;} [/code]

[quote="Prime factorization"]Do I put that in the new package, or in the experimental picture? I tried a variety of things with the line you suggested but no luck. Can you give me an example of a working package you have created?[/quote]

[b]mnm_pack.asy[/b]
[code]import cse5;
import geometry;
import geometry_dev;
import olympiad;

texpreamble ("\usepackage[d]{esvect}");

defaultpen (fontsize (10pt));

pen MNM_orange = cmyk (0.00, 0.69, 0.98, 0.00);
pen MNM_gray = cmyk (0.00, 0.00, 0.00, 0.60);

arrowfactor *= 0.75;
perpfactor *= 0.5;
real MNM_angleradius = 16;
[/code]

[b]translation.asy[/b]
[code]import "mnm_pack.asy" as mnm_pack;

size (9cm, 0);

triangle P; P.init ((0.0, 0.0), (3.0, -2.0), (3.75, 0.5));
triangle PP = shift ((4.5, 1.25) - P.A) * P;

draw ("$\vv{v}$", centroid (P) -- centroid (PP), S, dashed + MNM_orange, Arrow);
draw (P, dot (Fill (white)));
draw (PP, MNM_gray, dot (MNM_gray, Fill (white)));

label ("$P$", "$Q$", "$R$", P);
label ("$P'$", "$Q'$", "$R'$", PP, MNM_gray);
[/code]

[b]mnm_pack.asy[/b] is in the same folder as [b]translation.asy[/b].

Thanks. My other problem is still in tact, though:

[quote]I have a new one which concerns with creating a particular function in asymptote. Particularly, I'm looking for a rule $ f$ such that:

$ f(1) \equal{} 0.2, f(2) \equal{} 0.6, f(3) \equal{} 0.75, f(4) \equal{} 0.8,$

$ f(5) \equal{} 0.5, f(6) \equal{} 0.3, f(7) \equal{} 0.5, f(8) \equal{} 0.2, f(9) \equal{} 0.85$[/quote]

Right now I'm thinking of creating a pseudo-random function that produces all these values, though I'd much rather find out if asymptote can define piecewise functions somehow.


Show that $\forall x,y,z \in R \neq0$ we have:
$ \sum( \frac{x^2}{y^2} + \frac{y^2}{x^2})^2 \geq \sum( \frac{x}{y} + \frac{y}{x})^2$

cheers! :D :D :cool:

Another one:
  sum (x/y+y/x)^2<=2 sum (x^2+y^2)=sum (x^2/y^2+y^2/x^2)^2.
   Can't you find please more interesting problems? Or, at least ,post them in the proposed section please (you do have solutions for thm, they aren't unsolved). Thank you.


Inegalitatea rearanjamentelor spun ca daca avem {X1,X2,...Xn} si {Y1,Y2,....Yn} doua n-uple la fel ordonate atunci avem ca X1Y1+X2Y2+....XnYn este cea mai mare suma dintre toate sumele de perechi XiYj . Intrebarea mea este daca putem generaliza aceasta inegalitate la n n-uple adica putem demonstra spre exemplu ca a^5+b^5+c^5 >= a^2b^3+b^2c^3+c^2a^3 folosind acest principiu al rearanjamentelor?

Păi tu nu ai generalizat nimic, ba chiar dimpotrivă, ai particularizat pentru $n=3$ inegalitatea, şi da asta este o demonstraţie :) (dacă toate numerele sunt pozitive).

Pai inegalitatea de care spui la sfarsit nu este nimic altceva decat inegalitatea rearanjamentelor pentru sistemele la fel ordonate $ (a^2,b^2,...)$ si $ (a^3, b^3,...)$ . Sau nu am inteles eu intrebarea.

Daca stii ruseste, vezi si [url=http://kvant.mccme.ru/1985/12/uporyadochennye_nabory_chisel.htm]pagina asta[/url]

Este un caz particular al inegalitatii rearanjamentelor! Bineinteles, inegalitatea este adevarata :lol:


...are online:  http://www.wocomal.org/statistics/olympiad/200708/200708o_level2.html

Congratulations to all who placed in the top 20.

Rick Yanco
Worcester Academy

Darn. Only a few points away...

Probably because I forgot to show the thingies were relatively prime on the pythagorean one (I thought we only had to show it was a right triangle). Ah well...

Huh, I failed worse than I thought...This wouldn't have happened had I not thought the chair one was the classic Fibonacci problem... :roll:

[quote="CatalystOfNostalgia"]Wow as I predicted, Noah won and I got a 64 (actually I predited a 66 but whatev) with my stupid mistakes. I didn't go, btw.[/quote]

Mr Roos was predicting Noah first and Stephen second, at least he got one prediction right.

Congrats everyone. Personally, considering I wasn't even a semifinalist last year I like my finish here.  :yup:

i'm never spending 1.5 hours on one problem again....   :oops:

yea...i got owned

I put the other 73 scores on the Olympiad L2 online, listed by your ID number.  So if you remember your ID number and you weren't in the top 20, you can find your score.

If you don't remember your ID number--sorry, I don't have the list (yet?).

http://www.wocomal.org/statistics/olympiad/200708/200708o_level2.html

What's with all the seniors getting owned?  :o

Haha, so I looked at the geometry problem, and I was like, eww, geometry, gonna work on the socks problem. Big mistake. *sigh*

So Noah got what I got last year =)

Mr. Yanco -- do you, by any chance, have a problem-by-problem analysis of scores?

[quote="The Zuton Force"]do you, by any chance, have a problem-by-problem analysis of scores?[/quote]
Here is the highest score, 75th percentile, median, 25th percentile, and lowest score earned for each question.  (Students who didn't submit anything for the question aren't counted in these statistics.)

P1. 15/12/9/7/3
P2. 15/12/8/4/1
P3. 15/13/11/9/2
P4. 15/5.5/5/5/0
P5. 15/15/6/3/0
P6. 10/10/1.5/0/0
P7. 15/2/1/1/0

For each question part:

P1a. 4/4/4/1/0
P1b. 3/3/3/3/0
P1c. 4/3/2/1/0
P1d. 4/3/1/0/0
P2a. 2/2/2/2/1
P2b. 3/3/3/2/0
P2c. 1/1/1/0/0
P2d. 2/2/2/0/0
P2e. 7/4/0/0/0
P3a. 2/2/2/2/1
P3b. 5/5/4/2/0
P3c. 3/3/3/3/0
P3d. 5/3/2/2/0
P4a. 5/5/5/5/0
P4b. 10/2/0/0/0
P5a. 1/1/1/1/0
P5b. 2/2/2/2/0
P5c. 3/3/3/0/0
P5d. 4/4/0/0/0
P5e. 5/5/0/0/0 *
P6. 10/10/1.5/0/0
P7a. 1/1/1/1/0
P7b. 4/0/0/0/0
P7c. 10/0/0/0/0

*Since answering P5e was enough to answer the entire fifth question, the grader put "15" (or 14, or 13, or 12) in this column when P5e was proven.  For these stats I filled the points in starting with P5a, with any points lost deducted from P5e.  So a 13 is shown here as 1-2-3-4-3.

Hope this helps.

Rick Yanco
Worcester Academy


Hey guys,

After returning from the three-week long IMO trip (training + competition), I have been working on a journal, telling my story of the journey (with photos). It is my goal to describe the IMO experience from the eyes of a [i]contestant[/i] (as opposed to the usual official descriptions or Leader's Reports).

I'm still in the process of writing the third (and most significant) part of the journey (as well as refining the rest of the work), but in the meantime, I've put on the web the parts I have completed. They can be (found at this URL:

http://web.mit.edu/yufeiz/www/imo2006/

I hope you enjoy it.

(I wasn't originally intending on showing this to the public until it's done, but Victoriya repeatedly urged me to present it to the AoPS community. So here it is. As well, if you want to write your own IMO story, retelling us your unique perspective, then I'd be happy to include it in my website.)

I love how Yufei keeps quoting Peng over and over again...

'Peng had a very fitting description of our food – there were only two kinds: "crap soup," and "stuff."'

Yufei, this is some really good stuff.  Can't wait for part 3!

That was awesome, nice to know something about the olympiad besieds the scores.

If rem is reading this, he better make some of these over the next few years!

All this reminds me: if all of the team is graduating, where are you all going next year?

[quote="rem"][quote="rzsolt"]If rem is reading this, he better make some of these over the next few years!
[/quote]
LOL. I will, if I get to IMO...

The pictures and the whole story are soooooooooooooooooo awesome!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

From these pics IMO looks even more fun. I SO WANT TO GET THERE. these pictures are such great motivation!

[quote="rzsolt"]All this reminds me: if all of the team is graduating, where are you all going next year?
[/quote]
Yufei is going to MIT and David is going to UW. I don't know about the rest though.[/quote]
Peng is going to Duke, Richard is coming to UW, Farzin is going to UBC, and Victoria is going to UT

Too bad you're not all comming to UW, you guys looked like alot of fun, and I had only met Richard and Peng briefly.

[quote="Kileel"]It's all about Peng and his [b]SICK[/b] dancing skills.[/quote]

Hi Joe! Good to see you here!

OMG... i have to face 2 IMOs in the math class this fall.

Wow Yufei what a great journal!!!  Definitely waiting for the next part.

EDIT:  LOL...you'll be able to tell who David is.  He's the one with the REALLY REALLY loud laugh.  :lol:

great stuff! very insightful. i think it's great for those who were not able to experience imo.

[quote="J-I-M-B-O"] i think it's great for those who were not able to experience imo.[/quote]
u bet it is. :!:
This journal portrays how fun it actually is to do math, and the amazing reward u get for training for many hours to get to imo. :wink:
Canmoo now has a can!!! YAY!

Wow, Yufei.  That's quite a trip.  Thanks for letting us all share the experience!

Geoffrey posting on AoPS? The world is coming to an end!

Hello Taotao!  This is an awesome site.  Thanks for suggesting it.


Does anyone know when is the AMC cutoff score (top 1% for AMC-10 and 5% for AMC-12) for AIME to be announced?

I think it's probably whenever they get all the tests corrected for the A and B exams, which is typically 1 or 2 weeks after the B.  BTW, the cutoff for the AMC12 is by rule never higher than 100, so if you scored in the 3 digits, you're automatically in.

[quote="Derek"]Last year's statistics on cutoff score on A tests were posted on March 3rd last year, so this year's should be around that date. Not sure though.[/quote]

But the AIME test is on March 8 this year.  It does give much time for test takers.


1,Find limit:
$ A\equal{}\lim_{x\to0}\frac{\sqrt[n]{1\plus{}\alpha x}.\sqrt[m]{1\plus{}\beta x}\minus{}1}{x}$ with $ \alpha\beta \neq 0$ and $ m, n$ is positive integers

2,Let $ P(x)\equal{}a_1x\plus{}a_2x^2\plus{}...\plus{}a_nx^n$, with $ n$ is a positive integer . Find limit :$ \lim_{x\to 0}\frac{\sqrt[n]{1\plus{}P(x)}\minus{}1}{x}$
 :lol:

[quote="Yuriy Solovyov"]1) $ (1 \plus{} ax)^n \equal{} 1 \plus{} anx \plus{} O(x^2)$ when $ x\to 0.$
Answer: $ \frac {\alpha}{n} \plus{} \frac {\beta}{m}$
2) $ P(x) \equal{} a_1x \plus{} O(x^2)$ :)
Answer: $ \frac {a_1}{n}.$[/quote]
[b]Can you solve clearly and more easy to understand?[/b]
Thanks so much!

It is  quite easy.the simple way is use Lopitan ' order


Find the radius of the inscribed circle of trapezoid $ABCD$ if $AB=5$, $CD=10$, $BC=AD=7.5$.

[quote="frt"]I think there is more than one possible inscribed circle... In order for quadrilateral $ABCD$ (in general) to have a unique inscribed circle, we must have $AB+CD=BC+DA$... or am I missing something? :?[/quote]

Okay, I have no idea how to do these, so I was just making something up. Fixed, i guess.

[hide="Hint"]
First, find the area of the trapezoid. It shouldn't be so hard to find the height. :)

Then use $K=sr$, where $K$ is the area, $s$ is the semiperimeter, and $r$ is the radius of the inscribed circle. This formula is true for any polygon with inscribed circle.[/hide]

[quote="frt"][hide="Hint"]
First, find the area of the trapezoid. It shouldn't be so hard to find the height. :)

Then use $K=sr$, where $K$ is the area, $s$ is the perimeter, and $r$ is the radius of the inscribed circle. This formula is true for any polygon with inscribed circle.[/hide][/quote]

Isn't $s$ the SEMI-perimeter?  I know it is with triangles.  I didn't know that formula could be extended to other polygons, though.

Yeah, you're right. Sorry for providing a wrong hint :P Edited.

just find the height of the trapezoid with they pythaogorean theorem and divide by 2
i got $\dfrac{5\sqrt{2}}{2}$

Well, since it is an iscoceles trapezoid, then each of the two right triangles would have a base of $\dfrac{10-5}{2}$ and the hypotenuse would be $7.5$, now just use pythagorus and divide by two to get the radius, which I got as $\dfrac{5\sqrt{2}}{2}$.

Oh yeah... no need for $K=sr$ :roll:

I got $\dfrac{5\sqrt{2}}{2}$ as well.

Okay, thanks....another related question: how do you [b]construct[/b] the incircle? Like for a triangle, you draw the angle bisectors and find their intersection to find the incenter. What about something like a trapezoid? :?

EDIT: http://mathworld.wolfram.com/TangentialQuadrilateral.html
It only works for quadrilaterals with only one unique incircle.

What if there is more than one incircle??? :(

[quote="chess64"]

What if there is more than one incircle??? :([/quote]

Then try breaking it up into triangles and using the $rs$ formula, and if that doesn't work, then just try to use the best geometry skills you have to try to find it

Here's an example, you have an equilateral triangle, say of sidelength of $2$ and you have a circle inscribed in it, then you have three more circles tangent to two sides of the triangle and the bigger circle, then find the radius of the smaller circles.

[quote="chess64"][quote="frt"]Then use $K=sr$, where $K$ is the area, $s$ is the semiperimeter, and $r$ is the radius of the inscribed circle. This formula is true for [b]any[/b] polygon with inscribed circle.[/quote]

Are you [b]sure[/b] about that? I can't seem to find a proof on mathworld or anything. :( Even for a quadrilateral...[/quote]

He means when breaking it down to triangles because that formula is only true for triangles

[quote="chess64"]:huh: What are you talking about? :huuh: I don't think that's what he meant...[/quote]

Wow, I think he is right, I've used some examples, for instance your own problem and that works out  :o

The diagram below should explain why it works :)

Thanks, frt! That's a really useful formula! :)


Is there a way to set line spacing uniformly? The default mode is too compact for my liking.

Clarification: I'm asking if there is a way to set line spacing in paragraphs, as in single space, 1 1/2 space, double space, or even setting tenths of inches between lines, etc.

Help appreciated!

In the preamble put \renewcommand{\baselinestretch}{1.2}) which gives a 20% increase in line spacing. 1 is the default and you can adjust the vaue to suit.

Wow, I never knew that.

Thanks a lot! Also, is there a way to draw a graph?

Like a table?

Tables: [[LaTeX:Layout#Tables]]
Graphs: [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_(Vector_Graphics_Language)]Asymptote[/url] and ask questions in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=519]Asymptote Forum[/url].


geometry
3D geometry
algebra unsolved
algebra

Prove that can exist the real numbers $\ x,y$ such that $\ x^2+y^2=3,x^3+y^3=4$.

Let $s = x + y$ and $p = xy$. Then $s^2 - 2p = 3$ and $s(3 - p) = 4$. Solving the system we get $p = -1$ and $s = 1$. Hence the solutions of $x^2 - x - 1 = 0$, which are reals, are a pair of solutions.


modular arithmetic
number theory solved
number theory

Prove that if a random square number has an odd number of digits in it the last digit is $6$
$x^2=\overline{a_na_{n-1}a_{n-2}....a_1}$
Prove that if
$n=2k+1$ then $a_1=6$

I think it can be modified to:

If $a_2$ is an odd number, then $a_1 = 6$.

Well, let me give a slightly more elegant solution.

Let $x^2 = \overline{a_ka_{k-1}\ldots a_1a_0}$ be the decimal representation.
We first note that $a_1a_0$ must be even. This can be easily proven (See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55743]www.mathlinks.ro/Forum/viewtopic.php?t=55743[/url]). Thus if $a_1$ is odd, then $a_0$ can only be $0, 4$, or $6$, by taking $\mod 10$. Now the case of $a_0 = 0$ is immediately ruled out, since it implies $a_1 = 0$. If the units digit is $4$, then $x \equiv \pm 2 \pmod{10}$, so letting $x = 10y \pm 2$, we have $x^2 = 100y^2 + 20y + 4$, which implies $a_1$ is even. This forces $a_0$ to be $6$.

Done.


geometry
geometric transformation
rotation
calculus
calculus computations

given the region bounded by $y=x^2$ and $y=4x$, create a volume of rotation by rotating that about $y=2x$, what is the volume of this solid?


calculus
integration
limit
calculus computations

hint:
for big $ n$
$ \frac{1}{n\plus{}\alpha}\equal{}\frac{1}{n}\minus{}\frac{\alpha}{n^{2}}\plus{}o(\frac{1}{n^{2}})$


Please read the announcement post.

Making posts consisting of "???" is uncalled for.

this should be moved to puzzles/brainteasers

[quote="NoSoupForYou"]Apparently, this guy is posting the same problem in every forum.[/quote]

It's in Getting Started as well. That's probably where it belongs. Thanks for the tip.

Topic closed.


Let a, b, c be given positive real numbers.
Prove that
$(a+b+c)^9abc \geq 27(ab+bc+ca)^6$

It is correct?
Let $a\ge b \ge c$, and c is very small. Then we have $(a+b)^9abc\ge27(ab)^6$ Contradition with very small c.

Sorry I 'm wrong.
Try this one instead:

a,b,c > 0.
Prove that:
$(a+b+c)^2(ab+bc+ca)^2+(a+b+c)^3abc \geq 4(ab+bc+ca)^3$

Your solution is the same as mine. I post it here for finding an more beautiful solution. Do you have one?

Well, I do not know yet a simpler solution... I will try to find. :)

Here is an easy solution:
Let $s: =a+b+c$ and $q: =ab+ac+bc$. We have to prove $s^2q^2+s^3abc\ge 4q^3 \Leftrightarrow s^3+\frac{s^4}{q^2}abc\ge 4sq$.
Now we have $s^2\ge3q$, square that and plug it in:
 we have to show $s^3+9abc\ge 4sq$ but this is Schur.  :D  :D

Oh. It's really a beautiful solution. Thank you, spanferkel


$a+b+c=p+1.$ Hence $1\leq a<p,1\leq b<p,1\leq c<p.$
$(a+b+c)^3=a^3+b^3+c^3+3(a+b+c)(ab+ac+bc)-3abc,$
$a+b+c\equiv1(modp),a^3+b^3+c^3\equiv1(modp).$
Hence, $1\equiv1+3\cdot1\cdot(ab+ac+bc)-3abc(modp).$
Hence, $abc-(ab+ac+bc)\equiv0(modp).$
Hence, $abc-(ab+ac+bc)+a+b+c-1\equiv0(modp).$
Hence, $(a-1)(b-1)(c-1)\equiv0(modp).$
Hence, $a-1\equiv0(modp)\vee b-1 \equiv0(modp) \vee c-1\equiv0(modp).$
Hence, $a=1$ $\vee$ $b=1$ $\vee$ $c=1.$ :)

This is problem 7 from [b]current[/b] Polish MO due November 7.


x,y,z are positive real numbers and n,m are integers n \geq m prove:
x^n/((y+z)^m)+y^n((z+x)^m)+z^n/((x+y)^m) \geq 1/2^m(x^(n-m)+Y^(n-m)+z^(n-m))

Suppose $z \leq y \leq x$
$\frac{x^n}{(y+z)^m+y^n/(x+z)^m+z^n/(x+y)^m} \geq \frac{3(x^n+y^n+z^n)}{(y+z)^m+(x+z)^m+(x+y)^m} \geq \frac1{2^m}\frac{3(x^n+y^n+z^n)}{x^m+y^m+z^m} \geq \frac1{2^m}{x^{n-m}+y^{n-m}+z^{n-m}}$
We used chebishev's inequality and $2^{m-1}(x^m+y^m) \geq (x+y)^m$ :)  :)


By variational principles I mean e.g. the Principle of Least action in Mechanics and Fermat's principle in Optics.
Which are the other such principles in Physics, Can our entire knowledge of physics be described in terms of just the variational principles???

I don't think that you can describe the fact that

$F_g=G\frac{m_1m_2}{r^2}$

(or similar ones) by any such principle. You probably can't describe any numerical fact at all by such principles.

I think you can using Lagrange's princple. Cant you???

I don't know what is Lagrange's principle, could you explain? However, I seriously doubt that Lagrange's principle contains value $6.673\times10^{-11} m^3 kg^{-1} s^{-2}$ anywhere in it.

Actaully my question is related to research in finding such universal principles for descibing Physics. HAs somebody done some research on this???

I believe the variational principles might be a good approach for unification for it seems to include the uncertainity of Quantum Mechanics as well as the preciseness etc. of Einsteins and Newton's theories!! This is because, say considering a photon. the "photon" will "think" of the possible paths and take the path by slightly varying it at every moment of time to sort of stick to the path as dictated my NEwton's laws!!!

What are the prospects of research in this field???


Let be $ A_m\equal{}\{x\in \mathbb{R}\ |\ |x\minus{}m|\plus{}|x\plus{}m|\le 2m\ ,\ m\in \mathbb{R}\}$ . Find $ A_m$ and :

$ A_1\cap A_2\cap A_3\cap ...\cap A_{2009}$

[hide="solution"]
If $ m<0$ then obviously there is no such $ x$.

If $ m\ge 0$,then by triangle inequality we have $ 2m\ge |m\minus{}x|\plus{}|m\plus{}x|\ge |(m\minus{}x)\plus{}(m\plus{}x)|\equal{}2m$,hence the equality must hold,which means $ m\minus{}x$ and $ m\plus{}x$ have the same sign,i.e. $ \minus{}m\le x\le m$,

$ A_m\equal{}\quad \left\{\begin{array}{l}\phi \;(m<0)\\\\
\{x\: |\: \minus{}m\le x\le m\} \;(m\ge 0)\end{array} \right.$.

hence $ A_1\cap A_2\cap A_3\cap ...\cap A_{2009}\equal{}\{x\: |\: \minus{}1\le x\le 1\}$[/hide]


Or you could have used the $Power$ $Mean$ $Inequality$.

[color=darkblue]$n\in N^{*}\ ,\ a>0\ ,\ b>0\Longrightarrow (a+b)^{n}\le 2^{n-1}\cdot \left(a^{n}+b^{n}\right)\ \ (1)\ .$
[b]Proof.[/b] Apply the elementary conditioned inequality $(x-y)(z-t)\ge 0\Longrightarrow (x+y)(z+t)\le 2(xz+yt)\ :$
$(a+b)^{2}\le 2(a^{2}+b^{2})\ .$
$(a+b)(a^{2}+b^{2})\le 2(a^{3}+b^{3})\ .$
$(a+b)(a^{3}+b^{3})\le 2(a^{4}+b^{4})\ .$
$\ldots\ldots\ldots\ldots$
$(a+b)(a^{n-1}+b^{n-1})\le 2(a^{n}+b^{n})\ .$
Multiply the above inequalities and obtain $(1)\ .$

[b]Generalization.[/b] If for any $i\in\overline{1,m}$ and $j\in\overline{1,n}$ we have $a_{ij}>0$ and $a_{i1}\le a_{i2}\le\ldots\le a_{in}$, then $\prod_{i=1}^{m}\sum_{j=1}^{n}a_{ij}\le n^{m-1}\cdot \sum_{j=1}^{n}\prod_{i=1}^{m}a_{ij}\ .$
[b]Particular case.[/b] If $a_{i}>0$ for any $i\in\overline{1,n}$, then $\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{m}\le n^{m-1}\cdot \left(a^{m}_{1}+a^{m}_{2}+\ldots+a^{m}_{n}\right)$ $\Longleftrightarrow$
  $\left(\frac{a_{1}+a_{2}+\ldots+a_{n}}{n}\right)^{m}\le\frac{a^{m}_{1}+a^{m}_{2}+\ldots+a^{m}_{n}}{n}$, i.e. $A^{m}(a_{1},a_{2},\ldots ,a_{n})\le A(a^{m}_{1},a^{m}_{2},\ldots ,a^{m}_{n})$, where $A$ is the arithmetical mean.

[b]Remark.[/b] See the Chebyshev's inequality at http://www.mathlinks.ro/Forum/viewtopic.php?t=7942 i.e.
$a_{1}\le a_{2}\le \ldots\le a_{n}\ \ \wedge\ \ b_{1}\le b_{2}\le\ldots \le b_{n}$ $\Longrightarrow A(a_{1},a_{2},\ldots ,a_{n})\cdot A(b_{1},b_{2},\ldots ,b_{n})\le A(a_{1}b_{1},a_{2}b_{2},\ldots ,a_{n}b_{n})\ .$[/color]

[quote="arkhammedos"]If $a,b$ are non-negative reals, such that $a+b=2$, prove that $a^{4}+b^{4}\ge 2$[/quote]

According to rem's solution $a,b\in\Re$ is ok too.

Another solution[but a laborious one] would be to consider \[a=1+x\]  \[b=1-x\] And after that it just depends on x


[color=darkblue]Let $w$ be the circumcircle of the triangle $ABC$ for which $b<a<c\ .$

Prove that exists (unique) a point $M\in [BC]$ so that $\frac{1}{PM}=\frac{1}{PB}+\frac{1}{PC}\ \ (*)\ ,$ where $\{A,P\}=AM\cap w$ and in this case $\frac{MB}{MC}=\frac{c(b-a)}{b(a-c)}\ .$
[hide="Proof."]
$\{\begin{array}{c}\triangle ABM\sim\triangle CPM\Longrightarrow \frac{MB}{c}=\frac{PM}{PC}\\\\ \triangle ACM\sim\triangle BPM\Longrightarrow \frac{MC}{b}=\frac{PM}{PB}\end{array}\ .$ Therefore, the relation $(*)$ is equivalently with the relation

$\frac{PM}{PB}+\frac{PM}{PC}=1$ $\Longleftrightarrow$ $\{\begin{array}{c}\frac{MB}{c}+\frac{MC}{b}=1\\\\ MB+MC=a\end{array}$ $\Longleftrightarrow$ $\{\begin{array}{c}MB=\frac{c(b-a)}{b-c}\\\\ MC=\frac{b(a-c)}{b-c}\end{array}$ $\Longrightarrow$ $\boxed{\ \frac{MB}{MC}=\frac{c(b-a)}{b(a-c)}\ }\ .$[/hide]

Show that : $\{\begin{array}{cccccccc}1.\blacktriangleright & \widehat{BAM}\equiv\widehat{CAM}& \Longleftrightarrow & b+c=2a & \Longleftrightarrow & PB=PC=2\cdot PM & \Longleftrightarrow & OI\perp AI\ ;\\\\ 2.\blacktriangleright & MB=MC & \Longleftrightarrow & \frac{1}{b}+\frac{1}{c}=\frac{2}{a}& \Longleftrightarrow & OI\perp AM\ ; & & \\\\ 3.\blacktriangleright & AM\perp BC & \Longleftrightarrow & \frac{a}{b+c}+\frac{r}{R}=1 & \Longleftrightarrow & \cos B+\cos C=1 & \Longleftrightarrow & OI\parallel BC\ .\end{array}$

[b]Remark.[/b] If the triangle $ABC$ is equilateral, then the relation $(*)$ is verified for any point $M\in (BC)\ .$ See http://www.mathlinks.ro/Forum/viewtopic.php?t=117564 [/color]

This should be moved to a Geometry section.

It is beyond the scope of high school -- it is akin to a whole test paper.


linear algebra
matrix
vector
linear algebra unsolved

Show that if A is a hermitian matrix, then <Au,u> is real and deduce that the generalized eigenvalue problem: Au = lambda Bu where A and B are hermitian has only real eigenvalues.

1) $ <Au,u>\equal{}<u,Au>\equal{}conj(<Au,u>)$.
2) $ <Au,u>\equal{}<\lambda{Bu},u>\equal{}\bar{\lambda}<Bu,u>$.
a) If $ <Bu,u>\not\equal{}0$ then $ \bar{\lambda}$ is real and $ \lambda$ also.
b) If $ <Bu,u>\equal{}0$ then the result can be false.
example: $ B\equal{}\begin{pmatrix}1&0\\0&\minus{}1\end{pmatrix},u\equal{}[1,1]^T,A\equal{}\begin{pmatrix}0&\minus{}i\\i&0\end{pmatrix}$. Then $ Au\equal{}\minus{}iBu$.

Assuming additionally that $ B>0$, i.e. that $ B$ is positive definite, then the claim is true: Given a generalized eigen vector $ v$ to eigen value $ \lambda$, $ Av\equal{}\lambda Bv$, then \[ \overline{\lambda}\langle Bv,v\rangle\equal{}\langle\lambda Bv,v\rangle\equal{}\langle Av,v\rangle\equal{}\langle v,Av\rangle\equal{}\langle v,\lambda Bv\rangle\equal{}\lambda\langle v,Bv\rangle\equal{} \lambda\langle Bv,v\rangle\,,\]hence $ \lambda\equal{}\overline{\lambda}$ since $ \langle Bv,v\rangle\equal{}\langle v,Bv\rangle>0$.


algebra
function
domain
real analysis
real analysis unsolved

find all fns G(x,y) so that F=(ye^(2x)+ 3x^(2)e^(y))i + G(x,y)j is a gradient fireld.

normally this would be an easy question but how do i figure it out when one of the fucntions in unknown???

any help would be very handy cheers.

if $A(x,y)$ and $B(x,y)$ are defined on a simply connected domain and continuously differentiable there, then $A(x,y)i+B(x,y)j$ is a gradient field on that domain if and only if $\frac{\partial A}{\partial y}=\frac{\partial B}{\partial x}$.

Another approach that does not require smoothness of $G$ is to find whose gradient it may be. This means that you'll have to find all solutions of the equation $\frac{\partial \Phi}{\partial x}=ye^{2x}+ 3x^2e^y$ and then get $G$ as $\frac{\partial \Phi} {\partial y}$.


[quote="ProtestanT"][quote="bpms"][quote="davidlizeng"]A number ABCD is called an "interesting number" if it fulfills one or more of the following

1. A, B, C, and D are all even
2. A, B, C, and D are all odd
3. A<B<C<D
4. A>B>C>D

how many "interesting numbers" are there supposing A is not 0?[/quote]
Ooooh this looks like a load of casework plus PIE
[hide="1"]
$4*5*5*5=500$[/hide]
[hide="2"]
$5*5*5*5=625$[/hide]
[hide="3"]
$\binom{9}{4}=126$[/hide]
[hide="4"]
$\frac{9*9*8*7}{4!}=189$[/hide]
How to account for the overlap, is the question...[/quote]
why is number 3 $\binom{9}{4}$?[/quote]
[hide]Clearly the numbers are all different. For every set of four numbers there is exactly one arrangement that works. Since 0=no go, $\binom{9}{4}$[/hide]

[hide="hint"]Consider which properties can overlap[/hide]

[hide="solution"]clearly, 1 and 2 can't overlap. neither can 3 and 4. Not caring about overlap, there are 1440 #s. 1 and 3 overlap with only 1 #(2468), 1 and 4 5 ways(6420, 8642, 8640, 8620, and 8420), and the 1-4 and 2-4 combos having 5, in the same way. Thus, our final # is 1424.[/hide]

[quote="davidlizeng"]1424 is incorrect....
i totally do not get this problem.....
the answer is 1445, which is greater than 1440, so bpms and perfect628 must have both undercounted.[/quote]

I believe this is the reason:

[quote="bpms"][/quote][quote="ProtestanT"][quote="bpms"][quote="davidlizeng"]A number ABCD is called an "interesting number" if it fulfills one or more of the following

1. A, B, C, and D are all even
2. A, B, C, and D are all odd
3. A<B<C<D
4. A>B>C>D

how many "interesting numbers" are there supposing A is not 0?[/quote]
Ooooh this looks like a load of casework plus PIE
[hide="1"]
$4*5*5*5=500$[/hide]
[hide="2"]
$5*5*5*5=625$[/hide]
[hide="3"]
$\binom{9}{4}=126$[/hide]
[hide="4"]
$\frac{9*9*8*7}{4!}=189$[/hide]
How to account for the overlap, is the question...[/quote]
why is number 3 $\binom{9}{4}$?[/quote][quote="bpms"]
[hide]Clearly the numbers are all different. [b]For every set of four numbers there is exactly one arrangement that works.[/b] Since 0=no go, $\binom{9}{4}$[/hide][/quote]

(Click on the last "Click to reveal hidden content")
What about the set $\{1,2,3,3\}$?

For the fourth case have you considered 10, choose 4?

bpms broke the problem into four cases.  The first three are correct, the fourth should be combinations of 4 selected from a set of 10 which is 210.  Combine that with the work of perfect628 in counting overlaps and you’ve got it.

how would you do complimentary counting on a problem like this??
There are too many things to take into consideration. i don't think that's a great idea.


search
algorithm
number theory
algebra solved
algebra

1. x and y are two natural numbers such that 3*x*x + x = 4*y*y +y . Prove that x - y is the square of a whole nnumber.*=multyping symbol
  Rate the problem's difficulty from 0 to 10 Personally i put 5.

I would say 3.

By the way, it has been posted at least 1000 times before.

A few days ago a more general problem about the equation $ax^2 + x = by^2 + y$ appeared on the forum...

Please search for the topics I am talking about, shouldn't be hard. (Use "Iran" and "square" as keywords.)

Arne is true . I can  tell that you can find it in unsolved problem of number theory As square...
but I give you a hint: [b]If $ab=k^2$ and $(a,b)=1$ then $a,b$ both are perfect square[/b]
and I think it is obvious ;)

You can see it at
http://www.mathlinks.ro/Forum/topic-30699.html

because x, y are both natural # so x-y is also a natural #
then we dont have to consider the case x-y=c with c<0
set x-y=c with c > 0
rewrite with x = c+y
3(c+y)^2 + (c+y) = 4y^2 + y
simplefi: y^2 -6cy - (3c^2 + c) = 0
then we have  y = 3c + squ(c(12c + 1))
as given y is a natural # so squ(c(12c + 1)) must be a integer.
therefore c(12c +1) is a square.

now consider about 12c+1 = 4(3c)+1
by one of the theorem related to division algorithm says:
"Every square interger is of the form 4k or 4k+1, where k is an integer." the proof of this one is easy.
so now we know 12c+1 is a square. rewrit 12c+1 = k^2

c*k^2 is a square ==> c=x-y  is a square


White makes a random move on a chess board, followed by a random move by black and another random move by white. What is the probability of the piece(s) white moved still being on his side of the board (rows [i]1[/i] through [i]4[/i])?

I assume an equal probability selection of all possible legal moves. Any way you interpret it though, it sounds like a bunch of incredibly disgusting case analysis, especially once you consider bishop moves and Black being able to move a pawn to prevent one legal move (i.e. 1. e4 b5 and 2. Ba6 is not possible). This throws enormous complications into the calculations.

Is this problem supposed to have any elegance to it?

[quote="MellowMelon"]Is this problem supposed to have any elegance to it?[/quote]
Probably not, because I just made this problem up.

@ JBL - You randomly choose a legal move.

[quote]Is this problem supposed to have any elegance to it?[/quote]
If anyone [i]did[/i] find an elegant solution to this, I'd be very impressed. And a bit scared.

I'd say that the best method on this problem would be to Monte Carlo it using a computer program.  The casework would be disgusting, whereas writing a program would be fairly simple.

I don't even think you need Monte Carlo... just simulate all possible moves with a program. It's not that many for a computer.

[quote="i_like_pie"]@ JBL - You randomly choose a legal move.[/quote]There are lots of ways of making a random choice.  In particular, it's quite natural in the context of chess to consider choosing a piece at random and then choosing a move for that piece at random.  This is "random," but it's uniform across pieces instead of across moves.

paladin8 is certainly correct about the best possible approach to this problem.

Oh, I see what you mean. I meant uniform across moves. Sorry, I didn't understand what you meant before. :oops:

And I also agree with paladin's approach being the best. :)

[hide="Possible Crosses"]All possible ways for White to get to a piece across the Black's side (i.e. rows 5 through 8):

Move any Pawn twice (two spaces on the first move (8 ways) (5% chance of being blocked)

Move any of Pawns B through G twice (two spaces on first move and capture opposing Pawn (also moved two spaces and in either capture zone)) (12 ways) (90% chance of not have a Pawn to capture)

Move either Pawn A or Pawn H twice (two spaces on first move and capture opposing Pawn (also moved two spaces and in the capture zone)) (2 ways) (95% chance of not having a Pawn to capture)

Move either Knight twice (inward on the board first move (i.e. King's Knight moving in front of King's Bishop, or Queen's Knight moving in front of Queen's Bishop) then either forward move) (4 ways) (not stoppable)

Move either Knight twice (outward on the board first move (i.e. King's Knight moving in front of King's Rook, or Queen's Knight moving in front of Queen's Rook) the moving the only move) (2 ways) (not stoppable)

All ways from this point forward involve moving a Pawn forward (any amount) and [i]then moving another piece[/i] to the other side of the board. All ways are (have already been) multiplied by 2 because it does not matter which of the Pawn's 2 moves it makes. The "Total # of ways" discount the Pawn's own ways of getting across.

Pawn A,B,C,F,G, or H forward (Total # of ways: 0):

There are no pieces that can use this to get across.

Pawn D forward (Total # of ways: 4):

Queen's Bishop may move to G5. (2 ways) (not stoppable)

Queen's Bishop may move to H6. (2 ways) (5% chance of being stopped by Black's Pawn on G5)

Pawn E forward (Total # of ways: 6):

The Queen may move to H5. (2 ways) (not stoppable)

King's Bishop may move to B5. (2 ways) (not stoppable)

King's Bishop may move to A6. (2 ways) (5% chance of being stopped by Black's Pawn on B5)[/hide]

And there you have it, all the ways to get a piece across. It's hidden because it's long.

I was going to put everything, but BBCode doesn't seem to like have 3 "hides." So here's a start.

Now I shall list all possible moves for White (assuming non-blocking by Black).

[hide="Possible Moves"]Pawn A forward one square allows 19 different, subsequent moves, none of which can cross. $ \boxed{\frac{0}{19}}$

Pawn B forward one square allows 21 different, subsequent moves, none of which can cross. $ \boxed{\frac{0}{19}}$

Pawn C forward one square allows 21 different, subsequent moves, none of which can cross. $ \boxed{\frac{0}{21}}$

Pawn D forward one square allows 27 different, subsequent moves, 2 of which can cross. $ \boxed{\frac{2}{27}}$ Assuming that Black's move was not to put a Pawn of G5, in which case 26 different, subsequent moves, 1 of which can cross. $ \boxed{\frac{1}{26}}$

Pawn E forward one square allows 30 different subsequent moves, 3 of which can cross. $ \boxed{\frac{3}{30}}$ Assuming that Black's move was not to put a Pawn of B5, in which case, 29 different, subsequent moves, 2 of which can cross. $ \boxed{\frac{2}{29}}$

Pawn F forward one square allows 20 different, subsequent moves, none of which can cross. $ \boxed{\frac{0}{20}}$

Pawn G forward one square allows 21 different, subsequent moves, none of which can cross. $ \boxed{\frac{0}{21}}$

Pawn H forward on square allows 19 different, subsequent moves, none of which can cross. $ \boxed{\frac{0}{19}}$

Queen's Knight forward and to the left (i.e. to A3) allows 18 different, subsequent moves, 1 of which can cross. $ \boxed{\frac{1}{18}}$

Queen's Knight forward and to the right (i.e. to C3) allows 20 different, subsequent moves, 2 of which can cross. $ \boxed{\frac{2}{20}}$

King's Knight forward and to the left (i.e. to F3) allows 20 different, subsequent moves, 2 of which can cross. $ \boxed{\frac{2}{20}}$

King's Knight forward and to the right (i.e. to H3) allows 18 different, subsequent moves, 1 of which can cross. $ \boxed{\frac{1}{18}}$

Pawn A forward two spaces allows 21 different, subsequent moves, 1 of which can cross. $ \boxed{\frac{1}{21}}$ Assuming that Black's move was not to put a Pawn on A5, in which case, 20 different, subsequent moves, none of which can cross. $ \boxed{\frac{0}{20}}$ Assuming that Black's move was not to put a Pawn on B5, in which case, 22 different, subsequent moves, 2 of which can cross. $ \boxed{\frac{2}{22}}$

Pawn B forward two spaces allows 21 different, subsequent moves, 1 of which can cross. $ \boxed{\frac{1}{21}}$ Assuming that Black's move was not to put a Pawn on B5, in which case, 20 different, subsequent moves, none of which can cross. $ \boxed{\frac{0}{20}}$ Assuming that Black's move was not to put a Pawn on either A5 or C5, in which case, 22 different, subsequent moves, 2 of which can cross. $ \boxed{\frac{2}{22}}$

Pawn C forward two spaces allows 20 different, subsequent moves, 1 of which can cross. $ \boxed{\frac{1}{20}}$ Assuming that Black's move was not to put a Pawn on C5, in which case, 19 different, subsequent moves, none of which can cross. $ \boxed{\frac{0}{19}}$ Assuming that Black's move was not put a Pawn on either B5 or D5, in which case, 21 different, subsequent moves, 2 of which can cross. $ \boxed{\frac{2}{21}}$

Pawn D forward two spaces allows 28 different, subsequent moves, 3 of which can cross. $ \boxed{\frac{3}{28}}$ Assuming that Black's move was not to put a Pawn on D5, in which case, 27 different, subsequent moves, 2 of which can cross. $ \boxed{\frac{2}{27}}$ Assuming that Black's move was not put a Pawn on either C5 or E5, in which case, 29 different, subsequent moves, 4 of which can cross. $ \boxed{\frac{4}{29}}$ Assuming that Black's move was not to put a Pawn on G5, in which case, 27 different, subsequent moves, 2 of which can cross. $ \boxed{\frac{2}{27}}$

Pawn E forward two spaces allows 31 different, subsequent moves, 4 of which can cross. $ \boxed{\frac{4}{31}}$ Assuming that Black's move was not to put a Pawn on E5, in which case, 30 different, subsequent moves, 3 of which can cross. $ \boxed{\frac{3}{30}}$ Assuming that Black's move was not put a Pawn on either D5 or F5, in which case, 32 different, subsequent moves, 5 of which can cross. $ \boxed{\frac{5}{32}}$ Assuming that Black's move was not to put a Pawn on B5, in which case, 30 different, subsequent moves, 3 of which can cross. $ \boxed{\frac{3}{30}}$

Pawn F forward two spaces allows 20 different, subsequent moves, 1 of which can cross. $ \boxed{\frac{1}{20}}$ Assuming that Black's move was not to put a Pawn on F5, in which case, 19 different, subsequent moves, none of which can cross. $ \boxed{\frac{0}{19}}$ Assuming that Black's move was not put a Pawn on either E5 or G5, in which case, 21 different, subsequent moves, 2 of which can cross. $ \boxed{\frac{2}{21}}$

Pawn G forward two spaces allows 21, different, subsequent moves, 1 of which can cross. $ \boxed{\frac{1}{21}}$ Assuming that Black's move was not to put a Pawn on G5, in which case, 20 different, subsequent moves, none of which can cross. $ \boxed{\frac{0}{20}}$ Assuming that Black's move was not put a Pawn on either F5 or H5, in which case, 22 different, subsequent moves, 2 of which can cross. $ \boxed{\frac{2}{22}}$

Pawn H forward two spaces allows 21, different, subsequent moves, 1 of which can cross. $ \boxed{\frac{1}{21}}$ Assuming that Black's move was not to put a Pawn on H5, in which case, 20 different, subsequent moves, none of which can cross. $ \boxed{\frac{0}{20}}$ Assuming that Black's move was not put a Pawn on G5, in which case, 22 different, subsequent moves, 2 of which can cross. $ \boxed{\frac{2}{22}}$

[/hide]There are now eight cases of probability which must be tested in order to determine the total probability.

Almost done, just two more posts, the second one's the answer :D .

P.S. Stupid "Consecutive posts not allowed for less than 10 minutes." Bah!

There are now eight cases of probability which must be tested in order to determine the total probability.

[hide="Black Annoyance"]
If Black did not put any Pawns on Row 5 (60% chance):
$ \frac{0\plus{}0\plus{}0\plus{}2\plus{}3\plus{}0\plus{}0\plus{}0\plus{}1\plus{}2\plus{}2\plus{}1\plus{}1\plus{}1\plus{}1\plus{}3\plus{}4\plus{}1\plus{}1\plus{}1}{19\plus{}19\plus{}21\plus{}27\plus{}30\plus{}20\plus{}21\plus{}19\plus{}18\plus{}20\plus{}20\plus{}18\plus{}21\plus{}21\plus{}20\plus{}28\plus{}31\plus{}20\plus{}21\plus{}21}\rightarrow\frac{6\cdot0\plus{}8\cdot1\plus{}3\cdot2\plus{}2\cdot3\plus{}4}{2\cdot18\plus{}3\cdot19\plus{}5\cdot20\plus{}6\cdot21\plus{}27\plus{}28\plus{}30\plus{}31}\rightarrow\frac{8\plus{}6\plus{}6\plus{}4}{36\plus{}57\plus{}100\plus{}126\plus{}27\plus{}28\plus{}30\plus{}31}\rightarrow\frac{24}{435}\rightarrow\frac{8}{145}\rightarrow .05517$
There is a 5.5% of White crossing if Black does not put a Pawn on Row 5.

If Black put a Pawn on either A5 or H5 (10% chance):
$ \frac{0\plus{}0\plus{}0\plus{}2\plus{}3\plus{}0\plus{}0\plus{}0\plus{}1\plus{}2\plus{}2\plus{}1\plus{}0\plus{}2\plus{}1\plus{}3\plus{}4\plus{}1\plus{}1\plus{}1}{19\plus{}19\plus{}21\plus{}27\plus{}30\plus{}20\plus{}21\plus{}19\plus{}18\plus{}20\plus{}20\plus{}18\plus{}20\plus{}22\plus{}20\plus{}28\plus{}31\plus{}20\plus{}21\plus{}21}\rightarrow\frac{7\cdot0\plus{}6\cdot1\plus{}4\cdot2\plus{}2\cdot3\plus{}4}{2\cdot18\plus{}3\cdot19\plus{}6\cdot20\plus{}4\cdot21\plus{}22\plus{}27\plus{}28\plus{}30\plus{}31}\rightarrow\frac{6\plus{}8\plus{}6\plus{}4}{36\plus{}57\plus{}120\plus{}84\plus{}22\plus{}27\plus{}28\plus{}30\plus{}31}\rightarrow\frac{24}{435}\rightarrow\frac{8}{145}\rightarrow .05517$
The same as Case 1, interesting, but the other cases will not be so similar.

If Black put a Pawn on B5 (5% chance):
$ \frac{0\plus{}0\plus{}0\plus{}2\plus{}2\plus{}0\plus{}0\plus{}0\plus{}1\plus{}2\plus{}2\plus{}1\plus{}2\plus{}0\plus{}2\plus{}3\plus{}3\plus{}1\plus{}1\plus{}1}{19\plus{}19\plus{}21\plus{}27\plus{}29\plus{}20\plus{}21\plus{}19\plus{}18\plus{}20\plus{}20\plus{}18\plus{}22\plus{}20\plus{}21\plus{}28\plus{}30\plus{}20\plus{}21\plus{}21}\rightarrow\frac{7\cdot0\plus{}5\cdot1\plus{}6\cdot2\plus{}2\cdot3}{2\cdot18\plus{}3\cdot19\plus{}5\cdot20\plus{}5\cdot21\plus{}22\plus{}27\plus{}28\plus{}29\plus{}30}\rightarrow\frac{5\plus{}12\plus{}6}{36\plus{}57\plus{}100\plus{}105\plus{}22\plus{}27\plus{}28\plus{}29\plus{}30}\rightarrow\frac{23}{434}\rightarrow .053$
A little different, a 5.3% chance of White crossing.

If Black put a Pawn on G5 (5% chance):
$ \frac{0\plus{}0\plus{}0\plus{}1\plus{}3\plus{}0\plus{}0\plus{}0\plus{}1\plus{}2\plus{}2\plus{}1\plus{}1\plus{}1\plus{}1\plus{}2\plus{}4\plus{}2\plus{}0\plus{}2}{19\plus{}19\plus{}21\plus{}26\plus{}30\plus{}20\plus{}21\plus{}19\plus{}18\plus{}20\plus{}20\plus{}18\plus{}21\plus{}21\plus{}20\plus{}27\plus{}31\plus{}21\plus{}20\plus{}22}\rightarrow\frac{7\cdot0\plus{}6\cdot1\plus{}5\cdot2\plus{}3\plus{}4}{2\cdot18\plus{}3\cdot19\plus{}5\cdot20\plus{}5\cdot21\plus{}22\plus{}26\plus{}27\plus{}30\plus{}31}\rightarrow\frac{6\plus{}10\plus{}3\plus{}4}{36\plus{}57\plus{}100\plus{}105\plus{}22\plus{}26\plus{}27\plus{}30\plus{}31}\rightarrow\frac{23}{434}\rightarrow .053$
Different equation but the same as Case 3.

If Black put a Pawn on C5 (5% chance):
$ \frac{0\plus{}0\plus{}0\plus{}2\plus{}3\plus{}0\plus{}0\plus{}0\plus{}1\plus{}2\plus{}2\plus{}1\plus{}1\plus{}2\plus{}0\plus{}4\plus{}4\plus{}1\plus{}1\plus{}1}{19\plus{}19\plus{}21\plus{}27\plus{}30\plus{}20\plus{}21\plus{}19\plus{}18\plus{}20\plus{}20\plus{}18\plus{}21\plus{}22\plus{}19\plus{}29\plus{}31\plus{}20\plus{}21\plus{}21}\rightarrow\frac{7\cdot0\plus{}6\cdot1\plus{}4\cdot2\plus{}3\plus{}2\cdot4}{2\cdot18\plus{}4\cdot19\plus{}4\cdot20\plus{}5\cdot21\plus{}22\plus{}27\plus{}29\plus{}30\plus{}31}\rightarrow\frac{6\plus{}8\plus{}3\plus{}8}{36\plus{}76\plus{}80\plus{}105\plus{}22\plus{}27\plus{}29\plus{}30\plus{}31}\rightarrow\frac{25}{436}\rightarrow .05734$
A 5.7% chance of White crossing.

If Black put a Pawn on E5 (5% chance):
$ \frac{0\plus{}0\plus{}0\plus{}2\plus{}3\plus{}0\plus{}0\plus{}0\plus{}1\plus{}2\plus{}2\plus{}1\plus{}1\plus{}1\plus{}1\plus{}4\plus{}3\plus{}2\plus{}1\plus{}1}{19\plus{}19\plus{}21\plus{}27\plus{}30\plus{}20\plus{}21\plus{}19\plus{}18\plus{}20\plus{}20\plus{}18\plus{}21\plus{}21\plus{}20\plus{}29\plus{}30\plus{}21\plus{}21\plus{}21}\rightarrow\frac{6\cdot0\plus{}7\cdot1\plus{}4\cdot2\plus{}2\cdot3\plus{}4}{2\cdot18\plus{}3\cdot19\plus{}4\cdot20\plus{}7\cdot21\plus{}27\plus{}29\plus{}2\cdot30}\rightarrow\frac{7\plus{}8\plus{}6\plus{}4}{36\plus{}57\plus{}80\plus{}147\plus{}27\plus{}29\plus{}60}\rightarrow\frac{25}{436}\rightarrow .05734$
How odd that this keeps on being the same, oh well, easier for me, :lol: .

If Black put a Pawn on F5 (5% chance):
$ \frac{0\plus{}0\plus{}0\plus{}2\plus{}3\plus{}0\plus{}0\plus{}0\plus{}1\plus{}2\plus{}2\plus{}1\plus{}1\plus{}1\plus{}1\plus{}3\plus{}5\plus{}0\plus{}2\plus{}1}{19\plus{}19\plus{}21\plus{}27\plus{}30\plus{}20\plus{}21\plus{}19\plus{}18\plus{}20\plus{}20\plus{}18\plus{}21\plus{}21\plus{}20\plus{}28\plus{}32\plus{}19\plus{}22\plus{}21}\rightarrow\frac{7\cdot0\plus{}6\cdot1\plus{}4\cdot2\plus{}2\cdot3\plus{}5}{2\cdot18\plus{}4\cdot19\plus{}4\cdot20\plus{}5\cdot21\plus{}22\plus{}27\plus{}28\plus{}30\plus{}32}\rightarrow\frac{6\plus{}8\plus{}6\plus{}5}{36\plus{}76\plus{}80\plus{}105\plus{}22\plus{}27\plus{}28\plus{}30\plus{}32}\rightarrow\frac{25}{436}\rightarrow .05734$
Same again, :D .

If Black put a Pawn on D5 (5% chance):
$ \frac{0\plus{}0\plus{}0\plus{}2\plus{}3\plus{}0\plus{}0\plus{}0\plus{}1\plus{}2\plus{}2\plus{}1\plus{}1\plus{}1\plus{}2\plus{}2\plus{}5\plus{}1\plus{}1\plus{}1}{19\plus{}19\plus{}21\plus{}27\plus{}30\plus{}20\plus{}21\plus{}19\plus{}18\plus{}20\plus{}20\plus{}18\plus{}21\plus{}21\plus{}21\plus{}27\plus{}32\plus{}20\plus{}21\plus{}21}\rightarrow\frac{6\cdot0\plus{}7\cdot1\plus{}5\cdot2\plus{}3\plus{}5}{2\cdot18\plus{}3\cdot19\plus{}4\cdot20\plus{}7\cdot21\plus{}2\cdot27\plus{}30\plus{}32}\rightarrow\frac{7\plus{}10\plus{}3\plus{}5}{36\plus{}57\plus{}80\plus{}147\plus{}54\plus{}30\plus{}32}\rightarrow\frac{25}{436}\rightarrow .05734$
That is it, thank goodness.[/hide]

Almost there...

My BBCode doesn't seem to be working... Could someone tell me why, it's exactly the same as my other posts  :( .

EDIT: Fixed! Thanks AstroPhys.

You closing hide tag has the slash the wrong way and it does not allow an ' to be in the hide text.

So, there is a 70% chance of having a 5.5% chance, which is $ .03862$ or 3.9%.
There is a 20% chance of having a 5.7% chance, which is $ .01147$ or 1.1%.
There is a 10% chance of having a 5.3% chance, which is $ .0053$ or 0.5%.

In total this is $ .055388118734$ which is a 5.54% chance or $ \boxed{553,881,187,339,359\text{ in }10^{16}}$ chance.

Yes, it's done, it only took three days. I feel accomplished. And was all done without a comp. (except the dividing of fractions :D )

I would appreciate feedback on my solve.


PQR is a right angled triangle with PR = 3cm and QR = 4cm. The square STUV is inscribed within triangle PQR. What is the length in centimetres, of the side of the square?

[hide="Hint"]
PST, UVQ, TRU, and the whole triangle are all similar.[/hide]
[hide="Solution"]
So let SV=ST=UV=x, by similarity, PS=3x/4 and VQ=4x/3.
But PS+SV+VQ=5.
So x(1+3/4+4/3)=5, x=60/37.
You could have also used that PT+TR=PR, or RU+UQ=RQ.[/hide]

there are 2 ways to inscribe the square, which one is it

Dgreenb801 talked about 1 way, but I think the poster wanted the other way since none of the coordinates on the triangle match the coordinates on the square.

yes that is correct dgreenb801. A young friend of mine was doing this question and asked me to help him out, i wanted to see whether it was as easy for everyone as i thought.

Well it wasn't really easy for me, because I just remember the exact same problem from Mathcounts, with different numbers  :P . Was the poster a Mathcounts poster?


calculus
integration
LaTeX
function
logarithms
geometric series
calculus computations

[$\LaTeX$ tip: with "named" functions, use a \ before the name. So lnx becomes $lnx$ but \ln x (the space matters) becomes $\ln x,$ which looks better. The same with cos, sin, etc.]

Note that this integral is proper at 1, albeit nonobviously, and its size is such that it converges at both the 0 and $\infty$ ends. Also, the function being integrated is always negative, so the integral will be negative.

First step: the change of variables $y=\frac1x$ turns $\int_1^{\infty}\frac{\ln x}{1-x^2}\,dx$ into $\int_0^1\frac{\ln y}{1-y^2}\,dy.$ Thus we rewrite the original problem as

$2\int_0^1\frac{\ln x}{1-x^2}\,dx.$

Second step: we use the geometric series $\frac1{1-x^2}=\sum_{k=0}^{\infty}x^{2k}$ and the computation $\int_0^1x^n\ln x\,dx=-\frac1{(n+1)^2}.$

$2\int_0^1\frac{\ln x}{1-x^2}\,dx=2\int_0^1\sum_{k=0}^{\infty}x^{2k}\ln x\,dx$

$=2\sum_{k=0}^{\infty}\int_0^1x^{2k}\ln x\,dx=-2\sum_{k=0}^{\infty}\frac1{(2k+1)^2}.$

Finally we note that we have seen that sum before. By other means (closely related to calculations of $\zeta(2)$) we know that

$\sum_{k=0}^{\infty}\frac1{(2k+1)^2}=\frac{\pi^2}8.$

Hence the answer is $-\frac{\pi^2}4.$


geometry
circumcircle
geometric transformation
reflection
trigonometry
similar triangles
angle bisector

cam someone please define external symmedians  and  why they  turn out to be the tangents  at the vertices  of the triangle?

The point is the following result:

[color=blue][b]Theorem 1.[/b] Let the tangents to the circumcircle of a triangle ABC at the vertices B and C intersect each other at a point X. Then, the line AX is the A-symmedian of triangle ABC.[/color]

[i]Proof of Theorem 1.[/i] Here is one of the possible proofs:

(See Fig. 1.) We use directed angles modulo 180°. Since the line BX is the tangent to the circumcircle of triangle ABC at the point B, the tangent-chordal angle theorem yields < (BX; BC) = < BAC. In other words, < XBC = < BAC.

(See Fig. 2.) Let the perpendicular to the line CA at the point C meet the line AB at U, and let the perpendicular to the line AB at the point B meet the line CA at V. Then, since < UBV = 90° and < UCV = 90°, the points B and C lie on the circle with diameter UV. The center of this circle is the midpoint of the segment UV; denote this midpoint by X'. Then, by the central angle theorem, < X'BC = 90° - < CUB. Since $CU\perp CA$, we have < (CU; CA) = 90°, so that < X'BC = 90° - < CUB = < (CU; CA) - < (CU; AB) = < (AB; CA) = < BAC = < XBC. Thus, the point X' lies on the line BX. Similarly, the point X' lies on the line CX. But the lines BX and CX have only one point in common, namely the point X; thus, the point X' coincides with the point X. As we know that the point X' is the midpoint of the segment UV, we thus conclude that the point X is the midpoint of the segment UV.

(See Fig. 3.) Now, since the points B and C lie on the circle with diameter UV, we have < BUV = < BCV, what rewrites as < AUV = - < ACB. Similarly, < AVU = - < ABC. Thus, the triangles ABC and AVU are inversely similar. Now, if M is the midpoint of the segment BC, then the points M and X are corresponding points in these inversely similar triangles ABC and AVU (in fact, they are the midpoints of the corresponding sides BC and VU of these triangles). Since corresponding points in inversely similar triangles form oppositely equal angles, we thus conclude that < BAM = - < VAX, what rewrites as $\measuredangle\left(AB;\;AM\right)=-\measuredangle\left(CA;\;AX\right)$.

Now, if w is the angle bisector of the angle CAB, then $\measuredangle\left(AB;\;w\right)=-\measuredangle\left(CA;\;w\right)$. Thus,

$\measuredangle\left(w;\;AM\right)=\measuredangle\left(AB;\;AM\right)-\measuredangle\left(AB;\;w\right)=\left(-\measuredangle\left(CA;\;AX\right)\right)-\left(-\measuredangle\left(CA;\;w\right)\right)$
$=\measuredangle\left(CA;\;w\right)-\measuredangle\left(CA;\;AX\right)=\measuredangle\left(AX;\;w\right)$.

Thus, the line AX is the reflection of the line AM in the line w. Since the line AM is the A-median of triangle ABC (as the point M is the midpoint of the side BC), and the line w is the angle bisector of the angle CAB, we thus see that the line AX is the reflection of the A-median of triangle ABC in the angle bisector of the angle CAB; in other words, the line AX is the A-symmedian of triangle ABC. Theorem 1 is proven.

We have denoted by X the point where the tangents to the circumcircle of triangle ABC at the points B and C intersect, and shown that the line AX is the A-symmedian of triangle ABC. Similarly, if we denote by Y the point where the tangents to the circumcircle of triangle ABC at the points C and A intersect, and by Z the point where the tangents to the circumcircle of triangle ABC at the points A and B intersect, then the line BY is the B-symmedian and the line CZ is the C-symmedian of triangle ABC. This is why the lines YZ, ZX, XY (i. e. the tangents to the circumcircle of triangle ABC at the vertices A, B, C) are sometimes called external symmedians or exsymmedians of triangle ABC (though I find this naming misleading; there is no such relationship between usual and external symmedians like the one between internal and external angle bisectors!).

By the way, a side corollary of Theorem 1:

[color=blue][b]Theorem 2.[/b] Let the tangents to the circumcircle of a triangle ABC at the vertices B and C intersect each other at a point X, and let M be the midpoint of the side BC of triangle ABC. Then, $AM=AX\cdot\left|\cos A\right|$ (we don't use directed angles here).[/color]

[i]Proof of Theorem 2.[/i] In the proof of Theorem 1, we showed that the points M and X are corresponding points in the similar triangles ABC and AVU. Corresponding points in similar triangles form similar triangles themselves; thus, the triangles ABM and AVX are similar, so that $\frac{AM}{AX}=\frac{AB}{AV}$. But in the right-angled triangle ABV, we have $\frac{AB}{AV}=\left|\cos\measuredangle BAV\right|=\left|\cos A\right|$, so that $\frac{AM}{AX}=\left|\cos A\right|$. Thus, $AM=AX\cdot\left|\cos A\right|$. This proves Theorem 2.

  Darij

Here's another possible proof of the above [color=blue][b]Theorem 1[/b][/color]:

We work, as Darij, with the same directed angles modulo $180^\circ$.

Let the parallel through $X$ to $BC$ meet $AB,AC$ in $E,F$. Then, $\measuredangle CXB = 180^\circ-2 \measuredangle BAC$.

The triangle $BXC$ is clearly isosceles, so $\measuredangle XBC = \measuredangle BCX = \measuredangle BAC$.

By the fact that $BC \| EF$ we get $\measuredangle BXE = \measuredangle FXC = \measuredangle BAC$, $\measuredangle XEB = \measuredangle XCF = \measuredangle CBA$ and $\measuredangle EBX = \measuredangle CFX = \measuredangle ACB$.

Thus, $\triangle XEB \sim \triangle XCF \sim \triangle ABC \sim \triangle AEF$.

Hence, $\frac{BX}{EX}= \frac{AF}{AE}\, \, \Longleftrightarrow BX = \frac{EX \cdot AF}{AE}$ and $\frac{CX}{AE}= \frac{XF}{AF}\, \, \Longleftrightarrow \, \, CX = \frac{XF \cdot AE}{AF}$.

But $BX = CX$, so $\frac{EX}{XF}= \frac{AE^{2}}{AF^{2}}.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=theorem+steiner&t=58872]By Theorem 5 (actually by its converse)[/url], we see that $AX$ is the $A$-symmedian of $\triangle AEF$. This is clearly the same as the $A$-symmedian of $\triangle ABC$, so we're done.

[quote="darij grinberg"][color=blue][b]Theorem 1.[/b] Let the tangents to the circumcircle of a triangle ABC at the vertices B and C intersect each other at a point X. Then, the line AX is the A-symmedian of triangle ABC.[/color][/quote]

Here is one more proof of this theorem 1 (with projective geometry) :
Let ABC be a triangle, X the intersection of the tangents at B and at C, and Y the second intersection of AX with the circumcircle of the triangle. Since BC is the polar of X wrt the circle, BX, BC, BY, BA are a harmonic line batch, that is B, C, Y, A are a harmonic point series, and thus AB, AC, AY and the tangent at A are a harmonic line batch. Let M be the midpoint of BC. AB, AC, AM and the parallel to BC through A are a harmonic line batch. Thus, since AB, AC and the tangent at A are the images of AC, AB and the parallel to BC through A respectively, by reflection in the angle bissector of <BAC, AY must also be the image of AM by reflection in the angle bissector of <BAC.

Benjamin


How do you get parentheses to get bigger so it covers a fraction? Can you give me an example? I am just starting to use Latex and I am lost.

[quote="JBL"]I think \big makes characters bigger, and \left( and \right) can also work ... something like that.  Try a search for "\big" on the forum and the right thing should pop out.[/quote]
\left( and \right) are much better in this case.

[tex]z=\left(\left(x^2+3(y+7)\right)^4\right)[/tex]


modular arithmetic
number theory proposed
number theory

Let N be the set of all n's wich have the next property:

          $2^{n}$, ends in $n$.

Prove or disprove that:

Then N has an infinite number of elements.   :maybe:

[quote="N.T.TUAN"]True! An Example $n=2^{k}(k=1,2,...)$ :D[/quote]

jiji thanks, but i made a mistake in writing the problem. Now you can see  the correct problem.

Thanks anyway. :blush:  :blush:

[hide="a start..."]
First, $2^{n}$ and $n$ have the same units digit, so that
\[2^{n}\equiv n\pmod{10}\]
Clearly $n$ is even, so let $n=2m$. Now we have
\[2\cdot (-1)^{m-1}\equiv m\pmod{5}\]
Then $m\equiv 7,8\pmod{10}$ so $n\equiv 14,16\pmod{20}$. Since this rules out single digit $n$, $2^{n}$ and $n$ must now be congruent mod 100, so $n$ is a multiple of 4. This leaves the case $n\equiv 16\pmod{20}$. Let $n=4k$, $k\equiv 4\pmod{5}$, so that
\[4\cdot 16^{k-1}\equiv k\pmod{25}\]
But $k\equiv 4\pmod{5}$ implies $16^{k-1}\equiv 16^{3}\equiv-4\pmod{25}$, so $k\equiv 9\pmod{25}$ and $n\equiv 36\pmod{100}$.

So we see 36 belong to N, but for more digits, the computation becomes much more intense quickly, and I don't know whether it will end...
[/hide]

[quote="Rust"]If $2^{n}=n(mod \ 10^{k}, n<10^{k}$, then there are unique solution $2^{m}=m(mod \ 10^{k+1})$ and $m=2^{n}(mod \ 10^{k+1}$.
Therefore from $2^{36}=36(mod \ 100)$ we get $2^{36}=736(mod \ 10^{3})$ and $2^{736}=736(mod \ 10^{3})$.
$2^{736}=8736(mod \ 10^{4})$. It fgive $2^{8736}=8736(mod \ 10^{4})$, e.t.c.[/quote]

I had a lot of difficulties to understand Rust's demo.

Here are my explanations of his nice demo.

1) remember that $a^{\phi(n)}=1 (mod \ n)$ for any coprimes n and a. So $2^{\phi(5^{p})}=1 (mod \ 5^{p})$, and $2^{4*5^{p-1}}=1 (mod \ 5^{p})$ and hence $2^{10^{p-1}}=1 (mod \ 5^{p})$ for any $p>2$

2) If I have a solution of $2^{n}=n (mod\ 10^{k})$ with $k>1$ and $n < 10^{k}$, Let $m$ be the $(k+1)$ rightmost digits of $2^{n}$. We have $m=2^{n}(mod\ 10^{k+1})$.

n are the k rightmost digits of $2^{n}$. So $10^{k}|(m-n)$. But, from 1) above,  $2^{10^{k}}=1 (mod \ 5^{k+1})$ since $k>1$, so $2^{m-n}=1 (mod \ 5^{k+1})$
$5^{k+1}|(2^{m-n}-1)$ $\Rightarrow $ $10^{k+1}|(2^{m}-2^{n})$ (since $k>n$) $\Rightarrow $ $2^{m}=2^{n}(mod\ 10^{k+1})$ and, since $m=2^{n}(mod\ 10^{k+1})$, we have $2^{m}=m (mod\ 10^{k+1})$

It remains to say that :
a) we have a starting solution with $k>1$ : $2^{36}=36(mod\ 10^{2}$)
b) m may be equal to n (new digit = zero) but, in a finite set of steps, we are sure to have a new solution, since $2^{n}\neq n$.

Nice demo, RUST.

--
Patrick

It proof: There are infinetely many $10^{k-1}\le n<10^{k}, \ k\in N$, suth that last k digits of $2^{n}$ is $n$.
Interest problem: How many numbers $10^{k-1}\le n<10^{k}$, suth that first k digits of $2^{n}$ is n?
For example the $2^{72780032758751764}=72780032758751764....$

[quote="aviateurpilot"][quote="fermat3"]i think that this probleme is a nice application for this : show that existe infinitly many n for fixid m such $m|2^{n}-n$ and it s posted before and solved by harazi :rotfl:[/quote]
$S_{m}=\{n|\ 2^{n}-n\ (\ mod\ m)\}$
if there exist $a\in \mathbb{N}$ such that $2^{a}\equiv a\ (\ mod\ m)$.
we have $\forall k\in \mathbb{N},2^{a+km\phi(m)}\equiv 2^{a}\ (\ mod\ m)\equiv a+m\phi(m) \ (\ mod\ m)$
then $|S_{m}|\ge |\{a+km\phi(m)| k\in \mathbb{N}\}|=+\infty$.[/quote]
i don't know why i can't idt my post
$S_{m}=\{n|\ 2^{n}-n\equiv 0\ (\ mod\ m)\}$


geometry
3D geometry
advanced fields
advanced fields unsolved

Hi all,

I am trying to prove that if $ g: X \rightarrow Y$ is continuous, then the mapping cylinder $ M_g\equal{}(X \times I) \sqcup Y/(x,1) \sim f(x)$ contains $ Y$ as a deformation retract.  I'm sure you hear this a lot with these types of questions, but I can see that what one might try to do is take the points in $ X \times I$ and "slide" them along the cylinder to where they "attach" to $ Y$.  This map restricted to $ Y$ is the identity.  The problem, of course, is writing down an explicit deformation retract.  My professor (he wouldn't give any other hints) told me to look at a proof that the point of a cone $ CX$ is a deformation retract of $ CX$, but I can't find a proof of that anywhere, nor can I think of a way to do that.

Any help would be greatly appreciated.

Okay, well I proved that the cone $ CX$ is contractible for any space $ X$.  First, I prove that there is a deformation retract of $ X \times I$ onto $ X \times \{0\}$.  Define $ H: (X \times I) \times I \rightarrow X \times I$ by $ H((x,t),s))\equal{}(x,ts)$ and note that $ H((x,t),1)\equal{}(x,t), H((x,t),0))\equal{}(x,0)$, and $ H((x,0),0)\equal{}(x,0)$, and so $ H$ is a deformation retract.  We can construct from $ H$ the map $ G$ that makes $ CX$ contractible.  Define $ G(([x],t),s)\equal{}[H(((x,t),s))]$.  $ H$ is well-defined since $ H((x,0),s)\equal{}(x,0)$ and $ H((y,0),s)\equal{}(y,0)$ are from the same equivalence class.  To see that $ G$ is continuous, note that it is the composition of $ H$ with the identification map $ p: X \times I \rightarrow CX$.

I'm not really seeing how this helps me.  Maybe a map that resembles this one?

Does anyone have a hint as to how to do this?


FIBONACCI:Prove that for n≥3 we have
φ(F_n )≡0 (mod 4)
F_1=F_2=1

it has been posted before :
http://www.mathlinks.ro/viewtopic.php?t=262785


Find all polynomial P(x) in R[x] such that:
  P(a-b) + P(b-c) + P(c-a) = 2P(a+b+c)
a,b,c in R such that ab+bc+ca=0

Same comment(s) as in http://www.mathlinks.ro/Forum/viewtopic.php?t=108836 ...

http://www.mathlinks.ro/Forum/viewtopic.php?t=14021