To cancel a registration, I think you have to drop the class via the 'Drop Class' button on your class's page.

Also btw you can only drop a class during the first 3 weeks of the class if u want a full refund.

Yes, [b]before[/b] the third class begins.

[quote="AIME15USAMO"]How do I cancel my registration for an AoPS problem solving class?

Thanks  :)[/quote]

You click "Drop Class" on the My Classes tab. But remember you can only do this until the third class ends.

[b]BEFORE[/b] the third class [b]BEGINS[/b].

If I cancel a class that requires an AoPS textbook, is the textbook delivery also canceled?

Once you register, the book is en route (except for the two classes, calculus and precalculus, for which the book is not available yet -- if you drop those, the book order is canceled).


Find the maximum value of

$ f(x,y,z,w)\equal{}\dfrac{\alpha xy\plus{}\beta yz \plus{}\gamma zw \plus{}\delta wx}{x^2\plus{}y^2\plus{}z^2\plus{}w^2}$

In terms of x,y,z,w.

also take a look [url=http://www.mathlinks.ro/viewtopic.php?p=1201665#1201665]here[/url] for the same problem but for three variables.

[quote="mathwizarddude"][quote="mgao"]Find the maximum value of

$ f(x,y,z,w) \equal{} \dfrac{\alpha xy \plus{} \beta yz \plus{} \gamma zw \plus{} \delta wx}{x^2 \plus{} y^2 \plus{} z^2 \plus{} w^2}$

In terms of x,y,z,w.[/quote]

Do you mean in terms of $ \alpha, \beta, \gamma, \delta$ instead?
This shouldn't be hard using a little knowledge of multivariable calculus, but of course, I assume Olympiad problems aren't allowed to be solved using calculus; is it true?[/quote]
it depends on how much calculus you want to use! :D
anyway I don't think it would be inappropriate if you post your solution here...

nope no calculus
yes, in terms of $ \alpha , \beta , \gamma , \delta$
sorry for the typo


For reals $a,b,c$ and positive $x,y,z$, define $\displaystyle [a,b,c] = \sum_{sym}x^a y ^b z^c = x^a y^b z^c + x^a y^c z^b + x^b y^a z^c + x^b y^c z^a +   x^c y^b z^a + x^c y^a z^b$.

Prove that ${}[5,0,0] +  2[3,2,0]+ 4[3,1,1] \geq 4[4,1,0]+3[2,2,1]$.

Well, if you bring everything into the left hand side and factor, you get

$2\left( x^{2}+y^{2}+z^{2}-yz-zx-xy\right) \left( x^{3}+y^{3}+z^{3}+3xyz-yz\left( y+z\right) -zx\left( z+x\right) -xy\left( x+y\right) \right) \geq 0$.

The left bracket of this product is nonnegative by Cauchy, the right one by Schur.

  Darij


Find all the functions $f: Z\rightarrow{Z}$ such that
$f(m-n+f(n))=f(m)+f(n) \forall{m,n\in{Z}}$ :)


You have an 8x8 board divided into squares of 1x1 each. How many total squares of side lengths 1 to 8 inclusive can be made?

Bonus: develop a general formula for an n x n sized board.

Have fun.

[hide]64[/hide]Not sure about the formula though

You can derive the formula quite easily actually, you try to derive it though.

Do you mean derive it from the formula for the sum of a sequence of general polynomials (which would just be simple algebra) or derive it using some kind of reasoning?

not trying to derive it right now, the formula is [color=white]n(n+1)(2n+1)/6=sum of square number from 1 to n :^2: [/color] Try the problem again with how many rectangles of any dimension, assuming their sides fall on the lines of a NxN checkerboard?

Yeah, it's just the sum of all the squares up to n

For the last question: [hide]((n+1)C2)^2[/hide]

Quote:I'm sorry. What I meant was a general formula for the kind of sum you posted. However, thinking about it now, it's really not something that kind be derived by most people, if any. It's more of something some people just know.

Of course it can be derived! How many things in math are "just known"? There is a pretty cool way using combinatorics. Think binomial coefficients.


Given that $ 9a^{2} \minus{}8b^{2} \equal{} 1800$ and $ \frac{a}{b} \equal{} \frac {4}{3}$, what is the value of the product $ ab$?

wrong.

on the second~third line, you said $ 3^2\equal{}3$

$ 3a \equal{} 4b \implies 9a^2 \equal{} 16b^2$

$ 16b^2 \minus{} 8b^2 \equal{} 1800 \implies b^2 \equal{} 225 \implies b \equal{} 15$

$ 3a \equal{} 4b \implies 3a \equal{} 60 \implies a \equal{} 20$

$ ab \equal{} 300$

Answer ; 300

Substituing in a=4b/3

9(4b/3)^2-8b^2=1800

16b^2-8b^2=1800

From there the solution is the same as fantasys.

$ 9a^2 \minus{} 8b^2 \equal{}1800$.
Since $ b \equal{} \frac34a$ and $ a \equal{} \frac43b$, we can substitute to get
$ 12ab\minus{}6ab\equal{}6ab\equal{}1800\implies ab\equal{}\boxed{300}$.


I just exchanged some posts with ml2 in the Getting Started forum in the thread titled "AOPS Chapter 11."  After each time he posted, his post in the thread appeared unread.  After I posted and left the forum, every single post in the thread (including my most recent one) began to appear unread, even when I entered and exited the thread repeatedly (by using the various forum buttons, not by hitting "Back" on my browser).  However, when I clicked "mark all topics read," it stopped happening.  Any thoughts?

Update: his single post continues to appear unread to me, even though I've viewed it several times.

Further update:  [i]This[/i] post also appears to be unread.

I think its something with the site it happens sometimes, do this, go to forum and check if the time of last time you came has changed (so that it isn't like 10 minutes ago, since you already read my post a minute ago), if not, its just this thing that happens every so often, if it does change then i am out of ideas.

It's still happening, and now it's happening in Fun and Games, too.  Actually, just with the new announcement in Fun and Games, but not with any of the regular posts.

Close all browsers, come back to AoPS. If that doesn't work restart the computer.

Maybe you should delete your AoPS cookies?

I don't think it's a cookie/cache problem.  Tare is probably right that restarting my computer will do the trick -- have stuff open I don't want to close right now, though.  I'll try later and report back.


Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \sqrt{\frac{8a(b\plus{}c)\plus{}9bc}{(2b\plus{}c)(b\plus{}2c)}} \plus{}\sqrt{\frac{8b(c\plus{}a)\plus{}9ca}{(2c\plus{}a)(c\plus{}2a)}} \plus{}\sqrt{\frac{8c(a\plus{}b)\plus{}9ab}{(2a\plus{}b)(a\plus{}2b)}} \ge 5\]
:)

Can we tell, given any three non-negative real nos. $ a,b,c$, no two of which are zero, any one of them will be greater than the $ G.M.$ of the other two???
 :maybe:
Only then, the problem might be solved by mixing variables $ b$ and $ c$ to $ \sqrt {bc}$.

I thought of a different approach....but dunno whether it is true or not......infact, I could solve herds of ineq.s with this technique....but it has no concrete proof or argument.....
It is somehow alike mv technique.....

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \sqrt {\frac {8a(b \plus{} c) \plus{} 9bc}{(2b \plus{} c)(b \plus{} 2c)}} \plus{} \sqrt {\frac {8b(c \plus{} a) \plus{} 9ca}{(2c \plus{} a)(c \plus{} 2a)}} \plus{} \sqrt {\frac {8c(a \plus{} b) \plus{} 9ab}{(2a \plus{} b)(a \plus{} 2b)}} \ge 5
\]
:)[/quote]
I am waiting for a nice proof, my proof used only Cauchy Schwarz but it is not very nice.
Notice that the equality holds for $ a\equal{}b\equal{}c$ and $ a\equal{}b,c\equal{}0$.

[quote="tronghieu"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \sqrt {\frac {8a(b \plus{} c) \plus{} 9bc}{(2b \plus{} c)(b \plus{} 2c)}} \plus{} \sqrt {\frac {8b(c \plus{} a) \plus{} 9ca}{(2c \plus{} a)(c \plus{} 2a)}} \plus{} \sqrt {\frac {8c(a \plus{} b) \plus{} 9ab}{(2a \plus{} b)(a \plus{} 2b)}} \ge 5
\]
:)[/quote]
Because equality holds for a=b; cp=0 so we can use Cauchy-schwartz to solve it. But very ugly
 :|[/quote]
Dear tronghieu,
Dont spam! I just saw you spamed on the forum, if you have the solution, you can post it here, then we will discuss.

[quote=can_hang2007]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \sqrt{\frac{8a(b\plus{}c)\plus{}9bc}{(2b\plus{}c)(b\plus{}2c)}} \plus{}\sqrt{\frac{8b(c\plus{}a)\plus{}9ca}{(2c\plus{}a)(c\plus{}2a)}} \plus{}\sqrt{\frac{8c(a\plus{}b)\plus{}9ab}{(2a\plus{}b)(a\plus{}2b)}} \ge 5\]
[/quote]
Stronger  ;)
\[\sqrt{\frac{8a(b\plus{}c)\plus{}9bc}{(2b\plus{}c)(2c+b)}} \geqslant \frac{1}{3} \cdot \frac{17a^2-b^2-c^2+20a(b+c)+20bc}{a^2+b^2+c^2+4(ab+bc+ca)}.\]

[quote=ishfaq420haque]please send your solution[/quote]


geometry
geometric transformation
invariant

Consider the circumference $ C$ of center $ O$ and radius $ 2 \text{ cm},$ and a straight line $ t$ distant $ 3 \text{ cm}$ of the point $ O.$ From the point $ A$, in straight line $ t$, two tangent straight lines to circumference $ C$ are traced. The straight line that joins the points of contact of these tangents cuts, in a point $ B$, the diameter of the circumference that is perpendicular to straight line $ t$. Find $ \overline{OB}.$

I'm not sure, but i think the translation is very bad. I did the figure to make it clear.

I have the answer now, but i would like to know why $ \overline{OB}$ is invariant for any position of point $ A.$

Let the points of contact of these tangents cuts be $ M,N$ and $ OB$ cut the straight line $ t$ at $ H$
Denote the circle has diameter $ OH$ by $ (C)$
It easy to prove that B is the radical centre of these 3 circle.
Thus, B is fixed.
So it easy to see $ \overline{OB}\cdot\overline{OH} \equal{} 4$ which implies $ \overline{OB} \equal{} \frac {4}{3}$


algebra
polynomial
function
quadratics
search
complex numbers

Hi guys, I'm in a college algebra course and this is my first time posting. I wanted some help on this problem that I'm having a hard time getting through.

Basically, I'm supposed to take this function:

f(x) = 2x^3-5x^2+x+2

and do the following:

a) List all possible rational zeros

b) Test the possible rational zeros and find an actual zero

c) Find the remaining zeros of the polynomial function

So, for the first part, I got +/- 1, 1/2, and 2 as possible factors

Then for the second part, I plugged in 1 into the function and I got a zero. After that I used long division to divide the polynomial by x-1 and I got 2x^2+3x+2

Now, I'm supposed to find out find which of the other possible zeros, if any are actual zeros, but I don't know how to do that.

Do I just put each of the other zeros through long division and eliminate the ones that give me a remainder or is there some other, faster way to do this? Can I factor the zeros out of 2x^2+3x+2?

[quote="rofler"]Well, you just eliminated the rational roots, but you still have irrational and non-real roots.

$ 2x^2 \plus{} 3x \plus{} 2$ will contain the other 2 roots.

$ x \equal{} \frac { \minus{} 3\pm\sqrt {9 \minus{} 16}}{4} \equal{} \frac { \minus{} 3\pm i\sqrt {7}}{4}$ are the other 2 roots.

This was, of course, made under the assumption you know about complex numbers and the quadratic equation :P[/quote]

I do. If you put the polynomial through the quadratic equation, you get the irrationals and non-reals, right?

If that's the case, then I can definitely do that, but my question is how do I find the other real zeros? I know it has to be among these: +/- 1/2, 1, or 2. But how do I determine which ones are and aren't actual zeros?

There are no other zeroes, since it is of degree 3 it has exactly 3 zeroes, two are imaginary as shown in the post above, and 1 is the third zero, so these are all the zero.

And more generally, I believe you typically establish if the root actually works or not by plugging each possible value in and checking. Granted, there are smarter ways (as in this example, using the quadratic equation to deduce that there are no more zeroes), but exhaustive search will never fail to find all the rational roots.

There is actually a name for this, the rational roots theorem, see [url]http://en.wikipedia.org/wiki/Rational_root_theorem[/url]


Euler
function
number theory
totient function
superior algebra
superior algebra unsolved

Here's a little theorem for you :)

Let n be a (positive) natural number and let G be a group with order $\prod_{p\in\mathbb{P}\cap ]n,2n[}{p}$

Prove that G is cyclic.

It has been shown on the forum that a group of order $m$ such that $(m,\varphi(m))=1$ (where $\varphi$ is the Euler totient function) is cyclic. Clearly, that product is such a number, so this applies here.


leepakhin, you can simplify your answer...

[hide="my solution"]$x^2-\sqrt{3}x+1=0 \Rightarrow x=\frac{\sqrt{3}\pm\sqrt{3-4}}{2}=\frac{\sqrt{3}\pm i}{2}=\text{cis} \pm 30$
$\frac{360}{30}=12$ so $(\text{cis} \pm 30)^{2005}=((\text{cis} \pm 30)^{12})^{167}(\text{cis} \pm 30)=(1)(x+\frac 1x) =\boxed{\sqrt{3}}$[/hide]

At first I went for leepakhin's formula as well, but there is no such need  :)

[quote="leepakhin"][quote="warut_suk"]Let $x+\frac{1}{x}=\sqrt{3}$, find $x^{2005}+\frac{1}{x^{2005}}$[/quote]
[hide="perhaps wrong"]Let $S_n=x^n+\frac{1}{x^n}$
by the identity $(x+\frac{1}{x})(x^{n-1}+\frac{1}{x^{n-1}})=x^n+\frac{1}{x^n}+x^{n-2}+\frac{1}{x^{n-2}}$
we have $S_n-\sqrt{3}S_{n-1}+S_{n-2}=0$
thus the characteristic equation for $S_n$ is $r^2-\sqrt{3}r+1=0$
$r_1=\frac{\sqrt{3}+i}{2}, r_2=\frac{\sqrt{3}-i}{2}$
Let $S_n=Ar_1^n+Br_2^n$
Initial values: $S_0=x^0+\frac{1}{x^0}=2, S_1=\sqrt{3}$
then $S_0=Ar_1^0+Br_2^0=A+B, S_1=Ar_1^1+Br_2^1=Ar_1+Br_2$
Solving simultaneaous equations for A and B, we get $A=B=1$
Therefore the general formula for $S_n$ is $(\frac{\sqrt{3}+i}{2})^n+(\frac{\sqrt{3}-i}{2})^n$
So $x^{2005}+\frac{1}{x^{2005}}=S_{2005}=(\frac{\sqrt{3}+i}{2})^{2005}+(\frac{\sqrt{3}-i}{2})^{2005}$[/hide][/quote]

U r absolutely right ...just that ur final result can be further simplified ,on application of De-Moiver's theorem , to 2cos(2005*pi/6) = sqrt(3)  :D


inequalities
function
integration
real analysis
real analysis unsolved

Let f be a differentiable function on [0,1] satisfying f'(x) ≤ 1+ 3f(x) for all x belongs to [0,1] with f(0)=0. Show that, 3f(x) ≤ e^(3x) -1.


calculus
integration
limit
trigonometry
calculus computations

For a positive number $x$, let $f(x)=\lim_{n\to\infty} \sum_{k=1}^n \left|\cos \left(\frac{2k+1}{2n}x\right)-\cos \left(\frac{2k-1}{2n}x\right)\right|$

Evaluate

\[\lim_{x\rightarrow\infty}\frac{f(x)}{x}\]

$f(x)=\int_0^x |sin(t)|.dt $ :D  , so $\lim_{x\rightarrow\infty}\frac{f(x)}{x}=\frac 2\pi$
 :cool:


How do you solve this recurrence ? I only know how to solve it if the recurrence is liniar .Can you give me a link or can you post a solution ? Thank you :D

but how can you say that if it holds for x^4 , x^5 ,x^6 it too wud hold for x^2008

i got the actual explanation , since x_n+1 - x_n > 0 and x_4 >1 therefore definately, x_2008>1 i  i.e  0<1/x_2008<1 or  2<3- 1/x_2008<3,
therfore its integral part is 2.


Obviosly $a|c$ or $c=ak$.
There are infinetely many solutions $c=a$ (k=1).

Hello kimnimalar
[quote="kimnimalar"]find all triples (a,b,c) of positive integers such that
\[abc+ab+c = a^{3}\]
[/quote]
$a$ divides $c$ and we can write $c=ad$ and $abd+b+d = a^{2}$
This implies $bd<a$ (else $abd\geq a^{2}$)
$bd\leq(a-1)$ $\Rightarrow $ $b+d=a^{2}-abd\geq a^{2}-a(a-1)=a>bd$ $\Rightarrow $ $b+d>bd$
Then if $b\geq 2$, $b+d>bd$ $\Rightarrow $ $d>b(d-1)\geq 2(d-1)$ $\Rightarrow $ $d<2$
So, either $d=1$ and $b=a-1$, either $b=1$ and $d=a-1$

And the only $(a,b,c)\in\mathbb{N}^{3}$ solutions are :
$(a, a-1, a)$ with $a>1$ and we have $a^{2}(a-1)+a(a-1)+a=a^{3}$
$(a,1,a(a-1))$ with $a>1$ and we have $a^{2}(a-1)+a+a(a-1)=a^{3}$

(and, for RUST, $k=1$ is not the only solution : $k=a-1$ is also a different solution)

--
Patrick


[quote="Nea"]Prove $\forall n \in N$

$\left. 7 \right|3^{2n + 1} + 2^{n + 2}$[/quote]

[hide]

  $3^{2n+1} + 2^{n+2} = ( 3^{2n} \cdot 3^1 ) + ( 2^n \cdot 2^2 ) =$

  $(9^n \cdot 3) + (2^n \cdot 4)$

  Now replace the 9 with its mod 7 residue to obtain:

  $(2^n \cdot 3) + (2^n \cdot 4)$

  Factor out the $2^n$ to obtain:

  $2^n( 3 + 4 )$

  $2^n(7)$

  Evidently this equation is divisible by 7.   [/hide]


2 probleme facute de mine, deci va puteti astepta sa fie usoare spre banale :D

$1.$ aratati ca $\forall x$ din $(0;\pi/2)$ avem:

$tg^{2}x+cos(2x)+ctg^{2}x \geq (cosx+tgx)(cosx+\frac{1}{sinx})-2 \geq 1$

$2.$ s-ar putea sa fie o mica problema la enunt, adica datele sunt corecte, dar formularea scartzaie

fie $z_{1}$,$z_{2}$,$z_{3}$,$z_{4}$ numere complexe nenule afixele a $4$ puncte mobile $A,B,C, respectiv D$. $|z1|=|z4| z_{1}^{2}+z_{2}^{2}=0$ si $z_{3}^{2}+z_{4}^{2}=0$. Aratati ca daca $|z_{1}-z_{4}+z_{2}-z_{3}|$ este maxim atunci $A,B,C,D$ sunt varfurile unui dreptunghi

A invata cateva comenzi $\LaTeX$ este, cred, o dovada de respect fata de de cei care sunt activi pe forum. Nu e greu, si nu dureaza mai mult de o jumatate de ora...:)

Am incercat de jdemii de ori si nu merge. Atunci, in semn de respect, nu o sa mai scriu porbleme si nici solutii. O sa dau doar idei.

[quote="spix"]Ce anume nu merge?[/quote]
mersi de ajutor, am gasit pe cineva care s-a ofeit sa ma ajute pe private. sper ca data viitoare sa postez cu LATEX :D


geometry
geometric transformation
reflection
projective geometry
geometry proposed

Let A', B', C' be the midpoints of the sides BC, CA, AB of a triangle ABC, and let the incircle of triangle ABC touch the sides BC, CA, AB at the points X, Y, Z. The lines B'C' and YZ meet at A"; the lines C'A' and ZX meet at B"; the lines A'B' and XY meet at C".

There are several facts known about this configuration. For instance, the lines B"C", C"A", A"B" pass through the points A, B, C, respectively, and the lines XA", YB", ZC" pass through the point of tangency F of the incircle and the nine-point circle of triangle ABC.

Now the problem is to show that the lines A'A", B'B", C'C" concur at one point, and this point lies on the common tangent of the incircle and the nine-point circle of triangle ABC at their point of tangency F.

  Darij

[quote="darij grinberg"]
Now the problem is to show that the lines A'A", B'B", C'C" concur at one point, and this point lies on the common tangent of the incircle and the nine-point circle of triangle ABC at their point of tangency F.

Darij[/quote]

Here follow some reflections which could be usefull, togheter the other facts, to solve the problem: if I'm not wrong what follows is a little bit more general result of that claimed.

Let c be a circle and X,Y,Z and W four points on it. We define A?=XW^YZ, B?=XZ^YW and C?=XY^ZW. If tX,tY and tZ are the tangent at c in X, Y and Z respectively, we define A=tZ^tY, B=tZ^tX and C=tX^tY. Several facts are well known, for instance it holds that A is collinear with B? and C? and the line where they lie on is the polar of A? wrt c. Similarly B is collinear with A? and C? and C collinear with A? and B?. Another fact is that A?B? is the polar of C? wrt c and similarly for B? and A?.

Let R=AB^A?B? and  S=AC^A?C?.  Given that R lies on A?B? (the polar of C?) it results that the polar of R pass through C?, then, as R lies on tZ, ZC? is the polar of R wich means that RW is tangent in W to c. Similarly one can prove that SW is tangent  in W to c. Then R,W, S are collinear.

Now if A’ is a point on BC, let B’=AC^A’C? and C’=AB^A’B?. From Desargues, we have that A?,B’,C’ are collinear.
If  P=C’C?^B’B?, again by Desargues, it holds that R,P,S are collinear. This means that, as A’ varies on BC, P trace the tangent to c in W.

PS
Of course, also A’A? pass through P.
I hope there are no typo (I changed many times the name of points to match as much as is possible the original problem)
If it is necessary I can post some figures.

Wow! That was a cool proof. I didn't ever think the problem was so simple.

Anyway, here is another, much simpler problem on the same configuration (I mean the configuration of my original problem, not your generalization):

Prove that AA" || BB" || CC".

Actually this one is really not difficult. There is a solution using polars (by Lev and Tatiana Emelyanovy, the authors of IMO 2000/6), but there is also a rather simple solution without polars.

  Darij

[quote="darij grinberg"]

Prove that AA" || BB" || CC".

Actually this one is really not difficult. There is a solution using polars (by Lev and Tatiana Emelyanovy, the authors of IMO 2000/6), but there is also a rather simple solution without polars.

  Darij[/quote]

One proof colud be the following.
We know, for instance, that AA"C and C"CA are projectively correnspondets through B". Then as, AC'//C"A' and C'A"//A'C, AA"//C"C.

Is this one, one of the two (or more) you know?

[quote="sprmnt21"]Is this one, one of the two (or more) you know?[/quote]

No, it isn't. It is simpler than both of them!

Here are the two proofs I mentioned:

[i]Proof by Lev and Tatiana Emelyanovy (sketched).[/i]

Since the line AB" = B"C" is the polar of the point A" with respect to the incircle of triangle ABC, while the lines AC and AB touch this incircle, the lines AB" and AA" are harmonical conjugates with respect to the lines AC and AB. Hence, if the lines AA" and AC meet the line BB" at the points V and W, then the points B" and V are harmonical conjugates with respect to the segment BW. But the point B" is the midpoint of the segment BW (since the point B" lies on the midparallel C'A'); hence, its harmonically conjugate point V is an infinite point, i. e. we have AA" || BB". Similarly, BB" || CC". Proof complete.

[i]My proof.[/i]

This proof uses almost no preknowledge about the configuration.

We will use directed segments on the lines BC, CA, AB. Hereby, these lines will be directed in such a way that the segments BC, CA, AB will be positive (and hence, the segments CB, AC, BA will be negative). Let a = BC, b = CA, c = AB.

If $s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC, then BX = s - b. On the other hand, since A' is the midpoint of the side BC, we have $BA^{\prime}=\frac{a}{2}$. Thus,

$XA^{\prime} = BA^{\prime}-BX = \frac{a}{2}-\left(s-b\right) = \frac{b-c}{2}$

(after an easy calculation), and similarly

$ZC^{\prime} = \frac{a-b}{2}$.

Hence,

$\frac{XA^{\prime}}{ZC^{\prime}} = \frac{b-c}{a-b}$.

Since the points Z, X, B" lie on one line, the Menelaos theorem gives

$\frac{A^{\prime}B^{\prime\prime}}{B^{\prime\prime}C^{\prime}} \cdot \frac{C^{\prime}Z}{ZB} \cdot \frac{BX}{XA^{\prime}} = -1$.

Since the two tangents from a point to a circle are equal in length, ZB = BX, and thus, this becomes

$\frac{A^{\prime}B^{\prime\prime}}{B^{\prime\prime}C^{\prime}} \cdot \frac{C^{\prime}Z}{XA^{\prime}} = -1$.

In other words,

$\frac{A^{\prime}B^{\prime\prime}}{B^{\prime\prime}C^{\prime}} = - \frac{XA^{\prime}}{C^{\prime}Z} = \frac{XA^{\prime}}{ZC^{\prime}} = \frac{b-c}{a-b}$.

Now, if the line BB" meets the line CA at W, then, since C'A' || CA, Thales tells us

$\frac{CW}{WA} = \frac{A^{\prime}B^{\prime\prime}}{B^{\prime\prime}C^{\prime}}$.

Consequently,

$\frac{CW}{WA} = \frac{b-c}{a-b}$.

Similarly, if the line AA" meets the line BC at H, we have

$\frac{BH}{HC} = \frac{a-b}{c-a}$.

After some easy algebra, this becomes

$\frac{CB}{BH} = \frac{b-c}{a-b}$.

Hence, $\frac{CB}{BH} = \frac{CW}{WA}$. By Thales, this implies AA" || BB". Similarly, BB" || CC". And we are done again.

Thanks again for all of your proofs!

  Darij


What is the number of ways of going up n stairs, if we can take one or two steps at a time?

This is Fibonacci sequence or Lucas' sequence.


Given parallelogram $ABCD$ with $E$ the midpoint of diagonal $BD$. Point $E$ is connected to a point $F$ in $DA$ so that $DF=\frac{1}{3}DA$. What is the ratio of the area of triangle $DFE$ to the area of quadrilateral $ABEF$?
(A) $1:2$
(B) $1:3$
(C) $1:5$
(D) $1:6$
(E) $1:7$

the answer is
[hide="answer"]
$\text{C}$ $1:5$
[/hide]

Please don't post just an answer. We hope to learn from how you solved the problem. Show your steps and your thinking process.

yeah but I dont have any geosketch stuff can anybody tell me where to get it for free. :?  :(
I really face problems with drawing diagrams in paint. :mad:

I use Macromedia Fireworks. It's really fast and the diagrams look really nice.

is it free? can you tell me where to find it if it is???
I really need a geometry program. :?

[quote="riddler"]is it free? can you tell me where to find it if it is???[/quote]

go to [url]http://www.macromedia.com/[/url] and in the featured products section, click on Fireworks (it'll probably be Fireworks MX 2004), and it will give you the product information. You should see a free 30-day trial. But it's not forever!

But I dont want to give my information to another site... even if it is for free... :?  :(


inequalities
function
inequalities theorems

Hello.
For : $x,y,z>0$ we have:
$(\frac{x+y+z}{3})^{3}\leq \frac{x^{3}+y^{3}+z^{3}}{3}$
Can we generalise this inequalitie for n variables :
with proof please.

hi FOURRIER :D
i suppose all the variables are positive .
then apply jensen's inequality to the convex function $f(x)=x^{n}$


geometry
inradius
ratio
incenter
circumcircle
arithmetic series
geometric series

[hide]
i'm not sure..
the triangle must be an equilateral with a common difference
of 0 and a common ratio of  1 between the sides.
that way ot forms both ari. and geo. progressions.
so the inrad. is 2 so the median is 6 since the the inradius is 1/3 of the median.
that means the circumrad. is 4, this can be easily seen if you draw the circle out.
thus the area is 16pi.[/hide]

A triangle has side lengths that form both an arithmetic and geometric sequence. If the triangle's inradius is 2, find the area of the circumcircle.
Let triangle has sides a,b,c then
$(a+b)=2b=2\sqrt{ab}=2b$
then
$a=b=c$  Equilateral Triangle
then
$A(ABC)=\frac{P}{2}*r=\frac{abc}{4R}$
from here $a=b=c=4\sqrt{3}$ and $R=4$
[hide]answer
$16\pi$[/hide]

since the sides are in an arithmetic series, say that they are

a-b,a,a+b, now this is a geometric series, so

a^2=(a-b)(a+b) so b=0

so all the sides are equal, then using a 30-60-90 triangle, (note that the incenter and the circumcenter of an equilateral triangle are the same point), R=2(r) so R=4, then we get $4^{2}\pi=16\pi$


Using math symbols and exactly four four's, come up with a way to write 33

for example you would write 2 as. 2 = (4/4) + (4/4)

Am I correct in assuming that you can't use any othere digit? ex. you can't do

34 - 4^(4-4)?

I remember one solution given by Paul M Dirac (Nobel prize in physics - I  did happen to take a course taught by him) to a similar problem. I am stealing his idea to give a solution for any n and i will use only  three 4's.  (Well  if you insist using all the  four fours .. just try     "4+29"  and use my method to write 29 using 3 4's)

 I am not using latex (some one can help me here.. and put the solution )  .. so here it goes ...

- log (to the base of sqrt(4) ) of
  log  (to the base of 4 ) of
  sqrt(sqrt(sqrt ........ 4)     (sqrt(sqrt... etc come "n" times )

1 = -log(base 2)( log(base 4) ( :sqrt: 4))
2 = -log(base 2) (log (base 4) ( :sqrt:  :sqrt: 4) )

etc...

Just use as many " :sqrt: " as you need.

Hope  you found it to be fun...

[quote="belenos"]Will you accept :

$\lceil -44\,\cos(4)+4 \rceil$

with $\lceil x \rceil$ = smallest integer $\geq$ x ?[/quote]

Nice one!!!

Sure if you accept mine:

4*4+4*4 - (i)*(i)  :D

Also:
Since  4^0  is (4*4*4 ...  "0 times")  but 0 times is no times at all ,,.. so it does not count ...and we have  33= 4*4+4*4+4^0  :lol:

[quote="Gyan"]
4*4+4*4 - (i)*(i)  :D
[/quote]

lol That's cheap! Using i you can get any integer.

[quote]Using i you can get any integer.[/quote]
  Oh!  get REAL!! <groan>  :D

(Okay - Before moderators kick me out ... I will ask a  cute question .. can you get 20 by using  2   3's ?  ( say use TI calculator and only number you can press is  ''3'" two times ... NO 'pi'  etc either)

Nice one.. but  I think, I said, "NO   pi Etc " .. (e  is even a letter in "etc"  :) ..... (well,  there are perhaps   more solutions.,,, I will give mine after I see a few other ones...)

Also    has any one  found other  cute solutions for the original problem  worth sharing ?

See on my website "la page 4" :
 [url]http://perso.wanadoo.fr/gery.huvent/page_4/the4page.htm[/url]
 :lol:

[quote="Gyan"]My solution was:  (A much simpler)    [b]3!/.3 [/b][/quote]

That's too easy.   ;)


Find all nonegative integers satisfying $3y^2=x^4+x$

[hide="Solution (correct?)"]$x=y=0$[/hide]

[quote="draperp"]That is probably the correct solution, but proving it would even better.[/quote]

Yeah, you are right, but I am too lazy to do it know. I will post it tomorrow! :)

Using the infinite descent method I guess? ;)

I let x=0,1,2,3...1000 and y=0,1,2,3...1000, and I found that yours was the only solution in that range.

(By the way, I think that even if you included all integers, the only additional solution would be x=-1, y=0.)

This shouldn't be that hard to prove...do you know any rules about what happens to the mod when a number is squared, cubed, etc.?

EDIT: Duh! The new mod is the mod of the the old mod raised to a power.  I don't think this helps though...sorry,my thinking's going around in circles right now.

Prove a stronger statement:

For any postive integer \[ a \neq 2  \]
the only positive integer solution of \[ ay^2 = x^4+x  \]
is \[ x=y=0  \]

(This is interesting: All positive integer values of [i]a[/i] work except for 2.  It has solutions (0,0), (1,1), and (2,3).)


Half way through a document I have restarted the chapter numbering.

The following code:

[code]
\setcounter{chapter}{0}
\renewcommand{\thechapter}{\arabic{chapter} }
\renewcommand{\chaptername}{Unit \thechapter: Assessment Sheet}
\chapter{Natural Numbers}
[/code]

produces,

[quote]
Unit 1: Assessment Sheet 1

Natural Numbers
[/quote]

I don't want the "1" after the word "Sheet" to appear. So basically I am trying to get the chapter number to appear within the chapter name only and not after it.

Any help would be greatly appreciated.

Shane

I've solved the problem. The correct code is,

[code]
\setcounter{chapter}{0}
\renewcommand{\thechapter}{\arabic{chapter} }
\renewcommand{\chaptername}[2]{Unit \thechapter: Assessment Sheet}
\chapter{Natural Numbers}
[/code]

It appears you need to use the arguments for the \renewcommand the [2]

Shane


Anne gave Betty as many cents as Betty had. Betty then gave Anne as many cents as Anne then had. At this pount, each had 12 cents. How much did Anne have at the beginning?

[hide]Since Anne ends up with 12 cents, Betty must have given her half of her current total, or 6.  This means that after the first step Betty had 18 cents.  Anne had given her half of her total, or nine cents.  Then Anne started with 6+9=15 cents.[/hide]

Anne=x
Betty=y

 2y+2(x-y)=24
 2x=12
 x=6
 so then y=18

 y=18 is double what y had been before
 so before x had bin 9+6=15

 Answer=15 :lol:


A $2004\times2004$ square array is drawn. Find the largest integer $n$ such that it is possible to draw a convex $n-$sided polygon whose vertices lie on the points of the array.

Actually, this is a problem from the traing of Chinese team for IMO(2004), the answer is 581, but the solution is too long and my English is very poor :blush: . I can only tell that Eular function is needed.


An ant starts from the bottom edge of a can and makes 2 complete revolutions to the top, ending up on the top of the can directly above where he started. If the can is 18 cm tall and has a radius of 3 cm, how far did the ant travel along the can?

[hide]
Imagine the cylinder is cut in half. Therefore the ant walks up a cylinder of height 9 and radius 3 twice.

Unrolling the the cylinder would give a rectangle and the length of the diagonal is $ \sqrt{(6\pi)^2 \plus{} 9^2}$

So total length is
$ 6\sqrt{4\pi^2 \plus{} 9}$[/hide]

[quote="TZF"]Is there any characterization of what kind of path the ant took?

For example, did the ant travel along the path of minimal distance?[/quote]
The original problem doesn't quite state that...:\ but I'm assuming when the cylinder is unrolled the ant should walk in a straight line.


limit
trigonometry
induction
geometry
calculus
algebra solved
algebra

Consider the sequence $x_1,x_2,...,x_n,...$ such that $x_1=\sqrt{2},x_{n+1}=\sqrt{2+x_n}$ for all $n>1$ . Find :

a) $\lim_{n \to \infty}x_n$

b) $\lim_{n \to \infty} 4^n(2-x_n)$ ;)

It is obviosly $x_1=2\cos{\frac{\pi}{4}}$, by induction $x_n=2\cos{\frac{\pi}{2^{n+1}}}$. Therefore
a) lim=1,
b) $\lim_{n\to \infty} 4^n(2-2\cos{\frac{\pi}{2^{n+1}}}=(\frac{\pi}{2})^2$

[quote="Rust"]It is obviosly $x_{1}=2\cos{\frac{\pi}{4}}$, by induction $x_{n}=2\cos{\frac{\pi}{2^{n+1}}}$. Therefore
a) lim=1,
[/quote]
No, $\lim_{n\to\infty}x_{n}=2$ :D

yes, $2$ is the correct answer, let that $a$ is what we're finding, so
$a= \sqrt{2+a}$ solve it we have $a=2$ or $a=-1$,but $a>0$ => $a=2$

a) $(x_{n})$ is bounded form above by 2 (easy inductive proof) and obviously increasing. Hence $(x_{n})$ is convergent and $x: =\lim x_{n}$ satisfies $x=\sqrt{2+x}$ and $\sqrt{2}<x\le 2$.
Therefore $x=2$.

i got that the lim for part is 2, but then how do you prove part b)? Do you sub in the limit 2 as x_n, but that gives you that the new limit is (infinity)*(0)

For (b), the only reasonable way seems to be to get a closed form using trigonometry, like what Rust does. You then get
\[x_{n}=2\cos \frac{\pi}{2^{n+1}},\ 2-x_{n}=2\sin^{2}\frac{\pi}{2^{n+2}}\]
From here, using $\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$ makes the limit $\boxed{\frac{\pi^{2}}{8}}$. Since you get a transcendental number, nothing using only algebraic methods will work; you either need geometry, trigonometry, or calculus beyond basic limits.


geometry
3D geometry
geometric transformation
rotation
analytic geometry
probability
parameterization

A particle starts at origin O in three-space. Think of the origin as centered in a cube 2 units on a side. One move in this walk sends the particle with equal likelihood to one of the eight corners of the cube. Thus, at every move the particle has a 50-50 chance of moving one unit up or down, one unit east or west, and one unit north or south. If the walk continues forever, find the fraction of particles that return to the origin.

Meaning return to the origin at some point?

Isn't it just a matter of reorienting and rescaling the axes so that they pass through the vertices of the cube?

[quote="JBL"]As far as I can tell, the Mathworld entry deals with a random walk where each step is 1 unit parallel to one of the axes, not what we have here.  I don't know how much (if at all) this would change the answer, but it certainly couldn't make it more ugly ;)[/quote]

I read the original problem very quickly, but my understanding is that this is essentially a walk on a diagonal lattice, which, under rotation and scaling, is essentially $\mathbb{Z}^{3}$.

I wrote the following before blahblahblah posted:

There are 8 vertices of a cube and 6 half-axes ;)

In fact, the two walks do behave fundamentally differently.  A "face-centered" (parallel to the axes) walk picks a random coordinate and increments it randomly +1 or -1.  This gives 6 possible steps, and we can group them into three pairs so that any path that returns to the origin must have the same number of both members of a pair.

In the "vertex-centered" walk we're given, all three coordinates are simultaneously incremented by +1 or -1.  This gives 8 possible steps, and there is no longer any such grouping.  For example, the steps $(1, 1,-1), (1,-1, 1), (-1, 1, 1), (-1,-1,-1)$ return you to the origin, and we didn't directly reverse any of our steps.

The structure really is quite different, and may be easier to work with.  Certain calculations are definitely easier; for instance, the probability of being back at the origin after $2n$ steps is just $\left(\binom{2n}{n}/ 8^{n}\right)^{3}$.

In response to blahblahblah: this problem is indeed walking on a diagonal lattice; I assumed the problem Mathworld dealt with was the first one I describe above, but going back and reading their entry, it's not clear to me that they indicate which type of walk they're describing (or if they do, they use notation I'm not familiar with).

Further addition: their post is about a walk on a $d$-D lattice -- I have no idea what that means.

[quote="JBL"]There are 8 vertices of a cube and 6 half-axes ;)[/quote]

True.   :oops:

[quote="JBL"]their post is about a walk on a $d$-D lattice -- I have no idea what that means.[/quote]

$d$-dimensional, i.e. $2$-D, $3$-D, $4$-D.

Oh!  It still doesn't explain what kind of walking they mean, but reading the page for 2-D random walk strongly suggests that they mean a non-diagonal path (i.e. the $2d$-directional kind, not the $2^{d}$-directional kind), in which case we have made exactly 0 progress ;).

Fair enough.  I know some of the theory of random walks in the lattice, and it turns out that simple random walks on the diagonal lattice in $\mathbb{Z}^{2}$ is isomorphic to the usual simple random walk in $\mathbb{Z}^{2}$ (this is how you prove that the $2-$dimensional simple random walk is recurrent); it seemed plausible to me that this would be true for three-dimensions as well.

Then again random walks behave very differently in 2 and 3 dimensions so I guess this is not that surprising!

Following JBL's earlier insight, the expected number of times we touch the origin (including the start) is $S=\sum_{n=0}^{\infty}\left(4^{-n}\binom{2n}{n}\right)^{3}$; this generalizes easily to any number of dimensions, and this sum converges by comparison to $\sum n^{-\frac32}$.
The probability of ever returning to the origin is $1-\frac1S$.

That leaves the question of whether we can actually evaluate $S$; I doubt that it has a closed form.

All of this generalizes easily to any number of dimensions.

Actually, Mathematica claims that sum is $\frac{\pi}{\Gamma(3/4)^{4}}$.  I haven't had enough time to think about why that is or why the expected value and probability are related in that way.

Moral of the story: A drunken walker  need not  count on luck in  two  dimensions... IOW  take a random walk and you will eventually go where you want to go (if you walk long enough) .... Not so in three dimensions...!

The number of times we touch the origin follows a geometric distribution with parameter equal to the probability that we don't come back; that's how the probability and the expected value are related.


number theory
prime numbers
number theory unsolved

250 numbers are chosen among positive integers not exceeding 501. Prove that for every integer $ t$ there are four chosen numbers $ a_1$, $ a_2$, $ a_3$, $ a_4$, such that $ a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \minus{} t$ is divisible by 23.

[i]Author: K. Kokhas[/i]

[b]Albanian Eagle[/b],
I think isn't. Lets took all integers looks like $ 1 \plus{} 23*i$, $ 2 \plus{} 23*i$, $ 3 \plus{} 23*i$, $ 4 \plus{} 23*i$, $ 5 \plus{} 23*i$, $ 6 \plus{} 23*i$ for all $ i$ bigger than $ \minus{} 1$ and lesser than $ 22$. All this numbers lesser than $ 501$, because $ 23*21 \plus{} 6 \equal{} 489$. It's easy to see that for $ t \equal{} 2$ we cann't choose $ a_1$, $ a_2$, $ a_3$, $ a_4$. And we choose $ 22*6 \equal{} 132$ - more than $ 126$ integers.
But if we use one theorem about prime numbers it will be obvious that something about 200 integers is enough.
Theorem: if we have $ 2$ set $ A$ and $ B$ of remainders at prime modulus $ p$ wiht $ k$ and $ l$ elements. Then set of sums pair of elements from $ A$ and $ B$ (Set of remainders $ a \plus{} b$, where $ a$ - element from $ A$ and $ b$ element from $ B$) have at least $ k \plus{} l \minus{} 1$ element. (or $ p$ elemets if $ k \plus{} l \minus{} 1 > p$).
But i suppose people from junior league at Tuymaada don't know this theorem. Maybe i am wrong.)

Sorry for my very bad english, I hope i didn't make a mathematical mistake.  :blush:

[quote="Akella"]
Theorem: if we have $ 2$ set $ A$ and $ B$ of remainders at prime modulus $ p$ wiht $ k$ and $ l$ elements. Then set of sums pair of elements from $ A$ and $ B$ (Set of remainders $ a \plus{} b$, where $ a$ - element from $ A$ and $ b$ element from $ B$) have at least $ k \plus{} l \minus{} 1$ element. (or $ p$ elemets if $ k \plus{} l \minus{} 1 > p$).
But i suppose people from junior league at Tuymaada don't know this theorem. Maybe i am wrong.)
[/quote]
It seems to me that this is [b]Cauchy-Davenport[/b] theorem.
I'm pretty sure that this theorem can not be used at Tuymaada olympiad,if one doesn't prove it during the contest.And I'm sure that 8 or 9 grader wouldn't be able to find its proof.
So let's try to find proof without using of this theorem...

[quote="Erken"]
It seems to me that this is [b]Cauchy-Davenport[/b] theorem.
[/quote]
Thank you. I always forgot names of theorems.  :roll:

And i think simple solution is something like this:
Obvious that we not need numbers and need only they remainders at modul $ 23$. Also every remainder we can take maximally $ 22$ times. (because $ 23*22 \equal{} 506 > 501$). Then obviously there is a remainder that we take at least 2 times. Let it be $ a$. And let our problem is wrong and $ t$ is the remainder such that for it we cann't choose $ a_1$, $ a_2$, $ a_4$, $ a_4$. Then let $ a_3 \equal{} a_4 \equal{} a$. We cann't choose $ a_1$ and $ a_2$ such that $ a_1 \plus{} a_2 \equal{} (t \minus{} 2a)$ (at modul $ 23$ off course). Then we take $ 12$ pairs of remainders such that sum of remainders in every pair is $ (t \minus{} 2a)$. (every remainder is element of only one pair, and one pair has two equate elements). In every pair we have only $ 1$ element. Then maximum number of elements that we can take is $ 11*22 \plus{} 1 \plus{} 2 \equal{} 245 < 250$. ($ 11$ elements from $ 11$ pairs with inequate remainders, $ 1$ element from pair with equate remainders and $ 2$ elements that we take at the beginnig of solution)

By the way the thorem of Chaushy Davenport has a quite nice and beautiful solution by using the so called Combinatorial Nullstellensatz  :P  :rotfl:

Modulo 23, there can be at most 22 equal remainders among the 250. Let's split remainders into "good", that is, ones that do not reoccur more than 4 times among the 250 numbers, and "bad" that occur 5 and more times. Say, there are $ n$ numbers that give good remainders, then there are at least $ \frac{250 \minus{} n}{22}$ bad remainders. Let's form sets $ A_1, A_2, A_3, A_4$ from the 250 remainders as follows: split reocurring good remainders into different sets, and let each bad remainder occur in all 4 sets. This can be done in a manner such that none of the 4 sets remains empty. Then $ |A_1| \plus{} |A_2| \plus{} |A_3| \plus{} |A_4| \geq n \plus{} 4\frac{250 \minus{} n}{22} \equal{} \frac{11n \plus{} 500 \minus{} 2n}{11} \geq \frac{500}{11}$, so $ |A_1| \plus{} |A_2| \plus{} |A_3| \plus{} |A_4| \geq 46.$

Let $ f(a_1, a_2, a_3, a_4) \equal{} (a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} t)^{22} \minus{} 1$

Then $ f(a_1, a_2, a_3, a_4) \equal{} \sum_{k_1 \plus{} k_2 \plus{} k_3 \plus{} k_4 \plus{} k_5 \equal{} 22}{\frac{k_1!k_2!k_3!k_4!k_5!}{22!}{a_1}^{k_1}{a_2}^{k_2}{a_3}^{k_3}{a_4}^{k_4}t^{k_5}} \minus{} 1$ (where $ k_i$ are nonnegative integers), and in particular,  coefficient of each monomial of the form $ {a_1}^{k_1}{a_2}^{k_2}{a_3}^{k_3}{a_4}^{k_4}$ with $ k_1 \plus{} k_2 \plus{} k_3 \plus{} k_4 \equal{} 22$,  is nonzero. Choose  $ k_1, k_2, k_3, k_4$ in a way that $ k_1 < |A_1|, k_2 < |A_2|, k_3 < |A_3|, k_4 < |A_4|$ and $ k_1 \plus{} k_2 \plus{} k_3 \plus{} k_4 \equal{} 22$ (this is possible since $ |A_1| \plus{} |A_2| \plus{} |A_3| \plus{} |A_4| \geq 26$ and none of the sets is empty). Apply now Combinatorial Nullstellensatz: there should exist $ a_1 \in A_1, a_2 \in A_2, a_3 \in A_3, a_4 \in A_4$ such that $ f(a_1, a_2, a_3, a_4) \neq 0$. That, in turn, means that  $ a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} t \equal{} 0$, q.e.d.

p.s. the proof actually shows that 143 numbers is always enough. There's a gap between 126 and 143... I wonder what's the least number for which the problem statement is still valid


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

If $ A \in M_{n}(\mathbb{K})$, where $ \mathbb{K}$ denotes a field of number

assume the characteristic polynomial of $ A$ is:

$ | \lambda I\minus{}A|\equal{}(\lambda\minus{}\lambda_{1})^{l_{1}}(\lambda\minus{}\lambda_{2})^{l_{2}}...(\lambda\minus{}\lambda_{s})^{l_{s}}$

where: $ \lambda_{i} \not\equal{} \lambda_{j}$, if $ i \not\equal{} j$

$ l_{1}\plus{}l_{2}\plus{}...\plus{}l_{s}\equal{}n$

[color=red][size=150]show that:[/size][/color]

if $ k \in \mathbb{K}$, and $ k \not\equal{} 0$

then the characteristic polynomial of $ kA$ is:

$ | \lambda I\minus{}kA|\equal{}(\lambda\minus{}k \lambda_{1})^{l_{1}}(\lambda\minus{}k \lambda_{2})^{l_{2}}...(\lambda\minus{}k \lambda_{s})^{l_{s}}$

if $ k\neq 0$, $ A\mathbf{x}\equal{}\lambda \mathbf{x} \Leftrightarrow kA\mathbf{x}\equal{}k\lambda\mathbf{x}$, so multiplying a matrix by a scalar multiplies all its eigenvalues by the same scalar.

That proof assumes there are eigenvectors, which there may not be, since the matrix may not be diagonalizable. There's one eigenvector for each $ \lambda_i$ but have we guaranteed that the other copies of the same factor will come along?

It would perhaps be better to use Schur's Lemma - available here because we are given that the characteristic polynomial does split over $ \mathbb{K}.$ That is, $ A$ is similar to an upper triangular matrices, and the elements on the main diagonal of that upper triangular matrix are the eigenvalues, listed with their algebraic multiplicities. If $ P^{\minus{}1}AP\equal{}T,$ then $ P^{\minus{}1}(kA)P\equal{}kT,$ and $ kT$ has the characteristic polynomial we want.

you're right. we can also see it by simply looking at how the determinant expands:

let $ f(\lambda)$ be the characteristic polynomial of $ A$, $ g(\lambda)$ the characteristic polynomial of $ kA$, and $ n$ the dimension of $ A$.

note that if $ c\lambda^j$ is the $ j$-degree term of $ f(\lambda)$, then the $ j$-degree term of $ g(\lambda)$ is $ ck^{n\minus{}j}\lambda^j\equal{}ck^n\left( \frac{\lambda}{k}\right)^j$, so $ g(\lambda)\equal{}k^nf\left( \frac{\lambda}{k}\right)$, from which the result is clear.

$ \det( \lambda I \minus{} kA) \equal{} k^{\dim A}\det((\lambda/k)I \minus{} A) \equal{} k^n \prod_{i \equal{} 1}^n (\lambda/k \minus{} \lambda_i) \equal{} \prod_{i \equal{} 1}^n (\lambda \minus{} k \lambda_i)$.

[quote="Kent Merryfield"]That proof assumes there are eigenvectors, which there may not be, since the matrix may not be diagonalizable. There's one eigenvector for each $ \lambda_i$ but have we guaranteed that the other copies of the same factor will come along?
[/quote]

we know , $ A$ can be diagonalizable if and only if there are $ n\minus{}$ linear independent eigenvectors. In this statement, [color=red]do you mean[/color] "If $ A$ can be
 diagonalizable, then the proof give above is correct"

[quote="Kent Merryfield"]It would perhaps be better to use Schur's Lemma - available here because we are given that the characteristic polynomial does split over $ \mathbb{K}.$ That is, $ A$ is similar to an upper triangular matrices, and the elements on the main diagonal of that upper triangular matrix are the eigenvalues, listed with their algebraic multiplicities. If $ P^{ \minus{} 1}AP \equal{} T,$ then $ P^{ \minus{} 1}(kA)P \equal{} kT,$ and $ kT$ has the characteristic polynomial we want.[/quote]

Schur's Lemma:
Assume $ A \in M_{n}(\mathbb{C})$ and for $ \forall$ $ G \in M_{n}(\mathbb{C})$
if $ AG\equal{}GA$, then we have $ A\equal{}kI_{n}$, where $ k \in \mathbb{C}$

[color=red]Do you mean this Lemma?[/color]

[quote="Kent Merryfield"]the characteristic polynomial does split over $ \mathbb{K}.$[/quote]

Does [color=red]"split"[/color] means the polynomial has the form: $ (\lambda\minus{}\lambda_{1})^{l_{1}}(\lambda\minus{}\lambda_{2})^{l_{2}}...(\lambda\minus{}\lambda_{s})^{l_{s}}$ ?

then you say: [color=red]It is similar to an upper triangular matrices[/color]

How to get this conclusion?

For my ignorance.......

Thank you very much :)

[quote]Do you mean this Lemma?[/quote]

No. Schur's lemma is also the name given to the following theorem: any matrix $ A \in M_{n} (\mathbb{K})$ is (unitarily) similar to an upper triangular matrix, as long as the characteristic polynomial of $ A$ can be factored in $ \mathbb{K}$.

[quote]Does "split" means the polynomial has the form:[/quote]

Yes, with $ \lambda_i \in \mathbb{K}$.


Is it safe to discuss the problems at this time?

No. You should have looked at the announcement.

It's up at http://www.unl.edu/amc/a-activities/a7-problems/putnamindex.shtml


linear algebra
matrix
vector
number theory
least common multiple

Given distinct arbitrary real numbers $ a_1, \dots, a_n$, prove or disprove the following:
There exist distinct integers $ b_1, \dots, b_n$ such that

$ \text{rank}\left(\begin{bmatrix}a_1 & b_1 & 1\\\vdots & \vdots & \vdots\\a_n & b_n & 1\end{bmatrix}\right)\equal{}2.$
Moreover, assuming that a sequence $ b_1, \dots, b_n$ exists, can you show that infinitely many such sequences exist for a given sequence of $ a_i$'s?

P.S. Do you think this classic problem is suitable for a second course on linear algebra? :)

Obviously, if $ n\equal{}1$ holds, no such sequence $ b_n$ exists. If $ n>1$, we always have infinitely many solutions. If the sequence $ a_i$ is constant, choose $ b_1\not\equal{}0$ all other $ b_i\equal{}0$. If the sequence $ a_i$ is non-constant, choose $ b_i\equal{}\beta a_i$ with arbitrary $ \beta$

I do not know what your second course is about. However, I would not pose this question in a math degree course, since its solution is a one-liner.

[quote="hjbrasch"]choose $ b_i \equal{} \beta a_i$ with arbitrary $ \beta$[/quote]
But the $ b_i$ are integers ;)

[quote="hjbrasch"]Obviously, if $ n \equal{} 1$ holds, no such sequence $ b_n$ exists. If $ n > 1$, we always have infinitely many solutions. If the sequence $ a_i$ is constant, choose $ b_1\not \equal{} 0$ all other $ b_i \equal{} 0$. If the sequence $ a_i$ is non-constant, choose $ b_i \equal{} \beta a_i$ with arbitrary $ \beta$

I do not know what your second course is about. However, I would not pose this question in a math degree course, since its solution is a one-liner.[/quote]
I am afraid your one-line solution is incorrect hjbrasch! Even if your solution were correct, I wouldn't agree with your opinion about the value of questions with one-line answers. At least in my math classes, I've seen many students without the ability of answering simple questions.
P. S. By second course in linear algebra I meant a course which is intended for students who have already studied some basics of linear algebra such as systems of linear equations, vector spaces, determinants, and eigenvalues in an elementary level.

You can see :$ \left|\begin{array}{cc}a_i & 1 \\
a_j & 1\end{array}\right| \equal{} a_i \minus{} a_j \neq 0,\forall i\neq j$ then $ \text{rank}\left(\begin{bmatrix}a_1 & b_1 & 1 \\
\vdots & \vdots & \vdots \\
a_n & b_n & 1\end{bmatrix}\right)\geq 2$
To $ \text{rank}\left(\begin{bmatrix}a_1 & b_1 & 1 \\
\vdots & \vdots & \vdots \\
a_n & b_n & 1\end{bmatrix}\right) \equal{} 2$
need $ \forall i \neq j$ ,$ \left|\begin{array}{cc}a_i & b_i \\
a_j & b_j\end{array}\right| \equal{} a_ib_j \minus{} b_ia_j \equal{} 0$,
then $ \forall i \neq j$ : $ a_ib_j \equal{} b_ia_j$

Okay, now with the requirement of distinct integers $ b_i$. I wonder, how I overlooked that, and I am sorry.

For $ n\equal{}1$ no solution exists. For $ n>1$ and $ a_i$ constant infinitely many solutions exist, just by choosing any sequence of distinct integers $ b_i$. For $ n>1$ and $ a_i$ non-constant, i.e. $ \bar{a}\not\equal{}\bar{0}$ we must have $ \bar{b}\equal{}\beta\bar{a}$ for some $ \beta\not\equal{}0$ to achieve $ rank \equal{} 2$. Apparently, we have either zero or infinitely many solutions, since if $ \beta\bar{a}$ is a solution, $ \alpha\beta\bar{a}$ will be a valid solution for any $ \alpha\in\mathbb{Z}^*$. Now we claim that we have no solution, iff either $ a_i$ are not distinct or $ \exists i,j: a_j\not\equal{}0,\frac{a_i}{a_j}\not\in\mathbb{Q}$. As for "$ \Longleftarrow$" note, that $ a_i$ not distinct implies $ b_i$ not distinct. Now for the second part observe the identity $ \frac{a_i}{a_j}\equal{}\frac{b_i}{b_j}\not\in\mathbb{Q}$ and $ \bar{b}$ cannot be a solution. For "$ \Longrightarrow$" assume the contrary, i.e. we have no solution and $ a_i$ distinct and $ \forall i,$ with $ a_j\not\equal{}0: \frac{a_i}{a_j}\in\mathbb{Q}$. Choose $ a_k\not\equal{}0$, then $ \forall i\not\equal{}k: a_i\equal{}\frac{q_i}{p_i}a_k$ with $ q_i,p_i\in\mathbb{Z}$ and $ p_i\not\equal{}0$. Set $ \beta\equal{}\frac{lcm\{p_i|i\not\equal{}k\}}{a_k}$ and $ \bar{b}\equal{}\beta\bar{a}\in\mathbb{Z}^n$ with distinct coefficients and we are finished by contradiction.

P.S. if you answer your own questions positively, do not ask them ! or maybe it is just that you take pleasure to argue the opposite answer  :wink:

[quote]For $ n\equal{}1$ no solution exists. For $ n>1$  and  $ a_i$ constant infinitely many solutions exist, just by choosing any sequence of distinct integers $ b_i$.[/quote]

What do you mean by "constant"? If you mean that $ \forall i,j, a_i \equal{} a_j$, then it is not the subject of the question, since it is declared that "Given [b]distinct[/b] arbitrary real numbers ..."

[quote]then it is not the subject of the question, since it is declared that "Given distinct arbitrary real numbers ..."[/quote]

[b]fesghelbahal[/b]  :?: I hope your recognize that $ a_i$ distinct is a special case of my treatment, making your comment redundant

[quote]we must have $ \bar{b}\equal{}\beta\bar{a}$[/quote]
Well, It is sufficient to choose $ \bar{b}\equal{}\beta\bar{a}$, but it is not necessary, since one can define  each $ b_i\equal{}a_i\plus{}1$, and $ a_i$ are distinct (or precisely speaking, there are $ i \ne j$ such that $ a_i \ne a_j$).

[quote]we have no solution, iff either $ a_i$ are not distinct or...[/quote]
There is also a solution when $ a$'s are not distinct: choose $ b$'s distinct!

You are right, I am very sloppy and have written a lot of nonsense and I seem to have confused everybody. Let $ \bar{e}\equal{}(1,\ldots,1)$ and $ n>1$.

If $ a_i$ are distinct, then the matrix has already rank $ \geq 2$. Therefore, the rank will equal $ 2$ with a valid $ \bar{b}$- vector, iff $ \exists\beta,\gamma\in\mathbb{R}: \bar{b}\equal{}\beta\bar{a}\plus{}\gamma\bar{e}$ and $ \beta\not\equal{}0$ (because the $ b_i$ have to be distinct) and $ \bar{b}\in\mathbb{Z}^n$. Clearly, a valid solution $ \bar{b}$ implies infinitely many valid solutions given by $ \mathbb{Z}\cdot\bar{b}$. The condition above is equivalent to $ \bar{a}\in\mathbb{R}\cdot\mathbb{Z}^n\plus{}\mathbb{R}\cdot\bar{e}$, because the representation $ \bar{a}\equal{}r_1\bar{z}\plus{}r_2\bar{e}$ with $ a_i$ distinct requires $ r_1\not\equal{}0$ and we can set $ \beta\equal{}r_1^{\minus{}1},\gamma\equal{}\minus{}r_2r_1^{\minus{}1}$


what operation can I do? choose an arbitrary point on a line, devide a segment in a rational ratioo...?

You just have to notice that the side of the big square must be $\sqrt{5}$, and that you can get the segment of the length ${\sqrt{5}\over 2}$ if you join a vertex and the midpoint of a non-adjacent side in a small square.

Here's the diagram:

I already knew this problem.
This is the easiest solution with only 4 cuts :D


geometry
geometric transformation
function
number theory
research

Can anyone help me in getting translated in English the  research paper " ASYMPTOTISCHE ABSCHATZUNGEN IN SHANNONS INFORMATIONSTHEORIE" by V. Strassen which is  in German.

Thanks.

Get a German dictionary-English dictionary or find someone who speaks German.  If you're at a university, the latter shouldn't be too difficult.

If you try and do it yourself, using a German to English mathematical dictionary (such as the one [url=http://www.uni-bonn.de/~manfear/mathdict.php?fromLang=de&toLang=en]here[/url]) might also be useful.  Fortunately, a good portion of mathematical German is filled with cognates (e.g. Asymptotische, Informationstheorie) so is not too hard to guess even without the dictionary.

Thank you very much JRav and kevinatcausa for your help; especially the math. dictionary suggested by kevinatcausa is a very useful one.  Actually I am from India and here is no German student in our institute.

[quote="kevinatcausa"]Fortunately, a good portion of mathematical German is filled with cognates (e.g. Asymptotische, Informationstheorie) so is not too hard to guess even without the dictionary.[/quote]
While a fair portion of words are cognates, German can quickly get difficult to read.  I know a little German and while I could probably read a little bit, I would definitely need some means of translation to help me.

I guess it depends somewhat on the material in question.  The one time I tried to translate a German paper it was something by Erdos in elementary number theory, and a combination of the cognates, a dictionary, and some half-remembered German from High School was enough to make it not too painful.  However, that may have been as much a function of the paper's subject matter and style (heavy on the equations and not too difficult to follow) as anything.

good luck with reading papers in german. claims that this is easy to do fluidly ("because mathematical german only has about 300 words" etc) are completely specious, as there are a lot of barriers to misunderstanding. For one thing, papers certainly do not limit themselves to a vocabulary of 300 words; even if that were true, the true difficulty with reading german is the grammer, where you have to recognize what is modifying what (i.e. verbs dativ or akkusitiv), and mingling this of course changes the meaning of the sentence. If you are reading a new area of math, this makes things really intractable.

As a solution, I would suggest getting a firm grasp of the grammer, first in a non math context (read trashy novels, newspapers such as ftd.de or sueedeutsche.de), and then try to understand the papers.

Again, my claim is not that reading german is not doable..you can read a lot of things with just a few week's worth of knowledge..but rather that it is quite easy to read things and mangle the meaning. I guess thats why people translate things into english.

Rather than using a goofy "german math" dictionary i would dict.leo.org which includes has math terms.


[quote="Virgil Nicula"]Ascertain the general term of the sequence $a_{1}=0\ ,\ a_{n+1}=a_{n}\sqrt 2+\sqrt{2+a^{2}_{n}}\ ,\ n\in N^{*}\ .$[hide="The partial answer :"]$a_{n+2}=2a_{n+1}\sqrt 2-a_{n}\ .$[/hide][/quote]
I think this really becomes trivial after the partial answer given above.

The roots of the characteristic equation associated with the recurrence are $\sqrt2 \pm 1$. So, $a_{n}= A(\sqrt 2+1)^{n}+B(\sqrt 2-1)^{n}.$ Then, it is just a matter of finding the reals $A$ and $B$ using the conditions $a_{1}= 0$ and $a_{2}= 4.$

After some algebraic manipulation, I got $A = 2(\sqrt 2-1)$ and $B =-2(\sqrt2+1 ).$

Thus, $a_{n}= 2[(\sqrt2+1)^{n-1}-(\sqrt2-1)^{n-1}]$ for all $n\ge 1.$

Now, I wonder if this has anything to do with Calculus!

We use [i]a classical method of solving recurrences[/i]:

Given something of the form $f \left( a_{n+1}, a_{n}\right) = g(a_{n})$ (in this case $a_{n+1}-a_{n}\sqrt 2 = \sqrt{2+a_{n}^{2}}$), we try to apply various methods (like squaring etc.) such that we get something of the form $h \left( a_{n+1}, a_{n}\right) = g \left( a_{n+1}\right)$. Then use the fact that $h \left( a_{n+1}, a_{n}\right) = f \left( a_{n+2}, a_{n+1}\right)$ (it's true since both are equal to $g(a_{n})$) to get some recurrence (if you're lucky, it's something you already know how to handle :) ).
In this case we are lucky and we obtain Virgil Nicula's partial answer.

$a_{n+1}= a_{n}\sqrt 2+\sqrt{2+a^{2}_{n}}$

$\, \Longrightarrow \, \, a_{n+1}-a_{n}\sqrt 2 = \sqrt{2+a^{2}_{n}}$

$\, \Longrightarrow \, \, a_{n+1}^{2}+2 a_{n}^{2}-2 a_{n}a_{n+1}\sqrt 2 = 2+a_{n}^{2}$

$\, \Longrightarrow \, \, a_{n+1}^{2}+a_{n}^{2}-2 a_{n}a_{n+1}\sqrt 2 = 2$

$\, \Longrightarrow \, \, 2 a_{n+1}^{2}+a_{n}^{2}-2 a_{n}a_{n+1}\sqrt 2 = 2+a_{n+1}^{2}$

$\, \Longrightarrow \, \, \left( a_{n+1}\sqrt 2-a_{n}\right)^{2}= 2+a_{n+1}^{2}$

$\, \Longrightarrow \, \, \left| a_{n+1}\sqrt 2-a_{n}\right| = \sqrt{2+a_{n+1}^{2}}$

Now just check the sign of $a_{n+1}\sqrt 2-a_{n}$ (it's $>0$), and then equate it with $a_{n+2}-a_{n+1}\sqrt 2$ (which is also equal to $\sqrt{2+a^{2}_{n+1}}$).
Thus, $a_{n+1}\sqrt 2-a_{n}= a_{n+2}-a_{n+1}\sqrt 2$, which we know how to handle.

And, yeah, this has nothing to do with calculus. Algebra or Pre-Olympiad would have been more appropriate.


$ {x,y,z \in R\plus{}}$ and $ {x^2\plus{}y^2\plus{}z^2\equal{}1}$; then find the mininum value of $ {\frac{yz}{x}\plus{}\frac{xz}{y}\plus{}\frac{xy}{z}}$.
Looking forward toyour nice solutions ,Dear mathlinkers! :)

hello, are you sure that a minimum exists?
Sonnhard.

let $ A\equal{}\sum \frac{xy}{z}$
then $ A^2\equal{}\sum \frac{x^2y^2}{z^2}\plus{}2(\sum x^2)\equal{}\frac{1}{2}*\sum (\frac{x^2y^2}{z^2}\plus{}\frac{y^2z^2}{x^2})\plus{}2\ge \sum x^2\plus{}2\equal{}3$
so A min= $ \sqrt{3}$

hello, that's correct, the searched minimum is $ \sqrt{3}$.
Thank you for your nice solution.
Sonnhard.

Using The lemma $ (a \plus{} b \plus{} c)^2 \ge 3(ab \plus{} bc \plus{} ca)$

we have $ (\frac {xy}{z} \plus{} \frac {yz}{x} \plus{} \frac {xz}{y})^2 \ge 3(x^2 \plus{} y^2 \plus{} z^2) \equal{} 3$

and so $ \min \equal{} \sqrt {3}$

tuandokim,Dr Sonnhard Graubner,Evariste-Galois,nice solutions that's right answer ,very easy problem ,I think :wink:

Yes, it is easy and famous problem. I think the first problem is Russian MO, afterwards French MO,for which I posted the solution in MathLinks.


Se considera unghiul XOY de masura 2a, 0<a<90
O sfera variabila in spatiu S(C;r) este tangenta la laturile unghiului dat in punctele A si B. Se noteaza cu H proiectia centrului C al sferei pe planul unghiului dat
a) Sa se arate ca (OH sin a)2 +(CH)2= (r)2
   (fiecare 2 dupa fiecare paranteza vrea sa fie la puterea a doua)
b) Sa se determine locul geometric al centrelor sferelor care au raza constanta r si care sunt tangente laturilor unghiului dat.

P.S. Multumesc anticipat celui care ma lumineaza si pe mine

n`am mai facut o problema de geometrie in spatiu de multa vreme, deci e foarte probabil sa ma insel...

[hide="solutie"]deoarece laturile unghiului sunt tangente la sfera, vor fi tangente si la un anumit cerc din sfera (si anume cercul determinat de sfera si planul unghiului), deci $H$ se afla pe bisectoarea unghiului $XOY$.
punctul a) este de fapt teorema lui pitagora aplicata in triunghiul $CHQ$, unde $Q$ este unul din punctele de tangenta :
$CQ=r$ si $HQ=OH \sin a$.
cum $CH$ perpendiculara pe planul $(XOY)$, rezulta ca planul $(CHO)$ este perpendicular pe planul $(XOY)$, deci $C$ se afla intr`un plan determinat.
in acest plan, fie axa $Ox$ bisectoarea unghiului dat. fie $C$ de coordonate $(h,k)$. atunci $H$ are coordonatele $(h,0)$.
relatia de la punctul a) devine :
$h^{2}\sin^{2}a+k^{2}=r^{2}$, sau
$\frac{h^{2}}{r^{2}\csc^{2}a}+\frac{k^{2}}{r^{2}}=1$, adica o elipsa centrata in origine, cu raza mare (asa`i zice?) egala cu $r\csc a$ si raza mica $r$.
din aceasta elipsa se "decupeaza" partea cu $h\leq 0$, adica ramane o "jumatate" de elipsa "deschisa" (adica fara capete).
deci locul geometric este aceasta "jumatate" de elipsa, care se afla in planul perpendicular pe $(XOY)$.[/hide]


conics
ellipse
ratio
geometry proposed
geometry

Let a ellipse of foci $F$ and $F'$ and let M any point of the ellipse. Traces for M two straight lines $MF$ and $MF'$ that they intercept the ellipse in $P$ and $P'$, respectively. Prove that the sum $\frac{MF}{FP}+\frac{MF'}{F'P'}$ is constant.

I got a long (really long) solution. Does anybody have another solution?

Let FF' meet the ellipse $\mathcal E$ at A, A', the points A, F, F', A' following in this order. Using a parallel projection, project the ellipse to a circle $\mathcal C$ centered at the midpoint O of FF', with diameter AA' and radius $r = OA = OA' = \frac{AA'}{2}$. $M, P, P' \in \mathcal E$ are carried into $M^{*}, P^{*}, P'^{*}\in \mathcal C$, such that $MM^{*}, PP^{*}, P'P'^{*}\perp FF' \equiv AA'.$ Parallel projection preserves segment ratios on a line, $\frac{MF}{FP}+\frac{MF'}{F'P'}= \frac{M^{*}F}{FP^{*}}+\frac{M^{*}F'}{F'P'^{*}}.$ The points F, F' have the same power $r^{2}-OF^{2}= r^{2}-OF'^{2}= \frac{AA'^{2}-FF'^{2}}{4}$ to the circle $\mathcal C,$

$M^{*}F \cdot FP^{*}= AF \cdot FA' = \frac{AA'^{2}-FF'^{2}}{4}= A'F' \cdot F'A = M^{*}F' \cdot F'P'^{*}$

Substituting for $FP^{*},\ F'P'^{*}$ in the sum of the ratios,

$\frac{M^{*}F}{FP^{*}}+\frac{M^{*}F'}{F'P'^{*}}= 4\ \frac{M^{*}F^{2}+M^{*}F'^{2}}{AA'^{2}-FF'^{2}}.$

$OM^{*}= r$ is the M-median of the triangle $\triangle M^{*}FF',$ hence

$r^{2}= OM^{*2}= \frac{2M^{*}F^{2}+2M^{*}F'^{2}-FF'^{2}}{4}$

$M^{*}F^{2}+M^{*}F'^{2}= \frac{4r^{2}+FF'^{2}}{2}= \frac{AA'^{2}+FF'^{2}}{2}$

$\frac{MF}{FP}+\frac{MF'}{F'P'}= \frac{M^{*}F}{FP^{*}}+\frac{M^{*}F'}{F'P'^{*}}= 2\ \frac{AA'^{2}+FF'^{2}}{AA'^{2}-FF'^{2}}$

which is constant regardless of the position of $M \in \mathcal E.$


geometry
cyclic quadrilateral
projective geometry
angle bisector
Parallel Lines

Dear Mathlinkers,
1. let Ta1 the tangent at A1
2. According to Pascal theorem applid to CBA Ta1 C1C, PM is parallel to BC
3. Same idea, Mq is parallel to BC
and we are done
Sincerely
Jean-Louis

I have done this using cyclic quadrilateral property. By the way guys, how many questions do you think one needs to do or the minimum marks required to get through the INMO?

Use Pascal's Theorem for hexagon $ ABB_1A_1C_1C$ to obtain $ P, M, Q$ collinear.
Next, $ \angle PAM \equal{} \angle PC_1A \equal{} \dfrac{A}{2}$. Hence, $ MAC_1P$ is cyclic.
Then $ \angle PMC_1 \equal{} \angle PAC_1 \equal{} \angle BCC_1$. So, $ PM \| BC$  and we are done.

$ \angle ACC_1 \equal{} \angle AA_1C_1 \equal{} \alpha , \angle ABB_1 \equal{} \angle AA_1B_1 \equal{} \beta$

$ \angle BCC_1 \equal{} \angle BA_1C_1 \equal{} \theta , \angle CBB_1 \equal{} \angle CA_1B_1 \equal{} \gamma$

$ \frac {|AP|}{|BP|} \equal{} \frac {|AA_1|}{|BA_1|}. \frac {sin\alpha}{sin\theta}$ and $ \frac {|AQ|}{|CQ|} \equal{} \frac {|AA_1|}{|CA_1|}. \frac {sin\beta}{sin\gamma}$

By Trigonometric Ceva in triangle $ ABC$ according to point $ M$;

$ \frac {sin\alpha}{sin\theta} \equal{} \frac {sin\beta}{sin\gamma}$

$ |BA_1| \equal{} |CA_1| \Longrightarrow$

$ \frac {|AP|}{|BP|} \equal{} \frac {|AQ|}{|CQ|} \Longrightarrow$

$ PQ \parallel BC$.

[hide=" Solution 1 (own)"] As given $ M$ lies on angle bisector of $ \angle A$, $ \angle BAA' \equal{} \angle BB'A' \equal{} \angle CAA' \equal{} \angle CC'A' \equal{} \frac {\angle A}{2}$
Also $ \angle PC'M \equal{} \angle PAM \equal{} \frac {\angle A}{2}$ so $ PMAC'$ is a cyclic quadrilateral.
And thus $ \angle AC'M \equal{} \angle APM$. Because $ \angle AC'M \equal{} \angle AC'C \equal{} \angle ABC$
$ \equal{} > \angle APM \equal{} \angle ABC$ and hence $ PM \parallel BC$

Using similar arguments we can easily prove that $ MQ \parallel BC$
As $ M$ is the common point in the two segments $ PM$ and $ MQ$ both of which are parallel to $ BC$, thus $ M$ lies on $ PQ$ and $ PQ \parallel BC$.
Hence Proved. :) [/hide]

[hide=" Solution 2 (Due to Pardesi)"] Clearly $ \angle MC'C \equal{} \angle BB'M \equal{} \frac {\angle A}{2}$.Join $ PM,QM$.Observe that quadrilateral $ AC'PM$,$ AB'PM$ and $ AC'MB'$ are cyclic.The first because of the previous angle equality the last one because of $ \Gamma$.Thus $ \angle AC'M \equal{} \angle AMQ$,and $ \angle AMP \equal{} \angle AB'M$.Thus $ P,M,Q$ are collinear and since $ \angle APM \equal{} \angle AC'M \equal{} \angle ABC$ we have $ PQ \parallel BC$ [/hide]

Excuse me now.
How can you post my solution as yours? It's shameful.
[quote]Two similar solutions...

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

Solution 1 (own)
As given  lies on angle bisector of ,
Also  so  is a cyclic quadrilateral.
And thus . Because
 and hence

Using similar arguments we can easily prove that
As  is the common point in the two segments  and  both of which are parallel to , thus  lies on  and .
Hence Proved.
[/quote]

dear viewers solutions to the given set are updated by HBCSE

http://olympiads.hbcse.tifr.res.in/uploads/inmo-2010

This problem can be solved by simple angle chasing :-D
The flow goes like this:
(1) First prove that  $AMPC_{1}$  and  $AMQB_{1}$  are cyclic and hence evaluate $\angle AMP$ and $\angle AMQ$.
(2) Then use (1) to show that $P, M, Q$  are collinear points.
(3) Finally by (1) again, show that  $PM || BC$  and hence conclude that  $PQ || BC$.

[quote=Wizard_32][quote=tenplusten]If $M$ is incenter. This problem is Azerbaijan JBMO 2016 TST.[/quote]

That means that INMO is harder than Azerbaijan's TST  :what?:[/quote]

But keep in mind that it was JBMO TST.

[quote=Wizard_32][quote=tenplusten]If $M$ is incenter. This problem is Azerbaijan JBMO 2016 TST.[/quote]

That means that INMO is harder than Azerbaijan's TST  :what?:[/quote]

Shouldn't it be? Student population of India is more than 300 million (while population of Azerbaijan is 10 million) and there must be hard problems to choose the best one  ;)

INMO [b]was[/b] really easy ;
$\angle BAM = \angle MAC$
Using cyclic quadrilaterals $ACA_1C$ and $ABA_1B$, $\angle BAM = \angle MAC = \angle MB_1Q = \angle MC_1P$
Also, $\angle PAC_1 = \angle PMC_1 = \angle BCC_1 \Longrightarrow PM \parallel BC$
         $\angle QMB_1 = \angle QAB_1 = \angle CBB_1 \Longrightarrow QM \parallel BC$
Thus, $PQ \parallel BC$

[quote=INMO 2010 P1]Let $ ABC$ be a triangle with circum-circle $ \Gamma$. Let $ M$ be a point in the interior of triangle $ ABC$ which is also on the bisector of $ \angle A$. Let $ AM, BM, CM$ meet $ \Gamma$ in $ A_{1}, B_{1}, C_{1}$ respectively. Suppose $ P$ is the point of intersection of $ A_{1}C_{1}$ with $ AB$; and $ Q$ is the point of intersection of $ A_{1}B_{1}$ with $ AC$. Prove that $ PQ$ is parallel to $ BC$.[/quote]
[b]One line proof[/b]
Pascal's Theorem on $C_1A_1B_1BAC$ gives, $P,M,Q$ are collinear and hence, $C_1PMA$ and $MQB_1A$ are concyclic with radical axis $AM\implies C_1B_1QP$ cyclic and thus by Reim's Theorem $PQ\| BC$.

easy.
Applying Pascal's theorem on the self-intersecting hexagon $ABB_1A_1C_1C$ we have $P,M,Q$ are collinear.
Also $$\angle BB_1Q=\angle BB_1A_1=\angle BAA_1=\angle A_1AC=\angle MAQ$$
$$\implies AB_1QM \text{is cyclic}$$.
By converse of Reim's theorem we have $MQ||BC \implies PQ || BC$.
EDIT: Umm.. my solution is actually exactly same as that of @above. Ok nvm it's a storage.


algebra
polynomial
algorithm
vector
linear algebra
matrix
real analysis

Hi,

I'm implementing the algorithm for finding roots of polynomial P(x) with the usage of Bairstow method. Inside Bairstow I use Newtons method. I've already implemented everything however it's not working properly so if someone could hgelp me find the reason I would be very gratefull. (The problem I think is inside Newtons method)

The bairstow method is described here: http://www.answers.com/topic/bairstow-s-method?cat=technology

Here is what I do:

I take some polynomial P(x) of degree at least 3, take some initial quess u0, v0
1) T(x) = x^2 - u0x - v0
    P(x) / T(x) = Q(x) +R(x)
2) someone clever has once proved that
Q(x) = b0x^(n-2)+b1x^(n-3)+...+bn-2
R(x) = bn-1(x-u0)+bn

knowing this I calculate vactor b
next I also know that
cj=bj+p*cj-1+q*cj-2 & c-2=c-1=0 & j=0,...,n-1
so I build  vector c
//those vectors are equivalent to c&d on web page to which I gave link above (c=b, d=c)
3) I use Newton method to calculate proper u & v.
4) I apply new u,v values to T(x) and divide P(x) by T(x)
   to obtain Q(x) & R(x). This is how I have 2 roots from T(x)=0 and I calculate next roots
5) P(x) = Q(x)
    and I repeat steps from 1) to 4) till Q(x) is of at lest 2nd degree
-------------
Newton
1)   x(k+1)[0] = new_u = x(k)[0] – p[0]
      x(k+1)[1] = new_v = x(k)[1] – p[1]

where
x(k)[0] = u0  x(k)[1] = v0

p[0] = F[0] * Fprime^(-1)[0][0] + F[1] * Fprime^(-1)[1][0]
p[1] = F[0] * Fprime^(-1)[1][0] + F[1] * Fprime^(-1)[1][1]

double denom = 1/(c[n-2])^2 - c[n-1]*c[n-3])

Fprime^(-1) [0][0] = c [n-2]/denom    Fprime^(-1) [1][0] = -c[n-1]/denom
Fprime^(-1) [0][1] = -c[n-3]/denom   Fprime^(-1) [1][1] = c[n-2]/denom

F[0]=b[n-1] F[1]=b[n]

I calculate new values for u &v till
k (iteration number) = 100 //in case method diverges

or

norm = p[0]^2 + p[1]^2 // vector norm
norm < precision (0.5*10^-8)

The method works however it almost always diverges or I get infinity or strange numbers for new_u & new_v.
I really don't know why so please help me to find my mistake

thx,
Dominika

Check all your formulae. This one
p[0] = F[0] * Fprime^(-1)[0][0] + F[1] * Fprime^(-1)[1][0]
p[1] = F[0] * Fprime^(-1)[1][0] + F[1] * Fprime^(-1)[1][1]
definitely looks very suspicious: the standard way to multiply a matrix by a vector would be
p[0] = F[0] * Fprime^(-1)[0][0] + F[1] * Fprime^(-1)[color=red][0][1][/color]
p[1] = F[0] * Fprime^(-1)[1][0] + F[1] * Fprime^(-1)[1][1]
Most likely, you just have one or more stupid mistakes like this in your code.


Let ABCD a concyclic quadrilateral.And O is the circumcenter.Line AB and CD meets at E.Line AD and BC meets at F.
The circle which diameter is EF meets the line which passes through O and perpendicular to EF at G and H.(OH<OG)
The line passes through H and perpendicular to GH meets the circumference of O at I and J.
Prove that GI and GJ are tangents to O.

First, I will rename the points as shown on the figure
Denote H is the intersection of AC and BD
It is well-known that H is the orthocenter of ⊿OEF => H,N,D,A   M,O,C,B   L,E,D,A  are concyclic respectively
From the above known result ,we have that FO*MO=FO^2-FM*FO=FO^2-(FO^2-R^2)=R^2

On the other hand , ∠EMF=∠ENF=90 => M,N lie on the ⊙O’ => MO*FO=OK*OG=R^2 =>
∠OIG=∠OJG=90
DONE!

A property of orthogonal circles, $ (EF)$ and $ (O)$ are orthogonal and so we can rewrite it as

Let $ (O_1),(O_2)$ be two orthogonal circles, and $ P$ is a point on $ (O_2)$, $ PO_1\cap (O_2) \equal{} M$, draw through $ M$ perpendicular of $ PO_1$ intersect $ (O_1)$ at $ K,L$ then $ PK,PL$ are tangents of $ (O_1)$.

It can be solved easily by harmonic property of two orthogonal circles.

Actually it has an extension as fllowing

Let $ (O_1),(O_2)$ be two orthogonal circles, and $ P$ is a point on $ (O_2)$, draw two tranversal $ PMN$ and $ PEF$ ($ M,N,E,F\in (O_1)$, this means $ P,M,N$ and $ P,E,F$ are collinear). Let $ ME\cap NF \equal{} K$, $ PO_1\cap (O_2) \equal{} L$ then $ PL\perp LK$.

[quote="mathson"][quote="plane geometry"]
It is well-known that H is the orthocenter of ⊿OEF
[/quote]

I did not understand this point clearly!! Can you explain it?[/quote]
That is Brokard's theorem.
Hints:EH is a polar of F wrt (O).=>$ EH \perp OF$.


Find all solutions in positive integers to $a^A = b^B = c^C = 1990^{1990}abc$, where$ A = b^c, B = c^a, C = a^b. $


function
algebra
domain
topology
real analysis
real analysis unsolved

Can somebody construct a continuous injective function $ f: \mathbb{R}^2 \to \mathbb{R}^2$ such that $ f^{\minus{}1}$ is discontinuous.Thank you

I honestly tried to construct such $ f$, but [url=http://en.wikipedia.org/wiki/Invariance_of_domain]Brouwer's invariance of domain theorem[/url] would not let me.  :(

Oh,I can't believe that...I see some theorem about the $ f^{ \minus{} 1}$ is continuous using particular condition.So I guess the general case will be wrong.But I am wrong finally.Thank you,Mlok.That's a wonderful theorem you offered(I use darij's way to enter the page.)Can you download the proof?

It's on page 172 of Hatcher's "Algebraic topology", available [url=http://www.math.cornell.edu/~hatcher/AT/ATpage.html]here[/url].

mlok   I looked p 172 there is nothing related to zhaobin problem

You may have a different version of the book, which is now in its 7th printing. Look for "invariance of domain" in the index. Also, page 172 of the book is page 181 of the pdf file.

Thanks,mlok.I find the proof in the book.But I should read more to understand the proof :blush:


geometry
number theory
least common multiple
greatest common divisor
analytic geometry
rectangle
ratio

Hello everybody. I am opening up a new game (this doesn't mean I'm not still hosting any others). There will be six players. Each round will consist of five math problems. The players will work together to answer the question. However the longer they take to answer the question, the more money will be taken away from their salary for that problem. If the players miss three problems then they will lose all their money that they had earned in the game. If they choose to stop answering questions then the six players win all the money that they had earned.

However, how will we decide how we award the money? Well, in the last stage of the game, the six players have to agree on a way of dividing the money amongst the six of them. For example they may have to award 30% to one player 25% to another player, 20% to another player, 10% to another player, 8% to another player, and 7% to the last player. However, while they are deciding, the money they are dividing amongst them is decreasing. So they have to come to an agreement quickly, before, they don't have any money left.  :o

There will be a scoreboard for the top money earners from playing, so try to win as much as possible!!!

Now there will be a queue for helping me write the problems, and there will be another sign-up thread for playing the game. Sign-ups begin now, and are always going on, even during a game, but you can only sign-up again if you are no longer in the queue.

Writing Problems;

Playing;

spaceguy524

Oh and another thing. You can't be writing problems and playing during the same game.

I join
The message is too small. Please make the message longer before submitting.

Writing Problems;

Playing;

spaceguy524
gogators
SuperNerd123
maybach

funny if Pythag will write the problems.

Or Mewto.

Problems.

/in

get pythag to write the problems...then we won't have to worry that last step  :rotfl:

Writing Problems;
abcak

Playing;

spaceguy524
gogators
SuperNerd123
maybach
zapi2007

Do you want me to ask pythag if he'll help out with this?

We'll start once we get one more person to play.

Writing Problems;
abcak

Playing;

Okay, if anybody else wants to play in a later game they can sign up too. Sign-ups to help write problems will be going on too. But for now these are the people playing;

spaceguy524
gogators
SuperNerd123
maybach
zapi2007
bluecarneal

Just since you've all been waiting, I'll give the first question. This question is for $ \$$10. Every three hours, you will lose 25 cents. So if you don't solve the problem in five days, then you don't earn anything. At least four people must bold their answer. As soon as four people have, that answer is locked in. You are allowed to change your answer if necessary until four people have given the same answer. This is the first question of round 1. Don't worry if the problem is too easy, because they will get much harder than this. The amount of money you earn will increase by a lot too.

[size=150][b]Question 1[/b][/size]

Simplify;

$ \sqrt{5 \plus{} \sqrt{300 \plus{} \sqrt{10,000}}}$

Aright so three people are still playing. I'm gonna send prods to the others. Sorry for this incredibly long delay. I'll post the next question (worth 75);

Circle 1 circumscribes Square 1. Circle 2 is inscribed in Square 1. Square 2 is inscribed in Circle 2. Given that this pattern continues, what is the ratio of the area of the 10th circle to the area of the first circle?

bluecarneal and gogators both quit unfortunately. I don't know about maybach though.

If anybody not in this game wants to replace in then they are welcome to and if anybody in this game wants to suggest this game to somebody. That would be good.

Since 2 is probably the majority I will lock in your answer. It was correct!!! You earned 48.75. The pot total is now at 184.15.

I'll wait for maybach before I post the next question.

Sorry.  I had to prepare for Science Olympiad tryouts so I haven't been on AoPS for like a week.

I'm done with them now.  But I think King! can replace me now, if he/she wishes to.

Or he could join us, but I'd like a majority of the people playing to agree with this. What do you guys think?


superior algebra
superior algebra unsolved

Let $R$ be a ring with 1 and for some $a \in R$ we have two distinct $b,c \in R$ such that $ab=ac=1$.
Prove that there are infinetly many $x \in R$ such that $ax=1$

Use the fact that $a$ has no left inverse to show that for $n=1,2,\ldots$, the right zero divisors $a^{n-1}-ba^{n}$ are all distinct.


Find all real functions f such that is |x-y|<|y-z|, then |f(x)-f(y)|<|f(y)-f(z)|.

Isn't this rather easy? Maybe I'm doing something wrong :D

I rewrote it like this:
|x-y|<|x-z| => |f(x)-f(y)|<|f(x)-f(z)| (it's obviously the same thing.

Now let's assume f(a)=f(b), so |f(a)-f(b)|=0, so |f(a)-f(x)|>=|f(a)-f(b)| for all real x, so |a-x|>=|a-b| for all real x, so a=b, so f is injective.

We now assume there exist x<y<z s.t. f(x)<f(y)>f(z). This means that f(y)-f(x) < |f(x)-f(z)|, so if f(z)>f(x) then f(y)-f(x)<f(z)-f(x) =. f(z)>f(y), contradiction with the assumption, so f(x)> f(z).

We then have |f(y)-f(z)| < |f(x)-f(z)|, so f(y)-f(z) < f(x)-f(z), so f(y)<f(x), which is another contradiction. We get something analogous if we assume there exist x<y<z s.t. f(x)>f(y)<f(z), so f must be strictly monotonous. It's obvious that every strictly monotonous function checks out, so this is the answer:

f is a strictly monotonous function from R to R. Think it's Ok?

Something is wrong. You say that every strictly monotonous functions is a solution. Take f(x)=x^3 and y=3, z=1 and x=4 in the ennounce. You will find a contradiction.

You're right! Sorry for my self-confidence (" isn't this rather easy?" :D). When I said that evey strictly monotonous function verifies this I didn't consider the case |x-y|<|y-z| and y between x and z.

I guess all we know up until now is that f must be strictly monotonous...

P.S. You might want to check the messages you're posting, just to make sure you don't post the same thing twice (or, who knows, even 3 times! :D ).

Sorry again!

Yup, I forgot about that :)

I might have a proof for the continuity of f, but I'm not sure and I can't post it because I'm kind of in a rush to catch a train.

I think the solution is all linear functions on R

Steps to prove this include
1. f is monotonic
2. f is continuous
3. f(x+k)-f(x)=f(x)-f(x-k)
4. f is linear

Step 1 is done.

To prove continuity, we see that, for example,
 0<|f(x+(1/3^n))-f(x)|<0.5*|f(x+(1/3^(n-1)))-f(x)|<...<((0.5)^n)*|f(x+1)-f(x)|, taking limit of n tends to infinity yields the result. ( I am not very sure if this proof is okay....)

Step 3 an 4 are obvious with step 2.

I did eventually reach the conclusion that f is linear, but looks like you beat me to it :D

Very nice problem, very nice solutions. Thank you for your ideas!


$ 8a \plus{} 3a \equal{} 12a \plus{} 9$

$ 11a \equal{} 12a \plus{} 9$

$ \minus{}a \equal{} 9$

$ a \equal{} \boxed{\minus{}9}$

8 times a # plus 3 times the # is the same as 9 more than 12 times the #. find the #

8#+3#=9+12#
11#=9+12#
-#=9
#=-9[/hide]

Don't post the exact same solution.

And don't revive.


Suppose that we have a polygon with $n$ vertices $A_{1},A_{2},...,A_{n}$ and we draw all of the diameters. We colour each of the sides and diameters with  one of the colours red, blue and green. Suppose that $r_{i},b_{i}$ and $g_{i}$ are the number of red, blue and green line segments connected to $A_{i}$ for every $1\leq i\leq n$. We say a triangle is "good" if its vertices are the vertices of the polygon and all of its sides are the same colour. Find the number of "good" triangles.

[quote="pbornsztein"]I assume that you are asking for the minimal number of 'good' triangles...
I solve the problem for two colors here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=triangles&t=5820
The method may be adapted for three colors but I did not check if it gives optimal bounds.

Pierre.[/quote]
No, I want the exact number of "good" triangles. Even when you use two colours, say red and blue, it can easily be proved that the number of "good" triangles equals ${n\choose 3}-\frac{1}{2}\sum_{i=1}^{n}r_{i}b_{i}$, where $b_{i}$ obviously equals $n-r_{i}-1$. However, I would be thankful for a lower bound on the problem for three colours.

[quote="pbornsztein"]In your example, I assume that $b_{i}$ is the number of blue edges from vertex number $i$ and $r_{i}$ for the red ones, am I right?
But the equality is incorrect for three points and two blue edges and one red... I think there is a $\frac{1}{2}$ missing on the sum...

Did I miss something?

Pierre.[/quote]

Yes, you're right. I will correct it. :)


My high school is going to offer independent study diff. eq. for a couple people next year, but we still need a book. Does anyone know a good one with tests, etc.?

Thanks,
Gabor D.

Elementary Differential Equations and Boundary Value Problems by BOYCE/DIPRIMA.

The book is a good introduction to DE`s.

How about the book "Differential Equations : A Modeling Perspective by Robert L. Borrelli, Courtney S. Coleman"? Does anyone here have any experience with that book? What are you opinions of it? Is it a good first introductory book to differential equations?

Questions of this sort have generally been handled in the "Other Problem Solving Topics" forum, so I'm sending it there.


linear algebra
matrix
complex numbers
superior algebra
superior algebra solved

A,B,C matrix of Mn(R), AB=BA,BC=CB,AC=CA
Proove that T= det(A+B+C-AB-BC-CA) \geq 0
---------------------------------------------------------
Here my solution if you have a different one i 'll take it.
Commutativity of the product two by two matrix A,B,C gives
A+B+C-AB-BC-CA
= (A+jB+jC)(A+jB+jC) where j=e^(2i \pi /3)
=|A+jB+jC| since j=j* complex conjugate of j, |z|=z.z*
T=|det(A+jB+jC)| \geq 0
Mr.Omarjee
STT math-teacher

yes I have a different one.
let us first prove that det A2 + B2 is non-negative. it is quite simple using the obivous decomposition A2 + B2 =  ( A + iB )( A- iB ) = CC' where C' is the conjugate matrix of C.

now let us use this lemma to prove that det ( A2+B2+C2 - AB-BC-CA ) \geq 0. We shall write A2+B2+C2 - AB-BC-CA =  1/2 ( (A-B)2 + (B-C)2+ (C-A)2)  as a sum of squares.

First let us denote A-B=X, B-C=Y and then C-A=-X-Y. So we have 1/2 X2 +1/2 Y 2 + 1/2 (X+Y)2 = X2 + Y2 + XY = (X+1/2Y)2 + ( \sqrt 3/2 Y)2 and we are done. :D


What is the sum of the measures of the angles in a parallelogram?

sum of the angles of all quadrilaterals is 360. i actually figured out an equation that will tell you the sum of the angle of every type of polygon.
$x=s-1\cdot 180$
where s=the number of sides of the polygon.
and x is the sum of the angles.

does that work?  it tried it and it didn't work.
lets try a quadrilaterall.
$x=4-1 \cdot 180
x=3 \cdot 180
x=540$
but its supposed to be 360.
is there something wrong or am i just doing it wrong?

[quote="236factorial"]sorry it's
(s-[b]2[/b]) * 180[/quote]
oh, that makes sense now

oh and if it has x perfectly equal sides, the exterior angle is 360/s so the interior angle is 180-360/s.  So if it has 10 sides the exterior angle is 18 degrees and interior is 162 degrees.

cool, i'm learning all these cool formulas.

what about the exterior angles  :D ? how does that quantity vary for polygons of different numbers of sides?

can anyone prove the general formula that sum of interior angles with n sides is:

S = (n-2) * 180?

Think about the number of triangles you can make in such a polygon.

[hide="click here if you need some hints on how to proceed"]draw a point inside the $n$-sided polygon. then draw lines from this point to the vertices of the polygon. now what did you get? how can you use the fact that the sum of the angles of any triangle is always $180^\circ$. start off with a square then try it with a pentagon and then try to generalize it. and lastly, dont just read my hints as if it was novel! take out a pencil and paper and start drawing![/hide]

Ok, so you can make n triangles from a central point by drawing lines to the vertices.  The sum of their angles is n*180.  But you have to subtract the 360 degrees about the central point so thats n*180 - 360 = 180(n-2). :!:


LaTeX
number theory unsolved
number theory

Show that there exist infinitely many different primes $ p,q$ such that $ q|(2^{p\minus{}1}\minus{}1)$ and $ p|(2^{q\minus{}1}\minus{}1)$.

posted before. For example $ m>3$ odd and $ p|\Phi_m(2)$ and $ q|\Phi_{2m}(2)$.

[quote="mto"]If (p,q) are two prime divisor of 2^2^m +1, then  ( p, q ) is a solution. ( Remark that p,q must have the form k.2^(m+1) +1 )[/quote]
Can you prove, that there are infinetely many composite numbers of form $ 2^{2^m}\plus{}1$?
When $ r>3$ prime we can chose $ p|2^r\minus{}1\equal{}|Phi_r(2)$ and $ q|\frac{2^r\plus{}1}{3}\equal{}\Phi_{2r}(2)$. It give $ pq|2^{2r}\minus{}1$ and $ 2r|p\minus{}1, \ 2r|q\minus{}1$.

If there are only a number finite of composite 2^2^m +1, then there exist M such that 2^2^m+1 is a prime number for all m>M

In this case, let's take p=2^2^m+1 and q= 2^2^(m+1) +1

[quote="mto"]

In this case, let's take $ p = 2{}^2{}^m + 1$ and $ q = 2{}^2{}^(m + 1) + 1$[/quote]
I think the first you should study typing Latex before post message .
Now it is an open problem .
Here is my solution :
$ F_n = 2{}^{2{}^n} + 1$
$ G_n = 2{}^{2{}^n} - 1$
[u]Lemma 1 [/u] If $ p|F(n)$ then $ p\equiv 1 (\mod 2^{n + 2})$
[u] Lemma 2 [/u] $ m > n$ then $ F_n|G_n$ and $ \gcd(F_m,F_n) = 1$
Now consider problem :
Call $ p,q$ is two prime divisor of $ F_n,F_{n + 1}$
From lemma we have $ p\equiv 1 (\mod 2^{n + 2})$ and $ q\equiv 1 (\mod 2^{n + 3})$
So $ F_{n + 1}|G_{n + 2}$ and $ G_{n + 2}|2^{p - 1} - 1$ so $ F_{n+1}|2^{p-1}-1$
Because $ q|F_{n+1}$ it gives that $ q|2^{p-1}-1$
Similar $ p|2^{q-1}-1$
It mean that $ (p,q)$ satisfy condition .
Problem claim.

In my solution, i first show that if (p,q) are two different prime divisors of 2^2^m+1 , then (p,q) satisfies requirements.

From here we have two cases:

1. If there exists infinitely (p,q) chosen like that, the problem is proved

2. If not, then there exist M such that for each m>M , 2^2^m+1 is the power of a prime. It's not hard to prove that 2^2^m+1 itself must

be a prime. That leads to the choice of couple (2^2^m+1, 2^2^(m+1) +1 )

I did not use at all the result of "open problem"

About LATEX: I do elementary math just for fun, to relive the time i did math at highschool, and math is not my career. Hence, i don't want to spend time on study LATEX, which is not very tough but not really useful to an IT engineer

[quote="TTsphn"]
[u]Lemma 1 [/u] If $ p|F(n)$ then $ p\equiv 1 (\mod 2^{n \plus{} 2})$
[/quote]

Why is this?

It is an well know lemma .
For $ n\geq 2$ then $ p\equiv 1 (\mod 2^{n \plus{} 1})$
It follow that $ (\frac { \minus{} 2}{p}) \equal{} 1$
So exist an integer $ a$ such that $ a^2\equiv 2 (\mod p)$
$ \Rightarrow p|a^{2^{n \plus{} 1}} \plus{} 1$
Easy to prove that $ p\equiv 1 (\mod 2^{n \plus{} 2})$


factorial
modular arithmetic
algorithm
number theory

Hi everybody,

I have discovered something strange and I do not have any explanation.
My knowledge in theory number is basic.
I do not understand why such equality hold.
Here is the picture.
Thank you for your help.

Wilson's, same as in the other threads.  $ a! b! \equiv a! (\minus{}a)(\minus{}a\minus{}1)...(\minus{}a\minus{}b\plus{}1) \equiv (\minus{}1)^b a (a \plus{} b \minus{} 1)! \equiv (\minus{}1)^{b\plus{}1} a \equiv (\minus{}1)^{b\plus{}1} b \bmod (a \plus{} b)$.

Look, if you're actually reading the responses I've been posting, you should be very familiar with this idea by now.  [b]Do you understand why this proof works, and do you understand why this is the same as the solution to all of your other conjectures?[/b]

I understand your explanation but it is not because it is derived from Wilson or more exactly from all the properties of congruences and factorials that it not [u]new.[/u] that it is not [u]helpful [/u] that it has [u]no consequences.[/u]

[quote="t0rajir0u"]It has [b]no consequences.[/b]  Let me state Wilson's theorem in a way that should make this clear.

[b]Definition:[/b]  A [i]complete residue system[/i] $ \mod m$ is a set of $ m$ integers $ a_0, ... a_{m \minus{} 1}$ such that for every integer $ n$ there is a unique $ a_i$ such that $ n \equiv a_i \bmod m$.

[i]Examples.[/i]  $ \{ 0, 1, ... m \minus{} 1 \}$ is a complete residue system.  If $ m$ is odd, so is $ \{ \minus{} \frac {m \minus{} 1}{2}, \minus{} \frac {m \minus{} 3}{2}, ... , \minus{} 1, 0, 1, ..., \frac {m \plus{} 1}{2} \}$, the [i]balanced[/i] system.

[b]Wilson's Theorem:[/b]  Let $ (a_i)$ be a complete residue system modulo a prime $ p$.  Then $ \prod_{i \equal{} 0}^{p \minus{} 1} a_i \equiv \minus{} 1 \bmod p$.

Every idea you've posted so far consists of choosing a different residue system.  In terms of computer memory, [i]it doesn't matter what residue system you pick - you haven't made computing the factorial any faster.[/i][/quote]

Ok.

Now I want to know if we combine [u]those 2 statements (see picture below) what we can achieve[/u]?
I'm not talking about slow or fast computing. [u]I'm not looking for efficiency or any kind of use of Wilson theorem as a primilaity test.[/u]

[quote="JBL"][quote="Farenhajt"]Some extra condition might be necessary since $ 2!3!\not\equiv 2,3\pmod{5}$[/quote]
$ 2! \cdot 3! \equal{} 12 \equiv 2 \pmod 5$  :maybe:[/quote]

Idiot me. In my book, $ 2!\equal{}4$  :roll:

Sorry all.

Hi,

If you have n=a+b and n=c+d and n prime

then

a!b! mod n = a or b
c!d! mod n = c or d

Now how can you solve
a! mod n = ?
b! mod n = ?
c! mod n = ?
d! mod n? = ?

If you solve the 4 equations above then you could write all the residue system without computing it

What do you think about that?
I want your thoughts.

[quote="t0rajir0u"]Wilson's, same as in the other threads.  $ a! b! \equiv a! ( \minus{} a)( \minus{} a \minus{} 1)...( \minus{} a \minus{} b \plus{} 1) \equiv ( \minus{} 1)^b a (a \plus{} b \minus{} 1)! \equiv ( \minus{} 1)^{b \plus{} 1} a \equiv ( \minus{} 1)^{b \plus{} 1} b \bmod (a \plus{} b)$.

Look, if you're actually reading the responses I've been posting, you should be very familiar with this idea by now.  [b]Do you understand why this proof works, and do you understand why this is the same as the solution to all of your other conjectures?[/b][/quote]

As someone signal it in another forum Wilson theorem is a particular case where a=1

p= 1 + (p-1)

1! (p-1)! mod (1+(p-1)) = p-1

With this as example

1!12! mod (13) =12
2!11! mod (13) =2
3!10! mod (13) =10
4!9! mod (13) =4
5!8! mod (13) =8
6!7! mod (13) =6

we can compute easily p-1! mod p

It is just question of generalizing the formula and creating an algorithm to handle it.
The question will be done.


In a class of 50 students, 18 take Chorus, 26 take Band, and 2 take both Chorus and Band.  How many students in the class are not enrolled in either Chorus or Band?

This should probably be posted in MathCounts. Its not really a high school thing.

[hide="use PIE"]$18+26-2+x=50$
so $42+x=50$
so $x=\boxed{8}$[/hide]


Can someone tell me what a harmonic sequence is?

Please use this website's wonderful resources!

For example, the AoPS Wiki article is
[[Harmonic Series]]

[[Harmonic series]] (series, not Series).

If that doesn't help, there's always [url=http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29]Wikipedia[/url].

ok..i just have a question about harmonic series. do they all tend to $ \infty$?

No. Take, for example, $ \sum_{k \equal{} 1}^{\infty}\frac{(\minus{}1)^{k\plus{}1}}{k}$. That series tends to $ \ln{2}$.

$ \ln {2}$ is basically $ \log_{e} 2$.  Logarithms are, however, more advanced than what mathcounts has


The notation $ a\equiv b\pmod{n}$, where $ n$ is a positive number, means that $ (a\minus{}b)$ is a multiple of $ n$. What is the smallest positive integer $ x$ such that $ 3x\equiv4\pmod5$?


You have pentagon ABCDE and <BAC=<CAD=<DAE and that <ABC=<ADE=<ACD. Show that the intersection point of BD and CE intersects the midpoint of CD.

I don't really remember the source since a friend gave it to me lol but I think its a G3 of some year

? The intersection point of BD and CE is on the interior of the pentagon, and the midpoint of CD is on the perimeter.  They can't be concurrent.


calculus
derivative
function
real analysis
real analysis unsolved

Can anyone prove the following?

A differentiable real-valued function on $R$ with bounded derivative is uniformly continuous.

What does the Mean Value Theorem tell you about $|f(b)-f(a)|?$

Thanks Kent, I actually got the proof right after I wrote it.

$\frac{f(x)-f(y)}{x-y}= f'(x_{0})$ if $x_{0}\in (x,y), x<y$. So we have $|f(x)-f(y)| \le M|x-y|$ and simply let $M = \frac{\epsilon}{\delta}$.

This is an outline.

That outline went (bounded derivative) implies (Lipschitz) and then (Lipschitz) implies (uniformly continuous). And, of course, (uniformly continuous) implies (continuous).

[Lipschitz means that there exists $M$ such that $|f(x)-f(y)|\le M|x-y|$ for all $x,y.$]

Can you give me these three examples, using functions on $\mathbb{R}?$

1. A continuous function that is not uniformly continuous?

2. A uniformly continuous function that is not Lipschitz?

3. A Lipschitz function that does not have a bounded derivative everywhere?

[A Lipschitz function does have to have a derivative almost everwhere, and that derivative is in $L^{\infty}.$ But that's not in the spirit of this question - and almost everywhere is not the same as everywhere.]

A continuous function that is not uniformly continuous would be $f(x) = e^{x}$.

I'm thinking about the other 2.

$f(x)=x^{\frac13}$ might be an example of 2


Hello all, how are your preparations for this RMO?
I am in IX th standard, & it's my second time writing the RMO. My preparations in Geometry can only be described using the word "awful"... :(
I am ready to encounter inequalities and Number Theory in the RMO, though. But I think I have to leave the Geometry part completely blank.
In case of combinatorics, I do not know. I might succeed in solving, might not... depends upon the situation of my head during the exam. :P

Please post how your preparations are in this topic. :) :)

hi.i'm aadil from 10th giving rmo first time.
my preperation i feel is ok in no theory & ineq.the rest i am doubtful

i dont understand why you have opened this thread when one already exists in the general discussion forum

[quote="Pocha"]Dat's coz he probably doesn't visit the general discussion much. :P[/quote]
In fact, I never do :D

I always hover round the Ineqs Proposed & Own Problems and the Pre Olympiad section.
[hide="P/S"] I do not like the Indian community (frankly) because people spam in it too much :no: . :( [/hide]
BTW, sorry for spamming and opening this useless thread.

I do not intend any offense to you Aadil, but I think it was clearly wrong of you to have first replied to Potla's topic regarding your preparations and then telling in another post that there already exists a similar forum in the general Discussion. It is worse than Potla's spam. Since you have been posting even there, I can take that you already know such a topic exists before posting your first message in this topic.

No offense to you anyway.

I don't want any quarrels to start, so I hope mod locks this topic.  :(
Mod please ;lock this topic.