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inequalities unsolved
inequalities

the ineq was Prove that for the positive $a_1,a_2,....,a_n$ holds that

$\frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+.....+\frac{a_n}{a_1+a_2} >\frac{n}{4}$

and the improvment is $\frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+.....+\frac{a_n}{a_1+a_2}\geq\frac{n}{2}K$

First of all prove the first ineq if it hasn't been posted again in the forum. And then find the better possible price of $K$ and the prove this ineq with the price of $K$ at the ineq :)

[quote="Arne"]I know that K = 0.989 works.

But why are you posting this here? It's way too hard, even for Mathlinks :D[/quote]

why do you say this? Someone can't post hard problems in this site ?

Use  search  for  ''Shapiro  inequality''

[quote="silouan"]Why do you say this? Someone can't post hard problems in this site ?[/quote]

You know what I mean... Post reasonable questions, please.


geometry
trapezoid
ratio
similar triangles

I have square inside isosceles triangle that height is 4 and width 2. I am trying to find
size of the square inside. Here's picture i made from my problem

[img]http://www.kotiposti.net/eeveri/problem.JPG[/img]

Sorry if this is wrong sub forum to ask, but I am new here and
getting pretty desparate with this problem.  :oops:

[hide]
$\displaystyle \frac{(x+2)x}{2}+\displaystyle \frac{1}{2}\times x \times \frac{4x}{2}=2\times4\times \displaystyle \frac{1}{2}$

$\displaystyle \frac{x^2+2x}{2}+x^2=4$

$3x^2+2x-8=0$

$x=\displaystyle \frac{4}{3}$
[/hide]

Thanks a lot, but i still dont quite understand it myself.  :blush:

Could you explain how you came up with this part? Specially the left side.
[quote="Adunakhor"][hide]
$\displaystyle \frac{(x+2)x}{2}+\displaystyle \frac{1}{2}\times x \times \frac{4x}{2}$
[/hide][/quote]

Here's my method, I'll give you an explanation along with it :)

Cut the large triangle in half, and we know that the base will be $1$ now. Consider the right triangle toward the right (inside the cut in half triangle) with height of $x$, it will have a base of $\frac{2-x}{2}$, now using similar triangles (since the small triangle is similar to the cut in half triangle), we have $\frac{x}{4}=\frac{(2-x)/2}{1} \Longrightarrow x=\frac{4}{3}$, so the area of the square is $\boxed{\frac{16}{9}}$

Heh, sorry for not posting an explanation :blush:

The area of the whole triangle must equal the sum of the area of the trapezoid with height $x$, base $2$, and top length of $x$, added to the area of the triangle that sits atop it. That triangle is similar to the larger triangle, so the ratio of the base of the larger triangle to the base of the smaller triangle has to equal the ratio of their heights. If we call the height of the triangle $h$, this gives us $\displaystyle \frac{x}{2}=\displaystyle \frac{h}{4} \Longrightarrow h=2x$.

Once you draw the altitude, use similar triangles to get what simplifies to a linear equation for x.


linear algebra
matrix
search
induction
function
vector
topology

Hi, for my first post on this forum, I propose you a problem :
(sorry i'm french i don't know yet what are all the names used in english to describe mathematical objects)

(E,(|)) euclidian. dim E>=1. f endomorphism  s.t. tr f=0.
Find x<>0 s.t. (x|f(x))=0.

(The purpose is to prove that there is a base B orthonormée in which the matrix of f has only 0 on its diagonal.)

I know the result when B is not necessarly orthonormée and it is easy to prove the result once it is proved it exists x<>0 s.t. (x|f(x)=0, but i don't see why :/.

Thx for your help.

already posted and proved many times on the forum recently.

Sorry i had already read posts about that exercice but not when E was euclidian.
I search with the search tool but it's pretty hard to find a particular exercice.
So do you have a link or if you remember how to prove it ?
Thx :).

sorry, it was not exactly this pb that was already posted.

Anyway I think one can conclude as follows:
  Suppose that $(x|f(x))$ is always non zero if $x \neq 0$. One can suppose for example that $(x|f(x))>0$ for any $x \neq 0$. (easy because of the continuity of $x \to  (x|f(x))$). Take any orthogonal basis $(e_1,..,e_n)$ and $A$ the matrix of $f$ in this basis. Then the coefficient on the diagonal of $A$ are precisely the $(e_i|f(e_i)>0$. But the sum of these coefficients must be zero ! contradiction.

And the conclusion of the pb follows by induction, as you have already noticed.

Here's a proof ,regarding the fact that $tr(f)=0$ so let $(e_1,....e_n)$ be an orthonormal basis of $E$ we have $\sum_{i=1}^n <e_i|f(e_i)> equal to 0$ let's even suppose that there is no $i$ such that $<e_i|f(e_i)>$ is null  thus there is $0<m,p\leq n$ such that $<e_m|f(e_m) >$ positive and $<e_p|f(e_p)> <0$ let's consider the function of real values $g$ such that $g(x)=<xe_m+(1-x)e_p|xf(e_m)+(1-x)f(e_p)>$ notice that $g$ is continuous and $g(0)<0$ meanwhile $g(1)>0$ thus $\exists \alpha in [0,1]$ s.t $g(\alpha)=0$ it's the vector $u=\alpha e_m+(1-\alpha)e_p$ what you're looking for ;).

I didn't notice that Alekk has sent a solution i guess he did while i was typing mine now u have two solutions that's better :D

Thx for your answers.

Meanwhile, I found another solution a bit longer than yours :
If f symetric, (e1,...,en) an orthonormal basis in which the f matrix is diagonal.
let x=e1+..+en, f(x)=a1e1+..+anen where a1,..,an are the eigenvalues.
(x|f(x))=a1+..+an=0 cqfd.
General case: f=s+a avec s symetric et a antisymetric and the conclusion follows.

[quote="Tsss"]Hi, for my first post on this forum, I propose you a problem :
(sorry i'm french i don't know yet what are all the names used in english to describe mathematical objects)

(E,(|)) euclidian. dim E>=1. f endomorphism  s.t. tr f=0.
Find x<>0 s.t. (x|f(x))=0.

(The purpose is to prove that there is a base B orthonormée in which the matrix of f has only 0 on its diagonal.)
[/quote]

This works for unitary spaces as well (i.e. spaces of the form $\mathbb C^n$, with the usual scalar product $(\cdot,\cdot)$). To restate it, let $M$ be a matrix. Then $M$ is unitarily equivalent to a matrix with all diagonal entries equal.

We may as well assume that $\mbox{tr}(M)=0$. Suppose we can find a unit vector $x\in\mathbb C^n$ such that $(Mx,x)=0$. We can then prove the result by induction on $n$. Let $P$ be the orthogonal projection onto the orthogonal complement $\mathcal H$ of $x$. Then the restriction of $PM$ to $\mathcal H$ has trace equal to $0$ (and for any $y\in\mathcal H$, we have $(My,y)=(PMy,y)$), as can easily be seen, so, by the induction hypothesis we can find an orthonormal system $x_2,x_3,\ldots,x_n\in\mathcal H$ such that $(Mx_i,x_i)=0,\ \forall i\in\overline{2,n}$. Then the matrix $U$ having $x_1,x_2,\ldots,x_n$ has the property that $U^*MU$ has all diagonal entries equal to $0$.

This means that the problem reduces, as Tsss said, to finding a non-zero (or equivalently, unit) vector $x$ such that $(Mx,x)=0$. I think this has been posted before (actually, I remember posting the problem, but there was no proof in that topic :)). Anyway, a quick way out would be using the Toeplitz-Hausdorff theorem, stating that given a bounded linear operator $A$ on a complex Hilbert space, the set $W(A)=\{(Ax,x)\ |\ \|x\|=1\}$ is convex ($W(A)$ is the so-called numerical range). We apply this to $M$: if $e_i$ is the usual orthonormal basis, then $\sum (Me_i,e_i)=0$, so $0$ belongs to the convex hull of $W(M)$, and thus to $M$, by the mentioned theorem.


Help! :) Please and thank you!

The hydrogen peroxide reagent contained 30.0 g H2O2 per liter.
a. You used 15 mL of the reagent. Approximately how many moles of H2O2 are present in 15 mL of the
reagent?

b. Write the balanced equation for the production of O2 from H2O2.

c. Based on the balanced equation in b. and the number of moles of H2O2 calculated in a., how many
moles of O2 would you expect to produce in your reaction, from 15 mL of reagent?

a. Use dimensional analysis. You're given 30g H2O2 /L and ther's 15mL of this solution. From this, you can find the number of grams of H2O2, and then convert that to moles.

[hide]$ \frac{30g H2O2}{L} *\frac{1L}{1000mL} *15mL$[/hide]

I hope the algorithm is ok with u

a. 1 liter reagent contains 30.0 gms $ \ce{H2O2}$ . So, find out how many gms $ \ce{H2O2}$ are present in 15mL i.e. 0.015 L. Once you know mass, u can easily find number of moles
$ number of moles \equal{} \dfrac{mass}{molar mass}$.

b. $ \ce{2H2O2} \minus{} > \ce{2H2O} \plus{} \ce{O2}$

c. 2 moles $ \ce{H2O2}$ produces 1 mole $ \ce{O2}$ from the above eqn. So, n moles $ \ce{H2O2}$ should give n/2 moles $ \ce{O2}$.


calculus
derivative
limit
trigonometry
function
complex analysis
complex analysis solved

I've just entered in analysis and it looks like pooling me into itself, and I'm finding really interesting things, following one is quiet simple:
I've started with searching the solution of the equation $f''(x)+f(x)=0$ $(I)$ then by using the definition of derivative we'll get that $f'(x+dx)-f'(x)=-f(x)\cdot dx$ and by the same way we'll get $f(x+2\cdot dx)-2f(x+dx)+(1+(dx)^2)f(x)=0$ let's make a sequence $a_{n}=f(n\cdot dx)$ and we'll get a recurrent formula of $a_n$ which is $a_{n+2}-2a_{n+1}+(1+(dx)^2)a_n=0$. Actually we know that the solution of $(I)$ is $f(x)=sinx$, when $f(0)=0$ and $f(dx)=dx$ that is $a_0=0$ and $a_1=dx$, let's elicit the formula of $a_n$. Let $\alpha$ and $\beta$ be numbers that satisfy to following equations: $\alpha \beta=1+(dx)^2$ and $\alpha +\beta=2$,so $a_{n+2}-\alpha a_{n+1}-\beta(a_{n+1}-\alpha a_{n})=0$ let $c_n=a_{n+1}-\alpha a_{n}$, so $c_{n+1}=\beta c_n$ it is easy to understand that $c_n=\beta^n c_0$ and $c_0=dx$ and after all we'll get $a_{n+1}-\alpha a_n=\beta^n dx$ $(II)$. As $\alpha$ and $\beta$ have equal rights then $a_{n+1}-\beta a_n=\alpha^n dx$ $(III)$ if we subtract $(II)$ and $(III)$ we'll see $a_n=\displaystyle\frac{\beta^n-\alpha^n}{\beta-\alpha} dx$ by solving system : $\alpha \beta=1+(dx)^2$ and $\alpha +\beta=2$ we'll get $\beta=1+i\cdot dx$ and $\alpha=1-i\cdot dx$ so $a_n=\displaystyle\frac{(1+i\cdot dx)^n-(1-i\cdot dx)^n}{2i}$ ,$f(n\cdot dx)=\displaystyle\frac{(1+i\cdot dx)^n-(1-i\cdot dx)^n}{2i}$ and so $sin(n\cdot dx)=\displaystyle\frac{(1+i\cdot dx)^n-(1-i\cdot dx)^n}{2i}$ so
$sinx=\displaystyle\lim_{n\to\infty} \displaystyle\frac{(1+\displaystyle\frac{ix}{n})^n-(1-\displaystyle\frac{ix}{n})^n}{2i}$. $sinx=\displaystyle\lim_{n\to\infty} \displaystyle\frac{(1+\displaystyle\frac{ix}{n})^{\displaystyle\frac{n}{ix}\cdot ix}-(1-\displaystyle\frac{ix}{n})^{-\displaystyle\frac{n}{ix}\cdot (-ix)}}{2i}$. And thus $sinx=\displaystyle\frac{e^{ix}-e^{-ix}}{2i}$ that a good formula... :) (but I suspect in some parts if you can help me please say.)

This is essentially correct, but there is some danger of assuming what you are trying to prove (what is "$\sin (x)$", for example --- is there a precise definition of this function)?.   If you write your results as a list of precise theorems with detailed proofs it will make it much clearer what has been done.

You it is not so hard to but I suspect in part where I'm getting $e$.

The formulas are all correct, but it can be a circular argument ([i]petitio principii[/i]) if there is no clear definition of $e^x$, $sin(x)$ and some other things in the argument, such as existence of solutions to second-order linear differential equation or linear recurrence.   This is why I suggest to write everything in detail and you (or we) can see what is correct.

[quote="zeus_three"] a [u]much[/u] easier way ... $cos\theta + isin\theta = e^{i\theta}$ [/quote]

Assuming the conclusion is not a simplification, unfortunately.


algebra
polynomial
function
real analysis
real analysis solved

Polynomial $P(x,y)$ satisty the following condition: $P(x+1,y+1)=P(x,y)$Prove that $P(x,y)=\sum a_k(x-y)^k$

Let $f(x)=P(x,0).$ This is a polynomial in one variable and can be written as $f(x)=\sum_ka_kx^k.$

Now fix $x$. Define the function $g(y)=P(x,y)$. Clearly, $g$ is a polynomial in one variable. But for each integer $n, g(n)=f(x-n).$ This is two one-variable polynomials that agree at infinitely many points. Two such polynomials that agree at a number of points greater than the degree of either must be identical as polynomials. Thus for all $y, g(y)=f(x-y).$ Since we can do this for any $x, P(x,y)=f(x-y)$ and we are done.


for solve, enter the equation, then a comma, then the variable you want to solve for, just like in Mathematica.

just put what you want to factor inside the factor parentheses.

I just got mine too so I don't any of the others.

[quote="Ktradsb1"]And how do you graph equations with more than 2 variables? like x^2+y^2=1 or something. How do you get the $y$ variable into the calculator? I have a TI-83+Se btw.[/quote]For your specific example, you have to solve for $y$...so you would put in the two functions $y=\sqrt{1-x^{2}}$ and $y=-\sqrt{1-x^{2}}$.

[quote="math92"]Does anyone here know how to use the algebra functions (such as solve(), factor(), expand(), zeros(), approx(), comDenom(), propFrac(), nSolve()) on a ti-89 titanium?  Any help is appreciated.  Thanks.  :)[/quote]

1.you can look it up in the manual
2.you can find it on the catalog thingy of your calc, press enter, and look at the bottom of the screen. There should be some description of the function.
 :D
[quote="Ktradsb1"]And how do you graph equations with more than 2 variables? like x^2+y^2=1 or something. How do you get the  variable into the calculator? I have a TI-83+Se btw.[/quote]
I dont think calculators does that kind of stuff(2 var, 1 equation)
but if there's 2 equations and 2 var, do this(for 89)
solve(x^2+y^2=1 and x+y=2,{x,y})
that gives you the solutions i think.
not sure about 83's tho.


[quote="nguoivn"][quote="hedeng123"][quote="nguoivn"]Given $ a, b, c > 0$. Prove that:
\[ a^6 \plus{} b^6 \plus{} c^6 \plus{} ab^5 \plus{} bc^5 \plus{} ca^5 \ge 2(a^5b \plus{} b^5c \plus{} c^5a)\]
:)[/quote]

the inequality does not hold    a counter-example is
$ a \equal{} \frac {3}{2},b \equal{} 1,c \equal{} \frac {1}{8}$

$ a^6 \plus{} b^6 \plus{} c^6 \plus{} ab^5 \plus{} bc^5 \plus{} ca^5 \minus{} 2(a^5b \plus{} b^5c \plus{} c^5a) \equal{} \minus{} \frac {156687}{262144}$.[/quote]
Sorry, I have edited it. Now, it's trues  :oops:[/quote]

sorry  the inequality   does not hold  also

 counter-example
$ a\equal{}\frac{2}{3},b\equal{}1,c\equal{}\frac{1}{17}$

[quote="nguoivn"]Are you sure, my friend ? Please check it again  :maybe:
[quote="nguoivn"]Given $ a, b, c > 0$. Prove that:
\[ a^6 \plus{} b^6 \plus{} c^6 \plus{} a^5b \plus{} b^5c \plus{} c^5a \plus{} 3abc(a \minus{} b)(b \minus{} c)(c \minus{} a) \ge 2(ab^5 \plus{} bc^5 \plus{} ca^5)\]
:)[/quote][/quote]
He is right, nguoivn. For $ a\equal{}2/3,b\equal{}1,c\equal{}1/17,$ the inequality does not hold.


algebra
polynomial
superior algebra
superior algebra unsolved

Find all the monic irreducible polynomials of degree $ \leq 3$ in $ \mathbb{F}_2[x]$, and the same in $ \mathbb{F}_3[x]$.

Start low and go up.  Remember that there are only finitely many such polynomials, and indeed, there aren't too many.  Also, if a polynomial of degree three is irreducible then at least one of the factors must be of what degree?


Let $ A,B,C$ be the 3 groups of people,each contains exactly $ n$ man.Each of them knows exactly $ n\plus{}1$ man from other groups.
Prove that it is possible to find three man,such that each of them knows other two.

http://www.mathlinks.ro/Forum/viewtopic.php?t=40982
http://www.mathlinks.ro/Forum/viewtopic.php?t=49347

The second link actually states something stronger. Any ideas on how to prove it?

  darij


How many ways can the word MATHEMATICS be partitioned so that each part contains at least a vowel ? (For example, MA-THE-MATICS, MATHE-MATICS, MATHEMATICS are such partitions).

Do the letters of the partitions have to be in the same order as the letters in MATHEMATICS, or can the letters be in any order?  For instance, would MMTHIC-AAECS be an acceptable partition?

It has to be in the same order as MATHEMATICS, so your example is not acceptable. Thx

[hide]the best way is to just 'count' them systematically :)
we see that we have 4 vowels
if we split it in one part, we obviously have 1 combination...
if we split the word in two parts, we have 3 combinations...
if we split the word in three parts, we have 3 combinations again...
if we split the word in four parts, we have 1 combination...

adding those, we get 8 as the answer

or we have $\sum_{n=0}^3\binom 3n=2^3$[/hide]

[hide="Answer"]One partition:

MATHEMATICS

Two partitions:

MA-THEMATICS
MAT-HEMATICS
MATH-EMATICS
MATHE-MATICS
MATHEM-ATICS
MATHEMA-TICS
MATHEMAT-ICS

Three partitions:

MA-THE-MATICS
MA-THEM-ATICS
MA-THEMA-TICS
MA-THEMAT-ICS
MAT-HE-MATICS
MAT-HEM-ATICS
MAT-HEMA-TICS
MAT-HEMAT-ICS
MATH-E-MATICS
MATH-EM-ATICS
MATH-EMA-TICS
MATH-EMAT-ICS
MATHE-MA-TICS
MATHE-MAT-ICS
MATHEM-A-TICS
MATHEM-AT-ICS

Four partitions:

MA-THE-MA-TICS
MA-THE-MAT-ICS
MA-THEM-A-TICS
MA-THEM-AT-ICS
MAT-HE-MA-TICS
MAT-HE-MAT-ICS
MAT-HEM-A-TICS
MAT-HEM-AT-ICS
MATH-E-MAT-ICS
MATH-EM-AT-ICS

Probably missed a few in there, but if I didn't, then the answer is $34$.[/hide]

I hope this works
[hide]
WE will consider each set of consecutive consanants and the vowels that separate them from the other.
Clearly the first M and A must be together.  Now we have four options for the next ATHE: partition: 1) ATHE, 2) ATH-E, 3) AT-HE, 4) A-THE. Now, independent of the previous choice, we have 3 choices for the EMA: 1)EMA, 2)EM-A, 3) E-MA.  Likewise, for ATI we have 3 choices, and then clearly te concluding CS must be in the partition with the A,  This gives $4 \cdot 3 \cdot 3=\boxed{36}$

(Mmm, this seems kind of big, my method is probably flawed)
[/hide]

I don't think you're wrong. I got $34$, and I wouldn't be surprised if I missed two.

Your listing looks perfect.  I probably overcounted a bit.

Well, alas, I was wrong about you being right.  I found (with the aid of a computer program) the two cases you missed.  You overlooked MATH-EM-A-TICS and MATH-E-MA-TICS.  I don't blame you though, it's a pretty daunting task to perform by hand.

[quote="Jan"]Well Psycho, that's what I had too, but then I realised that this is not correct. Look at the following:
MAT-HEMATICS
MA-THEMATICS
MATH-EMATICS

are all different, valuable combinations. I have no idea how to solve it though...[/quote]
woops, thought he meant syllables  :roll:
nevermind then :)


Could anybody give me a hint for solving the following problem?

Prove that all functions $ f: \mathbb{R}\to \mathbb{R}$ that satisfy $ |f(x+y)-f(y)|\le x^{2}$, for all real $ x,y$, are constant.

for all real $ x,y$
so we can just take $ x=x'-y$ for any $ x',y$

=> then the statement becomes

Prove that all functions $ f: \mathbb{R}\to \mathbb{R}$ that satisfy $ |f(x')-f(y)|\le (x'-y)^{2}$, for all real $ x',y$, are constant.

1) $ |f'(x)|=|\lim_{y\to 0}\frac{f(x+y)-f(x)}{y}|\le \lim_{y\to 0}y=0$
so $ f'(x)=0$ gives $ f$ is constante

2) $ {\forall x\in\mathbb{R}: \ \lim_{n\to+\infty}|f(x)-f(0)|\le \lim_{n\to+\infty}\sum_{i=0}^{n-1}|f(x-\frac{(i+1)x}{n})-f(x-\frac{ix}{n})|\le \lim_{n\to+\infty}\frac{x^{2}}{n}=0}$
so $ \forall x\in\mathbb{R}: \ |f(x)-f(0)|=0$ gives $ f$ is constante


geometry
3D geometry
tetrahedron
triangle inequality
IMO
IMO 1969

For each of $k=1,2,3,4,5$ find necessary and sufficient conditions on $a>0$ such that there exists a tetrahedron with $k$ edges length $a$ and the remainder length $1$.

Take a tetrahedron $ABCD$ which has $k$ edges of measure $a$ and the others of length $1$.
$\textbf{Switch (k)}\ \{$
$\textbf{Case 1:}$ WLOG $CD=a$ and the measure of the other edges is $1$. Then if $M$ is the midpoint of $AB$ we have $MC+MD>CD\iff \sqrt{3}> a$. So $a<\sqrt3$ is a necessary condition. It is easy to ascertain that this is also a sufficient condition for the existence of $ABCD$.
$\textbf{Case 2:}$ We have two subcases:
$\textbf{(a)}$ Two opposite edges measure $a$ and the other edges are of length $1$, say $AB=CD=a$. Since $ABCD$ is isosceles and its faces are isosceles triangles which are acute then $ABCD$ exists if and only if its faces exist, that is when $a<2$.
$\textbf{(b)}$ Two edges emanating from the same vertex have length $a$ and the others have length $1$, say $AD=BD=a$. Then $D$ lies on the circle $\omega$ with center $C$ which is perpendicular to the edge $AB$. Thus $AD=a$ attains its extremal values when $D$ attains its extremal values, that is when $D$ tends to one of the points of intersection of $\omega$ and the plane $ABC$. Hence we get $\sqrt{2-\sqrt3}<a<\sqrt{2+\sqrt3}$. Of course this is also a sufficient condition for the existence of $ABCD$.
Thus we conclude that the desired necessary and sufficient condition for $k=2$ is $a\in (0, 2)\cup\left(\sqrt{2-\sqrt3}, \sqrt{2+\sqrt3}\right)=\left(0, \sqrt{2+\sqrt3}\right).$
$\textbf{Case 3:}$ We have three subcases:
$\textbf{(a)}$ Three edges of length $a$ emanate from the same vertex, say $D$. Then $ABCD$ exists if and only if $a$ is greater than the circumradius of $\triangle ABC$, that is when $a>1/\sqrt3$.
$\textbf{(b)}$ There is an open broken line with $3$ edges of length $a$ between the edges of $ABCD$. WLOG $AD=DB=BC=a$. Then $ABCD$ exists if and only if its faces exist which is just $1/2<a<2$.
$\textbf{(c)}$ A face has all edges of length $a$. WLOG $AB=BC=CA=a$. This is the dual of the subcase $\textbf{(a)}$ so we get $a<\sqrt3$.
Thus we conclude that for $k=3$ $ABCD$ exists for $a\in(1\sqrt3, +\infty)\cup(1/2, 2)\cup(0, \sqrt3)=(0, +\infty)$.
$\textbf{Case 4:}$ This case is dual to the $\textbf{case 2}$ so the answer is $a\in\left(1/\sqrt{2+\sqrt3}, 1/\sqrt{2-\sqrt3}\right)\cup(1/2, +\infty)=\left(1/2, 1/\sqrt{2-\sqrt3}\right)$.
$\textbf{Case 5:}$ This case is dual to the $\textbf{case 1}$ so the answer is $a\in (1/\sqrt3, +\infty)$.
$\textbf{\}}$


function integration limit symmetry trigonometry

I've been trying to solve the 1D diffusion equation

$ D\frac {\partial^2 \phi}{\partial x^2} = \frac {\partial \phi}{\partial t}$, where $ \phi(x,t)$ is the "concentration" of a particular substance at position $ x$ at time $ t$, and $ D$ is the diffusion coefficient.

This particular problem is subject to the following:

1) $ x$ is limited to be between $ 0$ and $ L$ inclusive, and time $ t \geq 0$.
2) The Initial Condition is $ \phi(x,0) =$ the step function centered on $ \frac {L}{2}$, that is, for $ x\leq \frac {L}{2}$, $ \phi(x,0) = 0$, and for $ x > \frac {L}{2}$, $ \phi(x,0) = 1$.
3) $ \int_0^L \phi(x,t)dx = \frac {L}{2}$, i.e. the total amount of substance is constant, all mass must remain between $ 0$ and $ L$ for all time (Conservation of Mass).
4) The only boundary condition is $ \phi(\frac {L}{2},t) = \frac {1}{2}$; the value of $ \phi$ is always changing at the edges $ x = 0$ and $ x = L$.
5) The steady state solution is $ \lim_{t\rightarrow\infty}\phi(x,t) = \frac {1}{2}$.
6) If you were to subtract $ \frac {1}{2}$ from the solution, the result should be an odd function about $ x = \frac {L}{2}$.
7) I do not believe I can make any analogies between the symmetric semi-infinite version of this problem (which would be to start with an initial $ \phi$ as square pulse of width $ \frac {L}{2}$, and enforce a boundary condition at $ x = 0$ that $ \phi = 0$, because this would violate the conservation of mass.)
8) That said, perhaps one can enforce an additional boundary condition that mass does not flow across the boundaries $ x = 0$ and $ x = L$. I'm not quite sure how to represent this as a differential equation that would be useful in this context, but maybe with further thought and some tips I could. I think that this condition is equivalent to #3; maybe this would be helpful in reformulating the problem.

Finding a separable solution for $ \phi$ is only slightly enlightening. I believe I should be able to obtain eigenvalues and then form an eigenfunction basis in order to formulate the actual solution. Thus if I if assume a separable solution $ \phi(x,t) = X(x)T(t)$, then I can write

$ \frac {X''(x)}{X(x)} = \frac {1}{D}\frac {T'(t)}{T(t)} = - k^2$. It might be helpful to set up the problem so that it repeats periodically in $ x$ so that $ \sin(kx)$ and $ \cos(kx)$ eigenfunctions can be used: Then the eigenfunctions for $ X(x)$ would be $ \sin(kx)$ and $ \cos(kx)$. The solution for $ T(t) = Ce^{ - k^2Dt}$.

Then:
$ \phi(x,t) = C_o + \sum_{n = 0}^\infty (A_n \cos(k_n x) + B_n(k_n x))e^{ - k_n^2Dt}$, where $ C_o$ is either 0 or $ \frac {1}{2}$ depending on where we position the problem.

I need to find the coefficients and the eigenvalues, but I am stuck, unsure about how exactly I should utilize the conditions set in this problem to find them. The lack of boundary conditions I think makes this difficult. However, if this problem is made to periodically go through zero, this might help. I could shift the system so that initially at ($ t = 0$) $ \phi$ is an periodic odd square wave: from $ x = 0$ to $ x = L$, $ \phi = \frac {1}{2}$; from $ x = - L$ to $ x = 0$, $ \phi = - \frac {1}{2}$. This repeats. And then the region of interest would be between $ x = - \frac {L}{2}$ and $ x = \frac {L}{2}$.

I think this setup would take care of #3 and #8 due to symmetry, in which case perhaps it might be possible to solve this using an alternative method with the mentioned above setup as the initial condition and the Green's function for the diffusion equation, which isn't separable.

So overall, can anyone give me insight on whether or not my approach(es) is(are) correct in solving for $ \phi(x,t)$ for this particular case?

You're not taking advantage of an important boundary condition, which holds for all time: $ \frac {\partial \phi }{\partial x }\Big |_{x \equal{} 0} \equal{} \frac {\partial \phi }{\partial x } \Big |_{x \equal{} L} \equal{} 0$

This is a result of the ends of the container being closed. Convince yourself why this must be true!

Once you've got this down, you'll realize that you don't need the sine terms -- throw them out!

You now need to find a Fourier cosine series for $ \phi(x|t \equal{} 0)$. First, determine the fundamental wavenumber (spatial frequency) using the closed-ends boundary condition. Without calculation, first convince yourself that the condition you already knew, $ \phi (t | x \equal{} L/2) \equal{} 1/2$, will automatically be met as a result of the Fourier coefficients of $ \phi (x | t \equal{} 0)$ -- this is instructive. Then, calculate the components and answer the problem.

What book are you using?


How many zeros does $ f(x) \equal{} \cos(\log(x)))$ have on the interval $ 0 < x < 1$?

$ \textbf{(A)}\ 0 \qquad
\textbf{(B)}\ 1 \qquad
\textbf{(C)}\ 2 \qquad
\textbf{(D)}\ 10 \qquad
\textbf{(E)}\ \text{infinitely many}$

Are we going by radians or degrees?

Edit: Sorry, didn't think about it. :blush:

[hide="Solution"]$ \cos(\log(x)) \equal{} 0\implies \log(x) \equal{} 0$, which will never happen, or $ \log(x) \equal{} \frac {\pi}{2} \plus{} \pi n\implies x \equal{} 10^{\frac{\pi}{2}\plus{}\pi n}$, which is out of our range for $ n\geq 1$.

However, we have solutions for $ x \equal{} 10^{\frac {\pi}{2}\cdot n}$ for all $ n\in\mathbb{Z}$ such that $ n < 0$.

So the answer is $ \boxed{\textbf{(E)}\ \text{infinitely many}}$.[/hide]


Okay so say I have an array which is like this:

.data

values:
      .half -17
      .half  16
      .half 153

and I want to print it as

-17
16
153

How do I do this?? If it is stored in a 4 byte of integer then it would be very easy to do, but not it's stored as a 2-byte integer.

In a 4 -byte integer I just quadruple starting from the first array address and then print the array, but how about this one?? Can someone help me??

I'm not very familiar with MIPS processor and its instruction set but is there really no way to convert a 2-byte integer into 4-byte integer? If there isn't you can always use bit masks and shift instructions - (number AND 0x00FF) gives you the lower 2-byte integer and (number SHIFT RIGHT 16) the upper one.

Okay, I happen to fix this problem. Now the problem is how do I convert a 1-byte integer into an ASCII characters.. say that I have the integer 65 and I want to convert that to 'A'. What is the code to do that??

Uh, 1-byte integer already IS an ascii character. You just need to print it (or do whatever you want to do) using the right function.


Euler
complex analysis
complex analysis unsolved

For which values of $ a\in C$ does the sequence $ a, a^a, a^{a^a}, a^{a^{a^a}},\cdots$ converge?

For real a it converges for a in [exp(-e),exp(1/e)).
This is due to Euler (and is not very hard). For complex a, the problem is much more difficult,
and was completely solved only recently:
J. Hubbard, D. Schleicher and M. Shishikura, Journal of the Amer. Math. Soc., 22 (2009) 1, 77-117,
the result is stated on page 81.

Pym.


induction
combinatorics proposed
combinatorics

Find the sum $\sum_{k=1}^n \binom{k}{[\frac{k}{2}]}.\binom{n-k}{[\frac{n-k}{2}]}$
 :love:  :love:

[quote="kunterbunt"]It's not very natural to start this sum at 1 and not at 0. You can substract the term for $k=0$ in the end if you like.
To get rid of the floors, we break the sum in two sums according to the parity of $k$ and consider two cases according to the parity of $n$ which gives 4 sums to calculate.

1) Prove that $\sum_k \binom{2k}k \binom{2n-2k}{n-k}=4^n$

[hide="Hint"] Compare the coefficients of $x^n$ in  $(1-4x)^{-1/2}(1-4x)^{-1/2}= (1-4x)^{-1}$.[/hide]

2) Prove that $\sum_k \binom{2k+1}k \binom{2n-2k-1}{n-k-1}=4^n-\frac 12 \binom{2n+1}{n+1}$.

[hide="Hint"] Use the identity $\binom{2m+1}m= \frac 12 \binom {2m+2}{m+1}$ for $m\not= -1$ for both factors on the left hand side. Then use 1). [/hide]

3) The first two results together give the sum for even $n$ starting with 0. Odd $n$ works analogously.[/quote]
Thank you for your problem,[b]kunterbunt[/b].Here is my answer(proved by induction)
$\sum_{k=1}^n$ $\binom{k}{[\frac{k}{2}]}$ $. \binom{n-k}{[\frac{n-k}{2}]}= 2^{n+1}-\binom{n}{[\frac{n}{2}]}-\binom{n+1}{[\frac{n+1}{2}]}$
  :blush:


geometry
incenter
trigonometry
IMO Shortlist
trig identities
Law of Sines

In the triangle $ ABC,$ with $ \angle A \equal{} 60 ^{\circ},$ a parallel $ IF$ to $ AC$ is drawn through the incenter $ I$ of the triangle, where $ F$ lies on the side $ AB.$ The point $ P$ on the side $ BC$ is such that $ 3BP \equal{} BC.$ Show that $ \angle BFP \equal{} \frac{\angle B}{2}.$

Let $ K\in AB$ such that $ \angle BKC \equal{} \angle ABI$ (A is between B and K). I will prove that $ \frac {BF}{FK} \equal{} \frac {1}{2}$.

$ *120^{o} \equal{} \angle B \plus{} \angle C$.$ \Longrightarrow 60^{o} \equal{} \frac {\angle B}{2} \equal{} \frac {\angle C}{2}$. $ \Longrightarrow\angle ACK \equal{} \angle ICA$.

Draw $ AT\perp AK ( T\in KC)$ and $ IM\perp BA(M \in BA)$

We have $ \angle TAC \equal{} \angle CAI \equal{} 30^{o}$

$ \Longrightarrow \bigtriangleup IAC \equal{} \bigtriangleup TAC$.

$ \Longrightarrow TA \equal{} AI \equal{} 2IM$

on the other hand $ \bigtriangleup KAT \thicksim \bigtriangleup BIK$

$ \Longrightarrow \frac {BM}{AK} \equal{} \frac {IM}{AT} \equal{} \frac {1}{2}$.

Notice that $ \frac {MF}{FA} \equal{} \frac {MF}{FI} \equal{} \frac {1}{2}$.

Hence $ \frac {BF}{FK} \equal{} \frac {BM \plus{} MF}{FA \plus{} AK} \equal{} \frac {1}{2}$. q.e.d  :lol:

Let $Q$ be the foot of the perpendicular from $P$ to $AB$, and let $R$ be on $\overline{AQ}$ so that $QR=QB$. We have $PR=PB$, so $\angle FBI=\frac{1}{2}B=\frac{1}{2}\angle BRP=\frac{1}{2}(\angle BFP+\angle RPF)$. Hence, it suffices to show that $\angle BFP=\angle RPF$, i.e. $RP=RF$.

We have $BC=BI\cos\frac{B}{2}+CI\cos\frac{C}{2}$. But $CI\sin\frac{C}{2}=BI\sin\frac{B}{2}$, so \begin{align*}BC&=BI\frac{\cos\frac{B}{2}\sin\frac{C}{2}+\sin\frac{B}{2}\cos\frac{C}{2}}{\sin\frac{C}{2}}  \\ &=BI\frac{\sin (\frac{B}{2}+\frac{C}{2})}{\sin\frac{C}{2}} \\ &=BI\frac{\sin 60^{\circ}}{\sin\frac{C}{2}}\\ \end{align*} Hence $BI=2BC\sin\frac{C}{2}\frac{1}{\sqrt{3}}$. By the law of sines on $\triangle BFI$, we have \[\frac{BF}{BI}=\frac{\sin (60^{\circ}+\frac{B}{2})}{\sin 60^{\circ}},\] so \begin{align*}BF&=\frac{2}{\sqrt{3}}BI\sin (60^{\circ}+\frac{B}{2})\\ &=\frac{4}{3}BC\sin\frac{C}{2}\sin (60^{\circ}+\frac{B}{2})\\ \end{align*}

However, $2\sin\frac{C}{2}\sin (60^{\circ}+\frac{B}{2})=\cos (A+\frac{B}{2}+\frac{C}{2})=\cos B-\cos 120^{\circ}=\cos B+\frac{1}{2}$. It follows that $BF=\frac{2}{3}BC\cos B+\frac{1}{3}BC=2PB\cos B+PB=BR+RP$. But $BF=BR+RF$, so $RP=RF$, as desired. $\Box$

it can be solved by use of intersection point of $FI$ and $BC$(name it $D$) and of congruency of $BFD$,$ABC$ with ratio:$\frac{a(a+c)}{2p}$ and computing the $\frac{BP}{PD}$ with sines,...

[u][b]Notations[/b][/u]:$IB \cap FP=K,IB \cap AC=L,\angle{BFP}=\theta$

[u][b]Proof[/b][/u]$IF \parallel AL \Rightarrow \frac{FB}{c}=\frac{IB}{BL}=\frac{a+c}{a+b+c} \Rightarrow \boxed{BF=\frac{(a+c)c}{a+b+c}}$
We also see that $BP=\frac{a}{3}$ and $b^2+c^2-a^2=bc$...(1)(From the fact that $\angle{A}=60^{\circ}$)

Now we start our manipulations:

$\theta=\frac{B}{2}$

$\Leftrightarrow \frac{sin(B+\theta)}{sin\theta}=\frac{sin\frac{3B}{2}}{sin\frac{B}{2}}=3-4sin^2\frac{B}{2}=1+2cosB=\frac{a^2+c^2-b^2+ac}{ac}$

$\Leftrightarrow \frac{3ac+c^2}{a^2+ab+ac}=\frac{a^2+c^2+ac-b^2}{ac}$(The left side follows by applying sine rule in $\triangle{BPF}$)

$\Leftrightarrow 3a^2c^2+3ac^3=(a^2+c^2+ac-b^2)(a^2+ab+ac)=(b^2+c^2-bc+c^2+ac-b^2)(a+b+c)$(Substituting from (1))

$\Leftrightarrow 3c(a+c)=(2c+a-b)(a+b+c)=2ac+2bc+2c^2+a^2+ab+ac-ab-b^2-bc$

$\Leftrightarrow c^2=bc+a^2-b^2$

$\Leftrightarrow b^2+c^2-a^2=bc$

$\Leftrightarrow \angle{A}=60^{\circ}$

which is true...

Let $\angle{BFP}=x$

So, $\frac{BF}{BP}=\frac{sin(x+\angle{B})}{sinx}$

$2r=AI=\sqrt{3}AF=4Rsin(\frac{\angle{B}}{2})sin(\frac{\angle{C}}{2})$

So we obtain,

$2\sqrt{3}sin(\angle{C})-4sin(\frac{\angle{B}}{2})sin(\frac{\angle{C}}{2})=\frac{sin(x+\angle{B})}{sinx}$

$\implies 2\sqrt{3}sin(\frac{\pi}{3}+\angle{B})-4sin(\frac{\angle{B}}{2})cos(\frac{\pi}{6}+\frac{\angle{C}}{2})=\frac{sin(x+\angle{B})}{sinx}$

$\implies 3cos(\angle{B})+2sin^2(\frac{\angle{B}}{2})=\frac{sin(x+\angle{B})}{sinx}$

$\implies 3-4sin^2(\frac{\angle{B}}{2})=\frac{sin(x+\angle{B})}{sinx}$

$\implies \frac{sin(\frac{\angle{B}}{2}+\angle{B})}{sin(\frac{\angle{B}}{2})}=\frac{sin(x+\angle{B})}{sinx}$

$\implies cot(\frac{\angle{B}}{2})=cotx$

$\implies x=\frac{\angle{B}}{2}$

Synthetic proof is not at all difficult if we proceed along right lines.:)
Let $P'$ be on $BC$ s.t. $\angle{BFP'}=B/2$.Similarly take $E$ on $AC$ where $IE$ is parallel to $AB$,and $Q$ on $BC$ with $\angle{CEQ}=C/2$.Let $X$ be the intersection of $QE$ and $AB$.
It is easy to see that $BFI$ is congruent to $XFE$.So $BF=FX$.Also we can easily show that $FP'$ is parallel to $EQ$.So $P'$ is the midpt of $BQ$.Similarly $Q$ is the midpt of $P'C$.So $P=P'$ and we are done .;)

more solutions  [url=https://artofproblemsolving.com/community/c6h100938p570209]here[/url]

[color=#f00]edit,[/color] it's a pity old links not to work, e.g.
[quote=livetolove212]http://www.mathlinks.ro/viewtopic.php?t=313492[/quote]


What is the only pair of integers (x,y) for which twice the square of the first equals three times the square of the second?

integers include negatives and negs square equal poss square so their are 4pairs if they aren't 0s
2x^2=3y^2
root 2 x= root 3 y
0,0

another way to think about this: the prime factorization of every perfect square consists of primes raised to EVEN powers.  when you multiply a square such as y by 3, you are shifting the 3 to an ODD power.


Are symmetries counted in the total number of grids? For example, would the grid on the far right in the [url=http://usamts.org/Tests/USAMTSProblems_19_3.pdf]example[/url] be the same or different than the same grid rotated 90 degrees?

We cannot answer this question while the round is in progress.


Let $ x_1,....x_n$ be postive real numbers
prove that
$ \frac{2\sqrt{n}\minus{}1}{5\sqrt{n}\minus{}1} \leq \sum \frac{2\plus{}x_i}{5\plus{}x_i} \leq \frac{2\sqrt{n}\plus{}1}{5\sqrt{n}\plus{}1}$

[quote="FOURRIER"]Let $ x_1,....x_n$ be postive real numbers
prove that
$ \frac {2\sqrt {n} \minus{} 1}{5\sqrt {n} \minus{} 1} \leq \sum \frac {2 \plus{} x_i}{5 \plus{} x_i} \leq \frac {2\sqrt {n} \plus{} 1}{5\sqrt {n} \plus{} 1}$[/quote]
At left the  inequality is trivial and at  right it's false  :D
:cool:

[quote="Diogene"] at  right it's false
[/quote]

can u give a contrexample ?

for $ n \equal{} 1$, the RHS is 1/2 while the sum in the middle has one term, which approaches 1 as we choose larger and larger values for $ x_1$. e.g., $ (2\plus{}10)/(5\plus{}10) \equal{} 12/15 > 1/2$


ce problema interesanta !
ce`i drept cred ca e destul de muncitoreasca, insa enuntul e dragut.
un numar natural $n>1$ se numeste [i]perfect[/i] daca suma tuturor divizorilor sai (inclusiv $1$ si el insusi) este egala cu $2n$. aflati toate numerele naturale $n$ care au urmatoarele proprietati :
- $n$ este [i]perfect[/i]
- $n-1$ si $n+1$ sunt numere prime.

Problema a mai fost pe forum (Olympiad - Number Theory) si solutiile date foloseau urmatoarea lema (care cica ar fi cunoscuta):

[hide="Lemă"][i]Dacă $n$ este un număr perfect par, atunci $n = 2^{p-1}\left( 2^{p}-1 \right)$, unde $p$ şi $2^{p}-1$ sunt numere prime.[/i][/hide]

interesanta lema..dar cat de cunoscuta o fi oare ? :)
oricum, eu cred ca am rezolvat`o fara aceasta lema...


http://www.research.att.com/~njas/sequences/A003095

A Tip. If you want a sequence, a serie. etc. you can try to find it on this site. it have a lot!!

But not all.

but is a good site

bye

This is high school basics. This was probably a "trick" question or possibly a challenge homework problem.
In short, it's inappropriate to mention Sloane's.... :mad:

[hide="solution"]The sequence is: square the term and add 1.
i.e.: $a_{n+1}=a_{n}^{2}+1$
No nice closed form :wink: [/hide]

err...

enter de site

[quote]
a(n) = a(n-1)^2 + 1.
[/quote]

If this is inappropriate so i think that

[quote]
$a_{n+1}=a_{n}^{2}+1$
[/quote]

is inappropriate too..

oh man.. .tsk tsk...

[quote="PLANETX1970"]First 4 terms are 1, 2, 5, 26 respectively find the possible 5th and 6th and nth term.

I know the 5th and 6th terms are 677 and 458330. Can't work ou the nth term.[/quote]

After careful consideration...

[hide]The pattern is $1$, $2$, $5$, $26$,...

If you look closely, you can see that each number is one more than the one before's square.

So, by this pattern, the $5$th term would be $677$
The $6$th term would be $458330$

The way to find the $n$th term would be to take the term before $n$ (let's make it $x$, and do the following equation $x^{2}+1=n$[/hide]

I think I'm right

[quote="yagaron"][quote="me@home"]That's the point - its inappropriate to site Sloane's when you can just as easily write the recursion in a simple way in your post...[/quote]

I just telling that he dont need to make a post for all series he dont know

he can just search on internet

but people like to make posts without make a carefully search on web.

thats the life, aho[/quote]

But you dont need to call hm an aho...I mean just because you wont be getting any..

asdff

I dont call anyone idiot.

I'm talking in English not in Japanese

I ever use the expressions

Hmpft, hmm, hmp, ahm. etc... and similar

is not my guilt the o is near the p


geometry
incenter
geometric transformation
reflection
circumcircle
geometry proposed

The bisectors of angles $ B$ and $ C$ of triangle $ ABC$ intersect the opposite sides in points $ B'$ and $ C'$ respectively. Show that the line $ B'C'$ intersects the incircle of the triangle.

$ I \equiv BB' \cap CC'$ is incenter of $ \triangle ABC.$ $ AI$ cuts $ BC$ at $ A'.$ Incircle $ (I)$ touches $ BC, CA, AB$ at $ D, E, F.$ Inversion in $ (I)$ takes $ A', B', C'$ into $ A'', B'', C''.$ Perpendicular bisectors $ x, y, z$ of $ IA'', IB'', IC''$ pairwise intersect at $ X \equiv y \cap z, Y \equiv z \cap x, Z \equiv x \cap y.$ $ \triangle XYZ \sim \triangle DEF$ with parallel sides are centrally similar. Let $ P$ be 9-point circle center of $ \triangle DEF.$ As inversion image $ A''$ of $ A'$ is reflection of the midpoint $ D'$ of $ EF$ in the midline $ d' \parallel EF$ of $ \triangle DEF$ $ \Longrightarrow$ $ d(I, x) \equal{} d(P, EF)$ and cyclic exchange $ \Longrightarrow$ $ I$ is 9-point circle center of $ \triangle XYZ$ $ \Longrightarrow$ $ \triangle XYZ \cong \triangle DEF$ are centrally congruent with coefficient -1 $ \Longrightarrow$ $ P$ is circumcenter of $ \triangle XYZ.$ Let $ P_x$ be reflection of $ P$ in $ x \equiv YZ$ $ \Longrightarrow$ $ I$ is midpoint $ XP_x.$ The intouch $ \triangle DEF$ is always acute, so is the congruent $ \triangle XYZ$ $ \Longrightarrow$ $ \angle XPP_x$ is obtuse $ \Longrightarrow$ $ 2XI \equal{} XP_x > PX \equal{} ID$ $ \Longrightarrow$ $ (I)$ intersects circle $ (X) \equiv \odot (IB''C'')$ $ \Longrightarrow$ $ (I)$ intersects line $ B'C'.$


in his book Kedlaya defines and i quote
[quote="Kiran Kedlaya"]Let F1 and F2 be two figures and suppose $ f : F_{1} \to F_{2}$ is a bijection (one-to-one correspon-dence). We say that $ F_{1}$ and $ F_{2}$ are congruent (via $ f$ ), and write $ F_{1} \equiv F_{2}$ , if we have an equality of distances $ PQ \equal{} f(P)f(Q)$ for all $ P, Q \in F_{1}$[/quote]
He defines  rigid motion as
[quote="Kiran Kedlaya"] A rigid motion of the Euclidean plane is a map from the plane to itself which preserves distances; that is, if $ P$ maps to $ P'$ and $ Q$ to $ Q'$ , then we have $ PQ \equal{} P^{'}Q^{'}$[/quote]
Then he states the following theorem
[quote="Kiran Kedlaya"]Given two congruent figures, each not contained in any line, there is a unique rigid motion that maps one onto the other (matching corresponding points).[/quote]
Except for the uniqueness part doesn't this follow from the definition of congruent and rigid motion

i got it and it doesn't follow  :) .
thanks to those who may have spent their time on this


What is the smallest number of planes needed to divide a cube into congruent tetrahedra?

5 cuts give 6 tetrahedra.

Best regards,
sunken rock

I think that 3 planes, give 6 congruent tetrahedra.

Kostas Vittas.


limit
logarithms
function
calculus
integration
geometric series
calculus computations

Find a sequence $x_{n}$ of real numbers such that $\lim_{n \to \infty}x_{n}=1$ and $\sum_{n=1}^{\infty}\frac{1}{n^{x_{n}}}< \infty$.

Too easy?

Is this possible?

We have $\sum_{n=1}^{\infty}\frac{1}{n^{x_{n}}}= \sum_{n=1}^{\infty}\frac{1}{e^{\ln(n^{x_{n}})}}= \sum_{n=1}^{\infty}\frac{1}{e^{x_{n}\ln n}}$.

Let's ascertain the nature of this series with the Root Test:

$\lim_{n\to\infty}(a_{n})^{1/n}= \lim_{n\to\infty}\frac{1}{e^{\frac{x_{n}\ln n}{n}}}$

and for the series to converge this limit must be less than 1, which is equivalent to have

$\lim_{n\to\infty}\frac{x_{n}\ln n}{n}> 0$.

But this cannot be, because

$\lim_{n\to\infty}\frac{x_{n}\ln n}{n}= \lim_{n\to\infty}x_{n}\cdot \lim_{n\to\infty}\frac{\ln n}{n}= 1 \cdot 0 = 0$.

What am I missing?

Carcul, what you are missing is that the Root Test can yield a value of $1$ which means that it is inconclusive.  Gelbaum and Olmstead give examples of this.

The root test never had a chance of working. Such a series must be smaller than the harmonic series but larger than any $p$-series for $p>1.$ The root test only compares to geometric series, and this doesn't compare usefully to any geometric series.

Can we make $\frac1{n^{x_{n}}}=\frac1{n\ln^{2}n}$ ? Work that out, and that would require $x_{n}=1+\frac{2\ln\ln n}{\ln n}.$ As required, $x_{n}\to1$ as $n\to\infty.$

$\sum_{n=2}^{\infty}\frac1{n\ln^{2}n}$ does converge, so we have our example.

Edoo and Professor Merryfield, thanks for the clarification.

Professor Merryfield, how did you think in $\frac{1}{n \ln^{2}n}$? That is, how did you make the connection between "the series must be smaller than the harmonic series but larger than any p-series" and $\frac{1}{n \ln^{2}n}$?

[quote="Kent Merryfield"]and this doesn't compare usefully to any geometric series.[/quote]

How can we see that?

Here is (a sketch of) another argument you may find useful. Suppose that we have a sequence of positive convergent series $\sum_{k}a_{n,k}$, $n=1,2,3,\dots$. Then we can choose $N_{1}$ such that $\sum_{k>N_{1}}a_{2,k}<1/2$. Then choose $N_{2}>N_{1}$ such that $\sum_{k>N_{2}}a_{3,k}<1/{2^{2}}$, and so on. Conclude that the series $\sum_{k}b_{k}$ converges where $b_{k}=a_{n,k}$ for $N_{n-1}<k\le N_{n}$. Use this trick to show that if you have any sequence of positive functions $f(k,x)$ and the series $\sum_{k}f(k,x)$ converges for any $x>0$, then there is a sequence $x_{k}\to 0$ such that $\sum_{k}f(k,x_{k})$ converges. The initial problem is a special case of this for $f(k,x)=\frac 1{k^{1+x}}$.

[quote="Carcul"][quote="Kent Merryfield"]and this doesn't compare usefully to any geometric series.[/quote]

How can we see that?[/quote]

For any convergent geometric series (or anything reasonably comparable to a geometric series), there is a constant $p > 1$ and another constant $c \neq 0$ such that our terms are growing like $\frac{c}{p^{n}}$.  $p^{n}$ grows faster than $n^{d}$ for any constant $d$; in our sequence, we are not even growing that quickly.

[quote="Carcul"]Professor Merryfield, how did you think in $\frac{1}{n \ln^{2}n}$? That is, how did you make the connection between "the series must be smaller than the harmonic series but larger than any p-series" and $\frac{1}{n \ln^{2}n}$?[/quote]

You need to grow by something smaller than any power of $n$.  $\ln n$ is the most usual example of such a thing.  Unfortunately, $\sum \frac{1}{n \ln n}$ doesn't converge (integral test), but luckily $\sum \frac{1}{n \ln^{2}n}$ does.  Then you still have to check that it satisfies the requirement on the $x_{n}$, which it does.


[quote="t0rajir0u"][hide="Cheap way"] Every number except $\boxed{ A }$ is far too large.  Recall that $\pi \approx 3.14$ radians is a half-circle. [/hide][/quote]

thats not cheap.  [hide]3 and 4 are obviously too big.  notice 2 is a little over a quarter of a circle.  the arc of a quarter of a pizza is 2*6pi=3 pi which isway bigger than 4.2 so it must be 1[/hide]

[quote="Interval"]A wedge-shaped piece is cut from a circular pizza. The radius of the pizza is 6 inches. The rounded edge of the crust of the piece measures 4.2 inches. To the nearest tenth, the angle of the pointed end of the piece of pizza, in radians, is

(1) 0.7        (3) 7.0
(2) 1.4        (4) 25.2[/quote]
[hide="Expensive Way  :P "] The circumference is $12\pi$.
$s=\frac{\theta}{2\pi}C$
$4.2=\frac{A}{2\pi}12\pi\iff A=\frac{4.2}{6}=0.7$ [/hide]


Look at illcrowflu's explanation
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150077

okay, i will post this not as a solution, b/c one has already been given, but as a warning to sharkman: this is not the place to come and spam with homework problems.  i am not saying that you are, but when the first ten posts or so are from the same author it is kinda suspicious...


decimal representation
number theory
Divisibility
IMO
IMO 1962

Find the smallest natural number $n$ which has the following properties:

a) Its decimal representation has a 6 as the last digit.

b) If the last digit 6 is erased and placed in front of the remaining digits, the resulting number is four times as large as the original number $n$.

Let $n$ have $k+1$ decimal digits. If $n$ checks a) and b), obviously $k > 0$. So you're reduced to solve the equation $4\cdot(10u + 6) = u + 6 \cdot 10^k$ in positive integers, making $u$ as small as possible for a $k$-digit decimal number. You need $13v = 10^k - 4$, letting $v = u/2$ be an integer. You check $\mbox{ord}_{13}(10) = 6$ and $10^5 \equiv 4 \bmod 13$. Therefore $13 \mid (10^k - 4)$ iff $k = 6h + 5$, for any $h\in\mathbb{N}$. Then $u = 2 \cdot \frac{10^{6h+5}-4}{13}$, and you're just involved for finding out the least $h\in\mathbb{N}$ such that $u$ has exactly $6h+5$ digits. Thanks to God, $h = 0$ works and $n = 153846$.

This is an alphametric!
Set it up like an old-school paper and pencil multiplication problem.
_ _ _ _ _ _ _ _ 6
x4
--------------------
6 _ _ _ _ _ _ _ _
Where we have added extra blanks in case the number is long. Now,
                   2
_ _ _ _ _ _ _ 4 6
x                    4
--------------------
6 _ _ _ _ _ _ _ 4

                 1 2
_ _ _ _ _ _ 8 4 6
x                    4
--------------------
6 _ _ _ _ _ _ 8 4

              3 1 2
_ _ _ _ _ 3 8 4 6
x                    4
--------------------
6 _ _ _ _ _ 3 8 4

And eventually you get 153846.

Sorry the spacing is wrong. How do I fix it?

Here is an even cooler method, I remember from a USAMTS solution to a similar problem:
Let the answer be abcdef....
Consider the repeating decimal 0.abcdef....abcdef....abcdef.......=K
Then 4K=0.6abcdef....6abcdef....6abcdef.......
Note that 10(4K)-6=K
Solving, K=2/13
We find the decimal respresentation of 2/13 is .153846153846.....

[quote="Jaswinder"]my soln is somewat dirty

write the no in base 10 and then take mod10,100,1000 both sides
and done[/quote]
Jaswinder,  the postings should not be in hints but the complete solution by Zetax  (http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135914)
[b][u]Generalistation, a way of analysing it:[/u][/b]
let the digits of the originally $n$ digit number be $x$ , denoted by $ a_n-1, a_n-2........,a_1,a_0=6$
$x$ =$10^{n-1}a_n-1+ 10^{n-2}*a_n-2+....+10a_1+ 6 $
according to this problem
$ 4x=6*10^{n-1}+(10^{n-2}a_n-1+10^{n-3}a_n-2+....+10a_2+a_1 )$
It is equal to this binomial exp:
$ 4x=6*10^{n-1}+ \frac{x-6}{10}$

$40x=6*10^n +x-6$

$13x=2(10^{n}-1)$

$ \rightarrow  $  this shows that $10^{n}-1$ (a number containing only $9$) must be divisible by $13$

$999999 = 13*76923 \rightarrow x= 2*76923 = 153846 $
this solution is rather guessing out , and  it's obvious this solution doesn't fetch us answer to these kind of problems . I just posted it to make sure there was another  indirect solution  for this problem . Readers don't take this post important as i mentioned u above :)   :ddr:

getting our hands dirty,we easily get the number $153846$ and indeed is $153846*4=615384$.It is from Imo $1962$
http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=1962&sid=7d944a84d144d5ce1f9d027cb681a3ad

[hide=My solution]Let it be $\overline{a_1a_2a_3...a_n6}$. Then \[4\cdot\overline{a_1a_2a_3...a_n6}=\overline{6a_1a_2a_3...a_n}.\]Let $\overline{a_1a_2a_3...a_n}=k$. Then \[ 6\cdot 10^n+k=4(10k+6)\iff 2(10^n-4)=13k \]Since $13\mid 10^5-4=99996$, hence the desired number id $153846$. [/hide]

Salutations,
[pywindow]
for uvwxyz in range(100000,250000):
    z = uvwxyz % 10
    uvwxy = uvwxyz // 10
    zuvwxy = 100000 * z + uvwxy
    if zuvwxy == 4 * uvwxyz:
        print(uvwxyz)
[/pywindow]

Let $n=10^k a_k+10^{k-1}a_{k-1}+\dots +10a_1+6$ be the number.
Writing $n$ in the form $n=10N+6,$ where $10^{k-1}<N<10^k,$ we have from (b),
$$4(10N+6)=6\cdot 10^k+N\implies 13N=2(10^k-4) \implies (-3)^k \equiv 4 \pmod{13}.$$ We find $k=5$ as the least that satisfies. So $N=15384$ and $\boxed{n=153846}.$


function
limit
calculus
derivative
algebra
polynomial
functional equation

Find all continuous functions which satisfy \[ f(x+y)=f(x)+f(y)+xy(x+y),x,y\in \mathbb R. \]

The answer is $f(x)=f'(0)x+\frac{1}{3}x^3.$

What's the source of this problem, chess?

Well, a bit of analysis and guesswork goes.
[hide]Expanding the second term gives $x^2y+xy^2$.
These appear to be the middle two terms of a cubic equation. Since the coefficients are $1$, instead of $3$ from the binomial expansion series of a cubic, the scale factor is $1/3$, which inspires us to try $f(x) = x^3/3$, which works.[/hide]

[quote="agolsme"]how did you find that kunny?[/quote]

All right, agolsme.

First plugg $x=0$ into the given functional equation, we have $f(0)=0.$

For $y\neq 0,\ \frac{f(x+y)-f(y)}{y}=\frac{f(y)-f(0)}{y}+x(x+y),$ thus we have

$\lim_{y\rightarrow 0} \frac{f(x+y)-f(y)}{y}=\lim_{y\rightarrow 0}\frac{f(y)-f(0)}{y}+\lim_{y\rightarrow 0}x(x+y).$

$f'(x)=f'(0)+x^2,$ integrate both of the sides with respect to $x,$ we have $f(x)=f'(0)x+\frac{1}{3}x^3+C.$

By $f(0)=0\Longleftrightarrow C=0,$ we obtain $f(x)=f'(0)x+\frac{1}{3}x^3.$

[quote="pkothari13"]but doesnt taking the derivative assume the function is already continuous?[/quote]

Check the wording of the problem.

well, what i mean to say is that doesn't the question ask for a function that IS continuous? how can you be sure that kunny's function is continuous?

Every polynomial function is continuous.

This function is right but I think that the solution is wrong.
 We do not know that $f$ has derivative , so we can not say that

              $lim\frac{f(x+y) - f(x)}{y} = f'(x)$

 [i](The problem is the existence of the limit)[/i]

 I have seen somewhere tat to those problems we put g(x) = a function of f(x)  and the given relation leads to the Cauchy form ect.

 [u]Babis[/u]

This is from an AoPS class, so I don't think you're supposed to use calculus. :|

Let $g(x)=f(x)-\frac{x^3}{3}$.Since $f$ is continuous,then so is $g$.Substituting $f(x)=g(x)+\frac{x^3}{3}$ into the given FE,then
$g(x+y)=g(x)+g(y)$.Thus
$g$ satisfies the [b]Cauchy Equation[/b] and continuity.Hence
$\exists a\in \mathbb R$ s.t. $g(x)=ax(\forall x\in \mathbb R)$.Namely
$f(x)=\frac{x^3}{3}+ax(\forall x\in \mathbb R)$.
Conversely
$\forall a\in \mathbb R:f(x)=\frac{x^3}{3}+ax(\forall x\in \mathbb R)$ satisfies the condition.Therefore the answer is
$\boxed{\forall a\in \mathbb R:f(x)=\frac{x^3}{3}+ax}(\forall x\in \mathbb R)\blacksquare$ :coool:


What is the value of:

$ (1 \minus{} \frac {1}{4})(1 \minus{} \frac {1}{9})(1 \minus{} \frac {1}{16}) \cdot\cdot\cdot (1 \minus{} \frac {1}{4012009})$

I rewrote it as:

$ \displaystyle\prod_{k \equal{} 2}^{2003} (1 \minus{} \frac {1}{k^2})$

I've tried messing around with it... looking for a geometric sequence or something, but I have come up with nothing so far.

[hide="Hint"]$ \displaystyle\prod_{k\equal{}2}^{2003}{(1\minus{}\displaystyle\frac{1}{k^2})} \equal{} \displaystyle\prod_{k\equal{}2}^{2003}{\displaystyle\frac{(k\minus{}1)(k\plus{}1)}{k^2}}$

Evaluate that for small terms and you may see a pattern.[/hide]

[hide="Hint 2"]$ \left(1 \minus{} \frac1{n^2}\right) \equal{} \left(1 \minus{} \frac1n\right)\left(1 \plus{} \frac1n\right)$[/hide]


inequalities
integration
real analysis
real analysis solved

Prove the following inequality.

\[\frac{n-1}{2n+2}<\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots\cdots+\left(\frac{n-1}{n}\right)^n<\frac{n}{n+1}\ \ (n=2,3,\cdots)\ \]

Personally speaking, this problem is fairly easy.

Let $f(x) = x^n, x\in [0,1]$.

Divide $[0,1]$ into $n$ equal parts, and we have
$\left(f(0) + f(1) + \cdots + f\left(\frac{n-1}{n}\right)\right)\frac{1}{n} < \int_0^1 f(x) dx < \left(f(0) + f(1) + \cdots + f\left(\frac{n-1}{n}\right)\right)\frac{1}{n} + \frac{1}{2n}$
where $\frac{1}{2n} = \sum_{i=1}^n \Delta_i$ and $\Delta_i = \int_\frac{i-1}{n}^\frac{i}{n} f(x)dx - \frac{1}{n}f(\frac{i-1}{n})$, therefore
$(\frac{1}{n})^n + \cdots + (\frac{1}{n-1})^n < \frac{n}{n+1} < (\frac{1}{n})^n + \cdots + (\frac{1}{n-1})^n + \frac{1}{2}$

The conclusion follows easily.


Don't look in the spoiler yet, kiddos.

[hide]Sophie Germain's Identity:

a^4+4b^4 = (a^2 - 2ab + 2b^2)(a^2 + 2ab + 2b^2)[/hide]

This was talked about not too long ago.  Also, this should probably be in Intermediate.  So  Ill move it there.

Ai ya.  What was wrong with leaving it in Getting started?

[quote="3X.lich"]Find all integers $x$ for which $x^4+4$ is a prime number.[/quote]
x= +-1

Fine, it's equivalent to (x^2+2x+2)(x^2-2x+2); both factors are necessarily integral and > 1 if x > 1, so $x=\pm 1$ are the only solutions.


trigonometry
geometry
angle bisector
trig identities
Law of Sines
geometry proposed

Let $ ABC$ be a triangle with $ \angle ABC\equal{}80$. Let $ BD$ be the angle bisector of $ \angle ABC$ with $ D$ on $ AC$. If $ AD\equal{}DB\plus{}BC$, determine value of $ \angle A$.

[hide="beginning?"]Let $ BA=c$, $ BC=a$, $ BD=d$, $ DA=n$, and $ DC=m$. Then we're given $ \angle DBA=\angle DBC=40$ and $ n=a+d$. From law of sines: $ \frac n{\sin 40}=\frac d{\sin A}\rightarrow n=\frac{d\sin 40}{\sin A}$. Also, $ \frac a{\sin BDC}=\frac d{\sin C}\rightarrow\frac a{\sin(A+40)}=\frac d{\sin(A+80)}$, so $ a=\frac{d\sin(A+40)}{\sin(A+80)}$. Hence:
\begin{eqnarray*}
\frac{d\sin 40}{\sin A}&=&\frac{d\sin(A+40)}{\sin(A+80)}+d\\
\frac{\sin 40}{\sin A}&=&\frac{\sin(A+40)}{\sin(A+80)}+1
\end{eqnarray*}
From which I probably should be able to derive the answer, which I can see is 20. Problem is i can't. Maybe someone can help...
[/hide]


I know that stuff like mathcounts and the AMCs take more precedence than the SAT for some of us, but how did you guys place in the math section and what do you think of it?

I got a 790 because I read a question wrong. I thought it was really easy though. I took the ACT and got a 33.

I got 800... ;) If you can do AMC problems, SAT problems are no sweat. Just check your answers a few times.

I hadn't taken SAT yet, but I got 80 on PSAT.

I caught two silly mistakes...
If you can do  AMC stuffs, it's pretty easy.


geometry
rectangle
linear algebra
matrix
combinatorics proposed
combinatorics

Take the part of a $n\cdot n$ latice (integer coordinated system) upper to the second diagonal. Consider only the dots which are not on the diagonal and neither on the $(n-1)$ diagonal and neither the vertice, so only the rightangled triangle with $(n-2)$ points on every side, without the rightangled vertice. From these points mark as 1's only $(n-1)$ points. Show that there are three points marked as 1's such that for every two of them, one is situated upper and on the left of the other one, and the rectangle made by drawing paralels to the sides of the square is in the half latice.

can you make a drawing? or give some more wordings on what is the 'first' and 'second' diagonal? first = / and second = \ ? And the (n-1) diagonal, is that the one below? So you start the bottom left in 0,0 and not the center in 0,0?

The way you choose the origin is up to you, or you can consider it as a matrix or whatever. Take the leftbottom to be $(0,0)$
The second diagonal is "\".
The (n-1) diagonal is the one from $(1,n-1)$ to $(n-1,1)$. So the points which interest us are $(x,y)$, with $x+y\geq n+2$, and without $(n-1,n-1)$
I hope the rest is clear.


USAMTS
analytic geometry
graphing lines
slope
trigonometry
pigeonhole principle
Pythagorean Theorem

[quote="flierdeke"]When writing a proof, what types of theorems is it acceptable to assume that the reader doesn't need you to prove?  The Pythagorean Theorem would clearly fall under this category, but there are others that (while reasonably common) aren't quite so famous.  Can anyone recommend a rule of thumb for determining if a theorem is "well-known" or not, or perhaps a list of theorems it's okay to use?
thanks
Kim[/quote]

List all the theorems it's okay to use? That would be longer than most textbooks.

A well-known theorem is a theorem that you learned in school yet the teacher did not provide a name because it was so simple. Something like, "A line perpendicular to a line with slope m has slope -1/m."

If a theorem or property has a name, such as the Law of Cosines or the Pigeonhole Principle, it improves your solution to use the name. You do not have to provide a proof to any theorem you learned in school or from a reference book--the whole idea behind theorems is that they are proven once and then we mathematicians can simply trust the result thereafter.

An example of a theorem that is not well-known is Stewart's Theorem (Art of Problem Solving, Vol. II, p. 39). When we graders were grading problem 5/2/11 on the USAMTS a few years ago, some students solved the problem by saying they were using Stewart's Theorem. We four who were working on that problem asked all the graders in the room if they had heard of Steward's theorm, and no-one had. Fortunately, the fifth student to use Stewart's Theorem quoted it in its entirety. Once I knew the statement of the theorem I scribbled a proof up on a whiteboard in the conference room so that we graders could see what Stewart's Theorem was capable of proving.

[quote="ComplexZeta"]On the USAMTS, if you can cite a theorem (by name or book reference or something) and it doesn't trivialize the result, you may use it. Try not to use theorems that aren't in MathWorld too often.[/quote]

We don't mind if you use a theorem so powerful that the USAMTS problem becomes trivial. However, if you do so, quote the statement of the theorem full length, because we graders won't have heard of it beforehand. On the other hand, if the proof of the theorem requires the result that you are trying to prove, then that is circular reasoning and it won't be a valid solution.

And don't assume that because a theorem is in Mathworld that we have access to it. Grading is conducted in conference rooms because we need at least 25 graders. Those rooms don't have Internet access.

Erin Schram
USAMTS grader


3 1 2 1 3 2
In the above sequence of six digits: there is one digit between the two 1's, two digits between two 2's and three digits between the two 3's.

The same properties are true in the sequence 4 1 3 1 2 4 3 2 where, in addition, there are four digits between a pair of 4's.

Can you write down a sequence of ten digits, pairs of 1's, 2's, 3's, 4's, 5's having these properties and in addition 5 digits between the two 5's?

[i]{Please, do not answer as either "Yes, I can write down" or "No, I cannot write down".}[/i] :D

I keep getting so close... as in, one number is missing one space between the two instances, or I can arrange it just missing one number entirely... :P

Well, JesusFreak197, I haven't tried it yet, but it might be impossible (in which case you'd have to prove that :D)

It's impossible

[hide]
Just look at the parity of the positions of the digits.
The 2's and the 4's have a different parity, so one 2 and one 4 is on a even position. (= 2 even positions)
The 1's, 3's and 5's have the same parity, so none or two 1's (or 3's, or 5's) are on a even position. (= 0,2,4,6 even positions)
But we need 5 digits on a even position, which is impossible.
[/hide]

I get it.
[hide]
Number the positions:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

There will be 5 on even positions and 5 on odd positions. Look at the twos and fours. If one is on an odd position, the other will be on an even position and vise versa so one two and one four will be on an even position..
With 1, 3, and 5, if one is on an even position the other will be on an even position, so there are eitehr 0, 2, 4, or 6 on an even position. However, since there are 5 even positions to be filled, it is impossible.
[/hide]

Note: this is clarifying his solution. I take no credit for it.

Ah, okay. That would explain why I can get solutions where one number is one space off from the necessary number (e.g., 3 5 2 4 3 2 1 5 1 4, where the 4 is one space more than it needs).

Yes, it is impossible!!! But, I would like a generic proof :lol: though the given explanation is excellent!!!

Here follows another sequence with 7 pairs of digits!!!!
7 1 3 1 6 4 3 5 7 2 4 6 2 5
 :ninja:

That is a generic proof. You can't do anything more, unless you mean prove all of the impossible sequences to be impossible.

Yes, Nathan, you are correct. I was actually looking for a way to prove that all the impossible sequences to be impossible.  :lol:

This is the [url=http://mathworld.wolfram.com/SkolemSequence.html]Skolem Sequence[/url], which is well-studied.

A sequence will work with $N$ pairs of numbers if:

if $N$ is odd, $\lfloor \frac N2 \rfloor$ must be odd.

if $N$ is even, $\frac N2$ must be even.

This means that $N = 4k \text{ or } 4k-1$ where k is a natural number.

I don't feel like proving it right now.


My computer's 2gHz, but when all that spyware was in my computer, it felt like 0mHz.

mines only 300 mHz to start with! :mad:  its only an intel celeron!

ahh sorry! didn't mean it in that way lol...
just that 300 is NOT enough for me since i have like a bajillion programs on the comp and i tend to...multitask...a lot... lol :P

Mine's pretty fast...at least I think so...

i just got a 2.4ghz laptop; before that i had a 300mhz computer

I didnt vote on the poll because 'really fast' isnt really a description, because some people have dif views on whats fast. Also, you can have a 3.6 Ghz computer and just totally mess up the compu ter and itll be really slow. My usual comp is i think 2.8 ghz, which is pretty fast, but since i take good care of it, it goes as fast as a lot of other people's computers which are above 2.8ghz.

i don't know how fast my computer is in ghz, but its pretty fast

my computer's 2.8ghz, but luckily, i dont have a lot of stuff installed on it, making it feel like 3+ ghz. :)

i depends in reference to what for the poll i used to think my comp was fast now i think it is average

I have a 2.8GHz Pentium 4. My computer alternates between being really fast and being so slow I want to bang my head againstt the monitor. I wonder why...

Man, I want an AMD proccessor!! They're slightly more expensive that Intel, but super fast!

[quote="mwahahahaha"]lol wow...i was wondering how it came to be 50-50...
then i saw the "2 total votes"...oh well lol  :P

my comp SAYS 1 gigahertz...but in reality i think it's died down to like 300 mh after it's 6 years of usage lol[/quote]

Wow, your computer must have cost a lot; 1 gigahertz was really good back then.

My computer has 3 ghz which is good enough for me

My laptop is an HP 3ghz AMD Athlon... This computer i'm typing on right now is an HP intel Pentium 4.     3.1ghz...

Mine is a 2 or 3ghz(I can't remember and don't know how to tell), but it is really slow because it only has a 40gig hard drive and I'm using 30 of the gigs.

3 GHz, a 250-GB hard-drive, 1GB RAM, but my family has so much stuff on it that it still seems slow!

My other computer is 266MHz, 9GB hard-drive, and 96 MB RAM, but runs almost as fast as the first because I just clean-installed Win98!  Aaah, 5-second shutdowns....  :D

[quote]3 GHz, a 250-GB hard-drive, 1GB RAM, but my family has so much stuff on it that it still seems slow!
[/quote]

that's about as good as a computer gets. the best stats i ever heard of were from dell. 3.8ghz, 1024 gb hard drive, 4gb RAM. i think its the dell XPS generation 5.

Well no. The best so far are 3.4 ghz pentium d dual core, which means that there are 2 3.4 ghz processors, so the spepd will be 6.8ghz+.  :)

[quote="aznmonkey1992"]i was wondering since my computer's slow as heck! :mad:[/quote]

I have one computer in my house that is slow as heck, one that doesnt work, and one that is pretty fast that often loses its conection.

well, its good to see only 10% of users have a slow computer. :)

Weeellll, first of all, CPU gHz is not everything, they're basically how many times it does calculations in a second, how many calculations it does every time is important too. My important specs are:
Athlon 64 3200+
GeForce 6600GT (MSI)
1 gig of RAM
250gig SATA HDD

Next upgrades: 2 more gigs of RAM (in a year or so?), second 6600GT to run in SLI (a year as well? Maybe half of a year? Depends on which games come out at Christmas), More HDD space!!! (maybe a Raptor, dunno though, probably another 250 gigger and a Raptor if I feel rich)

I love building computers, prebuilt (not on eBay), this would cost ~1500 dollars, my price:

$850! Cha-ching!

[quote="ToBe"]Well, technically a 64 bit dual core would be ideal, w/ 4GB of DDR RAM or similar.. but unfortunately that costs a lot, and we dont even have software that supports that. On an additional note, RAM is more important than CPU speed; a 2.8 Ghz computer with  a gig of RAM is faster in almost all cases than a 3.4 Ghz with 512 MB RAM.[/quote]

CPU power and RAM are TOTALLY unrelated. The CPU does calculations, the memory stores variables. Different processes use both in different proportions. For example, Firefox is using around 32 megs of RAM right now, but, say, my prime factorizing program factoring a large number takes up oodles of CPU power (acutally, it eats up all available), but almost no memory. Bang. :D

speaking of speed, what type of video card does everyone have?

[quote]I have an X800XT and a X300.[/quote]

those are very good. :) i have an X300 as well.

[quote="sponge008"]

CPU power and RAM are TOTALLY unrelated. The CPU does calculations, the memory stores variables. Different processes use both in different proportions. For example, Firefox is using around 32 megs of RAM right now, but, say, my prime factorizing program factoring a large number takes up oodles of CPU power (acutally, it eats up all available), but almost no memory. Bang. :D[/quote]

Well alright in purpose they are unrelated, true, but unless you are using a program like yours that relies solely on the processor, you wont notice the difference. For most applications, a 2.8 ghz is just as good as a 3.4 ghz, if it has enough RAM. Although its worthwhile to note that you cant do the same w/ an 800 Mhz processor, the calculations would be too slow. But 2.8 ghz is plenty fast enough for most applications.

I have an ATI Radeon Something. and my isster has an Nvdia NForce

my computer is pretty slow, but after i deleted Mcacfee, it speeded up every thing about 5 times :o

yeah but w/out virus protection your computer can be slowed down to 4x slower than it was before. And w/ that virus you can lose valuable information to hackers. Or just ruin your computer, so in most cases Mcafee or Norton or whatever is worth keeping on your cmoputer.

samehere. my comp. came with mcaffe trial, and it took forever for the welcome screen to go off. (it was configured for autoboot). Then after i deleted mcaffe and added symantec antivirus, and avg antivirus, it sppedded up by 10x.

mine takes a while to get started, but then its really fast :)

u guys should try avg free. it's stronger than mcaffe and symantec, or norton.

[quote="stupidkid"][quote="sponge008"]Weeellll, first of all, CPU gHz is not everything, they're basically how many times it does calculations in a second, how many calculations it does every time is important too. My important specs are:
Athlon 64 3200+
GeForce 6600GT (MSI)
1 gig of RAM
250gig SATA HDD

Next upgrades: 2 more gigs of RAM (in a year or so?), second 6600GT to run in SLI (a year as well? Maybe half of a year? Depends on which games come out at Christmas), More HDD space!!! (maybe a Raptor, dunno though, probably another 250 gigger and a Raptor if I feel rich)[/quote]
Nice computer, why not get a new GeForce 7800 to replace your SLI 6600 GT, supposedly one 7800 can out perform two 6800s.[/quote]

Do you mean a 7800GT, or a 7800GTX? In either case, I'd rather spend $150 than $400/$600.


function
real analysis
real analysis unsolved

A function $ f$ is continoous and one to one on $ \mathbb R$ and $ f(2x \minus{} f(x)) \equal{} x$ $ \forall x \in \mathbb R$.If  there is a

$ x_0$ such that $ f(x_0) \equal{} x_0$ , prove that  :

 $ f(x) \equal{} x$ ,  $ \forall x \in \mathbb R$.

 Babis

Since f is injective you wish to show that $ f(2x \minus{} f(x)) \equal{} f(x)$. We can rewrite this as $ \forall x\forall \varepsilon > 0\exists \delta > 0|f(2y \minus{} f(y)) \minus{} f(y)| < \varepsilon$, where y is in the delta neighborhood of x.

On each closed interval, by the uniform continuity, we can say $ \forall \varepsilon > 0\exists \delta > 0\forall x(|x \minus{} f(x)| < \delta \Rightarrow |f(2x \minus{} f(x)) \minus{} f(x)| < \varepsilon )$

If x is a fixed point of f we have $ |(2y \minus{} f(y)) \minus{} y| \equal{} |f(y) \minus{} y| < \delta$, $ \delta$ as above, for each y in some neighborhood of x by continuity. f is uniformly continuous on any bounded neighborhood of x so we can apply the above.

From the above comments, it follows that the set of fixed points of f is open. Use connectedness of the space and continuity of f.

EDIT: Note that the general problem linked to reduces immediately to this one when we substitute f here with what the solver calls 'h'. In fact, we have proven that it is not necessary to require the connected topological space to be an ordered field (R).


There are 14 numbers, so the median is the AM of the 7th and 8th, which is

$ \frac{18 \plus{} 20}{2} \equal{} \boxed{19}$


[quote="Pineappleperson"]yea....i don't know where this would go...but

my teacher said that there was a story behind the derivative of e^x; It consisted of something about buddha, elephants and e^x...

does anyone here know what the story was?[/quote]

All I know is $ (e^x)'\equal{}e^x$  :lol:

[quote="ProtestanT"][quote="Pineappleperson"]yea....i don't know where this would go...but

my teacher said that there was a story behind the derivative of e^x; It consisted of something about buddha, elephants and e^x...

does anyone here know what the story was?[/quote]
try googling it...type in "buddha elephants e x derivative"

also, theres a yahoo answers question on it
http://sg.answers.yahoo.com/question/index?qid=20071026113616AAxxVuq[/quote]

LOL i bet it's the same person
it gives you dumb wiki article though

This reminds me, many stories teachers come up with came from their personal experiences or are made up by the teachers themselves. Therefore many of them might not bother putting them on the internet.


calculus
integration
LaTeX
calculus computations

The $LaTeX$ is very confusing.  If you wouldn't mind simply using $exp(x)$ in place of $e^x$ ...

Also, you use $\{ \}$ for multi-character exponents, not $[ ]$.

Nevermind, I get it.  It's due to x^-0.5 and x^-3/2 are symmetric through x = y. :lol:


AMC
AIME
geometry
parallelogram
perimeter
ratio
area of a triangle

In the triangle ABC, the segments have the lengths shown and x + y = 20. Find its area.
 See the attachment for picture, have fun!! It's a very good problem

A first step, maybe even to be described as a hint:

[hide]We know that 6/(6 + 6) + 3/(3 + 9) + y/(x + y) = 1, by examining the areas of the triangles ABO, BCO, and CAO, where O is the center point there (I can't read what it is actually labelled).  Thus, y = 5 and x = 15.[/hide]

That's pretty close to I did it too... [hide]Call the intersection point of AD, BE, CF point O. Call Area of triangle AFO a, FOB b, OBD c, ODC d, OEC e, AOE f. Because many of the triangles shares the same altitude, we have a+b=c, e+f=d, (e+f)/3=a, (c+d)/3=b,(20-x)(a+b)/x=f, and  (20-x)(c+d)/x=e. We have a+b=(e+f+c+d)/3=c, sub in d and c for a+b and e+f we have c=d. Then we have e+f=(a+b+c+d)(20-x)/x, substitution again and simplify it to 20-x/x=1/3, so x=5, and y=15. [/hide]

I did the same thing, but I might have done it "illegaly".

[hide]I applied ceva's to the lenghts associated with each segment emanating from the point of concurrency to the triangle and then the x+y=20 thing and got x=5 and y=15. Is this allowed, or did it just by accident come out correctly?[/hide]

What you did can be proven to always come out correctly by examining the areas of the triangles I described.  Try to do it yourself, by figuring out what fraction of the total area each of those three triangles occupies, and then adding to get 1.

Now, does anyone have a second step?

I used mass-point geometry to get x = 15 and y = 5 also.  Note that 2 * [BPC] = [ABC] because P bisects AD.  Looking at triangle BPC, PD is the median intersecting side BC at D.  We can easily find the length of BC by Stewart's Theorem or by forming a parallelogram and finding the lengths of the diagonals.  BC = 6 :sqrt: 13.  Using Heron's formula,  :sqrt: (12 + 3 :sqrt: 13)(12 - 3 :sqrt: 13)(3 :sqrt: 13 - 3)(3 :sqrt: 13 + 3) = 54 = [BPC].  54 * 2 = [ABC] =108 and that's our answer.

What I did was....[hide]since c=d, then BD=DC because c and d have the same altitude. The I used Heron's Formula, call the semi-perimeter of PDC s we have equations c=d, or s(s-6)(s-9)(15-s)=(s+3)(s-3)(s-12)(18-s). Multiply it out s^4-30s :^3: +279s :^2: -810s=s^4-30s :^3: +207s :^2: +270s-1944. Simplify it to 72s :^2:-1080s+1944=0, or s :^2: -15s+27=0. The root is (15+3 :sqrt:(13))/2. Sub into Heron's formula we have d=27. 27*4=108[/hide]

What is mass-point geometry and what's Stewart's theroem?

http://mathworld.wolfram.com/StewartsTheorem.html
http://mathworld.wolfram.com/BarycentricCoordinates.html

WarpedKlown, you didn't prove that D was the midpoint!

JBL, could you explain how you were able reason that 6/(6 + 6) + 3/(3 + 9) + y/(x + y) = 1? is it a fundamental of geometry that i missed?

its pretty fundamental, he took three smaller triangles which add up to the bigger one, BCP/BCA + ABP/ABC + CAP/CAB = 1.

oh ya, forgot, and you know they have the same bases so you can just take the length of the line to represent the triangles in ratios.

Let me just do this in general.  Look at the diagram below.  In the top picture, we have three lines in the triangle intersecting in one point.  We draw the two perpendicular lines shown, which give us two similar right triangles from which we get that a/(x + y) = b/y, or in other words that b/a = y/(x + y).  Now look at the second triangle.  The dark blue triangle shares a base with the whole (both colors of blue) triangle.  Thus, the ratio of their areas, [Dark Blue]/[Whole Triangle] = (height of dark blue)/(height of whole triangle) = b / a = y/(x + y).

Now, do the same thing around the other two sides.  In each case, we see that the fraction of the area taken up by those triangles is equal to the ratio of the segment between the middle-point and edge to the entire segment (sorry I didn't use letters, hope it's clear).  But when we add the three fractions for the area up, we get 1 (since together, they make up the entire area).  So when we add the fractions for the cevians up, we also get 1.

ah. thanks. i understand. that makes perfect sense. i need to brush up on some geometry before the aime.

Congradulations to all the people who solved it. If you solved it in a different way than us, please post it. So far including my way, I have seen three ways to do this.

I still haven't seen a complete proof yet -- no one has shown that D is in fact the midpoint, which is pretty fundamental to solving this.

By the way, was this from 2003?

I think I have a proof. This is from 1989.

I have a proof using mass-point geometry, it's really helpful on concurrency problems, but I can't post it now because I have to leave.

It is easy to show D is midpoint through mass point geometry.

Just let A = 2, B, C = 1. Tada, D must be a midpoint.

So far, I think warpedklown has the best solution...

[quote="JBL"]I still haven't seen a complete proof yet -- no one has shown that D is in fact the midpoint, which is pretty fundamental to solving this.[/quote]

Well, I pretty much used Menelaus on triangles APE, AFP, and DBF to get three equations relating the ratios between FA and FB, DB and DC, and EA and EC. After this hunk of mess, I substituted some stuff and used Ceva to show that D was the midpoint of BC. I then used Stewart's on triangle BPC to find BC and went from there (Heron's, etc..). It was quite a pain, but, hey, you do what you have to. ;)

[quote]By the way, was this from 2003?[/quote]

No, 1989.


$ABC$ is a triangle such that:$AB=9,BC=15,CA=16$. $D$ is a point in $AC$ such that $\measuredangle ABD = 2 \measuredangle DBC$. Find $cos \measuredangle ADB$

If you have a nice solution, please post it :D .

[quote="frt"]Hmm... I actually found the numerical value for $\angle ADB$... That's probably wrong since the problem asks for $\cos\angle ADB$, which implies that we can't find the numerical value of the angle.

Well, I'll post my solution anyway:
[hide]
Extend $BA$ to $E$ such that $AE=16$ ($BE=25$).
$AC=AE$, so we can let $\angle AEC=\angle ACE=\alpha$ and $\angle BAC=2\alpha$.
Note that $\triangle ABC\sim\triangle CBE$ by SAS.
Hence $\angle BAC=\angle BCE=2\alpha$.
But $\angle ACE=\alpha$, so $\angle ACB=\alpha$.
Therefore, $\angle BAC=2\angle BCA$.

Now let $\angle ABD=2\beta$ and $\angle DBC=\beta$.
We have $\angle BAC+\angle ACB+\angle ABC=3\alpha +3\beta=180^\circ \Longleftrightarrow \alpha +\beta =60^\circ$.
Then $\angle ADB=\angle DCB+\angle DBC =\alpha +\beta =60^\circ$.

Hence $\cos\angle ADB=\boxed{\frac{1}{2}}$
[/hide][/quote]

It is nice and I believe it is correct  :D


ABC is a triangle. BD is the an angle bisector. E, F are the feet of the perpendiculars from A, C respectively to the line BD. M is the foot of the perpendicular from D to the line BC. Show that angle DME =  angle DMF.

[quote="shobber"]Let DM and AE meet at N. Then B, E, N and M lie on the same circle. Hence <DNE=<DBM=<DBC=<DBA. Therefore B, A, D and N are concyclic. Therefore <DAN=<DBN=<EBN=<EMN=<EMD. Since $AE \parallel CF$, so <DAE=<DCF. Since D, F, C and M are concyclic, so <DMF=<DCF=<DAE=<DME.[/quote]
Nice proof shobber. :lol:


And what is usually the qualifying score?

Yes...a 70...I don't think I will make it...

Can they write the CMO afterwards, as well?

Actually, I meant "Can American students qualifying for CMO then take the CMO?"

From what I've seen, Americans can qualify for the CMO, which generally implies that you can take it.  ;)

I'm not sure if you can actually win prizes from the CMO though.

Canadians win stuff from the USAMO, so we should be able to win prizes in the CMO too  :wink:

One reason Canadians write the USAMO is because it is taken into account for Canada's IMO team selections.  However, USA IMO people do not qualify for the team through the CMO, which usually is much easier than the USAMO.  So I don't think Americans should write the CMO.  Plus, if they officially could, Canada would get embarassed.  :(

I do not know if in fact Americans can write it.

I am pretty sure the Americans cannot write CMO. This is kind of unfair, but like Kileel said, if they did, Canada would be veru embarassed cuz more than half the ppl writing cmo would be Americans. :rotfl:

On the contrary Alex, I am certain Americans can write the CMO.  But like I said, I'm not sure if they could actually win prizes or place on it.  John Zhou and Sunil Agrawal...both people from Michigan, wrote the CMO last year.

I don't think there would ever be a problem such that more than half of the CMO writers are American.  Most high schools in the United States simply don't have the administration power to administer the contest at their respective places.  Infact, I think only Michigan so far has the ability to administer the COMC; and it's not a high school that's doing so.  I think they've got this gigantic learning centre or something...

Americans are probably unofficial participants.

Of course, the CMS has to know that you're a Canadian citizen.  It will probably assume that you are an American citizen otherwise if you're taking the COMC in the United States.

Yeah, if you do random stuff like the MOCP or subscribe to CRUX, then they'll probably figure that you're Canadian.  If you've done other stuff before with Canadian Math, then they'll probably figure that you're Canadian.  But the CMS doesn't automatically assume that anyone's Canadian.

I never realized the COMC was this important to American Math.  Just take the AMCs and AIMEs jeez...

Yeah i remember when i qualified for cmo they asked me 3 questions:
1. are you under 20 yrs of age?
2. did you take putnam?
3. Are you a Canadian Citizen or Permanent Resident?
If the answers were yes, no, yes, then only you could write cmo. :!:

[quote="mathgeniuse^ln(x)"][quote="rem"]Sorry I did not know that...
Do you know if Americans are recognized as official particpants or not?[/quote]

If I am not mistaken...all you have to have is citizenship to be official...

For example, I was born in Montreal, Canada, so I have citizenship...so if I qualify for the CMO, I can take it officially, and win prizes...which is not going to happen.[/quote]

I don't think this is true (at least, I don't think it should be). If you are a permanent American resident, you shouldn't be able to represent Canada if you are a canadian citizen.

If you have dual citizenship...it should be okay.

Well I'm not aware that the US allows Canadian residents to attend MOP even if they have American citizenship, so I don't see why it would work the other way around.

Phelpedo brings up a point in that there's a difference between being defined Canadian, and explicitly Canadian.  Like the CMS would have to know that you're clearly a Canadian citizen before they invite you to other stuff like the APMO.

I mean seriously...just take the AMCs and AIMEs lol.  It probably shouldn't matter then whether you're an official CMO participant or not; an "official USAMO qualifier" is pretty darn cool as well.

And stop invading our boards.

Stupid Americans...

Jk.   :lol:

I'm mostly taking the COMC to qualify for the CMO to practice for the USAMO. No fair that you canadians get to practice on three olympiads before taking the USAMO :P  :wink:

lowest_reynolds: We are trying to save this forum.  So, do [b]NOT[/b] scold people who, by upping the post count, are trying to help us. :rotfl:

Phelpedo: COMC is not considered an olympiad.  Also, CMO and APMO are not merely practice for USAMO.  For Canadians, they are important too!

Comment: COMC is this wednesday!  OMG! :D

[quote="Kileel"]lowest_reynolds: We are trying to save this forum.  So, do [b]NOT[/b] scold people who, by upping the post count, are trying to help us. :rotfl: [/quote]

Haha, fair enough Kileel. :P


integration
topology
geometry
real analysis
real analysis theorems

Stokes Theorem states:
$\int_{\partial \Omega}d\omega = \int_\Omega\omega$

(1)  Could we somehow generalize the idea to $\Omega$ being a manifold of fractional dimension?
(2)  In particular, could we define a cuboid or simplex of fractional dimension?  :oops:

This reminded me of Manin's survey [i]The notion of dimension in geometry and algebra[/i] in Bull. Amer. Math. Soc. 43  (2006),  no. 2, 139--161.  Unless I am mistaken, Bulletin of the AMS is in open access - [url=http://www.ams.org/journal-getitem?pii=S0273-0979-06-01081-0]try this link[/url]. From p.143:
[quote]I do not know whether some subsets S are so topologically “good” as to deserve the name of D–dimensional manifolds. Is there a (co)homology theory with geometric flavor involving such fractional dimensional sets?[/quote]
I do not know either.  :|


I've recently shut down the Weekly Potpourri because few people were submitting answers. :(

However, I'd like to try running a competition one more time. :) Do you guys have suggestions for what you might be interested in? I'll gather this advice into a new competition :)

My thoughts:
1) Less problems at a time, more frequently
2) Multiple Choice?
3) A good name :P

Note to Fellow Mods: I will move this thread once the competition starts. I want to have it here right now because it will be hosted over this forum and not all people visit the "Other Problem Solving Topics" forum :P

This is about the time when everyone gets really busy with all sorts of things.  It's usually a time issue rather than an issue of the content of the competition.

Yeah, I really wanted to do the Weekly Popourri, but school and USAMTS usually take away all of my time. Maybe in the summer, though. I doubt I'd do any competition now.

You're probably right. I'll just get back to solving problems then...  :|

i think the problem is that finals are coming soon, and also, the time that i have to do math is unreliable, so i usually just look at a few problems in the forums before i have to stop


In a country 6 cities are connected two by two with round-trip air routes operated by exactly one of the two air companies in that country.

Prove that there exist 4 cities $A$, $B$, $C$ and $D$ such that each of the routes $A\leftrightarrow B$, $B\leftrightarrow C$, $C\leftrightarrow D$ and $D\leftrightarrow A$ are operated by the same company.

[i]Dan Schwartz[/i]

There is a company which operates at least $8$ routes, so let's show that in a graph with $6$ vertices and $8$ edges there is a $4$-cycle.

We count the pairs $(x,\{a,b\})$, where $x$ is connected to both $a$ and $b$. Let $d_i,\ i\in\overline{1,6}$ be the degrees of the vertices. By considering each vertex $x$, and for each vertex counting the pairs $a,b$, we find our number to be equal to $\sum\frac{d_i(d_i-1)}2$, which is $\ge 14$ (simple estimate obtained from $\sum d_i=16$ and $6\sum d_i^2\ge (\sum d_i)^2$).

Let's show that there are at least two pairs $(a_1,b_1),(a_2,b_2)$ which do not have any common neighbours, if our graph has no $4$-cycles. Assume the contrary. We can find $a,b$ which are connected s.t. they have a common neighbour $c$. Let the other three vertices be $d,e,f$. Now, since there are at least two edges from the group $d,e,f$ to the group $a,b,c$ and we assumed that there was at most one pair of vertices with no common neighbours, we may assume that $a,d$ are connected and have a common neighbour. This neighbour cannot be $b,c$, because then we would have a $4$-cycle, so assume $ad,de,ae$ are edges.

We can't add any more edges to the configuration $abcde$ without getting a $4$-cycle, so there are at least two edges from $f$ to this configuration, but we can easily see that this gets us, again, a $4$ cycle.

We have thus shown that if there's no $4$-cycle, then there are at least two pairs of vertices with no common neighbours. However, according to the counting argument above, there are at least $14$ pairs of the form $(x,\{a,b\})$ with $xa,xb$ edges, so there must be a pair $a,b$ which appears twice here (since there are totally $15$ pairs), meaning that $a,b$ have at least two common neighbours, and these, together with $a,b$, form a $4$-cycle.

it's not very rigurous.. but here goes...
We'll prove that every graph with 6 vertices and 8 edges has at least one cycle of length 4.(from this the conclusion follows).
Denote with $d_i$, $i=1,\ldots,6$ the degrees of the vertices.
case1. We have a vertex with degree 5. Let's say that $d_1=5$. Now we must connect the vertices $2,3,4,5,6$ with 3 edges. But this means that one will have degree at least 3 (let's say 2, and 2 is connected with 3 and 4). Then 1,3,2,4 is a cycle of length 4.
case2. We have a vertex with degree 4. Let's say $d_1=4$ and that 1 is connected with 2,3,4,5. Then between 2,3,4,5,6 there must be 4 more edges. If the vertex 6 has at least two edges with 2,3,4,5 then the 4-cycle is formed. Else, (let's say  (6,2) is an edge ) then  between 2,3,4,5 we must have 3 edges so at least one will have the degree 2 (with respect to the subgraph $\{2,3,4,5\}$ ) and analogue with the first case the 4 cycle is formed
case3. all vertices have their degree at most 3. If $k$ of them has the degree 3 then
Then $16 = \sum d_i\le 3k+2(6-k)=k+12$ so $k\ge 4$. Now I think the 4-cycle can be easily found.


1.  let [tex] f(x,y) = (e^x cos(y), e^x sin(y)). [/tex]  show that the function is locally invertible, but does not have a global inverse.

2.  let [tex] f(x,y) =( x^2 - y^2, 2xy). [/tex]  where is this function locally invertible?  does it have a global inverse?

for #1, i found that the determinant of f'(x,y) is equal to [tex]
e^2^x [/tex] which is never equal to 0.  so by the inverse function theorem the inverse should exist everywhere.  right?  but i'm not sure how to prove it does not have a global inverse.

for #2, i set determinant of f'(x,y) equal to 0, and got the equation [tex] x^2 + y^2 = 0, [/tex] so determinant is equal to 0 when x=0 and y=0.  so by the inverse function theorem the function is locally invertible everywhere except when x=0 and y=0.  am i on the right track?  i don't know how to answer the 2nd part of this question.  how do i find whether it has a global inverse?

To show that these functions lack global inverses, you merely need to show that they are not one to one.

In the first case, for every $n\in\mathbb{Z},\,f(0,2\pi n)=(1,0).$ In fact, more generally, $f(r,\theta)=f(r,\theta+2\pi n).$ This function is periodic in $y$ and fails to be be one-to-one. In fact, it is infinite-to-one.

In the second case, $f(1,0)=f(-1,0)=(1,0).$

More generally, if $(x,y)\ne(0,0),\,f(-x,-y)=f(x,y)$ and the function is two-to-one.

---

To anyone with the slightest familiarity with complex function theory, these functions should be extremely familiar.

The first one is $f(z)=e^z.$ Saying that it does not have a global inverse is saying that there is no uniquely defined complex logarithm.

The second one if $f(z)=z^2.$ Every nonzero complex number has two square roots.


quadratics
number theory proposed
number theory

Find all nonegative integers $ x,y,z$ such that $ 12^x\plus{}y^4\equal{}2008^z$

I participated in that competition, and I had this solution but I made stupid mistake:

[hide]It's easy to see that (x,y,z)=(0,0,0) is a solution. Prove that x,y,z>0. If $ x$ is odd, then $ \minus{}3$ is quadratic residue mod $ 251$, if $ x$ is even then $ \minus{}1$ is quadratic residue mod $ 251$ and that's it.
The mistake that I made was that I proved that $ \minus{}3$ really is quadratic residue mod $ 251$...  idiot[/hide]

Bye

Here is my solution:
x=0  =>  y=z=0.
Examine the problem when $ x\geq 1$
* If x is even, $ x \equal{} 2k$.
We have $ 12^x \plus{} y^4\equiv 0 (mod251)$, then $ (12^x)^{125} \plus{} (y^4)^{125}\equiv 0 (mod251)$.
It means that $ A \equal{} (12^k)^{250} \plus{} (y^2)^{250}\equiv 0 (mod251)$
Use Fermat theorem, with $ (12^k,251) \equal{} 1$ and $ (y^2,251) \equal{} 1$, we have $ (12^k)^{250}\equiv (y^2)^{250}\equiv 1 (mod 251)$, so $ A\equiv 2(mod 251)$.
* If x is odd. Let  $ y \equal{} 2^m. t$ ( $ t$ is an odd numbers, $ m\geq 1$).
At there, $ 2^{2x}.3^x \plus{} 2^{4m}.t^4 \equal{} 2^{3z}.251^z$
We can easy solve this case.
Answer: $ (x,y,z) \equal{} (0,0,0)$

I have a question, if my solution is correct.

Of course we first try trivial cases when one of $ {x,y,z}$ is $ 0$, which leads us to conclusion that all must be $ 0$. So, in the continuation we can assume $ x,|y|,z$ are all greater than $ 0$.

We can denote $ y\equal{}2^ay_1$, where $ y_1$ is odd integer and $ a>0$.

Hence the equation becomes:

$ 2^{2x}3^{x}\plus{}2^{4a}y_1^4\equal{}2^{3z}251^z$

If there exists $ min\{2x,4a,3z\}$, then on dividing the equation with $ min\{2x,4a,3z\}$ we obtain an odd number on one side and even on the other. Hence $ min\{2x,4a,3z\}$ does not exist. Hence at least two numbers from $ \{2x,4a,3z\}$ are the same. Furthermore if all three were the same, we would again get one odd and one even side. Therefore exactly two are the same.

Now we have three cases:

1.$ \textbf{2x\equal{}4a}$; $ \textbf{max\{2x,4a,3z\}\equal{}3z}$

$ \rightarrow a\equal{}\frac{x}{2}\rightarrow 2^{2x}(3^{x}\plus{}y_{1}^{4})\equal{}2^{3z}251^{z}\rightarrow 3^{x}\plus{}y_{1}^{4}\equal{}2^{3z\minus{}2x}251^{z}$

Now since $ 3z\minus{}2x>0, 2$ must divide $ y_{1}^{4}\plus{}3^{x}$. But since $ y_{1}^{4}\equiv1(mod 8)$ and $ 3^{2k}\equiv 1 (mod 8)$, $ 3^{2k\plus{}1}\equiv 3 (mod 8) \rightarrow y_{1}^4\plus{}3^{x}\equiv 2$ or $ 4 (mod 8)$.

If $ x\equal{}2k$ we can use Fermat`s theorem as mr.danh has already shown.

If $ x\equal{}2k\plus{}1 \rightarrow 4|y_{1}^4\plus{}3^{x} \rightarrow 3z\minus{}2x\equal{}2 \rightarrow z\equal{}\frac{2(x\plus{}1)}{3} \rightarrow 3|x\plus{}1 \rightarrow x\equal{}3x_{1}\plus{}1, z\equal{}2x_{1}$.

Now our equation becomes:

$ 2^{2}251^{2x_{1}}\equal{}y_{1}^{4}\plus{}3^{x}$, which is equivalent to
$ (2\cdot251^{x_{1}}\minus{}y_{1}^{2})(2\cdot251^{x_{1}}\plus{}y_{1}^{2})\equal{}3^{x}$
Therefore $ 2\cdot251^{x_{1}}\minus{}y_{1}^{2}\equal{}3^{l}$, $ 2\cdot251^{x_{1}}\plus{}y_{1}^{2}\equal{}3^{j} \rightarrow 3^{l}\plus{}3^{j}\equal{}3s\equal{}4\cdot251^{x_{1}}$. But since $ 3$ does not divide  $ 4\cdot251^{x_{1}}$ this leads us to contradiction.

2.$ \textbf{2x\equal{}3z}$; $ \textbf{max\{2x,4a,3z\}\equal{}4a}$

$ \rightarrow z\equal{}\frac{2x}{3} \rightarrow x\equal{}3x_{1} \rightarrow z\equal{}2x_{1}
\rightarrow 2^{2x}(251^{2x_{1}}\minus{}3^{2x_{1}})\equal{}2^{4a\minus{}2x}y_{1}^{4} \rightarrow
251^{2x_{1}}\minus{}2^{4a\minus{}2x}y_{1}^{4}\equal{}3^{2x_{1}} \rightarrow (251^{x_{1}}\minus{}2^{2(a\minus{}x)}y_{1}^{2})(251^{x_{1}}\plus{}2^{2(a\minus{}x)}y_{1}^{2})\equal{}3^{2x_{1}}$

But this again leads us to $ 2\cdot251^{x_{1}}\equal{}3s$, which is contradiction.

3.$ \textbf{4a\equal{}3z}$; $ \textbf{max\{2x,4a,3z\}\equal{}2x}$

$ \rightarrow z\equal{}\frac{4a}{3} \rightarrow a\equal{}3a_{1} \rightarrow z\equal{}4a_{1}
\rightarrow 2^{2x}3^{x}\equal{}2^{4a}(251^{z}\minus{}y_{1}^{4})$

Again $ 2^{2x\minus{}4a}3^{x}\equal{}(251^{2a_{1}}\minus{}y_{1}^{2})(251^{2a_{1}}\plus{}y_{1}^{2})$ and $ 2\cdot251^{2a_{1}}\equal{}3s$.

Therefore $ (0,0,0)$ is the only solution to this equation.

[quote="tmrfea"]
1.$ \textbf{2x\equal{}4a}$; $ \textbf{max\{2x,4a,3z\}\equal{}3z}$

$ \rightarrow a\equal{}\frac{x}{2}\rightarrow 2^{2x}(3^{x}\plus{}y_{1}^{4})\equal{}2^{3z}251^{z}\rightarrow 3^{x}\plus{}y_{1}^{4}\equal{}2^{3z\minus{}2x}251^{z}$

Now since $ 3z\minus{}2x>0, 2$ must divide $ y_{1}^{4}\plus{}3^{x}$. But since $ y_{1}^{4}\equiv1(mod 8)$ and $ 3^{2k}\equiv 1 (mod 8)$, $ 3^{2k\plus{}1}\equiv 3 (mod 8) \rightarrow y_{1}^4\plus{}3^{x}\equiv 2$ or $ 4 (mod 8)$.

If $ x\equal{}2k$ we can use Fermat`s theorem as mr.danh has already shown.

If $ x\equal{}2k\plus{}1 \rightarrow 4|y_{1}^4\plus{}3^{x} \rightarrow 3z\minus{}2x\equal{}2 \rightarrow z\equal{}\frac{2(x\plus{}1)}{3} \rightarrow 3|x\plus{}1 \rightarrow x\equal{}3x_{1}\plus{}1, z\equal{}2x_{1}$.

Now our equation becomes:

$ 2^{2}251^{2x_{1}}\equal{}y_{1}^{4}\plus{}3^{x}$, which is equivalent to
$ (2\cdot251^{x_{1}}\minus{}y_{1}^{2})(2\cdot251^{x_{1}}\plus{}y_{1}^{2})\equal{}3^{x}$
Therefore $ 2\cdot251^{x_{1}}\minus{}y_{1}^{2}\equal{}3^{l}$, $ 2\cdot251^{x_{1}}\plus{}y_{1}^{2}\equal{}3^{j} \rightarrow 3^{l}\plus{}3^{j}\equal{}3s\equal{}4\cdot251^{x_{1}}$. But since $ 3$ does not divide  $ 4\cdot251^{x_{1}}$ this leads us to contradiction.[/quote]
Actually since $2x=4a$ then $2|x$, therefore you don't have to consider the case $2 \nmid x$.


Let $P_1P_2P_3\dots P_{12}$ be a regular dodecagon. Show that \[\left|P_1P_2\right|^2 + \left|P_1P_4\right|^2 + \left|P_1P_6\right|^2 + \left|P_1P_8\right|^2 + \left|P_1P_{10}\right|^2 + \left|P_1P_{12}\right|^2\] is equal to \[\left|P_1P_3\right|^2 + \left|P_1P_5\right|^2 + \left|P_1P_7\right|^2 + \left|P_1P_9\right|^2 + \left|P_1P_{11}\right|^2.\]

Place the 12 points of the regular dodecagon on a circle.  Notice $P_1P_7$ is the diameter which means triangle $P_1P_7P_k$ for $k \in \left\{ 2,3,4,5,6,8,9,10,11,12\right\}$ is a right triangle all with $P_1P_7$ as the hypotenuse.
So we have
\[\left|P_1P_2\right|^2 + \left|P_1P_8\right|^2=\left|P_1P_3\right|^2 + \left|P_1P_9\right|^2\]
\[\left|P_1P_{4}\right|^2 + \left|P_1P_{10}\right|^2=\left|P_1P_5\right|^2 + \left|P_1P_{11}\right|^2\]
After cancelling out the above terms we are left with showing
\[\left|P_1P_6\right|^2 + \left|P_1P_{12}\right|^2=\left|P_1P_{7}\right|^2\]
which is true by the pythagorean theorem.

wow, that is an olympiad problem, another way would be to use complex numbers, but using the fact that chords that intercept the congruent arcs in congruent circles is much easier