inequalities
symmetry
linear algebra
matrix
calculus
topology
function

Let $ \mathbf x,\mathbf y\in\mathbb R^n$ be nonzero, define the [b]angle[/b] between $ \mathbf x$ and $ \mathbf y$ by

$ \angle(\mathbf x, \mathbf y) \equal{} \arccos \frac {\langle\mathbf x,\mathbf y\rangle}{\|\mathbf x\| \|\mathbf y\|}.$

We can check by Cauchy-Schwarz Inequality that it is well-defined.

The linear transformation $ T: \mathbb R^n\to\mathbb R^n$ is said to be [b]angle preserving[/b] if $ T$ is 1-1; and for any nonzero $ \mathbf x, \mathbf y\in\mathbb R^n$, we have

$ \angle(T(\mathbf x), T(\mathbf y)) \equal{} \angle(\mathbf x, \mathbf y).$

Now, let $ \{\mathbf v_1, \cdots, \mathbf v_n\}$ be a basis for $ \mathbb R^n$, and $ \lambda_1, \cdots \lambda_n\in\mathbb R$ satisfy

$ T(\mathbf v_i) \equal{} \lambda_i\mathbf v_i$ for all $ 1\le i\le n$.

Show that $ \lambda_1 \equal{} \cdots \equal{} \lambda_n$ if $ T$ is angle preserving.

[hide="HINT"]Consider the angle between $ v_i$ and $ v_i\plus{}v_j$, and try to find a limiting argument.[/hide]

1)   $ \{\mathbf v_1, \cdots, \mathbf v_n\}$ must be an orthonormal basis.

 2) $ \angle(\mathbf x, \mathbf y) \equal{}\angle(\mathbf y, \mathbf x)$ then the following symmetry is angle preserving:
$ T\equal{}\begin{pmatrix}1&0\\0&\minus{}1\end{pmatrix}$.

In an edition of Spivak's Calculus on Manifolds, the absolute values are present, i.e., you must prove that $ |\lambda_i| \equal{} |\lambda_j|\forall i,j$.  Unfortunately, I thought of the angle in a geometrical way (that is, $ \angle(x,y)\neq\angle(y,x)$), and so the counterexample that Loup Blanc provided, which was the same as mine, did not convince me. To prove the [i]cotrrect[/i] version, assume to the contrary that $ |\lambda_i|\neq|\lambda_j|$ for some $ i\neq j$. Assume without loss of generality that $ |\lambda_i| > |\lambda_j|$.  Then it follows that $ \angle(v_i,v_i \plus{} v_j) \equal{} \angle (v_i,v_i \plus{} \frac {\lambda_j}{\lambda_i}v_j) \equal{} \angle (v_i,v_i \plus{} \left(\frac {\lambda_j}{\lambda_i}\right)^2v_j) \equal{} \hdots \equal{} \angle(v_i,v_i) \equal{} 0$.  Thus, $ v_i$ and $ v_j$ are linearly dependent, which is a contradiction to the fact that $ \{v_i\}$ is a basis.  Notice that the above proof does hint that things can go wrong if $ \lambda_i \equal{} \minus{} \lambda_j$ for some $ i,j$.

Hi boxedexe,
your proof is correct bur your remarks are not:
1) "I thought of the angle in a geometrical way (that is,    $ \angle(\mathbf x, \mathbf y) \not\equal{}\angle(\mathbf y, \mathbf x)$)
Geometrically it is a nonsense:
For example, in $ \mathbb{R}^3$, you CANNOT define the signum of an angle between 2 vectors. An angle is a number in $ [0,\pi[$.
2) "Notice that the above proof does hint that things can go wrong if  $ \lambda_i\equal{}\minus{}\lambda_j$ for some $ i,j$.
It is false because necessarily $ <v_i,v_j>\equal{}0$.
   In fact this exercise is a very particular case of this one:
Let $ E$ be an euclidean space.
Let $ T$ be a linear function : $ E\rightarrow{E}$ s.t. if $ <u,v>\equal{}0$ then $ <Tu,Tv>\equal{}0$. Show that there exist a non negative  real $ a$ and an orthogonal application $ U$ s.t. $ T\equal{}aU$.

Replies:

1) By the first statement, I simply meant that I worked in $ \mathbb{R}^2$ in an intuitive way (calculating angles by taking orientation into consideration), and so that lead to misunderstandings.

2) By saying that the original statement may go wrong if $ \lambda_i\equal{}\minus{}\lambda_j$ was a way of saying that your counterexample may be thought of by taking $ \lambda_1\equal{}\minus{}\lambda_2\equal{}1$.  Also, I did not assume that $ \left< v_i,v_j\right>\equal{}0$.

I hope that the above replies are clearer than before (and that they make sense).

[i]PROOF[/i]: Since $ Tv_i\equal{}\lambda_i v_i$ is equivalent to $ T(\frac{v_i}{\|v_i\|})\equal{}\lambda_i\frac{v_i}{\|v_i\|}$, one can without loss of generality assume that $ \|v_i\|\equal{}1$. Now for $ i\not\equal{}j$, consider the verctors $ v_i\plus{}v_j$ and $ v_i\minus{}v_j$. One has $ \langle v_i\plus{}v_j,v_i\minus{}v_j \rangle\equal{}0$ by direct computation. Hence $ \langle T(v_i\plus{}v_j),T(v_i\minus{}v_j) \rangle\equal{}0$, since $ T$ is angle-preserving. It follows from the latter equation that $ \lambda_i^2\equal{}\lambda_j^2$ or $ |\lambda_i|\equal{}|\lambda_j|.\,\Box$


I believe he means:

[quote="ashrafmod"]find all real solution x
$x^{10}-x^{8}+8x^{6}-24x^{4}+32x^{2}-48=0$[/quote]

Well, it factors into $(x^{2}-2)(x^{8}+x^{6}+10x^{4}+24-4x^{2})$.

So that means two solutions are $\pm \sqrt{2}$.  Now if we look at the right side we see that any solutions would have to be imaginary because from $x=1$, the polynomial is strictly increasing, as they all have even powers.  Therefore, those are the only two real solutions.  This is still a little bit too easy for pre-olympiad.


Prove that if Ax=0 has unique solution (thus, the columns and rows of A are linearly independent), then A is left invertible.

Assume $ A\in\mathcal{M}_{n,m}: \mathbb{R}^m\rightarrow\mathbb{R}^n$. Here $ ker(A)\equal{}\{0\}$ that is the columns are linearly independent and $ n\geq{m}$. Then you can easily construct a matrix $ B\in\mathcal{M}_{m,n}: \mathbb{R}^n\rightarrow\mathbb{R}^m$ s.t. $ BA\equal{}I_m$.  $ B$ is the left inverse of $ A$ and is onto.


I've seen Cauchy's function equation in the form $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ with the addtional condition is continuity .
I'm wondering if there is a solution for the general case :
$ f(x_1\plus{}x_2\plus{}..\plus{}x_n)\equal{}f(x_1)\plus{}f(x_2)\plus{}..\plus{}f(x_n)$ and f is a continuous funtion

p/s : i'm a newbie at funtion equation , so if you guys have any books or anything thing related to it please let me know  :blush:

[quote="Mathias_DK"][quote="HTA"]I've seen Cauchy's function equation in the form $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ with the addtional condition is continuity .
I'm wondering if there is a solution for the general case :
$ f(x_1 \plus{} x_2 \plus{} .. \plus{} x_n) \equal{} f(x_1) \plus{} f(x_2) \plus{} .. \plus{} f(x_n)$ and f is a continuous funtion

p/s : i'm a newbie at funtion equation , so if you guys have any books or anything thing related to it please let me know  :blush:[/quote]
Well, let $ x_3 \equal{} x_4 \equal{} ... \equal{} x_n \equal{} 0$ then you have $ f(x_1 \plus{} x_2) \equal{} f(x_1) \plus{} f(x_2)$ :) (Obviously $ f(0) \equal{} 0$. Let $ x_1 \equal{} ... \equal{} x_n \equal{} 0$ then $ f(0) \equal{} n \cdot f(0)$...)[/quote]
oh right , silly me  :blush:

[quote="Poincare"]what is cauchy's function? :blush:[/quote]
One of his functions $ f(x) \plus{} f(y) \equal{} f(x \plus{} y)$. It's only solution is $ f(x) \equal{} ax$.
Proof:
$ x\equal{}y\equal{}0$ gives $ f(0) \equal{} 0$.
$ x \equal{} \minus{}y$ gives $ f(x) \equal{} \minus{}f(\minus{}x)$
Let $ f(1) \equal{} a$. By induction:
$ f(n) \equal{} na \quad \forall n \in \mathbb{N}_0$.
It is true for $ n\equal{}0,1$. And $ f(n\plus{}1) \equal{} f(n) \plus{} f(1) \equal{} an \plus{} a \equal{} a(n\plus{}1)$.
In the same way you can prove that $ f(nx) \equal{} nf(x)$
So $ f(n \cdot \frac{1}{n}) \equal{} f(1) \equal{} nf(\frac{1}{n})$ $ \iff$ $ f(\frac{1}{n}) \equal{} a\frac{1}{n}$
So $ f(m \cdot \frac{1}{n}) \equal{} m \cdot f(\frac{1}{n}) \equal{} a\frac{m}{n}$
Thus $ f(x) \equal{} ax \quad \forall x \in \mathbb{Q}_\plus{}$ Because of $ f(x) \equal{} \minus{}f(\minus{}x)$ we get $ f(x) \equal{} ax \quad \forall x \in \mathbb{Q}$.
Because of continuity we get $ f(x) \equal{} ax \forall \mathbb{R}$

Note however that continuity is not a must, and there are solution other than $ ax$ to Cauchy's equation. Actually constructing them involves treating $ \mathbb{R}$ as a $ \mathbb{Q}$-vector space, finding a base and defining arbitrary values for $ f$ on that base. Then, $ f$ can come arbitrarily close to any point in the plane; so the only "tame" solutions are indeed $ f(x)\equal{}ax$.


geometry
geometric transformation
function
invariant
real analysis
complex numbers
Functional Analysis

Consider $ q\in\mathbf{C}$, and
\[ T: f(x)\mapsto f(qx)
\]
where $ f$ is a function defined on a subset $ D\subset\mathbf{C}$ that is invariant under multiplication by $ q$
Then what is the spectrum of this operator $ T$?

on which banach space is $ T$ defined? functions on arbitrary $ D$ don't work ..

I am looking at the case for arbitrary $ D$
I believe different $ D$ will give different spectrum.

For example, if $ D\equal{}\{q^n|n\in\mathbf{Z}\}$, then the map is equivalent to a shift $ a_n\mapsto a_{n\plus{}1}$, and the spectrum will be the unit circle.

ok this does not answer my question ;). you want to describe the set of complex numbers $ \lambda$, such that $ T \minus{} \lambda$ is not invertible in WHICH ring? according to your example, it could be $ L(H)$ where $ H$ is a hilbert space with orthonormal basis $ D$.

Oops sorry~
Should take $ B(H)$ where $ H\equal{}L^2(D)$

Let's fix a measurable space D and consider $ H\equal{}L^2(D)$

I think what I really want is using the standard notion of a spectrum of any linear operators:
$ T\minus{}\lambda$ is invertible iff it has a bounded inverse.
This applies to both bounded and unbuonded operators (and unbounded operators do not even form a ring!)

Uh-oh. :roll: $ L^2$ with respect to [b]what measure[/b]?

I am given a measurable space $ D$, so it comes with a measure.

Probably I can start with a simple case, say $ q$ is real and $ D\equal{}\mathbf{R}$ with Lebesgue measure $ dx$?
or $ D\equal{}\{q^n: n\in\mathbf{Z}\}$ with discrete measure ($ q$ is not a root of unity)


parameterization
algebra proposed
algebra

Solve in $ \mathbb {C}$ the equation and determine parameter $ k  \in \mathbb {C}$
$ x^5\minus{}9x^4\plus{}kx^3\minus{}3x^2\minus{}\frac{92}{3} x\plus{}20\equal{}0$,we know has two roots for which sum  is 0.        ($ x_1\plus{}x_2\equal{}0$).

[quote="soruz"]Solve in $ \mathbb {C}$ the equation and determine parameter $ k \in \mathbb {C}$
$ x^5 \minus{} 9x^4 \plus{} kx^3 \minus{} 3x^2 \minus{} \frac {92}{3} x \plus{} 20 \equal{} 0$,we know has two roots for which sum  is 0.        ($ x_1 \plus{} x_2 \equal{} 0$).[/quote]

Writing the equations with $ x_1$ and $ \minus{} x_1$ and adding the results, we find $ 9x_1^4 \plus{} 3x_1^2 \minus{} 20 \equal{} 0$
So we have two possibilities for $ x_1^2$ : $ \minus{} \frac {5}{3}$ or $ \frac {4}{3}$

Dividing $ x^5 \minus{} 9x^4 \plus{} kx^3 \minus{} 3x^2 \minus{} \frac {92}{3} x \plus{} 20$ by $ x^2 \plus{} \frac {5}{3}$, we find :
 $ k \equal{} \minus{} \frac {251}{15}$ and $ x^5 \minus{} 9x^4 \plus{} kx^3 \minus{} 3x^2 \minus{} \frac {92}{3} x \plus{} 20$ $ \equal{} (x^2 \plus{} \frac {5}{3})(x^3 \minus{} 9x^2 \minus{} \frac {92}{5}x \plus{} 12)$
So a first solution : $ \boxed{k \equal{} \minus{} \frac {251}{15}}$ and roots $ \minus{} i\sqrt {\frac {5}{3}}$, $ \plus{} i\sqrt {\frac {5}{3}}$ and the three real roots of $ x^3 \minus{} 9x^2 \minus{} \frac {92}{5}x \plus{} 12 \equal{} 0$ that I've not computed but easy to find, I think, with Cardan's method, for example.

Dividing $ x^5 \minus{} 9x^4 \plus{} kx^3 \minus{} 3x^2 \minus{} \frac {92}{3} x \plus{} 20$ by $ x^2 \minus{} \frac {4}{3}$, we find :
 $ k \equal{} \frac {65}{3}$ and $ x^5 \minus{} 9x^4 \plus{} kx^3 \minus{} 3x^2 \minus{} \frac {92}{3} x \plus{} 20$ $ \equal{} (x^2 \minus{} \frac {4}{3})(x^3 \minus{} 9x^2 \plus{} 23x \minus{} 15)$ $ \equal{} (x^2 \minus{} \frac {4}{3})(x \minus{} 1)(x \minus{} 3)(x \minus{} 5)$
So a second solution : $ \boxed{k \equal{} \frac {65}{3}}$ and roots $ \minus{} \frac {2\sqrt {3}}{3}$, $ \plus{} \frac {2\sqrt {3}}{3}$ $ ,1,3$ and $ 5$


calculus
integration
LaTeX
logarithms
trigonometry
calculus computations

1. $ \int\frac{ x^2 \plus{} 2x \plus{} 3 }{ ( x\plus{}1) \sqrt{x^2 \plus{} 1 } } dx \equal{}  \int \frac{ ( x \plus{} 1 ) ^2 \plus{} 2 }{ ( x\plus{}1) \sqrt{x^2 \plus{} 1 } } dx$

$ \equal{} \int\frac{x}{\sqrt{x^2\plus{}1}} dx  \plus{}  \int\frac{1}{\sqrt{x^2\plus{}1}} dx  \plus{} \int{2}{( x\plus{}1) \sqrt{x^2 \plus{} 1 } }dx$

in the first one put $ \sqrt{ x^2 \plus{} 1} \equal{} t^2$ and proceed .

in the second one put $ x \equal{}tan\theta$  so the integral becomes  $ \int sec\theta d \theta  \equal{}  \log |sec\theta \plus{} tan \theta |$

in the third one again  $ put x \equal{} \tan \alpha$ and proceed .

2 . $ \int\frac { \sqrt{x^2 \plus{} 1 } } {x} dx$

multiply and divide by x so we get  $ \int\frac { x \sqrt{x^2 \plus{} 1 } } {x^2} dx$

put $ x^2 \plus{}1 \equal{} t^2$

so the integral becomes $ \int\frac { t }{ t^2 \minus{} 1 } dt$

again put $ t^ 2 \minus{}1 \equal{} y$ and proceed  .

3. $ \int x \sqrt {\frac { 1 - x^2 }{1 + x^2 } } dx$

put $ \frac {1 - x^2}{1 + x^2 } = t^2$ therefore $ xdx = \frac { - 4t }{ ( 1 + t^2 )^2 } dt$ hence the integral becomes  $ - 4 \int\frac { t^2 }{ (1 + t^2 ) ^2 } dt$ That can be written as $ - 4 \int\frac { t^2 + 1 - 1 }{ (1 + t^2 ) ^2 } dt = - 4 \int\frac { 1 }{ (1 + t^2 ) } dt + 4 \int\frac { 1 }{ (1 + t^2 ) ^2 } dt$

the first integral becomes $ \arc \tan ( t ) + c .$

the second integral can be evaluated by jus expanding the denominator .

easier to make the substitutions $ x^2 \equal{} t$ and then $ t \equal{} \cos 2 \theta$

ya thats also fine . but that is more difficult for a first timer

4. $ \int\frac{1\plus{} x^{\frac{2}{3} } } { 1\plus{} x^{\frac{1}{3} } }  dx.$

it can be written as $ \int\frac{1\plus{} x^{\frac{1}{3} } \plus{} x^{\frac{2}{3} } \plus{}x^{\frac{1}{3} } \plus{}\minus{}2x^{\frac{1}{3} } }{ 1\plus{} x^{\frac{1}{3} } }  dx.$

split the integrals .

on splittingit becomes $ \int 1 dx \plus{} \int\frac{ ( x^{\frac{1}{3} } ) ( x^{\frac{1}{3} } \plus{} 1  ) }{ 1\plus{} x^{\frac{1}{3} } }  dx.    \plus{}  \int\frac{ \minus{}2x^{\frac{1}{3} } }{ 1\plus{} x^{\frac{1}{3} } }  dx$

the first two integrals can be evaluated easily . for the last integral  multiply and divide by

$ x^{\frac{2}{3} }$ . so the integral becomes $ \int\frac{\minus{}2x}{  ( 1\plus{} x^{\frac{1}{3} } ) ( x^{\frac{2}{3} } )  }  dx$

now jus put $ x^{\frac{1}{3} }  \equal{} t$ and proceed .


I don't think we should jump the gun on deleting people's posts. You're right, that using a computer or CAS defeats the purpose of most of the problems posted here. But rather than deleting posts, it would be better to make a comment about it where you see it and maybe they'll edit it, or "spoilerize" it, or try the problem by hand.

On some problems it's completely okay to investigate using a computers or CAS - or to verify the solutions that you already got by hand. Learning to use a computer or CAS to solve problems is a valuable skill. It's good that people are learning how to use them. They just shouldn't use them in a way that obscures the mathematics behind a problem or defeats its purpose.

Using CAS to make problems trivial : bad.

Using CAS to help formulate conjectures: priceless ;).

Mm, although again a big part of problem solving is working out what you have to prove, ie formulating the conjecture without the computer..

mmm. .. potentially good point.

I used a CAS for finding out everything I possibly could about ordered partitions... it was quite useful though still took me about 6 weeks w/o looking stuff up.

[quote="ComplexZeta"]There have been an increasing number of good problems that people have been solving with computer algebra systems. In my opinion, this defeats the purpose. These problems are to be solved with only a brain. Therefore, in the future, solutions using a CAS will be deleted unless the problem writer specifically permits the use of them. I would like to encourage moderators in other forums to follow suit.[/quote]

I think that's a very good policy, and this kind of announcement seems to be an eminently fair way to announce such a policy. Individually replying to each offending post would only tend to junk up the board even more, and once a thread has received a reply, people can't delete their own posts--which is why it's a good thing that moderators can always delete posts in the forums they moderate.

I share the concern of JBL and other moderators that this site shouldn't be allowed to degenerate into a chat room for teeny boppers. JBL has evidently seen other examples of math discussion sites that started out well-only to decline as too many users ignorant of netiquette came on board. I, for one, strongly support what the site owner is doing here, and I will do my utmost to make sure that the board is welcoming to young people who desire to practice using their brains to solve mathematical problems--even if that means making it unwelcome to people who junk up the forums.

It seems very reasonable to me to treat most challenge problems posted on most forums here as problems to be solved under competition conditions, to the extent that computer algebra systems are not permitted to reach the solution, however useful those programs may be for certain kinds of pedagogy and mathematical investigation.


Question. In high school chemistry, do they generally get to pi and sigma hybrid bonds? Just asking cuz i want to skip chemistry.

Wow, AoPS is really going far from math.

yes they do in fact (at least at the chem course in my school)

Yep. In fact, our school's first year chemistry course went over it a little, mostly just to introduce the theory, and then second year (AP) chemistry went much more in depth.

If you're thinking about skipping, here is a simple list of topics that our pre-AP class covered:
Dimensional Analysis
Stoichiometry
Organic Chemistry
Nuclear Chemistry
Electrochemistry
Thermochemistry
Oxidation-Reduction Reactions
Kinetics
Solubility
Atomic Structure/Periodic Table
Bonding
Water and Aqueous Systems
VSEPR
Electrons in Atoms (Models, Quantum Numbers)
Gas Laws

Chances are, if you've made it to this site, you won't have any trouble with the math, but you may be a little behind in future classes if you skip, just because everyday exposure during the school year is worth a lot. Of course, if you already know the actual theory, go for it! :)

well, not in China.
I am taking Chem and Physic and Math contest course now and the teachers of Chem told us that $\omega$ and $\pi$ bonds are not to be taught to students who are not interested in Chemistry Competition.

definitely taught in AP; pretty sure it's taught in singapore at some point...

wow...you guys are lucky. my regular chemistry class covered sig. figs. and nuclear chemistry/radioactivity. i had to LEARN everything new this year in AP chem. i mean i knew most of the organic chem, thermochem from biology and college biology classes, but yeah the VSEPR, VB, MO theories were pretty new to me. actually, the diagrams looked a lot like balloons that clowns blow up at birthday parties.

heh. I never took regular chemistry. jumped straight into AP but am somehow surviving...  I guess I have to thank AP Bio for introducing me to basic chem concepts. =D

I know a few people who have no prior chemistry knowledge and are doing pretty well is AP.  As long as you have a good teacher, almost any subject is easy.

at my school, almost all of the people doing AP chemistry have never taken any sort of pre-req type of general chemistry.  I guess it varies from school to school, but I don't think that many people are dying in the course or anything.... though I must say I think some are.

Well, I believe I could have gotten at least a 4 if I took the AP test after my first year of chemistry. AP Chem has not really been that invigorating, because there are a few new things, but the rest is just review. Also, a lot of the work is busy work.

Yes I just learn it in my high school! About the orbitals overlap and hybridization...Arghhhhhh

Educational standards often differ from state to state, so you should probably consult your state standards (or aska teacher who would know them).


Let $ ABC$ is a triangle, with $ B, C$ are fixeds point and $ A$ is a change point. Construct out of triangle $ ABC$ two square $ ABMN$ and $ ACPQ$. Porove that when $ A$ change then $ MP$ always pass a fixed point.

:maybe:  I think the problem should be "When A change ,  MP pass though a fixed point".

See the figure when A is fixed ,we can prove the midpoint of MP is fixed and this problem will be a "real problem" .

H is the fixed point which is the midpoint
see my figure

Nice solution plane geometry ,your solution is much easier than mine . :lol:

Yes. It's nice and sort. Thank you very much. Dear CCMath1 and plane geometry, I have change my post, because it's wrong. Sorry.


number theory
relatively prime
least common multiple

For a natural number $n$, let $f(n)$ be the highest order $(\mod n)$ among all the numbers coprime with $n$. For all natural numbers $a,b$ such that $(a,b)=1$, show that $f(ab)=lcm(f(a),f(b))$

what do you mean by "highest order"? :blush:

The order of these $a$ coprime to $n$ is the smallest $k$ with $a^{k}\equiv 1 \mod n$.
Now take the biggest order.

Well, if you consider an element of $\mathbb{Z}_{a}$ which has order $x$ and an element of $\mathbb{Z}_{b}$ which has order $y$, and use the CRT to find the resultant residue, that must have order $[x, y]$ mod ab, again by the CRT.

[quote="malinger"]Well, please explain in more elementary terms, could you?  :)[/quote]

First, notice that if a and b are relatively prime, a number is 1 mod ab if and only if it is 1 mod a and 1 mod b.  The general form of this is called the Chinese Remainder Theorem. (CRT)

Now, notice that if $u$ has order $n$ mod $m$, then
$u^{k}\equiv 1\mod{m}\implies n|k$.

So suppose $q$ has order $x$ mod a, order $y$ mod b, and order $z$ mod ab.  Then
$q^{z}\equiv 1\mod{ab}$
$\implies q^{z}\equiv 1\mod{a}, q^{z}\equiv 1\mod{b}$
$\implies x|z, y|z$
$\implies [x, y] | z$

Also, notice
$q^{[x, y]}\equiv 1\mod{a}, q^{[x, y]}\equiv 1\mod{b}$
$\implies q^{[x, y]}\equiv 1\mod{ab}$
$\implies z|[x, y]$

If two numbers divide each other, they must be equal, so thus
$z=[x, y]$.

I think your solution is wrong, K8107. You just consider one number at a time ($q$). The problem asks you to consider all number coprime at the same time (and take the one with highest order...)

I suppose he does make the assumption that if $q$ has maximal order $\bmod ab$ then it follows that it also has maximal order $\bmod a, \bmod b$.  Is this obvious, or should it be proven?

Hmm... how is it obvious?

Also, if it has maximal order $\bmod a,\bmod b$, does it also have the maximal order $\bmod ab$?

So prove it for me please, t0rajir0u.  :)

I did, in fact, realize that I assumed that statement in order to prove the problem: it follows from the fact that given any order $x$ mod a, $x|f(a)$, and similarly for b.

To elaborate: suppose $q$ has order $f(ab)$ mod ab, and has orders $x, y$ mod a and mod b, respectively, where $[x, y]=f(ab)$.  There exist two such $x, y$ by the CRT, because there exists an element of $\mathbb{Z}_{ab}$ with order $f(ab)$.  Then
$x|f(a), y|f(b)$
$\implies [f(a), f(b)] \ge f(ab)$

But there exists an element of $\mathbb{Z}_{ab}$ with order $f(a)$ mod a and $f(b)$ mod b.  That element cannot have order exceeding $f(ab)$ mod ab, by assumption, and therefore, $[f(a, f(b)] \le f(ab)$.

Thus, $f(ab)=[f(a), (f(b)]$.

[quote="malinger"][quote="K81o7"] the fact that given any order $x$ mod a, $x|f(a)$
[/quote]

Hopefully my last question is, why is this true?[/quote]

$f(a)$ is the least common multiple of all of the orders $\bmod a$.  The lemma that K81o7 proved in his previous post tells you that an element of this order must exist, and clearly it has maximal order.

[quote="malinger"]I'm sorry, but which lemma are you talking about?[/quote]

I didn't actually prove that in $\mathbb{Z}_{m}$, there exists an element whose order is divisible by those of all others in the ring.  But to prove that, you can show that given any two elements, there is an element whose order is divisible by the order of both.  I proved it at some point, I think if the elements are $u_{1}$ and $u_{2}$ and have orders $a_{1}$ and $a_{2}$ such that $a_{1}=kb_{1}$ and $a_{2}=kb_{2}$, where $(b_{1}, b_{2})=1$, you can take $x, y$ such that $xb_{1}+yb_{2}=1$, and then you take, I believe $u_{1}^{y}u_{2}^{x}$, which has order $kb_{1}b_{2}$.  Maybe there are a few additional steps required, but that sketches the proof...

[quote="malinger"][quote="K81o7"]I believe $u_{1}^{y}u_{2}^{x}$, which has order $kb_{1}b_{2}$.  [/quote]

Really?  :o[/quote]

Did I mess it up?  As long as neither of $b_{1}, b_{2}$ is 1, that should be right.  If that element has a smaller order, it divides $kb_{1}b_{2}$, you should be able to show it's impossible.

Ah, come on. Just show me how to prove that...

Well, or there should be an easier solution...

There isn't a significantly easier solution.  It's not a trivial result.

Anyway, you can do it by contradiction.  The details escape me at the moment.


How can I prove Haruki's theorem, http://mathworld.wolfram.com/HarukisTheorem.html .

Use the angle version of Ceva's theorem.

I have moved this discussion out of the "Theorems and Formulas" section, since Haruki's theorem is not really an important theorem (I have never seen it being actually used, although I could imagine some few cases where it could occur). Here is the proof from Honsberger's "Episodes in Nineteenth and Twentieth Century Euclidean Geometry":

[color=blue][b]Haruki theorem.[/b] Let a, b, c be three circles. Assume that the circles b and c meet at two points A and A', that the circles c and a meet at two points B and B', and that the circles a and b meet at two points C and C'. Then, $\frac{BA^{\prime}}{CA^{\prime}}\cdot\frac{CB^{\prime}}{AB^{\prime}}\cdot\frac{AC^{\prime}}{BC^{\prime}}=1$.[/color]

[i]Proof.[/i] The lines AA', BB', CC' are the pairwise common chords of the three circles a, b, c; in other words, these lines AA', BB', CC' are the pairwise radical axes of the three circles a, b, c, and thus they concur at a point R.

We will use directed angles modulo 180°. Since the points A, B, A', B' lie on one circle (namely, on the circle c), we have < AB'B = < AA'B; in other words, < AB'R = - < BA'R. Also, trivially, < ARB' = - < BRA'. Thus, the triangles AB'R and BA'R are inversely similar, and hence we get $\frac{BA^{\prime}}{AB^{\prime}}=\frac{BR}{AR}$. Similarly, $\frac{CB^{\prime}}{BC^{\prime}}=\frac{CR}{BR}$ and $\frac{AC^{\prime}}{CA^{\prime}}=\frac{AR}{CR}$, and thus,

$\frac{BA^{\prime}}{CA^{\prime}}\cdot\frac{CB^{\prime}}{AB^{\prime}}\cdot\frac{AC^{\prime}}{BC^{\prime}}=\frac{BA^{\prime}}{AB^{\prime}}\cdot\frac{CB^{\prime}}{BC^{\prime}}\cdot\frac{AC^{\prime}}{CA^{\prime}}=\frac{BR}{AR}\cdot\frac{CR}{BR}\cdot\frac{AR}{CR}=1$,

and the Haruki theorem is proved.

  Darij

What is the angle version of Ceva's theorem? My geometry knowledgement is not very good.

Let $D$, $E$, $F$ be points on sides $BC$, $CA$, $AB$ of triangle $ABC$, respectively.  Then $AD$, $BE$, $CF$ concur if and only if
\[\frac{\sin \angle BAD \cdot \sin \angle ACF \cdot \sin \angle CBE}{\sin \angle CAD \cdot \sin \angle BCF \cdot \sin \angle ABE} = 1.\]
It follows from the regular statement of Ceva's and sine law.


I have attached several files.

I try to make an output that looks exactly like LaTeXpic.gif

I am satisfied with everything in LaTeXQ.pdf, which is compiled from the tex file, EXCEPT (sorry for bolding but this needs emphasis) the vertical line.

No luck in fixing it so far. Do you guys have any suggestion?

How does this look to you (assuming the labels aren't vital)?

$\begin{array}{llll} 1 & P \to Q & & A \\ 2 & Q \to R & & A \\ \multicolumn{1}{l|}{3} & P & & H \\ \multicolumn{1}{l|}{4} & Q & & 1,3 \to E \\ \multicolumn{1}{l|}{5} & R & & 2,4 \to E \\ 6 & P \to R & & 3-5 \to I \\ \end{array}$

[url=http://www.g32.org/latable/]LaTable[/url] makes this sort of thing easy.

lol it would be nice if I can use label, but this is okay as well.


This is an example problem from Aops book I (pg. 38 example 4-4).
Given that x is directly proportional to y and to z ,and is inversly proportional to w, and that x=4 when (w,y,z) = (6,8,5), what is x when (w,y,z) = (4,10,9)?

I understand that xw, x/y, and x/z is constant.  Does anyone know why you have to order the variables in this form xw/yz= (4)(6)/(8)(5) = 3/5   to get the answer.  why cant you order the variables this way xy/wz=32/30?

thanks  :)

Myc, you are missing a crucial fact here.

You see, we can simply the proportion to xw=yz ( from xw/zy). The only way to get y to the left side is to divide xw and so you get xw/y = z. Flip the w around and you will get x/y = z/w. Piece it together and you will get...

 x/y
 ___

 z/w

Use what you learnt in school to get....

xw/ zy...

From which we conclude that we can't change it.

[quote="myc"]This is an example problem from Aops book I (pg. 38 example 4-4).
Given that x is directly proportional to y and to z ,and is inversly proportional to w, and that x=4 when (w,y,z) = (6,8,5), what is x when (w,y,z) = (4,10,9)?

I understand that xw, x/y, and x/z is constant.  Does anyone know why you have to order the variables in this form xw/yz= (4)(6)/(8)(5) = 3/5   to get the answer.  why cant you order the variables this way xy/wz=32/30?

Because then you are implying that x is inversely proportional to y and directly proportional to both w and z, which is not what the problem is saying.
thanks  :)[/quote]


A point $ P$ is performing a random walk on the $ X$-axis. At the instant $ t\equal{}0$, $ P$ is at a point $ x_0$ ($ |x_0|\le N$, where $ x_0$ and $ N$ denote integers, $ N>0$). If at an instant $ t$ ($ t$ being a nonnegative integer), $ P$ is at a point of $ x$ integer abscissa and $ |x|<N$, then by the instant $ t\plus{}1$ it reaches either the point $ x\plus{}1$ or the point $ x\minus{}1$, each with probability $ \frac12$. If at the instant $ t$, $ P$ is at the point $ x\equal{}N$ [$ x\equal{}\minus{}N$], then by the instant $ t\plus{}1$ it is certain to reach the point $ N\minus{}1$ [$ \minus{}N\plus{}1$]. Denote by $ P_k(t)$ the probability of $ P$ being at $ x\equal{}k$ at instant $ t$ ($ k$ is an integer). Find $ \lim_{t\to \infty}P_{k}(2t)$ and $ \lim_{t\to \infty}P_k(2t\plus{}1)$ for every fixed $ k$.

it is essentially equivalent to study a standard random walk modulo $ 2N$.


Lockers are numbered with consecutive positive integers beginning with 1, and the digit 2 is used exactly 106 times. What is the number of the last locker?

in the numbers 1-99, we use the number 2 a total of 20 times (confirm this yourself)
in the numbers 100-199, we use the number 2 a total of 20 times as well

here is where it starts getting annoying...

we have already used the number 2 a total of 40 times, so we have to use it 66 more times
we know that every number in the 200's uses at least one 2
so now we start:

from the number 200-209, we use the number 2 a total of 11 times
we use the number 2 the same number of times when writing the numbers 210-219

now we have to use the number 2 a total of 44 more times
now we think about the numbers from 220 to 229
every number uses 2 2's except for 222, which uses 3
so we have 9 numbers which use 2 twice and one that uses 2 three times, for a total of 21 2's used

now we have to use 2 23 more times
we write the numbers 230-239, for another 11 uses
we write the numbers 240-249, for another 11 uses

now we need to use the number 2 1 more time, so the number on the last locker is $ \boxed{250}$

i really hope i did this right....

You got it right. Another way of solving this problem is to Google the following:
"digit 2 106 times"


http://www.newgrounds.com/portal/view/445688

I got a bat, but then I died a while later, cuz everything was missing my web, and then like 3 queen bees came. My final score was somewhere around 14k. Yeah, not very good.

[quote="Nerd_of_the_Ages"]My final score was somewhere around 14k. Yeah, not very good.[/quote]

 :o dang i suck...


induction
number theory proposed
number theory

Prove that for all positive integers $n\geq 1$ the number $\prod^n_{k=1} k^{2k-n-1}$ is also an integer number.

[i]Laurentiu Panaitopol[/i].

We seem to have seen this one twice before: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47653]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1779]here[/url].

[quote="Maxal"]\[ \prod_{k=1}^n k^{2k-n-1} = \dfrac{n!^{n-1}}{1!^2 2!^2\dots (n-1)!^2} = \prod_{i=1}^{n-1} {n\choose i}  \]
is a product of integers.[/quote]

can you give a brief proof without induction?


Find all real valued functions $ f(x)$ on the reals such that $ f( (x \minus{} y)^2) \equal{} x^2 \minus{} 2y f(x) \plus{} f(y)^2$.

Plug in $ x\equal{}y$ to get $ f(0)\equal{}((f(x)\minus{}x)^2$. In the case $ x\equal{}0$, $ f(0)\equal{}0$ or $ f(0)\equal{}1$.

If $ f(0)\equal{}0$, then $ (f(x)\minus{}x)^2\equal{}0$ so $ f(x)\equal{}x$ for all $ x$. If $ f(0)\equal{}1$, then for all positive $ a$, plug in $ x\equal{}\sqrt{a},y\equal{}0$ to get $ f(a)\equal{}a\plus{}1$, and then plug in $ x\equal{}\minus{}a,y\equal{}a$ to get
\[ 4a^2\plus{}1\equal{}a^2\minus{}2af(\minus{}a)\plus{}(a\plus{}1)^2\Rightarrow f(\minus{}a)\equal{}\minus{}a\plus{}1\]
This case gives $ f(x)\equal{}x\plus{}1$ for all $ x$. So the solutions $ \boxed{f(x)\equal{}x,f(x)\equal{}x\plus{}1}$.


inequalities
LaTeX
inequalities proposed

Prove the following inequality:

                 x^y+y^x>=1+xy if 0<x,y<1 or if x,y>1.

let's begin putting it into LATEX

$\ x^y + y^x  \ge \ 1+xy\ $                                   [color=Blue] if 0< x, y< 1  or if x,y >1[/color]

for c>=1: x^c >= 1 + c(x-1) and the ineq is reversed if c =< 1.  It is shown here;  http://mathworld.wolfram.com/ExponentialInequality.html ;; also easily proved.

Suppose x,y>=1.  Then $x^y + y^x >= 2xy + 2 - x - y >= 1 + xy$ by considering $(x-1)(y-1)$.  can do x,y<1 similarly


Is there some way to, in a tabular environment, create a vertical line between two cells within one row (as apposed to running down all rows)?

Do you mean like this? $\begin{tabular}{|ll|}\hline \multicolumn{1}{|l|}{1}& 2 \\ 3 & 4 \\ \hline \end{tabular}$
[code]\begin{tabular}{|ll|}
\hline
\multicolumn{1}{|l|}{1} & 2 \\
3 & 4 \\
\hline
\end{tabular}[/code]
The work is done by multicolumn but you don't have to learn the syntax as [url=http://www.g32.org/latable/]LaTable[/url] does it all for you - in fact I used it to generate the code above.

Thanks, that's exactly what I needed.

About Latable: Isn't the whole point of LaTeX that it's not wysiwyg?  That seems to be sort of cheating :P .

It's not really cheating as it helps you to write good Latex code which can then be compiled. You could regard it as a way of teaching you how to do complicated things with tables and once you've learnt you can write the code directly.

At the moment this is just curiosity, although I may want to act on it some time in the future, but is there anything similar to excel in LaTeX (talk about violating non-wysiwyg! :roll: )?

I haven't tried it but [url=http://www.gnome.org/projects/gnumeric/faq.shtml]Gnumeric[/url] will export its spreadsheet files into LaTeX format. [url=http://www.ctan.org/tex-archive/help/Catalogue/entries/excel2latex.html]Excel2Latex[/url] does what it says. I have used [url=http://www.ctan.org/tex-archive/help/Catalogue/entries/csvtools.html]csvtools[/url] for mail merging a CSV file in LaTeX and it works very well. There's also [url=http://pegasus.rutgers.edu/~elflord/unix/software/spreadsheet2latex/]Spreadsheet2latex[/url] and [url=http://latexss.sourceforge.net/]LaTeXss[/url].


I'm having trouble with synthetic division. Can someone help me out?

Do you have a specific problem?  Synthetic division is really just a compact form of long division.  However, the only time you usually use synthetic division is when the divisor is linear.

This is the easiest way IMO

[quote="Sir.Aaron"]This is the easiest way IMO

Let

    $P(x)=2x^3+3x^2-4\,\!$
    $Q(x)=x+1\,\!$

We want to divide P(x) by Q(x) using Ruffini's rule. The main problem is that Q(x) does not appear to be a binomial of the form x − r, but rather x + r. We must rewrite Q(x) in this way:

    $Q(x)=x+1=x-(-1)\,\!$

Now we apply the algorithm:

1. Write down the coefficients and r. Note that, as P(x) didn't contain a coefficient for x, we've written 0:

    |     2     3     0     -4
    |
 -1 |
----|----------------------------
    |
    |

2. Pass the first coefficient down:

    |     2     3     0     -4
    |
 -1 |
----|----------------------------
    |     2
    |

3. Multiply the last obtained value by r:

    |     2     3     0     -4
    |
 -1 |          -2
----|----------------------------
    |     2
    |

4. Add the values:

    |     2     3     0     -4
    |
 -1 |          -2
----|----------------------------
    |     2     1
    |

5. Repeat steps 3 and 4 until we've finished:

    |     2     3     0     -4
    |
 -1 |          -2    -1      1
----|----------------------------
    |     2     1    -1     -3
    |{result coefficients}{remainder}


I was wondering if anyone else is going to boarding school, because I'm considering going  to Exeter or Andover next school year. I'm also looking at private day schools in my town. Please tell me about applying to them or whats it's like to go.

my mom wants me to go to a boarding school *bleah*
she says you live with other people while going to school

I go to Detroit Country Day School.

It's an [b]optional[/b] boarding school. I personally don't board, but there are a bunch of people from England and all over Europe at the school.

They have so many cool accents!

Hahaha, Andover and Exeter. Brrrrrr, very cold in winter. And lots of snow on the ground. Definitely not fun when you have to walk a long way to your next class. Snowball fights are fun though, although I'm not sure if they allow them (kids in sports jackets throwing snowballs at each other and tackling each other, amusing image).

I'd reccommend going to one of those two. They're worth it a lot. Both are pretty darn good, even if it is a bit cold. You get to meet people from like a bajillion (ok, more like 20) countries and a lot of states too, like 120. And the teachers are usually much better than pub schools. I went to a good public middle school, and the teachers at Choate are just so much better at teaching the material. Also, you see them like at lunch and dinner, so you can ask them questions about assignments, reading, etc. Ask somebody else about boarding life, I'm still a day student. Maybe I'll force JSRosen3 to come online and answer you.

[quote="yif man12"]Hahaha, Andover and Exeter. Brrrrrr, very cold in winter. And lots of snow on the ground. Definitely not fun when you have to walk a long way to your next class. Snowball fights are fun though, although I'm not sure if they allow them (kids in sports jackets throwing snowballs at each other and tackling each other, amusing image). [/quote]

Eh, living in Minnesota I'm used to that.  :P

I'd reccomend applying to a couple others though (mainly because one of those under a couple others is my school, but that doesn't change the fact that I'm right :D ) It's useful to find which school suits you best, because that's what it comes down to in the end.

I'd also reccomend saying no to any school that doesn't allow snowball fights between classes. There's just too much temptation there. Although it can be a hassle when kids just randomly start chucking snowballs at you while you walk.

[quote="A+MATH"][quote="xtrememath8"]Suffield Academy[/quote]

Thanks for your reply. Have you been there very long? What kind of academics do they have there?[/quote]

This is my first year. I like it a lot, but the math program isn't prime.


geometry
inequalities proposed
inequalities

Let $m_a,m_b,m_c$ and $h_a,h_b,h_c$ the medians and heights of a triangle of sides $a,b,c$ and area $S$. Prove that

$a^2+b^2+c^2 \geq max(\frac{m_a}{h_a},\frac{m_b}{h_b},\frac{m_c}{h_c})4\sqrt{3}S$

Very old. Published by Mihaly Bencze in many many magazines.


Two probably identical problems:

[b]1.[/b]  An equilateral triangle is given.Prove that it is possible to cut this triangle into several parts and to construct a square using these parts.

_________________________________________________________________________________________________________________

[b]2.[/b]  Challenge: Find out what conditions [b]can[/b] triangle satisfy, so that it is possible to cut this triangle into parts and to construct a square using only these parts.

In the last problem,I doubt if it is possible to find all necessary conditions,but some sufficiently conditions can be found,do not forget to use the first problem as a hint.

These "sufficiently conditions" can be unclear,like "it is sufficiently to some point be inside of the initial triangle".

Best regards,Erken.

[quote="greenvert"]About part 2
[url=http://en.wikipedia.org/wiki/Bolyai-Gerwien_theorem]Bolyai-Gerwien theorem[/url][/quote]
Oops... :blush:   I guess I should add some conditions into the statement.You need to show in both problem how exactly can this cutting be done.
I mean solution must look like:
"Take a point on the side $ AB$ and construct a line through this point" and so on...
I doubt if the proof of [b]Bolyai-Gerwien theorem[/b] has solution in the form I've just mentioned.I'm sorry If I am mistaken.
On the other hand,I'm pretty sure that the proof of this theorem is much harder than the proof of the second problem,as this problem has very elementary  and easy solution(probably not so easy,but not really hard in any case).

The proof of Bolyai-Gerwien theorem is very constructive, and I will post it here I find or remember it.
About the first problem, it's the  [url=http://mathworld.wolfram.com/HaberdashersProblem.html]Haberdasher's Problem[/url] by Dudeney

You haven't even given a chance to think for anyone who wanted to solve this problem and who have already visited the links above.
For others I would suggest to not visit these links.Try to solve them by yourself.The solution posted in the link above is somehow similar to my solution (I haven't read details yet),so feel free to post your own solutions,of course if they are different from the one given in the link.

I just want to clarify that in the second problem you only need to find any generalizations of the 1st problem.I mean to consider not only equilateral triangles but maybe isosceles or just triangles that satisfy to some conditions.
Actually,I am wondering whether in the second problem we can make cuttings of a triangle,which do not satisfies to any conditions,but please do not write that this is particular case of [b]Bolyai-Gerwien[/b] theorem.


[size=150]A search plane has a cruising speed of 250 mph and carries enough fuel AT MOST 5 hours of flying.  If there is a wind that averages 30 mph and the direction of search is with the wind one way and against it the other, how far can the search plane travel?[/size]

[hide]
Say the plane can travel a distance of $x$ miles.  Since the speed it travels with the wind is 30+250=280 mph and the speed against the wind is 250-30=220 mph. we have $\frac{x}{280}+\frac{x}{220}=5.$   Solve for $x$ to get that $x=616$.  Therefore, the plane can travel $616$ miles. [/hide]

i guess it helps to know the answer  :P  but to finish ur solution you are looking for the total distance travelled which is

$2x=2(616)=\boxed{1232}$

[quote="SM4RT"]to finish ur solution you are looking for the total distance travelled[/quote]
We are? :huh:

I thought the question meant how far [b]out[/b] the plane can travel.

the question is pretty much word for word from what I know of ... but i dunno i guess its how u interpret it..

[quote="Interval"]A search plane has a cruising speed of 250 mph and carries enough fuel AT MOST 5 hours of flying.  If there is a wind that averages 30 mph and the direction of search is with the wind one way and against it the other, how far can the search plane travel?[/quote]
I think that it is asking one direction out because it says "how far can it travel..." and it seems like it's saying "...before it needs to come back"

K that would certainly make more sense...I think it should be stated in the original question..


Simplify, assuming that $ ab\neq0$ and $ a\neq b$:
$ \dfrac{\frac{1}{a} \minus{} \frac{1}{b}}{ \frac{1}{b} \minus{} \frac{1}{a}}$ .

$ \dfrac{\frac {1}{a} \minus{} \frac {1}{b}}{\frac {1}{b} \minus{} \frac {1}{a}}$

$ \implies \dfrac{\frac {b \minus{} a}{ab}}{\frac {a \minus{} b}{ab}}$

$ \implies \frac {b \minus{} a}{a \minus{} b}$

$ \implies \minus{} 1(\frac {a \minus{} b}{a \minus{} b}) \equal{} \boxed{ \minus{} 1}$

$ \dfrac{\frac {1}{a} \minus{} \frac {1}{b}}{\frac {1}{b} \minus{} \frac {1}{a}} \equal{} \dfrac{\minus{}\left(\frac {1}{b} \minus{} \frac {1}{a}\right)}{\frac {1}{b} \minus{} \frac {1}{a}} \equal{} \boxed{\minus{}1}$


Just the solution.. look only if you can't find it!
[hide=";)"]
= (x+y)(x-y)(x²+y²-4)[/hide]

[quote="MCrawford"]Please stick to the general level of difficult of the forums.  This problem is obviously much harder than the other problems in this middle school forum, so I'm moving it.[/quote]
In Macao, this kind of problems are very easy for middle school students to solve.

[quote="ifai"]Factorize $x^4-y^4-4x^2+4y^2$.[/quote]

you can use an application of the factor theorem, set it equal to zero, then solve for x and y, and you know that the thing factors in that way

so it is pretty obvious that
x=-y or x=y, so dividing those out, we get x^2+y^2-4

Or just group the way it is, as in group the first two terms together by difference of two squares and factor out a 4 in the last two and simplify.

[quote="Silverfalcon"][quote="MCrawford"][quote="ifai"]
In Macao, this kind of problems are very easy for middle school students to solve.[/quote]

Whether or not all Macao middle school students are solving problems equivalent to hard AMC/easy AIME problems, we still group problems into levels of difficulty and this is the easiest grouping for us.[/quote]

I'll add one more comment. Compared to other nations (for instance, Romania has very high standards on math), U.S. isn't very advanced on middle school math. The difficulty rises when the level reaches high school but even then, the problems are easier than other nations.

That doesn't mean U.S. is not strong at math.. It is in fact VERY strong but it takes longer time in basics before moving to advanced. So, please do not post the problems that are in level of middle school to our middle school forum. It will be more likely to be placed in Getting Started.

Thanks.[/quote]
ok. And may you please tell me your location?


conics
parabola
geometry
geometric transformation
special factorizations

This is suggested by a post in [url]http://www.mathlinks.ro/viewtopic.php?p=1519160#1519160[/url] which asked how to prove that $ a$ determines the concavity of the parabola $ f(x)\equal{}ax^2\plus{}bx\plus{}c$ without using calculus. So let's have a little 'proof' practice for beginners.

[quote="mavropnevma"]Strange question. Since $ y \equal{} ax^2 \plus{} bx \plus{} c \equal{} a \left (x \plus{} \frac {b} {2a} \right )^2 \minus{} \frac {b^2 \minus{} 4ac} {4a}$, with the translation $ y' \equal{} y \plus{} \frac {b^2 \minus{} 4ac} {4a}$ and $ x' \equal{} x \plus{} \frac {b} {2a}$, the equation becomes $ y' \equal{} ax'^2$. But everybody knows the concavity of $ Y \equal{} X^2$, so ...[/quote]Yea, I agree that it's a bit strange when I created it. You've got the idea that I was seeking: completing the square and we have the standard form $ y\equal{}a(x\minus{}h)^2\plus{}k$. $ (x\minus{}h)^2\ge0$ for every real value $ x$, so the sign of $ a$ determines the concavity of a parabola.


function
calculus
derivative
superior algebra
superior algebra unsolved

Let $ {\mathbb Q}_p$ (where $ p\geq 3$) the field of $ p$-adic numbers and $ {\mathbb Z}_p \equal{} \{x\in{\mathbb Q}_p: \;|x|_p\leq1\}$ all the $ p$-adic integers. For each $ x \equal{} d_0 \plus{} d_1p \plus{} d_2p^2 \plus{} \cdots\in {\mathbb Z}_p$ we define $ f(x): \equal{} d_0^2 \plus{} d_1^2p \plus{} d_2^2p^2 \plus{} \cdots$. Show that $ f: {\mathbb Z}_p\to {\mathbb Q}_p$ is countinuos but non-differentiable on $ {\mathbb Z}_p$.

You need $ p>2$ there- if $ p\equal{}2$, it's the identity map. The coefficients are all interpreted mod $ p$.

Hmm... you can do a lot of weird things. How about a function which is differentiable everywhere, with derivative identically zero- and it's injective?

Yes you're right and I have edited it, thank you.
I'm not sure what you mean. If you want to find a example about a function which is differentiable everywhere, with derivative identically zero- and it's injective we can chose for each $ x \equal{} d_0 \plus{} d_1p \plus{} d_2p^2 \plus{} \cdots$ we consider $ g(x) \equal{} d_0 \plus{} d_1p^2 \plus{} d_2p^{4} \plus{} d_3p^{6} \plus{} \cdots$ then $ |g(x) \minus{} g(y)|_p \equal{} |x \minus{} y|^2_p$.
But how it relates to my question?  :)

It isn't closely related- it's just another fun thing you can do with this totally disconnected space.


probability
AMC
AIME
number theory
prime factorization

x and y are positive integers.
$7x^{5}=11y^{13}$ The minimum possible value of x has a prime factorization of $a^{b}c^{d}$ What is $a+b+c+d$?

In a tournament, there are no ties. There are 7 teams and each team plays each other exactly once. Each team has a 50% chance of winning each game. The winner is awarded 1 point and loser no points. Total points are accumulated to decide the ranking of the teams. In the first game, Team A beats team B. The probablity that A finishes with more points than B is $\frac{m}{n}$. $m$ and $n$ are relatively prime. Find $m+n$

For the first problem, the minium value of $x$ is obtained only if $x$ is divisible by $7$ and $11$ and nothing else. Let $x=7^{b}*11^{d}, y=7^{k}*11^{l}$ then:
$13l+1=5d$ so minium value of $d$ is $8$
$5b+1=13k$ so ,minmium value of $b$ is $5$.
Hence the answer is $11+7+8+5=31$

I just did that second one, is it from an AIME? I think I saw it in the one from 1992, or the AIME2 from 2006.

[hide="2"]Each team plays six total games, so after the first game teams A and B each have five games remaining.  Let $W_{A}$ and $W_{B}$ be the number of games teams A and B win in their remaining five games, respectively.  For team A to finish with more points than team B, team A must win more games or the same number of games as team B.  We want $P(W_{A}\ge W_{B})$.  We have $P(W_{A}\ge W_{B})=P(W_{A}>W_{B})+P(W_{A}=W_{B})$.

Since teams A and B both have the same probability of winning each game, $P(W_{A}>W_{B})+\frac{P(W_{A}=W_{B})}{2}=P(W_{A}<W_{B})+\frac{P(W_{A}=W_{B})}{2}=\frac{1}{2}$.  Therefore, $P(W_{A}\ge W_{B})=\frac{P(W_{A}=W_{B})}{2}+\frac{1}{2}$.

The probability that teams A and B each win $k$ games is $\left(\frac{\binom{5}{k}}{32}\right)^{2}$.  Therefore, $P(W_{A}=W_{B})=\sum_{k=0}^{5}\left(\frac{\binom{5}{k}}{32}\right)^{2}=\frac{1}{32^{2}}\sum_{k=0}^{5}\binom{5}{k}^{2}=\frac{252}{32^{2}}=\frac{63}{256}$.  The desired fraction is $\frac{63}{512}+\frac{1}{2}=\frac{319}{512}$, and the answer is $319+512=831$[/hide][list][/list]


search Princeton college topology \/closed

I think that MathLinks can now...go to school, because this site has already SIX YEARS ! :rotfl: The childhood is over !

I wish to it Happy Many Others Years !

And thank you all of you , specially Valentin Vo[b]i[/b]nicu (sorry, Vo[b]r[/b]nicu !) and his "team" :clap2:  :clap2:  :clap2:

Yes, this thread is spam, but I would also like to congratulate mathlinks and AoPS on doing such a wonderful job with everything!  :)

[quote="AIME15"]Yes, this thread is spam[/quote]

Indeed; this thread is infinitely less useful than any other in this forum. What kind of twisted, demented people wouldn't consider appreciation spam?

Oh man I'm getting old. Still getting carded for looking under 21 though so it's all good:P

Wow, congratulations all around to admins!

For those who identify the site as AOPS, it's a little short of six years. The anniversary in that case is in late May.

[quote="jmerry"]The anniversary in that case is in late May.[/quote]
I was just looking at the join dates of the top 10 posters. rrusczyk is listed with a March, 2003 date - was he a member of Mathlinks before he started the AoPS site?

Actually, it's probably that the forum wasn't opened right away. I was going by posts on the forum rather than join dates.

Based on join dates for other users, the right date for AOPS would be May 24, 2003. The earliest posts I can find are May 25.

I actually wanted to a similar posting but rather on 14th February 2003 which Val seems to claim as the starting date of ML [url=http://www.mathlinks.ro/viewtopic.php?p=169946#169946]here.[/url] But his first posting on the forum dates from [url=http://www.mathlinks.ro/search.php?search_author=Valentin%20Vornicu&ppage=20&start=7300]4th February 2003.[/url] At some point I suggested a collaboration between ML and AoPS. But initially there was only love at second glance and the relationship was speeded up by possible legal threat of ML which led to kind of merger and acquisition of ML. ML was lucky enough to retain its "brand" name as within the URL and forum skin.

For example if you browse early posts of my profile you might guess I have joined some time after IMO in July 2003 but my registration date is not before 23rd December. Who else of you has posted under your user name before you registered it? Not that many active users these days I guess!  :D The reason is that the 23rd of December is my AoPS registration date but I have been posting on ML before. But when merging the date and users I got assigned the AoPS registration date for some reason.

- Valentin Vornicu (Basketball whiz): Now University of Bucharest alumni and AoPS employee.
- Darij Grinberg (German high-school student): Now university of LMU math undergraduate
- Hojoo Lee (Korean graduate student): Now really popular for his online problem collections.
- Le Viet Bao (full score in IMO 2003): Soon to be Cambridge alumni and grad student in the US
- Andrei Negut (two times IMO silver medalist, future Princetonian): Now Princeton alumni
- Fedor Petrov (gold medalist): Now postdoc at Petersburg Department of Steklov Institute of Mathematics RAS
- Pierre Bornstein: Now French IMO Deputy Team Leader (I was betting on this event but he did not believe my prediction!  :wink: )
- Gabriel Dospinescu aka harazi: Now he is a student at ENS in Paris, apparently his birthday is today. So happy birthday!  :)
- (Insider joke for Val: I retired as German team leader.  :P)

And we all hope Val gets a bit more responsive again. A guy whose user name suggests a connection to Riemann hypothesis asked for a topology sub-forum. ML has moved on and I suggested a few new people like April as mods.  :)

[quote="jmerry"]Based on join dates for other users, the right date for AOPS would be May 24, 2003. The earliest posts I can find are May 25.[/quote]
Richard seems to have posted the first [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=23214#23214]AoPS Challenge[/url] problem on May 23, 2003.  Those early days were fun.

The archives for the early days are weird, too. That thread you linked can't be found from the main listing of topics in the Pre-Olympiad forum.

Yeah, you're right.  It and a bunch of other posts seem to be sorted as if they have a date of approximately July 5, 2004.  I'm guessing that oddity has to do with the merging of AoPS and MathLinks.

[quote="orl"]And we all hope Val gets a bit more responsive again. A guy whose user name suggests a connection to Riemann hypothesis asked for a topology sub-forum. ML has moved on and I suggested a few new people like April as mods.  :)[/quote]We're working on quite a shocker update on the site so that's why I might have seem "less responsive".

Wow it's hard to think that all this was going on while I was in 1st grade...

*Mouth open in awe*

MATH PROBLEMS FOR EVERYBODY! :D
(If you watched Fiddler on the Roof you would get that phrase)

Thanks to Valentin and Richard for making ML and AoPS!

Most probably [url=http://www.mathlinks.ro/viewtopic.php?p=2#2]this[/url] is the very first post of Mathlinks.  :D
(Though it shows that it is post #2) Most probably 1st post is deleted somehow.

BTW congratulation to everyone related to Mathlinks :)


[quote="jacob8888888"][quote="Klebian"][hide]Just say that $0.0454545... = x$

Then, $10x = .454545...$
To calculate a repeating fraction, you take the repeating number (45) and divide it by 9, 99, 999, etc (the number of 9's being the number of digits in the repeating number).

So, $10x = \frac{45}{99}$, simplified into $\frac{5}{11}$

$10x = \frac{5}{11}$, divide by 10, $x = \frac{5}{110}$, or $\frac{1}{22}$.[/hide][/quote]

Do you divide the repeating fraction by 9 for every single repeating fraction or does the divisor change for every repeating number?[/quote]

If I'm interpreting what you're asking correctly, then every repeating number.

For example, .88888... = 8/9, but we could put it as 88/99, which simplified = 8/9.

.343434... = 34/99=3434/9999

.123412341324 = 1234/9999

And so on.
If that's not what you meant, feel free to ask.

Actually what I wanted to know was whether you diveded by 9 every time or does the number change. Such as, do you every divide by a three, or a 5 etc or is it always ? THanks :)

It's always $9$'s, although it may simplify to $1$'s or $3$'s if the repeating portion is divisible by $3$ or $9$. :) This is because of the general rule for repeating fractions. Say, for example, we have the repeating decimal $0.174\overline{174}$:

$x=0.\overline{174}$
$1000x=174.\overline{174}$
$1000x-x=999x=174$
$x=\frac{174}{999}=\frac{58}{333}$

That's why you divide by an $n$-digit number, with $n$ the number of digits that repeats, with all $n$ digits as $9$'s. ;)

How do put 56.89898989.....into a fraction? Such as when there is a number to the left of the decimal point.

hm... i know this might not relate to this topic, but i've always wonder, what is the exact definition of a common fraction? say like if the number is $1\frac{8}{9}$, would i leave it as $\frac{17}{9}$ or $1\frac{8}{9}$?


I had made a lab for measuring boiling point of Methanol ($CH_3OH$). Here is my steps of measuring it:

   Using a twist tie, attach  test tube to a thermometer. Place about 100 uL of Methanol inside the test tube. Take a melting point capillary tube (one end sealed). Insert it,  open end down, into the test tube. The open end of the capillary tube is under the surface of the liquid. Insert the assembly (thermometer with the test tube) inside a water bath held in a 10 mL beaker. Slowly heat the bath until the rapid and continuous stream of bubble comes out of the capillary. Remove the heat and allow the bath to cool. Record the temperature at which bubbles cease and liquid begins to rise in the capillary. This is the boiling point temperature of the liquid .

 My question is that
1. What is the bubble? ( I mean what made the buble?)
2. At the step :" The temperature at which bubbles cease and liquid begins to rise in the capillary is the boiling point?" I don't understand why the bubble is cease when the temperature is at boiling point.
Thank you!

This procedure appears to me to be overly complex and will not reliably produce the boiling point of the methanol.

I don't understand why you are using  a melting point capillary tube.

Let's consider what is happening.  As the temperature of the liquid (methanol) heats, it will heat the melting point capillary tube and the air trapped.  As this occurs, the air in the melting point capillary tube will expand and, having no other exit, will bubble out throught the methanol.  It stopps when the temperature of the liquid (methanol) stops rising, which should be the boiling point, If the temperature bath is higher than the boiling point.

The same thing can be done less expensively and more reliabily by reading the temperature at reasonable intervals and noting when the temperature pauses rising, or the temperature that the liquid reaches before all of it boils away.  I would start taking the temperature each 5 degrees then use that time interval for the rest of the record.

One can also watch for the boiling bubble (methanol vapor).

One might think that one wanted to consume melting point capillary tubes in your experiemental setup.

Hope this helps.


some interesting problems on tangents and normals or just mention the source please!!!

The following may be far too easy for you: http://www.mathcentre.ac.uk/students.php/all_subjects/differentiation/tangents/resources/335

So might this:
http://www.yteach.co.uk/index.php/resources/gradient_tangent_curve_normal_function.html


A world-record swimmer and a dolphin compete in a 400-meter race. The swimmer swims at a rate of 2 meters per second, and the dolphin swims at a rate of 8 meters per second. When the dolphin crosses the finish line, how many meters is the swimmer from the finish line?

Dolphin needs $ 400\div 8\equal{}50$ seconds, and swimmer crosses $ 50\times 2\equal{}100$ meters during that time.
He is $ 400\minus{}100\equal{}300$ meters from the finish line.


function
trigonometry
real analysis
real analysis unsolved

It is nicer than I thought!
[hide]We have equality as functions, so we can take a derivative.  Thus $2y'y'' + 2yy' = y'$, so either $y' \equiv 0$ or $y + y'' = \frac12$.  The first yields $y \equiv 0$ and $y \equiv 1$ while the second yields solutions of the form $y(x) = A\sin x + B\cos x + \frac12$, and substitution into the original equation will yield a relationship between $A$ and $B$.[/hide]


Let $n \geq 2$ be a positive integer.
Determine the maximum value of $k_1 + k_2 + ... + k_n$,
given that $k_1,\;k_2,\;...,\;k_n$ are nonnegative integers such that $k_1^3 + k_2^3 + ... + k_n^3 \leq 7n$.

I think Arne posted on both ML and AoPS before the merger.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13801


number theory
real analysis
real analysis unsolved

For each integer $ n$ let $ \sigma_{k}\left(n\right)$ be the sum of the $ k$th power of its divisors.  Prove that

\[ \sum_{n\equal{}1}^{\infty}\frac{\sigma_{2}\left(n\right)}{n!}\]
is irrational.

[quote="Albanian Eagle"]number theory, maybe...[/quote]
there exist simpl solution without number theory . ( also I don't know how you can  use number theory   :maybe:.    )


Given real different numbers $ a,$ $ b$ and $ c$. Prove that:
\[ \frac {(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)^3}{(a \minus{} b)(b \minus{} c)(c \minus{} a)}\left(\frac {1}{(a \minus{} b)^3} \plus{} \frac {1}{(b \minus{} c)^3} \plus{} \frac {1}{(c \minus{} a)^3}\right) \leq \minus{} \frac {405}{16}
\]
When the equality occurs?

Let $ x \equal{} (a \minus{} b)^2,\ y \equal{} (b \minus{} c)^2,\ z \equal{} (c \minus{} a)^2\ xyz\neq 0$, we can rewrite the inequality as $ \frac {(x^2 \plus{} y^2 \plus{} z^2)^3}{xyz}\left(\frac {1}{x^3} \plus{} \frac {1}{y^3} \plus{} \frac {1}{z^3}\right)\leq \minus{} \frac {405}{2}$.

Let $ x\plus{}y\plus{}z\equal{}p,\ xy\plus{}yz\plus{}zx\equal{}q,\ xyz\equal{}r$, we have $ \frac{(r^2\plus{}3r\minus{}3pq)(p^2\minus{}2q)^3}{r^3}\leq \minus{} \frac {405}{2}$.

well i used $ x \equal{} (a \minus{} b) y \equal{} (b \minus{} c) z \equal{} (c \minus{} a)$ and the inequality turn to be
for all different real x,y,z which $ x \plus{} y \plus{} z \equal{} 0$
$ \frac {(xy \plus{} yz \plus{} xz)^3}{xyz}(\frac {1}{x^3} \plus{} \frac {1}{y^3} \plus{} \frac {1}{z^3}) \geq \frac {405}{16}$
then we got that
$ (xy \plus{} yz \plus{} xz)^3 \equal{} \frac { \minus{} 1}{8}(x^2 \plus{} y^2 \plus{} z^2)^3$ and
$ \sum x^3y^3 \equal{} \frac { \minus{} 1}{6}(39x^2y^2z^2 \plus{} 3\sum_{sym}x^4y^2)$
then problem turn to be
$ \frac {(x^2 \plus{} y^2 \plus{} z^2)^3 (39x^2y^2z^2 \plus{} 3\sum_{sym}x^4y^2)}{48x^4y^4z^4} \geq\frac {405}{16}$
which can be easily proven by AM-GM since all of the degree in L.H.S is even  :lol:

I found it accidentally that the equality will held when x=y=-1 z=2 but i cant find it y may be it's more formal form...


function
integration
real analysis
calculus
derivative
real analysis unsolved

Suppose f is a measurable real-valued function defined on $ \mathbb{R}$. Suppose for all pairs of real
numbers $ a < b$, $ 0\leq \int_{a}^{b} f dx \leq b \minus{} a$. Show that $ 0 \leq f (x) \leq 1$ for almost all x.

For the question to make any sense, $ f$ needs to be locally integrable (integrable over every bounded interval) as well as merely measurable.

Assuming that, we have that $ 0\le \frac1{|I|}\int_If\le 1$ for all intervals $ I.$

The Lebesgue theorem on the differentiation of the integral says that the limit as a sequence of intervals $ I$ shrinks to a point $ x$ of $ \frac1{|I|}\int_If$ equals $ f(x)$ for almost every $ x.$ A quantity which is always in $ [0,1]$ can only have a limit that is in $ [0,1].$

ah, me and a friend spent some time working on this ill-posed problem. Thank you!


Diophantine equation
number theory
quadratics
equation
IMO Shortlist

Let $ a, b \in \mathbb{Z}$ which are not perfect squares. Prove that if \[ x^2 \minus{} ay^2 \minus{} bz^2 \plus{} abw^2 \equal{} 0\] has a nontrivial solution in integers, then so does \[ x^2 \minus{} ay^2 \minus{} bz^2 \equal{} 0.\]

If $(x_0, y_0, z_0, w_0)$ is the nontrival solution of $x^2-ay^2-bz^2+abw^2=0$, then

$(x_0z_0-ay_0w_0, y_0z_0-x_0w_0, z_0^2-aw_0^2)$ is the nontrival solution of $x^2-ay^2-bz^2=0$.

How to get to the solution above: We first have to understand numbers of the form $x^2-ny^2$. The key idea is that the product of two such numbers is also such a number. In particular, we have the identity $\left(p^2-nq^2\right)\left(r^2-ns^2\right)=\left(pr-nqs\right)^2-n\left(ps-qr\right)^2$.

With this in mind, the following solution seems slightly less magical.

First rearrange to get $b\left(z^2-aw^2\right)=x^2-ay^2$.

Now multiply both sides by $\left(z^2-aw^2\right)$.

So we have  $b\left(z^2-aw^2\right)^2=\left(x^2-ay^2\right)\left(z^2-aw^2\right)$.

Now we use the identity above to get $b\left(z^2-aw^2\right)^2=\left(xz-ayw\right)^2-a\left(xw-yz\right)^2$.

Simply bring all the terms to one side, and $\left(xz-ayw\right)^2-a\left(xw-yz\right)^2-b\left(z^2-aw^2\right)^2=0$.

This is a solution of the form we wanted, so we are done.

Note we could have factored out with $a$ instead of $b$, so we also get that $\left(xy-bwz\right)^2-a\left(y^2-bw^2\right)^2-b\left(xw-yz\right)^2=0$ is a solution.

[quote=IMO2019]Perhaps it is like cross product of corfficients.[/quote]

This idea originates from the properties of determinants. For example, using determinants you can prove that the product of two numbers of the form $a^3+b^3+c^3-3abc$ is also of this form.

A different solution:

If $x^2 + abw^2 = ay^2 + bz^2$, then we have:

$$a(xy+bzw)^2 + b(zx-ayx)^2 = (ay^2 + bz^2)^2.$$


geometry
3D geometry
algebra
polynomial
linear algebra
matrix
induction

Let $Q_n$ be the graph of the n-dimensionale cube; that is, the vertices are the sequences $a_1,...,a_n$ with $a_i\in \{0; 1\}$ for all $1\leq i\leq n$ and two vertices are connected by an edge iff the vertices differ in exactly one coordinate.

A subgraph $H$ which contains all the vertices such that every pair of vertices is joined by a path in $H$ and which contains no cycles is called a spanning tree.

Show that the number of spanning trees of $Q_n$ is ${4^{n\choose 2} 6^{n\choose 3} 8^{n\choose 4}}\cdots (2n)^{n\choose n}$.

For example, if $n=2$, $Q_2$ looks like a square and the spanning trees are those subgraphs where exactly one of the edges is missing. Of course, there are exactly 4 of those, as predicted by the formula.

Note: There is no purely combinatorial proof known yet.

We can use [url=http://en.wikipedia.org/wiki/Kirchhoff%27s_theorem]Kirchhoff's Theorem[/url]. All we need to do is find the characteristic polynomial of [url=http://en.wikipedia.org/wiki/Admittance_matrix]the admittance matrix[/url] of the graph $Q_n$, and it's easy to prove by induction on $n$ that this polynomial is precisely $x\cdot(x-2)^{\binom n1}\cdot(x-4)^{\binom n2}\cdot\ldots\cdot(x-2n)^{\binom nn}$.


algebra
function
domain
inequalities
difference of squares
special factorizations

These should be simple questions, but I don't know why I couldn't get the answers.  Thanks for helping.

Q1.
Simplify
2/(x^2 - 1) - 1/(x - 1)
the answer is
-1/(x+1) but I don't know how you get there.  (And I don't know how to type up the math formulas with the special program.)

---

Q2.  Is f(x) = -x^3 / (3x^2 -9) even or odd?  I graphed it and found the answer to be odd, but I don't get this answer with:

f(-x) = -(-x)^3 / [3(-x)^2 - 9] =? x^3 / (3x^2 -9) --> where is the mistake?

---

Q3. How can I restrict the domain of a graph on a TI-83 plus?

---

Q4.  How do you find the domain restriction for f (g (x)) with detailed algebra?  I know the answer is {x | x<-1 or x>1}  (from the back of the book). But  I was confused by my own math.

f(x)=(x+1)^(1/2)   g(x) = 2 / (x-1)

---

thanks.

Q1.
$\frac{2}{x^2 - 1} - \frac{1}{x - 1} = \frac{2 - (x + 1)}{(x + 1)(x - 1)} = \frac{- (x - 1)}{(x + 1)(x - 1)} = \frac{-1}{x + 1}$

Q2.
$f(x)$ is odd.  By definition a function $f$ is odd if $f(-x) = - f(x)$.  And your work shows it.

Q3.
In Y=, divide the function by the restriction.  For example, to graph $y = x, x > 1$, input $Y_{1} = x/(x>1)$.  The greater than, less than, etc relations are located in TEST, which is 2nd MATH.

Q4.
$f(x) = (x + 1)^{\frac{1}{2}} = \sqrt{x + 1}$

$g(x) = \frac{2}{x - 1}$

$f(g(x)) = \sqrt{\frac{2}{x - 1} + 1} = \sqrt{\frac{2 + (x - 1)}{x - 1}} = \sqrt{\frac{x + 1}{x - 1}}$

Since you have a fraction and a square root, the denominator cannot equal 0 and $\frac{x + 1}{x - 1} \ge 0$.

$x - 1 \neq 0$, so $x \neq 1$

$\frac{x + 1}{x - 1} \ge 0$, so $x \le -1$ or $x > 1$

Q1.
You must first find a common denominator.
$\frac{2}{x^2-1} - \frac{1}{x-1} \Rightarrow \frac{2(x-1) - (x^2 - 1)}{(x^2-1)(x-1)}$
Then you just simplify from there.
$\frac{2x-2-x^2+1}{(x^2-1)(x-1)}$
$\frac{-[x^2-2x+1]}{(x^2-1)(x-1)}$
$\frac{-[(x-1)(x-1)]}{(x^2-1)(x-1)}$
$\frac{-(x-1)}{x^2-1}$
The denominator can be factored as the difference of squares:
$\frac{-(x-1)}{(x+1)(x-1)}$
$\frac{-1}{x+1}$

Hmm dakyru's is way better for Q1, it would have been a good idea to factor the denominator first heh.

Thank you dakyru and SnowStorm.
I have more...

Q3.5
How can I restrict the domain of a graph between two numbers?

Q4.5
Please show in detail...
$\frac{x + 1}{x - 1} \ge 0$, so $x \le -1$ or $x > 1$

Q3.5
Do what dakyru says only divide by another restriction using and inbetween.  For example, to graph $y = x$ from 2 to 5, graph the function $y = x/(x<5 \mbox{ and } x>2)$.  The $and$ can be found under the Logic (2nd->Math->Logic) section on your calculator.

Q4.5
You want a positive number.  This means that both $x+1$ and $x-1$ must have the same sign, if not the quotient will be negative.  This will only occur when $x \leq -1$ or when $x > 1$.

[quote="SnowStorm"]Q3.5
Q4.5
You want a positive number.  This means that both $x+1$ and $x-1$ must have the same sign, if not the quotient will be negative.  This will only occur when $x \leq -1$ or when $x > 1$.[/quote]

Would you please prove this with algebra?  (Because I can't.)  Thank you.

(x + 1)/(x - 1) => 0

since u can only multiply a positive number to a inequality without changing the sign, u can only multiply (x - 1)^2 to both sides.

so therefore, u will cancel out one of the (x - 1) to get::::

(x + 1).(x - 1) => 0

there u are


Prove or disprove, for $ m,n,x,y \in \mathbb{Z}^\plus{}$ and $ m \le n,\mbox{ }x \le y$.
If \[ \frac{1}{m}\plus{}\frac{1}{m\plus{}1}\plus{}...\plus{}\frac{1}{n}\equal{}\frac{1}{x}\plus{}\frac{1}{x\plus{}1}\plus{}...\plus{}\frac{1}{y}\]
then $ m\equal{}x$ and $ n\equal{}y$.


Evaluate the sum:
\[ S=\sum_{i=0}^n \frac{4^i}{1+\cos{2^i}}  \]

[quote="bhzhan"]Very good problem indeed. I wonder where it came from...
[hide]
Simply add $\frac{1}{1-\cos1}$ to the front of the sum. Observe:
\[ \frac{4^i}{1-\cos{2^i}}+\frac{4^i}{1+\cos{2^i}}=\frac{4^{i+1}}{2-2\cos^2{2^i}}=\frac{4^{i+1}}{1-\cos{2^{i+1}}}  \]

Final answer: $\sum_{i=0}^n \frac{4^i}{1+\cos{2^i}}=\frac{4^{n+1}}{1-\cos{2^{n+1}}}-\frac{1}{1-\cos1}$.[/hide][/quote]

From a magazine.


Hi I drew a picture, it's a man with a tree. What do you think about it? :blush:
If you want to see more, visit christophsbaum.npage.de(it's written in german, but pictures haven't got a language)

This picture is beautiful, you are a really good painter.

Can somebody give me a tip how I can draw it better?

wait you really made it with paint? it looks more like pencil/ink. the picture's great, but it would be nicer with more details and contrast.

Yes, it looks like everything is drawn with cross-hatching...MESSY cross hatching.

I drew this picture with pencil and ink, I have never said that I made it with paint.

no, i just read what "bearkiller" wrote.

[quote="kabi"]picture is quite ordinary even then it has something very very special which made me to write a post.
That man standing near the tree.there is fire but standing away from it .May be he is having something else in his mind.May be he is thinking about something .
I wish u will continue this work.
all the best or in my language "tum accha kaam karte rahu"[/quote]

 :huuh:

i think it's good. maybe a bit more blending and detail would make it awesome =] Viel Glück!!! (i think that's good luck on german...idk got it from my friend that takes german =])

sometimes my interpretation to  things bounces from many people's mind.
Anyways I still think this sketch has something rear in it.
Just try to observe it like I did.

What you could really do is improve the shadow. I personally do not like the fire because it kind of contradicts the theme. You have a dark shady tree and a light background, and then you have a fire. A dark tree with a dark background and a fire would be better if you want to keep using the fire.


number theory
greatest common divisor
algebra unsolved
algebra

Let $ X$, $ Y$, $ Z$ be distinct positive integers having exactly two digits in such a way that:

$ X\equal{}10a \plus{}b$
$ Y\equal{}10b \plus{}c$
$ Z\equal{}10c \plus{}a$

($ a,b,c$ are digits)

Find all posible values of $ gcd(X,Y,Z)$

[quote="lambruscokid"]Let $ X$, $ Y$, $ Z$ be distinct positive integers having exactly two digits in such a way that:

$ X \equal{} 10a \plus{} b$
$ Y \equal{} 10b \plus{} c$
$ Z \equal{} 10c \plus{} a$

($ a,b,c$ are digits)

Find all posible values of $ gcd(X,Y,Z)$[/quote]
$ 11$ does not divide $ \gcd(X,Y,Z)$, else we would have $ a\equal{}b\equal{}c$ and $ X\equal{}Y\equal{}Z$
$ \gcd(X,Y,Z)|(X\plus{}Y\plus{}Z)\equal{}11(a\plus{}b\plus{}c)$ and so $ \gcd(X,Y,Z)|(a\plus{}b\plus{}c)$ and so $ \gcd(X,Y,Z)\leq 26$
$ 5$ does not divide $ \gcd(X,Y,Z)$, else we would have $ a\equal{}b\equal{}c\equal{}5$ and $ X\equal{}Y\equal{}Z$
$ 6$ does not divide $ \gcd(X,Y,Z)$, else we would have $ a,b,c$ even with pairwaise sums all equal to $ 6$ or $ 12$, which is impossible
$ 9$ does not divide $ \gcd(X,Y,Z)$, else we would have $ (a,b,c)\in\{(0,0,0),(9,9,9)\}$

So $ \gcd(X,Y,Z)\in\{1,2,3,4,7,8,13,14,16,17,19,21,23,26\}$, so just $ 14$ tests :

$ \gcd(11,12,21)\equal{}1$
$ \gcd(22,24,42)\equal{}2$
$ \gcd(33,39,93)\equal{}3$
$ \gcd(44,48,84)\equal{}4$
$ \gcd(14,49,91)\equal{}7$
$ \gcd(X,Y,Z)\equal{}8$ $ \implies$ $ a,b,c$ all even and so $ X,Y,Z\in\{24,48,64,88\}$ so no solution.
$ \gcd(13,39,91)\equal{}13$
$ \gcd(28,84,42)\equal{}14$
$ \gcd(X,Y,Z)\equal{}16$ $ \implies$ $ a,b,c$ all even and so $ X,Y,Z\in\{48,64\}$ so no solution.
$ \gcd(X,Y,Z)\equal{}17$ $ \implies$  $ X,Y,Z\in\{17,34,51,68,85\}$ so no solution.
$ \gcd(X,Y,Z)\equal{}19$ $ \implies$  $ X,Y,Z\in\{19,38,57,76,95\}$ so no solution.
$ \gcd(X,Y,Z)\equal{}21$ $ \implies$  $ X,Y,Z\in\{21,42,63,84\}$ so no solution.
$ \gcd(X,Y,Z)\equal{}23$ $ \implies$  $ X,Y,Z\in\{23,46,69,92\}$ so no solution.
$ \gcd(X,Y,Z)\equal{}26$ $ \implies$  $ X,Y,Z\in\{26,52,78\}$ so no solution.

Hence the answer : $ \gcd(X,Y,Z)\in\{1,2,3,4,7,13,14\}$

I hope there exists a more interesting solution ...

I've got a slightly different solution, but I wouldn't say it's interesting.

We have $ \gcd(X$, $ Y$, $ Z) \mid \alpha X \plus{} \beta Y \plus{} \gamma Z$ for some integers $ \alpha$, $ \beta$, $ \gamma$. Choosing $ ( \alpha$, $ \beta$, $ \gamma) \equal{} (1$, $ 1$, $ 1)$ one gets $ \gcd(X$, $ Y$, $ Z) \mid 11(a \plus{} b \plus{} c)$ and hence  $ \gcd(X$, $ Y$, $ Z) \mid a \plus{} b \plus{} c \le 26$, as $ \gcd(X$, $ Y$, $ Z) \mid 11$ would lead to $ a \equal{} b \equal{} c$, a contradiction.
Moreover, choosing $ ( \alpha$, $ \beta$, $ \gamma) \equal{} (1$, $ 100$, $ \minus{} 10)$, one gets $ \gcd(X$, $ Y$, $ Z) \mid 1001b \equal{} 7 \cdot 11 \cdot 13b$ and hence $ \gcd(X$, $ Y$, $ Z) \mid 7 \cdot 13 b$ and the same for $ a$ and $ c$. But now we can rule out some cases, because one of the numbers $ 5$, $ 6$, $ 8$, $ 9$ can't divide $ a$, $ b$, $ c$ simultaneously. Thus, $ \gcd(X$, $ Y$, $ Z) \mid l \cdot 7 \cdot 13$, where $ l \in \{1$, $ 2$, $ 3$, $ 4$, $ 7\}$. Hence $ \gcd \in \{ 1$, $ 2$, $ 3$, $ 4$, $ 7$, $ 13$, $ 14$, $ 21$, $ 26 \}$ and each of these values, except $ 21$ and $ 26$, gives a solution indeed.
Hence the anser is: $ \gcd(X$, $ Y$, $ Z) \in \{ 1$, $ 2$, $ 3$, $ 4$, $ 7$, $ 13$, $ 14 \}$.

[quote="Rafikafi"][quote="FelixD"]
...
But now we can rule out some cases, because $ a$, $ b$, $ c$ can't divide one of the numbers $ 5$, $ 6$, $ 8$, $ 9$ simultaneously. Thus, $ \gcd(X$, $ Y$, $ Z) \mid l \cdot 7 \cdot 13$, where $ l \in \{1$, $ 2$, $ 3$, $ 4$, $ 7\}$.
...[/quote]

You mean one of $ 5,6,7,8,9$ can't divide $ a,b,c$ simultainiously otherwise $ a \equal{} b \equal{} c$. Thus, $ \gcd(X$, $ Y$, $ Z) \mid l \cdot 7 \cdot 13$, where $ l \in \{1, 2, 3, 4\}$.[/quote]

Of course I mean $ 5$, $ 6$, $ 8$, $ 9$ can't divide $ a$, $ b$, $ c$ simultaneously. But cou can't rule out the $ 7$ IMHO, because we already have $ \gcd \mid 7 \cdot 13l$.

Yes you can but at this example it doesn't matter. $ l$ is nothing else than $ gcd(a,b,c)$ If $ l\equal{}7 \Rightarrow 7|a$ and $ b$ and $ c$ Therefore $ a\equal{}b\equal{}c\equal{}7$ This doesn't mean that $ 7$ cannot divide $ gcd(X,Y,Z)$ it only means that $ 49$ doesn't divide $ gcd(X,Y,Z)$


geometry
incenter
circumcircle
geometric transformation
homothety
trigonometry
search

Let $ ABC$ be a given triangle with the incenter $ I$, and denote by $ X$, $ Y$, $ Z$ the intersections of the lines $ AI$, $ BI$, $ CI$ with the sides $ BC$, $ CA$, and $ AB$, respectively. Consider $ \mathcal{K}_{a}$ the circle tangent simultanously to the sidelines $ AB$, $ AC$, and internally to the circumcircle $ \mathcal{C}(O)$ of $ ABC$, and let $ A^{\prime}$ be the tangency point of $ \mathcal{K}_{a}$ with $ \mathcal{C}$. Similarly, define $ B^{\prime}$, and $ C^{\prime}$.

Prove that the circumcircles of triangles $ AXA^{\prime}$, $ BYB^{\prime}$, and $ CZC^{\prime}$ all pass through two distinct points.

We start by observing that a positive homothety centre $ A'$ takes the circumcircle to $ K_a$, and that a positive homothety centre $ A$ takes $ K_a$ to the incircle. Thus the composition of these two homotheties takes the circumcircle to the incircle, is positive, and has homothetic centre $ K$ on the line $ AA'$ (it is well known that the fixed point of a composition of two homotheties lies on the line containing the fixed points of the two individual homotheties). But $ K$ is clearly the external centre of similitude of the incircle and circumcircle, since the composed homothety is positive and takes the circumcircle to the incircle, and so also (because this characterisation is symmetric in $ A,B,C$) $ K$ lies on lines $ BB'$,$ CC'$.

Having established this, we can define $ A',B',C'$ in terms of $ K$ and finish the problem with (not necessarily normalised) areal co-ordinates.
$ K$, the external centre of similitude of the circumcircle and incircle, is Kimberling centre $ X_{56}$ which is known to have areal co-ordinates $ (\frac {a^2}{b \plus{} c \minus{} a},\frac {b^2}{c \plus{} a \minus{} b},\frac {c^2}{a \plus{} b \minus{} c})$ (I shall attach a proof of this fact at the end if it is not considered sufficient to simply quote this from the Encyclopedia of Triangle Centers).

Thus $ AA'$ has parametrisation $ (\frac {a^2}{b \plus{} c \minus{} a},\frac {tb^2}{c \plus{} a \minus{} b},\frac {tc^2}{a \plus{} b \minus{} c})$ with $ A$ at $ t \equal{} 0$. Intersecting this with the circumcircle $ a^2yz \plus{} b^2zx \plus{} c^2xy \equal{} 0$, clearing denominators, and dividing by $ a^2b^2c^2$ gives:
\[ t^2(b \plus{} c \minus{} a) \plus{} t(c \plus{} a \minus{} b) \plus{} t(a \plus{} b \minus{} c) \equal{} t(t(b \plus{} c \minus{} a) \plus{} 2a) \equal{} 0
\]
.
Thus the value of $ t$ corresponding to $ A'$ is $ t \equal{} \frac { \minus{} 2a}{b \plus{} c \minus{} a}$.
This gives the co-ordinates of $ A'$ as (after multiplying through by common things and -1) $ A'( \minus{} a,\frac {2b^2}{c \plus{} a \minus{} b}, \frac {2c^2}{a \plus{} b \minus{} c})$.

It may now be easily verified that $ A,A',X$ ($ X(0,b,c)$ well known - equivalent to angle bisector theorem) all lie upon the circle
\[ \Gamma_A: a^2yz \plus{} b^2zx \plus{} c^2xy \plus{} a^2(x \plus{} y \plus{} z)(\frac {c^2(a \plus{} c \minus{} b)y \minus{} b^2(a \plus{} b \minus{} c)z}{(b^2 \minus{} c^2)(a \plus{} b \plus{} c)}) \equal{} 0
\]
By cyclic permutation of letters, we obtain the equations for other two circles in the question, containing $ B,B',Y$ and $ C,C',Z$ respectively.
\[ \Gamma_B: a^2yz \plus{} b^2zx \plus{} c^2xy \plus{} b^2(x \plus{} y \plus{} z)(\frac {a^2(b \plus{} a \minus{} c)z \minus{} c^2(b \plus{} c \minus{} a)x}{(c^2 \minus{} a^2)(a \plus{} b \plus{} c)}) \equal{} 0\]
\[ \Gamma_C: a^2yz \plus{} b^2zx \plus{} c^2xy \plus{} c^2(x \plus{} y \plus{} z)(\frac {b^2(c \plus{} b \minus{} a)x \minus{} a^2(c \plus{} a \minus{} b)y}{(a^2 \minus{} b^2)(a \plus{} b \plus{} c)}) \equal{} 0\]

It is known that the areal radical axis of two circles is obtained by subtracting the equations of the circles from one another and dividing by $ x\plus{}y\plus{}z$. Doing this for $ \Gamma_A,\Gamma_B$, we get that their radical axis is described by the equation

\[ \frac {a^2c^2(a \plus{} c \minus{} b)y \minus{} a^2b^2(a \plus{} b \minus{} c)z}{(b^2 \minus{} c^2)(a \plus{} b \plus{} c)}\equal{}\frac {b^2a^2(b \plus{} a \minus{} c)z \minus{} b^2c^2(b \plus{} c \minus{} a)x}{(c^2 \minus{} a^2)(a \plus{} b \plus{} c)}\]

This simplifies, clearing denominators and grouping coefficients on the RHS, to
\[ b^2c^2(c^2\minus{}b^2)(b\plus{}c\minus{}a)x\plus{}c^2a^2(a^2\minus{}c^2)(c\plus{}a\minus{}b)y\plus{}a^2b^2(a\plus{}b\minus{}c)(b^2\minus{}a^2)z \equal{} 0\]

But this line is symmetric in $ ((a,x),(b,y),(c,z))$, and so the same line would have been obtained by intersecting $ \Gamma_B,\Gamma_C$ or $ \Gamma_C,\Gamma_A$. Therefore $ \Gamma_A,\Gamma_B,\Gamma_C$ are coaxal, and since the point $ K$ is interior to all of them, they must all intersect one another at two points, so, being coaxal, they must all intersect one another at [i]the same[/i] two points, so they have two common points, and we are done. :)
-----------------------------------------------------------------------------

[u]Deriving the areal co-ordinates of $ K\equal{}X_{56}$[/u]

The homothety centre $ K$ factor $ r/R$ takes the line $ BC$ to a line tangent to the circumcircle at the midpoint $ L$ of the arc $ BC$. Let $ L$ projected onto $ BC$ be $ D$. Let the areals of $ K$ be $ (ax,by,cz)$ where $ x$ is the perpendicular distance from $ K$ to $ BC$, etc.
Then $ x \equal{} \frac{r}{R\minus{}r}LD\equal{}\frac{r}{R\minus{}r}\frac{a}{2}\tan{\frac{A}{2}}$
So $ x: y: z \equal{} a\tan{\frac{A}{2}}: b\tan{\frac{B}{2}}: c\tan{\frac{C}{2}}$

But $ \tan{\frac{A}{2}}: \tan{\frac{B}{2}}: \tan{\frac{C}{2}} \equal{} \frac{r}{s\minus{}a}: \frac{r}{s\minus{}b}: \frac{r}{s\minus{}c} \equal{} \frac{1}{b\plus{}c\minus{}a}: \frac{1}{c\plus{}a\minus{}b}: \frac{1}{a\plus{}b\minus{}c}$ considering the triangles formed dropping perpendiculars from the incentre.

Thus $ ax: by: cz \equal{} \frac{a^2}{b\plus{}c\minus{}a}: \frac{b^2}{c\plus{}a\minus{}b}: \frac{c^2}{a\plus{}b\minus{}c}$, yielding the areal co-ordinates required.

For brevity in writing: $ A^{\prime} \equal{} A_1$
Lemma1: $ A_1I$ is the angle bisector of $ BA_1C$ Proof:http://www.mathlinks.ro/Forum/viewtopic.php?search_id=657778495&t=22925
Let $ R_a$ the midpoint of the minimum arc BC of the circumcircle ABC, and $ S_a$, the midpoint of the arc BC containing A, and O the circumcenter, $ R_a,O,S_a$ are collinear. from the lemma  $ S_a,I,A_1$ are collinear, (wlog a>b>c)set $ X_a$, the point of BC such that $ X_aI$ is perpendicular to $ AI$, because $ < AXX_a \equal{} C \plus{} A/2$, then $ < IX_aX \equal{} B/2 \minus{} C/2$, and $ < BIC \equal{} C/2 \equal{} < ICX_a$
then $ X_aB.X_aC \equal{} (X_aI)^2 \equal{} (X_aT)(X_aR_a)$, where T is the intersection point of $ X_aR_a$ and the circumcircle BAC, then $ < ITR_a \equal{} 90$, implying: $ I,T,S_a$ are collinear, then $ T \equal{} A_1$, then:
$ (R_aX)(R_aX) \equal{} (R_aC)^2 \equal{} (R_aI)^2 \equal{} (R_aA_1)(R_aX_a)$, so $ XA_1X_aA$ is inscriptible.
Now use inversion with circle: Incircle of triangle BAC, this map the problem to another one:
In a triangle ABC(wlog a>b>c), let $ A_1,A_2,A_3,X_a$, the midpoint of BC, the foot of the perpendicular from O to the altitude passing throught $ A$, the foot of the perpendicular from A to the bisector of BC and the foot of the altitude throught A, I have to prove the circles $ A_1A_2A_3 \equal{} w_a$ concur in two points. Because $ A_1A_2A_3X_a$ is a rectangle the center of the circle $ w_a$ is in the bisector of $ AX_a$, then for simmetry, it pass througt a point in $ AX_a$ such that $ AM_a \equal{} X_aA_2 \equal{} RcosA \equal{} AH/2$ where H is the ortocentre of BAC and R the circumradius.
Then $ M_a$ is the midpoint of AH, because $ OA_1HM_a$ is a paralelogram $ M_AA_1$ intersect $ OH$ in W: center of the 9 point circle(w). Then the power of W respect to $ w_a$ is the same of $ W$ respect to $ w \equal{} \minus{} R^2/4$, similarly for the other circles. Then W is the intersections of the radical axis of this 3 circles: radical center(if they dont intersect in 2 points, either is just a point in the radical axis).
If I prove the center of this circles are collinear then supposing they dont have 2 points in common, then the 3 radical axis are parallel(perpendicular to the line joining the centers), and the dont have an intersection, but W is this point, a contradiction.
Finally in order to prove this: the center of $ w_a$ = $ O_a$ is in the line passing throught W and penpendicular to $ M_aA_1$, because this line is the perpendicular bisector of this segment,  let $ R_a$ the circuncenter of $ A_1X_aM_a$, then letting $ 2 < x$ the measure of the arc $ A_1M_a$ of $ w_a$, then $ 2R_asenx \equal{} R$, and $ O_aW \equal{} R_acosx \equal{} R/2.cotx$, but for simmetry in the rectangle $ x \equal{} < A_2A_1O \equal{} < X_aA_2A_1$ then by easy computacion $ cotx \equal{} cosA/sen(B \minus{} C) \equal{} l_a$, similarly $ l_b,l_c$ angle between l_a and l_b is $ 180 \minus{} 2C$, because l_a is perpendicular to $ M_aA_1$ then is perpendicular to $ AO$, similarly l_b, and $ < AOB \equal{} 2A$, similarly the angle between $ l_b,l_c$ is $ 180 \minus{} 2A$, then the angle between $ l_a,l_c$ is $ 2B$, now to prove the collineality, use areas, because if $ O_b$ is in the interior of triangle $ WO_aO_c$ the sum of the areas will be least, and if the $ O_b$ is outside the sum of the areas will be greater, so i have to prove:
$ R^2/4. \frac {cosAcosB}{sen(B \minus{} C).sen(A \minus{} C)}.sen2C \plus{} R^2/4. \frac {cosBcosC}{sen(A \minus{} C)sen(A \minus{} B)}.sen2A \equal{} R^2/4. \frac {cosAcosC}{sen(B \minus{} C)sen(A \minus{} B)}.sen2B$, which is equivalent to $ senC.sen(A \minus{} B) \plus{} senA.sen(B \minus{} C) \equal{} senB.sen(A \minus{} C)$, because $ sen2A \equal{} 2senAcosA$, finally this is equivalent to
 $ sen(A \plus{} B).sen(A \minus{} B) \plus{} sen(B \plus{} C).sen(B \minus{} C) \equal{} sen(A \plus{} C).sen(A \minus{} C)$, this is true because $ sen(x \plus{} y).sen(x \minus{} y) \equal{} sen^2x \minus{} sen^2y$


Find 2 values of k for which angle BAC is a right angle: A(k,5), B(5,4), and C(-1,-3).

[hide]
Use vectors (the dot product must be zero):

$(k+1)(k-5)+8=0$

$k=1, 3$[/hide]

different way to go about it.. definatly not the fastest, but it was fun for me to do :D
[hide]since there is more than one point on $y=5$ in order for A to be a 90 degree angle, then the points must lie on a circle which has a diameter of the distance between point B and C. Recall that if the base of a triangle is the diameter of its circumcircle then it must be a right triangle.

Find the diameter of this circle
$d = \sqrt{(-1-3)^{2}+(5-(-1))^{2})}= \sqrt{85}$
so we know that the circle has a radius of $\frac{\sqrt{85}}{2}$
in order to set up an equation for this circle, we must find the mid point of this line (the center of our circle) which is:
$\left(\frac{-1+5}{2}, \frac{-3+4}{2}\right) = \left(2, \frac{1}{2}\right)$
so we know the circle was shifted 2 units right and 1/2 unit up, so now we can create an equation for our circle:
$(x-2)^{2}+(y-\frac{1}{2})^{2}= radius^{2}= \frac{85}{4}$
our values of k (which is x) lie along $y=5$ so plugging in 5 for all the y values yields:
$(x-2)^{2}+(5-\frac{1}{2})^{2}= \frac{85}{4}$
$(x-2)^{2}= \frac{85}{4}-\frac{81}{4}= 1$
$x = 2 \pm \sqrt{1}$
Which gives us our answer of

$\boxed{k_{1}= 1, k_{2}= 3}$
[/hide]


If $y^2-x^2=1$, find the range of $\frac{y}{x+1}$.

[hide]The answer is $\left|\frac{y}{x+1}\right|\geq \frac{1}{\sqrt{2}}.$[/hide]


calculus derivative function LaTeX algebra polynomial limit

y(t) is a monotoniically increasing function, given by the sum of (-1)^n (a(n))(t^n)/n! for n = 1 to infinity so that dy/dt = the sum of (-1)^(n+1)(a(n+1))(t^n) from 1 to infinity. y(t) approaches Y as t goes to infinity. The question is: is it possible to express dy/dt as the sum of c(n) y^n, where, since dy/dt = 0 when y = Y, Y will be the first zero of this sum? Is there a general expression for the coefficients c(n)?

Please, this is unreadable.

[quote]Let $ y(t)$ be a monotonically increasing function, given by $ f(t) \equal{} \sum\limits_{n\equal{}1}^\infty (\minus{}1)^n a_n \frac{t^n}{n!}$ so that $ \frac{dy}{dt} \equal{} \sum\limits_{n\equal{}1}^\infty (\minus{}1)^{n\plus{}1} a_{n\plus{}1} t^n$. Now, we have $ \lim\limits_{t \to \infty} y(t) \equal{} Y \in \mathbb{R}$.
The question is: is it possible to express $ \frac{dy}{dt}$ as the sum of $ c_n y^n$, where, since $ \frac{dy}{dt} \equal{} 0$ when $ y \equal{} Y$, $ Y$ will be the first zero of this sum? Is there a general expression for the coefficients $ c_n$?[/quote]

This seems weird, as being able to express the derivative of $ y$ like that gives you what the $ a_n$ are. So then you would know what $ Y$ is. Etc.

Don't get the rest of the post. Please elaborate (using LaTeX, see my post to learn how). Is this a question from a book?

No, it's not from a book. Someone asked a mutual friend to ask me. It's connected with something he's working on, but I don't know what that is. I'll have a look at Latex.

$ f'(t) \equal{} \sum_{n\equal{}1}^\infty (\minus{}1)^n a_n \frac{t^{n\minus{}1}}{(n\minus{}1)!}$ so by comparison with the formulation of the derivative given we must have $ a_n\equal{}0$ for $ n \neq 1,2$.  The derivative given also excludes a constant term so $ a_1\equal{}0$ and the function simplifies to $ f(t) \equal{} a_2 t^2 /2$.  And no $ Y$ exists.

Thanks for your reply, facis. Your expression for the derivative of f(t) is correct; the one I was given is wrong as it omits the term in a(1). So I guess the comparison cannot be made. Any other ideas?

well I suppose we could just adjust the series you gave to start with $ n\equal{}0$ but in any case the problem, as stated, is inconsistent since the series you gave for the derivative makes the function into a simple 2nd degree polynomial, and no $ Y$ will exist in that case.  If you omit your condition on the derivative, it becomes a whole different story.  But I don't think plugging in your formula for your series is going to be enough.

So how do you want to change your problem?

Yes. The derivative dy/dt which I was given is simply wrong and should be as you stated.

well the condition that $ y(t) \equal{} \sum_{n\equal{}0}^\infty (\minus{}1)^n a_n t^n/n!$ is really just the statement that $ y$ can be represented by its maclaurin series, which is not particularly useful.  Monotonically increasing means that $ y'(t) > 0$, and the fact that $ \lim_{t \to \infty} y(t)\equal{}Y$ tells us that $ y(t)$ is bounded.

WE wish to know if there is a sequence $ c_n$ such that $ y'(t) \equal{} \sum c_n y^n$.  With an initial condition of the form $ y(a)\equal{}b$, we could clearly estimate a unique solution with such a differential equation using Euler's method.  But it is interesting to note that a slope field of this differential equation would not depend on $ t$.  But I think it is possible to write a differential equation with $ y'$ as a function of $ y$ for any specific function - just possibly not for more general forms.

Then if we have $ y'\equal{}g(y)$ for some function $ g$, and $ g$ converges to it's maclaurin series, then we can find the coefficients $ c_n$; otherwise, it is impossible.

beyond this general analysis, I really don't know what to do.


Παιδιά μήπως μπορείτε να με βηθήσετε με αυτήν την άσκηση?

1.Αν y≠0 και n=2k (άρτιος).
Δείξτε ότι: x^n + y^n = (x + y)^n
μόνο όταν x=0

2. Αν y≠0 και n=2k+1 (περιττός)
Δείξτε ότι: x^n + y^n = (x + y)^n
μόνο όταν x = 0 ή x = −y

Spine h askhsh anaferetai se x,y akeraious kai k akeraio megalutero h iso me to 1 etsi?

An nai, h lush pou skefthka xrhsimopoiei Teleutaio 8ewrhma Fermat(T8F)

H genikh idea exei ws e3hs:
Gia thn prwth periptwsh e3etazw periptwseis:k=1 kai k>1
H prwth dinei 2xy=0 =>x=0 afou y diaforo tou mhden
Gia k>1 an x diaforo tou mhden mporoume na paroume anti gia x,y,x+y tis apolutes times autwn (logw ths artias dunamhs) pou erxetai se anti8esh omws me to T8F. Opote x=0, pou epalh8euei kai thn sxesh

Gia thn deuterh periptwsh:
Pairnw tis periptwseis gia x,y,x+y 3exwrista:an x diaforo tou mhdenos:
1. x,y>0 =>x+y>0 atopo logo T8F
2.1.1 x<0 y>0 x+y>0 kai h dosmenh sunepagetai y^n=(x+y)^n+(-x)^n atopo
2.1.2 x<0 y>0 x+y<0 kai h dosmenh sunepagetai y^n=(-x-y)^n+(-x)^n atopo
2.1.3 x<0 y>0 x+y=0 pou sunepagetai x=-y kai pou plhrei thn dosmenh
2.2.1 x>0 y<0 x+y>0 kai h dosmenh sunepagetai x^n=(x+y)^n+(-y)^n atopo
2.2.2 x>0 y<0 x+y<0 kai h dosmenh sunepagetai x^n=(-x-y)^n+(-y)^n atopo
2.2.3 x>0 y<0 x+y=0 pou sunepagetai x=-y kai pou plhrei thn dosmenh
3. x,y<0 =>x+y<0 pou dinei (-x-y)^n=(-x)^n+(-y)^n pali atopo

Eukola vlepoume oti gia x=0 h dwsmenh epalh8euetai

Opote h askhsh lu8hke

UG

Sorry paidia gia ton tropo grafhs m alla den 3erw na grafw se latex!

3exasa na pw oti an n artios uparxei h periptwsh x+y=0 (pou den thn kaluptei to T8F) opote pairnoume:
x^2k+y^2k=0 =>x=y=0 atopo

Όχι salonike η άσκηση πρέπει να λυθεί στο R, συγνώμη που δεν το διευκρίνισα,
μόνο τo n είναι ακέραιος,εγώ διαίρεσα με ψ (αφού είναι διάφορο του 0)  τη σχέση και μετά έθεσα,όπου χ/ψ =ω και βγήκε
(1+ω)^n=ω^n+1 την έθεσα σαν συνάρτηση   f(ω) και προσπάθησα να αποδείξω ότι δεν έχει άλλες ρίζες εκτός τις προφανείς , με βάση την μονοτονία της,
έλαβα περιπτώσεις για το ω , απέδειξα ότι είναι γνησίως αυξουσα στο (0,οο) και στο (οο, -1) αλλά στο (-1,0) δεν ξέρω
για δες το και συ σαν συνάρτηση με πεδίο ορισμού το R , να μου πεις
Σε ευχαριστώ πολύ για τον χρόνο σου και τις συμβουλές σου!

[quote="spinos"]απέδειξα ότι είναι γνησίως αυξουσα στο (0,οο) και στο (οο, -1) αλλά στο (-1,0) δεν ξέρω
[/quote]

Spine nomizw oti sto (-oo,-1) den einai panta au3ousa. gia n artio sumfwnw alla gia n peritto nomizw pws einai f8inousa.
gia n=3: f(-2)=(1-2)^3-(-2)^3-1=-1+8-1=6
enw  f(-3)=(1-3)^3-(-3)^3-1=-8+27-1=18

mporeis na steileis ligo thn apodei3h s gia thn monotonia?

pantws gia n artio sto (-1,0) nomizw oti mporeis na apodei3eis oti den exei lush ws e3hs:
-1<w<0 dinei oti 1>w^n>0 ara -1<-w^n<0 kai 0<1+w<1 ara 0<(1+w)^n<1 ara -2=0-1-1<f(w)=(1+w)^n-w^n-1<1+0-1=0

[quote="spinos"]έχεις δίκιο solonike για n άρτιο αποδεικνύεται πως έιναι γνησίως αυξουσα παντού σε όλο το R γιατί η συναρτηση της f'(x) είναι n[(1+x)^n-1-x^n-1] που είναι πάντα θετική γιατί το αν θέσουμε μια άλλη συνάρτηση την φ(t)=t^n-1 ( με n άρτιο ) αυτή είναι γνησίως αυξουσα και,άρα αφου το 1+χ>χ συνεπάγεται (1+χ)^n-1>x^n
όμως για n=περιττό δεν μπορώ να τα αποδείξω[/quote]
Στην περίπτωση $ n\equal{}2k\plus{}1$ με $ x/y\equal{}z$ και θέτοντας $ f(z)\equal{}1\plus{}z^{2k\plus{}1}\minus{}(1\plus{}z)^{2k\plus{}1}$ εύκολα δείχνεις ότι $ f''(z) \neq 0$ άρα με δισδοχικά rolle η $ f$ μπορεί να έχει το πολύ 2 ρίζες.(διότι αν είχε  3 ... άτοπο) Αλλά $ f(0)\equal{}f(\minus{}1)\equal{}0$

Μπράβο μπορείς!
αυτή είναι η λύση ,έχεις δικιο σε ευχαριστώ!
σας ευχαριστώ όλους για την πολίτιμη βοήθεια σας και τον χρόνο σας!


I'm hoping that #24 is geometry and #25 is algebra... or better yet, #22 and #23 are geometry and #24 and #25 are algebra.

That way I have a small hope that I might be able to solve either #24 or #25.

[quote="Sunny"]CMO invites are out, btw.  March 30th is the date[/quote]

WHAT?!!!!

My math teacher never told me...
Obviously winter camp or COMC means nothing to him!!! :mad:

What books have you gotten for qualifying CMO the second and third time?

I got my book today, it says that I'm supposed to get [i]In Polya's Footsteps[/i], yet enclosed was another [i]CMO 69-93[/i]...  ;) .

I was about to ask...

I DID NOT GET ANY BOOK!!!!  :mad:

Btw, Yufei, why are you online?

[quote="jpark"]
I DID NOT GET ANY BOOK!!!!  :mad:

Btw, Yufei, why are you online?[/quote]

The book comes in a white envelope with your name on it. Ask your teacher for it.

I'm online since I'm on lunch break. You?

I have the white envelope, but I only had two pieces of papers, one the letter congratulating me on being able to write the CMO, and another the blue sheet.

I'm in computer science, pretty much bored to death.

I'm going to go talk to my math teacher right now.

Book eh?

Jenn, we're missing out!!  I do have In Polya's Footsteps and the old CMO book though, maybe that's all I can get, lol.  Hmm, I shoulda really get around to doing that book sometime...

[quote="jpark"]

I'm in computer science, pretty much bored to death.

[/quote]

Tsk tsk, not on task in school eh.  Whenever I'm caught off task in computers, 10 pushups.  It doubles every time your caught, lol.  AoPS isn't allowed cuz its all forums and forbidden.  Like I'm really going to be abducted by a math site.