geometry
circumcircle
projective geometry
geometry proposed

The quadrilateral $ ABCD$ is inscribed in a circle. The lines $ AB$ and $ CD$ meet at $ E$, while the diagonals $ AC$ and $ BD$ meet at $ F$. The circumcircles of the triangles $ AFD$ and $ BFC$ meet again at $ H$. Prove that $ \angle EHF\equal{}90^{\circ}$.

Are you sure $ AB$ and $ CD$ intersect? It wouldn't be a quadrilateral right?

[quote="livetolove212"][quote="moldovan"]The quadrilateral $ ABCD$ is inscribed in a circle. The lines $ AB$ and $ CD$ meet at $ E$, while the diagonals $ AC$ and $ BD$ meet at $ F$. The circumcircles of the triangles $ AFD$ and $ BFC$ meet again at $ H$. Prove that $ \angle EHF = 90^{\circ}$.[/quote]

Let (O,R) be the circumcircle of quadrilateral ABCD.
[b]Lemma: [/b]Let L be the intersection of the tangents of (O) which pass through A and D then E,F,L are collinear. (*)
This lemma is well-known :D
Let $ EF\cut (O) = \{M,N\}$ then applying lemma (*) we get $ (ADMN) = - 1$ so the tangents of (O) which pass through M, N, and AD are concurrent at J.
Let $ OJ\cut MN = \{K\}, JE\cut OF = \{H'\}$
We have $ JK.JO = JO^2 - OK.OJ = JO^2 - R^2 = JA.JD$
But $ JO\perp MN$ and it's easy to show that $ JE\perp OF$ too.
Thus $ JK.JO = JE.JH'$ we obtain $ JE.JH' = JA.JD$ so quadrilateral $ ADEH'$ is cyclic then $ H'\equiv H$
->QED[/quote]
Please be careful, you have misunderstood the problem.
[quote] $ AFD$ and $ BFC$ meet again at $ H$[/quote].
Not $ (AED)\cap (BED)\equiv H$.

[quote]The quadrilateral $ ABCD$ is inscribed in a circle. The lines $ AB$ and $ CD$ meet at $ E$, while the diagonals $ AC$ and $ BD$ meet at $ F$. The circumcircles of the triangles $ AFD$ and $ BFC$ meet again at $ H$. Prove that $ \angle EHF = 90^{\circ}$.[/quote]
[i]Proof:[/i]
Let $ F\equiv EB\cap CD$ and $ J\equiv AD\cap BC$. It is easy to notice that $ FJ$ is the polar of $ E$ wrt $ (O)$- the circumcircle of $ ABCD$. Further, $ E$ is also the orthocenter of $ \triangle OEF$. Let $ X,Y,H$ respectively be the projection of $ E$ onto $ OF,OJ,FJ$. Consider the inversion through $ O$, power $ R^2$, where $ R$ is the radii of $ (O)$, we get, $ \mathcal {I}(O,R^2): F\mapsto X$, $ A\mapsto A$, $ E\mapsto H$, $ D\mapsto D$. But $ A,D,E,X$ are concyclic hence $ A,D,F,H$ are concyclic also. With the same argument, we also have $ F,H,C,D$ are concyclic, which implies that $ H\equiv (AFD)\cap (BFC)$. The result is lead as follow.

[quote="mathVNpro"][quote="livetolove212"][quote="moldovan"]The quadrilateral $ ABCD$ is inscribed in a circle. The lines $ AB$ and $ CD$ meet at $ E$, while the diagonals $ AC$ and $ BD$ meet at $ F$. The circumcircles of the triangles $ AFD$ and $ BFC$ meet again at $ H$. Prove that $ \angle EHF \equal{} 90^{\circ}$.[/quote]

Let (O,R) be the circumcircle of quadrilateral ABCD.
[b]Lemma: [/b]Let L be the intersection of the tangents of (O) which pass through A and D then E,F,L are collinear. (*)
This lemma is well-known :D
Let $ EF\cap (O) \equal{} \{M,N\}$ then applying lemma (*) we get $ (ADMN) \equal{} \minus{} 1$ so the tangents of (O) which pass through M, N, and AD are concurrent at J.
Let $ OJ\cap MN \equal{} \{K\}, JE\cap OF \equal{} \{H'\}$
We have $ JK.JO \equal{} JO^2 \minus{} OK.OJ \equal{} JO^2 \minus{} R^2 \equal{} JA.JD$
But $ JO\perp MN$ and it's easy to show that $ JE\perp OF$ too.
Thus $ JK.JO \equal{} JE.JH'$ we obtain $ JE.JH' \equal{} JA.JD$ so quadrilateral $ ADEH'$ is cyclic then $ H'\equiv H$
->QED[/quote]
Please be careful, you have misunderstood the problem.
[quote] $ AFD$ and $ BFC$ meet again at $ H$[/quote].
Not $ (AED)\cap (BED)\equiv H$.

[quote]The quadrilateral $ ABCD$ is inscribed in a circle. The lines $ AB$ and $ CD$ meet at $ E$, while the diagonals $ AC$ and $ BD$ meet at $ F$. The circumcircles of the triangles $ AFD$ and $ BFC$ meet again at $ H$. Prove that $ \angle EHF \equal{} 90^{\circ}$.[/quote]
[i]Proof:[/i]
Let $ F\equiv EB\cap CD$ and $ J\equiv AD\cap BC$. It is easy to notice that $ FJ$ is the polar of $ E$ wrt $ (O)$- the circumcircle of $ ABCD$. Further, $ E$ is also the orthocenter of $ \triangle OEF$. Let $ X,Y,H$ respectively be the projection of $ E$ onto $ OF,OJ,FJ$. Consider the inversion through $ O$, power $ R^2$, where $ R$ is the radii of $ (O)$, we get, $ \mathcal {I}(O,R^2): F\mapsto X$, $ A\mapsto A$, $ E\mapsto H$, $ D\mapsto D$. But $ A,D,E,X$ are concyclic hence $ A,D,F,H$ are concyclic also. With the same argument, we also have $ F,H,C,D$ are concyclic, which implies that $ H\equiv (AFD)\cap (BFC)$. The result is lead as follow.[/quote]
I have a mistake in typo $ E\leftrightarrow F$. But my solution is correct.

Let $AD$ meet $BC$ at $S$, easy we get $H, F, S$ are collinear and $HS$ is polar line of point $E$ with $(O)$. Let $ED$ meet $HS$ at point $K$ $\Longrightarrow (ECKD)=-1$. But we have $\angle DHF=\angle DAF = \angle DBC = \angle FHC$ $\Longrightarrow HS$ is bisector of angle $\angle DHC$, hence $EH\perp HS$.


For $a,b,c$ real positive numbers prove

$(a-b+c)(a-b)^2 + (b-c+a)(b-c)^2 + (c-a+b)(c-a)^2 \geq 0 $

It is [b]very easy to solve[/b] since it's [b]not true[/b]! (Take $a$ close to 0 and  $b > 2c$,  for example, a=0.1, b=3, c=1).

In fact, you can use this to prove (I am sure this has been posted before):
[quote]
If $a$, $b$, $c$ are sides of an triangle, then
\[
2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq 3+ \left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right).
\]
[/quote]
since these two inequalities are algebraically equivalent  when $a,b,c > 0$. However the first one is clearly true when $a$, $b$, $c$ are sides of an triangle.

Sorry, I have confused letters

$(a+b-c)(a-b)^2 + (b+c-a)(b-c)^2 + (c+a-b)(c-a)^2 \geq 0 $

and if you like it you can also prove

$(a^2+b^2-c^2)(a-b)^2 + (b^2+c^2-a^2)(b-c)^2 +(c^2+a^2-b^2)(c-a)^2 \geq 0 $

Both are Schur's inequality for r=1 and 2,

$a^r(a-b)(a-c) + b^r(b-c)(b-a) + c^r(c-a)(c-b) \geq 0.$

In fact for any r, we have algebraic identity:

$(a^r + b^r - c^r)(a - b)^2 + (b^r + c^r - a^r)(b - c)^2 + (c^r + a^r - b^r)(c - a)^2$

$ = 2[a^r(a - b)(a - c) + b^r(b - c)(b - a) + c^r (c - a)(c - b)]$.

[quote="fuzzylogic"]In fact, you can use this to prove (I am sure this has been posted before):

If $a$, $b$, $c$ are sides of an triangle, then
\[
2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq 3+ \left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right).
\]
[/quote]

Indeed, it's posted before right here:

http://www.mathlinks.ro/Forum/viewtopic.php?p=106139


Suppose that $A=(a_{ij})_{n \times n} \in C^{n \times n}$ and $\delta A =(\delta a_{ij})_{n \times n} \in C^{n \times n}$ are two square matrices. $A=VDV^-1$, where $V$ is nonsingular matrix and $D$ is diagonal matrix.

Prove the following:

$(a)$ If $\lambda$ is an eigenvalue of $A$, then there is an $i (1 \leq i \leq n)$ which may depend on $\lambda$, such that $\vert \lambda - a_{ij} \vert \leq \sum_{j \neq i} \vert a_{ij} \vert$

$(b)$ If $\mu$ is an eigenvalue of $A + \delta A$ , then there is an       eigenvalue of $A$ denoted as $\nu$ which may depend on $\mu$, such       that $\vert \mu - \nu \vert \leq \kappa_\infty(V) \Vert \delta A \Vert_\infty$

(CANNOT use Bauer-Bike Theorem)

I think I know how to prove the part $(a)$.

Basically, $(b)$ is Bauer-Fike theorem, but I'm stuck with using $(a)$ to prove $(b)$, Please help.


probability
integration
function
linear algebra
matrix
advanced fields
advanced fields unsolved

Suppose that $ (X, \mathcal{F}, \mu)$ is a probability space and $ T$ is a continuous operator on $ L^\infty(X)$.

1) If $ (Tf(x))^2 \equal{} T(f(x)^2)$, what can we say about $ T$? In particular, must $ T$ be the identity?

2) What if we just have $ \int (Tf(x))^2 d\mu(x) \equal{} \int T(f(x)^2) d\mu(x)$? What can we say about $ T$ then?

Well, any change of variable $ f\mapsto f\circ\omega$ where $ \omega: X\to X$ is an arbitrary measurable map satisfies (1). I don't know if there are other examples, perhaps not...

[quote="fedja"]Well, any change of variable $ f\mapsto f\circ\omega$ where $ \omega: X\to X$ is an arbitrary measurable map satisfies (1).[/quote]

You run into trouble unless you impose the condition that the measure $ \mu\circ\omega^{ \minus{} 1}$ be $ \mu$-absolutely continuous, don't you? :? If there is a $ \mu$-null set $ A\in\mathcal F$ such that $ \mu(\omega^{ \minus{} 1}(A)) > 0$, then the characteristic function of $ A$, although zero almost everywhere and hence zero in $ \mathcal L^\infty(\mu)$, is taken by $ f\mapsto f\circ\omega$ into the characteristic function of $ \omega^{ \minus{} 1}(A)$ (which is non-zero in $ \mathcal L^\infty(\mu)$), so the map in question can't define a linear operator on $ \mathcal L^\infty$.

[quote="Xevarion"]Suppose that $ (X, \mathcal{F}, \mu)$ is a probability space and $ T$ is a continuous operator on $ L^\infty(X)$.

1) If $ (Tf(x))^2 \equal{} T(f(x)^2)$, what can we say about $ T$? In particular, must $ T$ be the identity?
[/quote]

There is a "characterization" of sorts, but I don't know if it's satisfactory. I'll try to describe it (although I'm not entirely sure it's correct :)), but I'll omit some details.

Let $ \mathcal A$ be the Boolean measure algebra formed by quotienting $ \mathcal F$ through the $ \mu$-null sets. Because $ T$ is multiplicative (i.e. takes products into products; this is immediate from the fact that it commutes with squares and from the identity $ fg \equal{} \frac12((f \plus{} g)^2 \minus{} f^2 \minus{} g^2)$) it induces a endomorphism (of Boolean algebras) of $ \mathcal A$: simply identify every element of $ \mathcal A$ with the corresponding characteristic function of a set in $ \mathcal F$, and see what $ T$ does to characteristic functions.  Conversely, given a Boolean algebra endomorphism of $ \mathcal A$, it defines the action of $ T$ on all characteristic functions of sets in $ \mathcal F$, and it should be a simple exercise to check that this can be extended through linearity to a bounded linear operator on $ \mathcal L^\infty(\mu)$. The conclusion is that such operators $ T$ are naturally in a one-to-one correspondence with the endomorphisms of the Boolean algebra $ \mathcal A$.

The examples given by fedja is of this sort: any measurable map $ \omega: X\to X$ having the property that $ \mu\circ\omega^{ \minus{} 1}$ is $ \mu$-absolutely continuous induces an endomorphism of $ \mathcal A$ by sending the class of $ A\in\mathcal F$ modulo $ \mu$-null sets into the class of $ \omega^{ \minus{} 1}(A)$. The question of whether all operators $ T$ are of the form $ f\mapsto f\circ\omega$ now translates to: is every endomorphism of the Boolean algebra $ \mathcal A$ induced by some measurable map $ \omega: X\to X$? I'm quite sure that there should be some pathological examples where this is not the case, although I don't have a counterexample. I think Halmos made some relevant comments in his book "Lectures on Ergodic Theory". I don't have the book close by to check though.

P.S.

By Boolean algebra homomorphism one usually means, among other things, that the largest element is taken into the largest element. In our case this corresponds to the requirement that $ T({\bf 1}) \equal{} {\bf 1}$, where $ {\bf 1}$ is the constant function equal to $ 1$. I took only such operators into consideration (fedja's example also has this property), but the results can be modified to encompass the general case.

P.P.S.

I'm clueless about (2) :).

Yes, grobber, you are right about some trouble with sets of measure $ 0$. :) This is actually quite funny: we can consider the space of all bounded measurable functions with $ \sup$-norm without identifying functions having the same value a.e. (this is [b]not[/b] $ L^\infty(X,\mu)$, of course, but the difference is not that great). Then, in this space, the mapping I suggested works perfectly...

I'm also clueless about (2) at the moment. It is clear that there are far more options here: for instance, if we are talking about complex-valued functions, we can take any linear functional from $ L^\infty(\mathbb T)$ to $ H^\infty_\plus{}(\mathbb T)$. Then both sides are always $ 0$. If we restrict ourselves to real-valued functions, take the 2-point space with measure $ 1/2$ assigned to each point and consider any linear operator with matrix $ \left[\begin{matrix}a & b\\ c& d\end{matrix}\right]$ satisfying $ a^2\plus{}c^2\equal{}a\plus{}c$, $ b^2\plus{}d^2\equal{}b\plus{}d$, $ ab\plus{}cd\equal{}0$ (they are plentiful: just pick any $ a,b,c,d$ satisfying the last equation with $ a\plus{}c,b\plus{}d\ne 0$ and multiply $ a,c$ and $ b,d$ by appropriate non-zero real numbers to satisfy the first 2 equations). So, there may be no reasonable description here.


Is it able to divide numbers 1,2,...,21 to groups such that in every group is greatest number equals to sum of others (of the group)? Every number can be just in one group.

I take it that every number has to be in a group, right?

yes, I surely mean that every number must be in one group.

No, it's not possible
[hide]
Make 2 other groups. One with all biggest terms of groups and one with other terms. From conditions these two groups have same sums, but it is not possible since sum $1+2+\ldots+21$ is odd number, so it cannot be divided by 2 :)
[/hide]

ok. it was to easy for you, wasn't it? :)


Let $ x,y,z$ be positive reals such that $ xyz\equal{}1$.
Prove that $ \frac{1}{x\plus{}y\plus{}1} \plus{} \frac{1}{y\plus{}z\plus{}1} \plus{} \frac{1}{z\plus{}x\plus{}1} \leq 1$

[quote="hendrata01"]Let $ x,y,z$ be positive reals such that $ xyz \equal{} 1$.
Prove that $ \frac {1}{x \plus{} y \plus{} 1} \plus{} \frac {1}{y \plus{} z \plus{} 1} \plus{} \frac {1}{z \plus{} x \plus{} 1} \leq 1$[/quote]
Use $ x\plus{}y\plus{}1\geq(\sqrt[3]x\plus{}\sqrt[3]y\plus{}\sqrt[3]z)\sqrt[3]{xy}.$ :wink:

But then we're left to show, for $ abc\equal{}1$,
$ \frac{1}{a\plus{}b\plus{}c} \left( \frac{1}{ab} \plus{} \frac{1}{bc} \plus{} \frac{1}{ca} \right) \leq 1$, which is not true.... if $ a \to 0$ then it doesn't hold.

When $ a \rightarrow 0$, if $ b$ is fixed, $ c \rightarrow \infty$.

Actually, the inequality is a equality.

I know another problem. It's from Bulgaria 1997.

$ a, b, c > 0$ and $ abc\equal{}1.$

Prove that
\[ \frac{1}{1\plus{}b\plus{}c}\plus{}\frac{1}{1\plus{}c\plus{}a}\plus{}\frac{1}{1\plus{}a\plus{}b} \leq \frac{1}{1\plus{}2a}\plus{}\frac{1}{1\plus{}2b}\plus{}\frac{1}{1\plus{}2c}.\]

Hint: Brute force!

[quote="hendrata01"]But then we're left to show, for $ abc \equal{} 1$,
$ \frac {1}{a \plus{} b \plus{} c} \left( \frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca} \right) \leq 1$, which is not true.... if $ a \to 0$ then it doesn't hold.[/quote]
note that from $ abc \equal{} 1$ we get that $ \frac 1{ab} \equal{} c  ,  \frac 1{bc} \equal{} a$ and $ \frac 1{ca} \equal{} b$ so your inequality is equivalent to:

$ \frac 1{a \plus{} b \plus{} c}(a \plus{} b \plus{} c)\leq 1$

which clearly holds...

[quote="chess"]I know another problem. It's from Bulgaria 1997.

$ a, b, c > 0$ and $ abc \equal{} 1.$

Prove that
\[ \frac {1}{1 \plus{} b \plus{} c} \plus{} \frac {1}{1 \plus{} c \plus{} a} \plus{} \frac {1}{1 \plus{} a \plus{} b} \leq \frac {1}{1 \plus{} 2a} \plus{} \frac {1}{1 \plus{} 2b} \plus{} \frac {1}{1 \plus{} 2c}.
\]
Hint: Brute force![/quote]

Let $ a \ge b \ge c$.  Then, by the upward concavity of $ \frac{1}{x}$,
$ \frac {1}{1 \plus{} c \plus{} a} \plus{} \frac {1}{1 \plus{} a \plus{} b} \le \frac{1}{1\plus{}2a}\plus{}\frac{1}{1\plus{}b\plus{}c}$
$ \frac{1}{1\plus{}b\plus{}c}\plus{}\frac{1}{1\plus{}b\plus{}c} \le \frac{1}{1\plus{}2b}\plus{}\frac{1}{1\plus{}2c}$
and you are done.

This is just special case of following inequality:

[url]http://www.mathlinks.ro/resources.php?c=1&cid=60&year=2004&p=133719[/url]


Let a,b,c be positive real numbers.Prove that:
   SUM {a^2(1+a^3)^2/bc(b+c)^3} >=3/2

[quote="tamtrantrong"]Let a,b,c be positive real numbers.Prove that:
   $\sum{\frac{a^2(1+a^3)^2}{bc(b+c)^3}} \ge \frac 32$[/quote]

Using $(1+a^3)^2\geq 4a^3$ by AM-GM, it suffices to prove
$\sum_{cyc}\frac{8a^5}{bc(b+c)^3}\geq 3$
By AM-GM Inequality again,
$\frac{8a^5}{bc(b+c)^3}\geq\frac{32a^5}{(b+c)^5}=(\frac{2a}{b+c})^5$
Now by Power Mean Inequality and Nesbitt Inequality respectively,
$(\frac{\sum_{cyc}(\frac{2a}{b+c})^5}{3})^{\frac15}\geq \frac{\sum_{cyc}\frac{2a}{b+c}}{3}\geq 1$
The conclusion follows. :D

I think this ineq is very easy, it isn't hard


[quote="Iversonfan2005"]Don't know if this is the right place to ask but when you go to the contests, where do you find the answers or solutions to the problems or are the moderators still working on this? :)[/quote]

The moderators/admins are still working hard on this project so there will be gaps for quite a while I imagine.  So try and bear with us :)


vector
inequalities
superior algebra
superior algebra solved

let $E$ be a vector space of finite dimension. Let f be an application (we don't know if it is linear)
from $E$ to $E$ such that:
1/ $|f(x)-f(y)|=|x-y|$ for all $x,y$
2/$f(0)=0$

what can we say about $f$ ?

May we assume that your vector space is a (real or complex) inner product space and that by $|x|$ you mean $\sqrt{x\cdot x}$ ? If that's not what you mean, then the whole thing falls apart, and we wouldn't be able to draw any useful conclusions about $f$.

sorry  :blush:  of course it is an inner product space with usual norm

f must be a linear isometry. It is clear that $ |f(x)|=|x|$. Then, square the given relation and see that in fact $ (f(x), f(y))=(x,y)$ (here $ (x,y)$ is the scalar product). It remains to show the linearity. It is not difficult to show, using the triangular inequality for norms that in fact the image of the midpoint of a segment is the midpoint of the images of the extremities of the segmet ( or just use the identity $ f(x)-f(\frac{x+y}{2})+f(\frac{x+y}{2})-f(y)=f(x)-f(y)$  and the triangular inequality). Thus, we have $ f(x)+f(y)=2f(\frac{x+y}{2})$. It easily follows that $f$ is in fact additive. Next, it follows that $ f(qx)=qf(x)$ for rational values of $q$ and then the continuity of $f$ implies that f is linear. Hope I didn't do stupid mistakes.  :blush:


Να βρεθούν όλες οι συναρτήσεις
$ f: R \to R$ ώστε $ (x\minus{}y)f(x\plus{}y)\minus{}(x\plus{}y)f(x\minus{}y)\equal{}4xy(x^2\minus{}y^2) , \forall x,y \in R$

[hide]Θετουμε $ t \equal{} x \minus{} y, u \equal{} x \plus{} y$ και η συναρτησιακη σχεση γραφετε:
$ tf(u) \minus{} uf(t) \equal{} ut(u^2 \minus{} t^2)$, με τα $ t,u$ να παιρνουν οποιαδηποτε πραγματικη τιμη (το συστημα $ x \minus{} y \equal{} t, x \plus{} y \equal{} u$, ειναι παντα συμβιβαστο).
Σταθεροποιουμε τη μεταβλητη $ u$ και λυνυμε ως προς $ f(t)$ και παιρνουμε $ f(t) \equal{} t^3 \plus{} kt$, οπου $ k$ ειναι μια πραγματικη σταθερα.
Μπορουμε τωρα με μια απλη αντικατασταση να δουμε οτι καθε συναρτηση της παραπανω μορφης επαληθευει τη συναρτησιακη εξισωση.[/hide]

Μια άλλη λύση
$ x\equal{}y\equal{}1$ δίνει $ f(0)\equal{}0$
Για $ x\minus{}y\equal{}a,x\plus{}y\equal{}b \Rightarrow af(b)\minus{}bf(a)\equal{}ab(b^2\minus{}a^2)$
Για $ a,b \ne 0 \Rightarrow \frac{f(b)}{b}\minus{}b^2\equal{}\frac{f(a)}{a}\minus{}a^2$
Αρα $ \frac{f(x)}{x}\minus{}x^2\equal{}k (Constant)$ και επειδή $ f(0)\equal{}0 \Rightarrow f(x)\equal{}x^3\plus{}kx$

Άλλη λύση, όχι τόσο ωραία όσο οι δύο προηγούμενες:
Θέτουμε x+y = t, x-y=a και F(t) = f(t)/t. Η σχέση γράφεται $ \frac {F(t) \minus{} F(a)}{t \minus{} a} \equal{} t \plus{} a$
Για a $ \neq 0$, παίρνοντας όριο t τείνοντος στο a, είναι F [b]'[/b](a) = 2a, άρα $ F(a) \equal{} a^{2} \plus{} c$. Οπότε $ f(a) \equal{} a^{3} \plus{} ca$. H περίπτωση a=0 και το αντίστροφο είναι απλά.

Τάκης


geometry
geometric transformation
reflection

Let $O$ be the circumcentre of $\triangle ABC$. If $O_A$ is the reflection of $O$ in $BC$, $O_B$ is the reflection of $O$ in $CA$, $O_C$ is the reflection of $O$ in $AB$, show that $AO_A, BO_B, CO_C$ are concurrent at the nine-point centre, $N$.

The quadrilaterals $AHO_AO$ a.s.o. (which have a common diagonal $HO$) are paralellograms and the center of the Euler's circle is the middlepoint of the segment $HO$.


logarithms
function
number theory unsolved
number theory

Show that there exists a natural number $n$ so that the first digits of $n!$ is $2001$

I'd very like to see a solution to this problem. I've seen similar problem but instead of 2001 there was another number and I couldn't solve it...  :(

The following is due to my friend Bruno Langlois, which will probably join us soon :)
It solves the given problem and more general ones.

Let $\mathbb {T}  = \mathbb {R} / \mathbb {Z}$ be the torus.
If $(s_n)$ is a real sequence, let $( \Delta s)_n = s_{n+1}-s_n$
Let $\Delta ^2 = \Delta o \Delta.$

Each of the four following claims is not difficult to prove (but quite boring to write) :

Claim 1 :
If $s$ is a real sequence which has limit $+\infty$ such that $\Delta s$ has limit $0$, then $s$ is dense in $\mathbb {T}$.

Claim 2:
If $s$ is a real sequence such that $\Delta ^2 s$ has limit $0$ and $\Delta s$ is dense in $\mathbb {T}$, Then $s$ is dense is $\mathbb {R}$.

Claim 3 :
If $w$ is a real sequence such that $\Delta w $ has limit $+\infty$ and $\Delta ^2 w $ has limit $0$, then $w$ is dense in $\mathbb {T}$.
(Just use claim 1 for $s = \Delta w$ to see that $\Delta w $ is dense in $\mathbb {T}$ and use claim 2)

Claim 4 :
If the sequence $(\log_{10} (u_n))$ is dense in $\mathbb {T}$ then for all positive integer $k$, there exists $n$ such that the decimal representation of $k$ begins with the decimal representation of $k$.

Thus, you just have to use it for $u_n = n!$, starting with claim 3 for $w_n = \log_{10} (u_n)$ and then claim 4.

Pierre.

Similar argument but a little more intutive -

we will prove that there exist a N such that  ( 10^(2N)) !     starts with 2001.. that is   n = 10^2N

All logs based 10:

(1/2) log (2pi) + (n+1/2) (2N) - n log (e) + (1/(12n+1)) log e  <log(n!) < (1/2) log (2pi) +(n+1/2)(2N) - nlog e + (1/12n)

We are only interested in the fractional part  (mantissa), and have to prove that  there exist a N where it is   between     log(2.001) and  log (2.002)

Since  (log e ) is irrational mantissa of  (n log e)  can  fall in any small range we desire..

[quote="pbornsztein"]The following is due to my friend Bruno Langlois, which will probably join us soon :)[/quote]

I was right  :D
http://www.mathlinks.ro/Forum/viewtopic.php?t=23528

Pierre.

[quote="Gyan"]All logs based 10:

(1/2) log (2pi) + (n+1/2) (2N) - n log (e) + (1/(12n+1)) log e  < n! < (1/2) log (2pi) +(n+1/2)(2N) - nlog e + (1/12n)

[/quote]

 Can you please explain ,Gyan??

 @pbornsztein : Nice generalization!!

Yes, there was a typo, I should have said log(n!) (Corrected in the original message)
but anyway that was  the famous  Stirlings Approximation
(I just did a google and found it here:

http://mathworld.wolfram.com/StirlingsApproximation.html

One of  the easy way to prove that, is  to write the  n! = int ( 0 to infinity) (x^n * e ^(-x) dx
If you take the function (x^n * e^(-x) )   as  (exp ( f(x)))  and expand f(x) near its maximum (which happens to be when x=n)   so you get something like

f(x-n) + (1/2)  f'' (x-n)^2) + ... Ignore the other terms and take the integral. etc..

[quote="pbornsztein"]
Let $\Delta ^2 = \Delta o \Delta.$[/quote]
What does this notation mean?

[quote="Gyan"]This problem has come to life again  [url=http://artofproblemsolving.com/Forum/viewtopic.php?p=285395#p285395]here[/url]

It is in Singapore Math Olympiad (for 9th-10th graders) .. and I  just realized a nice proof which does not require higher-level math.  If there is an interest, I may post the solution in the  other thread.[/quote]

Which thread are you referring to?


is there a formula for finding the number of different circular permutations of $m$ red beads and $n$ blue beads?

@Mimoide: no, check $n=m=2$.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=64797 for a similar problem.


inequalities
LaTeX
symmetry
algebra
system of equations
inequalities unsolved

Can Someone Please Post The Solution to The Following Inequality USING CAUCHY SCHWARZ INEQUALITY ONLY(to be specific):->
a,b,c,d, are any Positive Reals.
\

here is my soln

cyclic{(a^2)/(ab+2ac+3ad)}>={{cyclic a}^2}/4{(cyclic ab)}                  (cauchy)

now {3[cyclic a^2]}>=2{cyclic ab}                             (am, gm)

hence dividing by 3 and adding {2[cyclic ab]}

we get {(cyclic a)^2}>=8/3{(cyclic ab)}

now dividing by 4{cyclic ab}

we r done

the am gm result can be proved with cauchy also using a bit differently than in 3 variables we will be done

plz post some nice hard problems

these problems are pretty damn easy and lack application

i want hard inequalities

plz reply aman i am from 10th and just joined this forum so i am new so plz help i am from delhi india

[quote="Jaswinder"]here is my soln

cyclic{(a^2)/(ab+2ac+3ad)}>={{cyclic a}^2}/4{(cyclic ab)}                  (cauchy)[/quote]
The "4{(cyclic ab)" part isn't true. "ac" in the LHS isn't a term in here at all.
I've not checked the rest of the solution for the poor typing. Plz learn latex soon. It's very useful and not very hard to learn.

My solution is AM-GM only. :)  Hope you don't mind Aman. :!:

[hide]Let $ p\equal{}a\plus{}2b\plus{}3c$ and $ q,r,s$ similarly.

By AM-GM \[ \sum \frac{p}{q}\ge 4,\sum \frac{q}{q}\equal{}4,\sum \frac{r}{q}\ge 4,\sum \frac{s}{q}\ge 4\]

Hence\[ \sum \frac{p}{q}\minus{}5\sum \frac{q}{q}\plus{}7\sum \frac{r}{q}\plus{}\sum \frac{s}{q}\ge 4\minus{}5\cdot 4\plus{}7\cdot 4\plus{}4\equal{}16\]

But \[ \sum \frac{p}{q}\minus{}5\sum \frac{q}{q}\plus{}7\sum \frac{r}{q}\plus{}\sum \frac{s}{q}\equal{}\sum \frac{p\minus{}5q\plus{}7r\plus{}s}{q}\equal{}\sum \frac{24a}{q}\]

Hence \[ 24\sum \frac{a}{q}\ge 16\iff \sum \frac{a}{q}\ge \frac{2}{3}\]
$ QED$[/hide]

[quote="Jaswinder"]
plz post some nice hard problems

these problems are pretty - easy and lack application

i want hard inequalities

plz reply aman i am from 10th and just joined this forum so i am new so plz help i am from delhi india[/quote]

Well Jaswinder , I am new to Inequalities and have just started to solve some simple Inequalities, so you see, I can't Really help you . You can try some IMO Inequalities Problems if you want to have fun!!!Bye The Way even I am From delhi.I Am in XIth

First of all who said $ a,b,c,d$ are postive reals to you Akashnil?
And who said that $ a,b,c,d$ are non negative reals reals to you  Jaswinder?
Aman, give full details of the problem before posting one.

Anyways, Akashnil, your soln for positive reals is briliant.
Could you please tell us how you thought of breaking the LHS in such a manner? Just beautiful!!!

[u]Solution.[/u] The inequality is false in general; if $ (a,b,c,d) = (0,0, - 1,1)$, then the left hand side is negative.  Suppose $ a,b,c,d\geq 0$.  Notice that no more than $ 2$ of the numbers $ a,b,c,d$ can be zero.  Therefore,
\begin{eqnarray*}\sum_{\text{cyc}}\frac {a}{b + 2c + 3d} & \geq & \frac {(a + b + c + d)^2}{\sum_{\text{cyc}}(a(b + 2c + 3d))} \\
& = & \frac {(a + b + c + d)^2}{ab + 2ac + 3ac + bc + 2bd + 3ab + cd + 2ac + 3bc + ad + 2bd + 3cd} \\
& = & \frac {(a + b + c + d)^2}{4(ab + bc + cd + da + ac + bd)}\geq\frac {2}{3} \\
&\iff & 3\sum_{\text{cyc}}a^2\geq 2(ab + bc + cd + da + ac + bd) \\
& \iff & (a - b)^2 + (b - c)^2 + (c - d)^2 + (d - a)^2 + (a - c)^2 + (b - d)^2\geq 0\end{eqnarray*}
$ \Box$

It was quite obvious to assume them positive. since otherwise it is clearly wrong.

[quote="Aravind Srinivas L"]Could you please tell us how you thought of breaking the LHS in such a manner?[/quote]
[hide]I had in mind the system of equations:
$ \begin {cases} a\plus{}2b\plus{}3c\equal{}p \\ b\plus{}2c\plus{}3d\equal{}q \\ c\plus{}2d\plus{}3a\equal{}r \\ d\plus{}2a\plus{}3b\equal{}s \end {cases}$
where $ a,b,c,d$ are variables and $ p,q,r,s$ constants.
It could get messy with eliminating variables. So we must make use of the symmetry.
First thing to do, is of course adding the four:
$ a\plus{}b\plus{}c\plus{}d\equal{}\frac{p\plus{}q\plus{}r\plus{}s}{6}\cdots (i)$
And second is a bit more tricky:
$ r\minus{}q\equal{}3a\minus{}b\minus{}c\minus{}d\cdots (ii)$
Now it's easy:
Adding $ (i)\plus{}(ii)$ and dividing by $ 4$:
$ a\equal{}\frac{r\minus{}q\plus{}\frac{p\plus{}q\plus{}r\plus{}s}{6}}{4}\equal{}\frac{p\minus{}5q\plus{}7r\plus{}s}{24}$[/hide]

hey akashnil

here cyclic ab includes all the combinations possible of 4 variables taken 2 at a time

now i guess u must be clear that my proof is perfectly right

plz reply

[quote="Jaswinder"]here cyclic ab includes all the combinations possible of 4 variables taken 2 at a time[/quote]
It is generally called symmetric not cyclic the case you are referring to. Anyway it's understandable if you specify what you denote.

[quote="Jaswinder"]now i guess u must be clear that my proof is perfectly right[/quote]
Yes. Nice solution. :)

[quote="Aravind Srinivas L"]boxedexe I did the same thing. Nice solution. :lol:[/quote]
That's what Jaswinder did too.

Yeah. I happen to see that only now.

Akashnil, even though you got the equation relation $ a$ how did u  think of the sigma cyclics.

Even if Now Latex , then Using Math Type 6a would be a good idea too!!

Sorry!! I simple forgot to mention that a,b,c,d were positive reals.... :blush:  :huh:

The given inequality seems to be lack of a vital condition. :wink:

[quote]{3[cyclic a^2]}>=2{cyclic ab} (am, gm) [/quote]

but  i think {3[cyclic a^2]}>=3{cyclic ab} (am, gm)

isn't it ???

plz let me know if i'm wrong..

sfsreporter

hey sfs u r totally wrong :P

u need to look that the inequality which u r talking about holds for 3 variables but here the inequality is in 4 variables so u see!!!!!!!!!!!!!!!!!!!

plz don't mind for that emotion

i am new to this forum so i will always post these smilies just for sake of fun so plz don't mind

 :o  :wink:  :maybe:

i got ur pt Jaswinder yaar

actually u have confused the cyclic and symmetric expressions..

i got it cleared from Akashnil

u hav written cyclic but mean symmetric that is why i was confused..

in fact i am right if we see the cyclicity and wrong if we see symmetry

sfsreporter


geometry
geometric transformation
rotation

Consider a rotating spherical planet. The velocity of a point on its equator is $V$. The effect of rotation of the planet is to make  $g$  at the equator  $\frac{1}{2}$  of  $g$  at the pole. What is the escape velocity for a polar particle on the planet expressed as a multiple of $V$.

well, escape velocity is rad2 times the oribital velocity

[hide="A Solution"]The centripetal acceleration at the equator is:
$\frac{V^2}{R} = \frac{g}{2} = \frac{GM}{2R^2}$

For the escape velocity we have:
$\frac{1}{2}v_{esc}^2 = \frac{GM}{R}$

Solving for $\frac{GM}{R}$ in the first equation and substituting it in the second we get:
$\frac{1}{2}v_{esc}^2 = 2V^2$

Therefore $v_{esc} = 2V$ (Keep in mind that $v_{esc}$ includes the original velocity of rotation $V$)[/hide]


On August 4th Richard Rusczyk and I will be hosting a Math Jam to explain the new AoPS Independent Study program and answer all questions.  The Math Jam will begin at 7:30 PM ET/4:30 PM PT.

[quote="mathnovice"]Auguts 11 or 4? By the way, the classroom isn't working for me for some reason. I already deleted the cookies, but it still won't work.[/quote]

I forgot to edit this post, but we have finished the Math Jam.  Sorry that you were not able to attend.  I will be posting the transcript soon and you can email me at crawford@artofproblemsolving.com if you have questions that were not answered.  Also, email me about this problem you had getting into the classroom.  We'll see if we can work that problem out quickly.  It might depend on your browser type or something simple like that.

mathnovice - what time did you show up for the Math Jam?

And what happened when you tried to get it?

Thanks.


[quote="Allan Z"]Given triangle $ABC$ with $AB=37$, $AC=12$, $BC=35$. $P$ is any point on segment $AB$. Two perpendicular lines are drawn from $P$ to $AC$ and $CB$, intersects at $M$ and $N$ respectively. Find the minimum length of $MN$.[/quote]

Right triangle  :yup:
[hide]
$PNCM$ is a rectangle, let $PM=NC=a$ and $MC=PN=b$, then we have some similar triangles, $\frac{b}{12}=\frac{35-a}{35}$ or $35b+12a=420$.

Cauchy-Schwarz: $(a_{1}^{2}+a_{2}^{2})(b_{1}^{2}+b_{2}^{2})\ge (a_{1}b_{1}+a_{2}b_{2})^{2}$, so $(35^{2}+12^{2})(a^{2}+b^{2})\ge (35b+12a)^{2}$. We are minimizing $\sqrt{a^{2}+b^{2}}$ so then $a^{2}+b^{2}\ge \frac{(35b+12a)^{2}}{35^{2}+12^{2}}=\frac{420^{2}}{37^{2}}$ so then $\sqrt{a^{2}+b^{2}}\ge \boxed{\frac{420}{37}}$.[/hide]

[quote="delta"]Much more simple[hide]
As $\angle C=90^{o}, CMPN$ is rectangle. $MN=CP$ and $CP$ is minimum when $CP$ is the height $CH$.
From $CA.CB=CH.AB$ $CH=\frac{420}{37}$. :D [/hide][/quote]

yup.. if you realized $CMPN$ is a rectangle.. bingo! .. nice solution, sick =D


function
inequalities
integration
real analysis
real analysis theorems

In several sources, a property holds "almost everywhere" whenever it holds everywhere except possibly on a null set (i.e, on a set contained in a set with measure $0$). In other sources, the set where the property does not hold must have measure $0$.

The two definitions do not coincide (except in the case of complete measure spaces). Which one is the standard one ?

P.S: I tried to check Mathworld, but as usual, they contradict themselves:
http://mathworld.wolfram.com/AlmostEverywhere.html
http://mathworld.wolfram.com/AlmostEverywhereConvergence.html

[quote="mlok"]You owe an apology to Mathworld. The following statements are equivalent:

(1) the set $\{x\colon f_{n}(x)\not\to f(x)\}$ is contained in a null set;
(2) $f_{n}(x)\to f(x)$ for all $x\in Y$, where $Y$ is a measurable subset of $X$ such that $\mu(X\setminus Y)=0$.

Either of them is a standard definition of convergence a.e.[/quote]

Certainly not. A null set is not necessarily measurable, it's just a set contained in another set with measure $0$. If the measure is not complete, such a null set has got no reason to be measurable.

Edit: Ok, maybe there is a confusion in your post. In the definition given by Mathworld and most books, your set $Y$ (in (2)) is supposed maximal. If you meant that $Y$ is not necessarily maximal, then you're right about the equivalence. Otherwise, no.

If I may ask, what difference does it make? I mean, if the property does not hold on $Z \subset F$ with $\mu(F) = 0$, then also the property doesn't hold on $F$, unless you want the property to fail on every point in the set for some reason. And what are you going to use $F \setminus Z$ for anyway? It's also a null set, throw it out the window.

Following up on what Xevarion said ...

Let's look beyond the definition itself to ask how we would prove a theorem which has "almost everywhere" as part of its conclusion. A certain sequence converges almost everywhere ... a certain function is finite almost everywhere ... a certain probabilistic limit holds almost surely. How do we prove that?

We prove that by collecting the "bad" sets - sets on which this property has any chance of being false. And we show that for any $\epsilon>0$ we can fit our "bad" set inside a set of measure less than $\epsilon.$ At no point in the proof are we looking at precisely the points at which the property fails; rather, we are always looking at a set of points on which the property might fail.

A very simple example: a nonnegative integrable function must be finite almost everywhere. How to prove this? First prove Chebyshev's (Markov's?) inequality:

$m\{x: f(x)>M\}\le\frac{\int f}M.$

The set on which $f(x)$ is infinite must be contained in the set $\{x: f(x)>M\}$ for all $M.$ Hence it must be contained in a set of measure $<\frac{C}{M}$ for all $M.$ That means it is contained in a set of measure zero.

[quote="mlok"]If there is a confusion here, it's all yours. Mathworld does not say that $Y$ must be maximal.[/quote]

You are right, I didn't read carefully ! So Mathworld doesn't contradict itself. Nevertheless, my initial question still remains (example in Foundations of Modern Analysis, where A. Friedman says that "a property is true a.e if the set of points where it fails to be true has measure $0$").

I guess (logically) the standard is Mathworld's definition, though.

Edit: have a look at http://planetmath.org/encyclopedia/AlmostSurely.html
Here, they use the other definition (clearly assuming that the set on which the property doesn't hold has measure $0$).

[quote="julien_santini"]Edit: have a look at http://planetmath.org/encyclopedia/AlmostSurely.html
Here, they use the other definition (clearly assuming that the set on which the property doesn't hold has measure $0$).[/quote]
Hm? It says that Let $(X, \mathfrak B, \mu)$ be a measure space. A condition holds almost everywhere on $X$ if it holds ``with probability'' , i.e. if it holds everywhere except for a subset of $X$ with measure 0.

[quote="liyi"]Hm? It says that Let $(X, \mathfrak B, \mu)$ be a measure space. A condition holds almost everywhere on $X$ if it holds ``with probability'' , i.e. if it holds everywhere except for a subset of $X$ with measure 0.[/quote]

So this subset on which it does not hold has measure $0$, and is in particular measurable ? The first definition from Mathworld does not require measurability.

[quote="julien_santini"]So this subset on which it does not hold has measure $0$, and is in particular measurable ? The first definition from Mathworld does not require measurability.[/quote]
You have said that 'measure zero', so...it is measurable...otherwise how can we say its "measure"?

A subset of a set of measure 0 is not necessarily measurable. You know it will have exterior measure 0, but Caratheodory's theorem doesn't tell us [i]what[/i] $\sigma$-algebra we get when we turn an arbitrary outer measure into a measure. We might not get everything that we "should" (in the sense that subsets of measure 0 sets intuitively ought to have measure 0).

Most of the time, you can complete your measure (so that all subsets of sets of measure 0 are measurable with measure 0 also) with impunity. That's why I think this discussion is kinda pointless.

On the other hand, it is good to be aware of this kind of thing.

Well, I agree that the discussion is useless, and that the point Kent made is actually relevant. However, seeing those two definitions arbitrarily mixed in a book, I was wondering which one is the most common.


I'm sure these are really easy, but I can't get them...  Haha so I'll let you guys have fun. :-P  (Should this be in the Getting Started Forum?  I just put these here because well, of the word[i] Olympiad [/i].

[quote] 1. Show that if an (infinite) arithmetic progression includes a square, then it must include infinitely many squares. [/quote]

[quote] 2.  Show that √x + √y + √(xy) = √x + √(y + xy + 2y√x). Hence show that √3 + √(10 + 2√3) = √(5 + √22) + √(8 - √22 + 2√(15 - 3√22)).  [/quote]

[quote] 3.  Show that [(4 + √11)n] is odd for any positive integer n.
[/quote]

[quote]  4.  Show that there are infinitely many primes in the arithmetic progression 3, 7, 11, 15, ... [/quote]

Well, they look like fun induction problems... :-P

#3)---Problem is FALSE. Try n=2.

Do you mean [tex]\lfloor(4+\sqrt{11})^n \rfloor[/tex]

#1

[hide]If a^2 is in this infinite arithmetic progression, and the difference between two consecutive terms is d, then (a+n*d)^2 are also in this arithmetic progression, where n is a natural number. d|(a+n*d)^2-a^2. So there are infinite squares in that progression. [/hide]

#1: suppose there is some square c^2 in the progression {ax+b}.
      numbers of the form (c+ka)^2=
      c^2 + 2kac+k^2*a^2=ax+b+2kac+k^2*a^2=a(x+2kc+k^2*a)+b, where k
      is an integer, are also in the progression {ax+b}. this gives infinitely many
      squares, which is what we wanted to prove.
#4: proof by contradiction. suppose not. we have finitely many primes
      p1,p2 ... p(k) of that form. consider K=4(p1*p2 ... *p(k))-1. K is of the form
      4k+3. none of p1,p2 ... p(k) are prime factors of K obviously, and since there
      are no further primes of the form 4k+3, all (odd) prime factors of K are of the
      form 4k+1. But the product of some numbers of the form 4k+1 is of the form
      4k+1, a contradiction.

[edit: sorry, i didnt supr008's post because it was 'invisible'. sorry for repeating what he said.]

#3: I think [(4+:sqrt: 11)^n] is odd for all n because that is correct.
      consider (4+:sqrt: 11)^n+(4-:sqrt: 11)^n=a(n). After expanding (using
      binomial theorom) we find a(n) is always an integer, and further more that
      a(n) is even. But we also know (4-:sqrt: 11)<1, so (4-:sqrt: 11)^n < 1, so
      thus [(4+:sqrt: 11)^n]=a(n)-1. Since a(n) is always even, a(n)-1 is always
      odd, and thats we needed to prove.
more generally, the following is true: [(p +:sqrt: q)^n] is always odd when
p=[:sqrt: q  + 1].
about #2, the first identity is trivial (subtract :sqrt:x and square both sides). for the second identity i think you should just square everything   :D  :D and keep on squaring, until you get rid of all the square roots.   :)  :)

[quote="SingTheSorrow"]Beta...I wouldn't know, but that's what I'm guessing what they mean...I just copied and pasted.[/quote]

Did you get these from an English source, or did you translate the Spanish yourself?  i would be interested in seeing the problems in the original language if you have them..

4) [hide]Assume there are finitely many primes congruent to -1 mod 4, and let these be denoted [tex]p_1, \ldots, p_n[/tex]. Consider the number [tex](p_1 p_2\cdots p_n)^2-2[/tex]. This number is -1 mod 4, and therefore it must have a prime divisor that is also -1 mod 4 (since if they were all 1mod4 then their product would also be 1mod4). But this prime divisor can't be one of the p's since then it would divide 2, which is impossible. Thus, there is another prime congruent to -1mod4, which gives the necessary contradiction.[/hide]

Farmertim - Oh, no, I got this information from an English source...


$ \text{Simplify\: } \sqrt{75x^3y^{\minus{}4}.}$

Yea I thought so. Today in math class, my teacher told me it's
$ \frac{5|x|\sqrt{3x}}{y^2}$

Suppose we have $ \sqrt{x^2}$, if we have a negative value of $ x$, we obtain the result as $ \minus{}x$ and when $ x>0$ we obtain $ x$.

I assume that we assume that $ \sqrt{x^2}\equal{}x^{\frac{2}{2}}\equal{}x$ returns a positive value.

Usually the $ \sqrt{x}$ means the positive root (called the principal square root), so the absolute value is unnecessary.

[quote="SnipedYou"]Usually the $ \sqrt {x}$ means the positive root (called the principal square root), so the absolute value is unnecessary.[/quote]

Yea I'm aware of that, but the $ \sqrt{x^2}$ is different.

your teacher is correct to say that $ \sqrt{x^2} \equal{} |x|$.  However, the technicality here is pointless.  If $ x<0$ then you are taking the square root of a negative number anyway, which I'd bet you do not want to deal with yet.  So you are probably only interested in the case $ x \ge 0$ anyway, under which condition $ \sqrt{x^2} \equal{} x$ is always true.

I remember doing this in Alg I. Depending on the power you're taking the root of, you may or may not need to have the absolute value.

And even $ \sqrt{x^2}$ should just be x, right? The radical implies that it is just the principal square root.

If you substituted $ y$ for $ x^2$, I don't think anyone would argue that it isn't...
I agree with SnipedYou.

Oh ok thanks :p. It just caught me off guard when she did it on the board because I've always been using $ \sqrt{x^2}\equal{}x$.

Hmmm....I think the rule is:
In an expression, its the positive root. So $ \sqrt{9}\equal{}3$.
In an equation, you have to take the $ \pm$. This is because if we have $ x^2\equal{}a$, $ x^2\minus{}a\equal{}0 \Rightarrow (x\plus{}\sqrt{a})(x\minus{}\sqrt{a})\equal{}0$, so $ x\equal{}\pm a$. Thus, if $ x^2\equal{}9$, $ x\equal{}\pm 3$.

Technically, $ x < 0 \Longrightarrow \sqrt {x^{2}} \equal{} |x| \neq x$, so the absolute value is not completely unnecessary; however, we are normally trying to find $ x$, and not $ |x|$, and the value on the other side of the equation is usually constant, or a polynomial in terms of other variables.

Always take every case (on competitions you can lose points for not discussing some case).

$ \sqrt{x^2}\equal{}|x|$. Don't take only case $ \sqrt{x^2}\equal{}x$ because it's easier, it could happen like metafor said.

Yeah, I realized that as I went to bed. I incorrectly assumed that $ x\ge0$.

Sorry to keep this going, but I'd like to point out that $ x$ must be $ \ge 0$ in this problem otherwise the expression is imaginary.

But yes you guys are right that $ \sqrt{x^{2}} \equal{} |x| \ne x$

I think we were talking about it in general.

facis already pointed out that x must be nonnegative.


geometry
circumcircle
geometric transformation
homothety
ratio

Let $ ABC$ be a triangle with circumcenter $ O$ and centroid $ M$ such that $ OM\perp AM$. Let $ AM$ intersect the circumcircle of $ ABC$ again at $ A'$. Let $ D\equal{}BA'\cap AC$ and $ E\equal{}CA'\cap AB$. Prove that the circumcenter of triangle $ ADE$ lies on the circumcircle of $ ABC$.

Nice, straightforward problem!
[hide]
First, because $ OM \perp AA'$, $ M$ must be the midpoint of $ AA'$.  Because $ M$ trisects the median from $ A$, it follows that $ BC$ bisects $ MA'$.  But $ MA'$ bisects $ BC$.  Therefore, $ BA'CM$ is a parallelogram.  Therefore, $ MC \parallel{} BA' \parallel{} A'D$, so $ D$ is the image of $ C$ under a homothety of ratio $ 2$ about $ A$.  Similarly, $ E$ is the ratio of $ B$ under the same homothecy.  Therefore, $ \triangle{ADE}$ is the image of $ \triangle{ABC}$ under this homothety, and so the circumcenter of $ \triangle{ADE}$ is the image of $ O$ under this homothety: namely, the point on the circle diametrically opposite $ A$.[/hide]


geometry
rhombus
geometric transformation
homothety
geometry proposed

Let $ ABCD$ be a convex quadrilateral such that $ AB$ and $ CD$ are not parallel and $ AB\equal{}CD$. The midpoints of the diagonals $ AC$ and $ BD$ are $ E$ and $ F$, respectively. The line $ EF$ meets segments $ AB$ and $ CD$ at $ G$ and $ H$, respectively. Show that $ \angle AGH \equal{} \angle DHG$.

Let $ I\equal{}AB\cap CD, J,K$ be the midpoints of $ AD, BC$. A well-known result: $ JK$ is parallel to the bisector $ l$ of angle $ GIH$.
But $ JFKE$ is a rhombus then $ JK\perp EF\Rightarrow l\perp GH$. Thus $ IGH$ is the isosceles triangle.

It is easy to see PFQE is an rhombus
PQ⊥EF   =>  PQ is the bisector of ∠FPE
Denote IJ is the bisector of ∠BJCS  =>  IJ∥PQ  => IJ⊥EF  =>  JGH is an isosceles triangle

I used homotety of triangles FPE and GJH

Let $S$ be the point of intersection of lines $AB$ and $CD$.
Using Menelaus theorem on the line containing $E$,$F$,$G$ and $H$ with respect to $\triangle ASC$ we get $\frac{HS}{CH}\frac{CE}{AE}\frac{AG}{GS} = 1$. Using Menelaus again this time on $\triangle BSD$ gives $\frac{HS}{DH}\frac{DF}{BF}\frac{BG}{GS} = 1$. Now $CE = AE$ and $DF = BF$, so $\frac{HS}{GS}\frac{AG}{CH} = \frac{HS}{GS}\frac{BG}{DH}$.
Hence $\frac{AG}{BG} = \frac{CH}{DH}$, but $AB = CD$, which means $AG = CH$ and $BG = DH$. Now because $\frac{HS}{GS}\frac{AG}{CH} = 1$
we get that $HS = GS$, so the angles $\angle AGH$ and $\angle DHG$ are indeed equal.


Let $OAB$ be a triangle, $r$ be a real number that $r>OA,OB$. Let $A'$ be on the extension of $OA$ such that $A$ is between $O$ and $A'$ and $OA' = \frac{r^2}{OA}$. Define $B'$ similarly. Let $X$ be the circumcenter of $OA'B'$. Show that $OX$ is perpendicular to $AB$.

in wach triangle $ABC$, $OH,OA$ are isogonal conjugate.

if we use makoos inversion, we can easily find why they are prependicular.

how ever it is nt an easy problem at least for this forum.

it would be better if u post it in geometry section.

i can tell u the complete proof later. ;)

Can you post here? I know the proof anyway, it's not that hard ;)  :)

yes ofcourse... ;)

i thought its basic forum :P


find the smallest solution to the following system :
n=5k+1
4k=5l+1
4l=5m+1
4m=5n+1
4n=5p+1
4p=5q+1


geometry
probability
AMC
AIME
number theory

I will be at Moe's (by the Rave Theater) in Vestavia at noon on Monday with a subject-oriented set of problems for the first of four or five days of problem solving instruction.  Students will have the afternoon to work on the problems after which we will convene at Vestavia Hills High School to discuss the problems.

All Birmingham area high school students are welcome to join in these sessions.

More info to come -- I'll edit later.

Mrs. Tipton told the class about the details today, but we didn't get a count of how many are going to show up. We guessed around 10. I'll especially try to get my friends to do it. For the topic, I'd like anything from geometry as a lot of us are relatively weak in that.

Sounds like a solid number.  I'll make sure to include some good geometry.

I haven't gotten any of my friends to say they're going. :( The lack of motivation and/or communcation is disappointing... Maybe I'll convince some more people when I get back, but don't be surprised if 10 is a big overestimate.

That's fine.  I'm only interested in focusing my attention on students who are interested in seeing what the next level of problem solving looks like.  I think it will be more enjoyable and easier for the students if there are several who can work together and share ideas, but ultimately, but that would just be an added bonus.

Subjects to be covered by day:

Monday: Algebraic Topics
Tuesday: Geometry Topics
Wednesday: Counting Methods and Probability
Thursday: Number Theory

I'll also hand out a few more topics at the end of Thursday that don't fit nicely into these categories.  Nearly all of the topics will be outside of the range of material typically covered in Alabama math team and Mu Alpha Theta.  This is both about learning what we can over a given week and also letting you see what is out there that you don't normally get experience working on.  You will not learn but a fraction of the material in a single week and that's fine.  You will know what to look for in the future when researching hard AIME to Olympiad level problem solving.

Each day after Monday I will be at the Bean Coffee House from 11:00 AM to noon with each new day's materials, but on Monday we'll meet at Moe's at noon.

I'd love to make this but I'll be out of town until 10 PM on 23rd (effectively missing pretty much every day you are here). I was wondering if there's a chance I could get a copy of the material handed out there?

Oh and even though I wouldn't be there I suggest Matrices and Vectors. The things we learn in the normal class is more memorization of rules and not understanding the concepts itself behind vectors.

There will be extra copies of the problem packets I hand out.

People now started revealing to me that they're going, so I'll probably be picking up several extra copies for people who are out of town or will be getting them later from me, if that's all right.


Consider the equation x^2+y^2=z^2+18, where x,y,z are positive integers.

Prove there are an infinite number of solutions where z=y+1.

[hide] substituting in, we get x^2=2y+19.  Well, for all odd x, we just square, subtract 19 and halve to get y, so all solutions of this form will be (2m+1, 2m^2+2m-9, 2m^2+2m-8)[/hide]


geometry
3D geometry
sphere
integration
calculus
calculus computations

The density of the point $(x,y,z)$ on the solid sphere $x^{2}+y^{2}+z^{2}\le{1}$ is proportional to its distance from the sphere's center.
What is the total mass of the solid sphere?

Cut it into spherical shells. For a shell of radius $r$ we have that $dV=4\pi r^{2}\,dr.$  Since mass = volume * density and the density on this shell is equal to $kr,$ then $dm=kr\cdot4\pi r^{2}\,dr=4\pi kr^{3}\,dr.$ Integrate that.

so the answer should be

$\int_{0}^{1}4k\pi{r^{3}}dr=k\pi$?

Yes. More generally, if the sphere were of radius $R$ rather than radius $1,$ you would get a mass of $k\pi R^{4}.$


Find all $ f: \mathbb{R}\rightarrow\mathbb{R}$ functions for which \[ f(x)\plus{}f(x\plus{}f(y))\equal{}2x\plus{}y\] $ \forall x,y\in\mathbb{R}$.

Puting x=0 we get $ f(f(y)) \equal{} y \minus{} f(0)$  . from this we conclude that there exist a w such that $ f(w) \equal{} 0$

puting x=y=w in the initial we get 3w=0 so $ w \equal{} 0$ and so $ f(0) \equal{} 0$
Now put y=0 in the initial we get $ f(x) \equal{} x$ which is in fact a solution


The polynomial $ P(X)\equal{}aX^3\plus{}bX^2\plus{}cX\plus{}d$ takes integer  values for $ X\equal{} \minus{}1, 0, 1, 2$  Prove that $ P(X)$ is an integer for any integer $ X$

You'll want to look at the technique discussed [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=227941]here[/url] and in particular Lemma 1 [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=178381]here[/url].


Dear Mathlinkers,
let ABC be a triangle, (1) the incircle of ABC, DEF the contact triangle of ABC, D' the midpoint of EF, U, X the second points of intersection of  DD', AD with (1), C' the meetpoint of UX and AB, and E' the meetpoint of the parallel to DE through U with DF.
Prove : C'E' // AD.
Sincerely
Jean-Louis

Well, very nice question jayme  :D !Here is my solution :blush: :

Thanks a lot for providing us with this nice problem jayme  :lol:  ! This problem is actually not very difficult , but it is complicated, there are many segments or parallel in it , so maybe we can’t get the proof easily.
By the way, jayme, where did you find this problem?
 :)

Dear CCMath1,
I found this problem by toying with my geometric programm... and proved it synthetically.
Sincerely
Jean-Louis


geometry
3D geometry
sphere
tetrahedron
ratio
inradius
perimeter

I don't know if this is the right place for this topic but maybe someone can point me in the right direction.  It is very well known (at least to many in this forum) that the radius of a triangle's incircle can be found using r=2K/P where K is the area of the given triangle and P is the perimeter.  Stemming from this one can also go about showing that the radius of an inscribed sphere for a tetrahedron in given by r=3V/S where V is the volume of the tetrahedron and S is the surface area.  With this in mind I hypothesized that for a simlar setup in n-space it's possibe that r=nV/S where V is the "volume" equivalent (n dimensions) and S is "surface area" (n-1 dimensions).  If it helps I noticed this same ratios held when the shape was a square/cube.  Any ideas?

The fact that the inradius r of a triangle is

$r = \frac{2V_2}{S_1}$

where $V_2$ is 2-dimensional volume of the triangle (triangle area) and $S_1$ 1-dimensional surface of the triangle (triangle perimeter) stems from the fact that the triangle area $V_2$ is equal to

$V_2 = \frac 1 2 b_1 h$

where $b_1$ is its 1-dimensional base and $h$ its altitude. Indeed, if $I$ is the incenter of a triangle $\triangle ABC$, then

$V_2 = |\triangle ABC| = |\triangle ABI| + |\triangle BCI| + |\triangle CAI| = \frac 1 2 c r  + \frac 1 2 b r  + \frac 1 2 a r = \frac 1 2 (a + b + c) r$

$r = \frac{2|\triangle ABC|}{a + b+ c} = \frac{2V_2}{S_1}$

Similarly, the fact that the inradius r of a tetrahedron is

$r = \frac{3V_3}{S_2}$

where $V_3$ is 3-dimensional volume of the tetrahedron and $S_2$ 2-dimensional surface of the tetrahedron  stems from the fact that the tetrahedron volume $V_3$ is equal to

$V_3 = \frac 1 3 b_2 h$

where $b_2$ is its 2-dimensional base and $h$ its altitude. Indeed, if $I$ is the incenter of a tetrahedron $ABCD$, then

$V_3 = |ABCD| = |ABCI| + |ABDI| + |BCDI| + |CADI| =$

$= \frac 1 3 |\triangle ABC| r + \frac 1 3|\triangle ABD| r  + \frac 1 3 |\triangle BCD| r  + \frac 1 3 |\triangle CAD| r = \frac 1 3 S_2 r$

$r = \frac{3 V_3}{S_2}$

The formula for the triangle area is usually demonstrated by inscribing the triangle in a rectangle, but we will do it differently. Slice the triangle parallel to its base $b_1$ into n thin trapezoids of equal heights $\triangle x$ and consider the thin rectangles with bases $b_{1i}$ and equal heights $\triangle x$. From similarity of the triangles with the bases $b_1, {b_{1i}}$ and altitudes $h, h - x_i$, the base $b_{1i}$ and surface ${V_2i}$ of a rectangle lying between $x_i$ and $x_i + \triangle x$ is equal to

$b_{1i} = b_1 \frac{h - x_i}{h} = b_1 \left( 1 - \frac{x_i}{h} \right)$

$V_{2i} = b_{1i} \triangle x = b_1 \left( 1 - \frac{x_i}{h} \right) \triangle x$

The area of the triangle is approximately equal to the sum of the areas of all the thin rectangles, and if $\triangle x \rightarrow 0$, the sum becomes an integral and equality is exact:

$V_2 \doteq \sum_{i = 1}^n b_{1i} \triangle x = b_1 \sum_{i = 1}^n \left( 1 - \frac{x_i}{h} \right) \triangle x \rightarrow b_1 \int_0^h \left (1 - \frac{x}{h} \right) dx =$

$= b_1 h \int_0^1 y dy = b_1 h \left[ \frac{y^2}{2} \right]_0^1 = \frac 1 2 b_1 h$

This process can be repeated in 3 dimensions for a tetrahedron and even in any number of dimesions. Slice the tetrahedron parallel to its base $b_2$ into thin layers of the same thickness $\triangle x$ and consider the thin rectangular prisms with bases $b_{2i}$ and equal heights $\triangle x$. From similarity of the tetrahedrons with the bases $b_2, b_{2i}$ and altitudes $h, h - x_i$, the base $b_{2i}$ and the volume $V_{3i}$ of the prism between $x_i$ and $x_i + \triangle x$ is

$b_{2i} = b_2 \left( \frac{h - x_i}{h} \right)^2 = b_2 \left( 1 - \frac{x_i}{h} \right)^2$

$V_{3i} = b_{2i} \triangle x = b_2 \left( 1 - \frac{x_i}{h} \right)^2 \triangle x$

Again, the volume of the tetrahedron is approximately equal to the sum of the volumes of all the thin prisms, and if $\triangle x \rightarrow 0$, the sum becomes an integral and equality is exact:

$V_3 = b_2 \int_0^h \left (1 - \frac{x}{h} \right)^2 dx = b_2 h \int_0^1 y^2 dy = b_2 h \left[ \frac{y^3}{3} \right]_0^1 = \frac 1 3 b_2 h$

It is now clear that in 4 dimmensions, the hypervolume $V_4$ of a 4-dimmensional object with 5 vertices is

$V_4 = b_3 \int_0^h \left (1 - \frac{x}{h} \right)^3 dx = b_3 h \int_0^1 y^3 dy = b_3 h \left[ \frac{y^4}{4} \right]_0^1 = \frac 1 4 b_3 h$

which implies that the inradius r of a hypersphere inscribed in such 4-dimmensional object is

$r = \frac{4V_4}{S_3}$

Similarly, in n dimensions, the n-dimensional volume $V_n$ of an n-dimensional object with n + 1 vertices is

$V_n = b_{n-1} \int_0^h \left (1 - \frac{x}{h} \right)^{n-1} dx = b_{n-1} h \int_0^1 y^{n-1} dy = b_{n-1} h \left[ \frac{y^n}{n} \right]_0^1 = \frac 1 n b_{n-1} h$

which implies that the inradius r of an n-dimensional sphere inscribed in such n-dimmensional object is

$r = \frac{nV_n}{S_{n-1}}$


integration
calculus
calculus computations

hola  :D

show

$ \sum_{k\equal{}1}^\infty\;\dfrac{(\minus{}1)^{k\plus{}1}\cdot \textbf H_k}{k^4}\;\equal{}\;\boxed{\dfrac{59}{32}\zeta(5)\minus{}\dfrac{1}{2}\zeta(2)\,\zeta(3)}$

where $ \textbf H_k\;\equal{}\;\sum_{n\equal{}1}^k\;\dfrac{1}{n}$

$ S \equal{} \sum _{k \equal{} 1}^{\infty } \sum _{n \equal{} 1}^k \frac {( \minus{} 1)^{k \plus{} 1}}{k^4 n}$

Write is as:

$ f(x,y,z,w,t) \equal{} \sum _{k \equal{} 1}^{\infty } \sum _{n \equal{} 1}^k \frac {( \minus{} 1)^{k \plus{} 1} \left((x y z w)^k t^n\right)}{k^4 n}$

$ \frac {\partial ^5f(x,y,z,w,t)}{\partial x\, \partial y\, \partial z\, \partial w\, \partial t} \equal{} \sum _{k \equal{} 1}^{\infty } \sum _{n \equal{} 1}^k ( \minus{} 1)^{k \plus{} 1} t^{n \minus{} 1} (w x y z)^{k \minus{} 1} \equal{} \frac {1}{(w x y z \plus{} 1) (t w x y z \plus{} 1)}$

$ S \equal{} \int _0^1\int _0^1\int _0^1\int _0^1\int _0^1\frac {1}{(w x y z \plus{} 1) (t w x y z \plus{} 1)}dtdwdzdydx$

This integral can be simplified to:

Can be simplified to....

 :?:  Can be simplified to?!  :?:

hm... The rest I give as a homework...

Now Seriosly:

There is a way to simplify integrals like

$ \int _0^1\int _0^1\frac {1}{(\textbf{v} \plus{} 1) (t \textbf{v} \plus{} 1)}dt\textbf{dv}$

Where $ \textbf{v} \equal{} x.y.z \cdots$

I saw or actually did someting in relation about this, but do not remember where.

And the sum can be generalizeed. I will try it other day.  :D

[quote="EulerIntegral"]

$ S \equal{} \int _0^1\int _0^1\int _0^1\int _0^1\int _0^1\frac {1}{(w x y z \plus{} 1) (t w x y z \plus{} 1)}dtdwdzdydx$

[/quote]

un - huh ? :wacko:  :what?:


[quote="ydog"]What would happen if distilled water were substituted for buffer in either the chamber solution or a gel solution?[/quote]
I've never heard of a Agarose Gel Electrophoresis before......But I think that if distilled water is used, the solubility of the DNA or RNA may decrease or the PH change may affect the result?

The procedure will simply not remain stable--ofc we dont hav  to dissolve the DNA just the slightest of ph changes will lead to mixing  in absence of buffer :)


Let $x$, $y$, $z$ are real numbers and they can't all be $0$ (one of them may be $0$, $2$ of them may be $0$, just not all of them are $0$). What is the maximum value of:

$\frac{xy+2yz}{x^{2}+y^{2}+z^{2}}$

I'll post my solution if i'm right. Is it
[hide] $\frac{\sqrt{5}}{2}$ [/hide] ?

(edit: I forgot to do one step on paper...  :P  Is my maximum correct?)

[quote="t0rajir0u"][hide="Solution"] Homogeneous with degree $0$; WLOG let $x^{2}+y^{2}+z^{2}= 1$ and make the substitution

$x = \sin \theta \cos \phi, z = \sin \theta \sin \phi, y = \cos \theta$

So we wish to maximize

$\cos \theta \left( \sin \theta \cos \phi+2 \sin \theta \sin \phi \right) = \frac{1}{2}\sin (2 \theta) \left( \cos \phi+2 \sin \phi \right)$

We can maximize each factor separately.  The left factor has maximum $\frac{1}{2}$ while the right has maximum $\sqrt{5}$ (Cauchy).  Hence our maximum is $\boxed{ \frac{ \sqrt{5}}{2}}$. [/hide][/quote]

[hide="My solution (similar, i guess)"]
Put vector $\text b{V}= (x,y,z) = \ \parallel \text b{V}\parallel ( \cos \alpha , \ \cos \beta , \ \cos \gamma)$.

 Then,
 $(\frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}}) (\frac{x+2z}{\sqrt{x^{2}+y^{2}+z^{2}}})=\cos \beta \ \cos \alpha+2\cos \beta \ \cos \gamma \stackrel{\text{Cauchy}}{\leq}\sqrt{ (5 \cos^{2}\beta)(\cos^{2}\alpha+\cos^{2}\gamma)}$

$\sqrt{ (5 \cos^{2}\beta)(\cos^{2}\alpha+\cos^{2}\gamma)}= \sqrt{5 ( \cos^{2}\beta)(\sin^{2}\beta)}= (\sqrt{\frac{5}{4}})(\sin 2\beta) \leq \boxed{\frac{\sqrt 5}{2}}$[/hide]

yup its $\frac{\sqrt5}{2}$  :D

[hide="another method  :) "]
The question is equavalent to find the smallest k such that ${xy+2yz}\leq{k(x^{2}+y^{2}+z^{2})}$

by AM-GM,
$x^{2}+ay^{2}\geq2\sqrt{a}xy$
$(1-a)y^{2}+z^{2}\geq2\sqrt{1-a}yz$
Adding two equations yields
$x^{2}+y^{2}+z^{2}\geq2\sqrt{a}xy+2\sqrt{1-a}yz$

We are getting close. Let ${2\sqrt{a}=\sqrt{1-a}}\implies{a=\frac15}$

Now we get the $a$, things get a lot easier.
Let's go back...
${x^{2}+{\frac15}y^{2}}\geq{\frac{2}{\sqrt5}xy}$
${\frac{4}{5}y^{2}+z^{2}}\geq{\frac{4}{\sqrt5}xy}$

adding these two you get
${x^{2}+y^{2}+z^{2}}\geq{\frac{2}{\sqrt5}(zy+2yz)}$
or ${\frac{zy+2yz}{x^{2}+y^{2}+z^{2}}}\leq{\frac{\sqrt5}{2}}$
The equality holds when $x: y: z=1: \sqrt5: 2$[/hide]


First I will quote something from the net and then ask my question:

[quote]

In research to be presented at the annual American Association of Artificial Intelligence conference in Providence, R.I., on July 28, computer science Professor Richard Korf has found the first optimal solutions to Rubik's Cube. The median optimal solution appears to be 18 moves, and it is believed any cube can be solved in no more than 20 moves.

"To solve Rubik's cube, one needs a general strategy, which usually consists of a set of move sequences, or macro operators, that correctly position individual cubies without violating previously positioned ones," Korf said. "Such strategies typically require 50 to 100 moves to solve a randomly scrambled cube. This research shows that most scrambled cubes can be solved in no more than 18 moves," he said.

http://www.seas.ucla.edu/hsseas/press/1997/korfcube.html

[/quote]

I'm not debating on the quote. I'm just reffering to the fact that the cube DOES have an optimal 'N' amount of moves to a solution. So whatever 'N' is, here's my question

My question is, if a 3X3X3 Rubik's cube would only have TWO colors instead of 6, but the rules would be the same: Get each side to be a solid color of one of the 2...

1. Would it be mathematically correct to say, that no matter how deep you scramble the cube, it would take less moves to solve since we only have two colors?

2. If the above IS correct, would it be mathematically correct to say that if this "two color cube" didn't have 3 of one color and 3 of the other, but say 4 sides one color and 2 sides the second color, (or 5:1... five of one color and one another color) that it would take EVEN LESS moves to solve, since 4 sides have the same color?

Thank you!

1. I would say yes, but I don't see how the colors would be arranged. There would be more than 1 possible solution, and the cube would be pointless and boring.
2. Say 5:1,  it took m e 6 seconds to do 1 (permutaions don't matter)

I know a cube like that would be boring. I don't need this information for a cube but for an algorithm I'm going to use in a program I'm making.

Are you saying it will be less moves, "theoretically" or do you know this as a mathematical fact? (I need to know that this is mathematically true and not theoretically. I myself also think it's theoretically true and that's why I asked the question here)

5:1 I know for a fact is true, well still theoretically, but see I solve a 3x3 in like 30-40 seconds, I solve a 5:1 in 6 seconds. Most of this is because orientation doesn't matter, as the edges have no way of telling the difference. Same as the corners. So to solve a normal one, there is 1 solution for the 1st layer, but for the 5:1 cube, there are 24*24 ways to solve the first layer. That must mean there are less moves, becuase there are more ways

For the 5:1, you only need to solve one side. So the question is: How easy is it to solve one side (without worrying about corner or side orientation)?


Hello,
What is the difference between if and if and only if?
In other words, what is the difference between if and iff?

Thank you

Right, but what is the difference between if and iff?

here is what i think (i might be wrong)

$A$ if $B$ means that $A$ happens if $B$ happens, but $A$ can occure even if $B$ does not.

however,

$A$ iff $B$ mean that $A$ [b][size=150]only [/size][/b]happens if $B$ happens

This was discuss lat week in a high school topic. Please, search for it.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103869[/url]  :D

my topic.

A  number is even[b][color=red] If  [/color][/b]  the last (unit) digit of a number is $0$ .

A number is divisible $10$[b] [color=red]Iff [/color][/b] the last (unit) digit of a number is $0$

(Point is there are numbers which are even even if the last digit is not zeoro but , say 2 or 4 or 6)

Hope this helps.

You may also understand the difference better if you try to prove an "if" and an "iff" statement.

In an "if" statement, you must prove that $A\implies B$ by assuming the antecedent and proving the consequence.

In an "iff" statement, you must prove that $A\implies B$ as you did with the "if" statement, but you must also prove that $B\implies A$ by assuming the consequence and proving the antecedent.  This is why you may see an "iff" statement written as $A\iff B$.


This forum is for [b]proposed[/b] and [b]own[/b] problems. How come there are so many problems posted from various contests (recently, Vietnam TST, but also tons of others)? Why aren't those being posted in Unsolved Problems?

[quote="N.T.TUAN"]I see ''Problems that you have already solved and you are interested in second opinions or solutions.''[/quote]

i also!  :wink:

And I repeat for the 30th time that the names of the subfores are inconvenient... ;)

http://www.mathlinks.ro/Forum/viewtopic.php?p=330787#330787 <-- actual meaning of the subfores, at least the way I heard it from Valentin.

  dg

That's right! :P Name of them are inconvenient.


I just moved in from Kansas about a week ago and was wondering about some math contests (I am going to be a freshman).  I already know about AMCs, but what other ones are there?  Also I live in Dallas.  Thanks!

Welcome to Texas!

There are contests at a number of Texas colleges and universities (A&M, Rice, UNT, and probably others); from my experience, these are good contests with good problems.  There also exists a contest (with various levels based on the level of math you are enrolled in at school) for Dallas area math students called the GDTCM (Greater Dallas Council of Teachers of Mathematics) McNabb; these are similar in form to the AMC 8 but cover largely one particular level (i.e. algebra II, geometry, calculus).  Talk to your math team sponsor (or head of the mathematics department if your school does not have a math team) after school starts; they should be able to give you more information about what your school participates in.

Welcome!  There will be a high school math contest in the Fall sponsored by MEANT.

You are vey lucky to be in a city with two math circles.

Dr. Titu Andreescu's Metroplex Math Circle meets at UT Dallas in Richardson [url]http://www.metroplexmathcircle.org[/url]

Dr. Dimitar Grantcharov's Mid Cities Math Circle meets at UT Arlington [url]http://www.midcitiesmathcircle.org[/url]

If you have any questions PM me.

While not specifically for Texans, there are two others that I feel are worth mentioning. The Mandlebrot competition is a multi-round national contest thats pretty interesting. There is also some Oklahoma State competition which is completed and mailed to them (I guess it isnt just for Okies as my school will be competing this year).I would like to encourage you to take the aforementioned McNabb. My math professor is now in charge of the problems I believe and as of last year was encouraging harder ones. Also, as Texas' Gold Team Captain for ARML, it is my duty to encourage you to sign up for our ARML program. With Clements and Klein conflicts, our teams have been shrinking and we may not have enough to go to ARML this year. If you are interested in that please PM me with any AMC/AIME/USAMO scores etc.


quadratics
Euler
modular arithmetic
geometry
3D geometry

Okay, in a mod n, we have that there are at most $\frac{n}{2}$ quadratic residues (because of negatives).
How can one find mods n such that there are n cubic residues mod n, (two examples are 5 and 6)?

By Euler, we have $a^{\phi(n)}\equiv 1 \pmod{n}$. However, it only holds for $\gcd(a,n)=1$... and gcd(2,6) and gcd(3,6) are not 1... :ninja:

If $p$ is a prime congruent to 2 modulo 3, then for all $a$,
\[a\equiv a^{p}\equiv a^{2p-1}\equiv \left( a^\frac{2p-1}{3}\right)^{3}\pmod{p}\]
So everything is a cube.

More generally, if $n$ is a squarefree integer containing no prime factor that is 1 modulo 3, then everything is a cube modulo $n$.

[hide="Another generalization"] For $m$ such that $\mathbb{Z}_{m}$ has a primitive root (that is, $m = 1, 2, 4, p^{k}, 2p^{k}$ for $p$ an odd prime) we have the following result:

There are precisely $\frac{ \varphi(m) }{ gcd(k, \varphi(m)) }\left( \frac{ \sigma(m) }{m}\right)$  $k$-adic residues $\bmod m$.

(Possibly.  It's late at night, and this simplifies considerably depending on the case.)[/hide]


geometry
circumcircle
trigonometry
geometry unsolved

Give an acute-angled triangle ABC. Bisector in triangle is AD. $ (O_1),(O_2)$ is circumcircle of triangle ABD and ADC. Two tangent of $ (O_1)$ and $ (O_2)$ intersect AD at P,Q. Prove that:$ PQ^2\equal{}AB.AC$

Assume that $ PQ^2 \equal{} AB \cdot AC$ and let $ r_1$ and $ r_2$ the circumradii of $ \triangle ABD, \triangle ACD$

$ BD \equal{} 2r_1 \cdot \sin  \frac {A}{2}  \ \ (1) \ \ , \ \ CD \equal{} 2r_2 \cdot \sin \ \frac {A}{2}  \ \ (2)$

$ AD^2 \equal{} AB \cdot AC \minus{} BD \cdot CD \Longrightarrow \ PQ^2 \equal{} AD^2 \plus{} BD \cdot CD$

$ \Longrightarrow (AD \plus{} 2PA)^2 \equal{} AD^2 \plus{} BD \cdot CD  \Longrightarrow 4PA^2 \plus{} 4AD \cdot PA \equal{} BD \cdot CD$

$ \Longrightarrow 4PA^2 \plus{} 4(PD \minus{} PA) \cdot PA \equal{} BD \cdot CD  \Longrightarrow 4PD \cdot PA \equal{} BD \cdot CD$

But $ PD \cdot PA$ is the power of $ P$ to both circles, hence $ XY^2 \equal{} BD \cdot CD,$ where $\overline{ XY} $ is their common tangent segment. Now we express $ XY$ in terms of radii $ r_1,r_2$ by using the formulas $ (1)$ and $ (2)$

$ (O_1O_2)^2 \minus{} (r_1 \minus{} r_2)^2 \equal{} 4r_1r_2 \cdot \sin^2  \frac {A}{2} $

But note that $ \angle O_1DO_2 \equal{} \angle BAC.$ Hence, by cosine law in $ \triangle DO_1O_2,$ we obtain

$ {r_1}^2 \plus{} {r_2}^2 \minus{} 2r_1 r_2 \cdot \cos A \minus{} {r_1}^2 \minus{} {r_2}^2 \plus{} 2r_1r_2 \equal{} 4r_1r_2 \cdot \sin^2  \frac {A}{2} $

Therefore, $ 2\sin^2  \frac {A}{2} \equal{} 1 \minus{} \cos A,$ which is true and the proof is completed.


geometry
geometric transformation
reflection
parallelogram
trapezoid
geometry unsolved

Consider a triangle $ ABC$.Let $ D$ and $ E$ be the reflections of vertices $ B$ and $ C$ into $ AC$ and $ AB$,respectively.Let $ F \equal{} BE\cap{CD}$ and $ H_a$ be the projection of the altitude from $ A$ onto $ BC$.Denote by $ F_a,F_b,F_c$ be the projection of $ F$ onto $ BC,CA,AB$ respectively.Prove that $ F_a,F_b,F_c,H_a$ are concyclic.

It is easy to acquire FcFb∥MN   FcFb=1/2MN
Notice A is the exincircle of ⊿FBC  ＝＞  CHa+FC=FB+BHa
＝＞  CHa+VN=MB+BHa   ＝＞  Ha is the midpoint of MN
＝＞  HaFbFcM is a parallelogram    ＝＞ FaFc=FbHa=MFc
＝＞  HaFaFbFc is an isosceles trapezoid  ＝＞ concyclic


I am preparing tomorrow's Math Jam and my copy of the AMC 10 B is partially cut off.  Could somebody tell me the (E) answer choices for problems 18-25?


Find all polynomials $ P(x)\in \mathbb R[x]$ such that for all $x \in \mathbb R$,
\[ P(x\plus{}1)\equal{}P(x)\plus{}3x \plus{} 7.\]

[quote="shortlist"]Find all polynomical $ P(x)$ in $ R[x]$ satisfy :
  $ P(x \plus{} 1) \equal{} P(x) \plus{} 3x \plus{} 7$[/quote]

If $ n\equal{}degree(P)>2$, identification of $ x^{n\minus{}1}$ terms in the equation gives $ na_n\plus{}a_{n\minus{}1}\equal{}a_{n\minus{}1}$ and so $ a_n\equal{}0$, impossible.

So $ P(x)\equal{}ax^2\plus{}bx\plus{}c$ and the equation becomes $ a(x\plus{}1)^2\plus{}b(x\plus{}1)\plus{}c\equal{}ax^2\plus{}bx\plus{}c\plus{}3x\plus{}7$ and so $ \boxed{P(x)\equal{}\frac{3x^2\plus{}11x\plus{}u}2}$ for any real $ u$


geometry inequalities AMC AIME USA(J)MO USAMO symmetry

Every year, people will start asking questions on how can I do well on AIME/AMC/USAMO? Most experienced problem solvers know that only way you can get better is to keep doing problems. But since there are so many people asking these questions, I decided to form a thread for "How To."

Basically, I'm looking for people to post a good advice on "How To" approach problems. You don't have to make a big paragraphs (you can but don't need to) about any type(s) of math problems you do well.

I'll go off first.

[b]Sequence[/b]

There are basically two options:

1. Plug in some numbers to see the terms or if you don't, brute force (if possible)

2. If #1 does not work, try to see if you can find a general formula for the sequence. For instance, see if you can find general formula that uses $n$ for the sequence that is defined as $a_n = (a_{n-1})(a_{n-2}) + a_{n-3}$. That way, you don't need to find every terms before $a_n$ you want.

Geometry: draw large diagram and CAREFULLY examine all the givens, then follow intuition :)  :)

Inequalities : Know all the substitutions for special given statements, normalize, homogenize, mix variables, try everything that comes to your mind. Don't give up on a approach just because it doesn't seem fruitful; keep it in the back of your mind.

The AMC is about working problems fairly quickly.
Try to get through the first 15ish problems in around 15mins. Then spend the remaining hour on the hard problems.

The AIME is about whether you can find a solution without taking an hour on 1 problem.
Since you have 3 hours, take the time to brute-force something since every point counts.

The USAMO is how creative are you.
You have an average of 90 min per problem, so spend time on it. If one approach doesn't work, try a new approach or go to a different one.

For all of these:
Do practice tests under "test-like" conditions.
Don't quit untill your proctor says that time is up. You never know what you might get in the last minute. Especially if you are missing class to take the test.  :D
Entirely re-work all problems at least twice. I made 3 careless errors on the AIME and ended up with a 7. :(

Combinatorics:

The nice thing is there are not that many things to use in combinatorics, so it's all about doing what you can with a small set of techniques.

First recommendation is to get good at recursion, as it is the trickiest part of combinatorics, but if you are good at it you can just look at something and say, "Oh this is recursion, let's find a formula."

Second recommendation is symmetry can be very useful for you, but be careful when making symmetry arguments.

Third recommendation is to just do lots of combo problems, that's the best way to get better.

Fourth recommendation is in terms of visualization, etc. Nice ways to visualise are graphs (not descartian graphs, graph theory graphs), drawing circles, and defining functions (not really visualization but helpful nonetheless). Many times when you define functions recursive solutions can sneak up on you.

Try some smaller problems too if you can't get a combo problem right away.

If you see yourself writing a bunch of sums, you can almost always rephrase it in terms of the "n choose k" operation.

If you see yourself writing a bunch of binomials out, you can usually simplify it into subsets of a set, or something along those lines, that results in a number of the form 2^n, or 3^n, or something like that. (Note that (n 0) + (n 1) + (n 2) + ... + (n n) = 2^n, and also be on the lookout for binomial factorizations).

This leads to another nice technique, which is case analysis. It comes in two forms:

Form 1: Break a problem up into different cases, each of which are fairly workable on their own.

Form 2: Show that something must satisfy either Case 1, Case 2, Case 3, etc., and that all cases can be satisfied independently of each other, or something along those lines. Then you have a very nice closed form solution of k^n, where k is the number of cases (including, oftentimes, the "null case", i.e. doesn't belong to any of the above categories) and n is the number of things that are satisfying cases.

Yeah that last one was pretty convoluted, but hopefully you can all get what I'm saying.

w00t time for HW.

Quote:

[quote]All proofs should be written neatly and coherently in paragraphs of standard American English.
Mathematical symbols like $\equiv$ and $\ge$ should be used only in equations, not as verbs or
prepositions in a sentence. Do not write things like “all of the $\triangle$’s angles are $\le 90^\circ$”—this
should be “$\angle{A},\angle{B},\angle{C}\le 90^\circ$” or “all of the triangle’s angles are at most 90.” Avoid the
symbols $\wedge \vee \forall \exists$; instead, write out “and,” “or,”, “for all,” “there
exists.” Similarly, do not use $\rightarrow$ or $\leftarrow\rightarrow$ except as part of a sequence of equations.
• Write a statement using words rather than symbols unless this would be unnecessarily awkward.
Do not invent more notation than is necessary to explain your solution.
• Write true statements. Do not write something that is only partially true, and then say how
to fix it later. If you assume something in one of your statements, say clearly what you are
assuming. Define all terms you make up. If you use figures, graphs, tables, etc., explain
thoroughly what they represent.
• Use the following formats for common proof patterns:
• Proof by induction: (1) state the claim, (2) check the base case, (3) prove the induction
step, and (4) conclude with the words “induction is complete.”
• Proof by contradiction: (1) state the claim, (2) state that you are assuming the opposite,
(3) derive a contradiction (and say why it is a contradiction), and (4) conclude that the
claim follows.
• Proof by case analysis: (1) state the claim, (2) state the various cases, (3) say why they
exhaust all possibilities, (4) analyze the cases one by one, and (5) conclude that the
claim follows.
• Draw accurate diagrams with compass and straightedge for geometry problems. This is for
your own good as well as the graders’.[/quote]


function
real analysis
real analysis unsolved

Consider two following sets :
$M_{1}= \{x\in C_{[a,b]}: x \text{is injective}, 0\leq x(t)\leq 1 \forall t\in [a,b]\}$
$M_{2}= \{x\in C_{[a,b]}: x \text{is surjective}, 0\leq x(t)\leq 1 \forall t\in [a,b]\}$
prove that $M_{2}$ is closed and $M_{1}$ is not closed

Concerning $M_{1}$, consider linear functions of the form $x(t)=ct$, where $c$ is small.

To prove that the complement of $M_{2}$ is open, suppose that $x(t)\ne \alpha$ for all $t\in [a,b]$, let $\epsilon=\min |x(t)-\alpha|$, and go on from there...


Εστω τριγωνο οξυγωνιο $ ABC$. Εστω $ (BC)\equal{}a, (AB)\equal{}c, (AC)\equal{}b$. Εστω οι κυκλοι $ C1(A,a), C2(B,b), C3(C,c)$. Εστω $ P$ το ριζικο κεντρο των ανωτερω κυκλων. Εστω $ H$ ορθοκεντρο του $ ABC$, $ N$ μεσο του $ AH$ και $ M$ μεσο του $ BC$. Να δειχθει οτι το περικεντρο $ O$ του $ ABC$ ειναι κεντρο βαρους στο τριγωνο $ PMN$.

Μια ερωτηση, επηδη δεν ειχα αρκετο χρονο για να πολυασχοληθω με Γεωμετρια περα απο τη σχολικη. Τι ειναι το ριζικο κεντρο?

[quote="vittasko"][quote]Τι ειναι το ριζικο κεντρο?[/quote]
Οι ριζικοί άξονες τριών τυχαίων κύκλων λαμβανομένων ανά δύο, τέμνονται στο ίδιο σημείο που ονομάζεται ριζικό κέντρο αυτών των κύκλων.
[hide] :(  :mad: [/hide]
Κώστας Βήττας[/quote]

Ευχαριστω κυριε Βηττα :)

Θα κοιταξω λιγο την ασκηση και αν βρω κατι θα το ποσταρω


[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq\ 3. \sqrt [4]{\frac {a^7 \plus{} b^7 \plus{} c^7}{abc(a^4 \plus{} b^4 \plus{} c^4)}}$
 :)[/quote]
This is not true, for a counter-example, $ a\equal{}22,\ b\equal{}21,\ c\equal{}20.$

Very sorry, I had a mistake when writing. I edited my problem. Thank you very much  :oops:
Now, it's trues.
[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2} \geq\ 3. \sqrt [4]{\frac {a^7 \plus{} b^7 \plus{} c^7}{abc(a^4 \plus{} b^4 \plus{} c^4)}}$
 :)[/quote]

A natural question: Find the best $ k$ for this ineq is trues:
[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2} \geq\ 3. \sqrt [k]{\frac {a^7 \plus{} b^7 \plus{} c^7}{abc(a^4 \plus{} b^4 \plus{} c^4)}}$
 :)[/quote]
And the hint of this problem can solve the hard ineq:
$ \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a} \geq\ 3. \sqrt{\frac{a^4\plus{}b^4\plus{}c^4}{a^2\plus{}b^2\plus{}c^2}}$
 :)


Five different four-digit integers all have the same initial digit, and their sum is divisible by four of them. Find all possible such sets of integers.

$ M \equal{} \{ k\cdot 10^4 \plus{} x_1,k\cdot 10^4 \plus{} x_2, \dots, k\cdot 10^4 \plus{} x_5 \}$ with $ k\in \{1,2,\dots , 9\}$ and $ x_1,x_2,x_3,x_5,x_5$ being $ 5$ distinct three digit numbers. Assume that the four elements that divide the sum are $ x_1,x_2,x_3,x_4$

and for brevity write $ S \equal{} (k10^4 \plus{} x_1) \plus{} (k10^4 \plus{} x_2) \plus{} \dots \plus{} (k10^4 \plus{} x_5)$

First we show that $ k \equal{} 1$

[hide="Proof"]
Notice that $ \frac {5k \plus{} 1}{k \plus{} 1} < \frac {S}{k10^4 \plus{} x_i} < \frac {5k \plus{} 4}{k} \qquad(*)$

[hide="proof of above"]
$ \frac {5\cdot k 10^4 \plus{} x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \plus{} x_5}{k10^4 \plus{} x_1} > \frac {5\cdot k 10^4 \plus{} x_1}{k10^4 \plus{} x_1}$

$ > \frac {5\cdot k 10^4 \plus{} 1000}{k10^4 \plus{} 1000}$

$ \equal{} \frac {5k \plus{} 1}{k \plus{} 1}$

Also

$ \frac {5\cdot k 10^4 \plus{} x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \plus{} x_5}{k10^4 \plus{} x_1} < \frac {5\cdot k 10^4 \plus{} x_1 \plus{} 4000}{k10^4 \plus{} x_1}$

$ < \frac {5k \plus{} 4}{k}$[/hide]

Therefore, since all elements in $ M$ are distinct, the four terms $ \frac {S}{k10^4 \plus{} x_i}...$ are distinct integers.

If $ k \equal{} 1$, From $ (*)$, we have $ 3 < \frac {S}{k10^4 \plus{} x_i} < 9$

implying that $ \frac {S}{k10^4 \plus{} x_i} \in \{4,5,6,7,8\}$

if $ k > 1$ however there aren't enough integers to have $ 4$ distinct terms. For example, if $ k \equal{} 2$

$ \frac {11}{2} < \frac {S}{k10^4 \plus{} x_i} < 7$

Wich would imply that $ \frac {S}{k10^4 \plus{} x_i} \equal{} 6$ for $ i \equal{} 1,2,3,4$, which is impossible if $ x_1,x_2,x_3,x_4$ are distinct.

Hence $ k \equal{} 1$
[/hide]

So we have that $ k \equal{} 1$ and as a result of the proof above we have that $ \frac {S}{1000 \plus{} x_i} \in \{4,5,6,7,8\}$

Clearly we cannot have two $ x_i,x_j$ such that $ \frac {S}{1000 \plus{} x_i} \equal{} 4$ and $ \frac {S}{1000 \plus{} x_j} \equal{} 8$

Otherwise $ 1000 \plus{} 2x_j \equal{} x_i$ which is impossible since $ x_i < 1000$

This leaves two systems

[b]system 1[/b]

$ 5000 \plus{} x_1 \plus{} \dots \plus{} x_5 \equal{} 4(1000 \plus{} x_1) \equal{} 5(1000 \plus{} x_2) \equal{} 6(1000 \plus{} x_3) \equal{} 7(1000 \plus{} x_4)$

Letting $ y_i \equal{} 1000 \plus{} x_i$ and we get simply

$ y_1 \plus{} y_2 \plus{} \dots y_5\equal{} 4y_1 \equal{} 5y_2 \equal{} 6y_3 \equal{} 7y_4$

Now $ S \equal{} y_1 \plus{} \frac {4y_1}{5} \plus{} \frac {4y_1}{6} \plus{} \frac {4y_1}{7} \plus{} y_5 \equal{} 4y_4 \Rightarrow 101y_1 \equal{} 105y_5$

Similarly $ 101y_2 \equal{} 84y_5, 101y_3 \equal{} 70y_5, 101y_4 \equal{} 60y_5$

This gives $ (y_1,y_2,y_3,y_4,y_5) \equal{} (105m,84m,70m,60m,101m)$ with $ m\in \{17,18,19\}$

[b]system 2[/b]

$ y_1 \plus{} y_2 \plus{} \dots y_5 \equal{} 5y_1 \equal{} 6y_2 \equal{} 7y_3 \equal{} 8y_4$

And by a similar method as above we find

$ (y_1,y_2,y_3,y_4,y_5) \equal{} (168m,140m,120m,105m,307m)$ but there is no $ m$ such that $ 1000 \le 120m \le 2000$ and $ 1000\le 307m \le 2000$

So the only solutions are

$ (y_1,y_2,y_3,y_4,y_5) \equal{} (105m,84m,70m,60m,101m)$ with $ m\in \{17,18,19\}$  :)


limit
induction
function
calculus
calculus computations

note that $x_{n+1}-x_{n}= \frac{(x_{n}-1)^{2}}{2}$

Take also:
$x_{n+1}-1 =\frac{x_{n}^{2}-1}{2}$ so:
$x_{n}+1=2\frac{x_{n+1}-1}{x_{n}-1}$
If we take 2 consecutive terms in the summation given:
$\frac{1}{x_{n}+1}+\frac{1}{x_{n+1}+1}=\frac{x_{n}-1}{2(x_{n+1}-1)}+\frac{1}{x_{n+1}+1}=\frac{x_{n}x_{n+1}+x_{n}-x_{n+1}-1+2x_{n+1}-2}{x_{n+2}-1}=\frac{x_{n}x_{n+1}+x_{n}+x_{n+1}-3}{x_{n+2}-1}=\frac{(1+x_{n})(1+x_{n+1})-4}{4(x_{n+2}-1)}$ Summing it with the next term:
$\frac{(1+x_{n})(1+x_{n+1})(1+x_{n+2})-8}{8x_{n+3}-1}$ So by induction:
$\frac{2^{n-1}(x_{n+1}-1)-2^{n}}{2^{n}x_{n+1}-1}$ so for n tending to infinity, the limit is half.i dont have any shot to the first part.
I would be glad if someone check what i did, since i usuallyt mess up a lot the stuff, but at least for me, sims to be quite right and nice! :D

it all lies in the faactoration of $x_{n}-1$
Just taking the formula givenm by the problem, we get that:
$x_{n}-1=\frac{x_{n-1}^{2}-1}{2}=\frac{(x_{n-1}-1)(x_{n-1}+1)}{2}$ so $\frac{1}{x_{n-1}+1}=\frac{x_{n-1}-1}{2(x_{n}-1)}$ Now if if we add to this term, the next term of the sequence S_n we get:
$\frac{(x_{n-1}-1)(x_{n}+1)+2x_{n}-3}{2(x_{n}-1)(x_{n}+1)}$
now adding the stuff in the numeratorwe get that $\frac{(x_{n-1}+1)(x_{n}+1)-4}{4(x_{n+1}+1)}$ and if you continue this process you will have:
$\frac{(x_{n-1}+1)(x_{n}+1)...(x_{m-1}+1)-2^{m-n+1}}{2^{m-n+1}(x_{m}-1)}$ for generic m and n. But for our sequence we have a special case. Since what values is the function $x_{n}-1$and when we multiply it by $x_{n}+1$ we get  $2(x_{n+1}-1)$ The trick is the factoration. so realize that $x_{1}-1=1$ so we can multiply the $(x_{1}-1)(x_{1}+1)(x_{2}+1)...(x_{n}+1)=2^{n}x_{n+1}-1$ and putting this in the formula:
$\frac{2^{n}(x_{n+1}-1)-2^{n}}{2^{n}(x_{n+1}-1)}=1-\frac{1}{x_{n+1}-1}=1$ since x_n is an increasing function  so it tends to infinity.
i Just dont know how to evluate the terms of the sequence, but i am quite sure that it tends to 1( yes i did a mistake in the post above!) if not, could you post your solution tuan?

[quote="Brambila"]
now adding the stuff in the numeratorwe get that $\frac{(x_{n-1}+1)(x_{n}+1)-4}{4(x_{n+1}+1)}$ [/quote]
You are wrong :D But it is litle :P Now, from your solution we have
\[\frac{1}{x_{n}+1}=\frac{1}{x_{n}-1}-\frac{1}{x_{n+1}-1}\]
And that is all, you solved problem :)


function
induction
algebra
functional equation
complex numbers
real analysis
real analysis solved

I need help on this problem:
      Find all functions f:R->R that verify: f(f(x))-3f(x)+2x=0 for every real x.
            Thanks.

hmmm i belive it must be continuous. are you sure it's a plain function?

Sorry I forgot to say it's a continous function.

I came up with a solution to the problem, and it was a bit unexpected (and rather long, but once you get the point, it's easy). You probably noticed that f(x)=x and f(x)=2x+a are solutions. Well, there is a more general solution: f(x)=2x-a, for x<=a, f(x)=x, for a<=x<=b, and f(x)=2x-b, for x>=b, where a and b may be equal, infinite, anything you might come up with. (Try drawing a graphic to see the whole thing better)

Now I have to make a few observations on the function, but they're quite obvious, so I won't prove them.
First, the function is injective, and being continuous, it must be strictly monotone. It can't be decreasing, because then f(f(x))-3f(x)+2x=0 would be the sum of 3 increasing functions, so f must be increasing.
It can't be bounded, because then it would have a finite limit in plus infinite or in minus infinite (being increasing and all), and by passing the hypothesis to limit in infinite or minus infinite we would obtain a contradiction.
So f is unbounded, and having Darboux's property, it's surjective, and therefore bijective.

Now let's denote by fn(x) f composed with itself n times (that is f(f(f...(f(x)..) ). If n =0, then f0(x)=x, and if n<0 then we consider fn(x)=f-1-n(x).

By induction we can easily prove that fn(x)=f(x)+(2n-2)(f(x)-x), for all [u]integer[/u] n. Thusly, for all integers n we form a sequence (fn(x))n integer infinite in both directions (just like Z, the set of integers).

Now let A={x | f(x)<>x}. If A has an element x, and, say, f(x)>x, we would have fn(x)>fn-1(x) for all [u]integers[/u] n.

Now that means that the sequence (fn(x))n integer is increasing, and unbounded from above, and bounded from below (by passing to limit when n tends to minus infinite, the sequence has the limit f(x)+(0-2)(f(x)-x)=2x-f(x).

So for all y greater then 2x-f(x), we would have an integer n such that  fn(x)> y>fn-1(x). Therefore for all m, we would have fn+m(x)>fm(y)>fn-1+m(x). By passing to limit when m tends to minus infinite we obtain 2x-f(x)<=2y-f(y)<=2x-f(x), so therefore, 2x-f(x)=2y-f(y), so for all y>2x-f(x), we have f(y)=2y+f(x)-2x.

In the same way, if A contains an element x such that f(x)<x, the sequence would be decreasing, unbounded from below, and having the limit 2x-f(x) from above. Therefore for all y<2x-f(x) we would have f(y)=2y+f(x)-2x.

Now let's divide A into two sets, B and C, such that B={x|f(x)>x}, and C={x|f(x)<x}.  Let's assume now both sets non empty (if any of them were empty if would be simpler, we'll get back to this case).Now let m=inf B and M=sup C.

If m=minus infinite, then let's show that f(x)=2x+a. Assume that we have x and y such that f(x)=2x+a and f(y)=2y+b, with a<>b. Because m is minus infinite, we would have a z in B such that z<x and z< y, so therefore f(x)-2x=f(z)-2z=f(y)-2y, so a=b.

The same happens whem M is plus ifinite. Now let's suppose that m is finite. We can prove that we must have f(m)=m. If f(m) would be greater than m, due to continuity, there would be numbers x smaller than m such that f(x)>x, so m wouldn't be inf B. Likewise, if f(m)<m we would have numbers x (a whole interval of them, thanks to continuity) greater than m such that f(x)< x, so m wouldn't be inf B.
We can show that f(x)=2x+a for all x>=m (In the same way as when we proved for m=minus infinite). Because f is continuous, then f(m)=2m+a=m so a=-m, so f(x)=2x-m, for all x>=m...

Well, I could go on, but we have many cases, depending on m and M (which is greater, are they both finite etc.) But you can see that we obtain the solutions that I wrote at the beginning.

hey mate
imposing solution but damn good.just a query.what exactly is darboux's property?
greetings!

darboux property as it is called in romania is actually the mean value property, i.e. a function f:(a,b) \rightarrow R has this property if and only if for every subinterval (x,y), and for every \lambda \in (f(x),f(y)) there is at least one c \in (x,y) for which f(c)= \lambda

obviously any continuous function has PD ( "property of darboux" abreviation in romanian )

I found this problem from last year's proposal for the national olympiad , 11th grade : determine all continuous f such that f(f(x)) + f(x) = 2x + a . I solved it using a different method (but also by iteration) , but my proof can't be adapted for the general case .. And this one doesn't seem to be adaptable too easy either . Is there a general method for solving these equations ? I asked some ppl already and no-one knew anything about them ..

It seems that these kind of functional equations are
very popular lately in the romanian olympiads!

Look here :

1)Find all monotone functions f:R->R which verify the functional equation :

f(f(f(x))) - 3f(f(x)) + 6f(x) = 4x + 3
(The Gazeta Matematica Contest,1999)

2)Find all continuous functions f:R->R which satisfy:
f(f(x)) + f(x) = 2x + a
(Jury,Romanian Olympiad 2002,11th Grade)

3)Let f,g:R->R be bijective and continuous functions
such that f(g^(-1)(x)) + g(f^(-1)(x)) = 2x , for all x
in R. Prove that if there is an x0 in R such that
f(x0)=g(x0) , then f(x) = g(00x) , for all x.
(Romanian National Olympiad 1997,11th Grade)

These three problems are quite different but still they
are using techniques from the same area.

Later,
Andrei

Method for solving functional equation with iteration f(x), f(f(x)),
f(f(f(x))):

Definition:Let f:I-->I, the iterate of f is the sequence of functions define
by (f^n)_{n \geq 0} where,
f^0(x) = x, f^1(x) = f(x), and f^n(x) = f(f^(n-1)(x)) or
f^n(x) = fo...of(x) n-time f.

Probem: a,b>0, find f:R+ ---->R+ such that
(*) f(f(x))+af(x) =b(a+b)x for x \geq 0
Shorlist 1992

It could be interesting to introduce the sequence x_n = f^n(x) and
study the property of x_n.

First x_n  \geq 0

We introduce the sequence x_0 = x and x_(n+1)=f(x_n)

(*) gives x_(n+2)+ax_(n+1)-b(a+b)x_n = 0

Now you can use the following
-----------------------------------------------------------------------------
Theoreme:Let U(n) real or complex number, a,b complex numbers
we suppose U(n+2) = aU(n+1)+bU(n) for any n \geq 0

i) If the equation x 2 =ax+b has two distincts roots r,s, then
there exist two constants p,q such that U(n) = p.r^n + q.s^n

ii)If the equation x 2 =ax+b has a root r with a multiplicity 2,
then U(n) = (p.n+q)r^n
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Here the equation is x^2-ax-b(a+b)=0 its roots are b and -(a+b)

x_n = p.b^n + q.(-1)^n.(a+b)^n

If q=/=0 for large n x_ is negative wich contradict x_n  \geq 0
so q=0.

For n=0 and n=1 we have

x_0 = x=p and x_1 = f(x) =p.b

so f(x) = b.x

And we check easily f(x) = bx is solution of the problem. :D

Application:

Solve problem 1) and 2) in the post of iandrei

Hello. I have a great problem with this:

"Give an example to an f: R->R function that is continuous in x=0, but it's
inverse is not continuous in x=f(0)"

Please don't revive so old topics!

Plus, I believe you already posted this problem.. unless I'm mistaken  :?

Can someone give me an example to an f: R->R injective function that is continuous
at x=0, but it's
inverse is not continuous at x=f(0). Where x=f(0) is meant for the inverse
function, like the inverse function is not continuous at an 'x', where 'x'
is the value of f(0).

This problem appears in my book (see http://www.teora.ro - colectia [u]Lyceum[/u] ) [u]Analiza matematica pentru clasa a XI-a. Teorie, exercitii si probleme, Editura Teora, 1999, Bucuresti[/u] (pag. 442 & 487, ex.17.1).

The solution of this problem is the set of the all functions $f_{mn}: R\rightarrow R$, where $-\infty\le m\le n\le\infty$ and the correspondence is\[ \displaystyle f_{mn} (x)=\{ \begin{array}{cc} 2x-m,\ x<m\\ x,\ m\le x\le n\\ 2x-n,\ n<x.\end{array}  \]


linear algebra
matrix
vector
algebra
polynomial
inequalities
superior algebra

An n x n matrix A is called nilpotent if, for some positive integer k, A^k = 0, where 0 is the n x n zero matrix.

Prove that if A is nilpotent, then det A = 0.

Thanks in advance!

[quote="blahblahblah"]$\vert A^k\vert=\vert A\vert^k=0$

Now, can we have a nonzero number become zero if we raise it to some power?[/quote]

not a number, but a matrix.

$\det A=\vert A\vert$ is not a matrix, it's a number.

Another proof: if $A$ is nilpotent, we must have $Ax=0$ for some vector $x.$
Just consider the chain $v,Av,a^2v,\dots$, which eventually becomes zero. Write $A$ in a basis containing $x$ and it is clear that $\det A=0.$

Also, unless otherwise stated matrices are over some field- no funny business here with nilpotent entries.

Actually, it is not hard to prove something even more powerful. Not only $\det(A)=0$, but all the eigenvalues are 0, that is, if $p(x)$ is the caracteristic polynomial of a nilpotent matrix $A\in\mathfrak{M}_n(\mathbb{R})$, or any other field, then $p(x)=x^n$. Also, the minimal $k$ in the definition holds with the inequality $k\leq n$.

Best,