inequalities
linear algebra
linear algebra unsolved

I gave (a version of) this problem to my students in the class but no one could give a solution in a week. So I decided to post it here: Prove or disprove; Let $ V_1, V_2, V_3$ be finite dimensional subspaces of $ V$, then
$ \\
\text{dim}(V_1 \plus{} V_2 \plus{} V_3)\le \sum_{i \equal{} 1}^{3}\text{dim}V_i \minus{} \sum_{i < j}\text{dim}(V_i\cap V_j) \plus{} \text{dim}(V_1\cap V_2\cap V_3).$

Hei,here is my solution. :lol:

$ \text{dim}(V_1 \plus{} V_2 \plus{} V_3) \equal{} \text{dim}(V_1 \plus{} V_2) \plus{} \text{dim}(V_3) \minus{} \text{dim}((V_1 \plus{} V_2)\cap V_3)$
$ \text{dim}(V_1 \plus{} V_2) \equal{} \text{dim}(V_1) \plus{} \text{dim}(V_2) \minus{} \text{dim}(V_1\cap V_2)$
$ \text{dim}((V_1 \plus{} V_2)\cap V_3) \ge \text{dim}(V_1\cap V_3 \plus{} V_2\cap V_3) \equal{} \text{dim}(V_1\cap V_3) \plus{} \text{dim}(V_2\cap V_3) \minus{} \text{dim}(V_1 \cap V_2\cap V_3)$
Thus:
$ \text{dim}(V_1 \plus{} V_2 \plus{} V_3) \le \sum_{i \equal{} 1}^3\text{dim}(V_i) \minus{} \sum_{i < j}\text{dim}(V_i\cap V_j) \plus{} \text{dim}(V_1 \cap V_2\cap V_3)$
The only inequality in the solution is:
$ \text{dim}((V_1 \plus{} V_2)\cap V_3) \ge \text{dim}(V_1\cap V_3 \plus{} V_2\cap V_3)$
This is because:
$ V_1\cap V_3 \plus{} V_2\cap V_3 \subset (V_1 \plus{} V_2)\cap V_3$
for every $ x \plus{} y \in V_1\cap V_3 \plus{} V_2\cap V_3$
Such that:$ x \in V_1\cap V_3,y \in V_2\cap V_3$
We have:$ x \plus{} y \in V_1 \plus{} V_2,x \plus{} y \in V_3$

Corollary: My students have not bothered to think on the problem!!:D
Now, I add another question. Can we generalize this, in the sense that we gain an analogous inequality for the sum of more than three subspaces?


Over what?

It has no rational or real factoring.  The answer you're probably looking for is

$(x+iy)(x-iy)$

Where $i = \sqrt{-1}$.

Or you could express it as:

$(x+y)^{2}-2xy$

Not really a "factoring."  Just a rewrite.

[quote="b-flat"][quote="tridecalogism_lover"][quote="sharkman"]Factor x^2+y^2[/quote]

you can't factor it. period.  What you probably meant is what mysmartmouth said.[/quote]

No, as t0rajir0u said, it can be factored, just not in reals.  The factorization $a^{2}+b^{2}=(a+b\imath)(a-b\imath)$ is useful in trig and complex numbers.[/quote]
It's also useful in modular congruences.

[quote="tridecalogism_lover"][quote="sharkman"]Factor x^2+y^2[/quote]

you can't factor it. period.  What you probably meant is what mysmartmouth said.[/quote]
$\boxed{(x-yi)(x+yi)}$ is one way to factor it.  Another way of factoring it, like mysmartmouth was saying, was that $x^{2}+y^{2}=(x+y)^{2}-2xy=\boxed{(x+y-\sqrt{2xy})(x+y+\sqrt{2xy})}$ by difference of two squares.

[quote="t0rajir0u"]That's not a valid polynomial factoring over any field, since $\sqrt{2xy}$ isn't a polynomial.[/quote]
Eh, but it is helpful with deriving the sophie germain identity (I think)

the technique is the same, but the problem is different...


geometry
3D geometry
tetrahedron
geometry proposed

Let $ABCD$ be a tetrahedron.  Let $G$ be the centroid of face $\triangle BCD$.  Show that the length of the median $\overline{AG}$ is given by the formula
\[9 AG^{2}= 3 AB^{2}+3 AC^{2}+3 AD^{2}-BC^{2}-BD^{2}-CD^{2}\, . \]


[quote="8826"]Let a,b,c be satisfying $ \frac {1}{\sqrt {2}} < a,b,c < 1$ and
 $ a^4 \plus{} b^4 \plus{} c^4 \plus{} 4a^2b^2c^2 \equal{} 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)$
Prove that
 $ \sum_{cyc}{\frac {a^4}{1 \minus{} a^2}} \geq 4abc(a \plus{} b \plus{} c)$[/quote]
As $ a^2b^2\plus{}b^2c^2\plus{}c^2a^2 \ge abc(a\plus{}b\plus{}c)$, we can even prove the more stronger inequality still holds:

If $ \frac{1}{2} <a,b,c<1$, such that
$ 4abc\equal{}2(ab\plus{}bc\plus{}ca) \minus{}a^2\minus{}b^2\minus{}c^2.$
Then
\[ \frac{a^2}{1\minus{}a}\plus{}\frac{b^2}{1\minus{}b}\plus{}\frac{c^2}{1\minus{}c} \ge 4(ab\plus{}bc\plus{}ca).\]

This inequality follows from the following inequality
\[ 8(a^2\plus{}b^2\plus{}c^2)\minus{}(ab\plus{}bc\plus{}ca) \ge \frac{21abc(a\plus{}b\plus{}c)}{2(ab\plus{}bc\plus{}ca)\minus{}a^2\minus{}b^2\minus{}c^2},\]
if $ a,b,c$ are the sidelengths of a triangle. ;) :)


I'd need some help on picking the right university... My first choice was camebridge, but apparently I'm about 6 months too late for that. So does anyone else know any good european universities for which I may be in time? (preferrably in France or UK or so...)

Currently I'm in Ghent, but the student level here is tremendously low here... in a matter that I'm pretty much alone here in being interested in maths. [ I'm currently doing engineer and mathematics at once in a hope to get some challenge at least :( ]

Does anyone know a good college in there for which I would be in time? Or will I have to wait another year? ...

Well Peter,

All my information about European Schools is second hand information, but I can share it with you. Nevertheless, I make the remark  :P  :P . I have a friend who told me that the year he spend as an exchange student Harvard-Cambridge in Cambridge was the best of his life. Oxford, of course, is also the other big one, and I really don't know much beside that. Nevertheless, I have heard that the university Paul Sabatier is amazingly good, and also, Paris VI, VII and XIII. I have a few friends that are currently in Paris and Toulouse. If you are interested, let me know (in a PM, and with at least two of them, I hope you can defend yourself in either Spanish or French :D )

But those are not the only options. I have a few friends in Germany, and they say that the Humbolt University and the Berlin Technological University (I am not totally sure of the name of the second one) are just great. About Italy, I don't know much, but I can ask. And well, there are nice places on Spain (Carlos III, Universidad Autonoma de Barcelona, Universidad de Madrid, etc), and my office mate is visiting for a semester, he comes from the UAB, and I have a friend who just left, who is in the Carlos III. I mean, Germany, Italy and Spain probaly would represent to learn a new language, but there are plenty of options.

Finally, I do not know if you can transfer from Europe to the US easely, but there are good schools here too, or you can consider it for Grad School.

[quote="tokenadult"]I hope other participants who are familiar with European universities will answer your question. You would like to transfer in as a second-year or third-year student, right? That is difficult to do in the most selective universities in the United States (but easy to do at many second-tier universities in the United States), but I don't know how transfer admission works between differing European countries.[/quote]
I have no clue what level camebridge is exactly. I know it's "high", but one would lose sense of height sometimes... is what you can find here in the college playground? Knowing that I didn't really solve much here, would going from second year here to second year there be a good option? Or would that lose challenge again then?

[quote="djimenez"]Well Peter,
All my information about European Schools is second hand information, but I can share it with you. Nevertheless, I make the remark  :P  :P . I have a friend who told me that the year he spend as an exchange student Harvard-Cambridge in Cambridge was the best of his life. Oxford, of course, is also the other big one, and I really don't know much beside that. Nevertheless, I have heard that the university Paul Sabatier is amazingly good, and also, Paris VI, VII and XIII. I have a few friends that are currently in Paris and Toulouse. If you are interested, let me know (in a PM, and with at least two of them, I hope you can defend yourself in either Spanish or French :D )[/quote]
I won't say no to that. Spanish will be a problem, but french should be ok, especially written. Also in the university, Paris is ok but still I think camebridge is better. :)

[quote="djimenez"]But those are not the only options. (...) I mean, Germany, Italy and Spain probaly would represent to learn a new language, but there are plenty of options.[/quote] Hmm... that is true. But I wasn't mainly looking for "studies abroad", rather for a really good university. (like some people told me, if you want to be a researcher/professor at a normal university, it is better to go to a top-level univerisity then to get good points in the proper univ...)

[quote="djimenez"]Finally, I do not know if you can transfer from Europe to the US easely, but there are good schools here too, or you can consider it for Grad School.[/quote]Nah, US is too far anyhow. I don't know the laws, I don't know the customs, I know nothing about there. I'd prefer to stay in europe.
[I've always been pretty europe-minded, perhaps because my father works for the EU?]

Besides, [Cambridge->home] or [home->Camebridge] is like 4 hours :) Paris is 3, and I don't think there are many other universities that close? Oxford isn't that great for maths I heard, I dunno if they're better for engineering?

You might try poking around the national forums or the college playground -- I get the sense that this forum is American-centric and that probably a small percentage of the French or German or English college students visit here, and they're the people who really know anything about these schools.


Let be $\ n$ a natural number.Prove that $\ [\sqrt n+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}]$ is equal whith the number $\ [\sqrt{16n+23}].$Can this problem be generalised?And how?
     $\ [a]$ is the  closest integer number smaller than $\ a$.

Why confine $n$ to just natural numbers?

EDIT: (Whoops...sorry I just realized that there are some numbers (such as 1.65) for which it dosen't work.  When I first examined it, it seemed to work for all postive $n$.

 It is here

[url= https://artofproblemsolving.com/community/c6h377979p2088023] see [/url]

https://artofproblemsolving.com/community/c6h377979p2088023


abstract algebra
vector
group theory
superior algebra
superior algebra solved

Let $C$ be an abelian group such that $pC=0$ for $p\in \mathbb{Z}$ a prime number. There exist a free abelian group $F$ such that $C\cong F\diagup pF$?

uhm..
if i recall correctly (and i seem to be convinced so), $C$ is a $\mathbb{F}_{p}$ vector space: take a basis $\{x_\alpha\}$ as $\alpha$ varies in some set of indices, and then consider the free abelian group they generate.
it seems like calling this last group $F$ would work..

ps. i wonder if there's something like a topological solution to this problem.. any idea?

[quote="ma_go"]uhm..
if i recall correctly (and i seem to be convinced so), $ C$ is a $ \mathbb{F}_{p}$ vector space: take a basis $ \{x_\alpha\}$ as $ \alpha$ varies in some set of indices, and then consider the free abelian group they generate.
it seems like calling this last group $ F$ would work..
[/quote]

Yeah, that's right. That $ C$ is an $ \mathbb{F}_p$ vector space follows since the map $ \mathbb{Z} \to \mathrm{End}(C)$ has kernel $ (p)$ and so induces a map $ \mathbb{F}_p \to \mathrm{End}(C)$.


complex numbers
algebra unsolved
algebra

Let a,b,c be complex numbers satisfying:

(a+c)(a+b)=b,   (b+c)(b+a)=c,   (c+a)(c+b)=a.

Prove that a,b,c are in fact reals.

You asked this before, remember? [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=63277]www.mathlinks.ro/Forum/viewtopic.php?t=63277[/url]


Let $ \imath$ denote the imaginary unit.  Show that

$ \sum_{k \ge 0} {n\minus{}k \choose k} (2\imath)^{n\minus{}2k} \equal{} (n\plus{}1) \imath^n$.

[hide]
Using "Snake Oil", (so probably not the 'nice' solution you're after)

Let $ f_n \equal{} \sum_{k \ge 0} \binom{n\minus{}k}{k} y^{n\minus{}2k}$

Consider the generating function, $ F(x)$ of $ \{ f_n \}$

$ F(x) \equal{} \sum_{n \ge 0} x^n \sum_{k \ge 0} \binom{n\minus{}k}{k} y^{n\minus{}2k}$

Changing the order of summation, so:

$ F(x) \equal{} \sum_{k \ge 0} y^{\minus{}2k} \sum_{n \ge 2k} \binom{n\minus{}k}{k} x^n y^n$

and $ F(x) \equal{} \sum_{k \ge 0} y^{\minus{}2k} x^{k} y^{k} \sum_{n \ge 2k} \binom{n\minus{}k}{k} (xy)^{n\minus{}k}$

$ F(x) \equal{} \sum_{k \ge 0} x^k y^{\minus{}k} \frac{(xy)^k}{(1\minus{}xy)^{k\plus{}1}}$

$ F(x) \equal{} \sum_{k \ge 0} \frac{x^2k}{(1\minus{}xy)^{k\plus{}1}} \equal{} \frac{1}{(1\minus{}xy)} \frac{1}{1\minus{}\frac{x^2}{(1\minus{}xy)}} \equal{} \frac{1}{1\minus{}xy\minus{}x^2}$

When $ y \equal{} 2i$

$ F(x) \equal{} \frac{1}{1\minus{}2ix \minus{}x^2} \equal{} \minus{}\frac{1}{(x\plus{}i)^2} \equal{} \sum_{k \ge 0} (k\plus{}1)i^k$
[/hide]

Congratulations, you've figured out what the [url=http://mathworld.wolfram.com/FibonacciPolynomial.html]Fibonacci polynomials[/url] look like!  :)

Now, I am very curious as to whether there's a direct approach, so I will say the following:  $ {n \minus{} k \choose k}$ is the number of ways to tile a $ 1 \times n$ board with $ n \minus{} 2k$ tiles of size $ 1 \times 1$ and $ k$ tiles of size $ 1 \times 2$, so a direct combinatorial solution might need to play around with powers of $ 2$ in the $ 1 \times 1$ tiles and investigate some kind of period-$ 4$ phenomenon in $ n$.  It might be good to break the problem down into the cases $ n$ even and $ n$ odd and take real and complex parts, in which case you are really proving four (closely related) identities.

Fibonacci numbers were EXACTLY what I thought of first. Unfortunately, I couldn't work out how to work in a coefficient of y in there. (I did try to look for the combinatorial one, using a similar approach to you, but I'm pretty weak at combinatorial arguments once they get too complex and I find the words getting in the way)

Silly me.  Dividing by $ \imath^n$ turns this into a "real" combinatorial identity again:  we only need to prove that

$ \sum_{k \ge 0} ( \minus{} 1)^k {n \minus{} k \choose k} 2^{n \minus{} 2k} \equal{} n \plus{} 1$.

Edit:  The direct proof is quite nice; try to find it :)

Edit #2:  I can't resist, the proof is [hide="too cute!"] $ {n \minus{} k \choose k} 2^{n \minus{} 2k}$ is the number of ways to tile a $ 1 \times n$ board with $ n \minus{} 2k$ tiles of size $ 1 \times 1$ and $ k$ tiles of size $ 1 \times 2$ and then to color the $ 1 \times 1$ tiles red or blue.  Given such a tiling-and-coloring, find the first index at which there is either a $ 1 \times 2$ tile or a red $ 1 \times 1$ tile followed by a blue $ 1 \times 1$ tile.

- If there is a $ 1 \times 2$ tile, change it into a red $ 1 \times 1$ tile and a blue $ 1 \times 1$ tile.
- If there is a red $ 1 \times 1$ tile and a blue $ 1 \times 1$ tile, change it into a $ 1 \times 2$ tile.

This operation changes the parity of $ k$, so establishes an almost-bijection between the tiling-and-colorings with $ k$ even and $ k$ odd.  It fails (that is, has a fixed point) if and only if there are no $ 1 \times 2$ tiles and in addition no blue tile follows a red tile, which occurs if and only if there are $ i$ blue tiles followed by $ n \minus{} i$ red tiles.  This occurs in $ n \plus{} 1$ ways and $ k \equal{} 0$ in each of these cases.  :) [/hide]


algebra
polynomial
calculus
derivative
induction
function
linear algebra

Let $ F: R^n \rightarrow R$ be $ C^{\infty}$ such that the directional derivatives $ D_i^P F \equal{} 0$ for some $ P$ and for all $ 1 \leq i \leq n$; show that $ F$ is a polynomial.

This is straightforward for $ n\equal{}1$, but I'm having trouble with the induction.

Write a stronger base case: that if the $ x_i$ partials after order $ M$ are zero for $ i \equal{} 1,2,\dots,k$, we have $ F(x_1,x_2,\dots,x_k,y_{k \plus{} 1},\dots,y_n) \equal{} \sum_{I\in \{0,1,2,\dots,M\}^k}G(y_{k \plus{} 1},\dots,y_n)x^I$ (using multi-index notation here); this is polynomial in the $ x_1,x_2,\dots,x_k$ of degree no more than $ M$ each, with the coefficients dependent on the other variables in some unknown way.
The induction step looks at how each of these coefficients behaves.

I remember there is a problem states: let $ f: \mathbb{R}\to\mathbb{R}\in C^\infty$, for any $ x$, there is an integer $ m$ s.t. for any $ n > m$, $ f^{(n)}(x) \equal{} 0$, then $ f$ must be polynomial..

I'm curious about whether it's generalization holds.

are you sure that induction works?  It seems like the coefficients should be dependent on the index, $ I$, as well

Thus, when you take the $ k \plus{} 1$'th partial of the sum, and you plug in $ 1$'s for the $ x$s, you have that the sum of the coefficient functions is a polynomial in $ y_{k \plus{} 1}$, but not necessarily that each one is, also.

Thanks.

Edit: Is the key that the coefficients only depend on the last $ x_k$, so that making these $ 1$ gives the desired result?

Yes, the coefficients are also dependent on the index- make that $ G_I$. Part of the induction step is to infer that each of those coefficient functions is polynomial in the next variable from the fact that the weighted sum is.

So, the problem reduces to showing that there exist $ r_i \in R^n$, $ i = 1,...,M^k$, such that the $ M^k \times M^k$ matrix $ A_{i,I} = r_i^I$ for each $ I \in \left\{1,...,M\right\}^k$ (where $ r_i^I = r{}_i{}_1^{I_i} \cdot r{}_i{}_2^{I_2} \cdot ... \cdot r{}_i{}_n^{I_n}$) is non singular.

I'm a bit stuck, here.  Does letting $ r_i = (i,...,i)$ work?

No. You need more linear independence than that.

The most natural choice is the points of a lattice, as $ x_i$ ranges over $ M\plus{}1$ possible values for each $ i$.

I would guess that the primes, in sequence, should work.

Is there an easier way to do this problem?

Not the primes. It doesn't matter what you choose, as long as they aren't all roots of the same low-degree polynomial. The lattice idea uses the one-dimensional fact that a nonzero degree $ \le M$ polynomial has at most $ M$ roots to get there, so we can choose any $ M\plus{}1$ values for each dimension. You should end up with a combination of Vandermonde determinants.


A [i]snake[/i] on the m x n chessboard is a nonempty subset S of the squares of the board with the following property: Start at one of the squares and continue walking one step up or to the right, stopping at any time.  The squares visited are the squares of the snake.  Attached is an example of the 8 x 8 chessboard covered with disjoint snakes.

Prove that the total number of ways to cover an m x n chessboard with disjoint snakes is a product of Fibonacci numbers.

Obersve how the snakes cut the sides along a diagonal...

I can post a solution this afternoon...

My friends and I used this problem to be the subject of the science fair.
We had studied it for about a year.
We solved the problem in May,2006.
Our conclusion is in the attachment.


limit
analytic geometry
graphing lines
slope
real analysis
real analysis unsolved

Dear All,

I have a recursive problem, which I am not able to solve.

$ P_{n}\equal{}\frac{1}{1\plus{}c_{1}*10^{\minus{}[(P_{0}*\frac{A}{n})\plus{}(P_{1}*\frac{A}{n})\plus{}...\plus{}(P_{n\minus{}1}*\frac{A}{n})]}}$

the initial values of $ P_{0}$, $ c_{1}$, and A are known.  n is any number of iterations 1 to infinity.

I was wondering if anybody know how to solve this problem?

here is an example, where you solve for the case where $ n\equal{}2$

If you have the following initial conditions:

$ P_{0}\equal{} 0.2$
$ c_{1}\equal{} 4$
$ A \equal{} 0.8$

First solve for the first iteration in the following manner:

$ P_{1}\equal{}\frac{1}{1\plus{}4*10^{\minus{}[(P_{0}*\frac{0.8}{2})]}}$

so,
$ P_{1}\equal{} 0.231$

Then solve for the second iteration for $ P_{2}$ for the final answer:

$ P_{2}\equal{}\frac{1}{1\plus{}4*10^{\minus{}[(P_{0}*\frac{0.8}{2})\plus{}(P_{1}*\frac{0.8}{2})]}}$

plugging in the initial $ P_{0}$, and the previously solved $ P_{1}$, you get an answer for $ P_{2}$ of:
$ P_{2}\equal{}0.271$

Thanks for your help.

If n =3, then the you would have to change the value of n in the equations, and as a result the values of P0, P1, and P3 would change.
Let me know if there are any questions, or if the previous example is not clear.

Thanks again.

Best,
Prad

[quote="Prad01"]Dear All,

I have a recursive problem, which I am not able to solve.

$ P_{n}\equal{}\frac{1}{1\plus{}c_{1}*10^{\minus{}[(P_{0}*\frac{A}{n})\plus{}(P_{1}*\frac{A}{n})\plus{}...\plus{}(P_{n\minus{}1}*\frac{A}{n})]}}$

the initial values of $ P_{0}$, $ c_{1}$, and A are known.  n is any number of iterations 1 to infinity.
[/quote]
Set $ e^{\alpha}\equal{} 10^{A}$, then the limit $ x$ of the recursion (if it exists) must satisfy
\[ x \equal{}\lim_{n\to\infty}\frac{1}{n}\left(P_{0}\plus{}\dots\plus{}P_{n\minus{}1}\right)\]
and hence
\[ x \equal{}\frac{1}{1\plus{}c_{1}10^{\minus{}Ax}}\equal{}\frac{1}{1\plus{}c_{1}e^{\minus{}\alpha x}}\equal{}\frac{e^{\alpha x}}{e^{\alpha x}\plus{}c_{1}}\]
The right hand side is monotone increasing from 0 to 1 with slope < 1, since $ \alpha < 4$. So there is exactly one solution. It can be obtained more easily with the iteration
\[ P_{n\plus{}1}\equal{}\frac{10^{AP_{n}}}{10^{AP_{n}}\plus{}c_{1}}\, .\]
In your example, the limit is $ P^{*}\equal{} 0.3046998...$.

One still has to prove that the original iteration always converges.


proof that $\forall A$, $\forall \epsilon>0$
there is an open set like $O$ such that $A\subset O$,
$m^{*}O<m^{*}A+\epsilon$

be more precise ? Sometimes it is false ..

the statement is correct.

recall: for any $E$, $m^{*}E=\inf\sum l(I_{n})$, where $\cup I_{n}$ is a countable set of open interval covering $E$.

So, for $A$, there exists a countable set $\cup I_{n}$ of open intervals covering $A$ and $\sum l(I_{n}) < m^{*}A+\epsilon$. Let $O=\cup I_{n}$, and u can check the inequality.


problem 1(russia 2000)

find all function $f: \mathbb{R}\to \mathbb{R}\ , f(x+y)+f(y+z)+f(z+x)\leq 3f(x+2y+3z)$,$x,y,z \in\ \mathbb{R}$

problem 2(moldova 2000)

find all fuction $f: \mathbb{R}\to \mathbb{R}, f(x+y)-f(x-y)=2y(3x^{2}+y^{2})$,$x,y \in\ \mathbb{R}$

Problem 2:

Let $x=y$ to get $f(2x)-f(0)=(2x)^{3}$. So $f(x)=x^{3}+c$, where $c$ is a real number.

1. y=z=0 give $f(x)\ge f(0)$ x=y=-z give $f(0)\ge f(2x)$. Therefore $f(x)=f(0)$ for all x.


quadratics
induction
number theory unsolved
number theory

Let $ n \ge 2$ and $ k$ be any positive integers.
Prove that $ (n\minus{}1)^2|(n^k\minus{}1)$ if and only if $ (n\minus{}1)|k$

posted before i think---- not able to find it though.. Just some mod chasing with $ n^k\minus{}1$ and also use the fact that $ n^k\minus{}1 \equal{} (n\minus{}1)(n^{k\minus{}1} \plus{} .....\plus{} 1)$....... I am sleepy now to post the complete solution.

The claim is true for $ n\equal{}2$. So, let $ n>2\Longrightarrow (n\minus{}1)^2>n\minus{}1\Longrightarrow k>1$ from now on. Direct calculation gives $ n^k\minus{}1\equal{}(n\minus{}1\plus{}1)^k\minus{}1\equal{}\sum_{i\equal{}1}^k\left(k\atop i\right)(n\minus{}1)^i\equiv k(n\minus{}1)\bmod{(n\minus{}1)^2}$, which immediately yields the assertion.

though i dont think it will work, is induction possible?

$ n^k\minus{}1 \equal{} (n\minus{}1)(n^{k\minus{}1} \plus{} .....\plus{} 1)$, let $ P(n)\equal{}n^{k\minus{}1} \plus{} .....\plus{} 1$ we must have $ p(1)\equal{}1^{k\minus{}1}\plus{}...\plus{}1\equal{}k$ divisible by $ (n\minus{}1)$, done. So it's just remainder theorem.


abstract algebra
search
function
group theory
superior algebra
superior algebra unsolved

I saw this problem in Herstein:

Given a finite abelian group G such that there are no more than $ n$ solutions to $ x^n \equal{} e$ for all positive integer $ n$, prove that G is cyclic.

I managed to prove this without assuming that G is abelian. I just wanted to make sure, is it true even if we don't assume that G is abelian?

Hmm yeah oops  :blush:

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1998026785&t=90624


calculus
integration
trigonometry
logarithms
algebra
partial fractions
calculus computations

If we make the substiution $ u \equal{} x^{\frac {1}{2}}$, we get $ du \equal{} \frac {1}{2}x^{\frac { \minus{} 1}{2}}\ dx$ and so

$ \int_{0}^{\infty} \frac {\sqrt {x}}{x^{2} \plus{} 1}\ dx \equal{} \int_{0}^{\infty} \frac {2u^{2}}{u^{4} \plus{} 1}\ du$.

The factors of $ u^{4} \plus{} 1$ are $ (u^{2} \plus{} \sqrt {u} \plus{} 1)(u^{2} \minus{} \sqrt {u} \plus{} 1)$, so we can use partial fractions to get something we can see how to integrate.

I make the partial fractions:
$ \frac { \minus{} 1}{\sqrt {2}} (\frac {u}{u^{2} \plus{} \sqrt {2}u \plus{} 1}) \plus{} \frac {1}{\sqrt {2}} (\frac {u}{u^{2} \minus{} \sqrt {2}u \plus{} 1})$

This gives:

$ \frac { \minus{} 1}{2\sqrt {2}}\ \int_{0}^{\infty} \frac {2u \plus{} \sqrt {2} \minus{} \sqrt {2}}{u^{2} \plus{} \sqrt {2}u \plus{} 1}\ du \plus{} \frac {1}{2\sqrt {2}} \ \int_{0}^{\infty} \frac {2u \minus{} \sqrt {2} \plus{} \sqrt {2}}{u^{2} \minus{} \sqrt {2}u \plus{} 1}\ du$

$ \equal{} [\frac { \minus{} 1}{2\sqrt {2}} \ln(u^{2} \plus{} \sqrt {2} u \plus{} 1)]_{0}^{\infty} \plus{} \frac {1}{2}\int_{0}^{\infty} \frac {1}{(u \plus{} \frac {1}{\sqrt {2}})^{2} \plus{} \frac {1}{2}} \ du$

$ \plus{} [\frac {1}{2\sqrt {2}}\ \ln(u^{2} \minus{} \sqrt {2}u \plus{} 1) ]_{0}^{\infty} \plus{} \frac {1}{2}\int_{0}^{\infty} \frac {1}{(u \minus{} \frac {1}{\sqrt {2}})^{2} \plus{} \frac {1}{2}}\ du$

$ \equal{} [\frac {1}{2\sqrt {2}}\ \ln(x \plus{} \sqrt {2x} \plus{} 1) \plus{} \frac {\sqrt {2}}{2}\ \tan^{ \minus{} 1}(\sqrt {2x} \plus{} 1) \plus{} \frac {1}{2\sqrt {2}}\ \ln(x \minus{} \sqrt {2x} \plus{} 1) \plus{} \frac {\sqrt {2}}{2}\ \tan^{ \minus{} 1}(\sqrt {2x} \plus{} 1)]_{0}^{\infty}$

$ \equal{} [\frac {1}{2\sqrt {2}}\ \ln(\frac {x \minus{} \sqrt {2x} \plus{} 1}{x \plus{} \sqrt {2x} \plus{} 1}) \plus{} \frac {\sqrt {2}}{2}\ \tan^{ \minus{} 1}(\sqrt {2x} \plus{} 1) \plus{} \frac {\sqrt {2}}{2}\ \tan^{ \minus{} 1}(\sqrt {2x} \minus{} 1)]_{0}^{\infty}$

$ \equal{} \frac {\pi}{\sqrt {2}}$

Thanks very much, Andrew Tom.

$ \int_{0}^{\infty} \frac{x^3}{e^x\minus{}1}dx$

$ \int_{0}^{\infty }\frac{x^{3}}{e^{x}\minus{}1}dx$

$ \equal{}\sum_{k\equal{}0}^{\infty }\int_{0}^{\infty }x^{3}e^{\minus{}(k\plus{}1)x}dx$

$ \equal{}\sum_{k\equal{}0}^{\infty }\frac{1}{\left( k\plus{}1\right) ^{4}}\allowbreak \int_{0}^{\infty }u^{3}e^{\minus{}u}\,du$

$ \equal{}\sum_{k\equal{}0}^{\infty }\allowbreak \frac{6}{\left( k\plus{}1\right) ^{4}}\equal{}\allowbreak \frac{\pi ^{4}}{15}$


linear algebra
matrix
search
function
linear algebra unsolved

As a follow up from http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1454790#1454790
Let $ S$ the set of nonsingular square matrix $ A$ with entries in $ \mathbb{K}$, such that for every positive integer $ k$, there is a matrix $ X$ with entries in $ \mathbb{K}$ s.t. $ X^k \equal{} A$.
What can we say about $ S$ in the following case:
[list]1. $ \mathbb{K}\equal{}\mathbb{Q}$,
2. $ \mathbb{K}\equal{}\mathbb{A}$, the set of real algebraic numbers
3. $ \mathbb{K}\equal{}\mathbb{R}$,
4. $ \mathbb{K}\equal{}\mathbb{C}$,
[/list] 2 of the 4 questions are simple, one is more difficult and one is open I think.

Case 1: $ K\equal{}\mathbb{C}$. Then $ S\equal{}GL_n(K)$
Proof:we consider the Jordan form of an invertible matrix $ A\equal{}diag(a(I\plus{}N),\cdots)$ where $ N$ is nilpotent and $ a\not\equal{}{0}$. Let $ X\equal{}a^{p}(I\plus{}(_1^p)N\plus{}\cdots)$ where $ p\equal{}1/k$ (binomial formula). Then $ X^k\equal{}a(I\plus{}N)$.
Case 2: $ K\equal{}\mathbb{R}$. Then $ S\equal{}exp(\mathcal{M}_n(K))$.
Proof: Let $ A$ be an invertible matrix. Necessarily there exists a real matrix $ X$ s.t. $ X^2\equal{}A$; then $ A\equal{}e^U\in{exp(\mathcal{M}_n(K))}$. Conversely $ X\equal{}e^{U/k}$ satisfies $ X^k\equal{}A$.

That's right loup blanc, I know that you can also answer the case $ \mathbb{K}\equal{}\mathbb{Q}$, even though it is more difficult. What about $ \mathbb{K}\equal{}\mathbb{A}$?

Case 3: $ K\equal{}\mathbb{Q}$. Then $ A\in{S}$ iff $ A(A\minus{}I_n)^n\equal{}0$.
see http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=909877333&t=42444
I had forgotten ! I sent (with a colleague) to the french review "Revue de Mathematiques speciales" a solution using Mahler's measure.
Case 4: $ K\equal{}\mathbb{A}$. Then  ?????

In my last post, $ A\in{S}$ is not assumed to be invertible.
My question is:
if $ A\in\mathcal{M}_n(\mathbb{Z})$ is a rootable matrix then is $ A$ necessarily a projector ?

[quote="loup blanc"]In my last post, $ A\in{S}$ is not assumed to be invertible.[/quote]
So for  $ K = \mathbb{Q}$, then $ A\in{S}$ iff $ (A - I_n)^n = 0$.
[quote="loup blanc"]if $ A\in\mathcal{M}_n(\mathbb{Z})$ is a rootable matrix then is $ A$ necessarily a projector ?[/quote]
I think so.

4- $ K=\mathbb{A}$ is a non algebraically closed field.
By 2- $ S\subset{exp(\mathcal{M}_n(\mathbb{R}))=\{U\in{GL_n(\mathbb{R}); \forall\lambda\in{spectrum(U)}\cap\mathbb{R}^-,\forall{p}: U}}$ has an even number of Jordan blocks $ J_p(\lambda)\}$.
See: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=535973634&t=43500 and
 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190912
A partial result: we assume $ spectrum(U)\subset{\mathbb{R}}$.
Let $ Z=\{U\in{GL_n(K); \forall\lambda\in{spectrum(U)}\cap\mathbb{R}^-,\forall{p}: U}$ has an even number of Jordan blocks $ J_p(\lambda)\}$.
Then $ U\in{S}$ iff $ U\in{Z}$
Proof: of course $ S\subset{Z}$.
Let $ U\in{Z}$ and $ k$ be an integer. There exists $ P\in{GL_n(K)}$ s.t. $ U=Pdiag(a(I+N),\cdots)P^{-1}$ where $ a\in{K}^*$ and $ N\in{\mathcal{M}_n(K)}$ is nilpotent.
Case 1: $ a>0$ ; we reason as in 1- ($ a^{1/k}\in{K}$).
Case 2: $ a<0$ ; then from $ diag()$ we extract 2 blocks $ V=diag(J,J)$ where $ J=a(I+N)$.
$ V=\begin{pmatrix}0&I\\J&0\end{pmatrix}^2$ but it is not sufficient.
The following is a sketch of the proof using an example ( I think that the result is true in general case):
Let $ J=\begin{pmatrix}-1&1\\0&-1\end{pmatrix}$. Then $ V=e^W$ with $ W=\begin{pmatrix}0&-1&\pi&0\\0&0&0&\pi\\-\pi&0&0&-1\\0&-\pi&0&0\end{pmatrix}$.
 $ X=e^{W/k}$ satisfies $ X^k=V$. Moreover the coefficients of $ X$ are algebraic functions of the $ cos(u\pi/v)$ and $ sin(u\pi/v)$ where $ u,v$ are integers; these reals are in $ K$.  QED


[quote="talfohrrt"]a^2+b^2+c^2=3 prove:
5(a+b+c)+3/(abc)>=18
[/b][/quote]
Maybe $ a,$ $ b$ and $ c$ are positive numbers? If so, see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86904


AMC AIME USA(J)MO USAMO Alcumus inequalities geometry

I'm trying hard to get a score greater than 8 this year on the AIME I, cuz I only got a 129 on the AMC 10. I know I must score really high to make USAMO, and I'm dying hard to do so this year. What's the best way to study, along with doing past AIME tests? What formulas should I know? What problem-solving strategies should I utilize to tackle 10+ problems?

All I do is do past AMC and AIME contests.

Work on Alcumus!! It maybe out of your level (meaning that your too good for it), but it really helps! It may help you build confidence! Alcumus could be the basics for you since you're already in 10th grade? But if there is a fundamental that you have not learned, Alcumus is there for you. You should learn a thing or two on the way! :D

I believe you'll score higher if you are not so crazy about USAMO.

[quote="funtwo"]know many, many inequalities. although that's more important for usamo because it's a good proof tool[/quote]This is terrible advice. Classical inequalities almost never show up on USAMO (and even less often on AIME), so don't worry about them. You should be familiar with the basics (PM-AM-GM-HM, Cauchy, Rearrangement/Chebyshev), but other than that you'll be fine.

A 209=8*10+129 index for an underclassmen will almost definitely qualify you. But to improve your score, [b]the most important thing is to do past tests[/b]. I can't stress that enough. Do them under conditions similar to testing (timed, primarily). When you're done, read all the solutions. Even to the problems you solved. You might have missed something.

There are no all-important formulas to memorize; 90% of your skill will come from doing problems. They don't need to be AIME problems - as long as you are doing ones of sufficient difficulty, you will learn and be better in the future. Nothing you do will improve your score really fast. A good portion of competition math, especially at the short-answer levels, is pattern recognition. If you've seen a similar problem before, you can generally find the solution very quickly. This happened to me on AMC 12A #25 this year; I'd seen a similar use of the tangent-sum identity before, and so I solved the problem in maybe 2 or 3 minutes maximum.

[quote]I've tried reading the last few chapters of AoPS Volume 2 and I really, really, really, really, really suck at graph theory. I totally do not understand it at all! Do I have to know graph theory for AIME or USAMO? :maybe:[/quote]Graph theory is simply a way of representing certain situations. It essentially never shows up on AIME, and you don't *have* to know it for USAMO. However, it will help you make your proofs more rigorous and more clear.

Thanks for the advice. I also have a few more questions about whether to know these well:

Stewart's Theorem
Homothecy
Inversion
Diophantine Equations
Block Walking
Recursion

I just learned Stewart's Theorem today from 2005 AIME I and I totally don't get the concept behind it, how it works, or how to remember it easily. Also, I don't know if it'd be rare that it'll erupt on a problem #13, 14, or 15.

The only use of Stewart's that I've ever seen is to find the length of an angle bisector. Homothecy shows up a lot in advanced geometry. Inversion doesn't; there is usually a less high-tech solution if you find an inversive one. Diophantine equations show up plenty as well. Block walking is occasionally useful, and easy enough to master that it's not too difficult to learn. Recursion shows up often - there was at least one problem on last year's AMC/AIMEs that required recursion and generally there is one somewhere. It's a great tool when you're dealing with counting possible strings of data (be it a sequence of letters with restrictions or even a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=750082]"spacy set"[/url]).

Wow...that really helped me reduce the amount of anxiety and stress I usually had on AIME's...thanks! :D

[quote="gauss1181"][quote="suma_milli"]Er... I'm pretty sure guass is in 8th grade..

I've heard mu alpha theta is a good resource :D[/quote]
Lol, yes I am in 8th grade. All i want to do is make USAMO--and my life will be fulfilled. jonathonchou711, how would Alcumus help? It really doesn't have AIME problems, and most of the problems are way too elementary.

I've tried reading the last few chapters of AoPS Volume 2 and I really, really, really, really, really suck at graph theory. I totally do not understand it at all! Do I have to know graph theory for AIME or USAMO? :maybe:[/quote]
what do you mean tried?
The NT section:
just know FLT, wilsons, primitive roots, the phi function
D.E. section:
know how to solve the linear one
Pells Equation: find the convergents congruent to 0 modulo the period of $ \sqrt{d}$, also know that all the solutions can be generated from the smallest one


What is the number of units in the distance from the origin to the point $(-5, 12)$ in a coordinate system?

[hide]$5^2 + 12^2 = 13^2$

the distance is 13 units.[/hide]

[quote="majidou"]distance = (5/(sin(atan(5/12)))   =   13[/quote]

Whoa! YOu didn't need that trig! It's not appropriate for this forum.

Just use the distance formula to get 13.

By the way, this is a pythagorean triple.

If you had explained sin and atan in your post, it would have been fine. But when you think trig on a MC problem, try to find a non-trig way. Trig's fine for checking or presenting another method of solving(with explanation of the functions you use!), but always show a non-trig method.

[quote="Treething"][hide]
5^2 + (-12)^2 = 13^2

[b]13[/b]

or maybe it's -5 and 12, oh well, too lazy to scroll down
[/hide][/quote]

lol it wouldn't have made a difference because the square of a real number is always postitive.  :lol:

[hide="solution"]$5^2+12^2=13^2$
                        the answer is [b]13[/b].[/hide]
lol...postitive :D

[quote="airelemental135"][quote="majidou"]distance = (5/(sin(atan(5/12)))   =   13[/quote]

What is the general form for this distance formula? I am just curious.[/quote]

The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is:

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.


algebra
polynomial
trigonometry
function
real analysis
real analysis unsolved

Show that there is no nonconstant polynomial $ P(u,v)$ in two variables (with real coefficients) such that $ P(x,\cos x)=0$ holds for all real $ x$.

We make repeated use of the fact that a real polynomial in a single variable must either be identically 0 or have only finitely many zeros.

Fix k between -1 and 1.  Then P(u,k) is 0 for infinitely many values of u (all those for which cos(u)=k).  But P(u,k), considered as a function of u, is a polynomial in u, and hence must be identically 0.  Thus P(u,v) is 0 for all -1<v<1.

Let q be any fixed real number.  Then P(q,v) is 0 for infinitely many values of v.  But P(q,v), considered as a function in v, is a polynomial, and hence must be identically 0.  The result follows.

Here is my approach:
After writing the original polynomial in decreasing powers of $ x$,we get the first term in this form,say,$ Ax^{n}f(x)$,where $ A$ is a nonzero constant and $ f(x)$ is a $ 2\pi$-periodic function which is not constantly zero.Then we change $ x$ into $ x+2\pi$ and make a major substraction:
$ \begin{gathered}A(x+2\pi)^{n}f(x+2\pi)-Ax^{n}f(x)=A(x+2\pi)^{n}f(x)-Ax^{n}f(x)\\=A(2\pi n x^{n-1}+\cdot\cdot\cdot)f(x) \end{gathered}$.
Hence it's obvious that if we repeat this process,this term will finally become $ A'f(x)$,where $ A'$ is a nonzero constant.It is exactly what the original polynomial is reduced to and has to be zero.Thus the fact that $ f(x)$ is not constantly zero leads to a contradiction.


We're trying to get a sense for what percentage of our users are on a high-speed connection.  We're considering alternative technologies for our classroom, but some would only be notably better for people with high speed connections.

I have one of those Bright House modems that is always on.

It's high speed.  :P

EDIT: What kind of changes to the class room are you suggesting?

We just got DSL, and I noted that my class-speed was very much increased. To the point that I usually instead of occasionally saw the LaTeXed problem before the replys.

[quote="eryaman"]I would expect that a majority of American users are on high-speed, not sure about international ones though.[/quote]I think that it depends where you are (ie in the country or in a city). For example most of the people in Bucharest using the site are on high-speed (DSL or cable and some even Fiber Optics :D) however other residents from the country are most likely using dial-up because the services are not as broad there.

i have dsl. it is really good in my opinion.

[quote="Danbert"]Whoo... dial-up.  Allegedly 56Kbps; the fastest I've ever seen it connect is 26.0Kbps--once.  Usually around 21.6Kbps; download speed averages 3KB/s.
DSL, cable, etc are not offered where I live.   :([/quote]

Been there.  :(  We just got cable finally and it rocks.  I used dialup with speeds like yours (and worse) for many years.   I hope cable or DSL comes to your area soon.

For Richard, if anyone uses satellite, they will probably have a classroom experience more similar to a dialup user than a DSL/cable user due to latency issues.   Satellite is ok for big uploads/downloads, but there's a delay for each message (as it goes up to the satellite, then back to earth) that is quite noticeable when sending/receiving multiple small messages like one would in the classroom.  Images would probably load in a more reasonable time frame though.  So, depending what the "other high speed" people are using, they may not get a high-speed classroom experience anyhow.

There was a poll about this before. (See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47822]here[/url]) :)

I have dial-up. I wish I had a high-speed connection instead... :(

I have cable.

Roadrunner! beep beep!  :D

We have Time Warner Cable (using Earthlink ISP through those cables). We live too far from our local phone exchange to get DSL. I feel that AoPS is one of the slowest-responding sites we use.

Dial-up. We were going to get DSL but apparently it's not available where I live.

We used to have cable, but we moved... now I use 26k dial-up [waits for emoticons to fully load] :surrender:  :help:  :noo:

[quote="IntrepidMath"]i have cable that claims 256k, but on AoPS, each page takes about 10-15 seconds.[/quote]That's too much. I have a slow cable connection (16 kB/s) and have to cross the Ocean to get to the server and still all the pages load in 2 seconds maximum. Basically after the first page all the images are cached on your computer, so you do not (should not I mean :) ) download them again ...

For those who have trouble loading the AoPS page, can you tell us what sort of connection you have, and what sort of times it takes for the page to load?  I use cable at work and satellite at home, and the site is roughly on par with other high-information sites (amazon, yahoo, etc).

Also, please let us know if it is the Forum only that is slow, or the whole site (the forum obviously calls for much more infrastructure than the rest of the site).

To you dial-up users - what's the AoPS experience like (both in the forum and in the classroom)?

I was on dialup until our phone line died, and the site was slow to load, but so was everything else.  (I live so far from main phone lines that the phone guy claimed he was surprised our phone line ever worked.)  Now I have satellite, and like mathmom notes, stuff coming in is fast, but stuff going out is still slow.  Any application that has lots of back and forth is therefore still dial-uppy.

I have dial-up, and I find the pages fine.  Sometimes the pages take a little while to load, but about the same as other websites, and in any event, we've never had anything else.

I am not sure, but sometimes it seems that in the classes I get stuff a little slower.  Otherwise, the forums are ok, just takes a little longer on the other computer which has a slower graphic card.

I have cable, and if I ever had to go down to dialup, I would commit suicide with a dialup modem. Such is the life of an FPSer. :D  :gleam:

well, I used to have Dial up, but now i have DSL for one computer. On my mother's lap top, it has wireless internet from the DSL.

The initial page takes about 45 seconds to load; later pages take around 10 on a good day.
I've actually never used the Classroom, so I can't tell you what it's like for me...

I have cable and the longest one page has ever taken to load was like 10 seconds

I have DSL, and it rocks bigtime.
   Ashwath

When I had dialup, AoPS ran very slowly.  But now I have DSL (I moved) and it works fine.  As for speed, the first page on AoPS takes about 8-10 seconds to get and about 1 to load after it downloads from the server.  Then, about 3 or 4 seconds per page.  (I also live across the continent)

Dial-up, no cable tv, and no cell phone ... and my dad's a computer science professor :) . (yes; we live in a good-sized city.)

But the dial-up is actually really fast--the home page loads immediately; the forum takes about half a minute. The only problems I ever have are LaTeX--after a certain amount (which is completely variable) in one post, it goes blank. Sometimes nothing shows up, usually everything shows up. The other problem is pdf files (but this is true for every site); I think I can speak for all dial-up users in saying that WAY too many files are pdf. They take anywhere from 1 to 10 minutes to load.

i have adsl
1mbps  :huh:

the first page on aops can be completed in 5-6 seconds

Well, my cousin has blazing fast dialup. It seriously loads faster on his computer than on my home computer (comcast cable), except downloads at his house are insanely slow :P .

AoPS loads very quickly for me.

Update: I now have FiOS, at 15 mbps, which is almost 2 megabytes a second. Unless I am capped by the server, I get most web pages instantly.

You guys can try Google Web Accelerator. When viewing a page, it automatically stores certain information on your computer so that you can view it more quickly next time.

Of course, it won't work for classrooms, but it did reduce the 4 seconds I needed to see the AoPS home page to about 1.5 seconds.  :)

[quote="eryaman"]I use DSL.  I would expect that a majority of American users are on high-speed, not sure about international ones though.[/quote]

Thing is that USA is the worst-connected industrialized country. Somalia has better connectivity then the US. Rural Louisiana only got phone service a few years ago.

i use comcast cable for internet and TV.
its usually very fast. 11.0 mbps

[quote="LordoftheMorons"]Verizion FIOS.

Annoyingly enough, my laptop's internet is MUCH slower than my pc's.[/quote]

Hmm...do you use wireless (802.11b maybe?) for your laptop?

I got a high-speed one. I open AoPS in about 4 seconds.


geometry
3D geometry
number theory unsolved
number theory

How to prove this theorem: No cube can be written as a sum of two coprime n-th powers if n ≥ 3. I read from Wikipedia that this follows from Taniyama-Shimura theorem, but I don't see the connection between these two theorems.

What is the difficulty of this problem?
  And do you have any clue? :?


Where can I find books&materials about solid geometry? In detail, are there any courses about [b]axiomatic Euclidean [/b]geometry and [b]analytic[/b] geometry [b]in three-dimension[/b]? If not, how can we solve problems about it? I've seen solutions using "vectors" , but it seems like a very advanced concept...

any geometry book should have it
I think there's an AoPS one too
even AoPS vol 1 and 2 haveit

yeah..
i THINK theres an AoPS book called [u]Introduction to Geometry[/u]
and Volume 1 has a lot about circles, triangles + trig, quadrilateral, and more...
i have volume 2, but i haven't looked at that yet...
(partially because i'm not at that level... xD, i WANT to take precalc really badly, but the highest for 8th graders at my school is geometry, and we just learned how to do two-column proofs.... it was SO annoying and no one in the class got it except for three people. (one guy, me, and my best friend))...

getting back to topic,
so AoPS has a lot of decent books with geometry...


[quote="Interval"]If a is < or = to b, and c < 0, show that ac is > or = bc.

I want a step-by-step easy explanation.[/quote]

Okay, just multiply both sides by c, and we get

$ac \le bc$. But c is negative, so the sign switches

Hence $ac \ge bc$

This is what I mean.  I like easy explantions.

Can you do the same for the previous question?


Duality [tex]a=x+y , b=y+z , c=z+x[/tex]  works   fine also

[quote="kunny"]Let $a,b,c$ be the lengths of sides of a triangle and let $a+b+c=2p$.
Prove the following inequality.

\[\frac{a}{p-a}+\frac{b}{p-b}+\frac{c}{p-c}\geq 6\][/quote]

  Come on guys,this inequality is trivial and it's equivalent with this well-known which follows directly from AM-GM.

                      $\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\geq 3$.

  i even posted a better one: $\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\geq\frac{1}{2}(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}) $ which is very easy too. ;)

[quote="shobber"]I got it.
$\frac{a}{p-a}+1+\frac{b}{p-b}+1+\frac{c}{p-c}+1-3$

$=\frac{p}{p-a}+\frac{p}{p-b}+\frac{p}{p-c}-3$

Use AM-HM:
$\frac{3}{\frac{p}{p-a}+\frac{p}{p-b}+\frac{p}{p-c}} \leq \frac{\frac{3p-a-b-c}{p}}{3}=\frac{\frac{3p-2p}{p}}{3}=\frac{1}{3}$

Hence $\frac{p}{p-a}+\frac{p}{p-b}+\frac{p}{p-c} \geq 9$

Hence $LHS \geq 6$.[/quote]

Sorry for question but May you explain what is AM-HM?


When someone else posts a long LaTeX formula, I will click the LaTeX to get a little pop-up window that tells me what the LaTeX code is.  Lately, however, the code doesn't come up.  Instead, it gives a message

[code]function Copy() { ie=
[/code]

I can also quote the user to find the LaTeX code, but the clicking thing was much easier.  Is this a bug related to the update?

Browser: Mozilla Firefox, Latest (non-Beta) version

I can confirm this bug (also using Firefox).

A workaround: You can select and copy that section of the post, then paste into a text editor window.

I'll bet I forgot to copy over some scripts.  I'll take a look tomorrow when I'm in the office.

Another "bug": when I clicked on the headline on earlier days, a window popped up with the URL in it; now it's just a link to the topic...
The old one was much better (and the new one makes no sense: why linking a topic to itself).

Guys guys guys. It's not bugs (the topic link not poping up an url), it's my yet limited capacity of restoring the old tweaks we had on the forum. I am slowly doing them all.

As for the $\LaTeX$ bug read: http://www.mathlinks.ro/Forum/viewtopic.php?t=97542 to view to currently found bugs and the ones that we are working on :)

The LaTeX click-and-get-the-code is now fixed (at least on IE).

It appears to work on firefox 1.0.7 as well.


Who will win ATP World Tour?

See the poll:

Results so far:
Andy Murray (4) def. Juan Martin Del Potro (5) 6-3, 3-6, 6-2
Roger Federer (1) def. Fernando Verdasco (8) 4-6, 7-5, 6-1

Ah!! Roddick (6) pulls out because of an INJURY

Soon:
Rafeal Nadal (2) vs. Robin Soderling (9)
Novak Djokovic (3) vs. Nikolay Davydenko (7)

Current result: Murray-Federer 6:3 2:4 and Murray has 15:30 on Fed's serve... BTW, Soderling and Djokovic won...

Djokovic wins!

While I don't agree that Djokovic has a clear win, I agree he is probably the challenge to Federer.

Wow Nadal lost?? He definitely has to win both other games to make it to SF.

So how does he win??

You know Djokovic will be tough to Nadal...

Nadal is out... lost to Davydenko 1-6, 6-7

Soderling has 4-0 sets, and probably will be @ semi finals

Federer to play Del Potro tomorrow
Murray vs. Verdasco

...
Good bye Nadal and hello Federer (and Soderling)

[quote="Bugi"]Soderling is definitely in semis, just it's 1st or 2nd... If Davidenko beats him, Djokovic is out...[/quote]

I'm looking to Djokovic to make, don't know how hard Nadal will try...

WOW What HAPPENED TO NADAL THIS YEAR

and Soderling?

Whoopee, Murray - Verdasco 6-4, 6-7, 6-6 (3-2)

[quote="pinkmuskrat"]Federer has won 2009 ATP World Tour.

Del Potro faces Davydenko in the championship of the ATP Finals Tournament.[/quote]

Federer has lost 2009-2010 world tour finals.

GRRRR... Davydenko up to #6

My opinion:

Davydenko played just awesome. He was slightly injured, but won. Wow.

Del Potro was tired... I don't like tournaments which don't organize (for example) semifinals at the same day, because players who play earlier have advantage... But he still played the tournament very good

Federer... He isn't the best on the court at the moment. But he is in the top...

Soderling played awesome also, an underdog almost got into finals...

Djokovic was tired... And lost against Soderling stupidly... The first set was the decider, he was so angry afterwards...

Nadal is totally out of form... He isn't for no. 2 anymore

Davydenko had injury???

He definitely didn't look like so... :lol:

Do you think that Nadal will come back sometime? He did [i]ok[/i] at US open this year...

EDIT: and that is the best he's ever done at US open (except for 2008: SF)


To practically everything form sumset theory  :wink:
The most canonical one is http://www.mathlinks.ro/viewtopic.php?t=14797 .


Simplify $ \Big(\sqrt[6]{27} \minus{} \sqrt{6 \frac{3}{4} }\Big)^2$

[quote="MCNational!"]Simplify $ \Big(\sqrt [6]{27} - \sqrt {6 \frac {3}{4} }\Big)^2$[/quote]

Hm, these problems look familiar. Anyway:

$ \left(\sqrt[6]{27} - \sqrt{6\frac{3}{4}\right)^2 \implies \left(3^{\frac{1}{2}} - \frac{3\sqrt{3}}{2}\right)^2 \implies \left(3^{\frac{1}{2}}\right)^2 - 2\eft(3^{\frac{1}{2}}\right)\left(\frac{3\sqrt{3}}{2}\right) + \left(\frac{3\sqrt{3}}{2}\right)^2 \implies \boxed{\frac{3}{4}}}$

What do you mean, $ 6\frac{3}{4} \equal{} \frac{9}{2}$?

Oh, I see what you mean. In general $ a\frac{b}{c}$ indicates $ a \plus{} \frac{b}{c}$.


I was wondering if someone could help with this problem.

Let m be a positive integer, and let $f(m)$ be the smallest value of $n$ for which the following statement is true: given any set of $n$ integers, it is always possible to find a subset of $m$ integers whose sum is divisible by $m$.

Determine $f(m)$.

Thanks

the answer is $2m-1$ and i'm sure it has been posted here before, somebody will probably give you an old link... this is a very beautiful problem

OK, but I was wondering what the key methods you had to use to get that f(m)=2m-1.

And, what should I search for if it has already been posted before, for I typed in the whole question, and only mine showed up.

maybe this one: http://www.mathlinks.ro/viewtopic.php?search_id=985160729&t=31191  :P

Oh, thanks.

I just didn't rearrange the words in the question to find it on the search.


real analysis
advanced fields
advanced fields unsolved

Let $\mathcal{B}$ be a Borel $\sigma$-algebra on $\mathbb{R}$.
Is there an example of set A, such that:
1. $A\notin \mathcal{B}$
2. There is $B$, $A\subset B$, $B\in\mathcal{B}$, $\lambda(B)=0$, where $\lambda$ is Lebesgue measure??

"Not Borel" is not the same as "not measurable"; your $A$ has Lebesgue measure zero.

A totally nonconstructive argument: there are more sets than there are Borel sets, in terms of cardinality. The cardinality of the Borel algebra is only $c$ (the cardinality of the reals). There are $2^{c}$ subsets of the Cantor set (or your favorite measure-zero uncountable set), so some subset is not Borel.

Or, for the vague illusion of it being more of a "construction":

Let $A$ be your favorite example of of a nonmeasurable set in $[0,1].$ Let $f: [0,1]\mapsto[0,1]$ be the usual Cantor-Lebesgue "devil's staircase" function. Let $B=f^{-1}(A)\cap K,$ where $K$ is the Cantor set. $B$ will be a non-Borel set of measure zero.

Thanks a lot!

And how can i prove that cardinaltiy of Borel $\sigma$-algebra is only $c$?


limit
inequalities
calculus
calculus computations

Let $ a$ and $ b$ are positive numbers. $ {a_n}$ and $ {b_n}$ are sequences defined by the following way:
$ a_n \equal{} \begin{cases}a, & n \equal{} 0; \\
\frac {a_{n \minus{} 1} \plus{} b_{n \minus{} 1}}{2}, & n\geq1\end{cases}$ and $ b_n \equal{} \begin{cases}b, & n \equal{} 0; \\
\sqrt {a_{n \minus{} 1}b_{n \minus{} 1}} & ,n\geq1\end{cases}$.
a) Prove that these sequences are convergent.
b) If $ \alpha$ and $ \beta$ are their limits find:
$ \lim_{n\to\infty }(a_1 \plus{} a_2 \plus{} ... \plus{} a_n \minus{} n\alpha)$ and $ \lim_{n \to \infty }(b_1 \plus{} b_2 \plus{} ... \plus{} b_n \minus{} n\beta)$

I'm sorry for the bad tex ... if it is possible for some moderator to edit this I'll be very happy. I think the problem is interesting.

[color=green]Moderator says: Done; check that the result is what you meant :)[/color]

a) By AM-GM inequality, $ b_n\leq a_n$ for all $ n\in\mathbb{N}$. Therefore, $ b_{n \minus{} 1}\leq \sqrt {a_{n \minus{} 1}b_{n \minus{} 1}} \equal{} b_n$ for all $ n\geq 2$. Since $ (b_n)_n$ is nondecreasing (after finitely many elements), it suffices to prove that $ a_n$ is convergent. However, $ a_n \equal{} \frac {a_{n \minus{} 1} \plus{} b_{n \minus{} 1}}{2}\leq a_{n \minus{} 1}$ since $ b_{n \minus{} 1}\leq a_{n \minus{} 1}$ for $ n\geq 2$. So, $ (a_n)_n$ is nonincreasing (after finitely many elements) and since all the numbers are positive, $ (a_n)_n$ is convergent.

Just a note about b): $ \beta$ is nowhere to be found in the problem, so I somewhat doubt the problem goes exactly like that.

To mornik:
$ \beta$ is $ b_n$'s limit.
I'm sorry in b) there was an error in the second limit. $ b_n$ is also convergent with limit $ \beta$ it is a part of a) condition. I have a solution for a)... Leonard Euler's works describes some similar sequences. In my source b) part is an open problem.

To the moderator - yes now it is exactly what I wanted to say. I'm sorry I'm so lazy. I believe it is a good problem. Thank you for the excellent work.

I found a way to solve the problem and many interesting properties of AGM sequences.

Can you tell or (which is better) show us this method ?

This is called Arithmetic-Geometric Mean.
See http://mathworld.wolfram.com/Arithmetic-GeometricMean.html and http://en.wikipedia.org/wiki/Arithmetic-geometric_mean

These are excellent materials but how it helps us for b)?

:-) it seems this problem to be nice. I'm surprised no one solved it completely. Is it so hard?!


linear algebra
matrix
probability
vector

I have a 9x9 matrix to be specific this is a bond ratings transition matrix it shows the probability that a bond would move from one rating to other . the first column is for AAA rated bonds, second AA, third A, BBB ,BB,B,CCC,CC, and the last column is default.

I am interested in probabilities of default for AA, A, and BBB rated bonds which is the second, third, and forth rows in the last column. This matrix represents the 1 year probabilities. In order to calculate the 2 years probabilities, I multiply the matrix by itself and the last column is the 2 year default probabilities, for 3 years i multiply the matrix by itself 3 times etc..

I am running a program where i need for each rating to multiply this matrix around 400,000 times (I need to do so because the time is changing continuously and consequently i need to calculate the corresponding default probabilities for that period) consequently I need to spend 6 month to obtain my results. I am wondering if there is a way to reduce the computations. Since i need only the AA,A,and bbb ratings ( 2nd column is for AA bonds,third is for A bonds, forth is for BBB bonds, and last column is the default column where the 2nd,3rd and forth observations in that column are what I need)I am wondering if there is a way to reduce the matrix to a 4 by 4 or something else to make the computations much quicker. To rephrase the question: is there a way to obtain the n years default probabilities for the AA,A,and BBB bonds by reducing the size of the 9 by 9 matrix? U could save my life !!!

If you can diagonalise the matrix $A$ by a change of variable: that is, find $B$ such that $B^{-1}AB = D$ is diagonal, then $A^n = BD^nB^{-1}$ and computing $D^n$ is a lot faster than computing $A^n$.

I am extremely glad that a solution exists although I need some guidance in what u have proposed.
the matrix % is the following:(I hope it would be readable . If not please let me know how to post it or send it by emaqil if possible.

94.7	5	0.3	0	0	0	0	0	0
0.7	88.4	10.4	0.4	0.1	0	0	0	0
0	1.5	90.9	6.7	0.7	0.2	0	0	0
0	0.1	3.1	88.7	6.6	1.1	0.2	0	0.2
0	0	0.3	3.9	86.2	7.7	0.3	0	1.6
0	0	0.2	0.5	5.8	83.6	2.1	0.5	7.4
0	0	0	0	1.2	4.9	63.7	4.7	25.6
0	0	0	0	0	10.4	22.3	57.7	9.6
0	0	0	0	0	0	0	0	1

so, how to apply what u have proposed to it?

Ok, maybe I am not the right person to help here, but if you want to find a $n^{th}$ power of something, you could find it in $log(n)\cdot C$, where C is the time to multiply one of those thing (for example matrices ;) )
Just use this (some kind of pseudocode in C)
matrix power(int n,matrix X)
{
if(n==1) return X;
if(n%2) return (power(2,power(n/2,X))*X);
return (power(2,power(n/2,X)));
}
I hope it's basicly clear  :?

Log(n).C is not giving the right results.thx for ur contribution

Firstly, I'm guessing that the bottom right-hand corner of your matrix is 100 not 1 (as they are all percentages).
The eigenvalues of the matrix are 97.73119459, 95.18721346, 92.30768210, 86.95186123, 83.63715841, 78.01812593, 70.02656653, 50.04019774, 100.0 as percentages, so the entries on the matrix in diagonal form will be these eigenvalues.  If you raise that to the power 400,000, the entries less than 1 will become negligible and the only column worth worrying about will be the eigenvector corrsponding to the eigenvalue 1.  This column is 1.008383160, 1.008382854, 1.008384981, 1.008334686, 1.008348020, 1.010215239, 1.005081765, 1.005190588, 1 and A^400000 will be zeroes except for that vector as your last column.

If you use Maple it can do this matrix exponentiation in under a second.

Thx a lot rgep for your reply and sorry fpr the delay in my replay.
Actually, u misunderstood me. It is not to the power 400,000. I have 400,000 dates to calculate the default probability for and that is why the program is slow.
Anyway, I managed by playing with the codes to reduce the time spent significantly. Technically instead of having them in the optimization problem , I calculate them in advance and I save many calculations steps.
Thx a lot for your replies and contribution.
Take care
JF


[i]Fiind the digits $ a,b,c,d$ so that $ c^{\overline{ab}}\equal{}\overline{abcd}$ and $ d\equal{}2c$.[/i]
[b]Claudiu Mîndrilă, Gazeta Matematică-B 4/2008[/b]

if c was 5 or more, then d is more than one digit.

so c<5.

$ c^{10a\plus{}b}\equal{}1000a\plus{}100b\plus{}10c\plus{}d\equal{}1000a\plus{}100b\plus{}12c$

right side of the equation is even, so left side should be, too.

so c is 2 or 4.

1) c=2
$ 2^{10a\plus{}b}\equal{}1000a\plus{}100b\plus{}24$

its true when a=1 and b=0.

so we have a=1, b=0, c=2, d=4.

2) c=4
$ 4^{10a\plus{}b}\equal{}1000a\plus{}100b\plus{}48$
$ 2^{20a\plus{}2b}\equal{}1000a\plus{}100b\plus{}48$

theres only one 4-digit power of 2 ends with 48; 2048.

then a should be 2, and b should be 0.

...but that makes $ 2^{40}\equal{}2048$, which isn't right.

so answer:

a=1, b=0, c=2, d=4.

hmmm i think thats right

[quote="Mîndrilă Claudiu"]In fact, the problem is for middle school. Let me show you my proof:
Because $ \overline{ab} \leq 10$ and $ 3^10 \equal{} 59049$, it`s obvious that $ c \equal{} 2$, because $ c > 1$. From $ d \equal{} 2c$ we get that $ d \equal{} 4$.
Now the problem is to fiind $ a,b$ so that $ 2^{\overline{ab}} \equal{} \overline{ab24}$.
First, we denote that $ 9 < \overline{ab} < 14(*)$. Then, because the last digit of $ \overline{ab24}$ is $ 4$ we denote that $ \overline{ab} \equal{} M_4 \plus{} 1(**)$.
From $ (*)$ and from $ (**)$ we reach to $ \overline{ab} \equal{} 10$.
Answer: $ a \equal{} 1, b \equal{} 0, c \equal{} 2, d \equal{} 4$.[/quote]

I think I understand it now, but it might be better in High School Basics.  Not sure.  But I do get it. :thumbup:

Own means that i`ve proposed the problem  :D .

[quote="Mîndrilă Claudiu"]In fact, the problem is for middle school. Let me show you my proof:
Because $ \overline{ab} \leq 10$ and $ 3^{10} \equal{} 59049$, it`s obvious that $ c \equal{} 2$, because $ c > 1$. From $ d \equal{} 2c$ we get that $ d \equal{} 4$.
Now the problem is to fiind $ a,b$ so that $ 2^{\overline{ab}} \equal{} \overline{ab24}$.
First, we denote that $ 9 < \overline{ab} < 14(*)$. Then, because the last digit of $ \overline{ab24}$ is $ 4$ we denote that $ \overline{ab} \equal{} M_4 \plus{} 2(**)$.
From $ (*)$ and from $ (**)$ we reach to $ \overline{ab} \equal{} 10$.
Answer: $ a \equal{} 1, b \equal{} 0, c \equal{} 2, d \equal{} 4$.[/quote]

yes cuz everyone knows how to formulate that....

and everyone knows 3^10=59049

Yes, that meens. I`ve proposed the problem in "Gazeta Matematică"  :)

I`ve proposed this problem for the $ 5^{th}$ grade. [b]mz94[/b], what`s your grade?

never mind

i didnt read -______-

nice problem, but i still think its in teh wrong place

[quote="Mîndrilă Claudiu"][i]Fiind the digits $ a,b,c,d$ so that $ c^{\overline{ab}} \equal{} \overline{abcd}$ and $ d \equal{} 2c$.[/i]
[b]Claudiu Mîndrilă, Gazeta Matematică-B 4/2008[/b][/quote]

This does NOT belong here.  Somebody please move to high school basics.


Consider $ A\equal{}
\left(
\begin{array}{cc}
1 & 1 \\
1 & 0  \end{array}
\right)$.

$ a)$ Prove that $ A^n\equal{}\left(
\begin{array}{cc}
F_{n\plus{}1} & F_n \\
F_n & F_{n\minus{}1}  \end{array}
\right)$, for any $ n \ge 1$, where $ (F_n)_{n \ge 0}$ is the sequence of Fibonacci defined by $ F_0\equal{}0, F_1\equal{}1, F_{n\plus{}2}\equal{}F_{n\plus{}1}\plus{}F_n.$

$ b)$ Prove that $ F_{m\plus{}n\plus{}1}\equal{}F_{n\plus{}1} \cdot F_{m\plus{}1}\plus{}F_n \cdot F_m,$ for all $ m \ge 0 , n \ge 0$.

a) By induction
We have $ F_2 \equal{} 1$ With $ n \equal{} 1$ we have $ A \equal{} \begin{bmatrix}F_2 & F_1 \\
F_1 & F_0\end{bmatrix}$ is true
Suppose correct to $ n$ mean $ A^n \equal{} \begin{bmatrix}F_{n \plus{} 1} & F_{n} \\
F_{n} & F_{n \minus{} 1}\end{bmatrix}$
Then  $ A^{n \plus{} 1} \equal{} A^{n}A \equal{} \begin{bmatrix}F_{n \plus{} 1} & F_{n} \\
F_{n} & F_{n \minus{} 1}\end{bmatrix}\begin{bmatrix}1 & 1 \\
1 & 0\end{bmatrix} \equal{} \begin{bmatrix}F_{n \plus{} 1} \plus{} F_n & F_{n} \plus{} F_{n \minus{} 1} \\
F_{n} \plus{} F_{n \minus{} 1} & F_{n}\end{bmatrix} \equal{} \begin{bmatrix}F_{n \plus{} 2} & F_{n \plus{} 1} \\
F_{n \plus{} 1} & F_{n}\end{bmatrix}$
b) From $ A^{m \plus{} n} \equal{} A^mA^n$ or $ \begin{bmatrix}F_{n \plus{} m \plus{} 1} & ... \\
... & ...\end{bmatrix} \equal{} \begin{bmatrix}F_{m \plus{} 1} & F_{m} \\
F_{m} & F_{m \minus{} 1}\end{bmatrix}\begin{bmatrix}F_{n \plus{} 1} & F_{n} \\
F_{n} & F_{n \minus{} 1}\end{bmatrix} \equal{} \begin{bmatrix}F_{n \plus{} 1}F_{m \plus{} 1} \plus{} F_nF_m & ... \\
... & ...\end{bmatrix}$
 Hence $ F_{m \plus{} n \plus{} 1} \equal{} F_{n \plus{} 1}\cdot F_{m \plus{} 1} \plus{} F_{n}\cdot F_{m},$

Prove that $ F_{n\plus{}1}F_{n\minus{}1}\minus{}F_n^2\equal{}(\minus{}1)^n.$

[quote="Kent Merryfield"]Prove that $ F_{n \plus{} 1}F_{n \minus{} 1} \minus{} F_n^2 \equal{} ( \minus{} 1)^n.$[/quote]
 We have $ \det(A)\equal{}\minus{}1$ then  $ \det(A^{n})\equal{}\det\begin{bmatrix}F_{n\plus{}1}& F_{n}\\ F_{n}& F_{n\minus{}1}\end{bmatrix}\equal{} F_{n \plus{} 1}F_{n \minus{} 1} \minus{} F_n^2$
 Implies $ F_{n \plus{} 1}F_{n \minus{} 1} \minus{} F_n^2 \equal{} ( \minus{} 1)^n.$

Can see class general of sequence   of Fibonacci is $ A^2\equal{}A\plus{}I_2$ ,For $ A \in M_n{(\mathbb{C})}$ (see from equation $ \lambda^2\minus{}\lambda\minus{}1\equal{}0$ of $ F_n$)
 Prove eassy $ A^n\equal{}a_nA\plus{}b_nI_2$ then $ A^{n\plus{}1}\equal{}a_{n\plus{}1}A\plus{}b_{n\plus{}1}I_2\equal{}A^nA\equal{}a_nA^2\plus{}b_nA\equal{}(a_n\plus{}b_n)A\plus{}a_nI_2$ then implies $ a_{n\plus{}1}\equal{}a_n\plus{}b_n,b_{n\plus{}1}\equal{}a_n$
Then $ a_{n\plus{}1}\equal{}a_n\plus{}a_{n\minus{}1}$ , it is sequence   of Fibonacci $ A^n\equal{}F_nA\plus{}F_{n\minus{}1}I_2$
 Let $ A\equal{}\begin{bmatrix}a&b\\c&d\end{bmatrix}$ such that $ Tr(A)\equal{}a\plus{}d\equal{}1,\det(A)\equal{}ad\minus{}bc\equal{}\minus{}1$
 Here this problem $ a\equal{}1,d\equal{}0,b\equal{}c\equal{}1$ ,can many matrix  example $ a\equal{}2,d\equal{}\minus{}1,b\equal{}\minus{}1,c\equal{}1$ then $ A^n\equal{}F_nA\plus{}F_{n\minus{}1}I_2\equal{}\equal{}\begin{bmatrix}2F_{n}\plus{}F_{n\minus{}1}& \minus{}F_{n}\\ F_{n}& \minus{}F_{n}\plus{}F_{n\minus{}1}\end{bmatrix}\equal{}\equal{}\begin{bmatrix}F_{n\plus{}2}& \minus{}F_{n}\\ F_{n}&\minus{} F_{n\minus{}2}\end{bmatrix}$
 Here we again  $ F_{n\plus{}2}F_{n\minus{}2}\minus{}F_n^2\equal{}(\minus{}1)^{n\plus{}1}$ can many matrix from sequence ......  :)

Let $ F_{n \plus{} 1}F_{n \minus{} 1} \minus{} F_{n}^{2} \equal{} ( \minus{} 1)^{n}$ ,$ \forall n \in \mathbb{N}$
  $ F_n$ is the sequence of Fibonacci  with $ F_0\equal{}1,F_1\equal{}1$
Don't answer by it is true sorry


The sum of two numbers is 12 and their product is 35. What is
the positive difference between the numbers?

The numbers are $ 5$ and $ 7$.
$ 7 \minus{} 5 \equal{} \boxed{2}$.

Textangle solved that using brute force, but here's a different way.

$ x \plus{} y \equal{} 12$

$ xy \equal{} 35$

Rewrite the first equation as $ x \equal{} 12 \minus{} y$.

$ y(12 \minus{} y) \equal{} 35$

$ 12y \minus{} y^2 \equal{} 35$

$ \minus{} y^2 \plus{} 12y \minus{} 35 \equal{} 0$

We have a quadratic. (Note: If you do not know how to solve quadratics, click the link in my signature, and read the entries titled "Quadratics Part 1" through "Quadratics Part 5".)

Factoring yields:

$ (y \minus{} 5)(y \minus{} 7) \equal{} 35$

Thus our two solutions are $ 5$ and $ 7$. This means that $ x$ is either $ 7$ or $ 5$, depending what $ y$ is. However, it doesn't really matter which one is which - one is 7, and one is 5, so $ 7 \minus{} 5 \equal{} \boxed{2}$.

Adding on to what isabella and tex said, you should know that the factors of 35 are only 1, 5, 7, and 35.


435 mL of a 0.100 M HNO3 solution is mixed with 235 mL of 0.175 M Ca(OH)2. Which of the following statements are not true?

The solution is acidic
Water is a product
Some HNO3 remains unreacted
The solution is alkaline
Some Ca(OH)2 remains unreacted

$ 2\, \ce{HNO3}(aq) \plus{} \ce{Ca(HO)2}(aq) \longrightarrow \ce{Ca(NO3)2}(aq) \plus{} 2\, \ce{H2O}(l)$

$ n(" \ce{HNO3} ") \equal{} 4.35 \times 10^{\minus{}2}$

$ n(" \ce{Ca(HO)2} ") \equal{} 4.11 \times 10^{\minus{}2}$

from where we can see that the slacked lime is in excess. Therefore, the false statements are "The solution is acidic" and "Some HNO3 remains unreacted".


inequalities
LaTeX
rearrangement inequality
inequalities solved

Today, in the attempt of solving some old MathLinks problems, I thought about the [b]2nd Problem of the MathLinks Contest 1st Edition[/b]. It seems that the second of [url=http://www.mathlinks.ro/Forum/contest/1st%20Edition/1st%20round%20sol.pdf]the two proposed solutions[/url] proves a slightly different inequality than the one proposed in the problem. Hence, we really get two different inequalities and one solution for each of them!

Actually, the inequality intended in the problem was

[b](1)[/b]    $\frac{a+b}{a+c} + \frac{b+c}{b+a} + \frac{c+a}{c+b} \leq \frac bc + \frac ca +\frac ab$

for three arbitrary positive reals a, b, c. This was proven in the first solution.

But the second solution proved the inequality

[b](2)[/b]   $\frac{a+c}{a+b} + \frac{b+a}{b+c} + \frac{c+b}{c+a} \leq \frac bc + \frac ca +\frac ab$,

which is much simpler. Actually, the proof of [b](2)[/b] given in the second solution can be simplified: Using the notation $\sum$ for cyclic sums, the inequality [b](2)[/b] can be rewritten as

$\sum \frac{a+c}{a+b} \leq \sum \frac bc$.

With the notion S = a + b + c, this becomes

$\sum \frac{S-b}{S-c} \leq \sum \frac bc$,

or, equivalently,

$\sum \left(\frac{S-b}{S-c} - \frac bc\right) \leq 0$.

But $\frac{S-b}{S-c} - \frac bc = \frac{\left(S-b\right)c - \left(S-c\right)b}{c\left(S-c\right)} = \frac{S\left(c-b\right)}{c\left(S-c\right)}$. Hence, it remains to prove

$\sum \frac{S\left(c-b\right)}{c\left(S-c\right)} \leq 0$,    or simply    $\sum \frac{c-b}{c\left(S-c\right)} \leq 0$.

Bringing the term  $\sum \frac b{c(S-c)}$  into the right member, we get

$\sum \frac c{c\left(S-c\right)} \leq \sum \frac b{c\left(S-c\right)}$.

But this follows from the rearrangement inequality, since the sequences (a, b, c) and $\left(\frac{1}{a\left(S-a\right)},\;\frac{1}{b\left(S-b\right)},\;\frac{1}{c\left(S-c\right)}\right)$ are oppositely sorted (in fact, assume WLOG that $a\geq b$; then, it follows that $\frac 1{a(S-a)} \leq \frac 1{b(S-b)}$ because $a\left(S-a\right)\geq b\left(S-b\right)$ because $a\left(S-a\right)-b\left(S-b\right)=\left(a-b\right)\left(S-a-b\right)=\left(a-b\right)c\geq 0$). This completes the proof of [b](2)[/b].

I wonder if this is really so simple or there is some error in the above (while else did the second solution contain so much calculation?).

  Darij

ups, my bad in that solution writeup :)

PS I have modified your post to include latex tags :D it looks nicer and easier to understand :D:D

Thanks, this actually looks nicer...

I don't find the error that bad... because now we have two different inequalities proved instead of one!  :D

Anyway, I think there is a shorter solution for [b](1)[/b]:

The inequality [b](1)[/b] can be written as $ \sum\frac {c \plus{} a}{c \plus{} b}\leq\sum\frac {a}{b}$, or, equivalently, as $ \sum\frac {a}{b} \minus{} \sum\frac {c \plus{} a}{c \plus{} b}\geq 0$. But

$ \sum\frac {a}{b} \minus{} \sum\frac {c \plus{} a}{c \plus{} b} \equal{} \sum\left(\frac {a}{b} \minus{} \frac {c \plus{} a}{c \plus{} b}\right) \equal{} \sum\frac {1}{b \plus{} c}\cdot\frac {c}{b}\cdot\left(a \minus{} b\right)$,

so it remains to prove that $ \sum\frac {1}{b \plus{} c}\cdot\frac {c}{b}\cdot\left(a \minus{} b\right)\geq 0$, what I have done in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=408785#p408785]http://www.mathlinks.ro/Forum/viewtopic.php?t=47342 post #4[/url] (in a pretty short way, I believe, even if the lemma is counted).

[b]EDIT:[/b]
Another topic for problem [b](1)[/b]: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32718
Another topic for problem [b](2)[/b]: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119472

  Darij


analytic geometry
geometry
3D geometry
sphere
trigonometry
integration
calculus

Hi! As we wait for the contest poll to end, here's a simple question:

What is the moment of innertia of a thin spherical shell of mass $M$ and radius $R$? I think that it should be something like $I = \frac{2}{7}MR^2$, but when I try to find this expression (by integrating the moments of innertia of infinitesimaly thin 'rings') I always end up with something complicated and, most likely, wrong.

The coefficient should be 2/3, and for your integrating, you probably want to take into account that the "rings" have a curved outside.  I'll see about writing out a solution tomorrow... my head's all tired from today.

Well, of course -- you think of rings as completely normal objects whose moment of innertia is $mr^2$... But I don't know about the $2\over 3$ coefficient...

Using spherical coordinates $r = R, \phi \in [0, 2\pi), \theta \in [0, \pi)$ and denoting $\rho$ the distance of a point on the sphere with the coordinates $(R, \phi, \theta)$ from the z-axis, we have $\rho = R \sin \theta$. The moment of inertia of the spherical shell is

$J = \int_{r = R} \rho^2\ \text d M = \frac{M}{4\pi R^2} \int_0^{2 \pi} \int_0^\pi \rho^2 R^2 \sin \theta\ \text d \theta\ \text d\phi =$

(integrating over $\text d \phi$ and substituting $\rho = R \sin \theta$)

$= \frac{2\pi M R^4}{4\pi R^2} \int_0^\pi \sin^3 \theta\ \text d \theta = \frac{ M R^2}{2} \int_0^\pi \sin^3 \theta\ \text d \theta =$

(substituting $\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta,\ \sin^3 \theta = \frac{3 \sin \theta - \sin 3\theta}{4}$ and then $\alpha = 3 \theta,\ \text d \theta = \frac{\text d\alpha}{3}$)

$= \frac{M R^2}{8} \int_0^\pi (3 \sin \theta - \sin 3\theta)\ \text d \theta = \frac{M R^2}{8} \left (3 \int_0^\pi \sin \theta\ \text d \theta - \frac 1 3 \int_0^{3\pi} \sin \alpha\ \text d \alpha \right) =$

$= \frac{M R^2}{8} \left (3 \left[-\cos \theta \right]_0^\pi - \frac 1 3 \left[-\cos \alpha\right]_0^{3\pi}\right) = \frac{M R^2}{8} \left (3 \cdot 2 - \frac 2 3\right) = \frac 2 3 MR^2$

Great!!!

Thank you very much!

I found out where I've been wrong with my integration (incorrectly chosen $dx$ during integration which made me effectively lose one of the $\sin \theta$ multipliers in [b]yetti[/b]'s equation). I've tried it my way (integrating rings) now and also got the result to be $\frac{2}{3}MR^2$.

@[b]daveed[/b] and [b]yetti[/b]: I know it's off topic, but I hope that you'll join in when the physics contests are renewed (the problems will be posted next friday, after the poll ends).

[color=blue]And I think this is also another solution to your problem, [b]Djole[/b]

$dS=2\pi Rdx$

$dm=\sigma dS$

$dI=dm(R^2-x^2)$

$\large I=\int_{-R}^{R}dI=2\pi R\sigma\int_{-R}^{R}(R^2-x^2)dx=\frac{8}{3}\pi\sigma R^4$

Remember that $m=4\pi R^2 \sigma$

$\Longrightarrow I=\frac{2}{3}mR^2$[/color]


Let $k,m,n\in{\bf N}^*$ with $k,m<n$. Find
\[ \displaystyle\sum\limits_{\sigma\in S_n}{\rm Card\{i\in\{1,2,\ldots,n\}\;|\;|\sigma(i+m)-\sigma(i)|=k\}  }\]
where $\sigma(n+p): =\sigma(p)$, $\forall\;p\in\{1,2,\ldots,n\}$, $S_n$ is the set of ($n$-order) permutations, and ${\rm Card}\{\ldots\}$ is the cardinal of the set $\{\ldots\}$.

Prove that \[ \displaystyle\sum\limits_{\sigma\in S_n}{\rm Card}\{i\in\{1,2,\ldots,n\}\;|\;|\sigma(i+m)-\sigma(i)|=k\}=2n(n-k)(n-2)! \]


Because the shapes are similar, we can set up the proportion $ \frac {\text{original short side}}{\text{original long side}} \equal{} \frac {\text{new short side}}{\text{new long side}}$.  So we have $ \frac {x}{10} \equal{} \frac {5}{x}$.  Cross multiplying, we get $ x^2 \equal{} 50\implies x \equal{} \sqrt {50} \equal{} \boxed{5\sqrt2}$.


MATHCOUNTS
Pascal\u0027s Triangle
States

Seems much harder im probaly gonna get like a 30 lol.
SEND ME SOLUTIONS AFTER I SUBMIT

There was a slight error on problem 15 on the sprint round I meant "Tonya and Sally" not "Tonya and Tonya"

CRICKET229 got 25/30 on the sprint round!  Very fine job.  Let's give cricket a big round of applause.

Current scores:
CRICKET229-25 sprint, yet to turn in target. :10:

Heres my applause.  :clap2:  :clap2:   :clap2:

Also Maybach, why did you not post your sprint round with your original thread? Just wondering.

What is the cutoff?  I did pretty bad on the chapter.

The cutoff is still TO BE DECIDED.

And I posted the state round on a separate thread just in case people who wanted to join NOW wouldn't get led astray to this test.  Oops, and when I posted the target, I meant to post the edited sprint round.

Wait, so I can't take it until I get approval from you?

No, you can take it.  But if you don't make the cutoff(unlikely) than it would be unofficial.  But that probably won't happen :)

How do you make problems so fast Maybach? Oh and i might be sending you the chapter round this afternoon.

I've been working on the state round since this Tuesday.  The reason I can plug things out more quickly is that I don't have to deal with an intermediary.  See, Powerofpi and vallon22 need each other's approval for problems, which involves PMs, waiting, and delays.  Since this is purely sponsored by just me, I don't have to worry about that.  :lol:

I realized I made an answer key writing mistake with Pascal's Triangle, so CRICKET229's score goes up by 1.

UPDATE:
yaofan-32(18 sprint, 14 target)
CRICKET-26(PLEASE TURN IN TARGET)

CRICKET appealed and I discovered an answer key error.

NEW SCORES:
yaofan-33(also affected)
CRICKET-27(SUBMIT TARGET PLZ)

yes mi targeto willo beo turnedo ino tomorrowo

OK.  Here's the state round.  C'mon people, turn this stuff in.

OK.  This has been prematurely cancelled.  I just don't have the time and really need to get preparing for science olympiad.

OK...thanks for the great contest! It was fun! :lol:  :w00tb:

The questions seem to be a fair difficulty for MC (I think it's a bit easy though for State), but there are too many words for me to read.

EDIT: I didn't realize this was from 2009  :P

[quote=Slammers1506]This seems way to hard to be a State’s test. Nationals might be a better category[/quote]

Lol if it was made it 2009, it could have been like 2 times harder than a nationals

If this is from 2009, it's pretty obvious that it's going to be easier than 2020...

[quote=yofro]Super easy

[hide="Sol to the 'hardest' problem"]
$x^2-x-6=(x-3)(x+2)$
The larger root is obviously $3$. So $3^2+3^3+3^4=\boxed{117}$ ezy
[/hide][/quote]

Lol it was so easy

hello i am trying to get into my school mathcounts team, can you give me any tips to study for mathcounts?

[quote=Maybach]I decided for those who are bored to take my state sprint round now(those who have already received scores.  The test may change a little(not sure yet) but if you take it now you don't have to take the edited test(or you can take the edited test and choose your better score).  IF YOU HAVE NOT TAKEN THE CHAPTER ROUND YET, DO NOT TAKE THIS TEST.

Please, comment on the test.[/quote]
if this test was a state test for mathcounts
in every tier one state, there would be at least a 8 way tie for 1st place with a 46

[quote=MC413551]"Tonya has $25 more than Tonya"
Interesting...
Problem 15.
I think it's a typo.

Also, problem 22.
"mathcountss"
Also, problem 23.
"sureveyed"[/quote]

check Post #3


[quote="Peter"]Show that the set of positive integers which cannot be represented as a sum of distinct perfect squares is finite.[/quote]
http://www.kalva.demon.co.uk/short/soln/sh00n6.html


Mary has 6 identical basil plants, and three different window sills she can put them on. How many ways are there for Mary to put the plants on the window sills?

If window sills are allowed to have no plants, we have to use a modified version of stars and bars. We put all 6 plants out in a line. For the two bars, we can put them in 7 different spots. Start with the first one- It can go in 7 places. The second one can also go in 7 places. However, here, we overcount the times where the 2 bars are in the same location. Subtract 7 for these.

We get- $ (7*7)\minus{}7\equal{}42$ ways

I think- maybe? Really not sure at all!

Obvious application of stars and bars/ball and urns.
\[ \binom{n \plus{} r \minus{} 1}{n \minus{} 1}\rightarrow\binom{6 \plus{} 3 \minus{} 1}{5} \equal{} \boxed{56}
\]
However watch me mess up.

Edit: HAHA I DID MESS UP!

Ohh I though that there was only 1 plant per windowsill. That makes it very different.

Windowsills can be empty, i.e not have plants in them. So just line up the 6 plants (dots) and divide them with two dividers. Since there are no restrictions in the problem, this just reduces to the number of permutations of 2 dividers and 6 plants, i.e $\binom{8}{2}$

[quote][i]Blackish said:[/i]
                                 Using the hokey stick identity we get $\binom{6+3-1}{3-1}=\binom{8}{2}=\boxed{28}.$[/quote]

Thats not nice maybe someone don't j=know how to do \binom{8}{2}...

This is a real simple way. I know it looks like endless casework, but it's quite simple.

[hide="Easier than Hockey Stick Identity"]

This is a problem in which you have to put $6$ indistinguishable balls into 3 distinguishable boxes.

Case 1: Box #1 has 0 Balls: There are $7$ ways because Box 2 has to have one of the numbers 0-6. The number in Box 3 is then dependent.

Case 2: Box #1 has 1 Ball: There are $6$ ways because now you can't put $6$ as a number in for Box 2, as $1+6=7$, over the limit of $6$ balls.

Case 3: Box #1 has 2 Balls: There are $5$ ways because now you can't have $5$ or $6$ as an option for Box 2.

Continuing this, it is just the sum of the numbers 1 through 7, which is $\frac{7\cdot8}{2}=\boxed{28}$ ways.

Whats kind of bad about this problem, is that the hockey stick identity and what not has not been learned at this point in the C&P courses. Its supposed to be solved in a harder way but hockey stick identity (for those who've read the textbook in advance) is the easiest way.

[hide="Stars and bars"]
We are writing 6 as the sum of 3 numbers. By stars and bars, this is $\binom{6+3-1}{3-1}=\binom82=\boxed{28}$
[/hide]

you just add all of these and there's 28

Divide the plants into 3 sections with 2 dividers.
Then it is 8 choose 2 = 8*7/2 = 28

I would like to point out when you proceed with casework here an interesting observation is that they are all 3! (the amount of "boxes") / (the amount of repeated numbers, for example (1,1,1) would be 3)!.

For example, you could either say

"If all the plants are on one window sill, there are $3$ ways to choose which window sill they are on."

Or you could say, because there are 3 slots we have 6! as a preliminary number. but, because 0 is repeated twice, we have to divide it by 2! to get 3.

Is this right?

[quote name="ad333" url="/community/p20807068"]
wait what's the hockey stick identity the sol i saw with that was the same as stars and bars
[/quote]

${\displaystyle {\binom {n+1}{k+1}}={\binom {n}{k}}+{\binom {n-1}{k}}+{\binom {n-2}{k}}+\cdots +{\binom {k+1}{k}}+{\binom {k}{k}}}$

[hide=solution (Stars and Bars)]$\binom{6+3-1}{3-1}=\binom82=\boxed{28}$

Do we assume that the window sills are distinguishable?

[quote=Blackish][hide="Insanely Easy Way"] Using the hokey stick identity we get $\binom{6+3-1}{3-1}=\binom{8}{2}=\boxed{28}.$ [/hide][/quote]

I thought the hockey stick identity stated that

$\binom{r}{r}+\binom{r+1}{r}+\binom{r+2}{r} +\cdot \cdot \cdot + \binom{n}{r}=\binom{n+1}{r+1}$?
I know that fact was in the Hockey Stick Identity section of Introduction to Number Theory, and I'm pretty sure I'm wrong.

That was annoying to type  :sleeping:


Two bicyclists are 7/8 of the way through a mile long tunnel when a train approaches the closer end at 40 mph. The riders take off at the same speed in opposite directions, and each escapes the tunnel as the train passes them. How fast do they ride?

Source: Mandelbrot #3

[hide]I agree:
When the train enters the tunnel, the first biker still has $\frac78-\frac18=\frac34$ of a mile left to go.  He goes this $\frac34$ of a mile in the same amount of time a train traveling 40 mph goes 1 mile.  Set up porportions where $s$ is the speed of the bikers:
$\frac{\frac34}{s}=\frac{1}{40}$
$s=\frac34\times40$
$s=30\text{ mph}$[/hide]

[quote="surge"][quote="keeyounglee"][hide]
Hint: use the formula $d = \frac{r}{t}$[/hide][/quote]

err... that was a typo on my part. It's $d=rt$ or $r=\frac{d}{t}$. How did you make the same mistake?[/quote]

Arhg. My bad, I thought the formula was $d = \frac{r}{t}$. I looked it up on the internet just now and I copied it wrong from the internet.

-.- yes i didnt know about $d=rt$

one of my questions i had while i was posting this problem is: the bicycler riding towards the approaching train would get crushed if the train met him on the way, which i assume it would, since there is nothing saying how far away the train was from the enterance. How did u guys find the answer for this bicycler?

[quote="surge"][quote="buzzer11"] The riders take off at the same speed in opposite directions, and each escapes the tunnel as the train passes them. [/quote]

You assume "each escapes the tunnel as the train passes them" to be the train and the men meet $\approx$ at the end of the tunnel. And since the "riders take of at the same speed in opposite directions" it wouldn't matter.[/quote]

but where is the train in relation to this second bicyclist? is it at the enterance of the tunnel, or is it further away? if so, how far? if the train is at the enterance of the tunnel, then, how could this second bicyclist escape "as the train passes them"?

why do i feel like im the only one who doesnt get this?  :huh:

Let me explain it for you:

_________________________________________Bicyclists-->_______<--Train
|---------------------------7/8 mi.-----------------------| |---1/8 mi.---|

The bicyclists go off in the other direction.  So the bicyclists travel 7/8 of a mile while the train travels 1 mile at 40 mph.


Aqui estan los problemas, son 6 sin nivel asi que piensenlos todos:

1. [b](corregido)[/b] Sea $I$ el incentro del $\triangle ABC$  con $AB<BC$ , donde $M$  es el punto medio de $AC$  y $N$ punto medio del arco $ABC$ de la circunferencia circunscrita del triangulo. Probar que: $\angle IMA=\angle INB$.

2. [b](corregido)[/b] Sea $\triangle ABC$ un triangulo y sea $\triangle PQR$ otro triangulo [b]semejante al inicial[/b] con $P$ en $BC$, $Q$ en $AC$ y $R$ en $AB$. Pruebe que el ortocentro de $\triangle PQR$ es el circuncentro de $\triangle ABC$.

3. Sea $T_1$ y $T_2$  dos circulos en el plano tal que no se tocan asi mismos. Una tangente externa a los circulos tocan a  $T_1$ en $A$  y a $T_2$  en $C$ , y una tangente interna toca a las circunferencias $T_1$  en $B$  y a $T_2$  en  $D$.
Probar que la interseccion de $AB$ y $CD$  cae en la linea que resulta de unir los centros de las circunferencias.

4. Sea  $\triangle ABC$  un triangulo y $D$ el pie de la altura trazada desde $A$ . Sean $E$  y $F$ sobre la linea que pasa por  $D$ tal que $AE$ es perpendicular a $BC$ y $AF$ es perpendicular a $CF$,   y  $E$ y $F$  son diferentes desde $D$ . Sean $M$  y $N$  los puntos medios de los segmentos $BC$ y $EF$ , respectivamente. Probar que: $AN$ es perpendicular a $MN$.

5. Sean $T_1$ y $T_2$ dos circunferencias que se tocan en $A$ y $B$. Sea $PQ$ una recta por $A$ tal que $P$ esta en $T_1$ y $Q$ esta en $T_2$. Hallar el lugar geometrico del punto medio de $PQ$ al variar esta recta.

6. Los circulos $C_1$ y $C_2$ estan dentro de otro circulo grande $C$ y son tangentes a el en $M$ y $N$ respectivamente. $C_1$ pasa por el centro de $C_2$. La cuerda comun de $C_1$ y $C_2$ al extenderse corta a $C$ en $A$ y $B$. Las lineas $MA$ y $MB$ cortan a $C_1$ en  $E$ y $F$. Pruebe que $EF$ es tangente a $C_2$.

Ahora si!!!.

en la mañana imprimi esto y no tenia la correccion en el prob 02
la semejanza entre $\triangle ABC$ y $\triangle PQR$ salva el problema

la $k$ de proporcionalidad no puede ser 1 (semejanza), en este caso hablariamos de congruencia entre triangulos, ...
entonces esta bien..

FABIOLA       :pilot:   :w00t:

Cuando ponen la 6??

y cual es el tema de la siguiente semana??

ojala sea  combiinatorio

Fabiola

siiii tiren combinatoria la semana que viene.

si la semana que viene es combinatoria, pero no cualquier combinatoria, ya van a ver... :ninja:

jaja no cualquier combinatoria, suena bueno........ :D

[color=blue]bueno entonces estaremos esperando los problemas de combinatoria(me muero de impaciencia!) :10: [/color]

Saludo el entusiasmo de varios de los  participantes de esta comunidad ....

pero recuerden que ya existe un grupo de problemas de esta semana
GEOMETRIA ...........
pues a intentarlas .....
y a enviar sus soluciones..  ;)

Gracias ...

Carlos Bravo
Lima -PERU

Soluciones hasta el  viernes 4 de noviembre.

ya pueden discutirlos

Gracias a este tipo que resolvio todo y me ahorro mucho trabajo!!! :lol:  :lol:  Felicitaciones >Chen!

Muy buen trabajo Chen ..  :first:

espero  que sigamos recibiendo su aporte en esta comunidad ..

extiendo este saludo a todos los chicos de los talleres del local de Huanquica (Lima)
y esperamos una mayor participacion de alumnos y entrenadores. ..

Gracias

Carlos Bravo :)
Lima -PERU

[quote="Pascual2005"]1. Sea $ I$ el incentro del $ \triangle ABC$  con $ AB < BC,$ donde $ M$  es el punto medio de $ AC$  y $ N$ punto medio del arco $ ABC$ de la circunferencia circunscrita del triangulo. Probar que: $ \angle IMA \equal{} \angle INB.$[/quote]
La mediatriz $ MN$ de $ AC$ corta la bisectriz $ BI$ en $ P,$ estando $ P$ en la circunferencia circunscrita $ (O).$ Es sabido que la circunferencia $ \mathcal{K}$ con centro $ P$ que pasa por $ A,C$ pasa también por el incentro $ I.$ Ahora considerando la inversión respecto a $ \mathcal{K},$ ésta trasforma $ M \to N$ y como $ I$ es un punto doble, entonces la circunferencia $ \mathcal{K}' \equiv (IMN)$ es invariable, es decir $ \mathcal{K}'$ es ortogonal al círculo doble $ \mathcal{K}.$ Por consiguiente, $ PI$ es tangente a $ \mathcal{K}'$ en $ I$ $\Longrightarrow$ $ \angle NIB \equal{} \angle IMN.$ De aqui deducimos que $ \angle INB \equal{} \angle IMA$ en vista que $ \angle NBI$ y $ \angle$ $ NMA$ son rectos.

[quote="Pascual2005"]
2. Sea $ \triangle ABC$ un triángulo y sea $ \triangle PQR$ otro triángulo [b]semejante al inicial[/b] con $ P$ en $ BC,$ $ Q$ en $ AC$ y $ R$ en $ AB.$ Pruebe que el ortocentro de $ \triangle PQR$ es el circuncentro de $ \triangle ABC.$[/quote]
Las circunferencias $ \odot(ARQ),\odot(BPR)$ y $ \odot(CQP)$ concurren en el [url=http://mathworld.wolfram.com/MiquelsTheorem.html]Punto de Miquel[/url] $ M.$ Luego por relaciones angulares en cuadrilateros cíclicos tenemos:

$ \angle BMC \equal{} \angle BMP \plus{} \angle CMP \equal{} \angle PQC \equal{} \angle BAC \plus{} \angle RPQ \equal{} 2\angle BAC.$ Analogamente se tiene $ \angle AMC \equal{} 2\angle ABC$ $ \Longrightarrow$ $M$ es circuncentro de $ \triangle ABC.$

Si $ PM$ corta $ RQ$ en $ H,$ entonces se tendrá

$ \angle RHP \equal{} \angle RQP \plus{} \angle MPQ \equal{} \angle RQP \plus{} \angle MCA  \equal{} 90^{\circ}$ $ \Longrightarrow$ $ PM \perp RQ.$ Mutatis mutandi $ RM \perp PQ$ $ \Longrightarrow$ $M$ es ortocentro de $ \triangle PQR.$

[quote="Pascual2005"]
3. Sea $ T_1$ y $ T_2$  dos círculos en el plano tal que no se tocan asi mismos. Una tangente externa a los círculos tocan a  $ T_1$ en $ A$  y a $ T_2$  en $ C$ y una tangente interna toca a las circunferencias $ T_1$  en $ B$  y a $ T_2$  en  $ D.$ Probar que la interseccion de $ AB$ y $ CD$  cae en la linea que resulta de unir los centros de las circunferencias. [/quote]
Si denotamos $ O_1,O_2$ los centros de las circunferencias $ T_1,T_2$ y $ E \equiv BD \cap AC, $ $P \equiv AB \cap CD,$ el resultado es directo aplicando el teorema degenerado de Pappus. Note que $ EO_1 \parallel CD ,$ $EO_2 \parallel AB,$ $AO_1 \parallel CO_2.$  $CD,EO_1$ se cortan en el punto del infinito $ U$ de $ CD.$ $ EO_2,AB$ se cortan en el punto del infinito $ V$ de $ AB.$ $AO_1,CO_2$ se cortan en el punto del infinito $ W$ del gradiente perpendicular a $ AC.$ $ U,V,W$ están en la recta del infinito y $ A,E,C$ están alineados, así $ P,O_1,O_2$ son colineales.

[quote="Pascual2005"]
4. Sea  $ \triangle ABC$  un triangulo y $ D$ el pie de la altura trazada desde $ A.$  Sean $ E$  y $ F$ sobre la linea que pasa por  $ D$ tal que $ AE$ es perpendicular a $ ED$ y $ AF$ es perpendicular a $ FC$   y  $ E$ y $ F$  son diferentes desde $ D.$ Sean $ M$  y $ N$  los puntos medios de los segmentos $ BC$ y $ EF,$  respectivamente. Probar que: $ AN$ es perpendicular a $ MN.$[/quote]
Como los cuadriláteros $ AEBD$ y $ ADFC$ son cíclicos, resulta que $ \angle AEF \equal{} \angle ABC$ y $ \angle AFE \equal{} \angle ACB$  $ \Longrightarrow$ $ \triangle ABC \sim \triangle AEF.$ $A$ es pues centro de la semejanza espiral con ángulo de giro $ \angle EDB \equal{} \angle AEB \equal{} \angle CAF.$ Como los puntos medios de $ BC$ y $ EF$ son homologos en la semejanza, entonces el ángulo $ \angle ANM$ es igual al ángulo de giro, por tanto $ \angle ANM \equal{} \angle CDF$ $\Longrightarrow$ $ ANDM$ es cíclico $ \Longrightarrow$ $ \angle ANM \equal{} \angle ADM \equal{} 90^{\circ}.$

[quote="Pascual2005"]
5. Sean $ T_1$ y $ T_2$ dos circunferencias que se tocan en $ A$ y $ B.$ Sea $ PQ$ una recta por $ A$ tal que $ P$ esta en $ T_1$ y $ Q$ esta en $ T_2.$ Hallar el lugar geométrico del punto medio de $ PQ$ al variar ésta recta.[/quote]
Denotemos $ T_1 \equiv (O_1,r_1) ,  T_2 \equiv (O_2,r_2)$ y  $ M$  punto medio de $ PQ.$ Las potencias de $ M$ respecto a $ T_1$ y $ T_2$ son

$ \frac {_1}{^2}PQ \cdot AM \equal{} (r_1)^2 \minus{} (MO_1)^2 \ , \ \frac {_1}{^2}PQ \cdot AM \equal{} (MO_2)^2 \minus{} (r_2)^2 $

$\Longrightarrow \ (MO_1)^2 \plus{} (MO_2)^2 \equal{}( r_1)^2 \plus{} (r_2)^2$

El lugar geométrico de $ M$ es pues la circunferencia de centro en el punto medio de $ O_1O_2$ coaxial con $ T_1,T_2.$

[quote="Pascual2005"]
6. Los círculos $ C_1$ y $ C_2$ están dentro de otro círculo grande $ C$ y son tangentes a él en $ M,N$ respectivamente. $ C_1$ pasa por el centro de $ C_2.$ La cuerda común de $ C_1$ y $ C_2$ al extenderse corta a $ C$ en $ A$ y $ B.$ $ MA$ y $ MB$ cortan a $ C_1$ en  $ E$ y $ F.$ Pruebe que $ EF$ es tangente a $ C_2.$[/quote]
Denotemos $ O_1, O_2$ los centros de $ C_1,C_2,$ $ PQ$ la cuerda común y $ G$ el segundo corte de $ AN$ con $ C_2.$ $ M$ es centro de la homotecia positiva que transforma $ C_1$ en $ C,$ por consiguiente $ AB \parallel EF.$ Entonces los arcos $ EP$ y $ FQ$ son iguales y como los arcos $ O_2P$ y $ O_2Q$ también son iguales, deducimos pues que $ MO_2$ es bisectriz de $ \angle EMF.$ Por otra parte, la inversión con centro $ A$ y potencia igual a la que tiene $ A$ respecto a $ C_1,C_2,$ deja a éstas invariables y transforma a $ C$ en una tangente común que pasa por los puntos inversos $ E,G$ de $ M,N.$ Por tanto, $ EG$ es tangente a $ C_1$ en $ E$ $\Longrightarrow$

$ \angle GEO_2 \equal{} \angle EMO_2 \equal{} \angle FMO_2 \equal{} \angle FEO_2$

$\Longrightarrow$ $EF$ es la simétrica de $ EG$ respecto a $ EO_2$ $ \Longrightarrow$   $EF$ es tangente a $ C_2.$


86 State Target #1

b is the smallest positive integral base for which $301_{b}$ is a square integer.  What is the square root of $301_{b}$ expressed in base b?

11 in base b
 :D  :D  :D
oops :blush:
13 in base b

Is this really Mathcounts? It remindes me of logarithms.

[hide]Since $301_b$ = $3b^2 + 1$ and in base b you can only use digits less than b, b must me greater than 3.  So the smallest possible value of b would be 4.  Start substituting 4 for b and you get $301_b = 3(4^2)+1 = 49$  You are loking for a square number, so there you go.  The square root of 49 = 7 which is $13_4$[/hide]


conics
parabola
geometry proposed
geometry

Let C be a circle with chord AB. Find the locus of the center of the circle which contacts C and AB.
Sorry if I have a English mistake

Let $AB$ a chord of a circle $C$ with center $O$ and radius $r$ , denote $d$ the line parallel to $AB$ at distance $r$ from $AB$. The required locus is the arc $AB$ of the parabola with focus $O$ and directrix $d$. In fact if $(P)$ is a generic circle with touch $C$ and the chord $AB$ (in $K$ ) we have:

$PO= r - PK = PH =$ dist$(P,d)$

Leon

[quote="Zao1989"]what do you mean by using those trandescent phrase,are they in advanced section?Can you post some more simple solution?wait your reply.....[/quote]

I do not understand what you mean for trascendent phrases. I have only used the definition of parabola :

[i][color=blue]A parabola is the locus of all points in a plane equidistant from a fixed point, called the focus, and a fixed line, called the directrix.[/color][/i]

I do not know a more simple solution.

Leon


Find maksimum possible value of $ n$. Such that;
$ \exists x$, $ k<x^k<k\plus{}1$ for all $ k\equal{}1,2,..,n$.

[quote="pqrs"]Find maksimum possible value of $ n$. Such that;
$ \exists x$, $ k < x^k < k \plus{} 1$ for all $ k \equal{} 1,2,..,n$.[/quote]
Help me
please :(

Maximum $ n$ is $ 4$ (and an example to $ k$ is $ 1.49$).
If $ n \ge 8$, then $ x < \dfrac{9}{7}$, but this implies $ x^2 < \dfrac{81}{49} < 2$, contradiction, therefore $ n \le 7$.
But if maximum $ n$ is greater than $ 4$, $ x< \sqrt[5] 6$, but then it's easy to show that $ x^3 < 3$, so maximum $ n$ is less or equal than $ 4$.


function
algebra
polynomial
limit
algebra unsolved

Let $f(x)$ be a real function such that for each positive real $c$ there exist a polynomial $P(x)$ (maybe dependent on $c$) such that  $| f(x) - P(x)| \leq c \cdot x^{1998}$ for all real $x$. Prove that $f$ is a real polynomial.


trigonometry
inequalities
geometry unsolved
geometry

Points $D,E,F$ are chosen on the sides $BC,AC,AB$ of a triangle ABC,respectively such that
$DC+CE=EA+AF= FB+BD$
   Prove that
      $DE+EF+FD\geq 1/2(AB+BC+AC)$

Projecting the segments EF, FD, and DE onto BC, AC, and AB, respectively, it becomes sufficient to show that

$\sum (DC+EC) \cos C \le s$

where $s$ is the semiperimeter of ABC. Indeed, this inequality follows directly from

$\cos A+\cos B+\cos C \le \frac{3}{2}$

which can be shown in a variety of ways and is well-known.


integration
calculus
calculus computations

hello, your integral can not be expressed by the known elementary functions.
Sonnhard.