inequalities
function
algebra
polynomial
inequalities proposed

Hi!. this is my partial solution. I managed to prove this inequality for the case $ ab \plus{} bc \plus{} ca\leq \frac {1}{4}$. Actually this is easy case...
So,

$ \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {1}{a \plus{} b} \plus{} 6(ab \plus{} bc \plus{} ca) \ge \frac {13}{2}$

$ \frac {a \plus{} b \plus{} c}{b \plus{} c} \plus{} \frac {a \plus{} b \plus{} c}{c \plus{} a} \plus{} \frac {a \plus{} b \plus{} c}{a \plus{} b} \plus{} 6(ab \plus{} bc \plus{} ca) \ge \frac {13}{2}$

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} 6(ab \plus{} bc \plus{} ca) \ge \frac {7}{2}$

From Cauchy-Schwarz:

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \geq \frac {1}{2(ab \plus{} bc \plus{} ca)}$

It is enough to prove:

$ \frac {1}{2(ab \plus{} bc \plus{} ca)} \plus{} 6(ab \plus{} bc \plus{} ca) \geq \frac {7}{2}$

Let $ ab \plus{} bc \plus{} ca \equal{} q$. Observe that:

$ \frac {1}{2q} \plus{} 6q \minus{} \frac {7}{2} \equal{} \frac {12(q \minus{} \frac {1}{3})(q \minus{} \frac {1}{4})}{2q}\geq 0$

The last inequality is trivial for $ q\leq \frac {1}{4}$.

For the non-trivial case, ($ \frac {1}{4}\leq q \leq \frac {1}{3}$),

We have to prove:

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} 6(ab \plus{} bc \plus{} ca) \ge \frac {7}{2}$

Observe that:
[list]$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \equal{} \frac {{1 \minus{} 2q \plus{} 3r}}{{q \minus{} r}}$[/list]
We should prove:
[list]$ \frac {{1 \minus{} 2q \plus{} 3r}}{{q \minus{} r}} \plus{} 6q \geq \frac {7}{2}$[/list]
From here, i don't know how to continue.  :(

I tried $ r \geq \frac {4q \minus{} 1}{9}$, but didn't really succeed.

Thanks for having patient and reading my post.  :)

I also have a partial solution. The given inequality implies that at least two of the numbers must be nonzero (otherwise we would have $ \frac1{0\plus{}0}$). Hence $ a\plus{}b\plus{}c\equal{}1\ \Rightarrow\ a,b,c<1$, and the function $ f(x)\equal{}\frac1{1\minus{}x}$ is convex for $ x<1$.

Hence, Jensen’s inequality gives $ \frac1{a\plus{}b}\plus{}\frac1{b\plus{}c}\plus{}\frac1{c\plus{}a}\geqslant\frac92$.

Now $ a\plus{}b\plus{}c\equal{}1\ \Rightarrow\ 6(ab\plus{}bc\plus{}ca)\leqslant2$ by Cauchy–Schwarz.

Unfortunately the inequality is the wrong way round, so I can’t add it to the Jensen bit. :(

Happy new year every one  :)

My dear Can :
with some Algebraic manipulations , I found that your inequality  is equivalent to :
$ \sum{ \frac{a^2}{b\plus{}c}} \geq \frac{ 5\sum{a^2}\minus{}2\sum{ab}}{2\sum{a}}$

or : $ \sum{ g(a,b,c)} \geq 0$ where  : $ g(a,b,c)\equal{} (a^2\plus{}b^2\minus{}c^2)(a\minus{}b)^2(a\plus{}b)$

that is : $ g(a,b,c) \plus{}  c^2 ( (a\minus{}c)^2(a\plus{}c) \plus{}(b\minus{}c)^2(b\plus{}c)) \plus{} (a\plus{}b)(a\minus{}b)^2 ( a^2\plus{}ab\plus{}b^2\minus{}c(a\plus{}b\plus{}c))$ which is obviously true if : $ a\geq b \geq c$

P.S 1: I think that you have a more simpler non-standard solution   :)
P.S 2 :  I have two terribly ugly proofs for Iran1996 I like the first which used the weaker inequality : $ \sum{\frac{2a^2\plus{}5bc}{(b\plus{}c)^2}} \geq \frac{21}{4}$  :wink:

Regards..

Rewriting the inequality, it is \[ \frac{1\plus{}\sum bc}{\sum bc \minus{}abc}\plus{}6\sum bc\ge \frac{13}{2}\] Now if we let $ \sum bc\equal{}\frac{1\minus{}q^2}{3}$, $ q\in[0,1]$, we have the inequality:\[ r\ge \frac{(1\plus{}q)^2(1\minus{}2q)}{27}\] which holds iff $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\equal{}0$. If we rearrange the inequality, we get \[ r\ge \frac{1\minus{}3q^2\minus{}4q^4}{27\plus{}12q^2}\] Thus it suffices to prove \[ \frac{(1\plus{}q)^2(1\minus{}2q)}{27}\ge \frac{1\minus{}3q^2\minus{}4q^4}{27\plus{}12q^2}\] \[ \iff (2\minus{}q)(q^2)(2q\minus{}1)^2\ge 0\] $ q\equal{}0$ gives $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$. $ q\equal{}\frac{1}{2}$ and the equality for our upper bound on $ r$ gives $ a\equal{}b\equal{}\frac{1}{2}$, $ c\equal{}0$ and permutations.

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {1}{a \plus{} b} \plus{} 6(ab \plus{} bc \plus{} ca) \ge \frac {13}{2}.
\]
Equality holds if and only if $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$ or $ a \equal{} b \equal{} \frac {1}{2},c \equal{} 0$ and its cyclic permutations. :)

P/s: This inequality is not hard to solve but the interesting point of it is: From this result, the Iran 96 Inequality follows immediately while the Iran 96 Inequality seems harder to solve. :)

[b]MERRY CHRISTMAS TO MATHLINKS MEMBERS. NICE TO MEET ALL OF YOU.

A SPECIAL REGARD TO ALL OF MY FRIENDS: MANLIO, SANCHIR, MR. MIRCEA LASCU, ANAS, DUONG DUC LAM, ... [/b]

:)[/quote]
Thank you all guys. :) Let me explain the reason why from this inequality, the Iran 96 Inequality:
\[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}\plus{}\frac{1}{(a\plus{}b)^2}\right) \ge \frac{9}{4}\]
follows immediately. :)
Rewrite it as:
\[ \sum \frac{a}{b\plus{}c} \plus{}\sum \frac{bc}{(b\plus{}c)^2} \ge \frac{9}{4}\]
\[ (a\plus{}b\plus{}c)\left( \sum \frac{1}{b\plus{}c}\right) \plus{}\sum \frac{bc}{(b\plus{}c)^2} \ge \frac{21}{4}\]
Now, we normalize for $ a\plus{}b\plus{}c\equal{}1$, the above inequality becomes
\[ \sum \frac{1}{b\plus{}c} \plus{}\sum \frac{bc}{(1\minus{}a)^2} \ge \frac{21}{4}\]
Now, notice that for all $ x \in [0,1]$, we have that
\[ \frac{1}{(1\minus{}x)^2} \ge 9x^2\plus{}\frac{3}{4}x\plus{}1\]
Using this inequality and Schur's Inequality, we have
\[ \sum \frac{bc}{(1\minus{}a)^2} \ge \sum bc\left( 9a^2\plus{}\frac{3}{4} a\plus{}1\right) \equal{}\sum bc\plus{} \frac{45}{4}abc \ge \sum bc \plus{}\frac{5}{4}\left( 4\sum bc\minus{}1\right) \equal{}6\sum bc\minus{}\frac{5}{4}\]
So, all we have to do is to prove
\[ \sum \frac{1}{b\plus{}c} \plus{}6\sum bc \minus{}\frac{5}{4} \ge \frac{21}{4}\]
or
\[ \sum \frac{1}{b\plus{}c} \plus{}6\sum bc \ge \frac{13}{2}\]
which is just the above inequality. :)

It's eqiuvalent to:
$ \sum\ \frac {a}{b \plus{} c} \plus{} \frac {6(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2} \geq\ \frac {7}{2}$
SOS kills it easily.
I remember that I used it to solve an old problem:
$ \sum\ \frac {a}{b \plus{} c} \geq\ \frac {3}{2}. (\frac {3(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2})^{\frac {2}{3}}$
PS: But I'm so surprise because it's stronger than Iran 96. Thank you.
 :)

I think the following generalization holds.

If $ a_1,a_2,...,a_n$ are nonnegative real numbers such that $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} n$, then

$ n(n \minus{} 2)(\sum \frac {a_1}{a_2 \plus{} a_3 \plus{} ... \plus{} a_n} \minus{} \frac n{n \minus{} 1}) \ge a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2 \minus{} n$.

On the other hand, the following similar inequality holds for any positive real numbers $ a,b,c$:

$ \frac a{b\plus{}c}\plus{}\frac b{c\plus{}a}\plus{}\frac c{a\plus{}b} \minus{}\frac 3{2} \ge (\sqrt 3\minus{}1)(1\minus{}\frac {ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2})$.

Is it known? Can you generalize it for $ n$ numbers?

[quote="Vasc"]I think the following generalization holds.

If $ a_1,a_2,...,a_n$ are nonnegative real numbers such that $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} n$, then

$ n(n \minus{} 2)(\sum \frac {a_1}{a_2 \plus{} a_3 \plus{} ... \plus{} a_n} \minus{} \frac n{n \minus{} 1}) \ge a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2 \minus{} n$.
[/quote]
Nice generalization, vasc. :)
My proof need to consider 3 cases: $ n \equal{} 3$ (the sharpest case), $ n \equal{} 4$ and $ n \ge 5$. What is your proof, VasC? I hope that it is not EV or LCF-RCF. ;)

[quote="can_hang2007"] My proof need to consider 3 cases: $ n \equal{} 3$ (the sharpest case), $ n \equal{} 4$ and $ n \ge 5$. What is your proof, VasC? I hope that it is not EV or LCF-RCF. ;)[/quote]

Why $ n\equal{}3$ is the sharpest case ?
Of course, RCF method is very simple.

[quote="Vasc"]
On the other hand, the following similar inequality holds for any positive real numbers $ a,b,c$:

$ \frac a{b + c} + \frac b{c + a} + \frac c{a + b} - \frac 3{2} \ge (\sqrt 3 - 1)(1 - \frac {ab + bc + ca}{a^2 + b^2 + c^2})$.

Is it known? Can you generalize it for $ n$ numbers?[/quote]

So Lets consider the general problem: Find the biggest $ k$ (the best constant) so that the following inequality holds.

$ \frac a{b + c} + \frac b{c + a} + \frac c{a + b} - \frac 3{2} \ge k(1 - \frac {ab + bc + ca}{a^2 + b^2 + c^2})$

So lets test the "power"  of  direct cauchy schwarz. From Cauchy Schwarz inequality, we have:

$ \frac a{b + c} + \frac b{c + a} + \frac c{a + b} \ge \frac {(a + b + c)^2}{2(ab + bc + ca)} = \frac {a^2 + b^2 + c^2}{2(ab + bc + ca)} + 2$

So it is enough to prove that:

$ \frac {a^2 + b^2 + c^2}{2(ab + bc + ca)} + 2 - \frac {3}{2} \ge k(1 - \frac {ab + bc + ca}{a^2 + b^2 + c^2})$

$ \frac {a^2 + b^2 + c^2}{2(ab + bc + ca)} + \frac {1}{2} \ge k(1 - \frac {ab + bc + ca}{a^2 + b^2 + c^2})$

To make calculations easier, denote:

$ \frac {a^2 + b^2 + c^2}{ab + bc + ca} = x$   (notice $ x\ge 1$)

so all we have to do is, to find the greatest $ k$ in the following inequality:

$ \frac {x}{2} - \frac {1}{2} \ge k(1 - \frac {1}{x})$

equivalently, we have:

$ \frac {x^2 - x - 2kx + 2k}{2x} \ge 0.$

$ x^2 - x - 2kx + 2k \0$

To find the maximum k, i prefered to change variable again. Because $ x\ge 1$ , there is a positive real $ m$ such that $ x = m + 1$, so that clearly implies that $ m \ge 0$. Therefore to find maximum $ k$ is much easier. We have:

$ (m + 1)^2 - (m + 1) - 2k(m + 1) + 2k \ge 0$

$ m^2 + m - 2km \ge 0$

Now, since $ m^2\ge 0$, we left with: $ m - 2km \ge 0$. Last one clearly indicates that, $ k_{max} \le \frac {1}{2}.$

There, we proved that,[b]direct[/b] (i am saying direct, because Cauchy Schwarz can be used differently (ingeniously) to produce magnificient inequalities.. Can_Hang is an example  :D  ) [b]application of Cauchy Schwarz[/b], can only produce

$ \frac a{b + c} + \frac b{c + a} + \frac c{a + b} - \frac 3{2} \ge (\frac {1}{2})(1 - \frac {ab + bc + ca}{a^2 + b^2 + c^2})$.

But wait!!! In VasC's inequality  $ k_{max} = \sqrt {3} - 1 = 0.732050808 > \frac {1}{2}$. So we should try other approaches. Lets try pqr method.

let $ a + b + c = 1$ (we can normalize this inequality), $ ab + bc + ca = q$ , and $ abc = r$

Direct Computation will show that:

$ \frac a{b + c} + \frac b{c + a} + \frac c{a + b} - \frac 3{2} = \frac {2 - 7q + 9r}{2(q - r)}$

and

$ 1 - \frac {ab + bc + ca}{a^2 + b^2 + c^2} = \frac {1 - 3q}{1 - 2q}$
actually i tried to sharpen the constant (instead of finding best constant ----> which is very hard to me  :| )...
and you will see that, still i can't do it.  :noo:
Since $ k \leq \frac {1}{2}$ by CS above is already proven, I said, lets try for$ k = 0.6$ which is still weaker than Vasc's Inequality. So for $ k = \frac {6}{10}$, we should prove that:$ \frac {2 - 7q + 9r}{2(q - r)}\ge \frac {6}{10}(\frac {1 - 3q}{1 - 2q})$  ---->  :idea:

From Schur's inequality for a third degree we have:

$ 1 + 9r\ge 4q$ or equivalently $ r\ge \frac {4q - 1}{9}$ , so applying this inequality for the  :idea: , we have to prove

$ \frac {9 - 27q}{2(5q - 1)} \ge \frac {6}{10}(\frac {1 - 3q}{1 - 2q})$

Observe that:

$ \frac {9 - 27q}{2(5q - 1)} - (\frac {6}{10})(\frac {1 - 3q}{1 - 2q}) = ( - \frac {3}{10})\frac {(3q - 1)(40q - 13)}{(5q + 1)(2q - 1)} \ge 0$

So, is my result true? because, i don't know how to study "cases" ... Anyways sorry if i am boring.

 [quote]Now, notice that for all $ x \in [0,1]$, we have that
\[ \frac {1}{(1 - x)^2} \ge 9x^2 + \frac {3}{4}x + 1
\]
[/quote]

well, we have:

$ \frac {1}{(1 - x)^2} - 9x^2 - \frac {3}{4}x - 1 = \frac {1}{4}\frac {(5 - 4x)(3x - 1)^2}{(x - 1)^2}$

and that is clearly holds for $ x \in [0,1]$

But how could you think of that function? can you please explain this little mystery?

Thanks  :)

[quote="Vasc"]
On the other hand, the following similar inequality holds for any positive real numbers $ a,b,c$:

$ \frac a{b \plus{} c} \plus{} \frac b{c \plus{} a} \plus{} \frac c{a \plus{} b} \minus{} \frac 3{2} \ge (\sqrt 3 \minus{} 1)(1 \minus{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2})$.

Is it known? Can you generalize it for $ n$ numbers?[/quote]

The constant $ \sqrt3\minus{}1$ comes from the polynomial
$ (9k\plus{}27)a^4 \plus{}(\minus{}36\minus{}6k)a^3\plus{}(\minus{}8k\plus{}24)a^2\plus{}(\minus{}8\plus{}6k)a\plus{}(1\minus{}k) >0$    (1)
We wish (1) to be corrected for all of a in $ [0..\frac13]$


I need to find how many subsets, $S_{i}$, of $\{1,\ldots,30\}$ I can take, for whom holds $\sum_{s \in S_{i}}s > 232$. I was thinking about the generating function for partitions in different integers, none of which is greater than $30$, but it's a long way.

I really don't think so, because $2^{30}$ is the number of all the subsets of $\{ 1,2,3..30\}$ , and not all of the subsets have the property from the hypothesis.

Problems like this come up in examples of applications of probability such as Chebyshev's inequality all the time, and I suspect there is no good way to do it, other than simply running a computer program.

[hide="Would these help?"]
(1)  I'm not sure of it, but the number of partitions might be symmetrical around $\frac{ \frac{30 \cdot 31}{2}}{2}= 232.5$.

(2)  It might help to biject to finding subsets $S_{j}$ the set $S' = \{-14, \ldots, 15 \}$, but what will the second condition become?

 :)  And we can always just expand $\prod_{k=1}^{30}\left( 1+x^{k}\right)$.  :P [/hide]

[hide="non generalized solution"]
the sum of $S_{i}$ is $\frac{30(31)}{2}= 465$
half of this sum is $232.5$

each subset $s \in S_{i}$ is either $> 232.5$ or $< 232.5$

now if you invert any subset whose sum is $> 232.5$, you will get a new subset whose sum is $< 232.5$ and vice versa

that means that the subsets can be paired, so half of them are greater than 232.5, or in other words greater than 232

$\frac{2^{30}}{2}= 2^{29}$

I don't know how to generalize this though. It only worked because 232.5 was half the sum of the set
[/hide]

I think that some sort of double- or triple-summation can be used to generalize. But it will most probably get extremely ugly, very quickly.

[quote="chickendude"]
now if you invert any subset whose sum is $> 232.5$, you will get a new subset whose sum is $< 232.5$ and vice versa

that means that the subsets can be paired, so half of them are greater than 232.5, or in other words greater than 232

$\frac{2^{30}}{2}= 2^{29}$
[/quote]
What do you mean by "inverting any subset"?

Let the number of subsets of $\{1, 2, ..., k\}$ with sum equal to $n$ be called $f(n, k)$.  Then, for each subset, it could

contain $k$.  There are $f(n-k, k-1)$ such subsets.

not contain $k$.  There are $f(n, k-1)$ such subsets.

Thus, $f(n, k)=f(n, k-1)+f(n-k, k-1)$

I don't see any way to turn this recursive relation into an explicit one, because the index has two parameters...
It seems you would have to use a computer...

But we don't have to know that the sum equal to a exact number, so I think the complement solution before works...???

I think by inverting he means to take the complement of the subset.  For example, if the subset is A, then B is the complement of A if and only if A u B = the whole set and A intersec B = null

oops, wrong term

my vocabulary is not up to par


Prove that for all real numbers $x$ and integers $n>1$,

$\lfloor nx \rfloor = \lfloor x \rfloor+\lfloor x+\frac{1}{n}\rfloor+\lfloor x+\frac{2}{n}\rfloor+\lfloor x+\frac{n-1}{n}\rfloor$

it's supposed to read
\[\lfloor nx\rfloor =\sum_{k=0}^{n-1}\lfloor x+\frac{k}{n}\rfloor\]
[hide]
Let $x=q+\frac{r}{n}+f$ where $q$ is an integer, $r$ is an integer with $0\leq r\leq n-1$, $f$ is a real number with $0\leq f<\frac{1}{n}$. Then $\lfloor nx\rfloor =nq+r$. For the summation,
\[\lfloor x+\frac{k}{n}\rfloor =q+\lfloor \frac{r+k}{n}\rfloor=\{ \begin{array}{rl}q,&k\leq n-1-r\\ q+1,&k\geq n-r\\ \end{array}\]
where there are $n-r$ $q$'s and $r$ $q+1$'s, so the sum is also equal to $nq+r$. (We could drop the $f$ from the floor because $\frac{r+k}{n}+f$ is nonnegative, strictly less than 2, and at least 1 if and only if $\frac{r+k}{n}$ is.)
[/hide]

[quote="boxedexe"]What about it?  Both sides equal to $0$.  This is, I believe, Hermite's Identity.    Use periodicity to prove it.[/quote]

how do you use periodicity?

[quote="boxedexe"][b]Solution.[/b] Define the function $\Psi: \mathbb{R}\rightarrow\mathbb{N}$ as $\Psi(x)=\sum_{k=0}^{n-1}\left\lfloor x+\frac{k}{n}\right\rfloor-\lfloor nx\rfloor$.  Notice that $\Psi(x)=0\ \forall x\in\left[0,\frac{1}{n}\right)$.  Moreover, $\Psi\left(x+\frac{1}{n}\right)=\Psi(x)$, i.e. $\Psi$ has period $\frac{1}{n}$.  Ergo, $\Psi\equiv 0$.  And the conclusion follows.$\Box$[/quote]

woah, i never thought of doing that sort of thing. in what other situations is periodicity useful?


geometry
quadratics
number theory
greatest common divisor
least common multiple
function

well can anyone tell me how  python programming language is useful in mathematics

In the same way any other programming language can be useful.

What area of mathematics are you looking at?

well i am not too sure its use in mathematics nor in which area of mathematics it is useful.
well that only i exactly wanted to know.

You can use Python to write a program for using the quadratic formula...

[code]
import math
randomvariable=raw_input("-->")
f,g,h=randomvariable.split(",")
a=float(f)
b=float(g)
c=float(h)
print "x=", (-b+math.sqrt(b**2-4*a*c))/2*a, "or", (-b-math.sqrt(b**2-4*a*c))/2*a
[/code]

It's pretty simple with IDLE

You can also use it for basic programming stuff like tic-tac-toe, adding/subtracting/multiplying/dividing fractions and it comes built in with GCD, LCM, add/subtract/multiply/divide, roots of numbers, etc...

What were you hoping to do with it :?:

can python be used in the informatics olympiad?

nope, the only ones that are allowed at the ioi are C++, C (same thing), and pascal

one reason might be because java and python have bignum, which makes bignum problems trivial.
python also has a parser function and regexp, which might make parsing problems trivial.
a large reason is also because those two languages both use interpreters

I am currently taking my Bachelor of Science in Applied Mathematics, major in Mathematical Finance, and I am required to study Python for two semesters.

I'd say it's just as good as any other programming language for mathematics. I created a program that solved one of the CGMO 2009 problems (the one involving $ n\sqrt{5}$)

[quote="lynbrook4math"]one reason might be because java and python have bignum, which makes bignum problems trivial.[/quote]
But there is [url=http://gmplib.org/]GMP[/url] so bignums are pretty easy in C/C++.


trigonometry
algebra
polynomial
Vieta
algebra proposed

$ e^{i\, 2nx} \equal{} (\cos x \plus{} i \sin x)^{2n} \equal{} \sum_{k \equal{} 0}^{2n} {2n \choose k} \, i^k \, (\sin x)^k \, (\cos x)^{2n \minus{} k}$

Compare the real parts $ \cos(2nx) \equal{} \sum_{k \equal{} 0}^n {2n \choose 2k} \, ( \minus{} 1)^k \, (\sin x)^{2k} \, (\cos x)^{2n \minus{} 2k}$

Therefore $ \frac {\cos(2nx)}{(\cos x)^{2n}} \equal{} \sum_{k \equal{} 0}^n {2n \choose 2k} \left( \minus{} \tan^2 x\right)^k$

The LHS equals zero for $ x \equal{} \frac {(2k \plus{} 1)\pi}{4n} \,\, , \,\, k \equal{} 0,...,n \minus{} 1$

Hence the polynomial $ P(X) \equal{} \sum_{k \equal{} 0}^n {2n \choose 2k} X^k$ has the roots $ X \equal{} \minus{} \tan^2\left(\frac {(2k \plus{} 1)\pi}{4n}\right)$ .

After a theorem of Vieta the negative sum of all roots is the coefficient in front of $ X^{n \minus{} 1}$ , namely $ {2n \choose 2(n \minus{} 1)} \equal{} n\, (2n \minus{} 1)$ .


$9.$ Let
\[\left(1+x^{3}\right)\left(1+2x^{3^{2}}\right)\cdots\left(1+kx^{3^{k}}\right)\cdots\left(1+1997x^{3^{1997}}\right) = 1+a_{1}x^{k_{1}}+a_{2}x^{k_{2}}+\cdots+a_{m}x^{k_{m}}\]
where $a_{i}\ne 0$ and $k_{1}< k_{2}< \cdots < k_{m}$. Determine the remainder obtained when $a_{1997}$ is divided by 1000.
[hide]Base Three[/hide]

[hide]The $k_{i}$ are numbers in base $3$ with only 0s and 1s and ending in 0.  Then the 1997th is 1997 in binary with a 0 added at the end, then read in ternary.  That is $111110011010$.  The coefficient of that must then be $1*3*4*7*8*9*10*11 \equiv 3*4*7*8*(-10) \equiv 7*(40) \equiv 280$.[/hide]

[quote="Altheman"]i believe this was from a bulgaria contest...[/quote]
It is? It's really easy though. Or did it come from one of the qualifying rounds?


An infinite sheet of charge with uniform surface charge density pn= 10
C/m2 is located at z = 0 (x-y plane), and another infinite sheet ps2 = -10
C/m2 is located at z = 4 m, both in free space. Using Gauss's Law, determine
the electric field intensity, j? in all regions.

I can't visualize this scenario so it's hard to provide a hint.  Do you have a picture, or could you describe it in more detail?

[hide="Hint 1"]Start with one sheet.[/hide]
[hide="Hint 2"]Use a cylinder as your Gaussian surface.[/hide]


Two bicyclists are seven-eighths 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 the opposite directions and each escapes the tunnel as the train passes them. How fast did they ride?

[hide="Answer"]The bicyclist heading for the close end goes $\frac{1}{8}$ the length of the tunnel before he exits, and since the two bikers are going the same speed, the other one is $\frac{3}{4}$ of the way from the other end of the tunnel at the instant that the train enters the tunnel. Therefore, the speed of the bicyclists equals $40\cdot \frac{3}{4}=30$ mph.[/hide]


Let $ \mathbb{S}$ be a set with $ \mbox{n}$ elements. In how many different ways can one select two not necessarily distinct or disjoint subsets of $ \mathbb{S}$ so that the union of the two subsets is $ \mathbb{S}$? The order of selection does not matter. For example, the pair of subsets $ \{a,c\}, \{b,c,d,e,f\}$ represents the same selection as the pair $ \{b,c,d,e,f\}, \{a,c\}$

Firstly what does "two not necessarily distinct or disjoint subsets" mean?

And I've tried experimenting a bit. I wrote down the 16 different subsets for $ \{a,b,c,d\}$. There seems to be some pattern but I don't think brute force like this is the way to go...

So can someone explain an easier method? (Please don't leave out any steps cause I'm a beginner at combinatorics :P)

Hint: to write $ S$ as a union of two subsets, $ A$ and $ B$, of $ S$, for each element of $ S$ you have to make one of 3 choices: put it only in $ A$, only in $ B$, or in both $ A$ and $ B$.

Thanks pleurestique for all your help :)

I think I got it, is it correct?

Let $ \mathbb{S} \equal{} \{a,b,c,...n\}$

So each element can either be in $ A$ or in $ B$ or in $ A \cap B$.

So we can have something like:

$ A \equal{} \{a,b,c...n\}$ and $ B \equal{} \{\varnothing\}$

$ B \equal{} \{a,b,c...n\}$ and $ A \equal{} \{\varnothing\}$

Which is an overcount since we don't care about order.

This means each pair is counted twice rather than once.

But there is one exception: If we picked all the elements in $ \mathbb{S} \equal{} \{a,b,c,...n\}$ to be in $ A \cap B$ then we have only counted that once.

$ \therefore \frac{3^n\minus{}1}{2} \plus{} 1$

Thanks again  :lol:


Let $ x$ and $ y$ be two real numbers such that $ 4x^2 \minus{}5xy \plus{} 4y^2 \equal{} 5$. What is the sum of the reciprocals of the maximum and minimum values of $ x^2 \plus{} y^2$?

take $ (x,y)\equal{}(\frac{1}{\sqrt{2}}(x'\plus{}y'),\frac{1}{\sqrt{2}}(x'\minus{}y'))$
it gives $ 3x'^2\plus{}13y'^2\equal{}10$ (the rest is easy)


Use titles that describe the problems! Please edit all four titles you recently made into something better!

[quote="tdl"]Prove that there exist pairwise distinct positive integers $ a, b, c$ and $ d$, such that the sum of digits of the number $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2$ equals to $ 19981998$.[/quote]

Consider the number $ u_k\equal{}\sum_{i\equal{}0}^{k\minus{}1}10^{2^k}$. We have $ S(u_k^2)\equal{}k^2$

Consider that, for any positive integers $ x,y,z,t$ it is always possible to find integers $ i,j,k$ such that $ S(10^ix\plus{}10^jy\plus{}10^kz\plus{}t)$ $ \equal{}S(10^ix)\plus{}S(10^jy)\plus{}S(10^kz)\plus{}S(t)$ $ \equal{}S(x)\plus{}S(y)\plus{}S(z)\plus{}S(t)$

So, choosing appropriate integers $ i,j,k$, we have $ S((10^iu_{4470})^2\plus{}(10^ju_{31})^2\plus{}(10^ku_{11})^2\plus{}(u_{4})^2)$ $ \equal{}S(u_{4470}^2)\plus{}S(u_{31}^2)\plus{}S(u_{11}^2)\plus{}S(u_4^2)$ $ \equal{}4470^2\plus{}31^2\plus{}11^2\plus{}4^2$ $ \equal{}19981998$


geometry
trapezoid
incenter
geometry unsolved

Let $ ABCD$ be a trapezium of parallel sides $ AD$ and $ BC$ and non-parallel sides $ AB$ and $ CD$
Let $ I$ be the incenter of $ ABC$. It is known that exists a point $ Q \in AD$ with $ Q \neq A$ and $ Q \neq D$ such that if $ P$ is a point of the intersection of the bisectors of $ \widehat{ CQD}$ and $ \widehat{CAD}$ then $ PI \parallel AD$ Prove that $ PI\equal{}BQ$

AICP is an isosceles trapezoid with AI=CP => AC=IP
denote I' is  incenter  of triangle ACQ => P is the exincenter of triangle ACQ => $ \widehat{ I'QC}$=$ \widehat{ I'PC}$=$ \widehat{ I'AC}$  => AQCB is an isosceles trapezoid => AC=BQ

it suffices to prove $A,I,C,P$ are concyclic.
not a difficult problem,but divinely nice~


Let $ a,b,c$ be positive real numbers. Prove that
\[ \frac{(a\minus{}b)^2}{ab}\plus{}\frac{(b\minus{}c)^2}{bc}\plus{}\frac{(c\minus{}a)^2}{ca}\plus{}2 \ge \frac{2\sqrt{3(a^4\plus{}b^4\plus{}c^4)}}{ab\plus{}bc\plus{}ca}\]
:)

[quote="Rose_joker"]This problem is very hard...I failed to solve it. Can you show me a solution please Can_hang2007?? I really want to know how is it easy...[/quote]

SOS help me, Rose_joker    :)


absolute value
linear algebra
linear algebra solved

$A,B \in M_n(\mathbb{C})$,$AB=BA^2$, $A$ has no eigenvalues of absolute value $1$. Prove that $A$ and $B$ have a common(non-zero) eigenvector

Let $x$ be some eigenvector of $A$ corresponding to the eigenvalue $\lambda$. Then $Ax=\lambda x\Rightarrow A(Bx)=\lambda^2(Bx)$, so $\lambda^2$ is alo an eigenvalue, which means that $\lambda^{2^k}$ is an eigenvalue of $A$ for all $k$. Since unless $\lambda=0$ or $|\lambda|=1$ the set $\{\lambda^{2^k}\}_k$ is infinite, it means that $\lambda=0$ (i.e. the only eigenvalue of $A$ is $0$, or $A$ is nilpotent).

Let $\mathcal L$ be the space of eigenvectors of $A$. Since $Ax=0\Rightarrow A(Bx)=0$, we find that $B$ acts on $\mathcal L$ (i.e. $\mathcal L$ is a fixed subspace of $B$). As a linear map from $\mathcal L$ to itself, $B$ has an eigenvector in $\mathcal L$, which will also be one for $A$.


First the question aks to prove that if $x$ is a limit point in a metric subspace $A\subseteq X$ then each open ball centered at $x$ contains an infinite number of distinct points of $A$. OK, this went fine. Then I'm supposed to use this to show that every finite subset of $X$ is closed, which we have defined it to be if it contains all its limit points. How? It seems intuitively clear, but I'm supposed to use the result just proved.

This was my reasoning: It suffices to show that every finite subset $A\subseteq X$ contains all points $x$ such that every open ball centered at $x$ contains an infinite number of points of $A$. But this is true vacuously, right?

Let $A\subset X$ be a finite set. Suppose $A$ is not closed. Then there is a limit point $x$ which is not a member of $A$. Consider an open ball around $x$. This ball must contain an inifinitely many elements of $A$ (by your previous result). BUt it is impossible since $A$ it self only have finitely many elements.


is there anywhere on the net problems or articles of the russian magazine kvant???

hello, here is a link for the Kvant, but only in russian
http://kvant.mirror1.mccme.ru/
Sonnhard.


Let ABCD be a square. Let E and F be points on BC and CD respectively which trisect their length with E closer to B than to C, and F closer to D than to C. Let G and H be points on BC and CD, respectively, which bisect their length.

Find the tan <EAF - cot <GAH.

Judging from the answer I'd think there would be a nicer solution, but this is what I came up with. We start by writing EAF=90-2BAE and GAH=90-2BAG. So we are looking for tan(90-2BAE)-cot(90-2BAG)=cot(2BAE)-tan(2BAG). Then tan BAE=1/3, and tan BAG=1/2, and tan(2a)=2tan(a)/(1-tan^2(a)), so now we have cot(2 arctan(1/3))-tan(2 arctan(1/2))=(1-1/9)/(2/3)-1/(3/4)=0.


In a 7 game series between the Cubbies and the Marlins, how many ways can the Cubs win? (ie. CCCC, CCMMMCC, CMCCC etc.)

Please respond with solutions and in spoiler.

(Ignore the fact that they are tied 3-3 at the moment)

im really tired. but the marlins won the pennant.

TheSouthpaw wrote:Okay, I did this in a couple seconds and I'm pretty sure I'm wrong, but what the heck.

[hide]
7c4 = 35

[/hide]

I think that's right...

im still not awake yet, but it sounds like gravvvvvy to me

That can't be right because when doing 7c4, that includes possibilities like CCCCMMM, which isn't possible because they stop after a team has won four games...

Actually that's ok though. Suppose that after we get 4 Cs, we fill the unfilled ones (if there are any) with Ms.  That shouldn't change the number of ways to get 4 Cs out of no more than 7 (when we stop when we get to 4 of either). Thus we only have to consider the ones with 4 Cs and 3 Ms.

....or putting a slightly different spin on it, to get the actual sequence of game winners, such as CCCC, we just take all combinations of 4C's and 3M's and delete the M's at the end. So for example, the combination CCCCMMM would yield CCCC, which would be the desired sequence, while something like CCMCMCM would yield CCMCMC, another acceptable sequence.

yeah, if you tried to count it your way (as in emily's) then youd get the same numero. so dont worry bout it

blah. you know you'll have to explain  your answer a bit more. and just out of curiosity, may i ask why you didn't look at the previous responses?

sorry to bring up this old question but theres a cool insight that some of you may find interesting:

indeed the answer is (7c4) = 35 as has been discussed above...

but can any of you tell me why this is also (8c4)/2 = 35?  :D

It is also 3C3 +4C3+5C3+6C3, because the Cubs have to win the last game and this is the number of ways they can win in 4 games 5 games ...

math92, you really need to stop spammong, or take a vacation


ghablan matrah shode
vali dar har hal barabare kamani az dayerei ast ka bar do sagh dar noghate $B,C$ momas ast
va age kolle safhe ro dar nazar begirim ghesmati az hozlooli niz be makan hendesimoon ezafe mishe

[quote="Behsan"]Makan hendesi noghati manande M dar mosallase motosavi alsagheyn ra biabid tori ke haselzarb faselehaye oon az 2 sagh  2barabare faseleye an az ghaede basad[/quote]

mage mishe haasele zarbe do taa tool baraabare tool baashe? :huh:

[quote="BaBaK Ghalebi"]manzooresh in bood ke faselash az ghaede barabare miangine hendesie faseleash ta do sagh beshe[/quote]
ahaa  :roll:

[quote="ashegh"]in harfi keh mizanam rabti be bahs nadare vali badak nist...

mitoonim hasel zarb e $2$ tool ro keshid...[/quote]
bale albateh ke mishe keshid:
farz kon do pare khate $x,y$ ro darim angah rooye yek pare khat be andazeye $x+y$ joda mikonim.($AB=x,BC=y$) angah dayere ee be ghotre $AC$ zade va dar noghteye $B$ amoody bar pare khate $AC$ rasm mikonim ta dayereye rasm shode ra dar $M$ ghat konad.darim $BM^2=xy \to BM=\sqrt{xy}$.
az tarafi baraye yaftan pare kahti betoole $BM^2$ kafist dar raveshe bala jaye $BC$ ,$1$ va jaye $BM$ khodash ra ghar dahim angsh bedast mi ayad $AB \times 1=BM^2 \to AM+BM$ :D


[quote="stergiu"][quote="billakos"]Καλημέρα στο forum,
θα ήθελα να ρωτήσω αν διαθέτει κάποιος κανένα από τα ακόλουθα βιβλία σε ηλεκτρονική μορφή:

-A modern view of geometry, Blumenthal, Dover, Ν.Υ. 1961
-A survey of geometry, Eves H., Allyn of Bacon Inc, Boston 1974
-The Foundations of Euclidean Geometry, Forder H., Dover 1958
-Non-Euclidean Geometry, Bonola R., Dover 1955

Αν είναι εύκολο όποιος μπορεί να με βοηθήσει ας μου στείλει pm.
Ευχαριστώ εκ των πρωτέρων

Βασίλης[/quote]

Καλημέρα σε όλους σας !

 Βασίλη , δυστυχώς δεν θυμάμαι αν έχω κάποιο από αυτά. Ωστόσο , με την ευκαιρία να σημειώσω ότι τελευταία έχουν ξανακυκλοφορήσει δύο από τα καλύτερα βιβλία Ευκλείδειας γεωμετρίας στην αγγλική σε εκδόσεις Dover. Είναι του Johnson και Ν.Α.Court.
 Φθηνά και μοναδικά, αποτελούν τα βιβλία αναφοράς σε κάθε σημαντικό πρόβλημα που παρουσιάζεται ατο αγγλόφωνο κοινό.Το μόνο μειονέκτημα είναι ότι δεν έχουν ασκήσεις ( το ένα έχει μόνο προτεινόμενες) , αλλά από θεωρία είναι πλήρη και μοναδικά.

Καλό Σαββατοκύριακο - Μπάμπης[/quote]

Ευχαριστώ Κύριε Στεργίου. Θα ψάξω να βρω τα συγκεκριμένα που προτείνατε.


calculus
derivative
calculus computations

7(sec^2(t))-7(t^3)

and

antiderrivative of [img]http://i250.photobucket.com/albums/gg247/spikypirate/anti2.png[/img]

In attempts to not just give the answer away...For the first one. Do you know the derivatives of $ \tan(t)$ and $ t^4$? If so, what constants do you need in front of them so that the derivative equals what you have?

i.e. solve $ \frac {d}{dt}(A\tan(t) \plus{} Bt^4 \plus{} C) \equal{} 7\sec^2(t) \minus{} 7t^3$ ... (added later) I guess technically you should throw a third constant, $ C$,  in.

F(0)=0 is the constant

i think i got it thanks
 answer of the 2nd is [img]http://img329.imageshack.us/img329/404/13ue8.gif[/img] ?


Was this forum down a little earlier today?
I tried logging on, and a strange window came up, and when i entered my username and password, it said "You must be an administrator to continue" or something like that... =|

We had a forum update running for 10 minutes today.

[quote="buzzer11"]okay, thank you. and now my profile button at the top is not working...[/quote]Ok, this is a bug thread, I'll fix it.


Given circle $ (O)$ inscribed in quadrilateral $ ABCD, \angle A\equal{}90^o$. Let $ OB\cap AD\equal{}\{F\}, OD\cap AB\equal{}\{G\}$. Prove that $ FG\perp AC$

It is well-known that ⊙I1 and ⊙I2 are tangent at one point on AC
∠A=90 => AN=AQ=r
MN=AM-AN=(AC+AD-CD)/2-r
QP=AP-AQ=(AC+AB-BC)/2-r
ABCD has an incircle => AD+BC=CD+AB => MN=PQ => IL/tan(D/2)=IK/tan(B/2)
=> (r-r1)/(r-r2)= tan(D/2)/ tan(B/2)  => (r-r1)/(r-r2)*(sin(B/2)/sin(D/2))= cos(B/2)/cos(D/2)
=> II1/II2= cos(B/2)/cos(D/2)
Notice IF/IG=(r/cos(B/2))/(r/cos(D/2))= cos(D/2)/cos(B/2)
=> IF/IG=II2/II1  =>  I1I2//FG
From the lemma I mentoned in the beginning , the proof completes

Let $ H,I,J,K$ be the tangencies of $ AB,AD, CD, CB$ and $ (O)$, respectively. $ L \equal{} HI\cap BD\cap KJ$
Since $ AC$ is the polar of $ L$ wrt $ (O)$ we get $ OL\perp AC. (1)$
Let $ M$ be a point on $ LO$ such that $ BM//OD$
We have $ \frac {MB}{OD} \equal{} \frac {LB}{LD}, \frac {OG}{OD} \equal{} \frac {R}{AD \minus{} R} \equal{} \frac {R}{ID}$ then $ \frac {MB}{OG} \equal{} \frac {LB}{LD}.\frac {ID}{R}$
Applying Menelaus's theorem for triangle $ ABD$ with line $ (LHI)$ we obtain $ \frac {LB}{LD} \equal{} \frac {BH}{ID}$
Therefore $ \frac {MB}{OG} \equal{} \frac {BH}{ID}.\frac {ID}{R} \equal{} \frac {BH}{R}$
On the other hand, $ \frac {OB}{OF} \equal{} \frac {AB \minus{} R}{R} \equal{} \frac {BH}{R}$
Thus $ \frac {MB}{OQ} \equal{} \frac {OB}{OF}$ which follows that $ \Delta GOF\sim \Delta MBO$
$ \Rightarrow OM//FG (2)$
From $ (1)$ and $ (2)$ we are done.

See the artice:" A theorem on circumscribed quadrilaterals"  at
http://www.cip.ifi.lmu.de/~grinberg/geometry2.html :)

Dear Mathlinkers,
I rewrite the theorem IV mentioned by Darij Grinberg:
ABCD a quadrilateral, (1) the inscribed circle in ABCD, I the center of (1), E the point of intersection of the perpendicular to AB at A with BI, and F the point of intersection of the perpendicular to AD at A with DI.
Prouve : EF is perpendicular to AC.
The challenge : can we imagine a proof without ratio?
Sincerely
Jean-Louis


What is the convention for writing Re and Im in math mode? I don't mean \Re and \Im, which produce the fancy symbols, because I want it to look exactly like

Re z = a, Im z = b

but in math mode. (Akin to \sin and \lim.)

Thanks.

Just use a text-style font. There are several choices: \text is one, for $ \text{Re}(z)$.

I personally think the \Im and \Re are unnecessarily superfluous and sides, can't even tell it's an I and can barely make out the R. The \i makes another type of i.  Who uses that?  Wouldn't it be better if we defaulted the \Re, \Im, and \i to do:

$ \textbf{Re} f(z)$
$ \textbf{Im} f(z)$
$ z\equal{}x\plus{}\textbf{i} y$

I usually do a

\DeclareMathOperator{\Real}{Re}
\DeclareMathOperator{\Imag}{Im}

in the beginning of tex-documents.


I have 3 fabulous diagrams that I scanned into my computer and saved as pdf things, but how in the world do I put them into my latex file?  I read the instructions for it on the website, but my computer said "Missing number, treated as zero" as an error message, and there's this enormous blank space on the dvi file where the picture should be.  And I don't get how you figure out what numbers to put in the [viewport=100 200 500 650] <-- that thing.  Help!

yeah its frustrating to include pictures, but i dont know what kind of errors result in that error message, so.. all i can say is for the [viewport=100 200 500 650] part, experiment and trial/error.

um the following is only if you get no error message.. if the coordinates of the lower left corner and the upper right corner are $(x_1,y_1)$ and $(x_2,y_2)$, then substitute [viewport=0 0 1000 1000] or extremely large numbers for the second coordinate and that will help you determine where in the axis the picture is located.

from here i just narrow down the white space: if the picture appears too much to the left, decrease the x coordinate of $(x_2,y_2)$ so that the white space to the left of the picture can be cut off. do the same thing for the other dimensions. and then you have to rescale the picture to fit your needs... :huh: its hard
im not sure if this will help anybody

thanks, i figured out that you have to make your latex file into a pdf instead of a dvi, (whatever that means), and then change those numbers a whole bunch.


linear algebra
matrix
algebra
polynomial

If the matrix $A$ of size $n\times n$ is idempotent and is rank deficient i.e., $rank(A)=r$ where $r<n$, can we say that its all its $r$ singular values are $1$ ?

And also how can we prove that for idempotent matrix all the eigenvalues are either $1$ or $0$?

thank you

srihari.

If $A$ is idempotent, $A^2-A=0$, and the minimal polynomial is a divisor of $x^2-x$. Since this polynomial factors completely into distinct linear factors, $A$ is diagonalizaable, and its eigenvalues are all zeros of this polynomial.

can you be please relate your argument to my question?

thank you

sri hari


Find all positive integers $ x,y,z$ such that $ 1\plus{}4^x\plus{}4^y\equal{}z^2$.

wlog $ x\leqq$

than for every $ x$, $ y = 2x - 1$ is a solution since $ (2^{2x - 1} + 1)^2 = 4^{2x - 1} + 4^x + 1$

if $ y > 2x - 1$ we get $ (2^y)^2 = 4^y < 4^y + 4^x + 1 < 4^y + 2\cdot 2^y + 1 = (2^y + 1)^2$ therefore no solutions

if $ y < 2x - 1$ and by noticing that $ z$ must be odd we come to the conclusion that $ m(m + 1) = 4^x(4^{y - x} + 1)$ where $ z = 2m + 1$ which means, due to the fact that $ gcd(m,m + 1) = 1$ and $ gcd(4^x,4^{x - y} + 1) = 1$, that $ 4^x + 1\leq m\leq 4^{y - x} + 1$ which is contradict he assumption $ y < 2x - 1$

so solutions are $ (x,2x - 1)$ for every positive integer $ x$

Shouldn't it $ m(m \plus{} 1) \equal{} 4^{x\minus{}2}(4^{y \minus{} x} \plus{} 1)$ ?

Also, why do you obtain that kind of bound for $ m$? (i.e. $ 4^x \plus{} 1\leq m\leq 4^{y \minus{} x} \plus{} 1$)

i miscalculated a bit but the method's ok.

wlog $ x\leq y$
$ 4^y \plus{} 4^x \plus{} 1 \equal{} 4^x(4^{y \minus{} x} \plus{} 1) \plus{} 1$. denote $ z \equal{} 2m \plus{} 1$ for a non-negative integer $ m$. therefore $ z^2 \equal{} (2m \plus{} 1)^2 \equal{} 4m(m \plus{} 1) \plus{} 1$. this means $ 4^x(4^{y \minus{} x} \plus{} 1) \plus{} 1 \equal{} 4m(m \plus{} 1) \plus{} 1$ which is equivalent to $ 4^{x \minus{} 1}(4^{y \minus{} x} \plus{} 1) \equal{} m(m \plus{} 1)$
now because of $ gcd(m,m \plus{} 1) \equal{} 1$ and $ gcd(4^{x \minus{} 1},4^{y \minus{} x} \plus{} 1) \equal{} 1$, $ 4^{x \minus{} 1}$ has to devide either $ m$ or $ m \plus{} 1$ which in any case  means $ 4^{x \minus{} 1} \minus{} 1\leq m$, therefore (since $ 4^{x \minus{} 1}(4^{y \minus{} x} \plus{} 1) \equal{} m(m \plus{} 1)$ ),   $ 4^{y \minus{} x} \plus{} 1$ has to be devisable by the other of $ m;m \plus{} 1$ which means in any case $ m\leq 4^{y \minus{} x} \plus{} 1$

according to both conclusions we now get $ 4^{x \minus{} 1} \minus{} 1\leq m\leq 4^{y \minus{} x} \plus{} 1$ which cleary implies that $ y\geq2x \minus{} 1$


Also $ \frac {abab\ldots ab} {cdcd\ldots cd} \equal{} \frac {ab} {cd}$. unicorn, you may abuse the "star in circle" important icon.

I'm in high school and can't solve this  :(


[quote="AndrewTom"]Let f(n) = k where k is the minimum number of powers of 2 required to represent n with addition and subtraction ( f(7) =2 since 7 = 2^3 - 2^0). Prove that k does not increase if two powers of two can be equal (2^m and 2^n). Prove that k does not increase if two powers of 2 can be consecutive (2^n and 2^(n+1)) Prove that f(1 + 4 + 4^2 + ... + 4^m) >= m+1.[/quote]
A bit hard to understand.
Can some1 convert to latex?

[quote="AndrewTom"]Let f(n) = k where k is the minimum number of powers of 2 required to represent n with addition and subtraction ( f(7) =2 since 7 = 2^3 - 2^0). Prove that k does not increase if two powers of two can be equal (2^m and 2^n). Prove that k does not increase if two powers of 2 can be consecutive (2^n and 2^(n+1)) Prove that f(1 + 4 + 4^2 + ... + 4^m) >= m+1.[/quote]

Let $ f(n) \equal{} k$ where $ k$ is the minimum number of powers of $ 2$ required to represent $ n$ with addition and subtraction ,$ f(7) \equal{}2$ since $ 7 \equal{} 2^3 \minus{} 2^0$.

Prove that $ k$ does not increase if two powers of two can be equal $ 2^m$and $ 2^n$.

 Prove that $ k$ does not increase if two powers of 2 can be consecutive $ 2^n$ and  $ 2^{n\plus{}1}$.

 Prove that $ f(1 \plus{} 4 \plus{} 4^2 \plus{} ... \plus{} 4^m) \ge m\plus{}1$

 :lol:  :D

Proof of first two parts: If a restriction is lifted, then any solution that applied before the lifting is still valid, so the new minimum cannot be greater than the old minimum.

Seriously, I think you made an error in stating that somewhere.

[hide="third part"]
Each power of 4 is a distinct bit, and there are $ m$ of them; the initial 1 is another distinct bit. That comes to a Hamming weight of $ m\plus{}1$.[/hide]-

[hide="better third part"]Since the sum is currently a sum of powers of two, and there are $ m\plus{}1$ of them, then $ f(1\plus{}4\plus{}4^2\plus{}\dots\plus{}4^m)\leq m\plus{}1$. To prove that is GREATER than or equal to $ m\plus{}1$, we must prove that it is EQUAL to $ m\plus{}1$.

== Lemma ==
$ 2^{2m}$ must be included in the sum, and it is the greatest power of 2 in the sum.

Proof: Let us assume that no power of two is in the sum greater than or equal to $ 2^{2m}$ and find a contradiction.
Let us sum all the lesser powers of two. Once we do that, we get $ 2^{2m}\minus{}1$. But the sum is GREATER than $ 2^{2m}$. Thus there is a power of 2 greater than or equal to $ 2^{2m}$.

Now let us assume that there is a power of two GREATER than $ 2^{2m}$ and find a contradiction.
The sum $ 1\plus{}4\plus{}4^2\plus{}\dots \plus{}4^{2m}<1\plus{}2\plus{}2^2\plus{}\dots \plus{}2^{2m}\equal{}2^{2m\plus{}1}\minus{}1$, and thus the sum is LESS than all powers of two greater than $ 2^{2m}$. Thus there is at least one power of two greater than or equal to $ 2^{2m}$ and there is no power of two GREATER than $ 2^{2m}$, and thus $ 2^{2m}$ is included in the sum, and it is the greatest power of two in the sum.

== End of Lemma ==

We subtract $ 2^{2m}$, and we have the sum of powers of 4 less than or equal to $ 4^{m\minus{}1}$, and from a similar version of the lemma, no powers of 2 greater than or equal to $ 2^{2m\minus{}2}$ exist other than $ 2^{2m}$ that are in the sum, and $ 2^{2m\minus{}2}$ is in the sum. Going down the list, we finally arrive at the sum $ 1\plus{}2^2\plus{}2^4\plus{}\dots \plus{}2^{2m}$. This has $ m\plus{}1$ terms, and thus $ f(1\plus{}4\plus{}4^2\plus{}\dots \plus{}4^m)\equal{}m\plus{}1$. Hence $ f(1\plus{}4\plus{}4^2\plus{}\dots \plus{}4^m)\geq m\plus{}1$.[/hide]

[quote="1=2"][hide="better third part"]

== Lemma ==
$ 2^{2m}$ must be included in the sum, and it is the greatest power of 2 in the sum.
[/hide][/quote]

Is this lemma correct? You are allowed to subtract, after all.

Thanks, Darmani. In the first two parts, both occurrences of "if two powers of two" should read "if no two powers of two".


The instruction is this:

The mode of a list of values is the value that occurs most frequently within the list. A list can have multiple modes.

What I get from this statement is that say that I have a list

1, 2, 2, 2, 2, 2, 4, 6, 6, 7, 7, 10

then the mode is 2??

How is it possible for a list to have multiple modes?? if it has the same amount of values in that list??

Yes, the mode of that list is 2, and the modes of the list 1,1,2,2,3,4,5 are 1 and 2, and the modes of a list in which each term occurs the same number of times are the entire list.


Given a right triangle $ABC$ with $BC=a$, $CA=b$, $AB=c$, and let $I$ be the center of its inscribed circle. Let  $r_1$, $r_2$, $r_3$  and  $r$ be the radii of the inscribed circles of the triangles $\triangle IAB$, $\triangle IBC$, $\triangle ICA$ and $\triangle ABC$, respectively.

Prove that   $r>\frac{(a+b+c)r_1r_2r_3}{br_1r_2+cr_2r_3+ar_3r_1}$.

                                                                                kunny

[quote="kunny"]Given a right triangle $ABC$ with $BC=a$, $CA=b$, $AB=c$, and let $I$ be the center of its inscribed circle. Let  $r_1$, $r_2$, $r_3$  and  $r$ be the radii of the inscribed circles of the triangles $\triangle IAB$, $\triangle IBC$, $\triangle ICA$ and $\triangle ABC$, respectively.

Prove that   $r>\frac{(a+b+c)r_1r_2r_3}{br_1r_2+cr_2r_3+ar_3r_1}$.[/quote]

I can't understand. Is it great kunny's joke?  :lol:
\[ r(br_1r_2+cr_2r_3+ar_3r_1)=brr_1r_2+crr_2r_3+arr_3r_1> br_1r_2r_3+cr_1r_2r_3+ar_1r_2r_3=(a+b+c)r_1r_2r_3, \]
since $r>r_1,r_2,r_3$ for [b]any[/b] triangle and for any point $I$ inside of it.

Am I missing something?

No,you aren't. You are right. This problem is very easy!  :blush:
Thank you,Myth.

kunny

Kunny! What did you mean when you posted this problem?


function
limit
trigonometry
calculus
calculus computations

Evaluate:

$ lim_{x \rightarrow 0^\minus{}} \frac{2(2x\plus{}1) sin ^2 2x}{cos(2x \minus{} 1)}$

[hide="..."]This function is defined and continuous at 0, so substituting in 0, we get that our answer is $ \boxed{0}$. [/hide]

I suspect that if the O.P. were more competent, the question would have been
$ \lim_{x \to 0^ \minus{} } \frac {2(2x \plus{} 1) \sin^2 2x}{\cos(2x) \minus{} 1}.$*
It also would have been placed in the Calculus Computations and Tutorials forum, where it belongs.

* Though even in this case, why we should have $ x \to 0^\minus{}$ instead of $ x \to 0$ is beyond me.

Ahh... René Descartes " i think therefore i am " :wink:

[hide="Outline of what I did for the new limit"]Let $ u \equal{} 2x$. Multiply our given limit by $ \lim_{u \to 0} (\frac{u}{\sin u})^2 \equal{} 1$. Then use L'hopital's. [/hide]

[quote="Zhero"]Then use L'hopital's.[/quote]  $ \lim_{u \to 0} \frac{1 \minus{} \cos u}{u^2}$ is sufficiently important (think Taylor series, for example) that it's probably worth knowing on its own.

$ lim_{x\rightarrow 0^{\minus{}}}\frac{2(2x\plus{}1) sin^{2}2x}{cos(2x\minus{}1)}$
since numerator and denominator are continuous
$ limit\equal{}\frac{limit(numerator)}{limit(denominator)}$
$ \equal{}\frac{0}{cos(\minus{}1)}$
$ \equal{}0$

am i correct?

well i think that it has to do with squeeze theorem. now, the problem is, i don't know, what is the range of $ sin^2(2x)$. Is $ 0  \leq sin^2(2x) \leq 1$?


Fun!  :lol:

In triangle $ABC$, $AB=20cm$, $m\angle A=30$ degrees, and $m\angle C=45$ degrees. Express the number of centimeters in the length of $BC$ in simples radical form.

I would scan my diagram, but it has my answer/work on it in pen.

Its from some kind of a sprint round, don't know the date/level.

Golden State rocks! w00t (just came back from GS/Dallas game)

[hide]Draw the altitude from $B$ to $AC$. Let its intersection with $AC$ be $D$. The length of $AD$ is half the length of $AB$, or $10$ cm, and the length of $BD$ is $10\sqrt{3}$ cm (both found with 30-60-90 triangle $ABD$). The length of $BC$ is $10\sqrt{6}$ cm (found with 45-45-90 triangle $BCD$).

[color=red][size=75]<Trigonometric reference deleted by moderator.>[/size][/color][/hide]

[quote="madskillz1"]uhh I get 10sqr2.[/quote]

$\boxed{correctomoundo}$

For some reason I thought $\text{m}\angle A=60^{\circ}$...

[quote="madskillz1"]uhh I get 10sqr2.[/quote]

That's a completely useless post, you need to post a solution. Posting an answer only is helpful to absolutely no one.

My solution:

[hide]
Draw BD, the height of the triangle from B.
It forms a 45-45-90 triangle and a 30-60-90 triangle.
Using the 30-60-90 triangle (ABD), $BD=\frac{AB}{2}=10$
Then, with the 45-45-90 triangle (BCD), $BC=AB\cdot\sqrt{2}=10\sqrt{2}$

So my answer is $10\sqrt{2}$
[/hide]


function
modular arithmetic
quadratics
number theory
number theory open

Hi,
In his book "Solved and Unsolved Problems in Number Theory", D. Shanks provided some information about Mersenne numbers ($M_q=2^q-1$ , q prime) and also some "open problems".
One of these problems deals with the LLT (Lucas-Lehmer Test) and with using it for representating the relationships between all numbers $0 \leq x < M_q$ as a graph.
D. Shanks provided an example with q=5 and said: "Prove it if you can!"

I'm studying this and I've been able to find properties and to study properties of Mersenne numbers with q=5, 7, 11, 13, 17, 19, 31 .
But I still do not have any proof for the property described by Shanks and for the ones I've found.
Can you help ?

The LLT says that a Mersenne number is prime iff it divides $S_{q-2}$, where $S_0=4 , S_{i+1}=S_i^2-2$.

Let call llt the function: $llt: x \mapsto x^2-2$.
In the following, I will take the example q=5. But everything is (expected to be) true for q and Mq prime.
For simplifying, I will also take numbers x between 0 and (Mq-1)/2, since $(M_q-x)^2-2 \equiv x^2-2 \pmod{M_q}$.

The problem:
If you apply the llt function to numbers between 0 and $\frac{M_q-1}{2}$ and draw the relationships, you get:
- one tree (q=5: starting at: 4, 9, 10, 11, and finishing at 8-0-2)
- and several loops (q=5: 1-1 , 12-13-12 , 3-7-16-6-3).
This is summarized in this [url=http://tony.reix.free.fr/Mersenne/PrimesAndSymmetrySUMMARY.pdf]page[/url].

I know the following properties:
 - There are $2^{q-2}$ [i]roots [/i] of the tree.
 - The tree ends with: $2^{\frac{q+1}{2}}$.
 - The length of the loops is a divisor of q-1.

My question is:
[b]How to prove that the llt function generates 1 tree and these loops for all Mersenne primes ?[/b]
And which theory can help studying the properties of these tree and loops ?

As an example, how one can prove that the number of loops of lengh 30 for q=31 is 17894588 ?
Also, multiplying all numbers of a loop modulo Mq results as +/- 1 (q=5: 12*13=1 , 3*6*15*7=-1). Why ?
Also, multiplying all numbers in a column of the tree results as $\pm 2^{(q+1)/2}$ (q=5: 14*5=8, 4*9*10*11=-8). Why ?
And, if you multiply all numbers of the tree, except 0, but including 2^(q+1)/2 and 2, the result is also +/- 1 . Why ?
And also, multiplying 2 numbers of the same sub-branch in the tree gives a number in the associated sub-branch (q=5: 4*9=5, 10*11=14). Why ?

Tony

This property of Mersenne numbers appears in Chapter "Supplementary Comments, Theorems, and Exercises" of Daniel Shanks' book, page 215 (Edition 1962).
He said: "Develop a general theory for all prime $M_p$, proving the main theorems, [b]if you can[/b]". (I have highlighted "if you can").
I think the explanation of this property could be in Chapter II (The Underlying Structure), [b]but I do not have the book at home[/b] !
Tony

Hi,
I've been warned about a very interesting paper by Jeffrey Shallit:
     On the iteration of certain quadratic maps over GF(p)

His theorem 17 and Corollary 15 provide the formula for computing the number of cycles for Mersenne primes under LLT (x^2-2) .
With PARI/gp, the formula can be computed only up to q=127 .

A preprint version is available at:
   http://www.cs.uwaterloo.ca/~tmjvasig/papers/newvasiga.ps

Studying his Theorem 5 and Corollary 3, I've been able to find that the formula giving the number of cycles for Mersenne primes (2^q-1) under x^2 seems independent on q and seems already well-known.
I'll open another Thread in AOPS for people interested in proving that.

Regards,

Tony


What is the sum of the tens digit and the units digit in the decimal
representation of $ 9^{2004}$?

every even power of 9 results in the last digit equaling one.

also, the pattern for the tens digit is as follows:

9^2=8
9^4=6
9^6=4
9^8=2
9^10=0
9^12=8
...

it takes 5 turns to complete the cycle.

thus, u must first find the number of turns (2004/2)he remai

and find the remainder when divided by 5 (2).

go an extra two turns to find the tens digit of 6.

add that to 1, and the answer is [b]7[/b], i think...


trigonometry
geometry
3D geometry
algebra unsolved
algebra

[quote="jagdish"]Solve for  x and y If
$ x^3 \minus{} 3xy^2 \equal{} 18$ (1)
$ 3x^2y \minus{} y^3 \equal{} 26$ (2)[/quote]

Obviously $ x,y\neq 0$. Let then $ y \equal{} ux$ and the equations become :

$ x^3(1 \minus{} 3u^2) \equal{} 18$
$ x^3(3u \minus{} u^3) \equal{} 26$

And so $ 18(3u \minus{} u^3) \equal{} 26(1 \minus{} 3u^2)$ $ \iff$ $ 18u^3 \minus{} 78u^2 \minus{} 54u \plus{} 26 \equal{} 0$ $ \iff$ $ (3u \minus{} 1)(6u^2 \minus{} 24u \minus{} 26) \equal{} 0$ and so $ u\in\{\frac 13,2 \minus{} 5\frac {\sqrt 3}3,2 \plus{} 5\frac {\sqrt 3}3\}$

And since $ x \equal{} \sqrt [3]{\frac {18}{1 \minus{} 3u^2}}$ and $ y \equal{} ux$, we get the three solutions :

$ (x,y)\in\{(3,1),$ $ ( \minus{} \frac {3 \minus{} \sqrt 3}2, \minus{} \frac {1 \plus{} 3\sqrt 3}2),$ $ ( \minus{} \frac {3 \plus{} \sqrt 3}2,\frac {3\sqrt 3 \minus{} 1}2)\}$

[quote="hsbhatt"]If we are solving for real x and y, we could say:

$ (x \plus{} iy)^3 \equal{} 18 \plus{} 26i$ which means $ (x \minus{} iy)^3 \equal{} 18 \minus{} 26i$

So that $ x \equal{} \frac {\sqrt [3] {18 \plus{} 26i} \plus{} \sqrt [3] {18 \minus{} 26i} }{2}$ and $ y \equal{} \frac {\sqrt [3] {18 \plus{} 26i} \minus{} \sqrt [3] {18 \minus{} 26i} }{2i}$ and so on[/quote]
Rather nice, but this gives as solutions :

$ x_1\equal{}\sqrt{10}\cos(\frac{\arcsin(\frac{13}{5\sqrt{10}})}3)$ and $ y_1\equal{}\sqrt{10}\sin(\frac{\arcsin(\frac{13}{5\sqrt{10}})}3)$

$ x_2\equal{}\sqrt{10}\cos(\frac{\arcsin(\frac{13}{5\sqrt{10}})\plus{}2\pi}3)$ and $ y_2\equal{}\sqrt{10}\sin(\frac{\arcsin(\frac{13}{5\sqrt{10}})\plus{}2\pi}3)$

$ x_3\equal{}\sqrt{10}\cos(\frac{\arcsin(\frac{13}{5\sqrt{10}})\plus{}4\pi}3)$ and $ y_3\equal{}\sqrt{10}\sin(\frac{\arcsin(\frac{13}{5\sqrt{10}})\plus{}4\pi}3)$

And, to be honest, I prefer my result ...

So do I.  :D

Only if the numbers were nicer, we could have used DeMoivre theorem to arrive at the cube roots easily. In that case, this method would prove useful.

Otherwise we could just as well leave it in this form


Show that the integer part of $(3+\sqrt{5})^{n}$ is odd for all $n\in\mathbb{N}$.

Exactly as the solution of mine. I'm not very good to develop interesting problems.


logarithms
complex numbers
calculus
calculus computations

${\sum_{n=0}^{\infty }\alpha_{n}x^{n}=\sum_{n=0}^{\infty }\beta_{n}\gamma^{x n}}$

what is the relation of ${\alpha}$ and ${\beta}$ ?!

Using MacLaurin expansion on $\gamma^{xn}= e^{\ln(\gamma)nx}$:

\[{\gamma^{xn}= \sum_{i=0}^\infty \frac{(\ln(\gamma)n)^{i}}{i!}}x^{i}\]

Substituting this then yields

\[{\sum_{i=0}^\infty \beta_{i}\gamma^{xi}= \sum_{i=0}^\infty \beta_{i}(\sum_{n=0}^\infty \frac{(\ln(\gamma)i)^{n}}{n!}}x^{n})\]

Since $\beta_{i}$ is independent of the variable $n$ in the above equation, the right hand side is also equal to

\[{\sum_{i=0}^\infty (\sum_{n=0}^\infty \beta_{i}\frac{(\ln(\gamma)i)^{n}}{n!}}x^{n})\]

The latter can also be rewritten as (switching sums)

\[{\sum_{n=0}^\infty (\sum_{i=0}^\infty \beta_{i}\frac{(\ln(\gamma)i)^{n}}{n!}}x^{n})\]

and since $x^{n}$ is independant of $i$, we have

\[{\sum_{n=0}^\infty (\sum_{i=0}^\infty \beta_{i}\frac{(\ln(\gamma)i)^{n}}{n!}}) x^{n}\]

Because this has to be equal to

\[\sum_{n=0}^\infty \alpha_{n}x^{n},\]

we compare coefficients to obtain

\[{\alpha_{n}= \sum_{i=0}^\infty \beta_{i}\frac{(\ln(\gamma)i)^{n}}{n!}}\]

PS> I'm a bit in a hurry now, so im not sure whether this solution's correct or not (i.e. I havent had the time to check)

i thougth this solution. but....

${e^{x}=\sum_{i=0}^{\infty }\frac{x^{i}}{i!}}$

${x\to \log (\gamma ) n x}$

${e^{\log (\gamma ) n x}=\sum_{i=0}^{\infty }\frac{(\log (\gamma ) n x)^{i}}{i!}}$

${(\log (\gamma ) n x)^{i}\neq (\log (\gamma ) n)^{i}x^{i}}$

Because

${\left( \begin{array}{ll}\gamma & x \end{array}\right)\in \mathbb{C}\land n\in \mathbb{Z}}$

but thanks for your solution... but this is not a general solution

[quote="Thales418"]
${(\log (\gamma ) n x)^{i}\neq (\log (\gamma ) n)^{i}x^{i}}$
Because
${\left( \begin{array}{ll}\gamma & x \end{array}\right)\in \mathbb{C}\land n\in \mathbb{Z}}$
[/quote]
What do you mean? :? The non-negative integer power of the product of complex numbers is always the same as the product of the powers. The only weak point in the above solution is switching the sums around. It requires some justification, of course.

n is integer....

hauhua

forget it

the last post didnt exist


limit
real analysis
real analysis unsolved

the sequence $ (y_n)$ of reals from $ [0,1)$ satisfies
\[ \lim_{n\to\infty}\frac {1}{\sqrt {1 \minus{} y_n} \plus{} \sqrt {1 \plus{} y_n}} \equal{} \frac {1}{2}
\]
Prove that $ y_n$ is convergent.

Any convergent subsequence converges to 0, which means the whole sequence converges to 0


geometry
incenter
ratio
trigonometry
geometry proposed

Let $ABC$ be a triangle with $\angle A = 60^{\circ}$, and let $I$ be the incenter of triangle $ABC$. Denote by $P$ the point on $BC$ that satisfies $3\cdot BP = BC$. The point $F$ lies on $AB$ and $IF$ is parallel to $AC$. Show that $\angle BFP = \angle FBI$.

[i]Alternative formulation.[/i] Let $ABC$ be a triangle with $\angle A = 60^{\circ}$, and let $I$ be the incenter of triangle $ABC$. Denote by $P$ the point on $BC$ that satisfies $3\cdot BP = BC$. The point $F$ lies on $AB$ and $IF$ is parallel to $AC$. Show that $\angle BFP = \frac{\angle B}{2}$.

[quote="j555"][i]Alternative formulation.[/i] Let $ ABC$ be a triangle with $ \angle A \equal{} 60^{\circ}$, and let $ I$ be the incenter of triangle $ ABC$. Denote by $ P$ the point on $ BC$ that satisfies $ 3\cdot BP \equal{} BC$. The point $ F$ lies on $ AB$ and $ IF$ is parallel to $ AC$. Show that $ \angle BFP \equal{} \frac {\angle B}{2}$.[/quote]

Denote a = BC, b = CA, c = AB. Let L be the foot of the bisector BI of the angle $ \angle B$ and let $ IF \parallel CA$ cut BC at G. The incenter I divides the bisector the $ \angle B$ in the ratio $ \frac {BI}{LI} \equal{} \frac {c \plus{} a}{b}$, hence, the triangles $ \triangle FBG \sim \triangle ABC$ are similar with similarity coefficient $ \frac {c \plus{} a}{a \plus{} b \plus{} c}$. Let a cevian from F forming the angle $ \frac {\angle B}{2}$ with FB cut the opposite side BG at P'. Using the sine theorem for the triangle $ \triangle BP'F,$

$ \frac {BP'}{BF} \equal{} \frac {\sin \frac {\widehat B}{2}}{\sin \frac {3 \widehat B}{2}} \equal{} \frac {1}{3 \minus{} 4 \sin^{2}\frac {\widehat B}{2}} \equal{} \frac {1}{1 \plus{} 2 \cos \widehat B}$

Substituting for $ BF$ and $ 2 \cos \widehat B,$

$ BP' \equal{} \frac {c^{2}a(c \plus{} a)}{(a \plus{} b \plus{} c)(ca \plus{} c^{2} \plus{} a^{2} \minus{} b^{2})}$

Using the cosine theorem formula $ a^{2} \minus{} b^2 \equal{} c^{2} \minus{} bc$ for $ \angle A \equal{} 60^\circ,$

$ BP' \equal{} \frac {ca(c \plus{} a)}{(a \plus{} b \plus{} c)(a \plus{} 2c \minus{} b)}$

Multiplying through the denominator and using $ a^{2} \minus{} b^{2} \equal{} c^{2} \minus{} bc$ again,

$ (a \plus{} b \plus{} c)(a \plus{} 2c \minus{} b) \equal{} 3c(c \plus{} a)$

Consequently, $ BP' \equal{} \frac {a}{ 3}$, i.e., the points $ P' \equiv P$ are identical.

I have a synthetic nice solution:

Let $G$ be apoint on $AC$ such that $IG$ is parallel to $AB$. It is easy to prove that $IFG$ is equilateral triangle. Let $X$ and $Y$ be points symetric to $I$ by $AB$ and $AC$, repectively. Points $X$, $F$, $G$, $Y$ are colinear and $XY = 3XF$. We know also that $\angle XBC+\angle YCB = 180^\circ$, so $BX$ is parallel to $CY$. Thus, also $PF$ is parallel to $BX$, so $\angle BFP = \angle FBX = \angle FBI$. :D

[quote="j555"][i]Alternative formulation.[/i] Let $ABC$ be a triangle with $\angle A = 60^{\circ}$, and let $I$ be the incenter of triangle $ABC$. Denote by $P$ the point on $BC$ that satisfies $3\cdot BP = BC$. The point $F$ lies on $AB$ and $IF$ is parallel to $AC$. Show that $\angle BFP = \frac{\angle B}{2}$.[/quote]
[color=darkblue][b]Lemma.[/b] $A=60^{\circ}\Longleftrightarrow a^{2}=b^{2}+c^{2}-bc\Longleftrightarrow (a+b+c)(2a+b+c)=3(a+b)(a+c)\ .$

[b]Proof of the proposed problem.[/b] Denote $S\in (BC)$ so that $AS\parallel FP$. Prove easily that $BF=\frac{a(a+c)}{a+b+c}$ and $\frac{BP}{BS}=\frac{BF}{BA}$ $\Longrightarrow$ $SB=\frac{a(a+b+c)}{3(a+c)}$ $\Longrightarrow$ $SB=\frac{a(a+b)}{2a+b+c}$ and $SC=\frac{a(a+c)}{2a+b+c}$, i.e. $\frac{SB}{SC}=\frac{a+b}{a+c}$. Using the my problem from http://www.mathlinks.ro/Forum/viewtopic.php?t=105533 results $m(\widehat{BAS})=\frac{B}{2}$, i.e. $m(\widehat{BFP})=\frac{B}{2}\ .$[/color]


the sum of 2 #s is 24. 7 times the smaller # is the same as 5 times the larger #. find the smaller #

$ \alpha \plus{} \beta \equal{} 24$

$ 7\alpha \equal{} 5\beta$

$ \alpha \equal{} \frac{5}{7}\beta$

$ \alpha \plus{} \frac{5}{7}\beta \equal{} 24$

$ \frac{12}{7}\alpha \equal{} 24$

$ \alpha \equal{} \boxed{14}$

Izzy, you were supposed to subtract in your 4th step. This solution makes the smaller number higher that the bigger number!

Here's the real solution.

$ a\plus{}b\equal{}24$
$ 7a\equal{}5b$

$ b\equal{}24\minus{}a$

So $ 7a\equal{}5(24\minus{}a)$

$ 7a\equal{}120\minus{}5a$

$ 12a\equal{}120$

$ a\equal{}10$

Izzy, your solution actually just reverses the numbers.


geometry
trapezoid
analytic geometry
circumcircle

[quote="Go Around the Tree"]Find the radius of the circle which circumscribes polygon ABCD with coordinates A (0,2) B (2,2) C (4,0) D (-2,0)[/quote]
[hide]Too lazy to make sure this is cyclic, the distance from two opposite vertices is $2 \sqrt{5}$. So the radius is $\sqrt{5}$[/hide]

[hide]
Shift the co-ordinates so that the points (3, 0) and (1, 2) lie on the circle which is symmetric about the y-axis.

$x^{2}+(y-h)^{2}= r^{2}$
$9+h^{2}= r^{2}$
$1+(2-h)^{2}= r^{2}$

$h =-1, r = \sqrt{10}$
[/hide]

[quote="Go Around the Tree"]Find the radius of the circle which circumscribes polygon ABCD with coordinates A (0,2) B (2,2) C (4,0) D (-2,0)[/quote]
[hide]The circumcenter is equidistant from the vertices, so we use the distance formula to find it.

$\sqrt{(x-0)^{2}+(y-2)^{2}}=\sqrt{(x-2)^{2}+(y-2)^{2}}=\sqrt{(x-4)^{2}+(y-0)^{2}}=\sqrt{(x+2)^{2}+(y-0)^{2}}$

We can square everything and simplify.

$x^{2}+(y-2)^{2}=(x-2)^{2}+(y-2)^{2}=(x-4)^{2}+y^{2}=(x+2)^{2}+y^{2}$

From the first two, we know $x^{2}=(x-2)^{2},$ so $x=1.$ We substitue $x$ into the middle two to find $y=-1.$

The circumcenter is at $(1,-1),$ so $r=\boxed{\sqrt{10}}.$[/hide]

[quote="Farenhajt"]Who said that the diagonals bisect one another  :?:

[hide="Correct answer"]$R=\sqrt{10}$[/hide][/quote]
They don't have to, I made a mistake because I thought that the middle of the diagonal was the center of the circle since it was equidistant from two of the points :roll:

[quote="Go Around the Tree"]so basically, what you have to know is that the circumcenter is equidistant from the vertices.

So you just plug in distance formula and simplify and solve for x and y.

Then find the coordinates of the circumcenter.

Then wah-lah, the answer is there  :)[/quote]
That's exactly what I did, except I actually did the work.


search
inequalities proposed
inequalities

[quote="apollo"]a+b+c=1,a,b,c>0
2($\frac{b}{a}$+$\frac{c}{b}$+$\frac{a}{c}$) $\geq$ $\frac{1+a}{1-a}$+$\frac{1+b}{1-b}$+$\frac{1+c}{1-c}$[/quote]

Welcome to the forum, apollo!

Next time please check the resources section at the right top (on mathlinks.ro) and use the search engine before posting.

German TST 2005: http://www.mathlinks.ro/resources.php?c=64&cid=54&year=2005
Your Problem: http://www.mathlinks.ro/viewtopic.php?p=161786#161786


If x, y, z are nonnegative numbers with x + y + z = 3, prove that
√x + √y + √z ≥ xy + yz + zx.

it's really have been posted here:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=phuong&t=82512


[quote="Stewie"]Is it true that we can find in the decimal representation of $ \pi$ any finite decimal string?[/quote]

No, simply because $ \pi$ is not an element of $ Q$

FOURRIER, that answer doesn't make sense. For instance, the decimal created by appending the natural numbers to each other ($ .012345678910111213141516171819202122\ldots$) has the required property, but it is certainly not rational. As a matter of fact, any element with this property can't be rational.

I'm sort of inclinced to go with archimedes' answer, but I don't know if it's actually "random" enough.

[quote="archimedes1"]Doesn't that work for any random infinite sequence of digits?[/quote]

What does this mean, exactly?  In what sense is the decimal representation of a (very specific) real number "random"?  In fact, the proof of the answer to this question is almost certainly very difficult.  For a related (but substantially stronger) idea, see http://mathworld.wolfram.com/NormalNumber.html

Pi doesn't have random representations of digits, it's just infinite, so there's no end to more digits.

Anyways, but the first few digits of pi are: 3.141592657...

Just because a number is irrational doesn't mean any sequence of digits can be found in it. Take for example Mathnerd314's Number:

$ M\equal{}0.42442444244442444442444444244444442444444442...$

where each time you put one more $ 4$ before the $ 2$

This is irrational (because the decimal will never repeat), but it obviously consists of only $ 4$s and $ 2$s, so you'd never find the string $ 65537$ in the digits.

Maybe pi is in this category of irrational numbers. Or maybe pi is more like MellowMelon's number $ 0.123456789101112131415...$, where you can obviously find every string of digits somewhere down the line.

JBL, about the "normal number" link, how would you define "average frequency" for an infinite number of digits?

[quote="mathnerd314"]how would you define "average frequency" for an infinite number of digits?[/quote]

Merely the limit of the average frequency for $ n$ digits as $ n \to \infty$.  Naturally, this makes proofs difficult, and you'll note that the link mentions that the only numbers thus far proven to be normal have been artificially constructed to be so.


The Mathematical Definition (From Wikipedia):
In a more general mathematical sense, a histogram is simply a mapping that counts the number of observations that fall into various disjoint categories (known as bins), whereas the graph of a histogram is merely one way to represent a histogram. Thus, if we let N be the total number of observations and n be the total number of bins, the histogram hk meets the following conditions:
[img]http://upload.wikimedia.org/math/1/6/e/16ee15dea3e0386c47bb779d0065955b.png[/img]

 Help On/Question: Can someone help me on explaining the formula. Such as
 What the E-like shape stands for, how to work the formula, How to find the index over the bins.

Cumulative Histogram

A cumulative histogram is a mapping that counts the cumulative number of observations in all of the bins up to the specified bin. That is, the cumulative histogram Hk of a histogram hk is defined as:

[img]http://upload.wikimedia.org/math/7/5/1/75198c4effbe455c6978c0b6a3d75a10.png[/img]

Help On/Question: Again, How to work this formula, step by step?

$\sum$ is the summation symbol.  It takes the variable defined below it (here, $k = 1$), plugs that value into the expression immediately after the symbol (known as the "summand"), and increments $k$ until it reaches the value above it (here, $k = n$), adding those new values of the "summand" until the summation terminates.

In other words,

$\sum_{k=1}^{n}h_{k}= h_{1}+h_{2}+h_{3}+...+h_{n}$

The notation makes the whole thing appear far more complicated than it is.  Here is a concrete example:  Given the data

$\{ 1, 2, 2, 4, 4, 4, 6, 9, 9, 9 \}$

We can construct the histogram

$h_{1}= 1$ (because there is one appearance of $1$)
$h_{2}= 2$ (because there are two appearances of $2$)
$h_{3}= 0$ (because there are no appearances of $3$)
$h_{4}= 3$ (because there are three appearances of $4$)
$h_{5}= 0$ (because there are no appearances of $5$)
$h_{6}= 1$ (because there is one appearance of $6$)
$h_{7}= h_{8}= 0$ (because there are no appearances of $7$ or $8$)
$h_{9}= 3$ (because there are three appearances of $9$)

In this example, the total number of bins is $9$ because the data ranges from $1$ to $9$, and the total number of observations is $10$.  This histogram satisfies

$10 = \sum_{k=1}^{9}h_{k}= h_{1}+h_{2}+...+h_{9}= 1+2+0+3+0+1+0+0+3 = 10$

Now, the cumulative histogram, as far as it applies to this example, is just a measure of how many points in our data set are less than or equal to some point in the data.  The notation is very confusing because it uses $k'$, which is the index of the new summation, as well as using $k$, which is the index of the cumulative histogram $H_{k}$.  I wouldn't worry too much about this one, but in the interest of completeness:

$H_{1}= h_{1}= 1$
$H_{2}= h_{1}+h_{2}= 3$
$H_{3}= h_{1}+h_{2}+h_{3}= 3$
$H_{4}= ... = 6$
$H_{5}= ... = 6$
$H_{6}= ... = 7$
$H_{7}= ... = 7$
$H_{8}= ... = 7$
$H_{9}= ... = 10$

[quote="KawaiiZia"]I'm confused - - Can you explain it in a much more easier way that a 13 year old would understand xD

See, I had to do 6 Histograms of the features of the people in my class that were numbered from 1-24; Which would be the total number of observations? (24) On The X Axis; it was the length of the feature/etc in cm, and there were 4 bars in the x axis, so 4 categories?... Uhh..So how does the formula work with that?[/quote]

Ignore the formula, since it doesn't tell you anything useful about actually making histograms.  All it tells you is that when you're dividing up your observations into categories, the sum of the number of observations in each category should equal the total number of observations.  That's much simpler, right?


superior algebra
superior algebra unsolved

Let $G$ be a finite group and $\alpha\in Aut(G)$. Let $F=\{g\in G|g^\alpha = g^{-1}\}$ and $|F|>\frac34 |G|$. Prove that $F=G$ and $G$ is abelian. Can the constant $\frac34$ be replaced by a smaller one?
[i]($g^\alpha$ denotes the image of $g$ under $\alpha$)[/i]

This problem seems to be famous.
[url]http://mathforum.org/library/drmath/view/66879.html[/url]

It is a beautiful question is it not? :)


Putnam
arithmetic sequence
number theory unsolved
number theory

Let $ a_1,...,a_k$ be different positive integers and suppose that for every natural number n we have $a_1\cdot...\cdot a_k$ divides $ (n+a_1)\cdot...\cdot (n+a_k)$. Then prove that these numbers are in fact the numbers 1,2,...,k. This is the easy and nice part of the problem, given at an oral at X. The "little" extension: if we suppose only that the divisibility holds for prime values of n, what are the possible values of the a_i?

The first part comes from Putnam 1982.

Pierre.

Yes, part 1 is fairly obvious. In fact, it is not difficult to show that if
\[
\frac{(x+a_1) \cdots (x+a_k)}{a_1 \cdots a_k} \in \mathbb{Z}
\]
for $k+1$ consecutive terms $x$ of an arithmetic progression $an+b, (a,b)=1$, then it holds for all $x$ (and hence the $a_i$ must be a permutation of $1,\ldots,k$). Since, as was recently shown by Green and Tao, there exists an arithmetic progression of length $k+1$ that consists solely of primes, part 2 follows as well.   :P


inequalities
function
inequalities proposed

$ \bigstar$ Let $ a,b,c,d$ be positive real numbers such that $ a+b+c+d=4$. Prove that:
\[ a^{3}+b^{3}+c^{3}+d^{3}+10(ab+bc+cd+da+bd+ca)\le 64 \]
There is a simple solution exist  :wink: Let kill it  :P
Well-come mathlinker  :wink:

@mornik: I've tried to find an error in your solution, but I couldn't... But I have to say that this inequality seems pretty obvious... For innstance, after homogenizing it you get a sum of nonnegative sums on the right side, and a zero on the left side (If anyone is interested how does it look like, should just multiply those abs with (a+b+c+d) and 64 with the third power of (a+b+c+d)...

It is probably $a+b+c+d=4,$ as it yields an equality case and is still true. Indeed,
\[(a+b+c+d)^{3}-a^{3}-b^{3}-c^{3}-d^{3}-\frac{5}{2}(a+b+c+d)(ab+bc+cd+da+ac+bd)= \]

\[=\frac{1}{4}\sum_{\mathbf{symmetric}}ab(a+b)+bc(b+c)+ca(c+a)-6abc\geq 0. \]

@bodan: Yes, you're right... $a+b+c+d=4$ makes perfect sense.

[quote="hsiljak"]@bodan: Yes, you're right... $ a+b+c+d=4$ makes perfect sense.[/quote]
I am so sorry, I have made mistake in typing .... I have edited it  :wink:
My solution only use mixing variable method and another small technique . I didn't use homogenizing it  :roll:
Maybe above solution is depends upon computer  :P Have you got any solution for it?  :wink:

[quote="The soul of rock"]bodan's proof is too nice. It's simple than mixing variable!
Do you think mixing doesn't take a lot of calculating?[/quote]
Yes, I think so :D Notice that, the inequality is equivalent to:
\[ a^{3}+b^{3}+c^{3}+d^{3}+16\le 5(a^{2}+b^{2}+c^{2}+d^{2})\]

[b]Another solution.[/b]

[i]Case I.[/i] If $ 0\leq a,b,c,d\leq 3$ we have $ a^{3}-5a^{2}\leq-7a+3$ etc., which is equivalent to $ (a-1)^{2}(3-a)\geq 0.$ Sum up cyclically and get the desired result.

[i]Case II.[/i] If some number exceeds $ 3$, WLOG $ a\geq 3,$ we have $ a^{3}+b^{3}+c^{3}+d^{3}\leq a^{3}+(b+c+d)^{3}=a^{3}+(4-a)^{3}$ and $ a^{2}+b^{2}+c^{2}+d^{2}\geq a^{2}+\frac{(b+c+d)^{2}}{3}=a^{2}+\frac{(4-a)^{2}}{3}.$ Now the inequality follows from
\[ a^{3}+(4-a)^{3}+16\leq 5\left(a^{2}+\frac{(4-a)^{2}}{3}\right)\quad \Leftrightarrow \quad (4-a)(2a-5)\geq 0 \]
which is obvious.

EDIT: changed sign in first case

[quote="zaizai-hoang"]My solution is as same as you  :P  In case 1, this is also my solution but case 2, I use MV method.
In case 2 of you,  why $ (4-a)(2a-5)\ge 0$ with $ a\le 3$ ?[/quote]

I have made a typo: In the first case $ 0\leq a,b,c\leq 3$ and we have $ a^{3}-5a^{2}\leq-7a+3\Leftrightarrow (a-1)^{2}(3-a)\geq 0$ and not $ a^{3}-5a^{2}\leq-7a+3\Leftrightarrow (a-1)^{2}(a-3).$ In the second case we have $ 3\leq a\leq 4\Rightarrow (4-a)(2a-5)\geq 0.$

nice proof, bodan  :wink: I am interested in your solution. Then, I will post my proof.
We consider 2 case. Assume that: $ a=max[a,b,c,d]$
Case 1: $ a\le 3$ we are done :D
Case 2: we put
$ f(a,b,c,d)=5(a^{2}+b^{2}+c^{2}+d^{2})-16-(a^{3}+b^{3}+c^{3}+d^{3})$
Then we have:
$ f(a,b,c,d)-f(a,b,\frac{c+d}{2},\frac{c+d}{2})=(c-d)^{2}\left(\frac{10+3a+3b-12}{4}\right)$
After that, by SMV theorem we only prove that the following inequality:
\[ 5(a^{2}+3t^{2})-16-a^{3}-3t^{3}\ge 0\]
With $ a=4-3t$. We have, above inequality is equivalent to:
\[ 24t(t-1)^{2}\ge 0\]
We are done  :P

You don't need to use SMV. If one of the variables is $ \geq 3$ then if $ a^{3}+b^{3}+c^{3}+d^{3}-5(a^{2}+b^{2}+c^{2}+d^{2})$ is maximum, the others are equal because the function $ x^{3}-5x^{2}$ has a single inflexion point.

[b]A third proof:  :D (very similar to yours and the second one) [/b] If $ a,b,c,d\leq 3$ we are done as in the previous proof. If not, the function $ f(x)=x^{3}-5x^{2}$ has a single inflection point, thus the maxima occurs when at most two of the variables are different. As $ a\geq 3,$ other variables cannot be greater than $ 3$ so the rest of the variables are equal to $ \frac{4-a}{3}\leq 1/3.$ Now
\[ a^{3}+\frac{(4-a)^{3}}{9}+16-5a^{2}-\frac{5(4-a)^{2}}{3}=\frac{8}{9}(a-1)^{2}(a-4)\leq 0. \]

All your solutions are very nice, bodan. Thank you,


prove that at least 99% of these numbers are composite numbers :

$ 10^1$ $ \plus{}$ $ 1$ , $ 10^2$ $ \plus{}$ $ 1$,$ ...$, $ 10^{2000}$ $ \plus{}$ $ 1$

$ a+b$ divides $ a^n+b^n$ for all odd natural n
so odd numbers are all covered except for n=1 which is prime
write the number as $ (10^k)+1$
and if k=$ 2^m$*$ l$
where m is a natural number and where l is an odd natural
then u can write the number as $ (10^(2^m))^l+1$
so the only numbers which may be prime are $ 10^k$
where k is a natural number and a power of 2
from 1 to 2000 there are only 10 powers of 2(excluding 1)
so only for 11 n's the number is prime
so the number is composite $ 99.45%
$ of the times


Sean:

$x_1=0,12345678910111213\ldots$

$x_2=0,149162536496481100\ldots$

$x_3=0,182764125216343512\ldots$

$x_4=0,1168125662512962401\ldots$

... y así sucesivamente.

La expansión decimal de $x_n$ se obtiene escribiendo la n-potencia de cada natural, una al lado de la otra.

Existe, en esta sucesión, algún $x_i$ racional??

[quote="Guía Rojo"][quote="José"]Problema intrigante... de qué año es??? qué país lo propuso?[/quote]de 1986, y no tengo información sobre el país que lo propuso...[/quote]

OK

Me parece que la respuesta es no. Procediendo por contradicción, supongamos que  $x_k$ es racional, para algun $k$ natural. Entonces tenemos tres casos:

PRIMER CASO: $x_k$ es decimal exacto, esto no puede ocurrir, pues en la representación decimal de $x_k$ existen infintos digitos distintos de cero, basta considerar las potencias $k$-esimas de los numeros de la forma $10^n$, para cualquier n natural.

SEGUNDO CASO: $x_k$ es periodico puro, esto es a partir del digito que ocupa la posición $l$ en la expresión desarrollada de $x_k$ la secuencia se repite. Como la secuencia comienza en 1, en cualquier bloque de digitos de longitud $l$ de la expresión desarrollada de $x_k$ debemos encontrar al menos un digito 1. Además, los digitos de $10^l$, estan en la expresion de $x_k$, esto es hay un bloque de $lk$ ceros, este  su vez contiene un bloque de $l$ ceros consecutivos, esto es hay un bloque de $l$ digitos que no contiene ningun numero 1. Esto es una contradiccion.

TERCER CASO: $x_k$ es periodico mixto, similar al caso anterior, con pequeñas variaciones.


Find all pairs of integers $a,b$ such that $a+b$ and $ab+1$ are integers powers of two.

I think the only solutions are (1,$2^{n}$-1).


Show that if a prime number p is divided by 30 then the remainder is a prime or 1.

Let's ask ourselves what numbers are possible remainders when dividing a prime number with $ 30$. Let's suppose that $ p>30$ (if $ p<30$, it is obviously its own remainder). If the remainder from dividing $ p$ with $ 30$ isn't prime, we can denote it by $ n$. Now, either $ n\equal{}1$ (which is alright for us) or $ n\equal{}ab$ ($ a\geq b>1$). However, obviously $ b\leq 5$ (otherwise $ n\geq 36$, which is not possible). So, $ b\equal{}2$, $ b\equal{}3$ or $ b\equal{}5$. However, $ p\equal{}30k\plus{}ab$ and since $ 30\equal{}2\cdot 3\cdot 5$, $ b|30$. Since $ b|30k\plus{}ab$, $ b|p$, which is a contradiction since $ p$ is prime.
The conclusion is: either $ n$ is prime or $ n\equal{}1$.

Primes greater than 3 are of form 6k+1 or 6k-1
(6k+1) mod 30 = (6(kmod5)+1) mod 30
(6k-1) mod 30 = (6(kmod5)-1) mod 30
kmod5 = {0,1,2,3,4}

The remainder when primes divided by 30 can be 1,7,13,19,29,5,11,17,23.(All are primes or 1)
Here, remainder can't be 25 because prime will be 30c+25 which is not possible.
remainder is 5 only in the case when prime is 5.

Is there a good reason why the same problem was posted in two different forums by two different people at around the same time?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=211371]see here[/url]

I doubt it's some kind of contest or journal problem since it's both purely computational and a well-known lemma.

[hide="I mean..."] There are $ 7$ primes between $ 1$ and $ 30$ not including $ 2, 3, 5$ and $ \varphi(30) \equal{} 8$, so $ \{ 1, \text{the primes} \}$ is precisely the group of units $ \bmod 30$. [/hide]


when 3 dice are rolled, what is the probability that all three of them show the same number?

[hide]$1/36$ It doesn't matter what the first dice rolled is. The second dice has a $1/6$ chance of being the same as the first, and the third dice has a $1/6$ chance of being the same as the first two. $1/6*1/6=1/36$[/hide]

Find $\text{P(sum of 3 numbers is even or at least 2 of the numbers are the same)}$.


Is there a positive integer $ k$ such that none of the digits $ 3,4,5,6$ appear in the decimal representation of the number $ 2002!\cdot k$?

Yes. There is also a positive integer $ k$ such that none of the digits $ 2,3,4,5,6,7,8,9$ appear in the decimal representation of the number $ 2002!\cdot k$ because it has the form $ 11...100...0$.

Well, in the sequence $ 1, 11, 111, 1111, ...$ there are two numbers $ k, l$ such that $ k\equiv l(mod2002!)$. Their difference is divisible by $ 2002!$ and has the form $ 11...100...0$.


We have finite number of distinct shapes in plane. A "[i]convex Kearting[/i]" of these shapes is covering plane with convex sets, that each set consists exactly one of the shapes, and sets intersect at most in border.
[img]http://aycu30.webshots.com/image/4109/2003791140004582959_th.jpg[/img]
In which case Convex kearting is possible?
1) Finite distinct points
2) Finite distinct segments
3) Finite distinct circles

It's possible for points and disks, but not for segments.

First a counterexample for segments: let $ABC$ be a triangle, and consider points $D, E, F$ on the rays $(CB,(BA,(AC$ respectively such that $A$ lies between $B$ and $F,\ B$ between $C$ and $D$, and $C$ between $A$ and $E$. Our segments will be $AF, BF, CD$, which we move slightly so that they do not intersect anymore. A picture might be helpful.

For points, there is the [url=http://mathworld.wolfram.com/VoronoiDiagram.html]Voronoi diagram[/url]: the convex chamber containing the point $P$ in our collection is defined to be the closure of the set of all the points in the plane which are closer to $P$ than to any other point in the collection. It can also be described as the subset of the plane containing $P$ and bounded by all the perpendicular bisectors of segments of the form $PQ$, as $Q\ne P$ ranges through the points in our collection.

For disks, we modify the idea used in the construction of Voronoi diagrams: for each disk $D$ in the collection, define the convex set which contains $D$ to be the subset of the plane which contains $D$ bounded by the radical axes of the pairs $D,E$ as $E\ne D$ ranges through the disks in the collection.


right, good thing someone's updating the ping :P


trigonometry
limit
algebra proposed
algebra

prove   for  all n

 l sin 1 l +l sin 2 l + l sin 3 l +........+ l sin(3n -1) l +l sin 3n l  <

  8n\5


inequalities
graph theory
combinatorics unsolved
combinatorics

Does there exist a graph G such that

1. G is planar

2. Every vertex has degree at least 5

3. No 2 vertices of degree 5 are adjacent

4. No vertex of degree 6 is adjacent to more than one vertex of degree 5?

I think this problem was mentioned during a graph theory course for the IMO, although I'm not 100% sure it is the same problem. If it is, the answer is no and the key is some inequality on the reciprocals of the degrees, but I really can't remember it exactly.

Actually I think such a graph exists. I din't finish the example because it's pretty messy but I think my idea is correct.

We shall construct a planar graph having $60$ $6$-vertices and $12$ $5$-vertices and $140$ triangular faces. The vertices can be partitioned into $12$ "pentagons" consisting of a $5$-vertex $P_i$ and five $6$-vertices $A_i,B_i,C_i,D_i,E_i$ connected to it. Let also $A_i,B_i$ be connected and so on.

There is a graph with twelve faces having $18$ triangular faces and each vertex having degreee $5$ (I invite you to construct it yourself). Now replace each of the twelve vertices with a pentagon. The edge between two pentagons will mean three edges betwenn two pairs of consectuve vertices of the pentagons. For example, one could match $A_i,B_i$ with $C_j,D_j$ by three edges: $A_i,C_j$, $B_i,D_j$ and $A_i,D_j$. Since each pentagons has five pairs of consecutive vertices, I believe the edges can be handled in such a way that the condition is satisfied.


[quote="AndrewTom"]What is 1 + 1/3 + 1/5 + 1/7 + ... + 1/(2n-1) = ?[/quote]

You can use the definition of the $ \Psi$-function to write
$ \sum_{k\equal{}1}^{n}\frac{1}{2k\minus{}1}\equal{}\Psi\left (n\plus{}\frac{1}{2}\right )\plus{}\frac{\gamma}{2}\plus{}\ln 2.$

This sum doesn't have a (nice) closed form.  It's related to the [url=http://en.wikipedia.org/wiki/Harmonic_number]harmonic numbers[/url], which also do not have nice closed forms (except in terms of special functions).  M@ri@n refers to the [url=http://en.wikipedia.org/wiki/Digamma_function]digamma function[/url].


No doubt everyone has heard of the recent Google China incident to some degree. Heck, I even [url=http://excelexcel.wordpress.com/2010/01/17/china-google-needs-to-be-punished/]blogged about it[/url]. What are your thoughts on this issue? What will happen as a result? Would this in any way impact global attitudes towards net censorship?

I thought that Google was a pretty big industry there. But apparently it wasn't.

Anyone know why the withdrew? :?:

[quote="candyhyperalert"]Very few Chinese people will care about the loss of Google due to Baidu, and Google does not care much for China given that their profits there are not so astounding.

Here is the official Google Blog about this happening:
http://googleblog.blogspot.com/2010/01/new-approach-to-china.html[/quote]

?

Google had 29% market share in China before they quit, which is definitely not "Very few". And the market share was actually GROWING if I recall correctly.

http://www.digitaleastasia.com/2010/01/05/baidu-google-yahoo-and-bing-2009-was-a-battleground-in-chinese-search/

I'm pretty sure google pulled out of China mainly because of this: http://www.cnbc.com/id/34927194

I heard about it, but I never really knew what happened in China.  Enlighten me, please? (pythag, your link is broken. :) )

[quote="pythag011"][quote="candyhyperalert"]Very few Chinese people will care about the loss of Google due to Baidu, and Google does not care much for China given that their profits there are not so astounding.

Here is the official Google Blog about this happening:
http://googleblog.blogspot.com/2010/01/new-approach-to-china.html[/quote]

?

Google had 29% market share in China before they quit, which is definitely not "Very few". And the market share was actually GROWING if I recall correctly.

[/quote]
Despite the 29% market share, google.cn only brought a small fraction of Google's profits.  As for the "very few," I did not mean very few people use Google.cn.  What I meant was very few people will mind google.cn disappearing because there exists baidu in China.  I don't know if this is a good analogy, but one example would be if a less popular search engine disappeared from America (AOL, yahoo, etc).  People could easily use google.com for their new search engine.  The transition is easy to make.  In China, google.cn users could just switch to Baidu.  I was perhaps, a bit too extreme, in my remark.  Some people will miss google.cn, but the majority won't.

does anyone actually use Baidu who can say if it is as good as google?  Apart from having gmail accounts hacked, i think google may be saying that it doesn't believe in censorship so if the chinese gov't wants to censor them, they will have to do it themselves and take the blame, not google.

Update: http://news.bbc.co.uk/1/hi/business/8581393.stm

Apparently google decided they are gonna stop using censorship. I expect the Chinese government will shut them down within a week.

[quote="Quickster94"]Lots of people have been saying that Google used this as an opportunity to pull out of a market where it wasn't making very much, and win a lot of PR points in the rest of the world.  There was an interesting article about this, I forgot if it was BusinessWeek, Economist, or TIME, or maybe more than one...[/quote] I don't really think this would be one of the reasons Google would pull out. While they only get 2% of their money from China at the moment, China represents a rapidly expanding market that would be very important in the future.

For Google, I guess India is a better market (also very rapidly expanding) than China. The current fallout (and probable shutdown) exemplifies China's autocratic political system.

The government don't like to see that there are too many political-words on net.what a shamed "harmonious society"! :wallbash_red:

Why is the Chinese government so wary of its detractors?

I think a point worth observing is that the number of internet users in China is astronomical, but the revenue that Google got in China is tiny (despite its large "market share"). I think Google doesn't mind making a stand against the Chinese government on the censorship issue because Google knows that the internet advertising market isn't sufficiently mature in China right now.

I use Baidu occasionally, and I don't like it very much. This may be because my Chinese isn't good enough for me to recognize which results are relevant, but it seems to me that for almost any search, many high ranked results are crappy things like forum posts in the Baidu forums and articles (also on Baidu) that look to me like a mediocre (and censored, of course) online encyclopedia. On the other hand, my impression of Chinese internet users is that anyone who doesn't know how to get around the government's controls probably doesn't care about much other than the Baidu forums ;)

Well you also have to note that while the revenue Google got in China is tiny, China represents a huge new market that will inevitably expand and whoever controls it in the future will get a lot of money. So I think Google cared at least a significant amount about China's censorship or else they wouldn't have done this.

[quote="Ihatepie"]China represents a huge new market that will inevitably expand and whoever controls it in the future will get a lot of money. [/quote]
My guess is that Google thinks that it will be a long time (relatively speaking) before this is significant. They think they can afford to make a stand on principle now without sacrificing too much of the long term opportunity. If Google's wrong and what happens now really matters that much, then they've made a monumentally bad business decision by leaving the Chinese market to Microsoft and Baidu. It's just not all that plausible that a company that's been so successful until now would make such a catastrophic mistake -- after all, it's a rare individual and an even more rare for-profit corporation that would sacrifice its future to stand by its principles.

Apparently two other companies, godaddy and networksolutions llc, have pulled out of China. I know that at least the former is big.

But anyways, to answer what you said: According to your hypothesis, google is still sacrificing its future. It would just take a few more years for this to happen. So either way, Google is losing money on purpose.

Also, call me crazy, but I think that if there was any company that would stand up on principle and would sacrifice money in order to do so, it would be google.


[color=darkblue]Hi! Every one. I want you give me [size=150][u]some way [/u][/size]to done this problem:

Let $ a,b\ge 0$. Prove that: $ (\sqrt {a} \plus{} \sqrt {b})^8\ge 64ab(a \plus{} b)^2$[/color]

This is a way of me:
$ (\sqrt {x} \plus{} \sqrt {y})^8 \equal{} [(a \plus{} b) \plus{} 2\sqrt {ab}]^4\ge [2\sqrt {(a \plus{} b) . 2\sqrt {ab}}]^4 \equal{} 64ab(a \plus{} b)^2$

I am bad at inequalities, so tell me if I am wrong.  :blush:

WLOG, assume $ a\plus{}b\equal{}1$. Then we must prove:
$ (\sqrt{a}\plus{}\sqrt{b})^8\geq64ab$
This is equivalent to
$ (a\plus{}b\plus{}2\sqrt{ab})^4\geq64ab\\
(1\plus{}2\sqrt{ab})^4\geq64ab$
By Cauchy-Schwarz,
$ (1\plus{}2\sqrt{ab})(2\sqrt{ab}\plus{}1)\geq(2\sqrt{2}\sqrt{\sqrt{ab}})^2\equal{}8\sqrt{ab}$
Square this, and then we are done.

[hide][color=darkblue]Thank you very much.[/color][/hide]

assume  √a=x,√b=y
 the problem is reduced to prove
(x+y)^8≧64x^2y^2(x^2+y^2)^2
let x+y=1,
This is equivalent to
1≧8xy(x^2+y^2)
1≧8xy((x+y)^2-2xy)
1≧8xy(1-2xy)
(4xy-1)^2≧0
which is true.

[hide][color=darkblue]THANK YOU VERY MUCH.[/color][/hide]

[quote="thanhnam2902"][color=darkblue]Let $ a,b\ge 0$. Prove that: $ (\sqrt {a} \plus{} \sqrt {b})^8\ge 64ab(a \plus{} b)^2$[/color][/quote]

Hello [b]Thanhnam2902[/b]!

The following identity is true for all non-negative $ a,b$:
\[ {\left( \sqrt {a} \plus{} \sqrt {b}\, \right) ^{8} \minus{} 64\,ab \left( a \plus{} b \right) ^{2} \equal{} \left( {a}^{2} \plus{} {b}^{2} \plus{} 6\,ab \plus{} 12\,\sqrt {ab} \left( a \plus{} b \right) \right) \left( \sqrt {a} \minus{} \sqrt {b}\, \right) ^{4}}
\]
Thank you for the nice inequality!

Sergey Markelov
markelov@mccme.ru


function
real analysis
real analysis unsolved

Is it true that if $f(s)$- $2\pi$-period function has absulute convergent Fourier series $\sum_k c_k e^{iks}$, then
$\frac{1}{n} \sum_{m=1}^n f(\frac{2 \pi m}{n}-s)= \sum_{n | k}c_k e^{-i k s}$.
It seems to be easy, but i'm looking at it and can't realize why it is so.

If this were the discrete Fourier transform, I'd call the operator on the left the "group" (or "grouping") operator and the right hand side the result of applying the "decimate" operator to the Fourier transform ([i]i.e.[/i], Fourier coefficients) of the function. And in that context, it's well known that the Fourier transform intertwines the "grouping" operator with the "decimate" operator - and this is very, very useful in building the Fast Fourier Transform.

You're using the letter $k$ for more than one thing - that's kind of confusing, so I'm going to go into your post and edit in another letter.

[Returning to what I was thinking.]

It comes down to this: what is $\frac1n\sum_{m=1}^nf\left(\frac{2\pi m}n-s\right)$ when $f$ is the trigonometric monomial $f(s)=e^{iks}?$

Calculate that. The rest requires passing to a limit, but you've put a very strong condition (uniform convergence of the Fourier series) on this to help with the limit.


[hide]1000x=y where x is the amount of time and y is the job.
1000x/4=y/4 or 250x=y/4.
But for the next quarter, we have only 900 people or 250x+225x.
But for the next quarter, we have only 800 people or 250x+225x+200x.  That's 675x.  Now, in the last quarter, we need to complete 325x to finish the job.  That's kx/4.  Remeber, x is a null variable.  By multiplying, we get 1300.  We have 800 workers right now so that's 500 more workers to hire.  [/hide]

That's what I got too. Is anyone else getting that?

You all got tricked.  Read the problem again.

[hide]The first quarter:
t_1 = 1, the work was done on time.

Second quarter:
t_2 = 1000/900 = 10/9

Third quarter:
t_3 = 1000/800 = 10/8 = 5/4

Fourth quarter:
t_4 = ?

In order for the work to be done on time, t_1 + t_2 + t_3 + t_4 = 4.  That means t_4 = 23/36.

Workers needed for fourth quarter = [1000 * 36/23] = 1566.  1566 - 800 = 766 and that's our answer.
[/hide]

Why did you use 1000/900 instead of 900/1000?

The time it takes to finish a quarter and the number of workers are inversely proportional, meaning wt = k.  I let 1 be the time taken for the first quarter and there were 1000 workers, so k = 1000.

And then when there were 900 workers, solving for t we had k/w = 1000/900 = 10/9.

Understand now?

[quote="joml88"]In order to complete a large job, 1000 workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then 100 workers were laid off, so the second quarter of the work was completed behind schedule. Then an additional 100 workers were laid off, so the third quarter of the work was completed still further behind schedule. Given that all workers work at the same rate, what is the minimum number of additional workers, beyond the 800 workers still on the job at the end of the third quarter, that must be hired after three-quarters of the work has been completed so that the entire project can be completed on schedule or before?[/quote]

[hide="solution"]one person does $\frac{1}{1000}$ of the job in a year

1st quarter, .25 done

2nd quarter

900/1000*t=.25 done

took 5/18 years

3rd quarter
800/1000*t=.25

took 5/16 years

4th quarter
time left=1-.25-5/18-5/16=23/144

workers/1000*23/144=.25

workers=1565.21...

needed addtion $1566-800=\boxed{766}$

[/hide]

[hide="solution"]

t_1 is the number of days it took to finish the first quarter, t_2 days which took to finish the second quarter and t_3 days which took to finish the third quarter of job.  and let $n$ be the number of workers to finish each quarter of job.
the amount of work for each quarter is equal, thus:
$1000t_1=900t_2=800t_3=n$
the remaining time is:
$4t_1-(t_1+t_2+t_3)=3t_1-t_2-t_3$

from the first eqution we conclude:
$t_2=\frac{10t_1}{9},t_3=\frac{5t_1}{4}$
thus the remaining time is:
$3t_1-t_2-t_3=\frac{23}{36}t_1$
let $m$ be the total amount of workers
we have:
$\frac{23}{36}t_1\cdot m=n\Rightarrow m=\frac{36}{23}\frac{n}{t_1}$
originally we had: $1000t_1=n\Rightarrow \frac{n}{t_1}=1000$
so $m=\frac{36000}{23}$
so the number of additional workers would be $\lfloor{m-800}\rfloor=\boxed{766}$[/hide]

[hide=Solution]For the project to be on schedule the sum of the times for each quarter must be $4$, so $1+\frac{10}{9}+\frac{5}{4}+t=4$ where $t$ is the last time. This means $t=\frac{23}{36} \implies \left \lceil 1000 \cdot \frac{36}{23} \right \rceil - 800 = \boxed{766}$.[/hide]