analytic geometry
graphing lines
slope
vector

If there are two and only two points on circle $(x-3)^{2}+(y+5)^{2}=r^{2}$ such that the distance between the points and the line $4x-3y-2=0$ is $1$, then find the range of the $r$.

[quote="lotrgreengrapes7926"][hide] Is it $4<r<6$? :maybe: If it is, I will post my solution. [/hide][/quote]
I got the same result

I had an inelegant solution. You can post your solution.  :)

[hide]The center of the circle is $C=(3,-5)$.  Draw the perpendicular line $l_{1}$ from $C$ to the given line $l_{2}$.  The slope of $l_{2}$ is $\frac{4}{3}$, so the slope of $l_{1}$ is $-\frac{3}{4}$.  In point-slope form, the equation of $l_{1}$ is $y=-\frac{3}{4}(x-3)-5$.  The lines $l_{1}$ and $l_{2}$ intersect at $X=(-1,-2)$, which can be found by solving the system of the two equations.  We have $CX=5$.

If $r<4$, there are no points on the circle within one unit of $l_{2}$.  If $r=4$, there is one point on the circle within one unit of $l_{2}$.  We can draw two lines parallel to and one unit away from $l_{2}$.  If $r>4$, the first line intersects the circle twice.  The second line doesn't intersect the circle unless $r\ge 6$.  Therefore, $4<r<6$.[/hide]

[hide]

center of circle: $O(x_{o},y_{o})=(3,-5)$

line $k: \ 4x-3y-2=0$

$d(O,k) = \frac{|4x_{o}-3y_{o}-2|}{\sqrt{4^{2}+3^{2}}}= \frac{25}{5}= 5$

There are two lines $m,n$ which are parallel to $k$ and $d(m,k)=d(n,k)=1$

Let $m$ be the line which is closer to $O$.

$d(O,m)=4$
$d(O,n)=6$

We have only two points with distance $1$ from line $k$, so the circle should intersect only the line $m$ (at two points), which means $4<r<6$[/hide]

[hide]The center of the circle is (3,-5). We find the distance from that point to the line.
$4x-3y-2=0 \implies y=(4/3)x-2/3$. The line perpendicular to that has slope -3/4, or is y=(-3/4)x+z, and passes through (3,-5):
$-5=(-3/4)3+z \implies z=-11/4$. Solving for the intersection, we get (-1, -2). The distance from (3,-5) to (-1,-2) is 5, so the range of r is $4<r<6$.[/hide]

[quote="lotrgreengrapes7926"]Extension: Prove that the distance between the point $(p,q)$ and the line $ax+by+c=0$ is $\frac{|ap+bq+c|}{\sqrt{a^{2}+b^{2}}}$.[/quote]

[hide="Solution"] We will make use of some vector techniques.

Consider an arbitrary vector from $(p, q)$ to the line.  We pick some point on the line, say $(0, \frac{-c}{b})$ and consider the vector $\vec{V}= \left< p, q+\frac{c}{b}\right>$.

Next we consider the normal vector to the line, that is, $\vec{N}= \left< a, b \right>$.

Suppose we project $\vec{V}$ onto $\vec{N}$.  The resulting vector is perpendicular to our line and has precisely the length we desire!  (That is, it is the perpendicular distance from our point to our line.)

The length of this vector is given by

$\frac{ |\vec{V}\cdot \vec{N}|}{ | \vec{N}| }= \frac{| ap+bq+c | }{ \sqrt{a^{2}+b^{2}}}$

(You can go through the algebra to prove that this expression is the same for arbitrary $\vec{V}$, but I'd rather use the projection interpretation.) [/hide]


calculus
integration
limit
function
ceiling function
algebra

[hide="Solution"]
We will use partial fraction decomposition to decompose the summand by writing:
$ \frac{2k\plus{}1}{(4k\plus{}1)(4k\plus{}3)(4k\plus{}5)} \equal{} \frac{A}{4k\plus{}1} \plus{} \frac{B}{4k\plus{}3} \plus{} \frac{C}{4k\plus{}5}$

By multiplying the equation by $ (4k\plus{}1)(4k\plus{}3)(4k\plus{}5)$, we have:

$ (4k\plus{}3)(4k\plus{}5)A \plus{} (4k\plus{}1)(4k\plus{}5)B \plus{} (4k\plus{}1)(4k\plus{}3)C \equal{} 2k \plus{} 1$

Setting $ k \equal{} \minus{}\frac{1}{4}$ yields $ 8A \equal{} \frac{1}{2}$, so $ A \equal{} \frac{1}{16}$. Similarly, by setting $ k \equal{} \minus{}\frac{3}{4}$ and $ k \equal{} \minus{}\frac{5}{4}$, we have $ B \equal{} \frac{1}{8}$ and $ C \equal{} \minus{}\frac{3}{16}$ respectively. Thus:

$ \frac{2k\plus{}1}{(4k\plus{}1)(4k\plus{}3)(4k\plus{}5)} \equal{} \frac{1}{64k\plus{}16} \plus{} \frac{2}{64k\plus{}48} \minus{} \frac{3}{64k\plus{}80}$

The sum we want to find can be split up as follows:
$ \sum_{k\equal{}0}^{\infty} \left( \frac{1}{64k\plus{}16} \plus{} \frac{2}{64k\plus{}48} \minus{} \frac{3}{64k\plus{}80} \right) \equal{}
\sum_{k\equal{}0}^{\infty} \left( \frac{1}{64k\plus{}16} \minus{} \frac{1}{64k\plus{}80} \right) \plus{} \sum_{k\equal{}0}^{\infty} \left( \frac{2}{64k\plus{}48} \minus{} \frac{2}{64k\plus{}80} \right)$

We will start by evaluating the sum on the left.
$ \sum_{k\equal{}0}^{\infty} \left( \frac{1}{64k\plus{}16} \minus{} \frac{1}{64k\plus{}80} \right) \equal{} \frac{1}{16} \sum_{k\equal{}0}^{\infty} \left( \frac{1}{4k\plus{}1} \minus{} \frac{1}{4k\plus{}5} \right)$

$ \frac{1}{16} \sum_{k\equal{}0}^{\infty} \left( \frac{1}{4k\plus{}1} \minus{} \frac{1}{4k\plus{}5} \right) \equal{} \frac{1}{16} \left( \frac{1}{1} \minus{} \frac{1}{5} \plus{} \frac{1}{5} \minus{} \frac{1}{9} \plus{} \cdots \right) \equal{} \frac{1}{16}$

So, the first sum is equal to $ \boxed{\frac{1}{16}}$.

Now, we will go about evaluating the second sum.

$ \sum_{k\equal{}0}^{\infty} \left( \frac{2}{64k\plus{}48} \minus{} \frac{2}{64k\plus{}80} \right) \equal{} \frac{1}{8} \sum_{k\equal{}0}^{\infty} \left( \frac{1}{4k\plus{}3} \minus{} \frac{1}{4k\plus{}5} \right)$

$ \frac{1}{8} \sum_{k\equal{}0}^{\infty} \left( \frac{1}{4k\plus{}3} \minus{} \frac{1}{4k\plus{}5} \right) \equal{} \frac{1}{8} \left( \frac{1}{3} \minus{} \frac{1}{5} \plus{} \frac{1}{7} \minus{} \frac{1}{9} \plus{} \cdots \right)$.

I thought that this last expression looked like an infinite series for $ \pi$, and after checking online  :) , I found that
$ \frac{\pi}{4} \equal{} 1 \minus{} \frac{1}{3} \plus{} \frac{1}{5} \minus{} \frac{1}{7} \plus{} \cdots$

(See [url=http://en.wikipedia.org/wiki/Leibniz_formula_for_pi]this wikipedia article[/url])

Then, we have:
$ 1 \minus{} \frac{\pi}{4} \equal{} \frac{1}{3} \minus{} \frac{1}{5} \plus{} \frac{1}{7} \minus{} \frac{1}{9} \plus{} \cdots$

This tells us that the second sum is equal to $ \frac{1}{8} \left( 1 \minus{} \frac{\pi}{4} \right) \equal{} \boxed{\frac{1}{8} \minus{} \frac{\pi}{32}}$.

Now that we have evaluated the two sums we can compute the answer:

$ \sum_{k\equal{}0}^{\infty} \frac{2k\plus{}1}{(4k\plus{}1)(4k\plus{}3)(4k\plus{}5)} \equal{} \sum_{k\equal{}0}^{\infty} \left( \frac{1}{64k\plus{}16} \minus{} \frac{1}{64k\plus{}80} \right) \plus{} \sum_{k\equal{}0}^{\infty} \left( \frac{2}{64k\plus{}48} \minus{} \frac{2}{64k\plus{}80} \right) \equal{} \frac{6 \minus{} \pi}{32}$.

So, the answer is $ \boxed{\frac{6 \minus{} \pi}{32}}$.

[/hide]

Wow, this is an interesting problem.

According to wolframalpha, the answer is $\frac{58-15 \pi}{480}$...

How do you prove all those weird stuff that equal something in terms of pi? Calculus or something, i suppose.

That series for $\pi/4$ comes from the Taylor series for $arctan(1)=\pi/4$.  It is derived as follows:

$\frac{1}{1+x^2} = 1-x^2+x^4-x^6+ \ldots$ (converges for $x<1$)

but $\arctan(x) = \int \frac{1}{1+x^2} dx$ so

$\arctan(x) = \int 1-x^2+x^4-x^6+\ldots dx$ (because there is a theorem which says if you integrate a Taylor series term by term, you get a Taylor series (with the same radius of convergence) for the antiderivative of the original function).

So $\arctan(x) = C + x - x^3/3+x^5/5-x^7/7 + \ldots$ for $-1<x \le 1$

now $\arctan(0) = 0$ so $C=0$ and thus

$\arctan(x) = x - x^3/3+x^5/5-x^7/7 + \ldots$ for $-1< x \le 1$

Plugging in $x=1$ gives the result $\pi/4 = 1-1/3+1/5-1/7+ \ldots$ (which the alternating series test shows converges)

This is a very interesting problem. Is there any solution without Calculus?

@facis: am I being too nit-picky if I say that proof is a bit incomplete, since you're dealing with $x=1$, which is not in the radius of convergence - so it is necessary to show convergence of the series for $x=1$, which you did, but also to show that it converges to $\arctan 1$.

[quote="BarbieRocks"]According to wolframalpha, the answer is $\frac{58-15 \pi}{480}$...[/quote][quote="Dr Sonnhard Graubner"]${\sum_{k=0}^{\infty}\frac{2k+1}{(4k+1)(4k+3)(4k+5)}}=\frac{29}{240}-\frac{\pi}{32}$[/quote]
Your answers agree, but unless I'm mistaken, they are both equal to $\sum_{k=1}^\infty\frac{2k+1}{(4k+1)(4k+3)(4k+5)}$, where $k$ starts at 1 instead of 0.

Oh, you're right. I misread the equation...

:facepalm:

hsiljak: I suppose you have a point.  So I'll be a bit more rigorous - though I should note, the topics needed for this proof are really part of Real Analysis.  See http://www.mathcs.org/analysis/reals/funseq/uconv.html in particular.

Claim: $f_n(x) = x-x^3/3+x^5/5-x^7/7 + \ldots + (-1)^{n-1} x^{2n-1}/(2n-1)$ converges uniformly on $[a,1]$, for any $a$ close to, but greater than, $-1$.

Proof: Let $f(x) = \lim_{n \to \infty} f_n(x) = x-x^3+x^5/5-x^7/7 + \ldots$.  From the alternating series test, it should be clear that this series of functions converges pointwise on $[a,1]$.

Now note that

$|f_n(x) - f(x)| = |(x-x^3/3+ \ldots + (-1)^{n-1} x^{2n-1}/(2n-1))- (x-x^3 + \ldots)|$
$= |\sum_{k=n+1}^\infty (-1)^k x^{2k-1}/(2k-1)|$
$ < |x^{2n+1}/(2n+1)| < 1/(2n+1)$ (the difference is an alternating (decreasing) series, which converges to some number less than the leading term).

Thus for every $\epsilon > 0$, there exist $N$ such that for all $n>N$, $|f_n(x)-f(x)|< \epsilon$ - namely, $N=\lceil\frac{1}{2}(1/\epsilon -1) \rceil$.  Thus $f_n(x)$ converges uniformly.

All the $f_n$ are continuous (since they are polynomials), so by the uniform convergence, $f(x)$ is also continuous on $(-1,1]$.  $f(x) = \arctan(x)$ for $(-1,1)$ because $x-x^3/3 + \ldots = \int 1-x^2+x^4 - \ldots dx$ and the term by term integral of a Taylor series is the Taylor series of the integral of the original function.  Now continuity of $f(x)$ tells us that $f(1) = \lim_{x \to 1^-} f(x) =\lim_{x\to 1^-} \arctan(x) = \arctan(1)$.  QED.


Let $ a,b,c$ be the nonnegative real numbers. Prove that
$ \frac {1}{\sqrt{4a^2 \plus{} bc}} \plus{} \frac {1}{\sqrt{4b^2 \plus{} ac}} \plus{} \frac {1}{\sqrt{4c^2 \plus{} ab}} \ge \frac {4}{a \plus{} b \plus{} c}$

[quote="Mathpro"]Let $ a,b,c$ be the nonnegative real numbers. Prove that
$ \frac {1}{\sqrt {4a^2 \plus{} bc}} \plus{} \frac {1}{\sqrt {4b^2 \plus{} ac}} \plus{} \frac {1}{\sqrt {4c^2 \plus{} ab}} \ge \frac {4}{a \plus{} b \plus{} c}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=146809


The arithmetic mean of two numbers 12. Their geometric mean is 8.  Find the sum of the square roots of the two numbers.  I got it down to a^2 + b^2 = 448.  Don't know what to do next...do i just square root everything?

Just calculate $ (\sqrt{a}\plus{}\sqrt{b})^{2}$.

it's actually a really simple problem
[hide]since the geometric mean is 8, we have $ \sqrt{ab}\equal{}8$ since the arithmetic mean is 12, we have $ \frac{a\plus{}b}{2}\equal{}12\Longrightarrow a\plus{}b\equal{}24$
the expression we want to find is $ \sqrt{a}\plus{}\sqrt{b}$. squaring the term we want, we have $ (\sqrt{a}\plus{}\sqrt{b})^{2}\equal{}a\plus{}2\sqrt{ab}\plus{}b$ we know what $ \sqrt{ab}$ is and we know what $ a\plus{}b$ equals. so. $ (\sqrt{a}\plus{}\sqrt{b})^{2}\equal{}24\plus{}16\equal{}40$. taking the square root of that, we have $ \sqrt{40}\equal{}2\sqrt{10}$ that is our answer.[/hide]


Prove that, for all $n,a$ $\in\N}$$Z$, the number $x^{2}+3x+5$ is not divisible by $121$


linear algebra
matrix
inequalities
floor function
vector
abstract algebra
linear algebra unsolved

Let $ A$ and $ B$ be $ n\times n$ matrices. Find the maximum possible value of

$ \mathrm{rank}\left(AB\right) \minus{} \mathrm{rank}\left(BA\right)$.

Assume $ n$ is even

and $ A\equal{}\begin{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}&&&&\\&\begin{bmatrix}1&0\\0&0\end{bmatrix}&&&\\&&...&&\\&&&...&\\&&&&\begin{bmatrix}1&0\\0&0\end{bmatrix}\end{bmatrix}$

       $ B\equal{}\begin{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix}&&&&\\&\begin{bmatrix}0&1\\0&0\end{bmatrix}&&&\\&&...&&\\&&&...&\\&&&&\begin{bmatrix}0&1\\0&0\end{bmatrix}\end{bmatrix}$

then we have $ rank(AB)\minus{}rank(BA)\equal{}\frac{n}{2}$

so at least $ \frac{n}{2}$ :roll:

This one is nice. Corollary 2 [b](b)[/b] below (with $ A$ and $ B$ interchanged) shows that $ \frac {n}{2}$ is an upper bound (see also Corollary 2 [b](a)[/b] below for a stronger inequality), so, combined with post #2, it yields that $ \left\lfloor\frac {n}{2}\right\rfloor$ is the required maximum. First, an easy and probably known lemma:

[color=blue][b]Theorem 1.[/b] Let $ k$ be a field. Let $ U$, $ V$, $ W$ be three $ k$-vector spaces. Let $ X: V\rightarrow W$ and $ Y: U\rightarrow V$ be two $ k$-linear maps. Then,

[b](a)[/b] we have $ \dim\left( \mathrm{Ker}\left( XY\right) \right) \leq\dim\left( \mathrm{Ker}X\right) \plus{} \dim\left( \mathrm{Ker}Y\right)$.

[b](b)[/b] we have $ \mathrm{Rank}\left( XY\right) \leq \mathrm{Rank}X$.

[b](c)[/b] we have $ \mathrm{Rank}\left( XY\right) \leq \mathrm{Rank}Y$.[/color]

[hide="Proof of Theorem 1"]
[i]Proof of Theorem 1.[/i] [b](a)[/b] The first isomorphism theorem (applied to the $ k$-linear map $ Y\mid_{\mathrm{Ker}\left( XY\right)}: \mathrm{Ker}\left( XY\right) \rightarrow V$) yields

$ \left( Y\mid_{\mathrm{Ker}\left( XY\right) }\right) \left(\mathrm{Ker}\left( XY\right) \right) \cong\left(\mathrm{Ker}\left( XY\right) \right) \diagup\mathrm{Ker}\left( Y\mid_{\mathrm{Ker}\left( XY\right) }\right)$,

so that

[color=green][b](1)[/b][/color] $ \dim\left( \left( Y\mid_{\mathrm{Ker}\left( XY\right) }\right) \left( \mathrm{Ker}\left( XY\right) \right) \right) \equal{} \dim\left( \left( \mathrm{Ker}\left( XY\right) \right) \diagup\mathrm{Ker}\left( Y\mid_{\mathrm{Ker}\left( XY\right) }\right) \right)$.

But $ \left( Y\mid_{\mathrm{Ker}\left( XY\right) }\right) \left(\mathrm{Ker}\left( XY\right) \right) \equal{} Y\left( \mathrm{Ker}\left( XY\right) \right) \subseteq\mathrm{Ker}X$ (since $ X\left( Y\left( \mathrm{Ker}\left( XY\right) \right) \right) \equal{} \left( XY\right) \left( \mathrm{Ker}\left( XY\right) \right) \equal{} 0$ yields $ Y\left( \mathrm{Ker}\left( XY\right) \right) \subseteq X^{ \minus{} 1}\left( 0\right) \equal{} \mathrm{Ker}X$) yields

[color=green][b](2)[/b][/color] $ \dim\left( \left( Y\mid_{\mathrm{Ker}\left( XY\right) }\right) \left( \mathrm{Ker} \left( XY\right) \right) \right) \leq\dim\left( \mathrm{Ker}X\right)$,

and

$ \mathrm{Ker}\left( Y\mid_{\mathrm{Ker}\left( XY\right)}\right) \equal{} \left\{ v\in\mathrm{Ker}\left( XY\right) \ \mid\ \left(Y\mid_{\mathrm{Ker}\left( XY\right) }\right) \left( v\right) \equal{} 0\right\}$
$ \equal{} \left\{ v\in\mathrm{Ker}\left( XY\right) \ \mid\ Y\left(v\right) \equal{} 0\right\}$ (since $ \left( Y\mid_{\mathrm{Ker}\left(XY\right) }\right) \left( v\right) \equal{} Y\left( v\right)$ for every $ v\in\mathrm{Ker}\left( XY\right)$)
[color=green][b](3)[/b][/color] $ \equal{} \underbrace{\left\{ v\in U\ \mid\ Y\left(v\right) \equal{} 0\right\} }_{ \equal{} \mathrm{Ker}Y}\cap\mathrm{Ker}\left(XY\right) \equal{} \mathrm{Ker}Y\cap\mathrm{Ker}\left( XY\right) \equal{} \mathrm{Ker}Y$

(since $ \mathrm{Ker}Y\subseteq\mathrm{Ker}\left( XY\right)$, since $ \left( XY\right) \left( \mathrm{Ker}Y\right) \equal{} X\left(\underbrace{Y\left( \mathrm{Ker}Y\right) }_{ \equal{} 0}\right) \equal{} X\left(0\right) \equal{} 0$).

Hence,

$ \dim\left( \mathrm{Ker}X\right) \geq\dim\left( \left(Y\mid_{\mathrm{Ker}\left( XY\right) }\right) \left(\mathrm{Ker}\left( XY\right) \right) \right)$ (by [color=green][b](2)[/b][/color])
$ \equal{} \dim\left( \left( \mathrm{Ker}\left( XY\right) \right)\diagup\mathrm{Ker}\left( Y\mid_{\mathrm{Ker}\left( XY\right)}\right) \right)$ (by [color=green][b](1)[/b][/color])
$ \equal{} \dim\left( \mathrm{Ker}\left( XY\right) \right) \minus{} \dim\left(\mathrm{Ker}\left( Y\mid_{\mathrm{Ker}\left( XY\right)}\right) \right)$
$ \equal{} \dim\left( \mathrm{Ker}\left( XY\right) \right) \minus{} \dim\left(\mathrm{Ker}Y\right)$ (by [color=green][b](3)[/b][/color]),

so that $ \dim\left( \mathrm{Ker}\left( XY\right) \right) \leq\dim\left( \mathrm{Ker}X\right) \plus{} \dim\left( \mathrm{Ker}Y\right)$, and thus, Theorem 1 [b](a)[/b] is proven.

[b](b)[/b] We have $ \left( XY\right) \left( U\right) \equal{} X\left(\underbrace{Y\left( U\right) }_{\subseteq V}\right) \subseteq X\left(V\right)$, so that $ \dim\left( \left( XY\right) \left( U\right)\right) \leq\dim\left( X\left( V\right) \right)$ and thus

$ \mathrm{Rank}\left( XY\right) \equal{} \dim\left( \left( XY\right)\left( U\right) \right) \leq\dim\left( X\left( V\right) \right) \equal{} \mathrm{Rank}X$.

Thus, Theorem 1 [b](b)[/b] is proven.

[b](c)[/b] The first isomorphism theorem (applied to the $ k$-linear map $ X\mid_{Y\left( U\right) }: Y\left( U\right) \rightarrow W$) yields

$ \left( X\mid_{Y\left( U\right) }\right) \left( Y\left( U\right)\right) \cong\left( Y\left( U\right) \right) \diagup\mathrm{Ker}\left( X\mid_{Y\left( U\right) }\right)$,

and thus

$ \dim\left( \left( X\mid_{Y\left( U\right) }\right) \left( Y\left(U\right) \right) \right) \equal{} \dim\left( \left( Y\left( U\right) \right)\diagup\mathrm{Ker}\left( X\mid_{Y\left( U\right) }\right) \right)$
$ \equal{} \dim\left( Y\left( U\right) \right) \minus{} \underbrace{\dim\left(\mathrm{Ker}\left( X\mid_{Y\left( U\right) }\right) \right)}_{\geq0}\leq\dim\left( Y\left( U\right) \right) \equal{} \mathrm{Rank}Y$.

But $ \left( X\mid_{Y\left( U\right) }\right) \left( Y\left( U\right)\right) \equal{} X\left( Y\left( U\right) \right) \equal{} \left( XY\right) \left(U\right)$ yields

$ \dim\left( \left( X\mid_{Y\left( U\right) }\right) \left( Y\left(U\right) \right) \right) \equal{} \dim\left( \left( XY\right) \left( U\right)\right) \equal{} \mathrm{Rank}\left( XY\right)$.

Thus,

$ \mathrm{Rank}\left( XY\right) \equal{} \dim\left( \left( X\mid_{Y\left(U\right) }\right) \left( Y\left( U\right) \right) \right)\leq\mathrm{Rank}Y$.

Thus, Theorem 1 [b](c)[/b] is proven.
[/hide]

[color=blue][b]Corollary 2.[/b] Let $ k$ be a field. Let $ n\in\mathbb{N}$. Let $ U$ be a $ k$-vector space such that $ \dim U \equal{} n$. Let $ A: U\rightarrow U$ and $ B: U\rightarrow U$ be two $ k$-linear maps. Then,

[b](a)[/b] we have $ \mathrm{Rank}\left( AB\right) \geq 2\mathrm{Rank}\left( BA\right) \minus{} n$.

[b](b)[/b] we have $ \mathrm{Rank}\left( BA\right) \minus{} \mathrm{Rank}\left( AB\right) \leq\frac {n}{2}$.[/color]

[i]Proof of Corollary 2.[/i] The first isomorphism theorem (applied to the $ k$-linear map $ BABA: U\rightarrow U$) yields

$ \left( BABA\right) \left( U\right) \cong U\diagup\mathrm{Ker} \left( BABA\right)$,

so that

$ \dim\left( \left( BABA\right) \left( U\right) \right) \equal{} \dim\left( U\diagup\mathrm{Ker}\left( BABA\right) \right) \equal{} \dim U \minus{} \dim\left(\mathrm{Ker}\left( BABA\right) \right)$
[color=green][b](4)[/b][/color] $ \geq\dim U \minus{} 2\dim\left( \mathrm{Ker} \left( BA\right) \right)$

(since Theorem 1 [b](a)[/b] (applied to $ V \equal{} U$, $ W \equal{} U$, $ X \equal{} BA$ and $ Y \equal{} BA$) yields $ \dim\left( \mathrm{Ker}\left( BABA\right) \right) \leq \dim\left( \mathrm{Ker}\left( BA\right) \right) \plus{} \dim\left(\mathrm{Ker}\left( BA\right) \right) \equal{} 2\dim\left( \mathrm{Ker} \left( BA\right) \right)$).

Besides,

$ \dim\left( \left( BABA\right) \left( U\right) \right) \equal{} \mathrm{Rank}\left( BABA\right) \leq\mathrm{Rank}\left( BAB\right)$ (by Theorem 1 [b](b)[/b] (applied to $ V \equal{} U$, $ W \equal{} U$, $ X \equal{} BAB$ and $ Y \equal{} A$))
[color=green][b](5)[/b][/color] $ \leq\mathrm{Rank}\left( AB\right)$ (by Theorem 1 [b](c)[/b] (applied to $ V \equal{} U$, $ W \equal{} U$, $ X \equal{} B$ and $ Y \equal{} AB$)).

Also, the first isomorphism theorem (applied to the $ k$-linear map $ BA: U\rightarrow U$) yields

$ \left( BA\right) \left( U\right) \cong U\diagup\mathrm{Ker}\left( BA\right)$,

so that

$ \dim\left( \left( BA\right) \left( U\right) \right) \equal{} \dim\left( U\diagup\mathrm{Ker}\left( BA\right) \right) \equal{} \dim U \minus{} \dim\left( \mathrm{Ker}\left( BA\right) \right)$

and thus

[color=green][b](6)[/b][/color] $ \dim\left( \mathrm{Ker}\left( BA\right) \right) \equal{} \dim U \minus{} \underbrace{\dim\left( \left( BA\right) \left( U\right) \right) }_{ \equal{} \mathrm{Rank}\left( BA\right) } \equal{} \dim U \minus{} \mathrm{Rank}\left( BA\right)$.

Now,

$ \mathrm{Rank}\left( AB\right) \geq\dim\left( \left( BABA\right) \left( U\right) \right)$ (by [color=green][b](5)[/b][/color])
$ \geq\dim U \minus{} 2\dim\left( \mathrm{Ker}\left( BA\right) \right)$ (by [color=green][b](4)[/b][/color])
$ \equal{} \dim U \minus{} 2\left( \dim U \minus{} \mathrm{Rank}\left( BA\right) \right)$ (by [color=green][b](6)[/b][/color])
$ \equal{} 2\mathrm{Rank}\left( BA\right) \minus{} \underbrace{\dim U}_{ \equal{} n} \equal{} 2\mathrm{Rank}\left( BA\right) \minus{} n$.

Thus, Corollary 2 [b](a)[/b] is proven.

Now,

$ 2\mathrm{Rank}\left( AB\right) \geq\mathrm{Rank}\left( AB\right)$ (since $ \mathrm{Rank}\left( AB\right)\geq0$)
$ \geq2\mathrm{Rank}\left( BA\right) \minus{} n$ (by Corollary 2 [b](a)[/b]),

so that $ 2\mathrm{Rank}\left( BA\right) \leq2\mathrm{Rank}\left( AB\right) \plus{} n$ and thus

$ \mathrm{Rank}\left( BA\right) \leq\frac {2\mathrm{Rank}\left( AB\right) \plus{} n}{2} \equal{} \mathrm{Rank}\left( AB\right) \plus{} \frac {n}{2}$,

so that $ \mathrm{Rank}\left( BA\right) \minus{} \mathrm{Rank}\left(AB\right) \leq\frac {n}{2}$. Thus, Corollary 2 [b](b)[/b] is proven.

  darij

We can also use the Flander's theorem about the Jordan's forms of $ AB$ and $ BA$. See:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=299895
We consider nilpotent Jordan blocks of dimension $ \geq{2}$ of $ AB,BA$. The ranks of each such blocks differ at most by $ 1$. We have at most $ E(n/2)$ blocks (where $ E()$ is the integer part); then $ sup(rank(AB)\minus{}rank(BA))\equal{}E(n/2)$.

Interesting

[url= https://artofproblemsolving.com/community/c7h2576955_matrix_and_rank]see [/url]
https://artofproblemsolving.com/community/c7h2576955_matrix_and_rank


How many distinct planes pass through 3 or more vertices of a cube? (ARML #5)

[hide="Solution."]
=====
Orient the cube such that there is a bottom face. There are five cases:

CASE ONE: The plane contains no vertices on the bottom face. There is only $ 1$ -- the one that contains the top face.

CASE TWO: The plane contains exactly 1 vertex on the bottom face. There's only one such plane for each vertex, so $ 4$ of these.

CASE THREE: The plane contains exactly 2 vertices on the bottom face. There's two for each pair of adjacent vertices (four ways to choose these) and three planes for each pair of non-adjacent vertices (two ways to choose these), so there are $ 2(4) \plus{} 3(2) \equal{} 14$ of these.

CASE FOUR: The plane contains exactly 3 vertices on the bottom face. There are none of these.

CASE FIVE: The plane contains exactly 4 vertices on the bottom face. Clearly there is only $ 1$ such plane -- the plane containing the bottom face.

In all, we find $ 1 \plus{} 4 \plus{} 14 \plus{} 0 \plus{} 1 \equal{} \boxed{20}$ such planes.
=====
[/hide]

[hide="Another way to break this into casework"][b]1.[/b] An equilateral triangle, containing the 3 vertices adjacent to a particular vertex
[i]Since each corresponds to a particular vertex, there are 8 of these[/i]

[b]2.[/b] A face
[i]There are 6 faces[/i]
[b]
3.[/b] A rectangle consisting of diagonally opposite, parallel edges
[i]Each contains 2 face diagonals, and there are 6*2 face diagonals, so there are 6 of these[/i]

A total of 20.[/hide]

dont any 3 points determine a plane??

wouldn't it then be 8C3 ??

Yes, but some of those planes may be the same if more than 3 points are coplanar - see TZF's case 2 and case 3.


The surface area of a cube is 294 square centimeters. What is the ratio of the number of square centimeters in the surface area to the number of cubic centimeters in the volume of the cube?

If the surface area of the cube is $ 294$ square centimeters, then the side length is $ \sqrt\frac{294}{6}=7$.
So, the volume is $ 7^3=343$, and the ratio of surface area to volume is $ \frac{294}{343}=\frac{6}{7}$.


calculus integration function limit LaTeX algebra domain

Hi all!

Could you please help me with this problem: we have to find the integral from 0 to infinity of: (sqrt(t))/(1+t^2) using contour integration.

I'm thinking about integrating (sqrt(z))/(1+z^2), but I'm not sure which contour to choose. I tried the "closed semicircle" which goes from -R to R on the real axis, then counterclockwise across the semicircle with radius R and center 0. But when I get to the point of splitting this integral into parts (from -R to 0, from 0 to R, and on the semicircle) I get into trouble, as I have a real integral of (sqrt(t))/(1+t^2) on the interval -R to 0, where sqrt(t) is not defined.

So, should I choose a "keyhole" contour, instead (small circle around zero, line to R, large circle with radius R, line back to smaller circle - as in avoiding the log branch cut?)? Or should I use a different integrand function? Or should I try a change of variables (i.e. take s=sqrt(t) or something else)? I'm not quite sure all the concepts are clear to me.... Please help!

Any ideas, hints, and even solutions are greatly appreciated.

Thanks,

Sitora

[quote="Sitora"]I get into trouble, as I have a real integral of (sqrt(t))/(1+t^2) on the interval -R to 0, where sqrt(t) is not defined.[/quote]
This means that you fail to understand a few basic ideas of complex analysis. $ \sqrt t$ has a perfectly well-defined branch in the upper half-plane continuous up to the boundary and coinsiding wiht the usual positive square root on the right semi-axis, so you can use this branch in the residue formula. Of course, the contribution of the negative semi-axis won't have real value but it can be easily expressed in terms of the original integral (or you can just note that it is purely imaginary and, therefore, can be killed by just taking the real part of the full integral).

Thank you for the reply. Yes, I don't feel I understand the basic concepts of complex analysis, though I've been struggling with it for a while. Could you recommend a site where they are clearly explained?

Also, could you clarify a couple of points you made?

1. When integrating a complex function along a contour, we sometimes split the contour up into parts (in the above case - positive semi-axis (call the integral over this part I1)), semicircle on upper half-plane (I2), and negative semi-axis(I3)). It sounds like a stupid question, but could you tell me if I've got this right?:

A given integral Ii is real if its contour is on the real axis, and the integrand function is defined on that real interval. Otherwise, the integral is a complex one.

You said:

[quote]the contribution of the negative semi-axis won't have real value [/quote]

But why are we assuming that a "complex integral" (eg I3) has a purely imaginary value?

2. OK, from what I understood, this is what I'm supposed to do to solve the problem:

Consider (int=integral)

 int( $ \frac{\sqrt{z}}{1\plus{}z^{2}}$ )dz around the closed semicircular contour we've already discussed.

The only pole inside the contour is i, so by the residue theorem,

int(  $ \frac{\sqrt{z}}{1\plus{}z^{2}}$  )dz=2(pi)iRes(f,i)

Res(f,i)= $ \lim_{z\rightarrow i}\frac{{(z\minus{}i)}\sqrt{z}}{1\plus{}z^{2}}$ =sqrt(i)/(2i)

Now, which root of i am I supposed to take? I suppose the positive one?

Next, I break the integral into three parts, I1, I2, and I3 (as above), show that I2 goes to 0 as R goes to infinity, note that I3 is purely imaginary (as you suggested), note that I2 is real, and, equating the real and imaginary parts, get

Int (from 0 to infinity) ( $ \frac{\sqrt{x}}{1\plus{}x^{2}}$ )= 2(pi)(sqrt(2))

Could you tell me if this is right?

3. Finally, do these forums only use LaTeX for formulas? Is there a way to copy-paste from MathType? (though I tried the latter and failed.)

Thank you a lot for your help!

[quote="Sitora"]Finally, do these forums only use LaTeX for formulas? Is there a way to copy-paste from MathType? (though I tried the latter and failed.)[/quote]

Mathtype is able to convert equations into tex/latex language.

First, precisely define an analytic determination of $ \sqrt {z}$, the domain of which includes your integral.  Here's one way:
\[ f(z) \equal{} \sqrt {z} \equal{} r^{1/2}e^{i/2 \Theta},\quad \minus{} \pi/2 < \Theta\leq 3\pi/2\]
Now, $ f(z)$ is entirely analytic in the complex plane except along the negative imaginary axis and the origin which is a branch point.  Therefore, if I now want $ f(i)$, that's easy right?  It's just $ |1|e^{i/2(\pi/2)} \equal{} e^{\pi i/4}$.  Then if I wish the residue of the expression $ \frac {f(z)}{1 \plus{} z^2}$ at $ z \equal{} i$, that's also easy.  $ \mathop\text{Res}_{z \equal{} i}\left\{\frac {f(z)}{(z \plus{} i)(z \minus{} i)}\right\} \equal{} \frac {e^{\pi i/4}}{2i}$.
Now, if I use a contour which travels along the real axis, with an indentation around the origin, and then circling back to the starting point, we can show the integral over the large semi-circle and small semi-circle parts tend to zero and letting $ z \equal{} re^{ \pi i}$ on the negative real axis and $ z \equal{} r$ on the positive:
\[ i\int_0^{\infty} \frac {\sqrt {r}}{r^2 \plus{} 1}dr \plus{} \int_0^{\infty} \frac {\sqrt {r}}{r^2 \plus{} 1}dr \equal{} 2\pi i\left(\frac {e^{\pi i/4}}{2i}\right)\]
or:
\[ \int_0^{\infty}\frac {\sqrt {r}}{r^2 \plus{} 1} \equal{} \frac {1}{1 \plus{} i}2\pi i\left(\frac {e^{\pi i/4}}{2i}\right) \equal{} \frac {\pi}{\sqrt {2}}\]

[quote="Soarer"]

Mathtype is able to convert equations into tex/latex language.[/quote]

Thank you! It didn't occur to me to check if this was possible. But I've got a bit of a problem: when I try to copy-paste in this way, what I get is a lot of gibberish, with the equation at the end.

Here's an example with 2/3:

% MathType!MTEF!2!1!+-
% feaafaart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr% MathType!MTEF!2!1!+-
% feaafaart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbGeaGGipv0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca
% aIYaaabaGaaG4maaaaaaa!3795!
\[ {2 \over 3}\]

Could you help?

Just remove the gibberish part before you post.  :wink:
I have not used Mathtype for a long time so I'm not sure, but I remember that you can remove it somewhere.

[quote="youriens"]First, precisely define an analytic determination of $ \sqrt {z}$, the domain of which includes your integral.  Here's one way:
\[ f(z) \equal{} \sqrt {z} \equal{} r^{1/2}e^{i/2 \Theta},\quad \minus{} \pi/2 < \Theta\leq 3\pi/2\]
Now, $ f(z)$ is entirely analytic in the complex plane except along the negative imaginary axis and the origin which is a branch point.  Therefore, if I now want $ f(i)$, that's easy right?  It's just $ |1|e^{i/2(\pi/2)} \equal{} e^{\pi i/4}$.  Then if I wish the residue of the expression $ \frac {f(z)}{1 \plus{} z^2}$ at $ z \equal{} i$, that's also easy.  $ \mathop\text{Res}_{z \equal{} i}\left\{\frac {f(z)}{(z \plus{} i)(z \minus{} i)}\right\} \equal{} \frac {e^{\pi i/4}}{2i}$.
Now, if I use a contour which travels along the real axis, with an indentation around the origin, and then circling back to the starting point, we can show the integral over the large semi-circle and small semi-circle parts tend to zero and letting $ z \equal{} re^{\pi i}$ on the negative real axis and $ z \equal{} r$ on the positive:
\[ i\int_0^{\infty} \frac {\sqrt {r}}{r^2 \plus{} 1}dr \plus{} \int_0^{\infty} \frac {\sqrt {r}}{r^2 \plus{} 1}dr \equal{} 2\pi i\left(\frac {e^{\pi i/4}}{2i}\right)\]
or:
\[ \int_0^{\infty}\frac {\sqrt {r}}{r^2 \plus{} 1} \equal{} \frac {1}{1 \plus{} i}2\pi i\left(\frac {e^{\pi i/4}}{2i}\right) \equal{} \frac {\pi}{\sqrt {2}}\]
[/quote]

Thank you for this post! It answered all my questions clearly and precisely! (as I was wondering about where the branch-cut should be, as well as other things mentioned in your post)

[quote="Sitora"]
I've got a bit of a problem: when I try to copy-paste in this way, what I get is a lot of gibberish, with the equation at the end.

Here's an example with 2/3:

% MathType!MTEF!2!1!+-
% feaafaart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr% MathType!MTEF!2!1!+-
% feaafaart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbGeaGGipv0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca
% aIYaaabaGaaG4maaaaaaa!3795!
\[ {2 \over 3}\]
Could you help?[/quote]

Not sure what version of MathType you're using, but when you select the LaTeX translator, you need to [b][i]uncheck[/i][/b] the 2 checkboxes
[list]Include translator name in translation
Include MathType data in translation[/list]
If you want an inline equation, $ r \equal{} \sqrt {{x^2} \plus{} {y^2}}$, check the [b]Inline equation[/b] command in the View menu. If you want a display equation,\[ f(z) \equal{} \sqrt z  \equal{} {r^{1/2}}{e^{i/2\Theta }},\quad  \minus{} \pi /2 < \Theta  \le 3\pi /2,\]make sure the [b]Inline equation[/b] command is unchecked.

If you find an equation in a Forum post that you want to use elsewhere, open MathType and drag the equation over to the open MathType window (you don't need to select the translator first -- just drag it). Depending on your browser or your operating system, drag & drop may not work. If not, try copy & paste.


inequalities
inequalities solved
3-variable inequality
square root inequality
Symmetric inequality

Let $ x$, $ y$, $ z$ be positive real numbers. Prove that
\[\sum_{\text{cyc}}\frac{\sqrt{xy\plus{}4yz\plus{}4zx}}{x\plus{}y}\geq\frac92\]

Are you sure the inequality is right? Are there any restrictions to the variables?

Counterexample: let x=y=z

when you simply the LHS you get 9 \sqrt (x/2) and we can arbitrarily let x =  \varepsilon>0  so that 9 \sqrt ( \varepsilon /2) < 9/2   :(  :(  :(

[quote="3X.lich"]x=y=z .. you get 9 \sqrt (x/2)[/quote] Sure ?

hehe my bad, just overlooked the square root was on the numerator, i took it for the whole expression  :D
Anyways, I'm still working on it. Thx for pointing out my sillyness :)

[/quote]
$ (y \plus{} z)^{2}(zx \plus{} 4xy \plus{} 4yz)(xy \plus{} 4yz \plus{} 4zx) \minus{} 4(xy^{2} \plus{} xz^{2} \plus{} 2y^{2}z \plus{} 2yz^{2} \plus{} 3xyz)^{2}$

$ \equal{} xyz(x \plus{} 4y \plus{} 4z)(y \minus{} z)^{2}\geq0$.
[/quote]

how can we think of this part? i think it's reali difficult!

[quote="morheli"]$ (y + z)^{2}(zx + 4xy + 4yz)(xy + 4yz + 4zx) - 4(xy^{2} + xz^{2} + 2y^{2}z + 2yz^{2} + 3xyz)^{2}$

$ = xyz(x + 4y + 4z)(y - z)^{2}\geq0,$

how can we think of this part? i think it's reali difficult![/quote]Try to guess how to come to this effect estimation ([url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=273]Namdung[/url]):

$ \sqrt {(zx + 4xy + 4yz)(xy + 4yz + 4zx)}\geq\frac{2(xy^{2} + xz^{2} + 2y^{2}z + 2yz^{2} + 3xyz)}{y + z}.$

$ \left(\sum\frac {\sqrt {yz + 4zx + 4xy}}{y + z}\right)^{2}$

$ = 2\sum\frac {\sqrt {(zx + 4xy + 4yz)(xy + 4yz + 4zx)}}{(z + x)(x + y)} + \sum\frac {yz + 4zx + 4xy}{(y + z)^{2}}$

$ \geq4\sum\frac {xy^{2} + xz^{2} + 2y^{2}z + 2yz^{2} + 3xyz}{(y + z)(z + x)(x + y)} + \sum\frac {yz + 4zx + 4xy}{(y + z)^{2}}$

$ = \frac {12(x + y + z)(yz + zx + xy)}{(y + z)(z + x)(x + y)} + \sum\frac {yz + 4zx + 4xy}{(y + z)^{2}}$

${ = \frac {45}{4} + 4\sum{\frac {yz + zx + xy}{(y + z)^2}}} + \frac {3}{4}\prod{\frac {(y - z)^2}{(y + z)^2}}$

$ \geq \frac {45}{4} + 4\times\frac {9}{4} + 0 = \frac {81}{4},$

which is true by [url=http://www.mathlinks.ro/viewtopic.php?t=32470]Iran 96 inequality[/url] :

$ \frac {4}{ (y + z)^2 } + \frac {4}{(z + x)^2} + \frac {4}{(x + y)^2}\geq\frac {9}{yz + zx + xy}.$

[hide="Another proof."]$ \left(\sum\frac {\sqrt {yz + 4zx + 4xy}}{y + z}\right)^{2}$

$ = 2\sum\frac {\sqrt {(zx + 4xy + 4yz)(xy + 4yz + 4zx)}}{(z + x)(x + y)} + \sum\frac {yz + 4zx + 4xy}{(y + z)^{2}}$

$ \geq4\sum\frac {xy^{2} + xz^{2} + 2y^{2}z + 2yz^{2} + 3xyz}{(y + z)(z + x)(x + y)} + \sum\frac {yz + 4zx + 4xy}{(y + z)^{2}}$

$ = \frac {81}{4} + \sum{\frac {(y - z)^2\left\{(y + z - x)^2\left[31\left(x^2 + yz\right) + 33x(y + z)\right] + 64xyz(y + z)\right\}}{16(y + z)^2(z + x)^2(x + y)^2}}$

$ \geq\frac {81}{4}.$[/hide]

See also : http://www.mathlinks.ro/viewtopic.php?t=28175

[quote="Ji Chen"]
$ \sqrt {(zx \plus{} 4xy \plus{} 4yz)(xy \plus{} 4yz \plus{} 4zx)}\geq\frac {2(xy^{2} \plus{} xz^{2} \plus{} 2y^{2}z \plus{} 2yz^{2} \plus{} 3xyz)}{y \plus{} z}.$

[/quote]
It was created from:
$ \sqrt{(a\plus{}k^2b)(b\plus{}k^2a)}\ge k(a\plus{}b)\plus{}\frac{2ab(k^2\minus{}1)^2}{(a\plus{}b)(k\plus{}1)^2}$ for $ a,b>0$ and $ k \ge 1$


algorithm
combinatorics open
combinatorics

I have a problem that I dont know if someone created it or not before.

N sets are given: A_1,..,A_N. Is there any algorithm or theorem show a way to find a smallest set A such that: the intersection of A and A_i is not empty? for i from 1 to N.

greedy algorithm... take an element which is in the greatest number of $A_i$ possible, then take an element which is in the greatest number of remaining $A_i$ possible, and so on

That is the first algorithm appear in my mind. But could you show me (by mathematics) that your algorithm will work properly?


[quote="ayush_2008"]can someone will tell  me about the site which will provide me free notes of iit chapter wise.[/quote]Try your country's subforum Community India and ask there: link below

[url]http://www.mathlinks.ro/index.php?f=514[/url]

IIT is Indian Institute of Technology. There are 5 of them, and they are among the best engineering schools in the world.

*When you're an Indian, you know stuff like this...

I know what IIT is.  I just didn't know what you were referring to in this context.

Ayush posted something about IIT in the India section:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1133576#1133576


[color=purple]Prove that :
  $ a^2\plus{}b^2\plus{}c^2\ge 4\sqrt{4\minus{}sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}}S$ for any triangle ABC
 And$ a^2\plus{}b^2\plus{}c^2\ge 4\sqrt{4\minus{}cosAcosBcosC}S$ for any acute triangle ABC. :) [/color]

I like first inequality.
I will try to prove it :)

We know         sin(A/2)sin(B/2)sin(C/2) = r/4R  so  4-8r/4R >3 implies R>2r  so I ask myself if the product of sin(A/2) in square root must be multiplied by 8


let ABCD be a convex quadrilaterall.DA and CB intersect each other in points M.and ,BA and CD intersect each other

in point N.aline goes through M,and intersects AB and CD,in points P and Q,respectivly.

PD and QA intersect each other in point X.

PC and QB intersect each other in point Y.

prove that: X,Y,N are colinear.

By Manelaus' Theorem,
Consider $AXQ$ on $\triangle MPD$, we have $\dfrac{MA}{AD}\cdot\dfrac{DX}{XP}\cdot\dfrac{PQ}{QM}=1$
$\Rightarrow \dfrac{DX}{XP}=\dfrac{AD}{MA}\cdot\dfrac{QM}{PQ}$
Consider $BYQ$ on $\triangle MPC$, we have $\dfrac{MB}{BC}\cdot\dfrac{CY}{YP}\cdot\dfrac{PQ}{QM}=1$
$\Rightarrow \dfrac{PY}{YC}=\dfrac{MB}{BC}\cdot\dfrac{PQ}{QM}$
Consider $NAB$ on $\triangle MCD$, we have $\dfrac{MB}{BC}\cdot\dfrac{CN}{ND}\cdot\dfrac{DA}{AM}=1$
$\Rightarrow \dfrac{CN}{ND}=\dfrac{AM}{DA}\cdot\dfrac{BC}{BM}$
Therefore, $\dfrac{PY}{YC}\cdot\dfrac{CN}{ND}\cdot\dfrac{DX}{XP}$
$=(\dfrac{MB}{BC}\cdot\dfrac{PQ}{QM})(\dfrac{AM}{DA}\cdot\dfrac{BC}{BM})(\dfrac{AD}{MA}\cdot\dfrac{QM}{PQ})=1$
By the converse of Manelaus' Theorem, $N, X, Y$ are collinear.


Έστω δύο κύκλοι $ (O1, R)$, και $ (O2, r)$ με $ R>r$ που εφάπτονται εσωτερικά στο $ A$. Έστω ε1 μια ευθεία που διέρχεται από το $ A$ και τέμνει τους κύκλους $ C1$, και $ C2$ στα $ C$ και $ B$ αντίστοιχα. Έστω ε2 μία άλλη ευθεία που διέρχεται από το $ A$ και τέμνει τους $ C1$ και $ C2$ στα $ D$ και $ E$ αντίστοιχα. Αν το $ BCDE$ είναι εγγράψιμο σε κύκλο κέντρου $ O3$, να δειχθεί ότι η παράλληλη που άγεται από το $ O3$ προς την $ O1O2$ διέρχεται από το σημείο τομής των $ BD$ και $ CE$.

Ελπίζω ότι είναι σωστή και θα σας αρέσει. :)

[quote="Protonios"]Έστω δύο κύκλοι $ (O1, R)$, και $ (O2, r)$ με $ R > r$ που εφάπτονται εσωτερικά στο $ A$. Έστω ε1 μια ευθεία που διέρχεται από το $ A$ και τέμνει τους κύκλους $ C1$, και $ C2$ στα $ C$ και $ B$ αντίστοιχα. Έστω ε2 μία άλλη ευθεία που διέρχεται από το $ A$ και τέμνει τους $ C1$ και $ C2$ στα $ D$ και $ E$ αντίστοιχα. Αν το $ BCDE$ είναι εγγράψιμο σε κύκλο κέντρου $ O3$, να δειχθεί ότι η παράλληλη που άγεται από το $ O3$ προς την $ O1O2$ διέρχεται από το σημείο τομής των $ BD$ και $ CE$.

Ελπίζω ότι είναι σωστή και θα σας αρέσει. :)[/quote]

Katarxin thelo apo edo na doso sugxarhthria ston kosta gia tin epidosi tou ston arximidi alla kurios giati ematha oti eluse ti geometria me desqrgues. apo ti stigmi pou arxises na asxoleisai me geometria kai luneis ti geometria me geometria de se fovamai katholou!theos!

tora gia tin askhsh... otan ti diavasa pistepsa oti vgazei kati kalo alla telika fainetai kai diorthose me an kano lathos oti kapoia magia pigaines na kaneis stin kataskeui alla den prepei na sou vgike. alla etsi einai oi dikes mas kataskeues...pote mas vgainoun pote de mas vgainoun...carry on!

loipon to olo thema einai oti to A einai kentro omoiothesias ton duo kuklon epomenos to BCDE einai trapezio ek kataskeuis. Ara gia na einai eggrapsimo tha prepei na einai isoskeles pragma pou simainei oti ta $ O_3,O_2,O_1,A$  kathos kai to shmeio tomhs  ton diagonion einai suneutheiaka.
elpizo na einai sosta ...kai na mh grafo paparies.

Επειδη δεν ξερω ομοιοθεσια θα δωσω μια λυση βασισμενη σε ριζικους αξονες,Εστω η ευθεια ε κοινη εφαπτομενη των Ο1,Ο2 στο Α.Η ε ειναι
 τοτε ριζικος αξονας των κυκλων Ο1,Ο2.Ευκολα αποδεικνυεται οτι η ε ειναι παραλληλη στην $ DC$ που ειναι ο ριζικος αξονας των κυκλων Ο1,Ο3.
 Η  $ EB$ ειναι ριζικος αξονας των κυκλων Ο2,Ο3.Επειδη ομως οι τρεις ριζικοι αξονες των κυκλων ανα δυο δεν μπορουν να συντρεχουν στο ριζικο κεντρο αφου οι δυο ειναι παραλληλοι τοτε υποχρεωτικα ειναι και οι τρεις παραλληλοι ,αρα το  $ DCBE$ ειναι τραπεζιο και ως εγγεγραμενο ειναι ισοσκελες.Αρα το σημειο τομης των διαγωνιων του και το Ο3 βρισκονται στην μεσοκαθετη της  $ DC$ .

[quote="Protonios"]M....κία στην κατασκευή. :(

Θεώρησα ότι οι ριζικοί άξονες τέμνονται ενώ σωστά υποδεικνύει και o nik και ο Σάκης ότι είναι παράλληλοι. :wink:

Κρίμα. Είχε ένα δύσκολο argument μετά... :)

Nick thx για τα καλά σου λόγια πάντως... :P
Δεν τα επιβεβαίωσα με αυτή τη χαζοκατασκευή.[/quote]

kala tora xaideuesai...re einai polu simantiko to oti boreis na kataskeuazeis askiseis kai to oti polles apo autes vgainoun koukles...tha hsouna polu egoistis an itheles na eisai o monadikos anthropos ston kosmo o opoios petuxainei se oles tou tis kataskeues...

(sil exo adiko? esena sou evgainan panta sfixtes? allo tora pou xeis parei to kollaei!!!)

loipon kosta perimeno na do askisi pano stin idea pou eixes kai de sou vgike-exallou den apokalifthike i idea!


I'd like to see the  all possible solutions of problems from IMC.
Hence I'll begin from 1994 year.
also I will represent my own solutions:
Find
\[ \lim_{N \to \infty}\frac {\ln^{2} N}{N}\sum_{k \equal{} 2}^{N \minus{} 2}\frac {1}{\ln k \ln (N \minus{} k)}
\]
Solution : (Let $ N$ is sufficiently large )
\[ \sum_{k \equal{} 2}^{N \minus{} 2}\frac {1}{\ln k \ln (N \minus{} k)} \equal{} \sum_{k \equal{} 2}^{N \minus{} 2}\frac {1}{\ln k (\ln N \plus{} \mathcal{O}\left( k/N \right)} \equal{} \sum_{k \equal{} 2}^{N \minus{} 2}\frac {1}{\ln k\ln N}\left( 1 \plus{} \mathcal{O}\left(\frac {k}{N \ln N} \right) \right) \equal{} \frac {N}{\ln^{2}N} \plus{} \mathcal{O}\left( \frac {N}{\ln^{3} N}\right)
\]
thus answer is $ 1$.
any other possible ways ?

You can see the solution here. [url]http://www.imc-math.org/[/url]


group theory
abstract algebra
superior algebra
superior algebra unsolved

The very first part, when you say for $ H \equal{} [4]$ is what you are doing; only difference is you are doing it for $ H$ the whole subgroup.   So,

$ [0] \plus{} H \equal{} \{0 \plus{} 0, 4 \plus{} 0, 8 \plus{} 0, 12 \plus{} 0\}$ etc.

It seems as though you got a little confused and tried to take all subgroups of $ H$ whereas you were asked for the cosets of $ H$ (hence, no need to do it for the cyclic group generated by 8 or 12.

Thanks for the reply.  Actually the problem is written like H ={[0], [4], [8], [12]}.  I assumed the [ ] brackets meant it was a subgroup with a generating set (ie [8] is a subgroup that generates a set of size 2).  Am I understanding the problem correctly?  Thanks in advance for clarifying.

Also, if this is true,  and I need to include [12] how far do I carry it out to?

Thank you.  Please bear with me on this as I'm clearly confused on the terms.  I understand the elements of Z16 are 0, 1, ..., 15.  I think the cosets of the subgroup are [a]+H, where [a] iterates thru the elements until your back to the original subgroup H ={[0], [4], [8], [12]}.

So if I understand you correctly, your saying that H ={[0], [4], [8], [12]} is just indicating the case where the class is [4] and not a subgroup of the classes =[0], [4], [8], [12].   So the correct answer to this question is just:
[0] + {0,4,8, 12} = {0, 4, 8, 12}
[1] + {0,4,8, 12} = {1, 5, 9, 13}
[2] + {0,4,8, 12} = {2, 6, 10, 14}
[3] + {0,4,8, 12} = {3, 7, 11, 15}
?


The answer is 185 snakes because 10in:26snakes.  You want 5ft 10, so to revert that to inches, you do:12*5+10 which gives you 70.  Therefore you have 10/26=70/x (where x is the number of snakes), and due to the cross multiplication rule, 1820=10x, which x=182.  Now it has to be taller, so to get 1 inch, you divide 26 by 10 and get 2.6 snakes per inch.  YOu can't have .6 of a snake, so its 3, so 182+3=185.

Treething didn't say that the pillar had to be 1 inch taller did he? Treething just said it had to be taller, so wouldn't 1 snake more make it taller? :huh:

[hide="unsure"]The ratio is 67.6. Treething is 70 in. $\frac{x^{2}}{70}=67.6$
$x=26\sqrt{7}$
2.67 is about $\sqrt{7}$.
$\lceil 2.67 \cdot 26 \rceil=$
$70$[/hide]

No correct answers yet.

bpms is on the right track though.

[quote="Treething"]Matt wishes to make pillars out of snakes. If the height of the pillar is proportional to the square of the amount of snakes in it and a ten-inch tower requires twenty-six snakes, how many snakes would need to be in a tower in order for it to be taller than Treething? (Treething is 5 foot 10.)[/quote]
[hide]We have

$\frac{s^{2}}{h}=k$, where $k$ is constant

So $\frac{26^{2}}{10}=k=67.6$

So, now we have

$\frac{s^{2}}{70}=67.6$

So $s^{2}=4732$

So $s=26\sqrt{7}$

$26\sqrt{7}$ snakes is the answer

But we have to take the ceiling of that, which is 69.

What a surprise

$69$[/hide]

Actually, the pillar can be alined and made so that one snake will not be enough.  I just put one extra inch so to be safe, but it can be anywhere from 183-185 snakes.

NO, what I was doing was answering the other guys question.  Sorry.  I didn't read the ratio part, if you did it my way, it's because you didn't read the entire problem.  I will vouch its 69, my bad.

[quote="Treething"]Matt wishes to make pillars out of snakes. If the height of the pillar is proportional to the square of the amount of snakes in it and a ten-inch tower requires twenty-six snakes, how many snakes would need to be in a tower in order for it to be taller than Treething? (Treething is 5 foot 10.)[/quote]
[hide]
let $y$ be height, $x$ be snakes
$y=kx^{2}$
$10=k26^{2}$
$k=\frac{10}{26^{2}}$
$k=0.014792899408284023668639053254438$
$y=70$
$70=0.014792899408284023668639053254438x^{2}$
$x^{2}=4732$
$x=68.789534087679355353042009594621$ snakes
to be taller than treething, i round x up
$x=69$ snakes
[/hide]


Đổi cho anh đi, anh có tất cả bộ của Titu rồi, trừ cuốn này. Đây là email của anh n111ttuan_qn@yahoo.com.vn , bỏ đi 3 số 1.

Em nhớ là em đã down trên VMF , anh lên tìm thử  :D . Còn chuyện sách, có thì cho đi anh, trao đổi cái nỗi gì  :P

em cũng đang cần cuốn này . Nếu có thì cho em xin bản ebook với.
email: batigol_leon@yahoo.com

[quote="quangpbc"]Lạ quá, em nhớ là mình đã từng có cuốn này. Khi trưa lên VMF tìm lại không thấy .  :blush: Chắc là thầy em cho rồi  :D Để khi nào xin lại , share cho mọi người sau.  :wink: Thế nhỉ .

Anh Tuân nhớ là trao đổi nha. Lắm sách mà không chịu chia sẻ gì cả  :([/quote]
Chú Quang có thì up lên đi! Cho mọi người chiêm ngưỡng cái! Giữ làm của riêng xấu lắm! :D

Oái . Oan cho tôi quá ông ơi  :( . Có đâu mà up với chả kô up  :D . Để xin lại thầy đã . Mấy lâu nay thầy có cho đọc sách khác đâu toàn phải đọc 30-4 để thi chọn ĐT thi QG :P

Hinh nhu cuon nay chua co tren ant thi phai

Anh N.T.TUAN oi, xoa ho em topic Vmo 2008 voi! :(  Khong co ai ban luan ca de nhin thi xau lam! :lol:

O ngay duoi topic nay thoi anh a.
Ma hinh nhu co ket qua thi quoc gia roi thi phai!
Ai co thong tin gi thi lam on post len cho moi nguoi cung xem voi! :lol:  Co danh sach thi cang tot  :D . Cam on nhieu!

8.3% có giải : 0 giải nhất, 3 giải nhì, 10 giải ba và 20 giải KK
Đấy là mình xem trên diễn đàn toán học. Chắc ông Chức cũng xem rồi. Nhưng hình như chưa co kết quả chính thức đâu.

Các anh ơi thế cái cuốn Algebra đấy nó có ebook ko ạ? Nếu có thì các anh và các thầy up lên đây luôn có được ko ạ? :roll:  :blush: Em cảm ơn các anh và các thầy nhiều

em cũng đang cần cuốn này . Nếu có thì cho em xin bản ebook với.
email: yunxiu@yeah.net

em cũng xin quấn này với các anh ơi
email: duongnguyenvietanh@gmail.com

gửi cho mình với    hangocbinh@ymail.com

Gui cho em voi, tim mai chang thay ebook dau. :help:  :help:  :help:
long30395@yahoo.com

cuốn này có trên trang Mathscope.Bạn vào mục tài liệu là có thui.

[quote="N.T.TUAN"]Có bạn nào có cuốn sau không?

[color=brown]101 Problems in Algebra: From the Training of the USA IMO Team
Titu Andreescu and Zuming Feng[/color]

Mình đang cần cuốn này, giúp mình với!  :lol:  Có gì chúng ta trao đổi cũng được, theo hai kiểu:
1)Sách-Sách.
2)Sách-Tiền.[/quote]
Bạn vào đây để lấy nhé- Here: [code]http://www.mathdoi.com/2011/08/101-problems-in-algebra-from-the-training-of-the-usa-imo-team/ [/code]

Anh post lên luôn đi! Bọn em đợi mãi sốt ruột! :D

See here file:

[color=#FF0000][Mod: Do not post copyright material on the forum.][/color]

[quote="hangochoanthien"][quote="AnhIsGod"]See here file:[/quote]
Cho em hỏi cái này dùng phần mềm nào mới xem đc hả anh.[/quote]
Phần mềm đọc djvu.

[quote="mathematicalolympiads"]mathematicalolympiads.wordpress.com/ you can find any books you search[/quote]

Notice this is really old?


linear algebra
matrix
vector
linear algebra unsolved

Let $ S$ be a set of invertible matrices, ie $ S\subset GL_n(\mathbb F)$, with the property

that the sum of any two distinct elements of $ S$ is non-invertible. Prove that $ S$ is finite.

[quote="The Captain"][quote="berti"]What is the field? Arbitrary?[/quote]
Yeah. Otherwise I would have explicitly named it. (Please use PM to ask these stuff.)[/quote]

You're saying that all requests for problem clarification should be private?

Let $ A,B \in S$. Since $ A\plus{}B$ is non-invertible, so is $ A^{\minus{}1}(A\plus{}B) \equal{} I\plus{}A^{\minus{}1}B$. Thus $ det(I\plus{}A^{\minus{}1}B) \equal{} 0$ for all $ A,B$. This means that $ \minus{}1$ is an eigenvalue of $ A^{\minus{}1}B$ for all $ A,B$. Now suppose that

$ A^{\minus{}1}B_1v_1 \equal{} \minus{}v_1$
$ A^{\minus{}1}B_2v_2 \equal{} \minus{}v_2$
...
$ A^{\minus{}1}B_nv_n \equal{} \minus{}v_n$

and

$ a_1v_1\plus{}a_2v_2\plus{}\ldots\plus{}a_nv_n \equal{} 0$. Then

$ A^{\minus{}1}(a_1B_1v_1\plus{}a_2B_2v_2\plus{}\ldots\plus{}a_nB_nv_n) \equal{} 0$

from which we can conclude that

$ a_1B_1v_1\plus{}a_2B_2v_2\plus{}\ldots\plus{}a_nB_nv_n \equal{} 0$.

Maybe from this you can get some sort of claims of linear indepdence among eigenvectors? Not sure though. It seems like linear independence is the best way to go here, though.

[quote="JSteinhardt"]

Maybe from this you can get some sort of claims of linear indepdence among eigenvectors? Not sure though. It seems like linear independence is the best way to go here, though.[/quote]

I don't think this is so simple because one can easily find a family of $ n!$ matrices that satisfies the conditions (those with $ \plus{}1$ or $ \minus{}1$ on the main diagonal)

[quote="The Captain"][quote="berti"]What is the field? Arbitrary?[/quote]
Yeah. Otherwise I would have explicitly named it.[/quote]
Not quite. You should demand that if $ \mathbb F$ is infinite, then $ \operatorname{char}\mathbb F\ne 2$ at the very least ;)

As Fedja wrote, the statement is false if $ \mathrm{char}\ {\mathbb F} \equal{} 2$. So we assume below that $ \mathrm{char}\ {\mathbb F} \neq 2$.

Let $ V$ be the linear space of maps from $ \{1,\dots, n\}\times \{1,\dots, n\}$ to $ \mathbb F$. For every matrix $ A$, we define a map $ f_A\in V$ as follows: $ f_A$ maps a set $ S\subset \{1,\dots, n\}\times \{1,\dots, n\}$ to $ \prod_{(i,j)\in S} a_{ij}$. Observe that $ \det(A\plus{}B)$ can be expressed as $ \det(A\plus{}B) \equal{} Q(f(A), f(B))$, where $ Q$ is a bilinear form on $ V$. Then for every two distinct $ A, B\in S$, $ Q(f(A), f(B)) \equal{} \det(A\plus{}B) \equal{} 0$. On the other hand, for every $ A\in S$, $ Q(f(A),f(A)) \equal{} \det(A\plus{}A)  \neq 0$ (here, we use that  $ \mathrm{char}\  {\mathbb F} \neq 2$).

Hence all vectors $ f(A)|_{A\in S}$ are linearly independent. Thus $ |S| \leq \dim V \leq 2^{n^2}$ (the bound is far from being optimal).


Find the number of $ (a,b,c) \in N$ such that
$ abc|(a\plus{}1)(b\plus{}1)(c\plus{}1)$

[quote="∫FaiLurE∮"]check $ (1,6,14)$ :blush:[/quote]

but 1,6,14 doesn't satisfy, 14*6 doesn't divide 14*15.


induction
algebra
partial fractions
combinatorics unsolved
combinatorics
Binomial coefficient identity

Very nice.

This can also be solved with generating functions.  For example

\begin{eqnarray*}\sum_{k=0}^{n}\binom{n+k}{2k}2^{n-k}& = & \sum_{k=0}^\infty \binom{n+k}{2k}2^{n-k}\\ \ & = & \sum_{k=0}^\infty \binom{n+k}{n-k}2^{n-k}\\ \ & = & \sum_{k=0}^\infty [x^{n-k}](1+x)^{n+k}2^{n-k}\\ \ & = & [x^{n}] \sum_{k=0}^\infty x^{k}(1+x)^{n+k}2^{n-k}\\ \ & = & 2^{n+1}[x^{n}] \frac{(1+x)^{n}}{2-x-x^{2}}\end{eqnarray*}

This can be re-expanded a different way to get the answer, but the details of that are not important.  Also, if you want $a_{n}=[x^{n}]f(x)$ where $f(x)$ does [i]not[/i] depend on $n$, then consider

\begin{eqnarray*}\sum_{k=0}^{n}\binom{n+k}{2k}2^{n-k}& = & [x^{n}]\sum_{j=0}^\infty \sum_{k=0}^\infty \binom{j+k}{2k}2^{j-k}x^{j}\\ \ & = & [x^{n}]\sum_{k=0}^\infty \sum_{j=0}^\infty \binom{j+k}{2k}2^{j-k}x^{j}\end{eqnarray*}

Note that after interchanging the order of summation, the inner sum can be evaluated with Netwon's Binomial Formula for Negative Exponents.  This should lead to an equation in the form $a_{n}=[x^{n}]f(x)$ where $f(x)$ does not depend on $n$.

I have been away a long time, and I reviewed this old post of mine to re-learn how to typeset math on this forum.  I said

"This can be re-expanded a different way to get the answer"

Does anyone know what I might have been talking about?  Maybe factoring the whole denominator and using partial fractions?  Or is there something else I might have been talking about?

Reads to me like you meant partial fractions.  Probably not worth worrying too much about: the past is a foreign country: they do things differently there :P


If a,b,c,d,e are positive real numbers prove that

 \sum (cyclic)(abc/de)^4 \geq  \sum (cyclic)abbc

This one can be proved by Cauchy for eight numbers:
2(abc/de)^4+2(bcd/ae)^4+2(cde/ab)^4+(dea/bc)^4+(eab/dc) \geq 8bccd...... :)


quadratics
calculus
derivative
algebra
polynomial
number theory
relatively prime

prove that the equation $ x^2\plus{}x\plus{}1\equal{}py$ has a solution $ (x,y) \in Z^2$  for  an infinit number of prime integer $ p$.

Suppose that there were only finitely many primes $ p_1,p_2,...,p_n$.  Then, let $ x\equal{}p_1p_2\cdots p_n$ and $ x^2\plus{}x\plus{}1$ is relatively prime to all $ p_i$.

Another proof comes from the fact that there exists infinitely many primes of form $ 3k\plus{}1$.

This is in fact the easiest proof that there exist infinitely many primes of the form $ 3k \plus{} 1$.

$ p | x^2 \plus{} x \plus{} 1 \Leftrightarrow p \equal{} 3 \text{ or } p \equiv 1 \bmod 3$, so if we prove that infinitely many primes divide an integer of the form $ x^2 \plus{} x \plus{} 1$ then we have proven that infinitely many primes are congruent to $ 1 \bmod 3$.  This is most easily done by cosinator's argument, the classic "Euclidean" argument.

Of course, you can just cite Dirichlet's theorem, but that's a much deeper result than this one.

For sure, I know that we are supposed to prove that $ p | x^2 \plus{} x \plus{} 1 \Leftrightarrow p \equal{} 3 \text{ or } p \equiv 1 \bmod 3$. But I cannot see such an argument for this to happen.

Factoring mod $ p$, we have that there's a solution $ x$ only if $ \minus{}3$ is a quadratic residue mod $ p$. Quadratic reciprocity then tells us that this is true iff $ p$ is a square mod 3; i.e. if $ p \equal{} 3$ or $ p \equal{} 3k\plus{}1$.

We don't actually have to appeal to quadratic reciprocity; this approach can generalize in a [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=155461]different direction[/url].  $ p | x^2 \plus{} x \plus{} 1 \Leftrightarrow p \equal{} 3 \text{ or } x$ is an element of order exactly $ 3$ in the group of units $ \bmod p$, which is possible if and only if $ 3 | p\minus{}1$.

This approach generalizes to proving that there exist infinitely many primes congruent to $ 1 \bmod n$, whereas the problem with using quadratic reciprocity as $ n$ gets larger is that residue classes get lumped together.

Hey, I have a question about that proof in your blog.

[quote]Case: $ ord_p(a)>1$. Now, since the primitive $ d^{th}$ roots of unity are distinct for distinct $ d$, it's not the case that $ p$ divides $ \Phi_{d}(a)$ for any $ d<n,d|n$.[/quote]

Can you please elaborate on how you reached "it's not the case that $ p$ divides $ \Phi_{d}(a)$ for any $ d<n,d|n$." from the first part?

I believe I did that case backwards.  Here's how I'd handle it now:  if $ p$ doesn't divide $ n$, then $ x^n \minus{} 1 \bmod p$ is relatively prime to its formal derivative, hence has no repeated roots.  We are given that $ \Phi_n(a) \equiv 0 \bmod p$ and $ \Phi_n(a) | a^n \minus{} 1$, and since $ a$ is a root of multiplicity at most $ 1$ it follows that $ a$ is not a root of $ \frac{x^n \minus{} 1}{\Phi_n(x)}$.

This actually proves a stronger result, namely that if $ n | p\minus{}1$ then factoring $ x^n \minus{} 1$ into cyclotomic polynomials $ \bmod p$ separates the roots by their orders $ \bmod p$.


Let us define $a_{n}$ a sequence of real numbers such that \[ 0<a_{n}\leq a_{2n}+a_{2n+1}\forall n\geq 1.  \]
 Prove that the serie $\sum_{n=1}^{\infty}a_{n}$ is divergent.

Easily we obtain $a_{n+1} + \cdots a_{2n+1} \ge a_1$ for all $n$. This fails the Cauchy criterion.


number theory
greatest common divisor
number theory proposed

Find a formula that generates the solutions of the following triple variable equation:

If $ \frac{a}{b}= c \rightarrow \frac{a}{b-1}= c+1$

For instance, $ 12/4 = 3$ while $ 12/3 = 4$. How do you generate all possible solutions with this pattern?

Never mind, I found the solution:

We have $ n \cdot (n+1) \rightarrow n^{2}+n$. Therefore,

$ \frac{n^{2}+n}{n+1}= n$ and $ \frac{n^{2}+n}{n}= n+1$

In the general case, simply substitute $ 1$ by $ m$, giving us:

$ n \cdot (n+m) = n^{2}+mn \rightarrow \frac{n^{2}+mn}{n}= n+m$ and $ \frac{n^{2}+mn}{n+m}= n$

Might as well ask something else.

What is the solution set to the following equality:

$ (x+y)^{2}=$ GCD$ (x,y)$


factorial
floor function
number theory unsolved
number theory

Determine the highest power of $1980$ which divides
 $\frac{(1980n)!}{(n!)^{1980}}$.

How about a more general problem?

The exponent of a prime $p$ in $n!$ is $\sum_{k=1}^{\infty}\left [ \frac{n}{p^{k}}\right ].$ ($[.]$ is the floor function).
And since $11$ is the largest prime factor of $1980$....., we don't have to take into account $2,3,5$.

[quote="t0rajir0u"]Well, it's pretty clear it's just

$\sum_{k=1}^{\infty}\left\lfloor \frac{1980n}{11^{k}}\right\rfloor-1980 \left\lfloor \frac{n}{11^{k}}\right\rfloor$[/quote]

The question asks to find the largest power of 1980 dividing $\frac{(1980n)!}{(n!)^{1980}}$.  So, if you knew the sum above was unbounded, then this would be the answer.  But, what if the sum above is bounded?  Then, one could potentially find an integer value for the highest power of 1980 dividing $\frac{(1980n)!}{(n!)^{1980}}$.

NOTE:  For any prime $p \le N$, the $p$-adic valuation for $N!$ can also be given by $\nu_{p}(N!)=\frac{N-s_{p}(N)}{p-1}$ where $s_{p}(N)$ denotes the sum of the base $p$ digits of $N$.  So, also we have that $\nu_{11}\bigg( \frac{(1980n)!}{(n!)^{1980}}\bigg) =\frac{1980 \cdot s_{p}(n)-s_{p}(1980 \cdot n)}{10}$.

Lastly, I think a little argument is necessary to show that it suffices to look at just the prime 11 and not 2,3, or 5.  At first this may seem obvious, but in the case for $n < 11$ notice that $(n!)^{1980}$ is not divisible by 11 and could be divisible by a large power of 2 which may potentially cancel out the large power of 2 in $(1980n)!$.  Similarly for 3 and 5.  Just a note.

Use that $\nu_{p}\bigg( \frac{(1980n)!}{(n!)^{1980}}\bigg) =\frac{1980 \cdot s_{p}(n)-s_{p}(1980 \cdot n)}{p-1}$.

The base 11 representation for 1980 is [0, 4, 5, 1] or $(1540)_{2}$.  Assume that $n$ has base 11 representation $[d_{0},d_{1},\ldots,d_{k}]$ where the $d_{i}'s$ are the base 11 digits (i.e. chosen from 0,1,2,...10).  Then, it is not hard to see that \[1980n=1980 \sum_{0\le j \le k}d_{j}11^{j}= 1\cdot\sum_{0\le j \le k}d_{j}11^{j+3}+5\cdot\sum_{0\le j \le k}d_{j}11^{j+2}+4\cdot\sum_{0\le j \le k}d_{j}11^{j+1}.\]  Notice that the sum of the base  11 digits from each summation is $\ge$ the sum of the base 11 digits of $1980n$. So, we have that \[s_{11}(1980n) \le (1+5+4)\cdot s_{11}(n) = 10\cdot s_{11}(n).\]   It would follow that \[\nu_{11}\bigg( \frac{(1980n)!}{(n!)^{1980}}\bigg) = \frac{1980 \cdot s_{11}(n)-s_{11}(1980 \cdot n)}{10}\ge 197s_{11}(n)\] and for the right choice of $n$, this would tend to infinity.


太难了吧！微软是不是要求设计程序通过计算机解决呢？

:)  应该不是，这题刚看起来觉得比较容易，但越想越复杂，一时想不到好的办法，就贴此共享一下。 :blush:

oh buddy, this looks like P vesus NP problem
however, we can change the amount of beans into a comparatively small amount, like 12
then start demonstrating.

应该都死了，除非有人做雷锋 :P

now, we can simply generalize it, with
5个海盗 changes in to $n$个海盗, and 分100两黄金 into 分$m$两黄金

楼上两位好主意！ :roll:

我当时博弈论文没什么写，就把一般情况来讨论了，如果楼上感兴趣，我可以把我的文章发上来让你指正 :P

博弈论很有意思，我也学过，不过学业不精， :blush:

当然欢迎你的研究了，如果有其它的类似问题，也希望你能贴过来。

[quote="zhaobin"][quote="Dreamer"]5个海盗，分100两黄金，规定每人编号1-5，从1开始每人提出一种分黄金的办法，如果有一半以上的人反对，则提出方案的人就被推下海，
问：第一个海盗提出怎样的分法，可使他既不被推下海，又可以分得最多的黄金。

条件：每个海盗智商一样。 :huh:[/quote]
这道题目我做过的，最终答案应该是98，0，1，0，1 :)[/quote]
Why?I can't get it.

谢谢，这是我写的文章，不过不是很好，学的不精。

答dingdongdong：这里主要用的是倒推的思想。 :)

[quote="Dreamer"]经典，变态推理题之（自己考虑，试着回答，合理就行）
                  第一个故事：远航
　　A和B是非常好的朋友。一次，A，A的妻子和B一起驾船在海上远航，航行途中，由于食物不足，A的妻子饿死了。A也因悲痛和饥饿交加而病倒。在B的照顾下，A的身体渐渐有了好转。终于，船靠了岸，劫后余生的A和B高高兴兴的来到岸边的一家餐馆想饱餐一顿。但A吃了一口服务生端上桌的海鸥肉以后，立刻号啕大哭了起来，并杀死了B，为什么？[/quote]

在海上B把A妻子的肉给A吃,并且骗他是海鸥肉。A吃了真的过后发现不对...

这个有很多版本，有的是企鹅肉...

本打算把这几个经典的推理都发上来，没想到大家这么强。
算了，发几个图形的 :D
选择一个，答时可以先说明是哪楼的题，再写出答案。
[img]http://www.ali30.com/bbs/upload/fyzw_2004113022588.gif[/img]

:rotfl:   :welcome:
[img]http://www.ali30.com/bbs/upload/fyzw_200473082336.jpg[/img]

[quote="Soarer"][hide]B在照顧A時讓他食了他妻子的肉，還胡謅的說是海鷗肉
[hide]其實以前看過 :D 也真夠變態的[/hide][/hide][/quote]
这个问题好恐怖 :)

[quote="Dreamer"]囚犯活命问题

详细介绍：

5个囚犯，分别按1-5号在装有100颗绿豆的麻袋抓绿豆，规定每人至少抓一颗，而抓得最多和最少的人将被处死，而且，他们之间不能交流，但在抓的时候，可以摸出剩下的豆子数。问他们中谁的存活几率最大？提示：  :idea:

　　　　　　1，他们都是很聪明的人  :roll:
　　　　　　2，他们的原则是先求保命，再去多杀人  :o
　　　　　　3，100颗不必都分完  ;)
　　　　　　4，若有重复的情况，则也算最大或最小，一并处死  :?:[/quote]

应该都死掉了
先假设豆子无穷多
但每个人可以知道前面所有人抓豆子的总和
设第一个人抓了n个豆子，那么第二个人应该抓n-1或n+1个
这样才有更大的可能自己的豆子不是最大或最小
这样下来前四个人抓的豆子是四个连续的整数
这样考虑的话，所以第五个人肯定是最大或者最小，他必死
那么他一定会抓四个整数中靠中间的数，这样的话可以拉一个垫背的一起死
但是这样第四个人就必死了
他想到这一点，他也会拉个垫背的
抓前三个数中中间的那个数目的豆子
第三个虽然也想到了这一点，但是他无论他怎么选择，都是无力回天
所以最后情况抓的豆子数目是m,m,m+1,m+1,m+2或m,m+1,m+1,m+2，m+2

最后到本题，第一个人的豆子不能超过22，不能小于18
否则，自己必死
别人顺序抓下来就可以了
所以它只能选19，20，21三个数
剩下的可以保证上面的推理合理

全都死！！！！！

[quote="zhaobin"][quote="Dreamer"]再来一个：[hide]
有3个人下棋，2人之间下赢了的可以和另外一个人下；如果一+个人能连赢2局，那么他就获胜。
问：这个规则是否公平？[/hide][/quote]
我想显然不公平。
来个类似的：
1次抽奖活动，有三张奖卷，其中两张无奖，一张可以得到一辆汽车，你抽了其中的一张，然后公证人员把剩下两张里面的其中一张无奖的给扔了。然后你有两种选择的机会
1.还是要自己原来抽到的一张
2.换成剩下的那张没被扔也没被你抽的那个奖卷
你会选择哪个？[/quote]

呵呵，二题等价的前提是
第一题中三人棋力相同
谁胜谁负都可能

[quote="yace"][quote="Dreamer"]囚犯活命问题

详细介绍：

5个囚犯，分别按1-5号在装有100颗绿豆的麻袋抓绿豆，规定每人至少抓一颗，而抓得最多和最少的人将被处死，而且，他们之间不能交流，但在抓的时候，可以摸出剩下的豆子数。问他们中谁的存活几率最大？提示：  :idea:

　　　　　　1，他们都是很聪明的人  :roll:
　　　　　　2，他们的原则是先求保命，再去多杀人  :o
　　　　　　3，100颗不必都分完  ;)
　　　　　　4，若有重复的情况，则也算最大或最小，一并处死  :?:[/quote]

应该都死掉了
先假设豆子无穷多
但每个人可以知道前面所有人抓豆子的总和
设第一个人抓了n个豆子，那么第二个人应该抓n-1或n+1个
这样才有更大的可能自己的豆子不是最大或最小
这样下来前四个人抓的豆子是四个连续的整数
这样考虑的话，所以第五个人肯定是最大或者最小，他必死
那么他一定会抓四个整数中靠中间的数，这样的话可以拉一个垫背的一起死
但是这样第四个人就必死了
他想到这一点，他也会拉个垫背的
抓前三个数中中间的那个数目的豆子
第三个虽然也想到了这一点，但是他无论他怎么选择，都是无力回天
所以最后情况抓的豆子数目是m,m,m+1,m+1,m+2或m,m+1,m+1,m+2，m+2

最后到本题，第一个人的豆子不能超过22，不能小于18
否则，自己必死
别人顺序抓下来就可以了
所以它只能选19，20，21三个数
剩下的可以保证]上面的推理合理

全都死！！！！！[/quote]

我也认为全都死，但我认为这种证法有误。
对“4，若有重复的情况，则也算最大或最小，一并处死 ”的理解应为并列最大或最小者死，而不是并列者皆死。
比如：若最后情况抓的豆子数目是m,m,m+1,m+1,m+2，应理解为1、2、5死而不是全死。

我的解法如下：
若1抓不少于21个，则2通过摸出剩下的豆子数知道1抓了几个，那么2比1少抓一个即可保命。（因为3、4、5至少有一人抓少于20个，故2不是最少，也不是最多）
同理，3、4摸出剩下的豆子数并按照2的推理，只需抓与2同样多即可活命，于是1、5死。
所以1抓至多20个。对于3、4、5，他们抓的个数只有尽量接近前几个人抓的平均数（可以过摸出剩下的豆子数算出）活命的机会才更大。
因此，若2与1抓的个数差不小于2，则1、2死，3、4、5活。
设1抓了X个，2抓了Y个。X与Y的差小于2
若X=20,Y=21,则5取不到20与21，5的个数小于20，所以2、5死，1活，3、4不定。
所以若1抓20个，1不能抓21个。
若X不等于20或Y不等于21，则3、4、5个数皆介于X、Y之间。但X与Y的差小于2，故皆死！

工具：两根蜡烛，一盒火柴
已知：每根蜡烛可燃一小时
请求出15分钟

一位商人花70元购进一件衣服,加价12元售出.后发现购买者支付的那张一百元是假钞,商人大悲.
现在请你帮那个倒霉的商人算算,他在这件衣服上共损失多少钱 ？

一根同时点两头，另一根同时点一边，当第一根烧完时把第二根的另一头也点上，这样从第一根烧完到第二根烧完即15分钟。好像是一个很老的题目。

88元

[quote="Dreamer"]My conclusion:
Each person is dead.[/quote]

[quote="standforn"]应该都死了，除非有人做雷锋 :P[/quote]

[quote="Kalimdor"]那就让俺上吧! :D[/quote]

我简单想了一下，也得出相同的结论。大家会趋向于每人摸20颗。谁多了或少了，后面的就会察觉，更加会摸20颗。
就像Survivor Panama有一次把石头装在三艘船上，大家都会尽量使每艘的石头数平均，以防止石头最少的一组轻易取胜。

不过，就算你要做雷锋，帮某个其他人越狱什么的，你又能如何做呢？
除非两个人做雷锋，一个摸一颗，一个全部摸完。摸一颗那个必死无疑，全部摸完那个还要看前面有没有人比他摸得多。
两个人的牺牲，可以换来3个人的生存，而如果没人牺牲，5个都要死。这不仅是数学，更是现代道德的鲜明写照。

Do you have an English version? Do you have a link to the question? Here is my translation, see if it makes sense:
[u]
Prisoners Problem[/u]

There are five prisoners enumerated 1-5. There are 100 beans in a sack and the prisoners each must take beans from it. The rule is that everyone must pick out at least one bean but those who pick the most and the least amount will be put to death. They can not communicate among each other but while they are picking they can feel for the number of beans still left over. Question: Which one has the greatest chance to live?

Warning:
They are all very brilliant
The most important thing is to survive first then kill the others off.
The 100 beans don’t have to be completely picked away
If there is a repeat, it doesn’t matter the most or the least amount they all have to die!


limit
trigonometry
real analysis
real analysis theorems

We all know that the limit $ \lim_{h \minus{} > 0} sin(\frac {1}{h})$ is not exist.

HOWEVER, what is the limit $ \lim_{h \minus{} > 0} \frac {h}{sin(\frac {1}{h})}$.does it exist.if so,is it a simple zero :maybe: .

No, that limit doesn't exist. You're dividing by something that's zero infinitely often, so the quotient blows up near those points and is undefined at them. On every interval $ (\minus{}a,a)$, $ \frac{x}{\sin\frac1x}$ takes every nonzero real value infinitely often, and is also undefined infinitely often.


Diophantine equation
number theory proposed
number theory

solve in positive integers the equation

$ (x\plus{}2)(y\plus{}2)(z\plus{}2)\equal{}(x\plus{}y\plus{}z\plus{}2)^2$

[quote="Victory.US"]solve in positive integers the equation

$ (x \plus{} 2)(y \plus{} 2)(z \plus{} 2) \equal{} (x \plus{} y \plus{} z \plus{} 2)^2$[/quote]

Let $ a\in\mathbb Z$ and consider two solutions $ (x_1,y_1)$ and $ (x_2,y_2)$ of the Pell equation $ x^2 \minus{} ay^2 \equal{} 4$.

Then $ (x_1,x_2,\frac {x_1x_2\pm ay_1y_2}2)$ is solution of the required equation (and all solutions are in this form for some $ a$)

For example :

$ a \equal{} 0$ gives the unique solution $ (2,2,2)$

$ a \equal{} 2$ gives infinitely many solutions. For example :
$ (6,6,34)$, $ (6,34,198)$, $ (6,198,1154)$, $ (6,1154,6726)$, $ (6,6726,39202)$...
$ (2,34,34)$, $ (34,34,1154)$, $ (34,198,6726)$, $ (34,1154,39202)$, $ (34,6726,228486)$ ...

$ a \equal{} 3$ gives infinitely many solutions. For example :
$ (4,4,14)$, $ (4,14,52)$, $ (4,52,194)$, $ (4,194,724)$, $ (4,724,2702)$, $ (4,2702,10084)$

...

So, it seems to me very difficult to give a general solution.
Where is this problem coming from ?

And, please, could you give us your own general solution?

we have
$ 8\plus{}xyz\plus{}2 \sum xy\plus{}4 \sum x\equal{}\sum x^2\plus{}2 \sum xy\plus{}4\plus{}4\sum x \implies xyz\plus{}4\equal{}x^2\plus{}y^2\plus{}z^2$
this equation has a minimum solution $ (2,1,1)$.we call it minimum because $ x\plus{}y\plus{}z$ is obviously minimum.
also there are infinitely many solutions like $ (2,y,y)$.
for getting all other solutions do same as what is done for finding all solutions of $ x^2\plus{}y^2\plus{}z^2\equal{}3xyz$.

This problem coming from " Problem from the books" of titu adreescu. Topic "equation & beyond"


I'll start it off by talking about the two most popular genres in the teenage crowd, rock vs rap.

Personally, I feel that rap is more "intelligent". Although mainstream rap is mostly about bling bling, women, drugs, etc. I've been getting into some underground stuff recently, where the rappers talk about their past, political views, and some other IMO deep stuff. Also, I feel that the beats in hip-hop are better than a guitar and drums.

Please don't take this as a flame to rock or anything, I personally like all kinds of music, listening to classical, reggae, heck even the asian country my parents put on sometimes :D

Your thoughts?

I feel it is very hard to compare rock and rap because they are so very different.

Rap is generally based on the beats, and not the lyrics. Yet, I think, because it is so raw, and the rules for the beats are not set, there is 'lyrical freedom' - they can put in mostly whatever they want, while teh song still sounds good. The beats adjust to the lyrics, but the rhythm is by the beat.

Rock (for me) is all about the guitars and drums (and occasionally all other sorts of instruments) - it follows more of a tune than a beat, and because the instrumental part of the song is so powerful and is "what the song IS", the lyrics MUST follow it, or the song doesnt sound right. Because of that, rock compared to underground "intelligent" rap may seem less intelligent - which it might or might not be - because it is harder to work with the lyrics and so theyll make sense. In addition, there are MANY intelligent rock songs, some of them arent even noticed because a lot of people only listen to the guitar, and dont pay attention to the meaning behind the song, whereas in underground rap, if you listen to the song, its because of the lyrics.

Personally, I like both genres, but rock more so, because of the talent behind it. I dont mean to offend anyone who's favorite music is rap, but IMO rapping isnt very hard, once you have a beat. Rhyming constantly takes a certain level of skill/practice to be good at, but the actual rapping ("talking") will sound good regardless of who raps it, if there is a beat. (of course, some people DONT sound good rapping w/ or w/out a beat, but I think most people do). Rock on the other hand, I appreciate because of all the talent behind (well, in my kind of rock anyway, there are punk bands that arent very good) it - each person in the band will have spent years learning an instrument. And, I like the sound more anyway..

Then again, there is classical and jazz which is a LOT of talent, but.. then again the discussion hasnt gone there yet.

[url=http://www.wired.com/news/culture/0,1284,67970,00.html]Rap is for geeks too.[/url]

From what I've heard, rap actually takes quite a lot of talent.

The beat is there for you to flow to it. Some rap artists amaze me even with their lyrics. You can read it, and it seems like a nicely written poem. Also, just making the beat takes quite a lot of talent also, hence making some legendary producers such as Dr.Dre, that others just can't match up to. Also, rap culture rose up partly from freestyle battling. It is extremely hard to come up with stuff like that on the spot.

Don't you think this should be moved into the Music and Art Section?

[quote="G-UNIT"]To my knowledge, rap in essence, was created as form of poetry, coming from african american roots.  It was usually adopted as channel for african americans to express their "struggles" and hardships of society.  Just like any poet a rapper must write his own lyrics, choose his own beats, and find an image, real or theatrical, that he would like to be potrayed as.  So all in all, it does take quite a bit of intelligence and effort to become a rapper, not anyone off the street could become a rapper, nor are rappers necessarily extremely intelligent people.  The reason rap has always revolved around "thug life" or whatever you want to call it, is becuase of the roots it was found in.  And as times go by, we see rap is slowly growing into much more than that.  Religious groups have emerged into rap.  New rappers, [b]Kanye West[/b] for instance, are incorparating current political, and religious issues into their music.  As a result, new songs such as [b]"Jesus Walks" and "Diamonds are Forever" [/b] have hit the radios, that talk about religios hardships and the domestic wars in Africa that take place because of the diamonds bought oversea.  Over time we will mostly likely see rap being used to express messages much more profund, and will shape media with its strong influence over the public.  In my opinion, it's quite difficult for an idiot to transform the expeirences and hardships of his life into a song that millions of people will listen to.[/quote]

I agree with G-UNIT. There's an article about Kanye West in Time magazine although I don't remember which. He definitely stands himself out of the crowd by being a non-conformist.


Prove that for all $n\in\mathbb N$ holds:
$1^{1}\cdot2^{2}\cdot3^{3}\cdot...\cdot n^{n}\leq\left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}}.$
Good luck!

Old, nice and super easy! :)

Let us apply AM-GM inequality like this \[\sqrt [\frac{n(n+1)}{2}]{1\cdot 2\cdot 2\cdot 3\cdot 3\cdot 3\ldots n\cdot n\ldots n}\leq\frac{1+2+2+3+3+3+\ldots+n+n+\ldots n}{\frac{n(n+1)}{2}}=\frac{n(n+1)(2n+1)}{6}\cdot\frac{2}{n(n+1)}=\frac{2n+1}{3}.\] The conclusion follows. :wink:

[quote="FieryHydra"][quote="Elemennop"]This is simple AMGM on $1^{2}+2^{2}+3^{2}+\ldots$ with $n^{2}$ expanded out into n elements.[/quote]
I think you misread the problem.[/quote]

No, read my statement carefully. It's exactly what Cezar did.

[quote="FieryHydra"][quote="Elemennop"]This is simple AMGM on $1^{2}+2^{2}+3^{2}+\ldots$ with $n^{2}$ expanded out into n elements.[/quote]
I think you misread the problem.[/quote]
Never mind!


Since the other thread is just men's singles, who do you think will win the 2009 US Open Women's singles, Women's doubles, Men's doubles, and Mixed doubles? Also, who do you want to win?

Yeah,

I got an email from them yesterday.

I'm not sure if Federer is out yet, is he?


Find all primes $p$ such that for each prime $q < p$ , the remainder of $p$ when divided by $q$ is square-free.

I didnt quite understand , Can you pz elaborate???

A more detailed version of Rgep's solution can be found in http://www.mathlinks.ro/Forum/viewtopic.php?t=5904 .

  darij

Thanks!!!! I really appreciate your spending time on these problems!!!


calculus
integration
logarithms
trigonometry
calculus computations

Help me to solve the integral, thanks.

$ \text{*  }\int\limits_{0}^{1}{\frac{\log \left| y\plus{}1 \right|.dy}{\sqrt{1\minus{}{{y}^{2}}}}}$

try using a substitution of y as sin(theta).
then write the series exapnsion of log(1+x) and try eliminating highewr order terms and therby solving the integral....

hello, i think the answer is
$ \frac{1}{2}\left(4\text{Catalan}\minus{}\pi\log(2)\right)$
Sonnhard.

$ I: \equal{}\int_0^1 \frac{\log(1\plus{}y)}{\sqrt{1\minus{}y^2}} dy$ is after substitution $ y\equal{}\cos x$

$ \int_0^{\frac{\pi}{2}} \log(1\plus{}\cos x)\, dx$ . Replace $ 1\plus{}\cos x$ by $ 2\cos^2 \frac{x}{2}$ .

$ I\equal{}\int_0^{\frac{\pi}{2}} \log\left(2\cos^2 \frac{x}{2}\right)\, dx\equal{}\int_0^{\frac{\pi}{2}} \left[\log\left(\left(2\cos \frac{x}{2}\right)^2\right)\minus{}\log 2\right] dx$

$ \equal{}2\int_0^{\frac{\pi}{2}} \log\left(2\cos \frac{x}{2}\right) dx\minus{}\frac{\pi}{2}\log 2$ . Use the fourier series expansion

$ \log\left(2\cos \frac{x}{2}\right)\equal{}\sum_{k\equal{}1}^\infty \frac{(\minus{}1)^k}{k}\, \cos kx$ to obtain

$ \int_0^{\frac{\pi}{2}} \log\left(2\cos \frac{x}{2}\right) dx\equal{}\sum_{k\equal{}0}^\infty \frac{(\minus{}1)^k}{(2k\plus{}1)^2}\equal{}K$ .

Hence $ I\equal{}2K\minus{}\frac{\pi}{2}\log 2$ .

But my solution is different ?

$ \text{*  }\int\limits_{0}^{1}{\frac{\log \left| y\plus{}1 \right|.dy}{\sqrt{1\minus{}{{y}^{2}}}}}\equal{}\frac{\pi \minus{}2}{2},\,\,\int\limits_{0}^{1}{\frac{\arcsin y.dy}{y\plus{}1}\,}\equal{}\frac{\pi }{2}.\left( \log 2\minus{}1 \right)\plus{}1$

$ I\equal{}\int\limits_{0}^{\pi /2}{\log \left( 1\plus{}\sin x \right)dx}\equal{}\int\limits_{0}^{\pi /2}{dx}\left( \int\limits_{0}^{\sin x}{\frac{dy}{y\plus{}1}} \right)\equal{}\int\limits_{0}^{1}{\frac{dy}{y\plus{}1}}\int\limits_{\arcsin y}^{\pi /2}{dx}\equal{}\int\limits_{0}^{1}{\frac{\pi /2\minus{}\arcsin y}{y\plus{}1}\,dy}$

$ D\equal{}\left\{ \left( x,y \right)|0\le x\le \frac{\pi }{2},\,\,0\le y\le \sin x \right\}$

$ y\le \sin x\Rightarrow x\ge \arcsin y,\,\,x\equal{}\frac{\pi }{2}\Rightarrow y\equal{}1\Rightarrow 0\le y\le 1,\,\,\arcsin y\le x\le \frac{\pi }{2}$

$ I\equal{}\frac{\pi }{2}\int\limits_{0}^{1}{\frac{dy}{y\plus{}1}\,}\minus{}\int\limits_{0}^{1}{\frac{\arcsin y.dy}{y\plus{}1}\,}\equal{}\frac{\pi }{2}.\left[ \log \left| y\plus{}1 \right| \right]_{0}^{1}\minus{}\int\limits_{0}^{1}{\frac{\arcsin y.dy}{y\plus{}1}\,}\equal{}\frac{\pi }{2}.\log 2\minus{}\int\limits_{0}^{1}{\frac{\arcsin y.dy}{y\plus{}1}\,}$

$ \text{Put }u\equal{}\arcsin y\Rightarrow du\equal{}\frac{dy}{\sqrt{1\minus{}{{y}^{2}}}},\,\,dv\equal{}\frac{dy}{y\plus{}1}\Rightarrow v\equal{}\log \left| y\plus{}1 \right|$

$ I\equal{}\int\limits_{0}^{\pi /2}{\log \left( 1\plus{}\sin x \right)dx}\equal{}\frac{\pi \minus{}2}{2}$

$ \Rightarrow I\equal{}\frac{\pi }{2}.\log 2\minus{}\int\limits_{0}^{1}{\frac{\arcsin y.dy}{y\plus{}1}\,}\equal{}\frac{\pi }{2}.\log 2\minus{}\left[ \arcsin y.\log \left| y\plus{}1 \right| \right]_{0}^{1}\plus{}\int\limits_{0}^{1}{\frac{\log \left| y\plus{}1 \right|.dy}{\sqrt{1\minus{}{{y}^{2}}}}}$

$ \equal{}\frac{\pi }{2}.\log 2\minus{}\frac{\pi }{2}.\log 2\plus{}\int\limits_{0}^{1}{\frac{\log \left| y\plus{}1 \right|.dy}{\sqrt{1\minus{}{{y}^{2}}}}}\equal{}\frac{\pi \minus{}2}{2}\Rightarrow \int\limits_{0}^{1}{\frac{\log \left| y\plus{}1 \right|.dy}{\sqrt{1\minus{}{{y}^{2}}}}}\equal{}\frac{\pi \minus{}2}{2}$

$ \frac{\pi }{2}.\log 2\minus{}\int\limits_{0}^{1}{\frac{\arcsin y.dy}{y\plus{}1}\,}\equal{}\frac{\pi \minus{}2}{2}\Rightarrow \int\limits_{0}^{1}{\frac{\arcsin y.dy}{y\plus{}1}\,}\equal{}\frac{\pi }{2}.\left( \log 2\minus{}1 \right)\plus{}1$

$ \text{*  }I\left( 1 \right)\equal{}\int\limits_{0}^{\pi /2}{\log \left( 1\plus{}\sin x \right)dx}\equal{}\left( \frac{\pi \minus{}2}{2} \right)$

$ I\left( y \right)\equal{}\int\limits_{0}^{\pi /2}{\log \left( 1\plus{}y.\sin x \right)dx}\Rightarrow {{I}^{'}}\left( y \right)\equal{}\int\limits_{0}^{\pi /2}{\frac{\sin x.dx}{\left( 1\plus{}y.\sin x \right)}}$

$ \text{Put }t\equal{}\tan \frac{x}{2}\Rightarrow x\equal{}2\arctan t\Rightarrow dx\equal{}\frac{2dt}{1\plus{}{{t}^{2}}},\,\,\sin x\equal{}\frac{2t}{1\plus{}{{t}^{2}}}$

$ x\equal{}0\Rightarrow t\equal{}0,\,\,x\equal{}\frac{\pi }{2}\Rightarrow t\equal{}1\Rightarrow {{I}^{'}}\left( y \right)\equal{}\int\limits_{0}^{1}{\frac{2dt}{1\plus{}{{t}^{2}}}.\frac{2t}{1\plus{}{{t}^{2}}}.{{\left( 1\plus{}\frac{2t.y}{1\plus{}{{t}^{2}}} \right)}^{\minus{}1}}}$

$ \equal{}\int\limits_{0}^{1}{\frac{4dt}{{{\left( 1\plus{}{{t}^{2}} \right)}^{2}}}.{{\left( \frac{{{t}^{2}}\plus{}2t.y\plus{}1}{1\plus{}{{t}^{2}}} \right)}^{\minus{}1}}}\equal{}\int\limits_{0}^{1}{\frac{4dt}{\left( 1\plus{}{{t}^{2}} \right)\left( {{t}^{2}}\plus{}2t.y\plus{}1 \right)}}$

$ \text{Put }\frac{4t}{\left( {{t}^{2}}\plus{}1 \right)\left( {{t}^{2}}\plus{}2t.y\plus{}1 \right)}\equal{}\frac{A}{{{t}^{2}}\plus{}1}\plus{}\frac{B}{{{t}^{2}}\plus{}2t.y\plus{}1}$

$ \Rightarrow A\left( {{t}^{2}}\plus{}2t.y\plus{}1 \right)\plus{}B\left( 1\plus{}{{t}^{2}} \right)\equal{}4t\Rightarrow \left( A\plus{}B \right).{{t}^{2}}\equal{}0,\,\,2A.t\equal{}4t,\,\,A\plus{}B\equal{}0$

$ \Rightarrow A\equal{}2\text{, }A\plus{}B\equal{}2\plus{}B\equal{}0\Rightarrow B\equal{}\minus{}2$

$ \Rightarrow {{I}^{'}}\left( y \right)\equal{}\int\limits_{0}^{1}{\frac{4dt}{\left( 1\plus{}{{t}^{2}} \right)\left( {{t}^{2}}\plus{}2t.y\plus{}1 \right)}}\equal{}\int\limits_{0}^{1}{\frac{2dt}{\left( 1\plus{}{{t}^{2}} \right)}}\minus{}\int\limits_{0}^{1}{\frac{2dt}{\left( {{t}^{2}}\plus{}2t.y\plus{}1 \right)}}\text{,  Regarding }y\equal{}1$

$ \equal{}2.\left[ \arctan t \right]_{0}^{1}\minus{}2\int\limits_{0}^{1}{\frac{d\left( t\plus{}1 \right)}{{{\left( t\plus{}1 \right)}^{2}}}}\equal{}2\frac{\pi }{4}\minus{}2\left[ \frac{\minus{}1}{\left( t\plus{}1 \right)} \right]_{0}^{1}\equal{}\frac{\pi }{2}\plus{}2\left( \frac{1}{2}\minus{}1 \right)\equal{}\frac{\pi \minus{}2}{2}$

$ \Rightarrow I\left( 1 \right)\equal{}\int\limits_{0}^{\pi /2}{\log \left( 1\plus{}y.\sin x \right)dx}\equal{}I\left( 0 \right)\plus{}\left( \frac{\pi \minus{}2}{2} \right)\int\limits_{0}^{1}{dy}\equal{}0\plus{}\left( \frac{\pi \minus{}2}{2} \right)$

$ \Rightarrow I\left( 1 \right)\equal{}\int\limits_{0}^{\pi /2}{\log \left( 1\plus{}y.\sin x \right)dx}\equal{}I\left( 0 \right)\plus{}\left( \frac{\pi \minus{}2}{2} \right)\int\limits_{0}^{1}{dy}\equal{}0\plus{}\left( \frac{\pi \minus{}2}{2} \right)$


What value of  $x$ will give the maximum value for  $-6 x^{2}+3 x+3$? Express your answer as a common fraction in simplest form.

Really? i thought you could find the value of x for which the minimum/maximum value of a quadratic expression, in the form ax + by +c, is obtained by using the formula
[hide]-b/2a[/hide]. but i guess i might've made a mistake somewhere...

[hide]
$-6x^{2}+3x+3 = 0$
$-2x^{2}+x+1 = 0$
$(-2x-1)(x-1) = 0$
$x =-\frac{1}{2}~\mbox{or}~x = 1$

those are the zeros of the function, so the x value of the vertex should be the midpoint of those two points: $\frac{-\frac{1}{2}+1}{2}= \fbox{1/4}$

EDIT: haha.. i totally forgot about $-b/2a$
[/hide]

see? i'm not the only one who got my answer. i think it's right.

I put in 1/4 into the box and the saab program said i was wrong and that the answer was 1/3 :oops:

hmm idk.. i plugged in both 1/3 and 1/4 into the original equation and x = 1/4 returns the greater y value..

[quote="#H34N1"]No. That is not correct. Also, please provide solutions :)[/quote]

are u sure ur answer is right, cuz pretty much whatever undefined gets is right...the fact that i got it doesn't really count :( , but can u check it anyway?


AwesomeMath
summer program
Mathcamp
videos
FTW

The point of this game is to list two things and the next poster has to pick one and list two more things.

Me or U?

I love this game!!

Me

Chemistry or Biology?

Chemistry
the other one was a month old so i renewed it =)
Fall Out Boy or Panic! at the disco ?

Fall Out Boy.

AwesomeMath Summer Program or Mathcamp?

rocks (I can make prettier shapes on the ground with them  :wink: )

Glue bottle or glue stick?  :lol:

glue bottle

80G Video iPod or iPod Touch?

Anyone whho knows my real name knows which one i choose

6 or 28?

i (jk $ \pi$)

Greek letters or common letters (in math problems)

:ddr:   all the way!

Um... Ancient Rome or Ancient Greece?


In a circumference of a circle, $ 10000$ points are marked, and they are numbered from $ 1$ to $ 10000$ in a clockwise manner. $ 5000$ segments are drawn in such a way so that the following conditions are met:

1. Each segment joins two marked points.
2. Each marked point belongs to one and only one segment.
3. Each segment intersects exactly one of the remaining segments.
4. A number is assigned to each segment that is the product of the number assigned to each end point of the segment.

Let $ S$ be the sum of the products assigned to all the segments.

Show that $ S$ is a multiple of $ 4$.

Umm? .. take any pair of intersecting lines with vertices $ a,b,c,d$ in clockwise order, because they do not intersect any other lines the number of points between $ a$ and $ b$ is divisible by $ 4$ and so forth. Then $ b,d \equiv a,c \pm 1 \pmod{4}$, eg even numbers will be paired with even numbers and odd number with odds. The product of the even numbers has no effect on $ S$ $ \mod 4$, and the odd numbers can be partitioned into two sets of $ 2500$ elements leaving residues of $ 1$ and $ 3$ respectively $ \mod 4$. Trivially we find that the sum of the products of pairs of elements from these sets will be divisible by $ 4$ ..


I think it's uncertain. Take x,y ->0 we get l.h.s. -> 0, take x->0 y=z->1/2 then l.h.s. -> 1/8

It will be 1/27 if my remember is not wrong .

[quote="blahblahblah"]However, if we have $x\geq y\geq z\geq 0$ with $x+y+z=1$, then the inequality

$xy^2+yz^2+zx^2\leq \frac {1}{8}$ holds[/quote]
it is not hard ;)

the original questin is this(i think it was canadian math olympiad)i solved it by lagragne inequality,but there shoud be another solution :)
ifx,y,z>0 and x+y+z=1then prove that
$xy^2+yz^2+zx^2\leq \frac {4}{27}

discussed before...
Anyways, the easiest way to prove that maximum is 4/27 is this:
let x>=y>=z, then show that if you change y to y+d and z to z-d, the expression increases....
So the maximum is when z=0.


In LaTex, how do you label the equations by sections.  I used \begin{equation} and \end{equation}, but that gives me label like 1,2,3...  How can I label the stuff by 1.1, 1.2....? thanks

look in there:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16377&sid=e2e334e900464e07797396272b51ff7c]www.artofproblemsolving.com/Forum/viewtopic.php?t=16377&sid=e2e334e900464e07797396272b51ff7c[/url]

ledurt


Asymptote AoPSwiki articles geometry circumcircle LaTeX incenter

I've also became very interested in learning asymptote, and I can't start doing that because I have problems from the begining. I solved a lot of them, with help, almost exclusively from AoPSWiki, but this one I can't.
I downloaded that Olympiad package. In fact, I didn't know what to do with it, beacuse it can't be downloaded, I went on "save target as", and saved it in my Asymptote folder. But it doesn't seem to work, this is my code
[code]\documentclass{article}

\usepackage{asymptote}

\begin{document}

\begin{center}

\begin{asy}
size(100);
pair A,B,C,O,I;
A=(0,0); B=(10,0); C=(5,15);
draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
\end{asy}

\end{center}

\end{document}[/code]
It only draws a triangle, but not its circumcircle...
Do you know maybe what the problem is?

Bye

You have to tell Asymptote to use the Olympiad package with the line import olympiad; as in
[code]\begin{asy}
import olympiad;
size(100);
pair A,B,C,O,I;
A=(0,0); B=(10,0); C=(5,15);
draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
\end{asy} [/code]otherwise it won't understand what circumcircle means.

I tried usepackage[olympiad] and similar styff, bot not import. I will try it, thank you :)

[edit]: still nothing :(
It says unknown variable circumcircle

[edit] [edit] IT WORKS :)
Thanks!!

Again something doesn't work..... I am going mad.

What should I type in tools -> customize -> .... in argument nad should I paick LaTex => pdf or ....

I want to make pdf file, latex code with asymptote code within...

Help pls....

It's all in the [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration]Wiki[/url] assuming you've done the stuff [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Getting_Started/Windows/Interactive_Mode]here[/url].

One more thing, I have this problem now...

This is the code:

[code]
\documentclass{article}
\usepackage{asymptote}

\begin{document}
\begin{asy}
import olympiad;
size(200);
pair A,B,C,O,I;
A=origin; B=2*right; C=1.5*dir(70);
O=circumcenter(A,B,C); // olympiad - circumcenter
I=incenter(A,B,C); // olympiad - incenter
draw(A--B--C--cycle);
dot(O);
dot(I);
draw(circumcircle(A,B,C)); // olympiad - circumcircle
draw(incircle(A,B,C)); // olympiad - incircle
label("I",I,W);
label("O",O,S);
\end{asy}

\end{document}

[/code]

names test1
When I "asy" - it it says only test1_0_1.pdf, and when I open pdf file (not test1_0_1.pdf, there is no such) there is only a part of picture. When I change size that part becames smaller or disapears... I don't know what to do any more...

I tried the code and it works well. Save as some new name (so no conflict with anything else), let's say test2.tex. Then compile to pdf, then run asymptote and finally compile to pdf again. There should be test2.pdf with your picture in it.

As a check there should be the following in the test2.asy file that is automatically created:
[code]access settings;
if(!settings.multipleView)
 settings.batchView=false;
settings.tex="pdflatex";

eval(quote{
defaultfilename='LaTeX1_1';
size(345.0pt,0,Aspect);

import olympiad;
size(200);
pair A,B,C,O,I;
A=origin; B=2*right; C=1.5*dir(70);
O=circumcenter(A,B,C); // olympiad - circumcenter
I=incenter(A,B,C); // olympiad - incenter
draw(A--B--C--cycle);
dot(O);
dot(I);
draw(circumcircle(A,B,C)); // olympiad - circumcircle
draw(incircle(A,B,C)); // olympiad - incircle
label("I",I,W);
label("O",O,S);
});[/code]

I am missing this line in asy file:

size(345.0pt,0,Aspect);

When I delete label comands (put // which is $\Leftrightarrow$ as delete :) ) then it all works fine. Now I saw "problem with labels" thread so I am going to read it now.

Thank you!

[edit:] I found nothing there :(  So, problem is now localized, LABELS. I tried to compiled it twice, still nothing...

How exactly are you compiling the file? Are you using TexnicCenter? Then either run asymptote from the command line using:
"C:\Program Files\Asymptote\asy.exe" -batchView -tex "pdflatex" -align C -f "pdf" nameofasyfile
(replacing nameofasyfile by the name of your .asy file)
or add in a pause to the DOS window by following [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148557]Reading asymptote error messages[/url].
Either way, you need to see what error messages you get.

I am using TexnicCenter, and compiling it with hotkey I made (just like they advise on AOPSWiki). There is no special error message, it only says, like I said before, if my file is called test, it says
test_0_1.df
and that's all. When I delete LABEL comand, it all works fine... Without labels, there is no problem...

The problem is that you won't see any error messages as the window closes too quickly which is why I suggested using my method for showing error messages.
[quote="SpongeBob"]There is no special error message, it only says, like I said before, if my file is called test, it says test_0_1.df and that's all. [/quote]What says test_0_1.df and where?

I have now added instructions about seeing the error messages to the Wiki at [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration]Asymptote: Advanced Configuration[/url] near the bottom of that page.

You didn't understand me...
I have done that, I went in start->run->cmd and there, I typed what you said, because the window closes to fast, and now it stays like that. When I compile it in the way you said, there is no error message, it only says what I said it says...

Try this suggestion from the [url=http://asymptote.sourceforge.net/FAQ/section2.html#miktex]Asymptote FAQ[/url]
[quote]Question 2.13. Why does Asymptote freeze upon trying to draw a label with my MikTex installation under Microsoft Windows?
Likely, this means that latex and dvips are not in your default path. Try adding the following lines to your config.asy file:
import settings;
latex="C:\texmf\miktex\bin\latex.exe";
dvips="C:\texmf\miktex\bin\dvips.exe";[/quote] Obviously you'll need to put in your paths for latex.exe and dvips.exe.

Well there is no such file: config.asy; I'll just move miktex into program files. THANK YOU!

config.asy is in C:\Documents and Settings\Username\.asy.
Careful when you say "move" as you can't just copy and paste as miktex has various configuration files and registry entries. You would need to uninstall and reinstall. Having said that, the sensible place to install miktex is in c:\texmf then no latex related program will get upset about paths with spaces in them.


Prove that for all positive integers n>=3 except for n=4 there exists no regular n-gon with lattice points as vertices.

There are lots and lots of solutions for this famous problem, and here's one of them:

It's easy to see that the existence of a regular polygon with n sides and vertices in the lattice points implies the existence of $2$ numbers $x$ and $y$ in $\mathbb Z[i]$ s.t. $\frac xy=\cos\left(\frac {2\pi}n\right)+i\sin\left(\frac {2\pi}n\right)$. To prove this just move one of the vertices of the polygon in the origin.

We thus have $x^n=y^n$, and since $\mathbb Z[i]$ is a factorial ring, it means that the primes which divide $x$ are the same as the primes which divide $y$, and their exponents are the same, so $\frac xy$ can only be invertible in $Z[i]$, or, in other words, $\frac xy$ in $\{1,-1,i,-i\}$. $\frac xy=1$ or $-1$ yield absurd results (I assumed $x$ and $y$ to be vertices of a regular polygon separated by another vertex), and $\frac xy=i$ or $-i$ correspond to the case n=4 and the polygon$=$square.

I don't think you need to use rings for this problem. You should be able to play around with values of n so that cos(2 :pi: /n) and sin(2 :pi: /n) are rational and then use Pick's Theorem to finish it off.

I don't think it's a question of whether you [i]need[/i] to use rings; as grobber said, there are many ways to prove the assertion. But I think the proof he presented is very nice and plus it's constructive.


calculus
integration
function
limit
real analysis
real analysis solved

For any functions f, g: [0; 1] ->  [0; 1], prove that if f
increases monotonically, then

\int10 f(g(x))dx \leq \int10 f(x)dx + \int10g(x)dx.

Suppose for some c we have
f(c)-c> \int f(x)dx
But  \int(0 to 1) f(x)dx>= \int (c to 1) f(x)dx>=(1-c)f(c)
So f(c)-c>(1-c)f(c) => cf(c)>c false!

So f(c)-c<= \int f(x)dx.

Take n a positive integer
And a{i,n}=sup{g(x): (i-1)/n <=x <=i/n}

1/n(f(a{i,n})-a{i,n})<=1/n \int f(x)dx  , any i
Add all this for i=1 to n, and get:

1/n \sum f(a{i,n}) <= \int f(x)dx + 1/n \sum a{i,n}

But a{i,n}>=g(i/n) so f(g(i/n))<=f(a{i,n}) =>

1/n \sum f(g(i/n)) <= \int f(x)dx + 1/n \sum a{i,n}

 \lim n->00 and get  \int f(g(x))dx<= \int f(x)dx + \int g(x)dx


probability
geometry
rectangle
search
number theory
relatively prime
similar triangles

A Circle of radius 1 is randomly placed in a $ 15\times35$ rectangle $ ABCD$ so that the circle lies completelly withing the rectangle.  Given that the probability that the circle will not touch diagonal $ AC$ is $ \frac{m}{n}$, were $ m$ and $ n$ are relatively prime postive integers, find $ m\plus{}n$.

hmm my solution involved similar triangles i keep getting the wrong answer.

[url]http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45[/url]

Just go to whichever year and problem this is, and click on the number of the problem, to go to the thread where this has already been discussed. I would post that link, but I have no idea what year this is from.


induction
number theory proposed
number theory

For each positive integer $ n$, define $ f(n)$ as the exponent of the $ 2$ in the decomposition in prime factors of the number $ n!$. Prove that the equation $ n\minus{}f(n)\equal{}a$ has infinitely many solutions for any positive integer $ a$.

Note that $ n\minus{}f(n)$ is equal to sum of digits of $ n$ in base $ 2$ .

Using the well-known Legendre's formula for $ f(n)$ it is easy to see that $ n\minus{}f(n)\equal{}2n\minus{}f(2n)$ thus if there is a solution for $ n\minus{}f(n)\equal{}a$ than there are infinitly many. To find one solution apply induction on a. it's obvious that $ n_0\equal{}1$ is a solution when $ a\equal{}1$ . Suppose we have $ b\minus{}f(b)\equal{}a\minus{}1$ for some positive integer a. Let $ k$ be the minimum number such that $ b<2^{k}$. Then $ n\equal{}2^{k}\plus{}b$ is a solution for $ n\minus{}f(n)\equal{}a$ and we are done.

[color=#38761D]Claim:[/color]: $ n=2^a-1$ works.
[color=#0f0]Proof:[/color]Using Legrendre`s formula  $f(n)=[n/2]+[n/4]+...$ we obtain that $ f(n)=2^a-1-a$ .
Let the set ${2^a-1=2^k-1   ,    k\ge a} $. Easily this satisfies the conditions.