Find all composite numbers $n$ satisfying the following property:
if $1=d_1<d_2<\dots<d_k<d_{k+1}=n$ are all divisors of $n$, then $d_1+d_2$ and
$d_1+d_2+\dots+d_k$ are also divisors of $n$.

If $n$ is odd then so is $d_2$ and then $d_1+d_2 = 1+d_2$ is even thus it cannot divide $n$.
Therefore, $n$ is even and $d_2=2$ and $d_3 = 1+d_2 =3.$
The second condition gives that $d_1+ \cdots d_k + d_{k+1} = 2n$ (since $d_1 + \cdots +d_k > d_k$ and has to be one of the $d_i$'s) which means that $n$ is a perfect number. It is well-known that an even perfect number  has form $n=2^{p-1}(2^p-1)$ where $2^p - 1$ is prime.
Since $3$ divides $n$ it follows that $2^p-1 = 3$ so that $p=2$ and $n=6.$

It is easy to verify that $n=6$ is indeed a solution.

Pierre.

You can solve it without using that theorem.


group theory
abstract algebra
search
superior algebra
superior algebra solved

G is a group of order 3825, Prove that if H is a normal subgroup of order 17 in G then H$\leq Z(G)$

I think it has already been discussed (in the general case).. try to search ..


Show that in any triangle we have:

(b^2+c^2)\m_a+(c^2+a^2)\m_b+(a^2+b^2)\m_c <=12R.

I'd guess that m_a, m_b, and m_c are the medians to sides a, b, and c respectively, and R is the circumradius.

I'll just [tex]\TeX[/tex] this thing real quick:

[tex]\displaystyle \mbox{Prove that}[/tex]

[tex]\displaystyle \frac{a^2+b^2}{m_c}+\frac{b^2+c^2}{m_a}+\frac{c^2+a^2}{m_b} \le 12R[/tex]

where [tex]m_a[/tex], [tex]m_b[/tex], and [tex]m_c[/tex] are the medians to sides a, b, and c, respectively, and R is the circumradius.

[hide]

Let F be the midpoint of AB and X be the point on the circumcircle such that the points C, F and X are colinear. Then [tex]CF\cdot FX = AF\cdot FB[/tex], hence [tex]FX=\frac{c^2}{4m_c}[/tex]. We have [tex]a^2(\frac{1}{2}c) + b^2(\frac{1}{2}c) = c(m_c^2 + (\frac{1}{2}c)^2)[/tex] by Stewart's Theorem. So [tex]a^2+b^2 = 2m_c^2 + \frac{1}{2}c^2[/tex]. Then [tex]\frac{a^2+b^2}{m_c} = 2m_c + \frac{c^2}{2m_c}= 2(m_c + \frac{c^2}{4m_c}) = 2(CF + FX) = 2 AX \leq 4R[/tex]. Similarly,  [tex]\frac{b^2+c^2}{m_a} \leq 4R[/tex] and [tex]\frac{c^2+a^2}{m_b} \leq 4R[/tex]. Adding the three inequalities together to get the required result.

Q.E.D

[/hide]


search
combinatorics unsolved
combinatorics

The space can not be represented as a finite reunion of planes. (search didn't give anything)

Actually, grobber has proved [url=http://www.mathlinks.ro/viewtopic.php?p=624250#624250]here[/url] a generalization.
If one's not comfortable with the language, grobber suggests picking a plane $\pi$ that is different from all the planes that presumably cover the whole space. Now, each plane intersects $\pi$ along a line (either that, or it doesn't intersect it all). Hence, we have finitely many lines that cover the plane and we can use the same procedure to prove that this isn't possible.


An object falls freely from rest near the surface of Earth. What is the speed of the object after having fallen a distance at 4.90 meters?

(1) 4.90 m/s      (3) 24.0 m/s
(2) 9.80 m/s      (4) 96.1 m/s

[hide]
From the formula $v_{f}^{2}=v_{0}^{2}+2ad$, we find that $v_{f}=9.8\,\,\text{m/s}$.[/hide]

If you consider distance as the under a velocity graph, we have $d=v_{0}t+gt^{2}/2$, where $g=9.8\ m/s^{2},\ v_{0}=0$.  So $t=1$.  After 1 second, the velocity is $1(9.8)\ m/s =9.8\ m/s$

I can see this to be a physics question, right?

[quote="rnwang2"][quote="Interval"]I do not understand the formula provided.

Can you break it down piece by piece?

Thanks![/quote]
It is easily derived using some basic physics.[/quote]
Basic math.  :P

Starting with the equation
\[d = vt, \]
where $v$ is the average velocity of the object. If the object undergoes constant acceleration, the average velocity is then equal to the average of $v_{i}$ and $v_{f}$, the initial and final velocities. So the above equation can be rewritten as
\begin{eqnarray*}d &=& \left (\frac{v_{i}+v_{f}}{2}\right) t\\ &=& \left (\frac{v_{i}+(v_{i}+at)}{2}\right) t\\ &=& v_{i}t+\frac{at^{2}}{2}\end{eqnarray*}
Squaring both sides of the equation $v_{f}= v_{i}+at$,
\begin{eqnarray*}v_{f}^{2}&=& v_{i}^{2}+2v_{i}at+a^{2}t^{2}\\ &=& v_{i}^{2}+2a\left(v_{i}t+\frac{at^{2}}{2}\right)\\ &=& \boxed{v_{i}^{2}+2ad}\end{eqnarray*}

Guys, you complicate things too much...
Start with the law of energy conservation.
$E_{gp}=E_{k}$ (gravitational energy transforms into kinetical)
$mg\Delta h=\frac{mv^{2}}{2}$
$v=\sqrt{2g\Delta h}$
Plug in $\Delta h=4.9\ m \Longrightarrow v=9.8\ m/s$


I was just wondering:

if we denot R(A) be the reduced echelon form of the matrix A is it tru that:

R(A*B) = R(A)*R(B)

Lets Add one more:

lets say that A = B*C where A = m by n, B = m by w, and C = w by n (w is the least possible) than this holds true:

Rank(A) = Rank(B) * Rank(C)

Thanx!

Could anyone please help me with the second problem!

Isn't it obvious? Multiplying ranks can't possibly be right. We have the inequalities
$\text{rank }B+\text{rank }C-w\le\text{rank }A\le\min\{\text{rank }B,\text{rank }C\}$,
so we can only have $\text{rank }A=\text{rank }B\cdot\text{rank }C$ when one of $B,C$ is zero or when $\text{rank }B=\text{rank }C=\text{rank }A=1$.

On the echelon form:

Try $A=\begin{bmatrix}1&1\\1&1\end{bmatrix}$ and $B=\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$.

the matrix A that u gave will be decomposed into  [1]
                                                                        [1] * [1 1] = [1 1]
                                                                                [1 1] and thus the problem holds. I think there is a mistake in what u wrote. Thanx for the help though!

I am sorry i think that u are right!
I found that:
Rank(A) + Rank(B) >= Rank(A*B) (I think it is Sylvester Inequality)
well than we have :L
Rank(A) + Rank(B) >= Rank(A*B) = Rank(A)*Rank(B) or
Rank A =Rank B = 1,2.
And each case can be solved separately!


[quote="enndb0x"]Let $ n\in \mathbb{N}$ and $ n\geq 3$.Prove or disprove that between numbers of following secuences

$ 2n \minus{} 1,2n,2n \plus{} 1,...,3n \minus{} 2$

$ 2n,2n \plus{} 1,2n \plus{} 2,...,3n \minus{} 1$

$ 2n \plus{} 1,2n \plus{} 2,2n \plus{} 3 ,...,3n$

          $ .$
          $ .$
          $ .$

There exist a perfect square in each secuence.[/quote]

You want to prove or disprove if, given  $ n\in \mathbb{N}$ and $ n\geq 3$, exists a perfect square in $ [2n \plus{} k \minus{} 1,3n \plus{} k \minus{} 2]$ $ \forall k\geq 0$

Take for instance $ k \equal{} n^2 \minus{} 2n \plus{} 2$. We have $ n^2 < n^2 \plus{} 1 \equal{} 2n \plus{} k \minus{} 1 < 3n \plus{} k \minus{} 2 \equal{} n^2 \plus{} n < (n \plus{} 1)^2$ and so no perfect square in $ [2n \plus{} k \minus{} 1,3n \plus{} k \minus{} 2]$

(and, btw, quicker : this is wrong for $ n\equal{}3$ : the first sequence $ (5,6,7)$ contains no perfect square)


calculus
integration
logarithms
calculus computations

I know this may seem annoying, but can any one direct me to the thread which discussed the evaluation of this integral (or something similar, I may not be remembering exactly):

$ \displaystyle\int_0^1 \ln{(1\minus{}x)}\ln{x} dx$

I recall that the solution involved converting one of the ln's into a power series, so maybe my rough memory of this integral is flawed.  Any direction will be appreciated, thank you!

Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106770]here[/url]. It's always a good practice to keep track of your own posts.


We will be doing a Math Jam as soon as we can after each AMC test.  Unfortunately, the day after the AMC B coincides with one of our courses.  Which of the following times would you like to see us hold the AMC B review Math Jam?

The day after! I will die if I don't know the definitive answers =D.

are teachers not allowed to let you see the solutions manual or tell you the answers the day of the test?  I thought they were.

I am not positive, but I believe that a set of solutions is available to your teacher after the AMC.  I am not sure this is true for the AIME.

We hope we can add to the process by explaining how to think along the lines that will produce the solution.

I think you have to pay for solutions for the AIME, whereas you get them automatically with the AMCs.

We've opted for the Saturday slot.  The last one went 3 hours, and we can't ask folks on the east coast to stay up til midnight on Thursday...

Is it possible that someone could post the answers thursday like last time on the board?


Let G = (V,E) be a graph. Then the line graph L(G) of G has the edges of G as the vertex, and two edges are joined by a line if they are incident with a common vertex of G.

Prove that a connected graph is isomorphic to its line graph iff it is a cycle.
Then determine the triangles (3-cycles) in L(G) in terms of the number of triangles of G and the degrees of the vertices.

[quote="cooley"]Prove that a connected graph is isomorphic to its line graph iff it is a cycle.[/quote]

If $G=L(G)$,
	If $G$ is not a cycle,
		Consider the largest cycle $C$ of length $x$ in $G$
		Since $G$ is connected and is not a cycle, there exists a vertex $V$ on $C$ that has degree $\geq 3$
		The edges of $C$ form the vertices of a cycle $C'$ on length $x$ in $L(G)$
		The $2$ edges of $V$ on $C$ and $1$ other edge form the vertices of a triangle $T'$ in $L(G)$
		But $C'$ and $T'$ share an edge
		So $XOR(C',T')$ is a cycle in $L(G)$ with length $x+1$ $\Rightarrow\Leftarrow$
	Thus $G$ is a cycle
If $G$ is a cycle,
	$L(G)$ is obviously a cycle
Therefore $G=L(G)$

[quote]Then determine the triangles (3-cycles) in L(G) in terms of the number of triangles of G and the degrees of the vertices.[/quote]

For any triangle $X'Y'Z'$ in $L(G)$ corresponding to edges $X,Y,Z$ in $G$,
	Each pair $(X,Y)$, $(Y,Z)$, $(Z,X)$ shares a vertex
	So either $X,Y,Z$ share a vertex or $X,Y,Z$ are the edges of a triangle
For any vertex $V$ in $G$ with degree $d$,
	There are $\binom{d}{3}$ unordered triples of edges with common vertex $V$
Let $t(G)$ be the number of triangles in $G$
Therefore the $t(L(G))$ is
	$t(G)+\sum{\binom{d_{i}}{3}}$

Thanks for the reply. Actually, i knew the answer of the second question by trial and error, but could not have managed to prove it.Now it is clear by your explanaton. Also, after asking the first question, i noticed that it can be proved by getting use of the automorphism. Anyway, your help is appreciated.


Let $ S_N \equal{} \{X \equal{} (n_1,n_2,...,n_k)|\sum_i n_i \equal{} N, n_1\ge n_2\ge ...\ge n_k\ge 1\}$ set of Jung's diagram.
Let $ F: S_N\to S_n$ defined by for $ X \equal{} (n_1\ge n_2\ge ...n_{l \minus{} 1} > n_l\ge ... n_m > n_{m \plus{} 1} \equal{} .. \equal{} n_k \equal{} 1)$, were $ n_{l \minus{} 1} > k,n_l\le k,n_m\ge 2, n_{m \plus{} 1} \equal{} 1$
$ F(X) \equal{} (n_1 \minus{} 1,n_2 \minus{} 1,...,n_{l \minus{} 1},k,n_{l} \minus{} 1,...,n_m \minus{} 1).$ (all term $ n_i$ decrease $ n_i \to n_i \minus{} 1$ and add new term k)
Let $ h(N) \equal{} \minus{} [\frac {1 \minus{} \sqrt {8N \plus{} 1}}{2}], t(N) \equal{} \frac {h(N)(h(N) \plus{} 1)}{2} \minus{} N.$
Let $ F^{(n)}$ is n-th iteration of F ($ F^{(1)} \equal{} F, F^{(n \plus{} 1)} \equal{} F(F^{(n)})$).
Diagram X is in attractor if $ F^{(h(N))}(X) \equal{} X$. Let $ L(X)$ is minimal distance to attractor $ F^{(h(N) \plus{} L(N))}(X) \equal{} F^{(L(N))}(X)$ but $ F^{(h(N) \plus{} L(N) \minus{} 1)}(X)\not \equal{} F^{(L(N) \minus{} 1)}(X).$
1. Prove: if $ X_1 \equal{} (1,1,...,1)$ - (N times 1), then $ L(X_1) \equal{} N \plus{} 1 \minus{} h(N).$ If $ N \equal{} [(m^2 \minus{} 1)/2],m \equal{} h(N)$, then for any $ X\in S_N$ $ L(X)\le L(X_1)$.
2. Prove that $ L(X)\le h(N)(h(N) \minus{} 1) \ \forall X\in S_N$ and if $ N \equal{} \frac {m(m \plus{} 1)}{2},m \equal{} h(N), X_0 \equal{} (m \minus{} 1,m \minus{} 1,m \minus{} 2,m \minus{} 3,...,2,1,1)$
then $ L(X_0) \equal{} h(N)(h(N) \minus{} 1).$

What does $ n_i\t0 n_i - 1$ mean?

Besides, "Jung" should be "Young".

  Darij


geometry
geometric transformation
number theory

[quote="� αναγιώτης1"]Hi. I'm searching for a good number theory book for intermediate solvers. If anybody has one please tell me.   :|[/quote]

use engel or 103 problems in Number Theory by andreescu :)

Forget about Engel as concerns Number Theory.
But I can't say much until I know what is meant by "intermediate solver".

David M. Burton, [i]Elementary Number Theory[/i], 1980

is suitable for different levels, but I don't know whether it is freely avaliable online (there was a time when its German translation was online as PDF; I can send it to you if you wish).

  darij


Let X be a asset of 1+2^k points in the Euclidean plane. Prove there exist 3 points in X making a triangle whose largest angle has size at least 180*(1-1/k)

By Euclidean plane, what do you mean? An ordinary plane? :maybe:


There is a new math symbol invented, it looks like this @

2@=6
3@=12
5@=30
7@=56
what is 34@?

nice job gr8sk8r, you might be good enough to take a shot at the Relay race(click on my signature)

[hide]Yeah, I got that (1190) too! :) [/hide]

in general,

n@ = n(n+1)
so 34@ = 34(35) or [color=cyan]1190[/color].

[hide]so when a number multiplies by @, it is basically N(N+1)
so 34*35=1190[/hide]

[hide]Let x = any real number

x@ = x(x+1)

So, 34@ = 34(35)

(need a calculator for this :-D)

[b]1190. There's your answer.[/b][/hide]

[quote="math92"]wow, i didn't think that many people would get it.  Okay there is another new math symbol that looks like this: #
if 5#=6
   4#=4
   3#=2
  10#=16
what is 32#[/quote]

[hide] $x$#$=2x-4$

$32$#$=64-4=\boxed{60}$[/hide]

[hide]the first n=8
the second n=254[/hide]

[quote="G-UNIT"][quote="10000th User"]*New*

Use math92 @ symbol:
If I have n@@=5256, what is n?

Use math92 # symbol:
If I have n###=2004, what is n?[/quote]

  I dont quite understand, could you explain a little more thouroughly.[/quote]
that's strange to see that it is easy to proceed forward but then it is harder to understand when doing things backwards. for example, finding 3@@ is easy because 3@=12 and 3@@=12@=156, but finding n when n@@=156 is not understandable...

[quote="math92"]wow, i didn't think that many people would get it.  Okay there is another new math symbol that looks like this: #
if 5#=6
   4#=4
   3#=2
  10#=16
what is 32#[/quote]

[hide]n#=2n-4

So 32#=2*32-4
=60[/hide]

[quote="math92"][quote="10000th User"]*New*

Use math92 @ symbol:
If I have n@@=5256, what is n?

Use math92 # symbol:
If I have n###=2004, what is n?[/quote][hide="the first one"]i used guess and check and got [b]8[/b] as my answer :lol: [/hide][/quote]
 :thumbup:
Not too hard eh?
I did this way: n@@=n@(n@+1)=5256 and find n@ is about 72.4. How to do it? OK see my square root calculate
[code]     72.
5256
49
 356    14_x_ missing digit is 2 (3 is too large)
 284
  7200    144_x_ missing digit is 4 (5 is a little too big)
-----not important------[/code]
So n@ is about 72.4. if you check 72x73=72x70+72x3=5040+216=5256
Now n@=n(n+1)=72 and you can check easy n=8

math92 please solve second one :D

[quote="math92"][quote="10000th User"]
math92 please solve second one :D[/quote]
i'm working on it...[/quote]
Hi math92
[hide="hint"]OK equation is n###=2004. You know # do this: double number and subtract 4. Now do 'backward' way. So add 4 and then divide by 2... this helps? ;)[/hide]

[hide]thanks alot for the hint.  I got 254 as my answer. Is that correct?[/hide]

[quote="math92"][hide]thanks alot for the hint.  I got 254 as my answer. Is that correct?[/quote]
 :yup:  :clap:  :10:  :flex:  :jump:  :thumbup:  :wow:

[quote="math92"]whoa, do you do that everytime someone answers a problem? :P[/quote]
Only for you because you are the most special middle school student here ;)

[hide]

I started out by trying to get a pattern: 2@=6, 2*3=6; 3@=12, 3*4=12; the pattern is @=n+1.

So 34@ = 34*35 = $  \framebox{1190}  $[/hide]

[hide]

I looked for a pattern and I got 2n - 4 is #.

So $  32\# = \framebox{60}  $[/hide]


let  f:R->R be a functionwith the properties:
1.  f(x)f(x)<=1  x for any real
2. f'(x)f'(x)+f"(x)f"(x)<=1  for any x real

Show that f(x)f(x)+f'(x)f'(x)<=1 for any x real.

it was at procopiu contest at 12`th grade ? :)

Yes, didn't go to the contest because i don't know any phisics :D , could you give me an idea, please?

How come we can't have a point $x$ such that $f(x)=.9$, $f'(x)=.9$, and $f''(x)=-.1$?

$f(x)f(x)=.81\leq1$, $f'(x)f'(x)+f''(x)f''(x)=.82\leq1$, but $f(x)f(x)+f'(x)f'(x)=1.62\nleq 1$.

Or am I misreading the question?  I assumed by $f(x)f(x)$ you mean $f(x)$ times $f(x)$.


[quote="ashegh"]let A=(1,2,...,99) be a set.50 number are chosen from A,inwich the sum of each two number isnt equal to 99 or 100.

prove that:

               the 50 chosen number should be : 50,51,...,99[/quote]
i think if we suppose that the numbers are not 50,51,...,99 $\rightarrow$ contradiction. :D

its a good fact...

but i think it doesnt help us too... :(

[hide]Suppose we choose an element $k$ ($k\ne 99$) from $A$.  We know that $99-k$ and $100-k$ cannot be chosen also.  Therefore, the elements in $A$ can be paired up as: $(1,98), (2,97),\ldots, (49,50), (99).$  We can only take one element from each set of parentheses.  Thus, we must choose the element $99$ in order to end up with 50 numbers.  But then we cannot choose 1 or else 99+1=100.  Thus we have to choose 98.  Similarily, we can't have 2.  So we must take 97.  This logic continues all the way down thus forcing us to choose $50,51,\ldots 99.$[/hide]


quadratics
calculus
calculus computations

\[ x\left( {\frac{{dy}}
{{dx}}} \right)^2  \plus{} (y \minus{} x)\frac{{dy}}
{{dx}} \minus{} y \equal{} 0\]

[quote="Xaenir"]
\[ x\left( {\frac {{dy}} {{dx}}} \right)^2 \plus{} (y \minus{} x)\frac {{dy}} {{dx}} \minus{} y \equal{} 0\]
[/quote]

It is a quadratic in $ \frac{dy}{dx}$

$ \Longrightarrow\frac{dy}{dx}\equal{}\frac{x\minus{}y\pm\sqrt{(y\minus{}x)^2\plus{}4xy}}{2}$

                          $ \frac{dy}{dx}\equal{}\frac{x\minus{}y\pm(x\plus{}y)}{2}$

                          $ \frac{dy}{dx}\equal{}x,\minus{}y$

                          $ y\equal{}\frac{x^2}{2}\plus{}C$ and $ y\equal{}e^{\minus{}(x\plus{}c)}$


inequalities
trigonometry
geometry
inequalities open

[quote="Michael Niland"]In a triangle ABC prove that

1+4CosACosBCosC >=2(CosACosB +CosBCosC +CosCCosA)[/quote]

This is equivalent to:

$(\cos A+\cos B)^2 + (\cos B + \cos C)^2 + (\cos C + \cos A)^2 \leq 3$

For acute triangle, it was proved at:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14567

This is wrong for obtuse triangles.

Yes Fuzzylogic, you are right.
This inequality is only true for acute triangle and right triangle. If you rewrite it in the form :
$ (cosA+cosB+cosC)^2 \leq\ ((sinA)^2+(sinB)^2+(sinC)^2) $
It is Walker's inequality.

Walker's inequality has a nice form :
If H is the orthocenter of the non-obtuse triangle ABC then :
$ (HA+HB+HC)^2 \leq\ AB^2+BC^2+CA^2 $

Treegoner has a general inequality :
For non-obtuse triangle ABC and m,n,p be real number
$(mcosA+ncosB+pcosC)^2 \leq\ (M(sinA)^2+N(sinB)^2+P(sinC)^2) $

$M=p^2+n^2-m^2 $
$N=m^2+p^2-n^2 $
$P=n^2+m^2-p^2 $

I remember that Treegoner also posted his theorem and proof on mathlinks recently.


number theory
prime factorization
number theory unsolved

A positive integer $ m$ is called [i]good[/i] if there is a positive integer $ n$ such that $ m$ is the quotient of $ n$ by the number of positive integer divisors of $ n$ (including $ 1$ and $ n$ itself). Prove that $ 1, 2, \ldots, 17$ are good numbers and that $ 18$ is not a good number.

[hide="Solution"]If $ m$ is an odd prime, then we can let $ n \equal{} 8m$.
If $ m \equal{} 1$, $ n \equal{} 1$
If $ m \equal{} 2$, $ n \equal{} 8$
If $ m \equal{} 4$, $ n \equal{} 36$
If $ m \equal{} 6$, $ n \equal{} 72$
If $ m \equal{} 8$, $ n \equal{} 80$
If $ m \equal{} 9$, $ n \equal{} 108$
If $ m \equal{} 10$, $ n \equal{} 180$
If $ m \equal{} 12$, $ n \equal{} 240$
If $ m \equal{} 16$, $ n \equal{} 128$
Let the prime factorization of $ n$ be $ 2^{a_1}3^{a_2}5^{a_3}...p_n^{a_n}$
Note that for any prime $ p$, $ \frac {p^x}{x \plus{} 1}$ is minimized when $ x \equal{} 1$. So if $ m \equal{} 18 \equal{} 2 \cdot 3^2$, then since $ 18|n$, $ n$ would have to be at least $ \frac {2^1 \cdot 3^2}{2 \cdot 3} \equal{} 3$. Thus, no prime $ k$ greater than $ 11$ can be in the prime factorization of $ n$, or we would have to multiply this by at least $ \frac {k}{2}$ which would make the result greater than $ 18$. If $ 7$ or $ 11$ was in the prime factorization of $ n$, then one of the exponents on the other primes would have to be a multiple of $ 6$ or $ 10$, making the quotient too great. Thus, the quotient is of the form
$ \frac {2^{a_1}3^{a_2}5^{a_3}}{(a_1 \plus{} 1)(a_2 \plus{} 1)(a_3 \plus{} 1)}$
We easily see $ a_3$ is either $ 0$ or $ 1$, or the fraction is too large. If it is $ 1$, then one of the other exponents must be $ 4$, and we can easily check that there is no solution. Thus, the quotient is of the form
$ \frac {2^{a_1}3^{a_2}}{(a_1 \plus{} 1)(a_2 \plus{} 1)}$.
Since $ a_2$ is at least $ 2$, and also odd, it must be $ 3$, as $ 5$ would make the fraction too large. But we can easily check there is no solution in this case Thus, $ 18$ is not a good number.[/hide]
But I can't attain $ 14$ or $ 15$. Can anyone show how to get $ 14$ or $ 15$?

[quote="dgreenb801"] But I can't attain $ 14$ or $ 15$. Can anyone show how to get $ 14$ or $ 15$?[/quote]

$ 14\equal{}\frac{252}{18}\equal{}\frac{2^2}3\frac{3^2}3\frac 72$

$ 15\equal{}\frac{360}{24}\equal{}\frac{2^3}{4}\frac{3^2}3\frac 52$


20% of the male students at Carnage Middle School buy lunch in the cafeteria. The number of female students who buy lunch is 80% of the number of male students who buy lunch, but there are 20% more female students at Carnage than males. If 48 female students buy lunch, how many students attend Carnage Middle School?

[hide="Strange name for a school."]
You know that 48 females buy lunch.  You also know that 48 is 80% of the number of male students who buy lunch, so there are 60 lunch-buying males.

60 is 20% of the total males; therefore, there are 300 males attending the school.

If there are 20% more female students than male students, then there are 360 female students.

Thus, there are $ \boxed{660 \text{ students}}$ attending Carnage Middle School.[/hide]


Prove that for any prime p we can find an integer n such that n8 = 16 (mod p).

we must have $p|n^{8}-16=(n^{2}-2)(n^{2}+2)(n^{4}+4)$
now we know that if $p=8k+1,8k+7$ then there exist $a$ such that $a^{2}\equiv2$ (mod p) , and if $p=8k+3,8k+1$ then there exist $a$
such that $a^{2}\equiv-2$ (mod p) . now we must prove the problem for $p=8k+5$
$n^{4}+4=n^{4}+4n^{2}+4-4n^{2}=(n^{2}+2n+2)(n^{2}-2n+2)$ , now we must have $p|(n^{2}+2n+2)(n^{2}-2n+2)=[(n+1)^{2}+1][(n-1)^{2}+1]$
now we now that if $p=4k+1$ then there exist $a$ such that $a^{2}\equiv-1$ (mod p) . now put $a=n+1$ and proof is complete[because $p=8k+5=4(2k+1)+1$]
:)
is my solution true ?

[quote="khashi70"]is my solution true ?[/quote]
Why do You ask? You know it is true.
For instance to get decidibility of cogruence $x^{2}\equiv-2 \pmod{p}$ we have to culculate Legandre simbol

$( \frac{-2}{p})= (-1)^{\frac{p-1}{2}+\frac{p^{2}-1}{8}}=1 \iff \frac{(p-1)(p+5)}{8}$ is even $\iff p\equiv 1 \pmod{8}\vee p\equiv 3 \pmod{8}$.

The only missing case is $p=2$ which is trivial.


$ a>1$ is not an integer.

define $ f(n)\equal{}\frac{1}{n}\lfloor{na}\rfloor$.

Prove or disprove that $ f(n)>f(n\plus{}1)$ for infinitely many $ n$.

in other topic it was written it should be known - could anyone give a link or a hint how to solve this problem ?

maybe it's not a good problem for "Pre-Olympiad" section?
if so, please move it

$ n\in\mathbb{N}$ , $ n$ is a positive integer (as usually $ n$ is) ;)

so your hint is OK but only in the case $ a\in\mathbb{Q}$   :P

what about irrational $ a$ ?

It's trivial to prove that if it's true for $ a_1,a_2$ so that $ a_1 < a_2$, then it's true for $ a_3\in (a_1,a_2)$.

This hinges on showing that $ \frac {1}{n}\lfloor na_2\rfloor > \frac {1}{n \plus{} 1}\lfloor a_1(n \plus{} 1)\rfloor$, which is of course true because $ \frac {1}{n \plus{} 1}\lfloor a_1(n \plus{} 1)\rfloor < \frac {1}{n}\lfloor na_1 \rfloor \le \frac {1}{n}\lfloor na_2 \rfloor$, the last inequality being true because $ a_1 < a_2$.

[hide]Let $ a \equal{} m \plus{} \alpha$ where $ m \in \mathbb{Z}$ is the floor of $ a$ and $ 0 \leq \alpha < 1$ is the fractional part. Then $ f(n) \equal{} m \plus{} \frac{1}{n} \lfloor n\alpha \rfloor$. Because $ \alpha < 1$, there are infinitely many $ n$ such that $ \lfloor n\alpha \rfloor \equal{} \lfloor (n\plus{}1)\alpha \rfloor$ (*). Those $ n$ work.

[hide="Proof of *"]$ \alpha < 1$ means $ \alpha < 1 \minus{} \frac{1}{k}$ for some $ k$. Then $ \lfloor n \alpha \rfloor > \lfloor (n\plus{}k)\alpha \rfloor \minus{} k$, so there is one value in between $ n$ and $ n\plus{}k$ where it didn't increase.[/hide][/hide]

perhaps [url=http://en.wikipedia.org/wiki/Hermite's_identity]Hermite's Identity[/url] may help.

um, except that it doesn't and has nothing to do with the problem except that they both contain floors?

Looks fairly helpful to me. It reinterprets $ f(n)$ as the average of $ \lfloor a \rfloor, \lfloor a \plus{} 1/n \rfloor, \ldots , \lfloor a \plus{} (n\minus{}1)/n) \rfloor$, which means you can analyze where the fractional part of $ a$ lies relative to $ 0, 1/n, 2/n \ldots 1 \minus{} 1/n$. I think that leads to a solution.

I admit I didn't see that and apologize for being hasty (and overly harsh, I guess), but that seems rather laborious and needlessly difficult.


pen
integration
modular arithmetic
number theory
relatively prime
Divisibility Theory

Find all positive integers $n$ that have exactly $16$ positive integral divisors $d_{1},d_{2} \cdots, d_{16}$ such that $1=d_{1}<d_{2}<\cdots<d_{16}=n$, $d_6=18$, and $d_{9}-d_{8}=17$.

Let $ n\equal{}\prod_{i\equal{}1}^mp_i^{k_i}$, then $ \prod_i (k_i\plus{}1)\equal{}16$. Obviosly $ d_{8\minus{}k}d_{9\plus{}k}\equal{}n,k\equal{}0,1,..,7$.
$ 18\equal{}d_6|n$, therefore $ p_1\equal{}2,p_2\equal{}3,k_2\ge 2, 2\le m\le 3.$
If $ m\equal{}2$, then $ n\equal{}2^33^3$ or $ n\equal{}2*3^7$ For first cfse $ d_6\equal{}8\not \equal{}18$. Second is solution.
If $ m\equal{}3$, then $ n\equal{}2*3^3*p,p\ge 5$, $ d_1\equal{}1,d_2\equal{}2,d_3\equal{}3.$
If $ p<9$ $ d_6\equal{}9$. If $ p\ge 11$ $ d_4\equal{}6,d_5\equal{}9$ $ d_6\equal{}18\to p\ge 19$.
We had divisors 27, 54, p,2p. If p<54, then must be $ d_9\equal{}54,d_8\equal{}54\minus{}17\equal{}37\equal{}p$, if p>54 $ d_8\equal{}54,d_9\equal{}54\plus{}17\equal{}71\equal{}p$.
Therefore all solutions are $ n\equal{}2*3^7,n\equal{}2*3^3*37,n\equal{}2*3^3*71$.

Rust

The first solution is false because d8=54 and d9=81 then the difference is d9-d8=27, a contradiction.

Yes. You are rite. I don't chek $ d_9\minus{}d_8\equal{}17$ for these case.

Slightly different finish:

So as above, we get that $n=2 \cdot 3^3 \cdot p$ for some prime $p$.

Then, we have that there exists $k$ with $k(k+17)=n=54p$

Considering modulo $27$ gives $k \equiv 0, 10 \pmod {27}$

In the former case, letting $k=27a$, we have $a(27a+17)=2p$. Since the two things are relatively prime, for there only to be prime factors of $p,2$ we must have $a=1,2$ (else there are at least 3 prime factors: a 2, and one further in each parenthetical part) $a=1$ yields $p=22$, which isn't prime, but $a=2$ yields $71$ which works.

In the latter case, letting $k=27a+10$ yields $(a+1)(27a+10)=2p$. Again, we have $a+1 \leq 2 \implies a \leq 1$. But $a=1$ works, yielding $p=37$.

Thus the only possible working values of $n$ are $\boxed{n=1998, 3834}$ (54 times 37 and 71 respectively). It can easily be verified that they work, so we're done.

@Mewto

$k(k+17)=n=54p$  How did you know that $d_8,d_9$ don't have common prime divisors?


[i]Find all integer solutions x,y of the equations
1)$(x+y^2)(x^2+y)=(x+y)^3$;
2)$(x+y^2)(x^2+y)=(x-y)^3$
3)$(x+y^3)(x^3+y)=(x+y)^4$
4)$(x+y^3)(x^3+y)=(x-y)^4$[/i]
[i]1) 45-th Swedish Mathematical Competition 2005, Final Round (Lund, November 19)
2) Proposed by Dang Hung Thang in Vietnamese Mathematical Olympiad
3) Australian Mathematical Competition 2004
4) Just to fill the collection  :)
I solved only number 3). Help me  :)[/i]


Putnam
geometry
integration
logarithms
function
college contests
Putnam calculus

Fix an integer $ b \geq 2$. Let $ f(1) \equal{} 1$, $ f(2) \equal{} 2$, and for each $ n \geq 3$, define $ f(n) \equal{} n f(d)$, where $ d$ is the number of base-$ b$ digits of $ n$. For which values of $ b$ does
\[ \sum_{n\equal{}1}^\infty \frac{1}{f(n)}
\]
converge?

jmerry just tried to link to this from the discussion of a 2008 problem, but the topic didn't exist. Now it does, and I'll go back and fix his link.

I realize that this is ancient history, but since I have an answer written up, I might as well post it. This is a harder problem then 2008 A4, which is the problem that got directed here.

======================================

For the integer $ b\ge 2,$ we have the sequence $ f(n)$ defined by $ f(1) \equal{} 1, f(2) \equal{} 2,$ and after that, $ f(n) \equal{} nf(d),$ where $ d$ is the number of base-$ b$ digits of $ n.$ Then  $ \sum_{n \equal{} 1}^{\infty}\frac {1}{f(n)}$  converges for $ b \equal{} 2$ and diverges for $ b\ge 3.$ As a lemma in our proof, we will need to estimate $ \sum_{n \equal{} b^{k \minus{} 1}}^{b^{k} \minus{} 1}\frac {1}{n}.$ On the one hand, picturing this sum as the sum of the areas of boxes shows that $ \sum_{n \equal{} b^{k \minus{} 1}}^{b^{k} \minus{} 1}\frac {1}{n} > \int_{b^{k \minus{} 1}}^{b^{k}}\frac {1}{x}\, dx .$
Hence, $ \sum_{n \equal{} b^{k \minus{} 1}}^{b^{k} \minus{} 1}\frac {1}{n} > \ln b,$ and for $ b\ge 3,\ \ln b > 1.$ On the other hand, since
 $ \frac {1}{nb} \plus{} \frac {1}{nb \plus{} 1} \plus{} \cdots \plus{} \frac {1}{nb \plus{} b \minus{} 1} < \frac {1}{n},$  $ \sum_{n \equal{} b^{k \minus{} 1}}^{b^{k} \minus{} 1}\frac {1}{n}$  is a strictly decreasing function of $ k,$ and, in particular, for $ k\ge 3,\ \sum_{n \equal{} 2^{k \minus{} 1}}^{2^{k} \minus{} 1}\frac {1}{n}\le \frac14 \plus{} \frac15 \plus{} \frac16 \plus{} \frac17 \equal{} \frac {319}{420},$ which is less than $ 1.$
Assume that $ b\ge 3.$ Then  $ \sum_{n \equal{} 1}^{\infty}\frac {1}{f(n)} \equal{} \sum_{k \equal{} 1}^{\infty}\sum_{n \equal{} b^{k \minus{} 1}}^{b^{k} \minus{} 1}\frac {1}{f(n)} \equal{} \sum_{k \equal{} 1}^{\infty}\frac {1}{f(k)}\sum_{n \equal{} b^{k \minus{} 1}}^{b^{k} \minus{} 1}\frac {1}{n} > (\ln b)\sum_{k \equal{} 1}^{\infty}\frac {1}{f(n)}.$
Thus  $ \sum_{n \equal{} 1}^{\infty}\frac {1}{f(n)} > (\ln b)\sum_{n \equal{} 1}^{\infty}\frac1{f(n)}.$ Since $ \ln b > 1,$ this must mean that the sum diverges.

Now assume that $ b \equal{} 2.$

$ \sum_{n \equal{} 1}^{\infty}\frac {1}{f(n)} \equal{} 1 \plus{} \frac12 \plus{} \frac16 \plus{} \sum_{n \equal{} 4}^{\infty}\frac {1}{f(n)}$

$ \equal{} 1 \plus{} \frac12 \plus{} \frac16 \plus{} \sum_{k \equal{} 3}^{\infty}\frac1{f(k)} \sum_{n \equal{} 2^{k \minus{} 1}}^{2^{k} \minus{} 1}\frac1{f(n)}$

$ \equal{} 1 \plus{} \frac12 \plus{} \frac16 \plus{} \sum_{k \equal{} 3}^{\infty}\frac1{f(k)}\sum_{n \equal{} 2^{k \minus{} 1}}^{2^{k} \minus{} 1}\frac1n$

$ < 1 \plus{} \frac12 \plus{} \frac16 \plus{} \left(\frac {319}{420}\right) \sum_{k \equal{} 3}^{\infty}\frac1{f(k)}.$

So, $ \sum_{n \equal{} 3}^{\infty}\frac1{f(n)} < \frac16 \plus{} \left(\frac {319}{420}\right) \sum_{k \equal{} 3}^{\infty}\frac1{f(k)},$ or  $ \left(\frac {101}{420}\right)\sum_{n \equal{} 3}^{\infty}\frac1{f(n)} < \frac16$   and the sum converges.


Solve for $x$ where $\sin(20^{\circ}+x)=2\cos40^{\circ}\sin{x}$.

[hide="Solution"]\begin{eqnarray*}&& \sin(20^\circ+x)=2\cos 40^\circ\sin x\\ &\iff& \sin 20^\circ\cos x+\cos 20^\circ\sin x=2\cos 40^\circ\sin x\\ &\iff& \sin 20^\circ\cos x=\sin x(2\cos 40^\circ-\cos 20^\circ)\\ &\iff& \tan x=\frac{\sin 20^\circ}{2\cos 40^\circ-\cos 20^\circ}\\ &\iff& \tan x=\frac{\sin 20^\circ}{2\cos (60^\circ-20^\circ)-\cos 20^\circ}\\ &\iff& \tan x=\frac{\sin 20^\circ}{2(\cos 60^\circ\cos 20^\circ+\sin 60^\circ\sin 20^\circ)-\cos 20^\circ}\\ &\iff& \tan x=\frac{\sin 20^\circ}{\cos 20^\circ+\sqrt{3}\sin 20^\circ-\cos 20^\circ}\\ &\iff& \tan x={1\over\sqrt{3}}\\ &\iff& x=30^\circ+k\cdot 180^\circ, k\in\mathbb{Z}\end{eqnarray*}[/hide]

[color=green]Different solution[/color]
[hide]$\sin(x+20^{o})=\sin(x+40^{o})+\sin(x-40^{o})$
$\sin(x+20^{o})-\sin(x-40^{o})=\sin(x+40^{o})$
$2\sin30^{o}\cos(x-10^{o})=\sin(x+40^{o})=\cos(50^{o}-x)$
$x-10^{o}=50^{o}-x+k.360^{o}$
$x=30^{o}+k*180$
This equation comes from a famous geometry problem do you know it?[/hide]


arithmetic sequence
algebra unsolved
algebra

[quote="xxxxtt"]How many progressions with 4 elements have the elements in the set {1,2,...,n}? :)[/quote]

Your problem is not very precise and many questions may be asked :
- are you speaking about arithmetic progressions, geometric progressions or both ?
- are you speaking about "4 elements in progression" or "progressions with exactly 4 elements in the set" ($ \{2,3,4,5\}$ would match the first, but not the second) ?
- are you speaking about "progressions" or "elements in progression". In the first case $ \{1,2,3,4\}$ must be count twice (common difference $ \plus{}1$ and common difference $ \minus{}1$). In the second case, this set must be counted once.
...

In the case where the problem is "Count the number of distinct ordered subsets of positive integers $ \{a,b,c,d\}$ such that $ 1\leq a<b<c<d\leq n$ and these elements belong to at least one arithmetic progression" :

For a given lowest element $ 1\leq a\leq n$, the number of such subsets is the number of positive integers $ r$ such that $ a\plus{}3r\leq n$, so is $ [\frac{n\minus{}a}3]$.

So the required count is $ f(n)\equal{}\sum_{k\equal{}1}^n[\frac{n\minus{}k}3]$

So, $ f(1)\equal{}f(2)\equal{}f(3)\equal{}0$ and $ f(n\plus{}3)\equal{}n\plus{}f(n)$

So, if $ n\equal{}3p\plus{}s>0$, with $ s\in\{1,2,3\}$, $ f(n)\equal{}\frac 32p(p\minus{}1)\plus{}ps$

IMHO


geometry
inequalities
rearrangement inequality
geometry unsolved

Prove :For all triangular we have:
 $ \frac{1}{2Rr}\leq \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}$

We have $ \frac {(a \plus{} b \plus{} c)r}{2} \equal{} K \equal{} \frac {abc}{4R}$, where $ K$ is the area of the triangle. from this we get $ \frac {1}{2Rr} \equal{} \frac {a \plus{} b \plus{} c}{abc} \equal{} \frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca}$, so the inequality to prove is
\[ \frac {1}{ab} \plus{} \frac{1}{bc} \plus{} \frac{1}{ca}\leq \frac{1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\]
which is obvious by the Rearrangement Inequality (or also by AM-GM).

equality occurs iff $ a \equal{} b \equal{} c$.


hey guys,

another year is over, so we have to do the yearly joke of course ;)
and as usually there were some surprising changes in the German team.
First the team and after this a few words.

Peter Scholze (he did it this year^^, i think you know him)
Friedrich Feuerstein (was at the IMO a few years ago but not in the last 3 years)
Georg Schröter (aka Naphthalin, first participation)
Stephan Neupert (also first one)
Lisa Sauermann (the first girl after about 5 years, first participation)
Malte Lackmann (aka Malte Luck-Man, first one again)

Lisa Sauermann is just 14 years old and can particpate till 2011 so you can await something  :P
surprisingly, christian sattler (aka NoName, golg medal last year) wasn't able to qualify for the IMO. also 2 guys who would have been in the team after the first 6 of 7 exams were "thrown" out...
and also meru alagalingam who was in the successfull team last year ([url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88832]look here[/url]) couldn't qualify although he was 3rd in the pre-selection in december.

because i am too old for this IMO, i wish the members of the team a lot of fun.

 :spam:   :rotfl:

dg

Just one question: Are you darij grinberg? :D

[quote="NoName"]Can't believe Christian Sattler didn't make it into the team :o[/quote]

Yeah, too bad that current students from LMU Munich cannot participate in their 14th year of school.  :D

but this wasn't unforseeable. last year he was nearly disqualified from the IMO because he did some n00by things during the TST's. But I don't know whether I am allowed to tell this to you...

Naphthalin

[quote="Valentin Vornicu"][quote="NoName"]Can't believe Christian Sattler didn't make it into the team :o[/quote]I heard he was so noob that he couldn't solve $\sum \frac a{b+c}\geq \frac{3}{2}$. So, definitely the medal from last year was revoked by the IMO Committee and he was forbidden to ever participate in the IMO (including IMO 2009 in Germany, for which he will have to be in another country, not to disturb the students).

It is unfortunate that a good member of the AoPS-ML community has received such a harsh treatment, nevertheless we agree with the punishment in his case.[/quote]

Wait...he got his medal revoked because he couldn't solve some inequality...?

Then how did he get on the team in the first place? I don't actually know the rules for IMO, but it's not like I will even make it; I haven't even made it to the USAMO yet :P.

[quote="Zellex"]Wait...he got his medal revoked because he couldn't solve some inequality...?[/quote]Unfortunately, yes  :(

[hide=":whistling:"]You do realize this thread is a hoax, right? :)[/hide]

[quote="Zellex"]Wait...he got his medal revoked because he couldn't solve some inequality...?[/quote]
I heard his medal was revoked, because he tried to achieve a second medal at the same IMO as "GER7: Christian Stettle".

[quote="Valentin Vornicu"][quote="Zellex"]Wait...he got his medal revoked because he couldn't solve some inequality...?[/quote]Unfortunately, yes  :(

[hide=":whistling:"]You do realize this thread is a hoax, right?[/hide][/quote]

How sad... wish him good luck next time  :D

[quote="Valentin Vornicu"]I heard he was so noob that he couldn't solve... [/quote]

Sorry for the stupid question but... What means [b]noob[/b]?  (i never heard this before)

$Tipe$

[quote="Zellex"]noob is an internet term. I think it developed from games. It basically transferred from "newb", who is a new player, or beginner.[/quote]

and doesn't know so much --> makes many stupid mistakes

It actually comes from "newbie":http://en.wikipedia.org/wiki/Newbie
That punishment seems unfair to me.


Call the positive integer selfdivisible if it is divisible by each sum of its consecutive digits, in particular it is divisible by
each of its digits. Prove that the set of selfdivisible integers is finite.

Isn't it infinite because all of the numbers consisting of only 1's are selfdivisible?

Examples:  $1$, $1111$, $1111111111111111111111111111$

[quote="rogue"] if it is divisible [b]by each sum of its consecutive digits[/b], in particular it is divisible by
each of its digits. [/quote]

For example, $1111$ is not divisible by $1+1$.

Are we to assume that $0$ does not divide any integer?

[quote="t0rajir0u"]
Are we to assume that $0$ does not divide any integer?[/quote]

Yes.

[quote="cincodemayo5590"]i think its infinite....

$24$, $2222 4$, $2222222 4$, etc are divisible by $2+2=4$ and $2+4=6$. Just keep adding three $2$'s to the beginning and it's always divisbile by 4 and 6.

Unless you mean that the sums can e taken of three or more digits...[/quote]

[b]Every[/b] set of consecutive digits.  So, for example, $2 + 2 + 2 + 2 + 2 = 10$ does not divide $22222224$.


Consider a 6x6 grid.
A)Find the number of squares of all sizes in this grid. Then generalize your result to an n x n grid.
B) Find the number of rectangles of all sizes in this grid. Then generalize your result to an n x n grid.

Part b)

First start by doing small values for $ n$,

When $ n \equal{} 1$ there are $ 1$ squares.
When $ n \equal{} 2$ there are $ 9$ squares.
When $ n \equal{} 3$ there are $ 36$ squares.

Now we can start seeing a pattern:
$ 1 \equal{} 1^3 \equal{} 1^2$

$ 1 \plus{} 8 \equal{} 1 \plus{} 2^3 \equal{} 9 \equal{} (1 \plus{} 2)^2$

$ 9 \plus{} 27 \equal{} (1 \plus{} 2^3) \plus{} 3^3 \equal{} 36 \equal{} (1 \plus{} 2 \plus{} 3)^2$

From this we can start to assume that for the $ n$ x $ n$ grid the amount of squares of all sizes is equal to the sum of the first $ n$ cubes or the equivalent, sum of the first $ n$ natural numbers, squared. So when $ n \equal{} 6$, the amount squares of all sizes is $ 1^3 \plus{} 2^3 \plus{} 3^3 \plus{} 4^3 \plus{} 5^3 \plus{} 6^3 \equal{} (1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} 6)^2 \equal{} 21^2 \equal{} 441$.

And for the $ n$ by $ n$ grid it's:

$ 1^3 \plus{} 2^3 \plus{} 3^3 \plus{} \cdots \plus{} n^3 \equal{} (1 \plus{} 2 \plus{} 3 \plus{} \cdots \plus{} n)^2 \equal{} \left(\dfrac{n(n \plus{} 1)}{2}\right)^2$

Edit: oops thanks, tennisperson3.

Part b)

You have to choose 2 rows, for the edges of the rectangle, and there are $ n\plus{}1$ rows.  Same goes with the columns.  Thus the answer is $ \left(\frac{n(n\plus{}1)}{2}\right)^2$.  For 6 this gives 441.

tonypr:  Unfortunately, for part a, you actually did part b instead.

Part a)

First, we choose the side length for the square.  This gives $ n$ choices, $ 1\minus{}n$.  If we choose $ k$ for this, the first $ n\minus{}k\plus{}1$ rows and first $ n\minus{}k\plus{}1$ columns can have the top corner.  Thus we have $ (n\minus{}k\plus{}1)^2$ possibilities for a square of $ k$ length.  Taking $ k$ from 1 to $ n$, we have $ n^2\plus{}(n\minus{}1)^2\plus{}...\plus{}1\equal{}\frac{n(n\plus{}1)(2n\plus{}1)}{6}$.  For 6, this gives 91.

Chrispy might have been asking about a 6x6 grid of points, not of squares.
\[ .\ .\ .\ .\ .\ .\\
.\ .\ .\ .\ .\ .\\
.\ .\ .\ .\ .\  .\\
.\ .\ .\ .\ .\ .\\
.\ .\ .\ .\ .\ .\\
.\ .\ .\ .\ .\ .
\]

It turns out that the formula for the number of squares (not necessarily integer side lengths) in this grid is $ \frac{n^2(n^2\minus{}1)}{12}$. We can prove this by considering the number of squares inscribed in a 1x1, 2x2, etc. and using induction.

I don't know the answer to b) for such a grid of points.

If it's a grid of points then all you have to do is just subtract by one on the formula for part b, so it would be $ \left(\dfrac{(n \minus{} 1)(n)}{2}\right)^2$

That formula doesn't work for, for example, $ n\equal{}3$.

For squares at least, it's not too hard to also count if they don't need to be orthogonal:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=155463

Like ThinkFlow, I don't know what the answer should be for rectangles, but perhaps someone can extend the method from the thread I've linked?


Hello,

How could I make a footer with the page number and some text and how can I make the foot note and can I link the number in the foot note to the note down?

Thanks in advance

[quote="hasan4444"]
How could I make a footer with the page number and some text[/quote][[LaTeX:Packages#fancyhdr]] which includes \cfoot{\thepage} which puts the page numbering the footer

[quote="hasan4444"]
and how can I make the foot note and can I link the number in the foot note to the note down?[/quote]Put \usepackage[hyperfootnotes]{hyperref} in preamble (see [url=http://www.tug.org/applications/hyperref/manual.html]hyperref manual[/url])


MATHCOUNTS
geometry
geometric transformation

To all of you going to nationals, wear a t-shirt with your username on it. I'll have DeeVee7 on mine.  Just post what you will put so we can recognize each other.

Thanks,
Karan Batra

I'll go write bob123 on my AOPS t-shirt :)

You guys can make 4 soccer teams and have a small championship :D :D

[quote="kchande"]Bob123, are you from CA? I hope to see many AOPSers at nationals, 49 exactly with names on their shirts.[/quote]

I'm #2 from CA... #1 is piggypi314, #3 is NoSoupForYou, #4 is aznness

[quote="Treething"]I'll just tape it to my shirt or something like that.[/quote]

Y'know, that's a good idea...

You can put it anywhere as long as people can see it.
BTW: not under a shirt

I will either be wearing my AoPS shirt with my username embroidered on it, or my Reader's Digest National Word Power Finalist shirt. :P

Billy

will it have your username on it?
it should

[quote="09husbri"]You can bet I'll be wearing my Seinfeld shirt with a picture of Kramer on it...or my United Nations T-shirt...or my chapter MathCounts shirt from 6th grade...or my "Deny Everything" DC Spy Museum shirt...or - forget it.[/quote]
Good idea (The forget it part)

...says "NoSoupForYou"... :lol:

No way, man! Seinfeld is [i]forever![/i]

Alex, i like ur avatar. However, mine is better.

[quote="NoSoupForYou"][quote="09husbri"]You can bet I'll be wearing my Seinfeld shirt with a picture of Kramer on it...or my United Nations T-shirt...or my chapter MathCounts shirt from 6th grade...or my "Deny Everything" DC Spy Museum shirt...or - forget it.[/quote]

A seinfeld shirt? Keep up with the times man - Seinfeld is so 90's  :D[/quote]

Last I recall: "No Soup for You!" - Soup Nazi... from Seinfeld....
GOOD LUCK AT NATIONALS! I could've made it, but being from texas where Houston area schools pwn, it was a long shot... got 39th though, with a pathetic score (because of target round... missed four of those, did well on sprint) of 28! My lowest ever

[quote="shake9991"][quote="NoSoupForYou"][quote="09husbri"]You can bet I'll be wearing my Seinfeld shirt with a picture of Kramer on it...or my United Nations T-shirt...or my chapter MathCounts shirt from 6th grade...or my "Deny Everything" DC Spy Museum shirt...or - forget it.[/quote]

A seinfeld shirt? Keep up with the times man - Seinfeld is so 90's  :D[/quote]

Last I recall: "No Soup for You!" - Soup Nazi... from Seinfeld.... [/quote]

*sigh*  Does :D mean anything to you?

[quote="09husbri"][quote="NoSoupForYou"]No. Tengo arroz pero no tengo leche. No tengo gusto de la leche. Usted consiguió pwned.

Or in Dutch: Nr. Ik heb krullen maar ik heb geen melk. Ik houd van geen melk. U kreeg pwned.

Or French: Non. Je prends le riz mais je ne prends pas le lait. Je n'aime pas le lait. Vous avez obtenu pwned.

Or German: Nein. Ich habe Wellungen, aber ich habe nicht Milch. Ich mag nicht Milch. Sie erhielten pwned.

Maybe Italian: No. Ho arricciature ma non ho latte. Non gradisco il latte. Avete ottenuto pwned.

Perhaps Russian =): Нет. Я имею рис но я не имею молоко. Я не люблю молоко. Вы получили pwned.

Or Potuguese: Não. Eu tenho ondas mas eu não tenho o leite. Eu não gosto do leite. Você começou pwned.

Φόβος μεταφραστών οι δεξιότητές μου! (Phear my translatoring skillz!)[/quote]
 :o Tell me you used some sort of electronic translator or a dictionary - or [i]soemthing[/i]. If you knew two or maybe even three of those languages, it would be (semi-)normal, but - whoa!

PS Three years earlier and I could have given you the Hebrew translation, but I haven't been to Hebrew school in a loooong time.

PPS Why am I telling this to somebody with a Nazi in his avatar? ;) Just kidding. You don't hate Jews, of course. You just don't give them soup.

PPPS Don't take offense. I know you're not a Nazi.[/quote]

lol, NoSoupForYou is jewish...

BTW, Alex, where'd you get that picture, it looks cool :)

[quote="NoSoupForYou"]husbri: I used [url=http://www.worldlingo.com/en/products_services/worldlingo_translator.html]WorldLingo[/url].

bob: What picture?[/quote]

your avatar


logarithms
real analysis
real analysis unsolved

working on solving an unknown sequance equation...
i have the (X,Y) exact curve ....but it seems that it's formula is unknown yet..
need ..assistance....(f(x)=????)

Looks like $y=Cx^{-p}$ for some $C,p>0$. Try to draw $\ln y$ versus $\ln x$. If you get a straight line, the dependence is, indeed, power.


I've been wondering about this function recently:

f(n,x,m) = the number of x's in the nth row of Pascal's Triangle mod m. That is, take each entry in Pascal's Triangle and reduce it modulo m, go to the nth row, and count how many occurences of the residue x there are.

For example, the 5th row of Pascal's Triangle, 1-5-10-10-5-1, when reduced mod 3, becomes 1-2-1-1-2-1. Then

f(5,0,3) = 0
f(5,1,3) = 4
f(5,2,3) = 2

I found some research out there on this when m = 2, but I was wondering how to go about determining this function for other moduli. I think there may just be a recurrence for it, or something.

A similar challenging question: look at the so-called "shallow diagonals" or "Fibonacci diagonals" of Pascal's Triangle, and find the same type of information. That is, get a recursion or some sort of closed form for

g(n,x,m) = the number of x's in the nth Fibonacci diagonal of Pascal's Triangle mod m. For example, the 7th diagonal, 1-5-6-1, when reduced modulo 4, gives us 1-1-2-1, and thus

g(7,0,4) = 0
g(7,1,4) = 3
g(7,2,4) = 1
g(7,3,4) = 0

I am hoping someone knows how to determine at least some information about either of these functions, because I can't think of how to go about solving for them. I also think the Fibonacci diagonal one is harder, by the way.

I would really appreciate any help - Thanks!

Let $p$ be a prime. Some useful identities:
If $p\nmid k$, $\binom{pn}{k}\equiv 0\mod p$
If $0\le i,j<p$, $\binom{pn+i}{pk+j}\equiv \binom{pn}{pk}\cdot\binom{i}{j}\mod p$
$\binom{pn}{pk}\equiv \binom{n}{k}\mod p$

Try to prove these if you can- they aren't too hard.

The consequence: If $n=\sum_{i=0}^{r}a_{i}p^{i}$ and $k=\sum_{i=0}^{r}b_{i}p^{i}$ are written in base $p$ ($0\le a_{i},b_{i}<p$), $\binom{n}{k}\equiv \prod_{i=0}^{r}\binom{a_{i}}{b_{i}}\mod p$. Once we know the first $p$ rows, we can generate everything else fairly easily.

If we aren't looking at things mod a prime, it gets much more complicated and annoying.


If $ x,y,z>0$ , then :

$ \sum_{cyc}\frac{x^5}{yz}\plus{}2\sum_{cyc}\frac{x^2y^2}{z}\ge \sum_{cyc}x^3\plus{}\sum_{sym}\frac{x^4}{z}$

$ \sum_{cyc}\frac{x^{5}}{yz}\plus{}2\sum_{cyc}\frac{x^{2}y^{2}}{z}\ge\sum_{cyc}x^{3}\plus{}\sum_{cyc}\frac{x^{4}\plus{}y^4}{z}$

 is equivalent to

 $ \sum{(x^2\minus{}yz)^2(x\minus{}y)(x\minus{}z)}\plus{}\sum{(yz)^2(x\minus{}y)(x\minus{}z)}\ge{0},$

it is obvious  holds.


Is there a way to do this problem [b]without dividing the entire thing out[/b]? No calculators are allowed.

How many times does the digit 8 appear in the first 2007 digits of the decimal expansion of 1/19?

Ans
[hide]222[/hide]

The answer is incorrect, the decimal is 0.052631578947368421.... where the part I put keeps repeating so there will be an infinite number of 8s.

heh... oops...
I meant the first 2007 digits of the decimal expansion.

EDIT:
iniquitus, I said I need a way without long division, but thanks anyways...

Notice that $ \frac{52631578947368421}{999999999999999999} \equal{} \frac{1}{19}$ by long division; therefore, in every 18 decimal places there are two 8's.

$ 2007 \div 19 \equal{} 111 R9$, so there are $ 2\times 111 \plus{} 1 \equal{} 223$ 8's??

Long division doesn't take that long; in any case, if you really wanted to do it quickly, you already know that the digits will repeat every $ 18$ digits because of [url=http://en.wikipedia.org/wiki/Fermat%27s_little_theorem]Fermat's little theorem[/url] and you can guess that approximately $ 2$ of those digits will be eights.

[quote="iniquitus"]Notice that $ \frac {52631578947368421}{999999999999999999} \equal{} \frac {1}{19}$ by long division; therefore, in every 18 decimal places there are two 8's.

$ 2007 \div 19 \equal{} 111 R9$, so there are $ 2\times 111 \plus{} 1 \equal{} 223$ 8's??[/quote]

Don't you mean $ 2007 \div 18 \equal{} 111 R9$?  by the way, I think this is correct.


Let $ x,y>0$ with $ xy\geq 1$ with $ xy\equal{}1$ if and only if $ x\equal{}y$ and $ x\geq y$ . Prove or disprove the following inequality

$ 2x^2y^4\plus{}x^3y^2\geq 2x\plus{}y^2$

I think that this inequality is not true. For example taking $ y \equal{} \frac {5}{4x}$ we land in an inequality:
$ 2*(\frac {5}{4})^{2}*\frac {1}{x^{2}}\geq \frac {7}{16}x \plus{} (\frac {5}{4x})^{2}$ which is not true for sufficiently big x :wink:

If $ y\equal{} \frac{5}{4x}$ then $ xy \equal{} \frac{5}{4} \neq 1$

Yes, but what the problem states is that:
I)$ xy\geq 1$
II)$ x\geq y$
III)$ (xy\equal{}1)\iff (x\equal{}y)$
At least I understood it like that :wink:

[quote="polskimisiek"]Yes, but what the problem states is that:
I)$ xy\geq 1$
II)$ x\geq y$
III)$ (xy \equal{} 1)\iff (x \equal{} y)$
At least I understood it like that :wink:[/quote]
I think the problem says
III) $ (xy \equal{} 1)\iff ((x \equal{} y)$OR$ (x \ge y)$.

[quote="archimedes1"]
I think the problem says
III) $ (xy \equal{} 1)\iff ((x \equal{} y)$OR$ (x \ge y)$.[/quote]That just wouldn't make since. I think the first interpretation is correct.


Hello to all! We have a very difficult homework and I don't know to answer it. I thing it is about Quadratic Equation. Here is the word problem:

[b]Genelle and Eljen can paint a room together in 4 hours. Working alone, Genelle can paint the room in two hours less than Eljen can. Find how long it takes Eljen to paint the room alone.[/b]

Please, I am begging you… I want the answer and how to solve it……
Thank You Very much and your right answer could be greatly appreciated.
Thank You once again….

[hide="Solution"]
Let $ g$ be the rate at which Genelle paints and let $ e$ be the rate at which Eljen paints.
Let $ t$ be the time it takes Eljen to paint the room.

Then,
$ 4(g\plus{}e)\equal{}1$
$ (t\minus{}2)g\equal{}1$
$ te\equal{}1$

Solving the equations,
$ 4(\frac{1}{t\minus{}2}\plus{}\frac{1}{t})\equal{}1$
$ 4(t\plus{}(t\minus{}2))\equal{}t(t\minus{}2)$
$ 8t\minus{}8\equal{}t^{2}\minus{}2t$
$ t^{2}\minus{}10t\plus{}8\equal{}0$
$ t\equal{}5\pm \sqrt{17}$

If $ t\equal{}5\minus{} \sqrt{17}$ then $ t\minus{}2<0$
Thus, $ t\equal{}5\pm \sqrt{17}$

Hope this helps...
[/hide]

Looking back at the problem, $ 5\minus{}\sqrt{17}$ is obviously not a solution ($ t\ge4$). Assuming mismatchtea is correct, the only solution is $ 5\plus{}\sqrt{17}$.


AMC
AIME
AMC 10
geometry
email
AMC 10 B
trigonometry

The official scores have been released?  Did everyone make it into AIME? :)

well.. now that the cutoff is 110.. YES!! lol =)

have score reports come out? or do you just know the cutoff and assume that you made it?

[quote="dnbmathguy"]123 on the 12s.

I heard the cutoff for the 10's was 119?[/quote]

No, the cutoff for the AMC10 was 110.

It would be awesome if the cutoff really is 110.

No... They really did send out the official scores.. I got em yesterday!

ok. thanks. i just took tha amc-10 and i'm not allowed to discuss it right now but i don't think i made it.

yea i made it through 10A too... 139.5

i don't think i did as good on B though... was it just me or was 10B harder than 10A??

I think that the geometry problems (#24 and #20) on AMC 10B are harder than the geometry problems oan the A exam......

Think I did alright on the amc 12B
i made it to the aime...120.5 on the 12A i think
maybe a couple of points higher on the 12B

The last few problems in 12B were definitely harder than the last ones on 12A

[quote="tetrahedr0n"]amazing! the similarities continue mathfan: on 20, i took a ruler and measured the diagram, and my measurement yielded very close to the correct d. then i filled in e on my sheet like an idiot :(. I similarly brute-forced 24 and scored points on it. why is 25 easier than 24 i wonder?[/quote]

Interestingly enough, I tried to do #20 with a ruler and diagram, but I didn't get anywhere so I just guessed C.  I considered guessing D, but I figured I'd go with my first guess, which was C.  Bah.

And 25 was pretty easy compared to 20 and 24...hmm...

[quote="Syntax Error"]I still don't remember!  Someone post the question please.[/quote]

In triangle ABC, points D and E lie on BC and AC, respectively.  If AD and BE interesect at T so that AT/DT = 3 and BT/ET = 4, what is CD/BD?

what did you get on the 12 A mystic? though it was a lot easier i just did a lot worse

25 is often not as hard as the others -- two years ago, the AMC 12A had an immensely easy question 25, if you know anything about polynomials.  It was a 1-line solution.

[quote="tetrahedr0n"]amazing! the similarities continue mathfan: on 20, i took a ruler and measured the diagram, and my measurement yielded very close to the correct d. then i filled in e on my sheet like an idiot :(. I similarly brute-forced 24 and scored points on it. why is 25 easier than 24 i wonder?[/quote]

the AMC people probably ordered the problems in such way to fool students. notice that number 25 is on the last page, alone by itself........

I don't think they did that to fool students.  In general, the AMC isn't out to fool you (although many people do get fooled, nonetheless).  For instance, the answer choices often don't include answers that people could arrive at by making a stupid mistake.  I know I've been saved before by not seeing my final answer as one of the choices...

[quote]For instance, the answer choices often don't include answers that people could arrive at by making a stupid mistake.[/quote]

How untrue (thinks back to AMC 12A #1, AMC 12B #20). :(

well, I took about 8 minutes (while in time trouble) trying to solve #20 without a clue until I turned my attention to #25 and solved it in about a minute... I really think that #25 should replace #20's position...

[quote="Fierytycoon"]How untrue (thinks back to AMC 12A #1, AMC 12B #20). :([/quote]

Compare to the exams from 15-20 years ago (I don't remember exactly when they switched they answer strategies).  They hardly ever use answers that result primarily from simple errors like arithmetic mistakes, but generally more complicated subtle mistakes or fundamental misunderstandings.

In general they don't want someone to get a problem wrong because they made a very minor (and mathematically unimportant) error. If people get problems wrong, it should be because they are guessing or because they made more fundamental errors. The whole point of AMC12A #1 was to make sure you actually read the question rather than guess what it's asking. They want you to do that, and I think that's reasonable.

[quote="texas137"]Are people saying that their teachers have received the official scores and list of AIME qualifiers by email, or just that the answers and cut-offs are available and people know their scores unofficially? My group has not received official scores, although AMC has an email address to send them to. Now I'm worrying that the answer sheets got lost in the mail...... :-([/quote]

I believe they're saying the former.  My teacher received our scores via e-mail today at lunch.  Considering that some people (e.g. Ragingg) got their scores several days back, it's likely that some scores are being sent out.  You'll probably get yours soon.

[quote="mathfanatic"]My teacher recieved our scores via e-mail today at lunch.  Considering that some people (e.g. Ragingg) got their scores several days back, it's likely that some scores are being sent out.  You'll probably get yours soon.[/quote]

I remember back in November after the AMC 8 the scores were not all emailed on the same day. I don't know what the queueing protocol is, but there was a several day span for sending out official emails with AMC 8 score notifications. I wouldn't worry yet--but I might worry next week if I still hadn't seen an A score.

I'm trying to calm my worries about a summer program application I submitted for my son, which was supposed to result in a confirmation of receipt postcard by now. I have to keep telling myself, "No, it's not lost in the mail; it's just that the mail is real slow."

Aside to Mathfanatic: I recall that you desire to practice good grammar, so find your spelling mistake in the quoted text above, and correct it by editing your original post. I'll probably make a spelling mistake without noticing it in this post, just so we can tease each other about this issue.

Best wishes to everyone who is still in suspense about scores.

In response to questions about emailing of school reports:  The email reports are still going out. and probably will go out for a couple more (work) days.  The very first paper reports were prepared on Friday, and I believe they went in the mail on Friday afternoon.

Here is an an overview of the process:  It takes several days for your answer sheets to even reach us after the contest.  For the AMC A on Tues Feb 10, it was not until Friday Feb 13, that we had substantial numbers of contests in the building, and it was Tuesday, Feb 17 (remember Feb 16 was a federal holiday) that we finally had the majority.  We are still receiving some A packages from distant parts of the globe.  When I left the office on Friday, we had only received about 20 or so B packages.

After receiving the packages, they are organized and grouped by states, and then opened and checked off the registration lists to determine which we have not received packages from.  The teacher certification form is also checked to determine if all rules were followed and that no irregularities occurred.  With  3000-4000 packages this takes several days.

The answer sheets are then scanned.  Our scanner is very touchy, and we have had two (very expensive) visitis from the repairman, and that slowed us  down some.  We generally scan all results from a state together, and generally largest enrollment states first.  The schools within each state are in random order.  After scanning, we must still "edit" the data, since improperly  filled in dots on the answer sheets leave gaps in our data.  If a student does not bubble in the name data correctly, or uses an ink pen (not a lead pencil)  then we must fill in the gaps.

AfTempter editing, we run the data and start actual scoring.  This year, this started on Monday Feb 23.   We have to score a substantial number to determine the invitation levels.  Only after all tht can we send the school reports by email, and of course, only if the school and contest manager provided an email address.  Because of the chances of organization, scoring by state, the volume of editing, it seems we can get out about 600 reports by email per day.  Therefore, it takes 5-8 (work) days to get all the reports completed by email.  Packaging the paper reports is slow and painstaking because each package has individualized contents.

We make every effort to get your results to you as quickly as possible, if you have not received your results yet, rest assured that they are coming.

Steve Dunbar
AMC Director

to Steve Dunbar - thanks for posting this. I know I found it interesting and informative!


algebra
polynomial
complex numbers
greatest common divisor
IMC
college contests

Let $p(z)$ be a polynomial of degree $n$, $n \geq 1$ with complex coefficeints and $a_1, a_2,...,a_k$ be $k$, $k \geq 2$ distinct complex numbers.
Prove that exist at least $(k-1)n+1$ complex numbers $z$ for which $p(z)$ is $a_1$ or $a_2$ or ... $a_k$.

The number of distinct roots of a polynomial $P$ is $n-\deg(P,P')$, where $(f,g)$ denotes, as usual, the greatest common divisor of $f$ and $g$. This means that the number of distnct roots of $P+a_i,\ i\in\overline{1,k}$ is $\sum_{i=1}^k[n-\deg(P+a_i,P')]=kn-\sum_i\deg(P+a_i,P')\ (*)$.

Now since $P+a_i$ are coprime, so are $(P+a_i,P')$, and since each one of them divides $P'$, their product must also divide $P'$, which has degree $n-1$. This means that $\sum_i\deg(P+a_i,P')\le n-1$, so $(*)\ge kn-(n-1)=(k-1)n+1$, as desired.


geometry
incenter
circumcircle
trigonometry
inequalities
geometry proposed

Let $ I$ and $ O$ denote respectively the incenter and the circumcenter of a triangle $ ABC$. Given that $ \measuredangle AIO \equal{} 90^{\circ}$, prove that the area of the triangle $ ABC$ is less than $ \frac {3\sqrt 3}{4}AI^2$

$ \angle AIO \equal{} 90^\circ \ \Longleftrightarrow \ \frac{r^2}{\sin^2 {\frac{A}{2}}} \plus{} R^2 \minus{} 2Rr \equal{} R^2 \ \Longleftrightarrow \ r \equal{} 2R \sin^2 {\frac{A}{2}}$

$ S <\frac{3 \sqrt{3}}{4} AI^2 \ \Longleftrightarrow \ p r < \frac{3 \sqrt{3}}{4} \frac{r^2}{\sin^2 {\frac{A}{2}}} \ \Longleftrightarrow \ p < \frac{3 \sqrt{3}}{2} R$ that is know

Why?
[quote="¬[ƒ(Gabriel)³²¹º]¼"]$ \angle AIO \equal{} 90^\circ \ \Longleftrightarrow \ \frac {r^2}{\sin^2 {\frac {A}{2}}} \plus{} R^2 \minus{} 2Rr \equal{} R^2 $[/quote]
How can solve this [quote]$ p < \frac {3 \sqrt {3}}{2} R$ that is know[/quote]

[quote="every"]Why?
[quote="¬[ƒ(Gabriel)³²¹º]¼"]$ \angle AIO \equal{} 90^\circ \ \Longleftrightarrow \ \frac {r^2}{\sin^2 {\frac {A}{2}}} \plus{} R^2 \minus{} 2Rr \equal{} R^2$[/quote][/quote]
For pytagora on $ \triangle OIA$ : $ AO^2 \equal{} OI^2 \plus{}AI^2$

[quote="every"]How can solve this [quote]$ p < \frac {3 \sqrt {3}}{2} R$ that is know[/quote][/quote]

$ p \equal{} 4R \cos \frac{A}{2}  \cos \frac{B}{2}  \cos \frac{C}{2}$ so the inequality become: $ \cos \frac{A}{2}  \cos \frac{B}{2}  \cos \frac{C}{2} \le  \frac {3 \sqrt {3}}{8}$

now $ \frac{a}{b\plus{}b}\equal{}\frac{\sin A}{\sin B \plus{} \sin C} \equal{} \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{2 \sin \frac{B\plus{}C}{2} \cos \frac{B\minus{}C}{2}}\equal{}\frac{\sin \frac{A}{2}}{\cos \frac{B\minus{}C}{2}} \ge \sin \frac{A}{2}$ with equality when $ B \equal{} C$

so $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \le \frac{1}{8}$ with equality when $ A\equal{}B\equal{}C$

It's know also $ \sin^2 \frac{A}{2} \plus{} \sin^2 \frac{B}{2} \plus{} \sin^2 \frac{C}{2} \plus{} 2\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \equal{}1$ in every triangle.

So we obtain $ \sin^2 \frac{A}{2} \plus{} \sin^2 \frac{B}{2} \plus{} \sin^2 \frac{C}{2} \ge \frac{3}{4}$ with equality when $ A\equal{}B\equal{}C$.

that is equal to $ \frac{9}{4} \ge \cos^2 \frac{A}{2} \plus{} \cos^2 \frac{B}{2} \plus{} \cos^2 \frac{C}{2} \ge 3 \sqrt[3]{\cos^2 \frac{A}{2} \cos^2 \frac{B}{2} \cos^2 \frac{C}{2}}$ by AM-GM.

so $ \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \le \frac{3\sqrt{3}}{8}$ with equality when $ A\equal{}B\equal{}C$


parameterization Vieta quadratics algebra

Source: Bulgarian MO 1980
1. For a real parameter $p \neq 0$, let $x_1,x_2$ be the roots of the equation $x^2+px-\frac{1}{2p^2}=0$. Prove that $x_1^4+x_2^4\geq 2+\sqrt{2}$.

[quote="rem"]Now let's prove that $p^4+\frac{1}{2p^4} \ge \sqrt{2}$.
We have the ineq equivalent to:
$p^4-\sqrt{2}+\frac{1}{2p^4} \ge 0$
$(p^2)^2-2*p^2*\frac{1}{\sqrt{2}p^2}+(\frac{1} {\sqrt{2}p^2})^2 \ge 0$
$(p^2-\frac{1}{\sqrt{2}p^2})^2 \ge 0$
Therefore $p^4+\frac{1}{2p^4} \ge \sqrt{2}$[/quote]
You want to switch that around. To prove it, you take a true statement and then make it into the thing you want to prove.
you would say:
Now let's prove that $p^4+\frac1{2p^4}\ge\sqrt2$
We know $\left(p^2-\frac1{\sqrt2p^2}\right)^2\ge0$
so ${\left(p^2\right)}^2-2\cdot p^2\cdot\frac1{\sqrt2p^2}+{\left(\frac1 {\sqrt{2}p^2}\right)}^2\ge0$
then $p^4-\sqrt2+\frac1{2p^4}\ge0$
therefore $p^4+\frac1{2p^4}\ge\sqrt2$.

You can also use the fact that every ineq is equivalent to the one below it.eg:
"$a^b+b^2=2ab$ is equivalent to $a^2-2ab+b^2 \ge 0$ which is equivalent to $(a+b)^2 \ge 0$", which holds.QED
Sorry I didnot write the part "the ineq is equivalent to" in my solution. :blush:
btw the way "the ineq is equivalent to" is more effective, because if you are on a contest running out of time and have no time for rough work, you will use this method and hope you will arrive at the result insead of solving the ineq and doing it your way, which would take more time.
But if you have time, either way is fine. :)

OMG how close I was .......
I gave up after I got $p^4+\frac{1}{2p^4}\geq \sqrt{2}$.
Anyway, thanks

is not $p^4 + \frac{1}{2p^4} \geq \sqrt{2}$ by AMGM?

Here is my solution.
Since $p\neq 0,$
$x^4=\left(x^2+px-\frac{1}{2p^2}\right)\left(x^2-px+p^2+\frac{1}{2p^2}\right)-\left(p^3+\frac{1}{p}\right)x+\frac{1}{4p^4}+\frac{1}{2}.$

Since $x_i\ (i=1,2)$ is the solution of the given quadratic equation, we have $x_i^2+px_i-\frac{1}{2p^2}=0.$ Thus
$x_1^4+x_2^4=-\left(p^3+\frac{1}{p}\right)(x_1+x_2)+2\left(\frac{1}{4p^4+\frac{1}{2}}\right)$
$=-\left(p^3+\frac{1}{p}\right)(-p)+2\left(\frac{1}{4p^4}+\frac{1}{2}\right)\ \because{x_1+x_2=-p}$
$=p^4+\frac{1}{2p^4}+2.$
As sen pointed, we can use $A.M.-G.M.$ inequality.


algebra
polynomial
real analysis
real analysis unsolved

Define the trigonometric polynomials $ P_0=Q_0=1, P_{n+1}(t)=P_n(t)+e^{i\cdot 2^nt}Q_n(t), Q_{n+1}(t)=P_n(t)-e^{i\cdot 2^n t}Q_n(t)$. Also, define $ f_n=\frac{P_n-P_{n-1}}{2^n}$ and finally $f=\sum_{n\geq 1}{f_n}$. Prove that even if $f$ is $1/2$ holderian the partial Fourier series of $f$ does not converge normally (it can be proved-I posted it some time ago-that if $f$ is $k>1/2$ holderian then the Fourier series converges normally).


inequalities
function
calculus
derivative
inequalities unsolved

if $ a,b,c>0$ then

$ \sum_{cyc}\left(\frac{a^2}{b^2\plus{}c^2}\right) \ge \sum_{cyc}\left(\frac{a}{b\plus{}c}\right)$

[quote="Evariste-Galois"]if $ a,b,c > 0$ then

$ \sum_{cyc}\left(\frac {a^2}{b^2 \plus{} c^2}\right) \ge \sum_{cyc}\left(\frac {a}{b \plus{} c}\right)$[/quote]
The inequality is equivalent to $ \sum \frac {ab(a \minus{} b)^2(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)}{(b^2 \plus{} c^2)(a^2 \plus{} c^2)(b \plus{} c)(a \plus{} c)} \geq 0$ which is obviously true !

[quote="rachid"][quote="Evariste-Galois"]if $ a,b,c > 0$ then

$ \sum_{cyc}(\frac {a^2}{b^2 + c^2}) \ge \sum_{cyc}(\frac {a}{b + c})$[/quote]
The inequality is equivalent to $ \sum \frac {ab(a - b)^2(a^2 + b^2 + c^2 + ab + bc + ca)}{(b^2 + c^2)(a^2 + c^2)(b + c)(a + c)} \geq 0$ which is obviously true ![/quote]

yes that's it Rachid  :lol:

me I stoped here :

$ \sum_{cyc}(ab(a - b)(\frac {1}{(b^2 + c^2)(b + c)} - \frac {1}{(a^2 + c^2)(a + c)}))\ge 0$

Which is true obviously true !! because $ (a - b)$ and $ (\frac {1}{(b^2 + c^2)(b + c)} - \frac {1}{(a^2 + c^2)(a + c)}))$ have the same sign !!

the function $ f(x) \equal{} \sum\frac {a^x}{b^x \plus{} c^x}$ is increasing because of th first derrivative so.....................


geometry
circumcircle
geometric transformation
reflection
geometry proposed

Given a triangle ABC, let the internal and the external angle bisectors of the angle CAB meet the circumcircle of triangle ABC at the points L and M, respectively (with $L \neq A$ and $M \neq A$, of course). Then let P and Q be the reflections of the point A in the points M and L. The circumcircles of triangles AMQ and ALP intersect each other at the point A and at some other point, which we call S. Show that AS || BC.

  Darij

Actually a little bit more general result holds. If P and Q are points so that AM/AP = AL/AQ then AS//BC.

Proof.

AL and AM are orthogonals, therefore ML, PL and MQ are diameters, i.e. O_1 (center of c(PAL)) and O_2 (center of c(MAQ)) stay on PL and MQ respectively.

Given that ML//PQ (from ipothesys AM/AP = AL/AQ) then O_1O2//ML. But ML is orthogonal to BC, by simmetrie, and O_1O_2 is orthogonal to AS also, then the thesis.

The problem actually holds for any $ P$ and $ Q$ | $ P \in AM$, $ L \in AL$ and $ PQ$ || $ ML$, we shall get $ \triangle OO_1O_2 \sim \triangle AML$ and $ OO_1 \parallel AM$ , $ OO_2 \parallel AL$, hence $ O_1O_2 \parallel ML$, done.

Observation: $ PL$ and $ MQ$ concur on $ OA$.

Best regards
sunken rock


geometry
rectangle
3D geometry
search
number theory
prime numbers

I am taking a college math class to become an elementary school teacher. I am currently working on an assignment and am struggling with finding the "shortcut"
Here is the question:

1. Which numbers less than 100 have an odd number of factors? Explain how rectangles help to figure out the answer.
2. Which numbers less than 100 have EXACTLY three factors? Explain.
3. Explain how to find all numbers less than 1000 with exactly three factors. ( I just have to explain it, not write out the numbers themselves)
4. Which numbers less than 100 have EXACTLY four factors? Explain.
5. Explain how to find all numbers less than 1000 with exactly four factors. (same as #3)

I know how to solve this the long way by writing out all the numbers, but what is the trick here? Help!

Sorry mine is without rectangles.
1) Those are only squares, if you look at formula for the number of factors: $ d(n)\equal{}(x_1\plus{}1)(x_2\plus{}1)...(x_k\plus{}1)$ if $ n\equal{}p_1^{x_1}p_2^{x_2}...p_k^{x_k}$ where $ p_1,p_2,...p_k$ are different primes, which then gives you that all $ x_1,x_2,...,x_n$ are even.
2) The formula above gives you those are squares of primes
3) = 2)
4) Again, the formula above gives you cubes of the primes, or product of 2 different primes
5) = 4)

1. The numbers that have an odd number of factors are the ones that are the squares. This is because the square root of these numbers doesn't have a different number to multiply with to get the square.

2. The numbers that have three factors are the ones that have the factors; $ 1, x,$ and $ x^2$. $ x$ is prime though, because otherwise we could factor out $ x$ and get more factors. The numbers that are prime and less than ten are 2, 3, 5, and 7. The squares of these are 4, 9, 25, and 49.

3. Well we can start off by finding the largest perfect square less than 1000. This is 961 which is $ 31^2$. So we are looking for the primes less than or equal to 31. Once we have found these primes, then we have to square them to get our answers.

4. Alright, this is a tricky problem. So I now think it is time to introduce a good trick of finding the number of factors of a number. First you factor the number into the form $ 2^a 3^b 5^c....$. We do this, because we want to know the exponents of the primes. Now this is how we find the factors;

We add one to all the exponents of the primes, and then we multiply all these numbers together to get;

$ (a \plus{} 1)(b \plus{} 1)(c \plus{} 1)...$. This is the number of factors. So there are four factors so;

$ (a \plus{} 1)(b \plus{} 1)(c \plus{} 1)... \equal{} 4$

$ 4$ can factor into $ (1)(4)$, or $ (2)(2)$. For the first case, we know that if the exponent of one of the primes is $ x$, then
$ x \plus{} 1 \equal{} 1$, and if $ y$ is the exponent of one of the primes, then $ y \plus{} 1 \equal{} 4$. So $ x \equal{} 0$, and $ y \equal{} 3$. If we look at $ x$, we notice that anything to the power of $ 0$ except $ 0$ is 1. We don't need to worry about $ 0$ though, because $ 0$ is composite. So that means any number that has four factors has to be a perfect cube, if this is the case. Now lets look at the other case. $ (2)(2)$. $ x \plus{} 1 \equal{} 2$, $ y \plus{} 1 \equal{} 2$. So $ x \equal{} y \equal{} 1$. So this means that the number can be expressed as;

$ a^1 b^1$ or $ c^3$ if it has four factors. So it is either the product of two distinct prime numbers, or the cube of a prime number.

5. To find all the numbers less than 1000 that have four factors, you have to search for a while by plugging primes into
$ a^1 b^1$, and $ c^3$.

Hope that helped.


calculus integration logarithms limit function algebra domain

hello, $ \int_{0}^{\infty}\frac{(\ln(x))^2}{(1\minus{}x)^3}dx\equal{}\frac{1}{3}\pi^2\plus{}\pi$.
Sonnhard.

Hello Sonnhard.
Why do you think so? My calculations shows that answer is $ \pi^{2}$. (but I'm not sure, I solved it , but my solution is very big since  all complex analysis lays there  :(   :( )

Oops... my calculation shows that the answer is $ \frac {\pi^2}{3}$.

[b]Step 1)[/b] $ PV \int_{0}^{\infty} \frac {\ln^2 x}{(1 - x)^3} \, dx = PV \int_{\mathbb{R}} \frac {t^2 e^{t}}{(1 - e^t )^3} \, dt$.

[hide="Solution"]Put $ I_{\epsilon} = \int_{0}^{1 - \epsilon} + \int_{1 + \epsilon}^{\infty} \frac {\ln^2 x}{(1 - x)^3} \, dx = \int_{ - \infty}^{\ln(1 - \epsilon)} + \int_{\ln(1 + \epsilon)}^{\infty} \frac {t^2 e^{t}}{(1 - e^t )^3} \, dt.$

If we let $ f(t) = \frac {t^2 e^{t}}{(1 - e^t )^3}$, then $ \lim_{t\to 0} t f(t) = 1$. So $ f(t) \approx \frac {1}{t}$ near $ t = 0$. Note that

$ I_{\epsilon} = \int_{\mathbb{R} - ( - \epsilon, \epsilon)} f(t) \, dt + E(\epsilon)$

where $ E(\epsilon) = \int_{ - \epsilon}^{\ln(1 - \epsilon)} + \int_{\ln(1 + \epsilon)}^{\epsilon} f(t) \, dt$. Since $ |E(\epsilon)| \leq C \ln \left| \frac {\ln(1 - \epsilon)}{\ln(1 + \epsilon)} \right|$ for sufficiently small $ t$, $ E(\epsilon) \to 0$ as $ \epsilon \to 0$. This proves the claim.[/hide]

[b]Step 2)[/b] As you can confirm,

$ I = PV \int_{\mathbb{R}} \frac {t^2 e^{t}}{(1 - e^t )^3} \, dt = \int_{0}^{\infty} \frac {t^2 (e^{ - t} - e^{ - 2t})}{(1 - e^{ - t})^3} \, dt.$

Note that $ \frac {1}{(1 - x)^3} = \sum_{n = 1}^{\infty} \frac {n(n + 1)}{2} x^{n - 1}$ for $ |x| < 1$. Then

\begin{eqnarray*} I & = & \sum_{n = 1}^{\infty} \frac {n(n + 1)}{2} \int_{0}^{\infty} t^2 (e^{ - nt} - e^{ - (n + 1)t}) \, dt \\
& = & \sum_{n = 1}^{\infty} n(n + 1) \left( \frac {1}{n^3} - \frac {1}{(n + 1)^3} \right) \\
& = & \sum_{n = 1}^{\infty} \left( \frac {1}{n} + \frac {1}{n^2} - \frac {1}{n + 1} + \frac {1}{(n + 1)^2} \right) \\
& = & 2\zeta(2) \\
& = & \frac {\pi^2}{3}. \end{eqnarray*}

It looks like true  :lol: , I'll try to check my solution and (also your).

The integral equals $ \int_{0}^{\infty }{\frac{x^{2}e^{x}}{\left( 1\minus{}e^{x} \right)^{2}}\,dx},$ which gives sos440's result.

hello, yes, my result is wrong, the result $ \frac{\pi^2}{3}$ is correct.
Sonnhard.

here is what I do:
Consider function :
$ f(z) = \frac {(\ln(z) - \pi i)^{3}}{(1 - z)^{3}}$
and integrate it over the contour .
note that :
$ \int_{\gamma}f(z) = - 3\pi^{3}i - 6\pi^{2}$ by residue .
on big and small cicrcle integral is equal to $ 0$,
let $ A$ will be upper contour and $ B$ will be down .
thus we have
$ \int_{A}\frac {(\ln(z) - \pi i)^{3}}{(1 - z)^{3}} + \int_{B}\frac {(\ln(z) + \pi i)^{3}}{(1 - z)^{3}}$
=
$ \int_{A}\frac {(\ln(z) - \pi i)^{3}}{(1 - z)^{3}} - \int_{A}\frac {(\ln(z) + \pi i)^{3}}{(1 - z)^{3}}$
=
$ \int_{A}\frac { - 6i\pi\ln^{2}z + 2i\pi^{2}}{(1 - z)^{3}}$
$ \int_{A}\frac {2i\pi^{2}}{(1 - z)^{3}} = - i\pi^{2}$ standart integral.
and ${ - 6i\pi\int_A}\frac {\ln^{2}z}{(1 - z)^{3}}dz = p.v - 6i\pi\int_{0}^{\infty}\frac {\ln^{2}x}{(1 - x)^{3}}dx - 6i\pi\int_{M}\frac {\ln^{2}z}{(1 - z)^{3}}dz$
where $ M$ is small half circle.
then :
$ \int_{M}\frac {\ln^{2}z}{(1 - z)^{3}}dz = i \pi res_{z = 1}\frac {\ln^{2}z}{(1 - z)^{3}} = i\pi$ by half residue
thus :
 $ p.v - 6i\pi\int_{0}^{\infty}\frac {\ln^{2}x}{(1 - x)^{3}}dx = - 3\pi^{3}i + i\pi^{2} - 6\pi^{2} - 6\pi^{2} = - 2\pi^{3}i - 12\pi^{3}$
and
$ \int_{0}^{\infty}\frac {\ln^{2}x}{(1 - x)^{3}}dx = \frac {\pi^{2}}{3} - 2i\pi^{3}$
real part is $ \frac {\pi^{2}}{3}$ it's answer yes I agree with sos440, but Why I have $ Im$ ?? why $ - 2i\pi^{3}$ ??

nice integral! i try to come up with another solution

[quote="Extremal"]find :
$ p.v. \int_{0}^{\infty}\frac {\ln^{2}x}{(1 \minus{} x)^{3}}dx$[/quote]

hold on, i see people are posting solutions, but i don't think the integral even exists.

when you have an integral like $ \int_0^\infty f(x)\;\ln^2\,{x}\;dx$

this integral will exist only so long as $ f(z)$ has a double pole at $ z \equal{} 1$, that's because the principal branch of $ \ln\,z$ has a simple $ 0$ at $ z \equal{} 1$. this integral still has a meaning and we can  evaluate it in the complex domain by taking a detour at $ z \equal{} 1$ using a small semicircle.

but what you have is $ f(z)$ with a [u]triple[/u] pole at $ z \equal{} 1$, and the numerator is only $ \ln^{2}\,z$,  so the integral will blow up.

yes, one can do
$ \int_{0}^{\infty}\frac {\ln^{2}x}{(1 \minus{} x)^{2}}dx$
or $ \int_{0}^{\infty}\frac {\ln^{3}x}{(1 \minus{} x)^{3}}dx$

but not $ \int_{0}^{\infty}\frac {\ln\,x}{(1 \minus{} x)^{2}}dx$
or $ \int_{0}^{\infty}\frac {\ln^{2}x}{(1 \minus{} x)^{3}}dx$

Hello Misan. I agree with  you about pole, Therefore I wrote $ P.V.$ this mean principal value of integral  :) .
or more generaly it's same as to find principal value of
$ \lim_{\epsilon \to 0} \int_{\mathbb{R}^{\plus{}}\minus{}(1\minus{}\epsilon, 1\plus{}\epsilon)} ...$

[quote="Extremal"]
real part is $ \frac {\pi^{2}}{3}$ it's answer yes I agree with sos440, but Why I have $ Im$ ?? why $ - 2i\pi^{3}$ ??[/quote]

Know what, I really feel I have to apologize for injecting so much complex analysis in here but it's a beautiful problem and sides, Extremal's having problems with it and well . . . I'm helping him.  :D

First let $ log(z) = ln(r) + i\Theta,\; 0\leq \Theta < 2\pi$.  This is a branch of $ \text{log}$ with the branch cut along the positive real axis.  Now, referring to the diagram below, the contour $ \Gamma$ encloses no poles and so:
\[ \int\limits_{\Gamma}\frac {(\text{log}(z) - \pi i)^3}{(1 - z)^3}dz = 0
\]
As Extremal stated above, the integral along the outer circular contour goes to zero as this contour is expanded and likewise goes to zero as the contour around the origin decreases.  So we're left with:
\[ \int\limits_{A}\frac {(\text{log}(z) - \pi i)^3}{(1 - z)^3}dz + \int\limits_{B}\frac {(\text{log}(z) - \pi i)^3}{(1 - z)^3}dz = 0
\]
Now:
\[ \int\limits_{A}\frac {(\text{log}(z) - \pi i)^3}{(1 - z)^3}dz = \text{P.V.}\int_0^{\infty}\frac {(\ln(x) - \pi i)^3}{(1 - x)^3}dx + \lim_{\rho\to 0}\int\limits_{Green}\frac {(\text{log}(z) - \pi i)^3}{(1 - z)^3}dz
\]

\[ \int\limits_{B}\frac {(\text{log}(z) - \pi i)^3}{(1 - z)^3}dz = \text{P.V.}\int_{\infty}^{0}\frac {(\ln(x) + \pi i)^3}{(1 - x)^3}dx + \lim_{\rho\to 0}\int\limits_{\text{Blue}}\frac {(\text{log}(z) + \pi i)^3}{(1 - z)^3}dz
\]
We can use the Residue Theorem to calculate the integral over the Green and Blue contours:
\begin{align*} \lim_{\rho\to 0}\int\limits_{Green}\frac {(\text{log}(z) - \pi i)^3}{(1 - z)^3}dz & = - \pi i\mathop\text{Res}\limits_{z = 1}\left\{\frac {(\text{log}(z) - \pi i)^3}{(1 - z)^3}\right\} \\
& = - \pi i\bigg(3\pi i - 3\pi^2/2\bigg) = 6\pi^2 + 3\pi^3 i/2 \end{align*}

\begin{align*} \lim_{\rho\to 0}\int\limits_{Blue}\frac {(\text{log}(z) + \pi i)^3}{(1 - z)^3}dz & = - \pi i\mathop\text{Res}\limits_{z = 1}\left\{\frac {(\text{log}(z) + \pi i)^3}{(1 - z)^3}\right\} \\
& = - \pi i\bigg( - 3\pi i - 3\pi^2/2\bigg) = - 6\pi^2 + 3\pi^3 i/2 \end{align*}
therefore:
\[ \lim_{\rho\to 0}\int\limits_{Green}\frac {(\text{log}(z) - \pi i)^3}{(1 - z)^3}dz + \lim_{\rho\to 0}\int\limits_{Blue}\frac {(\text{log}(z) + \pi i)^3}{(1 - z)^3}dz = 3\pi^3 i
\]
Combining this result with the two principal-valued integrals:
\[ \text{P.V.}\int_0^{\infty}\frac {(\ln(x) - \pi i)^3}{(1 - x)^3}dx - \text{P.V.}\int_{0}^{\infty}\frac {(\ln(x) + \pi i)^3}{(1 - x)^3}dx + 3\pi^3 i = 0
\]
or:
\[ \text{P.V.}\int_0^{\infty}\frac {(\ln(x) - \pi i)^3 - (\ln(x) + \pi i)^3}{(1 - x)^3}dx = - 3\pi^3 i
\]
simplifying:
\[ \text{P.V.}\int_0^{\infty}\frac {2\pi^3 i - 6\pi i \ln^2(x)}{(1 - x)^3}dx = - 3\pi^3 i
\]
Then finally:
\[ \text{P.V.}\int_0^{\infty}\frac {2\pi^3 i}{(1 - x)^3}dx - \text{P.V.}\int_0^{\infty}\frac {6\pi i\ln^2(x)}{(1 - x)^3}dx = - 3\pi^3 i
\]

\[ - \pi^3 i - \text{P.V.}\int_0^{\infty}\frac {6\pi i\ln^2(x)}{(1 - x)^3}dx = - 3\pi^3 i
\]

\[ \text{P.V.}\int_0^{\infty}\frac {\ln^2(x)}{(1 - x)^3}dx = \frac {2\pi^3 i}{6\pi i} = \frac {\pi^2}{3}
\]

yes, :) . I get $ Im$ maybe I don't calculate right half residue.

[quote="shawtoos"]
We can use the Residue Theorem to calculate the integral over the Green and Blue contours:
\begin{align*} \lim_{\rho\to 0}\int\limits_{Green}\frac {(\text{log}(z) - \pi i)^3}{(1 - z)^3}dz & = - \pi i\mathop\text{Res}\limits_{z = 1}\left\{\frac {(\text{log}(z) - \pi i)^3}{(1 - z)^3}\right\} \\
& = - \pi i\bigg(3\pi i - 3\pi^2/2\bigg) = 6\pi^2 + 3\pi^3 i/2 \end{align*}

\begin{align*} \lim_{\rho\to 0}\int\limits_{Blue}\frac {(\text{log}(z) + \pi i)^3}{(1 - z)^3}dz & = - \pi i\mathop\text{Res}\limits_{z = 1}\left\{\frac {(\text{log}(z) + \pi i)^3}{(1 - z)^3}\right\} \\
& = - \pi i\bigg( - 3\pi i - 3\pi^2/2\bigg) = - 6\pi^2 + 3\pi^3 i/2 \end{align*}
[/quote]

I think this is a problem (not to mention combining two principal-value integrals further down in the analysis but I digress).  The particular theorem used to make those equalities  is only valid for a simple pole and those aren't simple.  I don't know why everything falls into place;  something ain't right I think.

just so you know, the contour you  are using will not work,  [u]where[/u] is your branch cut ? you cannot have your branch cut on $ [0,\infty)$ 'cause you have a pole there. it has to be defined  somewhere else  e.g. $ ( \minus{} \infty,0],\quad\text{or}\quad ( \minus{} \infty\imath, 0\imath],\quad\text{or}\quad [0\imath, \infty\imath)$

I see. therefore you need to choose such contour as mine. Choose my contour and then will be everything good.  :)  you will have simple pole.

no, your contour will not work. your contour is no different from the one that shawtoos used. in fact, there is no contour on this planet that will work for the $ f(z)$ you have posed to solve the integral.

No misan,  I don't need brunch cut since I integrate it on countour  and therefore I consider my function In area (inside my contour) .
I know that there is pole on $ [0,\infty]$ maybe I must explain well every my step.. if you try to write everything well then will be good.

Here is red area where I consider my function .I'm sorry about my picture, since it not looks good.   :blush:  )

For integrals like this, the standard contour is real axis from -R to R with two little semicircle around x = 0 and x= 1, and with a big semicircle around x = 0 with radius R. All semicircles are on the upper half plane.

extremal, look at your contour,  you should get
$ \oint _{|z| \equal{} \delta,\delta\to 0}\;f(z)\;dz\; \plus{} \;\int_0^\infty \frac {(\ln\,x \minus{} \pi i)^{3}}{(1 \minus{} x)^{3}} \;dx$

$ \plus{} \oint_{|z| \equal{} R,R\to\infty}\;f(z)\; dz\; \plus{} \int_\infty^0 \frac {(\ln(x\;e^{2\pi i}) \minus{} \pi i)^{3}}{(1 \minus{} x\;e^{2\pi i})^{3}}\;e^{2\pi i}\;dx \quad \equal{} \quad \minus{} 3\pi^{3}i \minus{} 6\pi^{2}$

 (contributions from the $ 2$ semi-circles at $ z \equal{} 1$ cancel each other out...)

now look   at the $ 4$ terms on the LHS, clearly one of them does [b]not[/b] go  to $ 0$.... can you see which one ?

I'd like to better understand this problem.  I'm not trying to force-fit it to my whims but only to understand precisely what the problem is.   Letting $ \displaystyle f_1(x)\equal{}\frac{(\ln(x)\minus{}\pi i)^3}{(1\minus{}x)^3}$ and $ \displaystyle f_2(x)\equal{}\frac{(\ln(x)\plus{}\pi i)^3}{(1\minus{}x)^3}$ then without regard to its relevance to the above problem, does the following limit exists:

\[ \lim_{\epsilon\to 0}\left\{\int_0^{1\minus{}\epsilon} f_1(x)dx\plus{}\int_{1\plus{}\epsilon}^{\infty}f_1(x)dx\plus{}\int_{\infty}^{1\plus{}\epsilon} f_2(x)dx\plus{}\int_{1\minus{}\epsilon}^0 f_2(x)dx\right\}\]

It looks to me the limit does exists in the Principal Value sense and approaches $ \minus{}3\pi^3 i$, that is,  the unbounded terms, while approaching infinity and minus infinity, nevertheless cancel.  Is this not true?

If it is true, then I believe an argument could be made to solve the original integral using the contour I posted.

I understand. I must explain every my step more clearly (I do it only once more, since Must prepare to my exams , and hence I have no free much time.)
Consider contour such as on picture (please Look at picture well  :(  ).

My picture is not very well (since I can't draw it well) I note that on the picture points $ (A,B,C,D,H,F,E,G)$ lie on same line $ (0,\infty)$
but also note that $ A \neq H$ but they are very close, such that  $ dist (A,H) \equal{} 0$  :lol: .
also for points $ (B,F)$ , $ (C,E)$ and $ (D,G)$. OK.
Now I try to explain other things on my contour.
 $ |AB| \equal{} |HF|$ or dist. is equal.
$ |CD| \equal{} |EG|$
let's point $ 1$ called point $ 1$  :lol:  and also point $ 0$ called point $ 0$.

and fix some small $ \epsilon > 0$ and big $ R > 0$   say :
$ |1F| \equal{} |1E| \equal{} |1B| \equal{} |1C| \equal{} |0H| \equal{} |0A| \equal{} \epsilon$
also
$ |0D| \equal{} |0G| \equal{} R$
every one understand that $ R$ is radius of big corcle , and $ \epsilon$ is radius of small circles.
now I can
 consider function :

$ f(z) \equal{} \frac {(\ln(z) \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}$ where $ \lim_{\delta \to 0} \ln (1 \plus{} \delta\i) \equal{} 0$ (or argument on upper line is equal to $ 0$.)
 in area of contour it's [b]well defined[/b] (I think every one understand: what does it  mean  "well defined". ) function .
lets' called this contour $ L$ and integrate on this contour our function $ f(z)$
$ \int\limits_{L}f(z)dz$
from complex analysis and residue theorem it's wellknown that integral will be equal to :
\[ \int\limits_{L}f(z)dz \equal{} 2 \pi i Rez _{z \equal{} 1}f(z)
\]
since only point $ 1$ is in area of our contour. OK>
Now consider our integral by next way (as sum of the contours.)
\[ \int\limits_{L}f(z)dz \equal{} \int\limits_{A,B,C,D}\frac {(\ln(z) \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \plus{} \int\limits_{Circle(D,G)}f(z)
\]

\[ \plus{} \int\limits_{G,E,F,H}\frac {(\ln(z) \plus{} 2\pi i \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \plus{} \int\limits_{circle(H,A)}f(z)
\]
first simple question: why I have :
\[ \int\limits_{G,E,F,H}\frac {(\ln(z) \boxed{ \plus{} 2\pi i} \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz
\]
Answer: since function $ ln (z) \equal{} \ln |z| \plus{} i arg(z)$
hence after contour $ Circle(D,G)$ argument of $ z$ will be $ 2\pi i$ hence I have $ \plus{} 2\pi i$
OK.
Now I use something simple, I say that : if I change orientation of contour then there will changed sign, or
\[ \int\limits_{G,E,F,H}\frac {(\ln(z) \plus{} 2\pi i \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \equal{} \minus{} \int\limits_{H,F,E,G}\frac {(\ln(z) \plus{} 2\pi i \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \equal{}
\]

\[ \minus{} \int\limits_{H,F,E,G}\frac {(\ln(z) \plus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz
\]
.
Ok, Now let's note that
\[ \int\limits_{Circle(D,G)}f(z) \to 0
\]
as $ R \to \infty$ or more exact it's something $ \equal{} O(\frac {\ln^{3} R}{R^{2}})$ for big $ R$
now note that :
\[ \int\limits_{circle(H,A)}f(z) \equal{} O(\epsilon \ln^{3} \epsilon)
\]
for small $ \epsilon$

(also note that $ \lim_{\epsilon \to 0} O(\epsilon \ln^{3} \epsilon) \equal{} 0$)
finally we have :
\[ .... \equal{} \int\limits_{A,B,C,D}\frac {(\ln(z) \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \plus{} O(\frac {\ln R}{R^{2}}) \minus{} \int\limits_{H,F,E,G}\frac {(\ln(z) \plus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \plus{} O(\epsilon \ln \epsilon)
\]
Let's denote by
\[ W(\epsilon,R) \equal{} O(\epsilon \ln^{3} \epsilon) \plus{} O(\frac {\ln^{3} R}{R^{2}})
\]
and note that
$ \lim\limits_{\epsilon \to 0, R \to \infty } W(\epsilon,R) \equal{} 0$
OK.
or we have
\[ \equal{} \int\limits_{A,B,C,D}\frac {(\ln(z) \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \minus{} \int\limits_{H,F,E,G}\frac {(\ln(z) \plus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \plus{} W(\epsilon,R)
\]
OK.
Now ()at first consider :
\[ \int\limits_{A,B,C,D}\frac {(\ln(z) \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz
\]
this integral .
Now I divide contour $ (A,B,C,D)$ by parts such that :
\[ \int\limits_{A,B,C,D}\frac {(\ln(z) \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \equal{} \int\limits_{A,B}\frac {(\ln(z) \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \plus{} \int\limits_{circle(B,C)}\frac {(\ln(z) \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \plus{} \int\limits_{C,D}\frac {(\ln(z) \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz
\]
note that
$ |AB| \equal{} (\epsilon,1 \minus{} \epsilon)$  and $ |CD| \equal{} ( 1 \plus{} \epsilon,R)$
thus we have
\[ \int\limits_{A,B,C,D}\frac {(\ln(z) \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \equal{} \int_{\epsilon}^{1 \minus{} \epsilon}\frac {(\ln(x) \minus{} \pi i )^{3}}{(1 \minus{} x)^{3}}dx \plus{} \int\limits_{circle(B,C)}\frac {(\ln(z) \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \plus{} \int_{1 \plus{} \epsilon}^{R}\frac {(\ln(x) \minus{} \pi i )^{3}}{(1 \minus{} x)^{3}}dx
\]
similary for:
\[ \minus{} \int\limits_{H,F,E,G}\frac {(\ln(z) \plus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \equal{} \minus{} \int_{\epsilon}^{1 \minus{} \epsilon}\frac {(\ln(x) \plus{} \pi i )^{3}}{(1 \minus{} x)^{3}}dx \minus{} \int\limits_{circle(F,E)}\frac {(\ln(z) \plus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \minus{} \int_{1 \plus{} \epsilon}^{R}\frac {(\ln(x) \plus{} \pi i )^{3}}{(1 \minus{} x)^{3}}dx
\]
after summation this two results we will have :
\[ \int\limits_{A,B,C,D}\frac {(\ln(z) \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \minus{} \int\limits_{H,F,E,G}\frac {(\ln(z) \plus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \plus{} W(\epsilon,R) \equal{}
\]

\[ \equal{} \int_{\epsilon}^{1 \minus{} \epsilon}\frac {(\ln(x) \minus{} \pi i )^{3}}{(1 \minus{} x)^{3}}dx \plus{} \int\limits_{circle(B,C)}\frac {(\ln(z) \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \plus{} \int_{1 \plus{} \epsilon}^{R}\frac {(\ln(x) \minus{} \pi i )^{3}}{(1 \minus{} x)^{3}}dx \minus{}
\]

\[ \minus{} \int_{\epsilon}^{1 \minus{} \epsilon}\frac {(\ln(x) \plus{} \pi i )^{3}}{(1 \minus{} x)^{3}}dx \minus{} \int\limits_{circle(F,E)}\frac {(\ln(z) \plus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \minus{} \int_{1 \plus{} \epsilon}^{R}\frac {(\ln(x) \plus{} \pi i )^{3}}{(1 \minus{} x)^{3}}dx
\]
Note that
\[ \frac {(\ln(z) \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}} \equal{} \frac {(\ln^{3}(z) \minus{} 3\ln^{2}(z) \pi i \minus{} 3\pi^{2}\ln (z) \plus{} i\pi^{3} )}{(1 \minus{} z)^{3}}
\]

\[ \frac {(\ln(z) \plus{} \pi i )^{3}}{(1 \minus{} z)^{3}} \equal{} \frac {(\ln^{3}(z) \plus{} 3\ln^{2}(z) \pi i \minus{} 3\pi^{2}\ln (z) \minus{} i\pi^{3} )}{(1 \minus{} z)^{3}}
\]
therefore :
\[ \int_{\epsilon}^{1 \minus{} \epsilon}\frac {(\ln(x) \minus{} \pi i )^{3} \minus{} (\ln(x) \plus{} \pi i )^{3}}{(1 \minus{} x)^{3}}dx \equal{} \int_{\epsilon}^{1 \minus{} \epsilon} \frac { \minus{} 6\ln^{2}(z) \pi i \plus{} 2i\pi^{3}}{(1 \minus{} x)^{3} }dx
\]
and
\[ \int_{1 \plus{} \epsilon}^{R}\frac {(\ln(x) \minus{} \pi i )^{3} \minus{} (\ln(x) \plus{} \pi i )^{3}}{(1 \minus{} x)^{3}}dx \equal{} \int_{1 \plus{} \epsilon}^{R} \frac { \minus{} 6\ln^{2}(z) \pi i \plus{} 2i\pi^{3}}{(1 \minus{} x)^{3}}dx
\]
finally we obtain :
\[ \equal{} \int_{\epsilon}^{1 \minus{} \epsilon} \frac { \minus{} 6\ln^{2}(x) \pi i \plus{} 2i\pi^{3}}{(1 \minus{} x)^{3}}dx \plus{} \int_{1 \plus{} \epsilon}^{R} \frac { \minus{} 6\ln^{2}(x) \pi i \plus{} 2i\pi^{3}}{(1 \minus{} x)^{3}}dx \plus{} \int\limits_{circle(B,C)}\frac {(\ln(z) \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \minus{} \int\limits_{circle(F,E)}\frac {(\ln(z) \plus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz
\]
since differenc between points $ (B,F)$ and also $ (C,E)$ is equal to $ 0$ then I can say that :
\[ \minus{} \int\limits_{circle(F,E)}\frac {(\ln(z) \plus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \equal{} \minus{} \int\limits_{circle(B,F)}\frac {(\ln(z) \plus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz
\]
and therefore :
\[ \int\limits_{circle(B,C)}\frac {(\ln(z) \minus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \minus{} \int\limits_{circle(F,E)}\frac {(\ln(z) \plus{} \pi i )^{3}}{(1 \minus{} z)^{3}}dz \equal{} \int\limits_{circle(B,C)}\frac { \minus{} 6\ln^{2}(z) \pi i \plus{} 2i\pi^{3}}{(1 \minus{} z)^{3}}dz
\]
note that :
\[ \int\limits_{circle(B,C)}\frac { \minus{} 6\ln^{2}(z) \pi i \plus{} 2i\pi^{3}}{(1 \minus{} z)^{3}}dz \equal{} \equal{} \int\limits_{circle(B,C)}\frac { \minus{} 6\ln^{2}(z) \pi i}{(1 \minus{} z)^{3}}dz \plus{} \int\limits_{circle(B,C)}\frac {2i\pi^{3}}{(1 \minus{} z)^{3}}dz \equal{} \minus{} \pi i Rez_{z \equal{} 1} \frac { \minus{} 6\ln^{2}(z) \pi i}{(1 \minus{} z)^{3}} \plus{} i\pi^{3}\frac {1}{(1 \minus{} z)^{2}}|_{1 \minus{} \epsilon}^{1 \plus{} \epsilon} \equal{} 6\pi^{2}
\]
after note that :
\[ \int_{\epsilon}^{1 \minus{} \epsilon} \frac { \minus{} 6\ln^{2}(x) \pi i \plus{} 2i\pi^{3}}{(1 \minus{} x)^{3}}dx \plus{} \int_{1 \plus{} \epsilon}^{R} \frac { \minus{} 6\ln^{2}(x) \pi i \plus{} 2i\pi^{3}}{(1 \minus{} x)^{3}}dx \equal{} \left( \int_{\epsilon}^{1 \minus{} \epsilon} \plus{} \int_{1 \plus{} \epsilon}^{R} \right) \frac { \minus{} 6\ln^{2}(x) \pi i}{(1 \minus{} x)^{3}}dx \plus{} \left( \int_{\epsilon}^{1 \minus{} \epsilon} \plus{} \int_{1 \plus{} \epsilon}^{R} \right) \frac {2i\pi^{3}}{(1 \minus{} x)^{3}}dx
\]
note that
\[ \left( \int_{\epsilon}^{1 \minus{} \epsilon} \plus{} \int_{1 \plus{} \epsilon}^{R} \right) \frac {2i\pi^{3}}{(1 \minus{} x)^{3}}dx \equal{} \minus{} \frac {i\pi^{3}}{(1 \minus{} \epsilon)^{2}}
\]
thus we have :
\[ \equal{} \left( \int_{\epsilon}^{1 \minus{} \epsilon} \plus{} \int_{1 \plus{} \epsilon}^{R} \right) \frac { \minus{} 6\ln^{2}(x) \pi i}{(1 \minus{} x)^{3}}dx \minus{} \frac {i\pi^{3}}{(1 \minus{} \epsilon)^{2}}
\]
finally we get :
\[ \int_{L}f(z)dz \equal{} 6\pi^{2} \plus{} \left( \int_{\epsilon}^{1 \minus{} \epsilon} \plus{} \int_{1 \plus{} \epsilon}^{R} \right) \frac { \minus{} 6\ln^{2}(x) \pi i}{(1 \minus{} x)^{3}}dx \minus{} \frac {i\pi^{3}}{(1 \minus{} \epsilon)^{2}} \plus{} W(\epsilon,R) \equal{} 2 \pi i Rez _{z \equal{} 1}f(z)
\]
Now let $ R\to \infty$ and $ \epsilon \to 0$ then
\[ \left( \int_{\epsilon}^{1 \minus{} \epsilon} \plus{} \int_{1 \plus{} \epsilon}^{R} \right) \frac { \minus{} 6\ln^{2}(x) \pi i}{(1 \minus{} x)^{3}}dx \minus{} \frac {i\pi^{3}}{(1 \minus{} \epsilon)^{2}} \plus{} W(\epsilon,R) \to P.V. \int_{0}^{\infty}\frac { \minus{} 6\ln^{2}(x) \pi i}{(1 \minus{} x)^{3}}dx \minus{} i\pi^{3}
\]
and finally :
\[ \int_{L}f(z)dz \equal{} P.V. \int_{0}^{\infty}\frac { \minus{} 6\ln^{2}(x) \pi i}{(1 \minus{} x)^{3}}dx \minus{} i\pi^{3} \plus{} 6\pi^{2} \equal{} 2 \pi i Rez _{z \equal{} 1}f(z)
\]
thus we obtain
\[ P.V \int_{0}^{\infty}\frac {\ln^{2}(x)}{(1 \minus{} x)^{3}}dx \equal{} \frac {2 \pi i Rez _{z \equal{} 1}f(z) \plus{} i\pi^{3} \minus{} 6\pi^{2}}{ \minus{} 6\pi i} \equal{} \frac {\pi^{2}}{3}
\]
:sleeping:

I yield.  With some effort on my part I was able to work through sos440's solution above.  It's quite elegant I believe (where's the emotiocon for "we're not worthy"?)   I'm sorry for cluttering-up this thread and the forum with mumbo-jumbo complex analysis.  Sorry Extremal and I'm sure you'll do fine on your test :)

[quote="shawtoos"]... I'm sure you'll do fine on your test :)[/quote]
thank you shawtoos  :roll:
sometimes explanations looks very hard (or very big)...  in this situation it easy to see solution then to try explain .  :|


search
function
LaTeX
number theory
number theory unsolved

If $a,b,c$ are positive integers and
$0<a^{2}+b^{2}-abc\leq c$

Show that $a^{2}+b^{2}-abc$ is a perfect square.

[quote="Hawk Tiger"]I agree it is trivial.
But How can I use seach function to find such a prolbem,ZetaX?
Which word should I input?"integer, or positive integer,or perfect square?I think all these word may not work because the result may be many.
I have little experience in Search,though I always seach and always fail.[/quote]

In fact, I had already searched "a^2+b^2-abc", but I couldn't find anything...

See http://www.mathlinks.ro/Forum/viewtopic.php?t=6233 .

The "trick" is to use good parts of the $\LaTeX$-code (here "abc"), since other signs like "+" or "^" don't work very well, and to include some other keyword (here "square).
Also don't forget to restrict the search to "Number Theory" if there are too many results.
And note that originally, there where >5 threads, I merged them now.

I see. I hope in the future I'll be a better searcher.  :D

But, in Myth's solution, case 2, why is $\frac{b^{2}-d}{a}<a$? I just can't see...