Solve by vector methods:
In a triangle, a line is drawn from A to BC intersecting BC at D, such that AD:DB = 2:1.This line intersects with the median from C to AB at P. Find AP:PD.

I think that this is not a physics problem  :maybe:

Let $\left\{ D \right\}= AP \cap BC, \, \left\{ E \right\}= BP \cap CA, \, \left\{ F \right\}= CP \cap AB$.

By [url=http://www.cut-the-knot.org/Generalization/ceva.shtml]Ceva's theorem[/url] we have that $\frac{\overline{BD}}{\overline{DC}}\cdot \frac{\overline{CE}}{\overline{EA}}\cdot \frac{\overline{AF}}{\overline{FB}}= 1$.
Since $F$ is the midpoint of $AB$, it follows that $\frac{\overline{AF}}{\overline{FB}}= 1$, so that $\frac{\overline{BD}}{\overline{DC}}\cdot \frac{\overline{CE}}{\overline{EA}}= 1$.
Hence, $\frac{\overline{BD}}{\overline{DC}}= \frac{\overline{EA}}{\overline{CE}}$.

By [url=http://planetmath.org/encyclopedia/VanAubelTheorem.html]Van Aubel's theorem[/url] we have that $\frac{\overline{AP}}{\overline{PD}}= \frac{\overline{AF}}{\overline{FB}}+\frac{\overline{AE}}{\overline{EC}}= 1+\frac{\overline{EA}}{\overline{CE}}= 1+\frac{\overline{BD}}{\overline{DC}}$.

Note that if we fix $B,D,C$, then the position of $A$ doesn't change the answer. All that matters is $\frac{\overline{BD}}{\overline{DC}}$.
Therefore, the hypothesis $AD : DB = 2 : 1$ is useless.


trigonometry
function
logarithms
inequalities unsolved
inequalities

Find max, min $ y\equal{}2^{|\sin x|}\plus{}2^{|\cos x|}$

[quote="Algadin"]Find max, min $ y \equal{} 2^{|\sin x|} \plus{} 2^{|\cos x|}$[/quote]

WLOG, we can assume $ x \in \left[ {0,\frac {\pi }{4}} \right]$ then we have
\[ 2^{\left| {\sin x} \right|} \plus{} 2^{\left| {\cos x} \right|} \equal{} 2^{\sin x} \plus{} 2^{\cos x} \equal{} 2^{\sin x} \plus{} \frac {1}{2}.2^{\cos x} \plus{} \frac {1}{2}.2^{\cos x} \ge 3\sqrt [3]{{2^{\sin x \plus{} 2\cos x \minus{} 2} }}
\]
And because $ x \in \left[ {0,\frac {\pi }{4}} \right] \Rightarrow \sin x \plus{} 2\cos x \minus{} 2 \ge 0$. Therefor,
\[ 2^{\left| {\sin x} \right|} \plus{} 2^{\left| {\cos x} \right|} \ge 3
\]
One more solution with deravative. We cansider the funtion $ f\left( x \right) \equal{} 2^{\sin x}  \plus{} 2^{\cos x}$ with $ x \in \left[ {0,\frac{\pi }{4}} \right]$. We have
\[ f'\left( x \right) \equal{} 2^{\sin x} \cos x\ln 2 \minus{} 2^{\cos x} \sin x\ln 2
\]
Now we consider the funtion $ g\left( x \right) \equal{} \frac{{2^x }}{x}$ with $ x \in \left[ {0,1} \right]$
\[ g'\left( x \right) \equal{} \frac{{2^x \left( {x\ln 2 \minus{} 1} \right)}}{{x^2 }} \le 0
\]
\[ \Rightarrow g\left( {\cos x} \right) \le g\left( {\sin x} \right)
\]
\[ \Rightarrow f'\left( x \right) \equal{} \ln 2\left( {2^{\sin x} \cos x \minus{} 2^{\cos x} \sin x} \right) \ge 0
\]
\[ \Rightarrow \left\{ \begin{array}{l}
 f\left( x \right) \ge f\left( 0 \right) \equal{} 3 \\
 f\left( x \right) \le f\left( {\frac{\pi }{4}} \right) \equal{} 2^{\frac{{\sqrt 2 }}{2} \plus{} 1}  \\
 \end{array} \right.
\]

[quote="Algadin"]Find max, min $ y \equal{} 2^{|\sin x|} \plus{} 2^{|\cos x|}$[/quote]
Max we can find from Caramata for $ f(x)\equal{}2^x$


geometry
parallelogram
similar triangles

Let a triangle  $ABC$  with  angle $A = 90$  ,  the altitude  $AD$ , the midpoint  $M$ of segment  $DC$  and a point  $E$  on the extension of $AB$  such, that  $BE=AB$. Prove that  $ED \perp AM$

 [u]Babis[/u]

 [i]I hope you will like this nice problem. In an old Crux issue I saw a metric solution but this morning I tried again and I found a very short and beautyfull solution. I shall share it with you , if the solutions that you will send are different from mine's . [/i]

my solution:[hide]$\Delta ABC$ is similar to $\Delta ADC$ so $\frac{AD}{CD}=\frac{AB}{AC}$.
we know that $CD=2CM$ and $AB=\frac{AE}{2}$ so we have $\frac{AD}{MC}=\frac{AE}{AC}$ and because $\angle BAD=\angle ACM$ so $\Delta ADE$ is similar $\Delta AMC$. Assume that the line prpendicular to $BC$ passing through $M$ intersects $AC$ at $P$, and let $ED$ intersect $AM$ at $Q$. because $\Delta ADE$ is similar $\Delta AMC$ so $\angle QDM=\angle BDE=\angle AMP$ and because $PM\|AD$ so $\angle QDM=\angle AMP=\angle DAM$, and this means $\Delta DQM$ is similar to $\Delta ADM$ and $\angle DQM=90$. [/hide]
Was this your solution?

[hide="My solution"]Extend $AD$ to $F$ such that $EF\perp AF$. Then $\triangle AFE\sim\triangle CDA$ (because $\angle DCA=\angle DAB$), and $ED$ and $AM$ are corresponding segments, since $D$ is the midpoint of $AF$, due to $BD\parallel EF$ and $BA=BE$. Since all the other pairs of corresponding segments in $\triangle CDA$ and $\triangle AFE$ are perpendicular, those two are perpendicular as well.

(Alternative explanation: $\triangle CDA\sim\triangle AFE\implies \triangle CMA\sim\triangle ADE\implies\angle CAM=\angle AED$, and since one pair of their sides is perpendicular, the other pair must be perpendicular too.)[/hide]

You(Sinajackson) wrote :

  [color=blue][b]''My solution[/b]
 $\Delta ABC$ is similar to $\Delta ADC$ so $\frac{AD}{CD}=\frac{AB}{AC}$.
we know that $CD=2CM$ and $AB=\frac{AE}{2}$ so we have $\frac{AD}{MC}=\frac{AE}{AC}$ and because $\angle BAD=\angle ACM$ so $\Delta ADE$ is similar $\Delta AMC ''$.  [/color]

[color=red] Your nice solution can become shorter:[/color]

  From the similar triangles $ADE, AMC$ we have

 $\angle AED = \angle MAC$  and so

 $\angle AED+\angle MAE = \angle MAC+\angle MAE= \angle CAE = 90$  and so  $ED \perp AM$.

[b] My solution[/b]

 IF  $K , L$ are the symmetric points of $A$ with respect to points $D , M$, then
    [b]a)[/b] points $E,K,L$ are collinear , $EKL // BC$ and  $AK \perp EL$ since         $AK \perp BC$ ,
     [b]b)[/b] $ACLD$ is a parallelogram , so  $LD // CA$ ,
         that is $LD \perp AE$.
   [b]c) [/b]Hence  $D$ is the ortocenter in triangle $AEL$ , so $ED \perp AM$.

[u] Babis[/u]

 [i]I had typed my solution , before having seen Fahrenhajt's(same solution!) , so I send it too.[/i]

[color=green]May me not too different but still[/color][hide]
Let $N$ be the midpoint of $AD$ then $MN \parallel AC \Rightarrow MN \perp AB \Rightarrow N$ is the orthocenter of $\Delta AMB \Rightarrow BN \perp AM$.
But $BN \parallel ED \Rightarrow ED \perp AM$.[/hide]


An ice cream cone has radius 1 inch and height 4 inches. What is the number of inches in the radius of a sphere of ice cream which has the same volume as the cone?

I just had ice cream :D  :D  :D  :D  :D  :D [hide]The formula for the volume of a cone is $v = \frac 13 \pi r^2h$. So the volume of the cone is $\frac 13 * \pi * 1 * 4= \frac 43 \pi$. The formula for the area of a sphere is $v = \frac 43 \pi r^3$. So $ \frac 43 * \pi = \frac 43 \pi r^3$. The two $ \frac 43 \pi$s cancel each other out leaving $r^3=1$. Cube root gives you $r = \boxed{1}$[/hide]

[hide]The base of the cone is $ \pi $. the area of the ice cream is $ \frac{4}{3}\pi $ or $ \frac{4}{3}\pi r^{3} $. So the radius is 1. My favorite flavor of ice cream is chocolate :lol:[/hide]


A Difficult Math

12 （21） 26

16 （30） 36

20 （38） 48

24 （54） 60

28 （ ? ） 84

Each numeral series stands on its own.
Ther relationship is among the first/second/third numbers.
The relationship is the same for each numeral series.
But each numeral series does not affect each other.

Define the function $f(x,y)$ on pairs of integers by:
$f(12,26) : = 21$
$f(16,36) : = 30$
$f(20,48) : = 38$
$f(24,60) : = 54$
$f(28,84) : = 23 \cdot 42$
and
$f(x,y) : = \frac{\zeta(x^2+y^2+\Gamma(\frac{1}{\gamma})) - \sigma(xy) }{\vartheta(e \cdot x+i\cdot(y^4+\pi))}$
otherwise.

So the number you are looking for is $23 \cdot 42 = 966$  :P

PS:
when it hasn't been sarcastic enough: from a mathematical point of view any number is possible and makes sense; this is the reason why I consider such sequence things as nonsense.
PPS: find the rule behind $1,2,4,8,16,32,64$ ;)

[quote="ZetaX"]PPS: find the rule behind $1,2,4,8,16,32,64$ ;)[/quote]
The sequence is the number of binary bit strings of length n with no block of 8 or more 0's (nonzero heptanacci numbers).  Or maybe it's a polynomial of degree 6...  maybe of degree 28...  not sure.  It looks like it might possibly be exponential, but that's a bit of a wild guess.

[quote="mathnerd314"]Hey, don't torture Vich. He's (she's?) new.

Any real solutions?[/quote]

966 is real :P

I'm going to say the blank is $\pi^{e^{\pi^{e^{\pi}}}}$. I'd post the function $f(x,y)$ that gives me this, but it's longer than I want to type.


the hmmt quagga or the ahsimc quagga?  :wink:  :roll:

I vote we should both be quagga to princeton and (if y'all go) to hmmt - ENDLESS HILARITY OMG

[quote="neelnal"]is quagga some sort of beast team?[/quote]

there are two different Quaggas (and yes, both of them are beast teams)

lol all these people coming with monster math buddies. i can only go with my school and we have no depth really other than me and another kid.

[quote="Pakman2012"]Who knows where the word Quagga comes from?  :lol: [/quote]

um me

[quote="Pakman2012"]Who knows where the word Quagga comes from?  :lol:

Speaking of Quagga... We're calling it for HMMT.... and probably Princeton.[/quote]

oh hm yea quagga does sound more intimidating than team carolina but w/e

[quote="Pakman2012"]I thought it was from a random GUTS round or something?

Its like a knight that can move 6x5 or something like that.[/quote]
Maybe they called the knight a quagga because of [url=http://en.wikipedia.org/wiki/Quagga]quaggas[/url].

[quote="Phelpedo"]Look at the last definition...I see you're going by the mantra, "If Wikipedia doesn't have it then it doesn't exist."[/quote]
huh i was talking about the last definition. [i]WHO [/i]is that last definition referring to?

Origin of Quagga:
Suggested by Ben Cunningham, the night before HMMT. We were sitting in the dorm room deciding the name for our team. Then we came across a Guts round from the previous year involving a hypothetical chess peice named Qugga. (moves at 3X5) We unamiously agreed that its pure coolness so we settled on that name... ;)

[quote="Iversonfan2007"][quote="junggi"]oh whoa what is up with this page
[url]http://en.wikipedia.org/wiki/Quagga_%28disambiguation%29[/url]
was there a quagga different from the quagga with arnav[/quote]

Hm, the quaggas that I know are:

The HMMT quagga, whose members are: Arnav and the Florida Math Team (Nimish, Yannik, Rui, etc)

The AHSIMC quagga, whose members are Arnav, Zhai, Yannik, Rui, and Shobhit.

This year the Florida Math Team will be named Quagga for Princeton, and the HMMT quagga will be Arnav+Florida

BTW SOMEONE NEEDS TO ADD QUAGGA TO AOPS WIKI[/quote]

HMMT Quagga = Quagga

Princeton Quagga = Quagga

that is all

Quagga transcends these petty discussions

[quote="joml88"]No, that's his team.

And that definition was added today...by an AoPSer I'd assume (not me).[/quote]

:whistling:

whoa that is some pretty advanced whistling going on there CHESS64

[quote="Cicatriz"]So we [i]are[/i] going to be Quagga after all?

I feel honored.[/quote]

who's "we", florida?

i think we should change the nc/sc princeton team's name to Quagga 2 or something b/c of arnav

dude no

I think we should BOTH BE QUAGGA

sheesh

think of the endless hilarity that would ensue

Hm, except you register online and it's all electronic.  I'm pretty sure you won't be able to make them [i]exactly[/i] the same...there's probably a way around it though so that they are basically the same...

aww the quagga entry got deleted :( (http://en.wikipedia.org/wiki/Quagga_%28disambiguation%29)

[quote]Revert. Last addition was no article link.[/quote]

come on... who wants to write a wikipedia article about quagga math team? :P

Quagga got 7th place in the HMMT for team A category last year.

Is this the same Quagga you guys are talking about?

yeah 3rd overall and individual 1st, 2nd, 5th places

HELL YEA

But that was just a fluke. TJ is going down this year. Exeter too.

this is true

well, probably

I don't actually know who's on the team and all that kind of stuff

See problem #37.  :)

http://web.mit.edu/hmmt/www/datafiles/problems/2003/pguts03.pdf


show there is bijection from N^N to R,
 where N -set of natural nos.
           R - set of real nos. :welcomeani:

[quote="Omid Hatami"]We know that $\mathbb R\sim(0,1)$ and each number in $(0,1)$ has the form $\overline{0.a_{1}a_{2}\dots}$. Easily we caould make a bijection.[/quote]
It give bijection $10^{N}\to (0,1)$ and it is not bijection numbers $\{0,9,9,9,...\}\in 10^{N}\ and \ \{1,0,0,...\}$ give same number in R. Bijection $N^{N}\to (0,1)$ can be given by continiosly fraction.


logarithms
limit
function
real analysis
real analysis unsolved

1.let's consider the equation \[ \ln x = (\cot x)^\alpha \] with $\alpha \in (0, +\infty)$ and $x \in (0,{\pi \over 2})$.
a. show that for any $\alpha$ there exist one and only one solution, and let us denote it with $s(\alpha)$.
b. show that $s$ is continuous.
c. find $\mathop {\lim }\limits_{\alpha \to 0^+ }x_\alpha $.

2. Find the cardinality of the set of the real continuous functions.

(2)

All constant functions are continuous. Hence the cardinality of continuous functions is $\geq c$

For continuous function $f$, define
$g(f) = \{ (x,y)\in\mathbb{Q}\times\mathbb{Q} : y\leq f(x)\}$
It can be shown that $g$ is an injection from the set of continuous functions to $\mathcal{P}(\mathbb{Q}\times\mathbb{Q})$. The cardinality of the latter set is $c$. Hence the cardinality of continuous functions is $\leq c$.

ok, one has gone :)
what about the other one?


geometry
parallelogram
geometry proposed

own (please don't laugh :rotfl:  :D )
consider a $ \Delta ABC$ and its inner napoleonic triangles $ ABC'$, $ BCA'$,  $ CAB'$
now we construct parallelograms $ BCB"B'$ and $ ACA"A'$
prove that $ \Delta A"B"C'$ is equilateral

We have △AA’C △BB’C are congruence
△A’BA △CBC’ are congruence
So CA''=CB''=CC' and ∠A''CC'' =∠ A''CB''=120
this means △A''C'B'' is a equilateral triangle


vector
integration
trigonometry
real analysis
real analysis solved

Prove that any finite set couldn't be complete in $L_2$ (I don't know whether the word "complete" is acceptable)

One question: $L^2$ of what? Of a bounded inteval? Of $\mathbb{R}$? Of the integers? (The answer is the same in all of those cases, but you probably should be specific.)

What I assume you mean by "complete" is this: set of elements of an inner product space is complete if and only if its span is dense in that inner product space.

Equivalent statement: $W$ is complete iff $(\text{span}(W))^{\perp}=\{0\}.$

It boils down to $L^2$ being an infinite dimensional vector space.

I'm sorry for being not correct: by $L_2$ i meant the set of all square-summable funtions on $[a,b]$, i.e. the set of all measurable fucntions s.t. $\int_{a}^{b}f^2(x)dx<\infty$
In my meaning "complete" set of fucntions means that the only elemnt, which is orthogonal to any $f\in{L_2}$ is $0$

So, $L^2$ of a bounded interval it is. And I assume in that last sentence you mean that the only element orthogonal to any $f$ in the set in question (not all of $L^2$) is zero. Assuming that, your definition of "complete" is the same as mine.

To actually prove that no finite set is complete, it suffices to exhibit an infinite orthogonal set. Let $f_n(x)=\sin\left(\frac{n(x-a)}{b-a}\right)$ for $n\in\mathbb{N}.$ This is an infinite orthogonal set. (Divide them all by the right constant and it's an infinite orthonormal set.) Now suppose $W$ is any finite set. Thus the span of $W$ is a finite-dimensional subspace. Since $\{f_n\}$, being orthogonal, is independent, there must be some $n$ for which $f_n\not\in\text{span}(W).$ (You can't have infinitely many independent vectors in a finite-dimensional vector space.) We can explictly construct $g\in\text{span}(W)$ such that $f_n-g$ is orthogonal to everything in $\text{span}(W).$ (The construction is a standard part of the Gram-Schmidt algorithm.) That proves that $W$ is not complete.


integration
calculus
geometry
calculus computations

Hello,

2 problems who can you integrate this: $\int \frac{e^x}{x^2} - \int \frac{2e^x}{x^3}$

Second find the integral $\int \int \int_{s} z \ dzdxdy$

where is the area in the first octant limited by a plane $y=0 \ \ z=0$ $x+y=2 \ \ 2y+x=6$ and the cilinder $y^2+z^2=4$

I think the best way to do thit is first makking a skecth

[img]http://img102.imageshack.us/img102/4025/pror5hc.jpg[/img]

then I find the integral:

$A= \int^2_0 \int^2_{2-x} \int^{\sqrt{4y-y^2}}_0 \ z \ dz \ dy \ dx=\frac{4}{3}$

Is this oké?

Then B is $B=\int^6_2 \int^{3-\frac{1}{2}x}_0 \int^{\sqrt{4-y^2}}_0 \ z \ dz \ dy \ dx =\frac{34}{3}$

Is this oké?
Then I compute and I get $\frac{4}{3} \ + \frac{34}{3} \ = \frac{38}{3}$

and this is not the same solution as in my book they get $\frac{26}{3}$

Where do I miss? greets.

First of all, the answer to the first integral is exp(x)/x^2 (check it by differentiating).  There's not much to explain (you can use integration by parts though, if you choose e^x as dv and x^-2 as u).

For the second one, your setups are correct.  I also checked them on my TI-89 (which can do multiple integrals) and I got 2 for A and 20/3 for B.  2 + 20/3 = 6/3 + 20/3 = 26/3, the answer you said the book gave.  So you should check for any mistakes in the integration.

[quote="Bert"]impossible?[/quote]

What you mean ? It is just integrating by Parts  ;)

[quote]

What you mean ? It is just integrating by Parts [/quote]

Yes i see that he use PI but I can't see who ye get the solution by this way??

Greets

[quote="shyong"][quote="Bert"]First problem who can you do that by PI ? I find it doing by a substitution $y = \frac{{e^x }}{{x^2 }} \Rightarrow dy = \left( {\frac{{e^x }}{{x^2 }} - \frac{{2e^x }}{{x^3 }}} \right)dx \Rightarrow \int {dy} = y + c \to \frac{{e^x }}{{x^2 }} + c$[/quote]

I guess he mean like this

$\int\frac{e^x}{x^2}dx=\frac{e^x}{x^2}-\int(-2)\frac{e^x}{x^3}dx=\frac{e^x}{x^2}+2\int\frac{e^x}{x^3}dx$

 ;)[/quote]

you are right.

I see that you can start so but finaly you need to find $\frac{e^x}{x^2}$ but who do you do that? do I need to do a PI again?

Greets.

[quote="shyong"][quote="Bert"]First problem who can you do that by PI ? I find it doing by a substitution $y = \frac{{e^x }}{{x^2 }} \Rightarrow dy = \left( {\frac{{e^x }}{{x^2 }} - \frac{{2e^x }}{{x^3 }}} \right)dx \Rightarrow \int {dy} = y + c \to \frac{{e^x }}{{x^2 }} + c$[/quote]

I guess he mean like this

$\int\frac{e^x}{x^2}dx=\frac{e^x}{x^2}-\int(-2)\frac{e^x}{x^3}dx=\frac{e^x}{x^2}+2\int\frac{e^x}{x^3}dx$

 ;)[/quote]

Then subtract 2*int(exp(x)/x^3)dx from both sides, and you're left with the original integral = exp(x)/x^2.  (Actually, I first just solved it in my head, since it looked like product/quotient rule, I tested exp(x)/x^2, which worked)


Find the maximum and the minimum values of the function 3x-y+5, subject to the restrictions y>=1, x<=y,and 2x-3y+6>=0.

This is just a linear programming problem. Graph the inequalities, find the satisfactory area, and take the points in the corners of your area. Plug them into the function, one of those must be the maximum.


[quote=pigfly]Let ABC be a triagle.Prove that:
$(a+b+c)^3\prod (p-a)\geq( \sum a^2)^3\prod cos A$[/quote]
Any one?

[quote=pigfly]Let ABC be a triagle.Prove that:
$(a+b+c)^3\prod (p-a)\geq( \sum a^2)^3\prod cos A$[/quote]

Nice!
In any triangle,have

$(a+b+c)^3\prod (p-a)\le( \sum a^2)^3\prod \sin{\frac{A}{2}} $

[quote=pigfly]Let ABC be a triagle.Prove that:
$(a+b+c)^3\prod (p-a)\geq( \sum a^2)^3\prod cos A$[/quote]

this is a hard!


ABC is a triangle. Show that there is a unique point P such that PA2 + PB2 + AB2 = PB2 + PC2 + BC2 = PC2 + PA2 + CA2.

a solution can be found at:

http://www.kalva.demon.co.uk/short/soln/sh95g2.html


The funniest amongst all... where does this belong ?

I don't know.
I found it on the internet.

Unless you stay in between the center rails.


geometry
algorithm
induction
graph theory
combinatorial geometry
combinatorics proposed
combinatorics

There are n points in the plane, and let d be the smallest distance between 2 of them. Prove that we can remove 3n/4 of the points such that no pair of the points we left is at a distance d.

Consider the graph $G$ whose vertices are the $n$ points, two joined by an edge if and only if they are at distance $d$ from each other.
It is a classical result that this graph $G$ is planar (see for exemple the course in graph theory at http://www.animath.fr ex. 5 chp. 2).

We will 'prove' that $G$ is 4-colorable :

Wlog we may assume that $G$ is connected, otherwise we only prove it for each of its components.
Since $G$ is planar then its dual graph $G^*$ is planar too. From the four-color theorem (this explain why I used the '' above), we know that any planar representation of $G^*$ is a 4-colorable map. It means that its own dual $(G^*)^*$ is 4-colorable.
Moreover, since $G$ is connected, it is also classical that $(G^*)^*$ is isomorphic to $G$, so that $G$ is 4-colorable.

Then, using a 4-coloration of $G$ we divide the vertex-set into 4 parts. We only have to remove the three of these parts with the least number of elements. In the remaining part, we have at least $\frac n 4$ of the points, and no two of them are at distance $d$ from each other by definition of the coloration, thus we are done.

Pierre.

Pierre you are very good in graph theory but I am not, so i ll be very thankful if you explained a little the results you used. As far as i see you showed one can color the vertices in 4 colors in such a way that every edge has verticces of different colors, right?...Anyway your proof is great.

However i Have a elementary proof.
As you said construct the same graph and worry about conected components. Let T be a conected component, we are going to show that we can remove 3 vertex from T that left unconected another vertex, and inductively the result follow.

Let l be a line not parallel to any line determined by 2 of the points, and sweep the plane with this line, let P be the first point from T it touches, then P is conected to at most 3 other vertices, because they all lie in half of the circunference with center P and radius, not including the endpoints. So remove these 3 vertexs and we get what we wanted.

Obviously yours is more direct, and i would like to understand it

Well, in fact there is not much to understand but you have to know about the four-color theorem (even you do not know its proof which is very very long. It uses computers to be achieved in a human time. In my particuliar situation I did not read it), which states that :
Every planar map on the plane is four-colorable. (1)

This is the most well-known way to express it, but as shown in my previous post, it can be also expressed as :
Every planar graph is 4-colorable. (2)

You are right, here a $k$-coloring is to color each of the points with one among $k$ available colors in such a way that every two points joined by an edge (here two points at distance $d$ from each other) have distinct colors.

Therefore, if you know about (2), the rest of my proof is easy (all the complicated stuff is to explain how to go from (1) to (2) ).

Anyway, I like your proof. I was thinking about something like that in the beginning  : any vertex has degree at most 6 and use the convex hull of the points, but I did not achieved it in that way :( and did not see that we may select the points to have degree at most 3 at each step... Then I saw the problem as a coloring-graph one, and I did not come back to the first approach. I'm really happy that there is such a nice way, which do not use the four-color theorem...

Pierre.

I agree that Pascual's solution is great.

Pascual! Did you participate in IMO?

Pierre! Is it true that four-colour theorem has only boring proof using numerous number of cases? At least, I heard about this.

That's what I heard too, even if it seems they have a bit simplified the proof but not as much as it would be to become attractive... :(

Pierre.

In another hand, Pascual' proof is 'just' an algorithm to prove that this particuliar planar graph is 4-colorable :
Indeed, use induction.
The induction steps is done by Pascual's argument to isolate one vertex, say $M$. Then, by the induction hypthesis, you may 4-color the induced subgraph (the one without $M$ and its edges). From Pascual's argument, $M$ has no more than $3$ neighbours, so it leaves at least one color for $M$, and we are done.

Pierre.

Thanks for your comments! I am really happy to receive such great comments from 2 great problem solvers. :D  :D

To Myth: Yes I took the IMO but i got 31 marks, a silver medal! Hope to do better next year!


Call a number $a-b\sqrt{2}$ with $a$ and $b$ both positive integers tiny if it is closer to zero than any number $c-d\sqrt{2}$ such that $c$ and $d$ are positive integers with $c<a$ and $d<b.$ Three numbers which are tiny are $1-\sqrt{2}$, $3-2\sqrt{2}$, and $7-5\sqrt{2}.$ Without using a calculator or computer, prove whether or not each of the following is tiny: \[ (a) 58-41\sqrt{2} , \hspace{1in} (b) 99-70\sqrt{2}. \]

I have an idea but would like to see proof/answer

[hide="See if it is correct??"]
We see that for $a-b\sqrt{2}$<$c-d\sqrt{2}$
$\Rightarrow\frac{1}{a+b\sqrt{2}} < \frac{k}{c+d\sqrt{2}}$
Which is true since $a>c$ and $b>d$ and for $k\ge1$
This implies $a-b\sqrt{2}$ is tiny if $(a-b\sqrt{2})(a-b\sqrt{2})=1$
$\Rightarrow a^2-2b^2=1 or -1$

$(i)$ $58-41\sqrt{2}$
  $58^2 - 2\cdot41^2=2$

$(ii)$ $99-70\sqrt{2}$
  $99^2 - 2\cdot70^2=1$
Hence only $(ii)$ is a tiny number.
[/hide]


Does anyone know where do I find the book Lightning- Physics and effect, by Rakov Vladimir and Martin Uman?
Please help me. I need it badly.


find the greatest commond divisor of natural numbers a and b satisfying
$(1+\sqrt{2})^{2007}=a+b\sqrt{2}$

[hide]Let . Then we have the recurrence  and . By the Euclidean algorithm, we have , so  is invariant, so we have.

[/hide]

[hide]
\[\left( 1+\sqrt{2}\right)^{2007}=a+b\sqrt{2}\Rightarrow \left( 1-\sqrt{2}\right)^{2007}=a-b\sqrt{2}\]
Multiplying the two,
\[a^{2}-2b^{2}=-1\]
Suppose $d$ is a (positive) common divisor of $a$ and $b$, $a=dx$ and $b=dy$ where $x$ and $y$ are positive integers. Then
\[d^{2}(x^{2}-2y^{2})=-1\]
In particular, $d$ divides 1, so $d$ must equal 1 and $\gcd (a,b)=1$.
[/hide]


Right so I made one like 2 days ago and posted it on my blog, and apparently its not AMC season, but who cares, if you want to take it, just pm me your answers by the 5th or 6th of July., and if you find any errors pm. i know i can't do anything about this, but please if you take it, do it like you would do a normal AMC 12. i think its a little easy for and AMC 12 and don't be surprised if some problems look farmiliar  :wink:.

why is it still named Mock AMC [b]10[/b].pdf?

:D

lol, idk. i guess i am still sleepy at 11:26 ? :|

yeah, its an old AIME problem, but i changed the numbers a bit.

I am making the deadline JULY 6th, since july 4th is july 4th and idk about july 5th.

dude look

aime level ~#9 $ \neq$ amc 12 level :P

i personally find AMC 12 problems harder than AIME problems for some reason. BTW the question that i used was a variation of an AIME #7. The problem itself is not that difficult. you'd be surprised at the difficulty of some problems that you do with out knowledge of placement. for example i was doing this semi-challenging problem once and i knew it was an AIME problem, i thought it was a #8 or 9, but it turned out to be a #14. had i known it was a #14, my mind would have told me, "you can't do this", and i would find the test alot more difficult.

@#H34N1 i remember once Silverfalcon made a Mock AMC 12 and a few of the problems were AIME #15 level according to a majority of the takers.

now there are 24 downloads, so if i don't get at least 10 by july 6th, then i am not going to give the AIME now, i will give it when i give my next AMC 12.

Several months later, I personally agree with cognos. The last AMC questions are harder than most AIME ones.


If he is just starting Algebra 1 then I say yes, AoPS Vol. 1 will be too hard.

[quote="bpms"][quote="Phoenix257"]Hi,

You might have seen my other topic about MC notes.  I am now wondering if it's ok to start my little brother (7th grade) with AoPS Vol 1.  He will just be starting algebra 1 this upcoming school year.  Do you think AoPS would be too hard for MC?

Thanks for your advice![/quote]
If you are going to teach him the materials then it is the perfect thing for him. If he is going to do it on his own then it might be too difficult, or it might not. One word of advice is don't use the sol. manual too much.[/quote]
I first read Vol 1 when I was in 7th grade. It is written in a quite easy to understand format. I picked up lot of terminologies. When you read on your own and try to understand the concepts, it registers better. There may be sections he needs to skip on the first time. May be you can suggest him what to skip initially. He can come back to the skipped ones later.

If your just starting algebra, you might have some difficulties.  I read it when i already had taken algebra.  I think its about the same level as mathcounts though.

I started it before I started Algebra 1. I could do some of it easily, but some sections were just too hard. It was much easier to understand some of the sections after I did Algebra though.

So, my advice would be to start, but not expect to be able to do everything and skip the sections that are too difficult/confusing.

Basically I'm just confirming what mthiyaga said.

uh i started the aops book 1 after i started pre-calc, meaning I forgot everything in algebra 1, so I'd say most of the chapters shouldn't be too hard for your little brother.

Make sure the stuff that you will be teaching is at your brother's level.  Then it will be suitable.  Okay.

[quote="Phoenix257"]Hi,

You might have seen my other topic about MC notes.  I am now wondering if it's ok to start my little brother (7th grade) with AoPS Vol 1.  He will just be starting algebra 1 this upcoming school year.  Do you think AoPS would be too hard for MC?

Thanks for your advice![/quote]

[hide]It depends. I thought it was just right when I was in Geometry/Algebra 2. It might be a bit boring though. It was for me[/hide]

It really doesn't matter what class he is taking, AoPS books and what happens in the classroom are almost parallel.

For example, my school is screwed up, so I took Pre-Algebra in 7th grade  :mad:  :mad:  :mad: but I went through about half of AoPS vol. 2, understanding almost everything. My experience with high school algebra and geometry is sadly, limited.

I had never heard of Algebra I when I started it.  It basically starts with the basics  :P  from what I know.  It should be good, it mostly emphasizes thinking rather than concepts.

aOPS Volume 1 is appropriate if your sister/brother wants to learn.

They also would probably want MathCounts Materials as well.

I don't even own AoPS volume 1, but from what I can see from the table of contents posted on the website and the amount of assumed knowledge in AoPS volume 2, the only prerequisites for volume 1 are a strong elementary school math backround, which hopefully your sister/brother has if he/she is taking Algebra I.

I in 7th grade and haven't taken an algebra 1 class, but am able to do most of the non-geometry in the book (I hate geometry! :diablo:).  I am not quite finished with the book but I really recommend it!

It is hard to do by yourself if you don't have the answers so if you do than you really have to want to study..

If you are unsure you can post a problem here where someone will solve it.


geometry
symmetry
geometric transformation
reflection
number theory

Does anyone know if there is an answer key or solutions to any part of PUMaC posted somewhere? Answers to questions you got would be great too.

I'ma get my tests and post what I rememberz.

OK. I took Algebra and Number theory, so here are those:

[hide="Algebra"]1. 7[/hide]

[hide="Number Theory"]1. 42
2. 1
3. 719
4. 6

6. 1111

8. 31[/hide]
I don't remember any other answers. :(

As for my own answers, I did algebra and combo so here are the ones I'm pretty much positive I got right...

ALGEBRA
1. 7
2. 455
4. 120
5. 8
7. 511

COMBO
1. 34
2. 8 (A lot of people got 10 or 11, but I've seen a similar problem before but with a different number than 2009 and the answer should be 8...The weights are 1, 3, 9, 27, 81, 243, 729, 2187; note that a weight can be placed on either side of the scale)
3. 72
4. 22
5. 9
6. 21
7. 606 (This is the answer if the problem said "fourth" or "fourth smallest" instead of "fourth largest," but I heard that it was a typo and 606 was counted as correct)

Number Theory:
1. 42
2. 1
3. 719
4. 6
5. 31
6. 1111
7. 595
8. 31

Combinatorics:
1. 34
2. 8
3. 72
4. 22
5. 9
6. 21
7. 606 (Intended answer. The actual wording gives an answer of 303, but 606 was the answer they used for grading)
8. 198

[quote="andre4177"]As for my own answers, I did algebra and combo so here are the ones I'm pretty much positive I got right...

ALGEBRA
1. 7
2. 455
4. 120
5. 8
7. 51
[/quote]

I'm pretty sure the Algebra Answer Key was:
1.  7
2.  455
3.  810
4.  120
5.  8
6.  3
7.  511
8.  125

For question #3 on Geometry A, I think that it could be argued that a concave 11-gon actually can have 2 symmetries (the do-nothing symmetry and the reflection symmetry).  Would that be a valid answer, or am I just missing something here?  The question was a bit ambiguous, so I think it could be argued either way.


if it`s $\frac{1}{n^{n}}$ and $n^{n}$ then AMGM in  both.....


Find all functions $f: \mathbb{Z}\to \mathbb{Z}$ such that for all $m,n\in \mathbb{Z}$: \[f(m+f(n))=f(m)-n.\]

I got that no function existed. :huh:

$ m \equal{} n \equal{} 0$ gives $ f(f(0)) \equal{} f(0)$, then $ m \equal{} 0, n \equal{} f(0)$ gives $ f(0) \equal{} 0$. $ m \equal{} 0$ gives $ f(f(n)) \equal{}\minus{}n$.

Suppose that $ f(1) \equal{} a$. Then $ f(a) \equal{}\minus{}1$, $ f(\minus{}1) \equal{}\minus{}a$, and $ f(\minus{}a) \equal{} 1$. We can see here that $ a$ can't be $ 0$, $ 1$, or $ \minus{}1$.

But now let $ m \equal{} 1, n \equal{}\minus{}a$. We get $ f(2) \equal{} 2a$. We can induct to show that $ f(n) \equal{} na$ for all $ n\in\mathbb{Z}$. But now consider $ n \equal{} a$. We get $ f(a) \equal{} a^{2}\neq\minus{}1$, contradiction.

[quote="MellowMelon"]I got that no function existed. :huh:[/quote]That is indeed correct. :)

Let $P(m,n)$ be $f(m+f(n))=f(m)-n$.It is easy to show that $f$ is bijective.

$P(0,0)\Rightarrow f(f(0))=f(0)$.From injectivity,$f(0)=0$.

$P(0,n)\Rightarrow f(f(n))=-n$.Then $f(m+f(n))=f(m)+f(f(n))$.From surjectivity,$f(m+n)=f(m)+f(n)(\forall m,n\in \mathbb{Z})$.This is [b]Cauchy equation[/b].Hence ∃$c\in \mathbb{Z}$ s.t. $f(m)=cm$.Substituting this into $P(m,n)$,$c^2n=-n$ which implies $c^2=-1$.This is absurd.Therefore [b]there is no such $f$[/b].$\blacksquare$

Let $P(m,n)$ denote the given assertion.
$P(0,0)\implies f(f(0))=f(0) \text{ and } P(0,f(0))\implies f(0)=0.$
$P(0,n)\implies f(f(n))=-n.$ So $f$ is surjective, let $f(z)=y.$
$P(x,z)\implies f(x+y)=f(x)+f(y).$ So the well known solution is $f(x)=kx,$ checking yields a contradiction. So no function exists.


search
number theory unsolved
number theory

Hi... I was wondering if anyone could help me prove the following:

If p is an odd prime, prove that

1^n + 2^n + 3^n + ... + (p-1)^n

is congruent to 0 (mod p) if (p-1) does not divide n
OR
is congruent to -1 (modp) if (p-1) divides n

I was thinking of using Dirichlet's theorem but then again I don't know how or where to start to start proving this. It would be appreciated if someone could help. Thanks in advance!

i guess this is  a standard result but forgot the  name.  :lol:

Someone please move this to the Pre Olympiad section.

No, someone rather locks it for being posted before (the newest one about two weeks ago). Search for "Chevalley Warning" (I posted a proof for that linking to this)
[hide=" :wink: "]http://www.mathlinks.ro/viewtopic.php?t=40171[/hide]


[b]Problem[/b]
Given $a, b, c , x , y , z > 0$.
Prove that $\frac{x^4}{a^3}+\frac{y^4}{b^3}+\frac{z^4}{c^3}\geq \frac{(x+y+z)^4}{(a+b+c)^3}$.

[b]Question[/b]
Besides using Holder's Inequality, is there any other method to solve the above problem?

I'm pretty sure expanding and using Muirhead would work well...

but why another solution? This is perfect example of Holder in Engel form

[quote="blahblahblah"][quote="molina"]This is a HOMOGENUOS inequation.

Then, we can do $\large{x+y+z\,=\,a+b+c\,=\,1}$

We need to prove that :

$\large{\frac{x^4}{a^3}+\frac{y^4}{b^3}+\frac{z^4}{c^3}\,\geq{1}}$

It`s a hint.[/quote]

I don't think that helps.[/quote]

  It works.
$\frac{x^4}{a^3}+a+a+a \geq 4 \sqrt[4]{x^4}=4x$
Sum up and use $a+b+c=x+y+z=1$ get
$\sum \frac{x^4}{a^3} \geq 1$

Very nice!
So, applying the same tecninque, the inequality can be extended to the form:
$\displaystyle {\sum_{i=1}^n \frac{x_i^{m+1}}{y_i^m}}\geq \frac{(\displaystyle\sum_{i=1}^n x_i)^{m+1}}{(\displaystyle\sum_{i=1}^n y_i)^m}$ for any positive integer $m$.

Well,  Chebychev's inequalitycan be used to solve too.
and AM-HM as the last step
 :)


Why on the TI-89, when graphing $x^{1/3}$ (or the odd root of any variable), the negative $x$ values are not graphed and considered imaginary? While on the TI-83, the negative $x$ values (of $x^{1/3}$) are graphed.

Hmmmm, the negative x values do appear on a TI-89. Maybe you changed your settings to include imaginary values. Try it on REAL format.


hey guys. I'm going to begin studying for the AMC's now,but I dont know where to begin. The problems are all just so hard for me to do, and in my area, I can't find any books that will teach me the material. Can you guys help me out?

Ditto. Work through the two AoPS books. They'll teach you the materials you need to know, and give you a decent amount of practice working with those kinds of problems. Supplement with old tests for practice.

Yeah thing is, no AoPSS books are in stock in Flushing New york. I checked... x.x"""

Or amazon.  Of course, always click a link from the aops website to get to amazon.  It helps out the aops people.

Just buy the AoPS books and if you are really motivated, got extra money, and are willing to spend sometime, buy one or two introductory books too.


Dickchimney mentionned the following problem in the topic : " sequence intersecting given segments" : if f is continuous and  f(nx)-> :inf: for all x>0 then f(x)-> :inf: .

Show that the following (more difficult?) result is also true : if f is continuous and for all x>0, the sequence f(nx) converges, then f(x) converges too.

[quote="alekk"]It has already been discussed on the forum. It's a consequence of Baire theorem  :)[/quote]

 Yes, it is one solution. Another solution which is suitable for highschool students is discussed here

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23115]Sequence Intersecting Intervals[/url]
 It is just few topics below this one :D


trigonometry
complex numbers
imaginary numbers

What is  :sqrt: (4+4i) ?

i= :sqrt: (-1)

Here's a general tip for finding square roots of imaginary numbers:

a + bi = :sqrt: (c + di)
(a2 - b2) + (2ab)i = c + di
a2 - b2 = c
2ab = d

And you can solve from there.

[quote="JMI"]What is  :sqrt: (4+4i) ?

i= :sqrt: (-1)[/quote]

Don't know of any "easy" ways, but:

[tex]\sqrt{4+4i} = \sqrt{4\sqrt{2}(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4})} = 2\sqrt[4]{2}(\cos \frac{\pi}{8} + i\sin \frac{\pi}{8})[/tex]

[quote="JMI"]Would the roots be  :rt2: -i :rt2: and - :rt2: +i :rt2: ?[/quote]

[tex](\sqrt{2}-i\sqrt{2})^2 = -4i[/tex]

[quote="probability1.01"]
Don't know of any "easy" ways, but:

[tex]\sqrt{4+4i} = \sqrt{4\sqrt{2}(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4})} = 2\sqrt[4]{2}(\cos \frac{\pi}{8} + i\sin \frac{\pi}{8})[/tex][/quote]

How do you know what goes where?

I think it's technically the nth root theorem, but yeah, whatever.  The basic idea is there.  Available at your closest math dealer.  Or, order online at AoPS for a limited time offer of $0.01!*

*[size=67]1000000 payments[/size]

Basically, multiply the following out and see what you get (a good trig exercise!).

[tex]r_1(\cos\theta+i\sin\theta)*r_2(\cos\phi+i\sin\phi)[/tex]

Notice how any imaginary number can be represented in this way on the imaginary plane.  (if you don't know what that is, google it)

Is the following De Moiveres Theorem?

[r(cos :theta: +i sin :theta: )]^n=r^n(cos n:theta: +i sin n:theta:)

Yes, but I encourage you to also consider what I posted earlier (a generalization).

I think your hints are sometimes a little too hard probability1.01. I think mine lately have been too. But a hint can't be harder than the original problem or too vague, or it isn't of much use. You know what I mean?

For this problem JMI, try to write 4 + 4i as r eix. Then use DeMoivre's Theorem (remember that there are 2 square roots also).

I just found something in a book and everything you have said makes a lot more sense now. Thanks for the help. (the stuff you said also helped make sense of what the book said.)


geometry
geometric transformation
rotation
reflection
vector
advanced fields
advanced fields unsolved

part i. Suppose $\sigma$ is an automorphism and $\tau$ is a rotation with oriented angle $\theta$ around a point $P$. Show that $\sigma\circ\tau\circ\sigma^{-1}$ is a rotation around $\sigma(P)$.
What is the angle of rotation?

part ii. Suppose $\sigma$ is an automorphism and $\tau$ is a reflection through a line $l$. Show that $\sigma\circ\tau\circ\sigma^{-1}$ is the reflection through $\sigma(l)$.

part iii. Suppose $\sigma$ is an automorphism and $\tau$ is translation by a vector $v$. Show that $\sigma\circ\tau\circ\sigma^{-1}$ is translation.
What is the displacement of this translation?

I dont understand the question.
Automorphism with what, a group? But with what group? Is it $\mathbb{R}^{2}$, i.e. Euclidean Plan under addition?

sorry it's automorphism in the Euclidean plane.

and i think it's probably safe to assume that it is $\mathbb{C}^{2}$


calculus
integration
function
conics
hyperbola
real analysis
real analysis unsolved

Calculate
$ \int\limits_0^1\int\limits_0^1 a^2\frac{(xy)^{a-1}}{ln(xy)}dxdy $ and
$ \int\limits_0^1\int\limits_0^1 a^2\frac{[(1+x)(1+y)]^{-a-1}}{ln(1+x)(1+y)}dxdy. $

I will only solve the first one as they are essentially the same.

as you can see the integrand is only a function of one variable namely u=xy .
so one might try to convert this double integral to a single integral. one way is a direct use of the jacobian method.
Note that the double integral represents the volume beneath the cure
f(x,y)=(xy)^a /ln(xy)
we will partition the square 0=<x<=1 and 0=<y<=1 by hyperbolas of the form xy=u where  0<=u<=1.
hence the same volume can be written in the form
Volume= integral[0 t0 1] [u^a/ln(u)] (ln(u))(1-1/u)du =integral u^a -u^a-1
and so the answer is -a/a+1.


linear algebra
matrix
vector
algebra
polynomial
number theory
least common multiple

Let $ p$ be a prime number and $ n\geq 2$ and $ A\in M_{n}(\mathbb{Z}_{p})$ be an invertible matrix. Prove that there exists $ k\in\mathbb{N}^{*}$ such that $ A^{k} \equal{} I_{n}$.

$ G\equal{}\{A \in \mathcal{M}_n(\mathbb{Z}_p) |\det A \neq 0 \}$ is a group.But since $ |G|< p^{n^2}$ it follows that $ G$ is a finite group.So for any $ A \in G$. there is $ k \in \mathbb{N}$ such that $ A^k\equal{}I_n$.

so now find the maximal order of an element of $ GL_n(\frac{Z}{pZ})$

[quote="alekk"]so now find the maximal order of an element of $ GL_n(\frac {Z}{pZ})$[/quote]

$ |GL_n(\mathbb{Z}_p)|\equal{}(p^n\minus{}1)(p^n\minus{}p)\cdots (p^n\minus{}p^{n\minus{}1})$

In order to deduce this formula, one can consider the columns of the matrix $ A$ as linearly independent vectors (since $ rank(A)\equal{}n$) and try to count the possibilities of choosing $ n$ linearly independent vectors over $ \mathbb{Z}_p$.

Over a field of characteristic zero, the answer is the largest value of $ m$ such that $ \varphi(m) \le n$, and this is a lower bound in the general case.  But in general the answer depends on how various cyclotomic polynomials split over characteristic $ p$, and I'm not certain there's a simple characterization.

Here are the numbers over $ \mathbb{Z}/p\mathbb{Z}$:
If $ m\equal{}p^r\cdot q$ with $ p\nmid q$, there is an element with order $ m$ if $ n\ge (p^{r\minus{}1}\plus{}1)\plus{}\sum s_i$ and $ q_i|p^{s_i}\minus{}1$ for some $ q_i$ with $ \mathrm{lcm} q_i\equal{}q$ ($ p^{r\minus{}1}\plus{}1$ is interpreted as $ 0$ here).

Essentially, we construct $ s\times s$ blocks corresponding to elements of the field of order $ p^s$, and large Jordan blocks for the $ p$-power elements. If we want $ (I\plus{}N)^{p^{r\minus{}1}}$ to be nonzero while its $ p$th power is zero, we need $ N^{p^{r\minus{}1}}$ to be nonzero.

The multiple $ s_i$ is also important; we can get order $ 2^4\cdot 5\cdot 13\equal{}1040$ at $ n\equal{}7$ this way when $ p\equal{}3$, for example.

For all of this complexity, the single largest order is easy to pinpoint as $ p^n\minus{}1$.

$ F_q$ is the field with $ q\equal{}p^r$ elements. We construct a matrix $ A\in{GL_n(F_q)}$ which is of order $ q^n\minus{}1$.
$ p$ is not a divisor of $ q^n\minus{}1$; then $ A$ is necessarily diagonalizable and $ A$ is of order $ q^n\minus{}1$ iff for all eigenvalues $ \lambda_i$, $ \lambda_i$ is a root of unity of order $ s_i$ with $ lcm(s_i)\equal{}q^n\minus{}1$. The product of all irreducible polynomials s.t. their degree is a divisor of $ n$ is $ x(x^{q^n\minus{}1}\minus{}1)$. Let $ F$ be the splitting field of $ x^{q^n\minus{}1}\minus{}1$. It is a finite field; let $ a$ be a generator of $ F^*$ and let $ P$ be the $ a$-minimal polynomial (of degree $ n$) over $ F_q$. Then $ P$ is a divisor of $ x^{q^n\minus{}1}\minus{}1$ and $ inf\{l; P$ is a divisor of $ x^l\minus{}1\}\equal{}q^n\minus{}1$. Then we take $ A$ as the companion matrix of $ P$.
Example: for $ n\equal{}4,q\equal{}5$ we can choose $ P\equal{}x^4\plus{}3x^3\plus{}3x^2\plus{}2$ or $ P\equal{}x^4\plus{}4x^3\plus{}3x^2\plus{}3$ for an order equal to $ 5^4\minus{}1\equal{}624$.
see also:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=116201

Alternately, find an irreducible $ n$th degree polynomial, and let $ B$ be a companion matrix for it. The polynomials in $ B$ form a field of order $ q^n\minus{}1$; let $ A$ be a generator of the multiplicative group of this field.


AMC
AMC 10
AMC 12
AIME
USA(J)MO
USAMO
MATHCOUNTS

Alright, so it looks like the AMC season is fast approaching, with the first A test (which I will not be taking) next Tuesday.

What are all of your goals for this year?

140+ on AMC10 (if I take it), 130+ on AMC12, 8+ on AIME, qualify for USAMO. Possibly MOP, but unlikely, unless I do noticeably better on USAMO than I'm expecting.

I'm aiming for 21 on USAMO and pretty much the best I can do on AMC/AIME

In fact screw the 21 for USAMO, I'm still aiming for the best I can do on USAMO, solve as many as I can and not lose proof-writing points.

Wow a 21 is pretty good. I'm expecting a 14 as of right now. I hope 140+ on AMC next Tuesday and probably a at least a 10 on the AIME. That's ok I think. Hopefully the index won't be like 240+ or something...

Well yeah 21 seems be unlikely right now, but if I keep working from now maybe I'll make it.

I BELIEVE!

Hmm, let me just throw out some random numbers for goals...

AMC-$135$
AIME- $12$
USAMO-$15$

Of course, it all depends on the difficulty of the tests.  I'm assuming 'average' tests.  I'll be happy with any improvement from last year.

Well, right now I'm hoping for an improvement on my AMC score (last year was 132.5), and then two plus the floor on the AIME (I'm guessing about nine-ish, depending on the difficulty of the test)

USAMO ... who knows.  I made it last year and got a 13, and would be thrilled to improve on that since I don't expect the test to be as easy as last year's.

JB

Hmmm let's see this is a fun topic to randomly post in so yeah time to randomly post:

AMC 12 -- 150
AIME -- 16 wait no maybe more like 12/13
USAMO -- ??? actually seriously hmmm well yeah I have no idea... but hopefully a high score ;)

amc 12 - 130
aime - 10+ but realistically more like an 8 :( :(
usamo - [i]if[/i] I even make it, anything to get me into MOP

Yes, they're all in high school, but those were some of the more well qualified problem solvers of the forums which is probably why they have such high goals.  Anyway my goal is USAMO qualification.

Well, most of the people who have already posted have made the USAMO before...so their goals will be higher than most people.

[quote="ch1n353ch3s54a1l"]Are most of the people in this thread in high school? Because from my standards, you people have pretty high goals. >.> which is why i won't post mine.  :)[/quote]

That's because people who posted above believes (or had believed) and you should believe too!

Actually maybe it's too late to believe now oh well. :(

Nice to see so many people setting high goals.  Good luck!

My goal is... to read all of your scores on Wednesday (or Thursday?)!  :P

boing ... I'll try to actually make mine this year - 130 on AMC

I don't know it shouldn't make that much difference. But yeah if you are taking both AMC's you can try that on AMC B.

Heh...now I will have three ways to qualify for the AIME: AMC12A, AMC12B, and USAMTS (if I qualify through USAMTS)

AMC 10A: 150 (More realistic 144, given the inevitable stupid mistake or annoying Geometry problem)
AIMEB: 7 or 8
USAMO: Would be happy to simply take the test... so how is this scored?

Is there any way we can take the AMC 10/12B outside of school, as mine doesn't offer the B test? I want to take the 12B just for experience sake, as I'm a freshman and have only taken the 10s once.

[quote="tokenadult"]Among the guys I have coached recently, caution gains more points than daring. For test-takers taking on both the "A" date and the "B" date, I recommend a conservative strategy to qualify for AIME for sure on the A date, and then if the test-taker wants to go for broke and throw caution to the wind, to do that on the B date.[/quote]

Exactly what I'm doing:

[quote="qweretyq"]A test I'm going to be a little bit more cautious and am not going to guess unless I'm fairly sure I'm correct. B test I'll be a bit more daring. [/quote]

another goal of mine is to get highest score in my school which is a bit of a reach considering my school usually gets 1st-3rd place

I hope to qualify for AIMEs next year and get around 4. I missed the cut-off this year by 1 question.

Does anyone know how the scoring system will be next year? It seems much tougher, with the 1.5 points for a blank deal. Does anyone know if correct answers are then worth more or if the cut off is lower? Anything to compensate for the decline in point reward?

I think I am going to work through Art of Problem Solving vol. 2 and have it finished by summer and just cultivate my skills from there.

[quote="JesusFreak197"][quote="chess64"]This is kind of random, but: for some people, it might be beneficial to do AMC's backwards (starting from #25). It raised my score a few points :P[/quote]

I actually wouldn't do that in general, especially not on the 12. If you happen to get caught up in a problem and spend too much time on it, you'll get behind. It's better to solve 15 problems or so in the first 30-35 minutes than to solve like 4 in that time. ;)[/quote]

Hmmm I don't know, for me it worked. I guess it's because I've been solving AIME problems, and the last few are often close to the difficulty of the first few AIME. When I go from the start to the end, I always make so many careless mistakes it's not even funny. When I do harder problems, I rarely make careless mistakes. *shrug*

[quote="breez"]Does anyone know how the scoring system will be next year? It seems much tougher, with the 1.5 points for a blank deal. Does anyone know if correct answers are then worth more or if the cut off is lower? Anything to compensate for the decline in point reward?[/quote]

I think the first 15 or so questions will decrease in difficulty.

[quote="tokenadult"]Among the guys I have coached recently, caution gains more points than daring.[/quote]
Well, on the 12B, I had essentially nothing to lose (9th-grader, made 123 on the 12A), and I acted pretty reckless, ending up with a 132.5.  On the other hand, I have a sort of obsessive-compulsive thing about checking my work, so "reckless" for me might not be the same as "reckless" for someone else.  And, of course, I would have been extremely cautious if I hadn't done well on the A.

I guessed on the last 5 on the B. I was hoping for a perfect, and I sorta just gave up after I couldn't do number 21...ironically, i guessed 21 right and all the other ones were wrong.

I'm think I got around a 100 on the AMC.  I'm in 9th grade right now.  Do u think I'll make it to the AIME in 10th grade?  Is that too high?  How many hours do people study per day to get in to the AIMe?

thanks

AIME- floor+
USAMO- not 0

To improve on the AIME, is the key learning new methods for solving problems, or just improving your ability to figure out methods? In other words, do people who qualify on the AIME immediately know now to solve every problem on the AIME and just need to be accurate and fast, or do they actually have to think up a method? Same for the USAMO.

Also, I just recently ordered art of problem solving vol. 1 and 2. Is vol.1  too easy for the AIME? I started it, but I'm seem to know most of what is in it, so should I start vol. 2?

Yes, start Vol. II.

I'm not sure what you mean by your other question. Just because someone scores the floor or above on the AIME doesn't mean that they automatically know how to do all of the problems and just ran out of time. In fact, most AIME problems are fairly simple if you just see the right way into them. For example, I can generally solve 7 or 8 of the first 10 problems in an hour or less, and I've also solved a #15 in five minutes, but I've never gotten a perfect score on an AIME; it's just because I saw the right way into them, and I've never yet seen the right way into more than maybe 11 or 12 problems, and that was on last year's test, one of the easier ones. However, if you do find the best approach, you're extremely unlikely to run out of time, just because once you do, there are never any unusually difficult calculations.

For solving the problems, some of it is due to practicing finding the right method, and some is due to learning new methods. For example, a lot of people don't normally think of using coordinate geometry to solve a problem, but in some cases, that's the fastest way. I don't think I've ever yet solved a problem with coordinate geometry which didn't specify coordinates to begin with. However, I know some other methods of approach that are quite useful at times, and because I know them, I'll immediately think of them when I see the problem. For example, the #15 that I solved so quickly was from 2003 AIME II. Here's the problem:

The distinct roots of the polynomial $x^{47}+2x^{46}+3x^{45}+...+24x^{24}+23x^{23}+22x^{22}+...+2x^2+x$ are $z_1,z_2,...,z_n$. Let $z_k^2$ have imaginary part $b_ki$. Find $|b_1|+|b_2|+...+|b_n|$.

I instantly factored the polynomial to $x(1+x+x^2+...+x^{22}+x^{23})^2$, so the roots are the $24^{th}$ roots of unity, which stems from complex numbers, which I quite enjoy. So, this problem fit my strengths perfectly, which is how I saw my way into each part of it so quickly.

Sometimes you'll see problems which are similar to those encountered, and sometimes problems which are extensions off previously done ones, but sometimes you'll see something totally new.  But you'd have learned STRATEGIES from problems of the same kind earlier...

[quote="123s"]Well... that was a good answer to my question. Basically what I was asking is whether most of the people who qualify for MOSP and USAMO solve most of the problems they do because they already know the trick/method needed, and are just encountering a variation, or whether they have to come up with a new method? Cause, learning how to do the latter is harder, and just involves doing more and more problems. Thanks.[/quote]

I think that in that case, it's just practice. However, that doesn't mean doing math for hours each day. I get on here and solve any AMC/AIME problems I find, and that's almost all that I've done to prepare for the AIME. However, last year I scored a 3 on the easiest AIME in recent years, and now I could probably score an 11 or 12 on the same test. So, any extra experience will help you be better prepared to solve problems when the test actually comes.


modular arithmetic
quadratics
number theory unsolved
number theory

Let $p$ be a prime number such that $p\equiv 1\pmod{4}$. Determine $\sum_{k=1}^{\frac{p-1}{2}}\left \lbrace \frac{k^2}{p} \right \rbrace$, where $\{x\}=x-[x]$.

I think that sum $= \frac{p-1}4$.

Lemma: if $p \equiv 1 \pmod 4$ then, for each $1 \le n \le \frac{p-1}2$ exists exactly one $1 \le m \le \frac{p-1}2$ such that $n^{2}+m^{2}\equiv 0 \pmod p$.
Proof: it is known that $-1$ is a quadratic residue mod p. There exist only two numbers, a and -a, such that $a^{2}\equiv (-a)^{2}\equiv-1 \pmod p$.
Then $m^{2}+n^{2}\equiv 0 \pmod p$
$\Leftrightarrow m^{2}m^{-2}+n^{2}m^{-2}\equiv 0 \pmod p$
$\Leftrightarrow (nm^{-1})^{2}\equiv-1 \pmod p$
$\Leftrightarrow nm^{-1}\equiv a \pmod p$ (or -a)
$\Leftrightarrow n \equiv am \pmod p$ (or -am)
But exactly one of am, -am lies between $1$ and $\frac{p-1}2$.

If $m^{2}+n^{2}= kp$ then $\frac{m^{2}}p+\frac{n^{2}}p = k$, k is an integer. The sum of the fractional parts is an integer. But it cannot be 0 ($\frac{m^{2}}p$ is not an integer) and it's less than two. Therefore:
$\{\frac{m^{2}}p\}+\{\frac{n^{2}}p\}= 1$.
Now, we have $\frac{p-1}2$ fractions in our sum. If we match them, we get a sum with $\frac{p-1}4$ pairs, each of them is 1. The game is done.  :wink:

You are right, I think , thank you  :wink:

I remember a nice problem from Korea in which $p$ is a prime number and $p \equiv 1\pmod{4}$.The problem was to find the value of $\sum_{k=1}^{p-1}{([\frac{2k^2}{p}]-2[\frac{k^2}{p}])}$ .This problem uses exactly the same idea as the original problem.The answer is $\frac{p-1}{2}$


A difficult thread where you make your own problems:

here's mine: Bobby is a gifted superhero who wants to take the superhero test and become a megahero. On the test he is asked what is the probability of him getting a super code with only one 5 if the code can be anything from 1000-5000?

hmm.. i thought there rnt's supposed to be a (sorta) marathon? :|

oh, right, you're correct... So this is no longer a marathon, but you just have to solve MY problem.

[hide="Solution"]
There are $ 5000\minus{}1000\plus{}1\equal{}4001$ total possibilities. Now we have four cases for the successful outcomes, one for every place the $ 5$ can be.

[hide="CASE 1 - Thousands digit - 1"]
We only have one case for this, and that is $ 5000$. [/hide]

[hide="CASE 2 - Hundreds digit - 324"]
We just do it digit by digit. We have $ 4$ choices for the first, since it can't be $ 0$ or greater than $ 4$. We have only one choice for the second ($ 5$). We can have anything but a $ 5$ for the third, so that is $ 9$. And the last is similar to the third, so we have $ 9$. Now, just multiply:
$ 4\times1\times9\times9\equal{}324$[/hide]

[hide="CASE 3 - Tens digit - 324"]
This is the exact same as CASE 2, except that the second and third digits are switched. Multiplying:
$ 4\times9\times1\times9\equal{}324$[/hide]

[hide="CASE 4 - Ones digit - 324"]
The same as CASE 2 and CASE 3.
$ 4\times9\times9\times1\equal{}324$[/hide]

Now we just add the total for all the cases:
$ 1\plus{}324\plus{}324\plus{}324\equal{}973$

The final step is to make the fraction. We put the total successes from our cases over $ 4001$, and since $ 4001$ is prime, our answer is:

$ \boxed{\frac{973}{4001}}$
[/hide]

correct...(I think... :lol: )

but you didn't have to take all the cases... You could have just multiplied 324 by 3 and then added 1... Then just put that over 4001....

but still, a good solution is one that explains all steps, so I guess you did do a good job!

Here's my problem:  At 2:38 in the afternoon, what is the degree measure between the hour and minute hands?  Express your answer to the nearest whole number.

[quote="cgyao15"]Sorry to mention this but did u read the previous posts? [sarcasm] it might have said this isn't a marathon [sarcasm][/quote]

Please refrain from YELLING and using [sarcasm] tags.

I think it is best to let this thread die, like all the other dead ones.

i wasn't yelling, i was inquiring  :wink: and sorry about the sarcasm


Suppose a_i, i=1...n, are positive reals with sum 1.
Prove:
a_1*a_2 + a_2*a_3 +... + (a_(n-1))*a_n <= 1/4

Nice solution.
Mine is somewhat different:
For n=3, a1*a2 + a2*a3 = a2*(1-a2) <= 1/4 * ((a2)+(1-a2))^2 = 1/4. For n=2, it is also easy.
Proof by induction.
Assume it works for k=n but not for k=n+1.
then a1*a2 + a2*a3 ... + a_n *a_(n+1) > 1/4 >= b1*b2 .. + b_(n-1)*b_n  (*),
for any set {b1 ... b_n} of non-negative reals such that b1 + b2 ... + b_n = 1.
Consider the n different sets of {b1, b2 ... b_n} = {a1 + a2, a3, a4 ... a_(n+1)}, {a1, a2 + a3, a4 ... a_(n+1)} ... {a1, a2 ... a_(n-1), a_n + a_(n+1)}.
Applying (*) for these sets, and cancelling terms, yields: a_i * a_(i+1) > a_(i-1) * a_(i+1) + a_i * a_(i+2), so a_i > a_(i-1) [2] , and a_(i+1) > a_(i+2) [3] .
Now listing inequalities [2] and [3] for different values of i, quickly gives a contradiction.
Thus our assumption is wrong, and inequality is also true for k=n+1.


geometry
circumcircle
complex numbers
geometry theorems

When working with complex numbers, is there any (easy) way to find the circumcenter of a triangle, when its 3 vertexes are given?

If there isn't, is there any formula when there are some ''special'' conditions? Like http://www.mathlinks.ro/viewtopic.php?p=1599930#1599930 post #25?

Hi, mathlinkers.
I derived the circumcenter formula for triangle $xyz$:
$p=\frac{xy \overline y - x\overline xy+zx \overline x - z \overline zx+yz \overline z-z y \overline y}{x\overline y-\overline xy+y\overline z-\overline yz+z\overline x-\overline zx}.$
Can somebody check is it true?


All of the dimensions of the American Flag are based on the $ hoist$, or width, of the flag. The width of the $ union$ (the star field) is $ \frac7{13}$ the hoist and the length of the union is $ .76$ times the hoist. The hoist of a given flag is 4 feet. How many square feet are in the area of the union? Express your answer as a decimal rounded to the nearest hundredth.

width of union = 7/13 * 4 = 28/13 feet
length of union = 19/25 *4 = 76/25 feet

Area = $ \frac{28\times 76}{13\times 25}\equal{}\frac{2128}{325}\approx 6.55 ft^2$


In triangle ABC, $3\sin A+4\cos B=6$ and $4\sin B+3\cos A = 1$. Then $\angle C$ in degrees is

(A) 30
(B) 60
(C) 90
(D) 120
(E) 150

[hide]Square the two equations and add them. We get

$(3\sin{A}+4\cos{B})^{2}+(4\sin{B}+3\cos{A})^{2}= 37$

$9(\sin^{2}{A}+\cos^{2}{A})+16(\sin^{2}{B}+\cos^{2}{B})+24(\sin{A}\cos{B}+\sin{B}\cos{A}) = 37$

$24(\sin{A}\cos{B}+\sin{B}\cos{A}) = 12$

$\sin{(A+B)}= \sin{A}\cos{B}+\sin{B}\cos{A}= \frac{1}{2}$

by the angle addition formula for $\sin$. So we must have $A+B = 150 \Rightarrow C = 30$ and not the other way around, because then the second equation is impossible.[/hide]

[quote]by the angle addition formula for $\sin$. So we must have $A+B = 90 \Rightarrow C = 90$[/quote]
Are you sure?  becuase in that case, $\sin A = \cos B$ etc ..
giving $\sin A = 6/7$ and $\cos A = 1/7$   etc  $\sin^{2}A+\cos^{2}A = 37/49$ ...$\implies$ typo somewhere  :wink:

paladin just made a small addition mistake.  It should say:

$9(\sin^{2}A+\cos^{2}A)+16(\sin^{2}B+\cos^{2}B)+24(\sin A\cos B+\cos A\sin )=37$

You had 12 instead of 24.  Continuing from there the same way you get:

$24(\sin A \cos B+\cos A \sin B) = 12\\ \sin (A+B)=\frac{1}{2}\\ A+B=30 \vee 150$

Some experimentation yields that $A+B=150\Rightarrow C=30$

-edit- you already fixed it.

But I'm pretty sure $C=30$, not 150.

[quote]Some experimentation yields that $A+B=150\Rightarrow C=30$[/quote]
One way to think about it, if   $A$ is less than $30^{o}$ than $\sin A < (1/2)$ and thus LHS in eqution 1 would be less than $(3/2)+4 = 5+(1/2)$ which is less than 6.  etc ... thus $A$ has to be greater than $30^{o}$  etc....


geometry
combinatorics unsolved
combinatorics

A connected graph has 1998 points and each point has degree 3. If 200 points, no two of them joined by an edge, are deleted, show that the result is a connected graph.

Hi,
   I cannot believe it!! I just posted the question as you! Can you please let me know where you found the question? Or where you can find a solution? THANK YOU!!

the solution that I found(in unknown document) is too long so I write some hint LEMMA: assume that G is a graph such that the degree of each vertices is at least 3and if we omit one vertice arbitrarily the graph will remain connected,and also we can draw G on the plane such that each point P out of the area of G( out of its cycles) can be connected to all the vertices of G such that we don't have intersection except vertices ( excuse me the proof of this LEMMA is your duty(because of being long))assume that K vertices of G are not not neighbor mutualy are omitted we name it G'( all its vertices has at least 3 neighbor) and also assume that omitting just one vertice make G' disconnected, we should show k>=200 suppose  we have f region(except the external) ,1998-k vertices,2997-3k edge, so f>=1-(1998-k)+(2997-3k)=1000-2k we can easily conclude that 3f<=1998-k so if u use the privious conclusion we have K>=200

[b]Lemma[/b] Suppose graph $G$ with $n$ vertices contains at least $\frac{4n}{3}$ edges. Then $G$ contains two intersecting cycles.

Proof.
Suppose contrary. Consider spanning tree of $G$ and let's add remaining edges of $G$ back to it. Each edges add at least one new cycle. If it add two cycles we are done. Since each cycle contains at least 3 vertices and we didn't get intersecting cycles, it follows that we obtained at most $n/3$ cycles, i.e. we added at most $n/3$ edges, but tree contains $n-1$ edges, so totally we obtain in graph at most $n-1+n/3<4n/3$ edges. Contradiction.

Back to the main problem.
If graph contains two intersecting cycles, then consider common vertex $O$, where these cycles disperse. It is clear $\deg O=3$. Delete $O$ and three edges $OA$, $OB$ and $OC$. $G\setminus \{O\}$ is connected graph again, since there is path, which connects $A$, $B$ and $C$!
Repeat operation while there are intersecting cycles.

How many vertices we removed? Denote this number as $x$. In the beginning graph contains $\frac{3N}{2}$ edges ($N=1998$) and each our operation removes 3 edges. Due to lemma we can remove edges while $\frac{3N}{2}-3x\geq \frac{4(N-x)}{3}$, i.e. $x\leq \frac{N}{10}$, therefore we can remove $[1998/10]+1=200$ vertices.


The problem is:

Find the fourth degree polynomial that has the roots 3 and 1-i.  There are more than one solutions.

Do the coefficients have to be real?
[hide="If yes"]Since the polynomial has a complex root, $ 1 \minus{} i$, another root must be its complex conjugate for the coefficients to be real. So the fourth degree polynomial has roots $ 3$, $ 1 \minus{} i$, and $ 1 \plus{} i$. The fourth root must be a real number since $ 3$ is a real number, say $ b$. Call the polynomial $ p(x)$.
We can write $ p(x) \equal{} a(x \minus{} 3)(x \minus{} (1 \plus{} i))(x \minus{} (1 \minus{} i))(x \minus{} b)$, with leading real coefficient $ a\ne 0$.
So, $ p(x) \equal{} a(x \minus{} 3)(x^2 \minus{} 2x \plus{} 2)(x \minus{} b)$ for all real numbers $ a$ and $ b$, $ a\ne 0$.[/hide]

if one of the root is $ 1 \minus{} i$, then one other root must be $ 1 \plus{} i$.

if last root is $ n$, then the polynomial would be $ (x \minus{} 3)(x \minus{} 1 \minus{} i)(x \minus{} 1 \plus{} i)(x \minus{} n)$.

$ (x \minus{} 3)(x \minus{} 1 \minus{} i)(x \minus{} 1 \plus{} i) \equal{} x^3 \minus{} 5x^2 \plus{} 8x \minus{} 6$

so there are infinite polynomials which have $ 3, 1 \minus{} i, 1 \plus{} i$, and $ n$ as the roots.

EDIT:

oops, fixed.


invariant
advanced fields
advanced fields unsolved

this is false for $ G\equal{}1$. do you assume that $ k$ is algebraically closed and consider only maximal ideals?

[quote="dran"]Let $ G\subset \text{GL}(n,k)$. How does one show that $ \text{Spec}(k[x_1,...,x_n]^G)\simeq k^n/G$ ?[/quote]

If you want to prove an isomorphism, you'd better know what the terms involved are. So, what is $ k^n/G$ as an algebraic variety (of course I want algebraic structure, not just set structure)? I am asking because some people would give the above as definition for the quotient.


Hi there =)
Well, my class is currently dealing with arenes (and their reactions), we've got some homework and I need help >.<

Actually, I have solved the main synthesis task, which, if I am correct, leads to azulene.
The latter reacts with a few other substances, yet I don't know how :|

1) Azulene reacts with tert-butanol, catalysed by an acid, in a substitution reaction to F.

2) With an alkali organyl or a Grignard reagent azulene reacts in an
addition reaction to form  a salt. This, after an aqueous preparation and oxidation with tetrachloro-para-benzo-quinone, becomes a respective alkyl substituted derivative of azulene.
Two different products (isomers) are possible. The first (G) emerges from the usage of methyllithium, the second (H)  from the usage of triphenylmethyllithium.

3) Also compounds of transition metal of azulene are synthesisable. If azulene reacts  with LiAlH4  and  Iron(II)-chloride, a red compound I is formed in the form of multiple stereoisomers.
Via careful hydrogenation of the isomers a single orange compound of J is formed with the empirical formula C20H26Fe.

My homework is to draw the chemical structure of F, G and H, as well as to draw all isomers of I and the chemical structure of J. But since I don't feel confident about the reactions or the products , I can't do it :(

I would be deeply grateful if you helped me :)

I believe F will be a tert-butyl substituent on azulene (with elimination of water), but as to where, I have no idea.  Since a cation is formed with the azulene... it should be stabler on the 7-member ring, yes?  Maybe?  I'm not very confident about this question.


Given an $ n\times n$ matrix $ B$, how do you determine if there exists an $ n\times m$ matrix $ A$ such that $ AA^T\equal{}B$. If there does exist such an $ A$, how would you construct it? What if $ A$ has to be a square matrix?

I would write $ A^TA$ instead, but that's just notation; I prefer the transposed part to come first. I also assume you're dealing with matrices over $ \mathbb{R}$. It's very important to specify the field in something like this.

There exists such an $ A$ if and only if $ B$ is symmetric and positive semidefinite. Squareness of $ A$ is irrelevant. An algorithm is described [url=http://en.wikipedia.org/wiki/Cholesky_decomposition]here[/url] in the positive definite case, which even makes $ A$ triangular.


You have $2$ columns of $11$ squares in the middle, in the right and in the left you have columns of $9$ squares (centered on the ones of $11$ squares), then columns of $7,5,3,1$ squares. (This is the way it was explained in the original thread, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=44430 ; anyway, i think you can understand how it looks)

Several rooks stand on the table and beat all the squares ( a rook beats the square it stands in, too). Prove that one can remove several rooks such that not more than $11$ rooks are left and still beat all the table.

[i]Proposed by D. Rostovsky, based on folklore[/i]

My idea is to reduce a rook to two reduced rook A and B, A beats only the row it stands and B beats only the column.

Call the two rows with only two squares the first and the 11th row. If every row has an A-rook, we are done. If not, let s=max{|i-6| | the ith row does not have an A-rook}. And let k=6+s or 6-s such that the kth row does not have an A-rook. Since the kth row must be beated, the (s+1)th~(12-s)th columns all have a B-rook. We can choose these B-rooks. Then the rest of the table can be seen as if it was reduced to size-(2s-1) (the original table is size-11) and we can choose 2s-1 more rooks. Finally the problem will be reduced to size-1 and we are done.
(Note that s>0 since the 6th row must have a rook)

BTW, please forgive my poor English.


function
quadratics
algebra
quadratic formula
set theory

Which among the following is/are incorrect?

1) (Infinity - Infinity) is a definite number
2) V4 takes exactly 1 value
3) -1 is a complex number
4) (Infinity + Infinity) = Infinity
5) more than 1 of the above
__________________

[quote="jnApp66"]Can more than one be right?[/quote]

check the 5th option :)

give reasons also ...

more than one of the ABOVE. I only know 1, so I'm afraid to step out and show it.  :oops:

[quote="JBL"]"Infinity" is not an element of any set with a conventionally defined addition or subtraction.[/quote]

So , you mean to say that inf + inf  is not defined ?

So wots ur ans from the choices given above ?

Any more takers ....??? ....

What is V4?

[hide="Every other option"] 1)  No.  Operations on infinity are not defined.
3)  Yes.  Every real number is a complex number.
4)  No.  See 1.

The answer depends on what V4 is. [/hide]

is V4 $\sqrt{4}$? (doubtful but still probable) then no it does not have exactly one value

[quote="t0rajir0u"]Probably.

[hide="In that case..."] 2) Yes.  The square root [b]function[/b] only takes on the [b] non-negative[/b] value.  So $\sqrt{4}= 2$.  It is [b]not[/b] correct to say that $\sqrt{4}=-2$, although it is correct to say that $x =-2$ is a solution to the equation $x^{2}= 4$.

Since 2) and 3) are both correct, the answer is $\boxed{5)}$. [/hide][/quote]

v stands for square root only ..

But ans cant be 2 , as by definition of square root it must always be +ve , although it can give 2 values  :)

Also , operations on infinity are definitely possible ...

Substracting one infinite quantity  from another is undetermined as we donno how much (and type) we r substracting from how much (and type).

Understanding infinity as a out of boundary entity on a refrence scale , when u add two infinities it will still be an out of bound entity hence giving it a value of infinity again .Hence it is very much defined ..

Now , the question is will thil value tend towards (->) infinity or ill it be equal to(=) ??

You can not perform operations on infinity.  How can you add more objects to something that already has infinitely many?

Also, which infinity are we talking about? Aleph 0 (the cardinality of integers)? Aleph 1 (the cardinality of reals)?

I suggest you read through some of the Math Jam sessions we've had about cardinalities and infinity before saying things such as:
[quote="Eccentric"]Also , operations on infinity are definitely possible ...

Substracting one infinite quantity from another is undetermined as we donno how much (and type) we r substracting from how much (and type).

Understanding infinity as a out of boundary entity on a refrence scale , when u add two infinities it will still be an out of bound entity hence giving it a value of infinity again .Hence it is very much defined ..

Now , the question is will thil value tend towards (->) infinity or ill it be equal to(=) ??[/quote]

[quote="Eccentric"]v stands for square root only ..

But ans cant be 2 , as by definition of square root it must always be +ve , although it can give 2 values  :)[/quote]

No.  The square root operator $\sqrt{x}$, as a [b]well-defined function[/b], can only give [b]one y-value[/b] for each of its $x$-values, so by definition and convention it only gives the non-negative one.  The negative value is called [b]a[/b] square root in the sense that $(-x)^{2}= x^{2}$, but it is not [b]the[/b] square root.  That's why the quadratic formula specifies $\pm \sqrt{b^{2}-4ac}$.

And regardless of whether or not operations on infinity make sense conceptually, they are not defined rigorously in the way you're using them.  Learn more about set theory or the theory of limits for [b]rigorous[/b] discussions of two different conceptions of infinity.


Show that the equation $ x^{3}+y^{3}+z^{3}=2$ has infinitely solutions in $ %Error. "matthb" is a bad command.
{Q}$.

[quote="flipper"]Show that the equation $ x^{3}+y^{3}+z^{3}=2$ has infinitely solutions in $ \mathbb{Q}$.[/quote]

[hide="An Idea"]
In the theory of elliptic curves, when we are given the curve,
$ y^{2}= x^{3}+c$ and $ (x_{0},y_{0})$ is a rational point on this curve.
Then there exists something called [b]Bachet's Duplication Formula[/b].
It is a really messy formula that gives [i]another[/i] rational point $ (x_{1},y_{1})$ which is on the curve. Repeating this process indefinitely we get $ \{(x_{0},y_{0}),(x_{1},y_{1}),...\}$ are infinitely many such points.

The way the Bachet's formula is found is as follows.
1)Pick a point rational point on the curve.
2)Draw the [i]tangent[/i] at that point.
3)The tangent intersects the curve at another point.
4)That other point is also rational.

My idea is to generalize this approach.

Given $ x^{3}+y^{3}+z^{3}=2$ let us show a stronger result that there exist infinitely many integers for $ x^{3}+y^{3}+(1-x)^{3}=2$. Meaning, I am showing a stronger result that not only there are infinitely many there are infinitely many with the relationship that $ z=1-x$. Note that $ (1,1,0)$ is a rational point on this curve.

Expand,
$ x^{3}+y^{3}+1-3x+3x^{2}-x^{3}= 2$
Thus,
$ y^{3}= 1+3x-3x^{2}$

I think the Duplication formula might work here, try it! Draw a tangent at the point $ (1,1)$ and see it intersects the curve at another rational point. If it does repeat this again and again. You got your formula.
[/hide]

oh, i'm sorry. i don't know nothing about elliptic curves :(
here si my ideea:
[hide="maybe a stupid ideea"]i thought at something involving extremal principle. let us consider the set of the solutions of the given equation be finite. so there exist a triple $ (a,b,c)$ so that this maximize on minimaize (did i spelt correctly?) either $ abc$ or $ a+b+c$. if we find a tripile that keep invariant $ abc$ or $ a+b+c$ the problem is solved. i guess that is easier to find a triple that keep invariant the sum $ a+b+c$, like $ (a-e,b-f,c+e+f)$, but i don't find $ e$ or $ f$.[/hide]

first, i fixed $ z=1-x$ as silow_theory did, but i didn't manage to show that there are infinitely many $ y$ so that $ y^{3}=-3x^{2}+3x+1$. :(

We can generalize in R

If $ k \neq 0$ is a real, show that

$ x^{3}+y^{3}+z^{3}= k$ admits infinite solutions in R.

Solution : $ (\sqrt[3]{\frac{k}{2}}+n^{3})^{3}+(\sqrt[3]{\frac{k}{2}}-n^{3})^{3}+(-n^{2}\sqrt[3]{6}\sqrt[9]{\frac{k}{2}})^{3}= k$ for every reals $ (k,n)\neq 0$

The solution in rationals is very nice, but the extension to reals is just silly:  if I choose any value for $ k, y, z$, there exists a unique real value of $ x$ such that $ x^{3}+y^{3}+z^{3}= k$.

Ok, thanks ^^ I generalized here http://www.mathlinks.ro/Forum/viewtopic.php?t=162514


medal talaye IMC 2006 ra be aghaye hatami tabrik migam.  :first:

[quote="Amir.S"]International Mathematic Competition(IMC) for Univerity Students[/quote]

hamoon olampiad e daneshjooyie digeh, na?

man ham tabrik migam, :wink:

motmaen bashin ke IMC oonghadra ham ke parobalesh midan etebaar nadare.ghol midam
hatta khodeshoonam ke raftano bargashtan tafavote imo va imc ro dark kardan va oonghadra ham barashoon fargh nakarde
albatte har chi bashe belakhare baraye sharif eftekhare :wink:

salam
manam tabrik migam be ishoon.
be omid moafaghaite ishoon dar aiande va hameie alaghe mandan. : :blush:  :)  :wink:

[quote="behzad"][quote="maryam rashin"]motmaen bashin ke IMC oonghadra ham ke parobalesh midan etebaar nadare.ghol midam
[/quote]
man aslan movafegh nistam! :wink:[/quote]

man aslan movafegh nistam. :D

parto jan, movafeghate mano to che tasiri dar koliate masale dare?

baghie chikar kardan? faghat yek taka dashtem?


calculus
integration
trigonometry
real analysis
real analysis unsolved

1) If $ I$ denotes the integral $ I \equal{} \int_0^\frac {\pi}{2} \sin^2 \sin x \plus{} \cos^2 \cos x \ dx$, then under substitution $ u \equal{} \frac {\pi}{2} \minus{} x$,
\[ I \equal{} \int_0^\frac {\pi}{2} \sin^2 \cos x \plus{} \cos^2 \sin x \ dx
\]

\[ \Rightarrow 2I \equal{} \int_0^\frac {\pi}{2}(\sin^2 \sin x \plus{} \cos^2 \sin x) \plus{} (\cos^2 \cos x \plus{} \sin^2 \cos x)\ dx
\]

\[ \equal{} \pi \Rightarrow I \equal{} \boxed{\frac {\pi}{2}}
\]
2) For any positive integer $ n$, since $ |\sin x|$ has period $ \pi$, we have
\[ \int_0^\pi |\sin nx|\ dx \equal{} \frac {1}{n}\int_0^{n\pi} |\sin x|\ dx \equal{} \int_0^\pi \sin x\ dx
\]
implying the integral in question evaluates to $ \boxed{0}$.

[quote="scorpius119"]1)
$ 2I \equal{} \int_0^\frac {\pi}{2}(\sin^2 \sin x \plus{} \cos^2 \sin x) \plus{} (\cos^2 \cos x \plus{} \sin^2 \cos x)\ dx$
$ \equal{} \pi \Rightarrow I \equal{} \boxed{\frac {\pi}{4}}$
[/quote]

Is it correct?


[b]DFT[/b], when dealing with Diophantine equations of this type a purely manipulative solution is impossible since you never use the restriction of the problem to integers.  It's almost never the case that a piece of information given in the problem is useless, so if you systematically ignore information then you won't solve the problem.

[hide="Partial solution that gives a few solutions"]
Solving the quadratic in $a$ we have
\[a = 12 \pm \sqrt{b^{4}-2b^{2}+4b+70}\]
Thus, $b^{4}-2b^{2}+4b+70=k^{2}$ for some integer $k$. Looking in $\bmod 4$ we see that $b$ must be odd, so we substitute $b=2m+1$:

\[16m^{4}+32m^{3}+16m^{2}+8m+73 = (4m^{2}+4m)^{2}+8m+73 = k^{2}\]
But $(4m^{2}+4m+1)^{2}= (4m^{2}+m)^{2}+8m^{2}+8m+1$, so for $m$ such that $8m+73 \geq 0$, i.e. that $m \geq-9$ we must have
\[8m^{2}+8m+1 \leq 8m+73 \implies m \in \left[-3, 3\right] \]
From here we test the few values, and find the solutions:
\[m=3 \implies b=7 \implies a= 12 \pm 49 \]\[m=-2 \implies b=-3 \implies a= 12 \pm 11 \]\[m=-3 \implies b=-5 \implies a= 12 \pm 25 \]
All of these give different values for $a^{2}+b^{4}$. And I don't even know if this is the full solution set, since for $m <-9$ it doesn't work out as nicely. Can anyone help? [/hide]

[hide]
$a=-13$ and $b=-5$ seems to work, which would make $a^{2}+b^{4}=794$
I'll work on showing it's true; that was basically guess and check.
[/hide]

[quote="me@home"]This is really interesting, but I think the only solution is along cincodemayo's lines ... and the solutions are weird too, for instance the first solution listed has $a=61,b=7$ and $a^{2}+b^{4}=6122=2(3061)$ and $3061$ is prime..!?[/quote]

Odd.  :lol:

Can anyone devise a better strategy with which to attack this problem?  :P

Well, I can tell you that the problem is mistyped and some conditions are missing... so the problem might be unsolvable.  :blush:

Anyway, if it is the correct one then your strategy will work, so you should keep trying this one!


Let $ a,b,c$ be positive reals such that $ a^2\plus{}b^2\plus{}c^2 \equal{}3$ then prove that

$ (abc)^2(a^3\plus{}c^3\plus{}c^3) \le 3$

It's weaker than the well-known: $ abc(a^2 \plus{} b^2 \plus{} c^2) \leq\ 3$  with $ a \plus{} b \plus{} c \equal{} 3$:wink:


calculus
integration
Ring Theory
search
geometry
3D geometry
number theory unsolved

[Edit] It's a solution to $y^2+2=x^3$ ...

$\mathbb{Z} (\sqrt{-2})$ is an UFD so for some $a,b$ integers we have:

$y+i\sqrt{2}=(a+bi\sqrt{2})^3=a^3-6ab^2+3a^2bi\sqrt{2}-2b^3i\sqrt{2}$ from which we obtain:

$y=a^3-6ab^2$ and $1=3a^2b-2b^3$ and hence we get $(a,b)=(1,1);(-1,1)$ and that yield $y=5$ or $y=-5$

[quote="Megus"]$\mathbb{Z} (\sqrt{-2})$ is an UFD so for some $a,b$ integers we have:

$y+i\sqrt{2}=(a+bi\sqrt{2})^3=a^3-6ab^2+3a^2bi\sqrt{2}-2b^3i\sqrt{2}$ from which we obtain:

$y=a^3-6ab^2$ and $1=3a^2b-2b^3$ and hence we get $(a,b)=(1,1);(-1,1)$ and that yield $y=5$ or $y=-5$ so there's no solution.[/quote]

what about $(x,y)=(2,2)$ and $(5,11)$？ :)

I solved $y^2+2=x^3$ by a mistake  :D

Solution to an original question goes the same - $\mathbb{Z}(i)$ is an UFD as well so we are to find solutions to:

$y=a^3-3ab^2$,    $2=3a^2b-b^3$ now there's some easy casework to do and we're done.

Somewhat offtopic but important:

@rohitsingh0812: please give your topics names that say something to the people; just calling them "Integeral solutions" and numbering them isn't good at all: one doesn't even have a clue what the topic contains; also searching for it gets a lot harder, one of the reasons why it's so difficult to search for problems here.
I have changed them now, you can change them to something else if you want.

[quote="rohitsingh0812"]Q.Find all integral solutions of the equation:-

    $y^2 + 4 = x^3$[/quote]
This problem was posted before(with complete solution):
[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=perfect+cube+%28and+++its+conjugate&t=74074[/url]


If $(1.0025)^{10}$ is evaluated correct to 5 decimal places, then the digit in the fifth decimal place is
\[ \text{(A)}\ 0\ \ \ \ \text{(B)}\ 1\ \ \ \ \text{(C)}\ 2\ \ \ \ \text{(D)}\ 5\ \ \ \ \text{(E)}\ 8 \]

[quote="white_horse_king88"]Were calculators allowed on this contest?[/quote]

No.  In fact, it's so old that there were very few calculators in the world as we think of them now.

Oops.  I wrote down 120 when i meant 210.  My answer is the same i just mistyped it.

[quote="Euler1001"][hide="Solution"]Well this probably isn't the best way but here goes.  If you rewrite the problem as (1+2.5 x 10^(-2))^10 and know the 10th row of pascal's triangle is 1,10,45,120,210,252,210,120,45,10,1 then i think you can solve it.

Since you are only look for the numbers that would contribute to 10^(-5) and     10^(-6) you could eliminate a lot of your answer.

You only have to look at 120(5/2)^4(10^-8) which gives 8.203125 x 10^-5

So my answer would be E - 8.[/hide][/quote]Just for clarification: (it took me a little while to see what you were doing..)

The expression can be written as $\left(1+2.5\cdot10^{-3}\right)^{10}$, which can then be expanded using the binomial theorem. Whether or not this is right I'll have to think about..


For any given integer $ m$, there is no polynomial $ p(x)$ with integer coefficients such that $ p(n)$ is prime for all integers $ n$ with $ n\ge m$.

[hide="?"] Let $ k$ be an integer such that $ p(k)\neq 0, 1,\minus{}1$.  Then $ p(k) | p(k\plus{}a p(k) )$ for arbitrary $ a$.  [/hide]

[quote="Mersenne"]I'm not sure about that :maybe:
My book shows a very long solution (aka 1.5 pgs)[/quote]

Whether or not that's true, T0r's solution works. (you book may have the same solution but perhaps very explicitly proves all of the steps, such as $ a\minus{}b|P(a)\minus{}P(b)$.)


function
limit
real analysis
real analysis unsolved

$f$-is continious funtion on R, s.t. $f(x+h)-2f(x)+f(x-h)\to0$ when $h\to+\infty$ for any real x. Prove that f-linear fucntion

We have $f((x+t)+(h-t))-2f(x+t)+f((x+t)-(h-t))\to$ when $h\to\infty$. It follows that $f((x-h)+2t)-f(x-h)=f(x+2t-h)-f(x-h)\to 2f(x+t)-2f(x)$, but $\lim_{h\to\infty}(f((x-h)+2t)-f(x-h))=\lim_{h\to-\infty}f(h+2t)-f(h)$. Therefore $f(x+t)-f(x)$ doesn't depend on $x$ for all $t$ !!!
From here using continuity of $f$ we can easily obtain:
1) $f(x+qt)-f(x)=q(f(x+t)-f(x)$, where $q\in\mathbb{Q}$,
2) so it is holds for $q\in\mathbb{R}$.
That's all.


calculus
integration
trigonometry
function
calculus computations

Find $ \int \frac{x^2}{\cos x\plus{}x\sin x} dx$,thanks a lot!

hello, the only idea that i have is this here
http://www-math.mit.edu/~djk/18_01/chapter24/section03.html
but i'm not sure if this will help you, sorry.
Sonnhard.

First of all, don't [url=http://www.mathlinks.ro/viewtopic.php?t=308062]doublepost[/url]. (EDIT: ok, you probably posted it by mistake  :)  )
[Moderator edit: Repost deleted.]

Second, an antiderivative of this function expressible in terms of elementary functions probably does not exist - why are we looking for it in the first place? Is this a problem proposed to you, or you just played with functions to make a hard calculus problem?

Sorry, I made a mistake so that posted doubly. This problem was from a friend of mine who needs to solve it. In fact, I strongly feels that this integral has no form of elementary functions.

I am afraid that using $ \tan \frac{x}{2}$ will not work.

[quote="sss"]Find $ \int \frac {x^2}{\cos x \plus{} x\sin x} dx$,thanks a lot![/quote]

@sss :
$ \int \frac{x^{2}}{\left( \cos x\plus{}x\sin x\right) ^{2}}\,dx$ would be solvable. Not your original question.


search
combinatorics unsolved
combinatorics

Let $ n$ be an integer greater than six, and let $ X$ be an $ n$-element set. Let $ A_1, A_2, \ldots , A_m$ be distinct $ 5$-element subsets of $ X$. Prove that for
\[ m > \frac{n(n\minus{}1)(n\minus{}2)(n\minus{}3)(n\minus{}15)}{600}
\]
there exist six distinct $ A_i$ whose union contains exactly six elements.

The solution in the book (Counting Strategies, by Andreescu and Feng) seems extraordinarily complicated. Using a standard counting in two ways estimation, I got the weaker bound
\[ \frac{n(n\minus{}1)(n\minus{}2)(n\minus{}3)(n\minus{}4)}{144}
\]
This is close enough to the one in the problem to make me hope that a simpler solution can be built upon it.

Have a look here
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=483433098&t=38391

Pierre.


A small variant of this interesting problem is :

1/3 + 1/5 + ... 1/(2n+1)  is not an integer n > 1.

I never like this term "well-known", because nothing is "well known" until you've seen it for the first time. And this is the reason people are here - for exposure to problems that they have not seen before.

Besides, even "well-known" problems can have new and interesting solutions.

Gauss202, the *only* purpose of my post was to take the opportunity to give a small variant of the original problem (which is well known to me, so I won't post any solution) not to claim that the problem was not interesting.

Okay. I understand what you meant now. Your problem is very interesting as well.

Well, this is how I did it:

1/n = (n+1)/n - 1

Since an integer minus another integer results in an integer, all we have to do is verify if (n+1)/n is an integer.

So, correspondingly, if this is true:

n+1 mod n = 0

n mod n = -1

0 = -1
Which is false, and therefore the sum cannot be an integer.

Anything wrong?

You showed that 1/n isn't an integer for all integers greater than 1 -- the questions asks you to prove that the [i]sum[/i] of those terms isn't an integer.

Scrambled, you need to write more clearly.  Explain what you are saying in more words.  What is the 5/6 doing there?  You stated an important step, but I don't understand how you related it to all the other parts.  Furthermore, I don't see at all how you showed it for all n.

[quote="Scrambled"][quote="fuzzylogic"]Suppose that  [tex]n>1[/tex] is an integer. Prove that the sum [tex]\displaystyle{1+{1\over2}+\cdots+{1\over n}}[/tex] is not an integer.[/quote]

group it into this manner.  1+ (1/2+1/3)+(1/4+1/5+1/6+1/7)...5/6,
these are each less than 1 and more than 1/2. (easily proven).  Then we can say that the since there is a prime between n and 2n, the base will be a multiple of a prime and so no base is the same.  the top of the fraction wont be divisible by any of the denominators.  I think thats conclusive enough.  Maybe i seriously screwd up[/quote]

You used Bertrand's Postulate.

As they said in PROMYS about this, "well, thats nice -- can you prove it?"

Try to find a solution that does NOT use this highly nontrivial theorem (it does exist, and its not too difficult to find.)

[quote]As they said in PROMYS about this, "well, thats nice -- can you prove it?" [/quote]

Yep :).

zabelman wrote:I could prove it [hide]if I looked up the proof![/hide]

If you have a TI-89 or some other symbolic processing device, look up the first ten terms or so and see if you can conclude anything.

I'll provide a Lemma.

We note that the can write the numerator of the sum of this series as:

n!/1+n!/2+n!/3+.......n!/n.

------------

Lemma: In the sequence of numbers 1 ...... n, n :ge: 2, exactly one of these numbers contains a power of 2 larger than any other number in the sequence.

Proof: Assume only two distinct numbers, 2^a, and 2^b are in the sequence 1 ..... n and both contain the largest power of 2 found in the sequence but no larger powers.  Let 2^a be the smallest number containing 2^a as a factor.  Then 2^b must be at least 2*2^a, which is a contradiction.  If 2^b = k(2^a), then 2^(a+1)  :le: 2^b, which is also a contradiction.

------------

Let k=2^t, where 2^t is the largest power of 2  :ge: n.  Consider n!/k, especially in relation to all the other terms in the numerator.

MAJOR HINT: [hide]If we factor all remaining powers of 2 out of n!/k, can we factor all the same power of 2 out of every other term in the numerator?  Can we factor these out of the denominator?[/hide]

EVEN BIGGER HINT: [hide]After factoring n!/k, consider the part which has no remaining powers of 2.  What is its parity?[/hide]

zabelman wrote:zabelman wrote:I could prove it [hide]if I looked up the proof![/hide]

I was talking about bertrand's postulate. I have already seen the question about [tex]\displaystyle \sum_{k=1}^n \frac{1}{k}\in\mathbb{N}.[/tex]

That makes sense, since you are better at problem solving than me.

I like Erdos' proof of Bertrand's Postulate a lot. It would take a long time for me to reproduce it without going back to look it up though.

[quote="Nukular"][quote="Scrambled"][quote="fuzzylogic"]Suppose that  [tex]n>1[/tex] is an integer. Prove that the sum [tex]\displaystyle{1+{1\over2}+\cdots+{1\over n}}[/tex] is not an integer.[/quote]

group it into this manner.  1+ (1/2+1/3)+(1/4+1/5+1/6+1/7)...5/6,
these are each less than 1 and more than 1/2. (easily proven).  Then we can say that the since there is a prime between n and 2n, the base will be a multiple of a prime and so no base is the same.  the top of the fraction wont be divisible by any of the denominators.  I think thats conclusive enough.  Maybe i seriously screwd up[/quote]

You used Bertrand's Postulate.
[/quote]

Actually, i have no idea what that is.

[quote="Nukular"][quote="ComplexZeta"]Chebyshev said it, and I'll say it again.
There's always a prime between n and 2n.[/quote]

I agree its a very nice and useful result, but using that on this problem is kinda like using rearrangement to show [tex]a^2+b^2 \ge 2ab[/tex][/quote]

Interestingly enough, since the proof of rearrangment hinges on factoring individual terms and showing that they are  :ge: 0 if they are ordered in a certain way, the rearrangment inequality applied to a^2+b^2  :ge: 2ab is equivalent to saying (a-b)^2  :ge: 0.

I'll (try to) pitch in a cent or two.

If we sum the powers of 1/2 in the series, we get something in the form

1 - 1/(2^k)

Now I need to prove that nothing can cancel out that 1 - 1/(2^k)...

[quote="Nukular"][quote]

Then we can say that the since there is a prime between n and 2n

[/quote]

How do you know this?

This is Bertrand's Postulate.[/quote]

oh.  I thought u were referring to something else :oops:


let $a; b; c > 0$ .find maximum value of the expression :
$P= \frac{a+3c}{a+2b+c}+\frac{4b}{a+b+2c}-\frac{8b}{a+b+3c}$