geometry
trapezoid
trigonometry
geometry proposed

In the convex quadrilateral  is chosen  point $M$.The rays $AB$ and $DC$ intersecting at the point $K$,the rays $BC$ and $AD$ at point $L$.It is known that $\angle{AMB}=70^{0}$
$\angle{BMK}=40^{0}$,$\angle{KMC}=\angle{CMD}=60^{0}$.Find $\angle{LMD}$.

Given are 5 rays $(MA, (MB, (MK, (MC, (MD$ such that $\angle AMB = 70^\circ,$ $\angle BMK = 40^\circ,$ $\angle KMC = 60^\circ,$ $\angle CMD = 60^\circ,$ $\angle DMA = 130^\circ$ and we have to find the direction of the 6th ray $(ML.$ Three of the points A, B, K, C, D can be chosen arbitrarily. We choose the vertex A arbitrarily and $AB \parallel MK,$, hence $\angle MBA = \angle BMK = 40^\circ.$ The angle $\angle MAB$ is then equal to $180^\circ - (\angle AMB + \angle MBA) = 40^\circ.$ The condition $AB \parallel MK$ sends the intersection $K \equiv AB \cap CD$ to infinity, so that our quadrilateral $ABCD$ is a trapezoid with $CD \parallel AB.$ This means that $\angle MDC = 180^\circ - (\angle KMC + \angle CMD) = 60^\circ.$ Thus the triangle $\triangle CMD$ has 2 angles equal to $60^\circ$ and it is equilateral. In addition, we choose the vertex D so that $\angle MAD = 10^\circ.$ Then $\angle MDA = 180^\circ - (\angle DMA + \angle MAD) = 40^\circ.$ Using the sine theorem for triangles $\triangle DMA,\ \triangle AMB,\ \triangle BMC,$

$\frac{DM}{AM} = \frac{\sin 10^\circ}{\sin 40^\circ},\ \ \frac{AM}{BM} = \frac{\sin 40^\circ}{\sin 70^\circ}$

$\frac{\sin \widehat{MBC}}{\sin \widehat{MCB}} = \frac{CM}{BM} = \frac{DM}{AM} \cdot \frac{AM}{BM} = \frac{\sin 10^\circ}{\sin 40^\circ} \cdot \frac{\sin 40^\circ}{\sin 70^\circ} = \frac{\sin 10^\circ}{\sin 70^\circ}$

In addition,

$\angle MBC + \angle MCB = 180^\circ - \angle BMC = 180^\circ - (\angle BMK + \angle KMC) = 80^\circ$

It follows that $\angle MBC = 10^\circ$ and $\angle MCB = 70^\circ.$ But then $\angle BAD = \angle MAB + \angle MAD = 80^\circ$ and $\angle ABC = \angle MBA + \angle MBC = 50^\circ,$ so that $\angle ALB = 180^\circ - (\angle BAD + \angle ABC) = 50^\circ.$ As a result, the triangle $\triangle ABL$ is isosceles with AB = AL. Since $CD \parallel AB$, the triangles $\triangle DCL \sim \triangle ABL$ are similar, hence DC = DL. From the equilateral triangle $\triangle CMD$, we have DC = DM. Consequently, DM = DL, so that the triangle $\triangle DML$ is isosceles with

$\angle MDL = \angle MDC + \angle CDL = \angle MDC + \angle BAD = 140^\circ$

$\angle DML = \angle DLM = \frac{180^\circ - \angle MDL}{2} = 20^\circ$


algebra
polynomial
complex analysis
complex analysis unsolved

[quote="mavropnevma"][hide="Hint"]Use Viete's relations after reducing the sum to degree $ 2$ [/hide][/quote]

Ha?? What is Viete's relations???
And how to reduce the sum to degree 2 ???
Please teach me^^

Since $ z_1,z_2,\ldots,z_{2009}$ are the roots of the equation $ 2z^{2009}\minus{}3z\plus{}1\equal{}0$, it follows $ 2z_k^{2010} \equal{} 3z_k^2 \minus{} z_k$ for all $ k\equal{}1,2,\ldots,2009$.
Then $ 2\sum_{k\equal{}1}^{2009} z_k^{2010} \equal{} 3\sum_{k\equal{}1}^{2009} z_k^2 \minus{} \sum_{k\equal{}1}^{2009} z_k$.
But $ \sum_{k\equal{}1}^{2009} z_k^2 \equal{} \left ( \sum_{k\equal{}1}^{2009} z_k \right )^2 \minus{} 2\sum_{1\leq i < j \leq 2009} z_iz_j$.

Now, for any polynomial $ P(z) \equal{} a_0z^n \plus{} a_1z^{n\minus{}1} \plus{} a_2z^{n\minus{}2} \plus{} \cdots \plus{} a_{n\minus{}1}z \plus{} a_n \equal{} a_0(z\minus{}z_1)(z\minus{}z_2)\cdots (z\minus{}z_n)$ one has $ \sum_{k\equal{}1}^n z_k \equal{} \minus{}a_1/a_0$ and $ \sum_{1 \leq i < j \leq n} z_iz_j \equal{} a_2/a_0$, by identification of coefficients (these and the others similar are Viete's relations).

With us, $ a_1 \equal{} a_2 \equal{} 0$, hence the final result is ... zero.

[quote="mavropnevma"]Since $ z_1,z_2,\ldots,z_{2009}$ are the roots of the equation $ 2z^{2009} \minus{} 3z \plus{} 1 \equal{} 0$, it follows $ 2z_k^{2010} \equal{} 3z_k^2 \minus{} z_k$ for all $ k \equal{} 1,2,\ldots,2009$.
Then $ 2\sum_{k \equal{} 1}^{2009} z_k^{2010} \equal{} 3\sum_{k \equal{} 1}^{2009} z_k^2 \minus{} \sum_{k \equal{} 1}^{2009} z_k$.
But $ \sum_{k \equal{} 1}^{2009} z_k^2 \equal{} \left ( \sum_{k \equal{} 1}^{2009} z_k \right )^2 \minus{} 2\sum_{1\leq i < j \leq 2009} z_iz_j$.

Now, for any polynomial $ P(z) \equal{} a_0z^n \plus{} a_1z^{n \minus{} 1} \plus{} a_2z^{n \minus{} 2} \plus{} \cdots \plus{} a_{n \minus{} 1}z \plus{} a_n \equal{} a_0(z \minus{} z_1)(z \minus{} z_2)\cdots (z \minus{} z_n)$ one has $ \sum_{k \equal{} 1}^n z_k \equal{} \minus{} a_1/a_0$ and $ \sum_{1 \leq i < j \leq n} z_iz_j \equal{} a_2/a_0$, by identification of coefficients (these and the others similar are Viete's relations).

With us, $ a_1 \equal{} a_2 \equal{} 0$, hence the final result is ... zero.[/quote]

wowowowowowow....
this is COOL xxxxxxxxxxxxxxxxxxxxxxxxxxx  :cool:  :clap:
thankyou^_^
but can you proof this Viete's relation theorem??

I did.
[quote] ... by identification of coefficients ...[/quote]
If you expand $ a_0\prod_{k\equal{}1}^n (z\minus{}z_k)$ and identify its coefficients with $ \sum_{i\equal{}0}^n a_iz^{n\minus{}i}$ you get Viete's relations.


calculus
derivative
logarithms
function
calculus computations

I really need this answer to be 100% correct, and my teacher claims there is something wrong in the last line. Can someone help. Please and Thank you!

1)Find the first derivative of y=xln(x)
Y=xln(x)
Ln(y)=ln(x)²
1/y•dy/dx=2ln(x)1/x
dy/dx=2(lnx)(1/x)(y)
dy/dx=2ln(x)/x•x^ln(x)
dy/dx=x^2ln(x)-1•2ln(x)

There is something wrong in the second line.  You actually have $ \ln y \equal{} \ln x \ln \ln x$.  Furthermore, logarithmic differentiation doesn't really help here; just use the usual product rule.  The correct first derivative is $ y' \equal{} \ln x \plus{} 1$.

O NO!! I am sorry i typed the question in wrong it should be x raised to ln(x) like this

y = x^ln(x)

Sorry!

I can read the last line in one of two ways, and in both ways it's wrong.  You can't simplify the derivative past $ \frac {2 \ln x}{x} x^{\ln x}$ unless you really wanted to write $ (2 \ln x) x^{\ln x \minus{} 1}$.

As t0rajir0u pointed out, you're making it overly complicated.  $ f(x)\equal{}x\ln{x}$ is just a product of two functions so the product rule should jump out at you.  $ f'(x)\equal{}\frac{d}{dx}(x)\ln{x}\plus{}x\frac{d}{dx}(\ln{x})\equal{}\ln{x}\plus{}\frac{x}{x}\equal{}\ln{x}\plus{}1$.

hello, if you really meant $ y=x^{\ln(x)}$ then we get $ \ln(y)=(\ln(x))^2$ and by differentiating with respect to $ x$ we get
$ \frac{1}{y}y^'=2\ln(x)\frac{1}{x}$ and from here we get $ y^'=x^{\ln(x)-1}2\ln(x)$.
Sonnhard.

I still think logarithmic differentiation makes it more difficult than it needs to be.

Let $ f(x)\equal{}x^{\ln{x}}$.  Then $ f(x)\equal{}e^{\ln{x}\ln{x}}\equal{}e^{\ln^{2}{x}}$.  Then $ f'(x)\equal{}\frac{2\ln{x}}{x}e^{\ln^{2}{x}}\equal{}2\ln(x)x^{\ln{x}\minus{}1}$, which is what Dr. Graubner got.


inequalities
trigonometry
geometry
angle bisector
trig identities
Law of Sines

In any triangle $ ABC$ show the inequality :

$ b \plus{} c \le a.csec(\frac {A}{2})$

geometrical solutions are better.

_________________
Tony Henderson.

Extend angle bisector of $ A$ to $ D$ on $ \overline{BC}$, law of sines on $ \triangle ACD$ gives $ b\sin \frac A2 \equal{} CD \sin \angle ADC \le CD$, similarly $ c \sin \frac A2 \le BD$ and adding gives the desired.

[if you want a purely geometric approach, drop perpendicular from $ C$ to $ AD$ at $ D'$ and $ b\sin \frac A2 \equal{} CD' \le CD$]


haha i just did one of the questions on the mathcounts forum #8 and my answer was off by 20 :rotfl:

but I also saw they posted some other question like what is the minimum integer value of (a+b/c)(d+e) for distinct integers a,b,c,d,e between 1-9.  [b]I think thats 8?[/b]

and some other question-#6 on target-I dont know what the question was asking, but it said minutes and hour hand make two 110 degress within an hour.  I got the two times to be [i]4:20/11[/i] and [i]4:460/11[/i]..but I dont know what its asking?  if its differnece in times, it should be 40 minutes..

just go ask lyndon (or richard if he comes) at the next arml thing in terre haute...which is way to far away i might add.

[quote="mathcrazed"]lol #5 and 6 on the target were SO ez...... we're really stupid -_-[/quote]

So what was the exact questoin?


videos Mafia AMC AIME USA(J)MO USAMO FTW

Okay, I made up a new type of game for FF. Basically, we start off with a list consisting of ten things that you like. Then the next person that posts has to choose the one that they like the least, and then replace it with something that they like more. It could be a movie, a TV show, some kind of activity, or something else. It can not be more than one thing. For instance it can't be a Lambrigini with video games inside it. It would have to be just a Lambrigini, or just video games. Try to be specific as possible, so instead of just saying a video game, say something like SSBB. (super smash brothers brawl). Try to keep the list in order from oldest thing on the list, to newest thing on the list.

This is an example;

*SSBB
*Brownies
*Lambrigini
*New England, Patriots
*Boston Red Sox
*Ping Pong
*Rubber Bands

Post #2

They took rubber bands away, and replaced it with balloons.

Post#3

They took away ping pong, and replaced it with a Wii.

Remember to take away the thing that you like the least. The thing you replace it with doesn't have to be your favorite thing, just something you like more.

Starting List;

*Nissan Altima
*SSBB
*Cheese
*Monkeys
*School
*Math
*Emoticons
*Mafia
*Unicorns :lol:
*Football

*Nissan Altima
*SSBB
*Cheese
*Monkeys
*School
*Math
*Emoticons
*Mafia
*Unicorns
[b]*Bicycling[/b]

*Nissan Altima
*SSBB
*Cheese
*Monkeys
*School
*Math
*Emoticons
*Mafia
*Bicycling
[b]*Dragons[/b]

*Nissan Altima
*Cheese
*Monkeys
*School
*Math
*Emoticons
*Mafia
*Bicycling
*Dragons
[b]*Dominions II (A computer game)[/b]

*Cheese
*Monkeys
*School
*Math
*Emoticons
*Mafia
*Bicycling
*Dragons
*Dominions II
*Pizza

*Monkeys
*School
*Math
*Emoticons
*Mafia
*Bicycling
*Dragons
*Dominions II
*Pizza
*AoPS

2 more things for me to get rid of... until I like everything on the list.

*Monkeys
*Math
*Emoticons
*Mafia
*Bicycling
*Dragons
*Dominions II
*Pizza
*AoPS
*Portal

[quote="herefishyfishy1"]*Monkeys
*Math
*Emoticons
*Mafia
*Bicycling
*Dominions II
*Pizza
*AoPS
*Portal
*AMC

YOU TOOK OFF DRAGONS!!!!

I hate you...

And how could you take off school!!!

*Math
*Emoticons
*Mafia
*Bicycling
*Dominions II
*Pizza
*AoPS
*Portal
*AMC
*Dragons

Can someone do me a favor and take off bicycling? Thanks...

I HATE DRAGONS!

And I didn't take off school. herefishyfishy1 did.

*Math
*Emoticons
*Mafia
*Bicycling
*Dominions II
*Pizza
*AoPS
*Portal
*AMC
*AIME

HOW COULD YOU HATE DRAGONS? WHY WOULD YOU TAKE OFF DRAGONS? DRAGONS ARE SUPERIOR TO JAMES'S! ARE YOU KIDDING? DRAGONS ARE AWESOME! BETRAYER JAMES!!!! (I could be angry enough to not call you james) ARE YOU SERIOUS JAMES? DO YOU REALLY HATE DRAGONS? DO YOU HAVE ANY CLUE WHAT DRAGONS ARE? YOU BETTER NOT TAKE IT OFF THE LIST (THAT INCLUDES EVERYONE ELSE AS WELL :D)! DRAGONS PWN BICYCLING! I COULD MAKE A LONGER SPEECH ABOUT HOW DRAGONS SHOULD STAY ON THE LIST AND SHOULD NOT BE TAKEN OFF THE LIST EVER! WHOEVER HATES DRAGONS SHALL PERISH AT YOUNG, RIPPED AND TORN BY A DRAGON'S PIERCING TALONS! IF NOT, HE OR SHE SHALL BE SCORCHED BY THE DRAGON'S FIERCE INFERNO! THEN THE INSULTER WILL BE CRUSHED BY ITS WEIGHT. AND THE DRAGON WILL TAKE OFF, LEAVING NO TRACES OF IT BEHIND, AS IT WHIPS UP A GALE WITH ITS MIGHTY WINGS AND VANISH INTO THE SKY. ANYWAYS, BACK TO MY SPEECH WHY DRAGONS SHOULD NOT BE TAKEN OFF THE LIST. JUST DON'T TAKE THEM OFF THE LIST! WHAT'S THE REASON IN TAKING IT OFF THE LIST? IS THERE A BENEFIT? WHY DON'T YOU TAKE BICYCLING OFF THE LIST? JAMES, I SWEAR I WILL GET BACK AT YOU IN SCHOOL!

*Math
*Emoticons
*Mafia
*Bicycling
*Dominions II
*Pizza
*AoPS
*AMC
*AIME
*Dragons (YOU BETTER NOT TAKE THIS OFF)

*Math
*Emoticons
*Mafia
*Bicycling
*Dominions II
*Pizza
*AoPS
*AMC
*AIME
*42

You and your dragon fetish.  :P

*Math
*Emoticons
*Mafia
*Bicycling
*Philosophy
*Pizza
*AoPS
*AMC
*AIME
*42

*Math
*Emoticons
*Mafia
*Bicycling
*Philosophy
*Pizza
*AoPS
*AMC
*AIME
*USAMO

Would people please bold the thing that they take away? Is that too hard?

*Math
*Emoticons
*Mafia
*Bicycling
*[b]Russian[/b]
*Pizza
*AoPS
*AMC
*AIME
*USAMO

*AoPS
*Math
*MathCounts
*Food
*The Round Table
* Mr. Green Mr. Green Mr. Green Mr. Green Mr. Green Mr. Green Mr. Green
*Emoticons
*The Game (which you just lost)
*More food
*FTW
* Mr. Green

*AoPS
*Math
*MathCounts
*Food
*India
*USA
*The Round Table
* Mr. Green Mr. Green Mr. Green Mr. Green Mr. Green Mr. Green Mr. Green
*Emoticons
*AIME
*The Game (which you just lost)
*More food
*Helitongue
*FTW
* Mr. Green

You guys went down to eleven. I brought back these fifteen items.

*AoPS
*Math
*MathCounts
*Food
*India
*USA
*The Round Table
 *Emoticons
*AIME
*The Game (which you just lost)
*More food
*Helitongue
*FTW
* Mr. Green
* Ninja Emoticon

*AoPS
*Math
*MathCounts
*Food
*India
*USA
*The Round Table
*Emoticons
*AIME
*The Game (which you just lost)
*More food
*:heli:
*FTW
* Mr. Green
* Ninja Emoticon

*AoPS
*Math
*MathCounts
*Food
*India
*USA
*The Round Table
*Emoticons
*AIME
*More food
* :heli:
*FTW
* Mr. Green
* Ninja Emoticon
*Monopoly

*AoPS
*Math
*MathCounts
*Food
*Spam
*USA
*The Round Table
*Emoticons
*AIME
*More food
*  :heli:
*FTW
* :D
* :ninja:
*Monopoly

*AoPS
*Math
*MathCounts
*Food
*USA
*The Round Table
*Emoticons
*AIME
*More food
* :heli:
*FTW
* Mr. Green
* Ninja Emoticon
*Monopoly
*Mafia

eggy, new ones go at the bottom

*OnlineMathCircle
*Math
*MathCounts
*Food
*Spam
*USA
*The Round Table
*Emoticons
*AIME
*More food
*
*FTW
*
*
*Monopoly

*OnlineMathCircle
*Math
*MathCounts
*Food
*Spam
*USA
*The Round Table
*Emoticons
*AIME
*More food
*Mafia
*FTW
*
*
*Monopoly

*OnlineMathCircle
*Math
*MathCounts
*Food
*Spam
*USA
*The Round Table
*Emoticons
*AIME
*More food
*Mafia
*FTW
* Alcumus
*
*Monopoly

*OnlineMathCircle
*Math
*MathCounts
*Food
*Spam
*USA
*The Round Table
*Emoticons
*AIME
*More food
*Mafia
*FTW
* Alcumus
* Pizza
*Monopoly

[list][*]OnlineMathCircle
[*]Math
[*]MathCounts
[*]Food
[*]Spam
[*]USA
[*]The Round Table
[*]Emoticons
[*]AIME
[*]More food
[*]Mafia
[*]FTW
[*]Alcumus
[*]Hot Dog
[*]Monopoly[/list]

■OnlineMathCircle
■Math
■MathCounts
■Food
■Spam
■USA
■The Round Table
■Emoticons
■AIME
■More food
■Mafia
■FTW
■Alcumus
■Hot Dog
■Monopoly
* Pokemon

■OnlineMathCircle
■Math
■MathCounts
■Food
■Spam
■USA
■The Round Table
■Emoticons
■AIME
■More food
■Mafia
■FTW
■Alcumus
■Hot Dog
■Sorry (board game)
■Pokemon

[list]OnlineMathCircle
Math
MathCounts
Food
Spam
USA
The Round Table
Emoticons
AIME
More food
Mafia
FTW
Alcumus
Hot Dog
Sorry (board game)
Pokemon[/list]
*The Hunger Games (look at my sig)(the trilogy)


let a,b,c and x,y,z be nonnegative real numbers. prove that
$ x(b\plus{}c)\plus{}y(c\plus{}a)\plus{}z(a\plus{}b)\ge 2\sqrt{(xy\plus{}yz\plus{}zx)(ab\plus{}bc\plus{}ca)}$

[quote="sxgfly"]let a,b,c and x,y,z be nonnegative real numbers. prove that
$ x(b \plus{} c) \plus{} y(c \plus{} a) \plus{} z(a \plus{} b)\ge 2\sqrt {(xy \plus{} yz \plus{} zx)(ab \plus{} bc \plus{} ca)}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=186636


geometry
parallelogram
analytic geometry
geometry proposed

Let M be the middlepoint of the side [BC]. Then

$BGCD-parallelogram\Longleftrightarrow MG=MD\Longleftrightarrow MA\cdot MD=MB\cdot MC\Longleftrightarrow m_a\cdot\frac 12 m_a =\left(\frac a2\right)^2\Longleftrightarrow b^2+c^2=2a^2\Longrightarrow 2a^2(b^2+c^2)=(b^2+c^2)^2\ge 4b^2c^2 \Longrightarrow$
$2a^2(b^2+c^2)\ge 4b^2c^2\Longleftrightarrow \frac{1}{b^2}+\frac{1}{c^2}\ge \frac {2}{a^2}.$

[quote="cezar lupu"]let $ABC$ be a triangle and $G$ is the centroid of the triangle $ABC$. Denote $D$ as the intersection point of median from $A$ with the circumcircle. Prove that , if $BGDC$ is a parallelogram , then $\frac{1}{AB^2}+\frac{1}{AC^2}\geq\frac{2}{BC^2}$ ;)[/quote]

[b][u]Generalization. [/u]I note [/b]$P(x,y,z)$ - [b]a point with the barycentrical coordinates [/b]$(x,y,z)$ [b]w.r.t. [/b]$\triangle ABC\ (x+y+z=1).$

$If\ P(x,y,z)\in (ABC),\ M\in BC\cap AP,\ \{ A,D\}\subset AP \cap C(O,R)$

$then\ MP=MD\Longleftrightarrow x(y+z)(yb^2+zc^2)=yz(1+x)a^2$

[b]and in this case [/b]$y\cdot \frac{1}{c^2}+z\cdot \frac{1}{b^2}\ge \frac {4x(1-x)}{1+x}\cdot \frac{1}{a^2}.$

[b][u]Remark. [/u]For G=P[/b]$\left( \frac 13,\frac 13,\frac 13\right)$ [b]we obtain the proposed [/b][b]problem.[/b]

[quote="levi"][quote="cezar lupu"]let $ABC$ be a triangle and $G$ is the centroid of the triangle $ABC$. Denote $D$ as the intersection point of median from $A$ with the circumcircle. Prove that , if $BGDC$ is a parallelogram , then $\frac{1}{AB^2}+\frac{1}{AC^2}\geq\frac{2}{BC^2}$ ;)[/quote]

[b][u]Generalization. [/u]I note [/b]$P(x,y,z)$ - [b]a point with the barycentrical coordinates [/b]$(x,y,z)$ [b]w.r.t. [/b]$\triangle ABC\ (x+y+z=1).$

$If\ P(x,y,z)\in (ABC),\ M\in BC\cap AP,\ \{ A,D\}\subset AP \cap C(O,R)$

$then\ MP=MD\Longleftrightarrow x(y+z)(yb^2+zc^2)=yz(1+x)a^2$

[b]and in this case [/b]$y\cdot \frac{1}{c^2}+z\cdot \frac{1}{b^2}\ge \frac {4x(1-x)}{1+x}\cdot \frac{1}{a^2}.$

[b][u]Remark 1. [/u]For G=P[/b]$\left( \frac 13,\frac 13,\frac 13\right)$ [b]we obtain the proposed [/b][b]problem.[/b][/quote]

[b]Remark 2. For I=P[/b]$\left(\frac as ,\frac bs ,\frac cs\right)$ [b]we obtain the relations [/b]

$(b+c)(b^3+c^3)=abc(b+c+2a)\ and\ a^2\ge (b+c)(s-a)\ because\ b^3+c^3\ge bc(b+c).$


limit
logarithms
calculus
calculus computations

$lim_{x\to 0+}\frac{ln^{44}xlnln^{34}x+(xlnx)^{1/17}lnln(1/x)}{(x^{3}e^{-1/x})^{1/5}}$
 :lol:  if somebody can't solve it then I put sulution. trust me it's not so hard how  it's looks.

The limit doesn't even make sense:

Think about your term Ln [Ln (1/x)]  when x->0. You start taking the Ln of a negative number. It doesn't even make sense for x = 1.

Where are you getting these problems from anyway? Are you making them up or are they from a book? I wish I had a problem book with problems such as these which also guides me through how to approach and solve them.

Cheers!

[quote="§outh§tar"]Think about your term Ln [Ln (1/x)] when x->0. You start taking the Ln of a negative number.[/quote]

As $x\rightarrow0^{+}, 1/x\rightarrow+\infty$.

[quote="§outh§tar"]I wish I had a problem book with problems such as these which also guides me through how to approach and solve them.[/quote]

$lim \frac{A+B}{C}=\frac{lim A+lim B}{lim C}$

if the last limit isn't an indetermination.

Hmm.. I just might not know but upon rewriting the

Ln [Ln 1/x] as Ln[-Ln x] things start to look hairy for certain values.

Maybe they can be ignored?  :ninja:

When you're taking $\lim_{x\to 0+}$, all you need to worry about is the stuff close to zero and positive, and $\ln x<0\ \forall x \in (0,1)$, which means that on that interval, $-\ln x>0$...

yes answer is $+oo$. Remembre when $x\to 0+$ then $lnln1/x\to+oo$.


calculus
derivative
group theory
abstract algebra
induction
superior algebra
superior algebra unsolved


Find the product of the real roots of x^2 + 18x + 30 = 2*Sqrt(x^2+18x+45).

I know the correct way of doing this problem. We can let u^2 = x^2 + 18x + 45, substituting to get (u-5)(u+3) = 0. For some reason, u can only equal 5, and not -3. Why is this? Anyways, following this method, we get the answer.

Also, supposed we instead let u = x^2 + 18x + 45. We then substitute to get u=25 and u=9. In this case, we are not alerted to the extraneous root. Why is this?

I think my two questions are similar (if not equal). So what's the deal with these extraneous roots in the problem?

Because, fiery, in ur original substitution, u was a square root, which under the reals cannot be negative

in ur second substituion, u was a quadratic
so its all good

But it wasn't really a square root--only the term u^2 was used. With this argument, letting u be negative is fine ( (-k)^2 = k). And how come we cannot detect the extraneous root using just u (and not u^2) in the substitution?

but fiery, if u let u^2 = x^2 + 18x + 45, then that means that the right side of the equation is 2*sqrt[u], which means u must be positive.

in ur second subsitution, u let u x^2 + 18x + 30, not u^2 = x^2 + 18x + 30.

That square makes the difference between extraneous/not-extraneous

No matter what you are doing, if you have squares or square roots or anything like that, you may be adding in extra solutions.. you just need to check that they fit the question afterwards.
In the first case, substituting u=5 works into the initial equation, but u=-3 gives -6=6, not true.
In the second case, substituting u=25 works, substituting u=9 does not.

You should always be alerted to extraneous roots, not just if you see a negative sign or something.

yea, if ur too lazy to understand algebraically whether a solution is extraneous,

just plug it in


How to make notes on the bottom part of the page, next to the page number.

For example, in page 32, I want a text like "32|Johan Gunardi" appear at the bottom. How to do it?

Use the fancyhdr package [[LaTeX:Packages#fancyhdr]] which will allow you to put whatever you want in the footer. To get what you want in the middle of the footer put
[code]\cfoot{\thepage|Johan Gunardi}[/code]If the vertical line comes out wrong then use
[code]\cfoot{\thepage$\mid$Johan Gunardi}[/code]instead.


calculus
derivative
algebra unsolved
algebra

Let  $ n > 1$ be a positive integer number. Prove that the equation :
                             $ 1 \plus{} 2\cdot x \plus{} 3\cdot x^2 \plus{} .... \plus{} n\cdot x^{n \minus{} 1} \equal{} n^2$ has rational root.

for $ x \neq 1$ we have :

$ 1 \plus{} x \plus{} x^2 \plus{} .... \plus{} x^{n} \equal{}\frac{x^{n\plus{}1}\minus{}1}{x\minus{}1}$

by differentiation we get :

$ 1 \plus{} 2x \plus{} 3.x^2 \plus{} .... \plus{} n.x^{n \minus{} 1} \equal{}\frac{nx^{n\plus{}1}\minus{}(n\plus{}1)x^{n}\plus{}1}{(x\minus{}1)^{2}}$

so the equation of the problem is tantamount to :

$ \frac{nx^{n\plus{}1}\minus{}(n\plus{}1)x^{n}\plus{}1}{(x\minus{}1)^{2}}\equal{}n^{2}$

an easy calculation shows that $ x\equal{}1\plus{}\frac{1}{n}$ is a rational solution.


A ball is thrown from point A.  It lands at point B.  Point B is 26 yards from point A.  The ball is in the air for 2.36 seconds.  Determine the speed at which the ball was thrown and the angle of the throw.

think of gravity: which is like 9.81 or something and use it along with some other things you figure out, using a velocity or distance formula.

It's just like throwing a baseball, just like pictures. That's how I always think of it.

We have these two equations:

$AB=v_{0}\cos\theta \Delta t$   and $v_{0}\sin\theta=\frac{g\Delta t}{2}$

Using these two equations to find the answer you want


geometry
circumcircle
MIT
college
trigonometry
geometry proposed

A circumscribed triangle ABC is also inscribed. M is midpoint of AC and N is midpoint of     arc AC (that consists point B).Prove that <BNI=<IMA(I is center of inscribed circle).  (please solve)

[quote="Ulanbek_Kyzylorda KTL"] [color=darkred]Let $ ABC$ be a triangle ($ b\ne c$) for which denote the circumcircle $ w = C(O,R)$ , the

 incircle $ C(I,r)$ , the midpoint $ M$ of $ [BC]$ and the point $ N\in w$ so that $ MN\perp BC$

and $ BC$ doesn't separate $ A$ , $ N$ . Prove that $ \widehat {ANI}\equiv \{\begin{array}{ccc} \widehat {IMC} & \mathrm{if} & b < c \\
\ \widehat {IMB} & \mathrm{if} & b > c\end{array}\ .$[/color] [/quote]

[color=darkblue][b][u]Proof I[/u].[/b] Suppose $ b<c$ . Denote the diameter $ [NS]$ of circle $ w$ , $ T\in BC$ , $ IT\perp IA$ and $ R\in NI\cap ST$ .

Prove easily that $ R\in w$ and the quadrilateral $ MITS$ is cyclically. Thus, $ IT\parallel NA$ and $ \widehat {IMC}\equiv\widehat {ISR}\equiv\widehat {ANI}\ .$[/color]

[color=darkblue][b][u]Proof II[/u].[/b] Suppose $ b < c$ . Denote $ D\in BC$ for which $ ID\perp BC$ . Observe that $ MD = \frac {c - b}{2}$

and $ \tan\widehat {IMC} = \tan\widehat {IMD} = \frac {ID}{MD} = \frac {2r}{c - b}$ $ \implies$ $ \boxed {\ \tan\widehat {IMC} = \frac {2r}{c - b}\ }\ .$

Since $ AN\perp AI$ obtain $ \tan \widehat {ANI} = \frac {AI}{AN} = \frac {b + c}{2p}\cdot\frac {2bc\cdot\cos\frac A2}{b + c}\cdot\frac {1}{2R\sin\frac {C - B}{2}} =$

$ \frac {bc\cdot\cos\frac A2\sin\frac A2}{2pR\sin\frac {C - B}{2}\cos\frac {B + C}{2}} =$ $ \frac {bc\sin A}{2pR(\sin C - \sin B)} = \frac {2pr}{p(c - b)}$ $ \implies$ $ \boxed {\ \tan\widehat {ANI} = \frac {2r}{c - b}\ }\ .$

In conclusion, if $ b < c$ , then $ \tan\widehat {ANI} = \tan\widehat {IMC} = \frac {2r}{c - b}$ , i.e. $ \widehat {ANI}\equiv\widehat {IMC}\ .$

You can prove analogously in the case $ b > c$ that $ \tan\widehat {ANI} = \tan\widehat {IMB} = \frac {2r}{b - c}$ , i.e. $ \widehat {ANI}\equiv\widehat {IMB}\ .$[/color]

Ray $AI$ cuts circumcircle $(O)$ of $\triangle ABC$ at the midpoint $P$ of the arc $AC$ not containing $B.$ Inversion through pole $P$ with radius $PA=PC=PI$ takes $M$ into $N$ and $I$ is double $\Longrightarrow$ Circle $\odot(IMN)$ is tangent to $PI$ through $I$ $\Longrightarrow$ $\angle INM=\angle MIP.$ Which implies that $\angle BNI=\angle IMA.$

See this post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32163[/url]


If [i]n[/i] has [i]k[/i] digits in its binary numeral, show that there are [u]at most[/u] 2^([i]k[/i]/2) numbers less than or equal to the sqrt of [i]n[/i].

Can you find an example where there are [u]exactly[/u] 2^([i]k[/i]/2)?

(i) Let $ n$ be a $ k$ digit integer when written in binary. Then, $ 2^{k\minus{}1} \leq n < 2^k$, since the smallest $ k$ digit integer is $ 1 0 \ldots 0_2 \equal{} 2^{k\minus{}1}$ and the largest is $ 1 \ldots 1 \equal{} 10 \ldots 0 \minus{} 1 \equal{} 2^k \minus{} 1$.

Hence, $ \sqrt{n} < \sqrt{2^k} \equal{} 2^{k/2}$. Thus, there cannot be more than $ 2^{k/2}$ naturals less than $ \sqrt{n}$.

(ii) The number of naturals less than or equal to $ \sqrt{n}$ is $ \lfloor \sqrt{n} \rfloor \leq \sqrt{n} < 2^{k/2}$, so there are no solutions.


There are $ 20$ pupils  in a back woods school. Any $ 2$ of them  have a common brother. Prove that there exist $ 14$ pupils having a common grandfather.

This doesn't seem like a trivial proof and therefore does not belong here.

[hide="My Thoughts"]
Seems like something perfect to use a graph for. Simply have a graph of $ 20$ vertices and make paths such that the vertices in every pair have a path to a common vertex. Not sure how to go from there though.
[/hide]

I don't know, you may need to specify it a little bit more (perhaps I'm wrong, but) If each pair has common brother, then they should all be brothers since they have a common brother, and therefore have the same grandfather. Can someone please help me understand what brother means in this case.

He's saying that if A has father F1 and mother M1, and B has father F2 and mother M2 where F1 != F2 and M1 != M2, then there exists a pupil who either {has father F1 and mother M2} or {has father F2 and mother M1}.


function
vector
algebra
domain
number theory
number theory unsolved

Suppose that a+bi divides c+di. Show that a^2+b^2 divides c^2+d^2.

Division in this case means using Gaussian integers, when the Gaussian integer a+bi divides the Gaussian integer c+di and results in another Gaussian integer e+fi.

I've been working on this for about 3 hours and I can't seem to get anywhere.

Let $x, y$ be Gaussian integers. We want to show that $x|y$ implies $N(x)|N(y)$. So, write $y=kx$. Taking the norm on both sides: $N(y)=N(kx)=N(k)N(x)$ which clearly gives us the result...

$\frac{c+di}{a+bi}= e+fi$.  Then

$(a+bi)(e+fi) = c+di$.  Norm both sides.

$(a^{2}+b^{2})(e^{2}+f^{2}) = c^{2}+d^{2}$.

[hide="A discussion of the complex norm"] First, we define the conjugation function $\overline{a+bi}= a-bi$.  Then we define

$N(z) = z \cdot \bar{z}$

This gives

$N(a+bi) = a^{2}+b^{2}$

This is called the "norm."  We usually use the square root of this to refer to the "length" of a complex vector, but when working with Gaussian integers, for number theory purposes, it is more useful to look at the norm, which is an integer.

[b]The relevant property:[/b]  The norm is multiplicative.  In other words,

$N(z \cdot w) = N(z) \cdot N(w)$

[b]Corollary:[/b]  If two Gaussian integers divide each other, then their norms divide each other. [/hide]

I haven't known "Gaussian integers" .It seem that the complex number $a+bi$ with $a,b$ integer?
Is there any other useful and interesting identity about Gaussian integers?
Thanks.

It's a unique factorisation domain (especially a Dedekind ring). Thus you have unique factorisation into primes.
Additionally, there is the norm mentioned already.

But it's just a special case of general number rings and all that stuff... I recomend getting a good knowledge in algebra, then to do algebraic number theory.

Thank you very much,ZetaX. I am learning algebra,But know little in algebraic number theory.
Sorry for the stupid question.


Find all positive integers $n$ such that $\frac{2^n+1}{n^2}$ is a integer.

My solution (I found it many years ago, when I was preparing to IMO :) ).
Suppose $n=p^aq^b...$, $p<q<...$.
Since $p\,|\,2^{2n}-1$ and $p\,|\,2^{p-1}-1$ we conclude $p\,|\,2^d-1$, where $d=(2n,p-1)$, but since $p$ is smallest prime number in composition of $n$, we obtain $d=2$. Thus $p\,|\,2^2-1$, i.e. $p=3$. So $n=3^km$, $m\geq 1$, $m$ is not divisible by 3.
Suppose $k>1$. Then $3^{2k}\,|\,2^{3^km}+1$. It is well known fact that $3^{k+1}\,|\,2^{3^k}+1$ and $2^{3^k}+1$ is not divisible by $3^{k+2}$. Using equality
\[2^{3^km}+1=(2^{3^k}+1)(2^{3^k(m-1)}-2^{3^k(m-2)+...+1})\]
we immediately obtain $2^{3^km}+1$ is not divisible by $3^{k+2}$, but $k+2\leq 2k$, so $2^{3^km}+1$ is not divisible by $3^{2k}$. Contadiction.
Therefore, $n=3m$.
Suppose $m>1$. Then $q\,|\,2^{2\cdot 3m}-1$ and $q\,|\,2^{q-1}-1$. It follows $q\,|\,2^k-1$, where $k=(2\cdot 3m,q-1)$. Since $k\neq 1$, we see that $k=6$, i.e. $q\,|2^6-1$. It means $q=7$, but $2^{3m}+1$ is not divisible by 7 (it is 2 modulo 7). Contradiction.

Finally, $n=3$.

Hope it is correct.

[quote] Hope it is correct.[/quote]

Considering that you won a gold medal at IMO after this training, I think we may hope it is  :)

Pierre.

Um here is a general version :
PROBLEM: Let be given $p$ a prime number. Find all $n$ positive integer such that
\[n^2\mid (p-1)^n+1\]

I solved it, so i leave it for you.

I can't quickly complete it with my method, so, please, show us your solution.

oh sorry i mean

Problem: Let be given $p$ is prime. Find all $n$ such that
1st) $n^{p-1}\mid (p-1)^n+1$
2nd) $n^k\mid (p-1)^n+1$ for $k>2$.

Now you can quickly do  :D . By the way I will give you one thounsand Vietnam dollards  :D  if you work out for the cases $k=1$ and $k=2$. :P

[quote="pluricomplex"]Problem: Let be given $p$ is prime. Find all $n$ such that
1st) $n^{p-1}\mid (p-1)^n+1$
2nd) $n^k\mid (p-1)^n+1$ for $k>2$.[/quote]
Is it one problem or two problems?
I think, it is impossible $n^k\mid (p-1)^n+1$, $k>2$. Right?

it's my careless typrest. They are two problems which could be proved by the same method as you used. Yes for $k>2$ no such $n$ exist!

[quote="pluricomplex"]Yes for $k>2$ no such $n$ exist![/quote]
Great! It is direct consequence of my approach.  :D

[quote]Great! It is direct consequence of my approach. [/quote]

yes but....false for $n=1$ uh? :D

:D  :D  :D
Of course, $n=1$ is not a "real" number  :P  :P  :P


[quote="Yosh..."]If $a,b,c \in R^{+}$ and $a+b+c=3$

Prove $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\geq a^{2}+b^{2}+c^{2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87919


blogs
limit
real analysis
real analysis solved

Compute the limit:

lim(x-->a) [ log_c(log_b(log_a(x^(b^c))))]^(1/(x-a)) where a,b,c are pozitive reals \{1}

cheers! :D :D

(i consider it as x^(b^c)).

Change variable: t = x-a \to 0.
Denote A = (b^c) ln (t+a) / ln a,
           B = ln A / ln b,
           C = ln B / ln c
The limit is  \lim e^(ln C / t) = e^(\lim (ln C / t)). We are going to find out \lim (ln C / t).
Use L'hospital rule, we need to compute
\lim 1 / ((a+t) ln (a+t) ln A ln B).
a+t \lim a
ln (a+t) \lim ln a.
A \lim b^c, ln A \lim c ln b
B \lim c, ln B \lim ln c

hence, the answer is e^(1/(a c ln a ln b ln c)).

Yes liyi, I have done it without l'Hopital, then when I posted it I didn't know l'Hopital, I have solved it with a lot of computation and the limit can be rewritten like so:

L=1/(ac ln(a)ln(b)ln(c)

cheers! :D


geometry
combinatorics unsolved
combinatorics

Let $ P$ family of squares in a plane such that any points on plane inscribed in the interior of $ \geq k$ squares in $ P$
Prove that $ P$ can be divided into $ 4k\minus{}3$ subfamilies such that in any subfamilies, every $ 2$ squares are disjoint

Are you sure that it is $ \geq k$ ?
Because if some point of the plane belongs to $ 4k\minus{}2$ of the squares then for any repartition into $ 4k\minus{}3$ parts, two of these squares will belongs to the same part... :wink:

Pierre.

Hmm...actually i'm not sure :blush: Because this is my problem :(
I'll check again


Let $a,b,c>0$ and $a+b+c=1$. Prove that:
$\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\leq2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)$

$a=\frac{x}{x+y+z}$;$b=\frac{y}{x+y+z}$; $c=\frac{z}{x+y+z}$;
So The inenquality $\Leftrightarrow \sum\limits_{cyclic} \frac{2x+y+z}{y+z} \leq 2 \sum\limits_{cyclic}\frac{x}{y}$
$\Leftrightarrow 2 \sum\limits_{cyclic}\frac{x}{y+z} +3 \leq 2 \sum\limits_{cyclic}\frac{x}{y}$

We have $3 \leq \sum\limits_{cyclic}\frac{x}{y}$ So we must prove :
$2 \sum\limits_{cyclic}\frac{x}{y+z}\leq \sum\limits_{cyclic}\frac{x}{y}$
$\Leftrightarrow 2 \sum\limits_{cyclic}\frac{1}{\frac{y}{x}+\frac{z}{x}}\leq \sum\limits_{cyclic}\frac{x}{y}$

We set $\frac{x}{y}=p; \frac{y}{z}=m; \frac{z}{x}=n$

The inenquality $\Leftrightarrow 2 \sum\limits_{cyclic}\frac{1}{\frac{1}{m}+\frac{1}{n}} \leq m+n+p$ for $mnp=1$
$\Leftrightarrow 2 \sum\limits_{cyclic} \frac{mn}{m+n} \leq \sum\limits_{cyclic} \frac{1}{mn}$

Apply: $\frac{1}{x}+\frac{1}{y} \geq \frac{4}{x+y}$, We have:
$\frac{1}{mn}+\frac{1}{np} \geq \frac{4}{mn+np}=\frac{4}{n(m+p)}=\frac{4mp}{m+p}$

We have some prroblems like this :

[b]1.[/b] Give  $a,b,c>0$
[i]i.[/i] Prove that $\frac{a+c}{b+c} + \frac{a+b}{a+c} +\frac{b+c}{a+b} \leq \frac{b}{a} + \frac{c}{b} + \frac{a}{c}$

[i]ii.[/i] Prove that  $\frac{a+c}{b+c} + \frac{a+b}{a+c} +\frac{b+c}{a+b} \leq \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$

[b]2.[/b] Give $a,b,c>0$ such as $a+b+c=1$

[i]i.[/i] Prove that $\sum_{cyclic} \frac{1+a}{1-a} \leq 2.(\frac{a}{b} + \frac{b}{c} + \frac{c}{a} )$

[i]ii.[/i] Prove that $\sum_{cyclic} \frac{1+a}{1-a} \leq 2. (\frac{b}{a} + \frac{c}{b} + \frac{a}{c} )$

Problems 2 i. and ii. are equivalent and discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=25780 ; problems 1 i and ii were solved at http://www.mathlinks.ro/Forum/viewtopic.php?t=5263 .

  darij

Oh, I see ! I have had more than  one proof for each problems !
Notice that all of them can be solved easily by SOS method !

I have a solution using Cauchy-Schwarz.
\[\;\frac{{1 + a}}{{1 - a}} + \frac{{1 + b}}{{1 - b}} + \frac{{1 + c}}{{1 - c}} \le 2\left( {\frac{b}{a} + \frac{c}{b} + \frac{a}{c}} \right)\]
\[ \Leftrightarrow \sum\limits_{cyc} {\frac{{2a + b + c}}{{b + c}}}  \le 2\sum\limits_{cyc} {\frac{b}{a}}  \Leftrightarrow \sum\limits_{cyc} {\left( {\frac{a}{c} - \frac{a}{{b + c}}} \right)}  \ge \frac{3}{2}\]
\[ \Leftrightarrow \sum\limits_{cyc} {\frac{{ab}}{{c(b + c)}}}  \ge \frac{3}{2} \Leftrightarrow \sum\limits_{cyc} {\frac{{{a^2}{b^2}}}{{abc(b + c)}}}  \ge \frac{3}{2}\]
By using Cauchy-Schwarz inequality, we have:
\[\sum\limits_{cyc} {\frac{{{a^2}{b^2}}}{{abc(b + c)}}}  \ge \frac{{{{(ab + bc + ca)}^2}}}{{2abc(a + b + c)}}\]
Easily we have: ${(x + y + z)^2} \ge 3(xy + yz + zx)$
So we have:
\[\sum\limits_{cyc} {\frac{{{a^2}{b^2}}}{{abc(b + c)}}}  \ge \frac{{{{(ab + bc + ca)}^2}}}{{2abc(a + b + c)}} \ge \frac{{3abc(a + b + c)}}{{2abc(a + b + c)}} = \frac{3}{2}\]


function
linear algebra
matrix
induction

Anyone have some non-inductive solutions to a few Fibonacci Identities? Or even some inductive solutions for the more complex ones.

(1) $ L_{k \plus{} 1}L_{k \minus{} 1} \minus{} {L_k}^2 \equal{} 5( \minus{} 1)^{n \plus{} 1}$

(2) $ L_n \equal{} F_{n \minus{} 1} \plus{} F_{n \plus{} 1}$

(3) $ 5F_n \equal{} L_{n \minus{} 1} \plus{} L_{n \plus{} 1}$

(4) $ F_{n\plus{}p\plus{}1}\equal{}F_{n\plus{}1}F_{p\plus{}1}\plus{}F_{n}F_{p}$

(5) $ (F_1)^2\plus{}(F_2)^2\plus{}(F_3)^2\plus{}...\plus{}(F_n)^2\equal{}F_{n\minus{}1}F_{n\plus{}1}$

(6) $ (L_n)^2\minus{}5(F_n)^2\equal{}4(\minus{}1)^n$

Eh, mostly I couldn't figure out the induction for the more complex ones, but I was wondering how to approach the cross Fibonacci/Lucas identities. I'm sure if I saw one or two clean examples I could work from there.

Could someone try a matrix proof of number 4?

[quote="t0rajir0u"]If by "non-inductive" you mean "bashing with Binet's Formula," then yes.  Otherwise, there are generating functions proofs and [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=175257]matrix[/url] proofs.[/quote]

You break my heart!  The Fibonacci numbers are, if nothing else, combinatorial in nature.  In particular, if we index $ F_0 \equal{} F_1 \equal{} 1$, $ F_n$ is the number of ways to cover a $ 1 \times n$ strip with $ 1 \times 1$ or $ 1 \times 2$ bricks.  Less well-known, the Lucas number $ L_n$ (indexed so $ L_1 \equal{} 1, L_2 \equal{} 3$) is the number of ways to cover a circle of $ n$ points by covering single points or pairs of consecutive points.  (I am indebted to user renatzu for pointing this out ([url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=109530[/url]).)

This makes results 2 and 4 (with corrected indexing) easy.  I haven't tried to derive the others, although I expect they will be harder in general.  I'm curious about (3), in particular.

(5) is incorrect (try it for any $ n \neq 2$).

Unless I've mucked something up, the Fibonacci number version of (1) can be generalized (using my chosen indexing) to the identity $ F_m F_{m \plus{} n} \equal{} F_{m \plus{} k}F_{m \plus{} n \minus{} k} \plus{} ( \minus{} 1)^m F_{k \minus{} 1}F_{n \minus{} k \minus{} 1}$.  I suspect a simlar (or joint?) generalization is possible for (1).

I tend to think of the Fibonacci numbers within the context of second-order linear recurrences as a whole, and while I recognize the combinatorial nature of such sequences I prefer to think of them analytically, so the analytical solutions are the first ones that occurred to me.  :P

[quote="Pakman2012"]I was wondering how to approach the cross Fibonacci/Lucas identities. I'm sure if I saw one or two clean examples I could work from there. [/quote]

(2) can be used to transform each problem into a problem based on the Fibonacci sequence alone, and is easy to prove if we consider a "general Fibonacci sequence" $ G_0 \equal{} a, G_1 \equal{} b, G_n \equal{} F_n b \plus{} F_{n\minus{}1} a$.  (3) follows straightforwardly from (2).  (4) also follows from an application of the "general Fibonacci sequence" idea; consider $ G_0 \equal{} F_{p}, G_1 \equal{} F_{p\plus{}1}$.

I have to agree. While the combinatorial proofs are beautiful, they are not a general approach. 'Binet Bashing' trivializes all of these problems.

Sure, but one could argue that a result that can only be proved using Binet is "merely arithmetical," while a combinatorial proof demonstrates something deeper.  (Whether one finds this compelling probably correlates with how much one appreciates combinatorics as a field.)

Also, the fix for number (5) is to replace $ F_{n \minus{} 1}F_{n \plus{} 1}$ with $ F_n F_{n \plus{} 1}$, if I'm not mistaken.

Finally, there are other types of proof for Fibonacci identities, e.g. proof based on other identities.  (3) follows more or less immediately from (2); I'm fairly sure that (5) can be proved from $ \sum_{i \equal{} 0}^n F_i \equal{} F_{n \plus{} 2} \minus{} 1$ (a result that can be proved combinatorially or lots of other ways).


quadratics
algebra
polynomial
pigeonhole principle
number theory
algebra unsolved

Consider a polynomial of degree 2 with integer coefficients that is not the square of a polynomial of degree 1.
For such a polynomial, what is the largest n such that P(1), P(2), P(3), ... and P(n) are all perfect squares?

Huge respect for that one who solves this.  :)

I have not got the solution but probably this is the first step in solving this problem:
Let $P(1)=a^2, P(2)=b^2, P(3)=c^2$ and $P(4)=d^2$ then you can easily get that
$3(c^2-b^2)=d^2-a^2$ and after some computations to get that :
$P(x)=x^2(\frac{a^2}{2}-b^2+{\frac{c^2}{2}}) + x(\frac{-5a^2}{2}+4b^2-\frac{3c^2}{2}) + (3a^2-3b^2+c^2)$.
And I pretty sure that the solution is $n=3$ and for it for example $P(x)=18x^2-42x+28$
Hope this helps  :lol:

heartwork,this comes from your message "this is kiddie?"in subforum Number theory Unsolved problem,isn't it?

 I also think the question is abitrarily large $n$,but i'm really sorry i still have no solution.

 Hey guys,could you possibly help me  :( ?

@heartwork:
Where on the web did you exactly find this problem? Was it part of some competition? Could you post a URL?

In fact "this is kiddie?" in subforum Number theory Unsolved problem came from this...
I strongly believe there is a nice elementary solution!  :?

Hey, we still don't know what is the answer of the problem, not to speak about a solution. And since we have already found that n=8 works, I really doubt that one can find an elemntary nice solution.

Finally, I think I solve it. The largest integer [i] n [/i] is indeed $8$.
Suppose there is a polynomial with [i]n[/i] $\geq 9$ such that all $P(1)$, $P(2)$, ..., $P($[i]n[/i]$)$ are perfect squares. Then at least $5$ consecutive of them are perfect squares such that each $2$ of them are distinct: $P(j+1)$, $P(j+2)$, $P(j+3)$, $P(j+4)$, $P(j+5)$, because $P$ is a polynomial of degree $2$.
For simplicity we may consider another quadratic polynomial, $Q(x)=P(x+j+3)$ that will be of the same type with integer coefficients and not the square of a polynomial of degree $1$: $Q(x)=ax^2+bx+c$.
Then $Q(-2)=k_{-2}^2$, $Q(-1)=k_{-1}^2$, $Q(0)=k_{0}^2$, $Q(1)=k_{1}^2$, $Q(2)=k_{2}^2$ and we get:
$c=k_{0}^2$, $b=\frac{k_{1}^2-k_{-1}^2}{2}=\frac{k_{2}^2-k_{-2}^2}{4}$, $a=\frac{k_{1}^2+k_{-1}^2-2k_{0}^2}{2}=\frac{k_{2}^2+k_{-2}^2-2k_{0}^2}{8}$ and the relations:
$2(k_{1}^2-k_{-1}^2)=k_{2}^2-k_{-2}^2$ and  $4(k_{1}^2+k_{-1}^2-2k_{0}^2)=k_{2}^2+k_{-2}^2-2k_{0}^2$.
It is quite easy to observe that both relations above cannot hold for some reasons of parity! First note that $k_{1}$, $k_{-1}$, respectively $k_{2}$, $k_{-2}$ are distinct and must have the same parity. If they are all even, we may divide all coefficients of $Q$ with $2$ so we can exclude this case. If they are all odd or $2$ are odd and the other $2$ are even, then we get a contradiction in each case.

[quote="heartwork"]
Suppose there is a polynomial with [i]n[/i] $\geq 9$ such that all $P(1)$, $P(2)$, ..., $P($[i]n[/i]$)$ are perfect squares. Then at least $5$ consecutive of them are perfect squares such that each $2$ of them are distinct: $P(j+1)$, $P(j+2)$, $P(j+3)$, $P(j+4)$, $P(j+5)$, because $P$ is a polynomial of degree $2$.[/quote]
Not sure I understand this. What if we have the following:
$P(1)=a^2$
$P(2)=b^2$
$P(3)=c^2$
$P(4)=d^2$
$P(5)=d^2$
$P(6)=e^2$
$P(7)=f^2$
$P(8)=g^2$
$P(9)=g^2$
$P(10)=h^2$
$...$

Do I miss something?

That was only an example, I'm still not sure how we can argue that there must be five consecutive distinct squares (by a pigeonhole principe there must be five distinct, but how do we know that they are consecutive)?

But it is a quadratic polynomial! :what?:

[quote="heartwork"]
Consider a polynomial [color=#FF0000]of degree 2[/color] with integer coefficients that is not the square of a polynomial of degree 1.[/quote]
I missed that part, sorry.


inequalities
trigonometry
geometry proposed
geometry

[color=purple] Let $ r_{a},r_{b},r_{c}$ are three exradii and $ l_{a},l_{b},l_{c}$ are three internal angle bisectors of triangle ABC.Prove that :
   $ \frac{2R}{r}\minus{}1\ge\frac{r_{a}}{l_{a}}\plus{}\frac{r_{b}}{l_{b}}\plus{}\frac{r_{c}}{l_{c}}\ge\sqrt{\frac{4R}{r}\plus{}1}$
  :D Nice or not ?  :blush: Can you make them stronger  :) [/color]

I think you misscalculated something since your RHS inequality is [b]wrong[/b]. Actually,
\[ \frac{2R}{r}\minus{}1\ge 3\geq\frac{r_{a}}{l_{a}}\plus{}\frac{r_{b}}{l_{b}}\plus{}\frac{r_{c}}{l_{c}}.\]
Moreover a non-cyclic one also holds, due to Ji Chen, see [url]http://www.mathlinks.ro/viewtopic.php?t=129850[/url].
\[ \frac{r_{a}}{l_{a}}\leq\frac{3}{\pi}A,\]
where $ A$ can be the minimum angle of the triangle. Hence one can deduce a refinement of the cyclic one, stronger than the one in the LHS,
\[ \frac{2R}{r}\minus{}1\ge 3\geq\frac{9}{\pi}\cdot\min\left\{A, B,C\right\}\geq\frac{r_{a}}{l_{a}}\plus{}\frac{r_{b}}{l_{b}}\plus{}\frac{r_{c}}{l_{c}}.\]

[quote="pohoatza"]I think you misscalculated something since your RHS inequality is [b]wrong[/b]. Actually,
\[ \frac{2R}{r}\minus{}1\ge 3\geq\frac{r_{a}}{l_{a}}\plus{}\frac{r_{b}}{l_{b}}\plus{}\frac{r_{c}}{l_{c}}.\]
[/quote]
   [color=purple] I don't know what do you want to say ? Do you think the following is wrong : ?

   $ \frac{2R}{r}\minus{}1\ge\frac{r_{a}}{l_{a}}\plus{}\frac{r_{b}}{l_{b}}\plus{}\frac{r_{c}}{l_{c}}.$
[/color] :)

No, I said that the following is wrong

\[ \frac{r_{a}}{l_{a}}\plus{}\frac{r_{b}}{l_{b}}\plus{}\frac{r_{c}}{l_{c}}\ge\sqrt{\frac{4R}{r}\plus{}1},\]

since $ \frac{r_{a}}{l_{a}}\plus{}\frac{r_{b}}{l_{b}}\plus{}\frac{r_{c}}{l_{c}}\leq 3$.

[color=purple]Are you sure pohoatza  :D ? I can show you the stronger :
   $ \frac{r_{a}}{m_{a}}\plus{}\frac{r_{b}}{m_{b}}\plus{}\frac{r_{c}}{m_{c}}\ge\sqrt{\frac{4R}{r}\plus{}1}$
  I can't prove the above but I know it true !  :D So my inequality is correct  :) ! Can you show me your disprove solution of you ? Complete please  :)
And if you like this problem try with this link :www.mathlinks.ro/Forum/viewtopic.php?t=166399[/color]

[color=darkblue]From where know you that is it truly ?!

For finding of the diverse inequalities, you can use

$ 1\blacktriangleright$ In the case of the [b]acute or right triangle[/b],

$ m_{a}\le R(1\plus{}\cos A)$, i.e. $ 2rm_{a}\le a(p\minus{}a)$ with equality iff $ b \equal{} c$.

For example, $ m_{a}\plus{}m_{b}\plus{}m_{c}\le r_{a}\plus{}r_{b}\plus{}r_{c}$.

$ 2\blacktriangleright$ [b]Generally[/b],

$ m_{a}^{2}\ge p(p\minus{}a)$ with equality iff $ b \equal{} c$ ;

$ b^{2}\plus{}c^{2}\le 4Rm_{a}$ with equality iff $ b \equal{} c\ \ \vee\ \ A \equal{} 90^{\circ}\ .$[/color]

[quote="Virgil Nicula"][color=darkblue]From where know you that is it truly ?![/color]

[/quote]
 [color=purple]What do you want to say Virgil ? I dont' know
 $ \sum \frac {r_{a}}{l_{a}}\ge\sqrt {\frac {4R}{r} \plus{} 1}$or$ \sum\frac {r_{a}}{m_{a}}\ge\sqrt {\frac {4R}{r} \plus{} 1}$
      :wink:[/color]

[quote="evarist"][color=purple][size=134][b]I can't prove the above but I know it true ![/b][/size][/color][/quote]I ask you again : from where you know it is truly ?!

[color=purple]Oh it's from my friend he's very good at geomettric inequality ! He said he couldn't prove it , too but he saw it true by computer ! Now what do you want to say ? [/color]

[color=darkblue]O.K. Is clearly : the computer cann't do all ! It is the my gladness. I can be lucky. I hope that you know what is a joke !? This inequality "tortured" me, little by little, all last week ...[/color]


Να λυθεί με την μέθοδο των διαδοχικών προσεγγίσεων η ολοκληρωτική εξίσωση
$ f(x)\equal{}\int_{0}^{x}\frac{1\plus{}f^2(t)}{1\plus{}t^2}dt$

Προφανώς αν αντικαταστήσουμε στη θέση του f(t) την αρχική συνάρτηση φο(t)=x τότε αποδεικνύεται όντως ότι f(x)=x,δεν εφαρμόζουμε όμως έτσι την μέθοδο των διαδοχικών προσεγγισεων.
Γι΄αυτό πρέπει να αντικαταστήσουμε στη θέση του f(x) μέσα στο ολοκλήρωμα την φο(t)=1 οπότε θα υπολογίσουμε το ολοκλήρωμα του
$ \int_{0}^{x}\frac{2}{1\plus{}t^2}dt$ που ως γνωστο είναι το 2τοξεφχ
και μετά θα αντικαταστήσουμε στη θέση της f(x) μέσα στο ολοκλήρωμα το 2τοξεφχ και θα υπολογίσουμε εκ νέου το καινούργιο ολοκλήρωμα... και ούτω καθ΄εξής,
όμως θα χρειαστούμε μου φαίνεται την δυναμοσειρά που εκφράζεται η εφχ
μήπως ξέρετε ποια είναι αυτή η σειρά?

[quote="spinos"]όμως θα χρειαστούμε μου φαίνεται την δυναμοσειρά που εκφράζεται η εφχ
μήπως ξέρετε ποια είναι αυτή η σειρά?[/quote]

Try [url=http://en.wikipedia.org/wiki/Trigonometric_functions#Series_definitions]this[/url]... δεν είναι όμορφο :(

Cheerio,

Durandal 1707


daca avem 2 functii f,g care au aceeasi convexitate, prin compunerea lor ce va rezulta? o functie cu aceeasi convexitate?

din manualul de Ganga clasa a XI-a analiza, la convexitate / concavitate, citire :

daca $g$ convexa , $f$ convexa CRESCATOARE, at $f \circ g$ ---> convexa

de retinut : $f$ , $g$ convexe nu implica numaidecat $f \circ g$ ---> convexa !!!

dar daca aplici definitia....nu da numaidecat?(intre timp am facut cu definitia...si..iese) :)

eu ti-am zis ce scrie in manual . si cred ca e bine ce scrie aici . depinde de la functii la functii, poti sa gasesti doua convexe ale caror compunere da o  functie care nu e convexa. feel free to search for a counterexample :)

[quote="spx2"]un exemplu ar putea fi
$f,g: [\pi,2\pi]\to R$ cu functiile $f(x)=x^2,g(x)=sin(x)$
avem in continuare $(f\circ g)(x)=sin^2(x)$ pe care o notam cu $h(x)$
se observa acum ca $h''(x)=2cos(2x)$ deci $Im(h''(x)) = Im(2cos|_{[2\pi,4\pi]})$
dar functia cos pe intervalul $[2\pi,4\pi]$ poate fi si convexa si mai putin convexa
deci avem contradictia pe macar o portiune a acestui interval amintit mai sus.
sper sa fie bine[/quote]

 Cum adik mai mai putin convexa?  :?  Si $\cos$ e concava pe $\left [0,\frac{\pi}{2}\right]$ din umilele mele cunostiinte. ;)

"mai putin convexa" adica concava  :)
si tocmai pentru ca $[2\pi+0,2\pi +\frac{\pi}{2}]\subset [2\pi,4\pi]$
deci functia cos e concava pe o portiune din $[2\pi,4\pi]$ ceea ce e suficient


analytic geometry
Pythagorean Theorem
geometry

Five unit squares are arranged in a T shape:

[ ]
[ ][ ][ ]
[ ]

(rephrased)
The figure is inscribed in a circle of radius r, find r.

Put the center of the circle in the middle of the leg of the T and move it up and down. The radius can be found by the Pythagorean theorem and when the two radii are equal, they will be minimized. Call $x$ the distance from the middle of the circle and the intersection of the leg and head of the T. The radius is the larger of $\sqrt{(3-x)^2+.5^2}$ and $\sqrt{x^2+1.5^2}$. Setting these equal and solving for $x$ yields $x=\frac43$. So the radius is $\sqrt{\frac{16}{9}+2.25}$ which is about $2.00693$. Is this right?
Edit: see other solution below, arithmetic error here.

No, that's not what I got.
[hide="Hint"]Put the figure in a coordinate plane[/hide]

Hokkage, that method works, but you should check your arithmetic.

Yeah, I subtracted 2.25 from 9.25 and got 8 somehow  :huh: ? Anyways, my answer revised: $x=\frac76$ so $\sqrt{x^2+2.25}=\sqrt{\frac{49}{36}+2.25}$ which is about $1.90029$

I was looking for an exact answer, which you got :)


How do I do a right arrow, not this kind $ \Rightarrow$ but the arrow with only one vertical line


A car travels on flat land 4 miles west and 3 miles south.  How many miles is the car from it’s starting point?

[hide]

[color=green]The answer is 5 miles.[/color]

We can draw a triangle to show the question.

Draw a point to signify the starting point.
Draw a line segment to the left and label it 4.
From the end of that line segment, draw another line segment down from it, perpendicular. Label this 3.  Now connect the end of [b]that[/b] line segment. You now have a right angle triangle. Use the Pythagorean Theorem: $3^{2}+4^{2}=25$
The square root of 25 is 5, which is your answer.

[/hide]

[hide="solution"]We can draw a 3 by 4 rectangle and draw a diagonal towards the ending point. We need the length of the hypotenuse, which is 5.[/hide]

Correct :D

EXTENSION: What if the car then travels 2 miles east and 2 miles south? :)

[hide="extension"]Using same methods as I did above, you must find $\sqrt{2^{2}+2^{2}}=\sqrt{8}=\sqrt{4\cdot2}=2\sqrt{2}$

$2\sqrt{2}$

EDIT: Latexed.[/hide]

[quote="ProtestanT"][hide="Solution"]
$\sqrt{2^{2}+5^{2}}=\sqrt{29}$
[/hide][/quote]
Not 5 miles.

[hide]$x=\sqrt{2^{2}+2^{2}}$

$x=\sqrt{8}$

$\boxed{x=2\sqrt{2}}$[/hide]

[quote="ProtestanT"]if the car then travels 2 miles east and 2 miles south, then isnt it 2 miles west and 5 miles south overall?[/quote]
I think you misinterpreted the extension. It is a different case, but the question is the same.

More Extension: A car travels 1 mile north, 2 miles east, 3 miles south, and 4 miles west. How far is it from its starting point? :D

[quote="i_like_pie"][quote="bpms"][hide] The car moves 2 net miles south and two to the east. $\sqrt{4+4}=2\sqrt{2}$[/hide]
Extension If a car moves 15 miles east and 36 miles north how far did it travel.[/quote]
I think you mean 2 west.

[hide="Answer to Extension 3"]$x=\sqrt{15^{2}+36^{2}}$

$x=\sqrt{1521}$

$x=39$[/hide][/quote]
Wrong. Reread the question ;).

[quote="anirudh"][hide]15+36=51 :rotfl: [/hide][/quote]
That is right. Also, another thing is that if you do pythagorean for say 90 and 120, you can remeber that that is similar to a 3-4-5 leading to the hypotenuse being 150...

[quote="bpms"][quote="anirudh"][hide]15+36=51 :rotfl: [/hide][/quote]
That is right.[/quote]
I was assuming that since this is an extension, it would have the same conditions. Apparently not. :|

[quote="alex__ib"][hide]$15+36=51$You should have said:"how far [u]from the starting point[/u]![/hide][/quote]
No that is what I meant...

[hide]5 because the end figure forms a right triangle, so that the angle is 90 degrees although that doesn't matter[/hide]


counting
combinatorics
Arithmetic Progression
Chessboard

Let $n\geq 3$ be an integer. Find the number of ways in which one can place the numbers $1, 2, 3, \ldots, n^2$ in the $n^2$ squares of a $n \times n$ chesboard, one on each, such that the numbers in each row and in each column are in arithmetic progression.

[quote="Arrange your tan"][quote="Rijul saini"][quote="Brut3Forc3"]Do the numbers have to be in the order of the progression? For example, is $ 1,7,4$ as a row okay?[/quote]
No Brut3Forc3, i don't think so....
Do you have the solution BTW?[/quote]

[quote="Rijul saini"]
anyone for this.....??[/quote]

Rijul saini,
because you are not the original poster of the message, you
should not be asking on others' behalf if they want to answer this.
Ask and/or comment to Rushil, the proposer, directly here if you
have a personal interest.  You are not to be a go-between for
other users who have decided for their own reasons to not post to
this (or any other) thread, at least for the time being.[/quote]

 :huh:
Excuse me, but this is a question of the INMO (Indian National Mathematical Olympiad), so naturally I would take an interest...I'm not reviving a thread just for the sake of it....

Also, about me asking Rushil, if you haven't noticed, then he hasn't come online since ages....

So, anyone got an answer to this?

This is a nice problem. Here is my solution. Its easy to verify that 1 must be placed in the (1,1) cell and$ n^{2} $ must be placed at one of the extreme positions of the last row. Now  observe that the maximum value of the common divisor is$ (n+1) $ . But here are total $ 2n $ number of rows and columns. Hence two of them must have same common difference.
CASE 1.
Let us assume that one row and one column have same common difference.Now look at the element at the intersection of these row and column. Observe that these row and column have many elements in common which is a contradiction. Hence case I is impossible.
CASE 2.
Hence the other possibility is that two rows or two columns must have same common  difference. Let $ n_{1}^{th} $ and $ n_{2}^{th} $ column have same common difference $ d_{1} $. Let the common difference of any  column is  d . Let the first element of the $ n_{1} $t h column is a. Then doing some calculations we will get $ d=d_{1} $. This is possible only when  if we put $ 1,2,3, ...,n^{2} $ in there usual order.
NOW REFLECTING THIS ARRANGEMENT WE WILL GET 8 SUCH ARRANGEMENTS.

[quote=Chandrachur]Sorry for reviving this but i don't understand how 8 solutions come by.
there are 2 solutions when the A.Ps are increasing from left to rt and frm top to bottom and 2 more for similar decreasing ones. What abt the other 4???[/quote]

Similarly, 2 cases when A.Ps  are decreasing from left to right and bottom to top and other 2 cases for increasing ones.


hello, do you meant $ 3\left(\frac{a^3}{b^3}\plus{}\frac{b^3}{c^3}\plus{}\frac{c^3}{a^3}\right)\geq(a^3\plus{}b^3\plus{}c^3)\left(\frac{1}{a^3}\plus{}\frac{1}{b^3}\plus{}\frac{1}{c^3}\right)$?
Sonnhard.

yes, I have made mistake in typing. Thank you  :wink:  Now, let kill it :)

Notice that for any $ n\in\mathbb{N},n\geq1$ we have: \[ 3\left(\frac{a^{n}}{b^{n}}\plus{}\frac{b^{n}}{c^{n}}\plus{}\frac{c^{n}}{a^{n}}\right)\geq\left(a^{n}\plus{}b^{n}\plus{}c^{n}\right)\left(\frac{1}{a^{n}}\plus{}\frac{1}{b^{n}}\plus{}\frac{1}{c^{n}}\right).\] Obviously $ a,b,c$ are the side-lenghts of a triangle.

[b]Proof[/b]. Denote $ x\equal{}a^{n},y\equal{}b^{n},z\equal{}c^{n}$. Then $ x,y,z$ are also the side-lenghts of a triangle. The inequality in question reduces to: \[ 3(\frac{x}{y}\plus{}\frac{y}{z}\plus{}\frac{z}{x})\ge 3\plus{}\displaystyle\sum_{cyc}\frac{x}{y}\plus{}\frac{y}{x}.\]
Using the Ravi substitutions, denote $ x\equal{}m\plus{}n,y\equal{}m\plus{}p,z\equal{}n\plus{}p$, with $ x,y,z>0$. The inequality becomes \[ \displaystyle\sum_{cyc}\frac{m}{m\plus{}p}\plus{}\displaystyle\sum_{cyc}\frac{m}{n\plus{}p}\ge 2\displaystyle\sum_{cyc}\frac{m}{m\plus{}n}.\] The last inequality follows easy from Rearrangement inequality.

[b]Remark.[/b] For $ n\equal{}2$ we get a well-known problem proposed in "The Mathematics Magazine" , vol. 43, nr. 4/1970, by A. W. Walker. For $ n\equal{}3$ we get the problem proposed by [b]zaizai-hoang[/b].

Finally, I wish you a Happy New Year :)

i don't think $ x,y,z$ are always the sides of a triangle. For instance if angle $ A$  is obtuse $ a^2, b^2 , c^2$ are not sides of a triangle

[quote="Mîndrilă Claudiu"]Notice that for any $ n\in\mathbb{N},n\geq1$ we have:
\[ 3\left(\frac {a^{n}}{b^{n}} \plus{} \frac {b^{n}}{c^{n}} \plus{} \frac {c^{n}}{a^{n}}\right)\geq\left(a^{n} \plus{} b^{n} \plus{} c^{n}\right)\left(\frac {1}{a^{n}} \plus{} \frac {1}{b^{n}} \plus{} \frac {1}{c^{n}}\right).
\]
Obviously $ a,b,c$ are the side-lenghts of a triangle.

[b]Proof[/b]. Denote $ x \equal{} a^{n},y \equal{} b^{n},z \equal{} c^{n}$. Then $ x,y,z$ are also the side-lenghts of a triangle. The inequality in question reduces to:
\[ 3(\frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x})\ge 3 \plus{} \displaystyle\sum_{cyc}\frac {x}{y} \plus{} \frac {y}{x}.
\]
Finally, I wish you a Happy New Year :)[/quote]

I do not get it. Why $ a,b,c$ are three sides of triangle then $ a^n,b^n,c^n$ are three sides of tringle.
For instance $ a \equal{} 2$ , $ b \equal{} 1$, $ c \equal{} 1.5$ are three sides of triangle because $ 2 \minus{} 1 < 1.5 < 2 \plus{} 1$
but $ x \equal{} 2^{100}$ , $ y \equal{} 1^{100} \equal{} 1$, and $ z \equal{} 1.5^{100}$ are not  :)  $ 2^{100} \minus{} 1 > 1.5^{100}$

The first, Mîndrilă Claudiu, can you tell me when does equality hold ? :D I don't think your solution is correct  :wink:

Let $ x\equal{}a^3$, a.s.o. WLOG $ z$ is the largest. Then we rewrite the inequality as \[ f(x,y,z)\equal{}\frac{1}{xyz}\cdot (z(x\minus{}y)^2\plus{}(2y\minus{}x)(z\minus{}x)(z\minus{}y))\]
$ \bullet$ If $ 2y\ge x$, we are done. Equality for $ x\equal{}y\equal{}z$.
$ \bullet$ If $ z\ge x\ge 2y\iff c^3\ge a^3\ge 2b^3$. So $ c\ge a>b$ so we only need the condition $ c\le a\plus{}b$.
$ \frac{d}{dz}f(x,y,z)\equal{}\frac{(x\minus{}2y)(xy\minus{}z^2)}{xyz^2}\le 0$. So we have two subcases:
$ \bullet$ if $ x\equal{}2y$, then $ f(2y,y,z)\equal{}\frac{1}{2}>0$.
$ \bullet$ If $ x>2y$. Then $ f$ is decreasing in $ z$, so $ f(x,y,z)\ge f(x,y,y)\equal{}\frac{(x\minus{}y)^2}{xy}\ge 0$ (now $ c\le a\plus{}b$ is trivially satisfied as $ b\equal{}c$) again $ x\equal{}y\equal{}z$.

Did I make a mistake? It seems too easy. [hide="The rewriting is correct..."]In[79]:= Factor[
 3 (x/y + y/z + z/x) - (x + y + z) (1/x + 1/y + 1/z) -
  1/(x y z) (z (x - y)^2 + (2 y - x) (z - x) (z - y))]

Out[79]= 0[/hide]

[quote="zaizai-hoang"]Let $ a,b,c$ be sides of a triangle. Prove that:
\[ 3\left(\frac {a^3}{b^3} \plus{} \frac {b^3}{c^3} \plus{} \frac {c^3}{a^3}\right) \ge (a^3 \plus{} b^3 \plus{} c^3)\left(\frac {1}{a^3} \plus{} \frac {1}{b^3} \plus{} \frac {1}{c^3}\right)
\]
[/quote]
The inequality is very hard  :blush: since it does not hold. Try $ a\equal{}3.90021020, b\equal{}3,c\equal{}1$ ;)


If $ a_1,a_2,...,a_n>0$ such that
\[ \left(a_1\plus{}\frac{1}{a_1}\right)\left(a_2\plus{}\frac{1}{a_2}\right)\cdots\left(a_n\plus{}\frac{1}{a_n}\right)\geq \left(\frac{a_1\plus{}a_2\plus{}...\plus{}a_n}{n}\plus{}\frac{n}{a_1\plus{}a_2\plus{}...\plus{}a_n}\right)^n\]
please find the value range of $ a_1\plus{}a_2\plus{}...\plus{}a_n$.

Could anyone post an answer...or with computer..

:maybe: It seems that it's quite hard for $ n$ variables....

fjwxcsl,how do you get the value of $ k_{0}$?


[quote="Ligouras"]Let $ a, b, c$ be positive real numbes such that $ \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \equal{} 1$. Prove or disprove that

$ \sum\frac {5a^2 \minus{} bc}{a \plus{} b} \ge 2$ .

_____________
 :)[/quote]

My solution. :)

[quote="nguoivn"]Yes, it's trues because of Cauchy Schwarts and rearrangement ineq  :)[/quote]

--------------
Interesting reasoning. Can I have a complete demonstration? thanks and good 2009....  :roll:


topology
real analysis
real analysis unsolved

Let $ X$ be a complete metric space.

We say $ E$ is an $ F_\sigma$ if $ E$ is the union of a countable collection of closed sets.

Suppose $ E$ and $ X\backslash E$ are both dense in $ X$. Show that at most one of them is an $ F_\sigma$.

Suppose both of them is $ F_{\sigma}$, Baire's theorem tells you that one of the closed set has nonempty interior. Say that closed set belongs to E, then that empty interior doesn't intersect X-E, contradicting the density of X-E.


WIthout looking at the labels, Adrien placed four CDs in four cases. What is the probability that exactly two of the CD's are in the wrong cases? Express your answer as a common fraction.

I don't know the answer, but I doubt that's the answer...

What year is it? It's either $1/21$, $1/91$, or $2/17$ depending on what year. I think it's 1997; is it? If so, it's $2/17$.

First of all, it is impossible to have exacly 3 in their original cases, trivially.
Then the probability, they are all in their cases is $\frac{1}{24}$: ABCD. The probability that none are in their cases is $\frac{9}{24}$: BADC DABC CADB BDAC CDAB DCAB BCDA CDBA DCBA.
The probability of having 1 in the original case is $\frac8{24}$,
So the probability of having 2 in the orriginal cases is $1-\frac1{24}-\frac{9}{24}-\frac8{24}=\frac{6}{24}=\frac14$

What if just one is in their correct cases?

[hide="my solution"]
Think of the cases as case A, B, C, and D. When he puts the CDs in, he has 4 CDs for case A, 3 for case B, 2 for case C, and 1 for case D. This means that there are $4 \cdot 3 \cdot 2 \cdot 1 = 24$ ways of putting the CDs in. For exactly 2 to be wrong, the two must have been switched. The ways in which this will occur are the combinations of 2 out of the 4 cases. The full list is:

AB
AC
AD
BC
BD
CD

which is 6 possibilities. Therefore, the probability is $\frac 6{24}$ or $\boxed{\frac {1}{4}}$

[/hide]

That is what I got,  :) I hope that's right.

[hide]

I got the same thing:

There are 4! ways to put the CD's in the cases. There are 4 cases and there are 2 cases that have the wrong CD in them. So the answer is $\frac{{4\choose 2}}{24} = \boxed{\frac{1}{4}}$[/hide]


[quote="chien than"]Prove that:
$a+b+c \geq ab+bc+ca$
where $a;b;c$ are positive numbers satisfying $a+b+c+abc=4$[/quote]
Let exist positive numbers $a,$ $b$ and $c,$ where $a+b+c<ab+ac+bc$ and $a+b+c+abc=4.$
Let $a=kx,$ $b=ky$ and $c=kz,$ for $k>0$ and $x+y+z=xy+xz+yz.$
Hence, $kx+ky+kz=k^{2}xy+k^{2}xz+k^{2}yz$ and $k>1.$
Hence, $4=a+b+c+abc=kx+ky+kz+k^{2}xyz>x+y+z+xyz.$
Thus, $4>x+y+z+xyz,$ where $x+y+z=xy+xz+yz$ and $x,$ $y$ and $z$ are positive numbers.
But this is contradiction since, $4\leq x+y+z+xyz$ $:$
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=83238

[b]Viet Nam MO 1996:[/b]
Let $a,b,c$ are positive numbers satisfying $ab+bc+ca+abc=4. $ [url=https://artofproblemsolving.com/community/c4h1157770p5499288]Prove that[/url]$$a+b+c \geq ab+bc+ca$$

[quote=chien than]Prove that:
$a+b+c \geq ab+bc+ca$
where $a;b;c$ are positive numbers satisfying $a+b+c+abc=4$[/quote]


quadratics function trigonometry Vieta algebra

Find all pairs (a,b) of real numbers such that whenever z is a root of the eqn x^2 + ax = b =0, z^-2 is also a root.

calcul the discriminant end you find the result in fonction of a and b
I don't understand the calcul with z
??

[hide="Solution"]
By using the condition repeatedly,we can see all $ z^{( \minus{} 2)^n} (n \equal{} 1,2,...)$ are roots.Then if $ |z|\ne 1$,their absolute values are distinct,so quadratic equation has infinitely many solutions,this is contradiction.so we have all its roots satisfy $ |z| \equal{} 1$.
1)When the equality has real roots.
possible pair of roots are $ (1,1)(1, \minus{} 1)( \minus{} 1, \minus{} 1)$ and easily check only $ (1,1)(1, \minus{} 1)$ work.

2)When it has complex roots.
we can let one of its roots $ z \equal{} \cos{w} \plus{} i\sin{w} (0 < w < \pi)$ and another $ \cos( \minus{} w) \plus{} i\sin( \minus{} w)$
by the condition $ z^{ \minus{} 2} \equal{} \cos( \minus{} 2w) \plus{} i\sin{( \minus{} 2w)}$ is also root,thus $ \cos(\minus{} 2w) \equal{} \cos{w}$ follows,then we have $ \cos{w} \equal{} \minus{} \frac {1}{2}$, i.e.$ z \equal{} \cos{2\pi/3} \plus{} i\sin{2\pi/3}$ and another is $ z \equal{} \cos( \minus{} 2\pi/3) \plus{} i\sin( \minus{} 2\pi/3)$,and this truely works.

finaly we have $ (a,b) \equal{} ( \minus{} 2,1)(0, \minus{} 1)(1,1)$.

[/hide]

the eqn x^2 + ax + b=0 (*)has 2 roots( if it has any roots), so if z and z^(-2) are two roots of the eqn, they must be 2 only roots
By using Vieta's( I don't know how exactly it's written), we have this equation system:
z+z^(-2)=-a;
1/z=b which mean two roots of (*) are 1/b and b^2, so a=-(1/b+b^2)


How many ways are there to color a 3x3 grid with red, blue, and green such that no two colors in each row are the same and no two colors in each column are the same?

Solution:

[hide]
Clearly each row must be some permutation of RBG. Suppose we set one row as such. Now the makeup of the other two rows is determined, keeping in mind there are two possible placements for those rows, i.e. they replace each other.

Thus there are 3! permutations of RBG for the first row and 2 placements of the next two rows for each unique permutation, thus there are (3!)2=$ \boxed{12}$ ways.
[/hide]


perhaps this morning began a day of competition  USAMO.
can anyone post problem of USA MO??

No, that's tomorrow. I don't believe it's supposed to be discussed anyway until after the second day.

[url=http://www.artofproblemsolving.com/Forum/index.php?f=133]This forum[/url] will probably be locked at some point tomorrow. When it is unlocked later in the week, the problems will be posted there almost immediately for certain, as well as a huge amount of discussion by people who took it.

the contest has ended,can anyone post problem of it?

math10 // the problems are in
http://www.unl.edu/amc/
which is official site of MAA

Problem 1: http://www.mathlinks.ro/viewtopic.php?t=274369
Problem 2: http://www.mathlinks.ro/viewtopic.php?t=274370
Problem 3: http://www.mathlinks.ro/viewtopic.php?t=274393
Problem 4: http://www.mathlinks.ro/viewtopic.php?t=274371
Problem 5: http://www.mathlinks.ro/viewtopic.php?t=274376
Problem 6: http://www.mathlinks.ro/viewtopic.php?t=274378


[quote="mathboy_cntHDVN"]for$(O;R_{1});(O;R_{2})\&(O;R_{3})$$A;B;C;D \in (O: R_{1})$
the ray $AB: BC: CD: DA \cap (O;R_{2})\equiv E;F;G;H$
the ray $MN;NP;PQ;QM \cap (O;R_{3})\equiv M;N;P;Q$
$ABCD$ and $MNPQ$ is similitude ?[/quote]
Oh ! I think $EF,FG,GH,HE \cap (O;R_{3})\equiv M;N;P;Q$
Is't it?

[quote]Is't it?[/quote]
yes it is
[quote]Now I thinks this pro is false  :maybe:  [/quote]
why ?

I don't understand the term "similitude" if it means similar then it is't see the figure

for$(O;R_{1});(O;R_{2})\&(O;R_{3})$$A;B;C;D \in (O: R_{1})$
the ray $AB: BC: CD: DA \cap (O;R_{2})\equiv E;F;G;H$
the ray $EF;FG;GH;HE \cap (O;R_{3})\equiv M;N;P;Q$
$ABCD$ is  similitude with 0$MNPQ$  ?


Circles k1 and k2 meet at two distinct points A and B. Two lines through
B meet k1 at C and D and k2 at E and F, respectively, so that B lies on
the segments CE and DF. Let M be the midpoint of CE and N that of
DF. Prove that the triangles ACD, AEF, and AMN are similar.


geometry
3D geometry
prism
analytic geometry
sphere
algebra
function

It's a very well known problem. You have an ant on a box of dimensions l,h,d that needs to travel from point A to point B, while staying on the surface. What is the minmum distance he must travel?

I know how to attack the problem, but is there a short way? The only wa I know requires testing each "possible path" and finding the smallest. Does anyone know of a quick, foolproof formula?

unfold the net of the prism and then draw a straight line from point A to point B on the net, that is the shortest route.

[quote]$\sqrt{l^2 + h^2 + d^2 + \frac{2lhd}{\max (l, h, d)}}$[/quote]

let me clarify a little bit.  I think the formula you gave me is if the ant starts at one corner and goes to the opposite corner. I want a [u]very[/u] generalized formula.
Let the box have side lengths of $(2a,2b,2c)$. Let  the ant start on one of the $ab$ sides. Also, let each side of the box be a rectangular coordinate system with the origin at the center (exactly $a$ or $b$ from each edge). Let the ant start at $(x_1,y_1)$, and let the ant be travelling along the box to the opposite side to coordinates $(x_2,y_2)$ (on the coordinate system for the other side). What is the minimum distance the ant must travel.

Also, as an aside, what about other shapes. A sphere is actually quite simple, but what about platonic solids? Or a torus, klein bottle, or even (god forbid) a projective plane?

The solution to the first one is most important to me, but I would like to see someones solution to the others.

Have fun.

The problem you're asking about is that of the [url=http://mathworld.wolfram.com/Geodesic.html]geodesic[/url], paths between two points on a surface which minimize distance (at least locally).  For smooth surfaces, this problem is in the domain of differential geometry.  For convex polyhedra, the problem is always solvable by unfolding the net of the polyhedron in question.  For concave polyhedra, it can get more complicated (as it is not always possible to unfold concave polyhedra without overlaps).  If you google "geodesics on a torus" you will find a paper which gives five (exhaustive) families of geodesics on the torus -- however, you'll need to be familiar with a fair amout of differential geometry to actually read the paper.


integration
trigonometry
calculus
calculus computations

Prove, for all real values of m, $\int^{\frac{\pi}{2}}_{0}\frac{sin^{m}x}{sin^{m}x+cos^{m}x}dx=\frac{\pi}{4}$. i don't even know what the substitution should be... any help is appreciated  :)

sorry, this is probably really stupid, but how do i proceed after getting $\int^{0}_{-\frac{\pi}{2}}\frac{-1}{1+tan^{m}\theta}}d{\theta$?

using substitution suggested by kunny , we have

$x=\frac{\pi}{2}-t$ so $dx=-dt$

so $\int^{\pi/2}_{0}\frac{\sin^mx}{\sin^mx+\cos^mx}dx=\int^{0}_{\pi/2}\frac{-\cos^mt}{\cos^mt+\sin^mt}dt$

Since $\int^{0}_{\pi/2}\frac{-\cos^mt}{\cos^mt+\sin^mt}dt =\int^{\pi/2}_{0}\frac{\cos^mx}{\cos^mx+\sin^mx}dx$

So we have $\int^{\pi/2}_{0}\frac{\sin^mx}{\cos^mx+\sin^mx}=\int^{\pi/2}_{0}\frac{\cos^mx}{\cos^mx+\sin^mx}dx$

So
 $\int^{\pi/2}_{0}\frac{\sin^mx}{\sin^mx+\cos^mx}dx=\frac{1}{2}(\int^{\pi/2}_{0}\frac{\sin^mx}{\sin^mx+\cos^mx}dx+\int^{\pi/2}_{0}\frac{\cos^mx}{\sin^mx+\cos^mx}dx)\\=\frac{1}{2}\int^{\pi/2}_{0}dx\\=\frac{\pi}{4}$


[quote="colorfuldreams"][color=red]Using Trigonometric Substitutions[/color]

$ xy \plus{} yz \plus{} zx\geq3 \plus{} \sqrt {1 \plus{} x^z} \plus{} \sqrt {1 \plus{} y^z} \plus{} \sqrt {1 \plus{} z^z}$[/quote]
I think It's \[ xy \plus{} yz \plus{} zx \ge 3 \plus{} \sqrt {1 \plus{} x^2 }  \plus{} \sqrt {1 \plus{} y^2 }  \plus{} \sqrt {1 \plus{} z^2 }\]

[quote="colorfuldreams"][color=red]Using Trigonometric Substitutions[/color]
Prove that
1) $ x,y,z$ be real numbers such that $ 0 < x,y,z < 1$ and $ xy + yz + zx = xyz$
Prove that
$ \frac {x}{1 - x^2} + \frac {y}{1 - y^2} + \frac {z}{1 - z^2}\geq\frac {3\sqrt {3}}{2}$

2) $ x,y,z$ be real numbers such that  $ xy + yz + zx = xyz$
Prove that
$ \frac {x}{\sqrt {1 - x^2}} + \frac {y}{\sqrt {1 - y^2}} + \frac {z}{\sqrt{1 - z^2}}\leq\frac {3\sqrt {3}}{2}$

3) $ x,y,z > 0$ and $ xy + yz + zx = xyz$
Prove that
$ xy + yz + zx\geq3 + \sqrt {1 + x^z} + \sqrt {1 + y^z} + \sqrt {1 + z^z}$[/quote]

$ xy + yz + zx = xyz$

substitution

$ x = \frac {a + b + c}{a},y= \frac {a + b + c}{b},z = \frac {a + b + c}{c}$ $ (a,b,c > 0)$

How about
$ z \equal{} \frac {1 \minus{} xy}{x \plus{} y}$

$ cot(x \plus{} y) \equal{} \frac {1 \minus{} tanxtany}{tanx \plus{} tany} \equal{} tanz$
?

[quote="hedeng123"]

$ xy \plus{} yz \plus{} zx \equal{} xyz$

substitution

$ x \equal{} \frac {a \plus{} b \plus{} c}{a},y \equal{} \frac {a \plus{} b \plus{} c}{b},z \equal{} \frac {a \plus{} b \plus{} c}{c}$ $ (a,b,c > 0)$[/quote]

$ xy \plus{} yz \plus{} zx \equal{} xyz$
=> $ x \plus{} y \plus{} z \equal{} 180^{\circ}$

but how about question 1?

[quote="colorfuldreams"][color=red]Using Trigonometric Substitutions[/color]
Prove that
1) $ x,y,z$ be real numbers such that $ 0 < x,y,z < 1$ and $ xy + yz + zx = xyz$
Prove that
$ \frac {x}{1 - x^2} + \frac {y}{1 - y^2} + \frac {z}{1 - z^2}\geq\frac {3\sqrt {3}}{2}$

2) $ x,y,z$ be real numbers such that  $ xy + yz + zx = xyz$
Prove that
$ \frac {x}{\sqrt {1 - x^2}} + \frac {y}{\sqrt {1 - y^2}} + \frac {z}{\sqrt{1 - z^2}}\leq\frac {3\sqrt {3}}{2}$

3) $ x,y,z > 0$ and $ xy + yz + zx = xyz$
Prove that
$ xy + yz + zx\geq3 + \sqrt {1 + x^z} + \sqrt {1 + y^z} + \sqrt {1 + z^z}$[/quote]

$ x,y,z$ be real numbers such that $ 0 < x,y,z < 1$ and $ xy + yz + zx = xyz$

$ xy + yz + zx = xyz \Rightarrow   xyz>xy \Rightarrow z>1$
but   $ 0 < x,y,z < 1$  ?????????

[quote="hedeng123"][quote="colorfuldreams"][color=red]Using Trigonometric Substitutions[/color]
Prove that
1) $ x,y,z$ be real numbers such that $ 0 < x,y,z < 1$ and $ xy + yz + zx = xyz$
Prove that
$ \frac {x}{1 - x^2} + \frac {y}{1 - y^2} + \frac {z}{1 - z^2}\geq\frac {3\sqrt {3}}{2}$

2) $ x,y,z$ be real numbers such that  $ xy + yz + zx = xyz$
Prove that
$ \frac {x}{\sqrt {1 - x^2}} + \frac {y}{\sqrt {1 - y^2}} + \frac {z}{\sqrt{1 - z^2}}\leq\frac {3\sqrt {3}}{2}$

3) $ x,y,z > 0$ and $ xy + yz + zx = xyz$
Prove that
$ xy + yz + zx\geq3 + \sqrt {1 + x^z} + \sqrt {1 + y^z} + \sqrt {1 + z^z}$[/quote]

$ x,y,z$ be real numbers such that $ 0 < x,y,z < 1$ and $ xy + yz + zx = xyz$

$ xy + yz + zx = xyz \Rightarrow xyz > xy \Rightarrow z > 1$
but   $ 0 < x,y,z < 1$  ?????????[/quote]
Yes, I think Colorfuldreams'Inequality is wrong, because the equality doesn't occur


Prove that the complex numbers a1,a2,..,an (n>=3) satisfy the following inequality :
| a1+a2*z+...+an*z^(n-1)|<=|a1+a2+...+an| , z from C with |z|<=1 if and only if there exist the pozitive real numbers b2,b3,..,bn (<>0) so that a1=b2*a2=...=bn*an.
Please help me!  I have been trying to solve this one for ages.. actually for 2 weeks and I haven't found a solution, yet! Would induction be the solution? If so how? Any other suggestions?

Thank you!
Louis.

Is that actually true?  I'm pretty sure it fails for a1 = 1, a2=i, a3=1, so that a1+a2*z+a3*z^2 = 1 + iz + z^2.  From there you can convert to polar z=r*e^(i*theta)  and use that to maximize the value of | a1+a2*z+...+an*z^(n-1)|^2 = a1+a2*z+...+an*z^(n-1) times its complex conjugate.  If I didn't make a stupid calcuation error, that maximum is achieved at z = 1, so the inequality is satisfied even though the coefficients are not real multiples of each other.

It's an interesting question, however.  The 'if' part should hold, even though the above gives a counterexample for the 'only if'.

--Alison

[quote="Lagrangia"]Prove that the complex numbers a1,a2,..,an (n>=3) satisfy the following inequality :
| a1+a2*z+...+an*z^(n-1)|<=|a1+a2+...+an| , z from C with |z|<=1 if and only if there exist the pozitive real numbers b2,b3,..,bn (<>0) so that a1=b2*a2=...=bn*an.

I believe he means, "and with equality if and only if blah blah blah..." Someone should give this a try. It's a nice problem.


[hide] Just set it up like:
$|x-(-1)| < 1$
So,
$x+1<1$
and
$x+1>-1$.

From the first, we get $x<0$

From the second, we get $x>-2$

Thus, $-2<x<0$.

You can do something similar for the second. [/hide]

[quote="Interval"]Express the fact that x differs from -1 by less than 1 as an inequality which has an absolute value.  Solve for x.

What about if the question stated that x differs from 2 by more than 2.

How do we solve each case?[/quote]
[hide]For the second part, you would set it up like so:
$|x-2|>2$
Then you have
$x-2>2$ and $x-2<-2$
$x>4$ and $x<0$
Combining, you get $0>x>4$.
(It looks like the picture)[/hide]

You are an amazing tutor.

I thank you for helping me with my preparation for the upcoming state exam.


How many ways can an 8x8 chessboard be marked so that no two marked squares are in the same column, row,  and none of the corners are marked?

I tried to solve it using combinations, but I got messed up with all the things I had to subtract.

[hide="My Guess"]$(7!-6!)^2$[/hide]
Think about the problem without the last restriction (the one about the corners) and think what happens when you add this restriction.

Although that seems like an astronomically huge number for a problem like this.

[hide]If we take this problem without the corner restriction, this means that there are 8 places to put the first piece (in the first row), 7 places for the second (in the second row), etc. making the total $8!=40320$. On the other hand, $(7!-6!)^2=18662400$!?

I'm not certain my answer is correct because, frankly, I am horrible at counting problems, but I'll give it a try.

Because you cannot place a piece in the corner, there are 6 choices for the first column. Then there are 5 choices for the last column. For the middle columns, there are 6 choices, 5 choices, 4 choices, etc, making my answer:

$30\cdot6!=21600$
[/hide]

It is actually very easy,

[hide]
Suppose that you want to choose first the position that you are going to take for the first and for the last row. As you can't take the corners, you have six posibilities on each... ups, no, you have 6 for the first one, 5 for the second one, and after that you have 36 options for the third, 25 for the fourth, and so forth, then, you have $6\cdot5\cdot6!\cdot6!=30\cdot(6!)$. Before I made the mistake to think that you had 6 for each of the first and last row, and hence, the resoult would follow as the original guess I did, but I made a mistake  :blush: .
[/hide]

Best,

[hide]Six for the first row, five for the last, and 6! for the next six. 30*6! is 30*720=21600[/hide]


vector
linear algebra
linear algebra unsolved

Prove that if $\beta=\{x_1,x_2,\ldots,x_r\}$ is a basis for the subspace $W$ and $\hat{\beta}=\{W+y_1,W+y_2,\ldots,W+y_s\}$ is a basis for $V/W$ then $\{x_1, x_2,\ldots,x_r,y_1,y_2,\ldots,y_s\}$ is a basis for $V$.

Think what a basis of $V/W$ implies. Here it implies that $\sum_{i=1}^s c_iy_i\in W\Longleftrightarrow c_i = 0$

Suppose $\sum_{i=1}^s c_iy_i+\sum_{i=1}^r d_ix_i = 0$, write it as $\sum_{i=1}^r d_ix_i = \sum_{i=1}^s (-c_i)y_i$. The left-handed side is in $W$, so is the right-handed side. Thus $c_i = 0$. Hence $\sum_{i=1}^r d_ix_i = 0$, thus $d_i = 0$. We have shown that $\{x_i\}\cup\{y_j\}$ are linearly independent.

It is not difficult to see that each vector of the space is a linear combinatorial of $x_i$ and $y_j$.

Ah, here was my solution:

Linear independent:
Let $\alpha_1x_1+\alpha_2x_2+\ldots+\alpha_rx_r+\gamma_1y_1+\gamma_2y_2+\ldots+\gamma_sy_s=0$.
Consider in $V/W$. (Can I say this?) Since $x_1, x_2, \ldots,x_r\in W$, $\alpha_1x_1+\alpha_2x_2+\ldots+\alpha_rx_r\in W$.
Thus, $W+\gamma_1y_1+\gamma_2y_2+\ldots+\gamma_sy_s=0$. But $\{W+y_1, W+y_2,\ldots,W+y_s\}$ is a basis for $V/W$, so $\gamma_1=\gamma_2=\ldots=\gamma_s=0$. Then $\alpha_1x_1+\alpha_2x_2+\ldots+\alpha_rx_r=0$. Since $\{x_1, x_2,\ldots,x_r\}$ is a basis for $W$, a subspace of $V$, $\alpha_1=\alpha_2=\ldots=\alpha_r=0$. Thus $\{x_1, x_2,\ldots,x_r,y_1,y_2,\ldots,y_s\}$ is a linearly independent set.

Spanning set:
Let $v\in V$. Then $v\in W$ or $v\not\in W$.
If $v\in W$, $V$ can be written as a combination of $x_1,x_2,\ldots,x_r$ since $\beta$ is a basis for $W$.
If $v\not\in W$, let $v=W+v_0$ for some $v_0\in V$.
Then $v$ can be written as a combination of $x_1,x_2,\ldots,x_r,y_1,y_2,\ldots,y_s$ since $\hat{\beta}$ is a basis for $V/W$ and $\beta$ is a basis for $W$.


geometry
circumcircle
incenter
trapezoid
geometry proposed

Let $ Fe,H,O,Na,X_{56}$ be the Feuerbach point$ ,$ orthocenter$ ,$ circumcenter$ ,$ Nagal point$ ,$ and the isogonal conjugate point of $ Na$ of a given triangle $ ABC,$  respectively$ .$Denote by $ M$ the intersection of line $ FeH$ with $ ONa.$ Prove that$ :$
 (I)$ :$The three points $ Fe,H,X_{56}$ are collinear$ .$
(II)$ :$ The point $ M$ lies on the circumcircle of triangle $ ABC.$

Dear Fang-jh and Mathlinkers,
your two results are known. I think you have made a little typo you say H but it is G the centroid and M is the point X100 in ETC.
Nice rediscovery
Sincerely
Jean-Louis

I do not think Fang-jh has a notation typo in his problem. If $ N$ is the 9-point circle center and $ I$ the incenter of the reference $ \triangle ABC,$ then the orthocenter $ H$ is the external similarity center of the circles $ (O), (N),$ the Feuerbach point $ Fe$ is the external similarity center of the circles $ (N), (I),$ and the isogonal conjugate $ X_{56}$ of the Nagel point $ Na$ is the extrernal similarity center of the circles $ (O), (I),$ so that $ H, Fe, X_{56}$ are collinear (Monge-d'Alembert theorem).

[hide="It is well known that the isogonal conjugate of Nagel point is the external similarity center of the circumcircle and incircle."] Angle bisectors $ AI, BI, CI$ cut $ (O)$ again at $ X, Y, Z.$ Let $ a^* \parallel BC, b^* \parallel CA, c^* \parallel AB$ be the tangents of $ (O)$ at $ X, Y, Z,$ forming $ \triangle A^*B^*C^*$ centrally similar to $ \triangle ABC$ with similarity center $ X_{56},$ the external similarity center of $ (O), (I).$ Let $ XX', YY', ZZ'$ be diameters of $ (O).$ Since $ YZ \perp A^*O, AX' \perp AI,$ and $ A^*O \parallel AI,$ it follows that $ AX' \parallel YZ,$ $ AX'ZY$ is cyclic isosceles trapezoid, and $ A^*X', A^*A$ are isogonals WRT the angle $ \angle C^*A^*B^*.$ $ A^*X'$ is the A-Nagel line of $ \triangle A^*B^*C^*,$ $ A^*X' \parallel ANa.$ Since $ A^*A$ goes through $ X_{56},$ the lines $ ANa, AX_{56}$ are isogonals WRT the angle $ \angle CAB.$ Similarly for $ BNa, BX_{56}$ and $ CNa, CX_{56}.$ [/hide]

Let $ A', B', C'$ be the midpoints of $ BC, CA, AB$ and let $ G$ be the common centroid of $ \triangle ABC, \triangle A'B'C',$ their similarity center and the internal similarity center of their circumcircles $ (O), (N).$ Let the ray $ (HFe$ cut $ (O)$ at $ M.$ The points $ M, Fe$ are homothetic with the similarity center $ H$ and coefficient +2. Let $ Fe' \in (N)$ be diametrically opposite point of $ Fe.$ Then the points $ M, Fe'$ are homothetic with the similarity center $ G$ and similarity coefficient -2 and so are the points $ O, N$ and the parallel lines $ OM \parallel NFe'.$ Since the line $ NFe' \equiv NFe$ goes through $ I,$ identical with the Nagel point of the medial $ \triangle A'B'C',$ the parallel line $ OM$ goes through the Nagel point $ Na$ of the reference $ \triangle ABC.$

Dear Fang-jh and Mathlinkers,
you are right, there is no typo in your problem. I made a confusion with another collinearity (Fe, G et X100, the anticomplément of Fe). Sorry.
I think that your collinearity was pproposed by Darij Grinberg. Perhaps Darij could gives more informations.
Thanks Darij in advance.
Sincerely
Jean-Louis

Sorry, but I don't really know what I have proposed over 4 years ago... Part (I) not a particularly hard collinearity anyway, since
- the Feuerbach point is the external center of similitude of the incircle and the nine-point circle;
- the orthocenter is the external center of similitude of the circumcircle and the nine-point circle;
- [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23668]the point $ X_{56}$ is the external center of similitude of the circumcircle and the incircle[/url];
- the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216735]Monge-d'Alembert theorem[/url] holds.
But I see Yetti already pointed this out.
I don't remember having thought about (II) ever, but thinking of it now, I get the same proof as Yetti, so I'm sorry, but I can't contribute much to this topic...

  darij


Between 2007 and 2008, the AMC Committee removed calculators from the AMC10 and AMC12. Did the test difficulty change to compensate or not?
This is for my information in practice scores. Thanks :)

didn't this happen in the amc 8 too??Well..not really, because for me, typing into the calculator takes me longer than me just writing it out by hand :)

The AMC8 was considerably easier, but I'm not sure about 10/12.

i dont believe it was any easier. All the problems on the AMC 10/12 can be solved(and are intended to be solved) without a calculator, so i dont think taking away calculators would change the difficulty at all.

Well, the loss of calculator opens up the possibility of careless errors in computation, but we're careful, right?  :D


logarithms
advanced fields
advanced fields unsolved

[size=150]1. Show that $ 2^x < \prod_{p \leq x} p < (13/4)^x$ for all sufficiently large $ x.$

[hide] Here I found that the product is equal to $ e^{\vartheta (x)},$ where
$ \vartheta(x) \equal{} \sum_{p \leq x} \log p.$
And my textbook says that $ \vartheta(x) \sim x,$ but I don't know if this is rigorous enough to conclude the assertion.[/hide]

2. Show that if $ n$ is a positive integer then $ T(n) \equal{} \log n!$. Here $ T(x) \equal{} \sum_{1\leq n \leq x} \log n.$ Show that
\[ \frac {2^{2n}}{2n} \leq \binom{2n}{n} \leq 2^{2n}.
\]
Deduce that
\[ (2 \log 2)n \minus{} \log 2n \leq T(2n) \minus{} 2T(n) \leq (2 \log 2)n.
\]
[hide] Here I found a solution for the first one, but don't know how to do the second and third identity. The teacher told us to expand $ (1 \plus{} 1)^{2n}$ using the binomial theorem.[/hide]

3. Prove that $ n! \equal{} m^k$ is impossible in integers $ k > 1, m > 1, n > 1.$
[hide] Here I know we are supposed to use Bertrand's postulate. It says that if $ x$ is a real number, $ x > 1$, then there exists at least one prime number in the open interval $ (x,2x).$
[/hide][/size]

Yes your argument to exercise 1 can be made rigorous. do you know how to prove your assertions?

I think I know how to do #1 and #2, but still couldn't figure out how to do #3 using Bertrand's postulate. If any one knows, please let me know.

The idea for #3 is to find a prime $ p$ such that $ p\mid n!$ but $ p^{k}\nmid n!$ for any $ k>1$.


calculus
integration
trigonometry
calculus computations

http://www.physicsforums.com/showthread.php?t=189310

Look at post 1, the first half of post 6 and post 9. Can anyone tell me what Im doing wrong? I think even the limit answer in post 9 is incorrect because it doesn't work numerically.

The integral in question is $ \int_0^{\pi}\frac {x}{1 \plus{} \cos^2 x}\,dx.$

I don't feel like working it myself, but you're much more likely to get responses if people don't have to follow a link just to see the problem.

First, note that
\[ \int_0^{\pi} \frac{x}{1\plus{}\cos^2 x} dx  \equal{} \int_0^{\pi/2} \frac{x}{1\plus{}\cos^2 x} dx \plus{} \int_0^{\pi/2} \frac{\pi\minus{} x}{1\plus{}\cos^2 (\pi \minus{}x)} dx \equal{} \int_0^{\pi/2} \frac{\pi}{1\plus{}\cos^2 x} dx.\]
Let $ y \equal{} \tan x$ (then $ dy \equal{} (1 \plus{} \tan^2 x) dx$) and $ z\equal{}y/\sqrt{2}$. We have
\[ \int_0^{\pi/2} \frac{dx}{1\plus{}\cos^2 x} \equal{} \int_0^{\pi/2} \frac{1 \plus{} \tan^2 x}{2\plus{}\tan^2 x} dx \equal{} \int_0^{\infty} \frac{dy}{2\plus{}y^2}\]
\[ \equal{} \frac{\sqrt{2}}{2} \int_0^{\infty} \frac{dz}{1\plus{}z^2} \equal{} \frac{1}{\sqrt{2}} \arctan z \Bigr|^{\infty}_0 \equal{} \frac{\pi}{2\sqrt{2}}.\]

[b]Answer:[/b] $ \frac{\pi^2}{2\sqrt{2}}$.

Can anyone give me a hint on how to get there from an alteration of my method?

EDIT: thanks yury


limit
probability
real analysis
real analysis unsolved

Show that $ \mathcal{L}^1$-almost every number in $ (0,1)$ contains every natural number in his decimal expansion infinitely many times.

Let $ A_j$ be the set of numbers for which the $ j$th digit is not $ 0.$ The measure of each $ A_j$ is $ \frac{9}{10}.$ Furthermore, the $ A_j$'s are probabilistically independent, so that $ \bigcap_{j\equal{}m\plus{}1}^{m\plus{}k}A_j\equal{}\left(\frac{9}{10}\right)^k.$

Taking the limit as $ k\to \infty,$ we see that $ B_m\equal{}\bigcap_{j\equal{}m\plus{}1}^{\infty}A_j$ has measure zero.

The set of all numbers with at most finitely many zeros in its decimal expansion is $ \bigcup_{m\equal{}1}^{\infty}B_m.$ That's a countable union of sets of measure zero, hence itself a set of measure zero. Hence, almost every number has infinitely many zeros in its decimal expansion.

Repeat this argument for each other possible digit. The set of numbers that takes on any one digit only a finite number of times is the union of $ 10$ sets of measure zero, hence itself of measure zero.

----

In fact, a much stronger statement can be made. As a consequence of the Strong Law of Large Numbers, we can say that almost every real number in this interval has the property that for each possible digit $ d,$ $ \lim_{n\to\infty}\frac{\text{number of }d\text{'s in first }n\text{ digits}}{n}\equal{}\frac1{10}.$

Actually, I misread that. I worked the problem of containing every [b]digit[/b] infinitely many times. Instead the problem was about containing every [b]natural number[/b] infinitely many times.

Suppose $ N$ is some particular $ r$-digit number. Block the decimal number into $ r$-digit chunks. (This is wasteful, as the number could occur in shifted frames - it doesn't matter.) As above, the probability that this number doesn't occur in $ r$-digit frames $ m + 1$ through $ m + k$ is $ (1 - 10^{ - r})^k.$ Take the limit as $ k\to\infty$ and we get that the probability that this number never occurs after the $ m$th frame is zero; take the union over all $ m,$ and we get that the probability that this number occurs finitely is a countable union of sets of measure zero, hence of measure zero.

We can do this argument for each natural number. Since there are countably many natural numbers, the measure of the set in which we omit any one of them is the countable union of sets of measure zero, hence of measure zero.


prove that every natural number can be written in the form  x^2+y^2-5z^2 :o

come in mathlinkers its a easy problem :ninja:  :blush:


geometry
3D geometry
algebra
factorization
difference of cubes
special factorizations
sum of cubes

I have a real hard time understading the difference of cubes.
However, there are several questions on this topic in the coming June state exam.

How do I play with this topic?

QUESTIONS:

(1) 64x^3 + 125

(2) 3y^3 - 81

Understood your notes.

Yes, I will post these types of question in the high school basics.


Let $ABC$ be a triangle,   $r_a$, $r_b$, $r_c$  the radii of its excircles and $p$ its semiperimeter.
Show that $(r_a+r_b+r_c)(\frac{r_a}{a}+\frac{r_b}{b}+\frac{r_c}{c})\geq9p$

 :)

I think the constant on the left might be $\frac{9}{2}$.

Write $K$ for the area, $s$ for the semiperimeter, and $t = s - a, u = s - b, v = s - c$. Then we know $r_a = \frac{K}{s-a} = \frac{K}{t}$, etc, so divide each factor of the LHS by $K$ and the RHS by $K^2$ and plug in for the exradii and the $K^2$ on the right to convert the inequality to: $(\frac{1}{t} + \frac{1}{u} + \frac{1}{v})(\frac{1}{t(u+v)} + \frac{1}{u(t+v)} + \frac{1}{v(t+u)} \ge \frac{9}{2tuv}$, which upon multiplying through the first factor of the LHS and the RHS by $2tuv$, and plugging in $x = tu + tv, y = tu + uv, z = tv + uv$, becomes: $(x + y + z)(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \ge 9$, which is true by AM-HM, Cauchy, or Chebyshev.


Is AOPS going to make a book on Calculus? And if they do, when will they?

[quote="rrusczyk"]
DPatrick is working on Calculus, which we expect to be out by the end of 2009.  I am working on Precalculus, which I hope to have finished by the end of the summer.[/quote]


analytic geometry
geometry
AMC
USA(J)MO
USAMO
vector
geometry unsolved

i dont think  always  coordinate  is the only alternative for geometry problems.  i  have found vector algebra to be equally useful.

[quote="gauss1181"]Generally speaking, what kinds of geometry problems would usually be best suitable for coordinate bashing? I know that it's some kind of brute force with usage of plane/space geometry (I think?) but isn't it often time consuming (as many others may think) on olympiad contests like the USAMO? Is this type of approach usually a good way to do a geometry problem if you can't think of any other efficient ways?

 :?:  :huh:[/quote]
What is coordinate bashing ? :oops:

Oh yay, I just happened to find this:

[url]http://activities.tjhsst.edu/vmt/pages/arml/lectures/0708/Choice2.pdf[/url]


limit
linear algebra
matrix
real analysis
real analysis unsolved

Let $a_n ,b_n$ such that :
$1)b_n$ is increasing and diverges to $\infty$
$2)\sum_0^{\infty} \frac{a_n}{b_n}$ converges
Show that $lim_{n\rightarrow \infty} \frac{\sum_0^n {a_k}} {b_n}=0$

If $S_n=\sum_{k=0}^n\frac{a_n}{b_n}$, after some algebraic manipulation this becomes $u_n=\frac{\sum_{k=0}^{n-1}b_{k+1}(S_{k+1}-S_k)}{b_n}\to 0\ (*)$. If $\ell$ is the limit of $(S_n)_n,\ (*)$ becomes $\lim_n\left(\frac{b_1}{b_n}S_0+\frac{b_2-b_1}{b_n}S_1+\ldots+\frac{b_n-b_{n-1}}{b_n}S_{n-1}\right)=\ell$.

We can prove a more general statement: let $(S_n)_{n\ge 0}$ be a sequence converging to $\ell$, and $(t_m^n)_{m,n\ge 0}$ be a matrix with the following properties:

$(1)\ t_m^n\ge 0,\ \forall m\ge n\ge 0$

$(2)\ \sum_{n=0}^mt_m^n=1,\ \forall m\ge 0$

$(3)\ \lim_mt_m^n=0,\ \forall n\ge 0$

Then $\sum_{n=0}^mt_m^nS_n$ goes to $\ell$ as $m\to\infty$.

The proof is a simple exercise, and we can apply this to our problem with $t_m^0=\frac{b_1}{b_{m+1}},\ t_m^n=\frac{b_{n+1}-b_n}{b_{m+1}},\ \forall n\ge 1,\ \forall m\ge 0$.

The quoted statement is called Toeplitz Theorem. I was thinking about following problem for a lot of time:

If $\sum \epsilon_i a_i$ converges, where $\epsilon_i \in \{\pm 1 \}$ and $a_n \to 0$ descending , then $(\sum_{i=1}^n \epsilon_i)a_n$ converges to 0.

Now I figured it out. Many thanks grobber.