Personally, I favor China.  :lol: (#1 in medal standings and # of golds for the moment, at least) :first:

Currently:

[code]			   Gold		Silver	Bronze		Total
China 		  22 	      8 		5 		  35
USA 			 10 	      9 		15 		 34
Germany 		7 		    2	 	3 		  12 [/code]

This will change when track & field starts. Usa will win, as usual.

YAY GO USA

China will win in future years though

I think they're starting a program (119) of focusing on the events which give the most medals

and all that

so blah

OH WELL

EVERYONE STILL KNOWS USA IS BETTER AT LIFE

I really do hope the US brings home more gold than anyone else

GO USA!!

I hope the US can win again. I think we will start winning more when track & field starts.

USA USA USA!!!

Go China!

起来！不愿做奴隶的人们！
把我们的血肉，筑成我们新的长城！
中华民族到了最危险的时候，
每个人被迫着发出最后的吼声。
起来！起来！起来！
我们万众一心，
冒着敌人的炮火，前进！
冒着敌人的炮火，前进！
前进！前进！进！

I think it is good to have a constructive way to have competition and interaction between countries.

It doesn't help anyone when cheating occurs, I believe the US had one of the Hamm brothers test out for drugs.  The age of some of the women/girl gymnasts seems inappropriate, but I admit I'm no expert in telling ages from looking at someone on TV.

I also do like to see a good effort and see continued effort in the face of adversity or difficulty result in victory, regardless of the nation the athlete represents.

[quote="MysticTerminator"]dangit what is with everybody rooting for china

I seriously don't get that

YO IF CHINA IS SO GREAT WHY DON'T YOU GO BACK THERE

no like seriously

are you planning to live out the rest of your life in china as soon as you're able to move out[/quote]
Actually yes...

China is gonna own. Last time they were down by a bit, and that's cuz they lost the fencing one due to unfair judge (who was banned from judging for 2 years, racist b****), diving from screw up, ping pong or badminton from screw up.

起来, 不愿做奴隶的人们!
把我们的血肉，筑成我们新的长城。
中华民族到了最危险的时候，
每个人被迫着发出最后的吼声。
起来！起来！起来！
我们万众一心，
冒着敌人的炮火，前进！
冒着敌人的炮火，前进！
前进！前进！进！
(tis the national anthem for those of you who were wondering)

Wow! China is beating China 7-0 in the poll!

(All right, that's just a silly bug. Is it connected to editing polls?)

[b]@ iostream.h:[/b] If they didn't do that, there would be a lot more angry athletes throwing away their bronze medals.

I don't think it is nice to use a weight-vector, since that would try to set the relative values of the medals. The only other reasonable measure is the Gold count, and that is already in use along-side the overall count.

I don't want to be against China, but now comes USA!( I don't favor them either)

[quote="iostream.h"]the real question is whether or not we weight all the medals equally when determining the global ranking.[/quote]

yeah I do

basically for two reasons

1. blah

2. I am also too lazy to want to take the dot product with some vector every time I want to brag about my country

China is winning (in golds), but after track starts, US will win. If US fails track, then of course China will win

It's not the gold medals that count, but how you get them. In the Cold War era the US consistently came in 2nd to the Soviet Union because we didn't take anyone who showed talent and put them in a training camp when they were little kids. Our athletes have always been the best becaus they want to, not because their country forces them to. We don't have great monetary incentives for doing well, so in a sense, the US embodies what the Olympics were meant for: A group of athletes who have worked hard to get where they are, not because of money, not because their country forces them to, but simply because they wanted to be the best in the world at their sport.

i agree that equal weighting makes the most sense, all things considered, especially in sports where the difference between medals is so much smaller than each athlete's $ \sigma$.

TZF: the gold medal count is a form of weighting.  you might think of it heuristically as bronze = 1, silver = infinity, gold = infinity^2.  indeed, it's probably the most extreme conceivable form of weighting.

mewto: are you kidding me?  the notion that athletes participate in the olympics because of some highfalutin' notion of amateurism has long been kicked out the window.  some athletes certainly do it for the love of the game, but they're ultimately professionals who pay their bills by winning contests.  government coercion, social pressure, an uncompromising thirst for fame and fortune: pick your poison.  at the end of the day, the olympics are just a part (albeit a large one) of the bread-and-circuses machine.

heck, if anything, i feel more admiration (or perhaps pity?) towards the athletes who succeed in the soviet-style system.  if your life was completely hijacked against your will and your future was always uncertain and contingent on performance, do you think you'd be able to keep it together at the olympic level?  i don't think i could.  imagine trying to be an east german gymnast in the bad old days, watching your friends drop like flies from your government's performance enhancing drug experiments.

well yeah, but you'd also be that much more souped up by the performance-enhancing drug experiments

even if you'd die a couple years later

dude also how about just using <gold, silver, bronze> \cdot <1, 0, 0>

that is somewhat simpler

[quote="serialk11r"]mewto5555:
What's this "government forcing" you speak of? The government doesn't force anyone to do sports. Kids who are picked at a young age to compete and train do so because they did well in some competition and want to continue it.[/quote]

Yeah I agree with that but why are you posting what mewto said?   :huh:

GO CHINA!!!! :D

mystic: but when they do gold medal count (e.g. on beijing's official website), they break ties by looking to silvers, then bronzes.

$ \langle 1, \varepsilon, \varepsilon^2 \rangle$

All these fuckers train in the U.S. and then go compete for other countries. So that should be accounted for. [color=green]Some trolling removed.[/color]

...

I don't think I even know how to respond to your posts anymore

like I don't think you're just a troll -- you'd like us to take your comments seriously. you're just making it fairly difficult for that to happen.

I guess I will try to respond to that post:

could you give some examples of people training in the US and competing for other countries?  Also, what's wrong with ping-pong and handball?  Even if you don't like them, insulting them with derogatory words isn't appropriate.  btw, in the case of ping, the athletes all train in China, then play for other countries.

[url]http://www.nbcolympics.com/tabletennis/news/newsid=223558.html#chinas+hand+downs[/url]

The best ping-pong player of the US, Gao Jun, was a silver medalist in doubles in 1992 for China, then moved to the US, and must train in China, with just a college team, b/c she can't find good training partmers/coaches in the states.

examples of people who trained in the US, then played for other countries include kirsty coventry (trained at auburn university, competes for zimbabwe), chris kaman (born in the united states, played in the united states his entire career, competes for germany), becky hammon (born/raised in united states, plays for russia), shawn redhage (born/raised in united states, playing for australia), j.r. henderson (went to UCLA, moved to japan and renamed himself "j.r. sakuragi"), davey johnson (coach for the new york mets, coached for the netherlands in 2004), etc. etc.

conversely, many athletes who were not born in america compete for us.  examples of such players like those are so easy to find that i won't bother listing them (just check out our track and field team.)

in any case, why do we care?

hah and did you hear the russians ripping on the georgian volleyball team for being brazilian

it was pretty intense yo

[quote="iostream.h"]many athletes who were not born in america compete for us. examples of such players like those are so easy to find that i won't bother listing them (just check out our track and field team.)[/quote]

But they don't go training in other countries... This is not "conversely" at all. Who are you trying to fool?

Also, I like ping pong (not handball though). But they're silly as Olympic sports. That's just obvious.

well, i guess it's not obvious to me, could you try to explain it?
I understand many american's view ping-pong as a recreational sport, but it's not like that in asia, ping-pong is a huge sport in, for eg, China.  Lots of sports r big in some countries, but not in others.  for eg, men's field hockey is non existant in the US, but pretty important in europe, and cricket is really popular in india, but almost no1 plays in the U.S.

[quote="durt"]But they don't go training in other countries... This is not "conversely" at all. Who are you trying to fool?

Also, I like ping pong (not handball though). But they're silly as Olympic sports. That's just obvious.[/quote]

from what i hear, in china, ping pong beasts are like sex symbols over there

it's like over here we have scantily clad movie stars and barack obama who teenagers put posters of up on their walls? in china, they put up scantily clad movie stars (yeah, we have things in common with chinese people!) and then also them ping pong players

and so i'd imagine that it's hard to call ping pong "silly" if it's business that serious across the pacific. open your mind bro

Dude I played with a professional coach in China for a few years, and it's no silly sport. It's damn serious and intense. If you play properly it's good exercise too.

[quote="jhredsox93"]hey no offense or anything, but how exactly is Russia not on that poll of yours :-P?[/quote]

When I put Russia in it said I had too many poll options.   :mad:

[quote="durt"]All these - train in the U.S. and then go compete for other countries. So that should be accounted for. Some trolling removed.[/quote]

lol, there are us athletes who have chinese and russian coaches
pwnt.

[quote="shobber"]It seems that the winner of 2008 Olympic Games is China.  :lol:[/quote]

DUDE NO I THINK WE HAVE PRETTY CONSISTENTLY LED

ALSO WE ARE MORE AWESOME

IT DIDN'T COUNT MY VOTE FOR JAPAN!!!  :cursing:


Draw the midpoints of PB and PC.  Then use congruent triangles.

Edit--specifically, letting the respective midpoints be X and Y, show that KXM and LYM are congruent.


Prove that the product of any four consecutive non-negative integers can be expressed as $ x^{4}\plus{}2x^{3}\minus{}x^{2}\minus{}2x$ for some $ x\in\mathbb{N}$.

let the numbers be $ n,n\plus{}1,n\plus{}2,n\plus{}3$
set $ n\plus{}1\equal{}x$ thus $ n\equal{}x\minus{}1,n\plus{}2\equal{}x\plus{}1,n\plus{}3\equal{}x\plus{}2$
so $ n(n\plus{}1)(n\plus{}2)(n\plus{}3)\equal{}(x\minus{}1)x(x\plus{}1)(x\plus{}2)\equal{}x^{4}\plus{}2x^{3}\minus{}x^{2}\minus{}2x$

I THINK...

Let be $ k$ a number

$ (k\minus{}1)k(k\plus{}1)(k\plus{}2)$

$ (k\minus{}1)k(k\plus{}1) \equal{}{k}^{3}\minus{}k$

$ (k\minus{}1)k(k\plus{}1)(k\plus{}2) \equal{} ({k}^{3}\minus{}k)(k\plus{}2)$

$ ({k}^{3}\minus{}k)(k\plus{}2) \equal{}{k}^{4}\plus{}2{k}^{3}\minus{}{k}^{2}\minus{}2k$

C'est  la meme chose pour $ x$

This is the only way i know to prove that


Two typists working together can type a paper in 5 house.  The faster typist can complete the paper in 9 hours.  How many hours will it take the slower typist to complete the paper alone? (Express your answer as a mixed number.)

[hide]Using the work formula, you find that 1/5-1/9=1/x where x is the slower typists speed. the formula 1/a-1/b=1/x is (ab)/(b-a). 45/4=[b]11 1/4[/b].[/hide]

[hide="Solution Written More Clearly"]  If it takes the faster worker 9 hours, than he types $\frac{1}{9}$ of the paper every hour.  To finish the paper in 5 hours, they must type at a rate of $\frac{1}{5}$ of a paper every hour.  Then the second typist must type at a rate of $\frac{1}{5}-\frac{1}{9}=\frac{9-5}{45}=\frac{4}{45}$.  Then he would take $\frac{1}{\frac{4}{45}}$ hours to type the paper alone, or $11\frac{1}{4}$ hours.[/hide]

Really, noneoftheabove, learn LaTex.


If a rectangle is formed by doubling the length of one side of a square and halving the other side of the square, what is the ratio, expressed as a common fraction, of the perimeter of the rectangle to the perimeter of the square?

Let the side length of the square be $ s$. Thus, the perimiter is $ 4s$.

The perimiter of the rectangle is $ 2(2s\plus{}\frac{s}{2})\equal{}5s$.

Thus, we have $ \frac{5}{4}$.


1. A hollow rubber ball has an outside diameter of 5.6 cm. The thickness o the rubber ball is 0.3 cm. Determine the amoutn of rubber used to construct the ball.

Can someone please explain this to me? Thank You.

Another question...

2. how much more ore less does a hemispherical tank hold compared to a cone if their diameters are 4.8 m and the height of the cone is 2.4 m.

Please show your work, I want to understand how you get the answer =)

first, we calculate the surface area of the rubber ball. the surface area formula is equal to 4pi r^2. the surface area of the ball is therefore 98.47cm^2 if we use 3.14 as pi. if we multiply this by the thickness of the ball, we get 29.54 cm^3 of rubber necessary to construct the ball.

for your next question, we start by calculating the volumes of the two figures. the volume of a sphere is equal to (4pi r^3)/3. this means that the volume of a sphere with the dimensions you provided would be 463.01 m^3. but, we're dealing with a hemisphere, so we divide that number by 2 to get 231.5 m^3. the volume of a cone is (pi r^2 h)/3, which is equal to 57.88 m^3 for the dimensions you provided. once again, we use 3.14 for pi. so, the difference in volumes is therefore 231.5-57.88, or,
173.62 m^3.

you can get a lot more of these formulas on mathworld.com

Do you think I have to memorize all these formulas for chemistry class?

These questions come from my summer chem packet, so if it's essential to memorize the formulas for chem, I think i should get started.

For the first one:

First find the area of the sphere when knowing that the radius is $2.8$, then the area of the sphere is $\frac{4}{3}(2.8)^3\pi=29.27\pi$. Now, you must find the area of the sphere which does not contain any clay, which has a radius of $\frac{5.6-.6}{2}=2.5$, now the area of this, using the formula is $20.83\pi$, now subtracting the areas, you find the amount of rubber, which is $\boxed{8.44\pi}$.

who's answer is correct? h78reg or Iversonfan2005 for #1

Are we looking for the volume or area of the rubber?

[quote]for your next question, we start by calculating the volumes of the two figures. the volume of a sphere is equal to (4pi r^3)/3. this means that the volume of a sphere with the dimensions you provided would be 463.01 m^3. but, we're dealing with a hemisphere, so we divide that number by 2 to get 231.5 m^3. the volume of a cone is (pi r^2 h)/3, which is equal to 57.88 m^3 for the dimensions you provided. once again, we use 3.14 for pi. so, the difference in volumes is therefore 231.5-57.88, or,
173.62 m^3.[/quote]

Oh, i think for the radius, you put the diameter, instead. Because in the question, they give you the diameter.. which means you have to divide it by 2. It's okay though.. I got the answer >> 14.48m^3

[quote]
2. how much more ore less does a hemispherical tank hold compared to a cone if their diameters are 4.8 m and the height of the cone is 2.4 m. [/quote]
Can i leave it in m^3? or Do i have to change it to... Liters?

If the other exercises in your packet ask to convert the result to liters, convert it. If they don't, you can just leave it as it is.

[quote="h78reg"]yikes! i think i miscalculated. either that or a typo. :blush:  iversonfan is correct.

i think iversonfan was looking for the volume though, judging by his formula. but, even so, i got a totally result for the volume of the sphere. but, after calculating the two volumes, iversonfan's answer is indeed correct.

third of all, i dont see why you need to know these formulas in chemistry class, but, they are very useful nonetheless.[/quote]
Why not? Volume formulas are very important in every science.

in simple dumb terms (for me) its just pretty much subtract the inner volume from the outer volume right?


geometry circumcircle inradius trigonometry

[color=darkred]Let $ ABC$ be a triangle with circumradius $ R$, inradius $ r$ and $ A$- exinradius $ r_{a}$. Prove that $ A \equal{} 60^{\circ}$ $ \Longleftrightarrow$ $ \frac{R}{r}\equal{}\frac{a}{p\minus{}a}$ $ \Longleftrightarrow$ $ OI\perp OI_{a}$.[/color]

[hide="Just the first part"]We have
$ \frac{R}{r}\equal{}\frac{a}{p\minus{}a}\Leftrightarrow\frac{R}{a}\equal{}\frac{r}{p\minus{}a}$
$ \Leftrightarrow\frac{1}{2sinA}\equal{}\tan\frac{A}{2}$
$ \Leftrightarrow\sin\frac{A}{2}\equal{}\frac{1}{2}$
And we get $ A \equal{} 60^{o}$.[/hide]

[hide="Some hints"] We have $ 1.\blacktriangleright$ $ OI^{2}\equal{}R^{2}\minus{}2Rr$, $ OI_{a}^{2}\equal{}R^{2}\plus{}2Rr_{a}$
$ 2.\blacktriangleright$ By the Thales's theorem, we have $ \frac{AI}{p\minus{}a}\equal{}\frac{AI_{a}}{p}\equal{}\frac{II_{a}}{a}$ $ \Rightarrow II_{a}\equal{}\frac{a}{p\minus{}a}\cdot AI$ $ \Rightarrow II_{a}^{2}\equal{}\frac{a^{2}}{(p\minus{}a)^{2}}\cdot\frac{bc(p\minus{}a)}{p}\equal{}\frac{4aRr}{p\minus{}a}$
Hence we have $ OI\perp OI_{a}\Longleftrightarrow 2R^{2}\plus{}2R(r_{a}\minus{}r)\equal{}\frac{4aRr}{p\minus{}a}$ $ \Longleftrightarrow R\plus{}\frac{pr}{p\minus{}a}\minus{}r\equal{}\frac{2ar}{p\minus{}a}\Longleftrightarrow\frac{R}{r}\equal{}\frac{a}{p\minus{}a}$[/hide]

[color=darkblue][b]Proof.[/b] $ A \equal{} 60^{\circ}$ $ \Longleftrightarrow$ $ \widehat{BIC}\equiv\widehat{BOC}\equal{} 120^{\circ}$ $ \Longleftrightarrow$ the quad. $ BIOC$ is cyclically $ \Longleftrightarrow$ the point $ O$ belongs to the circumcircle of $ \triangle BIC$ $ \Longleftrightarrow$
the point $ O$ belongs to the circumcircle (with the diameter $ II_{a}$) of the quadrilateral $ BICI_{a}$ $ \Longleftrightarrow$ $ OI_{a}\perp OI$.[/color]

adding to it if $ A\equal{}60$ then $ O,H,I,I_{a},B,C$ all lie in one circle


If $ x,y,z>0$ such that $ xyz\equal{}1$ then prove the following inequality:

    $ \frac{x^3\plus{}y^3}{x^2\plus{}y^2}\plus{}\frac{y^3\plus{}z^3}{y^2\plus{}z^2}\plus{}\frac{z^3\plus{}x^3}{z^2\plus{}x^2} \geq 3$

$ \frac{x^3\plus{}y^3}{x^2\plus{}y^2} \ge \frac{x\plus{}y}{2} \Leftrightarrow (x\minus{}y)^2(x\plus{}y) \ge 0 \Rightarrow  \sum \frac{x^3\plus{}y^3}{x^2\plus{}y^2} \ge \sum \frac{x\plus{}y}{2} \equal{} \sum x \ge 3\sqrt[3]{xyz} \equal{} 3$.

Thanks for your reply Felix.That's my solution too :)


[quote="Inspired By Nature"]Prove that for every natural n, the following inequality is held:
$(2n+1)^{n}\geq (2n)^{n}+(2n-1)^{n}$.[/quote]
This is equivalent to ${(1+\frac{1}{2n})}^{n}\geq 1+{(1-\frac{1}{2n})}^{n}.$

Just expand and collect all the terms, except 1, on the LHS. I think this was what [b]manuel[/b] meant in his post earlier.


A right triangle ABC is given in which F is the midpoint of the hypotenuse AB, and BC=3AC.  Let the points D and E divide the side BC in three equal segments. Prove that the triangle DFE is isosceles and right-angled.  (M&IQ 1992)

I proved triangle DFE is isos.   Can anyone give me a hint on proving triangle DFE is right?  I drew in line AD.  I know that the acute angles of triangle ACD are 45 degrees.  am I on the right track?

Thanks :)

[hide]
Let $FG \perp BC$ at $G$, then $FG=GD=GE$ yielding $\angle{DFE}=90^{o}$.[/hide]


limit
logarithms
trigonometry
calculus
derivative
function
calculus computations

Find the limits:
[b]1.[/b] $\mathop {\lim }\limits_{x \to 0} \frac{{\ln \sin ^2 ax}}{{\ln \sin ^2 bx}}$ $a \ne b$
[b]2.[/b] $\mathop {\lim }\limits_{x \to b} \frac{{a^x - a^b }}{{x - b}}$ $a>0$
[b]3.[/b] $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} tgx \cdot (\frac{\pi }{2} - x)$

[quote="Nea"]Find the limits:

[b]2.[/b] $\mathop {\lim }\limits_{x \to b} \frac{{a^x - a^b }}{{x - b}}$ $a>0$
[/quote]

By L'hopital rule
\[ {\lim }\limits_{x \to b} \frac{{a^x - a^b }}{{x - b}} = {\lim }\limits_{x \to b} \frac{a^x\ln a}{1} =a^b \ln a \]

[quote="Nea"]Find the limits:
[b]3.[/b] $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} tgx \cdot (\frac{\pi }{2} - x)$[/quote]

Also by L'Hopitales rule
\[ \lim_{x \to \frac{\pi }{2}} \frac{ (\frac{\pi }{2} - x)}{ctgx} = \lim_{x\to \frac{\pi }{2}} \frac{-1}{\frac{-1}{\sin^{2}x}} = \sin^2 (\pi/2) =1  \]

One can find the first limit with use of L'Hopital rule too...  :)

Steering clear of L'Hopital:

${\lim }\limits_{x \to 0} \frac{{\ln \sin ^2 ax}}{{\ln \sin ^2 bx}} = {\lim }\limits_{x \to 0} \frac{{\ln |\sin ax|}}{{\ln |\sin bx|}}$ is essentially $\lim_{x\to 0} \frac{\ln|ax|}{\ln|bx|} = \lim_{x\to 0^+} \frac{\frac{\ln |a|}{\ln x} + 1}{\frac{ln |b|}{\ln x} + 1} = 1$. It doesn't matter if $a = b$.

$\frac{a^x - a^b}{x - b} = a^b\ln a\cdot\frac{e^{\ln(a)\cdot(x-b)} - 1}{\ln(a)\cdot(x-b)} \to a^b\ln a$ as $x\to b$, by a standard limit.

$\left(\frac{\pi}{2} - x\right)\tan x = \sin x\cdot\frac{\pi/2 - x}{\cos x} \to 1\cdot 1 = 1$ as $x \to \pi/2$ (Substitution of $y = \pi/2 - x$ shows that $\lim_{x\to\pi/2}\frac{\pi/2 - x}{\cos x} = \lim_{y\to0}\frac{y}{\sin y} = 1$.

Always remember to never use L'Hopitals rule :)

[quote="anik_andrew"][quote="Nea"]Find the limits:
[b]3.[/b] $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} tgx \cdot (\frac{\pi }{2} - x)$[/quote]

Also by L'Hopitales rule
\[ \lim_{x \to \frac{\pi }{2}} \frac{ (\frac{\pi }{2} - x)}{ctgx} = \lim_{x\to \frac{\pi }{2}} \frac{-1}{\frac{-1}{\cos^{2}x}} = \cos^2 (\pi/2) =0  \]

One can find the first limit with use of L'Hopital rule too...  :)[/quote]
I don't understand your logic :huh:. Maybe there is somtheing wrong.

He just did the derivative wrong. It should be $\sin^2$ instead of $\cos^2$. Anyway, L'hopital's rule isn't any good. Especially for number 2 it is ridiculous (do you see why?).

[hide="Without the l-Hospital - rules."]
$f_1(x)\equiv \frac{\ln \sin ^2 ax}{\ln \sin^2 bx}=\frac{\ln \left|\frac{\sin ax}{ax}\right|+\ln |a|+\ln |x|}{\ln \left|\frac{\sin bx}{bx}\right|+\ln |b|+\ln |x|}\Longrightarrow \lim_{x\to 0} f_1(x)=1.$
$f_2(x)\equiv \frac{a^x-a^b}{x-b}=a^b\cdot \frac{a^{(x-b)}-1}{(x-b)}\Longrightarrow \lim_{x\to b} f_2(x)=a^b\cdot \lim_{t\to 0} \frac{a^t-1}{t}=a^b\ln a.$
$f_3(x)\equiv \left( \frac{\pi }{2}-x\right)\tan x=\frac{\left( \frac{\pi }{2}-x\right)}{\tan \left(\frac{\pi }{2}-x\right)}\Longrightarrow \lim_{x\to \frac{\pi}{2}} f_2(x)=\lim_{t\to 0} \frac{t}{\tan t}=1.$[/hide]

Before the point gets lost, what Kalle was saying about the second problem is this:

Let $g(x)=a^x.$ The the limit being asked for is, by the raw definition of the derivative, $g'(b).$

I agree with you that L'Hopitales rule is not a lovely method, (i don't use them very often too) but if you forgot your "standart limits", what do you will to do???
When function is not very "large" i think we ca use it...
But this is a matter of taste ;)

[quote="Nea"]:D  i don't like L'hopital's rule too. I don't know why everybody use it.[/quote]

me, too. :lol: L'Hopital's theorem isn't permitted to use for university entrance exam in Japan.

As for these problems, they are based on the definition of differential:$\lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}=f'(a)$ as shown by virgil Nicula.

[quote="Virgil Nicula"]Anik Andrew, the standard limits you can not prove (logically) with the l'Hospital 's rules (it is a vicious circle !), may by for finding them again.[/quote]
I'm not dispute. I think, when i will write such limit, no one will interesting how i get him. And L'hospital's rules is only helm me to remember the answer... :)


calculus
derivative
vector
linear algebra
matrix
real analysis
real analysis unsolved

Show that: $ \left(\frac{\partial z}{\partial y}\right)_{u}  \equal{}  \left(\frac{\partial z}{\partial x}\right)_{y} \left[ \left(\frac{\partial x}{\partial y}\right)_{u} \minus{} \left(\frac{\partial x}{\partial y}\right)_{z} \right]$

I have Euler's chain rule and "the splitter."  Also the property, called the "inverter" where you can reciprocate a partial derivative.

If I write Euler's chain rule, I only know how to write it when there are 3 variables, I usually write it in the form:
$ \left(\frac{\partial x}{\partial y}\right)_{z} \left(\frac{\partial y}{\partial z}\right)_{x} \left(\frac{\partial z}{\partial x}\right)_{y}  \equal{}  \minus{}1$

Where I can write x,y,z in any order as long as each variable is used in every spot.  However, I do not know how to work this chain rule if I have an extra variable (u in this case).

I also tried using the "splitter" to do something like writing:
$ \left(\frac{\partial z}{\partial y} \right)_{u}  \equal{}  \left(\frac{\partial z}{\partial x} \right)_{u} \left(\frac{\partial x}{\partial y}\right)_{u}$

However, I do not know what to do with this because I have the term
$ \left(\frac{\partial z}{\partial x} \right)_{u}$ , which doesn't appear in the original problem.

Any help would be appreciated.

Thanks in advance.

(This is for a thermodynamics course, but we are still in the mathematics introduction.)

Apparently, here you have two independent relations between $ x,y,z,u$, so you are talking about a $ 2$-dimensional surface in $ \mathbb R^4$. Now, you have $ 3$ vectors in the tangent plane to this surface: $ [(x_y)_u,1,(z_y)_u,0]$, $ [1,0,(z_x)_y,(u_x)_y]$ and $ [(x_y)_z,1,0,(u_y)_z]$. They should be linearly dependent, which, in particular, means that
\[ \det\begin{bmatrix} (x_y)_u & 1 & (z_y)_u \\
1 & 0 & (z_x)_y \\
(x_y)_z & 1 & 0 \end{bmatrix} \equal{} 0\,,
\]
whence the conclusion.
"Splitter" and "Euler's chain rule" are useless here because they are applicable only when you have a $ 2$-dimensional surface in $ \mathbb R^3$.


inequalities
trigonometry
inequalities unsolved

In any triangle ABC  show that
[tex] R-2r \geq l_a - h_a  [/tex]

[tex] l_a [/tex]  is  the  bisector  of  angle  A.
The other  symbols  are  as  ussual.
Thanks

While trying to solve this one I found that:

$\displaystyle R-2r \geq \sqrt{p(p-a)}-h_{a} \geq l_{a}-h_{a}.$

I will just prove the first inequality as the second is trivial:

\[1) \displaystyle R-2r=R(1- 4 \sin \frac{A}{2} \cos \frac{B-C}{2}+ 4 \sin^{2} \frac{A}{2},\]

\[2) \displaystyle \sqrt{p(p-a)}=2R \cos \frac{A}{2} \sqrt{ \cos^{2} \frac{B-C}{2} -\sin^{2} \frac{A}{2}},\]

\[3) \displaystyle h_{a}=2R( \cos^{2} \frac{B-C}{2}-\sin^{2} \frac{A}{2}).\]

From 1), 2) and 3) it follows that our inequality is equivalent to: $4t^{2}(y-t)^{2}+8yt(y-t)^{2}+(4yt-1)^{2} \geq 0.$

cheers! :D it is indeed cute! :)

Lagrangia, I see you  like geometric inequalities. :D

Your solution is very nice so I have another nice gift for you.

Prove that

$ \frac{R}{2r} \geq \frac{m_a}{h_a}  $

where $ m_a$  and $h_a$ are respectively the median and height of side $a$.

[quote="harazi"]I'm not sure, but I think that Laurentiu Panaitopol published an article where those stronger inequalities appear and if I'm not wrong, with the same solution. And as far as I know, this problem was created by Laurentiu Panaitopol.[/quote]

Hmmm.. which issue? (if you know)

I found it on INTERNET but I don't remember the web site.

I don't know the issue by heart and I can't check right now, since I'm in Athens. But I will check and tell you.

Manlio showed me a file with the proof of the above two problems and it is the same as I gave and now there it was the reference of L. Panaitopol and the issue number. So now I know, perhaps someone I know has that issue..

cheers! :D :D


complex analysis
function
real analysis
real analysis solved

Let p,q and r be non-constant non-vanishing entire holomorphic functions such that

p + q + r = 0.

Prove that there exists an entire function h such that p, q, r are constant multiples of h.

This proof is due to Pliviu, but he is not keen on posting, so here it goes:
Take f=-p/r and g=-q/r. Then f and g are holomorphic functions non-vanishing, and f+g=1. Because f does not take the value 0, g does not take the value 1. But g does not take the value 0.
So g skips 0 and 1 from it's image. This statement should lead to g=constant and from this f=constant and so p,q,r are constant multiples of one of them.
The statement that if a holomorphic function skips 2 points from it's image, then f is constant I think is due to Picard (NOT Patrick Stewart). I don't know how to prove it, but maybe Mindspa knows.


What is the relationship between the medians and sides in a triangle?

One thing that I learned was that the median to side c (where a,b,c are the sides of the triangle) has length: sqrt(a^2/2 + b^2/2 - c^2/4) adn you can use that to find the other median lengths too, just put the side the median is going to at the end.

Sorry, I don't know latex very well...

Err, nvm, I keep thinking mixing up the meaning of median of a triangle. Sorry!

As Original Pi Guy said, you can find median lengths given the sides of the triangle. A direct corollary to this is that the sum of the squares of the medians of a triangle are equal to $\frac{3}{4}$ the sum of the squares of the sides. This follows directly from the formula.

if the length of a median to side a is d then

d = 1/2 :sqrt: ( 2(b 2 + c 2 ) - a 2 )


geometry calculus integration conics parabola trigonometry rotation

Hello everyone. i'm having a little trouble solving these problems that deal with the area between curves.

1) The curve of $x^2=y^2*(4-y^2)$

2) The parabolas $x=y^2-5y$ and $x=3y-y^2$

3) The total area bounded by the cubic $x=y^3-y$ and the line $x=3y$

also, one definite integral problem that has me stumped  :huh:

4) $\int\limits_{1}^{2}\frac{dx}{\sqrt{4-x^2}}$

well for the last one sub in $x=2sin(\theta)$

ok this is kind of embarrassing... but i need to graph these also using my ti-89 titanium, so how would i solve these for y so that i can input them into the calculator's graphing mode?  :blush: thanks to anyone that can help me out and to those who have already helped!

for #2 and 3 you could just plug them in as is and then rename you axies.
what i mean is
put $Y_1=x^2-5x$ let it graph and then just rotate what it displays $90^o$ to the right.

and for number one just do the same thing except take the spquareroot first and get
$x=\pm\sqrt{(y^2)(4-y^2)}$


A right triangle has perimeter $ 32$ and area $ 20$. What is the length of its hypotenuse?

$ \textbf{(A)}\ \frac{57}{4} \qquad
\textbf{(B)}\ \frac{59}{4} \qquad
\textbf{(C)}\ \frac{61}{4} \qquad
\textbf{(D)}\ \frac{63}{4} \qquad
\textbf{(E)}\ \frac{65}{4}$

[hide]Let the legs of the triangle be $ a$ and $ b$ then,

Perimeter: $ a \plus{} b \plus{} \sqrt {a^2 \plus{} b^2} \equal{} 32$
Area: $ ab/2 \equal{} 20$

Simplifying the perimeter equation,
$ \sqrt {a^2 \plus{} b^2} \equal{} 32 \minus{} (a \plus{} b)$
$ a^2 \plus{} b^2 \equal{} 1024 \plus{} (a \plus{} b)^2 \minus{} 64(a \plus{} b)$
$ 1024 \plus{} 2ab \equal{} 64(a \plus{} b)$

From the area equation, $ 2ab\equal{}80$. Substituting,
$ 64(a \plus{} b) \equal{} 1104$

So the sum of the legs is $ a \plus{} b \equal{} 1104/64\equal{}69/4$. The length of the hypotenuse is 32 minus that: $ \sqrt {a^2 \plus{} b^2} \equal{} 32 \minus{} 69/4 \equal{} 59/4$.

Hence the answer is $ \boxed{B}$[/hide]


vector
invariant
combinatorics solved
combinatorics

Given a simple connected graph with an even number of edges, show that we can direct the edges s.t. the out-degree of each vertex is even.

[The problem doesn't mention connectivity, but it would be false if we allowed non-connected graphs. Just think of two disjoint $K_3$'s (triangles).]

start with an arbritary choice of directions. clearly the total number of vertices with an odd out-degree is even. thus, as long as it is not zero, choose two vectors, choose a way between them(assume the graph is connected) and reverse all directions in that path. the parity of the out-degree of all vertices of the path is invariant except for the endpoints, where it changes, which decreases the number of vertices with an odd out-degree by 2. continuing, we get a desired choice of directions.

Peter

Of course! :) I did exactly the same thing. Thanks!


i know this question seems so easy, but for some reason, everytime i try it i come out with the wrong answer..help?

a bank PIN number has four digits where the leading number cannot be zero, the four digits cannot be in sequence, and all four digits cannot be identical. what is the probability that a thief would guess the number in three attempts?

Define in sequence. Is $ 1234$ allowed? $ 1423$? $ 2468$?


limit
real analysis
real analysis solved

$x_1>0,\ x_{n+1}=\left| x_n -n\right| ,\ y_n=\frac {x_n}{n},\ n\in N^*\Longrightarrow \lim_{n\to\infty} y_n =\frac 12.$

Now it satisfies:

Take $x_n<n$,$\forall n=1,2,3...$ where $x_n = x_1$, for $n$ odd and $x_n = n-1-x_1$ for $n$ even. It verifies that $x_{n+1} = |x_n - n|$,$\forall n \in {\mathbb N}^*$ and it is clear that $\lim_{n \to \infty} y_n \neq \frac{1}{2}$.

Your example does not satisfy the recurrence. The original problem is true.

For some $k$, $x_k<k$, since the sequence decreases until this is true. Now for $n\ge k$, $x_n<n$ and $x_{n+2}=(n+1-(n-x_n))=x_n+1$.
Obviously, $\lim_{m\to\infty}\frac{x_{n+2m}}{n+2m}=\lim_{m\to\infty}\frac{x_n+m}{n+2m}=\frac12$.


[quote="hjbrasch"]$ x\equiv 123\bmod{280}$[/quote]

Could you give a solution?

[quote="anncar"]Solve the simultaneous congruences

2x = 1 (mod7)
x= 3(mod5)
x= 3(mod8)

those aren't supposed to be equal signs- they are three slashes(congruences[/quote]
[hide="Solution"]
We have that $ 2x\equiv 1\bmod 7$, so we can find that $ x\equiv 4\bmod 7$  Now, $ x\minus{}3$ is divisible by both $ 5$ and $ 8$ and thus by $ 40$, so $ x\equiv 3\bmod 40$, giving us that $ x\equiv 3\bmod 40$.

Now, we say that $ x\equal{}3\plus{}40a$ and $ x\equal{}4\plus{}7b$.  So, we obtain that $ 3\plus{}40a\equal{}4\plus{}7b$, giving that $ 7b\plus{}1\equal{}40a$, so $ 7b\plus{}1\equiv 0\bmod 40$ and therefore that $ 7b\equiv 39\bmod 40$.  Thus, we can show that $ b\equiv 17\bmod 40$.  We say that $ b\equal{}17\plus{}40c$ and plug it in to get that $ x\equal{}4\plus{}7b\equal{}4\plus{}7(40c\plus{}17)\equal{}123\plus{}280c$, so $ \boxed{x\equiv 123\bmod 280}$.  [/hide]

[hide]$ 2x\equiv 8 \implies x\equiv 4\pmod{7}$

$ x\minus{}3 \equiv 0 \pmod{40}$

40(7k+3)+3=280k+123[/hide]


limit
real analysis
real analysis unsolved

I need to prove this version of L'Hopitals theorem.

If limf(x)=limg(x)=infinity (as x goes to infinity) then lim[f(x)/g(x)]=lim[f'(x)/g'(x)] (as x goes to infinity)

Any help please?

Thank you

$ \lim_{x \to \infty} \frac{f(x)}{g(x)} \equal{} \lim_{u \to 0} \frac{f(1/u)}{g(1/u)} \equal{} \lim_{u \to 0} \frac{[f(1/u)]'}{[g(1/u)]'} \equal{} \lim_{u \to 0} \frac{f'(1/u)(1/u)'}{g'(1/u) (1/u)'} \equal{} \lim_{u \to 0} \frac{f'(1/u)}{g'(1/u)}$

$ \equal{} \lim_{x \to \infty} \frac{f'(x)}{g'(x)}$,

which means that to prove your version of L'Hôpital's rule you can use the proof for the version

$ \lim_{t \to 0} \frac{f(t)}{g(t)} \equal{} \lim_{t \to 0} \frac{f'(t)}{g'(t)}$

[quote="maxcat00"]If limf(x)=limg(x)=infinity (as x goes to infinity) then lim[f(x)/g(x)]=lim[f'(x)/g'(x)] (as x goes to infinity)[/quote]The statement is not true.  It is true that [i]if[/i] $ \lim_{x\to \infty}\frac {f'(x)}{g'(x)}$ and $ \lim_{x\to \infty}\frac {f(x)}{g(x)}$ exist, then they are equal, but this assertion is significantly weaker.

I still didn't manage to find this. Can anyone help please.

Thank you

Excuse me, but what are you not able to find? Can't you be more specific about your doubts?


[quote="Chalao"]Prove the following equality:

1/(cos0cos1)   +1/(cos1cos2)    +1/(cos2cos3)........ +1/(cos88cos89)= (cos1)/(sin^2(1))

Where angles are measured in degrees. Please explain it clear...thanks  :lol:[/quote]

A start might be...
1/(cos0sin89)   +1/(cos1sin88)    +1/(cos2sin87)........ +1/(cos88sin1)

I'm still trying to forge a full solution :D

Hi. I thought of leaving the first term (which is $ \frac{\cos 1}{\cos ^{2} 1}$), and then putting the other eighty-eight terms in pairs: second with last, third with second last, and so on; and then using $ \cos \theta \equal{} \sin (90 \minus{} \theta)$ and so on but haven't got it yet.

It's USAMO, 1992:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58210[/url]

Ah! VANCHAN123's solution is neat (except that $ \cot(n\plus{}1) \minus{} \cot(n)$ should read $ \tan(n\plus{}1) \minus{} \tan(n)$ and then the rest is correct).


search
Diophantine equation
number theory proposed
number theory

hi,
i need some help with the following problem:
let $ a,b\in\mathbb{N}$ and if $ ab | a^{2}\plus{}b^{2}\plus{}1$ then $ \frac{a^{2}\plus{}b^{2}\plus{}1}{ab}\equal{}3$.
can somebody give me a solution, please.

Look for PEN by Hojoo Lee. :wink: Problem number A5.Here is the link: http://www.mathlinks.ro/viewtopic.php?t=150373

But we can find all solution of solution
$ x^{2}\plus{}y^{2}\plus{}1\equal{}3xy$

[quote="TTsphn"]But we can find all solution of solution
$ x^{2}\plus{}y^{2}\plus{}1 \equal{} 3xy$[/quote]
I don't understand,what do you mean?

We can find all solution of$ (x,y)$ such that
$ x^{2}\plus{}y^{2}\plus{}1\equal{}3xy$

[quote="TTsphn"]We can find all solution of$ (x,y)$ such that
$ x^{2}\plus{}y^{2}\plus{}1 \equal{} 3xy$[/quote]
Yes,we can,and i think it is very easy  :wink:

[quote="TTsphn"]But we can find all solution of$ (x,y)$ such that
$ x^{2}\plus{}y^{2}\plus{}1 \equal{} 3xy$[/quote]

$ x^{2}\plus{}y^{2}\plus{}1\equal{}3xy\Leftrightarrow (2x\minus{}3y)^{2}\minus{}5y^{2}\equal{}\minus{}1$.

Use Pell equation  :)

[quote="quangpbc"][quote="TTsphn"]But we can find all solution of$ (x,y)$ such that
$ x^{2}\plus{}y^{2}\plus{}1 \equal{} 3xy$[/quote]

$ x^{2}\plus{}y^{2}\plus{}1 \equal{} 3xy\Leftrightarrow (2x\minus{}3y)^{2}\minus{}5y^{2}\equal{}\minus{}1$.

Use Pell equation  :)[/quote]
You are wrong.

This has been discussed time and time again - USE the search function.


I would imagine that this question will not be revealed.

[quote="iYOA"]are there people that are testing the algebra part of alcumus?[/quote] Not currently.

Apparently, they are working on putting the algebra stuff on ALCUMUS, or they are thinking about doing it next, I forgot where I've read it, but I've read it somewhere on the forums today.


conics
ellipse
symmetry
geometry
geometric transformation
reflection

This may be kinda trivial but why is it true that the midpoint of the foci of an ellipse is the center?

The definition of an ellipse is symmetric with respect to the foci.

Is that a restatement of the fact that $ F_{1}Z \plus{} F_{2}Z$ is constant for foci $ F_{1} , F_{2}$ and a point on the ellipse $ Z$?

I didn't learn conics well in class this year, but your definition is right mathemagician.  I think he meant, BY definition an ellipse is symmetric with respect to the foci.  But idk, depending on how "foci" are defined out of the context of an ellipse, he might have meant the other.

It's obvious if apply your definition of an ellipse.  Due to the equal weighting of the foci, the impact of foci are the same.  So, the image ought to be the same if you switched the two foci, which is equivalent to producing a mirror image of the original image.  The only way for this equivalence of the original image and the mirror image to occur is for a perpendicular running through the midpoint of the line containing the foci to be an axis of symmetry across which the left side can be reflected to produce the right side.

Also, cutting the ellipse through the line running through the two foci, why should the top and bottom look different?  So now we have 2 axis of symmetry.

Why not make the intersection of these two lines of reflection the center as it is the only point that fits both symmetries?


geometry
MATHCOUNTS
AMC
AIME
email
AMC 10
AMC 12

I can't tell...is there new material or did they just want to give it a new look?

I don't see any differences...I mean the last time I checked it looked...holy cow! It changed! (explanation: I forgot the [url=http://www.unl.edu/amc/index.html]AMC website[/url]'s URL so I searched for it via Google, so it linked me to one of the older pages. I just clicked on "Home")
They still seem to have the same information though.

As far as I can tell, the only new things are:
- a copy of the 50th anniversary AHSME on the site (I got it from a different site)
- links to various math competitions in the Problem Directory at [url]http://www.unl.edu/amc/a-activities/a7-problems/problemdir.html[/url].  As it turns out, the content on this page was previously on a different website (courtesy of Kiran S. Kedlaya), and only recently did the AMC add it to its own site

If you're feeling nostalgic for the old site, you can see it at [url]http://www.unl.edu/amc/1-archive/archive-2002-03/index2003.html[/url]

Other than that, everything is just better organized so it is more non-hacker*-friendly

P.S. According to the site, "The AMC Web Site was last updated on 9/23/2003"

*hacker=me, for having the 10,000th post

No, there are a few other things here and there like state winner pictures here: http://www.unl.edu/amc/e-exams/e1-archives/2002-2003/0203StateWinnersPic.html

I wish [i]my[/i] picture were on the net for the AMC, or that our state had a meeting with the governor and an awards reception... sigh... too bad I live in the middle of nowhere.

Random thought: whatever happened to andy17null?  Or rippledance?  Or akatookey?  Or jojobird?  Or ......?  Or ......?

[quote="mathfanatic"]*hacker=Tare or me, for having the 10,000th post and the 1000th topic, non-respectively.[/quote]
Myeh!!!*battle screech*
I am not worthy of being called a hacker just because of that, I mean I did that [b]legitimately [/b]for pete's sakes! It wasn't even funny! :lol:
Hacker means either a computer hacker or a practical joker. Example for the first would be to change your post count by hacking, the example of second would be hacking all the chairs in the classroom so it's glued up-side down See the difference? :lol:

If you want the old AMC site[b]s[/b] (They have 17 :shock: ), [url=http://web.archive.org/web/*/http://www.unl.edu/amc/index.html]lookey here[/url]

Maybe they were too shy to put up their picture...or maybe...*ponders* :?

I would like to lift the discussion back to the original comment and ask  several questions:

What would you like to see on the AMC website?

How would you like the AMC Website to be organized?  Is it feasible to navigate the AMC website to find what you want as of now?

The AMC is a project of the Mathematical Association of America.  If you go to the MAA website at http://www.maa.org, and choose Students -> High School from the menus at the left, the only subsequent link is back to the AMC website.  What other (pre-college) student services and information should MAA offer?

Personally, I think that the AMC website is great so far, but it would be nice if there were more things like Kiran Kedlaya's notes on geometry and inequalities.

[quote="TheSouthpaw"]I also think you should have a forum like the MathCounts board to discuss scores and answers and whatnot.[/quote]
they could always link to here  :wink:

[quote="mathfanatic"]Hehe.  You forgot the L in URL.  I changed my mind.  You aren't a hacker.  But you did basically change the post count (by spamming,  :wink: ) so does that count?[/quote]
er, no.

I would love to see all the past problems online, but other than that it's a great website.

[quote="AMCDirector"]I would like to lift the discussion back to the original comment and ask  several questions:

What would you like to see on the AMC website?[/quote]

Thank you for joining this forum, and for asking. I would like to see a lot more on the Web site by way of testimonials from participants about why the AMC tests are a valuable experience. I'm a homeschooling dad, always trying to get my son into "competition culture," and here in MN (where AMC doesn't even have a state director [sigh]) I have a lot of difficulty finding other parents who are interested in AMC programs. It would be great if when I direct local friends to the AMC Web site (as I do), I could point to a specific page of kids saying, e.g., "AMC 8 really motivated me to study a lot more math" (I think this would be appealing to some parents I know  :wink: ) or teachers saying, "I think the best prepared students in my school are the ones who take the AMC tests" or whatever quotations you could find. There is a wee bit of this kind of thing already on the site, I know, but there should be a lot of it, all grouped together in an easy link for newbies to the test to spot.

[quote="AMCDirector"]How would you like the AMC Website to be organized?  Is it feasible to navigate the AMC website to find what you want as of now?[/quote]

The current condition of the AMC Web site is so-so, I would say. The premier Web site about Web usability is Jakob Nielsen's Useit.com

http://www.useit.com/

and I would urge you, and especially your staff members who are specifically tasked with keeping up the Web site, to get familiar with the free advice on that site at your earliest opportunity. It used to be HARD to find good information on the AMC Web site, but it is getting better. Improving the Web site would doubtless improve participation in AMC programs.

[quote="AMCDirector"]The AMC is a project of the Mathematical Association of America.  If you go to the MAA website at http://www.maa.org, and choose Students -> High School from the menus at the left, the only subsequent link is back to the AMC website.  What other (pre-college) student services and information should MAA offer?[/quote]

The number-one information I want as a homeschooling parent is reviews of SCHOOL MATHEMATICS (that is, K-12 mathematics) books from the perspective of real-world mathematicians. It is apparent that most United States mathematics instruction is regrettably lousy--I note, as one datum, that my home state of Minnesota had NO perfect scorers on last year's AMC 8, while the distinctly average Chiang Tsui Junior High School in Taiwan, which my niece attends, had three perfect scorers. If the kids in Taiwan going to average schools are able to ace a test of mathematics given in the English language, while the kids in Minnesota can't, the schools in Minnesota aren't trying hard enough in math instruction. (I don't expect you to publicly reply to this comment, as that might be politically dangerous for you  :oops:  but this is why I feel impassioned about reform of math education.) I loved math as a kid, but I didn't get very good teaching in it (hence my signature tagline from W. W. Sawyer). I now feel it is less risky to mostly educate my children myself (the Chinese language major dad with the piano performance major mom teaching their children early elementary math) rather than to send my kids to school for math instruction.

So, to specifically answer your question, I think MAA ought to have a panel of expert reviewers review all the major K-12 math curricula, especially all those that are used among parents who homeschool or "afterschool" (supplement at home) and let parents know what the good textbooks are. Again, the problem here is political. And I acknowledge that not all learners thrive with the same kinds of materials. But I think too few parents know of the better materials (e.g., the Primary Mathematics series published in Singapore) and it would be good to have a "one-stop-shopping" place for this kind of information that is less tendentious than Mathematically Correct, less commercial than Singaporemath.com, and includes more different curricula described in PERMANENT pages than the excellent curriculum discussions at Sonlight's Math Forum page

http://www.sonlight.com/forums/math.cgi

Thanks for asking. Welcome to the board, and best wishes for much continued success in the AMC programs.

I'd like to point out he joined and started posting back in September...

I agree with tokenadult on the School Curricurum point because even though I live in a "good" city our education (especially math) is bad. We're not learning things systematically at school so we're on our own. I also think the list of resourses would be a great idea, not only for the AMC but for math in general, though that may be asking for too much...

If you could also write out all the categories in which the problems are in and/or give a few example for each category that'll really help me to study.

One small mistake on the AMC web site: On the What's When page, the column named "2003-2004 Tentative Dates" should be "2004-2005 Tentative Dates", I think.

Ya, i noticed a few things wrong with that section:

first of all, the year, obviously...

second, it says -

AIME --
Tuesday, March 8, 2005 or
Alternate AIME --
Tuesday, March 22, 2004

i am assuming you mean March 22, 2005

From the AMC 10 section:

There is also [b]aspecial[/b] AMC 10P Contest in Taiwan on January 5, 2002.

That's all :D

I'd like to see some practice problems or example problems of some sort on the site, both as a study guide and for people to see what's going to be on the test. You could either use a few questions from old tests, or make up a few, or at least list topics likely to be covered.

Suggested reading materials for learning the techniques necessary for the test would also be great - especially if some of it was available online for free.

Message boards or maybe a "problem(s) of the week/day" email (like the SAT one) would be cool. It might be a lot of work, but I think it'd be worth it for the kiddies.

There already are sample problems, you just have to look a little:

[url]AMC10[/url]
[url=http://www.unl.edu/amc/b-registration/b1-archive/2001-2002/02bro1012/02brohtml1012/1012bro2002quest.html]2002 Samples[/url]
[url=http://www.unl.edu/amc/a-activities/a7-problems/practice1012p/02practice.html]2002 Practice Questions[/url]
[url=http://www.unl.edu/amc/b-registration/b1-archive/2002-2003/03bro1012/03bro1012html/03bro-10que.html]2003 Practice Questions[/url]
[url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2002-3pub/03tm12/03amc1012tm.html]2003 Student Worksheet[/url]

[u]AMC12[/u]
[url=http://www.unl.edu/amc/b-registration/b1-archive/2002-2003/03bro1012/03bro1012html/03bro-12que.html]2003 Samples[/url]
[url=http://www.unl.edu/amc/b-registration/b1-archive/2001-2002/02bro1012/02brohtml1012/1012bro2002quest.html]2002 Samples[/url]
[url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/1999-12a/99ahsq/99ahsque.htm]1999 Samples[/url]
[url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/1998-12a/98ahsq/98ahsque.htm]1998 Samples[/url]
[url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/1997-12a/97hsquest/97ahsque.htm]1997 Samples[/url]

[u]AIME[/u]
[url=http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/1999-aa/99aimequest/99aimque.htm]1999 Samples[/url]
[url=http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/1998-aa/98aimeq/98aimque.htm]1998 Samples[/url]
[url=http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/1997-aa/97aimequest/97aimque.htm]1997 Samples[/url]

[url=http://www.unl.edu/amc/a-activities/a7-problems/problemarchive.html]USAMO[/url]

[url=http://www.unl.edu/amc/a-activities/a7-problems/problemarchive.html]1989-2003 Problems[/url]

[url=http://www.unl.edu/amc/a-activities/a7-problems/problemarchive.html]Team Selection Test[/url]

[url=http://www.unl.edu/amc/a-activities/a7-problems/problemarchive.html]IMO 1986-2003[/url]

[url=http://www.unl.edu/amc/a-activities/a7-problems/problemarchive.html]APMO[/url]

However, I agree that it'd be nice to have at least one full AMC test on the website so that people know exactly what the competition is like.

As for message boards, you could link to the AMC board at http://www.artofproblemsolving.com/wforum/.

The group of people who write the AMC 12 and 10 tests each year must have a large number of problems left over, probably enough to make assemble an entire test. It would be very nice if they could post these problems on the website.


It is said that there are 12 Chinese got gold in the 2004 Athens IMO,that's a good news,now how many Chinese got gold in the 2005 Mexico IMO,hope more than 12. :)

10

1 from Canada  :D
5 from China
1 from Hong Kong
3 from Taiwan


Prove that there is no integral triple satisfying $ x^3 \plus{} y^3 \plus{} z^3 \equal{} 2003$.

[hide="Hidden by mod"]take the equation mod 9

every cubic residue in mod 9 is either 0 1 or -1 but 2003 si 5 mod 9[/hide]

@stevenmeow: please hide your solutions, no matter how short. Other people would have liked to solved this problem.


Hurt=-2 and heal=+1

Sony - 21
Activision - 20
THQ - 20
EA Games - 20
LUCASARTS - 18

These obviously aren't all of them, just five that are common.

Another Hurt and Heal? Jesus - I might ban all Hurt and Heals in the Fun Factory if this keeps up...

And you forgot Bungie, which makes me mad. You don't want an angry mod.  :P

Therefore, I shall Hurt and Heal two companies-

Sony - 21
Activision - 22
THQ - 18
EA Games - 20
LUCASARTS - 18
Bungie - 22

(Activision for their Call of Duty 4, and Bungie for Halo, obviously).

Heal is only +1

Sony - 21
Activision - 21
THQ - 18
EA Games - 20
LUCASARTS - 18
Bungie - 21

;

Sony - 22
Activision - 21
THQ - 18
EA Games - 20
LUCASARTS - 16
Bungie - 21

[quote="7h3.D3m0n.117"]Another Hurt and Heal? Jesus - I might ban all Hurt and Heals in the Fun Factory if this keeps up...[/quote]
Could you? Please?

[quote="7h3.D3m0n.117"][quote="Brut3Forc3"]Could you? Please?[/quote]

Hmm, I'm having a bad day actually, so from here on out, I decree that any more Hurt and Heal variations will be deleted on the spot.[/quote]

Maybe a subforum?

[quote="mewto55555"]Maybe a subforum?[/quote]

Nope. The only reason why Mafia was a subforum was because it involves [url=http://dictionary.reference.com/search?q=strategy]strategy[/url].

:rotfl: An actual definition! THANK YOU! :D

I agree, the hurts and heals are getting to be overwhelming. If anyone can think of at least 5 things that would be in G&FF that are NOT Hurt & Heal, it would be appreciated.

P.S. Idea: Contain all the hurt+heal into one topic and do one at a time.

yes but that would most likely reduce to spams because people woul think their topic is better than the one currently up and just post theirs anyway. I say ban all hurt and heals because it doesn't encourage discussion. If you really want to know which is better or worse simply encourage a discussion.

Sony - 22
Activision - 21
THQ - 18
EA Games - 20
LUCASARTS - 16
Bungie - 21

We're kinda getting off topic. Please stick to the game. I didn't make any changes, because nobody else did.

This is really boring, along with the rest of the hurt-and-heal games.

Lock and finish it up.

Sony - 18
Activision - 21
THQ - 18
EA Games - 20
LUCASARTS - 16
Bungie - 22
Nintendo - 21

[quote="7h3.D3m0n.117"]But I like video games :|

Sony - 20
Activision - 21
THQ - 18
EA Games - 20
LUCASARTS - 16
Bungie - 22[/quote]

isnt that a bit biased....

I love video games too but sometimes you gotta let go of what you love  :(

Sony - 19
Activision - 21
THQ - 18
EA Games - 20
LUCASARTS - 16
Bungie - 20
Nintendo - 21

Shh!!! CCM, don't give him any bad ideas.

LOCK THIS UP!  LOCK THIS UP!  LOCK THIS UP!

Sony - 19
Activision - 21
THQ - 18
EA Games - 20
LUCASARTS - 16
Bungie - 20
Nintendo - 21

Shh!!! ernie, don't give him any bad ideas.

Sony - 19
Activision - 21
THQ - 18
EA Games - 20
LUCASARTS - 16
Bungie - 21
Nintendo - 19

[quote="Wickedestjr"]Sony - 19
Activision - 21
THQ - 18
EA Games - 20
LUCASARTS - 14
Bungie - 21
Nintendo - 20

Get rid of LUCASARTS, that's what I'm getting rid of.[/quote]

Well, is anybody going to help me finish this?

[quote="Wickedestjr"]Well, is anybody going to help me finish this?[/quote]

The vast majority of AoPS isn't involved in video games. Or if they are, they're Wii fans (ew).

[quote="crazychinesemaniac"]I agree XBox 360 and PS3 are way better. They have the good games. Wii games are usually fun to beguin with but end up getting boring after a while.[/quote]

True, but there are some games that last.

Please help me finish this hurt and heal, or we will assume that people like Sony the best.


Let $ a,b,c$ be nonnegative real numbers, no two of them are zero. Prove that:

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

$ \geq \frac {13}{4} \plus{} \frac {15(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{4(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2} \minus{} \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$

:)

It isn't the strongest. Here is the stronger, my friend:
$ (ab \plus{} bc \plus{} ca)(\frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \plus{} \frac {1}{(a \plus{} b)^2}) \geq\ \frac {13}{4} \plus{} \frac {15(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{4(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2} \minus{} \frac {64a^2b^2c^2}{(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2}$
We can solve it easily by expand  :)


1.a)Find with proof all pairs $(x,y)$ (positive integers) such that the last two digits of $x^{2}$ are reversed in $y^{2}$. i.e, $x^{2}=abcd$, $y^{2}=abdc$; where a,b,c, and d are the digits.
[hide="extension"]b)Find with proof all pairs $(x,y)$ (positive intgers) such that besides the last two digit of $x^{2}$ (which are reversed in $y^{2}$), the rest of the digits of $x^{2}$ are the same as $y^{2}$.[/hide]
[hide="another extension"]c)Find with proof all pairs $(x,y)$ such that x and y are positive consecutive integers.[/hide]
:) Thanks for the editting Gyan

There's some mistake in the statement c. Please review it.

No, ther is nothing wrong with it. Please elaborate.

[quote="Farenhajt"]To find all the pairs of positive consecutive integers? No forum is big enough to list infinitely many of them...[/quote]
Oh, thats what you meant. Keep in mind part a when answering the others. The base question is a, b and c are variations of it.

[quote="Gyan"][quote="Farenhajt"]To find all the pairs of positive consecutive integers? No forum is big enough to list infinitely many of them...[/quote]
:)
But even for "a" part - You still have this  problem of too many answers ...
(10,20,30,40 ....) take any two of them as (x,y) :)
Or (12,112,212,312, 56712 56612 ...) , for that matter.
May be its, time to restate the problem, explicitly stating, all the conditions.[/quote]
I forgot to mention the last two digits can't be the same. Thanks for catching that Gyan. :)

I thought the solution is clearly finite, since it is limited by x^2 being a 4-digit number, further the 1st 2 digits of x^2 and y^2 are the same. There is no problem of too many solutions but its pretty easy to single them out by exhaustive method.

[quote]I forgot to mention the last two digits can't be the same. Thanks for catching that Gyan[/quote]
Trouble is, even with that condition,  there are still infinite solutions . (and all fairly trivial) .... unless you reword the problem  to make it clear it will cause confusion.

[quote="Ahduhduh"]I thought the solution is clearly finite, since it is[b] limited by x^2 being a 4-digit number[/b],....[/quote]
I don't believe anything of the kind  is stated. ... the original problem:

[quote]Find with proof all pairs $(x,y)$ (positive integers) such that the last two digits of $x^{2}$ are reversed in $y^{2}$.
 [/quote]

(One could easily put something like "$(x^{2})$  is four digit number" as a  condition ... I know, the example shows    'abcd' .. but still it is not clear if  that it is just an example or  part of the statement of the problem. ..

Hope that helps....

[quote="Gyan"]
(One could easily put something like "$(x^{2})$  is four digit number" as a condition ... I know, the example shows 'abcd' .. but still it is not clear if that it is just an example or part of the statement of the problem...[/quote]
The 4 digits were just part of the example, the number it self can be however long.
[quote="Farenhajt"][quote="Gyan"][quote]I forgot to mention the last two digits can't be the same. Thanks for catching that Gyan[/quote]
Trouble is, even with that condition,  there are still infinite solutions.[/quote]
I don't think there are infinitely many solutions, and here's
[hide="why"]$x^{2}-y^{2}=\overline{cd}-\overline{dc}=9(c-d)$, but $c,d$ being digits, the set of possible values of $c-d$ is finite, hence after factorizing $(x-y)(x+y)$ we get only a finite number of solutions. The casework is tedious though.[/hide][/quote]
Thanks Farenhajt. I hope the question is clearer now.

Sorry to nit-pick  and point out the obvious, but for [b]part b[/b]
[quote]b)Find with proof all pairs $(x,y)$ (positive intgers) such that besides the last two digits, the rest of the digits are the same.[/quote]
( or other  such conditions) .....  may be  :)

But if you  just want to reverse last two digits (with no condition on other digits) ...you can have (31,4),(31,104),(31,204) ...(131,4).. as many as you please ...:)

IMO it is much easier to state the problem in such a way .. one can't even play these games..:) ...

May be add a  part like " ... except for the last two digits, which are reversed... all digits of $x^{2}$ are identical with $y^{2}$" ... or something like that...

(

$4^{2}=16$ and $31^{2}=916$
I stated that all numbers beside the last two are the same (which we know are reversed). we can write 16 as 016. Comparing to 961, 9 and 0 are obvious distinct. I'm not sure what you mean.

Maybe you can rewirte the whole problem, and i'll see if it still has the same meaning. :)

[quote][b]I stated that all numbers beside the last two are the same [/b][/quote]
I didn't see where you mentioned that, (As, I said, I   see it was stated  before - only for extension b) but not for a) )
In any case, I think   that, makes the problem   clear.
(Sorry for diversion, i will edit/delete  all my posts out ..  shortly as soon as you edit the original problem and add the line)  thanks.

[quote="Inspired By Nature"]
b)Find with proof all pairs $(x,y)$ (positive intgers) such that besides the last two digits, [b]the rest of the digits are the same[/b].[/quote]
Well I did write it, I guess it isn't clear enough.
[quote="Gyan"][quote][b]I stated that all numbers beside the last two are the same [/b][/quote]
I didn't see where you mentioned that, (As, I said, I   see it was stated  before - only for extension b) but not for a) )[/quote]
The condition [b]is[/b] only for b, not a.

Any other confusions, mistakes, errors, etc. Please tell me. :)


At Frank's Fruit Market, $ 3$ bananas cost as much as $ 2$ apples, and $ 6$ apples cost as much as $ 4$ oranges. How many oranges cost as much as $ 18$ bananas?

$ \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$

[quote="worthawholebean"]At Frank's Fruit Market, $3$ bananas cost as much as $2$ apples, and $6$ apples cost as much as $4$ oranges. How many oranges cost as much as $18$ bananas?

$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 18$[/quote]
[hide]Make equations. So we have $3b=2a$ and $6a=4o$. So $a=\frac{3}{2}b$. Plugging this in to the second equation we get that $9b=4o$. Since its asking for 18 bananas we doubled to get that so its B.[/hide]

[hide="Answer"]Comparing ratios, we get bananas:apples:oranges = 9:6:4. If the number of bananas is 18, 9:4  = 18:8, or B.[/hide]


function algebra domain parameterization conics parabola geometry

Hello,

In my Intermediate Algebra book, it says that "...variations of the square root function are of the general form, g(x) = a + b sqrt(cx + d)."

The square root function is f(x) = sqrt(x) so I would like to know where this variation came from.

Thank you.

I'm not sure I really understand your question.  Basically, the graph of a variation on a square root function is the graph of the square root function transformed in some way, like stretched or shifted, for example.

We have

$ y = a\sqrt{b(x-h)}+k$ where:

(h,k) is the vertex of the function
the value of a is a vertical expansion factor
the value b is a horizontal compression factor
the sign of a and b determine whether the function will increase or decrease according to this pattern:

if a > 0 and b > 0, the function increases over its entire domain. However, making either a < 0 or b < 0 reverses this pattern. Obviously, making both < 0 would make the function increase over its entire domain as well.

Does that clarify the issue? Keep in mind that any function can be written in the form

$ y = a \{ b(x-h) \}+k$

where $ \{ \}$ are the delimiters that define the function. For instance, the $ | \; |$ delimiters represent the absolute value function.

Hello there,

Thank you for your replies. I will rephrase my question.

I know that the square root function is f(x) = sqrt(x). The book says that "variations of the square root function are of the general form, g(x) = a + b sqrt(cx + d)".

My question is, where did g(x) = a + b sqrt(cx + d) come from?

That's the same form as I stated, just using different variables. The formula for a function can be transformed geometrically (as I mentioned before) which is where all the extra parameters come into play.

As an analogy, the square function $ f(x) = x^{2}$ has the variations $ f(x) = ax^{2}+bx+c$.  These are also known as parabolas.  The graph of any parabola is essentially the graph of $ f(x) = x^{2}$ shifted appropriately, and then stretched or squeezed horizontally or vertically as appropriate.

The "variations of the square root function" are, similarly, all the different shifts, stretches, and squeezes of the graph of $ f(x) = \sqrt{x}$ (which, by the way, if you graph both the negative and the positive square roots, is a sideways parabola).  The graphs of $ g(x) = a+b \sqrt{cx+d}$ are general sideways parabolas.  (And if you're wondering why there are four parameters in there, the parameter $ b$ is not necessary.)

In other words, the square root function is part of the inverse function of a parabola. According to the general form, simply interchange the x and y variables.

$ x= ay^{2}+by+c$

$ x-c = ay^{2}+by$

$ \frac{x-c}{a}= y^{2}+\frac{b}{a}y$

$ \frac{x-c+d}{a}= y^{2}+\frac{b}{a}y+d$ where $ d$ allows completion of the square. substitute $ \frac{1}{a}$ by $ g$

$ gx-gc+gd = (y+e)^{2}$ where $ e = \frac{b}{2a}$ now substitute  $ gc+gd$ by $ f$

$ gx+f = (y+e)^{2}$

$ |y+e| = \sqrt{gx+f}$

$ y+e = \pm \sqrt{gx+f}$

$ y = \pm \sqrt{gx+f}-e$ substitute $ -e$ by $ m$ and ignore one of the $ \pm$ signs to get a function instead of a relation and we have

$ \boxed{y = \sqrt{gx+f}+m}$

now just substitute the variables I used for the ones used in your book and you're done.

Thank you very much t0rajir0u and Prime factorization for your detailed, helpful replies.

I just have one last question: could you please state what each variable denotes? (a, b, c, and d)

Thank you again!

$ a$ is the vertical translation factor, $ \frac{d}{c}$ is the horizontal translation factor, $ b$ is the vertical expansion factor and $ c$ is the horizontal compression factor. Or, you could say that $ \frac{1}{c}$ is the horizontal expansion factor. I already covered in my first post on this topic what would happen if the sign of $ b$ and $ c$ are either positive or negative.

Hope this helps.

The vertex is located at the point $ \left( a,-\frac{d}{c}\right)$. That's what those variables are for.

The $ b$ variable is a little tricky. Basically, it controls the "squish" factor. For smaller values of $ b$, the graph will be very, very squished. Notice that if $ b$ is $ 0$, the graph is just a horizontal line! You can think of a line as having infinite "squishness". For larger and larger values of $ b$, the graph will become less and less squished (or, you can think of it as being more stretched). In fact, it will actually start approaching a vertical line (this is very hard to imagine, but a vertical line has infinite "stretchedness")!

You can actually click on the picture for more high resolution.

The picture is a good example. We have:
$ a=-3$
$ b=0.5$
$ c=2$
$ d=5$

Note that the vertex is $ \left(-3,\frac{5}{2}\right)$. The $ b=0.5$ is pretty small, so the graph is pretty squished.

Personally, I think the form $ y = a\sqrt{b(x-h)}+k$ is the simpler form because the vertex is simply $ (h,k)$ instead of $ \left(a,\frac{-d}{c}\right)$. You could also use the form $ y = a\sqrt{\frac{(x-h)}{b}}+k$, that way you know if b increases then your graph stretches horizontally.


How many feet longer is 1$ \%$ of a mile than 60$ \%$ of a yard?

$ 1\%$ of a mile is 528 feet.  $ 60\%$ of a yard is 1.8 ft.

Actually $ 1\%$ of a mile is: $ 0.1 \times 5280 \equal{} 52.8$.


Show that for every odd prime p, there exist positive integers u and v such that puv is divisible by 4uv-u-v.

I can prove a lot of cases.  A lot.  I've sat here for some time now just proving cases.

The simplest:

Well, the case where p :equiv: -1 mod 3 is simple.  Let u=1, let v=(p+1)/3.
4uv-u-v=4v-1-v=3v-1=p+1-1=p.  So (4uv-u-v) divides p, and hence divides puv.

Unfortunately, a full proof has escaped me thus far.

The case $3|p+1$ can, I think, be generalized to $p+1$ having a divisor of the form $M_4+3$, but there's still no general solution...

It can be, fairly easily.  That is, if M_4 means a multiple of 4.


Prove: $ \forall k>1$, the equation $ \tau (n)\equal{}k$ has countless solutions.

[hide="Proof"]There are infinitely many primes. (This is very well-known, so this will go without proof.)

All numbers of the form $ p^{k\minus{}1}$ for prime $ p$ has exactly $ k$ divisors. But, there are infinitely many primes, so there are infinitely such numbers for $ k>1$.

[/hide]


Post the image that you want transformed on this topic and I will transform it into a ipod ad avatar!
example: my avatar.
It would help if you come up with an i-thing for me (for example, iFly in my current one.)
Don't expect it instantly, though. I will PM you when it is done.

I guess I'm going to need a new avatar to do this...darn...I wish I still had sauron...

If you want, just tell me character andsource(game, anime, manga he or she is from)
It can be another source, IF you can find it with a solid background.

Solid background isn't an issue, you can always make it solid. If you are using photoshop, my suggestion would be cutting character using pen tool (because it's most precise method I know if you have patience) and placing it on solid bg. That way you can also easily paint your character in black making a shape out of it.

What method do you use, maybe I can help as I have lots of experience in photoshop.

I'm not making a profit off of it, am I?
And, come on, haven't you seen like THOUSANDS of these pics on the net?

I'm guessing you don't do animated avatars. xD

I probably could...
Tell me what you want. I might be able to make it.


algebra
polynomial
Olimpiada de matematicas

HOLA, ACA UN PROBLEMA KE NO ME SALE HACE YA UN MES, PORFAS PONGA LA SOLUCION.
$9^{x}-7^{y}=2$, hallar las soluciones enteras de x,y.

Bueno, aki te pongo mi solucion, aver k te parece:

Analicemos la escuacion modulo 8.
$9^{x}-7^{y}=2 (mod 8)$
$1^{x}-(-1)^{y}=2 (mod 8)$
$(-1)^{y}=-1 (mod 8)$
De esto ultimo sakamos que y es impar. $y=2n+1$
Ahora, analicemos la ecuacion modulo 5.
$9^{x}-7^{y}=2 (mod 5)$
$(-1)^{x}-2^{y}=2 (mod 5)$
Si $x es par$, entonces:
$2^{y}=-1 (mod 5)$
 Analizando las posibles congruencias de $2^{y}$ recordando que y es impar, tengo que ninguna cumple la condicion.
Entonces $x es impar$
Ya sabiendo que $x$ y $y$ son impares, tenemos que:
$9^{x}-7^{y}=(9-7)(9^{x-1}+...+7^{x-1})=2$
De esta ultima factorizacion, vemos que si $x,y>1$ el factor derecho $>1$ entonces la unica solucion es $x=y=1$

Oye, una disculpa en la solucion. No me habia fijado que la ultima factorizacion no es correcta!!! Todo lo demas esta bien.
Lo ice de otra manera, es muy bonita, te pongo la idea.
Observa que la ecuacion es igual a:
$9(9^{x-1}-1)=7(7^{y-1}-1)$
y de eso oserva qe pasa cuando $x, y>1$ sabiendo que $x, y$ son impares

SI $x,y$ son positivos mayores que 1 es claro que esto es mayor a dos, ahora si uno es positivo y el otro negativo entonces la cosa que me queda es negativa en el caso de $y$ positivo y $x$ negativo, ya que $9^{x}$ con $x$ negativo es menor a uno, tambien observamos que si $x$ positivo y $y$ negativo entonces la cosa se pasa de 2, por lo que $x,y$ son negativos o alguno de los dos igual a 1,0 por separado, analizando por separado el de 1,0 tenemos que la unica que lo cumple es $x=1,y=1$; ahora solo falta demostrar para negativos pero para negativos es claro ya que $9^{x}$ negativo$<1$ y $7^{y}$ negativo$<1$ por lo que $9^{x}-7^{y}$ con $x,y$ negativos
es menor a 1, por lo que las unicas soluciones son $x=1$ y $y=1$.
jeje :lol:

[quote="Tony2006"]SI $x,y$ son positivos mayores que 1 es claro que esto es mayor a dos[/quote]
mmm :wink: , tu tomas esa parte como obia, lo cual no es asi, podrias ser mas riguroso, y poner la demostracion de este paso ya q a mi parecer de eso depende el problema.
conejita no se si podrias corregir tu demostracion en esa factorisacion, yo tb intente poor modulos mas de una ves y no llegue a nada. :maybe:
chaufas :wink:

Oye pues usar modulos si te ayuda, en mi caso encontramos que tanto $x,y$ son impares!!!!  Y con eso, y usando la ultima formula que te di, ya sale!!!!  Te repito k lo unico incorrecto en mi solucion anterior, es la ultima factorizacion!!
Tony2006, no es tan obvio decir que como son p ositivos $>1$entonces es $claro$que la suma es $>2$!! Hay que demostrarlo, y alli se encuentra el problema!!!

hola conejitaa, tranki ok, me refiero a k yo no llegue a mucho, llegue a k
$x,y$ son impares lo cual es cierto, tambien llegue ke $Y$ es congruente a 1 modulo 24 ,  :rotfl:  :rotfl:  :rotfl:  aunke no paresca.
pues estoy esperando conejita, termina de postear tu solucion.
chaufas

:mad: bueno aca les va  la solución, http://www.mathlinks.ro/Forum/viewtopic.php?p=629498#629498, ta en ingles, la termine  :D
chaufas

Hey aev5peru si supiera tu nombre me referiria a ti por tu nombre .

Bueno en la solucion que posteas en el link la verdad que hay muchas cosas que afirmas y que estan mal .

Me gustaria saber en donde aparecio dicho problema porque la verdad que parece bastante dificil y dejame decirte que a mi aun no me ha salido . Por cierto no lo he intentado mucho tiempo .

Podria proponer dos problemas del mismo tipo pero con diferente tecnica de solucion .

Por ejemplo :

1) $3^{x}-2^{y}=1$ ( técnica de la diferencia de cuadrados )

Aqui la solucion es (1,1) y (2,3) es facil probar que para y mayores que 3 ; x debe ser par y con una diferencia de cuadrados se termina la solucion

2) $7^{x}-3^{y}=4$ ( tecnica de resto gaussiano y encontrar un $(pr+1)^{pt}-1$=multiplo de $p^{2}$ ; p es primo )

Aqui en la solucion mediante una serie de restos gaussianos se puede probar que : y = 6r+1 luego que x = 12s+1 y despues d eso que $7^{12s}-1$ es multiplo de 9 y ali termina la solcuion

3) $2.5^{x}-7^{y}=1$
Este problema no me ha salido pero la situacion si que es dificil
parace que la unica solucion es x=2 e y=2 .

Y es que cuando las unicas soluciones son x=2 e y=2 y a su vez es dificil buscar una diferencia de cuadrados solo me queda confiar en mi tecnica numero 2 que por cierto no se si pueda funcionar .

Y es que el problema $9^{x}-7^{y}=2$ es un caso particular del problema $3^{x}-7^{y}=2$ en donde la única solucion es x=2 e y=1 y a su vez es muy pareciod en la forma al problema # 3 que propuse .

Jaja que lio si alguien le da solucion a este problema por favor posteelo que ya me dio curiosidad .

[quote="DarthTenebrus"]
1.-Hey aev5peru si supiera tu nombre me referiria a ti por tu nombre .

2.-Bueno en la solucion que posteas en el link la verdad que hay muchas cosas que afirmas y que estan mal .

3.-Me gustaria saber en donde aparecio dicho problema porque la verdad que parece bastante dificil y dejame decirte que a mi aun no me ha salido . Por cierto no lo he intentado mucho tiempo .
.[/quote]HOLA ME GUSTA SER PUNTUAL ASI QUE:
1.-me llamo amilcar
2.- un favor ps, me podrias decir donde ta los errores.  :blush: porfa.
3.- el problema lo vi en un libro de math-team, que ahora es cjeo-peru.
chaufas y porfa posteme los errores te lo agradeceria.

Hey amilcar cula es tu apellido . y si em podria decir de donde eres osea donde estuidas bueno al grano

Tiene errores en la congruencia gaussiana que usas porque concluyes x=2m+1 ; y =2n+1 cuando solo se pdoria concluir
x=3m+1 ; y = 3n+1 pero eso es leve porque de oalgua manera se pdoria conseguir eos que dices pero lo que vien esta sinceramentte tirado de los cabellos

eso de que :

 be the polynomial $P(x)=7x^{2}+2x-9k^{2}-2k$,
so that it has rational roots => $7$it divides $-9k^{2}-2k$, of ..(1). this does not exist., x=n, $(\to <-)$.

no tiene sentido alguno . Bueno yo todavia sigo a la espera de la solucion correcta.

OE DISCULPA, :blush:  JAJAJA NO SE EN K TABA PENSANDO TALVES LA EMOCION NO??, JEJEE BUENO A LO Q ME REFIERO ES QUE DE AHI KEDA UNA CUADRATICA ENTONCES ANALISANDO DISCRIMINANTE TIENE K SER CUADRADO PERFECTO, DESPS ACOTALO, ENTRE 2 CUADRADOS, ESO FUE LO Q ISE EN EL CARRO , PERO NO RECUERDO COMO,  :(  BUENO GRACIAS AMIGO, NO SE KE ME PASO, FUE UN LAPSUS.
CHAUFAS. Y AVER SI LO TERMINO.

No creo que sea asi como dices ya que la suma de las raices en esa cuadratica es -2/7 mas bien si la discriminante seria un cuadrado entonces seria el cuadrado de un numero racional pero aun asi reparando eso quedaria algo horrible de analizar asi nomas bueno eso era todo lo que queria decir...bye

bueno, que paso con las soluciones de este problema, porfa, es un trauma, ya voy 4 meses intentando.
espero que la posteen
bye

[color=blue]Observen este sitio:

[url]http://groups.google.com.py/group/es.ciencia.matematicas/browse_thread/thread/365b0d6b82871c8e/754a5601eb1d82a6?hl=es#754a5601eb1d82a6[/url]

Saludos.[/color]


Let $ a,b,c>0$ .Prove that:
$ \frac {a}{{b \plus{} c}} \plus{} \frac {b}{{c \plus{} a}} \plus{} \frac {c}{{a \plus{} b}} \plus{} \frac {1}{2} \ge \frac {{{a^2} \plus{} {b^2} \plus{} {c^2}}}{{ab \plus{} bc \plus{} ca}} \plus{} \frac {{ab \plus{} bc \plus{} ca}}{{{a^2} \plus{} {b^2} \plus{} {c^2}}}$

[quote="leedt26"]Let $ a,b,c > 0$ .Prove that:
$ \frac {a}{{b \plus{} c}} \plus{} \frac {b}{{c \plus{} a}} \plus{} \frac {c}{{a \plus{} b}} \plus{} \frac {1}{2} \ge \frac {{{a^2} \plus{} {b^2} \plus{} {c^2}}}{{ab \plus{} bc \plus{} ca}} \plus{} \frac {{ab \plus{} bc \plus{} ca}}{{{a^2} \plus{} {b^2} \plus{} {c^2}}}$[/quote]

$ LHS \minus{} RHS \equal{} \frac {1}{2}\sum{\left(\frac {1}{(b \plus{} c)(c \plus{} a)} \minus{} \frac {1}{ab \plus{} bc \plus{} ca} \plus{} \frac {1}{a^2 \plus{} b^2 \plus{} c^2}\right)(a \minus{} b)^2}$

 :(

how to prove it $ \ge 0$? ... can S.O.S theorem sovle it?  :oops:

or it can give another SOS form? ... :maybe:

er... expanding can prove it

after expanding, ineq becomes

$ [4,2,0]\plus{}2[4,1,1] \ge [3,3,0]\plus{}[3,2,1]\plus{}[2,2,2],$

by Muirhead Inequality, we have $ [4,2,0] \ge [3,3,0]$ and $ [4,1,1] \ge [3,2,1]$ and $ [4,1,1] \ge [2,2,2]$, adding them together, get proved.  :huh:

[quote="leedt26"]Let $ a,b,c > 0$ .Prove that:
$ \frac {a}{{b \plus{} c}} \plus{} \frac {b}{{c \plus{} a}} \plus{} \frac {c}{{a \plus{} b}} \plus{} \frac {1}{2} \ge \frac {{{a^2} \plus{} {b^2} \plus{} {c^2}}}{{ab \plus{} bc \plus{} ca}} \plus{} \frac {{ab \plus{} bc \plus{} ca}}{{{a^2} \plus{} {b^2} \plus{} {c^2}}}$[/quote]
AM-GM works here  :)

We know by Nesbitts that: $ \sum\frac{a}{b\plus{}c} \geq\frac{3}{2}$

so, it can be written as $ 2\geq \frac{\sum a^2}{\sum ab} \plus{} \frac{\sum ab}{\sum a^2}$

Since $ \sum a^2 \geq \sum ab$, we can cancel out first term of RHS to get 1.

then, to prove is $ 1 \geq \frac {\sum ab}{\sum a^2}$     which is true from the same logic as above.

[quote="analytic"]We know by Nesbitts that: $ \sum\frac {a}{b \plus{} c} \geq\frac {3}{2}$

so, it can be written as $ 2\geq \frac {\sum a^2}{\sum ab} \plus{} \frac {\sum ab}{\sum a^2}$

Since $ \sum a^2 \geq \sum ab$, we can cancel out first term of RHS to get 1.

then, to prove is $ 1 \geq \frac {\sum ab}{\sum a^2}$     which is true from the same logic as above.[/quote]

YOU ARE WRONG

Try to find your faults....

Dimitris

[quote="analytic"]We know by Nesbitts that: $ \sum\frac {a}{b \plus{} c} \geq\frac {3}{2}$

so, it can be written as$ 2\geq \frac {\sum a^2}{\sum ab} \plus{} \frac {\sum ab}{\sum a^2}$

Since $ \sum a^2 \geq \sum ab$, we can cancel out first term of RHS to get 1.

then, to prove is $ 1 \geq \frac {\sum ab}{\sum a^2}$     which is true from the same logic as above.[/quote]
  $ 2\le \frac {\sum a^2}{\sum ab} \plus{} \frac {\sum ab}{\sum a^2}$ by AM$ \ge$GM
so i think ur solution is wrong  :oops:

Indeed, analytic's proof was wrong but of course, we can use AM-GM to solve this problem but with the another hint. You should try it again  :)

[quote="leedt26"]Let $ a,b,c > 0$ .Prove that:
$ \frac {a}{{b \plus{} c}} \plus{} \frac {b}{{c \plus{} a}} \plus{} \frac {c}{{a \plus{} b}} \plus{} \frac {1}{2} \ge \frac {{{a^2} \plus{} {b^2} \plus{} {c^2}}}{{ab \plus{} bc \plus{} ca}} \plus{} \frac {{ab \plus{} bc \plus{} ca}}{{{a^2} \plus{} {b^2} \plus{} {c^2}}}$[/quote]
Notice that
\[ \frac{a}{b\plus{}c}\minus{}\frac{a^{2}}{ab\plus{}bc\plus{}ca}\equal{}\frac{abc}{(b\plus{}c)(ab\plus{}bc\plus{}ca)}\geq \frac{abc}{(b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2})}.\]
Therefore, it suffices to prove that
\[ \sum \frac{abc}{(b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2})}\plus{}\frac{1}{2}\geq \frac{ab\plus{}bc\plus{}ca}{a^{2}\plus{}b^{2}\plus{}c^{2}},\]
or
\[ 2abc\left( \frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\plus{}\frac{1}{a\plus{}b}\right)\plus{}a^{2}\plus{}b^{2}\plus{}c^{2}\geq 2(ab\plus{}bc\plus{}ca).\]
By the Cauchy-Schwarz Inequality, we have
\[ \frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\plus{}\frac{1}{a\plus{}b}\geq \frac{9}{(b\plus{}c)\plus{}(c\plus{}a)\plus{}(a\plus{}b)}\equal{}\frac{9}{2(a\plus{}b\plus{}c)}.\]
So, it is enough to check that
\[ \frac{9abc}{a\plus{}b\plus{}c}\plus{}a^{2}\plus{}b^{2}\plus{}c^{2}\geq 2(ab\plus{}bc\plus{}ca),\]
which is Schur's Inequality.

[quote="can_hang2007"][quote="leedt26"]Let $ a,b,c > 0$ .Prove that:
$ \frac {a}{{b \plus{} c}} \plus{} \frac {b}{{c \plus{} a}} \plus{} \frac {c}{{a \plus{} b}} \plus{} \frac {1}{2} \ge \frac {{{a^2} \plus{} {b^2} \plus{} {c^2}}}{{ab \plus{} bc \plus{} ca}} \plus{} \frac {{ab \plus{} bc \plus{} ca}}{{{a^2} \plus{} {b^2} \plus{} {c^2}}}$[/quote]
Notice that
\[ \frac {a}{b \plus{} c} \minus{} \frac {a^{2}}{ab \plus{} bc \plus{} ca} \equal{} \frac {abc}{(b \plus{} c)(ab \plus{} bc \plus{} ca)}\geq \frac {abc}{(b \plus{} c)(a^{2} \plus{} b^{2} \plus{} c^{2})}.\]
Therefore, it suffices to prove that
\[ \sum \frac {abc}{(b \plus{} c)(a^{2} \plus{} b^{2} \plus{} c^{2})} \plus{} \frac {1}{2}\geq \frac {ab \plus{} bc \plus{} ca}{a^{2} \plus{} b^{2} \plus{} c^{2}},\]
or
\[ 2abc\left( \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {1}{a \plus{} b}\right) \plus{} a^{2} \plus{} b^{2} \plus{} c^{2}\geq 2(ab \plus{} bc \plus{} ca).\]
By the Cauchy-Schwarz Inequality, we have
\[ \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {1}{a \plus{} b}\geq \frac {9}{(b \plus{} c) \plus{} (c \plus{} a) \plus{} (a \plus{} b)} \equal{} \frac {9}{2(a \plus{} b \plus{} c)}.\]
So, it is enough to check that
\[ \frac {9abc}{a \plus{} b \plus{} c} \plus{} a^{2} \plus{} b^{2} \plus{} c^{2}\geq 2(ab \plus{} bc \plus{} ca),\]
which is Schur's Inequality.[/quote]

very nice!

my solution:
(i)if $ a^2\plus{}b^2\plus{}c^2\geq 2(ab\plus{}bc\plus{}ca)$,then $ \frac{1}{2}\geq\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}$,so it is suffice to show
\[ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\geq\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\]
By Cauchy we have
\[ LHS\geq\frac{(a^2\plus{}b^2\plus{}c^2)^2}{a^3(b\plus{}c)\plus{}b^3(c\plus{}a)\plus{}c^3(a\plus{}b)}\]
we need to prove that
\[ (a^2\plus{}b^2\plus{}c^2)(ab\plus{}bc\plus{}ca)\geq\sum{a^3(b\plus{}c)}\Longleftrightarrow abc(a\plus{}b\plus{}c)\geq 0\]
(ii)if $ a^2\plus{}b^2\plus{}c^2\leq 2(ab\plus{}bc\plus{}ca)$,By Cauchy
\[ \sum{\frac{a}{b\plus{}c}}\geq \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)}\]
it's easy to check the inequality is equalent to
\[ \frac{3}{2}\minus{}\frac{a^2\plus{}b^2\plus{}c^2}{2(ab\plus{}bc\plus{}ca)}\geq\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}\]
it's true with the assumption $ a^2\plus{}b^2\plus{}c^2\leq 2(ab\plus{}bc\plus{}ca)$.
Our problem is done. :)

Nice hint, Tourish :) . It's nicer than my proof. I used my old lenma (similar can_hang's hint)  :)
\[ \sum \frac{a}{b\plus{}c} \ge \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\plus{}\frac{4abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\]

[quote="Tourish"]my solution:
(i)if $ a^2 \plus{} b^2 \plus{} c^2\geq 2(ab \plus{} bc \plus{} ca)$,then $ \frac {1}{2}\geq\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}$,so it is suffice to show
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\geq\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\]
By Cauchy we have
\[ LHS\geq\frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b)}\]
we need to prove that
\[ (a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)\geq\sum{a^3(b \plus{} c)}\Longleftrightarrow abc(a \plus{} b \plus{} c)\geq 0\]
(ii)if $ a^2 \plus{} b^2 \plus{} c^2\leq 2(ab \plus{} bc \plus{} ca)$,By Cauchy
\[ \sum{\frac {a}{b \plus{} c}}\geq \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)}\]
it's easy to check the inequality is equalent to
\[ \frac {3}{2} \minus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{2(ab \plus{} bc \plus{} ca)}\geq\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\]
it's true with the assumption $ a^2 \plus{} b^2 \plus{} c^2\leq 2(ab \plus{} bc \plus{} ca)$.
Our problem is done. :)[/quote]

very good,  :D


I think it should be rather in the number theory section ;).

It's true for $p>3$.
In $Z_p$ we can divide all residues (besides 0) in distinct, pairwise disjoiny pairs $(a,b)$ such
that $a \neq b$ and $ab \equiv 1$ $(\mod p)$ except the pair $(1,p-1)$ we does not satisfie the second condition.
So the set $\{\frac{1}{2^2},\frac{1}{3^2},...,\frac{1}{(p-2)^2} (\mod p) \}$ consists all numbers $i^2$  $(\mod p)$ for $ 2 \leq i \leq p-2$  and $\frac{1}{1^2}=\frac{1}{(p-1)^2}=1^2=(p-1)^2 \equiv 1 (\mod p)$
so $\sum_{i=1}^{p-1} \frac{1}{i^2} = {\frac{p(2p+1)(p+1)}{6} \equiv 0}$  $(\mod p)$.


cảm ơn bạn ráy nhiều , nhưng nó ít quá .Bạn có thể pót thêm không


Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a^2\plus{}b^2\plus{}c^2\equal{}3.$ Prove that:
\[ \frac{a^2}{2a\plus{}b\plus{}c}\plus{}\frac{b^2}{2b\plus{}a\plus{}c}\plus{}\frac{c^2}{2c\plus{}a\plus{}b}\geq\frac{3}{4}\]

From CBS we get
$ [a(2a \plus{} b \plus{} c) \plus{} b(2b \plus{} a \plus{} c) \plus{} c (2c \plus{} a \plus{} b)][\frac {a^2}{2a \plus{} b \plus{} c} \plus{} \frac {b^2}{2b \plus{} a \plus{} c} \plus{} \frac {c^2}{2c \plus{} a \plus{} b}]\geq$
$ \geq(a^2 \plus{} b^2 \plus{} c^2)^2$
This means that $ \frac {a^2}{2a \plus{} b \plus{} c} \plus{} \frac {b^2}{2b \plus{} a \plus{} c} \plus{} \frac {c^2}{2c \plus{} a \plus{} b}\geq \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{2a^2 \plus{} 2b^2 \plus{} 2c^2 \plus{} 2ab \plus{} 2ac \plus{} 2bc}\geq$
$ \geq \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{4a^2 \plus{} 4b^2 \plus{} 4c^2} \equal{} \frac {3}{4}$

You was wrong ^^
$ [a(2a\plus{}b\plus{}c)\plus{}b(2b\plus{}a\plus{}c)\plus{}c\plus{}(2c\plus{}a\plus{}b)][\frac{a^2}{2a\plus{}b\plus{}c}\plus{}\frac{b^2}{2b\plus{}a\plus{}c}\plus{}\frac{c^2}{2c\plus{}a\plus{}b}]\geq (\sqrt{a^3}\plus{}\sqrt{b^3}\plus{}\sqrt{c^3})^2$
^^ :D

Using Cauchy - Schwarz Inequality
$ \sum\frac{a^2}{2a\plus{}b\plus{}c}\equal{}\sum\frac{a^4}{2a^3\plus{}a^2(b\plus{}c)}\geq\frac{(a^2\plus{}b^2\plus{}c^2)^2}{2(a^3\plus{}b^3\plus{}c^3)\plus{}\sum a^2(b\plus{}c)}$
Using S.O.S and by the hypothesis, we get
$ \frac{(a^2\plus{}b^2\plus{}c^2)^2}{2(a^3\plus{}b^3\plus{}c^3)\plus{}\sum a^2(b\plus{}c)}\geq\frac{3}{4}$

[b]Sudoku[/b], I don`t understand what do you want to say with [b]"S.O.S"[/b]  :!:  :?:

[quote="rachid"][quote="Sudoku"]
Using S.O.S and by the hypothesis, we get
[/quote]
what is the meaning of S.O.S?.[/quote]
It means Sums Of Square.
$ S_a(b \minus{} c)^2 \plus{} S_b(c \minus{} a)^2 \plus{} S_c(a \minus{} b)^2\geq 0$, where $ S_a,S_b,S_c$ are expressions of a,b,c. They can be not positive.  :)

Let $ a \plus{} b \plus{} c \equal{} p,\ ab \plus{} bc \plus{} ca \equal{} q,\ abc \equal{} r$, we have $ (a \plus{} b \plus{} c)^2 \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca)\Longleftrightarrow q \equal{} \frac {p^2 \minus{} 3}{2}$,
$ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)$
$ \Longleftrightarrow a^3 \plus{} b^3 \plus{} c^3 \equal{} 3p \minus{} pq \plus{} 3r$.
Thus $ a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b) \equal{} a^2(p \minus{} a) \plus{} b^2(p \minus{} b) \plus{} c^2(p \minus{} c)$
$ \equal{} p(a^2 \plus{} b^2 \plus{} c^2) \minus{} (a^3 \plus{} b^3 \plus{} c^3) \equal{} 3p \minus{} (a^3 \plus{} b^3 \plus{} c^3) \equal{} pq \minus{} 3r$.

$ \therefore \frac {a^2}{2a \plus{} b \plus{} c} \plus{} \frac {b^2}{2b \plus{} c \plus{} a} \plus{} \frac {c^2}{2c \plus{} a \plus{} b} \equal{} \frac {a^2}{p \plus{} a} \plus{} \frac {b^2}{p \plus{} b} \plus{} \frac {c^2}{p \plus{} c}$

$ \equal{} \frac {a^2(p \plus{} b)(p \plus{} c) \plus{} b^2(p \plus{} c)(p \plus{} a) \plus{} c^2(p \plus{} a)(p \plus{} b)}{(p \plus{} a)(p \plus{} b)(p \plus{} c)}$

$ \equal{} \frac {(a^2 \plus{} b^2 \plus{} c^2)p^2 \plus{} \{a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b)\}p \plus{} abc(a \plus{} b \plus{} c)}{p^3 \plus{} (a \plus{} b \plus{} c)p^2 \plus{} (ab \plus{} bc \plus{} ca)p \plus{} abc}$

$ \equal{} \frac {(q \plus{} 3)p^2 \minus{} 2rp}{2p^3 \plus{} pq \plus{} r}\ \Leftarrow 2q \equal{} p^2 \minus{} 3$

$ \equal{} \frac {p^4 \plus{} 3p^2 \minus{} 4rp}{5p^3 \minus{} 3p \plus{} 2r}$

It is suffice to prove that $ 4(p^4 \plus{} 3p^2 \minus{} 4rp)\geq 3(5p^3 \minus{} 3p \plus{} 2r)\Longleftrightarrow \boxed{4p^4 \minus{} 15p^3 \plus{} 12p^2 \plus{} 9p\geq 2(8p \plus{} 3)r}$.

Now $ (a \plus{} b \plus{} c)^3\geq 27abc\Longleftrightarrow 0 < r\leq \frac {1}{27}p^3$ and $ 3(a^2 \plus{} b^2 \plus{} c^2)\geq (a \plus{} b \plus{} c)^2\Longleftrightarrow 0 < p\leq 3$,

Hum, I'm stuck here. :(

There are 2 hints !
If $ 0 < p\le \frac {9(\sqrt {593} \plus{} 15)}{184}$ we are done. Remain case is:
\[ \frac {9(\sqrt {593} \plus{} 15)}{184}< p\le 3
\]
Second way:
If $ 0 < p\le \frac {\sqrt {1569} \plus{} 27}{28}$ we are done. Remain case is:
\[ \frac {\sqrt {1569} \plus{} 27}{28}< p\le 3
\]
I don't have enought time to check remain case. I don't think it is easy  :wink: I will try later.
The following inequality is also holds:
\[ \sum \frac {a^2}{a \plus{} 3}\ge \frac {3}{4},\forall a,b,c\ge 0, a^2 \plus{} b^2 \plus{} c^2 \equal{} 3
\]

[quote="zaizai-hoang"]
The following inequality is also holds:
\[ \sum \frac {a^2}{a \plus{} 3}\ge \frac {3}{4},\forall a,b,c\ge 0, a^2 \plus{} b^2 \plus{} c^2 \equal{} 3
\]
[/quote]
I think, it can be $ \sum \frac {a^2}{a \plus{} 3}\le \frac {3}{4}.$ :wink:

[quote="arqady"][quote="Sudoku"]Using Cauchy - Schwarz Inequality
$ \sum\frac {a^2}{2a \plus{} b \plus{} c} \equal{} \sum\frac {a^4}{2a^3 \plus{} a^2(b \plus{} c)}\geq\frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{2(a^3 \plus{} b^3 \plus{} c^3) \plus{} \sum a^2(b \plus{} c)}$
Using S.O.S and by the hypothesis, we get
$ \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{2(a^3 \plus{} b^3 \plus{} c^3) \plus{} \sum a^2(b \plus{} c)}\geq\frac {3}{4}$[/quote]
Sudoku, are you sure that your last inequality is true? :wink:[/quote]
I am not sure  :lol: I think so  :lol:  :)

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3.$ Prove that:
\[ \frac {a^2}{2a \plus{} b \plus{} c} \plus{} \frac {b^2}{2b \plus{} a \plus{} c} \plus{} \frac {c^2}{2c \plus{} a \plus{} b}\geq\frac {3}{4}
\]
[/quote]
My hint is to prove the stronger but "easier" :)
\[ \sum \frac{a^2}{2a\plus{}b\plus{}c} \ge \frac{3}{8}(\frac{a\plus{}b\plus{}c}{3}\plus{}\frac{a^2\plus{}b^2\plus{}c^2}{a\plus{}b\plus{}c})\]
the second step is to normalize that $ a\plus{}b\plus{}c\equal{}1$ and use Vornicu Schur. :)

[quote="can_hang2007"][quote="arqady"]Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3.$ Prove that:
\[ \frac {a^2}{2a \plus{} b \plus{} c} \plus{} \frac {b^2}{2b \plus{} a \plus{} c} \plus{} \frac {c^2}{2c \plus{} a \plus{} b}\geq\frac {3}{4}
\]
[/quote]
My hint is to prove the stronger but "easier" :)
\[ \sum \frac {a^2}{2a \plus{} b \plus{} c} \ge \frac {3}{8}(\frac {a \plus{} b \plus{} c}{3} \plus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c})
\]
the second step is to normalize that $ a \plus{} b \plus{} c \equal{} 1$ and use Vornicu Schur. :)[/quote]

Very nice  :lol: . Your hint is very nice. We can show it by S.O.S too.

$ \sum {\left( {\frac {{a^2 }}{{2a \plus{} b \plus{} c}} \minus{} \frac {a}{4}} \right)} \equal{} \sum {\left( {\frac {{a\left( {a \minus{} b} \right)}}{{4\left( {2a \plus{} b \plus{} c} \right)}} \plus{} \frac {{a\left( {a \minus{} c} \right)}}{{4\left( {a \plus{} 2b \plus{} c} \right)}}} \right)}$

$ \equal{} \sum {\frac {{\left( {a \minus{} b} \right)}}{4}\left( {\frac {a}{{2a \plus{} b \plus{} c}} \minus{} \frac {b}{{a \plus{} 2b \plus{} c}}} \right)} \equal{} \sum {\frac {{\left( {a \plus{} b \plus{} c} \right)}}{{4\left( {2a \plus{} b \plus{} c} \right)\left( {a \plus{} 2b \plus{} c} \right)}}\left( {a \minus{} b} \right)^2 }$

$ \frac {3}{8}\left( {\frac {{a^2 \plus{} b^2 \plus{} c^2 }}{{a \plus{} b \plus{} c}} \minus{} \frac {{a \plus{} b \plus{} c}}{3}} \right) \equal{} \frac {1}{8}\left( {a \plus{} b \plus{} c} \right)\sum {\left( {a \minus{} b} \right)^2 }$

It suffices us to show that $ \sum {x\left( {b \minus{} c} \right)^2 } \ge 0$

with

$ \begin{array}{l} x \equal{} \left( {a^2 \plus{} ab \plus{} ac \minus{} bc} \right)\left( {2a \plus{} b \plus{} c} \right) \\
y \equal{} \left( {b^2 \plus{} ba \plus{} bc \minus{} ac} \right)\left( {2b \plus{} a \plus{} c} \right) \\
z \equal{} \left( {c^2 \plus{} ca \plus{} cb \minus{} ab} \right)\left( {2c \plus{} a \plus{} b} \right) \\
\end{array}$

We can assume $ a \ge b \ge c$ then $ x,y \ge 0$ and $ y \plus{} z \ge 0$. So the hint is proved.

Using the hint and by AM-GM we get the result. Really nice, Can. :)

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3.$ Prove that:
\[ \frac {a^2}{2a \plus{} b \plus{} c} \plus{} \frac {b^2}{2b \plus{} a \plus{} c} \plus{} \frac {c^2}{2c \plus{} a \plus{} b}\geq\frac {3}{4}
\]
[/quote]
the inequality is equivalent to $ (a\plus{}b\plus{}c)(\sum {\frac{a}{2a\plus{}b\plus{}c}})\leq a\plus{}b\plus{}c \minus{}\frac{3}{4}$
$ \Leftrightarrow (a\plus{}b\plus{}c)(\sum {\frac{a}{2a\plus{}b\plus{}c}} \minus{}\frac{3}{4})\leq\frac{a\plus{}b\plus{}c \minus{}\sqrt{3(a^2\plus{}b^2\plus{}c^2)}}{4}$
$ \Leftrightarrow (a\plus{}b\plus{}c)[\sum {\frac{(a\minus{}b)^2}{(a\plus{}2b\plus{}c)(2a\plus{}b\plus{}c)}}]\geq\frac{\sum {(a\minus{}b)^2}}{a\plus{}b\plus{}c\plus{}3}$
$ \Leftrightarrow \sum {\frac{(a\plus{}b\plus{}c)(3\minus{}a\minus{}b)}{(a\plus{}b\plus{}c\plus{}3)(a\plus{}2b\plus{}c)(2a\plus{}b\plus{}c)}(a\minus{}b)^2}\geq 0$ which is trivial.
now we are done!