inequalities
trigonometry
triangle inequality
inequalities proposed

[quote="Yimin Ge"]My solution:
[hide]
The left inequality is equivalent to
\[ (a+b-c)(a+c-b)(b+c-a)>0  \]
which is true since $a,b,c$ are sides of a triangle

The right inequality is equivalent to
\[ (a+b-c)(a+c-b)(b+c-a)\leq abc  \]
Take $x=a+b-c,\ y=a+c-b,\ z=b+c-a$ which are positive by triangle inequality, hence the inequality is equivalent to
\[ xyz\leq \frac{x+y}2 \cdot \frac{x+z}2 \cdot \frac{y+z}2  \]
which is true by AM-GM
[/hide]

Any other solutions?[/quote]
The right inequality is equivalent to[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=346917][b]IMO1964[/b][/url]\[ a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \leq 3abc . \]
The right inequality is equivalent to
\[ a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)> 2abc   .\]

The right inequality is true for all positive real numbers  $a,b,c$ .
[b]Swiss 1983:[/b]
Let $a,b,c$ be the positive real numbers . Prove that\[ abc\geq (a+b-c)(a+c-b)(b+c-a) .\]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=53&p=2393126][b]sieubebuvietnam：[/b][/url]
Let $a,b,c$ be sides of a triangle. Prove that\[abc\geq \frac{1}{8}(\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}})(\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{b}})(\sqrt[3]{\frac{c}{a}}+\sqrt[3]{\frac{a}{c}}) (a+b-c)(b+c-a)(c+a-b)\]\[\geq (a+b-c)(a+c-b)(b+c-a) .\]

[quote="Yimin Ge"]Let $a,b,c$ be sides of a triangle. Prove that
\[ 2 < \frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} - \frac{a^3+b^3+c^3}{abc}\leq 3  \][/quote]

Let $a,b,c$ be sides of a triangle.
\[ \frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} - \frac{a^3+b^3+c^3}{abc}= 2\sum_{cyc}{\cos{A}}  \]

Note that

$$\sum \frac{a+b}{c} -\frac{a^{3}+b^{3}+c^{3}}{abc} =\frac{1}{abc}\sum a(b^{2}+c^{2}-a^{2})=2\sum \frac{b^{2}+c^{2}-a^{2}}{2bc}=2\sum cosA$$

Thus, $$2<\sum \frac{a+b}{c} -\frac{a^{3}+b^{3}+c^{3}}{abc} \le 3 \Leftrightarrow 1 < cosA+cosB+cosC \le \frac{3}{2} $$

The right side of the inequality is true by Jensen's inequality as

$$cosA+cosB+cosC \le 3cos\left(\frac{A+B+C}{3}\right)=3cos\left(\frac{\pi}{3}\right)=\frac{3}{2}$$

For the left inequality, note that
$$cosA+cosB+cosC=1+4sin\left(\frac{A}{2}\right)sin\left(\frac{B}{2}\right)sin\left(\frac{C}{2}\right) >1 $$

I have used the identity $cosA+cosB+cosC=1+4sin\left(\frac{A}{2}\right)sin\left(\frac{B}{2}\right)sin\left(\frac{C}{2}\right)$. This can be proved as:

$cosA+cosB+cosC=2cos\left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right)+1-2sin^{2}\left(\frac{C}{2}\right)$

$=2sin\left(\frac{C}{2}\right)cos\left(\frac{A-B}{2}\right)+1-2sin^{2}\left(\frac{C}{2}\right)$

$=1+2sin\left(\frac{C}{2}\right)\left(cos\left(\frac{A-B}{2}\right)-cos\left(\frac{A+B}{2}\right)\right)$

$=1+4sin\left(\frac{A}{2}\right)sin\left(\frac{B}{2}\right)sin\left(\frac{C}{2}\right)$

[quote=Yimin Ge]Let $a,b,c$ be sides of a triangle. Prove that
\[ 2 < \frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} - \frac{a^3+b^3+c^3}{abc}\leq 3  \][/quote]

The right inequality is equivalent to Schur.The left inequality is equivalent to

$xyz<3xyz(a=y+z,b=z+x,c=x+y)$

Since $x>0, y>0, z>0$, this obviously holds. :D

Let $a,b,c$ be the positive real numbers . Prove that$$a^2(b+c)+kb^2(c+a)+c^2(a+ b)\geq (2+4\sqrt k)abc$$
Where $k>0.$
$$a^2(b+c)+4b^2(c+a)+c^2(a+ b)\geq 10abc$$

[hide=Thanks.][quote=TYT][quote=sqing]
Let $a,b,c$ be the positive real numbers . Prove or disprove $$8abc\geq (a+2b-c)(b+2c-a)(c+2a-b)$$[/quote]

It is wrong for $a = 0, b = 1, c = 1.5$.[/quote][/hide]

Let $a,b,c$ be the positive real numbers . Prove that$$a^2(b+c)+kb^2(c+a)+k^2c^2(a+ b)\geq 2\left(k+(k+1)\sqrt k\right)abc$$
Where $k>0.$
$$a^2(b+c)+4b^2(c+a)+16c^2(a+ b)\geq 28abc$$
[b]#10[/b]

Let $a,b,c$ be the positive real numbers . Prove that
$$a^3(b+kc)+b^3(c+ka)+c^3(a+k b)\geq \left(k+1\right)abc(a+b+c)$$
Where $k>0.$

[quote=smy2012]The right inequality is equivalent to
$ \sum_{cyc} x(x-y)(x-z) \geqslant 0$
It's the famous Schur's Inequality.[/quote]

indeed

[quote=Yimin Ge]My solution:
[hide]
The left inequality is equivalent to
\[ (a+b-c)(a+c-b)(b+c-a)>0  \]
which is true since $a,b,c$ are sides of a triangle

The right inequality is equivalent to
\[ (a+b-c)(a+c-b)(b+c-a)\leq abc  \]
Take $x=a+b-c,\ y=a+c-b,\ z=b+c-a$ which are positive by triangle inequality, hence the inequality is equivalent to
\[ xyz\leq \frac{x+y}2 \cdot \frac{x+z}2 \cdot \frac{y+z}2  \]
which is true by AM-GM
[/hide]

Any other solutions?[/quote]

Effective

[quote=sqing]Let $a,b,c$ be the positive real numbers . Prove that$$ka^2(b+c)+(b^2+c^2)(a+b+c)\geq  (2+4\sqrt{2k})abc$$Where $k>0.$[/quote]
$$ka^2(b+c)+(b^2+c^2)(a+b+c)\geq 2ka^2\sqrt{bc} +2bc(a+2\sqrt{bc} )=2abc+2\sqrt{bc} (ka^2+2bc)\geq
(2+4\sqrt{2k})abc$$


geometry
circumcircle
incenter
geometry unsolved

$ \odot O$ passes the vertex $ B,C$ of $ \triangle ABC$ and meets $ AC$,$ AB$ at $ D$,$ E$ respectively. $ BD$ meets $ CE$ at $ F$, $ OF$ meets the circumcircle of $ \triangle ABC$ at $ P$

Prove:$ \triangle PBD$ and $ \triangle PCE$ have the same incenter.

P=⊙ABC ∩ OF  so P is the miquel point of complete quadrilateral AEBCQD
It is well-known if EDCB are concyclic then OF ⊥ AQ(http://www.mathlinks.ro/Forum/viewtopic.php?t=217333)
Let AM,AN are two tangent lines of ⊙O , then M,F,N are collinear =〉 A,M,O,N,P are concyclic     =〉  MF*FN=OF*FP=EF*FC=FD*BF    =〉 E,O,C,P   P,D,O,B are concyclic respectively  =〉 ∠EPO=∠CPO   ∠BPO=∠DPO  =〉 OP is the bisector of ∠EPC and∠BPD
Denote I is the intersection of ⊙O and OP
We have IO=EO=CO=OB=OD  so I is the incenter of △BPD  and △EPC
Done!

[quote="beparoa"]sir, can u teach me how to draw these types of geometrical figure?
thank you[/quote]

I think you need a software like [b]Geometer's Sketchpad [/b]or[b] GeoGebra[/b], then you can draw these figures easily

[quote="beparoa"]sir, can u teach me how to draw these types of geometrical figure?
thank you[/quote]
I use a tool called Sketchpad which is very powerful

i have searched a lot but didnt get the full version of sketchpad.can you please tell me where would i get this file to download?

thank you...


Find all $f:\mathbb R\to\mathbb R$ for any $x,y$, it is ture that $f(x^2+y+f(y))=2y+f^2(x)$.

[i]Edited by Myth[/i]

I have got that $f(0)=0$.Can it help you? :(

Sorry for delay,[b]Myth[/b](I have bad internet connection).
I have checked it and I found a mistake :blush:  :blush: .I will try to correct it.

[quote="Armo"]I have bad internet connection[/quote]
I understand you very well :(

I think it's as similar as "function from VietNam TST2004"(was solved by Harazi).Right?
The link: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=VietNam&t=5619


Fifteen chairs are evenly placed around a circular table on which are name cards for fifteen guests. THe guests fail to notice these cards until after they have sat down, and it turns out that no one is sitting in front of his own card. Prove that the table can be rotated so that at least two of the guests are simultaneously correctly seated.

[hide]The table can be rotated into 15 possible different positions.  Each guest will be matched up with his or her name in exactly one of these positions.  In the initial position, no one is matched up correctly.  Therefore, there are 15 "matchings" left for only 14 positions, and by the Pigeonhole Principlie, at least one of those positions must contain at least two matchings.[/hide]

Hi,

Another approach..

[hide]

Consider any person. The distance of his card from him (in one direction - consider clockwise distance throughout) is either 1, 2 upto 14.  Thus, there are 15 persons, and if you classify them according to their distances from their correct cards, then at least two people will have the same distance. So, the table can be rotated with this distance, as a result of which, at least two people will have the correct card.

[/hide]

***************************************************************************

Sambit


function Asymptote LaTeX algebra polynomial calculus conics

Alright, I need some tricks for graphing rational functions.  Here is what I know so far:

If $f(x)$ is a rational function where $f(x)=\frac{p(x)}{q(x)}$ and $q(x)\neq 0$:

The vertical asymptotes are on the zeros of $q(x)$.

If the degree of $p(x)$ is less than the degree of $q(x)$, the horizontal asymptote is at $y=0$.

If the degree of $p(x)$ is equal to the degree of $q(x)$, the horizontal asymptote is at $y=\frac{a}{b}$, where $a$ is the leading coefficient of $p(x)$ and $b$ is the leading coefficient of $q(x)$.

If the degree of $p(x)$ is less than the degree of $q(x)$, there is no horizontal astymptote.

Now...

How do you find the verteces of each part of the graph?

In functions similar to  $f(x)=\frac{2x^2}{x^2-4}$, how do you find the range? (I feel like an idiot asking this...I know you need to solve for $x$ in terms of $y$...but how!)

Something completely off subject (yet I have been wondering this aswell), how do you find the vertex of a logaritmic function and exponential function?

If there is anything else that you think would be good for me to know please feel free to tell!

I am just about finished with my Algebra II...  All I have left to learn are rational equation graphing, and conics... The book I am using absolutely sucks for learning graphing of any kind... it overall isn't very great either!  Thank God for AoPS!  :P

Umm.... I don't think they're actually asymptotes rather than boundaries of the range. Like for f(x)=1/(x^2+1), the range may be (0,1], but there's only one asymptote at y=0.

Also, you might want to know about oblique/slant asymptotes as well, but I can't really help you with that. I've got to look them up first.

EDIT: Uggghh. I wish I had latex, but this will have to do.

Basically the graph of a rational function can have a slant asymptote if the degree of the top polynomial is greater than the degree on the bottom. The slant asymptote can be expressed as y=(coeff. on top)/(coeff. on bottom)*x^((deg. of top)-(deg. of bottom))

Sorry it's so butchered up.

Is this what you mean:

Coeff. of $p(x)=a$
Coeff. of $q(x)=b$
Degree of $p(x)=y$
Degree of $q(x)=z$

$f(x)=\frac{p(x)}{q(x)}$

Asymptote = $\frac{a}{b} x^{(y-z)}$

You are pretty much out of luck for finding local extrema (vertices) unless you have calculus unless it is a parabola or something.  To find the range, I always just graphed the function and then inspected the graph.  Solving for x in terms of y would be pretty nasty in some of these.

If you have the function:
$y = \frac{(x-2)(x+3)}{(x-6)(x-8)}$

I would first draw in the asymptotes (x = 6, 8) and mark the zeros (2 and -3). Then I would find the intervals we are interested in.  For this graph they would be, $(-\infty, -3); (-3, 2); (2, 6); (6, 8); (8, \infty)$  (in interval notation).  I would then find the sign of the function in each interval.  So in the interval from negative infinity to negative 3, the value of the function is positive and I would draw that in appropriately on my graph.  Do this for every interval and you should have a rough sketch of the graph.  I do not think there is much more you can do without using calculus.

Yep that's all I was taught how to do in my Alg II, College Alg classes.


AMC
AMC 8
AIME
USA(J)MO
USAMO
AMC 12
AMC 10

[quote="batteredbutnotdefeated"]darn. I didn't make my AMC8 goals(25), but i'm looking forward to the AMC10. My goals: 138+/150 on AMC10. What are yours?[/quote]

I hope to get 120 and qualify for the AIME.    :)

It's in my signature but my goal for the AMC 10 is to get a 150  :lol:

150 would be nice; I've gotten em in practice.

145.5, 144, 139.5 are more realistic though...

I aim for 150, but I will probably get 144.

I've gotten 130-135s in practice, but the last time I took one was a while ago. Since then, I had been preparing for the AMC 8. Now that that is over, I am preparing for the AMC 10/AMC 12/AIME.

My goal is a 135, because all I want to do is qualify with a relatively high score for the AIME, and I am focusing more on that.

Man, if I could take the AMC 10 to get to USAMO, then I would probably be going for something like a 140+

Buuuuuuuuuut, I have to take the AMC12, so I'll be going for 132+ on that (as can be seen in my sig)

[quote="fwolth"]hey guys I have this urge to share personal math competition goals with all of you, without even a pretense of stating them except the apparent but by mutual agreement unstated one of showing off that I'm amazing enough at math to have relatively high goals.[/quote]
fwolth posted this in another goals thread, and I completely agree with him.

Anyways, I won't be taking the AMC10, since our school only offers the B date.

athunder those are high goals... but then again, me scoring a 9+ on the AIME is unreasonable. Sadly(surprisingly :?: ), I tend to do better on harder tests...  :huh:

My goal would be to get 120, but I'd probably get 119.5 cuz I'm fail like that.

I really want to see if I got 25 on my amc 8 but my math teacher hasn't gone over it yet.  :o

My goal is 150+  :)

i set high goals for myself...

My goal is 150... even though I got a 93 on it last year :lol:

I improved a lot...

My goals are 132+. That's pretty realistic for me considering last year I got a 112.5

lol a good bump tho
[s]stop stalking evan[/s] :P

My goal is to qualify for both the AMC 8 and AMC 10. I realize that this will be tough to achieve, but I will try my hardest and never give up.

My goal is to get 24 or 25 on AMC 8 and qualify for AIME through AMC 10 with a score of 120+.

[quote=franzliszt]I thought you don;t need to qualify for AMC 8 lol[/quote]

Yeah, you don't. You don't need to qualify for AMC 10 either. I think @Orangecounty2 means he wants to qualify for AIME by AMC 10.

Nice bump :coolspeak:

I am aiming for a 151.5.
sHoot fOr tHe mOon. eVeN iF yOu miSs, yOu'Ll LaNd aMoNg tHe sTaRs.

[quote=v_Enhance]Just passing the AMC 10 (120 points) will suffice for me..
In my sevie year I only got 87.... everyone else beat me like.... badly.... :oops:
So this year I have to make up for it.  :P[/quote]

a legend was born.....

[quote=AopsUser101][quote=v_Enhance]Just passing the AMC 10 (120 points) will suffice for me...[/quote]
[I]proceeds to win USAJMO...[/i][/quote]

You know, as a teenager, my parents always warned me to be careful what I said on the Internet, maybe I should have listened to them :P

[quote=mira74][size=50]Welp, it seems like it worked so I'll try it too[/size]

Just passing the AMC 10 (120 points) will suffice for me..
In my sevie year I only got 87.... everyone else beat me like.... badly.... :oops:
So this year I have to make up for it. :P[/quote]

I'll try too...oh wait, I can't even take AMC 10 :wallbash:

[size=50]same, haha[/size]

Just passing the AMC 10 (120 points) will suffice for me..
In my sevie year I only got 87.... everyone else beat me like.... badly.... :oops:
So this year I have to make up for it. :P

[size=50]Ok I guess I'm doing it too.
[/size]
Just passing the AMC 10 (120 points) will suffice for me..
In my sevie year I only got 87.... everyone else beat me like.... badly.... :oops:
So this year I have to make up for it. :P

[size=50]Yeah, Evan's onto something here
[/size]
Just passing the AMC 10 (120 points) will suffice for me..
In my sevie year I only got 87.... everyone else beat me like.... badly.... :oops:
So this year I have to make up for it. :P

[size=50]It can't hurt. Plus, it might work. Hopefully.[/size]

Just passing the AMC 10 (120 points) will suffice for me..
In my sevie year I only got 87.... everyone else beat me like.... badly.... :oops:
So this year I have to make up for it. :P

Wow, so many people got 87 in 7th grade.


modular arithmetic
number theory solved
number theory

$n=2^h+1, h>1$. Show that $n$ is a prime if and only if $3^\frac{n-1}{2}\equiv -1\pmod n$.

If $n$ is prime, then $n \equiv 1 \mod 4$ and (otherwise you would get $3|n$) $h=2a \implies n \equiv 2 \mod 3 $ and this gives
$ \left( \frac{3}{n} \right) = \left( \frac{n}{3} \right) = \left( \frac{2}{3} \right) = -1$, so $3$ is not a square $\mod n$.
But half of the prime residue classes $\mod n$ are squares, and there are $\phi ( \phi (n)) = \phi (2^h)=2^{h-1}$ primitive roots $\mod n$, which is also the half, so every integer that is not a square is a primitive root (because squares cannot be primitive roots), so $3$ is a primitive root.
This gives $3^{\frac{n-1}{2}} \equiv -1 \mod n$

If $3^{\frac{n-1}{2}} \equiv -1 \mod n$ then (in $\mathbb{Z}_n$) this gives $ord(3) | 2^h$ but $ord(3)$ doesn't divide $2^{h-1}$, so $ord(3)=2^h$, but we also have $n-1=ord(3) | \phi (n) <n \implies \phi (n)=n-1$ which gives that $n$ is prime.

q.e.d.

3 can be replaced by $k$ which satisfies $\left(\frac{k}{2^h+1}\right)=-1$.

[quote="pbornsztein"]This is known as Pepin's test (1877).

Pierre.[/quote]
Thanks, Bornstein.


[hide]expanding the equation gives $x^{2}-3x+2=0$, so $x=2,1$ and the minimum value is $1$[/hide]

[hide]The equation, implies another equation

$x^{2}-3x+2=0$, so $x=1,2$

$1$[/hide]

[quote="mathnerd314"][hide]
$(x+1)(x+2)=6x$
$x^{2}+3x+3=6x$
$x^{2}-3x+3=6x$
Therefore the vertex is at $\frac{-b}{2a}=\frac{3}{2}$
1.5
[/hide]

Edit: I think I did something wrong...[/quote]
That is for the vertex of a graph if it was y=....

[quote="mathnerd314"][hide]
$(x+1)(x+2)=6x$
$x^{2}+3x+3=6x$
$x^{2}-3x+3=0$
Therefore the vertex is at $\frac{-b}{2a}=\frac{3}{2}$
1.5
[/hide]

Edit: I think I did something wrong...[/quote]
Using $-\frac{b}{2a}$ gives the x value when $x^{2}-3x+3$ is at a minimum. We are looking for the least [b]root[/b] of $f(x)=x^{2}-3x+3$.

[quote="mathnerd314"]There should be a "D'oh" smiley.

I get it now.[/quote]
Also, you made a careless mistake: $(x+1)(x+2)=x^{2}+3x+\boxed{2}$.
(And don't worry, we all make those kinds of mistakes. It's nothing to be ashamed about.  :) )

[quote="bpms"]Better to err now than at states or even nats :ninja:[/quote]

Like I did last year... my 8th grade year... :blush:

[quote="Ignite168"]Wtf? I had a solution posted to this problem. Where'd it go?[/quote]
In the notes topic I spent like 20 min. to type up and derive a formula for $1^{4}+2^{4}+3^{4}+...+k^{4}$ but when I hit submit it gave me some wierd page. I hit backspace but my solution was deleted and I was too mad to retype it.

[quote="bpms"]In the notes topic I spent like 20 min. to type up and derive a formula for $1^{4}+2^{4}+3^{4}+...+k^{4}$ but when I hit submit it gave me some wierd page. I hit backspace but my solution was deleted and I was too mad to retype it.[/quote]
LOL, that's exactly what I'm thinking when I've got to retype a solution (PC problems, Automatic log out for some reason, Accidental backspace, etc.)  :P  Seriously, I'm really pissed off at those moments. I'm happy to see that I'm not the only one.  :D

Oh well, I also added how you could do it by casework if the problem said integers only in my solution.

[quote="Treething"](x + 1)(x + 2) = 6x

Find the minimum value of x satisfying this equation.[/quote]

[hide="Solution"]
$\\ (x+1)(x+2)=6x\\ x^{2}+3x+2=6x\\ x^{2}-3x+2=0\\ (x-2)(x-1)=0\\ x={1,2}\\$ The least value for x is 1, so the answer to the question is [b]1[/b].
[/hide]

[size=75][color=darkred]Please try not to type text in LaTeX when you can just use regular text. Thanks.
-nebula42[/color][/size]

[quote="mathnerd314"][quote="dr.silly"]x^2=3x-2
boxed{1}[/quote]

:welcome:

You have to put dollar signs around your LaTeX code.

Also, it's a good idea to put hide tags around your solution.[/quote]
You also have to put a slash before boxed \boxed{1} in latex is $\boxed{1}$

[quote="Treething"](x + 1)(x + 2) = 6x

Find the minimum value of x satisfying this equation.[/quote]
[hide]x^2+3x+2=6x
x^2-3x+2=0
(x-2)(x-1)
so lowest is 1?[/hide]


a)$a,b$ are positive reals and $m$ is natural.Prove that $(1+\frac{a}{b})^m+(1+\frac{b}{a})^m\geq 2^{m+1}$.

b)Prove that $a^2+b^2+\frac{1}{a^2}+\frac{b}{a}\geq \sqrt{3}$, where $a,b$ are reals not equal to $0$.

a),b)When does equality occur?

a) $(1 + a/b)^m + (1 + b/a)^m \geq 2.(2 + a/b +b/a)^{m/2} \geq 2^{m+1}$

(the first and the second by AM-GM).

[hide="b)"]

$b^2+\frac{b}{a}+\left(a^2+\frac{1}{a^2}\right) = \left(b+\frac{1}{2a}\right)^2+\left(a^2+\frac{3}{4a^2}\right) \ge \left(a^2+\frac{3}{4a^2}\right)$

So we maximize $\left(a^2+\frac{3}{4a^2}\right) \ge 2\sqrt{\frac{3}{4}} = \sqrt{3}$ by AM-GM with equality at $a = \sqrt[4]{\frac{3}{4}}$ and $b = -\frac{1}{\sqrt[4]{12}}$.[/hide]

if the first can use the Inductive method and Launch the binomial equation to solve :?:


abstract algebra
group theory
superior algebra
superior algebra unsolved

$ G$ - abelian group
$ Fin(G)$ - subgroup of all $ g\in G$  having finite rank.

give a few examples of abelian $ G$'s such that $ Fin(G)$ are not abelian

Subgroups of abelian groups are always abelian...


A box contains 2 pennies, 4 nickels, and 6 dimes. Six coins are drawn without replacement, with each coin having an equal probability of being chosen. What is the probability that the value of coins drawn is at least 50 cents?
$\text{(A)} \ \frac{37}{924} \qquad \text{(B)} \ \frac{91}{924} \qquad \text{(C)} \ \frac{127}{924} \qquad \text{(D)} \ \frac{132}{924} \qquad \text{(E)} \ \text{none of these}$

[hide]
We want to find total that work divided by total possible.  We know there are $\binom{12}{6} = 924$ possible ways to choose the 6 coins.  We see there are only two ways to get 50 cents - 5 dimes + 1 anything, or 4 dimes + 2 nickels.
For the first case there are $\binom 6 5 \cdot \binom 7 1 = 42$ possibilities.
For the second case there are $\binom 6 4 \cdot \binom 4 2 = 90$ possibilities.
This sums to 132, so the total probability is $\frac{132}{924} \implies \boxed D$
[/hide]

i don't think that we simplify because the answers are in that form...but it does simplify to [hide]$\frac{1}{7}$[/hide]

also, could you post some more of these amc level probability problems because i am a little shaky on them

The 5 dimes plus one of any other one includes the 6th dime, if I read there answers right.

[quote="Artzilla03"]Why can't we have 6 dimes?[/quote]

6 dimes is equal to 5 dimes + 1 dime, which is covered in case 2.

[quote="Chocoboom"][hide="I think the answer is C"]
Case 1: 4 dimes and 2 nickels, $\binom{6}{4}\binom{4}{2}= 90$ ways.
Case 2: 5 dimes and 1 non-dime coin, $\binom{6}{5}\binom{6}{1}= 36$ ways.
Case 3: 6 dimes, ONLY 1 WAY.
So therefore, the answer is $\frac{127}{924}$[/hide]
 :huh:  :huh: (sorry about bringing back an old old post)[/quote]

Yep, that is correct.

Because, we aren't imposing any restrictions on the 5 dimes chosen.

So, we have to consider 5 dimes by itself, and then 6 dimes.

What would the probability be if there's replacement?

[quote="mathwizarddude"][quote="Flame"]I also got $ C$ for answer.

Considering case with replacement:

There are still 3 cases: 6 dimes, 5 dimes + any other coin and 4 dimes +2 nickels
so the probability is:

$ \left(\frac {1}{2}\right)^6 \plus{} \binom 6 5 \cdot \left(\frac {1}{2}\right)^5 \cdot \left(\frac {1}{2}\right) \plus{} \binom 6 4 \cdot \left(\frac {1}{2}\right)^4 \cdot \binom 4 2 \cdot \left(\frac {1}{3}\right)^2 \equal{} \boxed{\frac {47}{64}}$[/quote]
I don't think we need the binomial coefficients here...[/quote]

Pretty sure you do.. you want all combinations of 5 dimes/1 other, 4 dimes/2 nickels, etc. not just the case where you choose 5 dimes in a row followed by 1 other coin.

[quote="tomwolf2008"][quote="Flame"]I also got $ C$ for answer.

Considering case with replacement:

There are still 3 cases: 6 dimes, 5 dimes + any other coin and 4 dimes +2 nickels
so the probability is:

$ \left(\frac {1}{2}\right)^6 \plus{} \binom 6 5 \cdot \left(\frac {1}{2}\right)^5 \cdot \left(\frac {1}{2}\right) \plus{} \binom 6 4 \cdot \left(\frac {1}{2}\right)^4 \cdot \binom 4 2 \cdot \left(\frac {1}{3}\right)^2 \equal{} \boxed{\frac {47}{64}}$[/quote]
I think the question said we don't consider replacement[/quote]

mathwizarddude asked for a solution with replacement

I don't think so... note that in this case there's replacement and if we are treating all the coins as distinct (despite whether they are dimes or nickels or pennies), then for the case of 5 dimes, there's a probability of $ 1/12$ of picking a particular dime each time and the 6 dimes can permute in groups of 5 with repetition in $ 6^5$ ways and $ 6^5 (1/12)^5\equal{}(1/2)^5$.
So no binomial coefficients required.

The binomial coefficient are needed because order in which we pick coins matters now. In case 1 there is only one order. In case 2 there are $ \binom 6 5$ different orders. In case 3 there are $ \binom 6 4$ different orders but there is a mistake, $ \binom 4 2$ is not needed as we already ordered nickels and total probability for any nickel is counted. The correct result would be: $ \frac{41}{192}$

Right; only the order of the different type of coins matters, but we do not need the binomial coefficient for the order of a particular coin.

[quote="4everwise"][hide="Solution"]Answer: $\text{(D)}$

Notice that there are two ways we can draw more than 50 cents. We can have 4 dimes and 2 nickels, or 5 dimes and any other coin.

There are $\binom{6}{4}\binom{4}{2}=90$ ways for the first case, and $\binom 65 \binom 71=42$ ways for the second.

Dividing by the total ways to draw 6 coins, $\binom{12}{6}$, our answer is $\frac{132}{924}$.[/hide][/quote]

In the second case, you overcount the six dimes case 5 times.  The answer is actually [hide]$C$.[/hide]


algebra
polynomial
function
calculus
derivative
superior algebra
superior algebra unsolved

1. Let $ F$ be a field of characteristic $ 2$, and let $ t^2 \plus{} bt \plus{} c \equal{} 0$ be an irreducible polynomial over $ F$. Find a necessary and sufficient condition on $ b$ and $ c$ that this be inseparable.

[hide]Is the answer simply $ b \equal{} 0$?[/hide]

2. Let $ F$ be the field of rational functions over $ \mathbb{F}_2$ in the variable $ x$, and let $ K$ be the splitting field of $ t^2 \minus{} x \equal{} 0$. According to the primitive element theorem, this has infinitely many intermediate fields because it is inseparable. What exactly are they? I can't seem to construct any.

1. Yes. The derivative is zero if $ b\equal{}0$ and the derivative is not zero if $ b\neq 0$, so you're answer is correct.

2. Why should it have infinitely many intermediate fields¿ You have: infinitely many intermediate fields <=> not generated by one element (but not: ... <=> inseparable). But here it is generated by a root $ t_0$ of $ t^2 \minus{} x$.
Everything gets different if you use two variables.


How many ways are there to shade 3 squares in a 3x3 array of squares such that none of the shaded squares are in the same row, column, and diagonal?

[hide="answer(might be wrong)"]
Since they are in different rows and columns and diagonals there are only these ways:(Where x means shaded and 0 is not shaded.) 4 Total
[hide="squares"]
1:
x|0|0
0|0|x
0|x|0

2:
0|x|0
x|0|0
0|0|x

3:
0|0|x
x|0|0
0|x|0

4:
0|x|0
0|0|x
x|0|0
[/hide]
[hide="further explination"]
An easy way to notice these is picking a corner, make the diagonal to the other side and switching the shaded squares by column(only the 2 squares that were added to make the diagonal not the corner you picked.)

For example in number 1: I picked the first corner

x|0|0
0|0|0
0|0|0

Then made the diagonal

x|0|0
0|x|0
0|0|x

And switched the last two squares added by their columns

x|0|0
0|0|x
0|x|0

Tada!
[/hide]
[/hide]

Place one of the squares-9 ways
There are 4 other places to place the second one
There are only one place to place the last one
Then you divide by 3! because we counted everything 6 six times, make sure you see why.
So 36/6=6

Or, the top row has 3 choices, the second row has 2 choices left and the last row has the forced square so 3*2=6

When you calculate the possibilities, you forget about the two ways in which they end up in a diagonal. So it's $ 3*2 \minus{} 2 \equal{} 4$

Oh yeah, I didn't see the word diagonal.


LaTeX
parameterization
function
algebra
polynomial
combinatorics unsolved
combinatorics

could you help me please.

i have the following pair of equations and need to find a(n) and b(n):

a(n+1)=a(n) - b(n)
b(n+1)=2*a(n) +b(n)+n

a(0)=2, b(0)=0

I went as far as M*(1 + i(sqrt(2))^n) + K*(1 - i(sqrt(2))^n) - (1/2)*n... and not sure what to do next.

n is of the form (Pn+T). I found Pn=-(1/2)*n, cannot find T. Without T, I cannot find M & K.

Thanx.

Since $ b(n)$ can be obtained by successive differences of $ a(n)$, it is enough to get a closed recursive equation for either
$ a(n)$ or $ b(n)$. I shall do this for $ a(n)$ , but I am not able to find a closed form for $ a(n)$ yet.

Equations given  were:
$ a(n\plus{}1)\equal{}a(n)\minus{}b(n)$            ... (1)
$ b(n\plus{}1)\equal{}2a(n)\plus{}b(n)\plus{}n$

Adding and subtracting we get,

$ a(n\plus{}1)\plus{}b(n\plus{}1)\equal{}3a(n)\plus{}n$
$ b(n\plus{}1)\minus{}a(n\plus{}1)\equal{}a(n)\plus{}2b(n)\plus{}n$

=> $ 2b(n\plus{}1)\equal{}4a(n)\plus{}2b(n)\plus{}2n$
=> $ b(n\plus{}1)\minus{}b(n)\equal{}2a(n)\plus{}n$             ....(2)

Using (1) and (2)
$ a(n\plus{}2)\minus{}a(n\plus{}1)\minus{}a(n\plus{}1)\plus{}a(n)\equal{}2a(n)\plus{}n$
=>$ a(n\plus{}2)\equal{}2a(n\plus{}1)\plus{}a(n)\plus{}n$
with $ a(0)$=2, $ a(1)$=2 [ $ a(1)$ was obtained by (1)]

Now both $ a(n)$ and $ b(n)$ may be recursively generated.

Thank you for your response.

I got a different a(n) formula though.

I'll show you my method and maybe you could tell me what you think:

the condition is:

a(n+1)=a(n)-b(n), so b(n)=a(n)-a(n+1), so b(n+1)=a(n+1)-a(n+2).

I put it into the second eqation: b(n+1)=2*a(n) + b(n)+n and got this: b(n+1) = a(n+1) - a(n+2)= 2*a(n) + (a(n) - a(n+1))+n =

-(a(n+2) - 2(n+1) +3(a))=n

The roots are: -(1+(sqrt(-2)) and -(1-(sqrt(-2))

So the homogeneous equation is M*(-(1+(sqrt(-2)))^n + K*(-(1-(sqrt(-2)))^n

I now need to find a particular solution for n for the equation: -(a(n+2) - 2(n+1) +3(a)=n where a(0)=2 and a(1)=2.

I am not sure how to do it. I wonder if you could help? Thank you, Bob

You are being really, excruciatingly sloppy in your writing.  Please, use the edit button to fix the numerous typos in what you've written.  (It would also be nice if you learned [[LaTeX]]; see also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]this topic[/url].)

Going back to your very first post: assuming that you're computations have been correct, you can find all remaining parameters by recursively generating a few values of the sequence and then solving a system of linear equations.  If you're looking for a particular solution, since you already know it has to be linear, substitute in $ a(n) \equal{} Pn \plus{} T$ and solve for $ P$ and $ T$.  (This should be an identity as functions of $ n$, so you should choose values of $ P$ and $ T$ so that all coefficients in $ n \plus{} (a(n \plus{} 2) \minus{} 2a(n \plus{} 1) \plus{} 3a(n))$ are $ 0$.

Thanks for the links.

I will now try to put it in a proper form

$ a(n\plus{}1)\equal{}a(n) \minus{} b(n)$

$ b(n\plus{}1)\equal{}2*a(n) \plus{}b(n)\plus{}n$

$ a(0)\equal{}2, b(0)\equal{}0$

This is my attempt to solve it:

$ a(n\plus{}1)\equal{}a(n)\minus{}b(n)$, so $ b(n)\equal{}a(n)\minus{}a(n\plus{}1)$, therefore $ b(n\plus{}1)\equal{}a(n\plus{}1)\minus{}a(n\plus{}2)$.

Putting this into the second equation:

$ b(n\plus{}1)\equal{}2*a(n) \plus{} b(n)\plus{}n$

I got the following:

$ a(n\plus{}1) \minus{} a(n\plus{}2) \equal{} 2*a(n) \plus{} (a(n) \minus{} a(n\plus{}1))\plus{}n$

So my difference equation is $ a(n\plus{}2) \minus{} 2a(n\plus{}1) \plus{}3a(n)\equal{} \minus{}n$

The roots for this difference equation are:

$ 1 \minus{} \sqrt{\minus{}2}$ and $ 1 \plus{} \sqrt{\minus{}2}$

So, the homogeneous equation is:

$ M*(1 \minus{} \sqrt{\minus{}2})^n$ + $ K*(1 \plus{} \sqrt{\minus{}2})^n$

For the particular expression, I substitute $ a(n)\equal{}Pn\plus{}T$ following your advice and get the following expression:

$ n \plus{} (P*(n\plus{}2) \plus{} T) \minus{} 2*(P*(n\plus{}1) \plus{} T) \plus{} 3*(Pn\plus{}T) \equal{} 0$

$ P\equal{}\minus{}(1/2)$

I still cannot find T..., and without T I cannot find M and K.

[quote="BobRBR"]$ n \plus{} (P*(n \plus{} 2) \plus{} T) \minus{} 2*(P*(n \plus{} 1) \plus{} T) \plus{} 3*(Pn \plus{} T) \equal{} 0$

$ P \equal{} \minus{} (1/2)$[/quote]

Surely you did not jump from the first of these equations to the second with nothing in between?  Rewrite the first equation as $ n(1 \plus{} 2P) \plus{} 2T \equal{} 0$.  In order for this to be identically true, we need $ P \equal{} \minus{}\frac{1}{2}$ and $ T \equal{} 0$.  Or, you could have followed my other suggestion and taken the equation $ a(n) \equal{} M\cdot r_1^n \plus{} K \cdot r_2^n \plus{} P\cdot n \plus{} T$ for any four values of $ n$ you like and solved the resulting linear equations.

I have solved it and checked it in Maple. There is still a bit missing but I am not sure what...

I've got

$ M*(1\minus{} \sqrt{\minus{}2})^n$ + $ K*(1\minus{} \sqrt{\minus{}2})^n$ - $ (1/2)*P$

So, from here I get

$ a(0)\equal{}2\equal{}M\plus{}K$, $ M\equal{}2\minus{}K$

$ a(1)\equal{}2\equal{}M*(1\minus{} \sqrt{\minus{}2})^1 \plus{} K*(1\plus{} \sqrt{\minus{}2})^1 \minus{} (1/2)*1$

$ 2\equal{}(2\minus{}K)*(1\minus{} \sqrt{\minus{}2})^1 \plus{} K*(1\plus{} \sqrt{\minus{}2})^1 \minus{} (1/2)*1$

$ K\equal{}\minus{}(1/8)* \sqrt{\minus{}2}*(1\plus{}4* \sqrt{\minus{}2})$

$ M\equal{}2\plus{}(1/8)* \sqrt{\minus{}2}*(1\plus{}4* \sqrt{\minus{}2})$

so,

$ a(n)\equal{}(2\plus{}(1/8)* \sqrt{\minus{}2}*(1\plus{}4* \sqrt{\minus{}2}))*(1\minus{} \sqrt{\minus{}2})^n) \minus{} (1/8)* \sqrt{\minus{}2}*(1\plus{}4* \sqrt{\minus{}2})*(1\plus{} \sqrt{\minus{}2})^n \minus{} (1/2)*n$

$ a(n)\equal{}(1\minus{} \sqrt{\minus{}2})^n \plus{} (1\plus{} \sqrt{\minus{}2})^n \minus{} (1/2)*n \plus{} (1/8)* \sqrt{\minus{}2}*(1\minus{} \sqrt{\minus{}2})^n \minus{} (1/8)* \sqrt{\minus{}2}*(1\plus{} \sqrt{\minus{}2})^n$

The solution in Maple is to follow.

The solution in Maple:

$ rsolve({n\plus{}a(n\plus{}2)\minus{}2*a(n\plus{}1)\plus{}3*a(n) \equal{} 0, a(1) \equal{} 2, a(0) \equal{} 2}, a(n))$

$ a(n) \equal{} (1\plus{}\sqrt{\minus{}2})^n \plus{}(1\minus{}\sqrt{\minus{}2})^n \minus{} (1/2)*n \minus{} (1/8)*(\sqrt{\minus{}2})*((\minus{}3/(\minus{}1\plus{}\sqrt{\minus{}2}))^n \minus{} ((1/8)*(\sqrt{\minus{}2})*(1 \plus{}(\sqrt{\minus{}2})) * ((\minus{}3/(\minus{}1\minus{}\sqrt{\minus{}2}))^n/ \minus{}1\minus{}\sqrt{\minus{}2}))$

So, not sure what I am doing wrong here, but some bit of the solution is missing...

Your answer is correct (although not, perhaps, written in the simplest possible way), so it doesn't interest me terribly what Maple says :).  In all likelyhood, the two expressions are equivalent up to some simple manipulations.  (I note, for example, that the expression from Maple contains $ \frac{\minus{}3}{\minus{}1\minus{}\sqrt{\minus{}2}}$; this is just equal to $ 1 \minus{} \sqrt{\minus{}2}$.)

$ \boxed{a_{n \plus{} 1} \equal{} a_n \minus{} b_n,\ b_{n \plus{} 1} \equal{} 2a_n \plus{} b_n \plus{} n,\ a_0 \equal{} 2,\ b_0 \equal{} 0}$ gives the following equations.

We are to determine $ \alpha ,\ p,\ q,\ r$ such that $ a_{n \plus{} 1} \plus{} \alpha b_{n \plus{} 1} \plus{} p(n \plus{} 1) \plus{} q \equal{} r(a_n \plus{} \alpha b_n \plus{} pn \plus{} q).$

Oh, my god. I have deleted my solution! :(

@ kunny: The solution is correct. If [1] and [2] can be obtained, I agree it is the matter of handling the equations to get the required closed forms. But how did you think of [1] and [2] ? Is there any general theory to solve problems of this kind?

[quote="srikanth"]Is there any general theory to solve problems of this kind?[/quote]
The same generating function theory that allows us to solve the question without the extra $ n$ works here.  The method the original poster suggests (split into a homogeneous and particular solution) works just as easily, since the particular solution associated to a polynomial is just another polynomial.


If $R_n = \frac{1}{2} (a^n + b^n)$ where $a = 3 + 2 \sqrt 2, b = 3 - 2 \sqrt 2,$ and $n = 0,1,2,..,$ then $R_{12345}$ is an integer. Its unit digit is

$\text{(A)} \ 1 \qquad \text{(B)} \ 3 \qquad \text{(C)} \ 5 \qquad \text{(D)} \ 7 \qquad \text{(E)} \ 9$

[quote="JesusFreak197"][hide="Answer"]Note that the radicals will always cancel. Next, we find a pattern:

$R_1=3$
$R_2=17$
$R_3=99$
$R_4=577$
$R_5=3363$

The pattern repeats every four numbers, so since $12345\equiv 1\bmod{4}$, then $R_{12345}\equiv 3\bmod{10}$. $\boxed{B}$[/hide][/quote]

Err, it actually doesn't repeat every four numbers... for example $R_6 = 19601$ and $R_7 = 114243$. Besides, without a calculator, how are you going to evaluate those...

Here's an interesting way to do it though....

[hide]
Considering writing the sequence as a recursion:

First note that $a+b = 6$ and $ab = 1$.

$R_{n+1} = \frac{1}{2}(a^{n+1}+b^{n+1}) = \frac{1}{2}[(a^n+b^n)(a+b)-ab(a^{n-1}+b^{n-1})] = 6R_n-R_{n-1}$ (since $a^n+b^n = 2R_n$ and $a^{n-1}+b^{n-1} = 2R_{n-1}$).

Now that's pretty cool. We can easily plug in values to find the repeat now (everything taken $\pmod{10}$):

$R_0 = 1$
$R_1 = 3$
$R_2 = 7$
$R_3 = 9$
$R_4 = 7$
$R_5 = 3$
$R_6 = 1$
$R_7 = 3$

Since we have $R_0 = R_6$ and $R_1 = R_7$ we know it repeats every six because the value of $R_n$ depends only on the two previous terms. So $12345 \equiv 3 \pmod{6} \Rightarrow R_{12345} \equiv R_3 \equiv 9 \pmod{10}$ so our answer should be $E$.[/hide]


inequalities
function
inequalities proposed

[quote="Hatucdao"]Prove that for all $ x,y,z\ge 0$ we have[/quote]
Step1 we prove that
$ A \equal{} x/(x \plus{} 2z) \plus{} y/(2x \plus{} y) \plus{} z/(2y \plus{} z) > \equal{} 1$
because $ (x \plus{} y \plus{} z)^{2} \leq A(x(x \plus{} 2z) \plus{} y(2x \plus{} y) \plus{} z(2y \plus{} z))$
Step 2
we see that $ x^{2} \plus{} y \plus{} 1 \geq 2x \plus{} y;...$
Step 3 Do by yourself

Great desmrhai! As you see, the problem is in fact homogeneous one  :D Could you please try two others?

[quote="Hatucdao"]Show that for all $ x,y,z\ge 0$ we have
\[ \frac {x}{x^2 \plus{} y \plus{} 1} \plus{} \frac {y}{y^2 \plus{} z \plus{} 1} \plus{} \frac {z}{z^2 \plus{} x \plus{} 1}\le 1
\]
and the equality holds iff x=y=z=1.

This problem seems strange because it is non-homogeneous and there is no relationship between variables. However, it is easy. Can you prove it?

If you can, please try with a harder one: Find the maximum, for all $ x,y,z\ge 0$, of the functions
\[ f(x,y,z) \equal{} \frac {x}{x^2 \plus{} y^2 \plus{} 1} \plus{} \frac {y}{y^2 \plus{} z^2 \plus{} 1} \plus{} \frac {z}{z^2 \plus{} x^2 \plus{} 1}
\]
and
\[ g(x,y,z) \equal{} \frac {x}{x^2 \plus{} \sqrt {y} \plus{} 1} \plus{} \frac {y}{y^2 \plus{} \sqrt {z} \plus{} 1} \plus{} \frac {z}{z^2 \plus{} \sqrt {x} \plus{} 1}.
\]
(Hint: the maximum is attained when x=y=z).[/quote]
2) \[ \sum \frac{x}{x^2\plus{}y^2\plus{}1} \le \frac{1}{2} \sum \frac{x}{\sqrt{x^2\plus{}y^2}} \le \frac{3}{2\sqrt{2}}\]
:)
I will think of 3), and post it if I can... :)

[quote="Hatucdao"]Hi Can, you are fast and correct ! :)  Similar to the first one, the main point in the second inequality is also of changing into the homogeneous cases. Such change is natural and always does not meet any risk because the original one (non-homogeneous) is in fact equivalent to the later one (homogeneous). Anyway, could you please explain some words about the estimate?
\[ \sum \frac {x}{\sqrt {x^2 \plus{} y^2}} \le \frac {3}{\sqrt {2}}
\]
I am looking forward to seeing your treatment for 3).[/quote]
I think that you prove $ \sum \frac {1}{\sqrt {1 \plus{} a^{2}}}$ for $ abc \equal{} 1$
Or $ \sum \frac {1}{\sqrt {1 \plus{} a}}$ for $ abc \equal{} 1$

[quote="Hatucdao"]Hi Can, you are fast and correct ! :)  Similar to the first one, the main point in the second inequality is also of changing into the homogeneous cases. Such change is natural and always does not meet any risk because the original one (non-homogeneous) is in fact equivalent to the later one (homogeneous). Anyway, could you please explain some words about the estimate?
\[ \sum \frac {x}{\sqrt {x^2 \plus{} y^2}} \le \frac {3}{\sqrt {2}}
\]
I am looking forward to seeing your treatment for 3).[/quote]

This is problem 113 in the book "Old and New Inequalities". I'll rewrite the proof later :)

[quote="Hatucdao"]Hi Can, you are fast and correct ! :)  Similar to the first one, the main point in the second inequality is also of changing into the homogeneous cases. Such change is natural and always does not meet any risk because the original one (non-homogeneous) is in fact equivalent to the later one (homogeneous). Anyway, could you please explain some words about the estimate?
\[ \sum \frac {x}{\sqrt {x^2 \plus{} y^2}} \le \frac {3}{\sqrt {2}}
\]
[/quote]
I proved it by using Cauchy Schwarz :)
\[ \left( \sum \frac{x}{\sqrt{x^2\plus{}y^2}}\right)^2 \le \left( \sum (x^2\plus{}z^2)\right) \left( \sum \frac{x^2}{(x^2\plus{}y^2)(x^2\plus{}z^2)}\right) \equal{}\frac{4(x^2\plus{}y^2\plus{}z^2)(x^2y^2\plus{}y^2z^2\plus{}z^2x^2)}{(x^2\plus{}y^2)(y^2\plus{}z^2)(z^2\plus{}x^2)} \le \frac{9}{2}\]
where the last inequality is valid by AM-GM Inequality. :)

[quote="dduclam"][quote="Hatucdao"]Show that for all $ x,y,z\ge 0$ we have
\[ \frac {x}{x^2 \plus{} y \plus{} 1} \plus{} \frac {y}{y^2 \plus{} z \plus{} 1} \plus{} \frac {z}{z^2 \plus{} x \plus{} 1}\le 1
\]
and the equality holds iff x=y=z=1.

.[/quote]
We have a question : Find the [b]minimum[/b] value of expression $ P \equal{} \frac {x}{x^2 \plus{} y \plus{} 1} \plus{} \frac {y}{y^2 \plus{} z \plus{} 1} \plus{} \frac {z}{z^2 \plus{} x \plus{} 1}$ if we add the condition $ x \plus{} y \plus{} z \equal{} 3$ ?
I think it's an interesting one.  :)[/quote]
If we want to find minimum, it is very easy. Notice that
\[ x^2\plus{}y\plus{}1 \le x^2\plus{}(3\minus{}x)\plus{}1 \le 3x\plus{}(3\minus{}x)\plus{}1\equal{}2x\plus{}4 \le 10\]
Hence
\[ P \ge \frac{x}{10} \plus{}\frac{y}{10}\plus{}\frac{z}{10}\equal{}\frac{3}{10}\]
We have equality for $ x\equal{}3,y\equal{}z\equal{}0$. Therefore $ \min P\equal{}\frac{3}{10}.$ ;) :)

[quote="can_hang2007"] I proved it by using Cauchy Schwarz :)
\[ \left( \sum \frac {x}{\sqrt {x^2 \plus{} y^2}}\right)^2 \le \left( \sum (x^2 \plus{} z^2)\right) \left( \sum \frac {x^2}{(x^2 \plus{} y^2)(x^2 \plus{} z^2)}\right) \equal{} \frac {4(x^2 \plus{} y^2 \plus{} z^2)(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2)}{(x^2 \plus{} y^2)(y^2 \plus{} z^2)(z^2 \plus{} x^2)} \le \frac {9}{2}
\]
where the last inequality is valid by AM-GM Inequality. :)[/quote]
Great work! This is exactly what I have waited for! I don't like the answers such as "the solution is in the book XYZ"  :D  I was surprised if you can find it in a short time (I have needed a long time :) ).

The problem of finding minimum when x+y+z=3 is homogeneous one, and it is not in this series. Now, try the third one, please!

[quote="Hatucdao"][quote="desmrhai"] I think that you prove $ \sum \frac {1}{\sqrt {1 \plus{} a^{2}}}$ for $ abc \equal{} 1$
Or $ \sum \frac {1}{\sqrt {1 \plus{} a}}$ for $ abc \equal{} 1$[/quote]
Yes, of course they are equivalent, but they are not easier :D. Could you give a simple solution for them?[/quote]
$ (\sqrt {\frac {2}{1 \plus{} a^{2}}} \plus{} \sqrt {\frac {2}{1 \plus{} b^{2}}})^{2} \leq 2(\frac {2}{1 \plus{} a^{2}} \plus{} \frac {2}{1 \plus{} b^{2}}) \leq \frac {8c}{c \plus{} 1}$
but
$ 2\sqrt {\frac {2c}{c \plus{} 1}} \plus{} \sqrt \frac {2}{c^{2} \plus{} 1} \leq 3$
Understand? And you? your soluton? Don't say that u don't know

[quote="Hatucdao"]
$ g(x,y,z) \equal{} \frac {x}{x^2 \plus{} \sqrt {y} \plus{} 1} \plus{} \frac {y}{y^2 \plus{} \sqrt {z} \plus{} 1} \plus{} \frac {z}{z^2 \plus{} \sqrt {x} \plus{} 1}.$
[/quote]
Using Am-Gm: $ \sum\ \frac {x^2}{x^4 \plus{} y \plus{} 1} \leq\ \sum\ \frac {x^2}{x^3 \plus{} x \plus{} y} \leq\ \sum\ \frac {1}{2} \sum\ \sqrt []{\frac {x}{x \plus{} y}} \leq\ \frac {3}{2\sqrt []{2}}$
But equality doesn't occurs  :wink:

[quote="nguoivn"] Using Am-Gm: $ \sum\ \frac {x^2}{x^4 \plus{} y \plus{} 1} \leq\ \sum\ \frac {x^2}{x^3 \plus{} x \plus{} y} \leq\ \sum\ \frac {1}{2} \sum\ \sqrt []{\frac {x}{x \plus{} y}} \leq\ \frac {3}{2\sqrt []{2}}$
But equality doesn't occurs  :wink:[/quote]
It seems you are trying to change the problem into a homogeneous one. This argument works for the first two inequalities but sometime it is impossible. For example, let me propose more a problem

[b]Problem 4)[/b] Prove that for all $ a,b,c\ge 0$ then
\[ \frac {a}{a^2 \plus{} \frac {3}{2}b \plus{} 1} \plus{} \frac {b}{b^2 \plus{} \frac {3}{2}c \plus{} 1} \plus{} \frac {c}{c^2 \plus{} \frac {3}{2}a \plus{} 1}\le \frac {6}{7}.
\]
The equality holds when $ a \equal{} b \equal{} c \equal{} 1$.

This problem seems similar to the first problem in this topic, but it is in fact much more interesting. One cannot using estimate $ a^2 \plus{} 1\ge 2a$ again, because now the supremum of
\[ \frac {a}{2a \plus{} \frac {3}{2}b} \plus{} \frac {b}{2b \plus{} \frac {3}{2}c} \plus{} \frac {c}{2c \plus{} \frac {3}{2}a}
\]
is 1, which is larger than $ \frac {6}{7}$. Now please try 4) before return to 3).

[quote="desmrhai"] Understand? And you? your soluton? Don't say that u don't know[/quote]
To desmrhai: Please be more polite before talking about mathematics.

[quote="Hatucdao"][quote="can_hang2007"] I proved it by using Cauchy Schwarz :)
\[ \left( \sum \frac {x}{\sqrt {x^2 + y^2}}\right)^2 \le \left( \sum (x^2 + z^2)\right) \left( \sum \frac {x^2}{(x^2 + y^2)(x^2 + z^2)}\right) = \frac {4(x^2 + y^2 + z^2)(x^2y^2 + y^2z^2 + z^2x^2)}{(x^2 + y^2)(y^2 + z^2)(z^2 + x^2)} \le \frac {9}{2}
\]
where the last inequality is valid by AM-GM Inequality. :)[/quote]
Great work! This is exactly what I have waited for! I don't like the answers such as "the solution is in the book XYZ"  :D  I was surprised if you can find it in a short time (I have needed a long time :) ).

The problem of finding minimum when x+y+z=3 is homogeneous one, and it is not in this series. Now, try the third one, please![/quote]
$ f(x)=\sqrt{x}$ is concave. Hence
$ \sum \sqrt{\frac{a}{a+b}}=\sum \frac{a+c}{2(a+b+c)}\cdot \sqrt{\frac{4a(a+b+c)^2}{(a+b)(a+c)^2}}\leq \sqrt{\sum \frac{2a(a+b+c)}{(a+c)(b+c)}}$. Hence it suffices to prove $ \sum \frac{2a(a+b+c)}{(a+c)(b+c)}\leq \frac{9}{4}$ which is equivalent to $ (a+b)(b+c)(c+a)\geq \frac{8}{9}(a+b+c)(ab+bc+ca)$ which is well known.

Hatucdao, i think your third inequality can be solved using calculator only.
\[ g(x,y,z) \equal{} \frac {x}{x^2 \plus{} \sqrt {y} \plus{} 1} \plus{} \frac {y}{y^2 \plus{} \sqrt {z} \plus{} 1} \plus{} \frac {z}{z^2 \plus{} \sqrt {x} \plus{} 1}
\]
As you said in your hint maximum is attained when $ x \equal{} y \equal{} z$ (unfortunately i don't know how to prove that, this inequality attains maximum at $ x \equal{} y \equal{} z$). Therefore, we should find maximum of
\[ g(x) \equal{} \frac {3x}{x^2 \plus{} \sqrt {x} \plus{} 1}
\]
which is attained at $ x \equal{} 1.3154751$ and the maximum value of the given function is [b]$ 1.0314267$[/b]

Can someone give elementary proof to the this question? Using Cauchy-Schwarz maybe?

Thanks!  :lol:

[quote="Hatucdao"][b]Problem 4)[/b] Prove that for all $ a,b,c\ge 0$ then
\[ \frac {a}{a^2 \plus{} \frac {3}{2}b \plus{} 1} \plus{} \frac {b}{b^2 \plus{} \frac {3}{2}c \plus{} 1} \plus{} \frac {c}{c^2 \plus{} \frac {3}{2}a \plus{} 1}\le \frac {6}{7}.\]
The equality holds when $ a \equal{} b \equal{} c \equal{} 1$.[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-32-p-t-nam.html


$ {x_n} : x_1=\frac{-5}{2} ; x_{n+1}=\frac{1}{2}x_n^2+x_n-2 $
 Prove that : $ x_n $ is converge and find $ \lim_{n->\infty}x_n $

Denote $y_{n}=x_n+1$, then $y_1=-3/2$ and $y_{n+1}=\frac{y_n^2}{2}-\frac{3}{2}$. And it is easy to show that $y_{n+1}>y_n$ for $y_n<-1$, moreover $y_n<-1$ implies $y_{n+1}<-1$. So $y_n$ increases and bounded by 1, those $y_n\to y\leq -1$. I.e. $y=\frac{y^2}{2}-\frac{3}{2}$, therefore $y=-1$.

$ y_n <-1 \to y_{n+1} <-1 $ : I think it's wrong !

We need more complicated arguments. Actually, $y_{2n+1}$ is increasing and $y_{2n}$ is decreasing, moreover $y_{2n+1}<-1$ and $y_{2n}>-1$. And we are to show that the only appropriate root of $y=\frac{\frac{y^2}{2}-\frac{3}{2}}{2}-\frac{3}{2}$ is $y=-1$.


The third term of an arithmetic sequence is 13 and the eighth term
is 38. What is the product of the first and tenth terms?

First you find the common difference.
38-13=25 and 8-3=5. So the common difference is 25/5=5
So the first term would be 13-5-5=3.
And the tenth term would be 38+5+5=48.
So it would be 144.

idk if theres a faster way...


algebra
polynomial
floor function
calculus
integration
function
domain

$P(x)\in\mathbb{Z}[x]$. Prove, that if polynomials $P(x)$ and $P(P(P(x)))$ have common real root, then they have common integer root.

[quote="ishfaq420haque"]what does it mean by $ P(x)\in Z\lfloor x rfloor$
CAN YOU elaborate your solution process[/quote]
$ P(x)\in \mathbb{F}[x]$ means that, this is a polynomial of $ x$ and all of the coefficients of this polynomial lie in the field $ \mathbb{F}$
and here it means that the coefficients lie in the integral domain $ \mathbb{Z}$


Hey guys,

I was just wondering how you guys study math.  For the past year I tried doing a lot of past contests for whatever contest was coming up, and although I learned some things, it never seemed like I made too much progress, and sometimes, I never even knew what was going on, by reading the Kalva solution.

Well part of the problem is that I haven't ever covered some important topics (like for example, I've never officially done any Inequalities seminar/problem set).  So i think before I do all the past contests I need to learn these basic things first.  So I was wondering what's the best way to get confortable with certain new topics (inequalities, transformational geometry, modular tables)...I don't even know what some of these topics that I need to learn may be.

I think it's a lot easier to learn them by doing a whole bunch of questions on that same topic at once instead of coming across one once over a lot of past contests.  So are there any good books/websites/services(ie.tutor) that have like lectures/whatever on these topics?  Like it teaches basic concepts, gives examples, gives problems, and gives solutions.  That's what I think is the best way to learn math.  Wondering how you guys all got so good :)

Thanks for your help.

EDIT: I also put this post in some other AOPS forum sections.  No one will mind, right?

The resources section on AoPS should give you enough to start with.


calculus
integration
calculus computations


last week i got The Art of Problem Solving volume 1. and when i looked through it, it seemed impossible. i tried the first few chapters and i had a lot of trouble with them. im in 7th grade and do u think this book is too hard for me?

The AoPS books give generally good explanations. If you still can't stuggle though the book reading line by line, very carefully, then you may want to look into some Introductory books before tackling Vol 1. Look in the bookstore for a big selection of books for a wide range of levels.


[quote="Fitim"]If $f: N\to N$, find functions
$f((m+n)+(m-n))=8m$

 :wink:[/quote]
f(m)=4m if m is even, f(m)=a_m if m is odd, where $(a_{m})$ is a sequence in $N$. :D

Sorry! I have a mistake, see now please! :wink:  :)

[quote="Rust"][quote="Fitim"]If $f: N\to N$, find functions
$f(f(m+n)+f(m-n))=8m$

 :wink:[/quote]
Can be $n\ge m$, what is $f(m-n)?$.
May be $f: Z\to Z$? (Solution is $f(n)=2n$.)[/quote]

$m\geq n$
Sorry again! :(

[quote="Fitim"]
$m\geq n$
Sorry again! :([/quote]
So $0\in\mathbb{N}$?

im pretty sure it would be over all integers and not just naturals, otherwise i cant see how to solve it immediately.

for Z->Z:

m=f(a+b) n=(a-b) for some a,b

=> f(8a+8b)=8f(a+b) [by subbing into original equation and then using the original equation to simplify]

=>f(0)=0

Subbing n=m into original equation:

f(f(2m))=8m

if f(f(x)) is bijective  => f(x) bijective

=> f(m+x) + f(m-x) = 2f(m) for integers x [from orig. eqn, and obtaining RHS from subbing n=0]

=> Varying x over positive integers shows f(m) is arithmetic progression.

subbing f(m)=cm into orginal equation:

f(2cm)=8m

=> c^2=4

$\Rightarrow f(n)= \pm 2n$


I would like to have full solution to these questions ^^

[b]Problem 1[/b]
Which of the following solution mixtures have the highest pH value?
A.  100ml 0.02M $H_{2} SO_{4}$ + 100ml 0.02M $NaOH$
B.  100ml 0.02M $HNO_{3}$ + 100ml 0.02M $NaOH$
C.  100ml 0.02M $HCl$ + 100ml 0.02M $NaOH$
D.  100ml 0.02M $CH_{3}COOH$ + 100ml 0.02M $NaOH$
E.   100ml 0.02M $HCl$ + 100ml 0.02M $NH_{3}.H_{2}O$

More problems coming soon!!
The problems are from a chemistry competition(I forgot its name).

Highest pH = most basic

A = pH less than 7 - insufficient base to neutralize diprotic sulfuric acid
B-C.  neutral salt of strong acid -strong base  pH 7
D.  salt of weak acid + strong base creates negatively charged carboxylate pH>7
E. salt of strong acid + weak base = positively charged ammonium ion pH < 7

[b]Problem 2[/b]
The set of correct stoichiometric coefficients for balancing the following reaction is
$Fe^{2+} + Cr_{2}O_{7}^{2-} + H^{+} \to Fe^{3+} + Cr^{3+} + H_{2}O$

A. 1,1,14 ; 1,2,7
B. 6.1.14 ; 6,1,7
C. 6,1,7  ; 6,2,7
D. 6,1,14 ; 6,2,7
E. none of the above

It's a redox reaction. Look at the reduction half: $Cr_{2}O_{7}^{2-} \to Cr^{3+}$

First, balance the Oxygens by addition of $H_{2}O$ to product. You thus have: $Cr_{2}O_{7}^{2-} \to 2 Cr^{3+} + 7 H_{2}O$
Now, add Hydrogens to products to balance Hydrogens and then add electrons to balance the charges:
$6 e^{-} + 14 H^{+} + Cr_{2}O_{7}^{2-} \to 2 Cr^{3+} + 7 H_{2}O$

Reduction is balanced. Now onto oxidation: $Fe^{2+} \to Fe^{3+}$
You just have to add an electron to balance. $Fe^{2+} \to Fe^{3+} + e^{-}$

You now have both oxidation reaction and reduction reactions balanced. Now, you need to multiply oxidation reaction by 6, so that net change in electrons is 0. so you have:
$6 e^{-} + 14 H^{+} + Cr_{2}O_{7}^{2-} + Fe^{2+} \to 2 Cr^{3+} + 7 H_{2}O + 6Fe^{3+} +6e^{-}$

The electrons cancel. You have 6, 14, 1 $\to$ 2, 7, 6 . thus, D is the answer


I don't understand a solution that I saw to the source problem above.

When dividing $ f(x)\equal{}ax^3\plus{}bx^2\plus{}1$ by $ Q(x)\equal{}x^2\minus{}x\minus{}1$, how would you know to put the cubic function in the following form:

$ ax^3\plus{}bx^2\plus{}1\equal{}(ax\plus{}(a\plus{}b))(x^2\minus{}x\minus{}1)\plus{}(2a\plus{}b)x\plus{}(a\plus{}b\plus{}1)$ to more easily divide?

[quote="aufha"]I don't understand a solution that I saw to the source problem above.

When dividing $ f(x) \equal{} ax^3 \plus{} bx^2 \plus{} 1$ by $ Q(x) \equal{} x^2 \minus{} x \minus{} 1$, how would you know to put the cubic function in the following form:

$ ax^3 \plus{} bx^2 \plus{} 1 \equal{} (ax \plus{} (a \plus{} b))(x^2 \minus{} x \minus{} 1) \plus{} (2a \plus{} b)x \plus{} (a \plus{} b \plus{} 1)$ to more easily divide?[/quote]

You know that $ a(x) \equal{} p(x)b(x)\plus{}r(x)$ with $ \deg{r}<\deg{a}.$ This implies that $ \deg a \equal{} \deg b \plus{} \deg p.$ So you just set $ ax^3\plus{}bx^2\plus{}1 \equal{} (cx\plus{}d)(x^2\minus{}x\minus{}1) \plus{} ex^2\plus{}fx\plus{}g$ and compare coefficients. You could also use long division (http://mathworld.wolfram.com/LongDivision.html).

[quote="aufha"]Thanks for explaining it to me.
Which method would you prefer and why?[/quote]

Well generally long division is the way to go for dividing polynomils. but if you have some other decomposition, comparing coefficients will almost alway work.


is magnetic susceptibility C/T or Cu/T, where is u is permittivity of free space?

Why don't u try wikipedia and don't double post.

the double post was by mistake, dunno how it happened.
wiki says it is C/T but M=CB/T where M is magnetisation and using this we get magnetic susceptibility is Cu/T(which agrees with ncert and contradicts wiki)

so i wanted to know which is correct


articles
geometry
geometric transformation
reflection
probability
expected value

$ N$ people play a game, each choosing a number between 0 and 100 inclusive. The person who has chosen the number closest to the mean of the $ N$ numbers times $ \frac{2}{3}$ wins. Assuming all who play are intelligent, what should each choose?

I think everyone choosing $ 0$ is a Nash equilibrium :?: ? I'm not sure how to show it's the only one though.

Oh, right.  Well, in any other collection of mixed strategies, someone's strategy has an expected value at least that of the mean, so that guy has something to gain from changing his strategy if he knows everyone else's.

A good lesson to learn from this game is that the optimal strategy is worthless if nobody plays optimally.

[quote="t0rajir0u"]It's pretty easy to see that this game has no Nash equilibrium, if that's what you're getting at.  When I play this game, I choose $ 100$ just to throw everyone else off :P[/quote]

 Then you'd never win because the highest mean is 100, and 2/3 of that makes 66. :P
I'd choose 33 if people are picking arbitrary numbers

"Throw everyone else off" was not intended to be compatible with "win."  For what it's worth, when I played this game the winning average was about 15.  It would have been about 13 if myself and another person hadn't chosen 100.

We played this game in my game theory class at CTY, and at first everyone guessed random numbers. The person who guessed close to 66 won the first round then the person who guessed 44 won then the person close to fifteen won. What was happening was that after the number close to 66 won, everyone picked a number close to 66 so that they could obtain the reward. When everybody did this 2/3 of the average dropped to 44. when it got to 15 the whole class noticed the pattern and the game regained its random selection. It seems that the best time to implement a strategy is the turn after no one else does. This game is like the prisoner's dilemma, when it is played as a single event there is no winning strategy. When the same players play the same game multiple times, however a winning strategy can be attained.  :D

In my experience, the number of times you multiply by $ \frac{2}{3}$ when playing this game for the first time is a rough measure of how far you think ahead (or at least, how the extent to which you think ahead measures up against your perception of how far everyone you're playing with thinks ahead).

I haven't taken any classes in game theory as I'm in 8th grade right now, but I've read about the traveler's dilemma. Looking at the wiki article, I'm sorta confused. HELP?!

Wiki:

In game theory, the traveler's dilemma (sometimes abbreviated TD) is a type of non-zero-sum game in which two players attempt to maximize their own payoff, without any concern for the other player's payoff.

The game was formulated in 1994 by Kaushik Basu and goes as follows:[1][2]

An airline loses two suitcases belonging to two different travelers. Both suitcases happen to be identical and contain identical antiques. An airline manager tasked to settle the claims of both travelers explains that the airline is liable for a maximum of $100 per suitcase, and in order to determine an honest appraised value of the antiques the manager separates both travelers so they can't confer, and asks them to write down the amount of their value at no less than $2 and no larger than $100. He also tells them that if both write down the same number, he will treat that number as the true dollar value of both suitcases and reimburse both travelers that amount. However, if one writes down a smaller number than the other, this smaller number will be taken as the true dollar value, and both travelers will receive that amount along with a bonus/malus: $2 extra will be paid to the traveler who wrote down the lower value and a $2 deduction will be taken from the person who wrote down the higher amount. The challenge is: what strategy should both travelers follow to decide the value they should write down?
Naively, one might expect a traveler's optimum choice to be $100; that is, the traveler values the antiques at the airline manager's maximum allowed price. Remarkably, and, to many, counter-inuitively, the traveler's optimum choice is in fact $2; that is, the traveler values the antiques at the airline manager's minimum allowed price.
For an understanding of this paradoxical result, consider the following rather whimsical proof.
* Alice, having lost her antiques, is asked their value. Alice's first thought is to quote $100, the maximum permissible value.
    * On reflection, though, she realises that her fellow traveler, Bob, might also quote $100. And so Alice changes her mind, and decides to quote $99, which, if Bob quotes $100, will pay $101!
    * But Bob, being in an identical position to Alice, might also think of quoting $99. And so Alice changes her mind, and decides to quote $98, which, if Bob quotes $99, will pay $100! This is greater than the $99 Alice would receive if both she and Bob quoted $99.
    * This cycle of thought continues, until Alice finally decides to quote just $2 - the minimum permissible price!
The $2 in this instance is the Nash equilibrium point for the game. However, when the game is played experimentally, most participants select the value $100 or a value close to $100, including those who have not thought through the logic of the decision as well as those who understand themselves to be making a non-rational choice. Furthermore, the travelers are rewarded by deviating strongly from the Nash equilibrium in the game and obtain much higher rewards than would be realized with the purely rational strategy. These experiments fail to show that either the majority of people use purely rational strategies, or that they would be better off financially if they did. This paradox has led some to question the value of game theory in general, while others have suggested that a new kind of reasoning is required to understand how it can be quite rational ultimately to make non-rational choices. Note that the $100 choice here is the optimal pure strategy under a different model of decision making called superrationality, which assumes that all logical thinkers must use the same strategy, so pure superrational strategies are restricted to the diagonal of the payoff matrix.
My thoughts:
Well, no matter what the other person picks (besides $2 or $100 in which case only one of the following results is possible) you will always get either $1 more than what they say or less For example, If they pick $44 the most you can get is $45 from picking $43. The least you can get (if you pick higher than them) would be $42. Therefore, the best choice would be to pick high, right? If you pick lower than them, you are bound to receive from $4 (not $4 more, $4 as what you actually get) to $1 more than what they chose(in this case, $43). If you pick higher than them, you are bound to get two dollars less than what they have picked. You have a much better chance of getting a reasonable amount if you pick higher than them. To clarify on my previous statement about having a better chance of getting a reasonable amount; if they chose $44 and you chose something beneath them, you could possibly get $1 more than that ($43), $0 more ($42), or anywhere from $1 less to $40 less ($41-$2). If you pick higher you will only get $2 less at most. Because you do not know what they are going to pick, the best choice would be $100 (possibly $99). I think that is fair reasoning for one hoping for the highest outcome.

So is the wiki article wrong?

Whenever your intuition disagrees with something that's written down, you should always assume you're wrong until proven right.  To do anything else would be incredibly arrogant.

The problem here is, I think, that "is a Nash equilibrium" means "is locally optimal" in some very precise technical sense, and this precise technical sense does not always correspond to the intuitive sense of what is optimal.  In particular, it is frequently the case in practice that two players playing two-player game (especially iterated) will cooperate and so increase their total welfare even when the "rational" thing to do (again in some precise technical sense) in any single play is to do something that will increase your personal payout at much larger cost to your opponent.  There are lots of possible explanations for this phenomenon, some of which are covered by the chunk of the article you quoted.  I think one very compelling one is that players in real life very rarely find the axioms of (this version of) game theory satisfied: you usually will have the chance to play again, or to retaliate against someone who takes personal benefit at the public expense.  Also, there's lots of evidence that people place a positive economic value on fairness, i.e., people will often irrationally sacrifice some level of personal well-being in order to achieve some level of fairness or justice.

If you aren't familiar with the precise meaning of Nash equilibrium, it's worthwhile to familiarize yourself with it -- to understand why the answer is what it is, you really have to know what the question is asking  :)

'kay, I've done some more research....Sorry if I was being arrogant or anything :maybe: , well I wasn't really saying the entire wiki article was wrong...just that the part about it being the most rational choice to pick $2...whatev, sorry.


Let $ c(t)$ be a curve in $ R^n$ parametrized by arc length with the property that $ |c(t)|^2$ has a local maximum at $ t_0$. Let $ p_0\equal{}c(t_0)$ and $ \rho^2\equal{}|p_0|^2$.
Show that
\[ \kappa(t_0)\geq \frac{1}{\rho}
\]
where $ \kappa(t_0)\equal{}|\ddot{c}(t_0)|$ (which is the curvature of $ c(t)$ at $ t_0$ if defined.)

Even hints are appreciated. Please help.

$ \left|c(t)\right|^2$ has a local maximum at $ t_0$ implies that $ \frac{d^2}{dt^2}\left|c(t)\right|^2|_{t_0}\equal{}2\dot {\vec c}(t_0)^2\plus{}2\vec c(t_0)\cdot \ddot{\vec c}(t_0)\leq 0$


I need an answer in standard form. Please explain in full steps. Thank you.

6i/3+i

[quote="JRav"]What do you mean "standard form"?[/quote]

What I mean is that I want you to write an answer in standard form. Of course, the problem is given below the first post here.

I think the question J Rav is asking, is "What is standard form". What is the standard form?

1) Use parentheses.  The answer to the question you've written is different than the answer to the question you were trying to write.

2) Define your terms, so other people know what you're talking about.

3) Do your own homework.

Standard form is probably the form $ a \plus{} bi, a, b \in \mathbb{R}$.  The question you intended to ask is probably $ \frac{6i}{3 \plus{} i}$.  Think conjugation.

I'll go based on t0rajir0u:
$ \frac{6i}{3\plus{}i}\left(\frac{3\minus{}i}{3\minus{}i}\right)\equal{}\frac{18i\plus{}6}{10}\equal{}\frac{9i\plus{}3}{5}\equal{}\frac{3}{5}\plus{}\frac{9}{5}i$, assuming I didn't mess up my multiplication.


Small but tricky Problem :

[color=blue]Problem.[/color]Find all of the pairs of natural numbers like $(n,a)$ such that : \[n^{2}< a^{3}< (a+1)^{3}<(n+1)^{2}\] [hide="My solution"]We will show that no such pair of numbers exist.Assume the contrary ,that there exist such a $(n,a)$ and  rewrite the equation in this way: \[(\frac{n}{a})^{2}<a <a+1 <(\frac{n+1}{a+1})^{2}\]  \[\frac{n}{a}< \frac{n+1}{a+1}\longrightarrow n < a \longrightarrow (n+1)^{2}<(a+1)^{3}\] Which clearly is a contadiction.

[/hide]

[quote="lomos_lupin"] \[(\frac{n}{a})^{2}<a <a+1 <(\frac{n+1}{a+1})^{2}\] [/quote]

Edit:  My apologies.  I didn't quite realize what you were doing; this step is valid after all.

[hide="Solution"] The problem condition requires $(a+1)^{3}-a^{3}< (n+1)^{2}-n^{2}$.  In other words,

$3a^{2}+3a+1 < 2n+1$

However, since $n^{2}< a^{3}$, it follows that

$n < a^{ \frac{3}{2}}$
$2n+1 < 2a^{ \frac{3}{2}}+1 < 3a^{2}+3a+1$

Hence no such $(a, n)$ exist.  Note that it is not necessary that $a$ or $n$ be integer.  The problem statement holds for real $a,n \ge 1$. [/hide]

[quote="killmesoft"]how do you guys write math symbols on the forum?[/quote]
Use latex. Put dollar signs around the math formulas. There is a forum on aops about Latex.

[quote="Elyot"]
Let $n = x^{3}$ and $a = y^{2}$
[/quote]
Creative substitution .


inequalities
abstract algebra
number theory
greatest common divisor
prime numbers

[hide="does this work?"] We know that $ (x^{2}\minus{}y^{2},2xy,x^{2}\plus{}y^{2})$ for $ (x,y)\equal{}1$ generates all Pythagorean Triples. Thus $ p^{2}\equal{}x^{2}\minus{}y^{2},q^{2}\equal{}2xy$. The second equation gives $ q\equal{}2,xy\equal{}2$, so $ x\equal{}2,y\equal{}1$. Therefore, $ p^{2}\equal{}3$, contradiction.[/hide]

Another approach:
If there is any one out of 3 variables which is divisble by 2, then it must be equal to 2.
If a=2 or b=2 we easily infer that the equation doesn't have any prime solutions!
If c=2 then we will have the inequality: a^4<=16 then easily to infer that a=2. But then b=0(contradicts to th assumption)
Now, if there is not any variables which are divisible by 2, then a^4+b^4 module 4 is equal to 2.
But c^4 module 4 is equal to 1(contradiction since a^4+b^4=c^4)
From 2 cases above, we are done!

[hide="One Line"]
Fermat's last Theorem, since all primes lie in the Natural Numbers
[/hide]
Is it valid in a contest? Absolutely  :icecream:

[quote]Fermat's last Theorem, since all primes lie in the Natural Numbers[/quote]

You are absolutely right but it was for 2.nd grades Maybe just a few students know about Fermat's last theorem.

[quote="t0rajir0u"][quote="CatalystOfNostalgia"]Thus $ p^{2} \equal{} x^{2} \minus{} y^{2},q^{2} \equal{} 2xy$. The second equation gives $ q \equal{} 2,xy \equal{} 2$, so $ x \equal{} 2,y \equal{} 1$. [/quote]

I'm not sure how you concluded this.  We can have $ x \equal{} 2a^2, y \equal{} b^2$ where $ \gcd(a, b) \equal{} 1$ and $ b$ is odd.  The proof that this equation has no solution is usually done by infinite descent.[/quote]
You're right, t0raijru! :) .But it isn't essential in this case since the restriction:"prime numbers" make the problem more easier than the original problem. That maybe is suitable for the level of student in that age.

[quote]I'd assume Fermat's Last Theorem wouldn't be accepted as it trivializes the problem.[/quote]

It could be interesting if someone solved problem by Fermat's Last theorem but I assume that nobody of contestants know about it ( 2.nd grade) Most probabaly solution based on this theorem would be accepted!

$ p^4\equal{}(r^2\plus{}q^2)(r\plus{}q)(r\minus{}q)$. Note that $ r^2\plus{}q^2>r\plus{}q>r\minus{}q$. Then $ r^2\plus{}q^2\equal{}p^3$, $ r\plus{}q\equal{}p$, $ r\minus{}q\equal{}1$. Then $ r\equal{}\frac{p\plus{}1}{2}$, $ q\equal{}\frac{p\minus{}1}{2}$, in the first equation, $ (p\minus{}1)(2p^2\plus{}p\plus{}1)\equal{}0$ which has no prime solution.

There is a solution in Aops volume 2:

[hide="outline"]pythagorean triple and infinite descent.[/hide]

A very easy approach of this problem:

If p and q are odd, then r is even, but if r is even, since r is prime, r=2.Here, it's easy to see that there aren't solutions for p^4+q^4.
So, at least one of this primes is even.Wlog, let p this prime, so p=2.Now, we have:16=(r²2+q²)(r²-q²) . Here, it's again easy to see that there aren't solutions.So we are done.

Bye bye :)   :P

Maybe I'm being stupid, but just Fermats last theorem????

The point is to prove this result independently.  The case $ n \equal{} 4$ was known even to Fermat, and the case for $ x, y, z$ prime is even easier.

1. Ok, fine.

2. This is a 2ND GRADE competition??

I can't really see how 2nd graders would be doing this.

Perhaps in Bosnia, Grade 2 does not equate to Grade 2 where you're from.
In California, Grade 2 is generally ages 7-8.
This does seem a bit difficult for that age bracket, but again, grade 2 in Bosnia is most likely seen differently.

[hide="easy..."]
Let $p< q<r$. Because $r$ and $q$ are odd, $p$ must be even, so $p=2$.
$16=(r^2-q^2)(r^2+q^2)$, so because $\text{gcd}(r^2-q^2,r^2+q^2)=2$, $r^2-q^2=2$ and $r^2+q^2=8$, witch give us solution $r^2=5$. Contradiction![/hide]

1)if $p,q>2=>r=2$ no solutions))
2)if either $p$ or $q$ equals $2=>16=(r-q)(r+q)(r^2+q^2)$ no solutions))
very easy one) :D

[quote="t0rajir0u"][quote="CatalystOfNostalgia"]Thus $ p^{2} \equal{} x^{2} \minus{} y^{2},q^{2} \equal{} 2xy$. The second equation gives $ q \equal{} 2,xy \equal{} 2$, so $ x \equal{} 2,y \equal{} 1$. [/quote]

I'm not sure how you concluded this.  We can have $ x \equal{} 2a^2, y \equal{} b^2$ where $ \gcd(a, b) \equal{} 1$ and $ b$ is odd.  The proof that this equation has no solution is usually done by infinite descent.[/quote]

Why do you think Pythagorean triples are wrong since one can take $a^2=p^4,b^2=q^4,c^2=r^4$ and get Pythagoras equation ?????

That method is perfectly ok!

$p, q, r\in\mathbb{P}$
$p^2=u^2–v^2$
$q^2=2xy$ not prime since $\sqrt{2}.k$
$r^2=u^2+v^2$ clearly not primes....

Work in modulo 6 and the answer will be revealed.

[quote=delegat]Prove that equation $ p^{4}\plus{}q^{4}\equal{}r^{4}$ does not have solution in set of prime numbers.[/quote]

We can prove by "descente infinite" that the equation $ p^{4}\plus{}q^{4}\equal{}r^{2}$ have not positive integer solution, so the equation $ p^{4}\plus{}q^{4}\equal{}r^{4}$ have not too positive integer solutions, even more prime solutions!

Yea somewhat like @rikabcd123456 says can't we simply work modulo $4$ ? $(p,q,r) \equiv \pm 1 \pmod 4 \implies p^4,q^4,r^4 \equiv 1 \pmod 4$.Then the LHS $\equiv 2 \pmod 4$ whereas RHS $\equiv 1 \pmod 4$.We are done.

@targo all primes are 4k+1 or 4k+3 except 2.

[quote=AYG]Yea somewhat like @rikabcd123456 says can't we simply work modulo $4$ ? $(p,q,r) \equiv \pm 1 \pmod 4 \implies p^4,q^4,r^4 \equiv 1 \pmod 4$.Then the LHS $\equiv 2 \pmod 4$ whereas RHS $\equiv 1 \pmod 4$.We are done.[/quote]

How you got $p,q,r\equiv \pm 1\pmod{4}$ ? ;)
You need to consider that , as , $p,q<r$ , we will have one of  $p$ or, $q$ equals to $2$,to maintain the parity.So, WLOG if you take $p<q$ then,$p=2$ .

Is this ok?

We can clearly see that $p=q=2$ is not a solution.

$q^4=(r^2+p^2)(r+p)(r-p)$
So if $p\neq 2 $ we get that no part of right side is equal to $1$ and no part is negative. Thus we have that the right side is "made" of $q^2, q, q$. So one part in brackets is equal to one of this. But that would mean that 2 of them would be equal which makes no sense because we have prime numbers which are greater than $1$. So that doesn't work. Now if we have $p=2$ we just switch $p$ and $q$ and we get the same equation.

[quote=delegat]Prove that equation $ p^{4}\plus{}q^{4}\equal{}r^{4}$ does not have solution in set of prime numbers.[/quote]

Lemma: Let $\left(a,b,c\right)$ be a Pythagorean triple. Then $abc$ is a multiple of $60$.
Proof: We begin by proving that one of $a$,$b$,$c$ is a multiple of $3$. Assume the contrary, then $c^2=a^2+b^2=2\left(\operatorname{mod}3\right)$. This is impossible hence one of $a$,$b$,$c$ is a multiple of $3$. Now take the numbers modulo $8$. If no $a$,$b$,$c$, then $c^2=1,4\left(\operatorname{mod}8\right)$ and $a^2+b^2=2,5,0\left(\operatorname{mod}8\right)$. Impossible, hence one of $a^2$, $b^2$, $c^2$ is a multiple of $8$ which implies one of $a$,$b$,$c$ is a multiple of $4$. Finally, take the numbers modulo $5$. Assume none are a multiple of $5$. Then
$c^2=1,4\left(\operatorname{mod}5\right)$ and $a^2+b^2=2,3,0\left(\operatorname{mod}5\right)$. Hence one of $a$,$b$,$c$ is a multiple of $5$.
$60=3\cdot4\cdot5$ hence $abc$ is a multiple of $60$.

From the lemma, $p^2q^2r^2$ is a multiple of $60$ and hence $pqr$ is a multiple of $30$. As they are primes, the only possible solution would be $\left(p,q,r\right)=\left(2,3,5\right)$. These do not work hence there are no solutions.

If $p, q, r$ are all odd primes, then there are no solutions by simply taking both sides mod 4 (LHS is $2 \pmod 4$ while RHS is $1 \pmod 4$). Hence at least one of $p, q, r$ is $2$. Clearly, it is not possible for $p=q=r=2$. It is also easy to check that we cannot have $2$ of the 3 variables be equal to 2 and the last one be an odd prime. This means that exactly one of the 3 variables must be equal to 2. If $r=2$, this is not possible as we get $p^4+q^4=16$, and the minimal value of the left side is $3^4+3^4=162 > 16$. So, WLOG let $p=2$. Therefore,
$$16+q^4=r^4 \implies 16=r^4-q^4 \implies 16=(r^2+q^2)(r+q)(r-q).$$ However, the minimal value of $r^2+q^2$ is $3^2+3^2=18$, so there are no solutions here. Since we've covered all the possibilities, we can conclude that there are no solutions.

Fermat's last theorem and directly it has no solution

Let $p, q, r$ are all $>3$ then must $p^4+q^4\equiv 2\pmod 6$ but $r^4\equiv 1\pmod 6$ done. Rest is case check of $\{2,3\}$.

Groupsolved with [b]IAmTheHazard, sixoneeight, CT17, CT18, megarnie, OronSH, MathLuis, pikapika007, Martin2001, naman12, IAteTheHazard, tienxion, Evil Chin Inc
[/b]

Follows immediately by Fermat's Last Theorem.

I think this was relatively simple - but maybe I did something wrong?
[hide=maybe solution]We will use FLT to solve this equation. By FLT, for any prime values of $a$, $a^4 \equiv 1 \pmod 5$. Taking mod 5 shows that $1+1 \equiv 1 \pmod 5$, or $1 \equiv 0 \pmod 5$. We can easily see that this is impossible, showing that this solution has no solution in a set of prime numbers.[/hide]

It follows immediately by Fermat's Last Theorem.

Or any 4th power of a prime is always 1 mod 3 besides if the prime is 3. Hence contradiction so one of the primes is 3. If r=3 no solutions. quick check yeilds no 4th powers have difference of 81.

[quote=qwerty123456asdfgzxcvb][quote=PaperMath]How do you instantly notice that any prime number to 4th power is always $1 \pmod 3$ expect for 3?[/quote]

any number to a 4th power mod 3 thats not 0 is equal to 1, since any number squared mod 3 = 1.[/quote]

Uhhhhh that doesn’t answer my question. That is literally what im asking, how people can identify those relationships better.

Can someone also check my solution? -
[hide=maybe solution]We will use FLT to solve this equation. By FLT, for any prime values of $a$, $a^4 \equiv 1 \pmod 5$. Taking mod 5 shows that $1+1 \equiv 1 \pmod 5$, or $1 \equiv 0 \pmod 5$. We can easily see that this is impossible, showing that this solution has no solution in a set of prime numbers.[/hide]

The mod $3$ argument is nearly identical to what you did for mod $5$. In general, if you reduce $a^k$ mod $p$, you'll get $1$ if $p-1\mid k$. So a good place to start is with primes $p$ for which $p-1$ is a factor of the exponent.

I just realized that my solution assumed that $gcd(a,5)=1$, which is obviously not true for $a=5$. So I will include the case $a=5$ here. [hide=a=5 case]we will continue to take mod 5 on the equation $p^4+q^4=r^4$. We can use mod 5 to deduce that either $p$ or $q$ (not both) is equal 5, or everything is equal to 5, which is obviously not true, since all the other cases lead to contridictions in mod 5. WLOG $p=5$ We can then deduce that there are no solutions since $r>q$, and $r$ must be prime, which is odd, $q$ must be even, or $2$, which lead to no cases.[/hide]

[quote=hsiljak]Grade 2 (more precisely, high school 2nd grade) in Bosnia is equivalent to grade 10, i.e. 15-16 year old students.[/quote]

I mean there are a significant amount of 10 graders that know fermat's last (like me)

[hide=sol using mods+alternate idea]By mod 16 we see that r is odd and p, q are different parities. Let p=2 WLOG. Then q^r and r^4 differ by 16. The smallest difference a^4 - b^4 for a,b >= 2 is 27-8=19. So obviously this is not possible, proven. Another method would be FLT to show that one of p,q is 5, and 2^4 + 5^4 = 641 which is not a prime's 4th power.[/hide]

[hide=solution using mod 6]
Claim: $p \equiv 1,5 \pmod{6}$ and $q \equiv 1,5 \pmod{6}$.
Proof: By the definition of a prime number, it is leaves a non-zero remainder when divided by $6$. If the remainder is $2$ or $4$, then $p$ or $q$ is even, a contradiction. If the remainder is $3$ then it is a multiple of $3$ contradiction. Therefore, the only possible remainders are $1$ and $5$, so the claim is proved.

Next, notice that $p^4 \equiv 1 \pmod{6}$, and $q^4 \equiv 1 \pmod{6}$. This means that the left side is congruent to $2 \mod{6}$. $r$ is also prime so it is congruent to $1,5 \mod{6}$. This means $r^4$ or the left side is $1 \mod{6}$. Contradiction.

If $p$, $q$, or $r$ equal 2 or 3, then there still are no solutions.

Therefore, there are no solutions.[/hide]

[quote=Awesome_Twin1][hide=solution using mod 6]
Claim: $p \equiv 1,5 \pmod{6}$ and $q \equiv 1,5 \pmod{6}$.
Proof: By the definition of a prime number, it is leaves a non-zero remainder when divided by $6$. If the remainder is $2$ or $4$, then $p$ or $q$ is even, a contradiction. If the remainder is $3$ then it is a multiple of $3$ contradiction. Therefore, the only possible remainders are $1$ and $5$, so the claim is proved.

Next, notice that $p^4 \equiv 1 \pmod{6}$, and $q^4 \equiv 1 \pmod{6}$. This means that the left side is congruent to $2 \mod{6}$. $r$ is also prime so it is congruent to $1,5 \mod{6}$. This means $r^4$ or the left side is $1 \mod{6}$. Contradiction.

Therefore, there are no solutions.[/hide][/quote]

Don't forget to account for the cases of $r=2,3$ as these are different!!


What is the best way to study (except for the AoPs books i have them already) for Mathcounts?
And plz don't tell me to go to the library it sucks(elementary schools  :roll: )

Work on old tests and do problems that are harder than what you'll be seeing.  If you can do the hard stuff, you can surely do the easy stuff, right?

Um.... Isn't the library for .... literature?

elementary kids can do mathcounts  :huh:

anyways, are you going to nats this year?

if so, what place/score/state?

[i]Some[/i] libraries have an extensive [i]mathematical[/i] literature collection, dynamo.  :)

I'm not really in a position to give advice on how to do well  :blush: , but do lots and lots of problems. And more problems. And yet more problems. And lots and lots of problems. And lots and lots and lots of problems. And be sure to do more problems. Also, doing more problems can be helpful.

Wish I had known that advice earlier, now there's no more Mathcounts for me  :noo: .

Dood, this question has been posted hundreds of times on this forum.  It gets really annoying, b/c there's no secret method.  If there was, no one would tell you, so... yeah.

HINT: Use the search function.

[quote="Poincare"]How much do the tests cost ?[/quote]
Each year's set probably is around 24ish dollars...(I think) They're rather expensive, though. Also, you can ask your coach if he/she has any past problems.


pigeonhole principle
inequalities unsolved
inequalities

This doesn't seem too bad . . . define $b_i = a_{i+1} - a_i, 1 \le i \le 99$. We are given $\sum_{i=1}^{99} b_i = 0$ and $|b_i| \ge 1996 \forall i, 1 \le i \le 99$. As there are $99$ of these, by pigeonhole, $45$ have the same sign. WLOG, say this sign is negative (if not, switch all signs). So now, the sum of the $b_i$ that are negative is at most $-45 \cdot 1996$, i.e. the sum of the positive $b_i$ is at least $45 \cdot 1996$. But then, again by pigeonhole, one of these has magnitude at least $\frac{45 \cdot 1996}{44}$, which can be worked out to be more than $2037$.

I had exactly the same train of thoughts, but when you apply the pigeonhole principle for the second time, there are 99-45=54 positive numbers (and not 44), so that your bound (1996*45/44) is not ok.
Suppose we have k positive bi and l negative bi, with k>l (wlog)(k+l=99). The sum of the positive bi is greater than 1996k, so the sum of the l negative bi is lower than -1996k which imply that one of them is lower than -1996k/l. In that case one can find a bi such that |bi|>1996*k/l. But $\frac{k}{l}=\frac{99-l}{l}=\frac{99}{l}-1\geq\frac{50}{49}$
So one can find an integer i so that $|a_{i+1}-a_{i}|>1996\frac{50}{49}=2036.73...$
if the "numbers" are in fact integers, then $|a_{i+1}-a_{i}|\geq 2037$.


Suppose p, q, r > 0 such that p + q + r = 1. Prove that

7(pq + qr + rp) <= 2 + 9pqr.

British Mathematical Olympiad 1999, quite simple.
I'm wondering why nobody did it not yet !
Is it too easy ? 8-)

[quote="Maverick"]A hint: homogenize the ineq like this:
7(pq+qr+rp)(p+q+r)<=2(a+b+c)^3+9abc[/quote]

Probably you mean : homogenize like this :

7(pq + qr + rp)(p + q + r) :le: 2(p + q + r) + 9pqr ;-)

I've got three different solutions again 8-)

No one wants to do this ? :wink: Too easy for you ?

Yeah...too easy :)

Anyway...i might as well post a solution since noone else has...

we want to prove that

7(pq + qr + pr)(p + q+ r)=< 2 (p + q + r)^3 + 9pqr

expanding, etc. we have

2(p^3 + q^3 + r^3) + 3 pqr >= (p + q + r)(pq + pr + qr)

2(p^3 + q^3 + r^3) >= p^2(q+r) + q^2 (p+r) + r^2 (p+q)

and this holds by the rearrangement inequality.

Another idea : expand (7 - 9p)(7 - 9q)(7 - 9r). This works just fine too :)


inequalities trigonometry calculus derivative

[quote="Virgil Nicula"][color=darkred]Prove that in any acute triangle $ ABC$ there is the chain [/color]

${ \frac 12\cdot\sum\cos (B - C)\le \sum\cos A\le \frac 32}$[/quote]
First, we have:
$ \frac 12\cdot\sum\cos (B - C)\le \sum\cos A$
$ \Leftrightarrow [cos A+ cos B-cos (A-B)]+[cos B+ cos C-cos (B-C)]+[cos C+ cos A-cos (C-A)] \ge 0$
Hence $ cos A+ cos B-cos (A-B)=2cos{\frac{A+B}{2}}cos{\frac{A-B}{2}}-2cos^2{\frac{A-B}{2}}+1$
$ =1+2cos{\frac{A-B}{2}}[cos{\frac{A+B}{2}}-cos{\frac{A_B}{2}}]$
$ =1-4cos{\frac{A-B}{2}}sin{\frac{A}{2}}sin{\frac{B}{2}}\ge 1-4sin{\frac{A}{2}}sin{\frac{B}{2}}$
Similarly, we have:
$ cos B+ cos C-cos (B-C) \ge 1-4sin{\frac{B}{2}}sin{\frac{C}{2}}$
$ cos C+ cos A-cos (C-A) \ge 1-4sin{\frac{C}{2}}sin{\frac{A}{2}}$
By Cauchy-Schwarz inequality:
$ 3(sin{\frac{A}{2}}sin{\frac{B}{2}}+sin{\frac{B}{2}}sin{\frac{C}{2}}+sin{\frac{C}{2}}sin{\frac{A}{2}}) \le (sin{\frac{A}{2}}+sin{\frac{B}{2}}+sin{\frac{C}{2}})^2$
$ \le \frac{9}{4}$
It follows that:
$ 3-4\sum sin{\frac{A}{2}}sin{\frac{B}{2}} \ge 0$
Or:
$ \frac 12\cdot\sum\cos (B - C)\le \sum\cos A$

[quote="Virgil Nicula"][color=darkred]Prove that in any acute triangle $ ABC$ there is the chain [/color]

$ \boxed {\frac 12\cdot\sum\cos (B \minus{} C)\le \cdot\sum\cos A\le \frac 32}\ .$[/quote]

For the second, $ \sum\cos{A}$ is maximized when all angles are acute, and Jensen gives the desired result.

here it is another unnecesary proof of $ \sum{\cos A}\le 3/2$

$ \sum{\{a^3\equal{}ab^2\plus{}ac^2\minus{}2abc\cos A\}}$ (the sum of this 3 equations) is equivalent to $ abc(3\minus{}2\sum{\cos A})\equal{}\sum{a^3\plus{}abc\minus{}ab^2\minus{}ac^2}$ wich is greater than zero by Schur. So we must have:
$ \sum{\cos A}\le \frac{3}{2}$

Second:
$ \sum cosA\equal{}2cos{\frac{A\plus{}B}{2}}cos{\frac{A\minus{}B}{2}}\plus{}1\minus{}2sin^2{\frac{C}{2}}$
$ \equal{}1\plus{}2sin{\frac{C}{2}}(cos{\frac{A\minus{}B}{2}}\minus{}sin{\frac{C}{2}}) \le 1\plus{}2sin{\frac{C}{2}}(1\minus{}sin{\frac{C}{2}}) \le 1\plus{}2.\frac{1}{4}\equal{}\frac{3}{2}$ (AM-GM)

[color=darkred]Very nice the your proof, [b]NguyenDungTN[/b] ![/color]

[color=darkblue]Here is interestingly only the [b]first[/b] inequality ([b]LHS[/b]).

The second is well-known and is truly in any $ \triangle\ ABC\ .$

Thus, $ \boxed {\sum\cos A = 1 + \frac rR}\le \frac 32$ because $ R\ge 2r\ .$ Indeed,

$ \sum \cos A = 1 - 2\sin^2\frac A2 + 2\cos\frac {B + C}{2}\cos\frac {B - C}{2} =$

$ 1 + 2\sin\frac A2(\cos\frac {B - C}{2} - \cos\frac {B + C}{2}) =$ $ 1 + 4\cdot\prod\sin\frac A2 =$

$ 1 + \prod\sqrt {\frac {(p - b)(p - c)}{bc}} =$ $ 1 + \frac {(p - a)(p - b)(p - c)}{abc} =$ $ 1 + \frac {pr^2}{4Rpr} =$ $ 1 + \frac rR\ .$[/color]

[color=darkblue][b]Remark I[/b] (see [b]NguyenDungTN[/b]).

$ \boxed {3 - 2(\cos A + \cos B + \cos C) = (2\sin\frac A2 - \cos\frac {B - C}{2})^2 + \sin^2\frac {B - C}{2}\ge 0}\ !$

[b]Remark II.[/b] Show analogously that for any $ \lambda > 1$ there is the inequality

$ \lambda^2\cos A + 2\lambda (\cos B + \cos C)\le \lambda ^2 + 2$ with equality iff $ B = C$ and $ \sin\frac A2 = \frac {1}{\lambda }\ .$
[hide="Proof."]
$ \boxed {(\lambda^2+2)-[\lambda^2\cos A+2\lambda(\cos B+\cos C)]=2[(\lambda \sin\frac A2-\cos\frac {B-C}{2})^2+\sin^2\frac {B-C}{2}]\ge 0}\ .$[/hide]
Here are some particular cases :

$ 1\blacktriangleright$ $ \lambda : = 2\implies$ $ \cos A + \cos B + \cos C\le\frac A2\ .$

$ 2\blacktriangleright$ $ \lambda : = \sqrt 2\implies$ $ \cos A + \sqrt 2 (\cos B + \cos C)\le 2\ .$

$ 3\blacktriangleright$ $ \lambda : = \frac {2}{\sqrt 3}\implies$ $ 2\cos A + 2\sqrt 3 (\cos B + \cos C)\le 5\ .$

We'll have equality iff $ B = C$ and $ \{\begin{array}{c} 1\blacktriangleright A = 60^{\circ}\ . \\
 \\
2\blacktriangleright A = 90^{\circ}\ . \\
 \\
3\blacktriangleright A = 120^{\circ}\ .\end{array}$ [/color]

Yet another method is calculus. Let

$ \mathrm{f}(A,B)\ \equal{} \ \cos{A} \plus{} \cos{B} \minus{} \cos{(A \plus{} B)}$

Now find $ A$ and $ B$ when $ \frac {\partial f}{\partial A}$ and $ \frac {\partial f}{\partial B}$ are both $ 0$, and use the second-partial-derivative test to verify that $ f$ has a maximum at these points.

[quote="Virgil Nicula"][color=darkred]Very nice the your proof, [b]NguyenDungTN[/b] ![/color]

[color=darkblue]Here is interestingly only the [b]first[/b] inequality ([b]LHS[/b]).

The second is well-known and is truly in any $ \triangle\ ABC\ .$

Thus, $ \boxed {\sum\cos A = 1 + \frac rR}\le \frac 32$ because $ R\ge 2r\ .$ Indeed,

$ \sum \cos A = 1 - 2\sin^2\frac A2 + 2\cos\frac {B + C}{2}\cos\frac {B - C}{2} =$

$ 1 + 2\sin\frac A2(\cos\frac {B - C}{2} - \cos\frac {B + C}{2}) =$ $ 1 + 4\cdot\prod\sin\frac A2 =$

$ 1 + \prod\sqrt {\frac {(p - b)(p - c)}{bc}} =$ $ 1 + \frac {(p - a)(p - b)(p - c)}{abc} =$ $ 1 + \frac {pr^2}{4Rpr} =$ $ 1 + \frac rR\ .$[/color]

[color=darkblue][b]Remark I[/b] (see [b]NguyenDungTN[/b]).

$ \boxed {3 - 2(\cos A + \cos B + \cos C) = (2\sin\frac A2 - \cos\frac {B - C}{2})^2 + \sin^2\frac {B - C}{2}\ge 0}\ !$

[b]Remark II.[/b] Show analogously that for any $ \lambda > 1$ there is the inequality

$ \lambda^2\cos A + 2\lambda (\cos B + \cos C)\le \lambda ^2 + 2$ with equality iff $ B = C$ and $ \sin\frac A2 = \frac {1}{\lambda }\ .$
[hide="Proof."]
$ \boxed {(\lambda^2 + 2) - [\lambda^2\cos A + 2\lambda(\cos B + \cos C)] = 2[(\lambda \sin\frac A2 - \cos\frac {B - C}{2})^2 + \sin^2\frac {B - C}{2}]\ge 0}\ .$[/hide]
Here are some particular cases :

$ 1\blacktriangleright$ $ \lambda : = 2\implies$ $ \cos A + \cos B + \cos C\le\frac A2\ .$

$ 2\blacktriangleright$ $ \lambda : = \sqrt 2\implies$ $ \cos A + \sqrt 2 (\cos B + \cos C)\le 2\ .$

$ 3\blacktriangleright$ $ \lambda : = \frac {2}{\sqrt 3}\implies$ $ 2\cos A + 2\sqrt 3 (\cos B + \cos C)\le 5\ .$

We'll have equality iff $ B = C$ and $ \{\begin{array}{c} 1\blacktriangleright A = 60^{\circ}\ . \\
 \\
2\blacktriangleright A = 90^{\circ}\ . \\
 \\
3\blacktriangleright A = 120^{\circ}\ .\end{array}$ [/color][/quote]
[color=blue]Nice solution, Mr [b]Virgil Nicula[/b]!
With second inequality, we can prove stronger inequality:
[b]Problem:[/b]
Prove that: $ \sum cosA \le 1+\frac{cos^2{\frac{B-C}{2}}}{2} \le \frac{3}{2}$
[b]Solution:[/b]
Set $ P=\sum cosA$, we have:
$ P=1-2sin^2{\frac{A}{2}}+2sin{\frac{A}{2}}cos{\frac{B-C}{2}}$(*)
$ \Leftrightarrow sin^2{\frac{A}{2}}-2cos{\frac{B-C}{2}}sin{\frac{A}{2}}+P-1=0$
$ \Delta '=cos^2{\frac{B-C}{2}}-2P+2 \ge 0 \Leftrightarrow P \le 1+\frac{cos^2{\frac{B-C}{2}}}{2}$
[u][i]Remark:[/i][/u]
With (*), we can prove the following inequalities:
$ \star \sum cosA \le 1+\frac{\sum cos^2{\frac{B-C}{2}}}{6}$
$ \star \sum cosA \le 1+\frac{cos{\frac{A-B}{2}}+cos{\frac{B-C}{3}}+cos{\frac{C-A}{4}}}{6}$[/color]


In an effort to make this forum more active, and to answer one of my questions:
I was wondering whether countdown was official in Ohio?

Edit: Whoops, I accidentally pressed the submit button twice. I didn't mean to make two threads, sorry.

Two years ago, it was not official.  And I have not heard anything to suggest it has changed.  I am pretty sure it is unofficial.  If you want to make sure, you can search the coordinator database for Ohio here: http://www.mathcounts.org/Volunteer/VolunteerSearch.asp like I did when I was in eighth grade.  And I think that all the chapters have to follow what the state does and have countdown unofficial...


Let ABC be a scalene triangle and k be its circumcircle. Let t(A), t(B), t(C)
be the tangents to k at A,B,C, respectively.

Prove that common points of
AB  t(C),
AC  t(B),
BC  t(A) exist, and that they are collinear.

Dear Mathlinkers,
the tangentiel triangle of ABC is perspective with his contact triangle i.e. ABC.
According to Desargues's theorem, the involving meet point are colinear.
Sincerely
Jean-Louis

Hi~
I think that applying Pascal theorem to $ AABBCC$ on the circle kills this problem.

[quote="77ant"]Hi~ I think that applying Pascal theorem to $ AABBCC$ on the circle [b]kills[/b] this problem.[/quote]

[b][u]@ 77ant & Jayme[/u].[/b] You killed a butterfly with an "atomic bomb". Maybe our user [b][u]matex (L)(L)(L)[/u][/b] is fighting for "the peace" as me.

[b][u]An easy proof[/u].[/b] Denote $ D\in AA\cap BC$ (I am understanding by $ AA$ - the tangent in a point $ A$ to the circumcircle of $ \triangle ABC$).

$ \triangle DBA\sim\triangle DAC\ \implies\ \frac {DB}{DA} \equal{} \frac {c}{b} \equal{}$ $ \frac {AD}{CD}\ \implies\ \left\|\begin{array}{c} DB \equal{} \frac cb\cdot DA \\
 \\
DC \equal{} \frac bc\cdot DA\end{array}\right\|\ (: \downarrow)\ \implies\ \boxed {\ \frac {DB}{DC} \equal{} \frac {c^2}{b^2}\ }\ .$ Show analogously that for

$ E\in BB\cap CA$ , $ F\in CC\cap AB$ we have $ \frac {EC}{EA} \equal{} \frac {a^2}{c^2}$ and $ \frac {FA}{FB} \equal{} \frac {b^2}{a^2}$ . Observe that $ \frac {DB}{DC}\cdot\frac {EC}{EA}\cdot\frac {FA}{FB} \equal{} 1\ \stackrel{\mathrm{(Menelaus' th.)}}{\implies}\ D\in EF$ .


Is x^y (x,y in R) which stands for the relation: x+y in Q (rational numbers)
an equivalence relation?

so 3^5 means: 3+5 in Q (which is correct since 8=8/1)

I think it is not an eq. relation, am I correct?

I don't need a proof, if someone can confirm it, it is ok

thanks guys


geometry
Euler
function
vector
analytic geometry
geometry unsolved

Can we find a suficient condition between the sides of a triangle  $ABC$( if there is one) , so that the incircle divides the median $AM$ into three equal parts?

   I have proved that a necessary condition is $BC = 2 AB$, but is that and suficient?

 [i] Babis[/i]

If G is the centroid of ABC and G' the midpoint of AG, the power of G wrt the incircle is
$p\left(G\right) = \frac{5}{36} \left(a^2+b^2+c^2\right) - \frac16 \left(ab+bc+ca\right)$ and the power of G' is
$p\left(G^{\prime}\right) = p\left(G\right) + \frac{1}{12} \left(a-2b\right)\left(a-2c\right)$.
Your problem is equivalent to p(G) = p(G') = 0 hence to
$\left(\frac{b}{a}=\frac12,\ \frac{c}{a}=\frac{13}{10}\right)$ or $\left(\frac{b}{a}=\frac{13}{10},\ \frac{c}{a}=\frac12\right)$.

Thank you very much!

[i]I' ll read carefully all steps of your solution.[/i]

  Babis

Jean - Pierre , please tell me :

  Have you any method  - and which  -  to find the power of $G , H$  and $O$  (the last needs Eulers formula , I guess , but what about the simplifications for $r$ ?) with respect to the incircle $( I, r)$ of the triagle , as a function of  $a , b ,c$ ? Your formula for  $P(G)$ is pretty nice. I think I need a [i]Hint[/i].

 Thank you - Babis

Consider the relations (where MN means the vector MN : I don't know how to put an arrow upon MN)
AG = (AB+AC)/3;AG' = (AB+AC)/6; AI = (b.AB+c.AC)/(a+b+c);
r^2 = (a+b-c)(b+c-a)(c+a-b)/4/(a+b+c);
AB^2 = c^2; AC^2 = b^2; AB.AC = 1/2(b^2+c^2-a^2) (dotproduct)
and compute
p(G) = (AG-AI)^2-r^2; p(G') = (AG'-AI)^2-r^2
Of course, the computation is easier if you know the barycentric equation of the incircle.

Very well!

  I was afraid to use vectors , but as I see ,  it is a good way. I must refresh the theory on Barycentric coordinates  as well.

 Thanks a lot !

  Babis


I filled out the entry form that was online, and then it told me that I needed to fill out another entry form, but that I didn't need to print it out now. I clicked on the link to the form, but it won't download onto my computer. It says to try saving it to a disk, so I inserted a floppy disk, but the form still won't download. What can I do to complete my registratrion?

What happens when you try to download?  Do you have Adobe Reader installed?

It says:

[quote="my stupid computer"]C:/DOCUME~`/EMILY/LOCALS~1/TEMP/USAMTSProblems_16_1.pdf could not be opened, because an unknown error occurred.

Sorry about that. Try saving to a disk first and then opening the file.[/quote]

That's what comes up when I attempt to download the problems for round one. Something very similar came up for the entry form.

I don't believe we have adobe reader installed on my home computer? Is that what is causing these problems?

Try downloading and installing Adobe Reader:

[url]http://www.adobe.com/products/acrobat/readstep2.html[/url]