MATHCOUNTS
geometry
probability
USAMTS
3D geometry

Come on guys, this thing is dead! (Copycatting off the Illinois forum)

Let's start discussing this years Mathcounts!

And other contests in other threads...

OK.  Since state is in 3 days I'll post.

Gah I need to get better at secants, tangents, 3D geometry, probability, counting paths, and dividing polynomials.  I'm not going to do very well.

I predict this will be a medium test year.

That's a lot of things to do. Perhaps you should focus on one at a time?

Why do you think this will be a medium test year?

Nooooo way.  The 2010 chapter test was way harder than the 2008 and 2009 tests....  Methinks 2010 is going to be a beast year.

Well, does anyone know where the 2009 rankings are?

Well, got 5th...

After your state, if you want to, post your ranking...

[b]DO not post your score or discuss the difficulty of the test![/b]

I believe I got mid 20's place but they never gave me my score.  I'm not too mad since there's next year and a bunch of pwnage people will leave :).

Oh yeah and here's my friend's belief why Honey Creek didn't get in the top 6 teams:

Well, the team was practicing for states, and an individual on our team, Flynn Lemonnier, noticed that everyone who was working with his pencil did worse than normal(this was about after 25 or so runs through).  So he believed the pencil was "cursed" and he gave it to a Honey Creek kid who didn't make CD. I don't believe it, but hey, it might put Wilbur Wright on the map!

Wow...
That is crazy

Any way good try Maybach you have another year

[quote="AwesomeToad"]Come on guys, this thing is dead! (Copycatting off the Illinois forum)
[/quote]

yes you are copycatting.

[quote="Mewto55555"][quote="resurrection"][quote="Maybach"]2008 was hard 2009 was easy 2010 will probably be medium.[/quote]

2008 wasn't hard. that's all I know.[/quote]

07 wasn't too bad either.

also i totally pwned all of you guys at state[/quote]

humble much?

[quote="cognos599"][/quote][quote="Mewto55555"][quote="resurrection"][quote="Maybach"]2008 was hard 2009 was easy 2010 will probably be medium.[/quote]

2008 wasn't hard. that's all I know.[/quote]

07 wasn't too bad either.

also i totally pwned all of you guys at state[/quote][quote="cognos599"]

humble much?[/quote]

no dude if you were there you would have seen it all go down

also if i could discuss i would say my score of 48 was slightly subpar but still acceptable

[quote="Demosthenes"]@mewto don't you live in missouri?[/quote]

Of course not, ignore the location on his profile.

dude i saw him at nats last year
and his usamts high score shows him as being from MO
so unless he moved in the last month, hes definitely from missouri

[quote="Maybach"]Well, I'm so glad ther's a next year for Mathcounts.  For now, I'm going to probably take a break because of sci oly....[/quote]

Thanks for making me feel better...
you know where I placed as 8th grade

 :roll: just kidding

[quote="Maybach"]Wait, were you at state sci oly? Your friend summai was.  Our team got second!!!!!![/quote]

Good job by the way. :) No I wasn't there.

[quote="Mewto55555"][quote="resurrection"][quote="Maybach"]2008 was hard 2009 was easy 2010 will probably be medium.[/quote]

2008 wasn't hard. that's all I know.[/quote]

07 wasn't too bad either.

also i totally pwned all of you guys at state[/quote]
2007 was pretty hard.  2009, 2006, and 2005 were all pretty easy.


a prufer sequence of a certain tree is a sequence of 7 numbers.  The numbers include only two labels, one of which occurs four times and the other three times.  For example, the sequence might be 1,2,1,2,1,2,2 or 6,6,6,3,3,3,3.  How many labled trees have this type of prufer sequence?

Are the lables intigers $ \in[0,9]$?
And which type of prufer sequence do you mean; $ abababb$ or $ aaabbbb$[/hide]

1) Do your own homework.
2) When you post problems, you should post enough background material/definitions so that someone reading the problem can actually know what it's about.


Hey guys,

Isn't there a method for extracting square roots from a number, which is basically an algorithm? What is it?

-rts2007

[url=http://www.itl.nist.gov/div897/sqg/dads/HTML/squareRoot.html]url[/url]

Hm. What is the government doing math for?


solve for x in R:
$ x^2  \plus{} 2x\sqrt x  \minus{} x \minus{} 2\sqrt x  \minus{} 24 \equal{} 0$

[hide="hint"]
let $ \sqrt{x} \equal{} a$[/hide]

Let $ x \equal{} y^2$ (such that $ y$ is positive or something of the form $ \minus{} ki$ for positive $ k$) so that we have

$ y^4 \plus{} 2y^3 \minus{} y^2 \minus{} 2y \minus{} 24 \equal{} (y \minus{} 2)(y \plus{} 3)(y^2 \plus{} y \plus{} 4) \equal{} 0\implies y \equal{} \{ \minus{} 3,2\}$

(since the solutions of $ y^2 \plus{} y \plus{} 4$ are not pure imaginaries, their squares can't be real).

However, since $ y$ must be positive, the only solution is $ y \equal{} 2\implies x \equal{} \boxed4$.


inequalities geometry circumcircle algebra function domain Euler

Let $ABC$ be an acute triangle such that $a \geq b \geq c$. Let $P$ be a point in the same plane such that $PA \geq PB \geq PC$. Show that $PA+PB+PC \geq 3R$, where $R$ is the circumradius of triangle $ABC$.

Nice problem treegoner :D
We have $R(\sum PA)=\sum OAPA\ge\sum\vec{OA}\vec{PA}=\sum(\vec{PO}+\vec{OA})\vec{OA}=\vec{PO}(\sum\vec{OA})+3R^{2}=\vec{PO}\vec{OH}+3R^{2}$thus we get $\sum PA\ge\frac{\vec{PO}\vec{OH}}{R}+3R$we will show when $PA\ge PB\ge PC$ and $a\ge b\ge c$ then $\vec{PO}\vec{OH}\ge 0$ indeed call $\Omega$ is domain such that $PA\ge PB\ge PC$ and $\Lambda$ is domain such that $\vec{PO}\vec{OH}\ge 0$ in the figure we see $\Omega$ is yellow domain (limited by the midline of $BC,AB$) and $\Lambda$ is right half plane of the line is through $O$ and perpendicular with $OH$(Euler line) to show $\Omega\subset\Lambda$ we only show $\angle HOA'\ge 90^{0}$ and $\angle HOC'\le 90^{0}$ but it is true from $a\ge b\ge c$ and $ABC$ is acute.


calculus
derivative
logarithms
function
calculus computations

can someone show me step by step how to do this

Y=(e^x)/(ln(x))

Go to [url=http://library.wolfram.com/webMathematica/Education/WalkD.jsp]Wolfram's Step-by-step derivative webpage[/url]. Into the first input field copy and paste  [code] exp[x]/ln[x] [/code]

then click "DO IT". :wink:

$ y=\frac{e^x}{\ln{x}}$
Taking the natural log of both sides we have
$ \ln{y}=x-\ln{(\ln{x})}$
Then we can use our logarithm rule (namely that $ \displaystyle \frac{dy}{dx}\left(\ln{x}\right)=\frac{u\prime}{u}$ where $ u$ is a differentiable function of $ x$).
Thus, $ \frac{y\prime}{y}=1-\frac{1}{x\ln{x}}\Leftrightarrow$
$ \begin{eqnarray*}\frac{dy}{dx}&=&y=\frac{y}{x\ln{x}}\\&=&\frac{e^x}{\ln{x}}-\frac{e^x}{x(\ln{x})^2}$


floor function
pigeonhole principle
algebra unsolved
algebra

Let $ \alpha$ be the positive root of the equation $ x^2 \minus{} 1989x \minus{} 1 \equal{} 0.$ Prove that there exist infinitely many natural numbers $ n$ that satisfy the equation:

\[ \lfloor \alpha n \plus{} 1989 \alpha \lfloor \alpha n \rfloor \rfloor \equal{} 1989n \plus{} \left( 1989^2 \plus{} 1 \right) \lfloor \alpha n \rfloor.\]

First, we should note that $ \alpha^{2}-1989\alpha-1=0$ Then,
$ \lfloor\alpha n+1989\alpha\lfloor\alpha n\rfloor\rfloor = 1989n+\left( 1989^{2}+1\right)\lfloor\alpha n\rfloor$
$\iff \lfloor(\alpha-1989)n+(1989 \alpha-1989^{2}-1)\lfloor\alpha n\rfloor\rfloor=0$
$\iff \lfloor \frac{\alpha n-\lfloor \alpha n\rfloor}{\alpha^{2}}\rfloor=0$
Finally, by pigeonhole principle, there must be infinitely $n$ such that $\alpha n-\lfloor \alpha n\rfloor$ is small than $\alpha^{2}$. The proof is complete.


Two bottles of equal volume contain alcohol and water, the ratios of the amount of alcohol to the amount of water being 3:1 and 2:1.  The contents of the two are mixed.  What will be the ratio of alcohol to water?

Do both bottles have the same amount of liquid in them?

[hide]Let x be the amount of liquid in each bottle.   The first bottle contains 1/4 x water and 3/4 x alcohol.  The second bottle contains 1/3 x water and 2/3 x alcohol.  Adding together the amounts of alcohol we get 3/4 x + 2/3 x = 9/12 x + 8/12 x=17/12 x.  Now adding together the amounts of water in each we get 1/4 x + 1/3 x = 3/12 x + 4/12 x = 7/12 x.  So the ratio of alcohol to water in the mix is 17/12 x to 7/12 x which equals 17 to 7.[/hide]

h_s_potter2002 wrote:[hide]The ratios are 3:1 and 2:1.  I would just average them to get 2.5:1 or 5:2.[/hide]

Ah, but this approach assume the 1 is the same unit in both cases, but it's not.  In one of the cases it represents 1/3 of the whole.  In the other it represents 1/4 of the whole.  You've got to be careful with these types of problems.

[quote="themonster"]Is my solution right?[/quote]

Yes it is.


Find all ordered pairs of integers (x, y) which satisfy x^2 + 4x + y^2 = 9.

Source:  MATC Mathematics Club

[hide]Completing the square and replacing x+2 with z, we are left with:

z^2+y^2=13

The only integer pairs that satisfy this will be where z or y=2 or -2 and the other equals 3 or -3.

Since we need x and not z, we plug x+2 back in for z and our solutions are:

[b](-5, 2), (-5, -2), (1, 2), (1, -2)[/b][/hide]

[hide="your solution missed a bunch"]
As before, we complete the square $(x+2)^2 + y^2 = 13$.

The only two square numbers that can add up to 13 are 4 and 9.

Therefore, we have a few sets of solutions:

${(x,y)|x+2=\pm2,y=\pm3}$

${(x,y)|x+2=\pm3,y=\pm2}$

If we find all ordered pairs in these sets, we have all the solutions:

$(0,3)(0,-3)(1,2)(1,-2)(-4,3)(-4,-3)(-5,2)(-5,-2)$

[/hide]

Oh, I'm doing these quickly, but thanks.


I got this from somebody's book, but I forgot what it was called.  There weren't any answers to exercises, and this problem has been bothering me for a while.

[img]http://i118.photobucket.com/albums/o105/urabillus/PROBLEM.jpg[/img]

Umm did I understand something wrong?  There is a flaw to the question I think.
It is true for n=1,2,4...
However, just in case I tried n=3 and n=5.
when
n=3 : $U_{3}=2U_{2}+1$ which equals 5.
As 5 is not a multiple of 3, the statement is false.
We can already see that this statement is only true for certain circumstances.

Not all of the values for $n$ satisfy the statement.  The problem asks which ones do.

[quote="max1546"]Umm did I understand something wrong?  There is a flaw to the question I think.
It is true for n=1,2,4...
However, just in case I tried n=3 and n=5.
when
n=3 : $U_{3}=2U_{2}+1$ which equals 5.
As 5 is not a multiple of 3, the statement is false.
n=5 : $U_{5}=4U_{4}+1$ which equals 57.
This is also false.
We can already see that this statement is only true for certain circumstances.[/quote]

And besides, $u_{4}=3u_{3}+1=16$, and $u_{5}=4u_{4}+1=65$, and that is a multiple of 5, so you should check your work next time.

[hide]Hint: Show that if $u_{n-1}\equiv 1(\mod n)$, then $u_{n}\equiv 0(\mod n)$

Then expand on that argument to figure out which $u_{k}$ satisfy these conditions and then figure out what do not ,and prove from there.[/hide]


Find all solutions of: $ (x^2 \plus{} 7x \plus{} 6)^2 \plus{} 7(x^2 \plus{} 7x \plus{} 6)\plus{} 6 \equal{} x$

[hide="Solution"]Observe that solutions to $ x^2 \plus{} 7x \plus{} 6 \equal{} x \iff x^2 \plus{} 6x \plus{} 6 \equal{} 0$ are also solutions to the given equation*.

We now wish to rewrite the equation in the form $ (x^2 \plus{} 6x \plus{} 6)(x^2 \plus{} ax \plus{} b) \equal{} 0$.

Equating the constant terms of the original equation and the factored form, we get $ 6^2 \plus{} 7 \cdot 6 \plus{} 6 \equal{} 6b \implies b \equal{} 14$
Similarly, equating $ x^3$ terms of both gives $ 2 \cdot 7 \equal{} 6 \plus{} a \implies a \equal{} 8$.

Thus, the original equation can be rewritten as $ (x^2 \plus{} 6x \plus{} 6)(x^2 \plus{} 8x \plus{} 14) \equal{} 0$

We can now solve this using the quadratic formula: $ \boxed{x \equal{} \minus{} 3 \pm \sqrt {3} , \; \; x \equal{} \minus{} 4 \pm \sqrt {2}}$

[i]* The left-hand form of this quadratic is a root equation for the given equation[/i][/hide]


What is the probability that the product of two numbers chosen randomly from the set of all positive integers is divisible by 2? Express your answer as a common fraction.

The two numbers chosen to multiply can be either both even, both odd, or one even and one odd. 2 of those cases give an even number so the probability is $ \frac{2}{3}$.

um the probability of there being one of each is actually twice the probability of the others, so it's actually $ \frac34$
in a more logical approach, we count how to get an odd product.
the probability that it's odd is the probability of the first being odd multiplied to the probability the second being odd.
that is $ \frac12\times\frac12\equal{}\frac14$
now subtract from one to get probability of an even product
$ 1\minus{}\frac14\equal{}\frac34$
answer is:
3/4


In the circle, $O$, with diameter, $AB=16$, there is an inscribed triangle, $\triangle{ABC}$, where the area of $\triangle{ABC}$ is maximized. Inside of $\triangle{ABC}$, there is a n inscribed square, $DEFG$, where D and E are on $AB$ and F and G are on $AC$ and $BC$, respectively.
What is the area of square $DEFG$?

[quote="mathgeniuse^ln(x)"][quote="vishalarul"][hide="Someone please check my answer..."]The area of $\triangle ABC$ will be maximized if it an isosceles right triangle. If you draw a picture, you see that triangles $EAF$ and $DBG$ are also isosceles right triangles. All of this implies that $AE = EF = ED = GD = DB$. Most importantly, we know that $AE = ED = DB = \frac{AB}{3}$. Thus, each side of square DEFG has length $\frac{16}{3}$. So the area of the square is $\boxed{\frac{256}{9}}$.

Am I right?[/hide][/quote]

[hide]How do we know that the right triangle maximizes it?  Because from this the area of the triangle is $64$, but the equilateral triangle has an area of $48\sqrt{3}$, which is greater than 64.[/hide][/quote]
how do you inscrib an equilateral triangle in a circle if one side is a diameter?

the triangle must be right, and if it iscoceles the area is maximized

[quote="mathgeniuse^ln(x)"][quote="vishalarul"][hide="Someone please check my answer..."]The area of $\triangle ABC$ will be maximized if it an isosceles right triangle. If you draw a picture, you see that triangles $EAF$ and $DBG$ are also isosceles right triangles. All of this implies that $AE = EF = ED = GD = DB$. Most importantly, we know that $AE = ED = DB = \frac{AB}{3}$. Thus, each side of square DEFG has length $\frac{16}{3}$. So the area of the square is $\boxed{\frac{256}{9}}$.

Am I right?[/hide][/quote]

[hide]How do we know that the right triangle maximizes it?  Because from this the area of the triangle is $64$, but the equilateral triangle has an area of $48\sqrt{3}$, which is greater than 64.[/hide][/quote]
In the original problem, the diameter is $AB$ and then the triangle is $\triangle{ABC}$, which shares A and B. You cannot inscribe an equilateral triangle if the side length is the diameter.

My Bad, I thought there were no restrictions.

Sorry.

And yeah, vishalarul's solution seems to be right then.

Hi !

I'd like to know how do you proove that the area of ABC is maximized when it is isosceles ?

I've tried to introduce the point K, middle of AB and considered the angle $\theta=\widehat{BKC}$.

With the Al-Kashi theorem (or "cosine rule" ?), I've found that area of ABC is equal to $64\sin\theta$. This is maximized when $\sin\theta$ is maximized.
Since $\theta\in]0;\pi[$, we found that $\theta=\frac\pi2$ maximizes ABC's area.
Then, we proove that ABC is isosceles.

[size=75]Excuse me for my bad English.[/size]

It is really not that complicated.

You can find the area by taking half the product of the hypotenuses and the altitude to that. In our case, the hypotenuse was constant, so all we needed to maximize was the altitude to the hypotenuse. This happens when $\triangle ABC$ is an isosceles right triangle.

[hide]let the leg lengths be $x$ and $y$. we wish to maximize $\frac{xy}{2}$. by AM-GM $\frac{x+y}{2}\ge \sqrt{xy}\rightarrow \frac{x^{2}+2xy+y^{2}}{4}\ge xy \rightarrow x^{2}+2xy+y^{2}\ge 4xy$ the quantity is maximized when both sides are equal. thus, $x^{2}+2xy+y^{2}=4xy\rightarrow (x-y)^{2}=0$

this occurs when $x=y$, or when the leg lengths are equal[/hide]

http://mathworld.wolfram.com/ProofwithoutWords.html
Hope that helps.

[quote="Quevvy"]What is AM-GM? :?:[/quote]
The arithmatic mean of a set of numbers is always greater than or equal to the geometric mean of that set. In math,

$\frac{x_{1}+x_{2}+x_{3}\cdots+x_{n}}{n}\ge\sqrt[n]{x_{1}x_{2}x_{3}\cdots x_{n}}$

[quote="13375P34K43V312"][quote="ragnarok23"][quote="Quevvy"]What is AM-GM? :?:[/quote]
The arithmatic mean of a set of numbers is always greater than or equal to the geometric mean of that set. In math,

$\frac{x_{1}+x_{2}+x_{3}\cdots+x_{n}}{n}\ge\sqrt{x_{1}x_{2}x_{3}\cdots x_{n}}$[/quote]

No, it's the n-th root on the right side, not the square root.[/quote]
oops I forgot to put that in lemme edit it

[quote="ragnarok23"]let the leg lengths be $x$ and $y$. we wish to maximize $\frac{xy}{2}$. by AM-GM
$\frac{x+y}{2}\ge \sqrt{xy}\rightarrow \frac{x^{2}+2xy+y^{2}}{4}\ge xy \rightarrow x^{2}+2xy+y^{2}\ge 4xy$ the quantity is maximized when both sides are equal.
thus, $x^{2}+2xy+y^{2}=4xy\rightarrow (x-y)^{2}=0$

this occurs when $x=y$, or when the leg lengths are equal[/quote]

Great !
Thanks for this idea  :lol: .


Let us forget about $1-a_{i}^{3}$ in the denominator.
It is enough to prove that
$\sum_{i=1}^{n}\frac{a_{i}}{2n+S}\leq \frac{1}{3}$ or
$3(a_{1}+a_{2}+...+a_{n}) \leq 2n+a_{1}^{3}+a_{2}^{3}+...+a_{n}^{3}$
That is true from the AM-GM inequality
$a_{i}^{3}+1+1\geq 3a_{i},$ and we are done.

Nice solution prowler  :)   here is mine :

Let $f(x)=\frac{x}{2n+1+S-x^{3}}$  , $x\in \left[ 0,1 \right]$

We can easily find that $f''(x)>0$  so $f$  is convex at $x\in \left[ 0,1 \right]$
.
So $f$ attains its maximum at the endpoints .
We find $f(1)>f(0)$  so the maximum is $f(1)$ so
$LHS\leq nf(1)=\frac{1}{3}$  and we are done .Did you like it?

Not really. We actually have that $n \cdot f(1) = \frac{n}{2n+S}\geq \frac13$, which is not particularly useful :maybe:

If I am not wrong $f(1)$ is the maximum so we have $\leq$

We have that $LHS \leq n \cdot f(1) = \frac{n}{2n+S}\geq \frac13$!!! And that doesn't prove anything! Clear now?

Yes  :blush:  . You are right . Sorry .It is wrong .

Firstly I misunderstood what you meant .
The mistake I did was :
Because $f(1)$ makes $a_{i}=1$ i take $S=n$ at the final realtion

[quote="silouan"]Nice solution prowler  here is mine :

Let  ,  $ f(x) = \frac {x}{2n + 1 + S - x^{3}}$

We can easily find that  $ f''(x) > 0$ so  is convex at  $ x\in\left[ 0,1\right]$
.
So  $ f$ attains its maximum at the endpoints .
We find  $ f(1) > f(0)$ so the maximum is  $ f(1)$ so
$ LHS\leq nf(1) = \frac {1}{3}$ and we are done .Did you like it? prowler[/quote]

you can take $ f(x) = \frac {x^{1/3}}{2n + 1 + S - x}$ and make Jensen
we can check that  $ f''(x) < 0$

After Jensen we have to prove $ n\frac {\left ( \frac {S}{n} \right )^{1/3}}{2n + 1 + S - \frac {S}{n}} \leq\frac {1}{3}$
or
$ \frac {2n + 1}{n} + \frac {n - 1}{n^2}S \geq 3 \left ( \frac {S}{n} \right )^ {1/3}$ (*)
AGM gives
$ \frac {n - 1}{n} + \frac {n - 1}{n} + \frac {n - 1}{n^2}S \geq 3 \left (\frac {S}{n} \right ) ^ {1/3} \frac {n - 1}{n}$   (**)

$ S = a_{1}^{3} + a_{2}^{3} + \ldots + a_{n}^{3} \leq n$ that's why
$ \frac {3}{n} \geq 3 \ \left ( \frac {S}{n} \right ) ^ {1/3} \frac {1}{n}$ (***)

(**)+(***) gives (*)


Method 1:

Simplify to
$ \int\frac {1}{(sinx \plus{} 1/2)^2 \plus{} 3/4}dx$
Now put $ Sinx \equal{} \frac{2tan(x/2)}{sec^2(x/2)}$

Method 2:
Put $ Sinx\equal{}t$
Multiply and divide by $ Cosx$
Your denominator is a Quadratic Root Quad. So Put $ \frac{1}{t} \equal{} z$.

I have to finish some stuff for maths exam so I'll solve these later if you want.

The methods u have gvn are rite bt very long
There is a simple method that takes u thru the sum like cakewalk :rotfl:  :rotfl:


[quote="Beat"]Suppose that real numbers $a, b, c, d$ satisfy $\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{d}+\frac{d}{c}+\frac{d}{a}+\frac{a}{d}=\frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b}+\frac{ac}{bd}+\frac{bd}{ac}+2$.
Prove that at least two of the numbers are equal.[/quote]

Multiply all this by $abcd$ and after some boring work we get $(a-b)(b-c)(c-d)(d-a)=0$. From here we have that at least two of the number are equal.


calculus
integration
trigonometry
function
calculus computations

Evaluate the indefinite integral:
\[ \int \frac{\sin^2 x\minus{} \cos^2 x}{\sin x} \ dx\]

hello, write $ \frac{\sin^2(x)\minus{}\cos^2(x)}{\sin(x)}$ in the form
$ \sin(x)\minus{}\frac{\cos^2(x)}{\sin(x)}$. The first integral
$ \int \sin(x)dx\equal{}\minus{}\cos(x)\plus{}C_1$ In the second one set $ t\equal{}\cos(x)$ and you will
get:$ \int\frac{t^2}{1\minus{}t^2}dt$.
Sonnhard.

[quote="123456789"]Evaluate the indefinite integral:
\[ \int \frac {\sin^2 x \minus{} \cos^2 x}{\sin x} \ dx
\]
[/quote]

$ \int \frac {\sin^2 x \minus{} \cos^2 x}{\sin x} \ dx \equal{}\int 2sin x dx\minus{}\int \frac{dx}{sinx}$
$ \equal{}\minus{}2cosx\minus{}\int \frac{d(cos x)}{1\minus{} cos^2 x}\equal{}\minus{}2cos x\plus{}\frac{1}{2}ln{\frac{1\minus{}cosx}{1\plus{}cosx}}\plus{}C$

How did you do this part?
\[ \equal{} \minus{} 2cosx \minus{} \int \frac {d(cos x)}{1 \minus{} cos^2 x} \equal{} \minus{} 2cos x \plus{} \frac {1}{2}ln{\frac {1 \minus{} cosx}{1 \plus{} cosx}} \plus{} C
\]What do I do from here?
\[ \int\frac {t^2}{1 \minus{} t^2}dt\]

[quote="kunny"]$ \frac {t^2}{1 \minus{} t^2} \equal{} \minus{} 1 \plus{} \frac {1}{1 \minus{} t^2} \equal{} \minus{} 1 \plus{} \frac {1}{2}\left(\frac {1}{1 \minus{} t} \plus{} \frac {1}{1 \plus{} t}\right)$.[/quote]How did you figure out to simplify $ \frac{t^2}{1\minus{}t^2}$ to $ \minus{}1\plus{}\frac{1}{1\minus{}t^2}$? And could you explain the other solution?

Just do Long division; divide $ t^2$ by $ 1\minus{}t^2$. :)

It's not a matter of doing long division, just write $ t^2\equal{}1\minus{}(1\minus{}t^2).$

[url=http://en.wikipedia.org/wiki/Polynomial_long_division#Example]See[/url], this is what I'd call a "long division." By modifyin' the numerator to make a quickly simplification, I wouldn't call it as a long division.

Write the long division out. It's one step. What is it? Why does long division work?

[quote="NguyenDungTN"]$ - 2cos x + \frac {1}{2}ln{\frac {1 - cosx}{1 + cosx}} + C$[/quote]
I get $ - 2cos x + \frac {1}{2}ln{\frac {1 + cosx}{1 - cosx}} + C$... I think it's correct because my function graphing program gets the same answer.

I found it more easy to substitute $ t=\cosh$ from $ \int \frac{\sin^2 x- \cos^2 x}{\sin x} \ dx=\int \frac{1-2\cos^2 x}{\sin x} \ dx$. Has anybody done the same thing?

Greetings
GioMott


Let V be a finite dimensional vector space over field F and let v belong to V,v not equal to 0.Then there exists some f in V* such that f(v) is not equal to 0.
(where V* is dual space of V)

We can essentially invoke the rank-nullity theorem. Let $ n\equal{}\dim(V).$ Since there exists $ v\ne 0,$ we have $ n\ge 1.$ In turn, $ \dim(V^*)\equal{}n.$ Now let $ W\equal{}\{f\in V^*: f(v)\equal{}0\}.$ $ W$ is a subspace of $ V^*$ of codimension 1, hence dimension $ n\minus{}1.$ That makes $ W$ a proper subspace of $ V^*$ and we can select an element of $ V^*$ that is not in $ W.$

I would hesitate to be any more concrete than that. Any attempt to write specific formula, depending on $ v,$ for the element $ f$ runs the risk of being invalid in some finite characteristic.

what class do you learn about this is in?

[quote="Kent Merryfield"]
I would hesitate to be any more concrete than that. Any attempt to write specific formula, depending on $ v,$ for the element $ f$ runs the risk of being invalid in some finite characteristic.[/quote]

Why? What if I extend $ v$ to a basis of $ V$, and set $ f$ to be the map that takes the coefficient of $ v$?

That works, of course, and is not what I meant to avoid. I was thinking of expressions based on a pre-existing basis; saying "extend $ v$ to a basis" is a little beyond a "specific formula."


geometry
rectangle
induction
combinatorics unsolved
combinatorics

Is it possible to write a natural number in every cell of an infinite chessboard in such a manner that for all integers $m,n>100$, the sum of numbers in every $m$x$n$ rectangle is divisible by $m+n$.

No, its not possible. Suppose (for sake of contradiction) that its possible. Note firstly that the condition "[i]for all $m, n>100$[/i]" is garabage. To see this, we construct another board with numbers as follows. Divide the infinite board into disjoint miniboards of $101$x$101$. Call the sum of the numbers of each miniboard the set of "minisums". Create another infinite board, with natural numbers filled in, such that each of the numbers is filled in with the set of minisums, and also such that "order" is preserved of the corresponding squares and minisums(e.g. square A of (original chessboard) dimension $101$x$101$ is adjacent to the right of square B (of original chessboard) of dimensions $101$x$101$ [b][i]implies[/i][/b], in the new board: square with the minisum of A is adjacent to the right to square with minisum of B). With this constructed infinite chessboard, it is easy to check the property "[i]it is possible to write a natural number in every cell of an infinite chessboard in such a manner that for all integers $m,n$, the sum of numbers in every $m$x$n$ rectangle is divisible by $m+n$.[/i] (*)" Thus we now consider the problem in terms of this new chessboard.

We prove by induction on $k$ on the claim: for all $k$, $2^{k}$ divides the number in each square. For $k=1$, this is obviously true. Suppose the claim works for $k=n$. We prove the claim for $k=n+1$.

Take any $2$x$(2^{n+1}-1)$ rectangle. We observe the sum of the rectangle. First tile it with two disjoint $1$x$(2^{n+1}-1)$ rectangles. By property (*), we have that both of them, and hence the entire $2$x$(2^{n+1}-1)$  rectangle's sum is divisible by $2^{n+1}$. On the other hand, stick a $2$x$(2^{n+1}-2)$ in the $2$x$(2^{n+1}-1)$. Since [i]its[/i] sum is divisible by $2^{n+1}$, the remainding $1$x$2$ rectangle's sum is also divisible by $2^{n+1}$. Since we chose any $2$x$(2^{n+1}-1)$ rectangle to begin with, we have the property: "[i]for [b]any[/b] $1$x$2$ square, its sum is divisible by $2^{n+1}$[/i]."(**)

Now we use the inductive hypothesis.  Using property (**) and the hypothesis, we deduce that either every single square is $0 mod (2^{n+1})$ or every single square is $2^{n}mod (2^{n+1})$. If the second is true, then consider any $1$x$(2^{n+1}-1)$ rectangle once again and deduce a contradiction. Thus the first is true, and thus the induction is complete.

Thus by the induction statement, all numbers filled into the chessboard are divisible by $2^{k}$, for all $k$, and the only possible way, if all the numbers are non-infinite, is iff all numbers are zero: not a natural number. Contradiction


logarithms
calculus
calculus computations

The mass of a substance is decreasing exponentially. Its mass now is $ 150$ grams and its mass, $ m$ grams, at a time $ t$ years from now is given by

$ m\equal{}150e^{\minus{}kt}$,

where $ k$ is a positive constant. Find, in terms of $ k$, the number of years from now at which the mass will be decreasing at a rate of $ 3$ grams per year.

[b]Solution.[/b]
$ dm/dt\equal{}150e^{\minus{}kt}(\minus{}k)\equal{}\minus{}3$...Rearranging, we get $ \frac{\ln{(50k)}}{k}$.. :)


A mathematical organization is producing a set of commemorative license plates. Each plate contains a sequence of five characters chosen from the four letters in AIME and the four digits in $2007$. No character may appear in a sequence more times than it appears among the four letters in AIME or the four digits in $2007$. A set of plates in which each possible sequence appears exactly once contains $N$ license plates. Find $\frac{N}{10}$.

[hide]I divided it up into 2 cases.
Case 1: There are 0 or 1 zeroes in the license plate
7C5*5! = 2520
Case 2: There are 2 zeroes in the license plate
6C3*5!/2! = 1200

$2520+1200=3720, \frac{3720}{10}=\boxed{372}$.[/hide]

[hide=Simple casework (similar to above)]Case 1: There are less than 2 $0$'s:
$\binom{7}{5} \cdot 5! = 2520$.

Case 2: There are exactly 2 $0$'s:
$\binom{6}{3} \cdot \frac{5!}{2!} = 1200$.

$2520 + 1200 = 3720$, so the answer is $\boxed{372}$.[/hide]

We do cases.

Case 1: $0$ or $1$ zeros.

There are $5!$ arrangements and $\binom{4+3}{5}=21$ ways to pick $5$ characters out of AIME and $2,0,7$, for a total of $5! \cdot 21 = 2520$ ways.

Case 2: $2$ zeros.

There are $\frac{5!}{2}$ arrangements and $\binom{4+2}{3}=20$ ways to pick $3$ characters out of AIME and $2,7$, for a total of $\frac{5!}{2} \cdot 20 = 1200$ ways. Summing gives $3720 \implies \boxed{372}$.


trigonometry
inequalities
inequalities unsolved

In every acute-angled triangle $ ABC$, prove that

$ \frac {\sqrt {\sin B\sin C\cos A}}{\sin A} \plus{} \frac {\sqrt {\sin C\sin A\cos B}}{\sin B} \plus{} \frac {\sqrt {\sin A\sin B\cos C}}{\sin C} > 2$

[quote="Ph-An"]This inequality is equivalently with the following inequality:

$\sum \frac 1a\cdot \sqrt{b^+c^2-a^2}\ge 2\sqrt 2.$

I think that $\ge 3$.[/quote]

think that a->1,b->1,c->$\sqrt{2}$

[color=blue][b]Problem.[/b] Let ABC be an acute-angled triangle. Prove the inequality

$\frac{\sqrt{\sin B\sin C\cos A}}{\sin A}+\frac{\sqrt{\sin C\sin A\cos B}}{\sin B}+\frac{\sqrt{\sin A\sin B\cos C}}{\sin C}>2$.[/color]

[i]Solution.[/i] Since the triangle ABC is acute-angled, the cotangents cot A, cot B and cot C of its angles A, B and C are all positive.

Now,

$\cot B+\cot C=\frac{\cos B}{\sin B}+\frac{\cos C}{\sin C}=\frac{\cos B\sin C+\cos C\sin B}{\sin B\sin C}=\frac{\sin\left(B+C\right)}{\sin B\sin C}$.

Since B + C = 180° - A (by the sum of angles in triangle ABC), we have sin (B + C) = sin A, and thus this becomes

$\cot B+\cot C=\frac{\sin A}{\sin B\sin C}$.

Thus,

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

Taking the square root, we arrive at

$\sqrt{\frac{\cot A}{\cot B+\cot C}}=\frac{\sqrt{\sin B\sin C\cos A}}{\sin A}$.

Similarly,

$\sqrt{\frac{\cot B}{\cot C+\cot A}}=\frac{\sqrt{\sin C\sin A\cos B}}{\sin B}$;
$\sqrt{\frac{\cot C}{\cot A+\cot B}}=\frac{\sqrt{\sin A\sin B\cos C}}{\sin C}$.

Thus, the inequality in question,

$\frac{\sqrt{\sin B\sin C\cos A}}{\sin A}+\frac{\sqrt{\sin C\sin A\cos B}}{\sin B}+\frac{\sqrt{\sin A\sin B\cos C}}{\sin C}>2$,

becomes

$\sqrt{\frac{\cot A}{\cot B+\cot C}}+\sqrt{\frac{\cot B}{\cot C+\cot A}}+\sqrt{\frac{\cot C}{\cot A+\cot B}}>2$.

But this is the inequality posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=46428 (and at http://www.mathlinks.ro/Forum/viewtopic.php?t=27481 and http://www.mathlinks.ro/Forum/viewtopic.php?t=48825 ), applied to the positive numbers cot A, cot B and cot C. Proof complete.

  Darij


So my sister and I got a letter in the mail asking us to apply for the Canada/USA Mathcamp, and i am sure if i got it then all of you ppl with the exception of mustit, probly got it too... i am wondering if anyone else is planning to apply and if they get admitted if they are planning to go.

nah i havent gotten it

but i dont even know what it is, i guess if you have free time in the summer and you are really dedicated at being better at math, maybe you should go (apparently awesomemath is better though)

meh, i dont have time to go to summer camps so i cant go

I think I'm going to awesome math... because they sent their recruitment letters first XD...

Mrs. Benjamin + Dr. Chi say Canada Math camp is good. awesomemath isn't. =P. and the research camps are good also.so w/e xD

[quote="cognos599"]yeah mrs. benjamin said canada usa is better and i got awesomemath too.... idk if ill go to either.[/quote]

i think you should since 7th to 8th grade is not that important of a summer (unless you are going back to india to visit relatives or something)

but for me, i have tons of crap to take care of.  and lyndon, tell your dad to stay off the street during the summer between 12:00-6:00 pm.

[quote="unimpossible"][quote="the future"][quote="cognos599"]im not going to india for a while, since i just went like last year. so i might apply, but if lyndon is going to awesomemath then, i might go cuz he's like the only person i know.[/quote]

actually im pretty sure you will know 2 other people that are going to awesomemath...dont know if they want to be mentioned though but you played cards with them last year at ARML.

but i think you need to pass a test to get into awesomemath or something like that.

i wish i went to some camp between 7th and 8th grade year...wouldnt of failed as bad as i did at mathcounts.[/quote]

possibly another person who played cards with you : me =P

i might go[/quote]

nah you were busy owning us at basketball.


Write the partial fraction decomposition for the rational expression.  Check your result algebraically.  Then assign a value to the constant a and check the result graphically.

$\frac{1}{a^{2}-x^{2}}$

[hide]
Answer: ${\frac{1}{2a}\left(\frac{1}{a+x}+\frac{1}{a-x}\right)}$
[/hide]

I finally understood them.
Thank you, tjhance!!

to write fractions, use [code]$\frac{a}{b}$[/code] to make $\frac{a}{b}$

as for the problem

[hide]
$\frac{1}{x^{2}-a^{2}}=\frac{1}{(x-a)(x+a)}=\frac{A}{x+a}+\frac{B}{x-a}$ for some numbers A and B

$A(x-a)+B(x+a)=1$
This should work for all x and a. Let $x=a$
$0+B(2x)=1$
$B=\frac{1}{2x}$
Now let $x=-a$
$A(2x)+0=1$
$A=\frac{1}{2x}$

So the answer is $\frac{1}{2x}\frac{1}{x+a}+\frac{1}{2x}\frac{1}{x-a}=\frac{1}{2x}\left(\frac{1}{x-a}+\frac{1}{x+a}\right)$

EDIT-I got the x and the a mixed up... you should still be able to understand the concept

[/hide]

[quote="1=2"][quote="tjhance"]
$\frac{1}{x^{2}-a^{2}}$...

Now let $x=-a$...[/quote]

That would make 1/0. :wink:[/quote]

I was a little worried about that myself... but the answer is correct:

$\frac{1}{2x(x+a)}+\frac{1}{2x(x-a)}=\frac{x-a}{2x(x+a)(x-a)}+\frac{x+a}{2x(x+a)(x-a)}=\frac{2x}{2x(x+a)(x-a)}=\frac{1}{x^{2}-a^{2}}$

I'm not quite sure why it works, but this is the way that AoPS volume 2 says to do it.

If your still nervous at doing it like that...

[hide]
$A(x-a)+B(x+a)=1$
$Ax-Aa+Bx+Ba=1$
$(B-A)a+(A+B)x=0a+1$
$B-A=0$
$(A+B)x=1$

Solve that system...

$B=A$
$2Ax=1$
$A=\frac{1}{2x}$
$B=\frac{1}{2x}$

[/hide]


At B.C. 440, hippocrates got pi by quadrature of the lune

Can I get some solutions of this question?

Please....

Hippocrates was trying to square the circle to get $ \pi$ but it is not possible because $ \pi$ is transcendental.
His best effort was to square a lune, but the construction does not give $ \pi$.

http://en.wikipedia.org/wiki/Lune_of_Hippocrates


Let $ a,b,c$ are positive

Prove
1) $ \sum_{cyc}\frac {3a^2 \plus{} ab}{a \plus{} b}\geq 2(a \plus{} b \plus{} c)$

2) $ \sum_{cyc} \frac {3a^2 \plus{} ab}{a^2 \plus{} b^2} \leq 6$   *true for  $ a \leq b \leq c$*

3)$ \sum_{cyc} \frac {3a^2 \plus{} ab}{a^3 \plus{} b^3} \leq \frac {6(a \plus{} b \plus{} c)}{ab \plus{} bc \plus{} ca}$

*Have been edited*

1)

[hide]
Each term on the LHS is $ \frac {2a^2}{a \plus{} b} \plus{} \frac {a^2 \plus{} ab}{a \plus{} b} \equal{} \frac {2a^2}{a \plus{} b} \plus{} a$. Hence it suffices to show that
\[ \sum\frac {2a^2}{a \plus{} b} \ge a \plus{} b \plus{} c
\]
By Cauchy,
\[ \sum\frac {2a^2}{a \plus{} b} \sum (a \plus{} b) \ge \left(\sum a\sqrt {2}\right)^2
\]
and dividing by $ 2a \plus{} 2b \plus{} 2c$ we have the desired result.
[/hide]

[quote="Gibbenergy"]Let $ a,b,c$ are positive

Prove
2) $ \sum_{cyc} \frac {3a^2 \plus{} ab}{a^2 \plus{} b^2} \leq 6$
[/quote]
Try $ a\equal{}4,$ $ b\equal{}1$ and $ c\rightarrow0^\plus{}.$ :wink:

[quote="arqady"][quote="Gibbenergy"]Let $ a,b,c$ are positive

Prove
2) $ \sum_{cyc} \frac {3a^2 \plus{} ab}{a^2 \plus{} b^2} \leq 6$
[/quote]
Try $ a \equal{} 4,$ $ b \equal{} 1$ and $ c\rightarrow0^ \plus{} .$ :wink:[/quote]

Well, it is actually wrong for the case where c is in the interval $ [0.5 ... 7]$
I have checked my sols , I can only prove for a case $ a \leq b \leq c$
Thanks Arqady for pointing it out


number theory
relatively prime
number theory proposed

let a and b are natural number such that b<a

prove that (a^2+b^2)/(a^2-b^2) isn't in N.

Ιt is not so hard. Set a+b=x and a-b=y .......

If $ (a,b) > 1$, we reduce the fraction by all common factors. So WLOG $ a$ and $ b$ are relatively prime.

$ \frac{a^{2}\plus{}b^{2}}{a^{2}\minus{}b^{2}}\equal{}\frac{2b^{2}}{a^{2}\minus{}b^{2}}\plus{}1$. So the problem is equivalent to proving $ \frac{2b^{2}}{a^{2}\minus{}b^{2}}$ is not a nonnegative integer.

Suppose it is. Then $ 2b^{2}\equal{} ka^{2}\minus{}kb^{2}$ for some nonnegative $ k$, so $ (k\plus{}2)b^{2}\equal{} ka^{2}$. Since $ a$ and $ b$ are relatively prime, $ k$ and $ k\plus{}2$ must both be multiples of squares.  No two squares have difference two, so $ k$ and $ k\plus{}2$ must both be double a square (if triple or more, the difference would be $ > 2$). Then we must have $ k \equal{} 0$, and thus $ b \equal{} 0$. Contradiction.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose $G$ is a nonabelian group such that all its proper subgroups are abelian. Prove that $G$ is not simple.


modular arithmetic
Diophantine equation
number theory proposed
number theory

Solve the equation:
\[3x^2-7y^2=-20,x,y\in\mathbb{N^*}\]

[quote="Myth"]Hmm... You can use standart reduction to Pell's equation...  :?[/quote]
No,I didn't use Pell's equation.If you solve it by Pell's equation,you're lovely :lol: :lol:

[quote="zscool"]looking mod 7, we get:

$3x^2 \equiv 1 \pmod 7$, but the only square residues mod 7 are:

{0,1,4,2} so 3x^2 is congruent to one of {0,3,5,6} so no solutions.[/quote]
Yes,you're right!
Now, replace $-20$ by $20$ .I think it's harder ;)

Myth show the way? i think! But may be it takes time to work out and to check the root sequences!

in the original equation, make the substitution:

$x = \frac{u}{2} + 7\frac{v}{2}$ and $y = \frac{u}{2} + 3\frac{v}{2}$ where $u, v \in \mathbb{Z}$ then we get:

$21v^2 - u^2 = 20$ or $u^2 - 21v^2 = -20$ a pell equation

with least solution (1,1), and this will give rise to all solutions for x and y.

I have a results about  problem :
if equation a.x^2 -b.y^2=1 (a,b,a.b don't square) have any roof
then (c,d) are satisfy Pell equation :X^2-a.b.Y^2=1, then
c-1/2a,c+1/2b are square


geometry
3D geometry
sphere
geometry open

Given a transparent sphere $ S$ with radius one.
Find the minimum number of black spheres with radius one such that:
Any of those spheres shares at most a point with $ S$
If you are standing in the center of $ S$ you see all black.

Clearly for the two-dimensional case, you need 6 circles for covering one central circle.

Similar to the kissing number problem, i think it wont be easy

I think that it is impossible to see all black, stading in the center of $ S,$ because between of any three black spheres as the problem states  $ ($ with radius one, tangent to each other and also tangent to the given transparent sphere $ S$ $ ),$ there is a hole.

Sorry if I misunderstood your problem.

Kostas Vittas.

PS. What is the ''kissing number problem'', dear [b][size=100]modeler[/size][/b]?

The problem does not state that the black sphere should be tangent to each other, nor they have to be tangent to the transparent sphere(it only states that they cannot have common volume).

wikipedia: [url]http://en.wikipedia.org/wiki/Kissing_number_problem[/url]
It is not directly related to the problem, but I guess their level of difficulty may be similar.


AMC USA(J)MO USAMO trigonometry articles LaTeX geometry

Here is USAMO 1996 #4:

D lies inside the triangle ABC. $\angle BAC = 50^{\circ}$. $\angle DAB = 10^{\circ}$, $\angle DCA = 30^{\circ}$, $\angle DBA = 20^{\circ}$. Show that $\angle DBC = 60^{\circ}$

Here is my solution, approximate scores and comments are appreciated!

Let BD, CD, AD intersect AC, AB, BC at F, E, G, respectively. Let $\angle DBC=\theta$.

Note that $\angle ADB = 150$, which implies $\angle BDG = 30 = \angle ADF$, so $\angle DFC = 70$, so $\angle FDC = 80$, so $\angle DCB = 180 - \theta$.

By Angle form of Ceva's Theorem, we have $\frac{\sin BAG}{\sin GAC} \cdot \frac{\sin ACE}{\sin ECB} \cdot \frac{\sin CBF}{\sin FBA} = \frac{\sin 10}{\sin 40} \cdot \frac{\sin 30}{\sin (80-\theta)} \cdot \frac{\sin \theta}{\sin 20} = 1$

Lemma: $\frac{\sin 10}{\sin 40} \cdot \frac{\sin 30}{\sin (20)} \cdot \frac{\sin 60}{\sin 20} = 1$

Proof: Substitue $\frac{1}{2}$ for $\sin 30$ and rearrange we have

$\frac{2(\sin 40)(\sin 20)}{\sin 10} = \frac{\sin 60}{\sin 20}$

Substitute $\sin 20 = 2(\sin 10)(\cos 10)$ and cancel out $\sin 10$ we have

$4(\sin 40)(\cos 10) = 4(\sin 40)(\sin 80) = \frac{\sin 60}{\sin 20}$

By Triple Angle formula, we have $\sin 60 = (\sin 20)(3 - 4\sin^2 20)$, cancel out the \sin 20, we have

$4(\sin 40)(\sin 80) = 3 - 4\sin^2 20$.
RHS= $1 + 2( 1 - 2\sin^2 20) = 1 + 2\cos 40$,
LHS= $2(\cos(80 - 40) - \cos(80 + 40)) = 2\cos 40 +1$ as desired.

By the lemma we have
$\frac{\sin 60}{\sin 20} = \frac{\sin \theta}{\sin 80-\theta}$
or
$\sin(20) \sin(\theta) = \sin(80-\theta)\sin(60)$

LHS = $\cos (20 - \theta) - \cos (20 + \theta)$,
RHS = $\cos (20 - \theta) - \cos (140 - \theta)$

Therefore $\cos (20 + \theta) = \cos (140 - \theta)$, Since $\theta$ is between 0 and 80,
$20 + \theta = 140 - \theta$, yields $\theta = 60$ as required. QED

Will there be deductions for an unelegant solution like this one?

Okay, here are my comments.

[quote="beta"]Let BD, CD, AD intersect AC, AB, BC at F, E, G, respectively. Let $\angle DBC=\theta$.

Note that $\angle ADB = 150$, which implies $\angle BDG = 30 = \angle ADF$, so $\angle DFC = 70$, so $\angle FDC = 80$, so $\angle DCB = 180 - \theta$.[/quote]

I'll borrow a quote from Messrs. Rusczyk and Crawford here ([url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php]this[/url] is the article I'm quoting):

[quote="How to Write a Solution"]...every time you make an experienced reader have to think to follow your solution, you lose.[/quote]

In following your solution, I had to go and chase those angles myself; you never give me a reason [i]why[/i] knowing one angle implies knowing another angle.

Furthermore, you got a little careless at the end.  $\angle DCB$ equals $80^\circ-\theta$, not $180^\circ-\theta$.

On an unrelated note, you could have gotten that last part a bit more easily: just using the fact that the sum of the angles of a triangle is $180^\circ$, we have $\angle ADB=180^\circ-10\circ-20^\circ=150^\circ$ and $\angle ADC=180^\circ-40^\circ-30^\circ=110^\circ$, and, since the sum of all the angles around a given vertex is $360^\circ$, we have $\angle BDC=360^\circ-110^\circ-150^\circ=100^\circ$, and $\angle DCB=80^\circ-\theta$ to follow.

[quote="beta"]By Angle form of Ceva's Theorem, we have $\frac{\sin BAG}{\sin GAC} \cdot \frac{\sin ACE}{\sin ECB} \cdot \frac{\sin CBF}{\sin FBA} = \frac{\sin 10}{\sin 40} \cdot \frac{\sin 30}{\sin (80-\theta)} \cdot \frac{\sin \theta}{\sin 20} = 1$[/quote]

Well stated, but you should give that big equation its own line and some space around it (one way to do so with $\LaTeX$ is to enclose your math with $\backslash$[ math stuff $\backslash$] instead of regular dollar signs-this even centers the math for you).

[quote="beta"]Lemma: $\frac{\sin 10}{\sin 40} \cdot \frac{\sin 30}{\sin (20)} \cdot \frac{\sin 60}{\sin 20} = 1$

Proof: Substitue $\frac{1}{2}$ for $\sin 30$ and rearrange we have

$\frac{2(\sin 40)(\sin 20)}{\sin 10} = \frac{\sin 60}{\sin 20}$

Substitute $\sin 20 = 2(\sin 10)(\cos 10)$ and cancel out $\sin 10$ we have

$4(\sin 40)(\cos 10) = 4(\sin 40)(\sin 80) = \frac{\sin 60}{\sin 20}$

By Triple Angle formula, we have $\sin 60 = (\sin 20)(3 - 4\sin^2 20)$, cancel out the \sin 20, we have

$4(\sin 40)(\sin 80) = 3 - 4\sin^2 20$.
RHS= $1 + 2( 1 - 2\sin^2 20) = 1 + 2\cos 40$,
LHS= $2(\cos(80 - 40) - \cos(80 + 40)) = 2\cos 40 +1$ as desired.[/quote]

There is a big style flaw here.

You proved the lemma by starting with what you want (the product of the three trig fractions is 1) and turning it into something obvious ($2\cos(40^\circ)+1=2\cos(40^\circ)+1$).  You want to do this [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWriteForward.php]backwards[/url], that is, start with $2\cos(40^\circ)+1=2\cos(40^\circ)+1$ and work toward the product of the trig fractions being equal to 1.  The way you did it could very well cost you points-you never state that the steps you take can be done in reverse.

[quote="beta"]By the lemma we have
$\frac{\sin 60}{\sin 20} = \frac{\sin \theta}{\sin 80-\theta}$
or
$\sin(20) \sin(\theta) = \sin(80-\theta)\sin(60)$[/quote]

You should state how you get that equation-setting both products of trig fractions equal to each other (since they are both equal to one) and cancelling.

On an unrelated note, you could have finished immediately after proving the lemma by noting that $\textstyle{\frac{\sin(\theta)}{\sin(80^\circ-\theta)}}$ is increasing over the interval $(0^\circ,80^\circ)$, so there can only be one solution (namely, $\theta=60$)

[quote="beta"]LHS = $\cos (20 - \theta) - \cos (20 + \theta)$,
RHS = $\cos (20 - \theta) - \cos (140 - \theta)$

Therefore $\cos (20 + \theta) = \cos (140 - \theta)$, Since $\theta$ is between 0 and 80,
$20 + \theta = 140 - \theta$, yields $\theta = 60$ as required. QED[/quote]

Not a bad finish, but you never explicitly state that $\theta=60^\circ$ is the only solution.  If there might be another solution, then the desired angle isn't necessarily $60^\circ$, and all you proved is that the angle [i]could[/i] be $60^\circ$.

I give it 5/7.  A few tweaks and it would be quite nice.  Good job.

P.S., I wouldn't say that this is inelegant.  It's long, but I think all this trig was intended.

The way the lemma was proven was kind of awkward I suppose, but generally it's still not a big deal. Same with the angle chasing - making the reader chase isn't a good idea but it wouldn't have been dealbreaking here. 7 is probably not the most likely score. 5 or 6 are more likely I suppose but I would personally would lean toward 6.

Of course, you would have included a diagram on a paper exam- that would make the angle-chasing easier to follow.

Actually, here's a purely synthetic argument, and it's not even very long.

Let $BE$ be the altitude from $B$, and let $F\in BE$ be such that $\angle FAB=20^\circ$. Then since $\angle FBA=40^\circ$, $D$ is on  the bisectors of angles $FBA$ and $BAF$, and $D$ is the incenter of $ABF$. Therefore $\angle BFD=\angle AFD=60^\circ$. If $G$ is the intersection of $BE$ and $CD$, $\angle CGE=60^\circ$, so $\angle BGD=60^\circ$ and $F=G$. Now $\angle FAC=\angle FCA=30^\circ$, so $FC=FA$ and $BE$ is the perpendicular bisector of $AC$. Finally, $\angle EBC=\angle EBA=40^\circ$ and $\angle DBC=\angle DBF+\angle FBC=20^\circ+40^\circ=60^\circ$.

I would include a diagram to make this angle-chasing clear, but I don't have a good drawing program.

[quote="darktreb"]The way the lemma was proven was kind of awkward I suppose, but generally it's still not a big deal.[/quote]

Oh, yes it is.  Since he didn't say that all the steps could be done in reverse, he would almost certainly lose points there, as he would if he proved an inequality going forwards instead of backwards.

[quote]he would almost certainly lose points there, as he would if he proved an inequality going forwards instead of backwards.[/quote]

Not always true.

I think they are 80% concerned with your ability to do the problem.  If it's obvious that you didn't just make some stuff up, I think they will give at least a 6.  After all, proof writing skills are generally not that hard to teach.  Last year I had some pretty poor proof writing skills, but they still gave me credit.

Eh...not always true. If your solution is hard to read, hard to follow, incomplete, written in poor format, etc. you will almost certainly lose points.


function
college
algebra
polynomial
complex analysis
real analysis
complex analysis unsolved

Here's a very easy one, but I liked it :):

If $f$ is a function analytic in the open unit disk, then there is a sequence $(z_n)_n,\ |z_n|<1$ s.t. $|z_n|\to 1$ and the sequence $(f(z_n))_n$ is bounded.

P.S.

The Berkeley preliminary exams from 1977 up to 2000 can be found here:

[url]http://math.berkeley.edu/~desouza/pb.html[/url]

Here's possibly a solution:

Suppose otherwise, then it follows that every sequence $z_n$ such that $|z_n|\to1$ satisfies $f(z_n)\to\infty$.
Then if we call $g = 1/f$, then we have that $g$ tends to zero on the boundry.
It might have poles, but we know that there are finitely many of them, since had $f$ had infinitely many zeros a subsequence of those tending to the boundry could be found.

So $g$ has finitely many poles. If we multiply $g$ by a polynomials with zeros at the places of the poles of $g$, we get a holomorphic function, continuous in $|z|\leq1$ and is zero on the boundry.
So $g$ must have been zero to begin with, but this surely cant be.

I hope that there are no errors.

That's the idea; thanks for the solution.

With the same idea, if u suppose for contradiction then $\, f\,$  have a finitely zeros in unit disc, then $\,f(z)=(z-\alpha_1)\cdot (z-\alpha_n)g(z)\,$, where $\,g\,$ dont have zero in unit disc (else, this zero is zero for $\,f\,$ ). Now, the no boundedness for  $\,f\,$, really is the non-boundedness of $\,g\,$. Define,  $\,h=1/g\,$ , $\,h\to 0\,$, then you apply the theorem of modulus maximus (with limsup), hence $\,h=0\,$ for all $\,z\in C\,$. Sorry about my english.

Maybe you should unclick the "Disable BBCode in this post" box.

That is the same proof, just with rearranged order.


geometry
3D geometry
algebra
difference of squares
special factorizations

Evaluate (:rt5: + 2)1/3 - (:rt5: - 2)1/3.

No calculators allowed. Put your answers in spoiler.


topology
geometry
3D geometry
sphere
function
advanced fields
advanced fields unsolved

Hello,
I want to ask some questions on differential topology that I have a real difficulty to
solve them...
Thank you very much (in advance) for the help...
the file is attached...

Michael

(1) just calculate the differential of point 0, that the inverse image is a torus is evident.
(2) the embedding is given by (1) already en... what is parametrisation? (shame...)
(3) you should follow the hint of (1), write down the equation and it is f(x,y,z)=... is an equation of R^n+1 and M is the inverse image of 0. Verify that the value of 0 is a regular value, and the proof is done. So the only problem is to write down the equation.

I will try to solve (4) later on, got something to do now, sorry.  :D

Thank you Victor,
but about that the inverse image is a torus is not so evident (to me)....
about the parametrisations, I think that it is similar to the regular parametrisation of the sphere in R^3 (an analog to the parametrisation of the circle in the plane R^2). Am I right?
About (iii), I think that the function here have to be of 3n variables, not just 3....
It is a product of k spheres....

Can you, please, help me with this problem?
I have a real difficulty with (ii),(iii),(iv)....


What does that mean? Is it like gold medalists choose first, silver medalists choose second, etc.? Or is it like the first person who calls it gets it?

[quote="eyefragment"]Assuming it works like it did last year, there will be a set of prizes to choose from.  Gold winners choose first (on April 17th, I think) and get to choose two prizes.  Silver winners are second (Apr 20), with two prizes.  Then come bronze winners (Apr 23) with one prize, and finally, Honorable Mentions (Apr 26) with one prize.  Last year, everyone also received a T-shirt.  Selection order matters because there are around 7 prizes to choose from, and some may run out if there is high demand, so gold winners generally get to pick from any of the prizes, while the other levels might find their first few choices unavailable.[/quote]

This is pretty much correct.  The staggered dates are due to the fact that we have limited supply of many prizes, so our gold winners get to choose first.  As with last year, everyone will get a Tshirt.

When you login to usamts, where do you choose the prizes?  :?

Thanks!

It should appear immediately above your scores on your "My USAMTS" page.  If it doesn't, it means your prize group hasn't been given the chance to select prizes yet.


Ok, now that the mock test is over, it's time to release the final results!!!!! :D

In 3rd place, with 34 points, is....

[hide]FantasyLover![/hide] :winner_third:

In 2nd place, with 35 points, is...

[hide][size=150]pythag011!![/size][/hide] :winner_second:

And the CHAMPION OF THE 2008 GAUSS1181 AOPS COMPETITION IS....

[hide][size=200]xpmath!!!![/size][/hide] :winner_first:

Now for the teams:

THE WINNING TEAM OF THIS CONTEST IS....

(drumroll playing, nerves shaking...)

[hide][size=200]Team 1337!!!!![/size][/hide] :winner_first:

Thanks to everyone for participating in this contest!!!!!

ACKNOWLEDGMENTS:

Thanks to pythag011, abacadaea, and xpmath for proofreading and helping me clarify errors/ambiguity in this mock test. Their help was gladly appreciated. :)

RESULTS:

Individuals:

1. xpmath (PHAIL): Sprint 27, Target 6, Total 39
2. pythag011 (1337): Sprint 25, Target 5, Total 35
3. FantasyLover (0): Sprint 22, Target 6, Total 34
4. aleph0 (0): Sprint 25, Target 4, Total 33
5. shentang (PHAIL): Sprint 22, Target 5, Total 32
6. AIME15 (1337): Sprint 23, Target 4, Total 31
7. ernie (1337)
8. Wickedestjr (I HATE SCHOOL): Sprint 19, Target 3, Total 25
9. isabella2296 (1337): Sprint 18, Target 1, Total 20

The rest of the participants didn't submit, so they got 0. (Those were Aylsero of Team 0, THECHAMP of Team 0, Poincare of Team I HATE SCHOOL, abacadaea of Team PHAIL, 584920...x of Team PHAIL, mewto55555 of Team MewPwns, math154 of Team MewPwns, CA Math of Team MewPwns, and nikeballa96 of Team MewPwns.)

Teams:

1. 1337: Team Round 5, Team Score 38.25 (Captain pythag011)
2. PHAIL: Team Round 5, Team Score 27.75 (Captain shentang)
3. 0: Team Round 0, Team Score 16.75 (Captain aleph0)
4. I HATE SCHOOL: Team Round 0, Team Score 0 (Captain Wickedestjr)
5. MewPwns: Team Round 0, Team Score 0 (sadly no one did the mock test :( ) (Captain mewto55555)

Rules and Procedures for the Final Countdown Match will be posted later. Those CDR participants will be pythag011 and xpmath. Good luck to them!

And it's possible that there might be a Masters Round featuring xpmath, pythag011, FantasyLover, and aleph0.

i did sprint+target in like 40 minutes thinking 'lets get this over with'

thats why i said are you serious xD

The only reason our team won is because everyone submittted.  YAY!

We have last, third last, fourth last, and second.  Phail.

WHOA I BEETED ERNIE AND IZZY!!!

Oh darn, 39...would've been nice to get 40+, but oh well :)

Good job everyone and thanks for the test. If both of our other members had participated, I'm pretty sure that we would've won, considering that they're both insanely beast.

As a side note, if anyone is too ashamed to have their scores posted, please let me know. That way we don't put people in tears :D

[quote="gauss1181"][quote="AIME15"]Kinda late now, isn't it?[/quote]
No, it's just that I feel that sometimes if people hate it when others see their score and disregard them, they can choose to not have me post their scores publicly.[/quote]

Sure, why not?  I don't want ppl to see my epic fail. :wink: Although it's probably too late.

Just noticed that we got $ 5$ on the team round :o

Can I make a discussion thread for the problems?  There's a couple I'd like to see a solution to.

Hmm.... interesting...

Apparently I have somehow gotten a lot faster in countdown in the last 3 months... hmm... just played a countdown yesterday...

I would like a master's round... :)

I will get killed in countdown. *gulp*.

Pythag, just pm me any time that you can countdown, and I'll say if I can or not.

@pythag: Yes, there will be a Masters Round, but after the CD round is done.
@Fantasy: *cough cough* read the above posts and the other thread about Mock Test CD Round :wink:
@xpmath: Dont' worry, people should cheer for you! :D Good luck both pythag011 and xpmath! Remember: winner is best of 3 games. Good luck again!

So, yeah.

[size=200][b][color=red]pythag011 vs. xpmath!!!!!!!! 2008 Gauss1181 MATHCOUNTS!!!! WHO WILL WIN? CHEER FOR YOUR FAVORITE MATHLETE!!!![/color][/b][/size]
 :starwars:

*everybody applauds*

ohh there was a cd round (read the first post thoroughly for the first time...)

blah missed it by 1 pt

hmm, xpmath or pythag, can you post the time here so that we can all watch? :P

Dude, sadly you missed the written portion of the mock test, but it's still good that you're here to watch the countdown. :D

And there will be a masters round later featuring xpmath, pythag011, FantasyLover, and aleph0, if all wish to participate. :)

Guys, and also look out for me! If I can, and the time is appropriate, I will be a spectator and arbiter. :)

@5849206328x: No way. I'm not going to be pwned when people are watching. I just hope I can win atleast one of the three.

Heh, in a way, I'm somewhat glad that 5849206328x didn't submit. I wouldn't have come first then. Then again, since he would probably get >44, we most likely would have won team...

there are a lot of distractions in my house... so probably not

EDIT: not in that way, xp...

Yo, five eight whatever, I'm pretty sure that you would, even with all the "distractions". In fact, I think you're probably even more beast than this one kid named Alan Chou I used to know - have you ever met him? I wonder why he's not on aops anymore...

C'mon, xp, let us watch you get pwned... I bribed pythag, y'know.

Oh snap? I actually won...weird. Darn, if that practice had been real...
Pythag, pm me a time that you can come on and I'll see if it works. Or just pm me when both of us are on.

So the first game was complete!

So far, 1/3 of the match's games were played!
xpmath won the first game, 4-1!

Let's say good luck again for the next 2 games!

Ok, everyone, the top 2 rivals of the gauss1181 Super Mathcounts Tourney have played it to the finish!!!!

Now it's time to announce the winner of the tourney!!!

[hide="CONGRATULATIONS!!!!"]In a straight 3-way win after 2 straight wins in a row, beating his opponent, pythag011, in two games 4-1 and 4-0...

xpmath is declared the winner of this tourney!!!!!! :first: [/hide]

Again, congrats xpmath and pythag011 for your hard work in this tourney. You guys battled it out to the end, and it was intense! I'm proud of you people.

Oh, and BTW there would be a Masters Round side event for the top four winners of this contest, and that'll feature (if none of them withdraw) xpmath, pythag011, FantasyLover, and aleph0!

Stay tuned for more details! Also in this thread feel free to congratulate the CDR participants or discuss anything you wish about the final match!

[quote="xpmath"]Good games, pythag. Heh, I lucked out. If pythag had actually NOT ENTERED random stuff for the second countdown, he would've won.

Who bets that pythag wins masters?[/quote]

I bet you that I DON'T.

Unless the masters round is something I know.

How do I see the questions I got right/wrong.

Aww man, I wanted 30+

[quote="xpmath"]Good games, pythag. Heh, I lucked out. If pythag had actually NOT ENTERED random stuff for the second countdown, he would've won.

Who bets that pythag wins masters?[/quote]

1. No I wouldn't have.

2. Of course I'm going to win masters, I bribed Temperal.  :)

[quote="pythag011"]
2. Of course I'm going to win masters, I bribed Temperal.  :)[/quote]

wrong master's round, pythag.  :wink:


function
algebra
polynomial
binomial theorem
combinatorics unsolved
combinatorics

I saw this problem that asks to earn this sequence with use from generating function
the sequence is

$ a_m \equal{}2 a_{m\minus{}1} \minus{} a_{m\minus{}2}$   that  $ a_0 \equal{}0 ,  a_1\equal{}\minus{}1$

 solution:

if f(x)=$ a_0 \plus{} a_1 x \plus{} a_2 x^2 \plus{}a_3 x^3 \plus{}..... \plus{}a_m x^m \plus{}...$ then

-2xf(x)=$ \minus{}2a_0 x \minus{}2a_1 x^2 \minus{}2 a_2 x^3 \minus{} 2a_3 x^4 \minus{}....\minus{}2 a_{m\minus{}1} x^{m\plus{}1}\minus{}....$

$ x^2 f(x)\equal{} a_0 x^2 \plus{} a_1 x^3 \plus{}a_2 x^4 \plus{} a_3 x^5 \plus{}....\plus{}a_m x^{m\plus{}2} \plus{}.......$

if we sum three frases above we have :

$ (1\minus{}2x\plus{}x^2 ) f(x)\equal{}a_0\plus{}(a_1 \minus{} 2a_0)x\plus{}(a_2\minus{}2a_1\plus{}a_0)\plus{}(a_3\minus{}2a_2\plus{}a_1)x^3\plus{}.....\plus{}(a_m\minus{}2a_{m\minus{}1}\plus{}a_{m\minus{}2} x^m\plus{}......$

    since $ a_2\minus{}2a_1\plus{}a_0 \equal{}0 and a_3\minus{}2a_2\plus{}a_1 \equal{}0 and   a_m\minus{}2a_{m\minus{}1}\plus{}a_{m\minus{}2}\equal{}0  \Longrightarrow$

$ f(x)\equal{} \frac {\minus{}x}{{(1\minus{}x)}^2} \Longrightarrow$

$ f(x)\equal{} \sum_{m\equal{}0}^{\infty} a_m x^m  \equal{} \frac {\minus{}x}{{(1\minus{}x)}^2} \equal{} \minus{}x {(\sum_{m\equal{}0}^{\infty} x^m)}^2$

in continuance what I have to do .I only want to earn $ a_m$ with use from generating function

The definition of polynomial multiplication gives $ \left( 1 \plus{} x \plus{} x^2 \plus{} ... \right)^2 \equal{} 1 \plus{} 2x \plus{} 3x^2 \plus{} 4x^3 \plus{} ...$.  Alternately, you can apply the general binomial theorem and write

$ \frac{1}{(1 \minus{} x)^2} \equal{} \sum_{n \ge 0} (\minus{}1)^n {\minus{}2 \choose n} x^n$

and note that $ {\minus{}2 \choose n} \equal{} \frac{(\minus{}2)(\minus{}2\minus{}1)...(\minus{}1\minus{}n)}{n!} \equal{} (\minus{}1)^n (n\plus{}1)$.


How good is this book if one is just starting to jump into Olympiad mathematics? Here is my standing:

I have AoPS 1 and 2 and have pretty much mastered 1 and 2 except for the weird geometry and a bit of number theoretical stuff in volume 2.

I can get about an 8 or 9 on the AIME.

I am joining WOOT this year.

If I get this book, will I be able to understand most if not all of it if I work at it consistently?

[quote="tim1234133"]You are talking about the book by Arthur Engel, I take it?

I suppose you should find it fine. However the book has not a great deal of actual theory, but instead tons of examples and tons of questions. Some of them are very hard, but if you start on a chapter (I find most chapters are independent of each other, each dealing with a different area e.g. functional equations) and work through the examples carefully, then you should be able to slowly work through the content... but to actually understand all of it and do all the questions would probably take months/years... and it takes some thorough work.

P.S.  I think perhaps this should be in the [url=http://www.mathlinks.ro/Forum/index.php?f=319]Books and Journals Forum[/url][/quote]

Oh whoops... I didn't see that forum. By bad. Thanks for your reply  :)


Εστω $ x_1,x_2,x_3$ ρίζες του πολυωνύμου $ f(x)\equal{}x^3\plus{}ax^2\plus{}bx\plus{}c$ και ισχύει
1)$ x_2\minus{}x_1\equal{}k>0$
2)$ x_3>\frac{1}{2}(x_1\plus{}x_2)$

Nα βρείτε τη μέγιστη τιμή του $ \frac{2a^3\plus{}27c\minus{}9ab}{k^3}$

Δημήτρη το k είναι σταθερό???

Οχι δεν είναι σταθερό! (εξάλλου αν ήταν σταθερό η ύπαρξή του στον παρονομαστή δεν θα είχε κανένα νόημα). Συγγνώμη πάντως που δεν το διευκρίνισα... :blush:

Ειναι $ f( \minus{} a/3) \equal{} ( \minus{} a/3 \minus{} x_1)( \minus{} a/3 \minus{} x_2)( \minus{} a/3 \minus{} x_3) \equal{} \frac {2a^3 \plus{} 27c \minus{} 9ab}{27}$ και απο $ Vieta$ εχουμε $ \minus{} a \equal{} x_1 \plus{} x_2 \plus{} x_3$ και απαλοιφοντας το $ a$ η ζητουμενη παρασταση γινεται $ \frac {(x_2 \plus{} x_3 \minus{} 2x_1)(2x_2 \minus{} x_1 \minus{} x_3)(2x_3 \minus{} x_1 \minus{} x_2)}{(x_2 \minus{} x_1)^3}$ και εδω (με το "χερι" του Sil) μετατρεπεται σε $ (1 \plus{} \frac {x_3 \minus{} x_1}{x_2 \minus{} x_1})(2 \minus{} \frac {x_3 \minus{} x_1}{x_2 \minus{} x_1})(2\frac {x_3 \minus{} x_1}{x_2 \minus{} x_1} \minus{} 1)$  συναρτηση μιας μεταβλητης και το ακροτατο ευκολη διαδικασια.....

Ωραία λύση! Για την ιστορία, το μέγιστο βγαίνει $ \frac{3}{2}\cdot{\sqrt{3}}$
[color=red]
[size=150]Καλή επιτυχία σε όλους όσους γράφουν αύριο!!![/color][/size]

Kαλη επιτυχια και απο μενα σε ολους τους διαγωνιζομένους για την ομαδα

Kali epitixia se oloys mas ston eswteriko.

Alitheia mesogeiada tha dwsoyme pali oloi? :?:


a,b,c>=0.Prove that:
     SUM (a+b)^2/a^2+b^2+2c^2\ <=3.
  anyone has the nice solution

Hmm... Can you write it better please? :)

$\sum\frac{(a+b)^2}{a^2+b^2+2c^2}\leq3\Leftrightarrow\sum\frac{a^2-bc}{2a^2+b^2+c^2}\geq0.$
Not better, but proof:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=57775

Ok, thanks Arqady :)
Well, in fact, at first, I thought there was a typo because of the $2c^2$  :blush:

You have another solution for my first problem,argardy or the problem 2 because it is not a topo

Of course, It's in my book, too.

Sadnees!

This problem is very difficult. I use SOS and easy to realize 2 is the best constant.

i have a solution by bunhiacosky.too short.And you,hungkhtn.Did you the same me
(Thuc ra nhung bai em post em deu da lam duoc nhung em muon tham khao kinh nghiem cac dai cao thu tren ml,neu vi pham ban quyen mong anh bo qua. :D .Neu anh ko cho phep em se ko post nhung bai tuong tu the nua.OK :D )

Here is my solution, though I am convinced there is a simpler one:

[color=blue][b]Problem.[/b] Let a, b, c be three positive reals. Prove the inequality

$\sum\frac{\left(b+c\right)^2}{2a^2+b^2+c^2}\leq 3$,

where the $\sum$ sign denotes cyclic sums.[/color]

[i]Solution.[/i] First, we equivalently transform the inequality in question:

$\sum\frac{\left(b+c\right)^2}{2a^2+b^2+c^2}\leq 3\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ 3-\sum\frac{\left(b+c\right)^2}{2a^2+b^2+c^2}\geq 0$
$\Longleftrightarrow\ \ \ \ \ \sum\left(1-\frac{\left(b+c\right)^2}{2a^2+b^2+c^2}\right)\geq 0\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ \sum\frac{2\left(a^2-bc\right)}{2a^2+b^2+c^2}\geq 0$
$\Longleftrightarrow\ \ \ \ \ \sum\frac{a^2-bc}{2a^2+b^2+c^2}\geq 0$.

Upon multiplication by $\left(2a^2+b^2+c^2\right)\left(2b^2+c^2+a^2\right)\left(2c^2+a^2+b^2\right)$, this becomes

$\sum\left(a^2-bc\right)\left(2b^2+c^2+a^2\right)\left(2c^2+a^2+b^2\right)\geq 0$.

Now, denote $T=\sum\left(a^2-bc\right)=a^2+b^2+c^2-bc-ca-ab$. Then, $T\geq 0$, since we know that $a^2+b^2+c^2\geq bc+ca+ab$ (this is a well-known inequality, holding for arbitrary reals a, b, c). Also, denote $S=\sum\left(2a^4+3b^2c^2\right)=2a^4+2b^4+2c^4+3b^2c^2+3c^2a^2+3a^2b^2$. Then, a short calculation yields

$\left(2b^2+c^2+a^2\right)\left(2c^2+a^2+b^2\right)=S-\left(a^4-2b^2c^2\right)$.

Hence,

$\sum\left(a^2-bc\right)\left(2b^2+c^2+a^2\right)\left(2c^2+a^2+b^2\right)=\sum\left(a^2-bc\right)\left(S-\left(a^4-2b^2c^2\right)\right)$
$=\sum\left(a^2-bc\right)\cdot S-\sum\left(a^2-bc\right)\left(a^4-2b^2c^2\right)=TS-\sum\left(a^2-bc\right)\left(a^4-2b^2c^2\right)$.

Now,

$\sum\left(a^2-bc\right)\left(a^4-2b^2c^2\right)=\sum\left(a^6-2a^2b^2c^2-a^4bc+2b^3c^3\right)$
$=\left(\sum a^6+2\sum b^3c^3\right)-\left(a^3+b^3+c^3\right)abc-6a^2b^2c^2$
$=\left(a^3+b^3+c^3\right)^2-\left(a^3+b^3+c^3\right)abc-6\left(abc\right)^2=\left(a^3+b^3+c^3-3abc\right)\left(a^3+b^3+c^3+2abc\right)$.

By a well-known identity,

$a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-bc-ca-ab\right)=\left(a+b+c\right)T$.

Thus, we obtain

$\sum\left(a^2-bc\right)\left(a^4-2b^2c^2\right)=\left(a+b+c\right)T\left(a^3+b^3+c^3+2abc\right)$.

Hence,

$\sum\left(a^2-bc\right)\left(2b^2+c^2+a^2\right)\left(2c^2+a^2+b^2\right)=TS-\sum\left(a^2-bc\right)\left(a^4-2b^2c^2\right)$
$=TS-\left(a+b+c\right)T\left(a^3+b^3+c^3+2abc\right)=T\cdot\left(S-\left(a+b+c\right)\left(a^3+b^3+c^3+2abc\right)\right)$.

But $T\geq 0$, and a computation shows that

$S-\left(a+b+c\right)\left(a^3+b^3+c^3+2abc\right)$
$=\frac32\left(\left(ca-ab\right)^2+\left(ab-bc\right)^2+\left(bc-ca\right)^2\right)$
$+\underbrace{a^2\left(a-b\right)\left(a-c\right)+b^2\left(b-c\right)\left(b-a\right)+c^2\left(c-a\right)\left(c-b\right)}_{\geq 0\text{ by the Schur inequality}}$
$\geq 0$.

Hence, $\sum\left(a^2-bc\right)\left(2b^2+c^2+a^2\right)\left(2c^2+a^2+b^2\right)\geq 0$, and this, according to the above, solves the problem.

The essential idea of the above solution is that if one tries to bunch the inequality, then it is basicly a 6th degree symmetric sum, but it factors into the 2-degree sum T and a 4-th degree sum. The above was an attempt to perform this factorization with as little algebra as possible, but 4-th degree calculations were still required.

  Darij

Please post your solution here. If it can be solved by Bunhiacopski, It's very good.

But my memory tells me that It's can be true?

[quote="tamtrantrong"]Apply Bunhiacovsky,we have
 (a^2/a^2+c^2)+(b^2/b^2+c^2)>=(a+b)^2.....
Sum up,we have the result.Too short.Am i true,hungkhtn?[/quote]
oh,nice solution,tamtrantrong.let me clear it for you.
by caughy(sorry,I understand "Bunhiacovsky" is caughy,maybe it is different in your country)
we have $\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2} \geq \frac{(a+b)^2}{a^2+b^2+2c^2}$
and there sum the the three similar...
nice solution.

thank zhaobin.
to hungkhtn:do you agree with my solution

BRAVO!!!!!
                       It's a  nice solution.
                 But I  think you have this solution first,then you propose this problem???????

Yes, It's out of my fancy...

But my ordinary problem is :

$\frac{c^2-ab}{a^2+b^2+kc^2}+\frac{b^2-ac}{a^2+c^2+kb^2}+\frac{a^2-bc}{b^2+c^2+ka^2} \ge 0$

For all real $k$ and $0 \le k le 2$.

Is it still vaild with the general?

A complementary inequality is:
$\sum\frac{2bc+3(b^2+c^2)}{2a^2+b^2+c^2}\geq6$.

Can you post your solution for your problem,Vasc? I think it very hard and i can't do that


A Senate committee consists of 5 Republicans, 6 Democrats, and 2 Independents. A subcommittee of 3 members is randomly chosen. What is the probability that the subcommittee consists of 1 Republican, 1 Democrat, and 1 Independent?

Overall, there are 13C3=286 ways to choose 3 people.
To choose one of each party, multiply the possibilities: 5*6*2=60
Final answer is
$ 60/286\equal{}30/143$


algebra
polynomial
number theory proposed
number theory

Does there exist a a sequence $a_{0},a_{1},a_{2},\dots$ in $\mathbb N$, such that for each $i\neq j, (a_{i},a_{j})=1$, and for each $n$, the polynomial $\sum_{i=0}^{n}a_{i}x^{i}$ is irreducible in $\mathbb Z[x]$?
[i]By Omid Hatami[/i]

Sure -- the simplest example I can think of is to take $a_{n}= p_{n}$ for a sufficiently fast increasing sequence of primes, so that all zeros of $\sum_{i=0}^{n}p_{i}z^{i}= 0$ lie inside $|z| < 1$ (for instance, it suffices that $p_{n}> p_{n-1}+\cdots+p_{0}$). Was this also your original solution? :) Of course, "most" sequences you take have this property, but for "most" explicit (closed-form given) sequences it should be very difficult if impossible to prove the irreducibility.

I propose this more difficult problem: is there an infinite set $A$ of positive integers such that for any pairwise different elements $a_{0},\ldots,a_{n}\in A$ the polynomial $\sum_{i=0}^{n}a_{i}x^{i}$ is irreducible?

Of course this came to my mind vess. I wanted to find all subsets $S\subset \mathbb N$ such that there exists a sequence $a_{0},a_{1},\dots\in S$ such that $\sum_{i=0}^{n}a_{i}x^{i}$ is irreducible for each $n$.
I proposed this problem as part (a) of the problem, but it was removed before the exam:
Find the size of smallest $S\subset \mathbb N$ such that the following sequence exists.
It is easily proved with Eisenstein that for $S=\{p,q\}$ ($p,q$ primes) the sequence $a_{0}=p$, and $a_{i}=q$ for $i\geq 1$ satisfies the problem conditions.
Of course more than this could be said. If $S$ has two square-free elements, then also with Eisenstein we  can prove that such sequence exists. But for the set $\{4,9\}$, I have no idea. I would be very happy if anyone finds all subset $S\subset\mathbb N$ such that for the the sequnce exists.

I can reproduce an idea of Ljunggren to prove that every two-element set $S$ has your property. Does this answer your question completely? Of course, Eisenstein can't be applied in this general case (e.g., for $\{4,9\}$). To get an idea how to do this, take a quick look at the slides http://www.math.sc.edu/~filaseta/seminars/SeminarNotes102901.pdf and http://www.math.sc.edu/~filaseta/seminars/michiganlect00.pdf and, even better, browse some papers of M. Filaseta on his webpage. (Obviously, one of the best places to look for irreducible polynomials :))

But the problem I intended was different, have you thought on this: construct an infinite set $A$ such that for any $n$ and [i] every[/i] (not just some!) sequence $a_{0},\ldots, a_{n+1}$ of $n+1$ pairwise different elements of $A$, the polynomial $\sum_{0}^{n}a_{i}x^{i}$ is irreducible (hence allowing arbitrary rearrangements of the sequence)?

I think it is same as the problem China TST 2006 problem

"Very nice idea, to use the fact that if all the complex roots are inside the unit circle, then the polynomial is irriducible in Z!"

Please, could someone explain, why is it true? For example, the polynom (2x-1)^2.

I think this works : We proceed inductively,taking a suitable base case $2+3x$ as a polynomial and then after having assumed a construction for $a_1,...,a_{n-1}$ we define $a_n$ as a number having no prime dividing any of the previous terms and $a_n$ much larger in absolute value than the sum of absolute values of $a_i's$ for $i=0,1,...,n-1$. Now, Perron's criterion for irreducibility nails it.

[quote = anantmudgal09] $a_n$ much larger in absolute value than the sum of absolute values of $a_i's$ for $i=0,1,...,n-1$. Now, Perron's criterion for irreducibility nails it. [/quote]

Unfortunately this is not true because Perron's criterion only applies when the coefficient in front of $x^{n-1}$ is very large. However the recursive construction can be salvaged. Suppose we have $a_0, a_1, \dots, a_{n}$ such that all polynomials $\sum_{t = 0}^{k} a_{i_t} x^t$ with $i_0, i_1, \dots, i_k$ distinct indices in $[0,n]$ are irreducible. We want to find $a_{n+1}$ so that the same holds if some $i_t$ equals $n+1$. Abstractly, we need a finite family of polynomials $\{P_{\alpha}(x) + a_{n+1} x^{t_{\alpha}}\} $ to be simultaneously irreducible. By Hilbert's irreducibility theorem, every polynomial in the family is irreducible for "almost all" $a_{n+1}$.  The result follows.

Sorry for reviving.
[quote=vess]Sure -- the simplest example I can think of is to take $a_{n}= p_{n}$ for a sufficiently fast increasing sequence of primes, so that all zeros of $\sum_{i=0}^{n}p_{i}z^{i}= 0$ lie inside $|z| < 1$ (for instance, it suffices that $p_{n}> p_{n-1}+\cdots+p_{0}$). [/quote]
I couldn't understand how this directly implies it's irreducible. Could someone please explain?

If you mean why the zeroes of $P(x):=\sum_{i=0}^{n}p_{i}z^{i}$ lie inside $|z|<0$ it's due to the Rouche's theorem ( $p_nz^n$ and $P(z)$ have the same number of roots inside $|z|<1$ providing $p_{n}> p_{n-1}+\cdots+p_{0}$ ). If the question is why $P$ is irreducible then here's why:
Suppose $P(x)=Q(x)R(x)$ and both $Q,R$ are with integer coefficients. Then one of them, say $Q$, is monic (with senior coefficient $1$) since all $p_i$ are primes. So, $Q$ is a monic polynomial with all roots lying inside $|z|<1.$ It means the constant term of $Q$ has absolute value less than $1$, hence it must be $0$. But $z=0$ is not root of $P$.

It would be nice if this works [hide=Soln]We claim that choosing $a_n>1+a_0+a_1+\dots a_{n-1}$,and all $a_i$ prime numbers
[b]Lemma 1[/b] If a polynomial $a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$ is irreducible so is $a_0x^n+a_1x^{n-1}+\dots a_n$
USing the following lemma and the classic problem of
[quote=Problem]If $f(x)$ is a polynomial with $a_0$ prime and $|a_0|>1+|a_1|+...|a_n|$ is irreducible[/quote]
we are done


Divide the total distance for a mile by the length of each step.  Then we see that Jim takes 2112 steps, and Jeffrey takes 1980.  Then the ratio is $ \frac{2112}{1980}\equal{}\frac{16}{15}$


geometry rectangle inradius incenter parallelogram quadratics invariant

Let $ABCD$ be a rectangle and $T$ a circle through $A$ and $C$. let $T_{1}$ be a circle with radius $r_{1}$ that is tangent to $AB$, $BC$ and $T$ and lies on the interior of $ABCD$ and in the exterior of $T$. Let $T_{2}$ be another circle with radius $r_{2}$ that is tangent to $AD$, $DC$ and $T$ and that also lies in the interior of $ABCD$, and in the interior of $T$.

Prove that $r_{1}+r_{2}=2r$ where $r$ is the inradius of $ABC$ and that one of the common interior tangents of $T_{1}$ and $T_{2}$ is parallel to $AC$

Discussed (and solved?) on http://www.mathlinks.ro/Forum/viewtopic.php?t=6490 .

  darij

[quote="Pascual2005"]Let $ABCD$ be a rectangle and $(T)$ a circle trough $A$ and $C$. Let $(T_1)$ be a circle with radious $r_1$ that is tangent to $AB$, $BC$ and $(T)$ and lies in the interior of $ABCD$ and in the exterior of $(T)$. Let $(T_2)$ be another circle with radious $r_2$ that is tnagent to $AD$, $DC$ and $(T)$ and that also lies in the interior of $ABCD$, and in the interior of $(T)$.

Prove that $r_1+r_2=2r$ where $r$ is the inradius of $ABC$ and that one of the common interior tangents of $(T_1)$ and $(T_2)$ is parallel to $AC$.[/quote]

[quote="darij grinberg"]Discussed (and solved?) on http://www.mathlinks.ro/Forum/viewtopic.php?t=6490.

  darij[/quote]
Right. This problem has been bothering me for some time. Moving it to the Solved Problems section was not justified, because in the original thread, the problem was solved using the following unproven lemma:

Let $I$ be the incenter of a triangle $\triangle ABC$, let $(P)$ be a circle passing through the vertices $B, C$ and $(Q)$ a circle tangent to the circle $(P)$ at point $T$ and to the lines $AB, AC$ at points $U, V$, respectively. The points $B, T, I, U$ are concyclic and the points $C, T, I, V$ are concyclic.

[color=blue][b](1)[/b][/color] Let $ABCD$ be a rectangle with diagonal intersection $X \equiv AC \cap BD$, opposite side lengths $a = AB = CD, b = BC = DA$ and diagonal lengths $c = AC = BD = \sqrt{a^2 + b^2}$. WLOG, assume that $a > b$. Let $(I_1), (I_2)$ be the incircles of the congruent right angle triangles $\triangle ABC \cong \triangle CDA$. Let $s$ be the semiperimeter of the triangle $\triangle ABC$. The radius $r$ of the congruent incircles $(I_1), (I_2)$ is then

$r = s - c = \frac{a + b - c}{2}$

The distance between the incenters $I_2, I_1$ is equal to

$I_2I_1 = \sqrt{(b - 2r)^2 + (a - 2r)^2} = \sqrt{(c - a)^2 + (c - b)^2}$

and the center line $I_2I_1$ of these 2 incircles forms with the line $AB$ the angle

$\phi = -\arctan{\frac{b - 2r}{a - 2r}} = -\arctan{\frac{c - a}{c - b}}$

Let $J_1, J_2$ be the tangency points of these 2 incircles with their common internal tangent, the rectangle diagonal $AC$. The segment length $J_2J_1$ of this common internal tangent between the tangency points is equal to

$J_2J_1 = \sqrt{I_2I_1^2 - 4r^2} = \sqrt{(c - a)^2 + (c - b)^2 - (a + b - c)^2} = \sqrt{c^2 - 2ab} = a - b$

and the common internal tangent $J_2J_1$ forms with the center line $I_2I_1$ the angle

$\theta = \arctan{\frac{2r}{J_2J_1} = \arctan{\frac{a + b - c}{a - b}}}$

[color=blue][b](2)[/b][/color] Let $E, H$ be 2 points on the rays $AD, CD$, such that $AE= CH = \frac d 2 \ge \frac c 2$. Let $AEA'F, CGC'H$ be 2 congruent squares with the side lengths $\frac d 2$ outside of the rectangle $ABCD$, the 1st square having the vertex $A$ and the 2nd squrae the vertex $C$ common with the rectangle $ABCD$. Obviously, $AA' = CC'$ and $AA' \parallel CC'$. Hence, the quadrilateral $ACC'A'$ ia a parallelogram and $A'C' = AC = C$. Let $\triangle A'C'O$ be an isosceles triangle outside of the parallelogram $ACC'A'$, with the base $A'C' = c$ and shoulder sides $A'T = C'T = \frac d 2$. Let $(T_1), (T_2)$ be 2 circles with radii $\rho_1, \rho_2$ intersecting at the point $T$, the circle $(T_1)$ tangent to the lines $AB, BC$ and the circle $(T_2)$ tangent to the lines $CD, DA$.

[color=red][b](2a)[/b][/color] First, assume that $d = c$ is minimum. The isosceles triangle $\triangle A'C'T$ then degenerates into the line segment $A'C'$ and $T$ is the midpoint of this segment. The median $\rho_1 = T_1T = \frac c 2$ of the triangle $A'C'T_1$ is equal to half of its opposite side $A'C' = c$, hence, the angle $\angle A'T_1C' = 90^o$ is right. Since $\rho_1 = AE = \frac c 2$, the lines $AB \parallel A'T_1$ are parallel, $\triangle A'T_1C' \cong \triangle ABC$ are congruent and $A'T_1 = a, C'T_1 = b$. Let $M_1, M_2$ be the midpoints of the sides $C'T_1, A'T_1$, i.e., $TM_1 = \frac{A'T_1}{2} = \frac a 2$, $TM_2 = \frac{C'T_1}{2} = \frac b 2$. Let the normal from the point $T$ to the line $AB$ intersect the normal from the point $T_2$ to the line $BC$ at a point $N_2$. Using Pythagoren theorem for the right angle triangle $\triangle T_2TN_2$ with the hypotenuse $T_2T = \rho_2$ and the legs

$T_2N_2 = \rho_2 - \left(\frac a 2 + \frac c 2 - a\right) = \rho_2 - \frac{c - a}{2}$

$TN_2 = \left(\frac b 2 + \frac c 2 - b\right) - \rho_2 = \frac{c - b}{2} - \rho_2$

we get a quadratic equation for the radius $\rho_2$ of the circle $(T_2)$:

$T_2T^2 = T_2N_2^2 + TN_2^2$

$\rho_2^2 = \left(\rho_2 - \frac{c - a}{2}\right)^2 + \left(\frac{c - b}{2} - \rho_2\right)^2$

$\rho_2^2 - [2c - (a + b)] \rho_2 + \frac{3c^2 - 2c(a + b)}{4} = 0$

Using $a^2 + b^2 = c^2$, The discriminant of this quadratic equation turns to be a full square:

${{D = 4c^2 - 4c(a + b) + (a + b)^2 - 3c^2 + 2c(a + b)}}{2} = (a + b - c)^2$

and the roots are

$\rho = \frac{2c - (a + b) \pm (a + b - c)}{2} = \frac c 2$ [color=white].[/color] or [color=white].[/color] $\frac{3c}{2} - (a + b)$

Since the radius $\rho_1 = \frac c 2$ and obviously, the radius $\rho_2 < \rho_1$, the 1st root is not acceptable and we have to take the 2nd root with the minus sign. Then

$\rho_1 - \rho_2 = \frac c 2 - \frac{3c}{2} + a + b = a + b - c = 2r$

The distance between the centers $T_2, T_1$ is equal to

$T_2T_1 = \sqrt{N_2M_2^2 + (T_2N_2 + T_1M_2)^2} = \sqrt{\left(b + \rho_2 - \frac c 2\right)^2 + \left(a + \rho_2 - \frac c 2\right)^2} = \sqrt{(c - a)^2 + (c - b)^2}$

and the center line $T_2T_1$ of these 2 circles forms with the line $AB$ the angle

$\phi' = -\arctan{\frac{N_2M_2}{T_2N_2 + T_1M_2}} = -\arctan{\frac{b + \rho_2 - \frac c 2}{a + \rho_2 - \frac c 2}} = -\arctan{\frac{c - a}{c - b}}$

Let $W_1, W_2$ be the tangency points of the circles $(T_1), (T_2)$ with their common external tangent not intersecting the rectangle $ABCD$. The segment length $W_2W_1$ of this common external tangent between the tangency is equal to

$W_2W_1 = \sqrt{T_2T_1^2 - (\rho_1 - \rho_2)^2} = \sqrt{(c - a)^2 + (c - b)^2 - (a + b - c)^2} =$[color=white].[/color]$\sqrt{c^2 - 2ab} = a - b$

and the common external tangent $W_2W_1$ forms with the center line $T_2T_1$ the angle

$\theta' = \arctan{\frac{\rho_1 - \rho_2}{W_2W_1} = \arctan{\frac{a + b - c}{c - b}}}$

Thus we obtained $T_2T_1 = I_2I_1,\ W_2W_1 = J_2J_1$ and $\phi' = \phi,\ \theta' = \theta$, which means that both the center lines $T_2T_1 \parallel I_2I_1$ and the common tangents $W_2W_1 \parallel J_2J_1$ are parallel to each other.

[color=red][b](2b)[/b][/color] Next, we repeat this process with $d > c$. Let $M$ be the midpoint of the segment $A'C'$ and $N$ the intersection of normals form the points $A', C'$ to the lines $BC, AB$, respectively. Let the normal from the point $T$ to the line $AB$ intersect the normals from the points $T_1, T_2$ to the line $BC$ at points $N_1, N_2$. From the right angle triangle $\triangle A'TM$ (or $\triangle C'TM$),

$TM = \sqrt{A'T^2 - A'M^2} = \frac{\sqrt{d^2 - c^2}}{2}$

Using Pythagoren theorem for the right angle triangle $\triangle T_2TN_2$ with the hypotenuse $T_2T = \rho_2$ and the legs

$T_2N_2 = \rho_2 - \left(\frac a 2 + \frac d 2 - a\right) - \frac b c\ TM = \rho_2 - \frac{d - a}{2} - \frac b c\ \frac{\sqrt{d^2 - c^2}}{2}$

$TN_2 = \left(\frac b 2 + \frac d 2 - b\right) + \frac a c\ TM - \rho_2 = \frac{d - b}{2} + \frac a c\ \frac{\sqrt{d^2 - c^2}}{2} - \rho_2$

we get a quadratic equation for the radius $\rho_2$ of the circle $(T_2)$:

$T_2T^2 = T_2N_2^2 + TN_2^2$

$\rho_2^2 = \left(\rho_2 - \frac{d - a}{2} - \frac b c\ \frac{\sqrt{d^2 - c^2}}{2}\right)^2 + \left( \frac{d - b}{2} + \frac a c\ \frac{\sqrt{d^2 - c^2}}{2} - \rho_2\right)^2$

$\rho_2^2 - \left[d - (a + b) + \frac{a + b}{c} \sqrt{d^2 - c^2}\right] \rho_2\ +$

$+ \frac{3d^2 - 2d(a + b)}{4} + \frac{d(a + b) - 2ab}{2c} \sqrt{d^2 - c^2} = 0$

Similarly, using Pythagoren theorem for the right angle triangle $\triangle T_1TN_1$ with the hypotenuse $T_1T = \rho_1$ and the legs

$T_1N_1 = \left(\frac a 2 + \frac d 2\right) + \frac b c\ TM - \rho_1 = \frac{d + a}{2} + \frac b c\ \frac{\sqrt{d^2 - c^2}}{2} - \rho_1$

$TN_1 = \left(\frac b 2 + \frac d 2\right) + \frac a c\ TM - \rho_1 = \frac{d + b}{2} + \frac a c\ \frac{\sqrt{d^2 - c^2}}{2} - \rho_1$

we get a quadratic equation for the radius $\rho_1$ of the circle $(T_1)$:

$T_1T^2 = T_1N_1^2 + TN_1^2$

$\rho_1^2 = \left(\frac{d + a}{2} + \frac b c\ \frac{\sqrt{d^2 - c^2}}{2} - \rho_1\right)^2 + \left( \frac{d + b}{2} + \frac a c\ \frac{\sqrt{d^2 - c^2}}{2} - \rho_1\right)^2$

$\rho_1^2 - \left[d + (a + b) + \frac{a + b}{c} \sqrt{d^2 - c^2}\right] \rho_1\ +$

$+ \frac{3d^2 + 2d(a + b)}{4} + \frac{d(a + b) + 2ab}{2c} \sqrt{d^2 - c^2} = 0$

Using $a^2 + b^2 = c^2$ repeatedly, the discriminants of both these quadratic equations turn to be full squares:

$D_2 = \left[2d - (a + b) + \frac{a + b}{c} \sqrt{d^2 - c^2}\right]^2 -\ 4\left[\frac{3d^2 - 2 d(a + b)}{4} + \frac{d(a + b) - 2ab}{2c} \sqrt{d^2 - c^2}\right] =$

$= \left(\sqrt{d^2 - c^2} + \frac{d(a + b)}{c} - c\right)^2$

$D_1 = \left[2d + (a + b) + \frac{a + b}{c} \sqrt{d^2 - c^2}\right]^2 -\ 4\left[\frac{3d^2 + 2 d(a + b)}{4} + \frac{d(a + b) + 2ab}{2c} \sqrt{d^2 - c^2}\right] =$

$= \left(\sqrt{d^2 - c^2} + \frac{d(a + b)}{c} + c\right)^2$

and their roots are

$\rho_2 = \frac 1 2 \left[2d - (a + b) + \frac{a + b}{c} \sqrt{d^2 - c^2} \pm \left(\sqrt{d^2 - c^2} + \frac{d(a + b)}{c} - c\right)\right] =$

$= \frac d 2 + \frac{a + b + c}{2c}\left(d - c + \sqrt{d^2 - c^2}\right)$ [color=white].[/color] or [color=white].[/color] $\frac d 2 - \frac{a + b - c}{2c}\left(d + c - \sqrt{d^2 - c^2}\right)$

$\rho_1 = \frac 1 2 \left[2d + (a + b) + \frac{a + b}{c} \sqrt{d^2 - c^2} \pm \left(\sqrt{d^2 - c^2} + \frac{d(a + b)}{c} + c\right)\right] =$

$= \frac d 2 + \frac{a + b + c}{2c}\left(d + c + \sqrt{d^2 - c^2}\right)$ [color=white].[/color] or [color=white].[/color] $\frac d 2 - \frac{a + b - c}{2c}\left(d - c - \sqrt{d^2 - c^2}\right)$

Since for $d \rightarrow c$, these roots approach

$\rho_2 \rightarrow \frac c 2$ [color=white].[/color] or [color=white].[/color] $\frac 3 2 c - (a + b)$

$\rho_1 \rightarrow \frac 3 2 c + (a + b)$ [color=white].[/color] or [color=white].[/color] $\frac c 2$

in both cases, we have to take the 2nd roots with the minus signs as before. Then

$\rho_1 - \rho_2 = \frac d 2 - \frac{a + b - c}{2c}\left(d - c - \sqrt{d^2 - c^2}\right) - \frac d 2 + \frac{a + b - c}{2c}\left(d + c - \sqrt{d^2 - c^2}\right) =$

$= a + b - c = 2r$

The distance between the centers $T_2, T_1$ is equal to

$T_2T_1 = \sqrt{N_2N_1^2 + (T_2N_2 + T_1N_1)^2}$

$T_2N_2 + T_1N_1 = \rho_2 - \frac{d - a}{2} - \frac b c\ \frac{\sqrt{d^2 - c^2}}{2} + \frac{d + a}{2} + \frac b c\ \frac{\sqrt{d^2 - c^2}}{2} - \rho_1 =$

$= \rho_2 - \rho_1 + a = c - (a + b) + a = c - b$

$N_2N_1 = TN_1 - TN_2 = \frac{d + b}{2} + \frac a c\ \frac{\sqrt{d^2 - c^2}}{2} - \rho_1 - \frac{d - b}{2} - \frac a c\ \frac{\sqrt{d^2 - c^2}}{2} + \rho_2 =$

$= \rho_2 - \rho_1 + b = c - (a + b) + b = c - a$

$T_2T_1 = \sqrt{(c - a)^2 + (c - b)^2}$

and the center line $T_2T_1$ of these 2 circles forms with the line $AB$ the angle

$\phi' = -\arctan{\frac{N_2N_1}{T_2N_2 + T_1N_1}} = -\arctan{\frac{c - a}{c - b}}$

The segment length $W_2W_1$ of their common external tangent between the tangency is equal to

$W_2W_1 = \sqrt{T_2T_1^2 - (\rho_1 - \rho_2)^2} = \sqrt{(c - a)^2 + (c - b)^2 - (a + b - c)^2} =$[color=white].[/color]$\sqrt{c^2 - 2ab} = a - b$

and the common external tangent $W_2W_1$ forms with the center line $T_2T_1$ the angle

$\theta' = \arctan{\frac{\rho_1 - \rho_2}{W_2W_1} = \arctan{\frac{a + b - c}{c - b}}}$

Thus we again obtained $T_2T_1 = I_2I_1,\ W_2W_1 = J_2J_1$ and $\phi' = \phi,\ \theta' = \theta$, which means that both the center lines $T_2T_1 \parallel I_2I_1$ and the common tangents $W_2W_1 \parallel J_2J_1$ are still parallel to each other.

[color=blue][b](3)[/b][/color] Consider the rectangle side lines $AB, BC, CD, DA$, the incircles $(I_2), (I_1)$, their common internal tangent $J_2J_1 \equiv AC$, the circles $(T_2), (T_1)$ and their common external tangent $W_2W_1$ directed. Assume that the common internal tangent of the incircles $(I_1), (I_2)$ is directed from the rectangle vertex $A$ to the rectangle vertex $C$, i.e., $\overline{W_2W_1} \equiv \overline{AC}$. The directed incircles must have the same directions at their tangency points with their common tangent, hence, the circle ${(I_2 \oplus})$ is directed counter-clockwise and the incircle $(I_1 \ominus)$ clockwise. This forces the rectangle side lines to be directed as $\overline{BA}, \overline{CD}, \overline{CB}, \overline{DA}$ and the circles $(T_2 \ominus), (T_1 \ominus)$ tangent to these directed lines have to be both direted clockwise and their common tangent as $\overline{W_2W_1}$. Radii of circles directed counter-clockwise are considered positive and radii of circles directed clockwise negative. Thus we have $\rho_2 < 0,\ \rho_1 < 0,\ \rho_2 > \rho_1,\ \rho_2 - \rho_1 = 2|r| > 0$. Furthemore, 2 directed circles have at most one pair of common directed tangents (the common external tangents of the undirected circles, if the 2 directed circles have the same direction, and the common internal tangents of the undirected circles, if the 2 directed circles have opposite directions). Finally, consider the point $T$ as a circle of zero radius (undirected, because its radius is zero - neither positive, nor negative). Perform an expansion transformation of the figure (the directed rectangle side lines $\overline{BA}, \overline{CD}, \overline{CB}, \overline{DA}$, the directed circles $(T_2 \ominus), (T_1 \ominus)$, and the point $T$ considered as a circle of zero radius) by the distance $\frac d 2 > 0$. The directed lines  $\overline{BA}, \overline{CD}, \overline{CB}, \overline{DA}$ are carried into directed lines  $\overline{B'A'}, \overline{C'D'}, \overline{C'B'}, \overline{D'A'}$ parallel to the original lines at the distance $\frac d 2$ right from the original lines when looking down their directions. Hence, they form a rectangle $A'B'C'D'$ congruent to the original rectangle $ABCD$, the vertices $A', C'$ coincident with the previously constructed vertices $A', C'$ of the parallelogram $ACC'A'$. The point $T$ and the directed circles $(T_2 \ominus, \rho_2), (T_1 \ominus, \rho_1)$ are carried into directed circles with the same centers, but with radii increased by the distance $\frac d 2$, i.e., into directed circles $(T \oplus, r = \frac d 2 > 0),  (T_2 \oplus, r_2 > 0), (T_1 \ominus, r_1 < 0)$,

$r_2 = \rho_2 + \frac d 2 = -\left[\frac d 2 - \frac{a + b - c}{2c}\left(d + c - \sqrt{d^2 - c^2}\right)\right] + \frac d 2 = \frac{a + b - c}{2c}\left(d + c - \sqrt{d^2 - c^2}\right)$

$r_1 = \rho_1 + \frac d 2 = -\left[\frac d 2 - \frac{a + b - c}{2c}\left(d - c - \sqrt{d^2 - c^2}\right)\right] + \frac d 2 = \frac{a + b - c}{2c}\left(d - c - \sqrt{d^2 - c^2}\right)$

$r_2 - r_1 = a + b - c = 2|r|,\ \ |r_2| + |r_1| = 2|r|$

The directed circle $(T \oplus)$ tangent to the directed circles $(T_2 \oplus, r_2), (T_1 \ominus, r_1)$ passes through the vertices $A', C'$ of the rectangle $A'B'C'D'$. The difference $\rho_2 - \rho_1 = r_2 - r_1$, the directed segment of the common tangent length and the directed angle between the common tangent of the circles $(T_1), (T_2)$ and the fixed direction $\overline{I_2I_1}$, identical with the fixed direction $\overline{T_2T_1}$, are all invariant with respect to the expansion transformation. This concludes the proof of the problem proposition.

I would appreciate a simpler proof either of the above lemma or of the problem proposition, because this proof too long. Besides, $ABCD$ can be just a parallelogram, not necessarily a rectangle, and the proposition still holds. The above proof can be adapted by replacing every use of Pythagorean theorem by the cosine theorem, but it becomes even longer and the formulas more complicated.

Yetti

Impressive, Yetti, really impressive....


Source: 360 PMC
Solve the equation
$\sin x \cos y+\sin y \cos z+\sin z \cos x = \frac{3}{2}$.

Using AM-GM $LHS\geq 3.\sqrt[3]{ \frac{sin(2x)sin(2y)sin(2z)}8}\geq \frac{3}{2}$, equality holds only if the sines of double argument are all 1 which means all x,y,z being of the form $\frac{(2k+1)\pi}4$.
However, for this estimate I need all the terms on LHS to be positive. If one is negative (wlog the first one), we,ll have
$sin z+cos z\geq3/2$, contradiction.


The police has captured n prisoners and is about to start interrogating them as follows. The prisoners will be taken to solitary cells, and no communication between them will be permitted. Police will chose one of the prisoners at will and take him to the interrogation room (same room for all prisoners at all times). In the interrogation room there are two switches - one blue and one red. The prisoner is required to switch exactly one of the switches, and is then taken back to his cell. The police will then choose a prisoner again and repeat this process. At any point any prisoner has an option of saying that all the prisoners has already visited the interrogation room. IF this claim is correct, all the prisoners go free, if not they are executed. Prisoners have some time before the first interrogation to agree on a strategy. Can they guarantee a successful escape?

(The police do not touch the switches. Prisoners have no sense of time. Assume that if all prisoners are silent, police will take each prisoner to the interrogation room arbitrary many times.)

This is a generalization of (I think) zabelman's problem in another post.

[quote="mathdool"]Will the prisoners know how many total prisoners there are?
Do the prisoners know how the lights start?[/quote]

Will the prisoners know how many total prisoners there are? [b]Yes, n prisoners[/b].

Do the prisoners know how the lights start? [b]No[/b].

[hide]The blue switch will be useless, the prisoners hit the blue switch when they choose to do nothing.

The first prisoner interrogated will turn the red switch on.

Whenever the red switch is off and a normal prisoner goes in that has never hit the red switch yet, he will hit it.

Make one of the prisoners the leader. Whenever the leader is interrogated(unless he is the first person in, then his job will be to turn the red switch off), he will turn the switch off if it is on. This will tell him that one new person has been in the switch room. When the leader has to turn the switch off for the n-1 time, he can declare an escape.

It will be best if the police don't know who the leader is.

Despite this method, if n is large enough, an escape can still not be guaranteed because most of the prisoners will be dead before they can escape.  [/hide]

Unfortunately, you don't know whether you're the first person, since you have no sense of time.

[quote="zabelman"]also, the red switch may be off or on at the start -- you dont know[/quote]

If the red switch is already the way its supposed to be, then the prisoner just does nothing (hits blue switch)

I see.

but if the red light starts on and the leader is the first person in (remember- he doesnt know hes the first) then he will flip the switch and THINK that someone has been in there to turn the light on. He will have miscounted!

[hide]We nominate one person (call this person ``Counter'') to turn the red switch off at the first visit. Then we give everyone two ``tokens'' to drop in the room. You can use a token by turning the red switch on if it is off. If you are not using a token, you must flip the blue switch. Every time Counter comes into the room, he will switch the red switch to the off position if it is on and will switch the blue switch if it is off. Counter starts counting how many times he switches off the red switch the second time he visits the room. When he reaches 2n-3, he declares that everyone has visited the room.[/hide]

Very Good! This is a problem from Berkeley Math Circle Monthly Contest 5 due Feb 8, 2004.


[quote="felipesa"]Solve x

Problem 1
$|x-1|-|x|+|2x+3|>2x+2$

[/quote]

For this one, you just have to divide cases $x\geq 1, 0 \leq x < 1, -\frac{3}{2} \leq x < 0, x < -\frac{3}{2}$

[quote="warut_suk"][quote="felipesa"]
Problem 2
$|x-2|x-3+|x-1|||=0$[/quote]

$|x-2|x-3+|x-1|||=0
\longleftrightarrow x-2|x-3+|x-1|| = 0
\longleftrightarrow x = 2|x-3+|x-1||
\longleftrightarrow x^2 = 4(x^2+9+x^2-2x+1-6x+2x|x-1|-6|x-1|)$

The rest should be easy...[/quote]

divide cases in x-2|x-3+|x-1||=0

(i)x >= 1 ,  x - 2|2x-4|=0  then get x

(ii)x<1 , x = 2 |x-3-x+1| = 4  contradict
 :?  :?  :?  :?

prob 1:

plot the graph of | 2x + 3 | - | x | + | x - 1 |

should get a \_/ type of graph. then plot 2x + 2 on the same graph. after which, find the range of x values where \_/ graph lies above the linear curve.

for prob 2 i dun really get the notations.. sry cant solve that :S

yes, the graph based solution is indeed the simplest and most accurate


AMC
AIME
geometry
arithmetic sequence
geometric sequence

0 < a < b < c < d are integers such that a, b, c is an arithmetic progression, b, c, d is a geometric progression, and d - a = 30. Find a + b + c + d.

I wrote it out as a=d-30, so then:

b=d-30+r
c=d-30+2r

So then (d-30+2r)/(d-30+r)=d/(d-30+2r).  because of geometry ratios.

Multiplying out:
3d(10-r)=4(r-15)^2

now I'm stuck, so r<10, but the rhs doesn't give anything because of trivial inequality.

help!

[hide]Let a, a+d, and a+2d be the first three terms that form an arithmetic progression. Thus, in order for the last three terms to form a geometric progression, the fourth term has to be (a+2d)^2/(a+d). Since the first and fourth terms differ by 30, (a+30)^2=(a+30)(a+d). Simplifying, we get 3a(10-d) = 2d(2d-15). Thus, 2d-15>0 and 10-d>0 or 2d-15<0 and 10-d<0. The latter has no solutions while the former yields d = 8,9. When d=8, the sequence isn't all integers, thus d=9 and the sequence is 18, 27, 36, 48. Their sum is 129.[/hide]

How do you know that the common difference is d?