geometry
circumcircle
geometric transformation
reflection
projective geometry
geometry unsolved

let ABC be a triangle with the circumcircle (C).a line passes through the area of the triangle and meet AB,AC,BC and (C) in the points X,Y,Z,E and F respectively such that EX=YF and let M be the midpoint of XY.if AM meet (C) in the point H. prove that ZH is tangent line. :lol:

i think its a nice problem because its solution is really short  :)

I think something is wrong here :huh: Perhaps is $ YH$ and not $ ZH$ ?

[quote="Erken"]He forgot to say that $ l$-line meets the sides $ [AB]$,$ [AC]$ and the line $ BC$ at points $ X,Y,Z$.[/quote]
 my problem is correct .i agree with Erken here. i forgot to say that L-line meets the circumcircle at the points E and F and also meets the sides AB,AC and the line BC at points X,Y,Z. :wink:

here is a hint: first use butterfly theorem after that use pascal theorem  :wink:

Let $ N$ be the midpoint of $ BC$
$ K$ is the reflection of $ A$ through $ M$
We have $ AX//YK$ then $ \angle YKA\equal{}\angle XAK\equal{}\angle BCH$
Moreover $ \angle CBH \equal{} \angle KAY$
Then triangles $ AYK$ and $ BHC$ are similar
Thus $ YK/HC\equal{}AK/BC\equal{}MK/NC$
Hence triangle $ MYK$ and $ NHC$ are similar
So $ \angle KMY \equal{}\angle HNC$
Therefore $ N,H,Z,M$ are cyclic $ (1)$
Also $ \angle ONZ\equal{}\angle OMZ\equal{}90$ ($ M, N$ are midpoint of $ EF,BC$)
Then $ O,M,Z,N$ are cyclic $ (2)$
From (1) and (2) $ O,M,Z,H,N$ are cyclic
Thus $ O,N,H,Z$ are cyclic
This means $ \angle OHZ\equal{}\angle ONZ \equal{}90$
Therefore $ ZH$ is the tangent of circle $ (O)$ $ q.e.d$

[quote="Huyền Vũ"]
 thank you "Huyền Vũ your proof is really nice :)

My solution:

Let the tangent of $ \odot (ABC) $ through $ A $ meet $ EF $ at $ V $ .
Let the tangent of $ \odot (ABC) $ through $ H $ meet $ EF $ at $ Z^* $ .

From Desargue involution theorem ( for quadrilateral $ AABC $ ) we get $ (V,X;E,F)=(Z,Y;F,E) $ ,
but $ M $ is the common midpoint of $ EF $ and $ XY \Longrightarrow M $ is also the midpoint of $ VZ $ .

From Desargue involution theorem ( for quadrilateral $ AAHH $ ) we get $ (M,V;E,F)=(M,Z^*;F,E) $ ,
so combine with $ ME=MF $ we get $ M $ is the midpoint of $ VZ^* \Longrightarrow Z \equiv Z^* $ . i.e. $ ZH $ is the tangent of $ \odot (ABC) $

Q.E.D


[quote="ifai"][hide]$x=\dfrac{3}{4}$[/hide][/quote]

Please show your work. And I don't think that's correct...

[quote="4everwise"]For what value of x does $1+x+x^2+x^3+x^4+...=4?$

[hide="hint"]$1=x^0$[/hide][/quote]
you mean this sequense is infinite?

[hide="complete solution"]Applying $\sum_{k=1}^n {aq^{k-1}}=a\dfrac{1-q^n}{1-q}$,
\[ 1+x+x^2+...+x^n=\sum_{k=1}^n{x^{k-1}}=\dfrac{1-x^n}{1-x}  \]
If $|x|\geq1$, then the sum = infinite. So $|x|<0$, and $x^n\rightarrow 0$ when $n\rightarrow\infty$.
Then $\sum_{k=1}^{\infty}{x^k}=\sum_{k=1}^{\infty}{x^{k-1}}=\frac 1{1-x}$. thus
\[ \frac 1{1-x}=4  \]
Therefore, $x=\frac34$[/hide]

[quote="math92"]ifai, thats a little too hard for the Basics forum.[/quote]

Actually it isn't that difficult, he just used some harder language for mathmatics that isn't too appropriate, but it's okay.

[hide="lets check the solution :D"]let $\frac{3}{4}+(\frac{3}{4})^2+(\frac{3}{4})^3+....=x$

$\frac{4}{3}x=1+\frac{3}{4}+(\frac{3}{4})^2+.......$.

$\frac{4}{3}x-1=x$

$x=3$

So back in the equation, $1+3=4$. [/hide]

I made an error when I said you weren't correct, sorry  :blush:

Don't mind and wise man would also make mistake.  :)

Please be careful to use the appropriate notation for the Basics forum.  Summation notation is not appropriate for the Basics forum.  ifai's solution will be impossible to read for nearly anyone in middle school.

[quote="boxedexe"]Well, ifai just restated the geometric progression formula which resulted into a decreasing one.  The decreasing geometric progression formula is $s=\frac{1}{1-x}=4\rightarrow 1-x=\frac{1}{4}\rightarrow x=1-\frac{1}{4}=\frac{3}{4}$.

Masoud Zargar[/quote]
i think you shouldn't use it here.it is the basics. :idea:

[quote="ashegh"]if u mean integer for X,i should say for non of the values of X . :)[/quote]

Well, if x's are integers, then the value of the equation up to $x^4$ will equal at least 5 if you exclude the values of x euqaling 0.  Besides, the $...$ indicates that it goes up to infinity.

Masoud Zargar

[quote="boxedexe"]Well, ifai just restated the [b]geometric progression formula[/b] which resulted into a decreasing one.  The decreasing geometric progression formula is $s=\frac{1}{1-x}=4\rightarrow 1-x=\frac{1}{4}\rightarrow x=1-\frac{1}{4}=\frac{3}{4}$.

Masoud Zargar[/quote]

What is that formula? I'm not really sure what you all mean. I haven't done much with geometric series. Can some one explain this to me?

I won't prove it to you because teh proof does not belong in the basics, however, I can tell you that you find the sum of $x^0+x^1+...x^n=a$ where $x^n$ is the value of an $x$ on a graph before they begin to have constant values of $y$.  If you want a full explanation, you can post your question in the high school forums. :)

Masoud Zargar


function
trigonometry
algebra
functional equation
calculus
calculus computations

Find all functions $ f: \mathbb{R}\to\mathbb{R}$ such that exists $ f'$ on $ \mathbb{R}$ and \[ f(x\plus{}y)\equal{}\frac{f(x)\plus{}f(y)}{1\minus{}f(x)f(y)}\;\;\;\forall x,y\in\mathbb{R}.\]

Let $ f(x) \equal{} \tan g(x)$.  The arctan function is onto, so this is permissible.  Then $ g(x \plus{} y) \equal{} g(x) \plus{} g(y)$, the Cauchy functional equation; the condition that $ f'$ exists gives $ g(x) \equal{} cx$.


$ a \geq b \geq c$
Let $f(a)=\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{4c^{2}}{a}+8(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})-6a-9b-15c$
We have $f'(a)=\frac{2a}{b}-\frac{4c^{2}}{a^{2}}+4\frac{\sqrt{b+\sqrt{c}}}{\sqrt{a}}-6 \geq 0$
(because by AM-GM: $\frac{a}{b}+2\frac{\sqrt{b}}{\sqrt{a}}\geq 3 \ and \frac{\sqrt{c}}{\sqrt{a}}\geq \frac{c}{a}(\frac{c}{a}\le 1) )$
It follows that: $f(a) \geq f(b) = \frac{b^{2}}{c}+\frac{4c^{2}}{b}+16\sqrt{bc}-6b-15c$
 $f'(b)= \frac{2b}{c}-\frac{4c^{2}}{b^{2}}+\frac{8\sqrt{c}}{\sqrt{b}}-6 \geq 0$
It follows that: $f(b) \geq f(c)= 0$

[quote="NguyenDungTN"]$a \geq b \geq c$
Let $f(a)=\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{4c^{2}}{a}+8(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})-6a-9b-15c$
We have $f'(a)=\frac{2a}{b}-\frac{4c^{2}}{a^{2}}+4\frac{\sqrt{b+\sqrt{c}}}{\sqrt{a}}-6 \geq 0$
(because by AM-GM: $\frac{a}{b}+2\frac{\sqrt{b}}{\sqrt{a}}\geq 3 \ and \frac{\sqrt{c}}{\sqrt{a}}\geq \frac{c}{a}(\frac{c}{a}\le 1) )$
It follows that: $f(a) \geq f(b) = \frac{b^{2}}{c}+\frac{4c^{2}}{b}+16\sqrt{bc}-6b-15c$
 $f'(b)= \frac{2b}{c}-\frac{4c^{2}}{b^{2}}+\frac{8\sqrt{c}}{\sqrt{b}}-6 \geq 0$
It follows that: $f(b) \geq f(c)= 0$[/quote]


inequalities
function
pigeonhole principle
Cauchy Inequality
inequalities proposed

Given $ a, b, c > 0$. Prove that: $ \left(\frac{a}{2a\plus{}b}\right)^3\plus{}\left(\frac{b}{2b\plus{}c}\right)^3\plus{}\left(\frac{c}{2c\plus{}a}\right)^3 \ge \frac{1}{9}$
I have seen it on the box "Inequalities Unsolved Problems". It exist a very nice solution by Am-Gm and Cauchy Schwarts  :)

The hint: Use the well-known result $ \frac{a\plus{}b\plus{}c}{ab\plus{}bc\plus{}ca} \ge \frac{a}{a^2\plus{}2bc}\plus{}\frac{b}{b^2\plus{}2ca}\plus{}\frac{c}{c^2\plus{}2ab}$  :)

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \left(\frac {a}{2a \plus{} b}\right)^3 \plus{} \left(\frac {b}{2b \plus{} c}\right)^3 \plus{} \left(\frac {c}{2c \plus{} a}\right)^3 \ge \frac {1}{9}$
I have seen it on the box "Inequalities Unsolved Problems". It exist a very nice solution by Am-Gm and Cauchy Schwarts  :)[/quote]
It seems the inequality should be $ \left(\frac {b}{2a \plus{} b}\right)^3 \plus{} \left(\frac {c}{2b \plus{} c}\right)^3 \plus{} \left(\frac {a}{2c \plus{} a}\right)^3 \ge \frac {1}{9}$.

The general problem:
Given $ a, b, c > 0$. Determine the best constant $ k$ for the following ineq still trues:
$ \left(\frac{a}{ka\plus{}b}\right)^{3}\plus{}\left(\frac{b}{kb\plus{}c}\right)^{3}\plus{}\left(\frac{c}{kc\plus{}a}\right)^{3}\ge\frac{3}{(k\plus{}1)^3}$

By Am-Gm and Cauchy Schwarts, we can prove this ineq with $ k \le 2$.
The ineq which you mention is a weak ineq (with $ k \equal{}\frac{1}{2}$).
With $ k\equal{}1$, we obtain VietNam TST 2005's problem  :)

Here is proof or nguoivn's hint:

http://www.mathlinks.ro/Forum/viewtopic.php?t=311079&search_id=848058871

but how to use is very misterious :maybe:

Dear manlio, I'll try to post full proof soon. Sorry for my late  :oops:

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \left(\frac {a}{2a \plus{} b}\right)^3 \plus{} \left(\frac {b}{2b \plus{} c}\right)^3 \plus{} \left(\frac {c}{2c \plus{} a}\right)^3 \ge \frac {1}{9}$
I have seen it on the box "Inequalities Unsolved Problems". It exist a very nice solution by Am-Gm and Cauchy Schwarts  :)[/quote]
[hide="@ Tran Quoc Anh"]chào anh,
ngoài lời giải lúc trước em em post trên ddbdt.co.cc với ý tưởng chưa thật sự sáng sủa
mới đây em đã tìm được 1 lời giải rất đẹp mắt chỉ với Cauchy và khỏa sát hàm số rất thú vị
mời anh tiếp tục thảo luận tại [url=http://ddbdt.co.cc/forum/ba-t-ae-ng-tha-c-chae-ca-la-oei-gia-i-thpt/61-ma-t-ba-i-chae-xa-la-i-gia-i-ca-qua-c-anh.html#post86]tại đây[/url][/hide]

Thank you. Wait for your solution, my friend  :)

i've not finished the solution, and it's not elegant, but it works...

we'll prove that for $ xy\geq 1$, the following ineq holds
$ \left(\dfrac{1}{2 \plus{} x^2}\right)^3 \plus{} \left(\dfrac{1}{2 \plus{} y^2}\right)^3\geq 2\left(\dfrac{1}{2 \plus{} xy}\right)^3$

suppose wlog that $ y\geq x$... after some algebraic manipulation and a division by $ (y \minus{} x)$ this is equivalent to
$ \dfrac{x\left((2 \plus{} xy)^2 \plus{} (2 \plus{} xy)(2 \plus{} x^2) \plus{} (2 \plus{} x^2)^2\right)}{(2 \plus{} x^2)^3}\geq \dfrac{y\left((2 \plus{} xy)^2 \plus{} (2 \plus{} xy)(2 \plus{} y^2) \plus{} (2 \plus{} y^2)^2\right)}{(2 \plus{} y^2)^3}$

it's easy to prove that $ x(2 \plus{} y^2)^3\geq y(2 \plus{} x^2)^3$ using that $ xy\geq 1$...

we're left to prove that $ \dfrac{x\left(4 \plus{} xy \plus{} x^2\right)}{(2 \plus{} x^2)^2}\geq \dfrac{y\left(4 \plus{} xy \plus{} y^2\right)}{(2 \plus{} y^2)^2}$...

after expanding and dividing by $ (y \minus{} x)$ it follows that we have to prove that
$ 16xy \plus{} 4xy(x^2 \plus{} xy \plus{} y^2) \plus{} x^2y^2(x^2 \plus{} xy \plus{} y^2) \plus{} x^2y^2\geq 16 \plus{} 4xy \plus{} 4(x^2 \plus{} xy \plus{} y^2)$,
which again follows from the fact that $ xy\geq 1$, and we have proved the result...

then it's enough to prove that $ 2\left(\dfrac{z}{2z \plus{} 1}\right)^3 \plus{} \left(\dfrac{1}{2 \plus{} z^2}\right)^3\geq \dfrac{1}{9}$...  :D

in fact, this last inequality can be rewritten as $ (z\minus{}1)^2(10z^7\plus{}8z^6\plus{}60z^5\plus{}39z^4\plus{}102z^3\plus{}15z^2\plus{}8z\plus{}1)\geq 0$, which is true, and we're done  :D

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \left(\frac {a}{2a + b}\right)^3 + \left(\frac {b}{2b + c}\right)^3 + \left(\frac {c}{2c + a}\right)^3 \ge \frac {1}{9}$
I have seen it on the box "Inequalities Unsolved Problems". It exist a very nice solution by Am-Gm and Cauchy Schwarts  :)[/quote]

[u][b]My Solution (NGUYỄN ĐÌNH THI).[/b][/u] Let $ x = \frac {b}{a},y = \frac {c}{b},z = \frac {a}{c}$ so $ xyz = 1$ and the inequality become
$ \frac {1}{{{{\left( {x + 2} \right)}^3}}} + \frac {1}{{{{\left( {y + 2} \right)}^3}}} + \frac {1}{{{{\left( {z + 2} \right)}^3}}} \ge \frac {1}{9}$
According to the Dirichlet principle number 2 of 3 $ \left( {x - 1} \right),\left( {y - 1} \right),(z - 1)$ same mint-mark. No loss of general, suppose $ \left( {x - 1} \right)\left( {y - 1} \right) \ge 0 \Rightarrow x + y \le xy + 1$. According Cauchy inequality we have
$ \frac {1}{{{{\left( {x + 2} \right)}^3}}} + \frac {1}{{{{\left( {y + 2} \right)}^3}}} \ge 2{\left( {\frac {1}{{\left( {x + 2} \right)\left( {y + 2} \right)}}} \right)^{\frac {3}{2}}} = 2{\left( {\frac {1}{{xy + 4 + 2\left( {x + y} \right)}}} \right)^{\frac {3}{2}}} \ge 2{\left( {\frac {1}{{3xy + 6}}} \right)^{\frac {3}{2}}} = 2{\left( {\frac {1}{{\frac {3}{z} + 6}}} \right)^{\frac {3}{2}}}$
Therefore we only need to prove
$ 2{\left( {\frac {1}{{\frac {3}{z} + 6}}} \right)^{\frac {3}{2}}} + \frac {1}{{{{\left( {z + 2} \right)}^3}}} \ge \frac {1}{9}$
Review $ f\left( z \right) = 2{\left( {\frac {1}{{\frac {3}{z} + 6}}} \right)^{\frac {3}{2}}} + \frac {1}{{{{\left( {z + 2} \right)}^3}}}$, we haveta có
$ {f^'}\left( z \right) = \frac {9}{{{{\left( {\frac {1}{{\frac {3}{z} + 6}}} \right)}^{\frac {5}{2}}}{z^2}}} - \frac {3}{{{{\left( {z + 2} \right)}^4}}}$
 $ {f^'}\left( z \right) = 0 \Leftrightarrow z{\left( {z + 2} \right)^8} = 27{\left( {2z + 1} \right)^5}$
Have
$ z{\left( {z + 2} \right)^8} - 27{\left( {2z + 1} \right)^5} = \left( {z - 1} \right)\left( {{z^8} + 17{z^7} + 129{z^6} + 577{z^5} + 833{z^4} + 465{z^3} + 97{z^2}} \right) = 0\Leftrightarrow {z_0} = 1$.
Set the table we see variation function $ y = f (z)$ changes sign from - to + when passing $ {z_0}$. Therefore $ f\left( z \right) \ge f\left( {{z_0}} \right) = \frac {1}{9}$
Allow proof is complete. Equality occurs if and only if $ x = y = z = 1$ or $ a = b = c$.

[hide="@Trần Quốc Anh:"]
anh có thể đọc rõ hơn bằng tiếng mẹ đẻ ở đây http://ddbdt.co.cc/forum/ba-t-ae-ng-tha-c-chae-ca-la-oei-gia-i-thpt/61-ma-t-ba-i-chae-xa-la-i-gia-i-ca-qua-c-anh.html
cũng mong sớm thấy lời giải của anh cũng ở trang này[/hide]

[size=150][quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \left(\frac {a}{2a \plus{} b}\right)^3 \plus{} \left(\frac {b}{2b \plus{} c}\right)^3 \plus{} \left(\frac {c}{2c \plus{} a}\right)^3 \ge \frac {1}{9}$
I have seen it on the box "Inequalities Unsolved Problems". It exist a very nice solution by Am-Gm and Cauchy Schwarts  :)[/quote]

I have problem LBQ107(a)(in 2000.6),Is that inequality.
At the same time, there are problem LBq107(b):
$ (\frac {x}{kx \plus{} y})^{k \plus{} 1} \plus{} (\frac {y}{ky \plus{} z})^{k \plus{} 1} \plus{} (\frac {z}{kz \plus{} x})^{k \plus{} 1} \geq \frac {3}{(k \plus{} 1)^{k \plus{} 1}}(k \geq 0)$.
BQ[/size]

[quote="VIMF"]
I think my solution not bad by Cauchy and Dirichlet.
Can you post your solution?[/quote]
Of course, I said that the first hints is nice but I only don't like the last step because it used computer. It's only my opinion.
[quote="VIMF"]
Have
$ z{\left( {z \plus{} 2} \right)^8} \minus{} 27{\left( {2z \plus{} 1} \right)^5} \equal{} \left( {z \minus{} 1} \right)\left( {{z^8} \plus{} 17{z^7} \plus{} 129{z^6} \plus{} 577{z^5} \plus{} 833{z^4} \plus{} 465{z^3} \plus{} 97{z^2}} \right) \equal{} 0\Leftrightarrow {z_0} \equal{} 1$.
[/quote]
[quote="VIMF"]
Can you post your solution?[/quote]
Not now because sometimes, I don't want to post them. Of course, Can_hang2007 checked my solution and I think it's nicer than your proofs. Besides, Can_hang also had had a proof made only by hand. I like the proofs which don't need to any help of computer like that  :)

[quote="VIMF"][quote="nguoivn"]Nice works, Campos and VIMF  :) . I think the idea mixing variables is very useful. Besides, it's only a point which I don't so like is the last step of the proofs above, we must to use computer.
  :oops:
[quote="xzlbq"]

I have problem LBQ107(a)(in 2000.6),Is that inequality.
At the same time, there are problem LBq107(b):
$ (\frac {x}{kx \plus{} y})^{k \plus{} 1} \plus{} (\frac {y}{ky \plus{} z})^{k \plus{} 1} \plus{} (\frac {z}{kz \plus{} x})^{k \plus{} 1} \geq \frac {3}{(k \plus{} 1)^{k \plus{} 1}}(k \geq 0)$.
BQ[/size][/quote]
Thank you very much, xzlbq. Indeed, we want to mention this problem with nice solutions in our new books but we don't know the author. Now, we know  :lol: .
PS: That's so special general problem. If it's trues, it's very amazing and strange with me  :maybe:[/quote]
I think my solution not bad by Cauchy and Dirichlet.
Can you post your solution?[/quote]

[size=150]I am sorry，I just Put forward a problem，
I have not given a proof.[/size]

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \left(\frac {a}{2a \plus{} b}\right)^3 \plus{} \left(\frac {b}{2b \plus{} c}\right)^3 \plus{} \left(\frac {c}{2c \plus{} a}\right)^3 \ge \frac {1}{9}$
I have seen it on the box "Inequalities Unsolved Problems". It exist a very nice solution by Am-Gm and Cauchy Schwarts  :)[/quote]
see [url]http://www.aoshoo.com/bbs1/dispbbs.asp?BoardID=48&ID=17380&replyID=&skin=1[/url]

Dear my friend, I don't see any solution in that link. Can you post it here ?  :maybe:

Also I cannot find any solution at that link!!! :(

Can you please post your proof, nguoivn?

Thank you very much. :)

I really cannot solve it!!!! :blush:  :blush:

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \left(\frac {a}{2a \plus{} b}\right)^3 \plus{} \left(\frac {b}{2b \plus{} c}\right)^3 \plus{} \left(\frac {c}{2c \plus{} a}\right)^3 \ge \frac {1}{9}$
I have seen it on the box "Inequalities Unsolved Problems". It exist a very nice solution by Am-Gm and Cauchy Schwarts  :)[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-52-b-q-liu.html

Thank you very much, my dear friend. :)

Looking at your solutions makes me really happy. :)


superior algebra
superior algebra unsolved

Can we characterize all those $ q$ such that $ 1/q$ has the same $ p$-adic expansion for all $ p > 2$.

There are none except $ 1$ (I assume you want integers $ q$).
For positive $ q$, consider the $ p$-adic expansion for some $ p \leq q$ and $ p>q^2$.
For negative ones, do the same for some $ p \leq \minus{}q$ and $ p>q^2$.


Bring glory to some of your favorite youtube videos! First, I'll start off with one of my own:

[youtube]wen1YwaxRiM[/youtube]

Add a little gamer comedy

[youtube]7oXx0qwe0wc[/youtube]

And top it off with some TAS love

[youtube]elYE0VdGiME[/youtube]

i already made a topic like this... it isn't exactly a month old... D:


logarithms
number theory proposed
number theory

The [i] radical[/i] $\mathrm{rad}(n)$ of a positive integer $n$ is the productg of its prime factors. Show that $\mathrm{rad}(2^{n}-1) > C^{\frac{n}{\log{n}}}$, $n > 1$, for some constant $C > 1$ (and find an explicit constant that works).

Can you remove the logarithmic factor? That is, can you show that $\mathrm{rad}(2^{n}-1) > C^{n}$ for some $C > 1$? I can't -- even though this must be true, by the $abc$-conjecture.

But the weaker bound with $n/\log{n}$ is surprisingly simple. Prove it :)

Hmmm... simple for you...  :blush:  :maybe:   Here is an idea, although I'm probably far from the solution. Suppose that $p$ is a prime divisor of $2^{n}-1$ with exponent $k$. Let $d$ the order of $2$ mod $p$ and let $2^{d}=1+y$, with $y$ a multiple of $p$. Remember the classical fact that of $p>2$ and $p|A-B$ and not both of them, then $v_{p}(A^{n}-B^{n})=v_{p}(n)+v_{p}(A-B)$. Thus, $v_{p}(2^{n}-1)=v_{p}((1+y)^{n/d}-1^{n/d})= v_{p}(y)+v_{p}(n/d)$. Moreover, $v_{p}(n/d)=v_{p}(n)$, because $d$ being a divisor of $p-1$ is cannot be a multiple of $p$. Thus $v_{p}(y)+v_{p}(n)=k$. But also $v_{p}(2^{p-1}-1)=v_{p}( (1+y)^{(p-1)/d}-1^{(p-1)/d})=v_{p}(y)$ because $(p-1)/d$ is not a multiple pf $p$. Thus $k=v_{p}(n)+v_{p}(2^{p}-2)$. And now... nothing. Note however that for most of the primes $v_{p}(2^{p}-2)=1$ (the Wiefrich primes are extremely rare) "so for most" primes $p$ dividing $2^{n}-1$ we have $k\leq \log_{p}(n)+1$. This would result in a good exponential lower bound for the radical of $2^{n}-1$ if we had some quantitative results about those $p$ with $v_{p}(2^{p}-2)>1$. By the way, is there a uniform upper bound for this object? I suspect yes, but I have no idea. Also, note that using this you cannot solve Vess's problem by employing the trivial estimate $v_{p}(2^{p}-2)=O(p/\ln p)$. So probably all this is completely useless, but since I have spent quite a long time on this problem, I thought it is better than nothing.


algorithm
number theory
greatest common divisor

What is the smallest integer $n$ for which $\frac{n-6}{5n+17}$ is non-zero and reducible?

[hide]
We want $(n-6,5n+17) \not = 1$
By Euclid's algorithm, this is equal to $(n-6, 47) \not = 1$
Since 47 is prime, the first integer such with gcd > 1 is 47, so $n=53$
[/hide]


LaTeX
inequalities proposed
inequalities

Let a,b,c,d are non-negative real number. Prove that
$ 3(a^{4}+b^{4}+c^{4}+d^{4})+4abcd\ge (a+b+c+d)(a^{3}+b^{3}+c^{3}+d^{3})+\frac{1}{3}\sum_{cyc}ab(a-b)^{2}$
And, $ \frac{1}{3}$ is the best constant here!

[b]fjwxcsl[/b] ,why don't you use $ LATEX$?.It's not hard and if you use $ \text{\LaTeX}$ your work will be better understood and better apreciated.


Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
\[a\sqrt{1+b^{3}}+b\sqrt{1+c^{3}}+c\sqrt{1+a^{3}}\le 5.\]

You are really excellent, VASC. When I created and then solved this problem, I used mixing variable with 2 pages solution.  :blush:

Yes ! When I can only find out this solution one week after  :lol:

[quote="manlio"]Dear Sunjee if $a+b+c=3$ how can you suppose that $a+2b=3$ I don't understand this step, sorry.  :wink:[/quote]
First sorry my bad english, Second watch my proof again. :lol:

[quote="hungkhtn"]Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
\[a\sqrt{1+b^{3}}+b\sqrt{1+c^{3}}+c\sqrt{1+a^{3}}\le 5. \]
[/quote]

For $0\leq a,\ b, \ c \leq 3$, we have $a\sqrt{1+b^{3}}\leq \frac{3}{2\sqrt{2}}ab+\frac{1}{2\sqrt{2}}a$ e.t.c.

$a\sqrt{1+b^{3}}+b\sqrt{1+c^{3}}+c\sqrt{1+a^{3}}\leq \frac{3}{2\sqrt{2}}(ab+bc+ca)+\frac{1}{2\sqrt{2}}(a+b+c)$

$\leq \frac{1}{2\sqrt{2}}(a+b+c)^{2}+\frac{1}{2\sqrt{2}}(a+b+c)=3\sqrt{2}\leq 5$.

[/quote]

$a\sqrt{1+b^{3}}\leq \frac{3}{2\sqrt{2}}ab+\frac{1}{2\sqrt{2}}a$ e.t.c.

[/quote]
I don't understand it. Can you post solution.

Sorry, my solution didn't make a sense.  :blush: The graph of $y=\sqrt{1+x^{3}}\ (x\geq-1)$ has an inflection point at $x=0$. I was trapped by the convexity.

[quote="manlio"]Sorry Sunjee, but I don't think you can set $2a=3-b$  :wink:[/quote]
At least one of following inequalities is true
$1.~~f(2a,b,0)\geq f(a,b,c)$\\

$2.~~f(0,2b,c)\geq f(a,b,c)$\\

$3.~~f(a,0,2c)\geq f(a,b,c)$\\

[quote="hungkhtn"]Let $ x,y,z$ be non-negative real numbers such that $ x \plus{} y \plus{} z \equal{} 3$. Prove that
$ x\sqrt {1 \plus{} y^{3}} \plus{} y\sqrt {1 \plus{} z^{3}} \plus{} z\sqrt {1 \plus{} x^{3}}\le 5$.[/quote]
[quote="Vasc"]Use $ \sqrt {1 \plus{} y^{3}}\le 1 \plus{} \frac {y^{2}}{2}$.[/quote]
$ [2 \plus{} y^{2}]^2 \minus{} 4(1 \plus{} y^3) \equal{} y^2(2 \minus{} y)^2\geq0$.
[quote="Sunjee"]How to prove: if $ x \plus{} y \plus{} z \equal{} 3$ then $ xy^{2} \plus{} yz^{2} \plus{} zx^{2}\leq 4$.[/quote]
$ 4(x \plus{} y \plus{} z)^3 \minus{} 27(xy^2 \plus{} yz^2 \plus{} zx^2)$

$ \equal{} (z \plus{} x \minus{} 2y)^2(y \plus{} 4z \minus{} 5x) \plus{} 9x(x^2 \plus{} y^2 \plus{} z^2 \plus{} 2yz \minus{} zx \minus{} xy) \geq 0$,

which is clearly true for $ x \equal{} \min\{x,y,z\}$.

[quote="Sunjee"][quote="Vasc"]Use
$\sqrt{1+a^{3}}\le 1+\frac{a^{2}}{2}$,
$\sqrt{1+b^{3}}\le 1+\frac{b^{2}}{2}$,
$\sqrt{1+c^{3}}\le 1+\frac{c^{2}}{2}$.[/quote]
How to prove: if $a+b+c=3$ then
\[ab^{2}+bc^{2}+ca^{2}\leq 4\][/quote]

We will prove stronger inequality.

Let  $a$, $b$, $c$  be three non-negative  real numbers  such that $a+b+c=3$ . Prove that
$3\le a \sqrt{1+b^3}+b\sqrt{1+c^3} +c \sqrt{1+a^3}\le5$
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=487447&p=2731584#p2731584
Let $a,b,c \in \mathbb{R^{+}}$ such that $a+b+c=3$.Prove the inequality\[a^2b+b^2c+c^2a+abc\le 4.\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=497887


Prove: $ \tau (n)$ is uneven if and only if $ n\equal{}t^2, t\in N$

Let's write $ n$ in its prime factorization:

$ n \equal{} p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$

$ \tau(n) \equal{} (\alpha_1 \plus{} 1)(\alpha_2 \plus{} 2)\cdots(\alpha_k \plus{} 1)$  for definition.

So, $ \tau(n)$ is odd if none of the $ \alpha_i$ is odd. This happens if and only if every exponent in $ n$'s prime factorization is even. So $ n$ is a (perfect) square.

[quote="Gabuntu94"]Let's write $ n$ in its prime factorization:

$ n \equal{} p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$

$ \tau(n) \equal{} (\alpha_1 \minus{} 1)(\alpha_2 \minus{} 2)\cdots(\alpha_k \minus{} 1)$  for definition.

So, $ \tau(n)$ is odd if none of the $ \alpha_i$ is odd iff all the exponents of $ n$'s prime factorization are even. So $ n$ is a (perfect) square.[/quote]

But, $ \tau (n)\equal{}\prod_{i\equal{}1}^{k}(\alpha_i\plus{}1)$ and "only if" be forgotten or easy.
Thanks!

Yeah

sorry I'm stupid... I meant that but wrote something completely nonsense!  :blush:

Now it's correct...


Two boxes are each filled with colored socks.  Box $ A$ has 2 red socks and 4 blue socks.  Box $ B$ has 4 red socks and 4 blue socks.  A box is chosen at random, and a sock is drawn from the box.  Given that the sock chosen is blue, what is the probability that it came from box $ A$?

[hide]You will choose blue 4/6 from A and 3/6 from B, and each box is equally likely to be selected... so 4 out of 7 times you draw a blue sock, it will be from A, right?[/hide]

This seems simpler... is it bogus?

I don't think anything's wrong with that.

@jbatterson
yeah, you're right.

copied from Solution page

Problem:
Two boxes are each filled with colored socks. Box A has 2 red socks and 4 blue socks. Box B has 4 red socks and 4 blue socks. A box is chosen at random, and a sock is drawn from the box. Given that the sock chosen is blue, what is the probability that it came from box A?

Solution:
First, we can find the probability of randomly choosing a blue sock from box A. The probability of choosing box A is 1/2, and the probability of choosing a blue sock from box A is 4/6, so the probability of doing both is: $\frac{1}{2}\cdot\frac{4}{6}=\frac{1}{3}$ Next, we can calculate the probability of randomly choosing a blue sock from either box. This will be the sum of the probabilities of choosing a blue sock from box A or box B. Similar to above, this probability is: $\frac{1}{2}\cdot\frac{4}{6}+\frac{1}{2}\cdot\frac{4}{8}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$ Therefore, the probability that the blue sock came from box A is: $\frac{\text{probability of drawing blue sock from box A}}{\text{probability of drawing a blue sock}}=\frac{\frac{1}{3}}{\frac{7}{12}}=\boxed{\frac{4}{7}}$


modular arithmetic
number theory
prime numbers
number theory unsolved

Let $n\in N$, n>2 and $a_1,a_2,...,a_n$ be some natural numbers, not all equal. Show that there is a infinite number of prime numbers p, such that it exists $k\in N$, with the propertie that $p|a_1^k+a_2^k+...+a_n^k$

It seems difficult for me, because I am not good at number theory, but pbornsztein will be solve it. :)

kunny

[quote="pbornsztein"]I'm quite sure it already appeared on ML, and was solved by Harazi or Grobber.

Pierre.[/quote]

In fact for $n=3$ [b]grobber[/b] solved longtime ago, but the method is okey for any $n>2$. I think the problem is true for $n=2$ and it deu to some man call Noimann, but i'm not sure. This seems inpisred by this ;)

[quote="pluricomplex"]
In fact for $n=3$ [b]grobber[/b] solved longtime ago, but the method is okey for any $n>2$.[/quote]

I was right  :)

The case $n=2$ is here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=57113

Pierre.

Some insignificant remarks:
   A nice problem is to prove that the sequence $f(n)/n$ where $f(n)$ is the greatest prime factor of $a^n+b^n$ is unbounded if $a,b$ are not equal.
    The problem posted by Frozen is a consequence of Polya's celebrated theorem about linear sequences that have a finite number of prime divisors. Does anyone have a solution for this result? I don't think Grobber's approach would work in this case.

I hope this works:

Assume WLOG that $(a_1,\ldots,a_n)=1$ and at least one of the $a_i$ is $>1$. Let $p_1,\ldots,p_s$ be some distinct primes, which we assume are all the prime factors of numbers of the form $a_1^k+\ldots+a_n^k$.

Choose a positive integer $t$, and take $k$ to be divisible by $\prod_{j=1}^s\varphi(p_j^t)$. Those $a_i$ which are not divisible by $p_j$ (for each $j$, at least one $a_i$ is in this situation) will satisfy $a_i^k\equiv 1\pmod{p_j^t}$. This means that for all $j$ we have $a_1^k+\ldots+a_n^k\equiv m_j\pmod{p_j^t}$, with $1\le m_j\le n,\ \forall j$. If $t$ is large enough, by our assumption, this will imply that $a_1^k+\ldots+a_n^k\le n!$, and this is obviously a contradiction (taking $t$ very large will also make $k$ very large, which, in turn, will make $\sum a_i^k>n!$, since at least one of the $a_i$'s is $>1$).

[quote="grobber"]will satisfy $a_i^k-1\equiv 1\pmod{p_j^t}$[/quote]
I think $a_i^k\equiv 1\pmod{p_j^t}$, but is just a typing error, you applied better later. I think the sollution is OK.

Yes, sorry about that, and thanks for noticing. I've modified it.

[quote="grobber"] If $t$ is large enough, by our assumption, this will imply that $a_1^k+\ldots+a_n^k\le n!$[/quote]

I don't understand this part  :blush:  - would anybody be so kind and explain it to me?

i was just reading over this problem, and i don't understand the part that megus specified either.
could someone please explain it clearly?

could someone please reply and explain why $a_1^k + a_2^k + \cdots + a_n^k \leq n$? Please?
and harazi, what is polya's celebrated theorem regarding prime divisors of linear sequences?


limit
logarithms
calculus
calculus computations

I'll give a guess based on intuition.  I don't have time right now to check with L'hospital's rule.

When $ x$ is huge, the $ 1$ is insignificant.  It is as if we just had $ (5e^x)^{\frac1x} \equal{} 5^{\frac1x}(e^x)^{\frac1x} \equal{} 5^{\frac1x}e$.  When $ x$ is huge, $ 5^{\frac1x} \approx 1$.  So I'm guessing the limit is $ e$.

Observe the estimation $ e \equal{} (e^x)^{\frac{1}{x}} < (1 \plus{} 5e^x)^{\frac{1}{x}} < (6e^x)^\frac{1}{x} \equal{} 6^{\frac{1}{x}}e$ for $ x > 1$. Then squeezing theorem directly applies, yielding $ e$ as the limit.

[quote="sos440"]Observe the estimation $ e \equal{} (e^x)^{\frac {1}{x}} < (1 \plus{} 5e^x)^{\frac {1}{x}} < (6e^x)^\frac {1}{x} \equal{} 6^{\frac {1}{x}}e$ for $ x > 1$. Then squeezing theorem directly applies, yielding $ e$ as the limit.[/quote]

Thank you, this is a good solution. It was on my BC final and I had to intuitively treat the one as negligible, but that's sloppy. I'll use the squeeze theorem more often now.  :)

Alternatively employ L'Hopital's Rule since this is a $ \infty^0$ case (although, this way, it's longer)

$ L \equal{} \lim_{x\to\infty} (1 \plus{} 5e^x)^{\frac {1}{x}}$

$ \ln L \equal{} \lim_{x\to\infty} \frac {\ln(1 \plus{} 5 e^x)}{x}$

$ \ln L \equal{} ^H \lim_{x\to\infty} \frac {5 e^x}{1 \plus{} 5 e^x}$

And that last limit can be evaluated using a technique of your choice.

$ \ln L \equal{} 1$

$ L \equal{} e$


Three fixed circles pass through the points A and B. X is a variable point on the first circle different from A and B. The line AX meets the other two circles at points Y and Z (with Y between X and Z). Prove that the ratio XY/YZ is independent of the position of X.

<edit: apparently the sysstem messed up my attachement... this one had nothing to do with the problem.>

Anyone who feels like posting a solution for this?

It's not hard, and it's a problem from some recent [hide]Canadian[/hide] olympiad.

I mean, apparently nobody felt like solving this, and this was posted 1.5 years ago :?

By the way, don't worry about Peters counterexample, it doesn't make sense to me.

Apparently the attachements from back then are mixed up, this drawing has completely nothing to do with the problem.

I wouldn't care too much though. I remember we discussed about it then... Probably either the detail was added or I was wrong... these are again posts from before I registred.  :roll:

[quote="probability1.01"]I'm pretty sure the problem is wrong... just draw some extreme cases and it seems to range anywhere from near zero to near infinity.[/quote]

Can you post a drawing, probability 1.01? I think this version looks correct...

The problem is correct. See http://www.mathlinks.ro/Forum/viewtopic.php?t=14550 and http://www.mathlinks.ro/Forum/viewtopic.php?t=20174 .

  darij


Given are many equilateral triangles, each joining at the midpoint of the one before .the sides of the Big triangle are of resistance $R$. the resistance of the smaller sides inside are proportional to their length. Find the resistance between points $A$ and $B$, on the given infinite fractile.

According to symmetry of the system,we can change the figure as in file.There $R_{x}$ is the net resistance between A and B.

$R_{1}=\frac{R}{2}+\frac{R}{2}=R$

$\frac{1}{R_{2}}=\frac{1}{R_{1}}+\frac{2}{R_{x}}$

$R_{2}=\frac{R_{x}R}{R_{x}+2R}$

$R_{3}=R_{2}+\frac{R}{2}+\frac{R}{2}$

$R_{3}=2R\frac{R_{x}+R}{R_{x}+2R}$

Because the net resistance is $R_{x}$,then  we can write like this,

$\frac{1}{R_{x}}=\frac{1}{R_{3}}+\frac{1}{R}$

We get the following equation:

$3R_{x}^{2}+2R_{x}R-2R^{2}=0 R_{x}=R\frac{\sqrt{7}-1}{3}$

I just do not understand how you simplify the circuit and state that the resistance of the triangles inside is R/2, Could you explain, please?


We have two lines, $d_1$ and $d_2$, and a fixed point $P$. Draw two distinct, variable lines $f_1,f_2$ through $P$. Let $\displaystyle \left\{ A^\prime \right\} = f_1 \cap d_1$, $\displaystyle \left\{ B^\prime \right\} = f_1 \cap d_2$ and $\displaystyle \left\{ A^{\prime \prime} \right\} = f_2 \cap d_1$, $\displaystyle \left\{ B^{\prime \prime} \right\} = f_2 \cap d_2$.

Find the locus of $\displaystyle \{ Q \} = A^\prime B^{\prime \prime} \cap A^{\prime \prime} B^\prime$.


geometry
3D geometry
frustum
calculus
integration
pyramid
ratio

Suppose that a and b each represents the edge lengths of the bottom and top squares, respectively, and h is height. Then how would you derive a formula for volume of that frustum by integrals?

To make the question easier, I'm going to give you the answer for the formula which is 1/3h(a^2+ab+b^2), however, what troubles me is that I don't understand how one can possibly derive the formula by using the values, namely, a, b, and h.

Thank you for anyone who would like to help me figure out the process of derivation to the formula.

The answer to this has been known since [url=http://en.wikipedia.org/wiki/Moscow_Mathematical_Papyrus]long before there was calculus[/url], although we have no clear idea how the ancient Egyptians came up with the factor of $ \frac13.$

To use calculus: slice through the pyramid at height $ y$ above the base. The cross-section will be a square whose side length $ s$ varies linearly from $ s\equal{}a$ when $ y\equal{}0$ to $ s\equal{}b$ when $ y\equal{}h.$ Our main task will be finding this linear formula - but it is linear. We should get that $ s\equal{}a\plus{}(b\minus{}a)\left(\frac yh\right).$ Square that to get the cross-sectional area and integrate that cross-sectional area to get the volume.

$ V\equal{}\int_0^h\left(a\plus{}(b\minus{}a)\left(\frac yh\right)\right)^2\,dy$

$ \equal{}\left.\frac13\cdot\frac h{b\minus{}a}\cdot\left(a\plus{}(b\minus{}a)\left(\frac yh\right)\right)^3\right|_0^h$

$ \equal{}\frac13\cdot\frac h{b\minus{}a}\cdot (b^3\minus{}a^3)\equal{}\frac h3\cdot\frac{b^3\minus{}a^3}{b\minus{}a}.$

The formula you quoted is an algebraic simplification of this.

[quote="Kent Merryfield"] The cross-section will be a square whose side length $ s$ varies linearly from $ s \equal{} a$ when $ y \equal{} 0$ to $ s \equal{} b$ when $ y \equal{} h.$ Our main task will be finding this linear formula - but it is linear.

how do you know that it is linear ???

You can see it, through triangle similarity, "cutting" the pyramid with the plane defined by the pyramid axis an one of the mentioned cross-section diagonal; the side $ s$ is proportional to the diagonal.

(a^3 – b^3) * tan- alpha / 6

a – width lower base

b – width upper base

alpha- angle between a lateral face and the base

* a regular pyramid with a square base

Note that this formula can be derived intuitively from a sequence of flower petals  (see Flower Power Graphic at dimleed.wordpress.com)

The sequence is 1...7...19...37... etc.,

The antiderivative of x^2 is,

         x^3 / 3    Applying into it the 2 values of x,
                                                          a = 4 and b = 2 and substracting,
                                                                                (4^3 - 2^3)/3 = 56/3

Now, for any pyramid the ratio of total height (H) to the length of it´s Base (a) yields the height of a small slice that corresponds to a change of (x) in 1 unit (also the height of the pyramidion of base 1).  Multiplying this height by the result of the previous integration,

56/3 x 12/4= 56 :

You found it. Problem 14 in the Moscow papyrus for a frustum of a pyramid solved by a sort of integration.

In a sequence that I mentioned before  1...7...19...37...etc.;        37 + 19 = 56 the frustum,

1 + 7 + 19 + 37 = 64  is the total volume of the pyramid in that same problem.

What did the Egyptians know, and when did they know it?


Find a recursion for the number of partitions of an n-set

[quote="eryaman"]Is partition = subset?[/quote]

Partition is breaking up a set into disjoint sets with union the whole set.

For example, $ \{1,2\}, \{3,4,5\}$ is partition for $ \{1,2,3,4,5\}$.

hmm i think i remember seeing this from Engel before.

[hide="Solution"]
Let $ P_{n}$ be the number of partitions of a $ n$-set. Consider another term $ n+1$. Place $ n+1$ in a partition of $ k$ terms from the $ n$-set, where $ k\in [0,n]$. These terms can be chosen in $ \binom{n}{k}$ ways. The remaining terms can be partitioned into $ P_{n-k}$ ways. So we have that
\[ P_{n+1}=\sum_{k=0}^{n}\binom{n}{k}P_{n-k}\]
[/hide]


This was a cool problem.

Gil is $17$ years older than Shelia. If his age were written after hers, the result would be a $4$-digit perfect square. The same statement can be made $13$ years from now. Find Shelia's present age.

Gil's  [hide="age:"]36[/hide] Shelia's [hide="age:"]19[/hide]

[hide]Let $s$ be Sheila's age. From the second sentence, $10\leq s\leq 99$ and $100s+s+17=101s+17$ is a perfect square ($y^2$). From the third, $100(s+13)+s+13+17=101s+1330$ is also a perfect square ($x^2$). Subtracting, $x^2-y^2=1313$. Factoring each, $(x+y)(x-y)=101*13$. Either $x+y=101*13$, $x-y=1$ or $x+y=101$, $x-y=13$. The first case will make $s$ too large. For the second case, subtract and divide by 2 to get $y=44$. Then $s=\frac{44^2-17}{101}=19$.[/hide]


A cheaTh game has the same rules as a chess game except each player makes 2 consecutive moves at a time.
Ex. W (to make move)-W-B-B-W-W...
Can anyone think up a non-losing strategy for WHITE?

(I have no idea about a simple proof. Maybe this proof can't be filled in one page)

[quote="rem"]What happens if white checks on the first of the two moves?[/quote]

Then white wins
In this case  1 2 3 4 5 6
                  ww b b ww
if white checks on the fifth move and black can't avoid a check on 5th by any 3rd and 4th moves, 1,2 by white is equivalent to the old checkmate.
This is my guess

See this thread:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=239228[/url]


inequalities
trigonometry
geometry
rectangle
analytic geometry
geometry proposed

I couldn't solve this problem and the only solution I was able to find was very unnatural (it was an official solution, I think) and I couldn't be satisfied with it, so I ask you if you can find some different solutions. The problem is really great one!

If $M$ is the centroid of a triangle $ABC$, prove that the following inequality holds: \[\sin\angle CAM+\sin\angle CBM\leq\frac{2}{\sqrt3}.\] The equality occurs in a very strange case, I don't remember it.

It was indeed hard problem. But my solution is not unnatural for me. Here is my idea:
I fix [tex]\angle{CBM}[/tex] and find out when does [tex]\sin{\angle{CAM}}[/tex] approach its maximum.
Let's define: [tex]\angle{CBM}=\varphi[/tex] and [tex]\angle{CAM}=\omega[/tex].
First let's discuss the case when [tex]\varphi<\frac{\pi}{2}[/tex]. We can also take [tex]BC=2[/tex] whithout loss of generality. Denote by [tex]K[/tex] the midpoint of [tex]BC[/tex]. Let [tex]O[/tex] be a point lying on the ray [tex]CB[/tex] so that [tex]OB=BC[/tex]. Let [tex]l[/tex] be a line passing through [tex]O[/tex] and parallel to [tex]BM[/tex]. All triangles that have [tex]\angle{CBM}=\varphi[/tex] are formed by taking any point A on the line [tex]l[/tex] on the same side as [tex]M[/tex] is. Now we must determine the point [tex]A[/tex] on line [tex]l[/tex] such that [tex]\sin{\omega}[/tex] is maximum. Now we take Decartese rectangle coordinate system (sorry if it is not the right word) with axis [tex]OX [/tex] lying on line [tex]l[/tex] such that points [tex]B,C[/tex] have positive coordinates. Axis [tex]OY[/tex] is perpendicular to [tex]OX[/tex] with positive direction such that points [tex]B,C[/tex] have positive coordinates. In this system we have [tex]B=(2,2\tan{\varphi}),K=(3,3\tan{\varphi})[/tex][tex],C=(4,4\tan{\varphi})[/tex]. The piont [tex]A[/tex] which is moving by the positive side of [tex]OX[/tex] has such coordinates [tex]A=(t,0)[/tex] where[tex] t>0[/tex]. Let's calculate [tex]\vec{AK}\cdot\vec{AC}=(3-t)(4-t)+12{\tan{\varphi}}^{2}=t^{2}-7t+\frac{12}{{\cos{\varphi}}^{2}}[/tex].
If [tex]\sin{\varphi}\leq\frac{1}{7}[/tex] then there is such positive value of t when [tex]\vec{AK}\cdot\vec{AC}=0[/tex] i.e there is such point A such that [tex]\angle{CAM}=\frac{\pi}{2}[/tex], thus the maximum of [tex]\sin{\omega}[/tex] is [tex]1[/tex] so [tex]{\sin{\varphi}+\sin{\omega}}\leq{1+\frac{1}{7}}=\frac{8}{7}<\frac{2}{\sqrt[2]{3}}[/tex].
If [tex]\sin{\varphi}>\frac{1}{7}[/tex] then [tex]\vec{AK}\cdot\vec{AC}>0[/tex] i.e. [tex]\angle{CAM}<\frac{\pi}{2}[/tex] so [tex]\sin{\omega}[/tex] approaches its maximum when [tex]\tan{\omega}[/tex] is maximum.
[tex]\tan{\omega}=\tan(\angle{MAX}-\angle{CAX})=\frac{t\tan{\varphi}}{t^2-7t+\frac{12}{{\cos{\varphi}}^2}}[/tex]
so [tex]\tan{\omega}[/tex] is maximum when [tex]t=\frac{12}{t{\cos{\varphi}}^{2}}[/tex] and [tex]\tan{{\omega}_{max}}=\frac{\sin{\varphi}}{4\sqrt{3}-7\cos{\varphi}}[/tex] and [tex]\sin{{\omega}_{max}}=\frac{\sin{\varphi}}{7-4\sqrt{3}\cos{\vaphi}}[/tex]. So it we need to prove that
                             [tex]\sin{\varphi}+{\sin{\varphi}}{7-4\sqrt{3}\cos{\varphi}}\leq\frac{2}{\sqrt{3}}[/tex].
The last one is equvalent to [tex]{(\sin{\varphi}+\cos{\varphi}-\frac{2}{\sqrt{3}})}^{2}\geq{0}[/tex] which is obvious and equality holds if only if  [tex]\sin{\varphi}+\cos{\varphi}=\frac{2}{\sqrt{3}}[/tex]. We are on.  :) . Cases  [tex]{\frac{\pi}{2}}\leq{\varphi}<{\pi}[/tex] are done in the same way.

Oh, thank you, Levon Nurbekian, your solution is excellent
(and it is also absolutely natural:)),
it was such a pleasure for me to read it.
In official solution they first consider the case when AB touches the circle AMC (in
that case it is much easier to prove the inequality), and after that they go on to prove
the general case using already proved case. I couldn't find any sensible explanation
for such a structure of the proof.

[color=darkred]Centroid $G$ of $\triangle ABC\ ,$ area $S=[ABC]\ ,\ \phi =m(\widehat {AGB})\ .$ Prove that
$\sin\widehat {CAG}+\sin\widehat {CBG}\leq\frac{2}{\sqrt3}\Longleftrightarrow \boxed{\frac{a}{m_a}+\frac{b}{m_b}\le \frac{2ab}{S\sqrt 3}}\Longleftrightarrow \frac{m_a}{a}+\frac{m_b}{b}\le \frac{\sqrt 3}{\sin \phi}\ .$
An angular inequality is equivalently with a linear (more accessible) inequality ![/color]

[u]See[/u] http://www.mathlinks.ro/Forum/viewtopic.php?t=83790


If $ x^{x^{x^{: }}}\equal{}4$ then $ x\equal{}?$ (there are infinitly x)

Hm. I might be wrong about this, but isn't the range of convergent tetration limits $ \left[\frac{1}{e},e\right]$, as proven here: http://www.mathlinks.ro/viewtopic.php?t=77942

so you couldn't possibly reach $ 4$...

$ x^{x^{x^{\ldots}}}\equal{}a\equal{}4 \\ 1\equal{}\log_4 4 \equal{} \log_4 a \equal{} a \cdot \log_4 x \equal{} 4 \log_4 x \\ \log_4 x \equal{} \frac{1}{4} \\ x\equal{}\sqrt{2}$

But when $ x\equal{}\sqrt{2}$ we have $ a>4$, so such x doesn't exist.

[quote="mayhem"]If $ x^{x^{x^{: }}} \equal{} 4$ then $ x \equal{} ?$ (there are infinitly x)[/quote]If your tower means the hopefully existing limit $ f(x)$ of the sequence with $ x_0\equal{}x$ and $ x_{n\plus{}1}\equal{}x^{x_n}$ then it holds $ f(x)\equal{}x^{f(x)}$ implying $ 4\equal{}x^4$. So there are only finitely many candidates to solve the equation and just a single one which is positive real ($ x\equal{}\sqrt{2}$).


algebra
polynomial
geometric series
arithmetic sequence
arithmetic series

Hello,

I am having some trouble with the following problems involving sequences and series and I'd appreciate some help. My work is shown below as well.

Thank you!

---

1. In a geometric series, $ t_5 + t_7 = 1500$ and $ t_11 + t_13 = 187,500$. Find all possible values for the first three terms.

2. If the interior angles of a pentagon form an arithmetic sequence and one interior angle is $ 90^{\circ}$, find all the possible values of the largest angle in the pentagon.

3. In a sequence, every term after the second term is twice the sum of the two preceding terms. The seventh term of the sequence is 8 and the ninth term is 24. Find the eleventh term of the sequence.

[u] My Work : [/u]

1.  $ t_5 + t_7 = 1500$
$ t_{11} + t_{13} = 187,500$

Let $ t_1$ be x:

$ xr^4 + xr^6 = 1500$
$ xr^{10} + xr^{12} = 187,500$

$ xr^4(r^2 + 1) = 1500$  ---- (1)
$ xr^{10}(r^2 + 1) = 187,500$ - (2)

Dividing (2) by (1): $ r^6 = 125$

$ r = \pm 2.2361$

Therefore:

$ t_1 = 2.2361$  or $ - 2.2361$.

Before I continued, I looked at the answer and what I got above was wrong.

2. Sum of the interior angles of a polygon = $ 540^{\circ}$

Formula for Arithmetic Series:

$ S_n = \frac {n(a_1 + a_n)}{2}$

Therefore:

$ 540 = \frac {5(a_1 + a_n)}{2}$

$ 1080 = 5(a_1 + a_n)$

$ a_1 + a_n = 216$

Since one of the angles is 90°,

$ 90 + a_n = 216$

One of the angles would be 126°.

When I looked at the answer, it says that [hide]144[/hide] is a solution as well. How would I calculate that?

3. The sequence in terms of $ x$s and $ y$s:

x, y, 2x + 2y, 4x + 6y, ...

Instead of writing out these polynomials up to the seventh and ninth terms, how would I solve this problem? I am not seeing the method.

[hide="3"]
We know that the seventh term plus the eighth term equals the ninth term. If we call the 8th term q (for no apparent reason), we have:
$ 8\plus{}q\equal{}24$ so
$ q\equal{}16$
Our series starts at 8, 16, 24, 40, and $ \boxed{64}$  [/hide]

[hide="1"]Hmm...  Let's see.  I agree that $ r^6=125$, so $ r=\pm\sqrt{5}$.  However, this is NOT the first term.  This is the common ratio.  We still need to solve for x.

Looking at equation 1,

$ xr^4(1+r^2)=1500
x(5)(5+1)=1500
x=10$

Thus the first three terms can either be  $ 10, 10\sqrt{5}, 50$ or $ 10, -10\sqrt{5}, 50$[/hide]

[hide="Correction for #3"]The ninth term is [b]twice[/b] the sum of the two preceding terms.

Let's try this again, still calling the 8th term $ q$.

$ 24\equal{}2(8\plus{}q)$
$ 24\equal{}16\plus{}2q$
$ 8\equal{}2q$
$ q\equal{}4$

So the 7th, 8th, and 9th terms are 8, 4, and 24, respectively.  The 10th term is 56 ($ 2(4\plus{}24)$), and the 11th term is [b]160[/b],  ($ 2(24\plus{}56)$).[/hide]

#2

540= (x+0r)+(x+r)+(x+2r)+(x+3r)+(x+4r)
540=5x+10r
540=5(x+2r)
108=x+2r

so we know that the middle term is 108
case 1
the first angle is 90 (alos can be the last one)
x=90 (or x+4r=90)
x+2r=108
r=9 or -9

so the largest angle will be 126

case 2 the second (or the fourth angle is 90)
x+r=90 or (x+3r=90)
x+2r=108
r= 18 or -18
lagrest angle = 144

Thank you very much to all those who replied! I appreciate the help!


Is there anyway to place text inside an align environment without it appearing bold and non-text like?  Also, when I try to place a long line of text using \intertext, it doesn't page break and also when I separate out the math codes, those seem not to display properly.  The short paragraph below the equations is what I'd like to place  where I've noted.  Also looks like the triple sum is too long for the display but I can work on that.

Any ideas?
Thanks,
Shaw

Let:
\begin{align*} \frac {z e^{\pi i/z}}{(1 - z)^2(z \pi i + 2)} & = \frac {1}{\pi i}\left(\frac {1}{(1 - z)^2}\right)\left(\frac {1}{z + \frac {2}{\pi i}}\right) z e^{\pi i/z} \\
& = \frac {1}{\pi i}\left[\sum _{n = 0}^{\infty } (n + 1) z^n \sum _{n = 0}^{\infty } \frac {\pi ^{n + 1} z^n ( - 1)^{n + 2} i^{n + 1}}{2^{n + 1}} \sum _{n = 0}^{\infty } \frac {(\pi i)^n}{n!} z^{1 - n}\right] \\
\intertext{\text{Combining the first two sums via Cauchy's Product: }} \\
\frac {1}{(1 - z)^2} \frac {1}{\left(z + \frac {2}{\pi i}\right)} & = \sum _{n = 0}^{\infty } \sum _{j = 0}^n \frac {(j + 1)z^j\pi ^{n - j + 1} z^{n - j} ( - 1)^{n - j + 2}(i)^{n - j + 1}}{2^{n - j + 1}} \\
& = \sum _{n = 0}^{\infty }\left( \sum _{j = 0}^n \frac {(j + 1)z^n\pi ^{n - j + 1}( - 1)^{k - j}(i)^{n - j + 1}}{2^{n - j + 1}}\right) \\
\intertext{\text{Combining these two with the third sum: }} \\
\frac {1}{\pi i}\left(\frac {1}{(1 - z)^2}\right)\left(\frac {1}{z + \frac {2}{\pi i}}\right) z e^{\pi i/z} & = \frac {1}{\pi i}\left(\sum _{n = 0}^{\infty } \sum _{k = 0}^n \sum _{j = 0}^k \frac {(j + 1) z^k \pi ^{k - j + 1} ( - 1)^{k - j}(i)^{k - j + 1}}{2^{k - j + 1}} \frac {(\pi i)^{n - k}}{(n - k)!} z^{1 - (n - k)}\right) \\
\intertext{\text{ - - - - add paragraph below here - - - - }} \\
\mathop\text{Res}\limits_{z = 0}\bigg[\frac {ze^{\pi i/z}}{(1 - z)^2(2 + \pi iz)}\bigg] & = \frac {1}{\pi i}\sum _{k = 0}^{\infty } \sum _{j = 0}^k \frac {(j + 1) \pi ^{k - j + 1}(i)^{k - j + 1}( - 1)^{k - j} (\pi i)^{k + 2}}{2^{k - j + 1}(2 + k)!} \end{align*}
Paragraph to enter above where indicated:
We only want the $ \frac {1}{z}$ terms and that happens when the power on $ z$ is $ - 1$ or $ k + 1 - n + k = - 1$ or $ n = 2k + 2$.  Substituting that into the triple sum, we can then remove the outer sum and are left with:

I think you're trying to do too much.  If you are typesetting in $ \text{\LaTeX}$ which I assume you are, then just type the text in regular mode.  Don't bother with math mode.  Otherwise, why are you using the \intertext command?  Why not just finish the line of math, enter a new line, and then use \mathrm or \text or \mbox?

Hey JRav.  In order to place regular text between the equations I have to leave math mode and  put the equations into separate \align environments right?.  But then the equal signs won't line up between multiple aligns.  I don't think this is overly complex for Latex; it's just a few equations with regular "paragraphs" between them with the equations lined up with the equal signs.  Perhaps though I should have used a simple example and dispensed with all those sums which clutter up the central problem, however I do wish to post that set here in AoPS for a problem I'm working on so I posted it as is.

Alright, maybe not mess with it and just post it as such.  Still though I think aligning everything with the equal signs is nicer.

Let:
\begin{align*}
\frac{z e^{\pi  i/z}}{(1-z)^2(z \pi  i+2)}&=\frac{1}{\pi  i}\left(\frac{1}{(1-z)^2}\right)\left(\frac{1}{z+\frac{2}{\pi  i}}\right) z e^{\pi  i/z} \\
&=\frac{1}{\pi i}\left[\sum _{n=0}^{\infty } (n+1) z^n\sum _{n=0}^{\infty } \frac{\pi ^{n+1} z^n (-1)^{n+2} i^{n+1}}{2^{n+1}}\sum _{n=0}^{\infty } \frac{(\pi  i)^n}{n!} z^{1-n}\right]
\end{align*}
Combining the first two sums via Cauchy's Product:
\begin{align*}
\frac{1}{(1-z)^2} \frac{1}{\left(z+\frac{2}{\pi  i}\right)}&=\sum _{n=0}^{\infty } \sum _{j=0}^n \frac{(j+1)z^j\pi ^{n-j+1} z^{n-j} (-1)^{n-j+2}(i)^{n-j+1}}{2^{n-j+1}} \\
&=\sum _{n=0}^{\infty }\left( \sum _{j=0}^n \frac{(j+1)z^n\pi ^{n-j+1}(-1)^{k-j}(i)^{n-j+1}}{2^{n-j+1}}\right)
\end{align*}
Combining these two with the third sum:
\begin{align*}
\frac{1}{\pi  i}\left(\frac{1}{(1-z)^2}\right)\left(\frac{1}{z+\frac{2}{\pi  i}}\right) z e^{\pi  i/z}&=\frac{1}{\pi i}\left(\sum _{n=0}^{\infty } \sum _{k=0}^n \sum _{j=0}^k \frac{(j+1) z^k \pi ^{k-j+1} (-1)^{k-j}(i)^{k-j+1}}{2^{k-j+1}} \frac{(\pi  i)^{n-k}}{(n-k)!} z^{1-(n-k)}\right)
\end{align*}
We only want the $ \frac{1}{z}$ terms and that happens when the power on $ z$ is $ -1$ or $ k+1-n+k=-1$ or $ n=2k+2$.  Substituting that into the triple sum, we can then remove the outer sum and are left with:
\[ \mathop\text{Res}\limits_{z=0}\bigg[\frac{ze^{\pi i/z}}{(1-z)^2(2+\pi iz)}\bigg]\&=\frac{1}{\pi  i}\sum _{k=0}^{\infty } \sum _{j=0}^k \frac{(j+1) \pi ^{k-j+1}(i)^{k-j+1}(-1)^{k-j} (\pi  i)^{k+2}}{2^{k-j+1}(2+k)!}\]

[quote="stevem"][quote="shawtoos"]Is there anyway to place text inside an align environment without it appearing bold and non-text like?  Also, when I try to place a long line of text using \intertext, it doesn't page break and also when I separate out the math codes, those seem not to display properly.  The short paragraph below the equations is what I'd like to place  where I've noted.  Also looks like the triple sum is too long for the display but I can work on that.[/quote]I don't understand this; \intertext uses the normal text font so won't be in bold or 'non-text like' whatever that means.
It's not clear where you want to use the code as the forum doesn't have page breaks. In a document LaTeX won't break the page with \intertext as it's unnecessary. If the text won't fit on a page then you can come out of maths mode since there is no point in aligning equations on two different pages.[/quote]

Hi Steve.  Saw your post after posting below.  I guess I meant a line-break.  Looks like \intertext does not line break with long sentences.  Also, the  first post I made has the \intertext in bold without having it specified as such in the code.  I don't wish to be a pain.  Just thought it would look nicer if all the equations are lined up with the equal signs.

[quote="shawtoos"]Hi Steve.  Saw your post after posting below.  I guess I meant a line-break.  Looks like \intertext does not line break with long sentences.  Also, the  first post I made has the \intertext in bold without having it specified as such in the code.  I don't wish to be a pain.  Just thought it would look nicer if all the equations are lined up with the equal signs.[/quote]\intertext does do line breaks:

\begin{align*}
x&=1\\
\intertext{this is intertext line 1 \\ this is line 2}
y&=2
\end{align*}

I can't see the bold text you mention using IE or Firefox.

Hey Steve.  I have Firefox 2.0.0.14 and it looks bold to me.  It's not in the same font weight as the other texts right?  I appreciate your help Steve and wish not to be a pain.  I posted it in the forum where the problem was and . . . well . . . I'm not satisfied with it.  It's worst than the last one I posted above since I had to include a line-break with that triple sum and it's not indenting, not to mention the equal signs aren't lining up . . . math is beautiful and should be illustrated as such  I think :)

I'm making progress.  This looks nicer using Fedja's suggestion in another post with \hspace{-50pt}:
\begin{align*} \frac {1}{\pi i}\left(\frac {1}{(1 - z)^2}\right)\left(\frac {1}{z + \frac {2}{\pi i}}\right) z e^{\pi i/z} = \\
& \hspace{ - 50pt} \frac {1}{\pi i}\left(\sum _{n = 0}^{\infty } \sum _{k = 0}^n \sum _{j = 0}^k \frac {(j + 1) z^k \pi ^{k - j + 1} ( - 1)^{k - j}(i)^{k - j + 1}}{2^{k - j + 1}} \frac {(\pi i)^{n - k}}{(n - k)!} z^{1 - (n - k)}\right) \end{align*}

Actually I think the whole post is fine now.  Really, the fact I need to indent via \hspace{-50pt} makes the point of lining up the equal signs moot now anyway.

Thanks guys for your comments :)


geometry
3D geometry
sphere
geometry unsolved

I've been presented with two similar questions which i just can't find an answer for.  The question is if you have a triangle on a sphere that has 2 right angles....What is the smallest interior angle sum and the largest interior angle sum possible?  I've thought and thought and just can't seem to put it together.  Anyone have any idea?

[quote="joe trinkley"]The question is if you have a triangle on a sphere that has 2 right angles....What is the smallest interior angle sum and the largest interior angle sum possible?[/quote]

Hmm, maybe I am misunderstanding something, but the problem seems to be trivial: Our spherical triangle has two 90° angles and a third angle, which we denote by $\gamma$. Then, the sum of the angles of our triangle is $90^{\circ}+90^{\circ}+\gamma=180^{\circ}+\gamma$. Now, the angle $\gamma$ lies between 0° and 180°, and in fact it can take any arbitrary value between 0° and 180° (in fact, imagine a triangle formed by two meridians and the equator of the globe; the two angles between a meridian and the equator are both 90°, while the angle between the two meridians can be anything between 0° and 180°). Thus, the sum of the angles of our triangle can take any value between 180° + 0° = 180° and 180° + 180° = 360°.

  Darij

I'd say that the third angle can actually be arbitrarily close to $2\pi$ so $\pi<\Sigma<3\pi$, with equality possible if degenerate cases are allowed.


function
inequalities unsolved
inequalities

If $f(x) = (\frac{x}{6} )^ \frac{5}{x}$, define a sequence $<f(r)>$ for every natural number $r$. Find the maximum value of $f(r)$ in the sequence and for what value of r this happens.

[mod. edit: normal size instead of 18 is readable enough ;)]

Seems not be an open problem ("Open Problems" are those for that no solutions is known to the world; you probably meant this as Unsolved Problem, those for that you don't know a solution but there should already be one).

Well for the problem: derivate the function.
Then you will see that there is some (real) number $s$ such that it's strictly increasing in $]0,s[$ and strictly decreasing in $]s,\infty[$.
Thus you have only to check the next integers to $s$, the smaller and the bigger one.


floor function
real analysis
real analysis unsolved

Let $ x_{n} = \{ {\sum_{k = 1}^{n}\frac {1}{\sqrt k} \}}$.Prove that $ x_{n}$ is dense on $ [0,1]$

RMO shortlist 2004 : Daniel Petriceanu !

$ x_n$ is an increasing sequence with $ x_1$ = 1. I cannot see how it can be dense in $ [0,1]$.

I believe

$ \left\{ x \right\}\equal{}x\minus{}\left\lfloor x \right\rfloor$

so, it's the fraction part!

take a sequence $ u_n$ such that $ u_n \to 0$ and $ \sum_n u_n \equal{} \infty$ and you will get the same conclusion.


Two circles of different radii $R$ and $r$ touch externally. The three common tangents form a triangle. Find the area of the triangle in terms of $R$ and $r$.

Let $r<R$.  Then the circle with radius $r$ is the inscribed circle.  Let that circle have center $A$ and the other one have center $B$.  We know that we have an isosceles triangle.  Let the vertex be $C$.  Then we can calculate for the height, $r+BC$ by the following...
Let's say that we inscribed another circle in this triangle, with radius $\frac{r^2}{R}$.  Then another with smaller radius.  So on.  Then we would have the height equal to
$r+AC=S=2(r+\frac{r^2}{R}+\frac{r^3}{R^2}+...)$
Then
$S\frac{R}{r}=S+2R$
$S=\frac{2R}{\frac{R}{r}-1}=\frac{2Rr}{R-r}$
Now we have the height of the triangle.  Now,
Draw the perpendicular $AE$ to $BF$, where $F$ is the point of tangency between the larger circle and one of the two common tangents (apart from the one which is tangent to both at one point; let that point be $D$). We know that $BE=R-r$, and $AB=R+r$.  Thus, by the pythagorean theorem, we have $AE=2\sqrt{Rr}$.  Let $G$ be the base vertex of our triangle which is on the same side of the triangle as $E$.  We know, by the fact that two tangents to a circle from one point have the same length, that $GF=GD=GH$, where H is the point of tangency for the smaller circle and the line (same side as $G$).  We also know that $HF=AE$.  Thus, $GD=\sqrt{Rr}$.  Now multiplying this by $S$, we get
$\frac{2Rr\sqrt{Rr}}{R-r}$
which is the area of the triangle.

this was one of the  SIMUW 2005, problems


what is the sum of the two perfect numbers between 2 and 30?

[hide]A perfect number is a number whose factors other than itself add up to exactly that number.

Six is the first perfect number, as $ 6=2\cdot 3$ and $ 1+2+3=6$.

28 is the other, since $ 1+2+4+7+14=28$.

Thus the sum of the two perfect numbers less than 30 is $ 6+28=\boxed{34}$.[/hide]

If you were cool enough to know that formula then you'd have the first two perfect numbers memorized already. :)

[quote="Math Geek"]A Mersenne prime number is any number of the form 2^2^n, where, n is an integer.[/quote]
Not  exactly. A Mersenne prime is a prime of the form $ 2^{n}-1$ where $ n$ is a positive integer.

[quote="archimedes1"]MathGeek, how can 2^anything be prime?[/quote]
$ 2^{1}$ is prime. $ 2^\log_{2}{2}$ is prime and so on.

[quote="archimedes1"]MathGeek, how can 2^anything be prime?[/quote]
I think he meant 2^32-1 wasn't prime

[quote="13375P34K43V312"]or you can be really cool and use the fact that $ 2^{n-1}(2^{n}-1)$ is a perfect number whenever $ 2^{n}-1$ is a mersenne prime

then just be like n=2,3 both make 2^n-1 prime so plug those in and those are both under 30[/quote]

i saw that on the wikipedia page for some random number, i was just typing in numbers and it came up.

[quote="Math Geek"]A Mersenne prime number is any number of the form 2^2^n, where, n is an integer.  It was originally conjectured by this guy named, well, Mersenne, that all Mersenne numbers are prime, but later, I think it was either Fermat or Euler who proved that wrong without a huge computer that 2^32 wasn't prime. :D[/quote]

that's a fermat prime
and since fermat was pretty cheap and made a bunch of random conjectures without proof, you can pretty much assume it was euler, because fermat didn't prove anything significant, other than Fermat's Little Theorem.

[quote="13375P34K43V312"]
that's a fermat prime[/quote]
No. A Fermat prime is a prime of the form $ 2^{2^{n}}+1$, not $ 2^{2^{n}}$.

[quote="Math Geek"]Same difference.  2^32-1 isn't prime, and puuhikki, you're wrong.  A Mersenne Prime is of the form 2^2^n-1.[/quote]
I'm not wrong in this case. Mersenne primes are primes of the form $ 2^{n}-1$, $ n\in\mathbb{Z}$. It can be proven that if $ 2^{n}-1$ is prime, then so is $ n$. Therefore there are not so many primes of the form $ 2^{2^{n}}-1$. There is a mistake in your book.

There's an awful lot of confusion in this thread, so to reiterate:

A [b]Mersenne prime[/b] is a prime of the form $ 2^{p}-1$, where $ p$ is prime.  For example, $ p = 2$ gives the Mersenne prime $ 3$, and $ p = 3$ gives the Mersenne prime $ 7$.  Not all prime $ p$ produce a Mersenne prime; $ 2^{11}-1 = 23 \cdot 89$ is the first counterexample.  But for any $ p$ such that $ 2^{p}-1$ is a Mersenne prime, the quantity

$ 2^{p-1}(2^{p}-1)$

Is a perfect number.  If you're familiar with the common method of finding the sum of the divisors of a number, this is not too hard to prove.  What is harder to prove (and also true) is that every even perfect number is of this form!

A [b]Fermat prime[/b] is a prime of the form $ 2^{2^{n}}+1$ (note the change in sign).  The only known Fermat primes are for the cases $ n = 0, 1, 2, 3, 4$.

[quote="Math Geek"]Same difference.  2^32-1 isn't prime, and puuhikki, you're wrong.  A Mersenne Prime is of the form 2^2^n-1.  If you don't believe me, check in the AOPS 1 Book.  It's one of those extras at the end of a chapter. :harhar:

Also, youdonotknowme, that has nothing to do with the topic, but if you insist, it just happened to be July 4th when I added an avatar to my profile :huh: .[/quote]

The AoPS book actually misprinted this.

How do you find the sum of a number's divisors? :blush:   If this seems like an extremely easy answer, try not to make me feel bad.  I don't memorize this kind of stuff.

I would just write them out... im not sure there's a formula. is there?

Guys, please use the search feature on top.  The formula has been posted in a fairly recent topic.


As tomorrow is the official date for the 2004 AMC8, I'd like to wish all the participants good luck.  May your brains be quick and your calculators not broken.

Additionally, please do not discuss any aspect of the test, including problems, difficulty level, etc., until Friday, November 26.  (That's not this Friday, but the one after that.)  This abnormally long time is because schools may take the AMC8 at any time between Tuesday, November 16th and Tuesday, November 23rd, as described [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2004-5pub/04TM8/html/04TM8-04Text.html#07-Policy]here[/url].

Any discussions mentioning the 2004 AMC8 prior to the 26th will be deleted mercilessly.  I know you're anxious to find out how you did, but discussing the test here could jeopardize the results.

Thank you.

(Props to cobra for pointing this out.  I had forgotten that the AMC8 was tomorrow and was unaware that the test could be taken at any time over the next week.)

It is tommorow! I've got to get a perfect score! :shock:

Hmph, I never really liked the AMC 8. For some reason, I would always get some silly error on problem 5 or something. I ended up getting the same score 6th through 8th grade.

Oh well, I don't want to be a parade rainer. Good luck everybody!

Thanks to mathfanatic for monitoring the conversations until after the AMC 8 contest period, November 16 through November 23, 2004.

Thanks to cobra for reminding mathfanatic.

And best of luck and happy problem solving to all the AMC 8 contestants.
I hope you enjoy the AMC 8.

Steve Dunbar
AMC Director

Let me add that we will be hosting an AMC 8 Math Jam on Tuesday, November 30th, at 7:30 ET / 4:30 PT, where we will be discussing the problems on this year's AMC 8.

This Math Jam is free and everyone is invited to attend!

The AMC's web site says that it should be no later than February 4th.  In practice it is probably somewhat sooner than that, like late December or early January.

Here is an approximate timeline for the scores.

Most schools gave the AMC 8 yesterday, Tuesday, November 16, but some will give the AMC 8 as late as November 23.  Since schools postal mail the contests to us, we will be receiving contest packages to score as soon as today and as late as November 30.  Don't forget that the Thanksgiving holiday is in that period too delaying both school mailings and the mail delivery itself.

We will begin scoring as soon as we have organized a sufficient number to run through our scanners.  I estimate that we can scan 10,000 to 15,000 forms per day, and we will have  about 150,000 forms to scan.  Therefore, it will take about 2 work-weeks or more to completely scan all results.

After enough answer forms have been scanned to confidently determine statistical results for awards, honors, etc. we begin emailing results to teachers who provided an email address.  We also begin printing and then stuffing the postal mailed results.  Mailings go out soon after that.  We aim to have that process completed by mid-December so that we can turn our attention  fully to the AMC 10 and AMC 12.

So you should expect to receive emailed results by mid-December, and postal mailed results by mid- to late-December.  If your school is not in session, you will probably have to wait until school resumes for your teacher to receive and report the results to you.

Steve Dunbar
AMC Director

Hm, we seem to have gone back in time.  Feel free to discuss the 2004 AMC8 now.

darn! i didn't get a perfect  :(  i made 2 stupid errors


MATHCOUNTS
geometry
3D geometry
Euler
perimeter
trigonometry
AoPS Books

So do you think that the 2009 chapter target ( and sprint too i guess) will be harder than the 2008. I know its gonna be AND THEN I"LL GET DOOMED AND GET LIKE 45th at chapter.

What you may not notice is that if it is harder, not only will you score lower, but EVERYBODY ELSE AT YOUR CHAPTER AS WELL!!!
Lol :D

Heres teh thing. There are really smart kids at my Chapter. So last year it was easy enough even for me. So I got lots of questions so tehy obviously did too. But when its harder and I don't know cause they know cause they're smarter.
[size=75]
[color=indigo]LadyKn1ght (mod edit): Since your caps lock appears to be stuck when you wrote this, I uncapitalized the extra capitals for you. Don't worry, I left the text as it originally was.[/color][/size]

Hm practicing and doing more hard problems might help. :)

Here are some suggestions for studying. :)

[b]1) AoPS Books[/b]

I recommend first reading the entire Introduction series if possible, and then reading AoPS Volume 1. Volume 1 is rather fast-paced and serves as a great review. Be sure you attempt all the exercises and at least most of the Review Problems. Make sure you understand the solutions, and not only read them when you answer a problem incorrectly, but if you got it right too - they may have a slicker and more elegant solution than you. You should definitely try a good portion of the Challenge Problems - and it's okay if you miss a bunch, because they are rather difficult.

[b]2) Old Tests[/b]

A few Google searches can yield a lot of old MathCounts tests. You should try them all, Chapter, State, and if you can get your hands on some Nationals, that's great! Most of the websites that give you the questions will have answers and solutions, and you must read them. You can never do too many MathCounts tests. Do lots and lots and lots and lots of problems. They don't even have to be MathCounts problems! Look at AMC 8's, 10's, and 12's! Dig up questions from every contest you can think of. Don't just do easy problems - these aren't going to help you. Do hard problems, problems that stump you. But don't quickly think for a minute and then glance over the solution. Think for a day, or more if you have to. The thrill of finally solving the problem will make it worth it. (Naturally, you do not want to take days at the actual MathCounts competition.)

[b]3) Formulas[/b]

There are so many things you have to memorize. I suggest memorizing squares up to 40, cubes up to 25, and all the useful formulas. Mr. Diaz's MathCounts Bible is great for memorizing these formulas. First memorize - later, when you've got them all down pat, learn how they are derived.

[b]4) Vedic Math[/b]

Go on to the Vedic math website and read [i]all[/i] the tutorials. The tricks are amazingly helpful. Again, first memorize - later you can figure out how and why they work.

If you're really motivated to do good, here's the list.

It's good to compile a record of ALL the useful or plain interesting formulas, as well as special identities to practice. I mean, everyone knows that the number of handshakes between 10 people with each person shaking hands with each other is $ \binom{10}{2}$, but hey, you save 1 sec knowing that it's $ \frac {n(n \minus{} 1)}{2}$.  :D

[hide="A short list"]
Remember:
[hide="Perfect Squares"]
At the very least, remember perfect squares under 25 and "special" perfect squares up to 45. By Special, I mean the perfect squares that are very close to 1000 or 2000, they are very helpful.
$ 31^2 \equal{} 961$
$ 32^2 \equal{} 1024$
$ 33^3 \equal{} 1089$
$ 44^2 \equal{} 1936$
$ 45^2 \equal{} 2025$.
Also remember, and this is the only bit of Verdic Maths I remember, how to quickly calculate perfect squares ending in 5.
[/hide]
[hide="Prime Numbers"]
VERY USEFUL for everything, especially checking for divisibility. Remember too, there are 25 perfect squares under 100 and the number of perfect squares between 0-10, 10-20, ..., 90-100, can be memorized by memorizing the following. $ 4 \minus{} 4 \minus{} 2 \minus{} 2 \minus{} 3 \minus{} 2 \minus{} 2 \minus{} 3 \minus{} 2 \minus{} 1$
[/hide]
[hide="Perfect 6th powers"]
These numbers are the numbers that are both perfect squares and perfect cubes, and very useful. The first few are 1, 64, 729, 4096.
[/hide]
[hide="Perfect numbers"]
[url=http://en.wikipedia.org/wiki/Perfect_number]http://en.wikipedia.org/wiki/Perfect_number[/url]
[/hide]
Also, Formulas for the max number of intersections between n points in a plane, What defines a plane, methods such as completing the square, up to how to calculate the sum of a harmonic series using the
[url=http://en.wikipedia.org/wiki/Euler-Mascheroni_constant]Euler-Mascheroni constant[/url].
Area Volume Perimeter Formulas, Sum of squares, sum of cubes, Pick's Formula, Law of Sines and Law of Cosines. Ask around how to do certain problems. If they says it comes from a formula, remember it.[/hide]

I'd say... Practice PRACTICE PRACTICE. Not kidding. Until you can perfect all your nationals. Until you can do every problem. Then practice some more. (I'm still trying to perfect states!)

Then, Study. Investigate why you got the problems you got wrong wrong. Keep a notebook or something, and write down notes and how to do certain types of problems. If you derive a formula that works faster than the usual method, remember it. Know how to do every problem thoroughly, and WHY the answer is what it is. Normally during a competition, I remember a formula but is unsure of why it works. So, even though you might not have the spare time, try to derive all formulas.

You might want to focus on State round, preparing for chapter. Always good to be one step ahead.
Find whatever [size=59]pilfered[/size] old MathCounts tests you can. Print them. Organize them into a nice binder. Keep them clean. (I feel lucky that I have a state coach to give me old tests...)

Don't go crazy in problem solving, though.

Finally, Verdic Math is blah. Do abacus for 6 years.  :D

And, yes, chapter round will be soooooo much harder because last year's chapter was sooooo easy.
You'll do fine.


The average age of a group of teachers and students is 20.  The average age of the teachers is 35, and the average age of the students is 15.  What is the ratio of teachers to students?  Express your answer as a common fraction.

hm....
[hide="Lucy only..hey that rhymes!!!"]
NS: number of studnets
NT: number of teachers
S: sum of students ages
T: sum of teachers' ages

so we named them all..

now let's convert the problem into equations

(S+T)/(NS+NT) = 20
T/NT = 35
S/NS = 15

and this is the problem we want to solve..

NT/NS = ?

well, we would know that
S + T = 20 NS + 20 NT
from the first equation

T = 35 NT
from the second

and
S = 15 NS
from the last...

then, we could change the S and T to
20 NS + 20 NT = 15NS + 35NT

from here, we get the equation,
15NT = 5NS
so,,, the ratio would have to be....
NT/NS = 5/15 = 1/3
[/hide]
I personally thought it was a tough problem for the Basics section...
hm.. should be between Mathcounts and Basics...
and...
 people!! post your answers with your works!!!

[quote="LuCy4EvA"]how would you do this?[/quote]
[hide="The teeter totter method works nicely for this problem"] The average of all of the students and teachers is 20.  That's the fulcrum of the teeter-totter.  If the average age of the students is 15 and the average age of the teachers is 35, how many of each do you need to pile onto the teeter-totter to keep it in balance?

The distance from the fulcrum on the left times the number of students on the left must equal the distance from the fulcrum on the right times the number of teachers on the right.

[img]http://www.artofproblemsolving.com/Forum/files/20050307_201.gif[/img]

If you think you understand this, solve the same problem, where the average age of teachers and students is 17 years old, and everything else stated in the problem remains the same.[/hide]

Thanks everyone! I get it now...
While I was doing RCV's new problem, I figured out a new way of doing it... or maybe it's the same as shinwoos:

[hide]x=number of teach
y=number of students

35x+15y/(x+y) = 17
so.....
35x+15y = 17x+17y
and when you solve that equation, you get
18x = 2y
so it's
9:1
[/hide]
hm... is that right?

Yes, Lucy, you seem to "get it".  Your answer to my new challenge is correct.


analytic geometry
invariant
function
algorithm
trigonometry
calculus
integration

Diffusion motion of particle in situation where a diffusion coefficient K is constant can be in 2 dimmensions described in Cartesian cordinates as

$ x_{n + 1} = x_n + dx$
$ y_{n + 1} = y_n + dy$

$ < x^2 > = dx = R_1 \sqrt {2K \Delta t}\hspace{20 mm}(1)$

$ < y^2 > = dy = R_2 \sqrt {2K \Delta t}\hspace{20 mm}(2)$

where R1,R2 are gaussian random numbers, delta t is time step.

How it will be in spherical coordinates ($ r, \theta$) ?

$ < r^2 > = dr = ?$
$ < \theta^2 > = d \theta = ?$

Thank you for answers-recomendations.

[b]EDIT : to give more precise question[/b] - what I am looking for is stochastic solution of equation

$ \frac {\partial f}{\partial t} = \frac {\partial }{\partial x_i}} \left( {K_{ij} \frac {\partial f}{\partial x_j} \right)$

solution in spherical coordinates.

I do not really understand your notation. Do you mean a diffusion like a $ 2$ dimensional brownian motion?
$ x_{n\plus{}1} \equal{} x_n \plus{} dx$
$ y_{n\plus{}1} \equal{} y_n \plus{} dy$
where $ dx,dy$ are $ 2$ independant normal random variables such that $ Var(dx) \equal{} Var(dy) \equal{} dt$. I think that if you want describe the same diffusion in polar coordinate, since this diffusion is isotropic (invariant under orthogonal transformations), you can easily see that this roughly equivalent to (with your notation):
$ r_{n\plus{}1} \equal{} r_n \plus{} dr$
$ \theta_{n\plus{}1} \equal{} \theta_n \plus{} d \theta$
where $ dr,d \theta$ are $ 2$ independant normal random variables such that $ Var(dr) \equal{} Var(r d \theta) \equal{} dt$.

Does it help you ?

[quote="alekk"]I do not really understand your notation. Do you mean a diffusion like a $ 2$ dimensional brownian motion?[/quote]

Yes, like brownian motion, there is many examples, principle is same.

[quote="alekk"]
$ x_{n \plus{} 1} \equal{} x_n \plus{} dx$
$ y_{n \plus{} 1} \equal{} y_n \plus{} dy$
where $ dx,dy$ are $ 2$ independant normal random variables such that $ Var(dx) \equal{} Var(dy) \equal{} dt$. I think that if you want describe the same diffusion in polar coordinate, since this diffusion is isotropic (invariant under orthogonal transformations), you can easily see that this roughly equivalent to (with your notation):
$ r_{n \plus{} 1} \equal{} r_n \plus{} dr$
$ \theta_{n \plus{} 1} \equal{} \theta_n \plus{} d \theta$
where $ dr,d \theta$ are $ 2$ independant normal random variables such that $ Var(dr) \equal{} Var(r d \theta) \equal{} dt$.
Does it help you ?[/quote]

No, it is obvious what you say. Anyway thank you - lets do it in this way - I have two solutions (my evaluation, what I need is check), here is first one:

It is in spherical coordinates, where system is axially symetric, so we have just $ dr, d\theta$ motion moments

1.   Solution where distribution function $ f$ of diffusion equation is substituted in following way $ f \equal{} r^2*sin{\theta} * g$. $ g$ is "new" distribution function, I must take it to account during simulation of course.

Solution:

$ < r^2 > \equal{} dr \equal{} R_1 \sqrt {2K \Delta t} \plus{} \frac {2K \Delta t}{r}\hspace{20 mm}(3)$

$ < \theta^2 > \equal{} d \theta \equal{} R_2 \sqrt {\frac {2K \Delta t}{r^2}} \plus{} \frac {cotg(\theta) K \Delta t }{r^2}\hspace{20 mm}(4)$

Is it correct?

Thanks for answers.

still do not get your notation, and do not really understand what you are trying to do ?

Say that you have a diffusion $ \partial_t f \equal{} \partial_i ( K_{i,j} \partial_j f)$, and to keep things simple (it is basically the same, just do a change of variables), say that this diffusion is simply: $ \partial_t f \equal{} \Delta f$ (Heat equation) with some Dirichlet boundary conditions $ f(t \equal{} T,x) \equal{} \Phi(x)$
So, what you are trying to do is to compute a numerical solution of this diffusion PDE through a stochastic simulation algorithm (Monte carlo etc ..), right ?
If this is the case, just use the Feynman Kac representation theorem that tells you that:
$ f(x_1, \ldots, x_n) \equal{} E[\Phi(B^{T}_1, \ldots , B^{T}_1)]$
where $ B \equal{} (B_1, \ldots, B_n)$ is a $ n$-dimensional brownian motion. (for your problem, take the appropriate SDE, etc..)

So now you want to work with polar coordinates: apply Ito formula (i.e. chain rule).
$ (x,y) \mapsto (\sqrt {x^2 \plus{} y^2}, \tan^{ \minus{} 1}(\frac {y}{x})) \equal{} (r, \theta)$
you should find something like (this is a $ 2$ dimensional bessel process):
$ dr \equal{} dW_1$
and for $ \theta$:
$ d \theta \equal{} \frac {1}{r} dW_2$

where $ (W_1, W_2)$ is a $ 2$ dimensional brownian motion (this is what was expected, isn't it)

Now it works basically the same with an arbitrary SDE ( but it is not a brownian motion anymore ), but the computations get nasty, as is expected. Nevertheless, if your coefficients $ K_{i,j}$ are constants, it should not be that bad. Is it finally what you meant ? If not, could you describe a little bit clearer what you are trying to solve ..

[quote="alekk"]
Say that you have a diffusion $ \partial_t f = \partial_i ( K_{i,j} \partial_j f)$, and to keep things simple (it is basically the same, just do a change of variables), say that this diffusion is simply: $ \partial_t f = \Delta f$ (Heat equation) with some Dirichlet boundary conditions $ f(t = T,x) = \Phi(x)$
So, what you are trying to do is to compute a numerical solution of this diffusion PDE through a stochastic simulation algorithm (Monte carlo etc ..), right ?[/quote]

Yes, you are right. Thanks. Let me explain it. I want understand one thing in Fokker–Planck equation solution by Monte Carlo method. To have it simple I work with drift equal zero.

So, Wiener process $ dX_t = dW_t$ (with drift term equal zero and diffusion coefficient equal 1/2) -> corresponding Fokker–Planck equation is

   $ \frac {\partial f(x,t)}{\partial t} = \frac {1}{2} \frac {\partial^2 f(x,t)}{\partial x^2}$

that is the diffusion equation.

I can solve it by Monte Carlo method in cartesian coordinates by equation (1) and (2) from my first post in this thread and it work in my opinion well.
I want do same in spherical coordinates. As a solution I get equations (3) and (4) from my second post in this thread  - it does not work well so I am currious what is wrong.

I start from

$ \frac {\partial f}{\partial t} = {\frac { - 1}{r^2}} {\frac {\partial }{\partial r}} \left( r^2 S_r \right) - {\frac {1}{r sin \theta}} {\frac {\partial}{\partial \theta}} \left(S_\theta sin \theta \right)$

where

$ S_r = - K \frac {\partial f}{\partial r}$ and  $ S_\theta = \frac { - 1}{r} K \frac {\partial f}{\partial \theta}$

finaly I get

$ \frac {\partial g}{\partial t} = - K {\frac {\partial^2 }{\partial r^2}[g]} - K {\frac {\partial }{\partial r}[\frac {2g}{r}] } + {\frac {\partial^2}{\partial \theta^2}[\frac {K g}{r^2}]} - K {\frac {\partial}{\partial \theta}[\frac {cotg \theta g}{r^2}]}$

and from this I see (3) and (4).

My problem is that in cartesian coordinates results does not depend on $ \Delta t$ (in (1) and (2)) what is ok. But in (3) and (4) results does not seems to be correct and moreover does depend on $ \Delta t$ what is in my opinion wrong.

[quote="fedja"]Note that a finite step instead of an infinitesimally small one leads to the right result in Cartesian coordinates equations (if the coefficients are constant, it doesn't really matter if you immediately compute the coordinates after time $ T$ using this formula or split $ [0,T]$ into several subintervals and do it in several steps; the final distribution will be the same) but it is no longer true for the above polar coordinate equations: they are valid only if $ \Delta t\to 0$.[/quote]

Thank you. What about situation when K is not constant? Are results after same for different $ \Delta t$ (with not constant $ K$) in cartesian coordinates?

What is reason for validity of results for polar coordinates only if $ \Delta t\to 0$ ?.

Thank you alekk and fedja.

[quote="alekk"]because in cartesian coordinates, if $ K$ is constant, this is basically a brownian motion and one exactly knows the distribution of $ B_t$: this is a multivariate gaussian variable. Nevertheless in polar coordinate $ (\theta_t, r_t)$ is not gaussian. If $ K$ is not constant, it does not work, even in cartesian coordinates (hence one has to use a Monte-Carlo like procedure ..).[/quote]

Mamy people do for many years MC simulations where $ K$ is not constant. How I should understand this?
To be sure that I understand what you wrote correctly. Do you wanna say that MC method for not constant $ K$ does not work?

Because all what I was asking was about MC (stochastic) types of solutions of FP (heat) equation.

[quote="4sidetriangle"]Thank you alekk and fedja.

Mamy people do for many years MC simulations where $ K$ is not constant. How I should understand this?
To be sure that I understand what you wrote correctly. Do you wanna say that MC method for not constant $ K$ does not work?

Because all what I was asking was about MC (stochastic) types of solutions of FP (heat) equation.[/quote]

No  :D  I was saying that if $ K$ is constant, the problem is essentially trivial but if $ K$ is not, then it becomes more interesting and one can use MC methods. Sorry for the misunderstanding ..

About validity only in cartesian coordinates. Solution should be a

1. do MC simulation in polar coordinations
2. do transformation from polar to cartesian coordinates before integration of SDE - so integration in cartesian

if I correctly understand - it means that

1. I evaluate $ dr$, $ d \theta$ - doing MC in polar coordinates

2.  evaluate new $ r$, $ \theta$
$ r_{n \plus{} 1} \equal{} r_n \plus{} dr$
$ \theta_{n \plus{} 1} \equal{} \theta_n \plus{} d\theta$

3.  evaluate now
$ x \equal{} r*cos(\theta)$
$ y \equal{} r*sin(\theta)$

4. at (x,y) position count-add to space bin appropriate for this position value 1 or some normalized value (depend on how was evaluated equations - transitions moments for polar coordinates)

Is it what transformation back to cartesian and integration in cartesian means?

Maybe stupid question, but better ask a stupid questions than do mistakes :)


trigonometry
calculus
symmetry
integration
function
calculus computations

Household electricity is supplied in the form of alternating current that varies from $ 155$ volts to $ \minus{} 155$ volts with a frequency of $ 60$ cycles per second. The voltage is thus given by the equation  $ E(t) \equal{} 155\sin(120\pi t)$, where t is the time in seconds. Voltmeters read the RMS (root-mean-square) voltage, which is the square root of the average value of  $ \left[E(t)\right]^2$ over one cycle. Calculate the RMS voltage of household electricity.

What is the average value of $ \sin^2$ over a period?  (It's clearly not $ 1$.)

Edit:  Once you've solved this using calculus, try to solve it with a symmetry argument.

I've assigned this problem as homework - I'm pretty sure the exact wording comes from Stewart's textbook. Except that was supposed to be $ E(t) = 155\sin(120\pi t).$ You've got a misprint: the correct statement agrees with "60 cycles per second" but what you've written doesn't answer to that description.

A pertinent observation and something I think is worth remembering for its own sake:
\[ \int_0^{\frac {\pi}2}\sin^2x\,dx = \int_0^{\frac {\pi}2}\cos^2x\,dx = \frac {\pi}4
\]
I like to get this by a symmetry argument and accompany it by a picture; because of the picture, I usually refer to this in class as the "yin-yang theorem." But the way I tell me students to remember the result is this: if you integrate $ \sin^2$ or $ \cos^2$ of a linear function and (this is very important) the interval of integration starts and stops at the ends of quadrants (in terms of the linear function on the inside), then the value of the integral is "half the box".

For example: $ \int_0^{10}\cos^2(\pi t)\,dt = \frac12(10 - 0) = 5.$ The theorem applies since both $ 0$ and $ 10\pi$ are ends of quadrants.

Another way to remember this is that the long-term average value (and hence average value over each quadrant) of $ \sin^2x$ or $ \cos^2x$ is $ \frac12.$

Thank you Dr. Merryfield. But how should I use this "yin-yang theorem" in my solution? Does it have an official theorem title that I can quote?

Why would you need to cite it?  The proof is very short:  whatever the average value of $ \sin^2$ over a period is, it should be the same as the average value of $ \cos^2$, but $ \sin^2 x \plus{} \cos^2 x \equal{} 1$, so their averages should also sum to $ 1$.

The "yin-yang" comes about when you graph $ y \equal{} \sin^2 x$ against $ y \equal{} 1$.  The space between the two graphs is $ 1 \minus{} \sin^2 x \equal{} \cos^2 x$.


Find $ x,y \in Z$ of this equation: $ x^2\plus{}2008x\plus{}2009y^2\plus{}y\equal{}xy\plus{}2009xy^2\plus{}2010$

$ (x\minus{}1)(x\plus{}2009\minus{}2009y^2\minus{}y)\equal{}1$. The rest is trivial.


Let $ a,b,c\geq0$ s.t. $ ab\plus{}bc\plus{}ca\equal{}3$.Prove that:
$ \dfrac{1}{a^2\plus{}1}\plus{}\dfrac{1}{b^2\plus{}1}\plus{}\dfrac{1}{c^2\plus{}1}\geq\dfrac{3}{2}$

I can't prove it by cauchy-schwarz :) .But I can prove it by  pqr
<=>$ p^2 \ge 12\plus{}3r^2\minus{}2pr$
If $ p^2 \ge 12$[b](1)[/b], easy to prove that :
$ RHS \le 12 \le LHS$ Done :)
If $ p^2 \le 12$ ,we have :$ r \ge \frac{p(12\minus{}p^2)}{9}$
<=>$ (12\minus{}p^2)(p^4\plus{}24p^2\minus{}27) \ge 0$
Done ;)

I am 100% sure that your solution is correct, but how do you reach to your last inequality?

Let $ p \equal{}a\plus{}b\plus{}c,q\equal{}ab\plus{}bc\plus{}c\equal{}3,r\equal{}abc$
\[ LHS \equal{} \frac{q^2\minus{}2pr\plus{}2p^2\minus{}4q\plus{}3}{(q\minus{}1)^2\plus{}(p\minus{}r)^2}\]
\[ \frac{q^2\minus{}2pr\plus{}2p^2\minus{}4q\plus{}3}{(q\minus{}1)^2\plus{}(p\minus{}r)^2} \geq \frac{3}{2}\] \[ \Longleftrightarrow \frac{2p^2\minus{}2pr}{4\plus{}(p\minus{}r)^2} \geq \frac{3}{2}\]
\[ \Longleftrightarrow 4p^2\minus{}4pr \geq 12\plus{}3(p\minus{}r)^2 \equal{} 12\plus{}3p^2\minus{}6pr\plus{}3r^2\]
\[ \Longleftrightarrow p^2\plus{}2pr \geq 12\plus{}3r^2\]
from Am-GM ; $ (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca) \geq 9abc$ $ \therefore p \geq 3r$
$ \therefore pr \geq 3r^2$
from Am-Gm ; $ p^2q\plus{}3pr \geq 4q^2 \rightarrow p^2\plus{}pr \geq 12$
$ \therefore p^2\plus{}2pr \geq 12\plus{}3r^2$
$ \therefore LHS \geq \frac{3}{2}$ #


[quote="nebula42"]How many positive integers less than or equal to $1000$ are divisible by $3$ or $5$ but not $7$?[/quote]

[hide]Um...
$\lfloor 1000/3\rfloor+\lfloor 1000/5\rfloor-\lfloor 1000/15\rfloor-\lfloor 1000/21\rfloor-\lfloor1000/35\rfloor+\lfloor1000/105\rfloor=\boxed{401}$?[/hide]

[hide]For this, we need to find how many numbers are divisible by 3 and 5 both, then subtract that from the numbers divisible by 3 plus the numbers divisible by 5.  There are 200 numbers for 5, 333 for 3, and 66 for 3 and 5 both.  $200+333-66 = 467$ numbers that are divisible by 3 or 5.

For the second part, we need to find how 3'numbers that are divisible by 7, how many 5'numbers that are divisible by 7, and then subtract how many (3 and 5)'numbers that are divisible by 7.  There are 47 for 3, 28 for 5, and 9 for 3 and 5.  $47+28-9 = 66$ numbers out of the 467 that are divisible by 7.  So, $467-66 = \boxed{401}$.[/hide]

[size=75][color=darkred]Please try to remember to use hide tags when posting a solution. Thanks.
-nebula42[/color][/size]

Yes, I've forgotten to put hide tags like 3 times in the past 2 days.

You are correct again  :) .

It's ok, for being away from AoPS for a while (i think).

To everyone: If you don't know this, use HIDE TAGS.  They are right above where I'm typing right now, in the same row as the bold, italic, underline, quote, code, list, img, and URL boxes.  Hide tags allow other people to do the problem without the answer staring at them.


Where are the transcripts available for previous classes for Olympiad Geometry? I missed the class yesterday, and I want to find out what I missed.

After logging in, click on the My Classes tab above.


A group contains elements a ,b ,c ,d ,e satisfying the equations ab=c , bc=d , cd=e , de=a and  ea=b.

Find the orders of these elements assuming that none of them is the identity.

Where are you stuck? Just keep substituting into equations.


This year to prepare for the AMC 10, I searched and searched and searched. I posted threads. Got replies. Followed the advice. Did all the AMC 10s. Tried some AIME's. Did HMMT geometry. Did all the intro books. Got started on Volume 2. After a year of work, I'm nowhere. I'm getting the exact same scores I was getting in practice a year ago (120's). Its extremely frustrating. Should I try more on the AIME's to help with me with the AMC 10s (I suck with the time limit)? My parents are suggesting for me to get a tutor for all of this (I don't think we'll find one that teaches AMC 10 stuff not school stuff), should I do that? Should I just keep doing problems (if so, which ones)?

well, first you should see which problems you are getting wrong.  then work on those problem especially much.  I also started out at around 120, but after working on some prob. and geo, I was able to consistenly get 130, and 138 on the 10B.

After that, work on doing it FAST.  Its probably easiest to speed up problems 1-15.

Hope this helps :).

First, I think you posted your score on your blog. If you've actually understood everything that you've learned from those books, you should definitely be able to get a very high score on the 10. The first thing you need to do is figure out why you're getting all those problems wrong. Are you making stupid mistakes on the computation and such? Or do you not know the needed knowledge.

The best way to fix those up is to do many problems. Also for computation and speed help, maybe you should take a look at some [url=http://texasmath.org/DL/NS/ns.html]number sense tests[/url].

I definitely improved my score by 22 points by just trying to learn stuff I didn't know. Redoing stuff you already know is useless, especially on the AMC. Keeping a notebook of all the problems you have missed on practice tests is a good idea, because you can kind of see a pattern of things you don't know.

@xpmath

I definitely understood everything in the intro books. I went did them again twice. I also had my mom quiz on random problems throughout the books.


I think that just doing many problems from previous olympiads is the best possible way. There is no secret formula I think. Also, you should do more problems on topics you are weak at, and that should then work out fine. :)

From what olympiads should I take these problems? Obviously things like the USAMO and Romanian one are going to be a little too hard at first, so could you reccomend 3-6 easier and medium level olympiads?

I think some of the easier, or at least not-so-hard, olympiads are Indian Regional Math Olympiad (RMO), British Math Olympiad (BMO) Round 1 and Irish Math Oympiad Paper 1. You should be able to find quite a lot of resources online on these. Else, you always have this forum to discuss problems you are having difficulty with. :wink:

France MO and Lithuania and Greece MO are easy, too
On Aops there are only a few years from which the olympiads have been posted, but maybe there is another link...

Hmm, I'm a descendent of a lithuaian :D . Thanks guys, are there any problem books you would reccomend besides what I posted above, and do you know of any more methods for training? Thanks a bunch.

As FieryHydra suggested, the Irish MO are comparatively easy. Kalva only goes up to 2003, you can find the rest [url=http://maths.ucd.ie/~dukes/imo.html]here[/url]. I have a hard copy of 2006 somewhere, and can scan if you're interested.

[quote]If one had, say, 3 months to study for a national math olympiad, and he had already read the usual books(PSS, AoPS, ACOPS, MoT, MM, etc plus some supplements for geometry), how would you reccomend he go about studying for olympiads? Is there a more efficient way than just doing more problems, or is that what is boils down to?
[/quote]

What is Mot and MM?

I believe they are Mathematical Olympiad Treasures and Mathematical Mosiacs, respectively.

Masoud Zargar


Come on, this is for High-School Basics.

Let $x$ and $y$ be the ages of Linh and Hoa. Then we have
$x-5 = \frac{2}{3}(y-5)$
$x+10 = \frac{5}{6}(y+10)$
After that, we can easily get $x=15 and y =20$

This should be in Classroom Math, I think.  :wink:

[quote="chien than"]Five years ago Linh was $\frac{2}{3}$ as old as Hoa.Ten years from now he will be $\frac{5}{6}$ as old as Hoa.How old are they now?[/quote]

I'd rather you not label a problem as "very very easy" when it is in fact "very very very easy". A more specific title would help.

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$(x-5) = \frac23(y-5)$
$6x-30 = 4y-20$
$6x = 4y+10$

$(x+10)) = \frac56(y+10)$
$6x+60 = 5y+50$
$6x = 5y-10$

$4y+10 = 5y-10$
$y = 20$
$x = 15$

Linh is 15 and Hoa is 20.
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