inequalities
inequalities proposed
China

Dear siuhochung

Please, rewrite your solution in more details.

Thank you very much.

[quote="hungkhtn"][quote="siuhochung"]This appeared in china team training 03. In my book, there are two solutions.
The first solution used mixing variables.
Let $f(a,b,c)=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}$, then $f(a,b,c) \ge f(0,a+b, \frac{1}{a+b})$. Then it is easy to finish it up.
The second one is quite ugly so i'm not posting it.[/quote]

It should be:

$f(a,b,c) \ge f(a,t,t)$

$t=\sqrt{(a+b)(a+c)}-a$[/quote]

what i wrote is correct.
My book writes
Let \[
f\left( {a,b,c} \right) = \frac{1}
{{a + b}} + \frac{1}
{{b + c}} + \frac{1}
{{c + a}}
\]
Let $a \le b \le c$, and then we have
\[
f(0,a + b,c*) \leqslant f(a,b,c)
\]
with $c*=\frac{1}{a+b}$.
In fact,
\[
\begin{gathered}
  f(0,a + b,c*) \leqslant f(a,b,c) \hfill \\
   \Leftrightarrow \frac{1}
{{a + b}} + \frac{1}
{{a + b + c*}} + \frac{1}
{{c*}} \leqslant \frac{1}
{{a + b}} + \frac{1}
{{b + c}} + \frac{1}
{{c + a}} \hfill \\
   \Leftrightarrow \frac{{a + b + 2c*}}
{{(a + b + c*)c*}} \leqslant \frac{{a + b + 2c}}
{{ab + bc + ca + c^2 }} \hfill \\
   \Leftrightarrow \frac{{a + b + 2c*}}
{{1 + c*^2 }} \leqslant \frac{{a + b + 2c}}
{{1 + c^2 }} \hfill \\
   \Leftrightarrow \frac{{a + b + 2c*}}
{{a + b + 2c}} \leqslant \frac{{1 + c*^2 }}
{{1 + c^2 }} \hfill \\
   \Leftrightarrow \frac{{2\left( {c* - c} \right)}}
{{a + b + 2c}} \leqslant \frac{{(c* - c)(c* + c)}}
{{1 + c^2 }} \hfill \\
   \Leftrightarrow \frac{2}
{{a + b + 2c}} \leqslant \frac{{c* + c}}
{{1 + c^2 }} \hfill \\
   \Leftrightarrow 2 + 2c^2  \leqslant (a + b + 2c)(c* + c) \hfill \\
   \Leftrightarrow 2 \leqslant (a + b)(c* + c) + 2cc* = 1 + (a + b)c + 2cc* \hfill \\
   \Leftrightarrow 1 \leqslant 1 - ab + \frac{{2(1 - ab)}}
{{(a + b)^2 }} \hfill \\
   \Leftrightarrow (a + b)^2 ab \leqslant 2(1 - ab) \hfill \\
\end{gathered}
\]
Since \[
2(1 - ab) = 2c(a + b) \geqslant 2\frac{{(a + b)^2 }}
{2}
\]
it is true. It remains to show that
\[
\begin{gathered}
  f(0,a + b,c*) \geqslant \frac{5}
{2} \hfill \\
   \Leftrightarrow \frac{1}
{{a + b}} + \frac{1}
{{a + b + \frac{1}
{{a + b}}}} + (a + b) \geqslant \frac{5}
{2} \hfill \\
\end{gathered}
\]
Denote $x=a+b+\frac{1}{a+b}$.
Then $x \ge 2$, and $x+1/x \ge 5/2$ since $x+1/x$ is increasing.
QED

This is  one  of  the  boring inequalities  that  Muirhed  kills .

Absolutely against this attitude! Could you show us a solution using Muirhead? I think it takes more time to make the computations than to find the smart solution.

By the way, does anyone have a solution for the following general problem:
  If $ x_1x_2+x_2x_3+...+x_nx_1=1$, what is the minimal value of $\frac{1}{x_1+x_2}+...+\frac{1}{x_{n-1}+x_n}+\frac{1}{x_n+x_1}$?

Well  Harazi , this is  the main  problem  using  Muirhead  :(  .Too much  boring  computations,,,
The  initial  ineq  becomes  [tex] (xy+yz+zx)(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x})^2\geq(\frac{5}{2})^2[/tex]
After  that  lets  get  our hands  dirty , and  find :
[tex](\sum{x^5y}-\sum{x^4y^2})+3(\sum{x^5y}-\sum{x^3y^3})+xyz(\sum{x^3}+14\sum{x^2y}+38xyz)\geq0[/tex]  (all  [tex]\sum[/tex]  are  symmetric) which is true  because :
1)The  third  term is clearly non-negative
2)The  first  term is  non  negative  since  [tex] (5,1,0)\succ(4,2,0) [/tex]
3).The  second  term is  non negative  since  [tex](5,1,0)\succ(3,3,0)[/tex]

I think  Muirhead  works  the same  with  n  variables,,,,but  i dont  dare  to  make  computations!!  Look  computations  for  3  variables,,,,and then  try  to make  for  n !!  :)

If there is something I really like is that Muirhead never works for n variables and inequalities with fractions. Just try it for 4 variables and we will discuss later... when you will have finished the computations.

[quote="nickolas"]In  3  variables  we have  equality  if-f  [tex] (x,y,z)=(1,1,0)  or (1,0,1)  or  (0,1,1) [/tex]  so  at the  general  ineq  i think  the  minimun value is  [tex] \frac{n+2}{2}[/tex][/quote]

what's the equality case?!

I don't think you're right. For 5 variables, I obtained values close to $5/2$ and I guess this is the answer for 5 variables. Anyway, it seems a very difficult problem.

[quote="nickolas"]At the initial  inequality  [tex]x,y,z\geq0 , xy+yz+zx=1  then  \sum\frac{1}{x+y}\geq\frac{5}{2}[/tex]  we have  equality  if-f  [tex](x,y,z)=(1,1,0) or (1,0,1) or (0,1,1) [/tex]  So i  just think  at  harazi's  problem maybe we have  the minimum  value  if [tex] (x_1,x_2,...,,x_n-1,x_n)=(1,1,...,1,0) [/tex] Its  just  a  thought  that  [tex] \frac{n+2}{2}[/tex]  is  the minimum  value  of  Harazi's  inequality.[/quote]
your $(x_1,...,x_n)$ does not fit in $x_1x_2+...+x_{n-1}x_n=1$.

You are right....sorry  for my mistake :blush:

[quote="hungkhtn"]You proof is very nice and short,Siuhohung! :)[/quote]

I think it's nice too, and I hope one day I can think of it. But now, I can't, and the solution is taken from my book which has chinese team 03 training problems :)

siuhochung,what about the second solution? ;)

[quote="ThAzN1"]
but does anyone have a nicer way?[/quote]

I have !!!  :lol:

WLOG $x\geq y\geq z$ if $x,y$ or $z=0$ $\Rightarrow$ $xy=1$ and our inequality <=> $1/x+1/y+1/(x+y)\geq 5/2$ <=> $(x+y)^{2}+1\geq 5/2(x+y)$, which is clearly true by  $A(m)\geq G(m)$.
For positive real $x,y,z$ I can prove stronger inequality

$Inequality$ $1$ (if new $\Rightarrow$ own)  :wink:

If $x,y,z$ are positive real numbers, such that $\sum{xy}=1$ then prove that
$\sum{\frac{1}{x+y}}\geq \frac{\sqrt{27}}{2}$.

$Proof$.

Let $x=tga/2$, $y=tgb/2$, $z=tgc/2$. Since $\sum{xy}=1$ $\Rightarrow$ $\sum{tga/2.tgb/2}=1$ $\Rightarrow$ $\sum{a}=\pi$.
So, $cyc(\frac{1}{x+y}=\frac{cosa/2.cosb/2}{sin{(a+b)/2}}=\frac{cosa/2.cosb/2}{cosc/2})$ $\Rightarrow$ we can rewrite our inequality as
$\sum{\frac{cosa/2.cosb/2}{cosc/2}}\geq \frac{\sqrt{27}}{2}$ <=> $\sum{\frac{cos^{2}a/2.cos^{2}b/2}{cos^{2}c/2}+2\sum{cos^{2}{a}/2}}\geq \frac{27}{4}$, using
Vietnam tst 2007 inequality http://www.mathlinks.ro/Forum/viewtopic.php?t=142763
it's enough to prove $\sum{cos^{2}a/2}\geq \frac{9}{4}$ <=> $\frac{3}{4}\geq \sum{sin^{2}a/2}$, which is obviously true by Jensen's inequality.

$Q-E-D$.
How is my solution ???  :lol:

I think, when we have the condition $xy+yz+xz=A$, for positive integers $A,x,y,z$ or even $cyc(xy/A>0)$, it's very useful to use Trigonometric Substitutions:
$cyc(\frac{x}{\sqrt{A}}=tg(a/2))$ $\Rightarrow $ $\sum{tg(a/2)tg(b/2)}=1$ $\Rightarrow $ $\sum{a}=\pi$.

 :wink:

$Inequality$ $1$ has been solved easily with this substitution, what other inequalities do you know, that can be solved with this kind of trigonometrical substitutions ?  :lol:  :wink:

I think inequalities $20$ and $21$, from Vasc's book can be solved with this trigonometrical substitution (but I think it's not the best way to do it   :lol:  ).

To answer Vardanch's message to me: SOrry I must say that Inequality 1 of your is wrong. You may have a mistake when using Iran 96 and Jensen in a not right way, moreover, you can not replace the best constant $\frac{5}{2}$ to the more best one $\frac{\sqrt{27}}{2}$  :lol: . Easy to find the counter example, just let $a\to 0$ and $b,c\to 1$.

WLOG $x\geq y\geq z$ if $x,y$ or $z=0$ $\Rightarrow$ $xy=1$ and our inequality <=> $1/x+1/y+1/(x+y)\geq 5/2$ <=> $(x+y)^{2}+1\geq 5/2(x+y)$, which is clearly true by  $A(m)\geq G(m)$.

For positive $x,y,z$
Let $x=tga/2$, $y=tgb/2$, $z=tgc/2$. Since $\sum{xy}=1$ $\Rightarrow$ $\sum{tga/2.tgb/2}=1$ $\Rightarrow$ $\sum{a}=\pi$.
So, $cyc(\frac{1}{x+y}=\frac{cosa/2.cosb/2}{sin{(a+b)/2}}=\frac{cosa/2.cosb/2}{cosc/2})$ $\Rightarrow$ we can rewrite our inequality as
$\sum{\frac{cosa/2.cosb/2}{cosc/2}}\geq \frac{5}{2}$ <=> $\sum{\frac{cos^{2}a/2.cos^{2}b/2}{cos^{2}c/2}+2\sum{cos^{2}{a}/2}}\geq \frac{25}{4}$, using
Vietnam tst 2007 inequality http://www.mathlinks.ro/Forum/viewtopic.php?t=142583
it's enough to prove $\sum{cos^{2}a/2}\geq 2$ <=> $1\geq \sum{sin^{2}a/2}$.
Let $cyc(a=90-A/2)$ $\Rightarrow$ $\sum{A}=\pi$ $\Rightarrow$ $A,B,C$ are angles of an acute angled triangle.
Since $cyc(sin^{2}a/2=cos^{2}A)$, our inequality can be rewritten as $1\geq \sum{cos^{2}A}$.
But $\sum{cos^{2}A}=1-2cosAcosBcos<1$ (since A,B,C<90) $\Rightarrow$ $1\geq \sum{cos^{2}A}$ is true $\Rightarrow$

$Q-E-D$.
How is my solution ???  :lol:, is it new ? Is it true ?

[quote="ThAzN1"]Let $ x,y,z$ be nonnegative real numbers such that $ xy \plus{} yz \plus{} zx \equal{} 1$.  Prove that
\[ \frac {1}{x \plus{} y} \plus{} \frac {1}{y \plus{} z} \plus{} \frac {1}{z \plus{} x} \geq \frac {5}{2}.
\]
Of course, there's a solution by homogenizing, expanding, applying Muirhead, etc... but does anyone have a nicer way?[/quote]
[u]my proof[/u]: this inequality is equivalent to
$ 2(x\plus{}y\plus{}z)^2\plus{}5xyz\plus{}2\geq 5(x\plus{}y\plus{}z)$
* if $ x\plus{}y\plus{}z>2$ it's trivial
*if $ \sqrt{3}\leq x\plus{}y\plus{}z\leq 2$,we use $ xyz\geq\frac{4(x\plus{}y\plus{}z)\minus{}(x\plus{}y\plus{}z)^3}{9}$so it's remain to prove that
$ 2p^2\plus{}5\frac{4p\minus{}p^3}{9}\plus{}2\minus{}5p\geq 0$ ($ p\equal{}a\plus{}b\plus{}c$ )
$ \Leftrightarrow (p\minus{}2)(5p^2\minus{}8p\plus{}9)\leq 0$ it's true because $ p\leq 2$ and $ 5p^2\minus{}8p\plus{}9\geq 0$
we are done!
p/s: I think it's nice proof :D

[quote="toanIneq"]
p/s: I think it's nice proof :D[/quote]
I agree with you! :lol:

Let $x,y,z$ be nonnegative real numbers such that $xy+yz+zx=1$. Prove that
\[\frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} \geq \frac{5}{2}+\frac{xyz}{2(x+y)(y+z)(z+x)}.\]
The following inequality is also true.
Let  $x,y,z$ be nonnegative real numbers with  $xy+yz+zx=1$. Prove that  \[ \frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}+ \frac{3}{x+y+z} \ge 4. \]

Generalization
Let $x,y,z$ be nonnegative real numbers such that$xy + yz + zx = 1$，$k \in R$ and $ - 1 \le k \le 3$. Prove that
\[\frac{1}{{x + y}} + \frac{1}{{y + z}} + \frac{1}{{z + x}} + \frac{k}{{x + y + z}} \ge \frac{{5 + k}}{2}\]
See http://blog.sina.com.cn/s/blog_547541ce0101545q.html

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=508311][b]here[/b]:[/url]
Let $a,b,c $ be non-negative real numbers such that $ bc+ca+ab=1$ , Show that\[
\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}+\frac{\lambda}{a+b+c}\ge\frac{5+\lambda}{2}.(-1\le \lambda \le3).\]
[hide="24 May 2014"]调整法证明的一道不等式:http://blog.sina.com.cn/s/blog_b9a401a40101ed3u.html[/hide]

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=248673

Let $x,y,z$ be nonnegative real numbers such that $xy+yz+zx=1$.  For ${\lambda} \ge 1$ , prove that\[\frac{1}{x^{\lambda}+y^{\lambda} }+ \frac{1}{y^{\lambda}+z^{\lambda}}+ \frac{1}{z^{\lambda}+x^{\lambda}} \geq \frac{5}{2}.\]

[quote="Soarer"]I think it's nice too, and I hope one day I can think of it. But now, I can't, and the solution is taken from my book which has chinese team 03 training problems :)[/quote]

Could you tell me what this book is called?

Let $x,y,z$be nonnegative real numbers such that$0 < xy + yz + zx \le 1$,$k \in R,\alpha  \in R$ and $\alpha  \ge 1, - 1 \le k \le 3$ Prove that
\[
\frac{1}{{x^\alpha   + y^\alpha  }} + \frac{1}{{x^\alpha   + z^\alpha  }} + \frac{1}{{z^\alpha   + x^\alpha  }} + \frac{k}{{x^\alpha   + y^\alpha   + z^\alpha  }} \ge \frac{{5 + k}}{2}
\]

Let $x,y,z$ be nonnegative real numbers .  Prove that
$$\frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} \geq \frac{5}{2\sqrt{xy+yz+zx}}.$$

Let  be nonnegative real numbers and   ，Prove that

Let $x,y,z$ be nonnegative real numbers and  $\alpha  \ge 1, - 1 \le k \le 3$，Prove that
$$\frac{1}{{x^\alpha   + y^\alpha  }} + \frac{1}{{y^\alpha   + z^\alpha  }} + \frac{1}{{z^\alpha   + x^\alpha  }} + \frac{k}{{x^\alpha   + y^\alpha   + z^\alpha  }} \ge \frac{{5 + k}}{{2\sqrt {\left( {xy + yz + zx} \right)^\alpha  } }}$$

[quote=xzlbq]In triangle

\[\sum{\frac{1}{\tan{\frac{B}{2}}+\tan{\frac{C}{2}}}}-\frac{5}{2}\geq (\frac{5}{2}\sqrt{3}-\frac{25}{6})\sum(\frac{r}{r_a+h_a-r})\geq  \left( 3\,\sqrt {3}-5 \right) {\it \sum} \left( {\frac {r}{{\it r_a}+{
\it r_b}}} \right)
\][/quote]

[quote=sqing]Let $x,y,z$ be nonnegative real numbers such that $xy+yz+zx=1$. Prove that
\[\frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} \geq \frac{5}{2}+\frac{xyz}{2(x+y)(y+z)(z+x)}.\]
[/quote]
Let $x,y,z$ be nonnegative real numbers such that $xy+yz+zx=1$. Prove that
\[\frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} \geq \frac{5}{2}+\frac{3xyz}{4(x+y)(y+z)(z+x)}.\]
$$\frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} \geq \sqrt{\frac{25}{4}+\frac{4xyz}{(x+y)(y+z)(z+x)}}.$$
$$\frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} \geq \frac{5}{8}\left(x+y+z+\frac{4}{x+y+z}\right)+\frac{xyz}{(x+y+z)(x+y)(y+z)(z+x)}.$$
[size=50](ZXJY(SX)  No.6(2019))[/size]

[quote=ThAzN1]Let $x,y,z$ be nonnegative real numbers such that $xy+yz+zx=1$.  Prove that

\[\frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} \geq \frac{5}{2}.\]

Of course, there's a solution by homogenizing, expanding, applying Muirhead, etc... but does anyone have a nicer way?[/quote]
[url=https://artofproblemsolving.com/community/c6h449840p2917008]2003 China TST[/url]
Let $x,y,z$ be nonnegative real numbers such that $(x+y)(y+z)(z+x)\neq 0. $ Prove that$$\frac{1}{x+y} +\frac{1}{y+z} +\frac{1}{z+x} \ge \frac{5}{1+xy+yz+zx}$$
$$\frac{1}{x+y} +\frac{1}{y+z} +\frac{1}{z+x} \ge \frac{5}{2 \sqrt{xy+yz+zx}}$$
Let $x,y,z \ge 0,$ [url=https://artofproblemsolving.com/community/c6h1932561p13276896]prove that:[/url]
$$\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x} \ge \frac{5+\frac{xyz}{xy+yz+zx}}{1+xy+yz+zx} $$

[quote=sqing]Let $x,y,z$ be nonnegative real numbers such that $xy+yz+zx=1$. Prove that
\[\frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} \geq \frac{5}{2}+\frac{xyz}{2(x+y)(y+z)(z+x)}.\]
[/quote]
Let $x,y,z$ be nonnegative real numbers such that $xy+yz+zx=1$. [url=https://artofproblemsolving.com/community/c6h449840p3617601]Prove that[/url]
\[\frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} \geq \frac{5+xyz}{2}.\]


[quote="mathcrazed"]1. A license plate consists of 6 digits 0-9 and the beginning digit may be 0.  Each license plate must be different in at least two spots to be considered different.  What is the largest possible number of license plates?

2. In acute triangle ABC, the altitude from C passes through points P and Q on the circle with diameter AB.  The altitude from B passes through points R and S on the circle with diameter AC.  Prove that P, Q, R, and S lie on a circle.[/quote]
Number two is a USAMO problem.  Look here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356630#p356630

The second is also a USAMO problem, and mountain.dew's solution is incorrect.

That's exactly what I thought also.. Do you mean the first?

After I posted it, I immediately got #2  :wink:  However, I'm still having trouble on number 1... I believe that mountain.dew's approach is correct, but there a ton of overcounts  :wink:

[quote="the future"][quote="mountain.dew"][hide="1"]The answer is just $ 10^6$ minus the ones where only 1 digit is different. So, the answer is $ 10^6 \minus{} 9\cdot6 \equal{} 999946$.[/hide]
edit: this is an example of an incorrect solution :D[/quote]
I think you have to multiply a $ 10^{5}$ to the second one because the second one will have $ 10^{5}$ different digits for the 5 places that are the same.
(like $ 10^6 \minus{} 9\cdot6\cdot10^{5}$)

However, then there would be some serious undercounts because after some initial subtractions there would be license plates that wont differ by 1 digit to another license plate but we've counted it as though the license plate it differs by 1 digit is still there (instead of being subtracted)-yea I ddint understand what I just said either...

Basically, my answer would be less than the real answer because lets say there are 3 license plates

ABCDEF
ABCDEH
AQCDEH

if we do what I did, then ABCDEH would not belong becuase there is a ABCDEF
Also, AQCDEH would not belong because there is a ABCDEH, but since ABCDEH is already subtracted, then AQCDEH would now belong....

If I would guess I would guess $ 10^{6} \minus{} 6*9*10^{5} \plus{} 5*10^{5}$
becuase there are 6 letters and for everyone plate that differs in 1 place is subtracted, 5 others that differed in the other 5 places from the subtracted plate that orignially wouldnt work would now work (since there is no longer a plate that now differs in 1 place from that plate)

Umm actually forget that answer...because its less than 0...

But I think I am on the right track[/quote]
I thought of that too, but then the answer was negative.. maybe I'll post something if I make progress on it.

Yep, it's a negative answer if you do that.  I also noticed that for every license plate, there are 54 other license plates which can be changed in one way to get to that license plate.  I don't know how, but I think it's possible to find the number of overcounts this way ~_-.

[quote="mathcrazed"]Well yes. But exactly how... It's kinda hard.[/quote]

you could also try using recursion?


Let $ a,b,c$ be posittive real numbers, such that $ a\plus{}b\plus{}c \ge \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}$
Prove that $ a\plus{}b\plus{}c \ge \frac{3}{a\plus{}b\plus{}c}\plus{}\frac{2}{abc}$

[quote="Zuko"]Let $ a,b,c$ be posittive real numbers, such that $ a \plus{} b \plus{} c \ge \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$
Prove that $ a \plus{} b \plus{} c \ge \frac {3}{a \plus{} b \plus{} c} \plus{} \frac {2}{abc}$[/quote]

Using supposition we have

$ a \plus{} b \plus{} c\ge \frac1{3}(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \plus{} \frac2{3}(a \plus{} b \plus{} c)\ge\frac {3}{a \plus{} b \plus{} c} \plus{} \frac2{3}(a \plus{} b \plus{} c)$

We only need prove that  $ a \plus{} b \plus{} c\ge\frac3{abc}$ <=> $ abc(a \plus{} b \plus{} c)\ge 3$

or $ xy \plus{} yz \plus{} zx\ge3$ for $ x \equal{} bc,y \equal{} ca,z \equal{} ab$.

We have  $ a \plus{} b \plus{} c \ge \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$<=>$ xy \plus{} yz \plus{} zx\ge x \plus{} y \plus{} z$

=>$ x \plus{} y \plus{} z \le xy \plus{} yz \plus{} zx\le\frac1{3}(x \plus{} y \plus{} z)^2\le\frac1{3}(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)$

=>$ 3\le xy \plus{} yz \plus{} zx$
done!

[quote="BaBaK Ghalebi"][quote="dduclam"]
We only need prove that  $ a + b + c\ge\frac3{abc}$ <=> $ abc(a + b + c)\ge 3$
[/quote]
there is no need to use substitutions to complete this,just note that:

\begin{eqnarray*}a + b + c\geq \frac 1a + \frac 1b + \frac 1c\Rightarrow (a + b + c)^2\geq (\frac 1a + \frac 1b + \frac 1c)^2\geq 3(\frac1{ab} + \frac1{bc} + \frac1{ca}) = \frac3{abc}(a + b + c)\end{eqnarray*}

thus:

$ a + b + c\geq \frac3{abc}$[/quote]
Oh,the solution is very nice because the problems is easy^^


the fulcrum of a lever is located at the 60 cm mark. the meter stick weighs 135g. how much force would it take to hold the mter stick parallel to the ground from the 100 cm mark? assume there is no friction.

[hide="i have gotten not so far"]i am getting that the 40 cm on either side of the 60 cm mark cancel out but different parts of the stick affect the force differently, so i need to find the amount of force the extra 20 cm exerts on the lever. i would think its just the amount exerted at the 10 cm point *20 but its probably not linear

this prkobably hasnt made any sense so just ignore it

i was wondering if it had anything to do with center of gravity[/hide]

just balance the torque about the fulcrum -->

the whole mass is concentrated at 50 cm mark ie at a distance 0.1 m from the fulcrum and this weight provides anticlockwise moment =

0.135g * 0.1

clockwise moment required to balance anticlockwise moment = F * (1 - 0.6) = F * 0.4

now anticlockwise moment = clockwise moment

hence

F*0.4 = 0.135g*0.1

taking g = 10

F = 0.3375 N

omg i actually got that.

idk if i have asked this before but is the density of water .98 of 1 g per $ cm^23$


What are the conditions for $ a,b,c$?

Is this an inequality problem? :maybe:

[quote="Eldest"]Find all pairs of integer $ m$ and $ n$, both greater than $ 1$, such that the following equality
$ \frac {a^{m \plus{} n} \plus{} b^{m \plus{} n} \plus{} c^{m \plus{} n}}{m \plus{} n} \equal{} \frac {a^{m} \plus{} b^{m} \plus{} c^{m}}{m} \equal{} \frac {a^{n} \plus{} b^{n} \plus{} c^{n}}{n}$[/quote]

It's a problem in [b]PROBLEMS IN THIS ISSUE[/b], Mathematics and Youth Magazine, Vietnam. So please lock it  :wink:


induction
limit
calculus
calculus computations

Suppose $ 0<a<b$. Define $ a_0\equal{}a, b_0\equal{}b$, and

$ a_{n\plus{}1}\equal{}\sqrt{a_n*b_n}, b_{n\plus{}1}\equal{}(a_n\plus{}b_n)/2$

for every $ n\in \mathbb{N}$. Show that $ (a_n)$ is an increasing sequence, that $ (b_n)$ is a decreasing sequence and that both converge to the same limit.

induction: $ a_n < b_n$

$ k\equal{}0: a_0 < b_0$ done

assume for $ k\equal{}n$, wts for $ k\equal{}n\plus{}1$:

$ 0<\frac{1}{2}(\sqrt a_n \minus{} \sqrt b_n )^2\equal{}\frac{a_n\plus{}b_n}{2}\minus{}\sqrt {a_n b_n }\equal{}b_{n\plus{}1} \minus{}a_{n\plus{}1}  \to a_{n\plus{}1} < b_{n\plus{}1}$

therefore $ a_n < b_n$

$ a_{n\plus{}1} \equal{} \sqrt {a_n b_n} > \sqrt {a_n a_n} \equal{} a_n$

$ b_{n\plus{}1} \equal{} \frac{a_n\plus{}b_n}{2} < \frac {b_n \plus{}b_n}{2} \equal{} b_n$

so now $ a_n$ increasing, $ b_n$ decreasing as desired

$ \lim a_n \equal{} a, \lim b_n \equal{} b \to a\equal{}\sqrt{ab}, b \equal{}\frac{a\plus{}b}{2}$

$ \to a\equal{}b$

This problem is convergence of AGM.

[url=http://en.wikipedia.org/wiki/Arithmetic-geometric_mean]Arithmetic-geometric mean[/url] - Wikipedia
[url=http://mathworld.wolfram.com/Arithmetic-GeometricMean.html]Arithmetic-Geometric Mean[/url] - MathWorld
[url=http://planetmath.org/encyclopedia/ArithmeticGeometricMean.html]arithmetic-geometric mean[/url] - PlanetMath

[url=http://planetmath.org/encyclopedia/ConvergenceOfArithmeticGeometricMean.html]convergence of arithmetic-geometric mean[/url] - PlanetMath


function
real analysis
real analysis solved

Let $c$ denote the continuum cardinality. Show that there are $2^c$ functions $f:\mathbb R\to\mathbb R$ which have the Darboux property (or the intermediate value property, if it sounds more familiar :)).

It's from [url=http://www.princeton.edu/~mathclub/puzzles4.html]here[/url]. Problem $4$ on that page.

I have invented a quite tangled construction  :?

Consider Hammel basis $H$ of $\mathbb R$ over $\mathbb Q$, $1\in H$. We know that $|H|=c$. Suppose $\{a_\alpha\}_\alpha\in \{1,2\}^c$, $\alpha\in\mathbb R$, $a_\alpha\in \{1,2\}$.

Define $f_a(0)=0$ and $f_a(x)=\sum_{i=1}^{k-1} a_{\alpha_i}x_{\alpha_i}h_{\alpha_{i}}$, where $x=\sum_{i=1}^{k} x_{\alpha_i}h_{\alpha_i}$, $x_i\in\mathbb Q$, $h_{\alpha_i}\in H$, $\alpha_1<...<\alpha_k$.

We see that $f_a(x)$ takes all real values on each segment $[u,v]$.

Oh, I realized: maybe Grobber asks to prove that there are exactly $2^c$ such functions?

Proving that there are at least that many functions suffices, of course. I think my construction was even more tangled, but I also used a Hamel basis and I also used it to construct a function which takes all real values on each interval. It's a cute problem, but I knew it wouldn't hang around for a long time... :)


I clicked on Stats this morning to check on my rankings and it looked like this.

Karma
Yours: 28 of 5035
21 7 15: 679 of 5035

Points
Yours: 60 of 5035
21 7 15: 1413 of 5035

Alcumus Score [?]
Yours: 4695 of 5035
21 7 15: 4478 of 5035

What does "21 7 15: 679 of 5035" mean?

21 7 15 is an AoPS user.  Apparently, he friended you, allowing you to see his stats.


Ok Mafia Game 8 is now starting.  The players particpating are as follows:

LadyKn1ght -confirmed
jli -confirmed
math92 -confirmed
dangvh -confirmed
Super Naterul -confirmed
h_s_potter2002 -confirmed
SuperNova -confirmed
CLee -confirmed
Silverfalcon
yif man12 -confirmed
Fuzzeh_Panda -confirmed
baseballfreak2 -confirmed
ambierona -confirmed
Lucky707 -confirmed
legolas -confirmed

There will be 4 mafia, 3 masons, 1 vigilante, 1 prosecutor, 1 miller, 3 citizens, 1 inspector, 1 doctor.

If you don't know what your jobs are go to mafia sign up to read about it.

PMs will be sent shortly.

elvenchamp777

all pms have been sent.  Please pm me back or post in this thread to confirm that you recieved the pm and are playing.

Here

I accuse whats-his-face for being the serial killer.

The game isn't starting til most people confirm their pm.

Are we going to hang someone, or go to sleep first?

I meant that if baseball is mafia, then someone else is miller, so if we lynch other people we always have the chance of killing the real miller and finding out about baseball. So killing someone other than baseball could give us more info than killing baseball. I'm hoping that if Fuzzeh_panda isn't mafia, then he's miller, because that would tell us that baseball is mafia.

Basically, I think we should keep lynching people other than baseball until one of them turns out to be miller or mafia. If they are mafia, we win. If they are miller, we lynch baseball.

[quote="zeb"]I'm hoping that if Fuzzeh_panda isn't mafia, then he's miller, because that would tell us that baseball is mafia.[/quote]

Wow that was extremely suspicous but I know you're a mason. If anyone else had said that I would be extremely suspicious.

Reason? If Fuzzeh_panda is mafia, it's already over.

Should I just tell everyone who I am so the game can move on? I mean, elven is taking his sweet time in calling me dead. I am dead, right?

[quote="zeb"]I'm only planning ahead for if Fuzzeh_Panda isn't mafia. There's no point in planning ahead for the case where Fuzzeh_Panda is mafia. What is so suspicious about that? Did you think I was talking about the case where Fuzzeh_Panda is mafia?[/quote]

I was saying it sounds suspicious because you HOPE fuzzeh_panda isn't mafia.

sorry, i went to the beach and got all sunburnt everyone :D heh... anyways...  fuzzeh_panda your DEAD and so are the rest of the mafia members cause he was the lone mafia member!

The town wins!

darkknight was citzen
h_s_potter2002 was prosecutor
xdl615x was citzen
baseballfreak2 was miller (that almost screwed you up :D)
ambierona and zeb were mason.

Woot I got the roles right!! Being a mason helped.. Yay town!

I did not mean for a pattern to appear so sorry if that messed up the game...

I just got a little peeved, it's my fault for being that mean...But only ambie and lucky saw it, and they stupidly had to go and tell everyone...grr... :P

The night LadyKn1ght was killed, I got all the roles EXCEPT inspector... It was either LadyKn1ght or SuperNova.  Darn...  Not that it would have made much of a difference, I guess.


[quote="pkothari13"]Nice! Just one quick question:

How can you go from $a_{1}+a_{2}+...+a_{n}= \frac{n(n+1)}{2}$?  Did you just assume it held true and stuck with it, or what?[/quote]

he used strong induction

Well it holds true for the first few terms...so I just assumed


Αποδειξτε οτι δεν υπαρχει πολυωνυμο P(n) με ακεραιους συντελεστες ωστε για καθε φυσικο n η τιμη του πολυωνυμου να ειναι πρωτος. :wink:

Mallon tha ennoeis mi stathero poliwnimo P(n) etsi? Dioti ean parw p.x. $P(n)=7, \ \ \forall \ \ x\in\mathbb{N}$ tote nomizw eimai Ok.

Apo ekei kai pera thewreis to $0$ stous fisikous 'h oxi?

Ean nai tote ean ipothesoume oti iparxei tetoio (mi stathero) poliwnimo, epeidi prepei $P(0)=$ [b]prwtos[/b], ara o statheros oros $a_{0}$ einai diaforos tou miden kai malista prwtos arithmos ara $a_{0}> 1$. Sinepws, $P(a_{0})=(\cdots+1)\cdot a_{0}=$[b]πολλ.[/b]$a_{0}$ ara to $P(n)$ einai sinthetos gia $n=a_{0}$ (O arithmos stin parenthesi einai $>1$ dioti afenos oi sintelestes tou poliwnimou einai ap'to $\mathbb{N}$ kai afeterou dioti den einai stathero poliwnimo kai sinepws iparxei enas akomi oros megaliterou vathmou apo tou midenikou tou opoiou o sintelestis einai diaforos tou miden).

Alexandros

http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html[url]
υπαρχει προβλημα με το λινκ .ειναι παντως γνωστο αποτελεσμα. αποδυκνυεται με τη χρηση ενος οριου που περιεχει την ακολουθια των πρωτων. μπορει να το δωσει καποιος ?[/url]

Καταρχην συγνωμη που δεν απαντησα αμεσως στις απορειες σας. :) Πραγματι ξεχασα να σας πω οτι το πολυωνυμο ειναι μη σταθερο και δεχεται τιμες >=1.DEGP(n)>o :wink:

Εχει κανεις καμια ιδεα? :lol:

[quote="fotis"][quote="cretanman"]Mallon tha ennoeis mi stathero poliwnimo P(n) etsi? Dioti ean parw p.x. $P(n)=7, \ \ \forall \ \ x\in\mathbb{N}$ tote nomizw eimai Ok.
[/quote]

Νομίζω ότι δεν εννοεί μόνο σταθερά πολυώνυμα γιατί κάπου έχω διαβάσει ότι η συγκεκριμένη πρόταση ισχύει για όλα τα πολυώνυμα.
πχ. Το πολυώνυμο [b]Q(x) = x^2 - 79x + 1601 [/b] δίνει πρώτους αριθμούς για x = 0 , 1 , 2 ........,79  αλλά όχι για x = 80.[/quote]

Σωστά! Επίσης το παράδειγμα που έδωσε ο Fotis είναι πολύ καλό για να το αποδείξετε. Υπάρχει και στη Θεωρία Αριθμών από την Ε.Μ.Ε.

Από εκεί μπορεί να το θυμάμαι και εγώ....Δεν είμαι σίγουρος

[quote="fotis"][quote="cretanman"]Mallon tha ennoeis mi stathero poliwnimo P(n) etsi? Dioti ean parw p.x. $P(n)=7, \ \ \forall \ \ x\in\mathbb{N}$ tote nomizw eimai Ok.
[/quote]

Νομίζω ότι δεν εννοεί μόνο σταθερά πολυώνυμα γιατί κάπου έχω διαβάσει ότι η συγκεκριμένη πρόταση ισχύει για όλα τα πολυώνυμα.
πχ. Το πολυώνυμο [b]Q(x) = x^2 - 79x + 1601 [/b] δίνει πρώτους αριθμούς για x = 0 , 1 , 2 ........,79  αλλά όχι για x = 80.[/quote]
μια παρατηρηση--> αν το P(n) δινει πρωτους το P(n-p) δινει και αυτο πρωτους αριθμους. αν θυμιθεις το πολυωνυμο  του Ουλερ n^2+n+41 και θεσεις n--->n-40 παιρνεις το πολυωνυμο αυτο αρα εχεις πρωτους για 40+40=80 διαδοχικες τιμες

Αφού το $P(n)$ παράγει πρώτους για κάθε $n \in \mathbb{N}^{*}$ τότε $P(1) = p \in \mathbb{P}$. Έυκολα αποδεικνύεται ότι το $P(p+1)$ θα είναι πολλαπλάσιο του $p$.

[quote="Αρχιμήδης 6"]Αν καποιοςαπο σας εχει αποδειξη παρακαλω ας την γραψη.Θα σας ειμαι ευγνωμων![/quote]

Nomizw oti apantisa amesws meta tin erwtisi sou.

Alexandros

(Cretanman δεν καταλαβα τι ακριβως εγραψες στο μηνυμα.Αν μπορεις πες μου γιατι δεν καταλαβα.)Ναι δεν διαφωνω cmad!Θα μου επιτρεψεις ομως να σου θεσω αλλιως το ερωτημα ωστε να δω αν καταλαβες την κεντρικη ιδεα.
Αν λοιπον το n επαιρνε τιμες μεγαλυτερες η ισες του 2007 πως θα το προσεγγιζες? :)

Estw oti iparxei tetoio poliwnimo $P(x)=a_{n}x^{n}+\cdots+a_{1}x+a_{0}$ me $n\geq 1$ kai $a_{n}\neq 0$ kai $a_{i}\in \mathbb{N}^{,}\ i=0,1,\ldots,n$.

Tote $P(0)=a_{0}$ ara $a_{0}=$prwtos ara $a_{0}\geq 1$ $P(a_{0})=a_{0}(a_{n}a_{0}^{n-1}+\cdots+a_{1})$ = sinthetos, atopo

Ara den iparxei poliwnimo pou na ikanopoiei tin dosmeni sinthiki.

Alexandros

Αλεξανδρε στην εκφωνηση του προβληματος πραγματι δεν ειχα βαλλει καποιον περιορισμο στον n αλλα μετα το διορθωσα και ειπα οτι πρεπει n>=1.Επειδη ομως ο cmad το ελυσε εθεσα μεγαλυτερο περιορισμο για n>=2007.Περιμενω την αποψη σου.... :)

Για να μη μου το γενικέψεις ξανά :lol: δε χρησιμοποιώ το 2007, αλλα έναν αυθαίρετα μεγάλο φυσικό Μ.

Κατ' αρχήν παρατηρούμε ότι για το $P$ πρέπει να ισχύει $\lim P(n) =+\infty$.

Πάρε έναν οποιονδήποτε ακέραιο $x > M$. Τότε $P(x) = p \in \mathbb{P}$. Αλλά αλλά και το $P(p+x)$ είναι πολλαπλάσιο του $p$ (και $p+x > x > M$). Και γενικά το $P(\alpha p+x)$ είναι πολλαπλάσιο του $p$ (και $\alpha \geq 1$). Για να σιγουρευτούμε ότι το $P(\alpha p+x)$ δεν ισούται με $p$ (έτσι θα ήταν πρώτος), παίρνουμε $\alpha \rightarrow+\infty$, οπότε το $P(\alpha p+x) \rightarrow+\infty$.

(Αυτό με το $\alpha \rightarrow+\infty$ έλειπε, κακώς, από την ιδέα μου με το $x = 1$)

στο mathworld λεει για ενα διαγωνισμο prime generating polynomials
http://www.recmath.org/contest/

Σωστα cmad.Αυτο ακριβως ηθελα. :)

Den boroume apla na poume oti $ P(2n)$ den einai prwtos $ \forall  n \in Z$ ?

Οχι φιλε basted.Εστω το πολυωνυμο $ P(X) \equal{} X^2 \plus{} 1$ .Προφανως δεν ισχυει οτι για καθε αρτιο $ X$ ο $ P(X)$ πρωτος... :)

Afto akribws elega! Gia ka8e polywnymo $ P(X)$ yparxei ena polywnymo $ P(2X)$.  Alla den to eixa skeftei ka8olou kala. Skefthka oti gia ka8e $ P(X)$ to diplasio 8a diaireitai me to 2. Alla profanws ekana la8os afou yparxoun st polywnyma oi sta8eroi oroi...


trigonometry inequalities geometry circumcircle function symmetry search

How is this obvious? Here is a solution to that inequality which I've seen before.
Since the circumcenter is the intersection of the perpendicular bisectors of a triangle, for any acute triangle, it will lie within the medial triangle (triangle whose vertices are the midpoints of the sides of ABC). Let M be the midpoint of AB, N the midpoint of AC, and O the circumcenter of ABC. Then we obviously have $ BM\plus{}MN\plus{}NC > BO\plus{}OC$, or $ \frac{a}{2}\plus{}\frac{b}{2}\plus{}\frac{c}{2} > 2R$, or $ a\plus{}b\plus{}c >4R$.

[quote="dgreenb801"]How is this obvious? Here is a solution to that inequality which I've seen before.
Since the circumcenter is the intersection of the perpendicular bisectors of a triangle, for any acute triangle, it will lie within the medial triangle (triangle whose vertices are the midpoints of the sides of ABC). Let M be the midpoint of AB, N the midpoint of AC, and O the circumcenter of ABC. Then we obviously have $ BM \plus{} MN \plus{} NC > BO \plus{} OC$, or $ \frac {a}{2} \plus{} \frac {b}{2} \plus{} \frac {c}{2} > 2R$, or $ a \plus{} b \plus{} c > 4R$.[/quote]

WOW! indeed very nice and clear proof. thanks...

[quote="dgreenb801"]How is this obvious? Here is a solution to that inequality which I've seen before.
Since the circumcenter is the intersection of the perpendicular bisectors of a triangle, for any acute triangle, it will lie within the medial triangle (triangle whose vertices are the midpoints of the sides of ABC). Let M be the midpoint of AB, N the midpoint of AC, and O the circumcenter of ABC. Then we obviously have $ BM \plus{} MN \plus{} NC > BO \plus{} OC$, or $ \frac {a}{2} \plus{} \frac {b}{2} \plus{} \frac {c}{2} > 2R$, or $ a \plus{} b \plus{} c > 4R$.[/quote]

that's precisely how it's obvious :)

Wow. I forgot to use the triangle inequality. What are good inequalities to know for geometry inequality problems in general?

There are a lot. Do you have the Art of Problem Solving's "Introduction to Geometry"? It has a lot of useful stuff in it. If you don't you should get it xD.

[quote="CornFlakes"]

Prove that sinA + sinB + sinC > 2[/quote]
Can we have a stronger inquality that this?

No because equality hold when with have a triangle with two 90 degree angles and one 0 degree angle (theoretically).

[quote="CornFlakes"][quote="Tiger100"]Can we have a stronger inquality that this?[/quote]Equality (almost) happens in the case of a really narrow isosceles triangle, so the sum of the sines can be arbitrarily close to 2.[/quote]

indeedably so, me lad oh me lad!

[quote="pluristiq"]the law of sines turns the inequality into $ a\plus{}b\plus{}c>4R$, where $ R$ is the circumradius, and this is pretty obvious by the triangle inequality.[/quote]
By the Problem 68 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1894316#p1894316]here[/url] and using $r_a + r_b + r_c = 4R + r$, it really proves to be a very weak inequality. :)

[quote="CornFlakes"]None of the angles in a triangle ABC exceeds 90 degrees.

Prove that sinA + sinB + sinC > 2

I've done it, but there are probably better ways.  What do you think?[/quote]

acute $\triangle ABC$, prove that $\sin A + \sin B + \sin C > 2$.

first I will prove that \[
\sin A + \sin B > 1 + \cos C .
\]
it's equivalent to

$2\sin \frac{{A + B}}{2}\cos \frac{{A - B}}{2} > 2\cos ^2 \frac{C}{2}$
$\Leftrightarrow \sin \frac{{A + B}}{2}\cos \frac{{A - B}}{2} > \sin ^2 \frac{{A + B}}{2}$
$\Leftrightarrow \cos \frac{{A - B}}{2} > \sin \frac{{A + B}}{2}$
$\Leftrightarrow 2\sin \frac{{\pi  - 2A}}{4}\sin \frac{{\pi  - 2B}}{4} > 0.$

obvious true.

so we get \[
\sin A + \sin B + \sin C > 1 + \sin C + \cos C = 1 + \sqrt 2 \sin \left( {C + \frac{\pi }{4}} \right),
\]
but $C + \frac{\pi }{4} \in \left( {\frac{\pi }{4},\frac{{3\pi }}{4}} \right) \Rightarrow \sin \left( {C + \frac{\pi }{4}} \right) > \frac{1}{\sqrt 2} $, so\[
\sin A + \sin B + \sin C > 2.
\]
done.

[quote="Merlinaeus"]Not exactly a rigorous proof, but thinking of the graph of sin(x) and using the fact that
$sin(x) \ge x/90$
where x is in degrees, and 0<x<90 gives:
$ sin(A)+sin(B)+sin(C) \ge (A+B+C)/90 = 180/90 = 2$

Merlin[/quote]

slick argument! though keep in mind that if you use degrees the graph will be different. in terms of radians, the inequality you want is $\sin x > \frac{2x}{\pi}$, which is true because $\sin x > x$ for $0 < x < \frac{\pi}{2}$

[quote="Potla"][quote="CornFlakes"]None of the angles in a triangle ABC exceeds 90 degrees.

Prove that sinA + sinB + sinC > 2
[/quote]
Let $x=\tan\dfrac A2, y=\tan\dfrac B2, z=\tan\dfrac C2$ such that $xy+yz+zx=1$ and rewrite this into
\[\frac{2x}{1+x^2}+\frac{2y}{1+y^2}+\frac{2z}{1+z^2}>2,\]
Which, on using $1+x^2=xy+yz+zx+x^2=(x+y)(z+x),$ rewrites into
\[\frac{x(y+z)+y(z+x)+z(x+y)}{(x+y)(y+z)(z+x)}>1,\]
Which is again equivalent with  $ \ 2 \ + \ xyz \ > \ x+y+z.$
But, note that since $\tan$ is an increasing function in $\left(0,\dfrac{\pi}{4}\right)$ we get $0<x,y,z<1.$ We consider two cases.
[b]Case I.[/b] $\boxed{x+y\ge 1}$
So $(1-x)(1-y)(1+z)>0$ leads to
$1+xy-yz-zx-x-y+z+xyz>0,$ implying
$xyz+2>x+y-z+z(x+y)-xy+1,$
Or, $xyz+2>x+y+z+2z[(x+y)-1]\ge x+y+z.$
Hence we are done when $x+y\ge 1.$
[b]Case II.[/b] $\boxed{x+y<1}$
Note that in this case, $1=xy+z(x+y)<xy+z\implies z<xyz+z^2;$
And $x+y<1$. Adding these two leads to our desired result, ie $x+y+z<xyz+z^2+1<xyz+2$, since $x,y,z<1.$
Our proof is complete. $\Box$
:)[/quote]
so interesting proof, Potla, but I think the prove can be more simpler because
$2+xyz-x-y-z=(1-yz)(1-x)+(1-y)(1-z)>0$
 :D

[quote="kuing"][quote="Potla"][quote="CornFlakes"]None of the angles in a triangle ABC exceeds 90 degrees.

Prove that sinA + sinB + sinC > 2
[/quote]
Let $x=\tan\dfrac A2, y=\tan\dfrac B2, z=\tan\dfrac C2$ such that $xy+yz+zx=1$ and rewrite this into
\[\frac{2x}{1+x^2}+\frac{2y}{1+y^2}+\frac{2z}{1+z^2}>2,\]
Which is again equivalent with  $ \ 2 \ + \ xyz \ > \ x+y+z.$
[/quote]
so interesting proof, Potla, but I think the prove can be more simpler because
$2+xyz-x-y-z=(1-yz)(1-x)+(1-y)(1-z)>0$
 :D[/quote]
Well, here's how I did it :
Let $f(x,y,z) = 2+xyz-x-y-z$.
Now, since $y,z \le 1 \implies yz \le 1$, therefore \[\frac{\partial f}{\partial x}  = yz-1 \le 0\]
Therefore, $f$ is concave (No need to check for $y,z$ due to symmetry), and thus by weierstrass theorem, $f$ assumes its minimum at its endpoints.
Thus, Finally, (Let $x \ge y \ge z$ WLOG)
\[min(f(x,y,z))= min(f(1,1,1),f(1,1,0),f(1,0,0),f(0,0,0)) = 2\], completing our proof ;)

[quote="pluristiq"] in terms of radians, the inequality you want is $\sin x > \frac{2x}{\pi}$, which is true because $\sin x > x$ for $0 < x < \frac{\pi}{2}$[/quote]
Well, the argument you gave is false, because $1 = \sin \frac{\pi}{2} > \sin 1^{c}  $, as sin is increasing. ($1^{c} = 1$ radian.)
In fact the result $\sin x > \frac{2x}{\pi}$ is true nevertheless, and is better known as 'Jordan's Inequality' :)

[quote="Rijul saini"]
Well, here's how I did it :
Let $f(x,y,z) = 2+xyz-x-y-z$.
Now, since $y,z \le 1 \implies yz \le 1$, therefore \[\frac{\partial f}{\partial x}  = yz-1 \le 0\]
Therefore, $f$ is concave (No need to check for $y,z$ due to symmetry), and thus by weierstrass theorem, $f$ assumes its minimum at its endpoints.
Thus, Finally, (Let $x \ge y \ge z$ WLOG)
\[min(f(x,y,z))= min(f(1,1,1),f(1,1,0),f(1,0,0),f(0,0,0)) = 2\], completing our proof ;)[/quote]
Sorry, I messed it up. Actually, all I wanted to write was that the function was linear in $x$ and thus assumes its minimum when $x$ assumes its extremes. And after that its simple.

[quote="hsbhatt"][hide="Karamata Inequality"]Since $\left(\frac{\pi}{2}, \frac{\pi}{2},0 \right) \succ (A,B,C)$ and $\sin x$ is concave in $[0, \pi]$ by Karamata Inequality, $\sin A + \sin B + \sin C \ge 2$[/hide][/quote]
[quote="hsbhatt"][hide="Karamata Inequality"]
Can better  explain this property
and give one or more examples related to this inequality

[quote="ayush_2008"]here Hsbhatt sir has used KARAMATA INEQUALITY.will some will define it .where it can be  used ?[/quote]
Google search reveals a lot of relevant results, you should try searching in google or AoPS. For details on how to use the AoPS Search function, look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=376080]here.[/url] A search for the keyword "karamata" leads to our desired search results. :)
However, the first few google results:
1. [url=http://en.wikipedia.org/wiki/Karamata's_inequality]Wikipedia[/url]
2. [url=http://elib.mi.sanu.ac.rs/files/journals/tm/14/tm813.pdf]Applications of Karamata, Schur, Muirhead etc.[/url]
3. [url=http://kheavan.files.wordpress.com/2010/06/karamata.pdf]Generalisation of Karamata Inequality.[/url]

:)


floor function
number theory
prime numbers
number theory unsolved

For every positive number $x$,let $a(x)$ denote the number of prime numbers not exceeding $x$ and for every positive integer $m$ ,let $b(m)$ denote the number of prime divisors of $m$.Prove that for every positive integer $n$, we have:
$a(n)+a(\frac{n}{2})+...+a(\frac{n}{n})=b(1)+b(2)+...+b(n)$

Just look at how often you count the prime number $p$:
On the LHS, it is counted as often as $\frac{n}{k} \geq p$ for $k=1,2,3,...$.
On the RHS, it is counted $\left\lfloor \frac{n}{p} \right\rfloor$ times.
But thats clearly the same.


find all functions $f : R\rightarrow R$ such that
$f(x^2-f(y))=1-x-y^2$ for all $x,y\in R$

$x \rightarrow -x$ :  $1+x-y^2 = 1-x - y^2$ $\forall x, y$

this is a major problem ...


inequalities
calculus
integration
function
inequalities proposed

Let $ f: [0;1]\rightarrow R$ has $ f"$ is a continue function and $ f"(x)>0\ \forall x\in[0;1]$. Prove that: $ 2\int_0^1 f(t)dt\geq 3\int_0^1 f(t^2)dt\minus{}f(0)$

My solution:
$ 2\int_0^1 f(t)dt\geq 3\int_0^1 f(t^2)dt \minus{} f(0)$
$ \Leftrightarrow 2\int_0^1 [f(t)\minus{}f(0)]dt\geq 3\int_0^1 [f(t^2) \minus{} f(0)]dt$
$ \Leftrightarrow 2\int_0^1 [f(t)\minus{}f(0)]dt\geq 3\int_0^1 [f(t) \minus{} f(0)]d\sqrt{t}$
$ \Leftrightarrow 2[f(t)\minus{}f(0)]t\Big|_0^1\minus{}2\int_0^1tf'(t)dt\geq 3[f(t)\minus{}f(0)]\sqrt{t}\Big|_0^1\minus{}3\int_0^1\sqrt{t}f'(t)dt$
$ \Leftrightarrow \minus{}f(1)\plus{}f(0)\plus{}\int_0^1(3\sqrt{t}\minus{}2t)f'(t)dt\geq 0$
$ \Leftrightarrow \int_0^1(3\sqrt{t}\minus{}2t\minus{}1)f'(t)dt\geq 0$ (Because $ f(1)\minus{}f(0)\equal{}\int_0^1f'(t)dt$)
$ \Leftrightarrow \int_0^1(1\minus{}\sqrt{t})(2\sqrt{t}\minus{}1)[f'(t)\minus{}f'(\frac14)]dt\geq 0$ (Because $ \int_0^1(3\sqrt{t}\minus{}2t\minus{}1).f'(\frac14)dt\equal{}0$)
Now the last inequality is true because $ (1\minus{}\sqrt{t})(2\sqrt{t}\minus{}1)[f'(t)\minus{}f'(\frac14)]\geq 0$


Putnam
abstract algebra
group theory
logarithms
college contests

Is there a finite abelian group $ G$ such that the product of the orders of all its elements is $ 2^{2009}?$

No, although I won't swear there isn't a nonabelian group.

First, for the product of orders to be a power of $ 2$, the order of each element must be a power of $ 2$, and the whole group has $ 2^n$ elements for some $ n$.

Let $ f(k)$ be the number of elements of order $ 2^k$. We have $ \sum_{k\equal{}0}^n f(k)\equal{}2^n$ by counting group elements and $ \sum_{k\equal{}0}^n kf(k)\equal{}2009$ from the product condition. By considering each cyclic subgroup of order $ 2^k$, $ 2^{k\minus{}1}$ divides $ f(k)$ for each $ k$. Therefore, $ \sum_{k\equal{}0}^n kf(k)\equiv 0\plus{}f(1)\plus{}2f(2)\equiv f(1)\mod 4$, and $ f(1)\equiv 2009\equiv 1\mod 4$.
Next, a useful fact: in an abelian group, the elements of order dividing $ m$ form a subgroup. In this case, that means that the elements of order $ 1,2$ form a subgroup, and the cardinality of this subgroup must be a power of $ 2$. Since that cardinality is $ 1\plus{}f(1)\equiv 1\plus{}1\equiv 2\mod 4$, this power of $ 2$ must be $ 2$ itself, and there is only one element of order $ 2$.
If an abelian 2-group is not cyclic, it decomposes as a product of two smaller groups. Each of those groups has at least one element of order $ 2$, so the full group has more than one element of order $ 2$. That doesn't happen in our case, so $ G$ is cyclic.
Finally, we test these counts on cyclic groups. The product of the orders of the elements in the cyclic group of order $ 256$ is $ 2^{1793}$, and the product in the cyclic group of order $ 512$ is $ 4097$. $ 2^{2009}$ lies between these, and is therefore not a possible value.

my conclusion was pretty much the same, but my argument was that abelian groups have generators (which I'm told is not true).  So idk what claiming this had to be cyclic will do to any score I may get.

So i said that if the products of the orders of the elements is 2^2009, then each element's order has to be of the form 2^k.

So really, we are looking at Z mod 2^k.  Simply calculating what the desired result is leads to Z mod 256 giving 2^1793 and Z mod 512 giving 2^4097.  Since 2^2009 is between these, it is impossible.  But it seems showing its cyclic is the crucial part.

Nice observation, jmerry!  I didn't make the corresponding observation until much later, so I ended up having to do more work:

By the structure theorem, there exist positive integers $ n_1 \ge ... \ge n_k$ such that $ G$ decomposes as the product of cyclic groups of order $ 2^{n_k}$.  Let $ a_n$ denote the number of elements of order dividing $ 2^n$; any such element is a direct product of elements of order dividing $ 2^n$, so

$ \log_2 a_n \equal{} \sum_{i \equal{} 1}^{k} \text{min}(n, n_i)$.

Let $ \lambda_j$ denote the number of indices $ i$ such that $ n_i \ge j$ (the conjugate partition); then I claim that $ \log_2 a_n \equal{} \lambda_1 \plus{} ... \plus{} \lambda_n$.  Moreover $ \lambda_1 \ge \lambda_2 \ge ... \ge \lambda_m$ where $ m \equal{} \text{max}(n_i)$ and we can recover uniquely the partition $ n_i$ given $ \lambda_i$, so forget about the $ n_i$.  Let $ b_n$ denote the number of elements of order exactly $ 2^n$.  Then $ b_n \equal{} a_n \minus{} a_{n \minus{} 1}$, so the product $ P$ of the orders of the elements of $ G$ satisfies

$ \log_2 P \equal{} \sum_{i \equal{} 1}^{m} i b_i \equal{} ma_m \minus{} a_{m \minus{} 1} \minus{} a_{m \minus{} 2} \minus{} ... \minus{} a_0$.

Suppose this is equal to $ 2009$.  One readily computes that we must have $ a_0 \equal{} 1, a_1 \equal{} 2$, but since this implies $ \lambda_1 \equal{} 1$ we then have $ \lambda_i \equal{} 1$ for all $ i$ and $ a_i \equal{} 2^i$.  (This is equivalent to jmerry's observation that $ G$ is cyclic.)  This gives $ m \ge 9$, but now it's easy to check that this is impossible $ \bmod 64$.

Someone I know did not see the word abelian and proceeded to tackle the more difficult challenge of constructing such a group. I hope his construction is posted in the official solutions so that his effort was not a complete waste.

Well, if he wants it to go up on the AMC-hosted solutions (which are not official), he'll have to put it somewhere Kiran can see it.

Hmm- I think I'll see if I can do it, too.

By the Fundamental Theorem of abelian groups, we can see this group is

$\mathbb{Z}_{2^{a_1}} \times \mathbb{Z}_{2^{a_2}} \times \cdots \times \mathbb{Z}_{2^{a_k}}$.

where $a_i\in \mathbb{N}$ for $1\le i\le k$

Consider a k-tuplet $(b_1,\cdots,b_k)$. The order of this k-tuplet is the lcm of order of $b_i$s in $\mathbb{Z}_{2^{a_i}}$, which is simply their max.

In $\mathbb{Z}_{2^t}$ the number of elements with order dividing $2^c$ is $2^{\min\{t,c\}}$. Therefore, there are $2^{\sum\limits_{i=1}^k \min\{a_i,c\}}$ elements with order dividing $2^c$. There are exactly $2^{\sum\limits_{i=1}^k \min\{a_i,c\}}-2^{\sum\limits_{i=1}^k \min\{a_i,c-1\}}$ elements with order exactly $2^c$.

Let $s_c=2^{\sum\limits_{i=1}^k \min\{a_i,c\}}$ be defined for $c\le \max a_i$. Note $s_c$ is strictly increasing. The product of our orders is $\prod\limits_{c=1}^{\max a_i} 2^{c(s_c-s_{c-1})}$

log 2 on both sides give $2009=\sum\limits_{c=1}^{\max a_i} \sum\limits_{i=1}^k cs_c-cs_{c-1}$

For $0\le c\le \max a_i-1,$ I added $cs_c$ but subtracted $(c+1)s_c$, so they contribute a total of $-s_c$. Thus, $2009=\max a_i 2^{\sum a_i} - \sum\limits_{c=0}^{\max a_i-1} s_c$

Note $s_0=2^0, s_1=2^k$, so $2^k\mid 2008=\max a_i 2^{\sum a_i} - \sum\limits_{c=1}^{\max a_i-1} s_c$

Thus, $k\le 3$. If $k<3$ then $s_1=s_2=k$, which implies $a_i=1$ which obviously fails. If $k=3$ then note $s_2\in \{3,4\}$. If $s_2=3$, then all $a_i$'s are equal to 1, impossible. Otherwise, two of the $a_i$'s are 1, say $a_1=a_2=1, $. We can see $s_c=c+2$ we have $2000=a_3 2^{a_3+2} - 2^{a_3+2} + 16$, which yields no solutions.


[quote="Michael Niland"]X and Y are real numbers such that

7(x^2) +3XY +3(Y^2) =1

Prove that the least possible value of [(X^2) +(Y^2)]/Y is 1/2[/quote]

Should $x$ and $y$ both be non-negative?  Otherwise, you can take $x = -1/5$, $y = -2/5$, to get $(x^2 + y^2)/y = -1/2 < 1/2$.

We have $7x^2 + 3xy + 3y^2 = 1$ and we need to prove that $2(x^2 + y^2) \geq y$.
This is equivalent to proving that \[4(x^2 + y^2)^2 \geq y^2(7x^2 + 3xy + 3y^2)\] or that \[4x^4 + 8x^2y^2 + 4y^4 \geq 7x^2y^2 + 3xy^3 + 3y^4\] or \[y^4 + x^2y^2 + 4x^4 \geq 3xy^3.\] But from AM GM it follows that \[\frac{y^4}{4} + \frac{y^4}{4} + \frac{y^4}{4} + \frac{y^4}{4} + x^2y^2 + 4x^4 \geq 3xy^3\] and we are done.

[quote="Arne"]We have $7x^2 + 3xy + 3y^2 = 1$ and we need to prove that $2(x^2 + y^2) \geq y$.
This is equivalent to proving that \[4(x^2 + y^2)^2 \geq y^2(7x^2 + 3xy + 3y^2)\][/quote]

2>-4 but 4<16

Of course I am working with positive reals...

[quote="gunes"][quote="blahblahblah"]You need numbers satisfying the condition, so your counterexample is meaningless.[/quote]
who told you " y is positive"?
its meaningful i think.[/quote]
well, concerning $a > b$, a is always positive, and b is not sure.
If b is negative, then it's true.
If b is positive, we can square both sides to see if it's still true. So there is no problem with Arne's solution.


Let S be the midpoint of side TH of sqaure MATH. Point P is inside the sqaure such that angle PMA= angle PAT= angle APS = x. find x.

[hide]
Let B be the midpoint of MA, and let Q be the midpoint of BS, or the center of the square.

$ \angle PMA \equal{} \angle PAT \equal{} x$ and $ \angle PAT \plus{} \angle PAM \equal{} 90^\circ$ so $ \angle APM \equal{} 90^\circ$.
So the circumcircle of $ \triangle PMA$ is centered at B.

Next, $ \angle PMA \equal{} x$ so obviously $ \stackrel{\frown}{PA} \equal{} 2x$. But $ \angle APS \equal{} x$, so PS must be tangent to the circle. Then $ \angle BPS \equal{} 90^\circ$, and the circumcircle of $ \triangle BPS$ is centered at Q.

Therefore, if the side of the square is 2 units, then BP = QP = 1. So obviously $ \triangle BPQ$ is equilateral, and $ \triangle MBP$ is isosceles, and $ \angle MBP \equal{} 30^\circ$, so $ x \equal{} 75^\circ$.
[/hide]


We need to be careful here, so as to not count a domino twice. Notice we can't overcount by $ 2$ either, since the doubles wouldn't be compatible with the rest. Thus, we could only resort to careful casework:

One side 0: $ 7$ dominoes (other side 0,1,2,3,4,5,6)
One side 1: $ 6$ more dominoes (other side 1,2,3,4,5,6; 0 already counted above)
One side 2: $ 5$ more dominoes (see above for explanation)
One side 3: $ 4$ more
One side 4: $ 3$ more
One side 5: $ 2$ more
One side 6: just $ 1$ extra

Adding them up, $ 1\plus{}2\plus{}3\plus{}4\plus{}5\plus{}6\plus{}7\equal{}\frac{7(7\plus{}1)}{2}\equal{}7(4)\equal{}\fbox{28}$.

Alternatively:

There are 7 doubles. Not considering doubles, there are 49 ordered pairs (x,y) 0<=x,y<=6. Add 7 to compensate for doubles being counted only once, and halve. (49+7=56, 56/2=28, our answer)


geometry circumcircle incenter trigonometry function inequalities inradius

The side lengths of the obtuse triangle $ABC$ are three consecutive odd integers and $11\cos \angle B=13\cos \angle C.$ Prove that $OI=\frac{2}{3}\sqrt{ab}$ ($O$ is the circumcenter and $I$ is the incenter of the triangle).

Since $\cos{\widehat B}, \cos{\widehat C}$ have the same sign and the cosine of the single obtuse angles is negative, the angle $\angle A > 90^o$ is obtuse and the opposite side $a = BC$ is the longest one. The angles $0 < \angle B, \angle C <90^o$ are both acute. Since $\frac{\cos{\widehat B}}{\cos{\widehat C}} = \frac{13}{11} > 1$ and cosine is a decreasing function for in $(0, 180^o)$, $\angle B < \angle C$ and $b = CA < c = AB$.
Let $b = 2n + 1, c = 2n + 3, a = 2n + 5$ be the consecutive odd integer triangle sides. Clearly, $n > 0$, otherwise it would be $a > b + c$, i.e, the triangle inequality would not hold. Since the triangle $\triangle ABC$ is obtuse, by Pythagorean theorem.

$a^2 > b^2 + c^2$

$(2n + 5)^2 > (2n + 3)^2 + (2n + 1)^2$

$4n^2 + 20n + 25 > 4n^2 + 12n + 9 + 4n^2 + 4n + 1$

$4n^2 - 4n - 15 < 0$

$\frac{4 - \sqrt{16 + 240}}{8} < n < \frac{4 + \sqrt{16 + 240}}{8}$

$-\frac 3 2 < n < \frac 5 2$

Thus we have only 2 possibilities, $n = 1$ or $n = 2$, i.e., $b = 3, c = 5, a = 7$ or $b = 5, c = 7, a = 9$. Calculating $\cos{\widehat B}, \cos{\widehat C}$, the 1st possibility $n = 1$ leads to

$\cos{\widehat B} = \frac{c^2 + a^2 - b^2}{2ca} = \frac{25 + 49 - 9}{70} = \frac{65}{70} = \frac{13}{14}$

$\cos{\widehat C} = \frac{a^2 + b^2 - c^2}{2ab} = \frac{49 + 9 - 25}{42} = \frac{33}{42} = \frac{11}{14}$

$\frac{\cos{\widehat B}}{\cos{\widehat C}} = \frac{13}{11}$

as required, while the 2nd possibility $n = 2$ leads to

$\cos{\widehat B} = \frac{c^2 + a^2 - b^2}{2ca} = \frac{49 + 81 - 25}{126} = \frac{105}{126} = \frac 5 6$

$\cos{\widehat C} = \frac{a^2 + b^2 - c^2}{2ab} = \frac{81 + 25 - 49}{90} = \frac{57}{90} = \frac{19}{30}$

$\frac{\cos{\widehat B}}{\cos{\widehat C}} = \frac{25}{19} \neq \frac{13}{11}$

Let $s = \frac{3 + 5 + 7}{2} = \frac{15}{2}$ be the triangle semiperimeter. Calculating the triangle area $\triangle$, inradius $r$ and circumradius $R$:

$\triangle = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{\frac{15}{2} \cdot \frac 1 2 \cdot \frac 9 2 \cdot \frac 5 2} = \frac{15 \sqrt 3}{4}$

$r = \frac{\triangle}{s} = \frac{\sqrt 3}{2}$

$R = \frac{abc}{4 \triangle} = \frac{105}{15 \sqrt 3} = \frac{7 \sqrt 3}{3}$

Using the relation between the circumradius, inradius and the distance $d = OI$,

$R^2 - d^2 = 2rR,\ \ d^2 = R(R - 2r) = \frac{7 \sqrt 3}{3} \left(\frac{7 \sqrt 3}{3} - \sqrt 3\right) = \frac{28}{3}$

$d = OI = 2 \sqrt{\frac 7 3} = \frac 2 3 \sqrt{21} = \frac 2 3 \sqrt{ab}$


geometry
circumcircle
parallelogram
cyclic quadrilateral
radical axis
power of a point
geometry proposed

Let $ ABCD$ be a cyclic quadrilateral. Diagonals $ AC$ and $ BD$ meet at $ E$. Let $ P$ be a point inside the quadrilateral. Let $ O_{1}, O_{2}, O_{3}$ and $ O_{4}$ be circumcenters of triangles $ ABP, BCP, CDP$, and $ DAP$, respectively. Prove that lines $ O_{1}O_{3}, O_{2}O_{4}$, and $ OE$ are concurrent.

Some previous problems deal with special cases when $ P$ coincides with $ E$ or $ O$:
http://www.mathlinks.ro/Forum/viewtopic.php?t=97487
http://www.mathlinks.ro/Forum/viewtopic.php?t=80818

Hey Yufei,

This problem is nice. To prove this problem, we need the following two lemmas

[b]Lemma 1 [/b] Let $ XYZT$ be a convex cyclic quadrilateral with circumcenter $ O$. Then $ O$ is the orthocenter of the triangle whose vertices are the intersections of opposite sides $ XY$ and $ ZT$, $ XT$ and $ YZ$, $ XZ$ and $ YT$.

This lemma is well-known, so I won't prove it.

[b] Lemma 2 [/b] Let $ PXY$ be a triangle and $ M$ a point in $ XY$. Denote by $ U$ the second intersection of the circumcircle $ PMX$ and $ YP$ and $ V$ the second intersection of the circumcircle $ PMY$ and $ XP$. If $ O'$ is the circumcircle of $ PUV$, then $ O'M$ is perpendicular to $ XY$.

[hide="Proof"] This is an application of Carnot's theorem. Let $ O_{1}$ and $ O_{2}$ be the circumcenters of $ PMX$ and $ PMY$. Then to prove this lemma, it is sufficient to show that the perpendiculars from $ M$, $ O_{1}$, $ O_{2}$ to $ XY$, $ PY$, $ PX$ respectively are concurrent. Now apply Carnot's theorem. [/hide]

Back to the problem, let $ X$ be the intersection of $ AD$ and $ BC$, $ Y$ be the intersection of $ AB$ and $ CD$. Call $ M$ the projection of $ O$ on $ XY$. It is not hard to see that we may ignore all the points $ A, B, C, D, O, E$ by lemma 1 and reduce the problem to lemma 2. Notice that $ U$, $ V$ in lemma 2 correspond to the intersections of the circumcircles of $ PAB$ and $ PCD$, $ PAD$ and $ PBC$ in the problem, and $ O'$ corresponds to the circumcircle of $ PUV$, which is the intersection of $ O_{1}O_{3}$ and $ O_{2}O_{4}$.

$ O_{1}O_{4}\ \text{and}\O_{2}O_{3}\perp AC$ and $ O_{1}O_{2},O_{3}O_{4}\ \perp BD$ hence $ O_{1}O_{2}O_{3}O_{4}$ is parallelogram.hecne $ O_{1}O_{3}$ and $ O_{2}O_{4}$ bisect each other.

$ \angle O_{2}EC=90^{\circ}-\angle DBC=90^{\circ}-\angle CAD\ \Longrightarrow EO_{2}\perp AD$ and $ OO_{4}\perp BC$
similar this we have $ O_{4}E\perp BC\ ,\ OO_{2}\perp BC$ so also $ EO_{2}OO_{4}$ is parallelogram,Now we are done.

[quote="Amir.S"]$ O_{1}O_{4}\ \text{and}\O_{2}O_{3}\perp AC$ and $ O_{1}O_{2},O_{3}O_{4}\ \perp BD$ hence $ O_{1}O_{2}O_{3}O_{4}$ is parallelogram.hecne $ O_{1}O_{3}$ and $ O_{2}O_{4}$ bisect each other.[/quote]

Why? This is not true.

My solution:

Let $ T=AB \cap CD, S=AD \cap BC $ .
Let $ X=\odot (O_1) \cap \odot (O_3), Y=\odot (O_2) \cap \odot (O_4), Z=O_1O_3 \cap O_2O_4 $ $ (X, Y \neq P) $ .

Easy to see $ Z $ is the center of $ \odot (PXY) $ .

Since $ T $ is the radical center of $ \{ \odot (O), \odot (O_1), \odot (O_3) \} $ ,
so it also lie on the radical axis $ PX $ of $ \{ \odot (O_1), \odot (O_3) \} $ .
Similarly, we can prove $ S \in PY $ .

Since $ \odot (PXY), \odot (O_1), \odot (O_3) $ have common radical axis $ PX $ ,
so $ T $ has equal power WRT $ \odot (PXY) $ and $ \odot (O) $ .
Similarly, $ S $ has equal power WRT $ \odot (PXY) $ and $ \odot (O) $ ,
so $ TS $ is the radical axis of $ \{ \odot (PXY), \odot (O) \} $ ,
hence we get $ OZ \perp TS $ . ... $ (1) $

Since $ TS $ is the polar of $ E $ WRT $ \odot (O) $ ,
so we get $ OE \perp TS $ . ... $ (2) $

From $ (1), (2) \Longrightarrow Z \in OE $ .

Q.E.D


The inhabitants of the island of Jumble use the standard Kobish alphabet (20 letters, A through T).  Each word in their language is 4 letters or less, and for some reason, they insist that all words contain the letter A at least once.  How many words are possible?

There are 20^4+20^3+20^2+20 possible words (excluding the A rule).  This is because there are 20 letters for 1-letter words, 20*20 for 2 letter words, etc.
This equals 168420.
Then, subtract 19^4+19^3+19^2+19 because those dont have A in them.
168420-137560=30,860 possible words.

This thing suck... i used permutation and i got way off....

[quote=firemonkey17]Don't quite understand Alcumus solution. Could someone explain? :([/quote]

Well Jongy's solution is the best

We want to find how many possible words there are and subtract how many letter that don't contain an A in them.
Either that or we could bash out a ton of cases

[quote=firemonkey17]Don't quite understand Alcumus solution. Could someone explain? :([/quote]

[hide = Solution] For this problem we want to use something called [b]"Complementary Counting"[/b] : Count all the cases you don't want then subtract it out from all the cases with no restriction!
[b]The restrictions in this problem are[/b] : "Each word in their language is 4 letters or less" & "All words contain the letter A at least once." -- The number 1 rule in counting problems is to deal with the restriction first! Now that we have a good idea about the problem we can start breaking it up into $4$ small cases!
[b]Case 1 :[/b] The word has 4 letters : (All the cases with no restriction) - (Cases you don't want) $\Longrightarrow$ $(20*20*20*20) - (19*19*19*19) = 29,679$
[b]Case 2 :[/b] The word has 3 letters : (All the cases with no restriction) - (Cases you don't want) $\Longrightarrow$ $(20*20*20) - (19*19*19) = 1,141$
[b]Case 3 :[/b] The word has 2 letters : (All the cases with no restriction) - (Cases you don't want) $\Longrightarrow$ $(20*20) - (19*19) = 39$
[b]Case 4 :[/b] The word has 1 letters : (All the cases with no restriction) - (Cases you don't want) $\Longrightarrow$ $(20) - (19) = 1$
Adding all these cases together we get : $29,679 + 1,141 + 39 + 1 = \fbox{30,860}$
 [/hide]
Hope this helped!  :)

I tried a way in which I made a case by the number of letters, and also how many a's. However, I got kind of off from the answer. I dont understand why. Here are my cases:
[hide=My wrong Solution please help me figure out what I did wrong]
Case 1: 1 letter
- 1A: [b]1[/b]
Case 2: 2 letters
- 1A: 2*19=[b]38[/b]
- 2A: [b]1[/b]
Case 3: 3 letters
- 1A: 3*19*19=[b]1083[/b]
- 2A: 3*2*19=[b]57[/b]
- 3A: [b]1[/b]
Case 4: 4 letters
- 1A: 4*19^3=[b]27436[/b]
- 2A: 4*3*19^2=[b]4332[/b]
- 3A: 4*3*2*19= [b]456[/b]
- 4A: [b]1[/b]
Then, I calculated the total of the bolded words to get $33406$
[/hide]

why are you multiplying by 4*3? it should be 6*19^2 because you are trying to find the number of ways where the two A's can be. So that is where you made the mistake. Hope you understood.  :blush:

Oh, thanks! I did 4*3 instead of 4 choose 3.  :blush:

Is it really necessary to calculate $19^4$?

[quote] Each word in their language is $4$ letters [b]or less[/b][/quote]

Wait what did I do wrong.
[hide]There is only 1 one letter word, then there are 40 two letter words since we can put the A in one of two places and then we have twenty options so it's 2*20, then similarly three letter words are 3*20^2 and four letters are 4*20^3. So adding it up its 1+41+1200+32000=33241. What am I over counting?[/hide]

The problem with @above's approach is that because you are using all 20 words in the 4 * 20^3, you count the same thing multiple times in different situations.

I did that at first too lol :p

Well actually, calculating all of that isn't too difficult, because 20^4 - 19^4, we can use the difference of squares. So it isn't really that difficult if you look at it that way.


The expression

$ \sqrt{104\sqrt{6}\plus{}468\sqrt{10}\plus{}144\sqrt{15}\plus{}2006}$

can be expressed in the form $ a \sqrt{2}\plus{}b\sqrt{3}\plus{}c\sqrt{5}$. Determine the value of the integer product $ abc$

Heh, I did this problem today.
[hide="outline"] Set the $ a\sqrt{2}\plus{}b\sqrt{3}\plus{}c\sqrt{5}$ equal to the radical, square, and compare like terms. You should then get four equations that involve a,b,c from which you can find what they are.[/hide]

...so we have $ 2a^2\plus{}3b^2\plus{}5c^2\plus{}2ab\sqrt6\plus{}2bc\sqrt{15}\plus{}2ca\sqrt{10}$, which gives us

$ 2a^2\plus{}3b^2\plus{}5c^2\equal{}2006$
$ ab\equal{}52\equal{}2^2\cdot13$
$ bc\equal{}72\equal{}2^3\cdot3^2$
$ ca\equal{}2\cdot3^2\cdot13$

Multiplying the last three and taking the square root, we find that

$ a^2b^2c^2\equal{}2^6\cdot3^4\cdot13^2\implies abc\equal{}\boxed{936}$.


1. yes
2. no

[hide="2"]There's the problem that some blocks are indistinguishable. So we divide 9! by the number of ways to arrange those. So $\frac{9!}{4!*3!*2!}=1260$[/hide]

[quote="chuchen"]Ten people have volunteered to help clean up the park.  Two people are chosen to pick up trash.  Two people are then chosen to plant flowers.  Two people are then chosen to mow the grass.  The rest of the people wash the playground equipment.  How many ways can this set of decisions be made?
Is the answer  10C2 * 8C2 *6C2 * 1

Kimmy the Kindergartener has 9 blocks.  Four are red, three are blue, and two are red.  In how many ways can she line up the blocks if you can not tell the difference between two blocks of the same color?  Is this answer 9![/quote]
Wait, did you mean that four are [b]red[/b], three are blue, and two are [b]red[/b]?


quadratics
trigonometry
advanced fields
advanced fields unsolved

Prove that [arccos((√5 - 1)/2)]/pi is irrational.

EDIT:  There's another (probably easier) way that does not use these advanced technique, but just some common knowledge of trigonometry.  :wink:

[quote="mathwizarddude"]Prove that $ \frac {\arccos(\frac {\sqrt {5} \minus{} 1}{2})}{\pi}$ is irrational.[/quote]

Dear Mathwizarddude,

The following theorem was proved by Jogn Jahnel:

If $ x$ is a quadratic irrationality and $ \frac {\arccos{x}}{\pi}$ is rational then either $ x \equal{} \pm\frac {\sqrt {2}}{2}$ or $ x \equal{} \pm\frac {\sqrt {3}}{2}$ or $ x \equal{} \pm\frac {\sqrt {5}}{4}\pm\frac {1}{4}$.

See page 4 of his work [url=http://www.uni-math.gwdg.de/jahnel/Preprints/cos.pdf]"When is the (co)sine of a rational angle equal to a rational number?"[/url]

Sergey Markelov
markelov@mccme.ru


group theory
abstract algebra
superior algebra
superior algebra unsolved

How many Sylow 5-subgroups of $ S_{5}$ are there?

The number of p-sylowgroups in $ S_p$ is exactly $ (p\minus{}2)!$. This can easily be seen if you consider the structur of these groups: Because they all have order p, they must be cyclic and be generated by an element of order p. This must be a p-cycle. One can choose this cycle s.t. it has the form (12abc...). There is exactly one cycle of this form in each p-sylowgroup and there are (p-2)! cycles of this form.

Thank you for the reply.

I understand your reasoning, but why isn't the cycle of the form (1abcd) so that there are (p-1)! different subgroups?

That gives $ (p\minus{}1)!$ total elements of order $ p$, but note that each subgroup of order $ p$ has $ p\minus{}1$ elements of order $ p$, and distinct subgroups of order $ p$ intersect in the identity. If you count this way, you get $ \frac{(p\minus{}1)!}{p\minus{}1}\equal{}(p\minus{}2)!$.

The above post says that there's a unique $ p$-cycle $ \alpha$ in each subgroup such that $ \alpha (1)\equal{}2$, and there are $ (p\minus{}2)!$ such $ \alpha$, that gives $ (p\minus{}2)!$ directly.


Find the largest $ 85$-digit integer which has property: the sum of its digits equals to the product of its digits.

Let $ S \equal{} x_1 \plus{} x_2... \plus{} x_{85} \equal{} x_1x_2...x_{85}$ and $ 9\ge x_1\ge x_2\ge...\ge x_{85}$.
Because $ 2^7 > 7*2 \plus{} (85 \minus{} 7)$ we get $ x_7 \equal{} x_8 \equal{} ... \equal{} x_{85} \equal{} 1$. Therefore $ 85 < S < 133$.
If $ x_6 > 1$, then $ x_2 \equal{} x_6$ - not solution, Therefore $ x_6 \equal{} x_{85} \equal{} 1$ and $ S\le 125$,
$ S \equal{} x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \plus{} x_5 \plus{} 80 \equal{} x_1x_2x_3x_4x_5\le 125$. If $ x_5 > 1$, then $ x_5 \equal{} 2,x_4 \equal{} 2$, because $ 3 \plus{} 3 \plus{} 3 \plus{} 2 \plus{} 2 \plus{} 80 \equal{} 93 < 3^32^2 \equal{} 108$ we get $ x_5 \equal{} x_4 \equal{} x_3 \equal{} 2$ and $ x_1x_2 \equal{} 12,x_1 \plus{} x_2 \equal{} 10$ had not solution. Therefore $ x_5 \equal{} 1$ and $ x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \equal{} x_1x_2x_3x_4 \minus{} 81$ and $ 85 < S\le 117$.
If $ x_1 \equal{} 9$, then $ 9 < R \equal{} x_2x_3x_4x_5\le 13$. Because $ R\not \equal{} 11,13$ we get $ R \equal{} 10$ or $ R \equal{} 12$.
If $ R \equal{} 10$ $ x_2 \equal{} 5,x_3 \equal{} 2,x_4 \equal{} 1$ - not solution. If $ R \equal{} 12$, then or $ x_2 \equal{} 4,x_3 \equal{} 3,x_4 \equal{} 1$ or $ x_2 \equal{} 3,x_3 \equal{} x_4 \equal{} 2$ and $ x_2 \plus{} x_3 \plus{} x_4 < 9*12 \minus{} 9 \minus{} 81 \equal{} 18$, Therefore $ x_1\le 8$ and $ S\le 120$. If $ x_1 \equal{} 8$ $ R \equal{} x_2x_3x_4 \equal{} 12 \ or 14$  and $ x_2 \plus{} x_3 \plus{} x_4 \equal{} 8*(R \minus{} 11) \minus{} 1$. $ R \equal{} 14$ is not solution, therefore $ x_2 \plus{} x_3 \plus{} x_4 \equal{} 7,x_2x_3x_4 \equal{} 12$ had not solution.
Therefore $ x_1\le 7$ and $ S\le 109$. Because $ x_3 > 1$ and $ x_1 \equal{} x_2 \equal{} x_3$ is not solution and all prime divisors of s less than 10 we get $ 90\le S\le 108$ If $ x_1 \equal{} 7$, then $ x_2x_3x_4 \equal{} 14\to x_2 \equal{} 7,x_3 \equal{} 2,x_4 \equal{} 1$. It give maximal solution:
$ 7721111...111$.
There are only 2 solutions, second $ 644111...111$.

[quote="Rust"]Let $ S \equal{} x_1 \plus{} x_2... \plus{} x_{85} \equal{} x_1x_2...x_{85}$ and $ 9\ge x_1\ge x_2\ge...\ge x_{85}$.

[color=red]Because $ 2^7 > 7*2 \plus{} (85 \minus{} 7)$ we get $ x_7 \equal{} x_8 \equal{} ... \equal{} x_{85} \equal{} 1$. Therefore $ 85 < S < 133$.[/color][/quote]

i dont get this...could you explain..? :)

ok. thats what i did..
it was very quick-written.. thanks very much..:)

[hide=My Solution]

After writing the Python program and trying out the problem myself, I can confirm that jgnr is right -- the answer is $\boxed{8322\underbrace{111 \cdots 1}_{81\text{ ones}}}$.

For the Python program, I used Rust's method to prove that $a_8, a_9, \cdots a_{85} = 1$ to narrow down the searching the computer needs.  It would take a much shorter time for the program to check the numbers.  I'd recommend using Idle to run the program (it still takes a while!).
[code]
number = [1,1,1,1,1,1,1]
for i in range(0,8888889): #Number of iterations
   number[6] += 1 #Adds one to a_7
   for j in range(0,6): #Regroups if necessary
      if number[6-j] == 10:
         number[6-j] = 0
         number[5-j] += 1

   s = 78 #Sum of numbers from a_8 to a_85
   for k in number: #Calculate sum
      s += k

   product = 1 #Product of numbers from a_8 to a_85
   for n in number: #Calculate product
      product *= n

   if s == product: #Checks if sum equals product
      print(number)
[/code]
For my written solution, I started like Rust, but then I went in all-out casework on the possible values of $a_1$.  Although it's a lot of casework, it demonstrates that there is no number greater than $\boxed{8322\underbrace{111 \cdots 1}_{81\text{ ones}}}$ that also works.  I wrote the solution at http://artofproblemsolving.com/wiki/index.php?title=2006_Indonesia_MO_Problems/Problem_8 .

[/hide]


Find all [tex]f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}[/tex] such that [tex](x+y)f(f(yf(x)))=(x^2)f(f(x)+f(y))[/tex].

I am not sure that my solution is totally correct  :lol:
We can easily show that : $\left( \forall x,y>0\right) :f(x)=f(y)\Rightarrow x=y$ (*)
take $x=0$ in the original equation we get : $\left( \forall y>0\right) :f(f(yf(0)))=0$ (1)
- Now suppose $f(0)>0$, hence by replacing $y$ by $1$ and by $f(0)$ respectivly in (1) we get : $f(f(f(0)))=f(f(f(0)^{2}))=0$
Since $f(f(0))>0$ and $f(f(0)^{2})>0$ by using (*) we get $f(f(0))=f(f(0)^{2})$
and again $f(0)>0$ and $f(0)^2>0$ and using (*) we get $f(0)^2=f(0)$
Since $f(0)>0$ then $f(0)=1$
replacing in (1) we get : $\left( \forall y>0\right) :f(f(y))=0$
Take $y=0$ and $x=1$ in the original equation we get $f(1)=f(f(1)+1)$
since $1>0$ and $f(1)+1>0$ using (*) we get $f(1)=0$
which means that $f(f(1))=f(0)$ so $0=1$. Impossible.
So $f(0)=0$
-Now by taking $y=0$ in the original equation we get: $\left( \forall x>0\right) :f(f(x))=0$
Suppose $\left( \exists x_{0}>0\right) :f(x_{0})>0$
By taking $x=x_0$ in the original equation we get $\left( \forall y>0\right) :f(f(x_{0})+f(y))=0=f(f(y))$
and one more time using (*) we get : $f(x_0)=0$ contradiction.
So $\left( \forall x>0\right) :f\left( x\right) =0$ and since $f(0)=0$
the only function is $f\left( x\right) \equiv 0$

How can you take $x=0$ in the original equation?  Since $f$ is only defined for positive $x$ I would assume the equation in question only holds for $x$ and $y$ positive.

Well as I know $\mathbb{R}^{+}$ or $\mathbb{R}_{+}$ is the set which contains all the positive reals including $0$, however $\mathbb{R}_{+}^{*}=\mathbb{R}_{+} - \{0\}$ is the set of all positive reals except $0$
and as it is said in the problem, the function $f$ is defined on $\mathbb{R}^{+}$
So it is possible to take $x=0$ in the functional equation.  :lol:

Well, R+ is the set of positive reals, So you can't take x=0.

[quote="erdos"]So you mean that $0$ is not a positive real ??
Sorry but maybe the notations are different from those we have in our programs  :lol:[/quote]
Of course 0 is not positive!
I think you have learnt long before 0 is not positive or negative.

In fact , what I have learnt is that $0$ is both positive and negative.  :blush:  :blush:
But now It is clear , thanks  :lol:

I have completely given up on this, and I really like to see a
solution for this one.

if by irmo tou mean iran mathematic olympiad then thats is not the correct form

the correct question is  $(x+y)f(yf(x))=(x^2)f(f(x)+f(y))$

it have been discussed before.

you can see it here:

http://www.mathlinks.ro/Forum/topic-35009.html

The problem is much simpler this way. Probably you are right.


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conics
ellipse
geometry
circumcircle
trigonometry
algebra
system of equations

Let $O$ be the midpoint of a chord $AB$ of an ellipse. Through $O$, we draw another chord $PQ$ of the ellipse. The tangents in $P,Q$ to the ellipse cut $AB$ in $S,T$ respectively. Show that $AS=BT$.

The problem conditions and the proposition are not affected by projecting the ellipse into a circle with a diameter identical with the ellipse main axis. Hence, it is sufficient to prove the proposition for a circle.

WLOG, assume that the points $A, P, B, Q$ follow on a circle $(C)$ in this order. The quadrilaterals $COPS$ and $COQT$ are both cyclic, the 1st one because the angles $\measuredangle COA \equiv COS = 90^o$ and $\measuredangle CPS = 90^o$ are both right, the 2nd one because the angles $\measuredangle COB \equiv COT = 90^o$ and $\measuredangle CQT = 90^o$ are both right. The circumcenters of these cyclic quadrilaterals are the midpoints $M, N$ of the segments $CS, CT$, respectively. Let $K \equiv PS \cap CO$ and $ H \equiv OS \cap CP$ be the intersections of these lines. The points $O, P$ are feet of altitudes $SO, CP$ of the triangle $\triangle CKS$ and their intersection $H$ the orthocenter of this triangle. It follows that the triangles $\triangle CKS \sim \triangle PKO$ are similar and the angles $\measuredangle CSP \equiv \measuredangle CSK = \measuredangle POK$ are equal. The angles $\measuredangle POK = \measuredangle COQ$ are vertical and the angles $\measuredangle COQ = \measuredangle CTQ$ are equal, because they span the same arc $CQ$ of the circle $(N)$. As a result, the right angle triangles $\triangle CPS \cong \triangle CQT$ are congruent by ASA, because $CP = CQ$ are radii of the circle $(C)$ and the angles $\measuredangle CSP = \measuredangle CTQ$ are equal. Consequently, their circumcircles $(M) \cong (N)$ are also congruent. These 2 circles intersecting at the points $C, O$ are symmetrical WRT the line $CO \perp AB$, hence $OS = OT$. Since $O$ is the midpoint of the chord $AB$, $OA = OB$ as well. As a result,

$AS = OS - OA = OT - OB = BT$.

As I expected, you posted first,yetti.

Here is my solution, not ellegant.

Let  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\ (a>0,b>0)$  be the equation of the ellipse and let  four points

$A,B,P,Q$ be $A(a\cos \alpha,\ b\sin \alpha),B(a\cos \beta,\ b\sin \beta),P(a\cos \gamma,\ b\sin \gamma),Q(a\cos  \delta,\ b\sin \delta)$,then the tangent at $P,Q$ are as follows.

The tangent at $P:\ b(\cos \gamma)x+a(\sin \gamma)y=ab$

The tangent at $Q:\ b(\cos \delta)x+a(\sin \delta)y=ab$

The line $AB$: $b(\sin \alpha-\sin \beta)x-a(\cos \alpha-\cos \beta)y-ab\sin (\alpha-\beta)=0$

The line $PQ$: $b(\sin \gamma-\sin \delta)x-a(\cos \gamma-\cos \delta)y-ab\sin (\gamma-\delta)=0$

Let $M$ be the midpoint of $AB$,

$M\left(\frac{a(\cos \alpha+\cos \beta)}{2},\ \frac{b(\sin \alpha+\sin \beta)}{2}\right)\in PQ$

$\Longleftrightarrow (\sin \gamma-\sin \delta)(\cos \alpha+\cos \beta)-(\cos \gamma-\cos \delta)(\sin \alpha+\sin \beta)-2\sin(\gamma-\delta)=0$
$\Longleftrightarrow \sin (\gamma-\alpha)-\sin (\beta-\gamma)+\sin (\delta-\alpha)+\sin(\beta-\delta)=2\sin (\gamma-\delta)\cdots\cdots\ (\heartsuit)$

Solving the system of equations between the line $AB$ and the tangent at $P$, we get the absissa of the point

$S:\ x_S=a\cdot \frac{\sin \alpha\cos (\gamma-\alpha)-\sin \alpha\cos (\beta-\gamma)-\sin \gamma\sin (\alpha- \beta)-\cos \alpha+\cos \beta}{\cos (\gamma-\alpha)-\cos (\beta-\gamma)}$.

Similarly $x_T=a\cdot \frac{\sin \alpha\cos (\delta-\alpha)-\sin \alpha\cos (\beta-\delta)-\sin \delta\sin (\alpha- \beta)-\cos \alpha+\cos \beta}{\cos (\delta-\alpha)-\cos (\beta-\delta)}$

Let $m$ be the slant of the line $AB$, then we have $\overline {AS}=\sqrt{1+m^2}|x_A-x_S|$.
Similarly,$ \overline {BT}=\sqrt{1+m^2}|x_B-x_T|$.Therefore using $(\heartsuit)$ we are to prove $|x_A-x_S|=|x_B-x_T|\Longleftrightarrow AS=BT$

kunny


Se stie ca daca $(m,n)=1$, atunci $\mathbb{Z}_m\times\mathbb{Z}_n$ e izomorf cu $\mathbb{Z}_{mn}$, unde grupurile sunt aditive. Intrebare: Ramane valabil si la inele? Intr-o problema de la Laurentiu Duican 2004, se zice direct ca da, din cauza lemei chieneze a resturilor. Mie insa nu mi se pare asa evident. Poate cineva justifica acest lucru?

Cred ca am reusit sa-i dau de cap :) Am gasit efectiv un izomorfism.

Găseşti şi [url=http://www.math.miami.edu/~ec/book/ch03.ps]aici[/url] o demonstraţie şi generalizarea (pag.50)


Relatively == percentile.

I tend to do better with harder aimes, with tons of geometry and number theory.

I do better with hard dones. I'm great at ruling things out.  :D

For me it really depends on geometry, if an AIME has lots of geometry problems then I do well (average score around a 3 or a 4) , if it mostly has number theory, I do horribly (around a 2 or a 3).

I'm amazing at number theory :D but suck at geometry :(

LOL, I'm "amazing" at geometry, but suck at number theory :P

easy! duh! :o but sometimes even though I suck at geometry I can do some of the harder ones by pounding it out.


MATHCOUNTS
inequalities
number theory
prime factorization

Here's a number theory question from Mathcounts:

For positive integer n such that n<10,000, the number n+2005 has exactly
21 positive factors. What is the sum of all the possible values of n?

Help!

[hide="Hint"] For every positive real number $ n$ with prime factorization $ n\equal{}{p_1}^{e_1}\times{p_2}^{e_2}\times{p_3}^{e_3}\times...\times{p_{i\minus{}1}}^{e_{i\minus{}1}}\times{p_i}^{e_i}$, the number of factors it has equals $ ({e_1}\plus{}1)\times({e_2}\plus{}1)\times({e_3}\plus{}1)\times...\times({e_{i\minus{}1}}\plus{}1)\times({e_i}\plus{}1)$.[/hide]

Let $ m \equal{} n \plus{} 2005 < 12005$. Since we want $ m$ to have 21 factors, it must be in the form $ x^2y^6$, for primes $ x$ and $ y$. (Note that $ x^{20}$ wouldn't work, because it's too big.) We know that $ n$ is a positive integer, so $ a\ge\sqrt{2005}< 45$ and $ a\le\sqrt{12005}< 110$. If $ a > 44$, then we have $ xy^3 < 110$, and so $ y \le 3$. If $ y$ was exactly $ 3$, then $ 44 < 27x < 110$, implying $ x \equal{} 2$. If $ y$ was instead $ 2$, then $ 45 < 8x < 110$, meaning $ x$ can equal $ 7, 11,$ and $ 13$.

This means $ a \equal{} 54, 56, 88,$ and $ 104$. $ n \equal{} a^2 \minus{} 2005$, so $ n \equal{} 911, 1131, 5739,$ and $ 8811$. Thus, their sum is $ \boxed{16592}$.

Just a variable that satisfies the restrictions in the inequalities I mentioned.


logarithms
real analysis
real analysis unsolved

Consider the sum $ \sum_{n \equal{} 1}^{\infty}\frac {1}{k^n n^k}$. Show that when k = 2 we have $ \sum_{n \equal{} 1}^{\infty}\frac {1}{2^n n^2} \equal{} \frac {\pi^2}{12} \minus{} \frac {1}{2}\log^2(2)$.

Can exact values for higher k be found?

[hide]I've done the k=2 case using the two series $ \sum_{n \equal{} 1}^{\infty}\frac {1}{n^2} \equal{} \frac {\pi^2}{6}$ and $ \sum_{n \equal{} 1}^{\infty}\frac {x^n}{n} \equal{} \minus{} \ln(1 \minus{} x)$. No ideas on the higher values of k though.[/hide]

What do you mean higher values of $ n$?  It's an infinite series so every natural $ n$ will be used.  Do you mean
$ \frac{1}{3^{n}n^{3}}$ in the sum?

Yes, sorry. The handwritten notes I was working off was using n's and k's. I think I fixed up the original post. I'll try and pay more attention next time :)

Well in the case that you have $ 1/n^{3}$, it's well known that it converges, but what it converges to, no one is sure.

as far as I know $ \zeta(3)$ is irrational :wink:


function
algebra
domain
trigonometry
algebra unsolved

Find all f such as :

$ f(x\plus{}y)\equal{}\frac{f(x)\plus{}f(y)}{1\minus{}f(x)f(y)}$
$ f: \mathbb{R}\minus{}\minus{}\minus{}>\mathbb{R}$
thanx.

[hide]Define $ g(x)$ such that $ f(x) \equal{} \tan(g(x))$ for all $ x \in \mathbb{R}$.

Then, the equation becomes: $ \tan(g(x \plus{} y)) \equal{} \frac {\tan(g(x)) \plus{} \tan(g(y))}{1 \minus{} \tan(g(x))\tan(g(y))} \equal{} \tan(g(x) \plus{} g(y))$.

Therefore, $ f(x) \equal{} \tan(g(x))$ is a solution for any function $ g(x)$ that is a solution of the Cauchy F.E.

If $ f(x)$ must be continuous, then the solutions are $ f(x) \equal{} \tan(kx)$ for some $ k \in \mathbb{R}$. [/hide]


geometry counting distinguishability Duke college probability

I'm not sure which category the problems fall into but i'm sure they are not advanced... so i'm posting it here...mods kindly let me know if it is not ok...

1. How many ways are there to choose 4 cards of different suits and different values from a deck of 52 Cards?
a.	13.12.11.10
b.	52C4
c.	13 ^ 4
d.	52.36.22.10

The answer given is 13.12.11.10 however i think it should be 52.36.22.10... can anyone help me crack the logic?

2.There are 5 Blue socks, 4 Red socks & 3 Green socks. In how many ways can 4 socks be selected?
a.	245
b.	120
c.	495
d.	60

This one i'm not sure how to solve it???

[hide="1."]
first, we need to choose 4 numbers out of 13, so we have $\binom{13}{4}= \frac{13!}{8!4!}$.  Then, label the cards A, B, C, D.  A can have 4 suits, B then have 3 suits, etc.  Therefore, multiply $\binom{13}{4}$ by $4!$ and after simplifying we get $A.$
[/hide]

One of the hardest thing to do with choosing is that one can unconsciously impose order on the objects chosen.  If you choose 52 as the first product, this is what you've done.  Always check your logic thoroughly so that you don't impose any unnecessary order.
[hide="2."]
let the number of blue socks be $x$, red be $y$, and green be $z$.  Then we have:
$x+y+z = 4$
with $0 \le x \le 5$, $0 \le y \le 4$, and $0 \le z \le 3$.
Then one can solve by simple case work since all the numbers are small.  a.k.a. if $z$ is 3, there "blah" solutions, if $z$ is 2, there's "blah" solutions, etc... then add them all up (I choose $z$ here because the upperbound is smaller).

One solves this general form by using inclusion-exclusion principle.  i.e. if one of the variable is greater than the upper bound, or two greater, three greater, etc.  But the generic formula is way too complicated and I forgot it...(I can derive it if you want)
[/hide]

Hi,

Thanks for the reply i'm gonna bother you more...

according to you

[13!/(4!8!)] x 4! is the answer but this will give me

13.12.11.10.9...???

or is the logic such that first i have 13 values to chose from then i have 12 values then 11 and finally 10???

As for the second one...whats the formula??? i can very do simple counting but what if the numbers where higher?

oops, I did the subtraction wrong...it suppose to be $9!$ below.

for the formula, it's VERY long (theoretically, but at some point all the rest of the terms are 0's) and it's WAY too complicated that I doubt anyone will know it by heart.  Just know how to derive it.

So, if the only restriction is non-negative integers, then by the balls in urn formula (or the bijection thing) the solution is $\binom{6}{2}$.  Then, if one of the variable (say $x$ first) is greater than 5, then we introduce a variable $x_{1}$ that : $(x_{1}+6)+y+z = 4$.  So the solution in this case will be $\binom{0}{2}$ (you see this term is actually 0).  if now $y$ is greater than 4, we again introduce a new variable $y_{1}$ that $x+(y_{1}+5)+z = 4$, solving we have the solution $\binom{1}{2}$.  Do the same thing for $z$ to get $\binom{2}{2}$.  Then, do the case that $x$ and $y$ are both greater than their bounds, then $y$ and $z$, then $x$ and $z$.  Finally, do the case where all three variables are greater than their bounds.

Now by PIE our answer is:

$\binom{6}{2}-(\binom{0}{2}+\binom{1}{2}+\binom{2}{2})+$ (the sum of all the cases where two variables are greater) $-$ (the sum of all the cases where three variable are greater) $+\dots$.

Obviously, this generalize to arbitrary numbers of variables and bounds.

[quote]The answer given is 13.12.11.10 however i think it should be 52.36.22.10... can anyone help me crack the logic?
[/quote]
You can think of this way, You have one spade, one heart, one diamond, and one club.
There are 13 choices for spade, now only 12 for heart, 11 for diamond and 10 for club so answer is ($13.12.11.10$)

(Another way, which I think you did, was to pick first card (52 choices), now pick second (36 choices only because there are 12 cards in each suit left, which can be picked up etc... so the answer would be $52.36.22.10$ but this answer has to be modified (because order does not matter - there are 24 way to arrange these cards) by $4!$)

Important point here, is not to forget the factor 4! .

hope this helps.
[hide=" for the second problem"]
For the second problem, if I am not missing something, none of the answer given is correct. Answer is $14$ (which is small enough to list all the cases) . Can you check the problem again.. thanks [/hide]

Thanks guys

as for the first problem my doubt is clarified... coming to the second one the given answer is 495... i'm unable to figure out how.... :o

one more requiring help(actually this is one of my weaker areas and so i have many more... :) )

3.There are 8 Consonants and 5 Vowels in a word jumble. In how many ways can we form 5 letter words having 3 consonants and 2 vowels?
a.	67,200
b.	8540
c.	720
d.	None of These

[hide="Remark on #2"]It seems that the answer of $495$ you are given is incorrect.  That would be true if there were $12$ [b]different[/b] colors, because then you would have \[\binom{12}{4}=495\] ways of choosing.  However, the answer is clearly less since you could choose the same sets multiple times due to repeated colors.[/hide]

[quote="fizsle"]3.There are 8 Consonants and 5 Vowels in a word jumble. In how many ways can we form 5 letter words having 3 consonants and 2 vowels?
a.	67,200
b.	8540
c.	720
d.	None of These[/quote]
[hide] There are $\binom{8}{3}$ ways to choose the consonants and $\binom{5}{2}$ ways to choose the vowels. Then, there are $5!$ ways to arrange these 5 letters. $\binom{8}{3}\binom{5}{2}5!=67200$[/hide]

IMHO, the wordings of these problems are  poor, and can cause misunderstanding. Unfortunately many books are written/edited poorly   and  some teachers   concentrate on formulas without really explaining the key points.  It is more important to understand the mathematics behind  these problems, than  to just   apply formulas.

(Of course, I am not picking on the person who put the problems here, rather on the book/teacher who originated the problems. And I will try to show how the problem could have been worded more clearly)

For example, the problem:

[quote]There are 8 Consonants and 5 Vowels in a word jumble. In how many ways can we form 5 letter words having 3 consonants and 2 vowels?
[/quote]
Does not specify if the consonants (and vowels) are all “distinct” or not. This is not a minor nit-pick, but really very important and  such a probelm    shows, (IMO) mathematical ignorance in the author’s part.

(Not to mention, “how many ways can we form” is pretty ambiguous. (I can   form a word, say “TO” by writing “T” followed by an “O” or putting “O” first and then putting a “T” before it .. are they counted as different ways to form?  Also does a sequence of letters “XQEAP” really count  as "word"?, if not, how does one count the “words”?

This problem could have been  more clear if stated as:
[quote]“There are 8 distinct consonants and 5 distinct vowels in a word jumble, How many different words can be formed  if one chooses 3 consonants and 2 vowels from the jumble to form a 5 letter word.  A word here is defined as any sequence of letters and need not be found in a dictionary”[/quote]

Then the answer given in one of the previous post  (67200) is correct.

(personally, instead of using  5 letter " word" I would have used "different licence plates" or something like that to be more clear)

Similarly  the other problem:

[quote].There are 5 Blue socks, 4 Red socks & 3 Green socks. In how many ways can 4 socks be selected?
(with answer  given as 495) .[/quote]

The key question here again is: “Are all blue (and red or green)  socks are identical or distinguishable?”. (I mean Tar heelers and  Blue Devils both have blue as their school color but Carolina blue is different than "real"  blue – at least that’s what my son, who goes to Duke, says :)   )

If the socks are distinguishable  (then there are 13 socks, and colors is just an irrelevant factor)  the answer is 495, if they are identical (except the difference in color)  then the answer is 14.

(Sorry for the long post, but I hope this is  helpful)

Well thanks for that... but unfortunately i dont have a way to question these authors.... not even a mail id...lol...how much i would love to debate answers with them.....

here is one more card problem....i'm confused....

8.From a pack of 52 cards, 3 cards are drawn at random. Find the probability of a king, a queen and a jack.
a.16/5525
b.1/13^3
c.1/14^3
d.1/15^3

the given solution is 16/5525

i get the answer as... 4/52 x 4/51 x 4/50      = 4/33150

Am i missing something in my logic??? also why not 1/13^3 ???
from the previous problem i can get the logic that each card has a probability of 1/13 and hence 1/13^3....Clarify...

You need to multiply your answer by $3!$ to account for the order in which the cards are chosen. You could either choose a  King, Jack, Queen (in that order), or King, Queen, Jack, or a Jack, King, Queen etc...

$ \left(\frac{1}{13}\right)^{3}$ would not count it properly because if the selection was done without replacement, in which case the pool of cards to choose from would decrease by $1$ each time (something you accounted for in your first approach).

Another way to look at the problem: You can choose either a King, Queen, or Jack for the first card, with a probability of $\frac{12}{52}$. For the second card, the pool of desired cards you want to choose is cut to $8$ since you no longer want to choose the type of card that was chosen the first time. And for the third card, the pool of cards is cut to $4$ for the same reason. So the desired probability is $\frac{12\times8\times4}{52\times51\times50}= \frac{16}{5525}$.


During a simple harmonic motion you have a mass of 350 grams attached to a spring which is hanging from the ceiling.  You draw the masses down by 42 cm below their resting position.  just before releasing, your frind drops an additional 20g of mass on the top of the original stack.  This 20g mass is not secured.  Will it lift off the stack anywhere in the motion? if so, where?

the constant of the spring must be given.depending of this constant there are 2 possible motions.

sorry,
the constant for the spring is 11.32 N/m.
can some one solve it now? :maybe:

It will lift off at 72.3 cm above the start position. This is the equilibrium position of the spring alone, which is where the force exerted by the spring will cause m1 to have a more negative acceleration than m2, resulting in m2 lifting off. The energy stored in the spring is enough to pull the masses there.


AMC
AIME
geometry
analytic geometry
graphing lines
slope

Triangle $ ABC$ lies in the Cartesian Plane and has an area of $ 70$. The coordinates of $ B$ and $ C$ are $ (12,19)$ and $ (23,20)$, respectively, and the coordinates of $ A$ are $ (p,q)$. The line containing the  median to side $ BC$ has slope $ \minus{}5$. Find the largest possible value of $ p\plus{}q$.

The third solution given for this problem on aops wiki considers the point $ A'$ as $ (17, 22)$, which lies along the line through M of slope $ \minus{}5$. How would one have seen that $ (17, 22)$ lies on that line? Why is the area of triangle $ A'BC$ $ \frac{1}{5}$ of our needed area?

$ M \equal{} \left(\frac{35}2, \frac{39}2\right)$, and then "travel" $ x \minus{} \frac{1}{2}, y \plus{} \frac{5}{2}$ to point $ A'$. What the solution does is that it finds that $ [A'BC] \equal{} 14$ (the solution doesn't describe exactly how to do that, but there's a bunch of different ways to get that), which is a particularily nice number because it divides the given area of $ [ABC] \equal{} 70$. So $ \frac{[A'BC]}{[ABC]} \equal{} \frac 15$.

Oh!!!!
Wow. Thanks.

That solution is pretty slick... how would one thought to find points along the median?


I noticed that too, but didn't know what to do after that either :(

My first guess was to expand $(x^2+x+1)(ax^3+bx^2+cx+d)$ and plug in values until it fits...but it wouldn't fit.

You could try synthetically dividing in both $\omega$ and $\omega^2$.

[quote="Elemennop"]Notice that if $\omega=e^{\frac{2\pi}{3}}$, then $\omega$ and $\omega^2$ are both roots of the equation...However, I can't factor this out.[/quote]

Huh? I'm getting $\omega^5+\omega^2+1=2\omega^2 +1$, and I can't figure out anything else that might be a factor.

Maybe it was something like $z^5+z+1$? Here $\omega$ does work as a root. Afterwards, you could either use synthetic division with $\omega$ and $\omega^2$, or set $z^5+z+1=(z^2+z+1)(z^3+az^2+bz+1)$ and figure out what the other coefficients are.

Jeez, how do people see these factorizations? o_0

Practice, practice, practice... which leads to recognizing patterns. Here "+z+1" can possibly be a part of "z^2+z+1", and trying that comes out nicely.

Or even after you concluded that $\omega$ and $\omega^2$ are the roots, you could just multiply $(z-\omega)(z-\omega^2)=z^2+z+1$ and divide $(z^5+z+1): (z^2+z+1)$ (I hope you're fluent in polynomial division :))


geometry
rhombus
number theory unsolved
number theory

Let $a,b,c,d$ be positive integers such that the number of pairs $(x,y) \in (0,1)^2$ such that both $ax+by$ and $cx+dy$ are integers is equal with 2004. If $\gcd (a,c)=6$ find $\gcd (b,d)$.

Let ax+by=m, cx+dy=n. Suppose ad>bc.
Solving the system we get 0<am-cn<ad-bc
0<dn-bm<ad-bc.
Draw the lines 0=ax-by, ad-bc=ax-by etc. They form a rhombus with lattice corners.
But the number of solutions is just the number of lattice points inside the rhombus.
Applying Pick's theorem we must get the result ( I dind't check it, but I think it's right)

This problem is very similar to following one (from Russian TST'2002):
Let $a,b,c,d$ be positive integers such that $ad-bc=k>0$, $\gcd(a,c)=\gcd(b,d)=1$. Prove that there are exactly $k$ pairs $(x,y)\in [0,1)^2$ s.t. both $ax+by$ and $cx+dy$ are integers.

[Edited: $\gcd(a,b)=\gcd(c,d)=1$ replaced with $\gcd(a,c)=\gcd(b,d)=1$]

But I cannot> You can proof? Please ! Thanksyou very much!

I don't understand you. iura's solution is absolutely correct, only replace "Draw the lines 0=ax-by, ad-bc=ax-by etc" with "Draw the lines 0=ax-cy, ad-bc=ax-cy etc" ;)


Since this is an intro physics thread I'm going to post something easy. I hope all of you w/o the help of people doing advance physics threads can solve this.

A train at a constant 60.0 km/h moves east for 40.0 min then in a direction 50.0 east of due north 20.min and than west for 50.0 min. What is the AVERAGE velocity during this trip

Hint: You should know that
 ${v_avg} \displaystyle\frac{x}{t} = \displaystyle\frac{x2-x1}{t2-t1}$