calculus
inequalities
inequalities proposed

I'm not sure;

$ f(a,b,c) \equal{} \sum_{cyc}\frac {ab}{c^{2} \plus{} 3}$

$ f(a,b,c)\leq f(\frac {a \plus{} b}{2},\frac {a \plus{} b}{2},c)$

$ \implies$  $ f(\frac {3 \minus{} c}{2},\frac {3 \minus{} c}{2},c)$

hello, with the help of calculus i have got the maximum as $ \frac{1}{24}(\minus{}45\plus{}11\sqrt{33})$ for
$ a\equal{}\minus{}3/2\plus{}\sqrt{33}/2$
$ b\equal{}6\minus{}\sqrt{33}$
$ c\equal{}\frac{1}{2}(\minus{}3\plus{}\sqrt{33})$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, with the help of calculus i have got the maximum as $ \frac {1}{24}( \minus{} 45 \plus{} 11\sqrt {33})$ for
$ a \equal{} \minus{} 3/2 \plus{} \sqrt {33}/2$
$ b \equal{} 6 \minus{} \sqrt {33}$
$ c \equal{} \frac {1}{2}( \minus{} 3 \plus{} \sqrt {33})$
Sonnhard.[/quote]

Yes ... you are right Dr Sonnhard Graubner :D

[i]Wow , this prob is really very hard

   I have just founded the folllowing solution

  Although it was witten in Vietnamese , I think all of you can understand it :wink:

[/i]

This is easier.

If $ a,b,c$ are nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}2$, then

$ \frac {bc}{a^2\plus{}1}\plus{}\frac {ca}{b^2\plus{}1}\plus{}\frac {ab}{c^2\plus{}1}\le 1.$

[quote="Vasc"][quote="Vasc"]
If $ a,b,c$ are nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 2$, then

$ \frac {bc}{a^2 \plus{} 1} \plus{} \frac {ca}{b^2 \plus{} 1} \plus{} \frac {ab}{c^2 \plus{} 1}\le 1.$[/quote]

Is this a difficult inequality?[/quote]
See here: http://canhang2007.wordpress.com/2009/11/15/inequality-67-p-k-hung/

[quote="Vasc"]Your solutions are nice, Can_hang.  :lol:
There is also a nice solution using the Cauchy-Schwarz inequality and the third degree Schur's inequality.
On the other hand, you are right, this inequality is due to Pham Kim Hung. It is problem 8.55( 8.37) in my Algebraic Inequalities book.[/quote]
Don't post it soon, please Mr Vasile. I want to try it.

[quote="Vasc"]Your solutions are nice, Can_hang.  :lol:
There is also a nice solution using the Cauchy-Schwarz inequality and the third degree Schur's inequality.
On the other hand, you are right, this inequality is due to Pham Kim Hung. It is problem 8.55( 8.37) in my Algebraic Inequalities book.[/quote]
I got now a solution based on Cauchy-Schwarz Inequality, but in the proof I must used the fourth degree Schur's Inequality, not the third one. Please take a look again the page: http://canhang2007.wordpress.com/2009/11/15/inequality-67-p-k-hung/ to view it.

Now,  I am really interested in the solution using the Cauchy-Schwarz Inequality and the third degree Schur's Inequality. Could you please post it, Mr Vasile?

[quote="Vasc"][quote="can_hang2007"]
Now,  I am really interested in the solution using the Cauchy-Schwarz Inequality and the third degree Schur's Inequality. Could you please post it, Mr Vasile?[/quote]
Sorry, I found a mistake in my proof with the third degree Schur's Inequality. But there is still one solution with the fourth degree Schur's Inequality (I am sure you know this), using
$ \sum\frac {a}{a^2 \plus{} 1} \ge \frac {(\sum a)^2}{\sum a(a^2 \plus{} 1)} \equal{} \frac 4{3abc \plus{} 10 \minus{} 6q}.$

Other nice similar inequalities are the following.

Let $ a,b,c$ be nonnegative real numbers.

(a) If $ a \plus{} b \plus{} c \equal{} 1$, then

$ \sum \frac {bc}{2a^2 \plus{} 1}\le \frac 3{11}$;

(b) If $ a \plus{} b \plus{} c \equal{} 3$, then

$ \sum \frac {bc}{a^2 \plus{} a \plus{} 6}\le \frac 3{8}$.[/quote]
I like (b), and here is my proof for it: http://canhang2007.wordpress.com/2009/11/21/inequality-75-v-cirtoaje/

[quote="nguyenvuthanhha"][i]Seems to be very familiar , but this problem is quite hard ( At least with me ) :blush:

   Let $ a ;b;c$ be nonnegative real numbers such that : $ a \plus{} b \plus{} c \equal{} 3$

   Find the maximum value of the following expression :

   $ \mathcal{A} \ \equal{} \ \frac {ab}{3 \plus{} c^2} \ \plus{} \ \frac {bc}{ 3 \plus{} a^2 } \ \plus{} \ \frac {ca}{3 \plus{} b^2 }$[/i][/quote]

I think for the same condition, we have
\[ \frac {ab}{3 \plus{} a^2} \ \plus{} \ \frac {bc}{ 3 \plus{} b^2 } \ \plus{} \ \frac {ca}{3 \plus{} c^2 }\le\frac3{4}\]

The following Inequality is true or false?
\[ \frac {ab}{1 \plus{} 2a^2} \ \plus{} \ \frac {bc}{ 1 \plus{} 2b^2 } \ \plus{} \ \frac {ca}{1 \plus{} 2c^2 }\le1\]

[quote="dduclam"]

I think for the same condition, we have
\[ \frac {ab}{3 \plus{} a^2} \ \plus{} \ \frac {bc}{ 3 \plus{} b^2 } \ \plus{} \ \frac {ca}{3 \plus{} c^2 }\le\frac3{4}\]

[/quote]

By Cauchy Schwarz we have : $ (\sqrt{a}b\plus{}\sqrt{b}c\plus{}\sqrt{c}a)^2 \leq (ab\plus{}bc\plus{}ca)(b\plus{}c\plus{}a) \leq \frac{(a\plus{}b\plus{}c)^3}{3}\equal{}9$

By AM-GM we have : $ \frac {ab}{3 \plus{} a^2} \ \plus{} \ \frac {bc}{ 3 \plus{} b^2 } \ \plus{} \ \frac {ca}{3 \plus{} c^2 } \leq \frac{\sqrt{a}b\plus{}\sqrt{b}c\plus{}\sqrt{c}a}{4} \leq \frac{3}{4}$


$\lfloor x \rfloor$ . $\lfloor y \rfloor$ = $x+y$

Let [х]=m,[y]=n, k=x-[x]+y-[y]. Then $(m-1)(n-1)=k+1.$
If x and y are integer, then k=0 and solution is $x=0,y=0$ or $x=2,y=2$.
Else k=1, solutions is $x=-1+t,y=1-t$ or $x=1-t,y=-1+t$ or $x=2+t,y=4-t$ or $x=3+t,y=3-t$, were $t\in (0,1)$.


function
integration
real analysis
real analysis unsolved

Let $f: [0,1]\to\R$ be a continous function such that $f(0)=f(1)$ . Prove that there exists $c\in (0,1)$ such that :
$\int_{0}^{c}f(t)dt=c\cdot f(c)$.


Let $ x,$ $ y$ and $ z$ are positive numbers. Prove that
\[ \frac{x}{\sqrt{(x\plus{}y)(y\plus{}z)}}\plus{}\frac{y}{\sqrt{(x\plus{}z)(y\plus{}z)}}\plus{}\frac{z}{\sqrt{(x\plus{}y)(x\plus{}z)}}\geq\frac{3}{2}\]

$ LHS \geq \sum \frac{2x}{x\plus{}2y\plus{}z} \equal{}2 \sum \frac{x^2}{x^2\plus{}2xy\plus{}yz}$
$ \geq 2\frac{(x\plus{}y\plus{}z)^2}{(x\plus{}y\plus{}z)^2\plus{}xy\plus{}yz\plus{}zx} \geq \frac{3}{2}$

Or
$ LHS\geq \sum \frac{2x}{2y\plus{}x\plus{}z}$ with a=x+y etc... We can write $ \sum \frac{a\plus{}c\minus{}b}{a\plus{}b}\geq \frac{9}{2}\minus{}2\sum \frac{a}{a\plus{}b}\geq \frac{3}{2}$. The last inequality can be proven by expanding.

[quote="Athinaios"]$ LHS \geq \sum \frac {2x}{x \plus{} 2y \plus{} z} \equal{} 2 \sum \frac {x^2}{x^2 \plus{} 2xy \plus{} yz}$
$ \geq 2\frac {(x \plus{} y \plus{} z)^2}{(x \plus{} y \plus{} z)^2 \plus{} xy \plus{} yz \plus{} zx} \geq \frac {3}{2}$[/quote]
Shouldn't it be $ 2 \sum \frac {x^2}{x^2 \plus{} 2xy \plus{} xz}$?


for positive reals $ a,b,c$
prove that

$ \frac {a \plus{} 3c}{a \plus{} 2b \plus{} c} \plus{} \frac {4b}{a \plus{} b \plus{} 2c} \minus{} \frac {8c}{a \plus{} b \plus{} 3c} \geq\ 12\sqrt 2\ \minus{} 17$

[quote="phymax"]for positive reals $ a,b,c$
prove that

$ \frac {a \plus{} 3c}{a \plus{} 2b \plus{} c} \plus{} \frac {4b}{a \plus{} b \plus{} 2c} \plus{} \frac {8c}{a \plus{} b \plus{} 3c} \geq\ 12\sqrt 2\ \minus{} 17$[/quote]
The following inequality is also true.
For positive reals $ a,$ $ b$ and $ c$ prove that:
\[ \frac {a \plus{} 3c}{a \plus{} 2b \plus{} c} \plus{} \frac {4b}{a \plus{} b \plus{} 2c} \minus{} \frac {8c}{a \plus{} b \plus{} 3c} \geq\ 12\sqrt 2\ \minus{} 17
\]

Original: LHS is positive, RHS is negative, end?

[quote="arqady"]See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=147142[/quote]
Denote x=a+2b+c,y=a+b+2c,z=a+b+3c.We have an equivalent inequality:2(y/x)+4(x/y)+4(z/y)+8(y/z)>=12.(2^(1/2)).It is obvious by AM-GM.We are done.


function
complex analysis
complex analysis unsolved

Find a bounded analytic function on $ \{z: |z|\leq 2\}\setminus [0,1]$ that does not extend to be analytic on $ |z|\leq 2$.

So my intuition was something like log, which would be analytic on the region minus $ [0,2]$, and wouldn't extend. So I tried manipulating the region a bit to get something in which I could get an analytic log. Unfortunately, I ended up having to map $ 1\mapsto \infty$ and so it wasn't bounded anymore.

An extra note, it can't be continuous across the segment, or else it will extend to be analytic.

Also, you know how to map $ \mathbb C\setminus [0,1]$ to the unit disk, don't you? ;)

Oh. That's silly.  :blush: $ z\mapsto z/(z\minus{}1)$ followed by $ z\mapsto \sqrt{z}$, followed by Cayley transform. So $ f: \mathbb{C}\setminus [0,1]\to \mathbb{D}$, and the map I want is just restricting to the disk of radius 2.


Krishanu and Shaunak each pick an integer at random between 1 and 10, inclusive. What is the probability that the product of their numbers is more than 10? Express your answer as a common fraction.

\[ \frac{10\plus{}5\plus{}3\plus{}2\plus{}2\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1}{100}\Rightarrow\boxed{\frac{27}{100}}\]
I hope it's correct.

I believe you forgot to take the complement?

$ 1\minus{}\frac{27}{100}\equal{}\boxed{\frac{73}{100}}$


inequalities
triangle inequality
inequalities unsolved

For $a,b,c$ positive real numbers prove that

$\frac{1}{\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}} \geq \frac{1}{3} +\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} $

For  $a = b = c = \frac{1}{2}$  Inequality is true ??????

I think instead of $\frac{1}{a+b} $ we must have $\frac{1}{a+1} $ and so on.

Sorry very much for my misprint  :blush:

Harazi, you are right.

Wel this reduces to
$\frac{abc+ab+bc+ca+a+b+c+1}{ab+bc+ca+2(a+b+c)+3} \geq $ $\frac{3abc+ab+bc+ca}{3(ab+bc+ca)}$.
Now note $s=a+b+c$, $p=abc$, $t=ab+bc+ca$.
Our inequality after obvious substitutions reduces to $2t^2+st \geq 6ps+9p$
which breaks to $2t^2 \geq 6ps$ and $st \geq 9p$ which are both obvious

And the general case? There is a very simple and nicesolution!

In n variables. Try to find the statement and the proof. If I'm not wrong, this problem was given in a Moldavian olympiad in the general case.

Indeed, this has a very cool generalization. I wanted to cry out Heureka out of luck when I found it (but I didn't since I was on a school lesson). The generalization has 2n variables. I think your n variables generalization, Harazi, is a special case of this.

In fact, the 2n variables generalization is:

If $ a_1$, $ a_2$, ..., $ a_n$, $ b_1$, $ b_2$, ..., $ b_n$ are positive numbers, then

$ 1\left/ \sum_{i \equal{} 1}^{n}\frac {1}{a_{i} \plus{} b_{i}}\right. \geq 1\left/ \sum_{i \equal{} 1}^{n}\frac {1}{a_{i}}\right. \plus{} 1\left/ \sum_{i \equal{} 1}^{n}\frac {1}{b_{i}}\right.$.

Clearly, this inequality is a triangle inequality for the $ L_{ \minus{} 1}$-norm in the positive "quadrant" of $ \mathbb{R}^n$. Of course, this interpretation is nice, but I am not at all sure whether it yields a proof of the inequality. Here is an elementary proof (admittedly written in a rather unpoetic way):

We denote $ A_i \equal{} \frac {1}{a_i}$ and $ B_i \equal{} \frac {1}{b_i}$, what transforms the above inequality into the form

$ 1\left/ \sum_{i \equal{} 1}^{n}\frac {1}{\frac {1}{A_{i}} \plus{} \frac {1}{B_{i}}}\right. \geq 1\left/ \sum_{i \equal{} 1}^{n}A_{i}\right. \plus{} 1\left/ \sum_{i \equal{} 1}^{n}B_{i}\right.$.

Of course,

$ \frac {1}{\frac {1}{A_{i}} \plus{} \frac {1}{B_{i}}} \equal{} \frac {1}{\left( \frac {A_{i} \plus{} B_{i}}{A_{i}B_{i}}\right) } \equal{} \frac {A_{i}B_{i}}{A_{i} \plus{} B_{i}}$.

Thus, we have to prove

$ 1\left/ \sum_{i \equal{} 1}^{n}\frac {A_{i}B_{i}}{A_{i} \plus{} B_{i}}\right. \geq 1\left/ \sum_{i \equal{} 1}^{n}A_{i}\right. \plus{} 1\left/ \sum_{i \equal{} 1}^{n}B_{i}\right.$.

After multiplication by $ \sum_{i \equal{} 1}^{n}\frac {A_{i}B_{i}}{A_{i} \plus{} B_{i}}\cdot \sum_{i \equal{} 1}^{n}A_{i}\cdot \sum_{i \equal{} 1}^{n}B_{i}$, this becomes

$ \sum_{i \equal{} 1}^{n}A_{i} \cdot \sum_{i \equal{} 1}^{n}B_{i}\geq \sum_{i \equal{} 1}^{n}\frac {A_{i}B_{i}}{A_{i} \plus{} B_{i}}\cdot \left( \sum_{i \equal{} 1}^{n}A_{i} \plus{} \sum_{i \equal{} 1}^{n}B_{i}\right)$.

Now, the left-hand side of this inequality is

$ \sum_{i \equal{} 1}^{n}A_{i} \cdot \sum_{i \equal{} 1}^{n}B_{i} \equal{} \sum_{1\leq i\leq n,\;1\leq j\leq n}\left( A_{i}B_{j}\right) \equal{} \sum_{i \equal{} 1}^{n}\left( A_{i}B_{i}\right) \plus{} \sum_{1\leq i < j\leq n}\left( A_{i}B_{j} \plus{} A_{j}B_{i}\right)$,

while the right-hand side is

$ \sum_{i \equal{} 1}^{n}\frac {A_{i}B_{i}}{A_{i} \plus{} B_{i}}\cdot \left( \sum_{i \equal{} 1}^{n}A_{i} \plus{} \sum_{i \equal{} 1}^{n}B_{i}\right) \equal{} \sum_{i \equal{} 1}^{n}\frac {A_{i}B_{i}}{A_{i} \plus{} B_{i}}\cdot \sum_{i \equal{} 1}^{n}\left( A_{i} \plus{} B_{i}\right)$
$ \equal{} \sum_{1\leq i\leq n,\;1\leq j\leq n}\left( \frac {A_{i}B_{i}}{A_{i} \plus{} B_{i}}\cdot \left( A_{j} \plus{} B_{j}\right) \right) \equal{} \sum_{1\leq i\leq n,\;1\leq j\leq n}\left( A_{i}B_{i}\frac {A_{j} \plus{} B_{j}}{A_{i} \plus{} B_{i}}\right)$
$ \equal{} \sum_{i \equal{} 1}^{n}\left( A_{i}B_{i}\frac {A_{i} \plus{} B_{i}}{A_{i} \plus{} B_{i}}\right) \plus{} \sum_{1\leq i < j\leq n}\left( A_{i}B_{i}\frac {A_{j} \plus{} B_{j}}{A_{i} \plus{} B_{i}} \plus{} A_{j}B_{j}\frac {A_{i} \plus{} B_{i}}{A_{j} \plus{} B_{j}}\right)$
$ \equal{} \sum_{i \equal{} 1}^{n}\left( A_{i}B_{i}\right) \plus{} \sum_{1\leq i < j\leq n}\left( A_{i}B_{i}\frac {A_{j} \plus{} B_{j}}{A_{i} \plus{} B_{i}} \plus{} A_{j}B_{j}\frac {A_{i} \plus{} B_{i}}{A_{j} \plus{} B_{j}}\right)$.

Hence it remains to prove the inequality

$ \sum_{i \equal{} 1}^{n}\left( A_{i}B_{i}\right) \plus{} \sum_{1\leq i < j\leq n}\left( A_{i}B_{j} \plus{} A_{j}B_{i}\right)$
$ \geq \sum_{i \equal{} 1}^{n}\left( A_{i}B_{i}\right) \plus{} \sum_{1\leq i < j\leq n}\left( A_{i}B_{i}\frac {A_{j} \plus{} B_{j}}{A_{i} \plus{} B_{i}} \plus{} A_{j}B_{j}\frac {A_{i} \plus{} B_{i}}{A_{j} \plus{} B_{j}}\right)$.

From this inequality, we subtract $ \sum_{i \equal{} 1}^{n}\left( A_{i}B_{i}\right)$ on both sides, and get

$ \sum_{1\leq i < j\leq n}\left( A_{i}B_{j} \plus{} A_{j}B_{i}\right) \geq \sum_{1\leq i < j\leq n}\left( A_{i}B_{i}\frac {A_{j} \plus{} B_{j}}{A_{i} \plus{} B_{i}} \plus{} A_{j}B_{j}\frac {A_{i} \plus{} B_{i}}{A_{j} \plus{} B_{j}}\right)$.

And this actually follows from the inequality

$ A_{i}B_{j} \plus{} A_{j}B_{i}\geq A_{i}B_{i}\frac {A_{j} \plus{} B_{j}}{A_{i} \plus{} B_{i}} \plus{} A_{j}B_{j}\frac {A_{i} \plus{} B_{i}}{A_{j} \plus{} B_{j}}$

holding for every i and j. To prove the latter, we see that for any four positive reals u, v, w, x, we have

$ ux \plus{} vw\geq uw\frac {v \plus{} x}{u \plus{} w} \plus{} vx\frac {u \plus{} w}{v \plus{} x}$,

since

$ ux \plus{} vw \minus{} \left( uw\frac {v \plus{} x}{u \plus{} w} \plus{} vx\frac {u \plus{} w}{v \plus{} x}\right) \equal{} \frac {\left( ux \minus{} vw\right) ^{2}}{\left( u \plus{} w\right) \left( v \plus{} x\right) }\geq 0$.

If somebody is still awake, let's put a $ \blacksquare$ at this place.

And actually, from the general inequality, the original one (posted by Manlio) follows if n = 3, $ a_1 \equal{} a$, $ a_2 \equal{} b$, $ a_3 \equal{} c$, $ b_1 \equal{} b_2 \equal{} b_3 \equal{} 1$.

  Darij

More generally,

If $x_1 ,x_2 , \ldots ,x_n $  are real numbers such that  $x_1 ,x_2 , \ldots ,x_n  > 0$
 prove that:
	\[
\frac{1}{{\frac{1}{{1 + x_1 }} + \frac{1}{{1 + x_2 }} +  \cdots  + \frac{1}{{1 + x_n }}}} \ge \frac{1}{n} + \frac{1}{{\frac{1}{{x_1 }} + \frac{1}{{x_2 }} +  \cdots  + \frac{1}{{x_n }}}}
\]

with equality iff $x_1  = x_2  =  \cdots  = x_n $

[u]Solution:[/u]
Here‘s my proof. We have successively:
\[
\frac{1}{{\frac{1}{{1 + x_1 }} + \frac{1}{{1 + x_2 }} +  \cdots  + \frac{1}{{1 + x_n }}}} \ge \frac{1}{n} + \frac{1}{{\frac{1}{{x_1 }} + \frac{1}{{x_2 }} +  \cdots  + \frac{1}{{x_n }}}} \Leftrightarrow
\]

\[
\Leftrightarrow n \cdot \left( {\frac{1}{{\sum\limits_{j = 1}^n {\frac{1}{{1 + x_j }}} }} - \frac{1}{{\sum\limits_{j = 1}^n {\frac{1}{{x_j }}} }}} \right) \ge 1 \Leftrightarrow
\]

\[
\Leftrightarrow n \cdot \left( {\sum\limits_{j = 1}^n {\frac{1}{{x_j \left( {1 + x_j } \right)}}} } \right) \ge \left( {\sum\limits_{j = 1}^n {\frac{1}{{x_j }}} } \right) \cdot \left( {\sum\limits_{j = 1}^n {\frac{1}{{1 + x_j }}} } \right) \Leftrightarrow
\]

\[
 \Leftrightarrow \frac{1}{n}\sum\limits_{j = 1}^n {\frac{1}{{x_j \left( {1 + x_j } \right)}} \ge \left( {\frac{1}{n}\sum\limits_{j = 1}^n {\frac{1}{{x_j }}} } \right) \cdot \left( {\frac{1}{n}\sum\limits_{j = 1}^n {\frac{1}{{1 + x_j }}} } \right)}
\]

Which is true, see Cebasev inequality. And our inequality is proven.

[quote="darij grinberg"]Clearly, this inequality is a triangle inequality for the $L_{-1}$-norm in the positive "quadrant" of $\mathbb{R}^n$. Of course, this interpretation is nice, but I am not at all sure whether it yields a proof of the inequality.[/quote]

Wouldn't this mean that the inequality follows directly from Minkowski?

[quote="billzhao"]Wouldn't this mean that the inequality follows directly from Minkowski?[/quote]

Rather from some generalization of Minkowski (actually, $L_{-1}$ is not a real norm, that's the problem). I haven't ever studied this subject so it would probably better to ask someone else...

  darij

In Kiran Kedlaya's A < B, he defined Minkowski's inequality as

Let $r > s$ be nonzero real numbers. Then for any positive real numbers $a_{ij}$,

\[
\left( \sum_{j=1}^{m} \left( \sum_{i=1}^{n} a_{ij}^{r} \right)^{s/r}
\right)^{1/s}
\geq
\left( \sum_{i=1}^{n} \left( \sum_{j=1}^{m} a_{ij}^{s} \right)^{r/s}
\right)^{1/r}.
\]

And he gives proof for it (though I haven't had the chance to go over the proofs). If this were correct, then Minkowski works for $L_{-1}$.


On sides BC CA AB in triangle ABC ( AB = BC) we take points X,Y,Z respectively.

Lines BY CZ interect in P. We know that: <BZX = <CXY = <BAC. Prove that quadrilatera A Y P Z is cyclic.

Denote P' be the intersection of circles (BXZ),(CXY),(AYZ).
$ \angle BP'X\equal{}\angle BZX\equal{}\angle C\equal{}180^o\minus{}\angle XP'Y$, thus B,P',Y are collinear.
$ \angle CP'X\equal{}\angle XYC\equal{}\angle B\equal{}180^o\minus{}\angle XP'Z$, so C,P',Z are also collinear.
These imply P' is coincide with P. hence result.

$ CAZX$ is cyclic (isosceles trapezoid), therefore $ \angle CAX \equal{} \angle CZX$. By construction, $ ABXY$ is cyclic too, so $ \angle CAX \equal{} \angle YBC$, consequently $ \angle CZX \equal{} \angle YBX$ and $ BXPZ$ is cyclic, with $ \angle BPZ \equal{} \angle BXZ$, hence $ \angle BAC \equal{} \angle BPZ$ and $ AZPY$ is indeed cyclic.

Best regards,
sunken rock


probability
real analysis
function
algebra
polynomial
calculus
derivative

I don't think there is an answer to your problem in closed form.

For small $ N$ you can use the convolution formula to calculate the density function of the sum $ x_1,\ldots,x_N$,
and then it is easy to calculate the precise probability of your event. However, this is getting more and more complicated as $ N$ increases.

For large $ N$ you can calculate the probability approximately using the central limit theorem.

I don't think there is much more you can do.

This cumulative distribution function is a polynomial spline; its derivative (the density) is also a polynomial spline.

I'm going to make things easier for the time being by assuming that $ a\equal{}0$ and $ b\equal{}1.$ We can rescale from that to get the more general answer.

Let $ B(x)\equal{}\begin{cases}1&x\in(0,1)\\0&\text{otherwise}\end{cases}.$

This is the density of a single uniform distribution. Note that the weak (distributional) derivative of $ B$ is $ B'(x)\equal{}\delta(x)\minus{}\delta(x\minus{}1).$

The density of the $ N$-fold independent sum of such distributions is $ f(x)\equal{}B*B*B*\cdots*B(x),$ the $ N$-fold convolution.

We calculate the $ N$th derivative by throwing one derivative onto each factor:

$ f^{[N]}(x)\equal{}(B')*(B')*\cdots*(B')\equal{} \sum_{k\equal{}0}^{N}\binom{N}{k}(\minus{}1)^k\delta(x\minus{}k).$

We used here that $ \delta(x\minus{}a)*\delta(x\minus{}b)\equal{}\delta(x\minus{}(a\plus{}b)).$

Now, the $ N$-fold integral (from the left) of $ \delta(x)$ is $ \frac{1}{(N\minus{}1)!}x^{N\minus{}1}H(x)$ where $ H(x)$ is the Heaviside step function that is $ 1$ for $ x>0$ and $ 0$ otherwise. (Note that we could have written $ B$ in the first place as $ B(x)\equal{}H(x)\minus{}H(x\minus{}1).$)

So I get that $ f(x)\equal{}\frac{1}{(N\minus{}1)!}\sum_{k\equal{}0}^N\binom{N}{k}(\minus{}1)^k (x\minus{}k)^{N\minus{}1}H(x\minus{}k).$

That's the density; you asked for the cumulative distribution function. For that, we merely need to take one more integral:

$ F(x)\equal{}\frac{1}{N!}\sum_{k\equal{}0}^N\binom{N}{k}(\minus{}1)^k (x\minus{}k)^{N}H(x\minus{}k).$

Let's illustrate this for $ N\equal{}4.$

For $ x>4,$ $ H(x\minus{}k)\equal{}1$ for all $ k\equal{}0,1,2,3,4$ and we have

$ F(x)\equal{}\frac1{24}\left(x^4\minus{}4(x\minus{}1)^4\plus{}6(x\minus{}2)^4\minus{}4(x\minus{}3)^4\plus{}(x\minus{}4)^4\right)\equal{}1$

For $ 3\le x\le4,$ we leave out the $ (x\minus{}4)^4$ term, getting:

$ F(x)\equal{}\frac1{24}(\minus{}x^4\plus{}16x^3\minus{}96x^2\plus{}256x\minus{}232)$

For $ 2\le x\le3,$ we leave out the $ \minus{}4(x\minus{}3)^4$ term:

$ F(x)\equal{}\frac1{24}(3x^4\minus{}32x^3\plus{}120x^2\minus{}176x\plus{}92)$

For $ 1\le x\le2,$ it becomes easier to add from the left:

$ F(x)\equal{}\frac1{24}(\minus{}3x^4\plus{}16x^3\minus{}24x^2\plus{}16x\minus{}4)$

Finally, for $ 0\le x\le1,$ we have $ F(x)\equal{}\frac{x^4}{24}.$

Differentiate that to get the density:

$ f(x)\equal{}\begin{cases}0&x<0\\ \frac16x^3&0\le x\le1\\ \frac16(\minus{}3x^3\plus{}12x^2\minus{}12x\plus{}4)&1\le x\le2\\ \frac16(3x^3\minus{}24x^2\plus{}60x\minus{}44)&2\le x\le 3\\ \frac16(\minus{}x^3\plus{}12x^2\minus{}48x\plus{}64)&3\le x\le 4\\ 0&x>4\end{cases}$

I'll leave it up to you to figure out how to rescale this for $ a$ and $ b$ other than $ 0$ and $ 1.$ There [i]is[/i] a closed form, but this is what it looks like. And as solyaris mentioned, this is fairly quickly approximated well by a normal distribution, according to the Central Limit Theorem.

[color=green][I don't guarantee the $ \color{green}N\equal{}4$ details above against arithmetical errors.][/color]

[quote="pk123"]integers  a, b, and S  are given

there are  N  events :

"choose real  number   x_i,  a <= x_i <= b  "    ( i = 1...N)
(uniform distribution )
and events are independent.

i'd like to know probabilty of :

x_1 + x_2  +  .... x_N  < S

thanks in advance for help.[/quote]

\[ P\left(\sum_{i \equal{} 1}^N X_i < S\right) \equal{} P\left(NX < S\right) \equal{} P\left(X < \frac{S}{N}\right)\]

If $ b < \frac{S}{N}$ then $ P\left(X <\frac{S}{N}\right) \equal{} 1$, but if $ \frac{S}{N} < a$ then $ P\left(X <\frac{S}{N}\right) \equal{} 0$; else, $ P\left(X < \frac{S}{N}\right) \equal{} \frac{1}{b \minus{} a}\int_{a}^{\frac{S}{N}} \mathrm{d}t$.

What Dr. No has said would only be true if the $ X_i$ were perfectly positively correlated - and that's not what the statement said. Instead, the statement was that the $ X_i$ are independent.

Suppose $ X_1$ and $ X_2$ are both uniform on $ [0,1].$ Let $ Y = X_1 + X_2.$ Dr. No has essentially claimed that $ Y$ is uniformly distributed on $ [0,2].$ I claim that the only way that this can be true would be for the entire (except possibly for sets of probability zero) distribution of the ordered pairs $ (X_1,X_2)$ be confined to the diagonal line $ X_1 = X_2.$

If instead, $ X_1$ and $ X_2$ are independent, then the ordered pairs $ (X_1,X_2)$ are uniformly (in area) distributed on the unit square $ [0,1]\times[0,1].$ Then for $ 0\le y\le 1,$ we have that $ P(Y\le y)$ is the area of the triangle with vertices $ (0,0), (0,y),$ and $ (y,0).$ This is easily computed: we get that $ P(Y\le y) = \frac {y^2}2.$ A similar area computation gets us that for $ 1\le y\le 2,$ we get $ P(Y > y) = \frac {(2 - y)^2}2,$ or $ P(Y\le y) = 1 - \frac {(2 - y)^2}2.$

Contrast this cumulative distribution function with the distribution function for the uniform $ [0,2]$ distribution, which would be ${ \frac {y}2}$ on $ [0,2].$

If we differentiate the correct cumulative distribution function, you get a "tent-shaped" density function:

$ f(y) = \begin{cases}y & y\in[0,1] \\
2 - y & y\in[1,2] \\
0 & \text{otherwise}\end{cases}.$

This is quite different from the constant you get for the density of the uniform distribution.

Proof of my claim that the sum of uniform distributions being uniform implies perfect correlation:

Suppose that $ X_1$ and $ X_2$ are both uniformly distributed on $ [a,b]$ and suppose that $ Y\equal{}X_1\plus{}X_2$ is uniformly distributed on $ [2a,2b].$ Then $ \text{Var}(X_1)\equal{}\text{Var}(X_2)\equal{}\frac{(b\minus{}a)^2}{12}$ and $ \text{Var}(Y)\equal{}\frac{(2b\minus{}2a)^2}{12}\equal{}4\cdot\frac{(b\minus{}a)^2}{12}.$

But $ \text{Var}(Y)\equal{}\text{Var}(X_1)\plus{}\text{Var}(X_2)\plus{}2\text{Cov}(X_1,X_2).$

Putting in the numbers above gives us $ \text{Cov}(X_1,X_2)\equal{}\frac{(b\minus{}a)^2}{12}.$

The correlation coefficient between $ X_1$ and $ X_2$ is $ \rho\equal{}\frac{\text{Cov}(X_1,X_2)}{\sqrt{\text{Var}(X_1)\text{Var}(X_2)}}\equal{}1.$

That's perfect correlation. Note that independence requires the random variables to be uncorrelated, and if $ Y\equal{}\sum_{i\equal{}1}^NX_i$ where $ X_i$ have the identical distribution to $ X$ and are uncorrelated, then $ \text{Var}(Y)\equal{}N\text{Var}(X).$ Which is not the same as $ \text{Var}(NX)\equal{}N^2\text{Var}(X).$

[quote="Kent Merryfield"]Proof of my claim that the sum of uniform distributions being uniform implies perfect correlation:

Suppose that $ X_1$ and $ X_2$ are both uniformly distributed on $ [a,b]$ and suppose that $ Y \equal{} X_1 \plus{} X_2$ is uniformly distributed on $ [2a,2b].$ Then $ \text{Var}(X_1) \equal{} \text{Var}(X_2) \equal{} \frac {(b \minus{} a)^2}{12}$ and $ \text{Var}(Y) \equal{} \frac {(2b \minus{} 2a)^2}{12} \equal{} 4\cdot\frac {(b \minus{} a)^2}{12}.$

But $ \text{Var}(Y) \equal{} \text{Var}(X_1) \plus{} \text{Var}(X_2) \plus{} 2\text{Cov}(X_1,X_2).$

Putting in the numbers above gives us $ \text{Cov}(X_1,X_2) \equal{} \frac {(b \minus{} a)^2}{12}.$

The correlation coefficient between $ X_1$ and $ X_2$ is $ \rho \equal{} \frac {\text{Cov}(X_1,X_2)}{\sqrt {\text{Var}(X_1)\text{Var}(X_2)}} \equal{} 1.$

That's perfect correlation. Note that independence requires the random variables to be uncorrelated, and if $ Y \equal{} \sum_{i \equal{} 1}^NX_i$ where $ X_i$ have the identical distribution to $ X$ and are uncorrelated, then $ \text{Var}(Y) \equal{} N\text{Var}(X).$ Which is not the same as $ \text{Var}(NX) \equal{} N^2\text{Var}(X).$[/quote]

First, the problem specified that the $ X_i$ are independent. This implies that $ \text{Cov}(X_j,X_k) \equal{} 0$ by definition.

Second, $ P(NX < S)$, was a poor choice of notation on my part.

\[ \sum_{i\equal{}1}^{N} X_i \doteq N\overline{X} \]

Therefore, I should have written $ P(\sum_{i\equal{}1}^N X_i < S) \equal{} P(N\overline{X} <S)$.

\[ \text{Var}(N\overline{X}) \equal{}N^2\text{Var}(\overline{X}) \equal{} N^2\cdot \frac{\text{Var}(X)}{N} \equal{} N \text{Var}(X)\]

\[ N\text{Var}(X) \equal{} \text{Var}(\sum_{i\equal{}1}^N X_i)\]

\[ P(\sum_{i\equal{}1}^N X_i < S) \equal{} P(\overline{X} < S/N)\]

Provided $ N$ is large, $ P(\overline{X} < S/N) \equal{} P(Z <\frac{S/N \minus{}\mu}{\sigma/\sqrt{N}}) \equal{} \text{erf}\left(\frac{S/N \minus{}\mu}{\sigma/\sqrt{N}}\right)$.

$ \mu \equal{} \frac12 (a \plus{} b)$

$ \sigma \equal{}\frac{1}{\sqrt{12}}\left|b \minus{} a\right|$.

[quote="Dr. No"]Provided $ N$ is large, $ P(\overline{X} < S/N) \equal{} P(Z < \frac {S/N \minus{} \mu}{\sigma/\sqrt {N}}) \equal{} \text{erf}\left(\frac {S/N \minus{} \mu}{\sigma/\sqrt {N}}\right)$.[/quote]
If you'll say $ \approx$ instead of $ \equal{},$ then I'll agree. (Well, I'm never sure exactly what the conventions around the definition of "erf" are, but the statement that $ \frac {S/N \minus{} \mu}{\sigma/\sqrt {N}}$ has approximately a standard normal distribution is certainly correct.)

This is a direct consequence of the Central Limit Theorem, which solyaris and I both mentioned.

Here's an interesting exercise for using a system with programming and graphing capabilities:

For $ N\equal{}1,2,3,..$ (stopping somewhere - maybe $ 5$ or $ 6$ - don't go too far) compute the explicit density of $ S$ as the polynomial spline given in my first post. Then provide the proper linear rescaling of this density to be the density of $ \frac {S/N \minus{} \mu}{\sigma/\sqrt {N}}.$ Plot those densities on the same set of axis. Then plot the density of the standard normal distribution on the same set of axes. Make the whole thing suitable for display, and show it.

[quote="Kent Merryfield"][quote="Dr. No"]Provided $ N$ is large, $ P(\overline{X} < S/N) \equal{} P(Z < \frac {S/N \minus{} \mu}{\sigma/\sqrt {N}}) \equal{} \text{erf}\left(\frac {S/N \minus{} \mu}{\sigma/\sqrt {N}}\right)$.[/quote]
If you'll say $ \approx$ instead of $ \equal{} ,$ then I'll agree.[/quote]

So stipulated. Provided $ N$ is large,
\[ P(\overline{X} < S/N) \approx P\left(Z < \frac {S/N \minus{} \mu}{\sigma/\sqrt {N}}\right)\]


Note that at some point, you must go along a diagonal, or else it will take at least $ 6$ moves. For the middle diagonal, there are $ 2$ ways to go to the left endpoint, and then after crossing the diagonal, $ 2$ more ways to finish. For the other two diagonals, there is $ 1$ way to go to the endpoint, and then $ 1$ way to finish. Thus there are $ 2\cdot2\plus{}2(1\cdot1)\equal{}\boxed6$ paths.


[quote="solafidefarms"]Chris and Joe have been put in charge of choosing the president and vice-president of the school's MOEMS club. There are 6 applicants for president, but there are only 4 applications for vice-president. 2 people applied for both president and vice-president, but a person cannot serve both roles simultaniously. Also, no person may be chosen for a position he/she did not apply for. How many ways can Chris and Joe choose the president and vice-president from the people who applied?

Billy[/quote]

er...i dont think this is it but w.e lol :lol: [hide]im thinking that you said those 2 people who applied for both are counted in the 6 for pres and 4 for pres. so lol lamely i just subtracted 1 from each and multiplied...being 15[/hide]

Let's say they choose President first, because it doesn't matter what order we go in.

Case 1: We get somebody who applied only for president.  Then there are $4\times 4=16$ ways.

Case 2: We get somebody who applied for both.  This takes one out for V.P., so we have $2\times 3=6$ ways.

Our answer is $16+6=22$ ways


Let $ X$ be a CW-complex and $ \pi : Y\to X$ be a covering space

a) show that $ Y$ is compact iff $ X$ is compact and $ \pi$ has finite degree

b) Assume that $ \pi$ has finite degree $ d$, show that $ \chi (Y) \equal{} d\chi(X)$

c) let $ \pi : \mathbb{RP}^{N}\to X$ be a covering map, show that if $ N$ is even, then $ \pi$ is a homemorphism

any idea will be highly appreciated. Thanks

i seem to remember pieces of this problem already appeared on this forum not so long ago... anyway:

a) this is a general fact about coverings (at least for sufficiently nice spaces): if $ Y$ is compact, obviously $ X$ is compact (image of a compact space) and $ \deg\pi<\infty$ (the fiber is discrete, hence finite).
the other way is a bit more delicate, i think, but it should just be standard "basic" topology (and i don't want to dive into details right now).(*)

b) this is really nice: the key fact is that $ \chi(C_*) \equal{} \chi(H_*(C))$ (here by i mean euler characteristic of the graduated abelian groups $ C_*$ and $ H_*(C)$).
now given a cover $ \pi$, assuming $ X$ is finite (which shouldn't be a consequence of requiring that $ H_*(X)$ is definitely trivial (*)), we can give $ Y$ a structure of CW-complex such that the pre-image of every cell of $ X$ is the union of $ d$ cells (of the same dimension) in $ Y$.
so $ \chi(Y) \equal{} \chi(H_*(Y))\equal{}\chi(C_*(Y)) \equal{} d\chi(C_*(X)) \equal{} d\chi(H_*(X)) \equal{} d\chi(X)$.

c) what's the euler characterstic of $ \mathbb{RP}^n$? it's half the characteristic of $ S^n$, which is $ 2$, so it's $ 1$, and by point a) there can be no non-finite cover, and by point b) there can be no finite nontrivial one.

(*) in our case (we're dealing with a finite-dimensional manifold), we can bypass both problems by talking about finite CW-complexes: the first point then becomes easy, since a finite cover of a finite CW-complex is just another finite CW-complex, which is then compact.

The other direction of (1) can be seen as follows: cover $ X$ by finitely many compact sets $ K$ such that $ p|_{p^{\minus{}1}(K)}: p^{\minus{}1}(K)\to K$ is a trivial covering. Then $ p^{\minus{}1}(K)$ is compact as the fibre is finite. Since $ Y$ is covered by the finitely many $ p^{\minus{}1}(K)$, it is compact.

Thanks Hanno,

the following is what I have : choose an open cover $ \{U_{i}\}_{i}$ for $ Y$, since $ p$ is covering map so $ \{p(U_i)_i\}_{i}$ is an open cover for $ X$,  $ X$ is compact so I have a finite subcover $ \{p(U_i)\}_{i \equal{} 1}^{n}$. Now I take the preimages of these open subsets then I have $ Y \equal{} \bigcup_{i \equal{} 1}^{n}p^{ \minus{} 1}(p(U_{i}))$ . Because the degree of $ p$ is finite so
$ p^{ \minus{} 1}(p(U_{i}))$  is finitely covered.  So I can find a finite cover for $ Y$. But   my problem is $ p^{ \minus{} 1}(p(U_{i}))$ [b]may not be equal[/b] $ U_{i}$  :( . Could you guy tell me what I should proceed from here. thanks


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

Let N and P $n\times$ matrix over R such that NP=PN and N nilpotent. Show that det(N+P)=detP.

I prove that det(N+I)=1 using caracteristic polinomio of N, maybe you can use this... ;)

There are infinitely many $\lambda$ for which $P\\lambda I_n$ is invertible. You have shown that $\det (M+I_n)=1$ for nilpotent $M$, right? Take $M=N(P+\lambda I_n)^{-1}$ in this assertion. Then, multiplying by $\det (P+\lambda I_n)$ on both sides, we find that $\det(N+P+\lambda I_n)=\det(P+\lambda I_n)$, for all $\lambda$ for which $P+\lambda I_n$ is invertible. Now, $\det(M+P+\lambda I_n)=\det(P+\lambda I_n)$ is a polynomial in $\lambda$ which takes the value $0$ for infinitely many $\lambda$, so it must be $0$ for $\lambda=0$ as well, which is what we want.


inequalities
calculus
integration
function
calculus computations

f is a continous function on [0,1] and $ \int_{x}^1 f(t)dt \geq \frac{1\minus{}x^2}2$ for all $ x \in [0,1]$. prove that:
$ \int_0^1 (f(x))^2dx \geq \int_0^1 xf(x)dx$

$ \int_x^1 f(t) \, dt \ge \frac{1\minus{}x^2}{2} \equal{} \int_x^1 t \, dt \Rightarrow f(t) \ge t \Rightarrow (f(t))^2 \ge t f(t) \Rightarrow \boxed{\int_0^1 (f(t))^2 \, dt \ge \int_0^1 tf(t) \, dt}$

are you sure about your first implication ?

for example, $ f(t) \equal{} 2t \minus{} \frac{1}{2}$, t in [0,1] verify the initial condition of the problem, but not $ f(t) \geq t$


[quote="Rust"][quote="parinaz.sa"]find all functions $f: Z \rightarrow R$such that for each  $m,n\in Z$we have this:
$f(m).f(n)=f(m+n)+f(m-n) , f(1)=\frac{5}{2}$[/quote]
m=1,n=0 give f(0)=2. Therefore
$f(m+1)-\frac{5}{2}f(m)+f(m-1)=0$. Solution is $f(m)=2^{m}+2^{-m}\ \forall m\in Z.$[/quote]
can you tel me some trick for how you understand what the solution is !!

Let $g(x)=f(x+1)-2f(x)$, and then $g(x)=\frac{1}{2}g(x-1)$.
The following is easy :)

[quote="parinaz.sa"]
can you tel me some trick for how you understand what the solution is !![/quote]
Let $f(m)=a_{m}.$ Then for the sequence ${a_{m},m\in\mathbb{Z}},$ we have $a_{m+1}-\frac{5}{2}a_{m}+a_{m-1}=0.$ Therefore $a_{m}=c_{1}t_{1}^{m}+c_{2}t_{2}^{m},$ where $t_{1,2}$ are the roots ot the equation $t^{2}-\frac{5}{2}t+1,$ which are $2$ and $\frac{1}{2}.$ Now it is easy to see that $c_{1}=c_{2}=1.$


This problems solved with one inequality(how is cauchy)!
You knows Zaimi-Marku inequality???

This is the special case of zaimi-marku inequality only..

For better click:
http://www.mathlinks.ro/Forum/viewtopic.php?p=618601#618601
 :wink:


[size=150]A recent survey asked 100 people if they read Sports Illustrated,Newsweek,or Time on a regular basis. Twelve people said that they read all 3, 15 read both SI and Newsweek, 14 read both SI and Time, and 16 read both SI and Newsweek. If 57 total people read SI,  56 total read Newsweek, and 51 total read Time, how many people in the survey only read one of the magazines on a regular basis?[/size]
[color=red]If the given answer is 38?,IF SO, HOW?[[/color]size=18][/size]

[quote="xixi"]A recent survey asked 100 people if they read Sports Illustrated,Newsweek,or Time on a regular basis. Twelve people said that they read all 3, [b]15 read both SI and Newsweek[/b], 14 read both SI and Time, and [b]16 read both SI and Newsweek[/b]. If 57 total people read SI,  56 total read Newsweek, and 51 total read Time, how many people in the survey only read one of the magazines on a regular basis?
If is given anwer is 38?[/quote]

Theres one obvious mistake in bold there. But I think the main problem is that where it says those three numbers about people who read X and Y, that should be ONLY  X and Y - ie, 14 people read both SI and Time but not Newsweek. That will give you the answer of 38.


Given two pairs of points $ \mathbf{a,b}$ and $ \mathbf{c,d}$ that define skew lines, find the volume of the resulting tetrahedron.

(Hint: You could treat the points $ \mathbf{a,b,c,d}$ as vectors.)

So just $ \frac {1}{6}\left|(\mathbf{b \minus{} a})\cdot\parallel{}(\mathbf{c \minus{} a}) \times ( \mathbf{d \minus{} a})\parallel{}\right|$


Four circles, none of which lies inside another, pass through the point P. Two circles touch the line L at P and the other two touch the line M at P. The other points of intersection of the circles are A, B, C, D. Show that A, B, C, D lie on a circle iff L and M are perpendicular.

Invert about P. L inverts to itself, and the two circles touching it at P invert to lines parallel to L. Similarly, M inverts to itself, and the two circles touching it at P invert to lines parallel to M. Hence ABCD is a parallelogram. But a parallelogram is cyclic iff it is a rectangle. So ABCD is cyclic iff L and M are perpendicular.

It's not the first time that I see a topic with both problem and solution being copied word-by-word (!!) from Kalva. This time the source is http://www.kalva.demon.co.uk/aus-pol/apsol/asol971.html . What's the use of doing this??

  Darij


AMC
AIME
trigonometry
AMC 12
geometry
probability
USA(J)MO

Which math topics make the most difficult AIME problems?

Trig: I'm currently in Algebra II  :D. So trig drives me NUTS because they usually lead to blank stares (at least at the AIME level.. AMC 12 is mostly intuitive.)

[quote="h_s_potter2002"]Haha, that's the same thing with me, cause I'm also in Algebra II. Except I don't even know enough trig to do a lot of the harder AMC 12 problems.[/quote]

Lol I am just teaching myself the double angle formulas and junk with the AoPS books, but for the AMC12s, I just rewrite everything as x/z and y/z or draw pictures and it usually works out. Plus the TI89 can't hurt....  :D

trig.......

I am in awe at all the ppl who did trigonometry before calculators were invented.

I should probably actually learn my trig formulas before next years AIME.  Other than trig, though, I LOVE the non-calculator aspect of the test.

Also, any problems involving unicorns and towers tend to be extremely hard to understand, due to ambiguous wording :mad: .

hmm...contrary to the poll, I LOVE geometry!!!!!!Weird enough, for me, geometry at AIME or lower level implies that I can solve it more than 80% of the time...

But even easy probability and counting kills me.....

hmm.. geo seems easier for me than trig, analyt, or complex algebra stuff even after i took those classes.....

Maybe memorizing lots of formulas is what I need......

#14 (a geometry problem) on this year's AIME didn't seem to require an elegant solution - I almost got the solution (small mistake near the end) just brute-forcing with Pythagorean Theorem.  :)

In my opinion, the hardest AIME problems are the combinatorics ones - ants on a tetrahedron, etc.

[quote="pieterminate"]But seriously, if you look carefully at what you're given, it  [geometry] becomes easy requiring only the basic formulas.  But when you see these complex algebraic problems, you won't even know where to start unless you had a super hard course.[/quote]

...Switch around geometry and complex algebraic, in my opinion :lol:

[quote="pieterminate"][quote="scorpius119"]If we're only looking at the last 5 or so problems on the AIME, I would have to say that the geometry is CRIPPLINGLY hard. Unless I see the super-elegant solution quickly enough (which rarely ever happens), I usually end up pulling something heavy in computation and usually end up making stupid mistakes all over the place.

I sure hope #1 and #4 on this year's USAMO won't be geometry...on most of those there aren't even bad alternatives.[/quote]

A simple and elegant solution is easy to spot by intuition :) !

But seriously, if you look carefully at what you're given, it becomes easy requiring only the basic formulas.  But when you see these complex algebraic problems, you won't even know where to start unless you had a super hard course.[/quote]

Ehhh... I can assure that no AIME problems are not doable with algebra I, school geometry, and basic combinatorics (basically, stuff that's "obvious"). However, solutions to some problems are considerably easier to see with the right math background.

Edit--I revise that statement: this does not include complex number problems, but those are generally not that hard.

Geometry, by far. It's not that I can't do it, I just don't have any geometric intuition whatsoever and always make mistakes. Three of the five problems I missed were geometry, one due to lack of intuition (the rhombus problem), one since I forgot the correct formulation of Stewart's Theorem (I remembered $bmb+cnc=dad+man$, but $b$ is NOT on the same side as $m$!), and one because I just...I don't know, messed it up counting (#10). What's really frustrating about these geometry problems is that I got 60-90% of each solution but messed up one little thing and got them wrong. With the algebra/combinatorics, iff I get them, I'm sure of it.

Sometimes it's just weird, like how I missed #1 (AIME I) last year; it's a geometry problem, but my error came in multiplying 314 by 3 (I got 642, that's why we need calculators). Either way, around 80% of the problems I've missed on AIME that I thought I got have been geometry.

Why no one thinks combinatorics (a.k.a. counting) is hard??? I can NEVER get any combinatorics problems right without luck (of course, I mean AIME level).  How do you guys learn combinatorics??

I don't know for sure, but I think MATHCOUNTS has a disproportionate number of combinatorics problems, so that might be it. I've always assumed combinatorics was just something I was good at, the opposite of geometry.

Yea, but the whole auxilary line junk is fairly useless when you revert a problem to a straight up Law of Sines and Cosines problem...that stuff is easy, if a bit messy, but I guess I have a naturally affinity towards messy solutions.

I believe counting and probability are the hardest.

[quote="chess64"][quote="pieterminate"][quote="chess64"]

Well you are right that you don't need to know many formulas for geometry. But you do have to have a lot of creativity. You have to like randomly draw lines in random places all the time.[/quote]

Really?  Is that what we need to do?  So draw a giant diagram and draw all the lines possible and we are (near) done, right?[/quote]

Yeah, you go and do that...[/quote]

It's not exactly random line drawing. Don't go on a spree, because you'll probably become frustrated as to which lines to use and you'll have a million angles floating around... not a very good thing  ;) I think the best way to approach geometry is to turn to "POWER OF A POINT MODE" when circles are present and draw multiple lines that [i]help[/i]. Such lines would especially be perpendicular lines and parallel lines to other important points. For example, a quadrilateral in a circle almost begs to have at least one of its diagonals drawn.

[quote="probability1.01"]Drawing lines is overrated. I would estimate that only one in five or six geometry problems has "drawing a line" as an important step in it. Even then, when you draw a line, usually you have a very specific purpose in mind already.[/quote]

I totally agree. I think the reason why "drawing random lines" is looked upon highly is because there are some people (for me, the best example of this is Ryan Ko) who are just able to insert seemingly random lines out of nowhere which end up becoming instrumental to a solution. Some people just have a "knack" for drawing "random" lines but even so, they are definitely drawing lines in randomly but rather have a very explicit purpose for the drawing of each line, which you may discover if you ask them about their thought process.

[quote="WarpedKlown1335"]Well what if you're one of those people who gets the right answer, but after seemingly an eternity.  For example, the ellipse problem on the AMC 12B this year, I got it using matrices but it took like 25 minutes.  How do you all see quick solutions to problems so well?[/quote]

You have to train yourself to look for them. It is very easy to brute force almost any problem, but it makes it extremely ugly and you are likely to 1) screw up and 2) take a lot of time. Just train yourself to look for nice solutions even if you see a long solution immediately and you will get into the habit of finding the faster solutions most of the time.

[quote="perfectnumber628"]i think trig is the hardest, only because i looked at the trig problem on the AIME this year, laughed, pretended to work on it, gave up, moved on to the one about sequences that i could figure out.  but later when i saw the solutions i was like, "oh that doesn't look that hard..."[/quote]

I think trig is one of the easier things to learn in all of the AIME topics and should generally be the most approachable problem on the test outside of the first few.  I'm really mad that I rushed through my work on number 12 and got the wrong answer because I didn't read the interval or something silly like that.  It doesn't take very long to learn and memorize the formulas or even better, learn to derive them.  Read AoPS Volume 2 Chapter what.. 3? and a precalculus textbook and you should be solid on approaching most of the problems for trig that show up on AIME.  Trig problems generally have relatively nice solutions as long as you're willing to spend the five minutes with the algebra and work through it.

[quote="WarpedKlown1335"] Well what if you're one of those people who gets the right answer, but after seemingly an eternity. For example, the ellipse problem on the AMC 12B this year, I got it using matrices but it took like 25 minutes. How do you all see quick solutions to problems so well?[/quote]

I strongly suggest you look at more problems that you see a long solution to and say, is there a faster way to do that, a more convenient way.  I know you decently well and I would venture that you could benefit more from that rather than trying to learn alot more new material.

And yeah..  I still don't get alot of the drawing of seemingly random parallel lines things.  Its beautiful when it works out though and I'm really proud of myself when I add a parallel line that makes everything work out so nicely.

My worst topic definitely appears to be a tie between combinatorics and geometry.

[quote="Pakman2012"]My worst topic definitely appears to be a tie between combinatorics and geometry.[/quote]

Yeah, that's why you won combinatorics at HMMT.

Well see, Saiem's from Florida, like me.  And Florida as of lately is becoming a math powerhouse :D .

I thought you always had good math students (with the 4th largest state population, go figure :P ).  It's just now you guys finally realize that there is a world outside of Mu Alpha Theta.

Anyway, that still doesn't explain how Saiem is horrible at combinatorics and yet won it at HMMT.

Maybe there's a difference in style of the HMMT and the AIME problems in combinatorics? Heh, I don't like probability either.  :D

I only got like 6 questions right out of 10! I was lucky I understood alot of the notation lol.

I mean, I'm okay at combinatorics I guess.. but I just don't like it very much.
I can usually see a pretty nice solution, largely thanks to alot of teaching from Ben Cunningham, senior at Rickards, who is beastly at Combinatorics.  He would've probably won if he hadn't made dumb mistakes.

And anyhow, I got 2nd.... stupid sudoku problem.

Like I forgot to multiply by two.  I spent like 15 minutes on that problem cause I wanted my damn 8 points for that question.

I'm still really happy with the way HMMT went... despite my 3 (or so) on Algebra.


In $ \Delta ABC$, $ AC \equal{} BC$, $ m\angle DCB \equal{} 40^{\circ}$, and $ CD \parallel AB$. What is the number of degrees in $ m\angle ECD$?

[asy]pair A,B,C,D,E; B = dir(-40); A = dir(-140); D = (.5,0); E = .4 * dir(40);
draw(C--B--A--E,EndArrow); draw(C--D,EndArrow);
label("A",A,W); label("C",C,W);label("B",B,E);label("D",D,E);label("E",E,E);[/asy]

$ \angle DCB = \angle ABC = \angle BAC = 40^\circ$.
So, $ \angle ACB = 180^\circ - (40^\circ + 40^\circ) = 100^\circ$.
Thus, $ \angle ECD = 180^\circ - (\angle ACB + \angle DCB) = 180^\circ - (100^\circ + 40^\circ) = \boxed{40^\circ}$.

We know $ m\angle ECD \equal{} \angle EAB$, and $ m\angle ECD \plus{} \angle EAB \equal{} 140^{\circ}$, so $ m\angle ECD \equal{} 70^{\circ}$.

Edit: Oops failed  :blush:

[b]Another Solution[/b]

$ \angle DCB \equal{} \angle ABC \equal{} \angle BAC \equal{} 40^\circ$.
So, $ \angle ECB \equal{} 40^\circ \plus{} 40^\circ \equal{} 80^\circ$.
Thus, $ \angle ECD \equal{} \angle ECB \minus{} \angle DCB \equal{} 80^\circ \minus{} 40^\circ \equal{} \boxed{40^\circ}$.

[b]$ \$\mathrm{LaTeX}\$$ like Solution[/b]

$ \angle DCB \equal{} \angle ABC \equal{} \angle BAC \equal{} 40^\circ$.
And we know $ \angle ECD \equal{} \angle EAB$, Thus $ \angle ECD \equal{} \boxed{40^\circ}$.


inequalities
function
inequalities proposed

For any three positive reals a, b, c, we have

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

I wonder if something similar to darij proof of inequality 9 could be done here but in this way, define in the same way $x,y,z$. we have then:

$a\sqrt{\frac{a^2+bc+bc}{3}}>x^2\frac{x^2+yz+yz}{3}$

Same as Inequality 9 you post before.Here is my proof.

By Chebychev Inequality.
\[ \displaystyle \sum_{cyc} a \sqrt{a^{2}+2bc} \geq \frac{\sum_{cyc} a \sum_{cyc} \sqrt{a^{2}+2bc}}{3} \]
since $f(x)=\sqrt{x}$ is convex function,By Jensen Inequality.
\[ %Error. "displaytex" is a bad command.
\frac{\sum_{cyc} \sqrt{a^{2}+2bc}}{3}\geq \sqrt{\frac{(\sum_{cyc} a)^{2}}{3}} = \frac{\sum_{cyc} a}{\sqrt{3}} \]
Thus $ \displaystyle \frac{\sum_{cyc} a \sum_{cyc} \sqrt{a^{2}+2bc}}{3} \geq \frac{(\sum_{cyc} a)^{2}}{\sqrt{3}} $
then easy to prove that $\displaystyle\frac{(\sum_{cyc} a)^{2}}{\sqrt{3}} \geq \sqrt{3}\sum_{cyc} ab $

[quote="Pascual2005"]i am also sure Siouchung has also a very cool solution using Shur-Vornicu and his well known technique :D[/quote]

 :( not really....
this time i get stuck..

mathpk, you are wrong. f(x)=sqrt(x) is concave..

this costs me alot of times!!! :D
maybe siuhochung have anice proof :)

and my proof:

using mathpk 's way:

chebysev:

$lhs\geq\frac{(a+b+c)^2}{3}\sum\frac{a}{\sqrt{a^2+2bc}}$

the last maybe easy:
$\sum\frac{a}{\sqrt{a^2+2bc}}\geq\sqrt{3}$

so we get:

$(a+b+c)^2\geq3(ab+bc+ca)$ :lol:

[quote="cuong"]the last maybe easy:
$\sum\frac{a}{\sqrt{a^2+2bc}}\geq\sqrt{3}$

[/quote]

Easy!!!! (not for me :( )

[quote="cuong"]
the last maybe easy:
$\sum\frac{a}{\sqrt{a^2+2bc}}\geq\sqrt{3}$
[/quote]

cuong, can you check this?
I believe it is false.
Transform the inequality into
$\sum \frac{1}{\sqrt{1+2x}} \ge \sqrt{3}$ with $xyz=1$. Take x tends to 0 and y,z tends to infinity you get the infremum = 1

i deduce this from:

$\sum\frac{a}{\sqrt{a^2+8bc}}\geq1$

this is from imo :(

i will try to complete it :blush:

sorry again!!! :(

i have two way to prove it but i only try to prove the firts:

i hope not to make mistakes again :D

our ineq is equivalent to:

$a^4+b^4+c^4+2\sum\sqrt{(a^4+2a^2bc)(b^4+2acb^2)}\geq3((ab)^2+(bc)^2+(ac)^2)+4abc(a+b+c)$

using bcs :

$2\sum\sqrt{a^4+2a^2bc)(b^4+b^2ac)}\geq2(\sum(ab)^2)+4abc(\sum\sqrt{ab})$

so our ineq remains:

$a^4+b^4+c^4+4abc(\sum\sqrt{ab})\geq4abc(a+b+c)+(ab)^2+(bc)^2+(ac)^2$

let $\sqrt{a}=x$......

so we get:

$\sum\frac{(x^4-y^4)^2}{2}-4(xyz)^2(\sum (x-y)^2)\geq0$

mixing variable we get:

$(x-y)^2A+(y-z)^2B+(z-x)^2C\geq0$

$A=(x^2+y^2)^2(x+y)^2-4(xyz)^2$.......

using cosi we get:

$A=4(xy)^2(x+y)^2-4(xyz)^2=4(xy)^2(x+y-z)(x+y+z)$

so
$(x+y+z)(\sum(x-y)^2(xy)^2(x+y-z))\geq0$

let x>z>y so:

$[(z-y)^2((xy)^2(x+y-z)+(yz)^2(y+z-x))+(zx)^2(z-x)^2(z+x-y)](x+y+z)\geq0$

computation:

we have $x^3y^2+x^2y^3+y^3z^2+y^2z^3\geq xyz(xy+yz)$

by muirhead

everything complete :D  :D  :D

[quote="cuong"]

we have $x^3y^2+x^2y^3+y^3z^2+y^2z^3\geq xyz(xy+yz)$

by muirhead

[/quote]

As $xy \geq yz$ and $x \geq z$ by rearrangement

$(x^2y^2)x+(y^2z^2)z \geq x^2y^2z+y^2z^2x= xyz(xy+yz)$

I think you proof is OK in my opinion.

It is nice!! :)

[quote="cuong"]
$\displaystyle (x+y+z)(\sum(x-y)^2(xy)^2(x+y-z))\geq0$

let x>z>y so:

$[(x-y)^2((xy)^2(x+y-z)+(yz)^2(y+z-x))+(zx)^2(z-x)^2(z+x-y)](x+y+z)\geq0$
[/quote]
There's one thing I didn't understand. The factor [tex](y+z-x)[/tex] could be either positive or negative. Is the reasoning true in both cases?

sorry because of making you confused

iwill edit it:

i let x>z>y

and we get x-y>z-y

so $(x-y)^2(xy)^2(x+y-z)\geq(z-y)^2(xy)^2(x+y-z)$

because $(z-y)^2=(y-z)^2 $

i obtain it:

$\sum a\sqrt{2a^2+bc}\geq\sqrt{3}(ab+bc+ca)$

maybe  this is more easy than the original problem

Actually, the ineq.

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

implies

$a\sqrt{a^2+pbc}+b\sqrt{b^2+pca}+c\sqrt{c^2+pab}\ge \sqrt{1+p}(ab+bc+ca)$

for any $0  \le p \le3$.

ah, that's strange.
I tried to apply the same method i used in http://www.mathlinks.ro/Forum/viewtopic.php?t=33515 but i didn't able to solve it. Maybe i made some mistakes..

ah, it really works fine here.
Let us post it.
\[
\begin{gathered}
  \sum {a\frac{{a^2  + 2bc - \frac{3}
{4}\left( {b + c} \right)^2 }}
{{\sqrt {a^2  + 2bc}  + \frac{{\sqrt 3 }}
{2}(b + c)}}}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {a\frac{{4a^2  + 2bc - 3b^2  - 3c^2 }}
{{\sqrt {a^2  + 2bc}  + \frac{{\sqrt 3 }}
{2}(b + c)}}}  \geqslant 0 \hfill \\
\end{gathered}
\]
Let us break it into two parts,
\[
\sum {a\frac{{a^2  + bc - b^2  - c^2 }}
{{\sqrt {a^2  + 2bc}  + \frac{{\sqrt 3 }}
{2}(b + c)}}}  \geqslant 0
\]
and \[
\sum {a\frac{{a^2  - bc}}
{{\sqrt {a^2  + 2bc}  + \frac{{\sqrt 3 }}
{2}(b + c)}}}  \geqslant 0
\]
denote $f(a)=
{\frac{{1 }}
{{\sqrt {a^2  + 2bc}  + \frac{{\sqrt 3 }}
{2}(b + c)}}}
$
If $a \ge b$,
how to prove f is increasing?
i cannot prove it without heavy computations..

[quote="zhaobin"]I think Vasc mean we can prove:$f(p)=\frac{a\sqrt{a^2+pbc}+b\sqrt{b^2+pca}+c\sqrt{c^2+pab}}{\sqrt{1+p}}
$
is disincreasing when $0 \leq p \leq 3$[/quote]

well, if the inequality mentioned by vasc is the strongest (p=3 case), I think my method which works for p=3 case also works for $p \le 3$.

Indeed, I found that $f'(p)<0$ for non-trivial case.


To each square of a $2007 X2007$ chessboard we put one of the numbers $1 ,-1$ . Let $Ai$ denote the product of the numbers in $i-$ row  and   $Bj$ denote the product of the numbers in  $j-$ column of the board, with  $i,j = 1,2,3 , ... , 2007$. Prove that the sum
$S = A_{1}+A_{2}+...+A_{2007}+B_{1}+B_{2}+...+B_{2007}$ is not equal to $0$.

[u] Babis[/u]

i will prove to all chessboards with odd side!
suppose (2n+1)(2n+1).
let $q_{a}$ the number of -1's in our $A_{i}$'s and $q_{b}$ the same for $B_{i}$'s.
so, we have that $q_{a}\equiv q_{b}(mod.2)$, because the product of all $A_{i}$ = product of all $B_{i}$.

now, our sum is $(2n+1-2q_{a})+(2n+1-2q_{b}) = 4n+2-2(q_{a}+q_{b}) \equiv 2 (mod.4)$


function
inequalities
algebra unsolved
algebra

Find all functions $f: \mathbb{N}\to\mathbb{N}$ which satisfy the inequality:
\[f(3x+2y)=f(x)f(y)\]
for all non-negative integers $x,y$.

Check [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26929]this[/url].

[color=#FF0000][moderator edit: and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=37&t=4994]this[/url].][/color]

[quote=jastrzab]I think the only such function is $f(n)=1$.Well the proof isn't relly complicated :) Our left side is symmettric with respect to $x$ and $y$. So the right side also has to be, which gives us \[ f(3x+2y)=f(2x+3y) \]. And we obtain \[ f(3x+2y)=f(3x+2y+(y-x)) \] By putting respectively $y$ equall to $x+1,x+2,x+3,x+4,x+5$ we may obtain equations: \[ f(5x+2)=f(5x+3) \] \[ f(5x+4)=f(5x+6) \] \[ f(5x+6)=f(5x+9) \] \[ f(5x+8)=f(5x+12) \] \[ f(5x+10)=f(5x+15) \] Every positive integer greater than $15$ can be written in one of those forms on the left. This means, that for $n>15$ we can write (after using our equations several times) : for $a\equiv 0(mod5) : f(n)=a$ and for $n\equiv 1(mod 5)$ or $n\equiv 4 (mod 5): f(n)=b$ and for $n\equiv2(mod 5)$ or $n\equiv 3(mod 5): f(n)=c$ By putting in our first equation $x\equiv y\equiv 0 (mod 5)$ we obtain $a^2=a$ so $a=1$ And by "plugging" $x=y$ we have \[ f(5x)=f(x)^2 \] so $a=b^2$,$a=c^2$ which gives together that for every $n>15$ (or something like that) $f(n)=1$. To obtain this result for $n\leq 15$ all we have to do is to take any $y>15$ Then we have $1=f(3x+2y)=f(x)f(y)=f(x)$ Finally $f(n)=1$ :lol:[/quote]

f(n) can be 0 too, other than that nice solution :)

[quote=pco][quote=tudor129]f(n) can be 0 too, other than that nice solution :)[/quote]
No, $0\notin\mathbb N$ (convention generally used in this international forum when no explicit definition is given in the problem statement)[/quote]

Ye, I'm afraid that the person who posted should have mentioned that 0 is valid tho, because in romania, 0 counts as part of N, if not stated otherwise. The official solution also actually gives f(n)=0 as a solution. But yeah, now it's all clear I think :) (Anyways, thanks for letting me know of the international convention)


real analysis
search
complex analysis
real analysis theorems

Hi everybody.

This semester I'm taking differential equations at my university. We are using this textbook

http://www.pearsoned.co.uk/Bookshop/detail.asp?item=248264
http://www.amazon.com/Differential-Equations-Computing-Modeling-Third/dp/0130673374/ref=pd_lpo_k2_dp_k2a_1_img/103-8926301-7533450

However, the book doesn't suit me at all. It is full of non-relevant stuff... Every theorem is followed by a minimum of 5 examples, which makes this book a 900 page heavy-weight. Almost all of the important theorems are not proved. Furthermore the authors really did a great job to disguise the use of Fourier theory, and the bit of complex analysis that is needed for some parts... Since I study pure mathematics, I'm convinced this book is more suited for engineers.

As a consequence I'm in need of a proper mathematical rigorous threatment of the subject. By self-study I already learned about hilbert/banach/fourier/complex analysis/measure [basicly rudin I+II]; which is not assumed as a prerequisite for this book (not sure if these subject matter in a rigorous threatment of DE). I hope some of the analysis experts can help me

Preferrably I'm interested in an abstract concise threatment [cf. Rudin] (including PDEs), where all theorems are proved and you really have [i]to do [/i]the mathematics.

Thanks

There is a nice little book by Richard Bellman [url=http://www.fetchbook.info/compare.do?search=9780486622101]"Stability theory of differential equations"[/url] which covers the basics of ODE. I imagine there are similar texts around; I mention Bellman's because I happen to have it myself.

[url=http://www.fetchbook.info/compare.do?search=9780821807729]"PDE" by Evans[/url] could be the best textbook ever published on the subject, but it's aimed at graduate students. (PDE is really a graduate-level subject anyway.) Its Chapter 2 is a condensed version of a typical undergraduate PDE course.

How about a rigorous introduction to ODEs? I'm using the 2nd ed of Diff. Eqs, Dynamical systems, and an introduction to chaos but would like some supplements. on the other hand i wouldn't mind worked examples alongside the definition-theorem-proof style. :)

Check out George F.Simmons: [i]Differential Equations with Applications and Historical Notes[/i] 2nd Edition. It is a very challenging text on ODEs.


trigonometry
calculus
integration
limit
calculus computations

The integral $I_{n}$ is defined by \[I_{n}=\int^{\pi}_{0}{\frac{(\frac{\pi}{2}-x)\sin{(nx+\frac{x}{2})}}{\sin{(\frac{x}{2})}}\,dx}\]

where $n$ is a positive integer.  Evaluate $I_{n}-I_{n-1}$, and hence evaluate $I_{n}$ leaving the answer in the form of a sum.

[hide="HINT"]When you get down to $I_{n}-I_{n-1}=(.)$, make the step that $\sum^{n}_{1}{(I_{r}-I_{r-1})}=\sum^{n}_{1}{(.)}$ and look out for cancellation[/hide]

Nobody want to try this?  I thought it would have been worth doing  :oops:

I think the answer is $\frac{2}{3}\pi^{2}.$

I didn't solve this problem at all, but
As you got the result $I_{2n-1}=4\sum_{k=1}^{n}\frac{1}{(2k-1)^{2}},$

Let $m=2k-1,$ we have $I_{m}=4\sum_{m=1}^{2n-1}\frac{1}{m^{2}}\longrightarrow 4\cdot \frac{\pi^{2}}{6}=\frac{2}{3}\pi^{2}\ (n\to\infty).$


symmetry
inequalities
triangle inequality
absolute value
advanced fields
advanced fields unsolved

Find the metric spaces in which all triangles are degenerate.

P.S.

Sorry it sounds so vague, but this is pretty much all one can say without giving away the answer, or a hint, or something, and ruining the problem :)

We could at least define "degenerate": the sum of two of the side distances is the third side.

[quote="jmerry"]Subsets of the real line (with the induced metric) clearly work. The question you should be asking yourself: What others are there?[/quote]
Take the union of a family of such subsets, each with a single distinguished point, and then identify these points.

No, that doesn't work.

It's been long enough- here's the full solution:
The possible metric spaces are the subsets of $\mathbb{R}$ and the four-point space $a,b,c,d$ with $\delta(a,b)+\delta(b,c)=\delta(a,d)+\delta(d,c)=\delta(a,c)$
$=\delta(b,c)+\delta(c,d)= \delta(b,a)+\delta(a,d)=\delta(b,d)$.
This also implies $\delta(a,b)=\delta(c,d)$ and $\delta(b.c)=\delta(a,d)$.

Let $X$ be our metric space.
Introduce the relation "between": $b$ is between $a$ and $c$, written $abc$, if $\delta(a,b)+\delta(b,c)=\delta(a,c)$. This relation has the following properties:
Symmetry: If $abc$, then $cba$. This is obvious.
Interpolation: If $abc$ and $acd$, then $abd$ and $bcd$; this is true because we have $\delta(a,b)+\delta(b,c)+\delta(c,d)=\delta(a,d)$, and the triangle inequality forces $\delta(b,d)=\delta(b,c)+\delta(c,d)$.
Trichotomy (specific to this problem): If $a,b,c$ are distince, then we have exactly one of $abc,bca,cab$. This holds because every triangle is degenerate.

We say that a set $S$ in $X$ is linearly ordered if there is an order $<$ on $S$ such that $abc$ if and only if either $a<b<c$ or $c<b<a$.
Lemma 1: If $S$ is linearly ordered, $S$ is isometric to a subset of $\mathbb{R}$.
Proof: Choose $x\in S$. Let $f(x)=0$, and $f(y)=\delta(x,y)$ if $x<y$, $f(y)=-\delta(x,y)$ if $y<x$. We will show that $f$ is an isometry in three cases:
Case 1: $x<y<z$. Then $\delta(x,z)=\delta(x,y)+\delta(y,z)$ and $\delta(y,z)=\delta(x,z)-\delta(x,y)=f(z)-f(y)=|f(z)-f(y)|$.
Case 2: $y<x<z$. Then $\delta(x,y)+\delta(x,z)=\delta(y,z)=-f(y)+f(z)=|f(z)-f(y)|$.
Case 3: $y<z<x$. Then $\delta(x,y)=\delta(y,z)+\delta(x,z)$, so $\delta(y,z)=\delta(x,y)-\delta(x,z)=-f(y)+f(z)=|f(z)-f(y)|$.
In all three cases $\delta(y,z)$ is as desired. If $z<y$, exchange the order first.

Lemma 2 (Extension of orders): If any collection $S_i$ of $X$ are linearly ordered by $<_i$ such that there exist three points $a,b,c\in \cap_i S_i$ with $a <_i b<_i c$ for each $i$, then there is a linear order $<$ on $\cup_i S_i$ which extends each $<_i$
Proof: By Lemma 1, each $S_i$ is isometric by $f_i$ to a subset of the real line, with $f_i(b)=0$, $f_i(a)=-\delta(a,b)$, and $f_i(c)=\delta(b,c)$. Also, since $f_i(x)<0$ iff $xbc$ and $f_i(x)>0$ iff $abx$, $f_i(x)=f_j(x)$ if $x\in S_i\cap S_j$. Now we can define $f$ uniquely by $f(x)=f_i(x)$ if $x\in S_i$. It remains to show that $f$ is an isometry from $S$ into $\mathbb{R}$.
Let $x\in S_i$ and $y\in S_j$. We have $\delta(x,a)=|f(x)-f(a)|$, $\delta(x,b)=|f(x)|$, and $\delta(x,c)=|f(x)-f(c)|$, while $\delta(y,a)=|f(y)-f(a)|$, $\delta(y,b)=|f(y)|$ and $\delta(y,c)=|f(y)-f(c)|$. Considering the triangle $bxy$, either $\delta(x,y)=|f(x)+f(y)|$ or $\delta(x,y)=|f(x)-f(y)|$. By the triangle $axy$, either $\delta(x,y)=|f(x)-f(y)|$ or $\delta(x,y)=|f(x)+f(y)-2f(a)|$ By the triangle $cxy$, either $\delta(x,y)=|f(x)-f(y)|$ or $\delta(x,y)=|f(x)+f(y)-2f(c)|$. Since $2f(a),0$ and $2f(c)$ are all distinct, the three distinct numbers $f(x)+f(y),f(x)+f(y)-2f(a),f(x)+f(y)-2f(c)$ cannot all have the same absolute value and $\delta(x,y)=|f(x)-f(y)|$.

Lemma 3: If $S$ has an "extreme" element $a$ such that either $axy$ or $ayx$ for each other $x,y\in S$, $S$ can be linearly ordered with $a<x$ for all $x\in S$.
Proof: By Lemma 2, we need only prove this for four-element subsets $a,b,c,x$ Assume WLOG $abc$. If $acx$, we also have $bcx$ and $abx$ by interpolation; this is enough to define $a<b<c<x$. If $axc$,we split into cases. If $axb$, we have $xbc$ by interpolation and $a<x<b<c$. If $abx$, we have $bxc$ by interpolation and $a<b<x<c$.

Lemma 4: Any five-element subset $S=\{a,b,c,d,e\}$ of $X$ can be linearly ordered.
Proof: Suppose not. If both $\{a,b,c,d\}$ and $\{b,c,d,e\}$ could be linearly ordered, we could order $S$ by Lemma 2. Therefore WLOG $\{a,b,c,d\}$ can't be linearly ordered.
WLOG, we have $abc$. We can't have $acd$ or $cad$, so we must have $cda$. Similarly, we must have $bcd$. Using this, we must have $dab$ and the betweenness relation is complete.
Now, what about $e$?
Case 1: $aeb$. Then we have $aec$ and $ebc$ by interpolation with $abc$, then $dae$ and $deb$ by interpolation with $dab$.
Case 1.1: $ced$. Then we have $eda$ by interpolation with $cda$, and this contradicts $dae$.
Case 1.2: $cde$. Then we have $dea$ by interpolation with $cea$, and this contradicts $dae$.
Case 1.3: $ecd$. Then we have $bde$ by interpolation with $bce$, and this contradicts $deb$.
Therefore, we don't have $aeb$. By symmetry, we don't have $bec,ced$, or $dea$ either.
Now, if $e$ is an extreme element, we have a linear order by Lemma 3. We don't, so either $aec$ or $bed$. WLOG, we have $aec$. If $eda$, we have $dec$ by interpolation, contradicting case 1. Therefore, we have $ead$. By symmetry, we also have $eab$, $ecb$, and $ecd$. Now, we have all of $eab,bad,dae$; any of $bde,deb,ebd$ generates a contradiction with these, and this case is impossible.
This completes the proof, as we have reached contradiction in all cases: the set has a linear order.

Theorem: If $X$ has at least five elements, $X$ is isometric to a subset of $\mathbb{R}$.
Proof: Choose three elements $a,b,c$. By Lemma 4, each five-element set containing $a,b,c$ has a linear order. By Lemma 2, there is a linear order on all of $X$ extending these orders. By Lemma 1, $X$ is isometric to a subset of $\mathbb{R}$.

I'll leave the details of the exceptional case as an exercise- this proof is already long enough.

I didn't read it, but the answer certainly was Ok, so I guess there's not much that could have gone wrong :). Thanks for the interest.


ratio
trigonometry
projective geometry
geometry proposed
geometry

Let $ABCD$ be a convex quadrilateral. $I=AC\cap BD$, and $E$, $H$, $F$ and $G$ are points on $AB$, $BC$, $CD$ and $DA$ respectively, such that $EF \cap GH= I$. If $M=EG \cap AC$, $N=HF \cap AC$, show that \[\frac{AM}{IM}\cdot \frac{IN}{CN}=\frac{IA}{IC}.\]

[quote="Jutaro"]  [color=darkred]Let $ ABCD$ be a convex quadrilateral and $ I \equal{} AC\cap BD$ , $ E\in (AB)$ , $ H\in (BC)$ , $ F\in (CD)$ , $ G\in (DA)$

such that $ I\in EF \cap GH$ . Denote $ M\in EG \cap AC$, $ N\in HF \cap AC$ . Prove  that $ \frac {MA}{MI}\cdot \frac {NI}{NC} \equal{} \frac {IA}{IC}$ .[/color] [/quote]

[color=darkred][b][u]Lemma 1[/u].[/b] Let $ ABC$ be a triangle and $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ and $ P\in AD\cap EF$ . Then $ \frac {PD}{PA}\cdot BC \equal{} \frac {EC}{EA}\cdot DB \plus{} \frac {FB}{FA}\cdot DC$ .[/color]

[hide="Proof."]
[color=darkblue] Denote $ X\in EF\cap BC$ and $ Y\in EF$ so that  $ AY\parallel BC$ . Suppose w.l.o.g. that $ B\in (XC)$ . Observe that $ \frac {PD}{PA} \equal{} \frac {XD}{AY}$

$ \frac {EC}{EA} \equal{} \frac {XC}{AY}\ \ ,\ \ \frac {FB}{FA} \equal{} \frac {XB}{AY}$ and $ XB\cdot DC \plus{} XC\cdot DB \equal{} XD\cdot BC$ because $ (XD \minus{} BD)\cdot DC \plus{} (XD \plus{} DC)\cdot DB \equal{} XD\cdot BC$ .

Therefore, $ \frac {EC}{EA}\cdot DB \plus{} \frac {FB}{FA}\cdot DC \equal{}$ $ \frac {XC}{AY}\cdot DB \plus{} \frac {XB}{AY}\cdot DC \equal{}$ $ \frac {XC\cdot DB \plus{} XB\cdot DC}{AY} \equal{}$ $ \frac {XD\cdot BC}{AY} \equal{}$ $ \frac {PD}{PA}\cdot BC$ .[/color][/hide]

[color=darkred][b][u]Lemma 2[/u].[/b] Let $ ABCD$ be a convex quadrilateral and $ I\in AC\cap CD$ , $ E\in (AB)$ , $ F\in (CD)$ such that $ I\in EF$ . Then $ \frac {EB}{EA}\cdot\frac {FC}{FD} \equal{} \frac {IB}{ID}\cdot\frac {IC}{IA}$ .[/color]

[hide="Proof."]
Observe that $ \left\|\begin{array}{c} \widehat {EIB}\equiv\widehat {FID} \\
 \\
\widehat {EIA}\equiv\widehat {FIC}\end{array}\right\|$ and $ \left\|\begin{array}{c} \frac {EB}{EA} \equal{} \frac {IB}{IA}\cdot\frac {\sin\widehat {EIB}}{\sin\widehat {EIA}} \\
 \\
\frac {FC}{FD} \equal{} \frac {IC}{ID}\cdot\frac {\widehat {FIC}}{\sin\widehat {FID}}\end{array}\right\|\bigodot\ \implies\ \frac {EB}{EA}\cdot\frac {FC}{FD} \equal{} \frac {IB}{ID}\cdot\frac {IC}{IA}$ .   [/hide]

[color=darkblue][b][u]Proof of the proposed problem[/u].[/b] $ \left\|\begin{array}{ccccc} I\in HG & \stackrel {(lemma\ 2)}{\ \ \implies\ \ } & \frac {HB}{HC}\cdot\frac {GA}{GD} \equal{} \frac {IB}{ID}\cdot\frac {IA}{IC} & \implies & \frac {GD}{GA}\cdot IB \equal{} \frac {HB}{HC}\cdot\frac {IC}{IA}\cdot ID \\
 \\
I\in EF & \stackrel {(lemma\ 2)}{\ \ \implies\ \ } & \frac {EB}{EA}\cdot \frac {FC}{FD} \equal{} \frac {IB}{ID}\cdot\frac {IC}{IA} & \implies & \frac {EB}{EA}\cdot ID \equal{} \frac {FD}{FC}\cdot\frac {IC}{IA}\cdot IB\end{array}\right\|$ $ \bigoplus\ \implies$

$ \frac {GD}{GA}\cdot IB \plus{} \frac {EB}{EA}\cdot ID \equal{} \frac {IC}{IA}\cdot\left(\frac {HB}{HC}\cdot ID \plus{} \frac {FD}{FC}\cdot IB\right)$ $ \stackrel{(lemma\ 1)}{\ \ \implies\ \ }$ $ \frac {MI}{MA}\cdot BD \equal{} \frac {IC}{IA}\cdot\left(\frac {NI}{NC}\cdot BD\right)$ $ \implies$ $ \boxed {\ \frac {MA}{MI}\cdot\frac {NI}{NC} \equal{} \frac {IA}{IC}\ }$ .[/color]

Dedicate to  [b]Armpist[/b] ...

[quote="Jutaro"]  [color=darkred]Let $ ABCD$ be a convex quadrilateral and $ I \equal{} AC\cap BD$ , $ E\in (AB)$ , $ H\in (BC)$ , $ F\in (CD)$ , $ G\in (DA)$

such that $ I\in EF \cap GH$ . Denote $ M\in EG \cap AC$, $ N\in HF \cap AC$ . Prove  that $ \frac {MA}{MI}\cdot \frac {NI}{NC} \equal{} \frac {IA}{IC}$ .[/color] [/quote]

[color=darkred][b][u]Lemma 1[/u].[/b] Let $ ABC$ be a triangle and $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ and $ P\in AD\cap EF$ . Then $ \frac {PD}{PA}\cdot BC \equal{} \frac {EC}{EA}\cdot DB \plus{} \frac {FB}{FA}\cdot DC$ .[/color]

[hide="Proof."]
[color=darkblue] Denote $ X\in EF\cap BC$ and $ Y\in EF$ so that  $ AY\parallel BC$ . Suppose w.l.o.g. that $ B\in (XC)$ . Observe that $ \frac {PD}{PA} \equal{} \frac {XD}{AY}$

$ \frac {EC}{EA} \equal{} \frac {XC}{AY}\ \ ,\ \ \frac {FB}{FA} \equal{} \frac {XB}{AY}$ and $ XB\cdot DC \plus{} XC\cdot DB \equal{} XD\cdot BC$ because $ (XD \minus{} BD)\cdot DC \plus{} (XD \plus{} DC)\cdot DB \equal{} XD\cdot BC$ .

Therefore, $ \frac {EC}{EA}\cdot DB \plus{} \frac {FB}{FA}\cdot DC \equal{}$ $ \frac {XC}{AY}\cdot DB \plus{} \frac {XB}{AY}\cdot DC \equal{}$ $ \frac {XC\cdot DB \plus{} XB\cdot DC}{AY} \equal{}$ $ \frac {XD\cdot BC}{AY} \equal{}$ $ \frac {PD}{PA}\cdot BC$ .[/color][/hide]

[color=darkred][b][u]Lemma 2[/u].[/b] Let $ ABCD$ be a convex quadrilateral and $ \underline{I\in AC\cap BD}$ , $ E\in (AB)$ , $ F\in (CD)$ such that $ I\in EF$ . Then $ \frac {EB}{EA}\cdot\frac {FC}{FD} \equal{} \frac {IB}{ID}\cdot\frac {IC}{IA}$ .[/color]

[hide="Proof."]
Observe that $ \left\|\begin{array}{c} \widehat {EIB}\equiv\widehat {FID} \\
 \\
\widehat {EIA}\equiv\widehat {FIC}\end{array}\right\|$ and $ \left\|\begin{array}{c} \frac {EB}{EA} \equal{} \frac {IB}{IA}\cdot\frac {\sin\widehat {EIB}}{\sin\widehat {EIA}} \\
 \\
\frac {FC}{FD} \equal{} \frac {IC}{ID}\cdot\frac {\widehat {FIC}}{\sin\widehat {FID}}\end{array}\right\|\bigodot\ \implies\ \frac {EB}{EA}\cdot\frac {FC}{FD} \equal{} \frac {IB}{ID}\cdot\frac {IC}{IA}$ .   [/hide]

[color=darkblue][b][u]Proof of the proposed problem[/u].[/b] $ \left\|\begin{array}{ccccc} I\in HG & \stackrel {(lemma\ 2)}{\ \ \implies\ \ } & \frac {HB}{HC}\cdot\frac {GA}{GD} \equal{} \frac {IB}{ID}\cdot\frac {IA}{IC} & \implies & \frac {GD}{GA}\cdot IB \equal{} \frac {HB}{HC}\cdot\frac {IC}{IA}\cdot ID \\
 \\
I\in EF & \stackrel {(lemma\ 2)}{\ \ \implies\ \ } & \frac {EB}{EA}\cdot \frac {FC}{FD} \equal{} \frac {IB}{ID}\cdot\frac {IC}{IA} & \implies & \frac {EB}{EA}\cdot ID \equal{} \frac {FD}{FC}\cdot\frac {IC}{IA}\cdot IB\end{array}\right\|$ $ \bigoplus\ \implies$

$ \frac {GD}{GA}\cdot IB \plus{} \frac {EB}{EA}\cdot ID \equal{} \frac {IC}{IA}\cdot\left(\frac {HB}{HC}\cdot ID \plus{} \frac {FD}{FC}\cdot IB\right)$ $ \stackrel{(lemma\ 1)}{\ \ \implies\ \ }$ $ \frac {MI}{MA}\cdot BD \equal{} \frac {IC}{IA}\cdot\left(\frac {NI}{NC}\cdot BD\right)$ $ \implies$ $ \boxed {\ \frac {MA}{MI}\cdot\frac {NI}{NC} \equal{} \frac {IA}{IC}\ }$ .[/color]

Notice that triangles $ \triangle AEG$ and $ \triangle CFH$ are perspective through $I,$ thus by Desargues theorem, the intersections $ X \equiv AD \cap BC,$ $ Y \equiv AB \cap DC$ and $ P \equiv EG \cap HF$ are collinear. Let $ Q \equiv AC \cap XY.$ Then by Menelaus' theorem for  $ \triangle MQP$ and $ \triangle NQP$ cut by the straight lines $ AB$ and $ CD,$ we have

$ \frac {YQ}{YP} \cdot \frac {PE}{EM} \cdot \frac {MA}{QA} \equal{} 1 \ \ , \ \ \frac {YQ}{YP} \cdot \frac {PF}{NF} \cdot \frac {NC}{QC} \equal{} 1$

$ \Longrightarrow  \frac {MA}{NC} \cdot \frac {PE}{EM} \cdot \frac {NF}{PF} \equal{} \frac {QA}{QC} \ (\star)$

But by Menelaus' theorem for $ \triangle MNP$ cut by $ \overline{FIE},$ we have $\frac {PE}{EM} \cdot \frac {NF}{PF} \equal{} \frac {IN}{IM}$

Combining the latter expression with $ (\star)$ yields $ \frac {MA}{NC} \cdot \frac {IN}{IM} \equal{} \frac {QA}{QC}$

Since $ (A,C,I,Q) \equal{} \minus{} 1,$ it follows that $ \frac {AM}{IM} \cdot \frac {IN}{CN} \equal{} \frac {QA}{QC} \equal{} \frac {IA}{IC}.$

Note that the problem statement is purely projective. Also, it is well-known that there is a projective transformation which maps $ABCD$ to a square $A'B'C'D'$. Let this transformation map $E,F,G,H,I,M,N$ to $E',F',G',H',I',M',N'$, respectively.

[asy]
defaultpen(fontsize(10pt));
size(150);

pair A=dir(135);
pair B=dir(225);
pair C=dir(315);
pair D=dir(45);
pair I=extension(A,C,B,D);
pair E=0.6*B+0.4*A;
pair H=0.8*B+0.2*C;
pair F=extension(E,I,C,D);
pair G=extension(H,I,A,D);
pair M=extension(E,G,A,C);
pair N=extension(H,F,A,C);

dot("$A'$", A, dir(A));
dot("$B'$", B, dir(B));
dot("$C'$", C, dir(C));
dot("$D'$", D, dir(D));
dot("$E'$", E, dir(E));
dot("$F'$", F, dir(F));
dot("$G'$", G, dir(G));
dot("$H'$", H, dir(H));
dot("$I'$", I, dir(270));
dot("$M'$", M, dir(185));
dot("$N'$", N, dir(0));

draw(A--B--C--D--A--cycle);
draw(E--G--H--F--E--cycle);
draw(A--C);
draw(B--D);
[/asy]

As projective transformations preserve cross-ratio and thanks to the central symmetry in a square, we have that

\begin{align*}
(A,I;M,C)&=(A',I';M',C')\\
&=\dfrac{\frac{A'M'}{M'I'}}{\frac{A'C'}{C'I'}}\\
&=\dfrac{\frac{C'N'}{N'I'}}{\frac{C'A'}{A'I'}}\\
&=(C',I';N',A')\\
&=(C,I;N,A)\\
\end{align*}

Finally, this implies that

\begin{align*}
\dfrac{\frac{AM}{MI}}{\frac{AC}{CI}}&=\dfrac{\frac{CN}{NI}}{\frac{CA}{AI}}\\
\dfrac{AM}{IM}\cdot\dfrac{IN}{CN}&=\dfrac{IA}{IC}\\
\end{align*}

We are done. $\square$


Compute the sum  ( 1 - 1/3)+( 1/2 - 1/4)+(1/3 - 1/5)+ ... +(1/168 - 1/170) . Express your answer to the nearest thousandths.

ANSWER [hide]1.488[/hide]

I couldnt find a pattern.  Can someone please help me?  thanks.

Note that all the terms appear twice except for $ 1,\frac{1}{2},\frac{1}{169},\frac{1}{170}$

They all cancel out except for those 4.  Do the arithmetic.

[quote="mathcrazed"]Note that all the terms appear twice except for $ 1,\frac {1}{2},\frac {1}{169},\frac {1}{170}$

They all cancel out except for those 4.  Do the arithmetic.[/quote]

How do you know that $ 1,\frac {1}{2},\frac {1}{169},\frac {1}{170}$ are the fractions that don't cancel?

Thanks.

there is a negative 1/3 and a positive 1/3, so those cancel

then there's a negative 1/4 and a positive 1/4, so those cancel.

Repeat, and you get those terms as the left over ones

[quote="happyme"]And the ones that don't cancel are either starting (as in the first term of that isn't in the sequence) or ending (as in the second one that would cancel comes after the sequence ends).[/quote]

I still don't get how the $ \frac{1}{169}$ and the $ \frac{1}{170}$ cancel out.

Thanks.


function
limit
real analysis
real analysis unsolved

Let $f$ be a continious function from the ball $B_n=(x\in\mathbb{R}^n |  ||x||<1)$ into itself.Assume $||f(x)||<||x||$ for all non-zero $x\in B_n$. Let $x_0$ be a nonzero point of $B_n$, and define the sequence $(x_k)$ by setting $x_k=f(x_{k-1})$.Prove that $\lim x_k=0$

$\|x_n\|$ is decreasing, so it tends to some real $0\le r<1$. Suppose $r>0$. From the sequence $(x_n)_n$ we can extract a subsequence $(y_m)_m$ converging to a point $x$ of norm $r$, but then $f(y_m)$ will have norm $<r$ for a large enough $m$, and this contradicts the fact that $\|f(x_n)\|>r,\ \forall n$.

[quote="grobber"] but then $f(y_m)$ will have norm $<r$ for a large enough $m$[/quote]
Why?

That's because $y_m=x_{n_m}\to x$, so $\|x_{n_{m+1}}\|\le\|x_{n_m+1}\|=\|f(x_{n_m})\|\to\|f(x)\|<r$.

P.S.

I hate it when people drag me back into old threads. It can be very distracting.

I am sorry, but I still do not understand.

Can't it be that $|x_{n_{m+1}}|$ and $|f(x_{n_m})|$ are both decreasing to $r$ and $|x_{n_{m+1}}|<|f(x_{n_m})|$.

And isn't $|f(x)|=r$?


I don't know if it is the correct place to ask if my solution is correct or not. (If it is not the correct place please tell me which one is :(  )

[b]Problem:[/b] If $ a$ and $ b > 2$ are any positive integers, prove that $ 2^a\plus{} 1$ is not
divisible by $ 2^b \minus{} 1$.
[b]My solution:[/b] We assume the opposite is correct.
If some $ d$ divides both of them,it must divide their sum.
So, $ d|(2^a\plus{}1)\plus{}(2^b\minus{}1)\equal{}2^a\plus{}2^b$
So, $ d|2$ which is a contradiction since both are odd and $ d>1$

Is it correct? Please help  :(

$ 2^a \plus{} 2^b$ has lots of odd divisors, so this isn't enough at all.  The relevant lemma is the following:  $ \gcd(2^x \minus{} 1, 2^y \minus{} 1) \equal{} 2^{\gcd(x, y)} \minus{} 1$.

Ah...now I see, I did a horrible mistake :(
Thank you for your reply.
But, how to solve this problem. :ninja:


algebra
polynomial
pigeonhole principle
IMC
college contests

Let $ \mathbb{Z}[x]$ be the ring of polynomials with integer coefficients, and let $ f(x), g(x) \in\mathbb{Z}[x]$ be nonconstant polynomials such that $ g(x)$ divides $ f(x)$ in $ \mathbb{Z}[x]$. Prove that if the polynomial $ f(x)\minus{}2008$ has at least 81 distinct integer roots, then the degree of $ g(x)$ is greater than 5.

I'm probably wrong, but this seems to be a very stupid problem:observe that $ g(x)$ is one of the $ 16$ integer divisors of $ 2008$ and so there must be at least 6 of the 81 integers giving the same value. Since $ g$ is nonconstant, its degree must be at least $ 6$.

[quote="harazi"]I'm probably wrong, but this seems to be a very stupid problem[/quote]I think you are right though. The idea of pigeonhole on roots of a polynomial keeps returning to the IMC, year after year after year.


Let $p_{k}$ be the probability of choosing a positive integer $n$ from the interval $[2^{k-1}, 2^{k}-1]$ such that the binary representation of $n$ does not contain three consecutive $1$'s. Evaluate the infinite summation $p_{1}+p_{2}+p_{3}+\cdots$.

[hide]$p_{k}$ will only be nonzero when $k\ge3$ for obvious reasons. $2^{k}= 2(2^{k-1})$, so the length of each interval is $2^{k-1}$. It is probably easier to calculate the probability that there are three consecutive $1$s. There will be $k-2$ numbers with three consecutive ones in each interval because the sequence of ones can only fit $k-2$ times. Therefore $p_{k}= \frac{2^{k-1}-k+2}{2^{k-1}}$. I don't know how to generate the infinite sum from this.[/hide]

I'm pretty sure you'll overcount like that. And $p_{k}$ will only be less than $1$ if $k \ge 3$ you mean.

[hide="hint"] Let $S_{k}$ be the number of $k$-digit binary numbers that don't contain three consecutive $1$'s. Show that we have the recursion
[hide="giveaway"]
$S_{k+3}= S_{k+2}+S_{k+1}+S_{k}$[/hide][/hide]

For anyone interested, I posted the solution on my blog [url=http://wangsblog.com/jeffrey/?p=303]here[/url]. It's a cool problem, really!


[url=http://img14.imageshack.us/i/solve.jpg/][img]http://img14.imageshack.us/img14/9594/solve.th.jpg[/img][/url]

can I use separation of variable to solve for this?

Yes, you can: [hide="solution"]

$ (2y\plus{}2)dy\equal{}(3x^2\plus{}2x\plus{}4)dx$

$ y^{2} \plus{}2y \equal{} x^{3} \plus{} x^{2} \plus{}4x \plus{}c$

Substituting $ x\equal{}0, y\equal{}\minus{}2$, $ c\equal{}0$.

Adding $ 1$ to both sides,

$ (y\plus{}1)^{2} \equal{} \minus{}(\sqrt{x^{3} \plus{} x^{2} \plus{} 4x \plus{}1} \plus{}1)$,

where we choose the negative square root because of the initial condion.

Hope this is helpful.

[/hide]


function
geometry
parallelogram
vector
complex analysis
complex numbers
complex analysis unsolved

Let $ f$ be an entire function on $ \mathbb C$ and $ \omega_1,\omega_2$ are complex numbers such that $ \frac {\omega_1}{\omega_2}\in{\mathbb C}\backslash{\mathbb Q}$. Prove that if for each $ z\in \mathbb C$, $ f(z) \equal{} f(z \plus{} \omega_1) \equal{} f(z \plus{} \omega_2)$ then $ f$ is constant.

This is immediate: the image of $ f$ is the same as the image of the parallelogram built on those complex numbers by double periodicity, so this image is bounded, therefore by Liouville the function is constant.

Yes, but in the case that $ \omega_1/\omega_2$ is in $ \mathbb R\backslash \mathbb Q$ then $ z$ is an accumulation point in $ f^{ \minus{} 1}(f(z))$ which proves that $ f$ is a constant function.

Indeed, I forgot the case when the two vectors are on the same line, and as you said this is again trivial by classical density result. Thanks for the observation.


benorin's proof is fine, but if you wanted something of $\epsilon$-$\delta$ kind, then set $\epsilon=f(a)/2$.

[quote="drunner2007"][quote="benorin"][quote="drunner2007"]Suppose that $U \subset R^{n}$ is open and $f: U \rightarrow R$ is continuous. If $a \in U$ and $f(a)>0$, prove that there is a $\delta >0$ such that $f(x)>0$ for all $x \in B(a,\delta)$.[/quote]

Define $V=\left\{ x\in R^{n}|f(x)>0\right\}$. $V$ is nonempty for certianly $a\in V$, and $V$ is open by the continuity of $f$ since $V$ is the preimage of the open set $f(V)=\left\{ f(x)\in R|f(x)>0\right\}$. Define $W=U\cap V$ which is open since the finite intersection of open sets is open. Thus $a\in W$ is an interior point of $W$, hence for every $a\in W$ there exists a $\delta >0$ such that $B(a,\delta)\subset W\subset V$, the required result.[/quote]

right on. that's how i proved it. would you buy chance be using the same multivariable text i got this from? (multivariable math. by theodore shifrin)[/quote]

I went off my memory of Rudin's Principles of Mathematical Analysis and looked-up continuous map on mathworld.com.


From the order axioms for R, show that the set of positive real numbers is closed under addition and multiplication.

That's very much a question of where you start, of what you take for the axioms. For instance, the form I usually use starts with the notion of the positive numbers and takes the closure of the positive numbers under addition and multiplication as axioms.

Obviously, there are other places to start. But, precisely because there are multiple reasonable versions, you can't just pose this problem without stating which version you are taking for the axioms. (And these would be the axioms of any ordered field, not just $ \mathbb{R}.$)


Prove that $ \tan 9^\circ\minus{}\tan 27^\circ\minus{}\tan 63^\circ \plus{}\tan81^\circ \equal{} 4$

Hi Obel1x. I don't know if I've got the same solution as you.

I regrouped the four terms on the left to get

$ (\tan 9 \plus{}\frac{1}{\tan 9}) \minus{} (\tan 27 \plus{} \frac{1}{\tan 27})$

Getting common denominators gives $ \frac{\sec^{2} 9}{\tan 9} \minus{} \frac{\sec^{2} 27}{\tan 27}$

This gives $ \frac{1}{\sin 9 \cos 9} \minus{} \frac{1}{\sin 27\cos27}$

$ \equal{}\frac{2}{\sin 18} \minus{} \frac{2}{\sin 54}$.

Now the values of $ \sin18$ and $ \sin 54$ are well known, e.g. see here: http://ecademy.agnesscott.edu/~lriddle/ifs/trig/trig2.htm

So we have $ \frac{8}{\sqrt{5}\minus{}1} \minus{} \frac{8}{\sqrt{5}\plus{}1}$

which simplifies to $ 4$.

You don't need to know the values of $ \sin 18$ or $ \sin 54$

We want to prove that :

\[ \frac{1}{\sin 18} \minus{}\frac{1}{\sin 54} \equal{}2\iff \sin 54 \minus{}\sin 18 \equal{}2\sin 18\sin 54\]

The last one is true since $ \sin 54 \minus{}\sin 18\equal{}2\sin 18\cos 36\equal{}2\sin 18\sin 54$

hello, using that
$ \tan\left(\frac{\pi}{20}\right)\equal{}1\plus{}\sqrt{5}\minus{}\sqrt{5\plus{}2\sqrt{5}}$
$ \tan\left(\frac{3\pi}{20}\right)\equal{}\minus{}1\plus{}\sqrt{5}\minus{}\sqrt{5\minus{}2\sqrt{5}}$
$ \tan\left(\frac{21\pi}{60}\right)\equal{}\minus{}1\plus{}\sqrt{5}\plus{}\sqrt{5\minus{}2\sqrt{5}}$
$ \tan\left(\frac{81\pi}{100}\right)\equal{}1\plus{}\sqrt{5}\plus{}\sqrt{5\plus{}2\sqrt{5}}$
we get $ 4$.
Sonnhard.

[quote="AndrewTom"]Hi Obel1x. I don't know if I've got the same solution as you.[/quote]
[i][hide="see my solution"]
[/i]
$ (\frac{\sin 81^\circ}{\cos81^\circ}\minus{}\frac{\sin 63^\circ}{\cos63^\circ})\plus{}(\frac{\sin 9^\circ}{\cos9^\circ}\minus{}\frac{\sin 27^\circ}{\cos27^\circ})$

$ \implies \frac{\sin 18^\circ}{\cos81^\circ \cdot\cos63^\circ} \minus{} \frac{\sin 18^\circ}{\cos9^\circ \cdot\cos27^\circ}$

$ \implies \frac{\sin 18^\circ\cdot \sin 54^\circ}{\cos81^\circ \cdot\cos63^\circ \cdot\cos9^\circ \cdot\cos27^\circ}$   $ \cdots (*)$

[i]Using[/i]
$ \cos x\equal{} sin(\frac{\pi}{2}\minus{}x)$ and $ \sin 2\alpha\equal{}2\sin \alpha\cos\alpha$

[i]Finally from[/i] $ (*)$ [i]we have[/i]

$ \frac{2\cdot2 \cdot\sin 9^\circ\cdot\cos9^\circ  \cdot\sin 27^\circ\cdot\cos 27^\circ}{\sin 9^\circ \cdot\sin27^\circ \cdot\cos9^\circ\cdot \cos27^\circ} \implies  \tan 9^\circ\minus{}\tan 27^\circ\minus{}\tan 63^\circ\plus{}\tan81^\circ \equal{} 4$[/hide]


How come if you have aops style you can put a source in the getting started forum, but if you have ML style you can't?

You can't put a source in AoPS style either. Are you sure it was the getting started? I just tested it, and it works as it should ... :?

This was posted in basics: http://artofproblemsolving.com/Forum/viewtopic.php?t=26822

And this in getting started: http://artofproblemsolving.com/Forum/viewtopic.php?t=26014

So what has that got to do with the style?  :?
As for why McCrawford and Silverfalcon can post sources, please read: http://www.mathlinks.ro/Forum/viewtopic.php?t=23778 :)


algebra
polynomial
number theory unsolved
number theory

$ a\geq 0,b\geq 2$,$ a,b\in\mathbb{Z ^{\plus{}} }$
Let $ P(x)$ has integer coefficients
$ P(0)\geq 0$
$ \forall n\in\mathbb{Z^{ \plus{} }}$ $ b^{n}|a^{n} \plus{} P(n)$  $ \implies$  $ b|a$ and $ P(x) \equal{} 0$

[quote="cnyd"]$ a\geq 0,b\geq 2$,$ a,b\in\mathbb{Z ^{ \plus{} } }$
Let $ P(x)$ has integer coefficients
$ P(0)\geq 0$
$ \forall n\in\mathbb{Z^{ \plus{} }}$ $ b^{n}|a^{n} \plus{} P(n)$  $ \implies$  $ b|a$ and $ P(x) \equal{} 0$[/quote]
It seems that this problem has been posted before . From the condition we have : $ b^n|a^n\plus{}P(n)$ and $ b^{n\plus{}1}|a^{n\plus{}1}\plus{}P(n\plus{}1)$ .
It is easy to see that $ b^n|a(a^n\plus{}P(n))\minus{}a^{n\plus{}1}\minus{}P(n\plus{}1)$ or $ b^n|P(n\plus{}1)\minus{}aP(n)$ . We knew that this fact holds iff $ P(n\plus{}1)\minus{}aP(n)\equal{}0$ for $ n$ is large enough . By the property of  polynomials we have :$ P(x\plus{}1)\minus{}aP(x) ,\forall x\in R$ . By comparing the leading coefficient , it is easy to show that $ P(x)\equiv 0$ .


Biff has 30 liters of pure ethanol, how much water must be added for the solution to be diluted to 60% ethanol solution?

how would you do these types of problem?

is there any quick method?

add 20 liters of water to get a fraction 30/50 equal to 3/5 which is 60 percent

just make the amount added be x and use x variable to solve

[hide]
Let s = volume of final solution in litres
$0.6s = 30$
$s = \frac{30}{0.6}= 50$
[/hide]

In general, the way to do these mixture problems is to think about the amount of pure solvent and the amount of pure solute.

I'll give another one for you to try.

You have 54 liters of 25% alchohol.
You want 34% alchohol.
How much 50% alchohol must you add to the solution to have 34% concentration?

[hide]
Set up the equation using the amnt. of pure solvent and solute:
$(0.25)(54)+(0.5)(x) = (0.30)(54+x)$
The reason for the $(54+x)$ is that we are adding some 50% alchohol, so we need to account for what we are adding.
I'll let you solve this on your own, and I'll also double embed it:
[hide]
Expand and collect like terms:
$0.5x+13.5 = 0.3x+16.2$
Subtract 13.5 from both sides:
$0.5x = 0.3x+2.7$
$0.2x = 2.7$
Multiply by 5:
$x = 13.5$
Therefore, you must add 13.5 liters of 50% alchohol to the 54 liters of 25% alchohol to get 67.5 liters of 34% alchohol.
Try doing this out to get the hang of the whole solvent-solute thing.

[/hide]
[/hide]


I was wondering if anyone knows how to prove the following:

Let $a_{1}$, $a_{2}$, $b_{1}$, and $b_{2}$ be all nonnegative, and
$a_{1}+a_{2}= b_{1}+b_{2}$ be given.
Suppose that $\min(a_{1},a_{2}) \le b_{1}\le \frac{a_{1}+a_{2}}2 \le b_{2}\le \max(a_{1},a_{2})$
Then prove that $a_{1}\cdot a_{2}\le b_{1}\cdot b_{2}$

$\text{\tiny{(2/21/2007 6{: }21 PM): \LaTeX ed by moderator, moved to Intermediate Topics}}$

you should post this in another forum as you probably won't get any replies here

Similar idea to standard proof of Am-Gm :wink:

[hide="A brutal solution"]From the given conditions we can easily get that $\left(a_{1},a_{2}\right)\succ\left(b_{1},b_{2}\right)$. Taking the natural logarithm of both sides of the inequality gives $\ln a_{1}+\ln a_{2}\leqslant \ln b_{1}+\ln b_{2}$. It is just the Karamata inequality for the concave function $\ln$, so we are done. [/hide]


School uniform parts are on sale. The $ \$$25 pair of slacks can be
purchased at a 15$ \%$ discount, and the $ \$$18 shirt can be
purchased at a 20$ \%$ discount. What is the total number of
dollars spent to purchase 1 pair of slacks and 1 shirt during the
sale? Express your answer to the nearest hundredth.


Let $ \mathsf{A}$ be a square matrix such that $ \mathsf{A}^m \equal{} 0$ for some integer $ m$. Show that every eigenvalue of $ \mathsf{A}$ is zero.

If $ \lambda$ is an eigenvalue of A there is a non-zero vector $ x$ such that $ Ax \equal{} \lambda x$: then, multiplying both sides by $ A^{m \minus{} 1}$ we have $ A^mx \equal{} \lambda^m x \equal{} > \lambda^m x \equal{} 0 \equal{} > \lambda^m \equal{} 0 \equal{} > \lambda \equal{} 0$. Such matrices are called [i]nilpotent matrices[/i].

I see - can such matrices be diagonalized?

Take a moment to think about that: if such a matrix could be diagonalized, what would the diagonal form have to look like? And what would that tell you about the original matrix?

A matrix $ A$ is diagonalizable iff there is a non-singular matrix $ P$ and a diagonal matrix $ D$ such that $ A \equal{} PDP^{\minus{}1}$ - in other words, if A is similar to a diagonal matrix. Due to the similarity relation, A and D share the same eigenvalues (among several other interesting properties). For $ A \neq O_{n \times n}$ nilpotent, whose all eigenvalues are zero, we would have $ D \equal{} O_{n \times n}$ and so $ A \equal{} O_{n \times n}$: a contradiction.

I knew [i]you[/i] knew that, Carcul. I was wondering if aidan could get there.

[quote="Carcul"]A matrix $ A$ is diagonalizable iff there is a non-singular matrix $ P$ and a diagonal matrix $ D$ such that $ A \equal{} PDP^{ \minus{} 1}$ - in other words, if A is similar to a diagonal matrix. Due to the similarity relation, A and D share the same eigenvalues (among several other interesting properties). For $ A \neq O_{n \times n}$ nilpotent, whose all eigenvalues are zero, we would have $ D \equal{} O_{n \times n}$ and so $ A \equal{} O_{n \times n}$: a contradiction.[/quote]

I understand what Carcul means here - I was actually trying to prove the original problem I posted here in this way although I abandoned it because I thought there were gaps:

If we assume that the matrix can be diagonalized, then $ \mathsf{A} \equal{} \mathsf{S}\Lambda\mathsf{S}^{\minus{}1}$, where $ \mathsf{S}$ is the eigenvector matrix of $ \mathsf{A}$ and $ \Lambda$ is the diagonal eigenvalue matrix. But the only way this can be zero is if the matrix $ \Lambda$ is the zero matrix. Is this correct? (but I suspect it isn't)

Why would you suspect not? If $ A \equal{} S^{ \minus{} 1}\Lambda S,$ then $ \Lambda \equal{} SAS^{\minus{}1}.$

Actually I meant $ \mathsf{A}^m \equal{} \mathsf{S}\Lambda^m\mathsf{S}^{ \minus{} 1}$, if the LHS is zero then the RHS is zero. But then the eigenvector matrix wouldn't exist if the eigenvalues are all zero, right? They would be arbitrary.

And if $ \mathsf{A} \equal{} \mathsf{S}\Lambda\mathsf{S}^{ \minus{} 1}$, $ \Lambda \equal{} \mathsf{S}^{ \minus{} 1}\mathsf{A}\mathsf{S}$ right?

Wikipedia also says "Some matrices are not diagonalizable over any field, most notably nonzero nilpotent matrices" but I can't quote that in my proof  :maybe:

Fixed my typo.

The ingredients are all here; you just have to think carefully about what you have.