Given $ a, b, c \geq\ 0.$ Prove that:
$ (\frac {a}{b \plus{} c})^3 \plus{} (\frac {b}{c \plus{} a})^3 \plus{} (\frac {c}{a \plus{} b})^3 \plus{} \frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq\ \frac {1}{2}. (\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca})^2$
It's stronger than a well-known:
$ (\frac {a}{b \plus{} c})^3 \plus{} (\frac {b}{c \plus{} a})^3 \plus{} (\frac {c}{a \plus{} b})^3 \plus{} \frac {5abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq\ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}$
PS: With this old ineq, we are easily to solve it by Am-Gm and...
Notice that: $ abc(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca) \geq\ 0$
 :)

[quote="nguoivn"]Given $ a, b, c \geq\ 0.$ Prove that:
$ (\frac {a}{b \plus{} c})^3 \plus{} (\frac {b}{c \plus{} a})^3 \plus{} (\frac {c}{a \plus{} b})^3 \plus{} \frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq\ \frac {1}{2}. (\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca})^2$
It's stronger than a well-known:
$ (\frac {a}{b \plus{} c})^3 \plus{} (\frac {b}{c \plus{} a})^3 \plus{} (\frac {c}{a \plus{} b})^3 \plus{} \frac {5abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq\ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}$
PS: With this old ineq, we are easily to solve it by Am-Gm and...
Notice that: $ abc(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca) \geq\ 0$
 :)[/quote]
I haven't tried with AM-GM as you said but the inequality is actually weak and we can prove it using Cauchy Schwarz + Schur as follow:
Applying Cauchy Schwarz Inequality, we have
\[ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b} \ge \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)}\]
It follows that
\[ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \le 2\left( \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b} \minus{}1\right)\]
And we can deduce our inequality to
\[ \frac{a^3}{(b\plus{}c)^3}\plus{}\frac{b^3}{(c\plus{}a)^3}\plus{}\frac{c^3}{(a\plus{}b)^3} \plus{}\frac{abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \ge 2\left( \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\minus{}1\right)^2\]
Setting $ x\equal{}\frac{2a}{b\plus{}c},y\equal{}\frac{2b}{c\plus{}a},z\equal{}\frac{2c}{a\plus{}b}$, then $ xy\plus{}yz\plus{}zx\plus{}xyz\equal{}4$ and our inequality becomes
\[ x^3\plus{}y^3\plus{}z^3\plus{}xyz \ge 4(x\plus{}y\plus{}z\minus{}2)^2\]
Now, we denote $ p\equal{}x\plus{}y\plus{}z,q\equal{}xy\plus{}yz\plus{}Zx,r\equal{}xyz,$ then $ q\plus{}r\equal{}4$ and our inequality is equivalent to
\[ p^3\minus{}3pq\plus{}4r \ge 4(p\minus{}2)^2\]
\[ p^3 \minus{}3p(4\minus{}r)\plus{}4r \ge 4(p\minus{}2)^2\]
\[ (p\minus{}4)(p^2\plus{}4) \plus{}(3p\plus{}4)r \ge 0\]
If $ p \ge 4$, it is trivial. If $ 4 \ge p \ge 3$, applying Schur's Inequality, we obtain $ r \ge \frac{p(4q\minus{}p^2)}{9}$, hence
\[ 4\equal{}q\plus{}r \ge q \plus{}\frac{p(4q\minus{}p^2)}{9}\]
It follows that
\[ q \le \frac{p^3\plus{}36}{4p\plus{}9}\]
and we obtain
\[ r\equal{}4\minus{}q \ge 4\minus{}\frac{p^3\plus{}36}{4p\plus{}9}\equal{}\frac{p(16\minus{}p^2)}{4p\plus{}9}\]
We have to prove
\[ (p\minus{}4)(p^2\plus{}4) \plus{}(3p\plus{}4)\cdot \frac{p(16\minus{}p^2)}{4p\plus{}9} \ge 0\]
\[ p(p\plus{}4)(3p\plus{}4)\minus{}(4p\plus{}9)(p^2\plus{}4) \ge 0\]
\[ 7p^2\minus{}p^3\minus{}36 \ge 0\]
\[ (p\minus{}3)(12\plus{}4p\minus{}p^2) \ge 0\]
which is obviously true because $ 4 \ge p \ge 3$.
This completes our proof. Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}b,c\equal{}0$ and its cyclic permutations. :)

[quote="nguoivn"][quote="nguoivn"]Given $ a, b, c \geq\ 0.$ Prove that:
$ (\frac {a}{b \plus{} c})^3 \plus{} (\frac {b}{c \plus{} a})^3 \plus{} (\frac {c}{a \plus{} b})^3 \plus{} \frac {5abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq\ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}$
 :)[/quote]
But with this old ineq, your method can't solve it. I tried.
I also see a long proof for it by SOS in Pham Van Thuan's book. But using Am-Gm, we have a very short solution  :)
And it's seems not weaker than mine  :oops:[/quote]
I haven't read Thuan's book, so I dont know that solution. Anyways, the following inequality is stronger than yours ;)
\[ \left( \frac{a}{b\plus{}c}\right)^3 \plus{}\left( \frac{b}{c\plus{}a}\right)^3 \plus{}\left( \frac{c}{a\plus{}b}\right)^3 \plus{}\frac{9abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \ge \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}.\]
;)

Yes, that's it  :P
Your proof is same to me  :oops:
I also used a nice lenma:
$ \sum\ \frac{a}{b\plus{}c} \geq\ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \plus{} \frac{4abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$
After expand, it's become: $ abc(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca) \geq\ 0$ (as I said)
And I think it's an useful lenma because notice that it's stronger than:
$ \sum\ \frac{a}{b\plus{}c} \geq\ \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)}$

[quote="nguoivn"]Yes, that's it  :P
Your proof is same to me  :oops:
I also used a nice lenma:
$ \sum\ \frac {a}{b \plus{} c} \geq\ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
After expand, it's become: $ abc(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca) \geq\ 0$ (as I said)
And I think it's an useful lenma because notice that it's stronger than:
$ \sum\ \frac {a}{b \plus{} c} \geq\ \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)}$[/quote]
We dont need to expand here, nguoi vn ;)
We rewrite it as follow:
\[ \sum \frac {a[a(b \plus{} c) \plus{} bc]}{b \plus{} c} \ge a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {4abc(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}
\]

\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc\sum \frac {1}{b \plus{} c} \ge a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {4abc(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}
\]

\[ \sum \frac {1}{b \plus{} c} \ge \frac {4(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}
\]
which is obviously true by Cauchy Schwarz because
\[ \sum \frac {1}{b \plus{} c} \ge \frac {9}{2(a \plus{} b \plus{} c)}
\]
and
\[ \frac {4(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \le \frac {9}{2(a \plus{} b \plus{} c)}
\]
;) :)

I meant:
$ \sum\ \frac {a}{b \plus{} c} \geq\ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
<=> $ \frac {(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) \minus{} abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq\ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}$ $ (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca) \minus{} abc \sum\ ab \geq\ [(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} abc](a^2 \plus{} b^2 \plus{} c^2)$
<=> $ abc(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca) \geq\ 0$
I think both of ways also short and nice  :)


inequalities
calculus
calculus computations

[url=http://bildr.no/view/489513][img]http://bildr.no/thumb/489513.jpeg[/img][/url]
[url=http://bildr.no/view/489514][img]http://bildr.no/thumb/489514.jpeg[/img][/url]
[url=http://bildr.no/view/489526][img]http://bildr.no/thumb/489526.jpeg[/img][/url]
4 pics are continual . because the forum only allowed 3 images. the last image is on the first reply post.
I can not understand the lines denoted by red line.

thanks

continued
[url=http://bildr.no/view/489516][img]http://bildr.no/thumb/489516.jpeg[/img][/url]

I have understood it .


What is the sum of the reciprocals of the first three positive even integers?

[hide]
the first three positive even integers are 2, 4, and 6.
The recipricols are:
$\frac{1}{2} + \frac{1}{4} +\frac{1}{6}=\frac{14}{12} =\frac{7}{6}=1 \frac{1}{6}$[/hide]

[quote="math92"][hide]
is it $\frac{11}{12}$[/hide][/quote]

That's better  :)


geometry
function
integration
logarithms
calculus
real analysis
real analysis unsolved

[quote="mayhem"]$y=f(x)$ is a differantiable function in $R$. $a<b$ and in  this interval $f(x)>0$.  Arc of this function from $a$ to $b$ is equal to area between $x=a , x=b , y=0 , y=f(x)$ Find all $y=f(x)$.[/quote]
$f: [a,b]\to R^{2}$ ?
If I understand Everything right then  ( if arc is $\int_{a}^{b}\sqrt{1+f'^{2}(x)}dx$) then
$(b-a)f(x)=\int_{a}^{b}\sqrt{1+f'^{2}(x)}dx$ and therefore
$f(x)=\frac{\int_{a}^{b}\sqrt{1+f'^{2}(x)}dx}{(b-a)}$

A little correction to Extremal:
$\int_{a}^{b}f(x)dx=\int_{a}^{b}\sqrt{1+f'^{2}(x)}dx$

Also, in tis question asked for all functions $f(x)$.

Oh, Then I was wrong.
I can represent some runctions for wich it will be true :

In first cours in my  universuty we still not begin to study How solve differential equations  therefore in my solution maybe will some simpl mistake  but I'll try    :oops:
if $f(x)=\sqrt{1+f'^{2}(x)}$ then our equation will be true
let's consider this case.

$f^{2}(x)-1=f'^{2}(x)$
note that $f'(x)$ can be $>0$ and $<0$
therefore consider two case
1. $f'>0$
$\sqrt{f^{2}-1}=f'$     $f' <>0$
$1=\frac{f'}{f^{2}-1}$
therefore
$\int=\int\frac{f'}{f^{2}-1}$
$x+C=\ln{(f+\sqrt{f^{2}-1})}$
$e^{x+C}=f+\sqrt{f^{2}-1}$
therefore
$f=\frac{e^{2x+2C}+1}{e^{x+C}}$
let check our conditions ( we do it for find constant $c$)

$(e^{x+c}+e^{-x-c})|_{a}^{b}=\int_{a}^{b}\sqrt{e^{2x+2c}+e^{-2x-2c}-1}dx$
you only need to find
$\int_{a}^{b}\sqrt{e^{2x+2c}+e^{-2x-2c}-1}dx$
=${\int_{a}^{b}\sqrt{2\cosh{(2x+2c)-1}}dx=\int_{2a-2c}^{2b-2c}\sqrt{2\cosh(y)-1}}dy$
And (if I remember right) last integral was discused in this forum few times ago)

[quote="yagaron"]extremal, i think that there is a small error here :) ${\frac{d\log \left(f(x)+\sqrt{f(x)^{2}-1}\right)}{dx}=\frac{f'(x)}{\sqrt{f(x)^{2}-1}}}$[/quote]
yes really I rewrite very fast
when I wrote
$f'=\sqrt{f^{2}-1}$
then we have
$1=\frac{f'}{\sqrt{f^{2}-1}}$
thanks for correction yagaron
 :wink:

no extremal. what I meant is that your solution is wrong.

You comit some mistakes. (actualy forgot to divide by 2  :oops:  )

and we can solve this ODE in a very easy way

[quote]
${\frac{d\cos (x)}{dx}=-\sin (x)}$,${\frac{d\sin (x)}{dx}=\cos (x)}$

${\left(\cos^{2}(x)+\sin^{2}(x)=1 \right)\Rightarrow \left( \cos^{2}(x)+\left(\frac{d\cos (x)}{dx}\right)^{2}=1\right)}$

${\frac{d\cosh (x)}{dx}=\sinh (x)}$,${\frac{d\sinh (x)}{dx}=\cosh (x)}$

${\left(\cosh^{2}(x)-\sinh^{2}(x)=1\right)\Rightarrow \left(\cosh^{2}(x)-\left(\frac{d\cosh (x)}{dx}\right)^{2}=1\right)}$

${\left(\sqrt{f'(x)^{2}+1}=f(x)\right)\Rightarrow \left(f(x)^{2}-f'(x)^{2}=1\right)}$

So, we can conclude ${f(x)=\cosh \left(x+c_{1}\right)}$ is a specific result.

we just need to proof uniqueness with some theorems, but i dont want to make this post more than it is
[/quote]


Circles $C_{1},C_{2}$meet at points $A,B$,A line through A is parrallel to the line through the centres of $C_{1},C_{2}$ and meets$C_{1},C_{2}$ again $C and D$respectively.Circle $C_{3}$ having $CD$as its diameter meets $C_{1},and C_{2}$again at $P$ and $Q$ respectively. Prove that lines $CP,DQ,AB$ are concurent and $CQ,DP,AB$ are concurent.

What an easy problem. Proof:

It's simple to see that $C, O_{1}, B$ line on the same line and so do $D, O_{2}, B$.

Let $I$ is the midpoint of the segment $CD$. We have: $IO_{2}$ is parallel to $CB$ [color=red](1)[/color]

As $Q$ lies on the circle whose diameter is $CD$, $QC$ is perpendicular to $QD$. On the other hand, $(I)$ and $(O_{2})$ intersect each other at $D$ and $Q$, thus $IO_{2}$ is perpendicular to $QD$. Therefore, $IO_{2}$ is parallel to $CQ$ [color=red](2)[/color]

From (1) and (2) we have $B$ lies on $CQ$.

Similarly, $B$ lies on $DP$


is there a way to extrude curved text (labelpath or texpath) ?
i'd be grateful for any tips
best
zb

[code]
import solids;
viewportmargin=(10,10);
size(200);
revolution sphere=sphere(0.75);
surface s=surface(sphere);
string text="some text";
real uoffset=-1.2;
real voffset=7.5;
real h=0.05;
draw(surface(xscale(0.4)*yscale(0.08)*text,s,uoffset,voffset,h),yellow);
draw(s.uequals(uoffset),red+dashed);
draw(s.vequals(voffset),blue+dashed);
draw(surface(s),palegray);

[/code]


inequalities
algebra
polynomial
algorithm
inequalities unsolved

Any type of problem in the form you posted can just as easily be solved by muirhead/am-gm, I am right?
What other problems have you "composed"? edit: this is my post #1000!!!!

[quote="me@home"]Any type of problem in the form you posted can just as easily be solved by muirhead/am-gm, I am right?
[/quote]
That is not in fact true. There was a user here not long ago [b]fjwxcsl[/b] who posted some solutions to symmetric polynomial inequalities that looked like pasted computer output. This is Shengli Chen of the Chinese Academy of Science, and he is the coauthor of a paper about inequalities of symmetric polynomials of three variables. In the paper, they present an algorithm which will efficiently split a polynomial into Schur-like forms:
$\sum_{cyc}x^{k}(x-y)(x-z)$
$\sum_{cyc}(yz)^{k}(x-y)(x-z)$
$\sum_{cyc}x^{k}(y+z)(x-y)(x-z)$
$\sum_{cyc}(yz)^{k}(y+z)(x-y)(x-z)$
All of these are positive; therefore, such a decomposition will always enable you to check the validity of a polynomial inequality. (For details about that, see their paper.)

However, I don't think it is true that any symmetric polynomial inequality can be decomposed in such a way that it is killed by AM-GM and Schur alone (certainly not AM-GM alone). See for example [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107851]here[/url].


induction
number theory
relatively prime

Prove that there are no integers $m,n \geq 1$ satisfying

$\sqrt{m+\sqrt{m+\sqrt{m+\ldots+\sqrt{m}}}}=n$

where there are $2006$ square root signs.

[hide="Hint"]
generalize (2006 isn't special)
[/hide]

[hide="Solution"]
Instead of solving this problem for 2006, we will solve it for $k$ nested radicals where $k$ is a natural number greater than one. We will proceed by induction.

The base case is where $k=2$, or:

$\sqrt{m+\sqrt{m}}=n$

Squaring both sides and subtracting by $m$ yields:

$\sqrt{m}=n^{2}-m \rightarrow m=(n^{2}-m)^{2}$

Therefore, $m$ is a perfect square. Let $m=x^{2}$:

$x=n^{2}-x^{2}\rightarrow n=\sqrt{x(x+1)}$

Since $x$ and $x+1$ are relatively prime, $n$ is not an integer, which is a contradiction, which closes the base case.

Now we will assume that the case with $k$ nested radicals does not yield an integer value for $n$ in order to prove the result for the case with $k+1$ nested radicals. After squaring both sides and subtracting $m$ from both sides in the equation with $k+1$ nested radicals, we find that:

$\sqrt{m+\sqrt{m+\sqrt{m+\ldots+\sqrt{m}}}}=n^{2}-m$

where the nested radical appears $k$ times. From our inductive hypothesis, the left side is not an integer and the right side is by the given information (when $m$ is a natural number), so we have a contradiction and induction is complete.
[/hide]

If you want to make your induction nice, do something like this:

Let $f^{n+1}(x)=f(f^{n}(x))$ for $n\in \mathbb{N}^{+}$ and $f(x)=\sqrt{m+x}$, $f^{1}(x)=\sqrt{m}$.


I could be wrong, but \[\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\ \geq\ \sqrt{x-1}-\sqrt{y-1}-\sqrt{z-1}.\] Therefore, I think the inequality posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=533927#533927]here[/url] implies this one.


Decide whether or not there is a function $f$ defined for all positive integers and taking positive integer values such that

$f(f(1)) = 5$

$f(f(2)) = 6$

$f(f(3)) = 4$

$f(f(4)) = 3$

$f(f(n)) = n+2$ for $n\geq5$

[hide]Let $y=f(1)$. Then $f(y)=5$. $y$ must be less than 5 or else $f(y)=y+2>y\geq 5$. Also $y>1$ because $y=1\Rightarrow f(y)=1$.

Then $y$ is 2, 3, or 4. But in either case, $f(y)=5$ so $f(5)$ is 6, 4, or 3 depending on whether $y$ is 2, 3, or 4. Since this is false, no such function can exist.[/hide]

I got what I think is a solution, even though the argument's shitty.

We're looking at a function $f$ which applied twice to $x$ gives $x+2$ for some values of $x$. What's more is that $f$ is defined for positive integers and have an image of positive integers. So all that is is $f(x) = x+1$. But that doesn't work for $x<5$. So, no, there isn't such a function.

Or is that too obvious to be true?

(Are you just going to post them all up Sen, or actually do some yourself?  :) )

how do we prove that $f(f(n)) = n + 2 \implies f(n) = n+1$ is the only function for f


There is a fly at point X on a vertex of a rectangular prism with height 6, length 4, and width 3 units. The fly wants to fly to point Y which is on the vertex farthest from point X. How far must the fly fly to get to point Y?

[hide] hint: use a 3-4-5 triange to find the distance from X to Y [/hide]

[quote="fireemblem13"]the title says "3-d pythagorean theorem." do u mean for us to find the space diagonal, but the fly can't go like that. otherwise, you would find the shortest distance on the surface of the prism from vertex to the other vertex.[/quote]

Why can't it go along the space diagonal? It can fly...
[hide]$\sqrt{6^2+4^2+3^2}=\sqrt{61}$[/hide]


geometry
rectangle
search
factorial
area of a triangle

1.  In rectangle [tex]ABCD[/tex], [tex]AD=10[/tex], and [tex]AB=8[/tex].  Point [tex]K[/tex] is chosen in [tex]DC[/tex] such that, when triangle [tex]ADK[/tex] is reflected in [tex]AK[/tex], the image of [tex]D[/tex] is on [tex]BC[/tex].  The length of [tex]DK[/tex] is

a) [tex]5[/tex] b) [tex]6[/tex] c) [tex]2\sqrt{5}[/tex] d)[tex]3\sqrt{3}[/tex] e)[tex]4\sqrt{2}[/tex]

2.  If [tex]P = 64 * 63 * 62 * ... * 3 * 2 * 1[/tex], then the largest positive integer [tex]n[/tex], such that [tex]P[/tex] is divisible by [tex]12^n[/tex] is

a) 5 b) 21 c) 28 d) 30 e) 32

3.  The number of integers between [tex]500[/tex] and [tex]600[/tex] whose digits have a sum of [tex]12[/tex] is

a) 4 b) 5 c) 6 d) 7 e) 8

4.  If the product of [tex](4^5)(5^{13})[/tex] is written as an integer, then the number of digits in the integer is

a) 12 b) 13 c) 18 d) 19 e) 24

5.  [tex]ABC[/tex] is a trangle with sides [tex]a,b,c[/tex].  If [tex]\measuredangle C = 90^\circ$[/tex], [tex]c= 4[/tex] and [tex]a+b=\sqrt{18}[/tex], then the area of triangle [tex]ABC[/tex] is

a) [tex]\frac{1}{2}[/tex] b) [tex]1[/tex] c) [tex]2[/tex] d) [tex]4[/tex] e) [tex]18[/tex]

For the degree sign it would probably be better to use \circ  so you would have 90^\circ instead of 90^0.

$90^\circ$

instead of

$90^0$

The latter might be confused to be equal to 1.

3.[hide]We know the first digit has to be 5, so the last two digits add up to 7.  So the questions is basically asking, how many pairs of integers (a, b), add up to 7 where a and b are both less than or euqal to 7, and less than or equall to 0.

Since 534 is different from 543, order does matter.  Let's list the pairs, since there is just a small number.

7, 0

6, 1

5, 2

3, 4

4, 3

2, 5

1, 6

0, 7

So the answer is e, 8.

If I'm correct, you can also use permutations.[/hide]

4.

[hide]

The product can be simpliefied to

 always has  digits, so  has  digits.

 = 125 which has 3 digits which is about

 which has 13 digits[/hide]

And for 5, is c supossed to be the hypotenuse?

Joml, thank you for the tip.

Hspotter, yes, c is the hypotenuse

5.

[hide]a 2  + b 2  can be written as (a+b) 2  - 2ab.  Thus, 18 2  - 2ab = 4 2  by Pythagorean Theorem.  So 324 - 2ab = 16, and ab = 154.  Area of a triangle will be ab/2, so ab/2 = 77.[hide][/hide][/hide]

4. If the product of (4^5)(5^13) is written as an integer, then the number of digits in the integer is ..?

[hide](4^5)(5^13) = (4^5)(5^5)(5^8) = (20^5)(5^8) = (10^5)(2^5)(5^8)

10^5 contributes 5 digits to (2^5)(5^8)

(2^5)(5^8) = (10^5)(5^3)

10^5 contributes another 5 digits to 5^3

5^3 = 5*5*5 = 125

So the integer is 125*10^10

which has 13 digits.[/hide]

For number 5, I made a typing mistake.  It was [tex]a+b=18[/tex] when it should have been [tex]a+b=\sqrt{18}[/tex].

The mistake has been corrected.

kool_dudy wrote:5.   is a trangle with sides .  If ,  and [tex]a+b=\sqrt{18}[/tex], then the area of triangle [tex]ABC[/tex] is

a) [tex]\frac{1}{2}[/tex] b) [tex]1[/tex] c) [tex]2[/tex] d) [tex]4[/tex] e) [tex]18[/tex]

[hide]We know that , and that  and  can be the height and width of the triangle respectively, since this is a right triangle.  That means that  is the area of the triangle.

We can also figure out that  and that .  We can both of these things together using an equation -

Since  is also the are of the triangle, the are of the triangle is a, or [/hide]

mikewd wrote:2. If P=64*63*62*...*3*2*1 , then the largest positive integer n, such that P is divisible by 12^n is ..?

[hide]64! = k.12^n

There are 5 factors of 12 from 1 to 64, so I'd say 5.[/hide]

How can you tell if something is a factor of 12?  Could there be any ways that you could produce more factors 12 than just the number 12,24,36...?  Heres a slight hint:  4! has one factor of 12.  Try to see if you can work with that.

hmm... 12 as a product of primes is 2*2*3. so maybe the way to find more is to search for how many 2*2*3 s there are there. Or actually, how many 2*2 s and how many 3 s, one will be the limiting factor?

Ok. how many times is something multiplied by 2 in 1*2*3*4*5*......*61*62*63*64  ?

Well at least 32 times, as there are 32 even numbers. so 64! can be written:

(1*3*5*7*...*61*63)(1*2*3*4*5*....*31*32)(2)

THen the middle bracket again can be half-factorised to get more even numbers out (and more 2 factors out):

32! = (1*3*5*7*...*31)(1*2*3*4*...*16)(2)

Now it's quite obvious that every time, the factorial we are half-dividing (dividing only the even multiples) gets smaller... first we took 32 2s out, then 16, then 8 then 4 then 2. then none left. add these to make 32+16+8+4+2 = 62 2s, or 31 (2^2)s.

 Now to find how many 3s, to see whether the (2^2)s or the 3s are the "weakest link"... :lol:

ok.. how many 3s in 64!... I guess the same method would work.

1*2*3*4*5*...*61*62*63*64 = (1*2*4*5*7*...*62*62*64)(3*6*9*...*60*63) = (1*2*4*5*7*...*62*62*64)(1*2*3*4...*20*21)(3)

Sucking the 3s out of the next part that has 3s (ones that used to be 9s, 18s, which have... oh yeah of course, we are just taking all 3s out and ounting them once, taking 9s out (but only counting them once because we've already counted them once in the previous step)

so skipping some of that, the highest multiple of 3 below 64 is 63, giving 21*3. 21 3s here. 7*3 is 21, so there are 7 more 3s here. the highest multiple of 3 below 7 is 6, and thats 3*2, so there are 2 more 3s here.

adding them up gives us 21+7+2 = 30. so there are 30 threes in 64!

ok now to go back and edit some working on the 2s.

As there are 31 2^2s, and 30 3s, the abundance is 2^2s and the limiting factor is the 3s. so it is therefore possible to divide 64! by 12 30 times, or divide it by 12^30, and n is 30.

crikey... thats probably the hardest maths I've ever had to actually think about. thanks for the hint :)


calculus integration geometry perimeter inequalities

[b]How many distinct isosceles triangles having sides of integral length and perimeter 113 are possible?[/b]

How do u approach this type of problem. Do we use an equation? i got 2x+y=113. but where do i go with tht? Do we use the fact tht the side of a triangle has to be smaller than the sum of the other two sides?

PLz show me a way where you do not plug numbers in if you can.

Yes.

$ 2x\plus{}y\equal{}113$

Also, $ 2x>y$

Now, assuming that $ 2x\equal{}y$ will give us the highest $ x$ which does not satisfy the conditions.

So, $ 4x\equal{}113$, $ x\equal{}\frac{113}{4}\equal{}28.25$
The smallest value of $ x$ that will satisfy the equation is $ 29$.

The largest is $ 56$.

$ 29\rightarrow56\equal{}28$ numbers

Triangle Inequality: a+b>c

so in this case, x+x>y


salam man taze in forume irano peida kardam vali hala mibinam ziad fa'aliat nadare. man ye soale rahat amma ghashang daram:

$ \frac {1}{\sqrt {a^2 \plus{} 1}} \plus{} \frac {1}{\sqrt {b^2 \plus{} 1}} \plus{} \frac {1}{\sqrt {c^2 \plus{} 1}}\geq\frac {1}{\sqrt {ab \plus{} 1}} \plus{} \frac {1}{\sqrt {ac \plus{} 1}} \plus{} \frac {1}{\sqrt {bc \plus{} 1}}$
$ a,b,c\geq1$

[color=red][Moderator edit : LaTeX formula corrected.][/color]

[quote="Kamyar1991"]
$ \frac{1}{\sqrt{a^2+1}}+\frac{1}{\sqrt{b^2+1}}+\frac{1}{\sqrt{c^2+1}}\geq\frac{1}{\sqrt{ab+1}}+\frac{1}{\sqrt{ac+1}}+\frac{1}{\sqrt{bc+1}}$

$ a,b,c\geq 1$
[/quote]
ino ghablanam post kardi,age kasi khas hallesho bebine bere inja:

[hide][url]http://www.mathlinks.ro/viewtopic.php?p=1102834#1102834[/url][/hide]

vase $ \text{\LaTeX}$ ham chize khassi lazem nis download koni,faghat kafie do tarafe ebaratet alamate dolar bezario ye seri chizaye payaro balad bashi ke inja tozih dade shode:

[url]http://www.mathlinks.ro/viewtopic.php?p=1135325#1135325[/url]

hi man koli ruye in soal dobare fekr kardam va farz bayad tabdil be $ a,b,c>3$ beshe az hamatun ham mazerat mikham.!! :blush:


Does there exist the greatest value of the expression $\boxed{\sqrt{x_1x_2}+\sqrt{x_2x_3}+\cdots+\sqrt{x_{n-1}x_n}}$, where $x_i >0, \sum_{i=1}^nx_i=1, i=\overline{1,n}$.

I think mixing variables, first for any consecutive pairs of $x_2,...,x_{n-1}$ and finally for $x_1,x_n$ shows that $max=\frac{n-1}n$ :)


Yeah!

So, um, I'm wondering what courses I should take (I'm in Pure Maths) and where I should live (which residence). Anyone who goes there have any comments?

Thanks in advance.

And if you're also going to Waterloo please share your excitement!

[quote="max_tm"]I see you just joined the UW forum (err, sorry for blowing your cover :P). So basically, you don't choose that major till second year. First year you would take two maths a CS and two electives. That means MATH 145 and 147 first semester (145 is a course in "classical algebra" and is more an introduction to formal mathematics than anything, and 147 is the calculus/analysis course) and 146 and 148 second semester (146 is Linear Algebra, 148 is the second half of 147). Then your choice of CS and electives. Matter of fact, you should be getting your first year in math book in a couple days, should answer most questions.

And yes, I'm soo excited for Waterloo in September![/quote]
Uhhh I've been trying to join that forum for a while but I haven't been able to register. Maybe I should sort that out.

I guess I will see you there!

ya i'm having trouble joining that forum too, it keeps saying i have an invalid UW ID..ya im going to UW Nanotech

[quote="max_tm"]Wait, so you're not the Steven on the forum who got a 94 on the Euclid?[/quote]
I am Steven and I got a 94 on the Euclid, but I did not make that post! Weird!

[quote="Scarfy"]I remebered you got 93, steven............[/quote]
Dude no, that's the UBC score, I got 94 after remarking at Waterloo.

lol...i dont even kno wut i got...
steven...how did u get a promise of scholarship from ubc if ur going to waterloo?...that was funny...

I'm going there in September too! Stream-4 (pure) math.  See you all soon ... :lol:

And if anyone has done math there already, how much homework is advanced math?  What are the cheesiest electives?  What is the more useful comp-sci course?

[quote="Ethanol"]
And if anyone has done math there already, how much homework is advanced math?  [/quote]

i heard advanced math was hard..idk if its completely true

[quote="max_tm"]I would say for a typical AoPSer, advanced math is exactly what they've wanted in a math course their whole lives but were not able to get in high school (obviously I've yet to do the course... I'm just basing this on what I know of the program and the fact that the reference textbook in MATH 147 is Spivak and the reference textbook in MATH 137 is Stewart... see the difference?) Most people don't take advanced math or find it difficult because they're accustomed to high school math and need that year to get used to a different style of instruction, different problem sets, etc. Not to mention that quite a few programs in the math degree aren't based around rigorous mathematics, so some people just aren't interested.

As for comp-sci courses... :P :

Two choices, Java-based and Scheme based:

Java-based, three entry points:
CS 125 - For the person who has never programmed, starts from scratch in Java.
CS 133 - For people with some programming experience, not necessarily in an OO language.
CS 134 - For people with 'extensive' programming experience in Java. This is actually the second semester course for 125 and 133 but can be taken in first semester.

Scheme-based, one entry point:
CS 135 - This course differs from the others in that it is accessible to any level (from never-evers to experienced programmer) because the focus is a little shifted; the course is centered on learning good programming principles and design, outside the context of a specific programming language (it uses Scheme, which is a functional programming language based off of Lisp, but the course isn't about 'Scheme programming' per-se). This course is relatively new, but it's been really successful and, starting fall '08, they're replacing the java based courses with variants of this course. IMO, this is the one to be in.[/quote]
Thanks!  As for residence, I'm hoping to get into a suite at MKV.  They seem nice.  What rez do you recommend?

I've joined that Adavanced Math Waterloo Facebook group, btu never bothered looking into it. :P
I guess I'm see many of you in Kitchener this coming September eh? cheers!

PS And since many german-speaking people are still living over there, all you need to know is to say "ich weiss nicht" when they try to talk to ye. hahaha.

Yeah, I just finished the first year of advance math at UW. If you're the kind of person who does math for fun, and shows up on aops, you'll have a blast in the 140's at UW. You /will/ work hard, but you'll enjoy the work.


algebra
polynomial
function
algebra proposed

Let $P(x)=2x^{2}+6x^{3}+....+n(n-1)x^{n}$  where n is a natural number $n\geq 2$ .Find $P(3)$ as a function of $n$ .


I'm writing up some lecture notes and want to refer back to a previously-defined equation, while still labelling it by its original equation number. To show you what i'm aiming at, the general scheme i've tried is:

[in the preamble]
  \newcounter{keepeqtn}

[in the text body]
  \begin{equation}
  a = b+c
  \label{eq:one}
  \end{equation}

[yadah yadah yadah]

  \setcounter{keepeqtn}{\value{equation}}
  \setcounter{equation}{\value{\ref{eq:one}}}
    \begin{equation}
    a = b+c
    \end{equation}
  \setcounter{equation}{\value{keepeqtn}}}

but this breaks at
  \setcounter{equation}{\value{\ref{eq:one}}}
with a "missing \endcsname" error

Any ideas?
[For info:  i'm using \documentclass{report}, and "\input"ing (not "\include"ing) separate "chapters".   I'm interested in the general case where i can refer back to equations in previous "chapters" - although the scheme above breaks even for equations previously defined in the same "chapter"]

Does \setcounter{equation}{\ref{eq:one}} instead of \setcounter{equation}{\value{\ref{eq:one}}} work?

Thanks for the suggestion, but no, it doesn't.

An alternative approach would just be something like
   a = b+c \hfill (\ref{eq: one})
but of course \hfill doesn't work in math mode so the alignment comes out all wrong (and \mbox-ing the \hfill doesn't work)

I've just realised what you want to do, in which case, providing that you have loaded amsmath (which you are likely to have done) you can do it with \tag and there's no need to use a counter. [code]\usepackage{amsmath}
...
\begin{equation}
a = b+c
\label{eq:one}
\end{equation}

[yadah yadah yadah]

\begin{equation}
a = b+c \tag{\ref{eq:one}}
\end{equation} [/code]

Steve - right on the button, thanks very much!


inequalities
geometry
circumcircle
geometry unsolved

$ AD$,$ BE$,$ CF$ are cevians in triangle $ ABC$(not necessarily concurrent), intersecting the circumcircle of triangle $ ABC$ in the points $ P$,$ Q$,$ R$.
Prove that $ \frac{AD}{DP} \plus{} \frac{BE}{EQ} \plus{} \frac{CF}{FR} \ge 9$

The bisectors $ AA_1, BB_1, CC_1$ ($ A_1,B_1,C_1$ lie on the sides) of the triangle $ ABC$ meets the circumcircle at again $ A_2,B_2,C_2$ respectively.
Denote $ H$ be the foot of the $ A$-altitude. We have that $ d(P;BC)\le d(A_2;BC)\Rightarrow \frac {AD}{DP} \equal{} \frac {AH}{d(P;BC)}\ge \frac {AH}{d(A_2;BC)} \equal{} \frac {AA_1}{A_1A_2}$
where $ d(X,l)$ is the distance from $ X$ to the line $ l$.
Similarly, $ \frac {BE}{EQ}\ge \frac {BB_1}{B_1B_2}$, $ \frac {CF}{FR}\ge \frac {CC_1}{C_1C_2}$.
The inequality $ \frac {AA_1}{A_1A_2} \plus{} \frac {BB_1}{B_1B_2} \plus{} \frac {CC_1}{C_1C_2} \ge 9$ can be get easily  :wink: .

[quote="keyree10"]$ AD$,$ BE$,$ CF$ are cevians in triangle $ ABC$(not necessarily concurrent), intersecting the circumcircle of triangle $ ABC$ in the points $ P$,$ Q$,$ R$.
Prove that $ \frac {AD}{DP} \plus{} \frac {BE}{EQ} \plus{} \frac {CF}{FR} \ge 9$[/quote]

See also:

http://www.mathlinks.ro/viewtopic.php?t=308544


inequalities
geometry
3D geometry
tetrahedron

Derive a necessary and sufficient condition for six positive reals $ a, b, c, d, e, f$ to be the side lengths of a tetrahedron.

I think the short answer is that this is complicated.  I haven't studied the [url=http://mathworld.wolfram.com/Cayley-MengerDeterminant.html]Cayley-Menger determinant[/url] in great detail, but it's possible that the determinant is positive if and only if there actually is a tetrahedron with given edge-lengths, so in principle you could just run through the 30 distinct permutations of the edges to see if any of the results are positive.

On [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=83093]this thread[/url] I asked some questions about when we can construct more than one tetrahedron of given side-lengths:

[quote="JBL"]1. Find constraints on positive reals $ a, b$ such that:
(a) There are three distinct tetrahedra with sides $ a, a, a, b, b, b$
(b) There are only two such tetrahedra
(c) There is only one such tetrahedron
(d) There are no such tetrahedra

2. What about if the tetrahedron is to have sides of length $ a, a, a, a, b, b$?[/quote]


calculus
derivative
inequalities
inequalities solved

Show that for any $x \in [0,1]$  we have $ 1+ \frac{x}{2}- \frac{x^2}{8} \leq \sqrt{1+x} \leq 1+ \frac{x}{2}$.

cheers! :D :D :cool:

sqrt(1+x)\leq 1+x/2 is easy to proof.. just square both sides..
can we use derivatives for the left inequality?

[quote="liyi"]can we use derivatives for the left inequality?[/quote]

well.. using derivates is kinda easy.. lets stick please to elementary methods!

cheers! thank you very much!

(1+x/2-x^2/8)^2 = 1+x-x^3/8+x^4/64
we need only to show that
-x^3/8+x^4/64 \leq 0
i.e.,
x^4\leq 8x^3
this holds obviously for x\in [0,1].


All the links in my post
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=198223
have been changed to some bullshiit by you righteous great Moderator. And I am not allow to edit it now.

Have you no shame?

Is this what you called freedom of speech?

I must thank you for moving my first post to
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=198129

Seems like you can not edit the post there.  :rotfl:

I don't think I ever said we are practicing freedom of speech here. This is mainly a [b]math and sciences[/b] forum. [b]It is NOT[/b] a forum where you complain about political situations.

Next post of yours in this tone will make me ban you.


How does the AMC/AIME committee choose names for characters in their problems?  Just curious...

Just because it's irrelevant doesn't mean it isn't fun. This might belong in G+FF, though.

[quote="gauss1181"]Lock this. No relevance at all.[/quote]
Relevant to what?  This is a new topic.   :)
Sure, a mod can move it wherever it needs to go.

THey used "Mr.Dunbar" like last year for a character name.

Maybe they use the names of people that work on the AMC/AIME

Or they pruposely make names that start with the same letter, so I have to denote them with different variables.

[quote="abacadaea"]there was this city in MC called IMAFREEZIN[/quote]

I remember seeing that.. and also i've seen

Alice, Bob, and Casey are ....

and a few more like that.

Al lot of the times on Canadian contests they'll do something like Andrew, Betty, Carlos  :lol: (ABC get it... okay not funny I know). Maybe this is sometimes done on the Aimes Amcs and stuff.
I remember one question where the names spelled out MATH

[quote="hadasah"]Look at problem 11 on the AMC 10B this year. Steve and LeRoy...Doesn't that sound like the names of two people involved with AMC? Hmmm...I wonder who they are...

HELLO!!!!! Steve as in Mr. Dunbar and LeRoy as in the other person on the back of the AMC cover...(sry, whoever you are, I forgot your full name).[/quote]
Hmm... [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1052015#1052015[/url]
Do you mean AMC 10A?

they go to the big fat book of [b]Choosing Names for Contest Problems[/b].

I like problems where they are called Albert, Bert, Carlos, etc. or something of the sort.  I distinctly remember, however, on the AMC this year a Brenda and Bobby problem.

As a totally unrelated comment, when I name people, I tend to use my own name, and the names of the people receiving the test. I always go to multicultural names, too. Like take the MATHCOUNTS test I'm writing. On one of the problems, my four people are Ahmed, Bhat, Carl, and Delfonso.

As an even more unrelated comment, on English vocab or grammar quizzes, when we have to make up sentences, I always name my people with stereotypical names, like Muhammad, Jean-Phillipe, Mr. Nguyen, Hsin-pu, Toyoko, Mthomba Mb'dbonouoe (MP 06 FTW), etc.

I can't wait for the day where Zuming is part of an AMC problem. If you want to find a winning strategy for one of Zuming and someone else on the USAMO, assume Zuming is going to win.  :D

Hmm, I don't usually think about these things.

@Catalyst: Lol, whenever I make names for my own problems, the only ones I ever use are Joe, Bob, and Samantha. Yes, I need more creativity.  :D

[quote="randomdragoon"]Personally, I would hope that the characters are chosen so that they have unique first letters. Beyond that I don't really care.

Can you just imagine a problem like this?

Billy-Bob, Bimbo-bob and billy-bill each have some number of cows. Billy-bob and billy-bill have seven more than billy-bill and Bimbo-bob combined... (etc.)[/quote]
ermm.. Bimbo-bob? ...

My names = Person A, B, C, etc. Amazing, I know

The best names to use are Albert, Joe, Bob, Tyler, Arlene, and Vladmir.

However, I don't remember any of the above names being used.

Sometimes is seems like they pick names that are from different languages and ethnicities just to seem like they're all multiculturalized and everything....

Lol, on [i]some[/i] Canadian Contests (the COMC), the names are always something like Alphonse and Beryl... Beryl usually wins...  :D

[quote="CatalystOfNostalgia"]I can't wait for the day where Zuming is part of an AMC problem. If you want to find a winning strategy for one of Zuming and someone else on the USAMO, assume Zuming is going to win.  :D[/quote]

*facepalms* I would laugh if I saw that.. but i don't think that Mr. Feng is going to allow that.. :)

[quote="CatalystOfNostalgia"]As a totally unrelated comment, when I name people, I tend to use my own name, and the names of the people receiving the test. I always go to multicultural names, too. Like take the MATHCOUNTS test I'm writing. On one of the problems, my four people are Ahmed, Bhat, Carl, and Delfonso.
[/quote]

You mean the one about the murder and you have to figure out who committed it?


Consider the following heterogeneous equilibrium:

$NH_{2}CO_{2}NH_{4}(s) \rightleftharpoons 2NH_{3}(g)+CO_{2}(g)$.

Discuss the effect that the addition of carbon dioxide at constant temperature and pressure might have on this equilibrium.

[quote="Carcul"]Consider the following heterogeneous equilibrium:

$ NH_{2}CO_{2}NH_{4}(s)\rightleftharpoons 2NH_{3}(g)\plus{}CO_{2}(g)$.

Discuss the effect that the addition of carbon dioxide at constant temperature and pressure might have on this equilibrium.[/quote]

Since no one tried this problem, then I'll give it a try. One way to look at this problem is to look at the gas equation PV = nRT. Since R is the universal constant and P and T are unchanging, the increase in the number of moles of carbon dioxide will increase the volume. According to La Chatelier's principle, when there is a change in the concentration, pressure, volume, or total pressure, the equilibrium will shift in such a way as to counteract this change. Thus, to decrease the volume, the equilibrium will shift to the reactant side with the solid. The amount of $ NH_{2}CO_{2}NH_{4}$ will increase, the amount of NH3 will decrease and the amount of CO2 will decrease but will still be more than before the additional CO2 was added.

well thsi may come handy the concentration of solid is 1 or nearly always the same

No, it doesn't have nothing to do with that.

[quote="Carcul"]That answer is only partially correct. On other words, that will only happen if a certain condition is met: otherwise, that will not happen.[/quote]

Do you mean that under constant temperture and pressure, the Kc would have to be more than 1 for this to happen. And that the number of molecules in the gas state on the product side would have to be more than the number of molecules in the gas state on the reactant side so the volume would be able to decrease.

At constant temperature and pressure Kc is constant. What I mean is, if a certain condition (to be found) is met, then adding carbon dioxide will shift the equilibrium to the right (producing more carbon dioxide!). If that condition is not met, then the usual prediction (the reaction will shift to the left) will happen. Now, what is that condition?

Just wondering, if you add more $ CO_{2}$ and keep the pressure constant, obviously you would need to increase the volume of the overall system. What if you increase the volume in such a way as to decrease the overall concentration of $ CO_{2}$ (i.e. you increase the volume so much that $ [CO_{2}]$ actually becomes smaller than it was originally).


At one hospital, there are 100 patients, all of whom have at least one of the following ailments: a cold, the flu, or an earache. 38 have a cold, 40 have the flu, and some number have earaches. If 17 have both colds and earaches, and 7 have all three, how many have earaches?

patients having earache=46.
I used Venn diagram to solve this. To see the diagram open the image file attached

yup ..............sets and stuff , venn 's diagram's that is ....... same what i did.......

isnt it that the venn's diagram's stuff that we do has been derived from algebra, itself? cos these sums you can do algebraically( possibly arithmetically too ) only venn's diagrams makes it a whole lot easier :P

[hide]i get 29.  i don't know what i'm doing wrong. Can somebody post a solution?[/hide]

I think that you counted some of the overlaps of the venn-diagram twice.


In the "Solved Reports" forum, the description says that you can't make new topics, but in the forum, there is a link that makes a new topic. Is this supposed to happen?

Click on it and see what happens.  :wink:

[quote]Only users granted special access can post topics in this forum.[/quote]


function
inequalities proposed
inequalities

Let a,b,c be positive number
Prove that
$ \sum_{cyc}\frac{a^2}{\sqrt{a^4\plus{}24b^3c^3}} \geq \frac{3}{5}$

[quote="8826"]Let a,b,c be positive number
Prove that
$ \sum_{cyc}\frac {a^2}{\sqrt {a^4 \plus{} 24b^3c^3}} \geq \frac {3}{5}$[/quote]
It's wrong. Try $ a\equal{}b\equal{}c\equal{}2.$ :wink:

[quote="8826"]Let a,b,c be positive number
Prove that
$ \sum_{cyc}\frac {a^2}{\sqrt {a^4 \plus{} 24b^3c^3}} \geq \frac {3}{5}$[/quote]
I suppose you mean:
$ \sum_{cyc}\frac {a^2}{\sqrt {a^4 \plus{} 24b^2c^2}} \geq \frac {3}{5}$, or equivalently:
$ P \equal{} \sum_{cyc}\frac {a}{\sqrt {a^2 \plus{} 24bc}} \geq \frac {3}{5}$.
Then Hölder gives us:
$ \left ( \sum_{cyc} a^3\plus{}24abc \right )P^2 \ge \left ( \sum_{cyc} a \right )^3$, so it suffices to prove

$ 25 \left ( \sum_{cyc} a \right )^3 \ge 9\left ( \sum_{cyc} a^3\plus{}24abc \right )$, which is obvious upon expanding...

If it is as as Mathias said...
Another solution is(dedicated to quykhtn-qa1):
$ \sum\frac {a^2}{\sqrt {a^4 \plus{} 24b^2c^2}} \equal{} \sum\sqrt {\frac {a^4}{a^4 \plus{} 24b^2c^2}} \equal{} \sum\frac {1}{\sqrt {1 \plus{} \frac {24a^2b^2c^2}{a^6}}}$
We consider the function
$ f(x) \equal{} \frac {1}{\sqrt {1 \plus{} \frac {24a^2b^2c^2}{x^6}}}$
$ f''(x)\geq 0$
So $ f(a) \plus{} f(b) \plus{} f(c) \geq 3f(\frac {a \plus{} b \plus{} c}{3})$
So the $ LHS \geq 3\sum\frac {1}{\sqrt {1 \plus{} \frac {3^6*24a^2b^2c^2}{(a \plus{} b \plus{} c)^6}}}$
So we have to prove that:
$ \sum\frac {1}{\sqrt {1 \plus{} \frac {3^6*24a^2b^2c^2}{(a \plus{} b \plus{} c)^6}}}\geq \frac {1}{5}\Leftrightarrow(a \plus{} b \plus{} c)^6 \geq 3^6 a^2b^2c^2$ which is true :)


1.Take away my first letter; take away my second letter; take away
all my letters, and I would remain the same. What am I?

2.The more you take the more you leave behind.

What are they?

3.Where is the only place that yesterday always follows today?

4.A certain large animal lives happily and thrives here on Earth.
One day, every single one of these critters is wiped out by a
mysterious disease which affects only this particular animal.
There are none left anywhere on earth -- they are all gone. Dead.
About a year or so later, they begin to reappear on Earth again.
How can this be?

The animals were not created artifically by humans

[hide] Hint on the fourth puzzle: Try to find out what kind of animal it is[/hide]

[quote="Hamster1800"][hide="4"]
are they mules? (mix between horses and donkeys, they can't actually breed among each other.)
[/hide][/quote]
[hide]Mules is the animal I was thinking of. Correct answer[/hide]

[quote="Hamster1800"][hide="4"]
are they mules? (mix between horses and donkeys, they can't actually breed among each other.)
[/hide][/quote]

well don't ligers have more issues? not only can they not interbreed, but they are larger and fit the description

[quote="box of rocks"]D3m0n ShaD0w,

your answer to the first problem

is equivalent to the one that I

ALREADY gave immediately following

the original post.[/quote]

oh sorry i didnt know i dont read hides unless i myself give a brainteaser


topology
advanced fields
advanced fields unsolved

Prove that in locally compact spaces, a subspace is of dimension 0 iff it is totally disconnected.

1. We talk about second countable metric spaces only.
2. dimension 0 means for any $p \in X$, there exists an arbitrary small neighborhood which is open and close.

In general, $\Rightarrow$ is true, but how to prove the reverse direction?

Take $X$ a totally disconnected compact space and $p\in X$. We must find arbitrary small open-closed neighborhoods, so take an open neighborhood $U$ of $p$. For each $y\in X-U$, there is an open-closed set $V_{y}$ such that $y\in V_{y}\subset X-\{p\}$. Cover $X-U$ by a finite number of $V_{y}$ (a closed set of a compact space is compact), the complement of the union of that finite number works as an open-closed neighborhood of $p$ inside $U$.

 :D

Please note that I am talking about LOCALLY compact spaces.


61, 52, 63, 94, 46, ?

[hide="Hint"]Quite familiar numbers.
Dont' need any calculation.[/hide]

[hide]LOL it's the two-digit squares backwards! The next one is 18[/hide]


I'm reading this book called Catch-22 by Joseph Heller now. It seems like a good book and I read about 60 pages but there is one question: What is the plot?? I mean, I read 60 pages so I'm hoping to see some actions but there isn't really anything going on. Maybe I'm not reading this well so perhaps someone who read this book can give me general outline of the story (don't tell how it will end or stuff like that). I know it's about war and something about U.S.O. (not sure what this is as well).

I appreciate this greatly.

[url=http://www.sparknotes.com/lit/catch22/]Sparknotes[/url] is your friend. As far as I remember, the plot was pretty much random throughout the whole book... but pretty funny anyway :P.

I'm reading the book too (about page 50ish) and it hilarious. You must remember that it's not one of those traditionally good books... at first it got very bad reviews until people realized the hilarity and greatness of the book.


1. The digits of a positive integer [i]n[/i] are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when n is divided by 37?

so it's like 1000x+100(x-1)+10(x-2)+(x-3)??? but how can I find out the remainder?

2. Set A consists of [i]m [/i]consecutive integers whose sum is 2[i]m[/i], and set B consists of 2[i]m[/i] consecutive integers whose sum is [i]m[/i]. The absolute value of the difference between the greatest element of A and the greatest element of B is 99. Find m.

Can any of you give me any clue, thanks.

The integers are $ 9876$, $ 8765$,...$ 4321$. The remainder when $ 4321$ is divided by $ 37$ is $ 29$. Each successively larger such $ n$ is $ 1111$ plus the previous $ n$. $ 1111 \equiv 1 \pmod{37}$, so the sum of the remainders is
\[ 29 \plus{} 30 \plus{} 31 \plus{} 32 \plus{} 33 \plus{} 34 \equiv \minus{} 8 \minus{} 7 \minus{} 6 \minus{} 5 \minus{} 4 \minus{} 3 \equiv \minus{} 33 \equiv \boxed{4} \pmod{37}.
\]

For number 2, let the smallest element of $ A$ be $ x$ and the smallest element of $ B$ be $ y$. Where can you go from there?

We want sum of the remainders, not mod 37 of sum of remainders :)

So answer is 189.


Let $ a, b, c$ be non-negative real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove or disprove that

$ \sum\frac{a^4(a\plus{}1) \plus{} b^5}{a^3 \plus{} b^3}\ge \frac{5}{6}$ .

_____________
 :?:

Yes it is true!
Follows from
1) $ \sum_{cyc}\frac{a^4}{a^3\plus{}b^3}\geq \frac{a\plus{}b\plus{}c}{2}$
2) $ \sum_{cyc}\frac{a^k\plus{}b^k}{a\plus{}b}\geq (a^{k\minus{}1}\plus{}b^{k\minus{}1}\plus{}c^{k\minus{}1})$ ($ k\geq 1$)

[quote="Albanian Eagle"]Yes it is true!
Follows from
1) $ \sum_{cyc}\frac {a^4}{a^3 \plus{} b^3}\geq \frac {a \plus{} b \plus{} c}{2}$
2) $ \sum_{cyc}\frac {a^k \plus{} b^k}{a \plus{} b}\geq (a^{k \minus{} 1} \plus{} b^{k \minus{} 1} \plus{} c^{k \minus{} 1})$ ($ k\geq 1$)[/quote]

used technique is very nice. Thanks

Aspect further resolutions....


known fact (*) $\prod a_{i}=1, 0<r<s \longrightarrow \sum a_{i}^{r}\le \sum a_{i}^{s}$
$\binom{n}{i}\frac{\sum a_{i}^{n}}{n}\geq \binom{n}{i}\frac{\sum a_{i}^{i}}{n}\geq \sum a_{\sigma(1)}a_{\sigma(2)}\cdots a_{\sigma(i)}$ sum these up and get:
$\sum a_{i}^{n}\geq \frac{n}{2^{n}}(1+a_{1})(1+a_{2})\cdots (1+a_{n})$
all we need to show now is $\sum a_{i}^{n+1}\geq \sum a_{i}^{n}$ which is easy from (*)


function
induction
linear algebra
matrix
algebra unsolved
algebra

Let a Mobius function be a function $f(z) = \frac{az + b}{cz + d}, cz + d \neq 0$.

Let $f_0(z) = z, f_{k+1}(z) = f(f_{k}(z))$, where $f(z)$ is a Mobius function.

Find, with proof, all $p$ such that some $f(z)$ exists such that $f_k$ is a sequence with period $p$.

(There is a solution in this issue somewhere, but I cannot find it and I would be interested in alternate approaches.  It's a very interesting problem!)

Anyway, this may be helpful.
If we define $A=\left(\begin{array}{cc}a & b\\c & d\end{array}\right)$, then, by induction,
$f_k(z)=\frac{a_k z+b_k}{c_k z+d_k}$, where $A^k=\left(\begin{array}{cc}a_k & b_k\\c_k & d_k\end{array}\right)$.
Should we find $p$ such that $A^p=nI$ ($n$ being any scalar, and $I$ being the identity matrix)?
But that would be a solution not dependent of $z$, and would only exist for particular values of $A$.
[Edit: added the $n$]

[quote="lordWings"]Is it $f: \mathbb{R}\to\mathbb{R}$ or $f: \mathbb{C}\to\mathbb{C}$?[/quote]

It doesn't matter.  And yes, you are on the right track :)


You can apply the know inequality $ |a|\plus{}|b|\le |a\minus{}b|$. It is not hard. :)

[i]Generalization[/i]
             $ S \geq 2(n \minus{}1)$ :wink:

Find the maximum $ F\equal{}\sum_{i<j} {|x_i\minus{}x_j|}$ ( $ x_{i,j}\in [0;1]$)


I've been thinking about this since a few weeks. Is the internet anarchical? When you first think about it, it just seems anarchist because you can put anything and everything you want on it because there is no "it". I mean, everything you see online is on a server, on a computer, you're just accessing the data on the computer so the internet isn't really a storage space that someone owns. But, then why do so many hacking and exploitation websites sink? I also don't understand why hackers and crackers discuss new hacking techniques and distribute tools in the depths of P2P and IRC?

The better description is that the internet is decentralized with a very flat hierarchical structure. Also note that the internet is fantastic, and this is [b]because of[/b] its "anarchistic" nature.

The internet is a parallel  world to each one . Everything there , i mean everthing is possible.

One thing I don't like about the internet is that you are at the whim of the website administrators.
For example, two of my accounts were deleted from language websites. Probably I was associated with spam by mistake. Moreover, I didn't get any notification- so I had to google my account to see if it existed.
Another way to be "dumped" is when a website cancels its service- for example Yahoo Photos, 360 blog or Geocities have been canceled.


geometry
calculus
integration
function
floor function
calculus computations

Find the area enclosed by the curve[IxI]+[IyI]=1where[] denotes greatest int function and I I denotes absolute value function

$ \lfloor |x| \rfloor \plus{}  \lfloor |y| \rfloor \equal{} 1$

$ \alpha \in R, \to \lfloor |\alpha| \rfloor \equal{} 0 , 1, 2, ..$

$ \to i) \lfloor |x| \rfloor \equal{} 1, \lfloor |y| \rfloor\equal{} 0$
or $ ii)\lfloor |y| \rfloor \equal{} 1, \lfloor |x| \rfloor \equal{} 0$

i) $ x\in (\minus{}2,\minus{}1] \cup [1,2), y\in (\minus{}1,1)$
ii)$ y\in (\minus{}2,\minus{}1] \cup [1,2), x\in (\minus{}1,1)$

$ A\equal{}8$ ?


Find all positive integers $ n$ such that $ 3|(2^n\minus{}1)$ and there exists an integer $ m$ satisfying $ \frac{2^n\minus{}1}{3}|(4m^2\plus{}1)$.

First condition equavalent to $ n \minus{} even$.
Second condition equavalent to $ \frac{2^n\minus{}1}{3}\equal{}\prod_i p_i^{k_i}$, were all $ p_i\equal{}1\mod 4$.
If $ m|n$ then $ 2^m\minus{}1|2^n\minus{}1$, if m -odd>1, then $ 2^m\minus{}1|\frac{2^n\minus{}1}{3}$. But $ 2^m\minus{}1\equal{}7\mod 8$ and exist prime divisor $ q\equal{}3\mod 4$, suth that $ q|2^m\minus{}1$.
Therefore second condition equavalent to $ n\equal{}2^k$, first $ k\ge 1$.


$ a_{1} \equal{} 1$, $ a_{k} \equal{} a_{k\minus{}1} \plus{} 3$ for every natural number $ k>1$. Find $ a_{n}$ in terms of $ n$.

Thanks dgreenb801.

[hide=" Proof by mathematical induction:"]

Basis step:

For $ n\equal{}1$, $ a_{n} \equal{} a_{1} \equal{} 1 \equal{} 3(1)\minus{}2$

Inductive step:

If $ a_{k} \equal{} 3k\minus{}2$, then $ a_{k\plus{}1} \equal{} a_{k} \plus{} 3 \equal{} 3k\minus{}2 \plus{} 3 \equal{} 3(k\plus{}1) \minus{} 2$

[/hide]

$ a_n \equal{} a_1 \plus{} \sum_{k\equal{}2}^n (a_k \minus{} a_{k\minus{}1}) \equal{} 1 \plus{} 3(n\minus{}1) \equal{} \boxed{3n \minus{} 2}$.
By [url=http://mathworld.wolfram.com/TelescopingSum.html]Telescoping Sum[/url]


inequalities
combinatorics unsolved
combinatorics

A graph has $n$ vertix.Prove that:
$\chi(G).\chi(\overline{G}) \geq n$

this is also trivial.since an independent set in $G$ becomes a clique in $\overline{G}$ it follows that $\alpha(G)\leq\chi(\overline{G})$. the inequality follows since $\chi(G)\geq n/\alpha(G)$.


Hello, everyone; I'm sorry to disturb you again, but: I found this formula/series which converges on $ \sqrt{n}$ for some non-negative integer "n". The first term of the series should always be 1, and the recursive formula of the series should be $ (x\plus{}ny)/(x\plus{}y)$ (Sorry I'm bad at LaTex), or $ 1\plus{}y(n\minus{}1)/(x\plus{}y)$. The idea is that the series should slowly converge to $ \sqrt{n}$. I tested this for a few examples (not many, just a few), and I was wondering whether someone here would be kind enough to either prove or disprove this (Probably this might be something simple, so I'm sorry if this is too easy for you)

Oh yeah, sorry  :( . x/y should be the term that precedes the new term (I'm sorry if the language is confusing, as in the first term of the series for $ \sqrt5$ is 1, so $ (1\plus{}5(1))/(1\plus{}1)$, or $ 6/2$ is the next term (well, it's faster if you don't simplify the fractions), and the term after that is $ (6\plus{}5(2))/(6\plus{}2)\equal{}16/8$, etc.

So you've got a sequence $ a_k$ defined by $ a_1 \equal{} 1, a_{k\plus{}1} \equal{} \frac{a_k \plus{} n}{a_k \plus{} 1}$.  The fixed points of the transformation $ q \mapsto \frac{q \plus{} n}{q \plus{} 1}$ are in fact $ \pm \sqrt{n}$, so if the sequence converges at all it must converge to one of these two numbers.

There are a couple of ways to do it from here.

[hide="Method 1"] $ a_{k\plus{}1} \equal{} 1 \plus{} \frac{n\minus{}1}{1 \plus{} a_k}$ gives a (non-simple) [url=http://en.wikipedia.org/wiki/Continued_fraction]continued fraction[/url], and there are some standard convergence theorems for these. [/hide]
[hide="Method 2"] Figure out what bound you need on $ a_k$ to apply the [url=http://en.wikipedia.org/wiki/Banach_fixed_point_theorem]contraction principle[/url] to $ f(x) \equal{} \frac{x \plus{} n}{x \plus{} 1}$. [/hide]
[hide="Method 3"] Explicitly show that the quantity $ \left| \sqrt{n} \minus{} a_{k\plus{}1} \right|$ is decreasing to zero, for example by computing it as $ \left| \sqrt{n} \minus{} \frac{a_k \plus{} n}{a_k \plus{} 1} \right|$. [/hide]

Thank you, though I can't really understand this because I'm not advanced (well probably, it is easy, but I have barely begun to even understand math yet). Can you please simplify this (is there any basic algebra-style method for this). Thanks for all your help.

Method 3 is probably the most elementary way to do it, but the argument doesn't seem to carry through unless you apply the recursion twice; this is because the sequence $ a_k$ approaches $ \sqrt {n}$ alternately from the top and bottom, and the error in the approximation isn't monotonic.

[hide="Details"] I claim that the inequality

$ \left| \sqrt {n} \minus{} a_{k \plus{} 2} \right| < \left| \sqrt {n} \minus{} a_k \right| \left( \frac {n}{n \plus{} 1} \right)$

holds.  Since $ \frac {n}{n \plus{} 1} < 1$, this implies convergence as $ k \to \infty$.

The proof is "basic algebra-style":  first write $ a_{k \plus{} 2} \equal{} \frac {a_{k \plus{} 1} \plus{} n}{a_{k \plus{} 1} \plus{} 1} \equal{} \frac { (n \plus{} 1) a_k \plus{} 2n}{ 2 a_k \plus{} (n \plus{} 1) }$ by applying the recurrence twice, then compute

$ \left| \sqrt {n} \minus{} \frac {(n \plus{} 1) a_k \plus{} 2n}{2a_k \plus{} (n \plus{} 1)} \right| \equal{} \left| \frac {(a_k \minus{} \sqrt {n})(\sqrt {n} \minus{} 1)^2}{2a_k \plus{} (n \plus{} 1)} \right| \equal{} \left| \sqrt {n} \minus{} a_k \right| \frac { (\sqrt {n} \minus{} 1)^2}{2a_k \plus{} (n \plus{} 1)}$

and then note that $ (\sqrt {n} \minus{} 1)^2 < n, 2a_k \plus{} (n \plus{} 1) > (n \plus{} 1)$.  The result follows. [/hide]

I should also mention that it is actually possible to write down closed forms for sequences of the form $ s_{k \plus{} 1} \equal{} \frac {as_k \plus{} b}{cs_k \plus{} d}$ using linear algebra.  The nature of these closed forms provides a good justification for why these convergence results are true of fractional linear transformations.

Also, Method 2 probably doesn't work.

$ \frac{q\plus{}n}{1\plus{}q} \minus{} q \equal{} \frac{n\minus{}q^2}{1\plus{}q}$. So, for $ \minus{}1 < q < \sqrt{n}$, $ q$ increases through the transformation, and for $ \sqrt{n} < q$, the transformation reduces $ q$.

[u]Here's a question I haven't tried:[/u] For what $ m$ does $ a_{n\plus{}1} \mapsto \frac{ma_n\plus{}n}{a_n\plus{}m}$ with initial value $ a_0$ converge to $ \sqrt{n}$?

[quote="TZF"]for $ \minus{} 1 < q < \sqrt {n}$, $ q$ increases through the transformation, and for $ \sqrt {n} < q$, the transformation reduces $ q$.[/quote]
This only proves boundedness, not convergence.

[quote="TZF"][u]Here's a question I haven't tried:[/u] For what $ m$ does $ a_{n \plus{} 1} \mapsto \frac {ma_n \plus{} n}{a_n \plus{} m}$ with initial value $ a_0$ converge to $ \sqrt {n}$?[/quote]
Again the fixed points are $ \pm \sqrt{n}$, so if the sequence converges at all it converges to one of the square roots.  The eigenvalues of $ \left[ \begin{array}{cc} m & n \\ 1 & m \end{array} \right]$ are $ m \pm \sqrt{n}$, so convergence clearly can't occur for $ m \equal{} 0$ but should occur for all real $ m > 0$.  Things get iffy for negative $ m$, since the denominator can blow up.


a convex $ 2m$-gon has vertecies; $ A_1,A_2,A_3,...A_{2m}$.  If a point, P, lies inside the $ 2m$-gon and does not lie on a diagonal, show that it lies inside an even number of triagnles, with vertecies at $ A_1,A_2...A_{2m}$.

Cute!

[hide]Choose a point very close* to edge $ A_{2m}A_1$.  Then this point is contained exactly in triangles $ A_{2m}A_1A_i$ for $ i \in \{2, 3, \ldots, 2m \minus{} 1\}$, an even number.  Now, it's clear that if we move a point around without crossing any diagonals, we don't change the number of triangles in which it lies.  So we just need to show that if we move from one side of a diagonal (and very close to that diagonal) to the other side (and still very close to the diagonal), we don't change the parity.  But it's easy to see that when a point is very close to diagonal $ A_iA_j$ and then crosses the diagonal, it stays in every triangle it was in except those of the form $ A_iA_jA_k$ and it stays out of every triangle it was out of except those of the form $ A_iA_jA_k$.  But since there are a total of $ 2m \minus{} 2$ triangles of this form, we're done.

* "Very close" means "close enough so there are no diagonals between the point and the line segment in question."

This method is essentially what Po-Shen Loh described as "brutal force" when he taught it at MOP, lo those many years ago.[/hide]


geometry
MATHCOUNTS
3D geometry
number theory
prime numbers

Hard puzzeling math problems that make you think out of the box.

[quote="goldendomer"]Do you want us to post "puzzle problems"?[/quote]

Yes, that is correct.

[quote="i_like_pie"]I thought we couldn't create marathons...[/quote]

I thought so too.

[quote="indianamath"]Marathon #2 has been goin on and i dont think a mod or administrator has said anything.[/quote]

Are you talking about Marathon 2 in the MATHCOUNTS forum? If you are, it's the only going in the MC forum. #1 was locked because it was way too big.

[quote="hairymonkey2"]its 4000pi[/quote]

Nope.

[hide]Hint: unfold the cone[/hide]

[quote="8parks11"]the answer is : 0 pi haha
sphericon[/quote]

Wrong. Hint: The answer does not include pi.

[quote="8parks11"]isnt it 0?
the pi was a joke[/quote]

The question is basically asking...

where did you get the name "Zoboop" anyway  :huh: ?

[quote="snipefreak"][quote="funcia"][quote="hrzdlake"]Isn't it just 8000 miles?
Or possibly 2000 pi?[/quote]
if they go through it is 8000[/quote]
He says that there is no pi in the answer...doesn't there have to be though?[/quote]
not if dig hole to china

[quote="pn12345"][quote="8parks11"]i understand haha
i thought this was a brain teaser
so i just said 0

shortest would be to take the triangular arc?
or the circumference? or the arc of the half of the circle?[/quote]

unfold ONE cone into a pie-like section with a piece missing. The draw a straight line between the two points on the section.
This is the shortest distance[/quote]
the shortest distance between two points is a direct straight line. it does not have to be above the surface. it can be just a direct line. the question did not ask for shortest traveling distance, just shortest distance.

[quote="pn12345"]Here is your fist problem:

The is a planet named Zoobop.
Zoobog is a peculiar planet, in the sence that it has two "hemispheres" that are shaped like right-circular cones.
The two cones are congruent, and are joined at their circular bases. The height of the cones are 3000 miles each, and they have a base radius of 4000 miles.
Two cities are located on the equator of Zoobop, 180 degrees apart.

What is the shortest distance between the two cities?[/quote]
Wouldn't you just double the height, giving you 8000 miles?

[quote="erdogankerem123"]Question 1: is it 10000 miles...oops, I didnt see the posts about 8000 miles, and explanations about... would 10000 miles be the shortest traveling distance, though?
Question 2:52?[/quote]
For question 2 I get 52 too.

[quote="pn12345"]Here is your fist problem:

The is a planet named Zoobop.
Zoobog is a peculiar planet, in the sence that it has two "hemispheres" that are shaped like right-circular cones.
The two cones are congruent, and are joined at their circular bases. The height of the cones are 3000 miles each, and they have a base radius of 4000 miles.
Two cities are located on the equator of Zoobop, 180 degrees apart.

What is the shortest distance between the two cities?[/quote]

I think it is the square root of ((1-cos144)(2)(5000^2)) which is about 9510.56516.  This is traveling in an arc across the lateral surface of the cone.  If you unfold the cone to get a section of a circle, you can compare the radius or the circle 5000 (lateral hight of the cone) to the radius of the base of the cone (4000, given) to find that the section represents 4/5 of the circle, which is 288 degrees.  one city is in the middle of the arc and one is on the end, so by dropping a line from the center of the circle to the second city and connecting the two cities with a strait line we form a triangle with two sides of 5000 and an included angle of 144 degrees.  Using law of cos we find the final side.

Here's a new problem:
3.The sequence 2, 3, 5, 6, 7, 10, … consists of all natural
numbers which are neither perfect squares nor perfect
cubes. Find the 75th term of this sequence.

[hide]The 75th number is 75, but 1^2, 2^2, 2^3, 3^2, 4^2, 5^2, 3^2, 6^2, 7^2, and 8^2 (or 4^3) are counted. This is 10 squares or cubes, so 75 is actually the 65th term in the sequence. 80 is the 70th term, then we have: 82, 83, 84, 85, 86. 86 is the 75th term.[/hide]

[quote="funcia"][quote="buzzer11"][quote="hrzdlake"]Isn't it just 8000 miles?
Or possibly 2000 pi?[/quote]

[quote="pn12345"]The answer does not include pi.[/quote][/quote]
spam?!
no, that person didn't read question[/quote]

i did NOT write that...  :maybe:

[hide]86. Wasn't this on a previous MOEMS test?[/hide]

[hide="Simple."]86[/hide]
This is a long thread.

[hide=" :arrow:  :roll:  :wink: "]That was weird. None, of course.[/hide]

[quote="xpmath"][quote="vamathletes"]$a$ is a prime number. In the set ${2a,3a,5a}$, how many prime numbers are there?
BTW, this thread isn't long if you compare  it to the problem solving thread in the Mathcounts Forum (>90 pages)[/quote]
[hide]Obviously, each will be divisible by $a$, and the coefficient, so no prime numbers.[/hide][/quote]

You should hide things in smilies.
[hide=":roll:"]It's cool.[/hide]

New problem?

[quote="Nerd_of_the_Ages"]what is the ones digit of 2007^2007?
obviously, you dont have a calculator with a screen big enough to fit that number.[/quote]
[hide]$2007^{2007}\equiv2007^{2007\mod4}\equiv2007^{3}\equiv7^{3}\equiv\boxed{3}\mod{10}$[/hide]

Bobby runs on a track at a two feet per second. If an object at his starting place is falling from 2 miles up starting at .1 feet a second (abnormally slow, I know), how many quarters of a lap will Bobby have run before the object hits the ground, to the nearest lap.

Are we assuming each lap is one mile? If so
[hide]Since the object falls .1 feet each second, and 1 mile is 5280 feet, it will take 10*5280*2= 105600 seconds.
Then, since Bobby travels 2 feet per second, he travels 211200 feet, which is 40 miles. So if each lap is one mile, he needs to do 40 laps?[/hide]

I'm really tired right now, so I probably messed up.

Are we assuming the object is falling at a constant rate, because in real life, there is acceleration, which changes almost the whole problem.

Well, I said [b]starting[/b] at that speed, implying acceleration.


Using the Binomial Theorem Part 3

We use our understanding of the Binomial Theorem to quickly find specific terms of expansions of powers of binomials, without finding the full expansion!


The "obvious" thing to do is solve the first equation for $ a$ and substitute it in to the second equation.  There're one "nice" and three "ugly" solutions (the roots of a cubic with integral coefficients; one irrational and two complex with irrational real and imaginary parts).

er... now i need help solving a 6th power:

$ a^6\plus{}2a^4\minus{}6a^3\plus{}a^2\minus{}3\equal{}0$

I thought you were solving for $ x$... :maybe:
[hide="Too lazy to show the numbers, sorry."]
Factor the second equation to get $ a(ax^2\plus{}a\minus{}2)\equal{}0$. Now we have $ a\equal{}0$ or $ ax^2\plus{}a\minus{}2\equal{}0$. The first case has $ x\equal{} \dfrac 12$. (Substitute $ a\equal{}0$ in the first equation.) The other case, $ ax^2\plus{}a\minus{}2\equal{}0$, when added with the first equation, yields $ \minus{}(x\plus{}\dfrac 12)\equal{}a$. Plugging that in the first equation eventually gives the cubic $ 2x^3 \plus{} x^2 \plus{}2x \minus{}2\equal{}0$, which has the "ugly" irrational and complex roots.
[/hide]

[quote="unimpossible"]On a harder problem, i have everything simplified down to a 2 by 2 quadratic, solving for x: Please help:

$ \minus{} ax^2 \plus{} 2x \plus{} a \equal{} 1$
$ a^2x^2 \minus{} 2ax \plus{} a^2 \equal{} 0$[/quote]
Wait, I'm confused...
[hide="Does this work?"]
We multiply the first equation by $ a$ to get that $ \minus{} a^2x^2 \plus{} 2ax \plus{} a^2 \equal{} a$.  Adding this to the second equation gives us that $ a^2 \plus{} a^2 \equal{} a\implies 2a^2 \equal{} a\implies a \equal{} \frac {1}{2}, 0$.  Then plug them in to get:

[b]Case 1: $ a \equal{} 0$[/b]
$ 2x \equal{} 1$ and $ 0 \equal{} 0$, so one solution is $ x \equal{} \frac {1}{2}$

[b]Case 2: $ a \equal{} \frac {1}{2}$[/b]
The first equation yields $ \minus{} \frac {x^2}{2} \plus{} 2x \minus{} \frac {1}{2} \equal{} 0\implies x^2 \minus{} 4x \plus{} 1 \equal{} 0\implies x \equal{} \frac {4 \pm \sqrt {16 \minus{} 4}}{2} \equal{} 2\pm \sqrt {3}$.  The second equation is now essentially the same thing as the first, so it will give the same solutions.

So, $ x \equal{} \boxed{\frac {1}{2}, 2\pm \sqrt {3}}$
[/hide]

The problem was originally written differently.  Your solution seems correct for the new problem.


Let a,b,c are non-negative numbers such that $ a^2\plus{}b^2\plus{}c^2\equal{}1$. Prove that
$ \sum_{sym}\frac {1}{2\minus{}3ab}\le 3$

[quote="thegod277"]Let a,b,c are non-negative numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. Prove that
$ \sum_{sym}\frac {1}{2 \minus{} 3ab}\le 3$[/quote]
Should it be $ \sum_{cyc}\frac {1}{2 \minus{} 3ab}\le 3$? $ a \equal{} b \equal{} c \equal{} \sqrt{\frac{1}{3}}$ gives:
$ \sum_{sym}\frac {1}{2 \minus{} 3ab} \equal{} 6 \frac{1}{2\minus{}1} \equal{} 6 > 3$

No.$ \sum_{sym}\frac {1}{2\minus{}3ab}\equal{}\frac {1}{2\minus{}3ab}\plus{}\frac {1}{2\minus{}3bc}\plus{}\frac {1}{2\minus{}3ca}$. So, if $ a\equal{}b\equal{}c\equal{}\frac {1}{\sqrt 3}$ we have $ \sum_{sym}\frac {1}{2\minus{}3ab}\equal{}3$

[quote="thegod277"]No.$ \sum_{sym}\frac {1}{2 \minus{} 3ab} \equal{} \frac {1}{2 \minus{} 3ab} \plus{} \frac {1}{2 \minus{} 3bc} \plus{} \frac {1}{2 \minus{} 3ca}$. So, if $ a \equal{} b \equal{} c \equal{} \frac {1}{\sqrt 3}$ we have $ \sum_{sym}\frac {1}{2 \minus{} 3ab} \equal{} 3$[/quote]
Maybe $ \sum_{cyc}\frac {1}{2 \minus{} 3ab} \equal{} \frac {1}{2 \minus{} 3ab} \plus{} \frac {1}{2 \minus{} 3bc} \plus{} \frac {1}{2 \minus{} 3ca}$ $ ?$ :wink:
By the way, $ \frac {1}{2 \minus{} 3ab} \plus{} \frac {1}{2 \minus{} 3bc} \plus{} \frac {1}{2 \minus{} 3ca}\leq3$ is wrong.
Try $ a\equal{}\frac{1}{\sqrt{10}}$ and $ b\equal{}c\equal{}\frac{3}{\sqrt{10}}.$ :wink:

[quote="arqady"][quote="thegod277"]No.$ \sum_{sym}\frac {1}{2 \minus{} 3ab} \equal{} \frac {1}{2 \minus{} 3ab} \plus{} \frac {1}{2 \minus{} 3bc} \plus{} \frac {1}{2 \minus{} 3ca}$. So, if $ a \equal{} b \equal{} c \equal{} \frac {1}{\sqrt 3}$ we have $ \sum_{sym}\frac {1}{2 \minus{} 3ab} \equal{} 3$[/quote]
Maybe $ \sum_{cyc}\frac {1}{2 \minus{} 3ab} \equal{} \frac {1}{2 \minus{} 3ab} \plus{} \frac {1}{2 \minus{} 3bc} \plus{} \frac {1}{2 \minus{} 3ca}$ $ ?$ :wink:
By the way, $ \frac {1}{2 \minus{} 3ab} \plus{} \frac {1}{2 \minus{} 3bc} \plus{} \frac {1}{2 \minus{} 3ca}\leq3$ is wrong.
Try $ a \equal{} \frac {1}{\sqrt {10}}$ and $ b \equal{} c \equal{} \frac {3}{\sqrt {10}}.$ :wink:[/quote]
Argady, I think when $ a \equal{} \frac {1}{\sqrt {10}}$ and $ b \equal{} c \equal{} \frac {3}{\sqrt {10}}.$
we have $ a^2\plus{}b^2\plus{}c^2\equal{}1.9>1$??? :maybe:  :maybe:

[quote="thegod277"][quote="arqady"][quote="thegod277"]No.$ \sum_{sym}\frac {1}{2 \minus{} 3ab} \equal{} \frac {1}{2 \minus{} 3ab} \plus{} \frac {1}{2 \minus{} 3bc} \plus{} \frac {1}{2 \minus{} 3ca}$. So, if $ a \equal{} b \equal{} c \equal{} \frac {1}{\sqrt 3}$ we have $ \sum_{sym}\frac {1}{2 \minus{} 3ab} \equal{} 3$[/quote]
Maybe $ \sum_{cyc}\frac {1}{2 \minus{} 3ab} \equal{} \frac {1}{2 \minus{} 3ab} \plus{} \frac {1}{2 \minus{} 3bc} \plus{} \frac {1}{2 \minus{} 3ca}$ $ ?$ :wink:
By the way, $ \frac {1}{2 \minus{} 3ab} \plus{} \frac {1}{2 \minus{} 3bc} \plus{} \frac {1}{2 \minus{} 3ca}\leq3$ is wrong.
Try $ a \equal{} \frac {1}{\sqrt {10}}$ and $ b \equal{} c \equal{} \frac {3}{\sqrt {10}}.$ :wink:[/quote]
Argady, I think when $ a \equal{} \frac {1}{\sqrt {10}}$ and $ b \equal{} c \equal{} \frac {3}{\sqrt {10}}.$
we have $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1.9 > 1$??? :maybe:  :maybe:[/quote]
It can be $ a \equal{} \frac {1}{\sqrt {19}}$ and $ b \equal{} c \equal{} \frac {3}{\sqrt {19}}.$ :D

[quote="arqady"][/quote][quote="thegod277"][quote="arqady"][quote="thegod277"]No.$ \sum_{sym}\frac {1}{2 - 3ab} = \frac {1}{2 - 3ab} + \frac {1}{2 - 3bc} + \frac {1}{2 - 3ca}$. So, if $ a = b = c = \frac {1}{\sqrt 3}$ we have $ \sum_{sym}\frac {1}{2 - 3ab} = 3$[/quote]
Maybe $ \sum_{cyc}\frac {1}{2 - 3ab} = \frac {1}{2 - 3ab} + \frac {1}{2 - 3bc} + \frac {1}{2 - 3ca}$ $ ?$ :wink:
By the way, $ \frac {1}{2 - 3ab} + \frac {1}{2 - 3bc} + \frac {1}{2 - 3ca}\leq3$ is wrong.
Try $ a = \frac {1}{\sqrt {10}}$ and $ b = c = \frac {3}{\sqrt {10}}.$ :wink:[/quote]
Argady, I think when $ a = \frac {1}{\sqrt {10}}$ and $ b = c = \frac {3}{\sqrt {10}}.$
we have $ a^2 + b^2 + c^2 = 1.9 > 1$??? :maybe:  :maybe:[/quote][quote="arqady"]
It can be $ a = \frac {1}{\sqrt {19}}$ and $ b = c = \frac {3}{\sqrt {19}}.$ :D[/quote]
oh, yes. I am wrong. Thank you argady, I will find a condition best for it.thank you.  :)
PS: A inequality $ \frac {1}{1-ab}+\frac {1}{1-bc}+\frac {1}{1-ca}\le \frac 92$ is right. So what is genelization inequality? :huh:


linear algebra
matrix
superior algebra
superior algebra solved

Let A,B be two matrices in M_2(R). Prove that det(I_2+ (2AB+3BA)/5)=det(I_2+ (3AB+2BA)/5)

Let P(x) = det(I2+xAB+(1-x)BA). Since it's well-known that det(!2+AB)=det(I2+BA) then P(0)=P(1) (*). P has degree 2, so P(x)=ax^2+bx+1. From (*) we get a+b=0, so P(x)=ax^2-ax+1. It's easy to check now that P(x)=P(1-x) and you apply this for x=3/5.

where can i find some documentation about the polinomyal form of a determinant??


I'm having difficulty with these problems - please help!!!

1.) y = (4x)^(1/2) from x = 0 to 8 about the x-axis
so in terms of x this is:
x = ((y-1)^(1/2))/2 from y = 1 to 257
so in the formula this comes out to:
2pi [int from 1 to 257] ((y-1)^(1/2)) / 2 * (1 + (1 /( 4(y - 1)^(1/2))) ^2) ^(1/2)
which comes out to be:
2pi* (1/192) (y-1)^(3/2) ((16y - 15) / (y - 1))^(3/2)
which is about 142
which is wrong! what have i done incorrectly?

the other is:
2.) x = 2e^2y from y = 0 to y = 1 about y-axis
in formula:
2pi * [Int from 0-1] 2e^(2y) (1 + (4e^(2y))^2)^(1/2)
which comes out to be:
2pi (1/8) ln((1 + 16e^(4))^(1/2)) + 4e^(2) + (1/2)e^(2) (1 + 16e^(4))^(1/2)
which is about 142
which is wrong! what have i done incorrectly?

And would you know how to solve:

Find the area of the surface obtained by rotating the curve:
$ y \equal{} 2e^{2x}$ from $ x \equal{} 0$ to $ x \equal{} 1$ about the $ x$ axis.

[quote="Beachic109"]I'm having difficulty with these problems - please help!!!

1.) y = (4x)^(1/2) from x = 0 to 8 about the x-axis
so in terms of x this is:
x = ((y-1)^(1/2))/2 from y = 1 to 257[/quote]

That doesn't hang together.

If $ y\equal{}\sqrt{4x}$ for $ 0\le x\le 8,$ then $ x\equal{}\frac{y^2}4$ for $ 0\le y\le 2\sqrt{2}.$

If $ x\equal{}\frac{\sqrt{y\minus{}1}}2$ for $ 1\le y\le 257,$ then $ y\equal{}1\plus{}4x^2$ for $ 0\le x\le 8.$

I can't even tell what problem you're working, and if I can't tell that, there's no way I can find what else might be wrong.


abstract algebra
superior algebra
superior algebra theorems

Please tell me what is erroneous about the following argument.

Say we have some homomorphism phi from group X to group Y. For some x in X, we have the equality (phi(x))^n = phi(x^n). From this we see that if the order of x is n, then phi(x^n) = phi(1) = 1 = phi(x)^n, and hence the order of phi(x) is n. (Note: Showing that n is minimal is simple; assume there exists some m less than n such that phi(x)^m = 1. This implies phi(x^m) = 1, however this contradicts than n is the order of phi(x), thus by contradiction n in minimal such that phi(x)^n = 1.)

However, this equality doesn't seem to make sense; for example if x comes from the kernel of phi, although the order of phi(x) is 1, that doesn't imply the order of x must be one. (A kernel can certainly contain more than just the identity.) So then what is wrong with the argument?

the argument uses its claim ...

you just can say, that the order of $ \phi(x)$ divides the order of $ x$.


Let $(a_{n})$ be a decreasing positive sequence and the sequence $S_{n}=a_{1}+a_{2}+...+a_{n}$ is convergent.
Show that $lim (na_{n})=0$

[quote="Ngông"]please write more details. I can't understand what you did[/quote]

$\\ S_{2n}-S_{n}= \sum_{k=n}^{2n}a_{k}\geq n a_{2n}>0$ since $a_{n}$ is decreasing.

Therefore $\lim_{n \to \infty }2n a_{2n}= 0$ from Squeeze theorem .

But since $a_{2n}\geq a_{2n+1}$ we have

$\\ \frac{2n+1}{2n}2n a_{2n}\geq (2n+1) a_{2n+1}\Rightarrow \lim_{n \to \infty }(2n+1) a_{2n+1}= 0$

Therefore $\lim_{n \to \infty }n a_{n}= 0$ .

I don't know Squeeze theorem .Please tell me more about it.

Suppose $(a_{n})$, $(b_{n})$, and $(c_{n})$ are three sequences of real numbers such that for all $i\in \mathbb{N}$ we have $a_{i}\leq b_{i}\leq c_{i}$. Suppose further that $\exists x_{0}\in\mathbb{R}$ such that $\lim_{n\to \infty}a_{n}=\lim_{n\to \infty}c_{n}=x_{0}$, Then squeezing therem states that $\lim_{n\to \infty}b_{n}=x_{0}$.


[quote="anirudh"]Does anyone have the answers for the following:
National 1991-1995 sprint and target
1996 target
1998 sprint and target
2003 sprint and target?

Thanks[/quote]
I have them somewhere. I need to dig them out.

Do you want us to pm, or should we post it like last time?

It seems like you want a lot of things.

It may be over the uploading quota, so just email them to me:
jdasarathy@yahoo.com

yes, that is for year 2003.

Thank You mad_skillz_aops.


Basicly, this is just tug of war, but now instead of moving left and right, you can move in 2 dimensions: left, right, up and down.  Each move increments or deincrements the x or the y value in the last position by 1.  Let's start:

[b](0,0)[/b]

uhhh ok apparently ccy is too tired to do this right.
so ignore the "let's start" part. we need teams anyways
So just sign up here I guess...

Not yet friends. Hold on a bit more! Tiem is not ripe. I'll decide the right moment for this. And I also have great plans ...

Keep guessing!

wtf bubka
I need more excuses so I can procrastinate on homework. Might as well start a 2d tow now

OMFG GUYS I BUMPED THE OLD ONE UP NO NEED TO HAVE ANOTHER THREAD

OMFG YOU BUMPED THAT ONE AFTER THIS GUY MADE THIS

OMFG SRRY BOUT TAT THE OLD ONE IS LOCKED NOW SO ALL IS GOOD

*BLAH*

[quote="mustafa"]YES..........[/quote]

UM IT WASN'T A YES OR NO QUESTION

YO GUYS WE NEED A MORE ORGANIZED START

SIGNUPS OVER HERE:

1)akalra01
2)ccy
3)junggi
4)bubka

[quote="Ignite168"]:roll: :roll:[/quote]
DOES THAT IMPLY THAT YOU SIGNED UP?

People you are doing nothing but turning this 2D tug of war into spam. Just be a bit more patient. I am making preparations.

And someone lock this.

What preperations do you need to make a tug of war game? :huh:

LOTS
[color=red]
WATCH [/color]AND LEARN


Prove that there exists $1977$ triangles $ABC$ satisfying the conditions:
$\frac{\sin{A}+\sin{B}+\sin{C}}{\cos{A}+\cos{B}+\cos{C}}=\frac{12}{7}$ and $\sin{A}\sin{B}\sin{C}=\frac{12}{25}$.

We have $\frac{4\prod cos\frac{A}{2}}{1+4\prod sin\frac{A}{2}}=\frac{12}{7}$ and  $8\prod sin\frac{A}{2}=\frac{12}{25}$.Thus $\prod cos\frac{A}{2}=0,6$ and $\prod \sin\frac{A}{2}=0,1$.
Notice that $sin\frac{C}{2}=cos\frac{A}{2}cos\frac{B}{2}-sin\frac{A}{2}sin\frac{B}{2}$ $\Leftrightarrow$ $sin^2\frac{C}{2}cos\frac{C}{2}=0,6sin\frac{C}{2}-0,1cos\frac{C}{2}$.
Let $cosC=t$,we have
$(1-t^2)^2=0,6\sqrt{1-t^2}-0,1t$
$\leftrightarrow$ $11t-10t^3=6\sqrt{1-t^2}$
Solved this equation,we have $cos\frac{C}{2} \in$ {$\sqrt{\frac{1}{2}},\sqrt{\frac{4}{5}},\sqrt{\frac{3}{10}}$}
therefore $sinC \in$ {$1;0,8;0,6$} and from here the conclusion follows.


Suppose that $ x\equal{}cos\angle{A}$, $ y\equal{}cos\angle{B}$, $ z\equal{}cos\angle{C}$, where $ \angle{A},\angle{B},\angle{C}$ are the angles of an acute triangle $ ABC$. Prove that  $ 2(xy\plus{}yz\plus{}zx)\plus{}3\geq{3(x\plus{}y\plus{}z)}$.

Sorry for my fault, the sign must be reversed. Hence, we need to prove $ 2(xy\plus{}yz\plus{}zx)\plus{}3\leq{3(x\plus{}y\plus{}z)}$.


[quote="revelation"]Here's a pretty neat contest problem that i found the other day...

If $(x-a)(x-10)+3 = (x+b)(x+c)$ is true for all $x$ and if $a, b,$ and $c$ are integers, then the number of possible values for $a+b$ is....

A) 0
B) 2
C) 3
D) 4
E) 6[/quote]

[hide]Let $x = a$. That forces us to have $3 = (a+b)(a+c)$, which is the product of 2 integers. Since $3$ is prime, it only has two factors (which can be positive or negative). Therefore, the answer is D. [/hide]

just expand it, and equate coefficients....

[hide]Expanding and rearranging, we get two equations:
$-10-a=b+c\Longrightarrow a+b=-10-c$
$10a+3=bc\Longrightarrow 10a+ac+3=ac+bc\Longrightarrow a(10+c)+3=(a+b)c$

Substituting in our value for $a+b$ and rearranging some more, we get:
$(10+c)(a+c)=-3$.
Because both of these factors are integers, $(10+c)$ can only be $1,3,-1, or-3,$ and then $a+b$ is also $\pm 1$ or $\pm 3$ and the answer is D.[/hide]

[quote="white_horse_king88"]And does it specify that $a+b$ must be positive?  :wink:[/quote]

WHY do i always miss small little details like this. :mad:


If $ r$ is a real number, prove that there must be at least one term in the sequence $ r, 2r, \cdots, 99r$ that is within $ \frac {1}{100}$ an integer.

By "within $ \frac{1}{100}$, you include numbers which are exactly $ \frac{1}{100}$ away from an integer right...

[hide="hint"]You know the difference between any two of them is also in the set.  So assume otherwise: now construct $ 98$ pigeonholes from the interval between two integers not including the sections which correspond to the desired property.[/hide]


On side $ BC$ of $ \triangle ABC$ is point $ D$, such that $ BD : DC \equal{} 2: 1$. What is the ratio in which the median from point $ C$ to side $ AB$ divides the segment $ AD$?

I'm also trying to figure it out without just looking at a solution, so a hint would be nice too.  :)

First hint:

[hide]Let the midpoint of AB be E. Draw the midsegment EF, where F is the midpoint of AC. Let the intersection of CE and AD be M, and let the intersection of EF and AD be N. See if you can find any interesting triangles? (and/or ratios involving the midsegment of the triangle?)
[/hide]

Second hint:

[hide]Indeed, triangles MNE and MDC are congruent. This means MN and MD are congruent. We can also note that EN is a midsegment of triangle ABD, so AN = ND.[/hide]

Wrapping it up:

[hide]MN=MD
MN+MD = ND
2MD = ND
MD = ND/2

So I'm not sure exactly what ratio you're looking for, but it would divide the median into parts with ratio 3:1[/hide]

I think. ;)

I think for this problem it's easier to use mass points.


calculus integration algebra function domain factorial algorithm

I saw posts about gaussian integer at

[url]http://www.mathlinks.ro/viewtopic.php?t=179715[/url]

but I can't understand it

So I want to know about the number's definition & concept.

And how to use it??

The set of Gaussian integers is $ \mathbb{Z}[i]$.  You can find lots of information about them on the internet, especially mathlinks.  I suggest that you get a book on abstract algebra.  :wink:

You may first want to look at my post http://www.mathlinks.ro/Forum/viewtopic.php?p=584714 for what rings are.
Then see http://www.mathlinks.ro/Forum/viewtopic.php?p=935293 for some more terms.

Gaussian numbers $ \mathbb Z[i] \equal{} \{ x \plus{} iy \in \mathbb C | x,y \in \mathbb Z \}$ are a special type of ring, namely they are:
a) commutative and unitary, thus $ ab \equal{} ba$ and there is $ 1 \in \mathbb Z[i]$ such that $ 1 \cdot a \equal{} a$ for all $ a,b \in \mathbb Z[i]$.
b) an integral domain: if $ ab \equal{} 0$ then $ a \equal{} 0$ or $ b \equal{} 0$. Or the other way around: the product of nonzero numbers is a nonzero number.
c) factorial: every such number can be written as a product of prime elements, and this is unique up to reordering and multiplication with units. See the second link.

Even more:
d) there is a norm: there is a mapping $ N: \mathbb Z[i] \to \mathbb Z$ given by $ N(x \plus{} iy) \equal{} (x \plus{} iy)\overline{(x \plus{} iy)} \equal{} (x \plus{} iy)(x \minus{} iy) \equal{} x^2 \plus{} y^2$ for $ x \plus{} iy \in \mathbb Z[i]$. It is multiplicative, thus $ N(a\cdot b) \equal{} N(a) \cdot N(b)$.
e) norm-eucidean: for $ a,b \in \mathbb Z[i]$, $ b \neq 0$, we always find $ q,r \in \mathbb Z[i]$ such that $ a \equal{} bq \plus{} r$ and $ N(r) < N(b)$. This is what the euclidean algorithm is based upon.

It can be shown that a), b) and e) imply c).
One example of usage is the thread you linked at. Some more problems:
- show that if $ x,y \in \mathbb Z$ are each sum of two perfect squares, then $ xy$ is a sum of two squares.
- let $ p$ be prime; show that $ x^2 \equiv \minus{} 1 \mod p$ has a solution iff $ p$ is a sum of two perfect squares. Show that this solution is unique (up to sign and permutation) and find all such $ p$.
- find the number of ways to write some $ n$ as a sum of two squares.

Well, and after concerning with Gaussian numbers, there are a lot of similar rings that deserve attention. Look for algebraic integers and number rings (number fields). I suggest some course on abstract algebra, then on algebraic number theory.


symmetry
geometry
similar triangles
geometry proposed

Let two circles (1),(2) intouches the circle (O). Two chords BC and DE of (O) are the external common tangents of (1),(2). A point A lies on the arc BC of (O), which does not contain D,E. AD,AE meets BC at M,N. Prove that if the A-median of triangle AMN touches (1),(2), then the second internal common tangent of (1),(2) passes through the midpoint of BC.

The first common internal tangent of the circles $ (1), (2)$ cuts $ BC \equiv MN$ at $ P$ and $ (O)$ at $ A$ on the arc $ BC$ which does not contain $ D, E.$ The 2nd common internal tangent of $ (1), (2)$ cuts $ DE$ at $ F$ and the circle $ (O)$ at $ Z,$ again on the arc $ BC$ which does not contain $ D, E.$ By the parallel tangent theorem, $ AZ \parallel BC \equiv MN.$ The $ \triangle AMN \sim \triangle ZED$ are oppositely similar, having equal angles: $ \angle MAN \equal{} \angle DAE \equal{} \angle DZE$ and $ \angle MNA \equal{} \angle MAZ \equal{} \angle EAZ \equal{} \angle EDZ.$ By symmetry, the angle $ \angle APM \equal{} \angle APB$ formed by the 1st common pair of the common external / internal tangents $ BC, AP$ of $ (1), (2)$ is equal to the angle $ \angle ZFE$ formed by the 2nd pair of their common external / internal tangents $ DE, ZF.$ Therefore $ P, F$ are corresponding points of the similar triangles $ \triangle AMN \sim \triangle ZED.$ Consequently, $ P$ is the midpoint of $ MN$ $ \Longleftrightarrow$ $ F$ is the midpoint of $ DE.$

[hide]With the above lemma, we are able to prove the problem at [url]http://www.mathlinks.ro/viewtopic.php?t=213165[/url][/hide]


Yes, that old game. Here's the drill. Find what "I" in each riddle or passage refers to. All are written by me unless listed otherwise.

[b][u]Submit your answer by private message[/u][/b]. On submission PMs, title (that means subject line) your submission [color=green]"WAI #[insert number]: [insert answer you're submitting]"[/color] for example [color=orange]"WAI #22: Seventy-four"[/color]. A scoreboard will be kept. [b][u]Do not, I repeat, do [i]not[/i], post your answers on the thread. [/u][/b]

Use the most [u]specific[/u] answer you can think of that fits [u]all[/u] clues given.

Okay, here's the first.

[size=200]Puzzle One[/size]

For five points, find the person who would be the "I" below.

A prodigy, a genius, a math wizard was me,
When I traveled to Britain it was plain to see.
I devised clever theorems when I was thirteen,
And I cranked out numbers fast as a machine.
I showed my peers math they never saw,
And they all stared at me in awe,
Although I must have went a little too askew,
Since I died when I was thirty-two.

There were [b]six[/b] submissions for [b]Puzzle One[/b].

[u]Current Scoreboard[/u]
tennis123 5
meewhee009 5
math_explorer 5
westiepaw 5
pacman2812 5
k00lperson 5

[size=200]Puzzle Two[/size]

For five points, find the geographical feature that would be the "I" below.

On Earth, I was a recent occurrence,
I show early mankind's appearance,
In fossils inside, Where the desert can hide.
And across my stretch over an entire continent,
Making a fault breaking ever so dissonant,
Are all these lakes,
Some are so great the seem fake,
And one so dangerous it puts a million people at stake.

I forgot to post the answer to Puzzle One. :wallbash: anyways, it's Ramanujan. From now on, I will only accept submissions for the currently active puzzle.
The answer to Puzzle Two was the Great Rift Valley, specifically.

There were [b]three[/b] submissions for [b]Puzzle Two[/b].

[u]Current Scoreboard[/u]
k00lperson 7
tennis123 5
meewhee009 5
math_explorer 5
westiepaw 5
pacman2812 5
PowerOfPi 5
ernie 2
ksun48 2

[size=200]Puzzle Three[/size]

For five points, find the scientific phenomenon that would be the "I" below.

I cause the west coasts to be warm,
I cause the cyclones to be formed,
I determine the winds that blow hilltops flat,
I am part of what makes the earth fat,
I impact climate (and maybe sensation),
And I am the result of the Earth's rotation.

HURRAH! I'm winning..
i probably am gonna start losing now.

Of course! You can.....

Like just pm him saying the answers

There were [b]twelve[/b] submissions (including [b]two[/b] [color=green]correct[/color] ones! =P) for [b]Puzzle Three[/b]. The answer was the [b]Coriolis Effect[/b]. There was a diverse variety of submitted answers, as can can see from the list of the twelve submissions (in chronological order of submitted time) below.

El Nino, El Nino, Centrifugal Force, Heat, Equator, Coriolis Effect, Gravity, El Nino, Tide, Coriolis Effect, Heat, Ocean Currents

Congratulations to [b]vahalla[/b] and [b]eqjj168[/b] who submitted correct answers!

[u]Also, start following the format. I'll start to take off points for the wrong submission format. See the first post. [/u]

[u]Current Scoreboard[/u]
k00lperson 7
math_explorer 6
vahalla 5
eqjj168 5
tennis123 5
meewhee009 5
westiepaw 5
pacman2812 5
PowerOfPi 5
ernie 3
abcak 2
ksun48 2
bluecarneal 1
dragon96 1

Puzzle Four will be up shortly.

darn!  i knew it was going to be the coroilis effect!

I learned that last year.

[size=200]Puzzle Four[/size]

For four points, find who or what the "I" in the following passage refers to.

Compared to my siblings, I am so small;
I can easily fit through the narrowest hall.
As to why I'm so small, well I don't know why,
I was cloned to get why, and the still they all sigh,
As I still don't know, and why still don't so.

There were [b]four[/b] submissions for [b]Puzzle Four[/b]. Indeed, "I" is the letter "I."

[u]Three out of four people were still docked off for the wrong format, despite my warning. Again, put WAI #[number]: [submitted answer] in the subject line. I know I'm a little picky here, but chances are one of your future teachers will be highly anal about strict following of directions. That having been said, let's move on. [/u]

[u]Current Scoreboard[/u]
vahalla 8
k00lperson 7
math_explorer 6
eqjj168 5
tennis123 5
meewhee009 5
westiepaw 5
pacman2812 5
PowerOfPi 5
aleph0 4
ernie 4
ksun48 3
abcak 2
bluecarneal 1
dragon96 1

[size=200]Puzzle Five[/size]

For five points, find the integer that fits the description of "I" below. The third line refers to the digits themselves, not the values of the digits.

I'm not too large, I'm not too small,
Just two digits, yup, that's all.
The ones is six more than the tens,
And I am prime! Win, win win!

Gah.  I put the format right, but I guessed a liger.

Contrariwise, ernie, the format was not right.  :wink:

Oh right, I got lazy and put the WAI in the subject.

[quote="PhireKaLk6781"]There were [b]twelve[/b] submissions for [b]Puzzle Five[/b], less than half of which had the right format. The answer is [b]seventeen[/b].

[u]Again, put WAI #[number]: [submitted answer] [b]in the subject line[/b] (without the brackets). It's for convenience reasons. [/u]

You are allowed to submit your own WAIs for use in this thread. In that case, you get the median score of the submissions.

[u]Current Scoreboard[/u]
vahalla 12
math_explorer 11
westiepaw 10
eqjj168 10
tennis123 10
k00lperson 7
ernie 7
ksun48 6
bluecarneal 6
musicnmath 5
meewhee009 5
pacman2812 5
PowerOfPi 5
Sabien5 5
aleph0 4
andrewjjiang97 3
monkeygirl13 3
abcak 2
dragon96 1

[size=200]Puzzle Six[/size]

For six points, find the musical form described as "I" below:

I was popular ten scores ago,
But still a little today,
My form has many different sections,
Between repeated section A.

There were [b]six[/b] submissions for [b]Puzzle Five[/b]. The answer is [b]rondo[/b].

[u]Current Scoreboard[/u]
vahalla 16
tennis123 15
eqjj168 12
math_explorer 11
westiepaw 10
ernie 8
k00lperson 7
ksun48 6
bluecarneal 6
zapi2006 5
musicnmath 5
meewhee009 5
pacman2812 5
PowerOfPi 5
Sabien5 5
aleph0 4
andrewjjiang97 3
monkeygirl13 3
abcak 2
dragon96 1

[size=200]Puzzle Seven[/size]

For four points, find the contemporary meme described below:

Pronunciation goes out of hand,
Like in a distant land.
And yes it happens in funny verse,
Because it's in reverse.
It isn't annoying as a rickroll,
But funny as bear and bull,
Overused? Maybe, but truth be said,
Google has over 9000 yet.

There were [b]four[/b] submissions for [b]Puzzle Seven[/b]. The answer is [b]In Soviet Russia[/b].

[u]Current Scoreboard[/u]
tennis123 16
vahalla 16
math_explorer 12
eqjj168 12
westiepaw 10
ernie 8
k00lperson 7
ksun48 6
bluecarneal 6
zapi2006 5
musicnmath 5
meewhee009 5
pacman2812 5
PowerOfPi 5
Sabien5 5
aleph0 4
andrewjjiang97 3
monkeygirl13 3
abcak 2
dragon96 1


integration
calculus
limit
function
complex analysis
calculus computations

That integral can be done in elementary terms- it's just ugly.

Actually, I like the more symmetric form $\lim_{N\to\infty}\sum_{n=N}^{N+1}\frac1{1+5n}$- it's a principal value.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=21316]here[/url].

[quote="Kent Merryfield"]Well, I can express it as

$\int_{0}^{1}\frac{1-x^{3}}{1-x^{5}}\,dx= 1-\int_{0}^{1}\frac{x^{3}+x^{4}}{1+x+x^{2}+x^{3}+x^{4}}\,dx.$

Whether that does any good is a different question.

Actually, I think I'd be better off trying to find a Fourier series than a power series.[/quote]
how are u able to do that?

Let $f(x)=\sum_{n=0}^{\infty}\left(\frac{x^{5n+1}}{5n+1}-\frac{x^{5n+4}}{5n+4}\right).$

Differentiate to find $f'(x),$ then note that we can compute $f'(x)$ as a geometric series.

did you find that by just evaluating the integral?

An identity: $\lim_{N\to\infty}\sum_{n=-N}^{N}\frac1{1+\frac{n}{z}}=\lim_{N\to\infty}\sum_{n=-N}^{N}\frac{z}{z+n}=\pi z\cot(\pi z)$ whenever $z$ is not an integer. The proof uses complex analysis; if we divide by $z$, we get an obviously periodic function with poles of residue 1 at every integer. There are several more steps to pin it down exactly, but that is doable- see also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=41504]here[/url].


function
inequalities
inequalities proposed

$ a,b,c,d\geq 0$ and at least two of them are small than 1.prove that
\[ (1\minus{}a)(1\minus{}b)(1\minus{}c)(1\minus{}d)\plus{}a\plus{}b\plus{}c\plus{}d\geq 1\]

Consider
\[ f(a) \equal{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d) \plus{} a \plus{} b \plus{} c \plus{} d \minus{} 1\]
as a linear function in which a is a variable lies in $ [0,1]$

Clearly, we can see that $ f(0)$ and $ f(1)$ are non-negative so we complete our proof here.

[quote="Tourish"]$ a,b,c,d\geq 0$ and at least two of them are small than 1.prove that
\[ (1 - a)(1 - b)(1 - c)(1 - d) + a + b + c + d\geq 1\]
[/quote]
It's a very nice Inequality.
Here my solution:
Expanding the Inequality , we have:
$ ab+bc+ca+bd+cd+ad+abcd\geq 1 \1$

[quote="Tourish"]$ a,b,c,d\geq 0$ and at least two of them are small than 1.prove that
\[ (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d) \plus{} a \plus{} b \plus{} c \plus{} d\geq 1\]
[/quote]
It's a very nice Inequality.
[b]Here my solution:[/b]
We may assume that $ a,b \le 1$
Expanding the Inequality , we have:
$ ab \plus{} bc \plus{} ca \plus{} ad \plus{} bd \plus{} cd \plus{} abcd \ge bc(a \plus{} d) \plus{} ad(b \plus{} c)$
or $ (a \plus{} c)(b \plus{} d) \plus{} ac \plus{} abcd \plus{} bd \ge ac(b \plus{} d) \plus{} bd(a \plus{} c)$
       $ \Leftrightarrow (b \plus{} d)(a \plus{} c \minus{} ac) \plus{} bd(ac \minus{} a \minus{} c) \plus{} ac \plus{} bd \ge 0$
       $ \begin{array}{l} \Leftrightarrow (b \plus{} d \minus{} bd)(a \plus{} c \minus{} ac) \plus{} ac \plus{} bd \ge 0 \\
\Leftrightarrow (b \plus{} d(1 \minus{} b))(a \plus{} c(1 \minus{} a)) \plus{} ac \plus{} bd \ge 0 \\
\end{array}$
Which is true because  $ a,b \le 1$
The equality occurs when $ a \equal{} b \equal{} c \equal{} d \equal{} 0$