Dear Mathlinkers,
some of you know that I like to work with the Reim's theorem which avoids an angle chasing.
see :  http://perso.orange.fr/jl.ayme  "à propos"
To day, I present you a nice generalization which avoid some calculation.

Let (1) and (2) two secant circles at C and D,
L a line intersecting (1) and (2) in this order A, P, B, Q (A, B on (1), P, Q on (2)).
Take two points, one on (1) and the other on (2) : for example A and Q.
Let La, Lq be two parallel lines passing through A, Q and intersecting CD at A', Q'.
we have : BQ' // PA'.

Thank to Kostas for opening my eyes…
Sincerely
Jean-Louis

What applications has this got? i've seen you do some problems (with great elegancy) and i don't think that you've used this form of reim.

P.S can you provide a proof of this?

Dear Mathlinkers,
see http://perso.orange.fr/jl.ayme  vol.5 le cercle de Hagge p. 51. for a proof and for an applicaztion p. 19-21.
Sincerely
Jean-Louis

thank you Jean-Louis. I'll check the link. Also, Do you know of an english book where reim's theorem occurs? I also sent a PM a while back, If you're not too busy would you please replu? I totally understand if you don't.
many thanks.

Dear Sincostan and Mathlinkers,
the complete study of this situation can be seen on my website in French
http://perso.orange.fr/jl.ayme  then "A propos"
Sincerely
Jean-Louis

Er..... Thanks. But i've been studying those figures already. I just don't understand how some of the problems you solve relate to them. Especially, Have you applied this generalized form of reim to any problems on ML?


Could anyone tell me picks fomula please? I have it in a book, but it dosnt appear to be right. :huh:
Thanks

the area of the smallest polygon determine by the lattice points in 1/2.
(i think thats picks formula)

Could you elaborate pleae?
Theres some kinda formula he derived for the area of a latice polygon, in my book it says:

$A=i+\frac{b+1}{2}$

Where i is the interior points and b is the boundry points... but this dosnt work for a 1 by 1 square...

[quote="WindSlicer"]isn't it $i + \frac{b}{2} - 1$?[/quote]

Yeah, I think this is right. Though I also think conventionally the $I$ and $B$ are capitalized... just being picky ;)


[hide]
Instead of looking at ways that she doesn't hear all of her favorite song, we look at the ways that she [i]does[/i] hear all of her favorite song. Either she hears it as the very first song, which occurs with probability $ \frac{1}{10}$; or she hears the 30-second song first and then her favorite song, which occurs with probability $ \frac{1}{90}$; or she hears the 1-minute song first and then her favorite song, which occurs with probability $ \frac{1}{90}$. Add 'em up: $ \frac{11}{90}$ and subtract from 1: $ \boxed{\frac{79}{90}}$.[/hide]


[quote="Anonymous52111"]I need your step-by-step here and need answer, of course. Thanks.

If 6a - 4 = - 2, what is the value of 12a - 4?[/quote]

Here is a slightly easier way:
[hide]Multiplying both sides by $2$, \[2(6a-4)=2(-2)\] Now, we may use the distributive property to get: \[2(6a)+2(-4)=2(-2)\] We may simplify this to get: \[12a-8=-4\] We may add the quantity $4$ to both sides to get: \[12a-8+4=-4+4\] Finally, simplifying this results in: \[12a-4=0\] Thus, the answer to your problem is just \[\boxed{0}\] Notice that we don't even have to find $a$![/hide]

i believe vishalarul's way only applys to certain problems

however, you should always find a first before continuing. (unless ur doing a competition then speed matters more)

[quote="Anonymous52111"]I need your step-by-step here and need answer, of course. Thanks.

If 6a - 4 = - 2, what is the value of 12a - 4?[/quote]
[hide]$6a-4=-2$

$6a=2$

$12a=4$

$12a-4=\boxed{0}$[/hide]


What shoe size do you wear? US shoe sizes please.

ahah its really sad how ppl shorter than me have bigger shoe sizes lol

umm i'm not sure what to vote for.................. i'm size 4 girls and size 6 womens... i guess i wear girls' shoes more. ok voting 4 size 4!! :D  :D

wow u ppls have big feet :rotfl: FIRST ONE TO VOTE FOR SIZE 1-4!!! GO ME!!!

my shoe is is like 9 or something i think. idk.

men's sizes and women's sizes are different - and both genders are voting, so...


Find f(2) if f(x)+xf(1-x)=120x for all real numbers x

We can also solve explicitly for $f(x)$ by solving the system of equations by plugging $x$ into the function equation and plugging $1-x$ into the functional equation.  We get $f(x)=\frac{120x^{2}}{x^{2}-x+1}$ from which we can see $f(2)=160$.

please could you explain your working to find the equation of the function?

thanks

[quote="Ravi B"]Technically, there is no function $f$ that satisfies the recursion for all real $x$.  That's because we get a contradiction when we plug in the two roots of $x^{2}-x+1$.[/quote]

But those roots aren't real numbers.


Prove that $\forall x,y,z \in R$ we have: $ \sum x^{2} - \sum xy \geq \frac{3}{4} (x-y)^{2}$.

cheers! :D :D

0<=3/2[sum(x^2)-sum(xy)]
=3sum(x^2)-3sum(xy)-[3/2[sum(x^2)-sum(xy)]]
=3sum(x^2)-3sum(xy)-3/4sum(x-y)^2 implying that
3sum(x^2)-3sum(xy)>=3/4sum(x-y)^2 (*), for all real x,y,z.

Due to complete symmetry, we may assume that (x-y)^2=min((x-y)^2,(y-z)^2,(z-x)^2), whence, after division by 3, (*) becomes

sum(x^2)-sum(xy)>=3/4(x-y)^2 as required.

Obviously, equality holds iff x=y=z.

Sorry Michael Lipnowski,but I don't think we can let
(x-y)^2=min((x-y)^2,(y-z)^2,(z-x)^2) :?  .If we can,how can we prove this one:
sum(x^2)-(sum(xy))>=3/4max((x-y)^2,(y-z)^2,(z-x)^2).
However,here is my solution:the inequality is equivalent to:
2(y-z) 2 +2(x-z) 2  \geq (x-y) 2 .Now let A=y-z,B=x-y,then we can easily see that:
   2A 2 +2B 2  \geq (A+B) 2
<=>(A-B) 2  \geq 0.
So the inequality is proved.The equality is x+y=2z :)


Prove that among any $39$ consecutive natural numbers it's always possible to find one whose sum of digits is divisible by $11$.

[hide]We will proceed by contradiction.  Assume there exists a set of $39$ natural numbers such that none of the $39$ numbers have a sum of digits divisible by $11$.  Let these numbers be $a_{1},a_{2},a_{3}\cdots a_{39}$.

If the last digit of $a_{n}$ is $0$, and $n\le 30$, then the sum of the digits of $a_{n}$ must be equivalent to $1\pmod{11}$.  Otherwise, the sum of the digits of one of the next nine numbers would be divisible by $11$.  There are exactly three numbers with ones digit $0$ among $a_{1},a_{2},a_{3}\cdots a_{30}$, and these three numbers are consecutive multiples of $10$.  Let these numbers be $10n,10(n+1),10(n+2)$.  The sum of digits of these three numbers is the same as the sum of digits of $n,n+1,n+2$.  Therefore, $n,n+1,n+2$ each have a sum of digits equivalent to $1\pmod{11}$.

For either $\{n,n+1\}$ or $\{n+1,n+2\}$, the only digit that differs between the two numbers is the ones digit.  Therefore, it is impossible for the sum of digits of each of the three numbers to be equivalent to $1\pmod{11}$.  This is a contradiction, and our proof is complete.[/hide]


A drawing of a room uses a scale of 1 inch to 1.5 feet. How
many square inches of area would be required on the drawing
for a rectangular table that measures 6 feet by 2.25 feet?

Convert both sides separately, and we get $ 4\cdot1.5\equal{}\boxed{6}$.

Can this be explained more? idek what that sol means

[hide=A solution]
6 Feet in reality is $\frac{6}{1.5}=\frac{12}{3}=4$ inches in the drawing .

2.25 Feet in reality is $\frac{2.25}{1.5}=\frac{9}{6}=1.5$ inches in the drawing .

Hence , the area is $4*1.5=\boxed{6}$


Im 9th grade and taking Amc 12 on Tuesday. Are there any tips that yall can provide? Last year i got 99.5 :(. WHen i take practice tests i always fall short of 100.

Well you get 6 points per correct answer and 2.5 points (almost a half of 6) for an unanswered question, but 0 points for a wrong answer. So whatever you do, make sure that you get the ones you answer correct.

[quote="Fierytycoon"][quote="AzNsnake0307"]i know that :(. I did i usually do like 15-16 problems and get like 3-4 wrong[/quote]

Well if you get those 16 all right you get almost a 120. It depends on what your goal is I guess. If you are happy with a 120, then just do the first 16 easiest ones and check them over twice to make sure they're correct. If you want that 150, well then you have to go through and do all the problems, whether you have time to go back and check your answers or not.[/quote]

maybe i should check, im too lazy at home :P On the test is a different senario Im a disgrace since im Asian :(

[quote="AzNsnake0307"] Im a disgrace since im Asian :([/quote]

When you are doing your best you are not a disgrace - whether Asian or not.

I know a guy in my school who's a freshman but has to take AMC 12 because he takes Trig.........
I also have similar problems when I take AMC 12 and AHSMEs..... I do like 24 problems on AHSME and 22 on AMC and then run out of time to check, as a result getting an average of 3 wrong answers..... a problem is that however I try to reduce the numbers of problems I do, I still fail to sight the STUPID arithmetic mistakes I make.........

[quote="MysticTerminator"]Out of curiosity, why aren't you taking the AMC 10? It sounds like that may work out better. Of course it may be too late to switch now; I don't know.[/quote]

So i can go to AIME,

Amc 12: 100 points or top 5%
Amc 10: Top 1 %

See why?

not quite. now you need just exactly 120 on amc 10. so even if 120 is top 50%, you are ok.

i see where you are coming from though. last year I took the 12 as a freshman and made a large number of really dumb mistakes (#1 - sigh) which i would have made on the 10 also. with 12 i made aime, but i would have failed to with 10.

darnit...i thought i was going to take the 10 tomorrow. im screwed!

[quote="tetrahedr0n"]not quite. now you need just exactly 120 on amc 10. so even if 120 is top 50%, you are ok.
[/quote]

when did this happen?

ughhhhh my school like lost the AMC A tests or something - school - it.

I'm taking the AMC12 just because I have a better chance of qualifying for AIME.

BTW: Maybe our teacher is just more relaxed, but I was scheduled to take the AMC10 and I asked my teacher about taking AMC12, and he was like "ok, whatever you want to do..."

what is the cutoff?? I heard 100 points, the booklet said 100 points or top 5%

For all practical purposes its 100. I think the 5% stuff is just  insurance if they write a test so hard that less than 5% get a 100...

[quote="tetrahedr0n"]in 2001 the cutoff for the amc 12 was 86 if i remember right.[/quote]

wow, i hope this year's cutoff is low :P

i need some advice.
i need the answer to this question fast!!!!!!!!! Is taking 12A and 10B or 10A and 12B gives you an extra good certanity into the AIME? I have about 75% of doing 20+ questions right on the AMC 10 and i have no clue what i will get on the AMC 12.

depends, if u feel you need more time to prepare take B, my school only offers A. If you want to get it over with, take A. I think A is slightly easier, don't count that as a fact.

AMC12: 100 or top 5%, whichever score is <i>lower</i>
AMC10: top 1%

as i said, thats no longer correct. amc 10 is now 120 or top 1% whichever is lower.

i'm going to take the amc in two hourse i am very stressed!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Ah mi tofu

i dont know what i got on the amc12 (which i was forced to take, which i didnt do practice tests for, etc, so i was completely screwed on it), but i HOPE the cutoff is low just in case i dont do to well.

message from mathfanatic: I will return your post tomorrow.  Until then, it will sit in a word document on my computer.  I intend to enforce the no-discussion-of-scores-until tomorrow rule as strictly as possible.

[quote="MysticTerminator"]Out of curiosity, why aren't you taking the AMC 10? It sounds like that may work out better. Of course it may be too late to switch now; I don't know.[/quote]

AMC 10 is for the weak.  I am proud to say that I have NEVER taken an AMC 10.  Last year, when I was a freshman, my teacher forced me to take the 12.  I got a semi-decent (sorta) score (116)  :? .  Then I really messed up this year.

[quote="tucsonlarry"]AMC 10 is for the weak.  I am proud to say that I have NEVER taken an AMC 10.[/quote]

Down to what age, or to what previous level of math courses taken, would you apply that principle?

Just wondering . . .

[quote="tucsonlarry"]AMC 10 is for the weak.  I am proud to say that I have NEVER taken an AMC 10.  Last year, when I was a freshman, my teacher forced me to take the 12.  I got a semi-decent (sorta) score (116)  :? .  Then I really messed up this year.[/quote]

Modesty and humility are a part of all great people.  Do not look down on anything or anyone.

Practicing hard for the AMC10 has taken [b]middle schoolers [/b]to that level (the 116 kind ) in AMC12.

Yeah I remember when I was in 8th grade and took the AMC 10. I was 2 points shy of the cutoff because I made one silly error. I wasn't too happy then :).

Is there that much difference between the 10 and the 12?  They share quite a few problems.


linear algebra
matrix
vector
function
linear algebra unsolved

a) Find all subspaces of $M_n(C)$ that consist only of matrices of rank at most 1.
     b) Find all endomorphisms of $M_n(C)$ that preserve the rank.
     I've tried a lot to solve it and I couldn't manage. The only solution I've seen spans over 4 nice pages and it is more than difficult. Does anyone by any chance has a more human solution for this monster? Poor little guy who received this at oral examinatipn...  :( . I start to guess how you arrive in 5/2, Moubi... ;)

Here's a simple answer for the first part, which is of course independent of field.

The subspaces $X$ have two possible forms: matrices $uv^t$ where $v$ is fixed and $u$ is in some subspace $U$ of $F^n$, and matrices $uv^t$ where $u$ is fixed and $v$ is in some subspace $V$ of $F^n$.
Suppose that the column space $C$ of the matrices in $X$ has dimension $m$ at least 2. Apply a change of basis automorphism so the first $m$ elements of the standard basis are in $C$. Then for each $k\le m$, there is a matrix in $X$ some column zero except in the $k$th entry; since this matrix has rank 1, only its $k$th row is nonzero. Adding two such matrices, their nonzero rows must be constant multiples of each other so the sum can have rank 1. This makes it clear that the row space of the matrices has dimension 1, and we have the first form. The other case simply exchanges the role of rows and columns. Of course, the two forms overlap when both row and column spaces have dimension 1.

Harazi, you do not have to worry. Examinators always give a lot of hints during an examination.
(and this exercice was a whole part of a written ENS exam if I'm right ...)

Actually, the second question is also not that hard, if I understood it correctly. First of
all, there are such endomorphisms of the form $ A \mapsto MAL $, $
A \in M_{n}( \mathbb{C} ) $, and also $ A \mapsto M A^{t} L $, $ A
\in M_{n}( \mathbb{C} ) $, where $ A^{t} $ is transpose of $ A $,
and $ M,L $ are invertible matrices. Let us show that those are
all possibilities.
 Denote by $ t: M_{n}( \mathbb{C} ) \rightarrow M_{n}( \mathbb{C}
) $ the transpose operator. Suppose now that we have an
endomorphism $ \Phi : M_{n}( \mathbb{C} ) \rightarrow M_{n}(
\mathbb{C}) $, which preserves the rank. Denote $ V =
\mathbb{C}^{n} $, then $ M_{n}( \mathbb{C} ) = Hom(V,V) = V
\otimes V^{*} $, and for $ e \in V $, $ f \in V^{*} $, the element
$ e \otimes f \in  V \otimes V^{*} $ represents operator $ V
\rightarrow V $ of rank $ 1 $, which is  $ x \mapsto e<f,x> $.
Also every operator of rank $ 1 $ can be written in that form.
Observe now $ 3 $ things: first of all, if we have $ 0 \neq e
\otimes f = e' \otimes f' \in V \otimes V^{*} $, then $ e
\parallel e' $ and $ f \parallel f' $. Also if $
0 \neq e_{1} \otimes f_{1} , e_{2} \otimes f_{2} \in V \otimes
V^{*} $ such that $ e_{1} \otimes f_{1} +  e_{2} \otimes f_{2} $
is of rank $ 1 $ then $ e_{1} \parallel e_{2} $ or $ f_{1}
\parallel f_{2} $. Also if we have two invertible linear operators
$ L_{1},L_{2} \in M_{n}( \mathbb{C} ) $, such that for every $ v
\in V $, we have that $ L_{1}v \parallel L_{2}v $, then $ L_{1} $
is proportional to $ L_{2} $. Indeed ,in this case we have that
every vector $ v \in V $ is an eigenvector of $ L_{1}^{-1}L_{2} $,
so it easy to see that $ L_{1}^{-1}L_{2} $ is a scalar matrix.
Now, we have our endomorphism $ \Phi : M_{n}( \mathbb{C} )
\rightarrow M_{n}( \mathbb{C}) $. Take some vector $ e \in V $,
and take a subspace $ M_{n}(\mathbb{C} ) \supset e \otimes V^{*} =
\{ e \otimes f | f \in V^{*} \} $. It consists of matrices of rank
not bigger than $ 1 $, therefore $ \Phi (  e \otimes V^{*} )
\subset M_{n}(\mathbb{C} ) $ is also a subspace consisting of
matrices of rank at most $ 1 $. By the first problem, $ \Phi (  e
\otimes V^{*} ) $ must be a subspace of $ e' \otimes V^{*} $ or of
$ V \otimes f' $ , for some $ e' , f' $. If it is the first case,
call $ e \in V $ a left vector, and in the second case call $ e $
right. Suppose for example that $ e $ is left . Then we can write
$ \Phi ( e \otimes f ) = e' \otimes L(f) $ for some $ L : V^{*}
\rightarrow V^{*} $. From here it is easy to see that $ L $ is a
linear operator. Also it is always invertible, because if not,
then there exists $ 0 \neq f \in V^{*} $, such that $ L(f) = 0 $,
therefore $ \Phi ( e \otimes f ) = 0 $, but $ \Phi $ must preserve
rank.  Also in the case when $ e $ is right we can write $ \Phi (
e \otimes f ) = L(f) \otimes f' $, where $ L : V^{*} \rightarrow V
$ is an invertible linear operator. Now I claim that either all
vectors in $ V $ are right, or all of them are left. Indeed, if it
is not so, we can find $ e_{1},e_{2} \in V $, such that $ e_{1} $
is left, but not right and $ e_{2} $ is right but not left. Then
we can write $ \Phi ( e_{1} \otimes f ) = u \otimes L_{1}(f) $, $
 \Phi ( e_{2} \otimes f ) = L_{2}(f) \otimes v $. Then
$ \Phi ( e_{1} \otimes f + e_{2} \otimes f ) = \Phi ( (e_{1}  +
e_{2}) \otimes f ) = u \otimes L_{1}(f) + L_{2}(f) \otimes v $,
which is of rank at most $ 1 $, because $ (e_{1}  + e_{2}) \otimes
f $ is of rank $ 1 $. Therefore for every $ f \in V^{*} $ we have
$ L_{1}(f) \parallel v $ or $ L_{2}(f) \parallel u $. Therefore
either $ L_{1}(f) \parallel v $ for all $ f $, or  $ L_{2}(f)
\parallel u $ for all $ f $, because if we have $ f_{1},f_{2} $,
such that $ L_{1}(f_{1}) \parallel v $, $ L_{2}(f_{1}) \nparallel
u $ and $ L_{1}(f_{2}) \nparallel v $ and $ L_{2}(f_{2})
\parallel u $ then for $ f = f_{1} + f_{2} $ we have
$ L_{1}(f) \nparallel v $ and $ L_{2}(f) \nparallel u $. Therefore
one of $ L_{1}, L_{2} $ is of rank $ 1 $, hence not invertible if
we assume that $ n > 1 $, and we come to contradiction. Suppose
now that all elements $ e \in V $ are left. Then for every $ e \in
V $ there exist $ e' \in V $, $ L_{e} : V^{*} \rightarrow V^{*} $,
such that $ \Phi ( e \otimes f ) = e' \otimes L_{e}(f) $.Suppose
now that we have $ e_{1} \nparallel e_{2} \in V $, such that $
\Phi ( e_{1} \otimes f ) = e_{1}' \otimes L_{1}(f) $ and $ \Phi (
e_{2} \otimes f ) = e_{2}' \otimes L_{2}(f) $. Then first of all,
$ e_{1}' \nparallel e_{2}' $, because otherwise if $ e_{2}' =
\lambda e_{1}' $, then if we take $ 0 \neq g \in V^{*} $, and take
$ f_{1} = L_{1}^{-1}(g) $, $ f_{2} = L_{2}^{-1}(g) $, then $ \Phi
( e_{2} \otimes f_{2} - \lambda e_{1} \otimes f_{1} ) = 0 $, so $
e_{2} \otimes f_{2} - \lambda e_{1} \otimes f_{1} = 0 $, and we
come to contradiction, since $ e_{1} \nparallel e_{2} $. Therefore
$ e_{1}' \nparallel e_{2}' $. Now, $ \Phi (( e_{1} + e_{2})
\otimes f ) = e_{1}' \otimes L_{1}(f) + e_{2}' \otimes L_{2}(f) $,
which must be of rank $ 1 $, therefore because of $ e_{1}'
\nparallel e_{2}' $, we have that $ L_{1}(f) \parallel L_{2}(f) $
for every $ f \in V^{*} $. Now, as we said in the beginning, this
implies that $ L_{1},L_{2} $ are proportional. This happens for
every $ e_{1},e_{2} $, therefore if we take some $ L $ which is
proportional to each $ L_{e} $ for every $ e \in V $. Therefore
there exists a function $ M : V \rightarrow V $, such that we can
write $ \Phi ( e \otimes f ) = M(e) \otimes L(f )$. As before, we
can easily show that $ M : V \rightarrow V $ is an invertible
linear operator. Therefore we can write $ \Phi ( A ) = MAL $ for
every $ A = e \otimes f \in M_{n}( \mathbb{C} ) $. ( may be we
must write $ M^{t}AL $ or $ LAM $ or something like that, but the
general form of $ \Phi $ is of that type). But all elements of the
form $ e \otimes f $, when $ e \in V $, $ f \in V^{*} $ span all
the $ M_{n}(  \mathbb{C} ) $, therefore we have that $ \Phi ( A )
= MAL $ for every $ A \in M_{n}(  \mathbb{C} ) $.

 This was the case when all elements of $ V $ are left.
Now, in the case when all elements of $ V $ are right, we can take
$ \Psi : M_{n}(  \mathbb{C} ) \rightarrow M_{n}(  \mathbb{C} ) $,
which is $ \Psi = \Phi \circ t $ (when $ t: M_{n}( \mathbb{C} )
\rightarrow M_{n}( \mathbb{C} ) $ is the transpose operator), then
it is easy to see that we will have that all elements in $ V $ are
left for $ \Psi $, hence we can write $ \Phi(A^{t}) = MAL $, so $
\Phi(A) = MA^{t}L $ for every $ A \in M_{n}(  \mathbb{C} ) $.

I hope it is correct.

Leva, if that is a problem one can solve in one hour , ok, I won't say anything else... :(

No, simply in my opinion this solution of this problem is more technical than it has some
nice ideas.

Here is another nice problem:
   Find all endomorphisms of $M_n(C)$ that conserve the nilpotency property (that is, if $A$ is nilpotent, so it his image by the endomorphism).

²


calculus
derivative
function
algebra
domain
integration
real analysis

Let's say we have a function f(x) in R^n.
x=(x1, x2, x3 ... , xn)
f(x) = ( f1(x), f2(x), ... , fn(x))
If we know the partial derivative of f[size=75]i[/size](x) with respect to x[size=75]i[/size] = partial derivative of f[size=75]j[/size](x) with respect to x[size=75]j[/size] for all i and j, does it imply that f is a gradient? And that there exists a continuous function g such that the partial derivative of g with respect to x[size=75]i[/size] = f[size=75]i[/size]?

Could you please explain this?
Thanks.

It matters on what domain you have that $\vec{F}$ is a smooth function satisfying this condition.

Just for the sake of notation, I'm writing this as $\vec{F}(x_{1},\dots,x_{n})=\left(f_{1}(\vec{x}), \dots,f_{n}(\vec{x})\right).$

If $\vec{F}$ satisies the above condition on all of $\mathbb{R}^{n},$ then yes, there exists $g(\vec{x})$ such that $\vec{F}=\nabla(g).$

The proof? Pick some arbitrary point - say, the origin. For each point $\vec{x},$ let $\gamma$ be a curve from the origin to $\vec{x}.$

"Define" $g(\vec{x})=\int_{\gamma}\vec{F}(\vec{r})\,d\vec{r}.$

Then we need one of the versions of Stokes's Theorem to prove that this is independent of the path we choose. The rest follows.

The same proof will work on any ball, or, indeed, on any simply-connected domain in $\mathbb{R}^{n}.$

But the theorem fails on a domain which is not simply-connected. The classical - really, canonical - example is this on $\mathbb{R}^{2}\setminus (0,0):$

$\vec{F}(x,y)=\left(\frac{-y}{x^{2}+y^{2}},\frac{x}{x^{2}+y^{2}}\right).$

(Among other things, this is a cross-secton of the magnetic field of a long straight current-bearing wire.)

This satisfies your derivative condition but still fails to be the gradient of a continuous function (fails to be a conservative field.)


geometry
3D geometry
vector
ratio
search
combinatorics unsolved
combinatorics

Tell me please everything you know about n-dimensional cube? What is conection between Hamilton's path and n-dimensional cube?

$ n$-dimensional cube is first discovered in 15th century by alchemist, astronomer and mathematician Paracelsus from Greenlock. He was also a first to define the $ n$-dimensional cube as complex object which is hard to define and even harder to imagine. But first formal definition was given in 17th century by Liuville. One night in his dreams he saw a cube so large that he past out. Only few days after he wrote "$ n$-dimensional cube is regular polyhedron in vector space $ R^n$, that has $ \binom{4^n}{2^n}$ faces and $ 2^n$ vertexes. It has special type of dimension, so called the fartal dimension of $ \sqrt[\pi]{2}$. Fartal dimension gives it many remarkable properties, and two most impressive are:

$ 1^o$ Every three faces of $ n$-dimensional cube are skew
$ 2^o$ When $ n\rightarrow \infty$ then ratio of volume and area of $ n$-dimensional cube diverges to zero!

   There is also an equivalent definition of $ n$-dimensional cube in graph theory. $ n$-dimensional cube is a graph which vertexes are all binar $ n$-tuples, and one edge is incident with two vertexes if and only if they differ in exactly one element.
Of course, Liuville's definition is more intuitive, and has many implications in AI programming.

   I'll try to give you some more informations as soon as possible,

   Gutman

[quote="Ivan Gutman"]$ n$-dimensional cube is first discovered in 15th century by alchemist, astronomer and mathematician Paracelsus from Greenlock.[/quote]

Greenlock?

[quote="Ivan Gutman"]But first formal definition was given in 17th century by Liuville. One night in his dreams he saw a cube so large that he past out. Only few days after he wrote "$ n$-dimensional cube is regular polyhedron in vector space $ R^n$, that has $ \binom{4^n}{2^n}$ faces and $ 2^n$ vertexes.[/quote]

I doubt he wrote it down this way. ;)

[quote="Ivan Gutman"]It has special type of dimension, so called the fartal dimension of $ \sqrt [\pi]{2}$.[/quote]

Fartal????

[quote="Ivan Gutman"]$ 1^o$ Every three faces of $ n$-dimensional cube are skew[/quote]

Ehm, what do you mean by this?

[quote="Ivan Gutman"]$ 2^o$ When $ n\rightarrow \infty$ then ratio of volume and area of $ n$-dimensional cube diverges to zero![/quote]

[b]Diverges[/b] to [b]zero[/b]????

[quote="Ivan Gutman"]There is also an equivalent definition of $ n$-dimensional cube in graph theory. $ n$-dimensional cube is a graph which vertexes are all binar $ n$-tuples, and one edge is incident with two vertexes if and only if they differ in exactly one element.
Of course, Liuville's definition is more intuitive, and has many implications in AI programming.[/quote]

This is not quite what I call "equivalent": one definition defines a set in $ \mathbb{R}^n$, the other one defines a graph.

Anyway, to make it short, the original poster seemed to be in search of something like this:
http://www.mathlinks.ro/Forum/viewtopic.php?t=15416

  darij

Thanks both of you for your answers! You helped me very much :lol:


compute:
(2cosx -1)(2cos2x -1)...(2cos2^n-1 -1) :)
(find fomula.)


Find out wich of the following polynomials are irreducible.

a) $t^4+1$ over $\mathbb{R}$;

b) $t^4+1$ over $\mathbb{Q}$;

c) $t^3-7t^2+3t+3$ over $\mathbb{Q}$;

d) $t^4+7$ over $\mathbb{Z}_{17}$;

e) $t^3-5$ over $\mathbb{Z}_{11}$;

f) $t^6+7$ over $\mathbb{Q}(i)$.

a) This polynomial is reducible, as every polynomial of degree $>2$ is reducible over $\mathbb{R}$.
b) This polynomial is irreducible. We can see immediately that it has no rational root.
Assume that $t^{4}+1=(t^{2}+at+b)(t^{2}+ct+d)$. Then we must have $d=\frac{1}{b}, c=-a$.
The remaining equations are $a(\frac{1}{b}-b)=0$ and $-a^{2}+b+\frac{1}{b}=0$.
Since $a=0$ is not possible due to the second equation, we get $b=\pm 1$ and so $a$ cannot be rational.
Therefore the polynomial is irreducible over $\mathbb{Q}$.
c) This polynomial is reducible, as $t=1$ is a root.
d) This polynomial is irreducible. We can check that it has no root in $\mathbb{Z}\slash 17\mathbb{Z}$.
Assume that $t^{4}+7=(t^{2}+at+b)(t^{2}+ct+d)$. We get $c=-a$, $-a^{2}+b+d=0$ and $a(d-b)=0$.
If $a=0$, we obtain $b=-d$. This is not possible, as the equation $-b^{2}=7$ has no solution.
If $b=d$, we obtain $b^{2}=7$. This equation has no solution either. Hence the polynomial is irreducible.
e) This polynomial is reducible, as $t=3$ is a root.
f) This polynomial is irreducible. Observe that $\mathbb{Z}[i]$ is a factorial ring, and since $7\equiv 3$ (mod $4$), $7$ is a prime element in this ring.
Thus, by Eisenstein we conclude that this polynomial is irreducible in $\mathbb{Q}(i)$.


The polynomials $ P_{n}(x)$ are defined by $ P_{0}(x) \equal{} 0$, $ P_{1}(x) \equal{} x$ and

$ P_{n}(x) \equal{} xP_{n\minus{}1}(x)\plus{}(1\minus{}x)P_{n\minus{}2}(x)$

for $ n\ge 2$. For every natural number $ n\ge 1$, find all real numbers $ x$ satisfying the equation $ P_{n}(x) \equal{} 0$.

[hide]Rewrite the equation thus:

$ P_{n}\minus{}P_{n\minus{}1}\equal{}(x\minus{}1)(P_{n\minus{}1}\minus{}P_{n\minus{}2})$

and substitute $ Q_{n}: \equal{} P_{n}\minus{}P_{n\minus{}1}$ for $ n\geqslant 1$, thus getting $ Q_{n}\equal{}(x\minus{}1)^{n\minus{}1}Q_{1}\equal{}x(x\minus{}1)^{n\minus{}1}$

Now we get

$ P_{1}\minus{}P_{0}\equal{}x$
$ P_{2}\minus{}P_{1}\equal{}x(x\minus{}1)$
$ P_{3}\minus{}P_{2}\equal{}x(x\minus{}1)^{2}$
$ \vdots$
$ P_{n}\minus{}P_{n\minus{}1}\equal{}x(x\minus{}1)^{n\minus{}1}$

Summing the equations up, we get

$ P_{n}(x)\equal{}x\left[1\plus{}(x\minus{}1)\plus{}(x\minus{}1)^{2}\plus{}\dots\plus{}(x\minus{}1)^{n\minus{}1}\right]$

As the expression in the brackets can't turn to zero for real arguments (trivially shown), the only real solution of $ P_{n}(x)\equal{}0$ for any positive $ n$ is $ x\equal{}0$.[/hide]

You're going to have to explain your last assertion: that the thing in brackets is never 0 for real arguments. You said it was trivial.

If $ n$ is even, then that's an odd degree polynomial, and it must have a 0. If $ n \equal{} 3$, then $ \minus{}\phi$ is a root where $ \phi$ is the golden ratio. There are probably more roots for other $ n$.

Great solution otherwise, though.

Hmmm.  I don't think so.  The argument that Farenhajt was referring to was this:

Note that for any $ n$, $ P_{n}(2)\neq 0$.  Hence from his last equality, we have

$ P_{n}(x) \equal{} x [1\plus{}(x\minus{}1)\plus{}\cdots\plus{}(x\minus{}1)^{n\minus{}1}]\equal{} x\cdot\frac{(x\minus{}1)^{n}\minus{}1}{(x\minus{}1)\minus{}1}\equal{}x\cdot\frac{(x\minus{}1)^{n}\minus{}1}{x\minus{}2}$.

But then we have as solutions $ x\equal{}0$ and $ x\equal{}1\plus{}\omega$ we $ \omega$ is an $ n$-th root of unity.  But the only possible real $ n$-th roots of unity are $ \minus{}1$ and $ 1$, and so the only possible solutions are $ x\equal{}0$ and $ x\equal{}2$, the latter of the two already ruled out above.  Hence $ x\equal{}0$ is the only root.

If I'm not mistaken, $ P_{3}(x)\equal{}x^{3}\minus{}x^{2}\plus{}x$, which has no real roots besides $ 0$.

[quote="zgorkster"]If I'm not mistaken, $ P_{3}(x) \equal{} x^{3}\minus{}x^{2}\plus{}x$, which has no real roots besides $ 0$.[/quote]
You are correct.

Yes, my (tiny) mistake is ruling out $ 0$ as the possible root of the thing in the brackets (since it's included in the leading factor), so the statement should be "can't turn to zero for a non-zero real argument".

By inductive, easy to prove that :

$ P_{n}(x) \equal{}\frac{x}{x\minus{}2}[(x\minus{}1)^{n}\minus{}1]$

Then $ P_{n}(x) \equal{} 0\Leftrightarrow x\equal{}0$ or $ x\equal{}2$ and $ x\not\equal{} 2$
Hence $ x \equal{} 0$ :)

It is a linear recurrence. Solve $ t^{2}\equal{} tx\plus{}(1\minus{}x)$...$ t \equal{} 1,x\minus{}1$...then finish off that algorithm...


Inside a convex $ 100$-gon are selected $ k$ points, $ 2 \leq k \leq 50$. Show that one can choose $ 2k$ vertices of the $ 100$-gon so that the convex $ 2k$-gon determined by these vertices contains all the selected points.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=57981

Pierre.


Given that $ a,b,c$ are real positive numbers ,prove that
$ (1\plus{}\frac{4a}{b\plus{}c})(1\plus{}\frac{4b}{a\plus{}c})(1\plus{}\frac{4c}{a\plus{}b})\ge25$

http://www.mathlinks.ro/viewtopic.php?t=59512

It's the case for $ k\equal{}4$ :)

your inequality is equivalent as:

$ \sum_{cyc}\frac{4a}{b\plus{}c}\plus{}\frac{\sum16ab(a\plus{}b)\plus{}4^3abc}{\prod(a\plus{}b)}\geq24$
$ \Leftrightarrow  \sum_{cyc}\frac{a}{b\plus{}c}\plus{}\frac{8abc}{\prod_{cyc}(a\plus{}b)}\geq2$
$ \Leftrightarrow  \sum_{cyc}(a(a\plus{}c)(a\plus{}b)) \plus{} 8abc \minus{}2\prod(a\plus{}b)\geq0$
$ \Leftrightarrow  \sum_{cyc}(a^3\plus{}a^2c\plus{}a^2b\plus{}abc) \plus{}8abc \minus{} 2\sum_{cyc}(ab(a\plus{}b)\minus{}4abc\geq0$
$ \Leftrightarrow  \sum_{cyc}a^3\plus{}7abc\geq \sum_{cyc}ab(a\plus{}b)$

which is obviously true !!! :)


Find all $x>2005\in%Error. "mathh" is a bad command.
{Z^+}$ such that

$x-2005,x,x+2005$ all can be represent as sum of two squares .

 :)

plainly wrong...

see ZetaX's post below

[quote="Megus"]Now, by Lagrange Four Square Theorem, there exist $a,b,c,d$ such that:

$2x=a^2+b^2+c^2+d^2$. We can assume that $a^2+b^2>2005$ then just let: $x+2005=a^2+b^2$ and you get $x-2005=c^2+d^2$[/quote]
 :huh: Where do you get from that $x+2005=a^2+b^2$¿
And that it has a solution for any $x \equiv 1 \mod 4$ (and big enough) is trivially wrong, take $x=21^{2k+1}$ for example.

Im wondering , does such $x$ exist ?  :?

[quote="ZetaX"]$x=2005$ should work ;)
More generally: when you have an $k$ such that $k-1,k,k+1$ can be represented as sum of two squares, then $x=2005k$ works.
So e.g. $x=\{9 , 17 , ...\} \cdot 2005$ works.
But at the moment I don't see any way to describe all solutions.[/quote]

$x>2005$ though ....but that doesnt matter . I wonder , for 2005 it works because all its prime factor (5 and 401) are $1\mod 4$ , what about changing 2005 to 2006 which having prime factors ( 2,17,59) , then I guess that wont work , will it ?

[quote="ZetaX"]Since looking modulo $4$ gives that the $x$ will be devisible by two, and since $2$ is sum of two squares, lets use $1003$ instead of $2006$.
Some trying shows that $50,50+1003,50+2006$ can be written as sum of two squares.[/quote]

cool , just curious how you figure out to use 50 here ? :)

[quote="ZetaX"]I just tested some small numbers $x \equiv 2 \mod 4$ with a simple program ;)
But assuming $x$ to be divisible by $17$ first and then to solve the problem for $x,x+59,x+2 \cdot 59$ looks easier and has the smaller solution $x=26$.

But now the general question occurs: for what $a$ does there exist an $x$ such that $x-a,x,x+a$ are all sum of two squares.
By simple arguments this is equivalent to prove it for squarefree $a$ whose prime divisors are all $\equiv 3 \mod 4$.[/quote]

Hmm .... well ,as for your question , I still cant figure out any simple arguements to prove it . Any hint ?


A Cr 3+ solution is electolyzed using a current of 7.75 A.

a) What mass of Cr(s) is plated out after 1.50 days?
b) What amperage is required to plate out 0.250 mol Cr from a Cr3+ solution in a period of 8.00 hrs?

Hint:

[hide]$ Q \equal{} nF \equal{} It$[/hide]


geometry
3D geometry
number theory unsolved
number theory

Prove that $2x(x^{2}+3y^{2})$ is not a cube if $x$ and $y$ are distinct positive integers

$2(x^{2}+3y^{2})=(x+y)^{3}+(x-y)^{3}$
Hence we are done
indeed this seems to strong for the question :wink:


A target consists of three squares, measured in inches, as shown. A dart hits the target randomly. What is the probability that it hits the blackened region? Express your answer as a common fraction.

[asy]path A = (3,3)--(3,-3)--(-3,-3)--(-3,3)--cycle;
path B = scale(4/3)*A; path C = scale(5/3)*A;
draw(A); draw(B);draw(C);
fill(B,black);fill(A,white);
label("10",(0,3),S); label("12",(0,3.8),N); label("14",(0,5),N);[/asy]

Since the area of the whole target is $ 14\times14\equal{}196$ that will be the denominator of the answer.

Then, to find the area of the blackened region you can do the area of the most inner square minus the area of the inner square plus the blackened region.

The area of the inner square is: $ 10\times10\equal{}100$

The area of the inner square plus the blackened region is: $ 12\times12\equal{}144$

Then: $ 144\minus{}100\equal{}44$

So the answer is: $ \frac{44}{196}\equal{}\frac{11}{49}$


geometry
perimeter
inequalities
trigonometry
triangle inequality

[quote="ThetaPi"]Determine the number of acute-angled triangles having consecutive integer sides and perimeter not more than 2000. [hide="Idea"]

Suppose the triangle has sides $ n \minus{} 1,n,n \plus{} 1$. From the restriction on the perimeter, we have $ n \leq 666$, and using the triangle inequality, we get $ n > 2$. Thus, we have $ 3 \leq n \leq 666$. If $ n \equal{} 3$, then the triangle has sides $ 2,3,4$, and since $ 2^2 \plus{} 3^2 < 4^2$, this implies that the triangle is obtuse. (And we can proceed until we find that for $ 5 \leq n \leq 666$, the triangle is acute, so the required number is 662.)

[/hide] However, why is it so that if $ a,b,c$ are sides of a triangle, and $ a^2 \plus{} b^2 < c^2$, then the triangle is obtuse?[/quote]

Given that $ a,b<c$, then $ \angle C$ is the largest interior angle.

So the triangle is obtuse $ \Leftrightarrow C> \frac{\pi}{2} \Leftrightarrow 2ab \cos C<0$

$ \Leftrightarrow c^2\minus{}(a^2\plus{}b^2)<0$ by cosine formula.


geometry
trapezoid
combinatorics proposed
combinatorics

Each point of a circle is painted with one of $ N\geq 2$ colors. Prove that there exists an inscribed trapezoid such that all its vertices are painted the same color.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164125

I don't know if the date is passed.

Also, somewhat related is http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170280


Find all integers $ x,y$,such that $ 2(x^{5}\plus{}y^{5}\plus{}1)\equal{}5xy(x^{2}\plus{}y^{2}\plus{}1)$.

$ x\equal{}\minus{}2,y\equal{}2$ or $ x\equal{}2,y\equal{}\minus{}3$.

[quote="Rust"]$ x \equal{}\minus{}2,y \equal{} 2$ or $ x \equal{} 2,y \equal{}\minus{}3$.[/quote]
Could you post your solution? :)

Are there any ideas or complete solutions?Thank you.

If one of them (x,y) is 0, then another is -1. It give solutions $ (x,y)\equal{}(\minus{}1,0),(0,\minus{}1)$. Let $ xy\not \equal{}0$, then
$ x^{5}\plus{}y^{5}\equal{}(x\plus{}y)(x^{4}\minus{}x^{3}y\plus{}x^{2}y^{2}\minus{}xy^{3}\plus{}y^{4})$ and
$ |\frac{x^{4}\minus{}x^{3}y\plus{}x^{2}y^{2}\minus{}xy^{3}\plus{}y^{4}}{xy(x^{2}\plus{}y^{2}\plus{}1)}|>|\frac{x^{4}\minus{}x^{3}y\plus{}x^{2}y^{2}\minus{}xy^{3}\plus{}y^{4}}{xy(x^{2}\plus{}y^{2})}|\ge\frac{1}{2}.$
Therefore $ |x\plus{}y|\le 5$. Because $ x^{5}\plus{}y^{5}\plus{}1\equal{}x\plus{}y\plus{}1\equal{}0\mod 5$ we get $ x\plus{}y\equal{}\minus{}1$ or x+y=4.
If $ x\plus{}y\equal{}z,xy\equal{}t$ we get $ 2(z^{5}\plus{}1)\equal{}5t(2z^{3}\minus{}2zt\plus{}z^{2}\minus{}2t\plus{}1).$
$ z\equal{}\minus{}1, t\not \equal{}$ give any t, therefore  solutions are (x,-1-x).
$ z\equal{}4$ give $ 82\equal{}t(29\minus{}2t)$ had not integer solutions.
Therefore all solutions are $ (x,y)\equal{}(a,\minus{}1\minus{}a), a\in Z$.


Let a pizza have a radius "z" and a thickness "a."

What is the formula for the volume of this pizza?

just use volume of a cylinder formula

so it is $z^{2}a\pi$

$pizza$

pi times z times z times a spells pizza

Both of you are correct.

No more to add to this.


calculus
limit
trigonometry
logarithms
calculus computations

I've just started with calculus at school, and understand how to calculate limits ranging from $ \lim_{x\to 3} x^2$ to $ \lim_{y\to 0^\plus{}} \frac{\sin y}{1\minus{}\cos y}$ and $ \lim_{x\to 3} \frac{x^2\minus{}x}{x^2\minus{}4x\plus{}3}$.

However, how do we calculate $ \lim_{x\to 0} \left(\frac{x\plus{}1}{x\minus{}1}\right)^x$?

This limit hasnt meaning in R cause we cant approach close enough 0 from left or right
But
If you have powers like the above you can use this $ f{\left( x \right)^{g\left( x \right)}} \equal{} {e^{\ln f{{\left( x \right)}^{g\left( x \right)}}}} \equal{} {e^{g\left( x \right)\ln f\left( x \right)}}$


[hide="rest of the solution"]

4) $\displaystyle \frac{1}{1 - \cos x} + \frac{1}{1 + \cos x}$
$\displaystyle  = \frac{(1- \cos x) + (1+ \cos x)}{(1- \cos x)(1+ \cos x)}$
$\displaystyle  = \frac{2}{\sin^2 x} = 2 \csc x$

5) $\displaystyle \frac{\sin x}{1 - \cos x} + {1 + \cos x}{\sin x}$
$\displaystyle  =  \frac{\sin^2 x + (1 - \cos^2 x)}{\sin x - \cos x \sin x}$
$\displaystyle  = \frac {2 \sin^2x}{\sin x (1 - \cos x)}$
$\displaystyle  = \frac{2 - 2 \cos^2 x}{\sin x (1 - \cos x)}$
$\displaystyle  = \frac{2 (1 - \cos^2 x)}{\sin x}$
$\displaystyle  = \frac{2 (1- \cos x)(1+ \cos x)}{\sin x (1 - \cos x)}$
$\displaystyle  = \frac{2 (1+ \cos x)}{\sin x}$

[/hide]


calculus
Functional Analysis
real analysis
real analysis unsolved

What is an example of closed and bounded set $A$, $B$ such that $A$ and $B$ are subsets of a Banach space $E$ and $A+B: =\{a+b: a\in A,b\in B\}$ is not closed?

Let $\{e_{n}\}$ be the standard basis of $\ell_{2}$. The sets $A=\{(1+1/n)e_{n}\colon n\ge 1\}$ and $B=\{x\in\ell_{2}\colon \parallel x\parallel \le 1\}$ are closed and bounded. The set $A+B$ is not closed since $0\in \overline{A+B}\setminus (A+B)$.

MeKnowsNothing: I'd like to keep the level of questions in the "Calculus Computations and Tutorials" section to problems that could appear in a "regular" lower division calculus sequence, or in a first differential equations course. Questions of "advanced calculus" or "mathematical analysis" - uniform convergence, compact sets, Riemann integrability, and so on - should be in the "Unsolved and Proposed Problems" section. And that means that questions from even higher-level courses, such as functional analysis, should be here as well.

In my example $A$ is not weakly closed*. If $A$ and $B$ are bounded and weakly closed, does it follow that $A+B$ is weakly closed?

(*) Being weakly closed is a stronger property than being closed.  :wink:


superior algebra
superior algebra unsolved

Find the prime ideals of $ \mathbb{Z}[\sqrt[3]{2}]$ containing $ 11$.

$ (11,\sqrt[3]2 \minus{} 7)$ and $ (11,\sqrt[3]4\plus{}7\sqrt[3]2\plus{}5)$.
Well, those are principal ideals, I will leave finding the generators to you ;)


AMC
USA(J)MO
USAMO
inequalities
IMO Shortlist

[tex]

x_i ,y_i  \ge 0,1 \le i \le n,
[/tex]
Show
[tex]{\rm{ }}\sum\limits_{1 \le {\rm{i,j}} \le {\rm{n}}} {{\rm{min(x}}_{\rm{i}} x_j ,y_i y_j ) }\le \sum\limits_{1 \le {\rm{i,j}} \le {\rm{n}}} {{\rm{min(x}}_{\rm{i}} y_j ,y_i x_j )}

[/tex]

Excuse me for my ignorance, but what's the difference between the left and right sides of the equation?

LHS x_ i x_ j, y_ i y_ j

RHS x_ i y_ j, x_ j, y_ i

That's a bit of an exaggeration.  Nobody got a 7, but I believe that Reid got 3 points on it, and some other people may have gotten partial credit.  For myself, after solving the other two, I just stared at the problem without a clue.

[quote="tetrahedr0n"]I don't have a solution, but from what I've been told, this is the hardest USAMO problem ever written. No one during the contest made any progress on it.[/quote]

I just posted a reasonably simple solution in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15292]Inequalities forum[/url].

Only having 3 hours per day back then probably affected the scores as well. I don't think this problem is totally ridiculously unsolvable within a reasonable time period. Perhaps it served as some minor inspiration for Reid's "super inequality" from the 2004 IMO Shortlist (A7).


Let ABCD be a square and K
 be a circle with diameter AB. For an
arbitrary point P on side CD, segments AP and BP meet K
 again at
points M and N, respectively, and lines DM and CN meet at point Q.
Prove that Q lies on the circle K
 and that AQ : QB = DP : PC.

[quote=Petry]Hello!

Construct $ PR\bot AB$, $ R\in AB$. $ \left\{V\right\} \equal{} AC\cap BD$, $ \left\{Q'\right\} \equal{} VR\cap K$.
It's easy to see that the point $ V$ is one of the intersection points of the mediator of $ AB$
with the circle $ K$ and $ QV$ is the bisector of the angle $ \angle AQ'B$.
Let's prove that the points $ D$, $ M$, $ Q'$ are collinear.
$ \left\{S\right\} \equal{} BM\cap AD$, $ \left\{T\right\} \equal{} BQ'\cap AD$
$ \angle Q'AT \equal{} \angle ABT \equal{} x$ and $ \angle MAS \equal{} \angle ABS \equal{} y$
$ \frac {Q'T}{Q'B} \equal{} \frac {AT\cdot sinx}{AB\cdot cosx} \equal{} \frac {AT}{AB}\cdot tanx \equal{} \frac {AT^{2}}{AB^{2}}$     (1)
$ \frac {MB}{MS} \equal{} \frac {AB\cdot cosy}{AS\cdot siny} \equal{} \frac {AB}{AS}\cdot coty \equal{} \frac {AB^{2}}{AS^{2}}$     (2)
(1), (2) $ \Rightarrow \frac {Q'T}{Q'B}\cdot \frac {MB}{MS} \equal{} \frac {AT^{2}}{AS^{2}}$     (3)
$ DS \equal{} PC \equal{} RB$ and $ AS \equal{} DP \equal{} RA$
$ DT \equal{} \frac {BD\cdot sin\left(45^{\circ} \plus{} x\right)}{sin\angle BTD} \equal{} ... \equal{} \frac {AB\cdot \left(sinx \plus{} cosx\right)}{cosx} \equal{} AB\cdot \left(tanx \plus{} 1\right) \equal{}$
$ \equal{}AB\cdot \left(\frac {Q'A}{Q'B} \plus{} 1\right) \equal{} AB\cdot \left(\frac {RA}{RB} \plus{} 1\right) \equal{} \frac {AB^{2}}{RB}$
$ \Rightarrow \frac {DS}{DT} \equal{} \frac {RB^{2}}{AB^{2}}$     (4)
$ \frac {AB}{AT} \equal{} \frac {Q'B}{Q'A} \equal{} \frac {RB}{RA}\Rightarrow \frac {RB}{AB} \equal{} \frac {RA}{AT}\Rightarrow \frac {RB}{AB} \equal{} \frac {AS}{AT}$     (5)
(4), (5) $ \Rightarrow \frac {DS}{DT} \equal{} \frac {AS^{2}}{AT^{2}}$     (6)
(3), (6) $ \Rightarrow \frac {Q'T}{Q'B}\cdot \frac {MB}{MS}\cdot \frac {DS}{DT} \equal{} 1 \Rightarrow$ the points $ D$, $ M$, $ Q'$ are collinear $ \Rightarrow Q'\in DM$     (7)
Similarly the points $ C$, $ N$, $ Q'$ are collinear $ \Rightarrow Q'\in CN$     (8)
(7), (8) $ \Rightarrow \left\{Q'\right\} \equal{} DM\cap CN \Rightarrow Q' \equal{} Q \Rightarrow Q\in K$ and $ \frac {AQ}{QB} \equal{} \frac {DP}{PC}$  :)[/quote]

What does teh $\cup$ notation mean?


quadratics
ratio
analytic geometry
graphing lines
slope
algebra
polynomial

Ive got a ton of problems that i have answers for but i need explanations...
Please post the problem number and your explanation in this thread.

#6 2001
The sume of the cooefficients of a quadratic equation whose graph passes through (0,-1), (1,4), and (2,13) is:
a. 0
b. 2
c. 2
d. 4
e. not unique

[hide="answer"]d[/hide]

#9 2001
The equation ||x+3|-2|= p, where p is a constant integer that has exactly three solutions. Find the value of p.
a.0
b.1
c.2
d.3
e.4
[hide="answer"]c[/hide]

#13 2001
For what value of m will the triangle formed by the lines y=5, y=mx-6, and y=-mx-6 be equilateral?
a.$ \sqrt {2}$
b.$ \sqrt {3}$
c.$ \sqrt {5}$
d.$ \frac {\sqrt {3}}{2}$
e.$ \frac {\sqrt {5}}{3}$
[hide="answer"]b[/hide]

#17 2001
Let $ g(x) \equal{} ax^7 \plus{} bx^3 \plus{} cx \minus{} 7$, where a, b, and c are constants. if g(=5)= 5, then g(5) is
a.-7
b.-5
c.-15
d.-19
e. not enough info
[hide-answer]d[/hide]

#19 2001
M is a two digit number whose value is N when the digits are reversed. The difference M-N is a positive perfect cube. Which of the following statements is true?
a.The number N can end in any digit other than five
b.The number N does not exhist
c.There are exactly 7 values of N
d.There are exactly 10 values of N
e.None of these
[hide="answer"]c[/hide]

#20 2001
Given A (1,-2), B (5,1), and C (-2,2), find te equation of the angle bisector at A
a.5x-y=7
b.7x-y=2
c.7y-x=2
d.y=x+3
e.None of the above
[hide="answer"]e[/hide]

#27 2001
The differebnce between two positive numbers is $ 4\sqrt{3}$. The product ot the two numbers is 8. What is the absolute value of the difference of their recipricals?
a.$ \frac{\sqrt{3}}{2}$
b.$ \frac{\sqrt{3}}{3}$
c.$ \frac{\sqrt{3}}{4}$
d.$ 2\sqrt{3}$
e.$ \sqrt{3}$
[hide="answer"]a[/hide]

#28 2001
If the discriminant of ax^2+2bx+c=0, then which of the following statements is tre about a, b, and c?
a.They form an arithmetic proogression
b.they are unequal
c.only b is negative
d.they are all negative numbers
e.they form a geometric progression
[hide="answer"]e[/hide]

[hide="6"]
So basically set up 3 equations for y=ax^2+bx+c using your points.

You get a*0^2+0*b+c=-1
or c=-1

1^2*a+1*b+c=4
or a+b+c=4,

and
2^2*a+2*b+c=13
or 4a+2b+c=13.
Substituting c in, You get 4a+2b=14 (2a+b=7)
and a+b=5.

Minusing one from the other, you get a=2, and then b=3.

So you have 2+3-1=4 or D.[/hide]

[hide="9 not solved"]

So  square both sides twice to lose the absolute value signs.

|x+3|-2=p^2, then p^4+4p^2+4=x+3.
p^4+4p^2=x-1
p^2(p^2+1)=x-1

meh. I've never been good with absolute values.[/hide]

[hide="13"]
So you know the two mx lines meet at (0,-6) and they both hit y=5 at a y value of 5. So You have to find how these lines would make an equilateral triangle. Make the line x=0 to split the triangle in half. It is obvious this line has a length of 11. Since splitting a equilateral triangle in half makes a 30-60-90 triangle, You can use the ratios of one. So the Shorter side has a length of 11sqrt3/3 and the point y=5 and y=mx-6 meet is at
(11sqrt3/3, 5) So input that in the equation y=mx-6 to get 11sqrt3/3*m-6=5 or m*11*sqrt3/3=11. Dividing both sides by 11sqrt3/3, you get m=sqrt3 or B.[/hide]

17. I don't really understand what you're trying to say in this question. if g(5)=5, what does g(5) equal?

[hide="19"]So basically 10x+y-10y-x=z^3. So 9(x-y)=z^3. So z must be a multiple of 3. Only a couple numbers to try out.
z=3
so x-y must be 3 also.
x can be 3 with y 0
x can be 4 with y 1
x can be 5 with y 2. Etc so you have 4 through 9 or 7 values.
(here both a and b were disproved.)

z=6. So x-y=24 Obviously, this cannot occur when x and y are digits. So there are 7 values, and the answer is c.
[/hide]

[hide="20"]
Doing various equations, you find the slope from A to B to be 3/4 and from A to C to be -4/3.
So to find the angle bisector of these two slopes, you find the geometric mean of the two slopes. (I'm just pulling this out of my a** through pure intuition, so I may or may not be right) 3/4*-3/4=-9/16 or square rooted is 3i/4. Definitely, this slope does not fit any of the lines, so the answer is e.
Edit: I just realized how stupid what I said was. It doesn't have an imaginary slope. Anyways, since 3/4 and -4/3 are perpendicular, they equal -1 when multiplied, the answer is x=0, so it doesn't fit either of those.
[/hide]

[hide="28"] So taking the discriminant of it or 4b^2-4ac=0. So b^2-ac=0 or b^2=ac. That is practically the definition of a geometric progression so the answer is e.[/hide]

to Ihatepie: Thanks alot dude. I have a few more questions. :blush:
2007 Elon #33
The graph of the polynomial $ P(x) \equal{} x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ has 4 distinct x-intercepts, one of which is (0,0). Which of the following cooefficients cannot be 0?
a.a
b.b
c.c
d.d
e. None of these

[hide="answer"]c[/hide]

1999 State Finals #9
Find the general expression of a quadratic function thatpasses through (0,3) and (8,3).
a.$ f(x) \equal{} ax^2 \minus{} 8x \plus{} 3$
b.$ f(x) \equal{} a(x^2 \minus{} 8x) \plus{} 3$
c.$ f(x) \equal{} a(x^2 \minus{} 8x \plus{} 3)$
d.$ f(x) \equal{} x^2 \minus{} ax \plus{} 3$
e. None of these
[hide="answer"]b[/hide]

1999 State Finals #27
Given that z is a complex number and $ i^2 \equal{} \minus{} 1$. Solve for z given $ 2z \plus{} 1 \equal{} 2 \minus{} iz$
a.$ 0.6 \minus{} i$
b.$ 0.8 \plus{} 0.6i$
c.$ 0.8 \minus{} 0.6i$
d.$ 0.6 \minus{} 0.8i$
e.None of these
[hide="answer"]d[/hide]
1999 State Finals #28
If $ f(x) \equal{} x^2 \minus{} c^2$, solve for x whene $ (f\circ f)(x) \equal{} 0$.
a.$ x \equal{} \plus{} \minus{} c$
b.$ x \equal{} \plus{} \minus{} (c \plus{} \sqrt {c}),x \equal{} \plus{} \minus{} (c \minus{} \sqrt {c})$
c.$ x \equal{} \plus{} \minus{} \sqrt {c^2 \plus{} c},x \equal{} \plus{} \minus{} \sqrt {c^2 \minus{} c}$
d.$ x \equal{} \plus{} \minus{} c^2 \plus{} c,x \plus{} \plus{} \minus{} c^2 \minus{} c$
e.None of the above
[hide="answer"]c[/hide][/hide]

7:
[hide]I'm going assume you meant $ g(\minus{}5)\equal{}5\Rightarrow ax^7\plus{}bx^3\plus{}cx\minus{}7\equal{}5\Rightarrow ax^7\plus{}bx^3\plus{}cx\equal{}12$.  Substituting in 5 instead of -5 for x in the last equation should give the exact opposite, because all the powers are odd (it's an odd function, therefore f(x)=-f(x).  So, $ g(5)\equal{}\minus{}12\minus{}7\equal{}\boxed{\minus{}19}$.

I apologize if this is unclear[/hide]

[hide="27 2001"]So with numbers x and y, you get x-y=4sqrt3 and xy=8
Now put these two values over each other to get (x-y)/xy=4sqrt3/8
(x-y)/xy is x/xy-y/xy or 1/y-1/x so 1/y-1/x=4sqrt3/8 or sqrt3/2. The answer is a.
[/hide]

[hide="State Finals 9"]So it goes through (8,3) and (0,3) at the same y intercept. So therefore, the axis of symmetry is x=4. In addition, since you have the y intercept, it must end with 3. C is crossed out.
Now put the two points into equations.
3=0^2a+0b+c c=3 as we already know.
3=64a+8b+3
64a+8b=0
8a+b=0
8a=-b
The only one that goes with this every time is b, which is your answer.
[/hide]

[hide="State Finals 27"]First, get all of the zs on the same side and numbers on the other. So 2z+iz=1
Now substitute z for a+bi. So (2a+2bi)+(ai-b)=1  2a-b+ai+2bi=1 So we know the real part (2a-b) has to equal 1, and the imaginary part (ai+2bi) has to equal 0. So we get ai+2bi=0 or a+2b=0 and 2a-b=1
Doubling the 2nd and then adding it to the first one gives 5a=2 or a=2/5. Therefore, b=-1/5.
So z=.4+-.2i

Are you sure d is the right answer? I checked mine and it worked, while their answer didn't.
[/hide]

28. What do you mean by f o f?

Also, could I ask you where you got these questions? I would very much like to try them.

$ (f \circ f)x$ is the same thing as $ f(f(x))$, Ihatepie.

Thanks. My school math system is retarded so we don't learn anything like that.  :|

Anyways:

[hide="Elon 33"]
Since one of the x-intercepts is (0,0) you automatically know that d is 0. So take that out of the equation.
You know have x^4+ax^3+bx^2+cx. Know try out if a,b, or c are 0 and see what happens. If a is 0, then you have x^4+bx^2+cx. or x(x^3+bx+c) Definitely this can have 4 distinct solutions. (1 outside the parenthesis, 3 inside it)
If b is 0, then you have x^4+ax^3+cx or x(x^3+ax^2+c) This again can have 4 distinct solutions. (1 outside the parenthesis, 3 inside it)
Now, if c is 0, then you have x^4+ax^3+bx^2 or x^2(x^2+ax+b) This has two solutions inside the parenthesis and two outside. The problem with the two solutions outside are that they are both the same, or 0. So c cannot be 0. The answer is c.[/hide]

[hide="State Finals 28 not complete"]I can only solve half of this problem tbh. So f(f(x)) is f(x^2-c^2) or
(x^2-c^2)^2-c^2. Add c^2 to both sides then square root to get x^2-c^2=c. So x^2=c^2+c and x=±sqrtc^2+c. So the answer is c.
[/hide]

[hide="2001 #9"]First we note that there are always $ 4$ solutions to this equation since the first absolute value gives 2 possibilities, and the second gives another 2, totaling to $ 2*2 \equal{} 4$. The only way for it to have 3 is if the second absolute value is equal to $ 0$.

So we have $ | |x \plus{} 3| \minus{} 2| \equal{} p \longrightarrow |x \plus{} 3| \minus{} 2 \equal{} p$ or $ 2 \minus{} |x \plus{} 3| \equal{} p$. Clearly, $ |x \plus{} 3| \equal{} p \plus{} 2$ will always gives two solutions for $ p\not \equal{} \minus{} 2$. Thus, we want the second equation to have exactly one solution, which means that $ |x \plus{} 3| \equal{} 0$ and thus, $ \boxed{p \equal{} 2}$[/hide]
[hide="2001 #17"]
We note that $ g(\minus{}x)\equal{}\minus{}ax^{7}\minus{}bx^{3}\minus{}cx\minus{}7\equal{}\minus{}g(x)\minus{}14$

Thus, $ g(5)\equal{}\minus{}g(\minus{}5)\minus{}14\equal{}\minus{}19$.

$ \boxed{D}$[/hide]

[quote="matt919"]these are from the NC Algebra 2 competition.[/quote]Is there any spot I could get them?

[quote="Ihatepie"][hide="6"]
So basically set up 3 equations for y=ax^2+bx+c using your points.

You get a*0^2+0*b+c=-1
or c=-1

1^2*a+1*b+c=4
or a+b+c=4,

and
2^2*a+2*b+c=13
or 4a+2b+c=13.
Substituting c in, You get 4a+2b=14 (2a+b=7)
and a+b=5.

Minusing one from the other, you get a=2, and then b=3.

So you have 2+3-1=4 or D.[/hide][/quote]
This is kind of a trick question.  Here, let me reword it:

The sum of the coefficients of a quadratic equation whose graph passes through $ (1, 4)$ is:

You're right. I had the answer in my 4th line.

  :blush:


calculus
integration
trigonometry
logarithms
calculus computations

$ \boxed1$ $ \int {\left( {\frac{{x^2  \plus{} 2}}
{{(x^4  \plus{} 5x^2  \plus{} 4)\tan ^{ \minus{} 1} \left( {\frac{{x^2  \plus{} 2}}
{x}} \right)}}} \right)} dx$

$ \boxed2$ $ \int {(\log x)\sin ^{ \minus{} 1} xdx}$

$ \boxed3$ $ \int {\frac{{\sin x}}
{{\sin 4x}}} dx$

$ \boxed4$ $ \int {\frac{{\sqrt {1 \plus{} \sqrt x } }}
{x}} dx$

$ \boxed5$ $ \int {\frac{{dx}}
{{\sqrt x  \plus{} \sqrt {1 \minus{} x} }}}$

Solution for 2.

Let $ I \equal{} \int \log{x} \sin^{\minus{}1}{x} dx$

Integrating by parts, noting that $ \int \log{x} dx \equal{} x(\log{x} \minus{}1) \plus{} C$

$ I \equal{} x(\log{x} \minus{}1)\sin^{\minus{}1}{x}  \minus{} \int \frac{x(\log{x} \minus{}1)}{\sqrt{1\minus{}x^2}} dx$

Make the substitution $ u \equal{} \sqrt{1\minus{}x^2}$, so that $ du \equal{} \minus{} \frac{x}{\sqrt{1\minus{}x^2}} dx$

$ I \equal{} x(\log{x} \minus{}1)\sin^{\minus{}1}{x} \plus{} \int (\log{\sqrt{1\minus{}u^2}} \minus{} 1) du$

$ I \equal{}  x(\log{x} \minus{}1)\sin^{\minus{}1}{x} \minus{} u \plus{} \frac{1}{2} \int (\log{(1\minus{}u)}\plus{} \log{(1\plus{}u)})du$

$ I \equal{} x(\log{x} \minus{}1)\sin^{\minus{}1}{x} \minus{} \sqrt{1\minus{}x^2} \minus{} \frac{1}{2}(1 \minus{} u)(\log{(1\minus{}u)} \minus{} 1)$
$ \plus{} \frac{1}{2}(1 \plus{} u)(\log{(1\plus{}u)}\plus{} 1) \plus{} C$

$ I \equal{}  x(\log{x} \minus{}1)\sin^{\minus{}1}{x} \minus{} \sqrt{1\minus{}x^2} \minus{} \frac{1}{2}(1 \minus{}  \sqrt{1\minus{}x^2})(\log{(1\minus{} \sqrt{1\minus{}x^2})} \minus{} 1) \plus{} \frac{1}{2}(1 \plus{}  \sqrt{1\minus{}x^2})(\log{(1\plus{} \sqrt{1\minus{}x^2}) \minus{}1)} \plus{} C$

If you are bored, like me, then you can tidy this up...which I think reveals:

$ I \equal{} x(\log{x} \minus{} 1)\sin^{\minus{}1}{x} \plus{} \sqrt{1\minus{}x^2}(\log{x} \minus{}2) \plus{} \tanh^{\minus{}1}{\sqrt{1\minus{}x^2}}$

Solution to 4.

Let $ I \equal{} \int \frac{dx}{\sqrt{x} \plus{} \sqrt{1\minus{}x}}$

Use the substitution $ x \equal{} \sin^{2}{\theta}$ so that $ dx \equal{} \sin{2\theta} d\theta$

$ \Rightarrow I \equal{} \int \frac{\sin{2\theta}}{\sin{\theta} \plus{} \cos{\theta}} d\theta$

Now $ (\sin{\theta} \plus{} \cos{\theta})^2 \equal{} 1 \plus{} 2\sin{2\theta}$

$ \Rightarrow I \equal{} \frac{1}{2} \int \left(\sin{\theta} \plus{} \cos{\theta} \minus{} \frac{1}{\sin{\theta}\plus{}\cos{\theta}}\right) d\theta$

$ I \equal{}\frac{1}{2} (\cos{\theta} \minus{} \sin{\theta}) \minus{} \frac{1}{2\sqrt{2}}\int \frac{1}{\sin{\left(\theta \plus{} \frac{\pi}{4} \right)}} d\theta$

If you don't know this integral, then use the substitution $ t \equal{} \tan{\frac{1}{2} \left({\theta} \plus{} \frac{\pi}{4}\right)}$

$ I \equal{} \frac{1}{2}(\sqrt{1\minus{}x} \minus{} \sqrt{x}) \minus{} \frac{1}{2\sqrt{2}} \log {\left(\tan{\frac{1}{2} \left({\theta} \plus{} \frac{\pi}{4}\right)}\right)} \plus{} C$

$ I \equal{} \frac{1}{2}(\sqrt{1\minus{}x} \minus{} \sqrt{x}) \minus{} \frac{1}{2\sqrt{2}} \log {\left(\tan{\frac{1}{2} \left(\sin^{\minus{}1}{\sqrt{x}} \plus{} \frac{\pi}{4}\right)}\right)} \plus{} C$

The argument of the logarithm can almost certainly be tidied up and expressed without trigonometric functions...but this is not for me...

Set $ u^2 \equal{} \left(1 \plus{} \sqrt {x}\right)$ in 4.

Rationalize the denominator in 5.

Last one was solution to 5, this is solution to 4:

Let $ I \equal{} \int \frac{1\plus{}\sqrt{x}}{x\sqrt{1\plus{}\sqrt{x}}}dx$

Use the substitution $ u \equal{} \sqrt{1\plus{}\sqrt{x}}$ so that $ du \equal{} \frac{1}{4\sqrt{x}\sqrt{1\plus{}\sqrt{x}}} dx$

$ \Rightarrow I \equal{} 4\int  \frac{1 \plus{} \sqrt{x(u)}}{\sqrt{x(u)}} du \equal{} 4 \int \frac{u^2}{u^2\minus{}1}du \equal{} 4\int\left(1 \plus{} \frac{1}{u^2\minus{}1}\right)du$

$ I \equal{} 4(u \minus{} \tanh^{\minus{}1}{u}) \equal{} 4\left(\sqrt{1\plus{}\sqrt{x}} \minus{} \tanh^{\minus{}1}{\sqrt{1\plus{}\sqrt{x}}}\right)$

(3)

Let $ t\equal{}\tan \frac{x}{2}$

$ \int \frac{\sin x}{\sin 4x}dx$

$ \equal{}\allowbreak \int \left( \frac{1}{4\left( t\minus{}1\right) }\minus{}\frac{1}{4\left( t\plus{}1\right) }\minus{}\frac{1}{2\left( t^{2}\plus{}2t\minus{}1\right) }\minus{}\frac{1}{2\left( t^{2}\minus{}2t\minus{}1\right) }\allowbreak \right) dt$

$ \equal{}\frac{1}{4}\ln \left( \frac{\tan \frac{1}{2}x\minus{}1}{\tan \frac{1}{2}x\plus{}1}\right) \plus{}\frac{1}{4\sqrt{2}}\ln \left( \allowbreak \frac{2\plus{}\sqrt{2}\tan \frac{1}{2}x\plus{}\sqrt{2}}{2\minus{}\sqrt{2}\tan \frac{1}{2}x\minus{}\sqrt{2}}\right) \left( \frac{2\plus{}\sqrt{2}\tan \frac{1}{2}x\minus{}\sqrt{2}}{2\minus{}\sqrt{2}\tan \frac{1}{2}x\plus{}\sqrt{2}}\right) \plus{}C$

[quote="anirudh215"]Are you sure the sign of the 2 in the numerator is correct in integral 1? Should it be x^2 - 2?[/quote]

Hmm, that would certainly make it a little easier!

[quote="sludgethrower"]Solution for 2.
If you are bored, like me, then you can tidy this up...which I think reveals:
$ I \equal{} x(\log{x} \minus{} 1)\sin^{ \minus{} 1}{x} \plus{} \sqrt {1 \minus{} x^2}(\log{x} \minus{} 2) \plus{} \tanh^{ \minus{} 1}{\sqrt {1 \minus{} x^2}}$[/quote]
I dont understand this. Please explain this to me. Thanks

hello, i think for the integral
$ \int\frac{x^2\plus{}2}{(x^4\plus{}5x^2\plus{}4)\arctan\left(\frac{x^2\plus{}2}{x}\right)}\,dx$ we will get no solution.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, i think for the integral
$ \int\frac {x^2 \plus{} 2}{(x^4 \plus{} 5x^2 \plus{} 4)\arctan\left(\frac {x^2 \plus{} 2}{x}\right)}\,dx$ we will get no solution.
Sonnhard.[/quote]

I think there is a way to solve it. I dont hvae much time right now to explain more, but what got into my mind was that $ \left(\tan^{ \minus{} 1}u\right)^{\prime} \equal{} \frac {1}{1 \plus{} u^{2}}$. and fortunately we have $ \frac {1}{1 \plus{} \left(\frac {x^{2} \plus{} 2}{x}\right)^{2}} \equal{} \frac {x^{2}}{x^{4} \plus{} 5x^{2} \plus{} 4}$


How do you create a vertical dashed line in a table?  I found a package which apparently will do it, called [b]arydshln[/b] (http://www.ctan.org/tex-archive/help/Catalogue/entries/arydshln.html)  I don't know how to interpret the instructions, though-- they say "Definitions of horizontal and vertical dashed lines for the array and tabular environment. Horizontal lines are drawn by \hdashline and \cdashline while vertical ones can be specified as a part of preamble using ‘:’. The shape of dashed lines may be controlled through style parameters or optional arguments."  What does "vertical ones can be specified as part of a preamble using `:' mean in terms of the actual commands I need to use?

You need to look at the manual linked on the page you gave. The first example on page 3:
[code]\begin{tabular}{|l::c:r|}\hline
A&B&C\\\hdashline
AAA&BBB&CCC\\\cdashline{1-2}
\multicolumn{2}{|l:}{AB}&C\\\hdashline\hdashline
\end{tabular}[/code]shows that : is a single dashed vertical line and :: is a double dashed vertical line. They are in the 'preamble' of the table which means the first line where you give the type of table that you want. It's very similar to the normal LaTeX usage where you can use | for solid vertical lines.
There are only 9 pages in the manual so it's worth looking through it first for the type of table that you want.


modular arithmetic
linear algebra
matrix
number theory unsolved
number theory

Find all positive integers $x>1, y$ and primes $p,q$ such that $p^{x}=2^{y}+q^{x}$

p>q, thus $p^{x}-q^{x}>1$(because x>1.) Thus we have $y\ge1$, i.e. p,q are both odd. \[p^{x}-q^{x}=(p-q)(p^{x-1}+\dots+q^{x-1}).\] Let r be any common prime divisor of both $p-q$ and $T=p^{x-1}+\dots+q^{x-1}$. Then (r,q)=1(otherwise r divides both q and p=(p-q)+q.)
$p\equiv q\pmod r\Rightarrow T\equiv xq^{x-1}\pmod r$, but $T\equiv0\pmod r$, and, since (q,r)=1, r divides x.
Let us remember that $p^{x}-q^{x}=2^{y}$, thus only possibility for r is r=2. If x is odd, we have thus (p-q,T)=1, but $(p-q)T=2^{y}$ and p-q is even $\Rightarrow$ T=1, but T>1. We proved, that x is even, x=2t. \[(p^{t}-q^{t})(p^{t}+q^{t})=2^{y}.\] If t>1 it follows from here that t is even, but that implies $p^{t}+q^{t}\equiv2\pmod4$, but $p^{t}+q^{t}>2$ and $(p^{t}-q^{t})(p^{t}+q^{t})=2^{y}$ is impossible.
So, x=2 and $(p-q)(p+q)=2^{y}$. Thus \[\left\{\begin{matrix}p+q=2^{\alpha},\\ p-q=2^{\beta}. \end{matrix}\right.\] $p=2^{\alpha-1}+2^{\beta-1}\qquad\Rightarrow\qquad\beta=1$ and $\alpha-1$ is power of 2(otherwise $2^{\alpha-1}+2^{\beta-1}$ is not prime.)
$q=2^{\alpha-1}-2^{\beta-1}\qquad\Rightarrow\qquad\beta=1$ and $\alpha-1$ is prime.
Thus $\alpha-1=2$, [b]p=5,q=3,x=2,y=4.[/b]
Excuse me for my bad English.


Same idea as the weakest area one.

Voted number theory.

Combinatorics and number theory for me. I voted combinatorics.

Combinatorics would go under counting/probablity?

Actually counting/probability go under combinatorics, combinatorics also include binomial, Pascal's Triangle, sets,

Eh, combinatorics is just a fancy way to say counting.

[quote="beta"]I am bad at everything.[/quote]

That's a lie and you should know it.  You have some of the most elegant solutions on the entire board.

[quote="beta"]I am bad at everything.[/quote]

Is this kind of comment really necessary? Self-deprecation will not accomplish anything. Especially when it's not even true.

As for strength...I voted Trigonometry. I'm pretty good at Algebra as well, although I'm fairly weak in Number Theory and Geometry.

Can't say I'm 'strongest' in Geometry, but I do have the most experience with it, out of all the discplines in math.

Which is why you see me participating in geometric threads a LOT more often than I do in other subjects.

Combinatorics!!

Since I just solved IMO 04/3 flawlessly :D

That's like beta's comment, maybe you should think of yourself as the last choice. I believe thinking your're the best is better than thinking your're the worst.

[quote="RC-7th"]I believe thinking your're the best is better than thinking your're the worst.[/quote]

Depends... if you're not, that's pretty stupid :P

No, I think it's good just to know what you're good at exactly, and try as much as possible only to show that talent instead of your weaker areas.

[quote="Peter VDD"]No, I think it's good just to know what you're good at exactly, and try as much as possible only to show that talent instead of your weaker areas.[/quote]

I've fooled people into thinking I'm smart with this very strategy.

I'm better at number theory and algebra than at anything else I suppose  :)

Inequalities and general topology... - I love Jensen's Inequality...

Well, I was going to use this post to brag about how amazingly good at math I am and be a pompous jerk, but then I realized that nobody likes people who are egotistical, and that talking about how good you are at math won't do anything to improve your mathematical skills.  Note to the slow:  that means stop bragging.


How do I show that the medians split up the triangle into 6 regions of equal area (without using the fact that the centroid trisects a median in a 2:1 ratio)?

Assuming we can still use the fact that the medians are concurrent:
[hide="Proof"]
Label the triangle $ ABC$, and let $ M$,$ N$,$ O$ be the midpoints of the respective opposite sides. Let $ G$ be the point of intersection of the three medians. By Same-Base-Same-Altitude, we get $ |BAM|\equal{}|CAM|$, $ |BGM|\equal{}|CGM|$, $ |AGN|\equal{}|CGN|$, and $ |AGO|\equal{}|BGO|$.  Subtracting the first two equations yields $ |AGB|\equal{}|AGC|$, which in turn implies $ |AGO|\equal{}|AGN|$.  Following the same argument on each corner of the triangle shows that all six little triangles have the same area.
[/hide]

[quote="cyberspace"]Assuming we can still use the fact that the medians are concurrent:
[/quote]

and if you are not yet sure of that just use Ceva´s theorem and crush it!


Sooo I have this problem of the week called Paving Patterns. These two people are placing tiles in a 2ft by 20ft area, the question is how many 1ft by 2ft tile combinations are there? so its definitely 1ft by 2ft tiles, it was late i read the problem wrong haha thanks guys. the tiles can be placed either horizontally or vertically

Does the order the tiles are placed in matter?  are they different colored or something?

err... 2 x 20 = 40 ft^2, and 1 x 3 = 3 ft^2, but since 3 doesnt divide 40 it is impossible to fill that area with those tiles. Does this make sense or am I reading this wrong?

Are you sure it's 2 by 20 ft and 1 by 3ft? If so, the tiles won't cover the area completely.


I've recently been trying to solve this problem, but I didn't get the solution :(

1st problem :

1 - [cos^2(a)/(1+cos(a))] = sin(a)

2nd problem :

If a,b and g are angles of a triangle, than prove:

sin2 :alpha: + sin2 :beta: + sin2 :gamma:  = 4sin :alpha: sin :beta: sin :gamma:

Thanx
Fis

This clearly is not an olympiad level problem and it should be moved to the appropriate level.

[b]1.[/b] Solve for a:

$1 - \frac {\cos^2{a}}{1 + \cos a} = \sin a$

$1 + \cos a - \cos^2{a} = \sin a + \sin a \cos a$

$\sin^2{a} + \cos a = \sin a + \sin a \cos a$

$\cos a (1 - \sin a) = \sin a (1 - \sin a)$

$(\cos a - \sin a)(1 - \sin a) = 0$

Either $\sin a = 1$, $a = 90^o + n \cdot 360^o$ or $\sin a = \cos a$, $a = 45^o + n \cdot 180^o$.

[b]2.[/b] Since $a + b + c = 180^o$,

$\sin{2c} = \sin{2[180^o - (a + b)]} = - \sin{2(a + b)}$

$\sin{c} = \sin{[180^o - (a + b)]} = \sin{(a + b)}$

Use the following trigonometric identities:

$\sin x + \sin y = 2 \sin{\frac{x + y}{2}} \cos{\frac{x - y}{2}}$

$\cos x - \cos y = -2 \sin{\frac{x + y}{2}} \sin{\frac{x - y}{2}}$

which can be easily verified by putting $x = \frac{x+y}{2} + \frac{x - y}{2}$ and $y = \frac{x+y}{2} - \frac{x - y}{2}$. Hence,

$\sin{2a} + \sin{2b} + \sin{2c} = \sin{2a} + \sin{2b} - \sin{2(a + b)} =$

$= 2 \sin{(a + b)} \cos{(a - b)} - 2 \sin{(a + b)} \cos{(a + b)} =$

$= 2 \sin{(a + b)} [\cos{(a - b)} - \cos{(a + b)}] = 2 \sin c \cdot \left(-2 \sin {\frac{2a}{2}} \sin{\frac{-2b}{b}}\right) =$

$= 4 \sin a \sin b \sin c$


For real numbers $ a$ and $ b$, define $ a\$b\equal{}(a\minus{}b)^2$. What is $ (x\minus{}y)^2\$(y\minus{}x)^2$?

$ \textbf{(A)}\ 0 \qquad
\textbf{(B)}\ x^2\plus{}y^2 \qquad
\textbf{(C)}\ 2x^2 \qquad
\textbf{(D)}\ 2y^2 \qquad
\textbf{(E)}\ 4xy$

[hide="Answer"]
$ (x \minus{} y)^2 \equal{} (y \minus{} x)^2 \equal{} z$. We want the value $ (x \minus{} y)^2 \$ (y \minus{} x)^2 \equal{} z \$ z \equal{} (z \minus{} z)^2$, hence the answer is $ \boxed{\textbf{(A)} \; 0}$
[/hide]

omg was this number 7 on AMC 12????? it was like the same on 10.


AMC
AIME
AMC 12
AMC 8
AMC 12 A
USA(J)MO
USAMO

The AIME invitation score for the 2004 AMC 10 A is 110.0 (the top 1% level.)

The AIME invitation score for the 2004 AMC 12 A is 100.0.

AMCDirector
Steve Dunbar

hm, cats... I won the bet. I'll be seeing you at ARML...;)

Im sure that i don't know what you're talking about.

[quote="AMCDirector"]The AIME invitation score for the 2004 AMC 10 A is 110.0 (the top 1% level.)

The AIME invitation score for the 2004 AMC 12 A is 100.0.[/quote]
Is that low for AMC?

The 110.0 score is low; the 100 score is normal.

WOW.... I think I made it....if I didn't put my answers in the wrong places..... very shocking... (I have 110, at least that's what I think is my score) but I'll redeem myself on Wed. anyway...(you know, AMC B)

Wow, 110! i'm gonna make AIME. There is a god!

wow talk about low scores... or mebbe just a lot of ppl taking it this year. but still... 110... you'll never see a 10 pt difference on the AMC 12.

I think the qualification score in 2001 was 84.

Yes... How strange that the cutoff score is 100...

there is god and i will succeed on the AMC 10 for i have been plagued by failures this week. that means god is devoting my full  strength towards the AMC.

serviam

fine. be that way :D ok, i got it, never will the AMC12 be 20 pts under! hah! prove me wrong now! (actually... dont, even if you can :P)

When I looked the AMC-12 A, it said that the cut off is 120.  Do you know why it did not say 110?

a.) I believe that's for the AMC 10
and
b.) If the cut off score is above 120 then the cut off score will be automatically lowered to 120. If the cut off score is lower than 120 then it stays that way.

W00T!!!!! 113!!!!! Unless another scantron error like on the AMC8, where 1 mark wasn't clear so i got 23 instead of 24.

[quote="Tare"]What exactly would I need to know to get a 9?[/quote]

A lot.  :mrgreen: Ok, that was a bad response, but hey, you gotta admit, the question was pretty vague. :D

Anyways, I think you oughta take a couple practice AIME's first and see how you do. Then you will know what you need to work on.

[quote]um, isn't that what I basically said? I was just putting a range for what I thought would be the AIME floor judging from the past years.[/quote]

yeah, actually it is. sorry. i was just trying to say that tare did not seem to know whether or not qualifying depended on amc 12/10 score for him/her. so i was just trying to specify that fact. anyway, does anyone think the floor could possibly be  9? that seems like ridiculously high.

I highly doubt it, but they might surprise us with a very easy AIME to make up for what was supposed to be a hard amc 12 and stuffs.

I mostly included it in the range just to be safe.

i dont think they will "make up" for the slightly increased amc 12 difficulty. in 2001, when the amc 12 was quite difficult, the aime was still pretty hard, floor being 6. last year, the amc 12 was the easiest ever, but so was the aime. hopefully extremely hard aime this year.

[quote="tokenadult"][quote="Tare"]I got a 112.5, so how much do I need to score on AIME to make USAMO? :lol:[/quote]

Tare, are you a United States permanent resident or United States citizen? If not, you have to find a different country to test in to make into the IMO selection process. I thought I read a post of yours months ago that indicated you would be going back overseas. I DON'T know what your citizenship or visa status is.[/quote]
Well...let's just say I'm screwed if they check for my citizenship during USAMO :lol:

Oooh, that reminds me, I forgot to sign up for that AIME cram session! Stupid me :)

Hmmm, I've tried the first couple of questions from 2003 AIME A and yet some of them don't come out integral between 0 and 999...(the answers on Kalva doesn't either :?:)

Hmmm, it seems I need to work on geometry, logarithms, trigonometry, and serieses (sp? :lol:) I'm pretty much screwed I guess :lol:

[quote]Hmmm, I've tried the first couple of questions from 2003 AIME A and yet some of them don't come out integral between 0 and 999...(the answers on Kalva doesn't either [/quote]

John Scholes reworded all of the problems. If you take an AIME from his site, the answers may not be integral.

[quote="alc"]I qualified on both the AMC10 and AMC12, so which one will they use for my combo score along with the AIME?[/quote]

The higher one.

[quote="alc"]I qualified on both the AMC10 and AMC12, so which one will they use for my combo score along with the AIME?[/quote]

We have been practicing with OLD AIME tests, and in those days you wrote in on the AIME answer sheet what your (then) AHSME score was. Can someone else confirm that what is done today is to indicate either an AMC 10 score or an AMC 12 score? See this year's [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2003-4pub/04AIME/html/04-04aime-text.html]AIME instructions[/url] from the AMC Web site. You would, of course, choose whatever score gives you a better index for qualifying for the USAMO.

I hope someone with AIME experience will correct me if I am mistaken in this. I would like the AIME answer sheet to be available for downloading on the AMC Web site, as well as the instructions.

the AIME answer sheet has a place to bubble in which prior exam you took and your score. If you took two exams, you should enter the one in which you scored higher.

The AIME instructions are available in the teacher's handbook on the AMC website.
http://www.unl.edu/amc/d-publication/d1-pubarchive/2003-4pub/04USAMO/04aimeusamoTM.html

The AMC-10A, 10B, 12A, and 12B Grand Statistics are now posted at [url]http://www.unl.edu/amc[/url]

If my calculations are correct:

AMC-10A:  1.38% made AIME cutoff
AMC-10B:  2.07% made cutoff
AMC-12A:  8.14% made cutoff
AMC-12B: 19.28% made cutoff

Disclaimer:  I have no connection with CAMC.

[quote]I would like the AIME answer sheet to be available for downloading on the AMC Web site, as well as the instructions.[/quote]

Looks like your wish was granted, tokenadult.

[url]http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2004-aa/04aimeanswerforms.html[/url]

[quote="JSRosen3"]Looks like your wish [that sample AIME answer sheets be publicly posted] was granted, tokenadult.[/quote]

[url]http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2004-aa/04aimeanswerforms.html[/url]

Thanks for letting me know. I've printed out a couple.

That .pdf of a single page is simply HUGE though (32-ish mbs)


algebra
function
domain
trigonometry
calculus
calculus computations

Hi i need help for this problem.

Find the formal solution of Dirichlet problem which statisfy:
i) Equation: $\frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{ \partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}=0$ in domain $1<r<2$ and $- \pi \leq \theta \leq \pi$
ii) $u(1, \theta) = \sin \, 4 \theta - \cos \, \theta$ , $- \pi \leq \theta \leq \pi$
iii) $u(2, \theta) = \sin \, \theta$, $- \pi \leq \theta \leq \pi$.

thanks you.

Have you tried using separation of variables?

I have a general solution for this problem with arbitrary continuous boundary functions and in domain $\rho < r < 1$. It was homework in my Fourier analysis class. I can post it if you want. But I skip a lot of simple steps in the beginning of the solution since we had already solved the problem in class for the Dirichlet problem in the unit disc. This especially includes the part that DVO gave you a suggestion on -- separation of variables to find the general form of $u(r,\theta)$.


A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. THe remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths $ 15$ and $ 25$ meters. What fraction of the yard is occupied by the flower beds?
[asy]unitsize(2mm);
defaultpen(linewidth(.8pt));

fill((0,0)--(0,5)--(5,5)--cycle,gray);
fill((25,0)--(25,5)--(20,5)--cycle,gray);
draw((0,0)--(0,5)--(25,5)--(25,0)--cycle);
draw((0,0)--(5,5));
draw((20,5)--(25,0));[/asy]$ \textbf{(A)}\ \frac18\qquad
\textbf{(B)}\ \frac16\qquad
\textbf{(C)}\ \frac15\qquad
\textbf{(D)}\ \frac14\qquad
\textbf{(E)}\ \frac13$

[hide="Answer"]
Easy to see that the upper side is missing $ 10$ meters thus the sides of the triangle must be $ 5$ meters. Combine the flower beds into a square to get $ 5 \cdot 5\equal{}25$ meters squared. The yard is $ 5 \cdot 25$ and $ \frac{25}{5 \cdot 25}\equal{}\frac{1}{5} \implies \boxed{(C)}$
[/hide]

There's a slight typo in the problem statement:

"[b]THe[/b] remainder of the yard has a trapezoidal shape"

I may be being picky, but could someone please fix this?


[quote="pardesi"]why does hydride transfer in the intermediate of cannizaro reaction with benzaldehyde require mono-deprotonation while with formaldehyde bideprotonation[/quote]

I don't really get your question... Can you explain? How does Bideprotonation occur in HCHO?

i meant in benzaldehyde the $ \ce{OH}^{\minus{}}$ attacks the carbonyll carbon atom hence an inermediate is  formed in whihc the $ \ce{C}$ is attache dto a negatively charged O and OH  is formed which transfers the hydride to another molecule.
but in case of formaldehyde. the tranfer only takes place after the $ \ce{OH}$ also has been deprotonated..

My book doesnt say so (and so doesnt your book) and i think only after the $ \ce{H\minus{}}$ attacks the HCOOH , It becomes $ \ce{CH3O\minus{}}$ and then only a proton transfers  :maybe: .
Atleast that's what the book says  :D

no my book(Sykes) says what i said :D and the rate law is $ r \equal{} k[\ce{OH}^{ \minus{} }]^{2}[\ce{HCHO}]^{2}$
and for benzaldehyde it is $ r\equal{}k[\ce{OH}^{ \minus{} }][\ce{PhCHO}]^{2}$

Then wat is the mechanism?  :D  Bcos how can a proton attack if there is no O- for it to attack (or in other words what drives the rn according to Sykes?)  :maybe:

sorry the question should have been why benzaldehyde is not deprotonated and formaldehyde monodeprotanated....
actually that doesn't change our discussion
see the book say the $ \ce{OH}^{\minus{}}$ captures the hydrogen attached to $ \ce{OH}^{\minus{}}$ first then the hydogen from that carbon attacks the other carbon

Wait i dont understand what you mean? Can you post ur full mechanism? For instance can you explain this atleast("see the book say the  captures the hydrogen attached to  first then the hydogen from that carbon attacks the other carbon")?  :maybe:

see [url=http://en.wikipedia.org/wiki/Cannizaro_reaction]this[/url]
here the formaldehyde loses a proton from $ \ce{OH}$ before it attacks while benzaldehyde doesn't

I dont see where in the link it is mentioned abt formaldehyde?  :maybe: . And also do you mean OH- loses a proton? How can that happen? And also isnt the tetrahedral intermediate more stable?

see the last reaction diagram replace R by H

[quote="pardesi"]no my book(Sykes) says what i said :D and the rate law is $ r \equal{} k[\ce{OH}^{ \minus{} }]^{2}[\ce{HCHO}]^{2}$
and for benzaldehyde it is $ r \equal{} k[\ce{OH}^{ \minus{} }][\ce{PhCHO}]^{2}$[/quote]

You are perfectly right about this even my chem prof said that when he did canizzaro reaction.

actually this supports the involvemnt of
$ \ce{OH}^{\minus{}}$ twice hence the doubly charged ion's hydrogena ttacking
but the question is why

A doubly charged ion does not make sense on this reaction...

it's not exactly doubly charged but something like the last reaction species in the link i provided to madness

Yes, I saw the wikipedia link. The dianion seems to form only under strongly basic conditions. Under normal Cannizaro reaction conditions, there is no dianion and hydride transfer occurs from an intermediate R-C(O-)(OH)-H.

thanks ...but isn't it strange....the active species has to wait till it becomes hyperactive


rotation
symmetry
calculus
calculus open

Show that if a set in R^n has strictly positive but finite measure, and is rotationally symmetric about two axes, these two axes must have a pt in common.

If anyone can help, i would really appreciate it.

Suppose two such axes $L_{1}$ and $L_{2}$ are at a positive distance $d$ from each other. Let $a_{k}\in L_{k}$ be two points such that $|a_{1}-a_{2}|=d$. Rotate $L_{1}$ about $L_{2}$ by $180$ degrees. The result is a line $L_{3}$ that is at distance $2d$ from $L_{1}$. This new line is also an axis of symmetry for the set. Continuing this process, you will obtain a sequence of axes $L_{k}$ such that the distance from $L_{k}$ to $L_{1}$ goes to infinity.

Since $E$ has finite positive measure, almost all of it is contained in some ball $B$. But there is $L_{k}$ that is disjoint from $B$, and rotating $E$ around $L_{k}$ by $180$ degrees, you will arrive at a contradiction.


geometry
probability
expected value
national olympiad

1.  How many subsets of $\{1, 2, ... , 2n\}$ do not contain two numbers with sum $2n+1$?

2.  The diagonals of a convex pentagon divide it into a small pentagon and ten triangles. What is the largest number of the triangles that can have the same area?

3.  $p \geq 5$ is prime. The sequence $a_0, a_1, a_2, ...$ is defined by $a_0 = 1$,$a_1 = 1$,$...$,$a_{p-1} = p-1$ and $a_n = a_{n-1} + a_{n-p}$ for $n \geq p$. Find $a_{p^3} \mod p$.

4.  The positive reals $x_1, x_2, ... , x_n$ have harmonic mean $1$. Find the smallest possible value of $x_1 + \frac{x_2 ^2}{2} + \frac{x_3 ^3}{3} + ... + \frac{x_n ^n}{n}$.

5.  An urn contains $n$ balls labeled $1, 2, ... , n$. We draw the balls out one by one (without replacing them) until we obtain a ball whose number is divisible by $k$. Find all $k$ such that the expected number of balls removed is $k$.

6.  $PA$, $PB$, $PC$ are three rays in space. Show that there is just one pair of points $B', C'$ with $B'$ on the ray $PB$ and $C'$ on the ray $PC$ such that $PC' + B'C' = PA + AB'$ and $PB' + B'C' = PA + AC'$.

\[\text{Day 1}\]
1. http://www.mathlinks.ro/Forum/viewtopic.php?p=352191#p352191
2. http://www.mathlinks.ro/Forum/viewtopic.php?p=352192#p352192
3. http://www.mathlinks.ro/Forum/viewtopic.php?p=148269#p148269
\[\text{Day 2}\]
4. http://www.mathlinks.ro/Forum/viewtopic.php?p=352194#p352194
5. http://www.mathlinks.ro/Forum/viewtopic.php?p=352195#p352195
6. http://www.mathlinks.ro/Forum/viewtopic.php?p=314072#p314072


This is humorous youtube. I laughed :laugh:

Check this out..  http://video.google.com/videoplay?docid=4241480654150026320

haha :laugh:
Is this material an existing thing from old days?
It was not in Japan to have looked :o

[quote="7h3.D3m0n.117"][quote="math92"][url]http://www.youtube.com/v/f2Re073wpls[/url]

 :o[/quote]

 :rotfl:  lol im sending that to my math teachers[/quote]

I second that.  Abbott and Costello did the same thing:
[youtube]pUmVe9qnw7E[/youtube]

Another video for your math teachers (this is from my school):
[youtube]P42VkspMqsI[/youtube]


rotation
geometry
parallelogram
trapezoid
perpendicular bisector
geometry proposed

Given 2 circles $(O_{1})$ and $(O_{2})$ and their 2 common points $A$, $B$. 2 arbitrary parallel lines passing through $A$, $B$ cuts 2 circles at $M$, $N$, $P$, $Q$, respectively. Denote $I$, $K$ are the midpoints of $MN$, $PQ$. Find the locus of the midpoint $H$ of $IK$ when $MN$ and $PQ$ rotate round $A$ and $B$.

In other words, since it's easy to see that $MNQP$ is a parallelogram, we want the locus of the center of this parallelogram.

The locus of the midpoint of $MN$ is a circle $\mathcal C$ centered at the midpoint of $O_1O_2$ and passing through $A,B$, and this is also the locus of the midpoint of $PQ$. To show this, consider a line through $B$ parpendicular to $AB$. Let this line cut the circles in $X,Y$, and let $P$ be the midoint of $XY$. An angle chase shows that $\angle PIA=\frac\pi 2$, and that's it.

As for our locus, it's the locus of the midpoint of a chord of a fixed circle $\mathcal C$ which has constant length ($IK=AB$). This locus must be a circle also centered at the midpoint of $O_1O_2$ and touching $AB$.

Here is a different solution, using a tactic (well, not really a tactic, rather an idea) which I use to call "midpoints chasing" (similarly to "angle chasing"). Mainly, in the following, I will define so many midpoints that it will be helpful to have a special notation for them: In the following, for any two points X and Y, we will denote by $M_{XY}$ the midpoint of the segment XY.

Now a remark to the problem statement itself:

[quote="Huynh Anh Hao"]Given 2 circles $(O_1)$ and $(O_2)$ and their 2 common points $A$, $B$. 2 arbitrary parallel lines passing through $A$, $B$ cuts 2 circles at $M$, $N$, $P$, $Q$, respectively. Denote $I$, $K$ are the midpoints of $MN$, $PQ$. Find the locus of the midpoint $H$ of $IK$ when $MN$ and $PQ$ rotate round $A$ and $B$.[/quote]

From your sentence "2 arbitrary parallel lines passing through $A$, $B$ cuts 2 circles at $M$, $N$, $P$, $Q$, respectively", it is not completely clear to me what exactly the points M, N, P, Q are assumed to be: for instance, should the points M and N lie on one and the same of the 2 parallel lines and on different circles or on different lines and on the same circle? (For instance, in Grobber's solution, the points M and N seem to be the intersections of the line through A with the circles $\left(O_1\right)$ and $\left(O_2\right)$; but I am not completely sure since Grobber seems to denote two different points by P in his solution, although his solution can be still understood quite well.) Anyway, this doesn't really matter since your definition of the point H as the midpoint of the segment $IK=M_{MN}M_{PQ}$ shows that this point H is actually the centroid of the four points M, N, P and Q, and the notion of centroid is completely symmetric in all involved points. So we can label our points of intersection of the parallel lines with the circles $\left(O_1\right)$ and $\left(O_2\right)$ in any order we want; the point H will always be the same. Hence I take the freedom to rewrite your problem as follows (I also rename the circles and their centers, since I don't really like expressions with three indices inserted into each other :D ):

[i]Given two circles c and d with centers C and D, and the two common points A and B of these two circles. Let a and b be two parallel lines through the points A and B, respectively. The line a intersects the circles c and d at the points M and N, respectively (of course, apart from the point A), and the line b intersects the circles c and d at the points P and Q, respectively (apart from B). Let H be the centroid of the four points M, N, P and Q. Find the locus of the point H when the direction of the parallel lines a and b varies.[/i]

We will show that this locus is the circle with center $M_{CD}$ and radius $M_{CD}M_{AB}$. In fact, in order to show this, it will clearly suffice to prove $M_{CD}H=M_{CD}M_{AB}$.

We will use directed angles modulo 180. Since the points A, B, M and P lie on one circle (namely on the circle c), we have < AMP = < ABP, or, equivalently, < (a; MP) = < (AB; b). Similarly, < (a; NQ) = < (AB; b). Thus, < (a; MP) = < (a; NQ), so that MP || NQ. Also, MN || PQ (this is just a reformulation of a || b). Hence, the quadrilateral MNQP is a parallelogram, and thus, the centroid H of the points M, N, P and Q is just the center of this parallelogram, i. e. the midpoint of its diagonal PN. Thus we have proven that the point H is the midpoint of the segment PN. In other words, $H=M_{PN}$.

Since AN || BP (this is again a rephrase of a || b), the quadrilateral ABPN is a trapezium (even if it is self-intersecting in the usual case). Since the midparallel of a trapezium is parallel to its bases, we thus have $M_{AB}M_{PN}\parallel AN\parallel BP$. On the other hand, since the segment BP is a chord in the circle c, the perpendicular bisector of this segment BP passes through the center C of the circle c; this yields $CM_{BP}\perp BP$. Since $M_{AB}M_{PN}\parallel BP$, we thus have $CM_{BP}\perp M_{AB}M_{PN}$. Similarly, $DM_{AN}\perp M_{AB}M_{PN}$. Thus, $CM_{BP} \parallel DM_{AN}$. This shows that the quadrilateral $CM_{BP}M_{AN}D$ is a trapezium. Since the midparallel of a trapezium is parallel to its bases, we therefore have $M_{CD}M_{M_{BP}M_{AN}} \parallel CM_{BP}$. But the point $M_{M_{BP}M_{AN}}$ is the centroid of the points A, B, P and N, and as such, it coincides with the point $M_{M_{AB}M_{PN}}$. Thus, we have $M_{CD}M_{M_{AB}M_{PN}} \parallel CM_{BP}$. Since $CM_{BP}\perp M_{AB}M_{PN}$, this yields $M_{CD}M_{M_{AB}M_{PN}}\perp M_{AB}M_{PN}$. Since $H=M_{PN}$, this can be rewritten as $M_{CD}M_{M_{AB}H}\perp M_{AB}H$. Thinking about it, this relation means nothing but just the fact that the point $M_{CD}$ lies on the perpendicular bisector of the segment $M_{AB}H$, so that $M_{CD}H=M_{CD}M_{AB}$. Proof complete.

  Darij


On the avatar panel, could there be a spot like flag for gender? It would eliminate some confusion...

Hm...that could help, but it shouldn't really matter that much...in terms of talking to people and what you think of them

If someone is that concerned about confusion, they should put their gender in their signature.  I don't recommend that.  What's wrong with saying "he/she" anyway?

The ambiguity might be offense to some, and most people don't remember to put the latter ".../she" and you can see how gender confusion could happen.


ye soale  bahal vase bache haye bahal:
dar yek jame $n$ nafare,har do nafar daghighan ye dooste moshtarek darand,prove that ye nafar hast ke ba hame dooste.(doosty yek rabeteye do tarafe ast) :D

pigeon hole principle.(asl e laneh kabootari)

This wonderful theorem has a very simple commonsense formulation.
Namely, given a society in which any two people have exactly one friend in
common, there must be a “host,” who is everybody’s friend. Of course, this
is a graph-theoretic theorem and in order to prove it we must express it in
graph-theoretic terms.
Theorem 1 (Friendship Theorem). Suppose that G is a graph such that,
if x and y are any two distinct vertices of G, then there is a unique vetex z
adjacent in G to both x and y. Then there is a vertex adjacent to all other
vertices.
We will prove this theorem in several steps.1 We will assume that a
counterexample G to the Friendship theorem does exist and will be working
with this counterexample until we get a contradiction. Then the theorem
will be established.
Definition 1. A sequence of vertices x0, x1, . . . , xn will be called a path of
length n, if xi−1 is adjacent to xi for all i = 1, 2, . . . , n. These vertices need
not be all different, i.e. going along this path we may visit a certain vertex
several times. Any path x0, x1, . . . , xn−1, x0 is called a cycle of length n.
Lemma 1. G does not have any cycles of length 4.
Proof. If we had a cycle x0, x1, x2, x3, x4, x0 of length 4, then x0 and x2
would have at least two neighbors in common, namely x1 and x3, which is
not possible.
Definition 2. The degree of a vertex is the number of other vertices adjacent
to it. The graph is called regular, if all its vertices have the same degree.
Lemma 2. Any two nonadjacent vertices of G have the same degree.
Proof. Let x and y be two nonadjacent vertices and let z be their unique
common neighbor. Then x and z will have a unique common neighbor u
and y and z will have a unique common neighbor v.
Now let u1, u2, . . . , us be all other vertices adjacent to x. For each
i = 1, 2, . . . , s, let vi be the unique common neighbor of ui and y. By
inspection we check that vi is different from any of the x, z, y, u, v, ui (every
such assumption lead to the existence of a 4-cycle). Also, no two vertices vi
and vj can coincide for i 6= j, according to the same reason.
Thus, we see that the degree of x is not greater than the degree of y. But
the situation is symmetric, i.e. we can also prove that the degree of y is not
greater than the degree of x. Hence, these two degrees coincide.
Lemma 3. G is regular.
Proof. Let d(x) denote the degree of the vertex x. Suppose that G is not
regular and that there exist two vertices a and b such that d(a) 6= d(b).
Then a and b must be adjacent by Lemma 2. There is a unique common
neighbor c of a and b. Since either d(c) 6= d(a) or d(c) 6= d(b), or both,
we may assume that the former is true and d(c) 6= d(a). Now let x be any
other vertex. Then x is adjacent to one of a or b, for otherwise by Lemma 2
d(a) = d(x) = d(b), contrary to the assumption that d(a) 6= d(b). Similarly,
x is adjacent to either a or c. But x cannot be adjacent to both b and c, as a
is their unique common neighbor, hence x must be adjacent to a. This now
shows that all vertices of G are adjacent to a and G is not a counterexample.
Hence G is regular.
Let m be the degree of G.
Lemma 4. m is an even number.
Proof. Let v1, v2, . . . , vm be the vertices adjacent to v. Let us consider v1.
Together with v it must have a vertex which is adjacent to both. Since
v1, v2, . . . , vm are all vertices adjacent to v, this third vertex must be among
v1, v2, . . . , vm. Let it be v2. No other vertex among v1, v2, . . . , vm can be
adjacent to v1 or to v2. Thus v1 and v2 form a pair. This way we can pair
off the vertices adjacent to v which implies that m is even. We show that
the neighborhood of every vertex v looks like a “windmill.”
Lemma 5. Let N be the number of vertices of G. Then N = m(m − 1) + 1.
Proof. Let v be any vertex and v1, v2, . . . , vm be the vertices adjacent to v.
We know that the neighborhood of v looks like a “windmill.” Without loss
of generality we assume that the vertices are paired off so that v1 is adjacent
to v2, v3 is adjacent to v4 and finally vm−1 is adjacent to vm. Every vertex
different from v and v1, v2, . . . , vm must be adjacent to one of the vi since
it must have a common neighbor with v. Each vi will have exactly m − 2
neighbors of this kind. In total we then have N = 1 + m + m(m − 2) =
m(m − 1) + 1 vertices.
Let us now note that m > 2, or else G is just a triangle which is not
a counterexample. Let p be any prime which divides m − 1. Since m − 1
is odd, p is also odd. In the following two lemmas we will consider the set
S of all cycles v0, v1, . . . , vp−1, v0 of length p with the fixed initial point v0.
This means that the same cycle with two different initial vertices will be
considered as two different elements of S. Let us agree that if the cycle is
written as v0, v1, . . . , vp−1, v0, then v0 is chosen as its initial point. Note that
we again do not require that all vertices in the cycle are different.
We shall compute the cardinality |S| of S in two ways.
Lemma 6. |S| is a multiple of p.
Proof. Every cycle of length p with the fixed initial point
v0, v1, . . . , vp−1
gives us p − 1 other cycles by changing the initial point of it: v1, v2, . . . , v0,
and v2, v3, . . . , v1, and so on. No two of such sequences are the same, assuming
the opposite will contradict to the primeness of p (see the solution
to Exercise 9 of the assignment ”Many faces of mathematical Induction”).
Since in every cycle of length p we can choose p initial points, and hence get
p different elements of S, it is clear that |S| is divisible by p.
Proof of the Friendship Theorem. Now we will prove that |S| is NOT divisible
by p which will give us a contradiction and the proof will be therefore
complete.
First, we will count the number of vertex sequences v0, v1, . . . , vp−2, such
that vi is adjacent to vi+1 for all i = 0, 1, . . . , p − 2. There are two types
of such sequences: 1) those for which v0 = vp−2 and 2) those for which
v0 6= vp−2. Let K1 and K2 be the number of sequences of the first and the
second type, respectively. Then K1 + K2 = Nmp−2. Indeed, we can choose
v0 in N different ways, and having chosen v0, v1, . . . , vi, we can choose vi+1 in
m different ways. Now we will return to cycles with fixed initial vertices from
S. Each of them can be obtained from a sequence of one of the above types
by adding a vertex vp−1 which is adjacent to v0 and vp−2 and considering
v0 as the initial vertex of this cycle. If v0 = vp−2, then we can choose vp−1
in m different ways, while if v0 6= vp−2, then such vp−1 will be unique. Thus
|S| = mK1 + K2. But now
|S| = (m − 1)K1 + (K1 + K2) = (m − 1)K1 + Nmp−2  Nmp−2 (mod p)
But N  1 (mod p), and m = (m − 1) + 1  1 (mod p), thus |S|  1
(mod p) which is a contradiction.
Therefore the Friendship theorem is proved.


See if you can get anywhere by considering either the sequence $ a_{k \plus{} 1} \minus{} a_k$ or the sequence $ \frac {a_{k \plus{} 1}}{a_k}$.  What will indicate when the function has reached a local maximum?