linear algebra
matrix
function
LaTeX
parameterization
linear algebra unsolved

Hi,
I want to show that if A(nxn) is symmetric then it is orthognally equvalent to a symmetric matrix .

let $S$ be a symetric matrix. let $O(S)$ be the set of
matrices that are othogonaly similar to $S$. We want to
prove that there exists $M \in O(S)$ whose elements on the
diagonal are equal.

Here what I did:
1/ the case $n=2$. We only have to prove it when $S=diag(a,b)$ is diagonal.
  It is easy, It suffices to take $P=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}})$.
  Then $PMP^{t}=diag(\frac{a+b}{2} , \frac{a+b}{2})$

2/ with a similar proof we can prove the case $n=2^k$

3/ for the general case I can't find a similar proof :-(

Then I considered $f: O(S) -> R$, $f(M)=\prod M_{k,k}$, the product of
elements on the main diagonal.
$f$ is a continuous function and achieved its maximum on the compact set
$O(S)$. Let $M_0$ be such that $f(M_0)$ is maximum. I think we can prove
that $M_0$ works.

Let $M_1 \in O(S)$ be such that $(M_1)_{i,i} \neq (M_1)_{j,j}$.
Consider $P \in O_n(R)$ defined as follows:
*$P_{k,k}=1$ if $k \neq i$ and $k \neq j$
*$P_{i,i}=P_{i,j}=\frac{1}{\sqrt{2}}$
*$P_{j,i}=-P_{j,j}=\frac{1}{\sqrt{2}}$
Then $(M_2)=P(M_1)P^{t}$ has the same diagonal as $M$, except
$(M_2)_{i,i}=(M_2)_{j,j}=\frac{(M_1)_{i,i} + (M_1)_{j,j}}{2}$.

Hence we can define a sequence of matries $M_k \in O(S)$. Let $M \in O(S)$
be an adherence point of this sequence $M_k$.
$M$ works (if i don't made stupide calculation mistakes)

Hi,
The point is the elements on the two diagonals should be equal.for example in case 2x2 it is like :[a b;b a]

What I found is that in case 2x2 if orthognal matrix is Q= [cos(teta) -sin(teta);sin(teta) cos(teta)] and A=[a b;b c] then for teta=.5*atan(c-a)/2b Q'AQ=[ f h;h f].

in fact I did make two big mistakes yesterday !

1/ if $M_0$ is such that $f(M_0)$ is maximum, it does
not implie that the elements on the diagonal of $M_0$ are equal.
It should be interesting to compute this maximum !!

2/ There was a computation mistake in my poor proof.
So I will write properly the end of what I meant.
Let $M_1 \in O(S)$,
$i$ such that $(M_0)_{i,i}=min{(M_0)_{k,k}}$
$j$ such that $(M_0)_{j,j}=max{(M_0)_{k,k}}$

Consider $P(\theta) \in \O_n(R)$ defined as follows:
i/ $P( \theta )_{i,i} = P(\theta)_{j,j}=cos( \theta )$
ii/ $-P( \theta )_{i,j} = P(\theta)_{j,i}=sin( \theta )$
iii/ $P( \theta )_{k,k} = 1$ for $k \neq i$ and $k \neq j$
iiii/ $P( \theta )_{a,b}=0$ everywhere else.
Then
$( ^{t} P( \theta ) (M_0) P( \theta ) )_{i,i} = cos( \theta )^{2}(M_0)_{i,i} + 2cos( \theta )sin( \theta )(M_0)_{i,j} + sin( \theta )^{2}(M_0)_{j,j}$
$( ^{t} P( \theta ) (M_0) P( \theta ) )_{j,j} = cos( \theta )^{2}(M_0)_{i,i} - 2cos( \theta )sin( \theta )(M_0)_{i,j} + sin( \theta )^{2}(M_0)_{j,j}$
$( ^{t} P( \theta ) (M_0) P( \theta ) )_{k,k} = (M_0)_{k,k}$ for $k \neq i$ and $k \neq j$.
Hence we can find $ \theta $ such that:
$( ^{t} P( \theta ) (M_0) P( \theta ) )_{i,i} = \frac{(M_0)_{i,i} + (M_0)_{j,j}}{2}$
$( ^{t} P( \theta ) (M_0) P( \theta ) )_{j,j} = \frac{(M_0)_{i,i} + (M_0)_{j,j}}{2}$
$( ^{t} P( \theta ) (M_0) P( \theta ) )_{k,k} = (M_0)_{k,k}$ for $k \neq i$ and $k \neq j$.
Then we can define $M_1 = (^{t} P( \theta ) (M_0) P( \theta ))$.
Let $M \in O(S)$ be an accumulation point of the sequence $M_k$.
$M$ works.(I hope)

Rq:
1/I think that with the same idea (if my proof is correct), we can prove
that any matrix $M$ is (orthogonally) similar to a matrix whose diagonal entries
are equal ?
2/let $A$ be an arbitrary matrix. What are the sets $D=(d_1,..,d_n)$ such
that there exists a matrix $M$ similar to $A$ (not nescessary orthogonaly similar)
whose diagonal is $D$ ?

Hi,
Thank you so much for spending your time on this case.Let me give you some hints which I got from your first reply and second reply :

1-We have to find a general form (nxn) for orthognal matrices and i thin it works :

 [cos(teta)  1  -sin(teta)  cos(teta)    1   -sin(teta) .............(just repeat that).....;
  sin(teta)   0   cos(teta)  sin(teta)    0    cos(teta) .............(just repeat that).....;
  0             0       0               0         0       0          the rest all zero                  ;
                         The rest are zero
                                    .
                                    .
                                                                                ]
Then show that B=Q'AQ is a matrix with equal elements on the diagonal.
The define f:R(nxn)----->R   f(x)=SUM(yii-yjj)  , Y=X'AX    and note that f is a continuos function.And then we have to prove that f has a global minima Q* in O(n) and for that f(Q*)=0

2-Sice A is symmetric then it can be written as A=V'YV such that Y=V'AV and Y is a diagonal mat with eighen value on its digonal and then same story for Y which is easier to find a Q matrix and then define a function f same as above and ....

By the way how can I use the special math notation in my text?

Thank you for your reply.

Sorry I think the general form of  matrices is not right.

One way which I am thinking about is that since A is symmetric then you can write A=V'DV and D=digonal(l1,l2......ln) ,l=eighenvalues,then it could be easier to prove.

Suppose that $ A : \mathbb{R}^{n} \rightarrow \mathbb{R}^{n} $ is symmetric.
In fact, we have to find an orthogonal basis
$ e_{1},e_{2}, \ldots , e_{n} \in \mathbb{R}^{n} $, such that
$ (Ae_{i},e_{i}) = \frac{tr(A)}{n} $, for $ i=1,2, \ldots , n $.
For this define $ B = A - ( \frac{tr(A)}{n}) I $, so now we know that
$ tr(B) = 0 $ and we have to find an orthonormal basis
$ e_{1},e_{2}, \ldots , e_{n} \in \mathbb{R}^{n} $, such that
$ (Be_{i},e_{i}) = 0 $, for $ i=1,2, \ldots , n $.
For this take arbitrary orthonormal basis
$ f_{1} , f_{2} , \ldots , f_{n} \in \mathbb{R}^{n} $. We know that
$ tr(B) = 0 $ , so $ (Bf_{1},f_{1}) + (Bf_{2},f_{2}) + \ldots + (Bf_{n},f_{n}) = 0 $.
If all $ (Bf_{i},f_{i}) = 0 $ then we are done. Otherwise there exist some
$ i,j $ such that $ (Bf_{i},f_{i}) > 0 $ and $ (Bf_{j},f_{j}) < 0 $.
Take a straight line segment joining $ f_{i} $ and $ f_{j} $.
Because of continuity there exists some $ v_{1} $ on this segment
such that $ (Bv_{1},v_{1}) = 0 $. Of course $ v_{1} \neq 0 $, so define
$ e_{1} = \frac{v_{1}}{|v_{1}|} $. Complete now $ e_{1} $ to an orthonormal basis
$ f_{1} , f_{2} , \ldots , f_{n} \in \mathbb{R}^{n} $
(I will denote it by the same letters). So $ f_{1} , f_{2} , \ldots , f_{n} $
is orthonormal and $ f_{1} = e_{1} $. Now once again,
$ (Bf_{1},f_{1}) + (Bf_{2},f_{2}) + \ldots + (Bf_{n},f_{n}) = 0 $, so
$ (Bf_{2},f_{2}) + \ldots + (Bf_{n},f_{n}) = 0 $. Once again, if all
$ (Bf_{i},f_{i}) = 0 $ then we are done. Otherwise
there exist some $ i,j > 1 $ such that $ (Bf_{i},f_{i}) > 0 $ and $ (Bf_{j},f_{j}) < 0 $.
Take a straight line segment joining $ f_{i} $ and $ f_{j} $.
Because of continuity there exists some $ v_{2} $ on this segment
such that $ (Bv_{2},v_{2}) = 0 $. Define now $ e_{2} = \frac{v_{2}}{|v_{2}|} $.
Then $ (Be_{1},e_{1}) = (Be_{2},e_{2}) = 0 $. Complete now again
$ e_{1},e_{2} $ to an orthonormal basis $ f_{1} , f_{2} , \ldots , f_{n} \in \mathbb{R}^{n} $
and do the same thing, and so on.
 At the end we will get an orthonormal basis
$ e_{1},e_{2}, \ldots , e_{n} \in \mathbb{R}^{n} $, such that
$ (Be_{i},e_{i}) = 0 $, for $ i=1,2, \ldots , n $.

[quote="akscola"]

how can I use the special math notation in my text?

[/quote]

Top of the screen there is a circle with $\sum$, $\pi$ Latex Help

Thank you for your reply but there is one mistake:the elements on the two diagonal should be equal.
For example for 2x2 [d e;e d]

In fact, I didn't observe that it is true for every matrix,
that you can make all elements on diagonal to be equal.  :blush:
But you want that also on the second diagonal all elements to be
equal, right?

THanks for your answer.Yes both diagonals.

Hi Gender,
Thanks for your reply. I've thaught about your solution and I need some clarifications:

1-To find :theta: you have two equations and one unknown  :theta:  .Are you sure they reach to one solution.
2- Waht do you mean by accumulation point.How did you prove there exist M such that the diagonals are same.

what is the exact statement of the problem ? I admit that i did not understand what the second diagonals have to be ?

I know it's been long ago that this topic has been "ended", but can anyone explain me what the hell it is about? Proving that every symmetric matrix is orthogonally equivalent to a symmetric matrix with all entries on the main diagonal equal?

  darij

It appears to involve both the main diagonal and whatever you call the diagonal that runs from top-right to bottom-left.  (Main anti-diagonal?)

The Leva1980's result is true (I haven't read its proof):
if $ A\in\mathcal{M}_n(\mathbb{R})$, then $ A$ is orthogonally similar to a matrix wich has  constant diagonal. Now $ A$ is symmetric and we want also to obtain a constant antidiagonal: the result is obvious for $ n\equal{}2$; for $ n\equal{}3$ it seems possible because we are in front of 3 parameters but I haven't really thought about that.

We assume that we work in $ \mathcal{M}_n(\mathbb{R})$.
Question: Let $ A$ be a symmetric matrix; is $ A$ necessarily orthogonally similar to a matrix with constant diagonal and antidiagonal ?
If $ n\equal{}3$ then the answer is no.
Proof: $ A$ must be orthogonally similar to a matrix in the form $ B\equal{}\begin{pmatrix}a&b&a\\b&a&c\\a&c&a\end{pmatrix}$. If $ trace(B)\equal{}0$ then necessarily $ det(B)\equal{}0$, that is a contradiction.     QED.


Prove that:

cos ( cos (x) ) >= sin(x)

for every real x.

The proof I am thinking about is quite elegant.

[hide="this is wrong"]
rewrite this as $\cos{x} \ge \arccos{(\sin{x})}$
write $\sin{x} = y$ for simplicity. Then $\arccos{\sin{x}}$ is $\sqrt{ 1 - y^2}$ if you draw a triangle. this is equivalent to $\sqrt{\cos^2{x}}$ which is obviously always $\le$ to $\cos{x}$

[/hide]

How is $\arccos{\sin{x}} = \sqrt{ 1 - y^2}$ ? I've tried for different x's and I don't think that's true. I think this is true: $\arccos{\sin{x}} = \arcsin\sqrt{ 1 - y^2}$

The problem doesn't make sense in degrees.  It's been posted before, and yes, there is a very neat solution.

You could always expand the Taylor series of both sides :P

[quote="t0rajir0u"]
You could always expand the Taylor series of both sides :P[/quote]

Not very elegant. :(


Construct a right triangle with given hypothenuse c such that the median drawn to the hypothenuse is the geometric mean of the two legs of the triangle.

$l_m=\frac{c}{2}$, and $ab=2S=hc$, so $h=\frac{ab}{c}=\frac{c}{4}$. Thus we know it is a 15-75-90 triangle.

Let $c=AB$. Construct a circle with centre $A$ and radius $c$ and a circle with centre $B$ and radius $c$. Let the intersection of the two circles be $C$. Thus $\triangle{ABC}$ is an equilateral triangle. Now construct the circumcircle of $\triangle{ABC}$. Let the midpoint of $AB$ be $M$. Bisect $BAM$ and let the intersection of the bisector and the circumcircle be $D$. $\triangle{ACD}$ is the desired result.

[quote="shobber"]Thus we know it is a 15-75-90 triangle.[/quote]

I don't think so... :)

$\triangle{ABC}$, $M$ is the midpoint of $AB$, $C=90^o$, $CH \perp AB$.

In right triangle $CMH$, $CM: CH=2: 1$, hence $\angle{CMH}=30^o$, hence $\angle{A}=15^o$.

Is there any problems?  :)

By the way, it's good to see you again, Arne.  :lol:

Oh, seems like I got a bit confused with your notation :)


SQL requests not achieved

DEBUG MODE

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SQL Request: DELETE FROM phpbb_weblog_contributors WHERE weblog_id = 673 AND user_id IN ( )
Line : 261
File : class_blog_options.php

that's what it says.

Yes, this has been posted before, several times (one of which was me). The admins are working on the blog problem.

I can't blame you for wanting to delete anirudh! When my blog was unpopular, I just added whoever I could to my contributors, but now I am overflowed with contributors. I guess I will just have to be more careful next time I add someone.

I didn't even post a single thing :roll:


For this question, I got 2011, around November, but the answer key says
[hide]2012[/hide]

3. According to the U.S. Census Bereau, the world's population passed 6 billion people on July 18 1999. Further, an average of 4.2 births and 1.7 deaths occur every second. If these rates were to remain constant, in what year will the population reach 7 billion?

[hide]after doing the division and blah blah blah.... you get an answer of :
12.6839168.... years....
well... It's in July so there are only like 6 months of days left.... that's half a year or .5 which is less than .6. So you add 1999+12+1=2012!!![/hide]

I must of miscalculated, I got around 12.3


inequalities
function
floor function
number theory unsolved
number theory

For any real $ x$, let $ \lfloor x\rfloor$ be the largest integer that is not more than $ x$. Given a sequence of positive integers $ a_1,a_2,a_3,\ldots$ such that $ a_1>1$ and
\[ \left\lfloor\frac{a_1\plus{}1}{a_2}\right\rfloor\equal{}\left\lfloor\frac{a_2\plus{}1}{a_3}\right\rfloor\equal{}\left\lfloor\frac{a_3\plus{}1}{a_4}\right\rfloor\equal{}\cdots\]
Prove that
\[ \left\lfloor\frac{a_n\plus{}1}{a_{n\plus{}1}}\right\rfloor\leq1\]
holds for every positive integer $ n$.

[hide="Solution"]

Assume the contrary. Then $ \left\lfloor\frac{a_{1}\plus{}1}{a_{2}}\right\rfloor>1$, or $ a_1\plus{}1\ge 2a_2$, $ a_2\plus{}1\ge 2a_3$, ...

So we get a non-increasing sequence of positive integers, but that sequence can't be infinite. The only case is when some two numbers are equal, but then the floor is exactly 1. Contradiction, so the inequality holds.

[/hide]

[hide="My solution"]Assume the contrary, let
\[ \left\lfloor\frac {a_n \plus{} 1}{a_{n \plus{} 1}}\right\rfloor > 1\]
holds for any $ n\in\mathbb{N}$.
We get:
\[ \frac {a_n \plus{} 1}{a_{n \plus{} 1}} > 1\]

\[ a_n \plus{} 1 > a_{n \plus{} 1}\]
Since $ a_n,a_{n \plus{} 1}\in\mathbb{N}$, we can sharpen the inequality into:
\[ a_n \plus{} 1\geq a_{n \plus{} 1} \plus{} 1\]

\[ a_n\geq a_{n \plus{} 1}\]
Since the inequality holds for any $ a_n\in\mathbb{N}$, we get:
\[ a_1\geq a_2\geq a_3\geq\cdots\]
From Infinite Descent, because $ a_n\in\mathbb{N}\forall n\in\mathbb{N}$, then there exist $ k\in\mathbb{N}$ such that $ a_k \equal{} a_{k \plus{} 1} \equal{} \ldots$.
If $ a_k \equal{} a_{k \plus{} 1} \equal{} \ldots > 1$ for a $ m\in\mathbb{N}$, we get:
\[ 1 < \left\lfloor\frac {a_k \plus{} 1}{a_{k \plus{} 1}}\right\rfloor \equal{} \left\lfloor\frac {a_k \plus{} 1}{a_k}\right\rfloor \equal{} 1\]
A contradiction. Hence, $ a_k \equal{} a_{k \plus{} 1} \equal{} \ldots \equal{} 1$.

Lemma:
If $ a_k \equal{} a_{k \plus{} 1} \equal{} \ldots \equal{} 1$, hence $ a_{k \minus{} 1} \equal{} a_k \equal{} 1$.

Proof:
We get:
\[ \left\lfloor\frac {a_{k \minus{} 1} \plus{} 1}{a_k}\right\rfloor \equal{} \left\lfloor\frac {a_k \plus{} 1}{a_{k \plus{} 1}}\right\rfloor \equal{} \left\lfloor\frac {1 \plus{} 1}{1}\right\rfloor \equal{} 2\]
Because $ a_k \equal{} 1$,
\[ \lfloor a_{k \minus{} 1} \plus{} 1\rfloor \equal{} 2\]

\[ a_{k \minus{} 1} \plus{} 1 \equal{} 2\]

\[ a_{k \minus{} 1} \equal{} 1\]
QED

Then, applying the lemma over and over again, we get $ 1 \equal{} a_k \equal{} a_{k \minus{} 1} \equal{} a_{k \minus{} 2} \equal{} \ldots \equal{} a_1 > 1$. A contradiction!

Hence,
\[ \left\lfloor\frac {a_n \plus{} 1}{a_{n \plus{} 1}}\right\rfloor\leq1\]
holds for any $ n\in\mathbb{N}$. QED.[/hide]

[quote="wangsacl"]Assume the contrary, let
\[ \left\lfloor\frac {a_n \plus{} 1}{a_{n \plus{} 1}}\right\rfloor > 1\]
holds for any $ n\in\mathbb{N}$.
We get:
\[ \frac {a_n \plus{} 1}{a_{n \plus{} 1}} > 1\]

\[ a_n \plus{} 1 > a_{n \plus{} 1}\]
Since $ a_n,a_{n \plus{} 1}\in\mathbb{N}$, we can sharpen the inequality into:
\[ a_n \plus{} 1\geq a_{n \plus{} 1} \plus{} 1\]

\[ a_n\geq a_{n \plus{} 1}\]
Since the inequality holds for any $ a_n\in\mathbb{N}$, we get:
\[ a_1\geq a_2\geq a_3\geq\cdots\]
From Infinite Descent, because $ a_n\in\mathbb{N}\forall n\in\mathbb{N}$, then there exist $ k\in\mathbb{N}$ such that $ a_k \equal{} a_{k \plus{} 1} \equal{} \ldots$.
If $ a_k \equal{} a_{k \plus{} 1} \equal{} \ldots > 1$ for a $ m\in\mathbb{N}$, we get:
\[ 1 < \left\lfloor\frac {a_k \plus{} 1}{a_{k \plus{} 1}}\right\rfloor \equal{} \left\lfloor\frac {a_k \plus{} 1}{a_k}\right\rfloor \equal{} 1\]
A contradiction. Hence, $ a_k \equal{} a_{k \plus{} 1} \equal{} \ldots \equal{} 1$.

Lemma:
If $ a_k \equal{} a_{k \plus{} 1} \equal{} \ldots \equal{} 1$, hence $ a_{k \minus{} 1} \equal{} a_k \equal{} 1$.

Proof:
We get:
\[ \left\lfloor\frac {a_{k \minus{} 1} \plus{} 1}{a_k}\right\rfloor \equal{} \left\lfloor\frac {a_k \plus{} 1}{a_{k \plus{} 1}}\right\rfloor \equal{} \left\lfloor\frac {1 \plus{} 1}{1}\right\rfloor \equal{} 2\]
Because $ a_k \equal{} 1$,
\[ \lfloor a_{k \minus{} 1} \plus{} 1\rfloor \equal{} 2\]

\[ a_{k \minus{} 1} \plus{} 1 \equal{} 2\]

\[ a_{k \minus{} 1} \equal{} 1\]
QED

Then, applying the lemma over and over again, we get $ 1 \equal{} a_k \equal{} a_{k \minus{} 1} \equal{} a_{k \minus{} 2} \equal{} \ldots \equal{} a_1 > 1$. A contradiction!

Hence,
\[ \left\lfloor\frac {a_n \plus{} 1}{a_{n \plus{} 1}}\right\rfloor\leq1\]
holds for any $ n\in\mathbb{N}$. QED.[/quote]

You can have a short solution like this:
[hide="My solution"]
Since $ S \equal{}${$ a_{i}$} is a sub-set of $ N*$, we can asume that: $ a_{k}$ is the smallest one
Hence $ \left \lfloor \frac {a_{k} \plus{} 1}{a_{k \plus{} 1}} \right \rfloor \leq 1$ or $ a_{k} \equal{} a_{k \plus{} 1} \equal{} 1$
The second case gives the fact that :$ a_{i} \equal{} 1 \forall i \in N*$( Contradiction!!!)
The first case gives us what we need [/hide]


The earth's galaxy milky way with its '[b] spiral[/b] galaxy[b] '[/b] is about 100,00 light[b] years[/b] across[b] - and up to 5,000 light[/b] years[b] thick[/b] . If assuming that  tachyonic[/b] particle[b] exists-and travels at the speed of 186,212 miles per[/b] fraction[b] of[/b] a[b] second[/b] ,'[b] HOW[/b] '[b] many  years will it takes for a tachyonic[/b] particle[b] to cross both ends of the mikly way galaxy? Please explain.

                                                                                Respectfully,
                                                                                A.Eddington[/b]

[quote="arthur eddington"]travels at the speed of 186,212 miles per[/b] fraction[b] of[/b] a[b] second[/b][/quote]

That would depend on what the fraction is, wouldn't it?

By the way, excessive use of bold is annoying and doesn't add anything to your message.

In fact, Arthur doesn't add anything to the forum ;)

First of all, it's not “inter-galactic travel” if the particle only travels through [i]one[/i] galaxy ([i]intra[/i]-galactic is more accurate).

Second, you haven't specified two things: what a “fraction of a second” refers to, and what course is the particle travelling through (we'll assume it's travelling a distance equal to the diameter of the galaxy).

To solve: Note that a tachyon is a hypothetical type of particle that travels at superluminal speed, so the particle would take, at [i]most[/i], 100,000 years (travelling at the speed of light). Hypothetically speaking, if the particle's speed exceeds that of light, then the particle would be able to travel back in time.

(Now, if, by any chance, the problem suggests that the particle travels at subluminal speed, then the ‘tachyon’ would be, instead, a ‘tardyon,’ or ‘bradyon,’ and we would calculate the time by converting light years into miles, and then using the formula $ time \equal{} \frac {distance}{speed}$.)

The problem in the original post [i]seems[/i] to mean "186,212 miles per second".
Question for the question - time relative to what?  Considering that the particle is moving [i]quite[/i] fast, relativity would need to be taken into account.  Once that is answered the question becomes easy.

I agree with LydianRain , the excessive bold is unnecessary and quite intrusive.

Also, I kind of agree with ZetaX  :( , the original post does not really add to the Astronomy forum.

metafor - Your approach is just a bit off for the above reason.  If the particle was moving slower, your approach to the problem would be more "accurate".

[quote]Your approach is just a bit off for the above reason.[/quote]

Which part? :P

[quote="metafor"]
(Now, if, by any chance, the problem suggests that the particle travels at [b]subluminal[/b] speed, then the ‘tachyon’ would be, instead, a ‘tardyon,’ or ‘bradyon,’ and we would calculate the time by converting light years into miles, and then using the formula $ time \equal{} \frac {distance}{speed}$.)
[/quote]

The use of the "subluminal" is not completely correct, that is what I meant by it being "a bit off".  If something is traveling "subluminally" [i]just[/i] under the speed of light then, relativity needs to be taken into account.  If it is traveling at a very slow speed relative to light the use of $ d\equal{}vt$ will result in a minuscule, unnoticeable  error when compared to using the relativity formulae.

True... thanks for the qualification.  xD

well according to me this is a case pure hypotheticality and of course frame of reference is needed in these questions??????


trigonometry
trig identities
Law of Cosines
geometry
Law of Sines

There is a triangle which is $ a \equal{} 30,b \equal{} 14,C \equal{} m \angle 29^\circ$
So by using the law of cosines we know that $ c \equal{} 19$
Then if we use the law of sine to know $ \angle B$ we would get $ B \equal{} 21.1^\circ$
But if I use it for $ \angle A$ I find it come to $ A \equal{} 50^\circ$
Then $ 21 \plus{} 50 \plus{} 29 \equal{} 100^\circ$
Is there is something wrong with the calculations  :(  or there is some concepts that I don't know  :wink: . Please till me  :)

$ sin130^\circ \equal{} sin50^\circ$

It's the "ambiguous case" of the law of sines and you have to pick the angle that best fits the triangle; in this case, 130.

EDIT:  Yea, geotrig is right.  I never bothered to check the rest of the triangle when I posted my reply :roll:

When you go through the math, there's still going to be the same issue with the angles, so just make sure you know which angle to make obtuse.

Sorry $ a\equal{}30$ it was a typo  :blush:

$ c^2\equal{}a^2\plus{}b^2\minus{}2ab\cos C$
$ c\equal{}\sqrt{14^2\plus{}30^2\minus{}2(14)(30)\cos 29^\circ}$
$ \boxed{c\equal{}19}$

Just use the Law of Cosines for the problem...you should get the right angle measurements!!!


How come when I try to send a private message, the message goes to my outbox but never gets sent?

The other guy isn't checking his inbox. Some participants here don't have PM notification by email turned on, and they forget to look in their inboxes once in a while.

Are the sent messages supposed to go to the "sentbox"?  They aren't appearing there.

They go to "sentbox" when the person reads it.


function
trigonometry
search
real analysis
real analysis unsolved

Given the equation $f(n+1)+f(n-1)= w f(n)$, with $n$ an integer and $w$ a complex number, what are the solutions $f(n)$. $f(n)$ has to be an analytic function.

I found two solutions by trail and error: $f(n) = e^{c n}$ and $f(n) = a n+b$ with $a,b,c$ all complex numbers. Of course combinations are also possible.

1. I'm wondering if there is a general way to solve such an equation? Or is it possible to proof that the two solutions stated above are the only two.
2. Can you call this a discrete differential equation or is there better name for it?
3. Possible new solutions are also welcome.

Because you have no constraints anywhere but the integers, I can do all sorts of wacky things to the function as a real function, e.g adding multiples of $\sin 2\pi n$ or multiplying by $\cos 2\pi n$ or other things.  If you're only interested in the function as a function on integers, however, that equation (called a [i]difference equation[/i]) has a unique solution if you give it any two values of $f$.  Choosing $f(0) = f_{0}$ and $f(1) = f_{1}$ arbitrarily, you can solve the difference equation uniquely for the values $\{f(n)\}_{n \in \bf Z}$.  Any method which works to prove Binet's formula for the Fibonacci numbers will work here -- try a search of the forum, or read the beginning section or two of [url=http://www.math.upenn.edu/~wilf/gfology2.pdf]Generatingfunctionology[/url] (.pdf).

There is a general formula for solutions of linear [url=http://en.wikipedia.org/wiki/Recurrence_relation]recurrence relations[/url] with constant coefficients.

Thanks for the links guys, there are of great help.

The general solution is given in: [url]http://eqworld.ipmnet.ru/en/solutions/fe/fe1203.pdf[/url]
Just put $x=n-1$ and $b=1$ and the equation stated in the link is the same as mine.


[quote="alkachi"]what about final fantasy??[/quote]

8 was awesome and I think it was the best

I have no idea how anyone can keep track of all of them.  Too confusing for me.

I liked 7 and 10 the best... hmm about FF 8, i didnt like it that much because I didnt like the story line that much... maybe there was too much of it i dunno lol. The ending wasnt too spectacular lol Squall, Zell, Rinoa, Quistis, Selphie, and lol that guy with the shotgun ( i forget his name) just run about tryin to get back to their right time lol

I also like 7 and 10 the best. My brother has beaten 10 many times and it's always full of new suprises. The first time he had trouble with the final boss and the last time he beat him barely getting hurt. The ultimate weapons are a challenge to get but it's fun to try. 7 was great even though i've only seen it beaten 1 or 2 times. THe ending was ok.

wow the movie was....interesting....
My favorite Final Fantasy game is 7, its great

[quote="MithsApprentice"]FFX! Man they need to release it for PC. *mutters*[/quote]

Haha. Don't have a PS2?... Yeah, FFX is my favorite. People that don't have PS2's complain about not having the old FF's (7, 8, X) released for the computer.

Yea. Most uncool that is. Guy can't even get a game on his comp these days. What is the world coming to...


Prove that a three  collored  $G_{80}$   contains a monochromatic quadrilateral.

Assuming $G_{80}$ means $K_{80}$, this isn't hard.

Select a random vertex $v$.  This vertex is connected to at least 27 edges of the same color.  Let $H$ be an induced subgraph $G$ consisting of 27 vertices that are connected to $v$ by edges of the same color, say blue.  Note that $H$ cannot have two connected blue edges, since that would form a blue quadrilateral using $v$.  So we can select an induced subgraph $K$ of $H$ with at least 14 vertices, with no blue edges. Now take a random vertex $x$.  $x$ is connected to at least seven vertices by edges of the same color, say red. Let $M$ be the induced subgraph with those seven vertices.  $M$ has no connected red edges, so we can select four vertices connected by only the third color.

Of course this can be improved a lot... For example, I think $K_{6}$ has to have a monochromatic quadrilateral, so we can reduce 80 at least to 32, probably more.


[quote="zaizai-hoang"]Let $ a,b,c,d\ge 0$ such that $ a \plus{} b \plus{} c \plus{} d \equal{} 4$. Find the maximum value of:
\[ V \equal{} a^2b \plus{} b^2c \plus{} c^2d \plus{} d^2a
\]

\[ A \equal{} a^2b \plus{} b^2c \plus{} c^2d \plus{} d^2a \plus{} abcd
\]

\[ N \equal{} a^kb \plus{} b^kc \plus{} c^kd \plus{} d^ka
\]
[/quote]
Along time ago, I solved on this forum the following problem (posted by hungkhtn)
If $ a,b,c,d \ge 0,a\plus{}b\plus{}c\plus{}d\equal{}3$ then
\[ ab(b\plus{}c)\plus{}bc(c\plus{}d)\plus{}cd(d\plus{}a)\plus{}da(a\plus{}b) \le 4\]
Using this problem, we can find the maximum of $ V$ and $ A$ :)

Can, can you post your solution?  :wink: I think the generalization is really very very hard. So I hope someone will prove this one.

I don't understand N... what are the restrictions on k? If there is no upper bound to it we can just set it as large as we want, set, for example, $ (a,b,c,d) \equal{} (3,1,0,0)$ and we have $ N \equal{} 3^k \plus{} 3$ which also has no upper bound.

Or do you want the max expressed in terms of k?

[quote="zaizai-hoang"]Yes, extractly. I want to find out the best constant $ k$ such that the inequality is correct. But I think it is very hard to solve simply by elementary method ... without using Lagrange's multiplier. Anyway, the first thing what I want to see is the maximum of V and A. Who can post solution? Please  :wink:[/quote]
Don't you see my hint?
Lemma. If $ a,b,c,d \ge 0$ then
$ ab(b\plus{}c)\plus{}bc(c\plus{}d)\plus{}cd(d\plus{}a)\plus{}da(a\plus{}b) \le 4\left( \frac{a\plus{}b\plus{}c\plus{}d}{3}\right)^3$
[url=http://www.mathlinks.ro/viewtopic.php?t=147852]Here is my proof[/url]
This inequality holds for all $ a,b,c,d \ge 0$ hence, it also holds for the sequence $ (c,b,a,d)$, that is
$ bc(a\plus{}b)\plus{} ab(a\plus{}d)\plus{}ad(c\plus{}d)\plus{}dc(b\plus{}c) \le  4\left( \frac{a\plus{}b\plus{}c\plus{}d}{3}\right)^3$
$ \Leftrightarrow \sum a^2b\plus{} \sum abc \le \frac{4^4}{3^3}$
Moreover
$ V \le A \le \sum a^2b\plus{} \sum abc$
we get
$ V \le A \le \frac{4^4}{3^3}$
Equality hold for $ a\equal{}2b,c\equal{}d\equal{}0$. :)


Since the points B and P lie on the circle $C_1$, whose center is O, we have OB = OP.

Since M is the midpoint of the segment AB, the segment PM is a median in the triangle PAB; thus, by the well-known formula for the length of a median in a triangle, we have $PM^2=\dfrac{2\cdot\left(PA^2+PB^2\right)-AB^2}{4}$.

Let N be the midpoint of the segment OM. Since the segment OM is a diameter of the circle $C_2$, it thus follows that this point N is the center of the circle $C_2$. Since the line PT is tangent to this circle at the point T, we thus have $PT\perp NT$. Hence, < PTN = 90º, so the triangle PTN is right-angled at T, and hence, by Pythagoras, $PT^2=PN^2-NT^2$.

Now, since N is the midpoint of the segment OM, the segment PN is a median in triangle POM; hence, by the well-known formula for the length of a median in a triangle, we have $PN^2=\dfrac{2\cdot\left(OP^2+PM^2\right)-OM^2}{4}$. On the other hand, since the point T lies on the circle $C_2$, whose center is N and whose radius is $\dfrac{OM}{2}$ (since OM is a diameter of this circle), we have $NT=\dfrac{OM}{2}$.

Finally, the center of a circle always lies on the perpendicular bisector of any chord of the circle; applying this to the circle $C_1$ with its center O and its chord AB, we obtain that the point O lies on the perpendicular bisector of the segment AB; in other words, $OM\perp AB$, since M is the midpoint of the segment AB. Thus, < OMB = 90º, so that the triangle OMB is right-angled at M, and thus Pythagoras yields $OM^2=OB^2-MB^2$. Now, remember that OB = OP; also, since M is the midpoint of the segment AB, we have $MB=\dfrac{AB}{2}$. Thus, $OM^2=OB^2-MB^2=OP^2-\left(\dfrac{AB}{2}\right)^2$.

Altogether, we get

$PT^2=PN^2-NT^2=\dfrac{2\cdot\left(OP^2+PM^2\right)-OM^2}{4}-\left(\dfrac{OM}{2}\right)^2=\dfrac{OP^2+PM^2-OM^2}{2}$
$=\dfrac{OP^2+\dfrac{2\cdot\left(PA^2+PB^2\right)-AB^2}{4}-\left(OP^2-\left(\dfrac{AB}{2}\right)^2\right)}{2}=\dfrac{PA^2+PB^2}{4}$,

so that $PA^2 + PB^2 = 4 \cdot PT^2$, and the problem is solved.

  darij

here is my solution:
let $S$ be the midpoint of $OM$ we have:
$PS^2=\dfrac{1}{2}PM^2+\dfrac{1}{2}PO^2-\dfrac{1}{4}OM^2$ and we have $PM^2=\dfrac{1}{2}PA^2+\dfrac{1}{2}PB^2-\dfrac{1}{4}AB^2$ and we have know that $OM \bot AB \rightarrow BO^2=OM^2+MB^2$ and $MB=\dfrac{1}{2}AB,BO=AO=OP=R$ and with these we R done.(there is only calculate) :D

i think this problem was nice but there was no new idea and there was only calculating...! :D

[quote=Rushil]In a circle $C_1$ with centre $O$, let $AB$ be a chord that is not a diameter. Let $M$ be the midpoint of this chord $AB$. Take a point $T$ on the circle $C_2$ with $OM$ as diameter. Let the tangent to $C_2$ at $T$ meet $C_1$ at  $P$. Show that $PA^2 + PB^2 = 4 \cdot PT^2$.[/quote]
We use vectors.
Let $O'$ be the midpoint of $OM$.
We first translate the figure so that $O$ is the origin. Let $\vec{OA}=\vec{a}$, $\vec{OB}=\vec{b}$, $\vec{OP}=\vec{p}$, $\vec{OT}=\vec{t}$. We can easily see that $\vec{OM}=\frac{\vec{a}+\vec{b}}{2}$ and $\vec{OO'}=\frac{\vec{a}+\vec{b}}{4}$.
Now, we dilate the figure so that the radius of the circle $C_{1}$ is 1, i.e.  $|a|=|b|=|p|=1$. Thus, $O'T \bot PT$ gives us
$$\left(\frac{\vec{a}+\vec{b}}{4}-\vec{t}\right).\left (\vec{p}-\vec{t}  \right)=0 \Leftrightarrow 4|t|^{2}-(\vec{a}+\vec{b}).\vec{t}=4\vec{p}.\vec{t}-\vec{p}.(\vec{a}+\vec{b})$$
Now, it suffices to show $PA^{2}+PB^{2}=4PT^{2}$, i.e.
$$(\vec{p}-\vec{a})^{2}+(\vec{p}-\vec{b})^{2}=4(\vec{p}-\vec{t})^{2} \Leftrightarrow 4-2\vec{p}.(\vec{a}+\vec{b})=4(1+|t|^{2}-2\vec{p}.\vec{t} )$$
$$\Leftrightarrow 2|t|^{2}=4\vec{p}.\vec{t}-\vec{p}.(\vec{a}+\vec{b})=4|t|^{2}-(\vec{a}+\vec{b}).\vec{t}$$
$$\Leftrightarrow 2|t|^{2}=(\vec{a}+\vec{b}).\vec{t} \Leftrightarrow  |t|^{2}=\left (\frac{\vec{a}+\vec{b}}{2} \right).\vec{t}$$
$$\Leftrightarrow |OT|^{2}=\vec{OM}.\vec{OT}=|OM||OT|\cos \angle TOM$$
$$\Leftrightarrow \cos \angle TOM=\frac{|OT|}{|OM|}$$
which is clearly true as $OTM$ is a right triangle.

[hide=Median bash]
Set \(N\) to be the midpoint of \(\overline{OM}\). Then,
\begin{align*}
4PT^2 &= 4PN^2 - OM^2 \\
&= 2OP^2 + 2PM^2 - 2OM^2 \\
&= 2OP^2 - 2\left(OP^2 - \frac{AB^2}{4}\right) + \frac{1}{2}\left(4PM^2\right) \\
&= \frac{AB^2}{2} + \frac{1}{2}(2PA^2 + 2PB^2 - AB^2) \\
&= PA^2 + PB^2
\end{align*}
[/hide]


Solve the eqs:

1) tan{x}*tan[x]=1
2) x^5-[x]-5=0

bye bye!!!

PS: thank you [b]Lagrangia[/b] for the solutions you posted for the other 4 problems! I really needed fast answers!

here is another fast one! :D :D :D

Solution:

for the 1st one : write the eqs like: cos[x]*cos{x}=sin[x]*sin{x} which is equivalent to:
cos([x]+{x})=0 which is cos x=0 => x={pi/2+k*pi; k integer}

cheers!! :D :D

okay...

Solution

for the 2nd one:

for x>=2 we get that x^5>=x^4*[x]>=16[x]=[x]+15[x]>=[x]+30>[x]+5

for x<1 we get that x^5<x^4([x]+1)<=[x]+1<[x]+5.

so the solutions are inside (1,2) so [x]=1 and the eqs becomes: x^5-6=0, d.h. x=6^(1/5)

cheers! :D :D

If am I not wrong the 2nd one was given at a maths contest this year "Dan Barbilian" for the 9th grade!
Or at least a simmilar one! :D :D


function
ceiling function
combinatorics proposed
combinatorics

There are $ 11$ rows of seats and each row has $ 199$ seats. $ n$ students are coming and it is only known that at most $ 39$ students are from the same school. If students from the same school must sit in the same row, determine the maximum number of students such that all the students will be seated.

Generalize this to $ r$ rows and $ s$ seats, with at most $ d$ students from the same school.

If there's exactly one student from each school, then we can fill up all the seats.  :maybe:  Perhaps you meant something else?

Students from different schools can fill up the same row of seats as long as the students in the same school sit in the same row. Is that your concern?

That was my understanding.  But then it is possible that $ n \equal{} 11 \cdot 199$, with exactly one student from each school, and we can always fill up the entire grid in this way.  I'm wondering whether there are some typos or some missing conditions in the statement of the problem to avoid this trivial solution.

For your $ n$, if we consider 25 students from each school (with one more school that is the remainder when $ n$ is divided by 25), then we can only seat 7 schools in each row (not considering the remainder), but $ \frac{11\cdot199}{25\cdot7}>11$.

Is this problem really that hard to understand?  :maybe:

Okay, I had misunderstood the quantifiers.  (This should work for *every* assignment of students to schools.)  Well, the answer is strictly less than 1900 because with 55 groups of size 34 and one group of size 30, we can't do it.  (Try placing the groups of size 34 first: we must have five in each row, so each row has 170 students when we try to add the group of size 30.)  The number 34 is the best possible that we get from a bound of "make all groups the same size" -- specifically, the function $ 11a \cdot \left\lceil \frac{199}{a}\right\rceil \plus{} \left(200 \minus{} a \cdot \left\lceil \frac{199}{a}\right\rceil\right)$ is minimized for $ a \in [1, 39]$ at $ a \equal{} 34$.  Similarly, in the general case we know that the answer must be strictly less than
\[ \min_{a \in [1, d]} ra \cdot  \left\lceil \frac{s}{a}\right\rceil \plus{} \left(s \plus{} 1\minus{} a \cdot \left\lceil \frac{s}{a}\right\rceil\right).\]
But this doesn't tell us about whether this value is really the truth (I would guess not, in general) -- we would still need to show that (in this case) 1899 is always possible, and I don't immediately see an easy way to do that.


the reduction potential values for the foll are:
 Br2 + 2e- → 2Br-              1.06
 O2 + 4H+ + 4e- → 2H2O    1.23

yet, when we add bromine to water, HBr is formed along with HOBr, which means bromine oxidises the water.
This contradicts the reduction potential values.

Am i missing something here?

There is no half rection in your post that shows bromine to be oxidized to HOBr.


geometry
trigonometry
angle bisector
geometry proposed

Let $ABC$ be a triangle. $M_a$ and $M_b$ are midpoints of the sides $a$ and $b$ respectively. $N_a$ and $N_b$ are the points where the incircle of $\triangle ABC$ touches the  sides $a$ and $b$ respectively. Suppose $AN_a \cap BM_b = {K}$,  $AM_a \cap BN_b = {L}$.  If $C,K,L$ are on same line $l$, prove that the line $l$ is the angle bisector of $\angle BCA$

Since K,L,C are collinear, $\frac{\sin\angle ACK}{\sin\angle BCL}=\frac{\sin\angle ACL}{\sin\angle BCL}$, and then apply the trigo form of Ceva. I have not carried out the computations, but I'm sure it works :)

I have another solution:

Lemma: Let $ABC$ be a triangle. $M$ is midpoint of $BC$. $P$ is a point on $[AM]$. $BP$ and $CP$ meets the sides $D$ and $E$. $ED$ is parallel to $BC$.

[hide="Proof"]Proof:
We know from Ceva Theorem:
$\frac {AD}{DC} \cdot \frac {CM}{MB} \cdot \frac {BE}{EA} = 1$
So  $\frac {AE}{EB} = \frac {AD}{DC} $. This proves $ED \| BC$

[/hide]

Let $CK$ meets $AB$ at $D$.
Using LEMMA we get $DN_b \| BC$ and $DN_a \| AB$. And $CN_aDN_b$ is a paralleogram.
And also we know $CN_a = CN_b$.  So $CN_aDN_b$ is a rhombus. This means $CD$ bisects $\angle N_bCN_a$


Even for $x, y, z > 0$, the problem is still wrong!

For example, let $0 < x, y, z < 1$, so they satisfy $x+y+z \leq 3$, but $x^2$, $y^2$, $z^2$, $xyz$ are all less than 1, and their sum is less than 4.

[quote="siuhochung"][quote="ehku"]Let $a^2+b^2+c^2+abc=4$. Prove that
$a+b+c \ge 3 $
Who have a short solution for this problem ?[/quote]
sign reversed, put $a=2cosA$ and so on and then $\sum cosA \le \frac{3}{2}$[/quote]
    Trigonometric proof is the best but this problem also has an algebraic proof  :)
   Put $ A= a+b+c $ ;
 The hypothesis is equivalent to $ (A-2)^2 = (2-a)(2-b)(2-c) \geq 0 $
  This infers $ (A-2)^2 \leq (\frac{6-a}{3})^2 \iff A \leq 3 $  ;)


I made  a mistake typing it the first time.  Here is the right problem.  Thank you.

[img]http://i227.photobucket.com/albums/dd269/joedowdle/prob2a.jpg[/img]

The original problem has a typo.  The top of the RHS should be $ 2n \plus{}1$, not $ 2n\plus{}2$.  Open problems is a completely inappropriate place for class problems -- problems of this difficulty probably should be placed in the PreOlympiad or Intermediate Topics fora.


trigonometry
geometry
ratio
analytic geometry
trig identities
Law of Sines

I have two good questions about angle ina triangle. And I can not solve. Please help. I am a student in middle school.

the second problem i get that x satisfies

$\frac{\sin 68}{\sin x}*\frac{\sin150-x}{\sin 82}=\frac{\sin 78}{\sin 146-x}*\frac{\sin x-68}{\sin 16}$

which is like ewwwww, so says my calc... $x\approx 81.0511264...$

the first one, i suspect, lacks a nice solution as well, the only method i have for that one is heavy analytic geometry, AECF is nicely defined,make EC=1, then length brute force with the law of sines...nasty

explain your second one, that is obviously  incorrect (look at the diagram or plug and chug angles)
I suspect both have nice solutions

oh wow haha, the answer is 4, i guess i messed my law of sines expression, basically use law of sines to express the ratio of the diagonal in two ways


probability
Putnam
geometry
ratio
irrational number
number theory

Here's a problem that I found at summer camp but never solved.

Pick 2 random numbers between 0 and 1. Call them m and n. What is the probability that the closest integer to (m/n) is even?

-Isaac

I think m/n is all positive reals.  Therefore the probability is 1/2?  It seems too simple though....

-interesting_move

I believe this problem was on the Putnam once, but I was unable to find it, so I solved it again. First note that it is possible to scale m and n so that they fit onto the unit square. Now we can find the areas of each little piece (i.e. the ones that make m/n closest to 1, 3, 5, etc.). The toughest case is the one that makes m/n closest to 1. This is a quadrilateral with vertices (0,0), (1, 1/2), (1,1), and (2/3, 1). Its area is (1/2+1/3)/2=5/12. For each odd number n>1, the area of the little triangle is (2/(2n-1)-2/(2n+1))/2. Thus the sum of all the pieces is 5/12+1/2(2/5-2/7+2/9-2/11+...). Furthermore, we know that 1-1/3+1/5-1/7+...=:pi:/4, so we add and subtract 1-1/3 to get 5/12+:pi:/4-2/3=:pi:/4-1/4.

Aah, thanks. I don't completely understand your solution but I get the general idea. Thanks Simon.

-Isaac

ON THE PUTNAM?  Wow, I was too quick to answer.  Maybe it would be 1/2 if m and n were had the range of (0,  :inf: )?

-interesting_move

They do have the range (0,:inf:). The neat thing about this problem is that by tweaking it so slightly from picking a random number a and seeing which integer it is closest to (which isn't a particularly interesting problem), we choose two numbers and see what their ratio is closest to. The point is that we are more likely to get 1 than any other integer this way, and it is especially very unlikely that we will get a really big number, such as 271828 (just a random example). Also whether or not a number is rational or not is rather academic in this case since we are concerned with area, and any irrational number can be approximated as closely as we want by infinitely many rational numbers.

Yes, it was on the Putnam, but since it wasn't on any of the Putnams that were on the AMC website, it was probably from a 1950s or 1960s Putnam, and those may have been easier. Then again, don't be intimidated by Putnam problems. A1 and B1 are usually not too hard, and A2 and B2 are often quite doable also. Furthermore, you may have a strength doing some problem that others find nontrivial, so you may be able to get a problem #4 or 5.

droooooooool.

Simon,  where did u learn that technique.?!

I don't know. All I can tell you is that I solve a lot of problems, and every problem I solve helps me in all aspects of mathematics.

I was thinking about this a bit more. I remembered that there are essentially four types of mathematics: algebra, analysis, number theory, and topology. Note that geometry is not one of them. Geometry is usually classified as analysis. If you remember that, you may remember to use pseudo-analytic processes on geometry problems, such as I did on this one.


Prove that:

$ \sum_{k\equal{}1}^{\frac{p\minus{}1}{2}}\frac{1}{k(p\minus{}k)} \equiv 0 \ \ (mod p)$

[quote="Rust"]$ \sum_{k \equal{} 1}^{\frac {p \minus{} 1}{2}}\frac {1}{k(p \minus{} k)} \equal{} \minus{} \sum_{k \equal{} 1}^{(p \minus{} 1)/2}\frac {1}{k^2} \equal{} \minus{} \frac 12 \sum_{k \equal{} 1}^{p \minus{} 1}\frac {1}{k^2} \equiv 0 \ \ (mod p)$[/quote]

 $ \ni j \equal{}k^{\minus{}1}\equal{}\minus{}(p\minus{}k)^{\minus{}1}$ or  $ j\equal{}\minus{}k^{\minus{}1}\equal{}(p\minus{}k)^{\minus{}1}$ because  $ (\mathbb Z/p\mathbb Z)$ is field

so  $ \sum_{k\equal{}1}^{\frac{p\minus{}1}{2}}\frac{1}{k(p\minus{}k)} \equal{}\sum_{k\equal{}1}^{\frac{p\minus{}1}{2}} j^2 \equal{}kp$


geometry
circumcircle
geometric transformation
dilation
geometry unsolved

Given a convex pentagon $ ABCDE$ with $ DC \equal{} DE$ and $ \angle BCD \equal{} \angle DEA \equal{} \frac {\pi}{2}$
let $ F$ be the point on segment $ AB$ such that $ \frac {AF}{BF} \equal{} \frac {AE}{BC}$

Prove that $ \angle FCE \equal{} \angle FDE$ and $ \angle FEC \equal{} \angle BDC$

May I ask you a favor? Thank you

[quote="77ant"]
Prove that $ \angle FCE \equal{} \angle FDE$ and $ \angle FEC \equal{} \angle BDC$
[/quote]
Dear 77ant, I am sure that you posted incorrectly. My proof (if true) implies $ \angle FCE\equal{} \angle ADE$, not $ \angle FDE$

$ Thanks$, mr.dahn
I think that you are right , not $ \angle FDE$ but $ \angle ADE$  :roll:

[hide="Solution"]
Let $ P$ be the intersection between $ AE$ and $ BC$, let $ Q$ be the intersection between the circumcircles of $ \triangle PAB$ and $ \triangle PEC$ and let $ R$ be the intersection between $ AD$ and $ CF$.  Now, since $ \angle PED + \angle PCD = 90 + 90 = 180$, we have that $ PEDC$ is cyclic, so $ Q$ lies on the circumcircle of $ PEDC$.  Now, $ \angle QEA = \angle QCB$ and $ \angle QAE = 180 - \angle QAP = 180 - \angle QBP = \angle QBC$ since $ Q$ is the intersection between the circumcircles of $ \triangle PAB$ and $ \triangle PEC$.  It follows that $ \triangle QAE\sim \triangle QCB$, so $ \frac {FA}{FB} = \frac {AE}{BC} = \frac {AQ}{QB}$, from which it follows that $ QF$ bisects $ \angle QAB$.  Now, we also have that $ PEDC$ is a kite, so $ PD$ bisects $ \angle EPC$.  Furthermore, since $ \triangle QAE\sim \triangle QBC$, we see that a spiral similarity maps $ \triangle QAB\sim \triangle QEC$, so $ \angle AQB = \angle EQC = \angle EPC$.  Thus,\[{ \angle DQC = \angle DPC = \frac {\angle EPC}{2} = \frac {\angle QAB}{2} = \angle AQF}\]Furthermore, $ \angle QAF = 180 - \angle QPB = \angle QDC$, so $ \triangle QDC\sim \triangle QAF$, so a spiral similiarity maps $ \triangle QAD\sim \triangle QFC$.  Thus, $ \angle RCQ = \angle RDQ$, so $ R$ lies on the circumcircle of $ \triangle QDC$.  Thus, $ R$ lies on the circumcircle of $ \triangle EDC$, so $ \angle RDE = \angle RCE$, from which we conclude that $ \angle FCE = \angle ADE$.  Similarly, $ \angle BDC= \angle FEC$. [/hide]

[quote="thaithuan_GC"]What is a spiral similiarity ?[/quote]
It is a dilation combined with a rotation.  Look here: http://mathworld.wolfram.com/SpiralSimilarity.html

I used it in the following form:
If $ \triangle PAB\sim \triangle PCD$, then $ \triangle PAC\sim \triangle PBD$

This can easily be proven with SAS similarity.


Let $ s$ be any arc of the unit circle lying entirely in teh first quadrant. Let $ A$ be the area of the region lying below $ s$ and above the $ x$-axis and let $ B$ be the area of the region lying to the right of the $ y$-axis and to the left of $ s$. Prove that $ A\plus{}B$ depends only on the arc length, and not on the position of $ s$.

Sorry, I didn't understand. Please, maybe you or someone can present a draw. Thank you.

Look here at page 6, problem 25: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84848&search_id=1493827583&start=100

Thank you, [b]The Quatto Master 6000[/b] ! It is very simply.


Euler
topology
geometry
3D geometry
sphere

A polyhedron (a solid with polygonal faces, not a regular one) has an equal number of quadrilaterals and pentagons as its faces. Exactly three faces meet at each vertex. How many faces does this polyhedron have?

[hide="Hint"]Don't forget, for any polyhedron $F+V-E=2$.[/hide]
[hide="Solution"]We know $F+V-E = 2$ for any polyhedron.

The solid has $n$ quadrilateral faces and $n$ pentagon faces. Altogether it has $2n$ faces, $\frac{4n+5n}{3}= 3n$ vertices, and $\frac{4n+5n}{2}= 4.5n$ edges.

Putting it into the equation above gives

$2n+3n-4.5n = 2$
$0.5n = 2$
$n = 4$

Thus the polyhedron has $8$ faces ($4$ pentagons and $4$ quadrilaterals)[/hide]

That is the correct answer, of course, and that is our intended method of solution.

We had this as question #1 on our test. It was a multiple-choice problem, and there was a "guessing penalty". A number of students said that they would have expected question #1 to be easier. We put this one as #1 because we had already given everyone T-shirts that, in honor of the Year of Euler, had "$V-E+F=2$" written on them. (Questions #2 and #3 were very, very easy.) Our result from this question is that the majority of students left it blank, although of the substantial minority who did answer it, most got it right.

What about non-convex polyhedra? I thought the formula only worked for convex polyhedra. Obviously, this problem was about convex solids... but in general, isn't it possible for that not to be valid?

As I just learned what this was, I'm not really somebody to be explaining it, but to the best of my understanding, the formula works for any polyhedron that is homeomorphic to a sphere (pretend the polyhedron is made of play-doh, if you could mould it into a sphere without tearing or ripping or closing holes up, it is homeomorphic (in the least mathematical way possible  :P).)


algebra
polynomial
number theory open
number theory

proof that the all  numbers $ 1001,1001001,1001001001,....$ are composite.

let $ a_n$ be that sequence.

$ 11| a_1$
and $ 1001|$ all odd members of that sequence.

$ 3|$ all $ a_{6k\plus{}2}$

Still pending to analyze all $ a_{6k\plus{}4}$

$ 1001001001....$(n times 1)$ \equal{}1\plus{}10^3\plus{}10^6\plus{}...\plus{}10^{3(n\minus{}1)}\equal{}\frac{10^{3n}\minus{}1}{10^3\minus{}1}\equal{}\frac{1000^{n}\minus{}1}{3^3.37}$
Let $ n\equal{}p_1^{\alpha_1}.p_2^{\alpha_2}...p_k^{\alpha_k}$. If the statement is true for some of the $ p_i$ (WLOG $ q/(1000^{p_1}\minus{}1)$) then is true for $ n$ (beacause $ q/(1000^{p_1}\minus{}1)/(1000^{n}\minus{}1)$). So it's sufficient to prove that $ 1000^p\minus{}1$(for every prime p) has a prime divisor (different from $ 3,37$)(for example $ 31/1000^5\minus{}1$). But THIS i cant prove! HELP!

[b]Lemma:[/b]  If $ p$ is a prime factor of $ \Phi_n(a)$, then either $ p | n$ or $ p \equiv 1 \bmod n$ where $ \Phi_n(x)$ is a [url=http://mathworld.wolfram.com/CyclotomicPolynomial.html]cyclotomic polynomial[/url].

Because $ \Phi_{p}(10) | 10^p \minus{} 1 | 1000^p \minus{} 1$ there exists a prime factor of $ 1000^p \minus{} 1$ that either divides $ p$ or is congruent to $ 1 \bmod p$.  So the result should be true for all but at most finitely many $ p$ and then it reduces to casework.

[hide]$ 1\plus{}10^3\plus{}10^6\plus{}...\plus{}10^{3n\minus{}3}\equal{}\frac{10^{3n}\minus{}1}{10^3\minus{}1}\equal{}\frac{(10^n\minus{}1)(10^{2n}\plus{}10^n\plus{}1)}{999}$. As $ 10^n\minus{}1$ is always divisible by 9, we merely need to prove that $ 111|(10^{2n}\plus{}10^n\plus{}1)$. In the original equation, if there are n 1s, the number is clearly divisible by 3, so n is not divisible by 3.

We can divide this into two cases where n=3m+1 or n=3m+2. It can be shown that $ 10^{3k}$ is congruent to 1 (mod 111), which can easily be used to show that the two cases result in $ 10^{2n}\plus{}10^n\plus{}1$ is divisible by 111.

Having done that, we have shown that the number can be written as the product of two integers.[/hide]

I don't think that I showed exactly why it can be written as the product of two integers though.

[quote="xpmath"][hide]$ 1 \plus{} 10^3 \plus{} 10^6 \plus{} ... \plus{} 10^{3n \minus{} 3} \equal{} \frac {10^{3n} \minus{} 1}{10^3 \minus{} 1} \equal{} \frac {(10^n \minus{} 1)(10^{2n} \plus{} 10^n \plus{} 1)}{999}$. As $ 10^n \minus{} 1$ is always divisible by 9, we merely need to prove that $ 111|(10^{2n} \plus{} 10^n \plus{} 1)$. In the original equation, if there are n 1s, the number is clearly divisible by 3, so n is not divisible by 3.

We can divide this into two cases where n=3m+1 or n=3m+2. It can be shown that $ 10^{3k}$ is congruent to 1 (mod 111), which can easily be used to show that the two cases result in $ 10^{2n} \plus{} 10^n \plus{} 1$ is divisible by 111.

Having done that, we have shown that the number can be written as the product of two integers.[/hide]

I don't think that I showed exactly why it can be written as the product of two integers though.[/quote]
Didn't you just show that $ \frac {(10^n \minus{} 1)(10^{2n} \plus{} 10^n \plus{} 1)}{999}$ is a natural number?

$ 10^{2n}\plus{}10^n\plus{}1$ is greater than $ 10^n\minus{}1$, which is greater than 999 for all n>3, so we're left with another factor after dividing $ 10^{2n}\plus{}10^n\plus{}1$. Hence, our expression is composite, as long as the number is more than 1001001, which, along with 1001, can easily be verified to be composite.

I don't think I'm saying that correctly or rigorously, sorry.

[quote="xpmath"]$ 10^{2n} \plus{} 10^n \plus{} 1$ is greater than $ 10^n \minus{} 1$, which is greater than 999 for all n>3, so we're left with another factor after dividing $ 10^{2n} \plus{} 10^n \plus{} 1$. Hence, our expression is composite, as long as the number is more than 1001001, which, along with 1001, can easily be verified to be composite.

I don't think I'm saying that correctly or rigorously, sorry.[/quote]
Ahh.. Now I understand you :) Nice solution! :thumbup:


Here's an algebra and a geometry problem (Source: Seton Hall Competition):
1.  April can do a certain task alone in the same number of hours that Charles can do the task alone and in 2/3 of the number of hours that it takes Bob to complete the task alone.  April worked on the task alone for a few hours, then Bob continued alone for 3 times as long as April had worked, and the task was finished by Charles alone in half the time that Brad worked on it.  The entire task CAN be completed by all three working on it only 1/4 hour longer than it took Charlie to finish the task. In how many hours can April and Bob working together complete the entire task?

2.  In triangle ABC, altitude CD is 8 feet long and is perpendicular to side AB at D (with D between A and B).  Line segment CE intersects side AB at E and CE bisects angle ACB and line segment DE is 1 foot long.  If side AB is 448/15 feet long, find the perimeter of triangle ABC.

[hide="Answer for 1"]
Let April be able to do the task in x hours. Thus, she (and Charles) can do the task at a rate of 1/x per hour, and Bob  2/(3x) per hour.

If April worked on the task for y hours, we know that y/x+(6y)/(3x)+(3y)/(2x)=(9y)/(2x)=1. We also know that [(3/2)y+1/4][1/x+1/x+2/(3x)]=1=[(3/2)y+1/4]8/(3x)=(12y+2)/3x.

The equations can be manipulated to become 9y=2x and 3x=12y+2. Solving, we get x=6, y=4/3. Thus, April and Bob have a combined rate of 5/18, so it will take [b]18/5 hours[/b].
[/hide]


algebra
polynomial
search
induction
limit
algebra unsolved

P(x) is a polynomial with rational coefficient. degP>1. The infinity rational series $q_{i}$ satisfy that:
$q_{i}=P(q_{i+1})$
Prove that $q_{i}$ is repeating.(That is, there exists positive integer N s.t. $q_{i}=q_{i+N}$ for all natural number i)

Search for  '' 1990 shortlisted problems'' you will find  your problem for $\deg{P}=3$. And...  :roll:

Or you try to solve it yourself.  :P

(1) there exists $m,n\in\mathbb{Z}\setminus \{0\}$ with $nq_{1}\in\mathbb{Z}$ and $mnP({1\over n}x)\in\mathbb{Z}[x]$ monic (i.e. leading coefficient $1$):
First we can find $\tilde{n}\in\mathbb{N}$ with $\tilde{n}q_{1}\in\mathbb{Z}$ and $\tilde{m}\in\mathbb{N}$ with $Q(x)=\tilde{m}\tilde{n}P({1\over \tilde{n}}x)\in\mathbb{Z}[x]$.
Now if $a\in\mathbb{Z}\setminus \{0\}$ is the leading coefficient of $Q$ and $d=\deg Q=\deg P\geq 2$,
then $a^{d-1}Q({1\over a}x)\in\mathbb{Z}[x]$ is monic, and so we can take $m=a^{d-2}\tilde{m}$ and $n=a\tilde{n}$.

By replacing $P(x)$ with $nP({1\over n}x)$ and $\{q_{i}\}_{i\in\mathbb{N}}$ with $\{nq_{i}\}_{i\in\mathbb{N}}$,
we may assume $q_{1}\in\mathbb{Z}$ and $mP(x)\in\mathbb{Z}[x]$ is monic for some $m\in\mathbb{Z}\setminus \{0\}$.

(2) $q_{i}\in\mathbb{Z}$ for all $i\in\mathbb{N}$:
This follows by induction on $i$.
We have $q_{1}\in\mathbb{Z}$, and for all $q_{i}\in\mathbb{Z}$ we have, that $q_{i+1}$ is
a rational root of $mP(x)-mq_{i}\in\mathbb{Z}[x]$, which is monic.
It follows $q_{i+1}\in\mathbb{Z}$. ($\mathbb{Z}$ is integrally closed in $\mathbb{Q}$)

(3) $\{q_{i}\}_{i\in\mathbb{N}}$ is bounded:
$\lim_{x\rightarrow\pm\infty}\left|{P(x)\over x}\right|=\infty$, as $\deg P>1$.
So there exists some $C>0$ with $|q_{1}|<C$ and
$\left|{P(x)\over x}\right|\geq 1$ for all $|x|\geq C$.
Then for all $|x|\geq C$ we have also $|P(x)|\geq |x|\geq C$, and so
$|q_{1}|<C$ and for all $i\in\mathbb{N}$ we have $|P(q_{i+1})|=|q_{i}|<C\Rightarrow |q_{i+1}|<C$.
So $|q_{i}|<C$ for all $i\in\mathbb{N}$ by induction and $\{q_{i}\}_{i\in\mathbb{N}}$ is bounded.

Because of (2) and (3) $\{q_{1},q_{2},\ldots\}$ is finite.
Take $p\in\mathbb{N}$ minimal with $\{q_{1},\ldots ,q_{p}\}=\{q_{1},q_{2},\ldots\}$.

And now
(4) $q_{n}=q_{n+p}$ for all $n\in\mathbb{N}$:
Let $n\in\mathbb{N}$ be minimal with $q_{n}\not =q_{n+p}$.
Then $\{q_{k},\ldots ,q_{k+p-1}\}=\{q_{k+1},\ldots ,q_{k+p}\}$ for all $1\leq k<n$.
So $q_{n+p}\in\{q_{1},\ldots ,q_{p}\}=\cdots=\{q_{n},\ldots ,q_{n+p-1}\}$
and $q_{n+p}=q_{m}$ for some $n\leq m<n+p$.
But $q_{n}\not =q_{n+p}$ and so $m\geq n+1$ and
$q_{p}=\underbrace{P(P(\cdots P(q_{n+p})\cdots ))}_{n\text{ times}}= \underbrace{P(P(\cdots P(q_{m})\cdots ))}_{n\text{ times}}=q_{m-n}\in\{q_{1},\ldots ,q_{p-1}\}$.
But then $\{q_{1},q_{2},\ldots\}=\{q_{1},\ldots ,q_{p}\}=\{q_{1},\ldots ,q_{p-1}\}$. Contradiction!


pigeonhole principle
combinatorics proposed
combinatorics

A set contains $ 372$ integers from $ 1,2,...,1200$ . For every element $ a\in S$, the numbers $ a\plus{}4,a\plus{}5,a\plus{}9$ don't belong to $ S$. Prove that $ 600\in S$.

The set
$ \{1,2,3,4, 14,15,16,17, 27,28,29,30, 40,41,42,43, ...\}$
is good (contains exactly $ 372$ numbers) and contains $ 600$.

I think this is the only good set, but I don't have the proof. Has anyone?

Yes, your example is in fact the only good set.

We count the number of pairs $ (x, I)$, where $ x\in S$ and $ I$ is an interval of 13 consecutive integers containing $ x$.

Method 1: For any interval $ I\equal{}\{i\plus{}1, i\plus{}2,\dots i\plus{}13\}$, $ S$ can contain at most 1 number from each of

$ (i\plus{}1,i\plus{}5,i\plus{}10)$
$ (i\plus{}2,i\plus{}6,i\plus{}11)$
$ (i\plus{}3,i\plus{}7,i\plus{}12)$
$ (i\plus{}4,i\plus{}8,i\plus{}13)$
So at most 4 numbers from $ I$.

However, not every interval contain exactly 4 numbers from $ I$.   For example, the interval $ (1200, 1201, 1202,\dots 1212)$ can only contain one number from $ S$, and in reality we have $ |I\cap S|\leq\min(|I\cap\{1\dots 1200\}|, 4)$

There are 1212 intervals intersecting $ S$, including 3 "short" intervals on each side whose intersection is less than 4.  Adding up over all these intervals gives that there are at most $ 1\plus{}2\plus{}3\plus{}4*1206\plus{}1\plus{}2\plus{}3\equal{}4836$ pairs.

Method 2: There are 372 integers in $ S$.  Each one lies in 13 intervals.  Thus there are $ 372*13\equal{}4836$ pairs.

This implies that equality must hold in our calculations in step 1, and in particular that each interval must contain exactly $ \min(|I\cap\{1,\dots 1200\}|, 4)$ elements of $ S$.

This implies that $ \{1,2,3,4\}\subset S$ (by looking at the interval $ \{\minus{}8,\minus{}7,\dots 4\}$), and that for $ a\leq 1187$ that $ a\in S$ iff $ a\plus{}13\in S$ (compare the number of elements of $ S$ in $ \{a,a\plus{}1,\dots a\plus{}12\}$ to the number in $ \{a\plus{}1,\dots a\plus{}13\}$).

I will use [i]Dirichlet Principle[/i] when I will inshallah prove your identity. Let 600 is not an element of S. b is the first element of S and so b + 4, b + 5, b + 9 are not. The others between them all are not known that whether or not they are elements. So in 4, we have only one good sample. Then we would have only 300 elements in such manner. But we have 372 elements. Then it is guaranttee that there are 72 elements which are compansed each other. So then we can again have 300/4 = 75 and we have 375. 375 - 372 = 3. So let 600 be the one of these 3. But we cannot have. Proven. It is certainly true, if you're claiming that I'm wrong thus you haven't understood the proof.

[quote="madshiet"]It is certainly true, if you're claiming that I'm wrong thus you haven't understood the proof.[/quote]
Well dang, if Poincare had just said that when he gave his (not correct) proof to his conjecture maybe it wouldn't be a Millennium Problem today.

...

There's two things wrong with that statement
1. You're not infallible. Neither is anyone else.
2. A proof isn't any good if no one understands it.

As far as the second point is concerned...
[quote]But we have 372 elements. Then it is guaranttee that there are 72 elements which are compansed each other. So then we can again have 300/4 = 75[/quote]
I have absolutely no idea what's going on here.


What's wrong with this picture?

If you've already seen this one, don't spoil it.

Hint: [hide]Try actually drawing this on graph paper.[/hide]

Ragingg wrote:--

???

[hide]the 2 triangles can't be arranged on the lattice points like that. 5x2, 8x3 and 13x5 are not similar triangles.[/hide]

[quote="yangyang2000"]The slopes are different...is this yasha from mc2k3?[/quote]

Yes, it is I. Are you coming back next year? I certainly am.

Yang Wang from 2K3? Hi! Sadly I can't go back this year - if you are going, have fun.


abstract algebra
number theory
graph theory
absolute value
combinatorics open
combinatorics

Do there exist two finite graphs such that their [url=http://en.wikipedia.org/wiki/Tensor_product_of_graphs]tensor product[/url] is isomorphic to their [url=http://en.wikipedia.org/wiki/Cartesian_product_of_graphs]Cartesian product[/url]?

(This is really a number theory question in disguise.)

Well, some stupid observations: the graphs need the same chromatic number, since the chromatic number of the Cartesian product is the maximum of the chromatic numbers, and the chromatic number of the tensor product is no larger than the chromatic numbers.  Also, by counting edges we need that $ 2e(G_1) \cdot e(G_2) \equal{} v(G_1) \cdot e(G_2) \plus{} v(G_2) \cdot e(G_1)$, where $ v(\cdot)$ and $ e(\cdot)$ count vertices and edges, respectively.  This suggests something like the product of two odd cycles -- does some simple example like that work?  (I mean, even $ K_3 \times K_3$ is plausible, but I'm not going to try to construct an isomorphism between them by hand :))

It looks like the direct product of odd cycles works, which is interesting - part of the reason I asked this question is that the Cartesian product adds eigenvalues while the tensor product multiplies them, so I thought a positive answer would be unlikely, but $ C_p$ has eigenvalues $ \zeta^k \plus{} \zeta^{ \minus{} k}, k \equal{} 0, 1, ... p \minus{} 1$ and

$ (\zeta^a \plus{} \zeta^{ \minus{} a})(\zeta^b \plus{} \zeta^{ \minus{} b}) \equal{} (\zeta^{a \plus{} b} \plus{} \zeta^{ \minus{} a \minus{} b}) \plus{} (\zeta^{a \minus{} b} \plus{} \zeta^{b \minus{} a})$

so the eigenvalues match if and only if $ (a, b) \mapsto (a \plus{} b, a \minus{} b)$ is invertible in $ \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$, which is the case if and only if $ p$ is odd.  That's a necessary condition, not a sufficient one, but it seems to work for $ p \equal{} 3$ and I believe that it'll work for larger $ p$.  Incidentally, I don't feel like I have a good idea of why the above identity is true (other than the isomorphism of the tensor and Cartesian product), which is the complex version of the product-to-sum identity.

The other reason I asked this question is that there's an obvious choice of infinite graphs for which this is true, which is two copies of the Cayley graph of $ \mathbb{Z}$, and I was wondering if there were finite approximations to this result.  The cycle graphs are natural, so I probably should've looked at those first.

Edit:  Now, of course, my question is whether there are any other examples.  By the above, the graphs $ G_1, G_2$ can't both have eigenvalues with absolute value greater than $ 2$.

Let $ d_G$ and $ d_H$ be the minimum degree in $ G$ and $ H$ respectively, and let $ i_G$ and $ i_H$ denote the number of isolated vertices in each graph.

The Cartesian product has $ i_G i_H$ isolated vertices, while the tensor (if I understand it correctly) has $ i_G |V(H)|\plus{}i_H |V(G)|$.  Comparing, we see that $ I_G\equal{}i_H\equal{}0$.

The Cartesian product has minimum degree $ d_G\plus{}d_H$, while the Tensor has minimum degree $ d_G d_H$.  Comparing (and using our knowledge that both minimum degrees are positive), we see that both $ d_G$ and $ d_H$ are $ 2$.  On the other hand, we know that the average degree is a lower bound on the largest eigenvalue.  So using the eigenvalue observation of t0rajir0u, we see that in fact both graphs must be $ 2\minus{}$regular (a disjoint union of cycles).  I guess it may just be doable by brute force at this point.

Are you assuming that the graphs are undirected?  Sorry, I should've been clearer - I'm allowing both directed edges and multiple edges.


probability
linear algebra
matrix
inequalities
integration
probability and stats

1) Show that if [b]u[/b] is a probability measure, and [b]v [/b]is a probability measure that has density f with respect to[b] u[/b], then:
2 [i][b]d_tv [/b][/i]([b]u[/b],[b]v[/b]) = || f-1||_[i]1[/i] < ||f-1||_[i]2[/i]

where [i][b]d_tv [/b][/i]is the Totaly Variation Distance
||.||_[i]1 [/i]is L1 norm
||.||_[i]2[/i] is L2 norm

2) Let K be a Markov chain on 0, 1, ..., n-1 which at each step jumps to a uniform random site different from its current position. Find the eigenvalues and the spectral gap of K.

I'm not sure about the wording here, does it mean K is a nxn matrix with 0 at the diagonals and 1/(n-1) at all other places????
And to find the eigenvalues, is there other smarter ways or I really have to calculate the determinant directly from K?????

For the first one, you may want to think of $ f\minus{}1$ as $ 1(f\minus{}1)$.  Do you know any inequalities that give an upper bound on the $ L_1$ norm of a product?

For the second one, another way to think about your matrix is $ \frac{1}{n\minus{}1} (J\minus{}I)$, where $ J$ is the matrix of all ones and $ I$ is the identity matrix.

If you know (or are able to figure out) the eigenvalues (with multiplicity) of $ J$, you can use those eigenvalues to get the eigenvalues of $ \frac{1}{n\minus{}1} (J\minus{}I)$.

"think of  f-1 as 1(f-1)" ...do you mean "the Expectation of 1(f-1)"???

"inequalities that give an upper bound..."....no, I don't know.

I mean think of $ f\minus{}1$ as the product of $ f\minus{}1$ and 1.  And the inequality I would use here is the Cauchy-Schwartz inequality:
\[ (\int fg)^2 \leq (\int f^2)(\int g^2)\]
If you replace $ \int$ by $ \sum$, you get a version of the inequality you may be more familiar with, but they're really the same thing.

Thank you for the hint on the eigenvalue question.

But I still don't see how the L1 and L2 question works..\\

The idea is to apply the Cauchy-Schwartz inequality I gave above, with a "nice" choice of $ f$ and $ g$.  We choose $ f(x)\equal{}|x\minus{}1|$, $ g(x)\equal{}1$.  Plugging in to the C-S inequality gives

\[ (\int |f(x)\minus{}1|)^2 \leq (\int |f(x)\minus{}1|^2)(\int 1)\]

The left hand side is the square of the $ L_1$ norm of $ f\minus{}1$, while the right hand side is the square of the $ L_2$ norm of $ f\minus{}1$.


I have attached the question.
Could anyone solve this? If so, please try your best to provide a reason or explain how you got to the conclusion.
Thanks

Chances that someone will prove it here are low cause it seems to be the Collatz conjecture, thus open. For the second part, just "go back".
And please post this in the appropriate forum (this is not combinatorics I think) and don't use attachments in the way you did. Write the text down instead.


How many prime values of p are there such that p^2+17 has four positive different divisors.

Davron

If p>3 then $p^2+17$ have divisors 1,2,3,6,$p^2+17$. Therefore p=2 or p=3.


What is the sum of the number of faces, vertices and edges in a prism with a base having 7 sides?

8sides?
13sides?
10000 sides?
n sides?

[hide]
The number of sides in the base of a prism is related to the number of Faces, vertices, and edges in that shape.

Supposing that a prism's base has $n$ sides,The number of faces in that prism is $2+n$.The number of vertices in that prism is $2n$, located at either end of the shape. And the number of edges in that shape is $3n$.

Therefore, the sum of the number of Faces, vertices, and edges in a prism with a base having $n$ sides is $6n+2$

---
The sums are $44$, $50$, $80$, $60002$, and $6n+2$ respectively.[/hide]

Good job, you were the only replier  :roll:
The formula was the important part of this.


inequalities
trigonometry
algebra unsolved
algebra

we  can easily  infer  that the equality occurs  at  $ \pi/4$.

now the  other cases; i couldnt find  any other better way then  by a case bash:

for  all x  in $ \left(2n\Pi ,2n\Pi \plus{}\frac{\Pi }{4} \right)$ , we  have  the  above reducing  as

$ sinx <a sin(bx\plus{}c)<cosx$

now looking at the RHS  of the inequality , we have

       $ asin(bx\plus{}c)<cosx$

that means  a\$a sin(bx+c)\epsilon $ \left[1,\frac{1}{\sqrt{2}} \right)$

                    we  have  a sinc=1     $ asin(b\Pi /4\plus{}c)\equal{}\frac{1}{\sqrt{2}}$

                 now  again  u can apply  similarly  for (2npi+pi/4 ,2npi+pi/2)            and proceed.

[quote="HTA"]Find all conditions for a,b,c such that :

min{$ sinx,cosx$} $ \leq asin(bx \plus{} c) \leq$ max{$ sinx,cosx$} $ \forall x \in R$[/quote]

Wlog say $ b\geq 0$ and $ c\in[ \minus{} \pi, \plus{} \pi)$
We have the trivial solutions $ \sin(x)$ and $ \cos(x)$

Let's look for non trivial solutions :
$ x \equal{} \frac {\pi}{4} \plus{} k\pi$ $ \implies$ $ a\sin(b\frac {\pi}{4} \plus{} c \plus{} kb\pi )\equal{} \frac {( \minus{} 1)^k\sqrt 2}{2}$ $ \forall k\in\mathbb Z$

So $ a\sin(bx \plus{} c)$ is not a constant (so $ a\neq 0$) and its period $ T$ is an odd divisor of $ 2\pi$ and so $ b \equal{} 2k \plus{} 1$ for some integer $ k$

$ \forall x\in (0,\frac {\pi}{2})$, $ \sin x > 0$ and $ \cos x > 0$ and so $ \frac {T}{2}\geq \frac {\pi}{2}$ and so $ b \equal{} 1$.

So we have $ a\sin(\frac {\pi}{4} \plus{} c) \equal{} \frac {\sqrt 2}{2}$ and, as a consequence $ a\sin(\frac {\pi}{4} \plus{} \pi \plus{} c) \equal{} \minus{} \frac {\sqrt 2}{2}$

The first positive root of $ a\sin(bx \plus{} c)$ need to be in $ (\frac {\pi}{2},\pi)$ (remember we excluded the two trivial solutions) and so $ c\in(0,\frac {\pi}{2})$

So we have the following mandatory constraints for non trivial solutions :
$ b \equal{} 1$
$ c\in(0,\frac {\pi}{2})$
$ a\sin(\frac {\pi}{4} \plus{} c) \equal{} \frac {\sqrt 2}{2}$

In order to verify that these constraints are enough, we just have to say that $ f(x) \equal{} a\sin(x \plus{} c)$ is such that :
The inequality rule is respected for $ x \equal{} 0$ and $ x \equal{} \pi \minus{} c$
$ f(x) \equal{} \sin(x)$ may have at most 2 roots in $ [0,2\pi)$ and we know them.
$ f(x) \equal{} \cos(x)$ may have at most 2 roots in $ [0,2\pi)$ and we know them.
Continuity now allow us to conclude.

[u][b]Synthesis[/b][/u] :

Basic solutions are :
Two trivial one : $ (1,1,0)$ and $ (1,1,\frac {\pi}{2})$
A lot of non trivial : $ \forall c\in(0,\frac {\pi}{2})$ : $ (\frac {\sqrt 2}{2\sin(\frac {\pi}{4} \plus{} c)},1,c)$

And all elaborated solutions built from basic one :
$ (a,b,c)\to(a,b,c \plus{} 2k\pi)$
$ (a,b,c)\to( \minus{} a, \minus{} b, \minus{} c)$
$ (a,b,c)\to( \minus{} a, b, c \plus{} (2k \plus{} 1)\pi)$


The sales tax rate in Bergville is $ 6\%$. During a sale at the
Bergville Coat Closet, the price of a coat is discounted $ 20\%$ from
its $ \$90.00$ price. Two clerks, Jack and Jill, calculate the bill
independently. Jack rings up $ \$90.00$ and adds $ 6\%$ sales tax,
then subtracts $ 20\%$ from this total. Jill rings up $ \$90.00$,
subtracts $ 20\%$ of the price, then adds $ 6\%$ of the discounted price
for sales tax. What is Jack's total minus Jill's total?  Express
your answer to the nearest dollar.

Jack

$ 90.00 \cdot 1.06 \cdot 0.8 \equal{} 76.32$

Jill

$ 90 \cdot 0.8 \cdot 1.06 \equal{} 76.32$

$ 76.32 \minus{} 76.32 \equal{} \boxed{0}$

Aha! When we are doing discounts and taxes, we only multiply by quantities. But multiplication is communative, so it doesn't matter what order the events occur in.


calculus
integration
algebra
function
domain
superior algebra
superior algebra unsolved

when one talks about the characteristic of a group does that imply that the group be
cyclic?
if not what example of group that is not cyclic altough it has some characteristic(not 0)
and why does $char(Q)=0$ (where Q is rational group) ?
can we talk of characteristic of other algebraic structures as well ?

yes , we can talk of the characteristic of some other algebraic structures as well. for example, integral domains and fields. characteristic of a field is the smallest positive integer 'm' such that $1_m=1+1+... m times=0$ . if no such integer exists then characteristic of the field is said to be 0 . for example the field $Z/pZ$ has char=p. it is an easy excersise to prove that a field which has non zero characteristic, has characteristic p, where p is some prime.  i hope this is what you were looking for


logarithms
inequalities
inequalities proposed

Let a,b,c positive numbers such that abc=1. Then

 $ \frac{ab}{1\minus{}c\plus{}c^2}\plus{}\frac{bc}{1\minus{}a\plus{}a^2}\plus{}\frac{ca}{1\minus{}b\plus{}b^2}\ge 3$

[quote="Marius Mainea"]Let a,b,c positive numbers such that abc=1. Then

 $ \frac {ab}{1 \minus{} c \plus{} c^2} \plus{} \frac {bc}{1 \minus{} a \plus{} a^2} \plus{} \frac {ca}{1 \minus{} b \plus{} b^2}\ge 3$[/quote]
$ \sum_{cyc}\frac {bc}{1 \minus{} a \plus{} a^2}\geq3\Leftrightarrow\sum_{cyc}\left(\frac {1}{a(1 \minus{} a \plus{} a^2)}\minus{}1\plus{}2\ln a\right)\geq0.$ :wink:

by power mean,
\[ \frac {\frac {1}{a \minus{} a^ \plus{} a^3} \plus{} \frac {1}{b \minus{} b^ \plus{} b^3} \plus{} \frac {1}{c \minus{} c^ \plus{} c^3}}{3}\geq\ (\frac {a \minus{} a^2 \plus{} a^3 \plus{} b \minus{} b^2 \plus{} b^2 \plus{} c \minus{} c^2 \plus{} c^3}{3})^{ \minus{} 1}\]
i.e.
\[ \frac {\frac {1}{a \minus{} a^ \plus{} a^3} \plus{} \frac {1}{b \minus{} b^ \plus{} b^3} \plus{} \frac {1}{c \minus{} c^ \plus{} c^3}}{3}\geq\ (\frac {a \plus{} b \plus{} c}{3} \minus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{3} \plus{} \frac {a^3 \plus{} b^3 \plus{} c^3}{3})^{ \minus{} 1}\]
by am-gm, each term on RHS is greater than $ 1$.
so equivalently,
\[ \frac {\frac {1}{a \minus{} a^ \plus{} a^3} \plus{} \frac {1}{b \minus{} b^ \plus{} b^3} \plus{} \frac {1}{c \minus{} c^ \plus{} c^3}}{3}\geq\ (1)^{ \minus{} 1}\]
proved.

We can use RCF-Corollary.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216604]see here[/url]

[quote="phymax"]
\[ \frac {\frac {1}{a \minus{} a^ \plus{} a^3} \plus{} \frac {1}{b \minus{} b^ \plus{} b^3} \plus{} \frac {1}{c \minus{} c^ \plus{} c^3}}{3}\geq\ (\frac {a \plus{} b \plus{} c}{3} \minus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{3} \plus{} \frac {a^3 \plus{} b^3 \plus{} c^3}{3})^{ \minus{} 1}\]
by am-gm, each term on RHS is greater than $ 1$.
so equivalently,
\[ \frac {\frac {1}{a \minus{} a^ \plus{} a^3} \plus{} \frac {1}{b \minus{} b^ \plus{} b^3} \plus{} \frac {1}{c \minus{} c^ \plus{} c^3}}{3}\geq\ (1)^{ \minus{} 1}\]
proved.[/quote]

Just because $ \frac {a \plus{} b \plus{} c}{3}, \frac {a^2 \plus{} b^2 \plus{} c^2}{3}, \frac {a^3 \plus{} b^3 \plus{} c^3}{3} \ge 1$ does NOT mean that:

$ \left(\frac {a \plus{} b \plus{} c}{3} \minus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{3} \plus{} \frac {a^3 \plus{} b^3 \plus{} c^3}{3}\right)^{\minus{}1} \ge 1$.

For instance, $ (a,b,c) \equal{} \left(2,2,\frac{1}{4}\right)$ yields $ \left(\frac{17}{12}\minus{}\frac{43}{16}\plus{}\frac{1025}{192}\right)^{\minus{}1} \equal{} \dfrac{192}{781} \not \geq 1$.

Another solution is to expand the original inequality
It follows to use Chebyshev synthetically

$ \sum_{cyc} ab(a^3\plus{}b^3) \ge \frac{2}{3} ( ab\plus{}bc\plus{}ca)(a^3\plus{}b^3\plus{}c^3)$

By the way, Arquady solution is much better in the way of using one variable and analyze its properties.  :)


logarithms
function
algebra unsolved
algebra

A={1，2，3, ...},B={[e^i]}(i=1,2,3, ...)(e is natural logarithm),C=A-B={a(n)}(n=1，2，3, ... ), then please give the formula of each a(n)(n=1，2，3, ...)(use n to express a(n)).

This Problem is from the lecture of Huang Yumin(Chinese should know him) in Jiangsu Province.
It is based on an easier Problem:
   Find the largest integer k, so that k^l-k<n.(l,n are integers)
First, guess k=[(n+n^(1/l))^(1/l)].
Then Prove it, that is not hard.
Then another problem is out:
   A={1,2,3,...}, B={1^l,2^l,3^l,...},C=A-B={a(1),a(2),a(3),...},give the formula of each a(n)(use n to express a(n)).
First guess a(n)=n+[(n+n^(1/l))^(1/l)].When n=k^l-k, a(n)=k^l-1,when n=k^l-k+1, a(n)=k^l+1.The proof is easy.

So, the problem I last posted is out, but unfortunately, Huang Yumin did not give the solution. Who would like to solve it? Tell some views about it is OK even have no idea?

I would be very thankful if anyone posts reply even have no idea.

For small value of $m$ this can be calculated directly.
Note that there are $f(x)=[\ln (x+1)]$ elements from $A$ $\leq x$.
Now $x$ is from $A$ iff $f(x)>f(x-1)$.
Hence $x=a_m$ iff $f(x)=x-m$ and $f(x)=f(x-1)$ hence
$f(x)=f(x-1)=x-m$.
Let $x-m=y$. Then $[\ln(y+m)]=[\ln(y+m+1)]=y$.
The function $g(x)=[ln(x+m)]-x$ is decreasing.
We notice $g([\ln m])\geq 0$ but $g([\ln m]+2)<0$.
Hence $y=[\ln m]$ or $y=[\ln m]+1$.
Now we can have $y=[\ln m]+1$ iff
$[\ln (m+[\ln m])]=[\ln m]+1$ or between $m$ and $[ m+\ln m]$ there is a natural power of $e$, let it be $e^k$.
Hence $m<e^k$ but $m+\ln m>e^k$. We easily see that this happens only for $m \in [e^k-k+1, \cdots e^k]$.
Hence the final answer is:
$m+[\ln m]+1$ for $m \in \bigcup [e^k-k+1, \cdots e^k]$ and
$m+[\ln m]$ otherwise.
I remind you: we analysed only big value of $m$, for example $m>6$.

[quote="iura"]Hence the final answer is:
$m+[\ln m]+1$ for $m \in \bigcup [e^k-k+1, \cdots e^k]$ (1)and
$m+[\ln m]$ otherwise.(2)
I remind you: we analysed only big value of $m$, for example $m>6$.[/quote]

If we combine (1) and (2), I think we could get a(n)=n+[ln(n+1+[ln(n+1)])], is that right?
Otherwise, thanks for your help. :D


function
trigonometry
factorial
trig identities
complex analysis
complex analysis unsolved

Hello, help me please to find the residue of the function $ \frac{z^2}{sin^3(1/z)}$ in $ {\infty}$. It should be the coefficient at $ 1/z$ in the Loran series, but i have some troubles with deciding how many terms of the sin series i should take.

If you need to find the Laurant series of a function like that, you can turn it into a product of functions you know how to work with.  So, find the Laurant (Taylor) series of $ \sin^{3}\frac{1}{z}$.  Assume the Laurant series you desire has the form
$ \cdots+\frac{a_{-2}}{z^{2}}+\frac{a_{-1}}{z}+a_{0}+a_{1}z+a_{2}z^{2}+\cdots$.
Now multiply this by your Laurant series for $ \sin^{3}\frac{1}{z}$.  You know it has to equal $ z^{2}$, so just match coefficients.  Then take coefficient you need, in this case, ${ a_{-1}}$.

Hmmm... I wonder JRav if you can show us the details of your approach :roll:

As to the original question, you want to determine the fraction $ \frac{z^2}{\sin^3\frac 1z}$ near $ \infty$ with an absolute error of order $ \frac{1}{z^2}$ or less, which is relative error $ z^{\minus{}7}$ (the fraction itself is of order $ z^5$). The numerator is exact, so you need to know $ \sin\frac 1z$ with the same relative error $ z^{\minus{}7}$, i.e., you can ignore the powers starting with $ z^{\minus{}8}$ in the [color=red]Laurent[/color] series of $ \sin\frac 1z$ near $ \infty$.

The whole thing is terribly messy anyway (just think of multiplying those factorials with pen and paper and adding the fractions that arise  :o). Do you need it for anything real or is it just a homework assigned by some overzealous TA?

It's a homework. Are there any other approaches here?
P.S who is TA?

The algebra is a little simpler with some trig identities: $ \sin 3t\equal{}3\sin t\minus{}4\sin^3 t$, so $ \sin^3\frac1z\equal{}\frac14\left(3\sin \frac1z\minus{}\sin \frac3z\right)$. That avoids multiplication of series, although we still have to do one series inversion.

[quote="iks i grek"]P.S. who is TA?[/quote]
Sorry, it is the standard American university abbreviation for "Teaching Assistant" (usually a graduate student who runs "practice sessions" for some course).

I've gotten similar questions on this forum before.  Do schools outside the US not have TA's or something or do they just have different titles?  TA's also usually grade the homework assignments and help the professor grade exams.

In Germany, we have "tutors" (give extra lessons for better understanding), trainers (more directly translated, the are called "exercise leaders"; they present and discuss the solutions of the exercise sheets), correctors (correct the exercise sheets). There is also a TA (directly translated called "teaching assistants", too), who primarily does any of the above (varies) and organizes all that.

Just for the record, here in Bosnia & Herzegovina (and rest of former Yugoslav republics, as far as I know), we have 'tutors' (formerly known as 'demonstrators', mostly senior undergraduate or graduate students) who take care of exercises and grading homework, while the 'assistants' (mostly M.Sc. and (rarely) Ph.D.) are in charge of preparing and grading exams, organizing the courses, and/or doing the tutors' job in some cases (obviously, there isn't a big difference in what I stated above and what ZetaX said about the system in Germany.)

So to answer my previous question, it seems like everyone has something similar to a TA, just with a different title.

Indeed. In the Netherlands we generally call them "student assistants" (which could be undergraduates, graduates, or PhD's).

You actually hav, e people with Ph.D's helping?  In the US, TA's are grad students who have yet to earn their doctorate.

Here it's mixed: it's done by both (PhDs and non-PhDs).

[quote="JRav"]You actually hav, e people with Ph.D's helping?[/quote]Yep. In fact, they are obliged to do so: it's part of their contract with the university.

So then they stay with the university even after they earn their degree?

I guess my formulation was a bit confusing, sorry. I meant PhD students (PhD candidates).

Yes, that's what I thought.  In the US, you remain a TA until you get your degree but I don't think anyone stays after they get their Ph.D.


trigonometry
ratio
geometry proposed
geometry

Let $O$ be a point in the plane of a triangle $ABC$. The perpendiculars to the lines $AO$, $BO$, $CO$ at the point $O$ meet the lines $BC$, $CA$, $AB$ in the points $X$, $Y$, $Z$ respectively. Prove that these points $X$, $Y$, $Z$ are collinear.

Suppose the perp. thru O to AO intersects AC and AB at $A_c$ and $A_b$ respectively(a pretty odd notation, isn't it :P ). $B_a,B_c,C_a,C_b$ are defined in a similar manner.
These lines are collinear is equivalent to $\triangle AA_bA_c$ and $\triangle OC_bB_c$ are perspective. Thus it suffices to show that $AO,A_bB_c,A_cC_b$ are concurrent. This clearly follows by Ceva.

I remember Grobber posted a solution to this problem nearly one year ago :?

[quote="prowler"]I think this was already posted before, anyway:
     Let $P$ be one point from the plane of triangle $ABC$. Perpendiculars in $P$ to the lines
$AP,\ BP,\ CP$ meet lines $BC,\ AC,\ AB$ in $X,\ Y,\ Z$ respectively. Prove that $X,\ Y,\  Z$ are collinear.[/quote]
[b]The first method (trigonometric). [/b]From the product of the relations

$\frac{XB}{XC}=\frac {PB}{PC}\cdot \frac{\sin {\widehat {XPB}}}{\sin {\widehat {XPC}}}=\frac{PB}{PC}\cdot \frac{\cos {\widehat {APB}}}{\cos {\widehat {APC}}},\ \frac{YC}{YA}=\frac{PC}{PA}\cdot \frac{\cos {\widehat {BPC}}}{\cos{\widehat{BPA}}}$ $,\ \frac {ZA}{ZB}=\frac {PA}{PB}\cdot \frac{\cos {\widehat {CPA}}}{\cos {\widehat {CPB}}}$

results the relation $\frac{XB}{XC}\cdot \frac{YC}{YA}\cdot \frac {ZA}{ZB}=1,\ i.e.\ Z\in XY.$

[b]The second method (vectorial). [/b] I note $\overline {XB}=m\cdot \overline {XC},\ \overline {YC}=n\cdot \overline {YA},\ \overline {ZA}=n\cdot \overline {ZB}\Longrightarrow$

$\overrightarrow {AB}-m\cdot \overrightarrow {AC}=(1-m)\cdot \overrightarrow {AM},\ \overrightarrow {PB}-m\cdot \overrightarrow {PC}=(1-m)\cdot \overrightarrow {PX}$.

From the relation $\overrightarrow {PA}\cdot \overrightarrow {PX}=0$ results that $\overrightarrow {PA}\cdot (\overrightarrow {PB}-m\cdot \overrightarrow {PC})=0,\ i.e.\ \overrightarrow {PA}\cdot \overrightarrow {PB}=m\cdot \overrightarrow {PA}\cdot \overrightarrow {PC}.$

Similarly, we obtain the relations $\overrightarrow {PB}\cdot \overrightarrow {PC}=n\cdot \overrightarrow {PB}\cdot \overrightarrow {PA},\ \overrightarrow {PC}\cdot \overrightarrow {PA}= p\cdot \overrightarrow {PC}\cdot \overrightarrow {PB}.$

From the product of this relations results that mnp=1, i.e. $Z\in XY.$

[b]The third method (metric).[/b] From the Stewart's relation applied to the line AX in the triangle ABC and to the line PX in the triangle BPC we obtain the relations:

$AB^2\cdot \overline {XC}+AC^2\cdot \overline {BX}=AX^2\cdot \overline {BC}+\overline {XB}\cdot \overline {BX}\cdot \overline {BC}$

$PB^2\cdot \overline {XC}+PC^2\cdot \overline {BX}=PX^2\cdot \overline {BC}+\overline {XC}\cdot \overline {BX}\cdot \overline {BC}$.

$AX^2=PA^2+PX^2\Longrightarrow AB^2\cdot \overline {XC}+AC^2\cdot \overline {BX}=PA^2\cdot \overline {BC}+PB^2\cdot \overline {XC}+PC^2\cdot \overline {BX}$.

I note $\overline {XB}=m\cdot \overline {XC},\ i.e.\ \frac {\overline {XB}}{m}=\frac {\overline {XC}}{1}=\frac {\overline {BC}}{1-m}$. We obtain

$(1-m)\cdot PA^2+PB^2-m\cdot PC^2=c^2-mb^2$.

Similarly, for the points Y and Z. Therefore,

$m=\frac {PA^2+PB^2-c^2}{PA^2+PC^2-b^2},\ n=\frac {PB^2+PC^2-a^2}{PB^2+PA^2-c^2},\ p=\frac {PC^2+PA^2-b^2}{PC^2+PB^2-a^2}$.

From the relation mnp=1 results $Z\in XY$.

For a very elegant proof of this beautiful result by Jacques Hadamard, see the note "Problem: The ortho-intercepts line" on [url=http://de.geocities.com/darij_grinberg]my website[/url].

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=1036 .

  darij

[quote="darij grinberg"]For a very elegant proof of this beautiful result by Jacques Hadamard, see the note "Problem: The ortho-intercepts line" on [url=http://de.geocities.com/darij_grinberg]my website[/url].[/quote]
A truly fantastic and unbelievable proof!
I wonder if this method can be applied to the other nasty problems(here I mean the problems dealing with ratios among segments only).

Nice problem  :wink: , here's my proof. It's very ugly  :(  :o
Let $\frac{\bar{XB}}{\bar{XC}}=\frac{x}{y}$ so $\vec{OX}= y\vec{OB}-x\vec{OC}$. We have:
$OX\perp OA \Leftrightarrow \vec{OX}.\vec{OA}= 0 \Leftrightarrow \vec{OA}(y\vec{OB}-x\vec{OC}) = 0 \\ \Leftrightarrow x\vec{OA}.\vec{OC}= y\vec{OA}.\vec{OB}\Leftrightarrow \frac{x}{y}= \frac{\vec{OA}.\vec{OB}}{\vec{OA}.\vec{OC}}\Leftrightarrow \frac{\bar{XB}}{\bar{XC}}= \frac{\vec{OA}.\vec{OB}}{\vec{OA}.\vec{OC}}$
We prove $\frac{\bar{YC}}{\bar{YA}}= \frac{\vec{OB}.\vec{OC}}{\vec{OB}.\vec{OA}}$ and $\frac{\bar{ZA}}{\bar{ZB}}= \frac{\vec{OC}.\vec{OA}}{\vec{OC}.\vec{OB}}$ similarly. Q.E.D


Hi guys

i really really want to go to the Waterloo seminar, however, i only got 91~93 on euclid, im a grade 11 girl in ontario, is it possible for me to be invited? (i did really bad on fermat)

thank you !

well ur a [b]grade 11[/b] and ur a [b]girl[/b] and u live in [b]ontario[/b] so chances are you will get invited..

lol thanks SM4RT

honestly, that is what i'm counting on as well =(

what do you guys think?

sure, I think you will get invited. I'm a guy, and last year I got only a 90 on euclid, and I still got invited. I think they atcually try and get girls involved more in math contests, so ya. Good Luck!

[quote]
A girl at my school with the same stats as you got invited. I don't know how well she did on Fermat though.
[/quote]
Her name is Ming. she got 91 on Euclid, 134 on Fermat.

I was talking about this with SM4RT the other day... i think if you are a girl in Ontario in grade 11 your chances are really high

[quote="zhuyihan"]hi guys
did MIng get accepted this year? or last year?

i have similar fermat as she did LOL

thanks[/quote]

she got accepted this year..

[quote="Effervescence"]A girl at my school with the same stats as you got invited. I don't know how well she did on Fermat though.[/quote]
Who is coming to Seminar from WCI?

Shalev and Ming? I think that's it.

Although the last time I talked to Ming she said she didn't know if she'll go to Seminar cause of summatives and the cost ($150 I think?)

Make the school pay! :ninja:
My school pays for everything (registration for math contests, seminar, going to this one chess tournament that cost 170 bucks). It's very nice. :)

Hmm.. I asked my school to help fund me the Seminar, and they offered to pay $\$100$ for Seminar... which really helped. So yeah, just ask!

[quote="blootzboy"]i'm a grade 10 boy in ontario who got 138 on fermat, proly around 80 on euclid, and like prolly around 35 on hypatia would my scores be too low for seminar? or is possible i may get invited[/quote]
probably too low :(
[quote="rem"]Make the school pay! Ninja[/quote]
oh i should have... maybe i'll go ask the school with ming

[quote="rem"]Yeah, proly too low. They invite gr 10s iff  they get perfect on cayley/fermat or pwn the olympiads.[/quote]

Well I don't really know much about how they invite, but they invited me based on the Galois... :|

I'm not really sure... my teachers don't really want to check the Waterloo site... but I think that I would have gotten 40 (maybe 1 point off if I like forgot to show a step or something... but I'm not really sure...) :lol:

Ah. We'll just have to wait until the offical results come out.

Well the results were out a long time ago and are available to your teachers, except they still havent posted the results booklet..

hmm pwn olympiads mmm 11 on usamo? :D lol :( aww thx tho

hey guys

i got my euclid mark today--93 =)

here are some of my other marks:

COMC-64
fermat-134 (@_@)
AMC12 --127.5
AIME--6

so, as a gr 11 ontario girl, how likely is it that i'm invited to seminar?

also when would i know for sure?

thanks!

Your chances are ok. and btw they don't look at AMC, AIME, or COMC.

However, I think everyone already got their seminar invitations, but I am not sure.

Approximately what level of math do you guys do at the seminar?

awww
everyone got theirs already?

well i knwo someone from my school is going, but she got perfect on fermat

so only1 person/school?

It depends on the school. In some schools nobody is invited, in some 2 or 3 ppl might get invited.

At Seminar, ppl are divided into 3 groups each of which does a certain level of math. The level is like Intermediate--Pre-Olympiad.

Hmm, no seminar for me this year >< UW is such a lolicon...

[quote="rem"]Interesting, maybe they just decided to invite some grade 10s...
Last yr we had like 4 or 5 grade 10s, a really low number.[/quote]

I don't quite agree with that...I think last year Gr.10s was the majority. Among 55 ppl, there were 15 seniors, 15 intermediate and and [color=red]25[/color] juniors.

Senior -- Gr.12 and below
Intermediate -- Gr.11 and below
Junior -- Gr.10 and below

In other words, if you really pwn in Olympiads, you can be in the senior group as a gr.10. However, you cannot be in a lower group than your grade in school.

In other words...for those 25 people who were in the junior group, they're either gr.9 or gr.10. And as far as I know, gr.9s are very rare.

so the number of gr.10 students should be around 25 (because the number of gr.9s in junior group is canceled out with the number of gr.10s in higher groups)

So...as a result, if you're a girl in gr.10 who lives in Ontario, sometimes you can make the seminar for no reason...like nothing else but a 142 Cayley...

Oh yes, I remember. I kind of did not notice that. :blush:

i didnt get invited...lol...maybe cuz i live in bc...and i mistakenly took cayley instead of fermat...but it overlaps with my school exams anyways... :rotfl:

btw...i luv sushi!!!

I wish I could go there again this year....However I'm not a girl, I'm not gr.11 and below, and altho I did all 3 olympiads i didn't pwn in any of them ><

hey guys
i didnt get invited=(

thx for all ur replies tho =P

i guess i'll never be able to go to waterloo seminar, but i might still have a shot at university of waterloo next year XD

Have they sent out all the invitations by now already?


The sum of 3 numbers $x,y,z$ is 100.  Find such numbers such that $S=xyz$ is a maximum.  You may use any method.

Masoud Zargar

do you mean positive numbers, use am-gm, otherwise, the product can be infinite, 101+k,-1,-k

So would [hide]100/3 100/3 100/3[/hide] be right?

[quote="t0rajir0u"][hide="Positive reals, or otherwise its infinite..."]
$\frac{a+b+c}{3} \ge \sqrt[3]{abc}$
$\left( \frac{100}{3} \right)^3 \ge abc$

Hence the maximum occurs when $a = b = c = \frac{100}{3}$.  This is by far the most straightforward method.  The calculus is ridiculously ugly for such a simple problem.
[/hide][/quote]

yep, just use AM GM

Also, the geometric solution. Since a cube has the most volume per perimeter (of the edges), if all the numbers are the same, the product (volume) will reach a maximum.
Wait... this is pretty much directly AM-GM. Whatever  :P


calculus
derivative
logarithms
calculus computations

To compare the acidity of different solutions, chemists use the pH. The pH is defined in terms of the concentration, x, of hydrogen ions in the solution as

$ pH \equal{} \minus{}\log x$.

Find the rate of change of pH with respect to hydrogen ion concentration when pH is 2.

[hide="I got"]
$ d/dx (\log x) \equal{} 1/x$
$ pH \equal{} \minus{}\log x$
$ (pH)' \equal{} \minus{}1/x$

It doesn't make sense when I plug $ 2$ into $ pH$[/hide]

Sure. You should plug $ x$ instead, to which end you should find it first ;)

Except for one thing ...

That particular use of $ \log,$ in its proper chemical context, is not the natural logarithm. It's the base 10 logarithm.

$ \frac{d}{dx}\log_{10}x\equal{}\frac{d}{dx}\,\frac{\ln x}{\ln 10}\equal{}
\frac1{\ln 10}\,\frac1x.$

So $ (pH)'(x)\equal{}\minus{}\frac1{x\ln 10}.$

For $ pH\equal{}2,$ that would be $ x\equal{}10^{\minus{}2}.$

$ \minus{}\frac1{10^{\minus{}2}\ln 10}\approx \minus{}43.4$ The units for that are in the reciprocal of moles per liter.

Chemists do use base 10 logarithms frequently. The number $ \ln 10$ which appears frequently for that reason is approximately $ 2.303.$

[quote="Jeez123"]The pH is defined in terms of the concentration, x, of hydrogen ions in the solution[/quote]

That's not strictly correct. The present definition of pH is actually an operational definition, where the pH of a solution S, $ pH_S$, is defined as

$ pH_S \equal{} pH_R \plus{} \frac{F(E_S \minus{} E_R)}{RT \ln(10)}$

where $ pH_R$ is the pH value for a reference solution, and F, R, and T have the usual meaning. The $ E_S$ and $ E_R$ are the emf's of the cells composed by a hydrogen electrode dipped in an aqueous solution S or R and a saturated calomel electrod. Even this expression is approximate, since we are considering the junction potentials to be roughly the same.
However, when solving theoretical pH problems we obviously don't use this definition (because it envolves an actual experimental measurement with a cell) but instead the thermodynamic definition, $ pH \equal{} \minus{} \log[a(H^\plus{})]$, where $ a(H^\plus{})$ is the activity of hydrogen ions (in fact, the $ pH_R$ values on the first definition are choosen in order to match as close as possible the values given by the thermodynamic definition). This quantity is related with the molar concentration by the expression $ a(H^\plus{}) \equal{} \gamma(H^\plus{}) \cdot [H^\plus{}]$, where $ \gamma(H^\plus{})$ is the activity coefficient of hydrogen ions. This last quantity can be calculated through the Debye-Hückel equation, which relates the activity coefficient with the ionic strenght of the solution. However, in most applications it is employed the approximation of a reasonably diluted solution, for which the activity coefficient is roughly unity, and so $ a(H^\plus{}) \approx [H^\plus{}]$: this then leads to the usually employed expression $ pH \equal{} \minus{} \log([H^\plus{}])$.


A record makes 1.2 revolutions per second. Find the angular speed.

My work:

Angular speed is the way the turning rate of an engine is described.

1.2 revolutions/seconds times 2pi radians/revolutions

1.2/seconds times 2pi radians

2.4pi radians/second = angular speed of record.

Is this right?  If not, please explain how to find the right answer.