complex numbers
algebra unsolved
algebra

Z is a unit root.PRoof that the real part of Z donnot have the form of (k+1)^1/2 -(K)^1/2

I once want to use trigonometric formulae to do the problem but not sucess

I think he means the following problem (I remember seeing it in some MOSP homework):

Prove that the real part of a root of unity cannot have the form $ \sqrt {k \plus{} 1} \minus{} \sqrt {k}$ for an integer $ k > 1$.

  darij

[color=red][See this : http://www.mathlinks.ro/viewtopic.php?t=113831 -mathmanman][/color]


modular arithmetic
combinatorics open
combinatorics

Determine if it's possible to arrange the numbers 1,1,2,2,...,n,n, such that there are j numbers between two j's, 1<=j<=n, when n=2000, n=2001, and n=2002.

Let $a_1,a_2,\ldots, a_n$ be the position in the sequence of the first $1$, the first $2$, and so on, respectively. We have $1+2+\ldots+2n=2(a_1+\ldots+a_n)+(2+3+\ldots+(n+1))$. This means that $n$ and $\frac{(n+1)(n+2)}2$ have different parities. A quick case analysis reveals that this implies $n\equiv 0$ or $3\pmod 4$, so $2001,2002$ are out.

I think such an arrangement exists for $n=2000$, but I don't really have the time to think about a construction right now.

Tanks Alex! Check this one: Using your idea,

Let's say: a permutation is good if it satisfies the above conditions. In general, there is a good permutation of the numbers 1,1,2,2,...,n,n if and only if n=4k or 4k-1 for some positive integers k.First we show there is a good permutation of numbers 1,1,2,2,...,n,n if n=1 or 2 modulo 4. We aproach indirectly by assuming that a good permutation a_1, a_2,...,a_2n exists. For each number k, let (i_k, j_k),where i_k < j_k. Denote the positions of two occurrences of k.Then  The sum of i_k + the sum j_k =1+2+...+2n=n(2n+1)=S_1 which is odd if n=1 (mod4) and even if n=2(mod 4).
On the other hand j_k - i_k = k+1, so
........The sum of i_k + the sum j_k =2+...+(n+1)=n(n+3)/2 =S_2 which is even if n=1 (mod 4) and odd
if n=2 (mod 4).Obtained 2(the sum of j_k = S_1 + S_2, an even number equal to an odd number, which is impossible...


function
inequalities
number theory
prime numbers
number theory unsolved

Find every positive integer $ n$ such that:

$ 1. \sigma (n) \vdots n\plus{}1$
$ 2. n\minus{}1 \vdots \varphi (n)$

where $ \varphi(n)$ is the Euler's function and $ \sigma (n)$ is the sum of positive divisors of $ n$ including itself.

[quote="Mashimaru"]Find every positive integer $ n$ such that:

$ 1. \sigma (n) \vdots n + 1$
$ 2. n - 1 \vdots \varphi (n)$

where $ \varphi(n)$ is the Euler's function and $ \sigma (n)$ is the sum of positive divisors of $ n$ including itself.[/quote]
Do you mean that n satisfy two conditions ? I will post my solution for this case . The first if $ n > 2$ ,from the condition it is not hard to show that $ n$ must be an odd  square free number  .
Let $ n = \prod_{i = 1}^{r} p_i$. We will show that $ r = 1$ (which means that only prime numbers satisfy these conditions ) .If not ,assume $ r\req 2$ . Two conditions can be  rewritten in the form :
\[ p_1...p_r + 1|(p_1 + 1)...(p_r + 1)
\]

\[ (p_1 - 1)...(p_r - 1)|p_1...p_r-1
\]
Because $ r\geq 2$ so $ 4|p_1...p_r - 1$ ,therefore $ 2\parallel{}p_1...p_r + 1$ . Because all $ p_$ are greater than 2 so we can write $ p_i + 1 = 2t_i$ and $ t_i\geq 1$ .
The condition i means that $ p_1...p_r + 1|2^r.\prod_{i = 1}^rt_i$ .By the fact that $ 2\parallel{}p_1...p_r + 1$ we have $ p_1...p_r + 1|2t_1...t_r$ or $ (2t_1 - 1)...(2t_r - 1)\leq 2t_1...t_r$ .This inequality is equivalent to $ (2 - \frac {1}{t_1})...(2 - \frac {1}{t_r})\leq 2$ .
Because  $ r\geq 2$ and $ t_i\geq 2$ so  $ \prod_{i = 1}^r (2 - \frac {1}{t_i})\geq (\frac {3}{2})^r > 2$
The converse is obvious because all prime numbers satisfy two conditions .
Hence  solutions are all prime number .

[quote="TTsphn"][quote="Mashimaru"]Find every positive integer $ n$ such that:

$ 1. \sigma (n) \vdots n \plus{} 1$
$ 2. n \minus{} 1 \vdots \varphi (n)$

where $ \varphi(n)$ is the Euler's function and $ \sigma (n)$ is the sum of positive divisors of $ n$ including itself.[/quote]
Do you mean that n satisfy two conditions ? I will post my solution for this case . The first if $ n > 2$ ,from the condition it is not hard to show that $ n$ must be an odd  square free number  .
Let $ n \equal{} \prod_{i \equal{} 1}^{r} p_i$.[/quote]
Sorry for my stupidness, but can you explain more clearer about the above property? :maybe:

If $ n$ has a divisor of form $ p^m$ and $ m>1$ then $ p|\varphi(n)$ but $ p\not |n\minus{}1$

Yes, I meant the two statements both have to be satisfied. This is a problem of teacher Đặng Hùng Thắng.

@[b]TTsphn[/b]: Are you Mr. Thọ Tùng? If not, I am sorry. If right, since you are very good at Number Theory and I am very bad at this kind, could you please share with me some exp of yourself :blush: ?

[i]A good solution , Tung . Congratulation :lol: [/i]

[quote="Ronald Widjojo"]What is exactly the meaning of

$ \vdots$

?[/quote]
It means "is divisible by".


calculus
integration
trigonometry
limit
calculus computations

The sequence $\{a_{n}\}$ is defined as follows. \[a_{1}=\frac{\pi}{4},\ a_{n}=\int_{0}^{\frac{1}{2}}(\cos \pi x+a_{n-1})\cos \pi x\ dx\ \ (n=2,3,\cdots)\] Find $\lim_{n\to\infty}a_{n}$.


i type
[color=red]

\documentclass[11pt]{article}
\begin{document}
Hello, world!
\end{document}[/color]

and compile, but when i open  'Hello.dvi' ,  i can only see some quotation marks
the basic page told me that i should see [color=red]Hello, world![/color]

can anyone tell me what the problem is?

[i][ Admin Edit please do not use [size=184]this size[/size] of font when asking support ][/i]

first of all, use normal characters to display your problems.
second of all, I just pasted that code into a new .tex file and compiled it, and I obtained a "hello world!" text on the .dvi file, so I can only assume that either

1) you haven't compiled it correctly
2) you haven't installed miktex correctly

please check for both.

[quote="Valentin Vornicu"]
1) you haven't compiled it correctly
2) you haven't installed miktex correctly

[/quote]

that takes time. i think i would rather do that later, coz my PC is very slow.

i want to ask u a question:
it seems that u are here everywhere (maybe becoz of your face)
how long do u stick online per day??

well, i made the fonts larger becoz i am still not used to read so many small english at a time. i simply didnt want anyone to get tired, esply for your eyes.

writing normally doesn't tire anyone. however, changing the color of the text in red, and making it larger, make tires me (at least). so please, do not do that again.

re-install your latex, and re-read the documentation carefully. these two should do the trick ;)


How many two-digit positive integers $ N$ have the property that the sum of $ N$ and the number obtained by reversing the order of the digits of $ N$ is a perfect square?

$ \textbf{(A)}\ 4\qquad
\textbf{(B)}\ 5\qquad
\textbf{(C)}\ 6\qquad
\textbf{(D)}\ 7\qquad
\textbf{(E)}\ 8$

[hide]If the first number is ab, then the reversal is ba, so the sum is 10a+b+10b+a=11a+11b=11(a+b).  For this to be a perfect square, a+b must equal 11* a perfect square.  But this means it must equal 11, for 11*4=44, which is too large for a sum of digits.  So a+b=11.  It is easy to check that the possibilities for N are 92, 83, 74, 65, 56, 47, 38, 29, so there are 8 possibilities. E.[/hide]

[hide]$10x+y+10y+x=11x+11y$. So $11(x+y)=k^2$ so therefore, $x+y=11$, so you can have $29, 38, 47, 56, 65, 74, 83, 92$.[/hide]

By the way, how do you add problems to the AMC archive thing? Do you just post the problems and give the year, etc, and the moderators do it for you?

Guys, please hide your answers.

As for your question, only specific moderators who were given permissions and administrators are allowed to add to the archives. I don't think you should simply post random problems and ask moderators to put them in the archive.

What I think is a good idea is if you have years below, send the PDF or HTML document to me and I'll do the job. Please send the choices and exact wording if possible.

Years 1990 through 1994
Years 1950 through 1979

[hide]

Let $ N\equal{}10a\plus{}b$, where $ a$ and $ b$ have only one digit. So, the sum of $ N$ and the reverse of $ N$ is $ 10a\plus{}b\plus{}a\plus{}10b\equal{}11a\plus{}11b\equal{}11(a\plus{}b)$. Clearly, this is a square if $ a\plus{}b\equal{}11x^2$ for some integer $ x$. However, if $ x\equal{}2$, $ 4(11)\equal{}44$ is too large to obtain with the sum of two single digit numbers, so $ a\plus{}b\equal{}11$. There are $ 8$ ways of doing this. $ \mathrm{ (E) \ }$ [/hide]

[hide="Solution"]$ N\equal{}\overline{ab}\equal{}10a\plus{}b$.
$ 10a\plus{}b\plus{}10b\plus{}a\equal{}11a\plus{}11b\equal{}11(a\plus{}b)$.
This is a perfect square iff $ a\plus{}b\equal{}11$ or $ a\plus{}b\equal{}11n^2$.
$ a\plus{}b\equal{}11$ has $ 8$ solutions.
That's already $ 8$, so the answer is clearly $ \boxed{\textbf{(E)}\ 8}$.

However, for the sake of completeness...
$ a\plus{}b\equal{}11n^2$ for $ n\in\mathbb{Z}$, $ n>1$.
Suppose $ n\equal{}2$. $ a\plus{}b\equal{}44$ has no integral solutions $ 10\leq\overline{ab}\leq99$. Clearly, for $ n>1$, there are no good solutions, so we are done.[/hide]


group theory
abstract algebra
analytic geometry
vector
algebra
function
domain

Hi all,

I'm an incoming high school senior, and took abstract algebra this past year. Obviously this isn't really general high school curiculum and it was only the second time that the teacher had taught the class. Anyway, I was working on isomorphism and there was something in the book about the complex plane with zero removed (punctured plane) being isomorphic to the group of all complex numbers with a magnitude of one (the unit circle). I have searched and searched for the proof to this and was wondering if anyone here had an idea. I have found other sources saying that they are isomorphic and the book even gave reference to the journal / edition in which the proof was published.

So it seems as though this isomorphism does exist, however elusive that it is. I also worked with my teacher for a few days trying to come up with the proof on our own, and had no such luck. It just doesn't make sense simply because the unit circle has one degree of freedom and the punctured plane has two, so it becomes very complicated trying to figure out how to create a one to one mapping.

Anyway, if anyone can help, it would be much appreciated.

Chris

Some hints.

Let $\mathbb C^{*}=\mathbb C\setminus 0$ and let $U=\{z\in\mathbb C\colon \mid z\mid =1\}$.

1) Prove that $\mathbb C^{*}\cong U\times\mathbb R^{+}$, where $\mathbb R^{+}$ is the group of positive reals with multiplication.

2) Prove that $\mathbb R^{+}\cong(\mathbb R,+)$ (reals with addition).

3) Prove that $U\cong(\mathbb R/\mathbb Z,+)$.

4) Prove that $\mathbb R=\mathbb Q\oplus\Bigl(\bigoplus_{i\in I}A_{i}\Bigr)$, where each subgroup $A_{i}$ is isomorphic to $\mathbb Q$.

5) Prove that $\mathbb R/\mathbb Z\cong\mathbb Q/\mathbb Z\oplus\Bigl(\bigoplus_{i\in I}A_{i}\Bigr)$

6) Use the above results to prove that $\mathbb C^{*}\cong U$.

There's a more simple way.

The homomorphism theorem tells us $S^{1}\cong \mathbb{R}/ \mathbb{Z}$, and the polar coordinates show $\mathbb{C}^{*}\cong \mathbb{R}\times \mathbb{R}/ \mathbb{Z}$. So, our aim is $\mathbb{R}/ \mathbb{Z}\cong \mathbb{R}\times \mathbb{R}/ \mathbb{Z}$. Since $\mathbb{R}$ is a infinite-dimensional vector space over $\mathbb{Q}$, the vector spaces $\mathbb{R}$ and $\mathbb{R}\times \mathbb{R}$ are isomorphic. We can even find an isomorphism $f$ mapping $1$ to $(1,0)$ (basis extension etc.). Then $g : \mathbb{R}/ \mathbb{Z}\to \mathbb{R}\times \mathbb{R}/ \mathbb{Z}, r+\mathbb{Z}\to (y,x+\mathbb{Z}) ,~ f(r)=(x,y)$ is an isomorphism, which can be easiliy checked.

Besides, a similiar proof shows  $\mathbb{R}/ \mathbb{Z}\cong \mathbb{R}\times \mathbb{Q}/ \mathbb{Z}$. Of course, these isomorphisms aren't continuous.

Thank you for the quick response.

It seems to me that lofar's way is pretty much a round about way of  doing -oo-'s way, but anyway, that all makes sense (its 4:30 in the morning and im about to get on a plane, so I am not going to try to work through the proofs at the moment) except that how can you properly make the isomorphism from R to R x R? (sorry, I don't know mathBB) It seems like any mothed that you would use wouldn't be able to produce a 1-1 corrolation.

I have to go, but thanks a lot for your response

Chris

It is well-knwon that $\kappa+\kappa = \kappa$ for infinite cardinals $\kappa$. So, if $V$ is a infinite-dimensional vector space, $\dim(V \times V)=\dim(V)+\dim(V) = \dim(V)$ and hence $V \times V \cong V$.

Sorry my message got cut off, I had to leave for the airport...

This makes sense in proof, but I am trying to understand it in practice. I have never really had a great concept of infinity as it is mostly counter intuitive at first glance. Anyway it seems that if you have a one dimmensional set of real numbers it will have an infite order, and the same is true for a two dimmensional, but if you take any given sample domain there will always be twice as much data in the two dimmensional set as the one. I am just trying to understand how this mapping can take place such that you can say that R is isomorphic to R^2. If this seems a little trivial or whatnot, its fine, don't bother to respond, but if you have the time I would very much appreciate it.

Thanks

Chris


What is the remainder when 9^1995 is divided by 7?

[hide] we can change this into

9^1995 (mod 7) which is

2^1995 (mod7)which is

8^665 (mod 7) which is

1^665 (mod 7) which is

1.[/hide]


linear algebra
matrix
linear algebra unsolved

Here's a really nice problem: what is the maximal number of 1 in a nilpotent order $n$ matrix over the reals?

The answer is n^2-2  times "1" (incredible!)
We take A=[Aij] with Aij=1 except Ann=-n+1 (trace(A)=0) and A1,n=z.
Then A^3=(z+n^2-n-1)A and A^2<>0 for all z.
Thus A is nilpotent iff z=-n^2+n+1.
We cannot do better than n^2-2 because trace(A)=0 and because the precedent "iff".

Nice example loup blanc.

However I understand harazi's question, he asked for "nilpotent order $n$ matrix", [i]i.e.[/i] $A^n=0$, but $A^{n-1}\not=0$.

Then this example does not answer the question :?:

Hi JC_math,
perhaps you are right, but your problem seems more difficult.
Harazi, what is really the question?

Both of them.  ;)  Really, the first one I intended to put on the site, since it was the one I could solve. But indeed, the second one is much more interesting, so why not asking it too?

Now we want also A^(n-1)<>0.
My preceding proof shows that if n=2,3,4 the answers are 2,7,<=14.
Let n=4. One cannot carry out the value 14:
2 cases are considered (with Maple):
a) Exactly one  of the Aij<>1 is on the diagonal. Then if A^n=0 then A^(n-1)=0.
b)  The Aij<>1 are both on the diagonal. Then A is never nilpotent.
Fact: the answer is 13 as the following example shows:
A is a n,n matrix s.t. aij=1 except
Ann=-n+1;A1,n-1=-n^2+2n+1; An-1,n=-n+1.
Then A is nilpotent and A^3<>0.
    Perhaps the general answer is n^2-n+1 ????? (it's true for n=3,4 thus it's true for all n).


geometry
circumcircle
trigonometry
power of a point
radical axis
geometry proposed

[quote="April"]Let $ ABC$ be a triangle and $ A_1$, $ B_1$, $ C_1$ the mid-points of the sides $ BC$, $ CA$, $ AB$, respectively. Let $ H$ be the foot of altitude dropped from $ A$, and $ K$ the second intersection of the circumcircles of triangles $ BHC_1$, $ CHB_1$. Prove that $ AK$ is the $ A$-symmedian line of triangle $ ABC$.[/quote]
Nice problem!
[hide="Solution"]
Notice that
\[ \angle C_1KB_1 \equal{} 360 \minus{} \angle C_1KH \minus{} \angle B_1KH \equal{} 360 \minus{} (180 \minus{} \angle ABC) \minus{} (180 \minus{} \angle ACB) \equal{} \angle ABC \plus{} \angle ACB \equal{} 180 \minus{} \angle B_1AC_1
\]
so it follows that $ C_1KB_1A$ is cyclic.  Now, since $ C_1$ is the midpoint of the hypotenuse of right $ \triangle AHC$, we see that $ HC_1 \equal{} C_1C$.  Similarly, $ HB_1 \equal{} B_1B$.  Now, since $ B_1$, $ C_1$, and $ A_1$ are midpoints of $ AC$, $ AB$, and $ BC$, we have that $ B_1C_1\parallel BC$, $ A_1C_1\parallel AC$, and $ A_1B_1\parallel AB$.  Therefore, $ \angle CKC_1 \equal{} \angle C_1HC \equal{} \angle ACB \equal{} \angle AC_1B_1 \equal{} \angle AKB_1$.  Furthermore, $ AC_1KB_1$ is cyclic, so $ \angle KC_1C \equal{} \angle AB_1K$, so $ \triangle AKB_1\sim \triangle CKC_1$.  Since $ AC_1 \equal{} CC_1$ and $ AB_1 \equal{} BB_1$, we have that\[ \triangle AKB\sim \triangle CKA\implies \frac {AK}{CK} \equal{} \frac {AB}{AC} \equal{} \frac {AB_1}{B_1A_1}\]Furthermore,\[ \angle AB_1A_1 \equal{} 180 \minus{} \angle BAC \equal{} \angle AB_1C_1 \plus{} \angle AC_1B_1 \equal{} \angle AKC_1 \plus{} \angle ACB \equal{} \angle AKC_1 \plus{} \angle C_1HC \equal{} \angle AKC_1 \plus{} \angle C_1KC \equal{} \angle AKC\]Thus, it follows that $ \triangle AB_1A_1\sim \triangle AKC$ and so $ \angle A_1AB \equal{} \angle KAC$, which means that $ AK$ the $ A$-symmedian of $ \triangle ABC$.  [/hide]

Nice solution too. I have a different approach.

Let $ A^{\prime}B^{\prime}C^{\prime}$ be the tangential triangle of $ ABC$ (i.e. $ A^{\prime}$ is the intersection of the tangents at $ B$, and $ C$ to the circumcircle of $ ABC$, and so on). It is known that $ AA^{\prime}$ is the $ A$-symmedian of triangle $ ABC$, and therefore it is suffice to prove that the second intersection $ K$ of circumcircles of triangles $ BHC_{1}$ and $ CHB_{1}$ lies on the line $ AA^{\prime}$.

Consider the inversion $ \Psi$ with pole $ O$, the circumcenter of $ ABC$, with power $ R^{2}$, where $ R$ is the circumradius of triangle $ ABC$. The images of the midpoints $ A_{1}$, $ B_{1}$, $ C_{1}$ are $ A^{\prime}$, $ B^{\prime}$, $ C^{\prime}$, and the image $ H_{i}$ of $ H$ under $ \Psi$ is the second intersection of the circumcircle of triangle $ A^{\prime}BC$ with the circumcircle of $ A^{\prime}B^{\prime}C^{\prime}$. Hence, the circles $ BHC_{1}$ and $ CHB_{1}$ are mapped into the circles $ BH_{i}C^{\prime}$, and $ CH_{i}B^{\prime}$, and since $ H_{i}$ is the Miquel point of the quadrilateral $ BCB^{\prime}C^{\prime}$ (this is because the circumcircles of triangles $ A^{\prime}BC$ and $ A^{\prime}B^{\prime}C^{\prime}$ intersect for the second time at $ H_{i}$), we conclude that the circumcircles of $ BH_{i}C^{\prime}$, and $ CH_{i}B^{\prime}$ pass through the intersection of the lines $ BC$ and $ B^{\prime}C^{\prime}$, i.e. the intersection $ T$ of the tangent at $ A$ to the circumcircle of $ ABC$ with the sideline $ BC$. On other hand, $ T$ is the image of $ K$ under $ \Psi$ (since $ AA^{\prime}$ is mapped into the circle $ AOA_{1}T$, and circle $ A^{\prime}BC$ into the line $ BC$), and therefore $ K$ lies on the line $ AA^{\prime}$. This proves our problem.

Thank you dear [b][size=100]April[/size][/b] for a nice result, new to me.

$ \bullet$ It is easy to show that the points $ A,\ C_{1},\ K,\ O,\ B_{1},$ where $ O$ is the circumcenter of $ \bigtriangleup ABC$ are concyclic and let $ (O_{1})$ be, their circumcircle.

$ ($ From $ OB_{1}\perp AC$ and $ OC_{1}\perp AB$ $ \Longrightarrow$ $ AC_{1}OB_{1}$ is cyclic with diameter $ AO$ and from

$ \angle AB_{1}K \plus{} \angle AC_{1}K \equal{} \angle KHC \plus{} \angle KHB \equal{} 180^{o}$ $ \Longrightarrow$ $ AC_{1}KB_{1}$ is cyclic and so, $ AK\perp KO$ $ ).$

We will prove now that the quadrilateral $ BKOC,$ is also cyclic.

Let be the point $ D\equiv (O_{1})\cap CO$ and because of $ \angle KOD \equal{} \angle KB_{1}D$ $ ,(1)$ it is enough to prove that $ \angle KB_{1}D \equal{} \angle KBC.$

$ \angle KB_{1}D \equal{} \angle KB_{1}A \minus{} \angle AB_{1}D \equal{} \angle KB_{1}A \minus{} \angle ACO \minus{} \angle B_{1}DO$ = $ \angle KB_{1}A \minus{} 2\angle CAO$  $ \Longrightarrow$ $ \angle KB_{1}D \equal{} \angle KB_{1}A \minus{} 2\angle CAO$ $ ,(2)$

$ \angle KBC \equal{} \angle KHC \minus{} \angle BKH \equal{} \angle KB_{1}A \minus{} \angle BC_{1}H \equal{} \angle KB_{1}A \minus{} 2\angle BAH$ $ \Longrightarrow$ $ \angle KBC \equal{} \angle KB_{1}A \minus{} 2\angle BAH$ $ ,(3)$

From $ (2),$ $ (3)$ and because of $ \angle CAO \equal{} \angle BAH,$ we conclude that $ \angle KB_{1}D \equal{} \angle KBC$ $ ,(4)$

From $ (1),$ $ (4)$ $ \Longrightarrow$ $ \angle KOD \equal{} \angle KBC$ and so, we have that $ BKOC$ is cyclic and let $ (O_{2})$ be, its circumcircle.

$ \bullet$ In the configuration now, of the three circles $ (O),\ (O_{1}),\ (O_{2}),$ we have that the tangent line of $ (O)$ at point $ A$ and the line segments $ KO,\ BC,$ as their radical axis, taken per two of them, are concurrent at one point, so be it $ S.$

We denote as $ A',$ the intersection point of the tangent lines of $ (O)$ at points $ B,\ C$ and so, the line segment $ AA',$ is the $ A$-symmedian of $ \bigtriangleup ABC,$ as well and it is enough to prove that $ AK\equiv AA'.$

But it is true, because of the line segment $ AK,$ as the polar of $ S$ with respect to $ (O)$ $ ($ because of $ SA$ tangents to $ (O)$ at $ A$ and $ SO\perp AK$ $ ),$ passes through the point $ A',$ since the line segment $ BC,$ as the polar of $ A'$ wrt $ (O),$ passes through the point $ S.$

Hence, the line segment $ AK$ is the $ A$-symmedian of $ \bigtriangleup ABC$ and the proof is completed.

$ \bullet$ This proof is dedicated to [b][size=100]Dimitris Papadimitriou[/size][/b].

Kostas Vittas.

[quote="April"][color=darkred]Let $ ABC$ be a triangle and denote the midpoints  $ M$ , $ N$ of the sides $ [AB]$ , $ [AC]$ respectively.

Let $ D\in BC$ be the point for which $ AD\perp BC$ . Denote be the second intersection $ P$ between

the circumcircles of $ \triangle BMD$ , $ \triangle CND$ . Prove that the point $ P$ belongs to the $ A$-symmedian in $ \triangle ABC$.[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ S\in AP\cap BC$ . Prove easily that the quadrilateral $ AMPN$ is cyclically (or it is well-known).

Thus, $ \{\begin{array}{ccccccc} \|\begin{array}{c} \widehat {MPB}\equiv\widehat {MDB}\equiv\widehat {ABC} \\
 \\
\widehat {MPA}\equiv\widehat {MNA}\equiv\widehat {ACB}\end{array}\| & \implies & \frac {MB}{MA} = \frac {PB}{PA}\cdot\frac {\sin\widehat {MPB}}{\sin\widehat {MPA}} & \implies & \frac {PB}{PA} = \frac cb \\
 \\
\|\begin{array}{c} \widehat {NPC}\equiv\widehat {NDC}\equiv\widehat {ACB} \\
 \\
\widehat {NPA}\equiv\widehat {NMA}\equiv\widehat {ABC}\end{array}\| & \implies & \frac {NC}{NA} = \frac {PC}{PA}\cdot\frac {\sin\widehat {NPC}}{\sin\widehat {NPA}} & \implies & \frac {PC}{PA} = \frac bc\end{array}\|$ $ \implies$

$ \boxed {\ \frac {PB}{PC} = (\frac cb)^2\ }$ . But $ \{\begin{array}{c} m(\widehat {SPB}) = 180^{\circ} - m(\widehat {BPM}) - m(\widehat {MPA}) = 180^{\circ} - B - C = A \\
 \\
m(\widehat {SPC}) = 180^{\circ} - m(\widehat {CPN}) - m(\widehat {NPA}) = 180^{\circ} - C - B = A\end{array}\|$ $ \implies$

$ \widehat {SPB}\equiv\widehat {SPC}$ . Therefore, $ \frac {SB}{SC} = \frac {PB}{PC}$ $ \implies$ $ \boxed {\ \frac {SB}{SC} = (\frac cb)^2\ }$ , i.e. the line $ AP$ is the $ A$ - symmedian in $ \triangle ABC$ .[/color]

[quote="Virgil Nicula"][color=darkred][b][u]An easy extension.[/u][/b] Let $ ABC$ be a triangle and denote the midpoints  $ M$ , $ N$ of the sides $ [AB]$ , $ [AC]$ respectively.

For a point $ D\in BC$ denote $ \|\begin{array}{c}
m(\widehat {BDM})=x\\\
m(\widehat {CDM})=y\end{array}$ , the second intersection $ P$ between the circumcircles

of $ \triangle BMD$ , $ \triangle CND$ and $ S\in AP\cap BC$ . Prove that $ \frac {SB}{SC}=(\frac cb)^2\cdot \frac {\cot C+\cot x}{\cot B+\cot y}$ .[/color][/quote]

It can be solved with inversion also.
Let $ P$ be the second intersection point of $ AB_1C_1$'s circumcircle and symmedian from $ A$. We will show that $ BHPC_1$ and $ CHPB_1$ are circumscribable. Make an inversion centered at $ A$ with radius $ \sqrt {AB\cdot AC}$. (inversion turnes point $ X$ to $ X'$) Now $ B'$, $ C'$ and $ P'$ are respectively the midpoints of $ AC_1'$, $ AB_1'$ and $ B_1'C_1'$. Also $ P'$ is the second intersection of $ AB'C'$'s circumcircle and the line that passes through $ A$ and $ AB'C'$'s circumcenter. Therefore $ P'$ is the circumcenter of $ AC_1'B_1'$. $ B'H'P'C_1'$ and $ C'H'P'B_1'$ are circumscribable. So are $ BHPC_1$ and $ CHPB_1$.


Ok, well there haven't been any new problems for a little while, so here's some ones that are semi-intermediate (easier) from the Furman Senior test:

1. What is the thousands' digit of 101^29?

2. Four positive integers a, b, c, d are given. There are exactly 4 distinct ways to choose 3 of a, b, c, d. The mean of each of the 4 possible triples is added to the fourth integers, and the four sums 29, 23, 21, 17 are obtained. What are the original integers?

3. Sin(a)=2/3 and Sin(b)=3/4 and a is in Quadrant II and B is in Quadrant IV, then sin(a - b) = ?

4. The four digit number 2pqr is multiplied by four and the result is the 4 digit number rqp2. It follows that p+q is...

[hide] 3. since sin(a-b) = sin(a+(-b)) we can use the identity to get:

sin(a)cos(-b) + sin(-b)cos(a) and because cos(-x) = cos(x) and sin(-x) = -sin(x) we get:

sin(a)cos(b) + -sin(b)cos(a) = (2/3)( :sqrt: (7)/4) - (3/4)( :rt5: /3) =  :sqrt: (7)/6 -  :rt5: /4.[/hide]

[hide] 4.

First, we know that r = 8, because 9*4 (mod 10) is not 2.

x is the floor value function.

So, we set up these equations for p and q. (4q+3)(mod 10) = p and 4p + [[(4q+3)/10]] = q.

Solving the second equation for p we get p = q - [[(4q+3)/10]]. Plug back into first equation and get:

(4q+3)(mod 10) = q - [[(4q+3)/10]]

By the "Theory of Guess-and-Check"  we find that 7 works. Plug it back into the first equation we get p = 1.

Therefore, p+q = 8.[/hide]

1. [hide]i think the hundredth grows by 1 each time, thus 101^29 ends with 2901. thousands digit is 2.[/hide]

2. [hide]adding up the means of 3 and the last, we get each integer twice , thus their sum is 45. multiplying each equation by 3 then derving everything is terms of 'a' for example, we get a=21 (if a is biggest). then b=12, c=9 and b=3.[/hide]

1. I disagree - [hide]thousands digit is always 0?[/hide]

Update: I'm an idiot. This is just further reaffirmation... I should've started getting suspicious when it started getting asymmetric at the 5th power.

hmm no try 101^12 ends in [hide]1201, use comp calculator if you don't belive me.[/hide]

Ml2=1 and 2 are right.

Paladin8: You got #4 right (Nice solution, btw. I did it differently.)
I don't think you got number three right...well maybe you did...could you rationalize that please?

well its rationalized but on a common denominator it would be:

[hide] (4 :sqrt:7 - 6 :rt5:)/24 [/hide]

That's incorrect. Try drawing triangles in those quadrants and see what you get. Besides, I think you used the wrong identity...it's sin(a - b) = sin(a)cos(b) - sin(b)cos(a).

MagnusRex wrote:3. Sin(a)=2/3 and Sin(b)=3/4 and a is in Quadrant II and B is in Quadrant IV, then sin(a - b) = ?

Perhaps you meant sin(b) = -3/4? If b is in Quad IV, sines are negative.

Anyhow I got [hide]cos(a) = - :rt5: /3, cos(b) =  :sqrt:7/4 .

sin(a-b) = (2 :sqrt:7 - 3 :rt5: ) / 12 [/hide]

ACK! I'm so sorry about that one. Here was what the problem was supposed to be:

3. Sin(a)=2/3 and cos(b)=3/4 and a is in Quadrant II and b is in Quadrant IV, then sin(a - b) = ?

But you don't have to solve it if you don't want to...


It is pretty well known that if you have a complex root of a polynomial with real coefficients that the conjugate is also a root.  i.e.  if f(a+bi)=0 then f(a-bi)=0.

I had never seen a short easy proof of it.

It's a nice problem, but if you just want to see the proof, here is a sketch:

[hide]

Let f(a+bi)= a-bi

Show that f(a+bi) + f(c+di) = f(a+c +(b+d)i)

Show that f(a+bi) * f(c+di) = f((a+bi) * (c+di))

Now prove that for any polynomial g with real coefficients

f(g(a+bi)) = g(f(a+bi)).

Then we get the desired result as a corollary.

[/hide]

--Dan

Is this right?

[hide]The following are true of all complex numbers and their conjugates:

c * (a + bi) is the conjugate of c * (a - bi)

(a + bi) + (c + di) is the conjugate of (a - bi) + (c - di)

(a + bi)^n is the conjugate of (a - bi)^n

Now consider two complex numbers w and z such that w is the conjugate of z.

From the first and third statements we know that c *z^n is the conjugate of c * w^n.

Now using the second we know that c_0 * z^n + c_1 * z^(n-1) + ... + c_n * z^0 is the conjugate of c_0 * w^n + c_1 * w^(n-1) + ... + c_n * w^0.

This means that P(a + bi) is the conjugate of P(a - bi) so if one is zero then the other must be.[/hide]

I assume you mean that c is real in the first statement. If that is the case, it looks good!


function
inequalities
algebra
domain
combinatorics proposed
combinatorics

Let $\mathcal{F}$ be the set of all the functions $f : \mathcal{P}(S) \longrightarrow \mathbb{R}$ such that for all $X, Y \subseteq S$, we have $f(X \cap Y) = \min (f(X), f(Y))$, where $S$ is a finite set (and $\mathcal{P}(S)$ is the set of its subsets). Find
\[\max_{f \in \mathcal{F}}| \textrm{Im}(f) |. \]

[quote="Valentin Vornicu"]Let $\mathcal{F}$ be the set of all the functions $f : \mathcal{P}(S) \longrightarrow \mathbb{R}$ such that for all $X, Y \subseteq S$, we have $f(X \cap Y) = \min (f(X), f(Y))$, where $S$ is a finite set (and $\mathcal{P}(S)$ is the set of its partitions). Find
\[\max_{f \in \mathcal{F}}| \textrm{Im}(f) |. \]
[/quote]

Sorry, your terminology is confusing.

Is $\mathcal{P}(S)$ the power set of $S$? i.e the set of all subsets of $S$?

If  $\mathcal{P}(S)$ is the set of partitions of $S$ and If $X \subseteq S$ then $X \not\in \mathcal{P}(S)$, so talking about $f(X)$ would not make sense.

$P(S)$ is indeed the set of subsets of $S$, and yes, you are right, it is a typo in the enuntiation, $X, Y \in P(S)$, instead of $X, Y \in S$.

${\mathcal P}(S)$ is the powerset of $S$, i.e. the set of its subsets, but he [i]does[/i] mean $X, Y \subseteq S$ (which is equivalent to $X, Y \in{\mathcal P}(S)$), not what pohoatza said.

[hide="Answer but no proof"]If $S$ has $n$ elements, then we can always achieve $n+1$ values, as follows: choose any chain $\emptyset = X_{0}\subset X_{1}\subset \ldots \subset X_{n}= S$.  Then define $f(Y) = \max_{i}(X_{i}\subseteq Y)$.  Then for any $Y, Z$,
\begin{eqnarray*}f(Y \cap Z) &=& \max_{i}(X_{i}\subseteq(Y\cap Z)) \\ &=& \max_{i}(X_{i}\subseteq Y\textrm{ and }X_{i}\subseteq Z) \\ &=& \min\{\max_{i}(X_{i}\subseteq Y),\max_{i}(X_{i}\subseteq Z)\}\\ &=& \min\{f(Y), f(Z)\}, \end{eqnarray*}
as desired.[/hide]

Yes JBL, I meant $X,Y \in P(S)$, anyway your answer is correct.

Here is the proof we cannot do better:

[hide]Let $|S| = n$, and consider the values on the $n-1$-element subsets of $S$, $X_{1}, X_{2}, \ldots, X_{n}$.  Choose any set $X \subset S$, $|X| = n-i$, and without loss of generality let $X_{1}, X_{2}, \ldots, X_{i}$ be the $n-1$-element supersets of $X$ in $S$.  Also without loss of generality, assume $f(X_{1}) \leq f(X_{2}) \leq\ldots\leq f(X_{i})$.  Then
\[f(X) = f(X_{i}\cap(X_{i-1}\cap\ldots(X_{2}\cap X_{1}))) = f(X_{i-1}\cap\ldots(X_{2}\cap X_{1})) = \ldots = f(X_{2}\cap X_{1}) = f(X_{1}), \]
where the leftmost equality follows from the equality of the sets involved while the other inequalities follow from right to left by the given property of $f$.  Thus, the values of $f$ on all sets other than $S$ itself are among these $n$ values, so $f$ can take at most $n+1$ different values.[/hide]

i think the value for the question is  positive infinte. As i haven't find the way send my prove, so pls wait me..

The domain of $f$ is finite, so I'm not sure how you expect to make its range infinite.  In any event, I have already posted a complete proof.


$a_{4}=8$, so from there on vishalarul is not correct.

[hide="Hint"] The problem is the same as showing that $a_{n+4}\cdot a_{n+5}\equiv{_{a_{n+3}}}7$   $\forall n \in \mathbb{N}$.[/hide]

Would you mind showing us how please? :)

[hide="hint that gives answer"]
mod 7
[/hide]

[hide="answer"]
Obviously the first three terms are 1 mod 7, evaluating the equation for $a_{n+3}$ in mod 7 gives $\frac{1 \cdot 1+7}{1}\equiv 1 \pmod 7$, therefore all terms of the sequence of 1 mod 7, which make all of them integers by the prior congruence.
[/hide]

[quote="deej21"]
[hide="answer"]
Obviously the first three terms are 1 mod 7, evaluating the equation for $a_{n+3}$ in mod 7 gives $\frac{1 \cdot 1+7}{1}\equiv 1 \pmod 7$, therefore all terms of the sequence of 1 mod 7, which make all of them integers by the prior congruence.
[/hide][/quote]

Your logic is incorrect.  If we have two numbers $a,b\equiv 1\pmod{7}$, $b$ does not necessarily divide $a$.  A simple counterexample is $a=15$, $b=8$.

What about proving that $a_{n+4}= 16 a_{n+2}-a_{n}$ for all $n$? This is an easy exercise (induction).

But this alternative definition for the sequence shows immediately that all terms are integers.


Given that $ p$ is prime, prove that $ _{p}C_j$ is divisible by $ p$ where $ j\in [1,p\minus{}1]$.

Because $ \binom{p}{j}\equal{}\frac{p!}{j!(p\minus{}i)!}\in\mathbb{N}$ and $ p\not |j!(p\minus{}j)!,p|p!$ we're done.

[quote="countdownmath"]Given that $ p$ is prime, prove that $ _{p}C_j$ is divisible by $ p$ where $ j\in [1,p \minus{} 1]$.[/quote]
The following is also true:
$ \binom{kp}{j}$ is divisible by $ p$ when $ k$ is an integer, $ gcd(j,p)\equal{}1$ and of course you define $ j<kp$.


Compute all functions $f: \mathbb R \to \mathbb R$ such that $ f(0) \equal{} 223$ and $ f(f(x)) \equal{} 3x \plus{} 1$ for all real $x$.

$ f(0) \equal{} 223$. Going back to edit.

[quote="darkdieuguerre"]Compute the function $ f(x)$ such that $ f(0) \equal{} 223$ and $ f(f(x)) \equal{} 3x \plus{} 1$.

(Original post incorrectly asserted $ f(0) \equal{} 1$.)[/quote]

Please, try to copy correctly your ten words problems !!!!!
This problem cant be a correct problem found in a real contest / book  / class.
Some informations are missing :

Is $ f(x)$ a function from $ \mathbb R\to\mathbb R$?
Is $ f(x)$ a function from $ \mathbb N\to\mathbb R$?
Is $ f(x)$ a function from $ \mathbb Z\to\mathbb R$?
...
For example, there exists infinitely many such functions from $ \mathbb R\to\mathbb R$, even continuous... .


integration
advanced fields
advanced fields unsolved

hi!
I solved this (simple?) problem, but unfortenetly my answer is'n corect (?) so I'm hopping someone can check it

$y'(t)+4y(t)+5\int^{t}_{0}y(u)\,du=e^{-}t;$ $y(0)=0; t\ge0$
[b][i]since[/i][/b]
$\int^{t}_{0}y(u)\,du=\frac{F(s)}{s}=\frac{sY(s)-y(0)}{s}$
[b][i]we have[/i][/b]
$sY(s)-y(0)+4Y(s)+5\frac{sY(s)-y(0)}{s}=\frac{1}{s+1}$
$sY(s)-y(0)+4Y(s)+5(\frac{sY(s)}{s}-\frac{y(0)}{s})=\frac{1}{s+1}$
[b][i]inserting value[/i][/b] $y(0)=0$
$sY(s)-0+4Y(s)+5(\frac{sY(s)}{s}-\frac{0}{s})=\frac{1}{s+1}$
$sY(s)+4Y(s)+5\frac{sY(s)}{s}=\frac{1}{s+1}$
$(s+9)Y(s)=\frac{1}{s+1}$
$Y(s)=\frac{1}{(s+1)(s+9)}$
[b][i]inverse laplace *can't find commad for it...[/i][/b]
$y(t)=-\frac{e^{-9t}-e^{-t}}{8}$
[b][i]i guess this is my misstake. but why can't I do so?becouse of [/i][/b]$y(0)=0$
$5\frac{sY(s)-y(0)}{s}=5(\frac{sY(s)}{s}-\frac{y(0)}{s})$

Er.  Forgive me if this problem looks simpler than you make it out to be.

Let $g(t) = \int_{0}^{t}y(u) \, du$.  Then the equation is second-order in $g$.  Specifically,

$g''+4g'+5g = e^{-t}$

The homogeneous solution is $g = Ae^{t}+Be^{4t}$ (by characteristic equation).  The nonhomogeneous solution is of the form $g = Ce^{-t}$, giving us

$(C-4C+5C) e^{-t}= e^{-t}\Leftrightarrow C = \frac{1}{2}$

Hence our solution is

$g = Ae^{t}+Be^{4t}+\frac{1}{2}e^{-t}$

Since $g(0) = \int_{0}^{0}y(u) du = 0$ we have

$A+B =-\frac{1}{2}$

Differentiating,

$y = Ae^{t}+4Be^{4t}-\frac{1}{2}e^{-t}$

And $y(0) = 0$ gives

$A+4B = \frac{1}{2}$
$3B = 1$
$B = \frac{1}{3}$
$A =-\frac{5}{6}$

So our final answer is

$\boxed{ y =-\frac{5}{6}e^{t}+\frac{4}{3}e^{4t}-\frac{1}{2}e^{-t}}$

As to the Laplace transform, however, I am of no use.  :(

[b]t0rajir0u[/b] thanx for the input, you are right about solving this way, but I kind of need solving with laplance trasforms.
Eather way unfortenutly yuor soliution isn't the same as in the book soliution... and not as mine eather... so I just wondering that is going on.. of cource I can by wrong.. and may book is wrong as well...
eny way the correct (?) answer is :

$\boxed{y(t) = \frac{1}{2}(3e^{-2t}sin(t)+e^{-2t}cos(t)-e^{-t});\;for\;t\ge0}$

[b]V.V.[/b]
this is true only if $y(0)=0$ nad I use it in my soliution

$\int_{0}^{\infty}e^{-st}\int_{0}^{t}y(u)\,du\,dt=\int_{0}^{\infty}\int_{u}^{\infty}e^{-st}y(u)\,dt\,du$

$=\int_{0}^{\infty}\frac1se^{-su}y(u)\,du=\frac{Y(s)}{s}.$

It doesn't matter what $y(0)$ is. V.V. is correct.


$ a,b,c$ are distinct real numbers such that
\[ a^2 \minus{} b \equal{} b^2 \minus{} c \equal{} c^2 \minus{} a
\]
Then find the value of:
\[ \left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)
\]

$ b^2\minus{}c\equal{}c^2\minus{}a\implies c\minus{}a \equal{} b^2 \minus{} c^2 \equal{} (b\minus{}c)(b\plus{}c)$

$ a^2\minus{}b\equal{}b^2\minus{}c\implies b\minus{}c \equal{} a^2 \minus{} b^2 \equal{} (a\minus{}b)(a\plus{}b)$

$ a^2\minus{}b\equal{}c^2\minus{}a\implies a\minus{}b \equal{} c^2 \minus{} a^2 \equal{} (c\minus{}a)(c\plus{}a)$

$ (c\minus{}a)(b\minus{}c)(a\minus{}b)\equal{}(b\minus{}c)(b\plus{}c)(a\minus{}b)(a\plus{}b)(c\minus{}a)(c\plus{}a)$

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


function
linear algebra
matrix
algebra proposed
algebra

WE have a sequence $ { \{a_n\} \in R_\plus{}}$,$ {a_1\equal{}a}$,$ {a_2\equal{}b}$, and for $ {m\plus{}n\equal{}p\plus{}q}$ with $ {m,n,p,q \in N_\plus{}}$; we also have :$ {\frac{a_m\plus{}a_n}{(1\plus{}a_m)(1\plus{}a_n)}\equal{}\frac{a_p\plus{}a_q}{(1\plus{}a_p)(1\plus{}a_q)}}$.
(a). when $ {a\equal{}\frac{1}{2}}$, $ {b\equal{}\frac{4}{5}}$, find $ {\{a_n\}}$ (use $ {n}$ to express $ {a_n}$);
(b).prove that :for any $ {a}$, there exists $ {\lambda}$ , such that :for every $ {n \in N_\plus{}}$, we have $ {\frac{1}{\lambda}\le a_n \le \lambda }$ .  [use $ {a}$ to express $ {\lambda}$]
[hide]This is a difficult problem from National Matriculation Examination[we call it 高考 in Chinese and it is just like SAT in The USA ]   in China  for jiangxi province in 2009 . students use 2 hours to solve 22 questions and this problem is the last one , because students don't have enough time ,so nearly no one solved it [I think the time for the examination is too short] ,Dear mathlinkers can you solve it ?  :wink: ][/hide]

[quote="CCMath1"]WE have a sequence $ {\{a_n\} \in R_ \plus{} }$,$ {a_1 \equal{} a}$,$ {a_2 \equal{} b}$, and for $ {m \plus{} n \equal{} p \plus{} q}$ with $ {m,n,p,q \in N_ \plus{} }$; we also have :$ {\frac {a_m \plus{} a_n}{(1 \plus{} a_m)(1 \plus{} a_n)} \equal{} \frac {a_p \plus{} a_q}{(1 \plus{} a_p)(1 \plus{} a_q)}}$.
(a). when $ {a \equal{} \frac {1}{2}}$, $ {b \equal{} \frac {4}{5}}$, find $ {\{a_n\}}$ (use $ {n}$ to express $ {a_n}$);
(b).prove that :for any $ {a}$, there exists $ {\lambda}$ , such that :for every $ {n \in N_ \plus{} }$, we have $ {\frac {1}{\lambda}\le a_n \le \lambda }$ .  [use $ {a}$ to express $ {\lambda}$]
[hide]This is a difficult problem from National Matriculation Examination[we call it 高考 in Chinese and it is just like SAT in The USA ]   in China  for jiangxi province in 2009 . students use 2 hours to solve 22 questions and this problem is the last one , because students don't have enough time ,so nearly no one solved it [I think the time for the examination is too short] ,Dear mathlinkers can you solve it ?  :wink: ][/hide][/quote]

I think this problem is to be solved without computing first the general expression for $ a_n$ but I will :)

Fist, if $ a_k \equal{} 1$ for some $ k$, the equation, using $ (m,n,p,q) \equal{} (2t \minus{} k,k,t,t)$ implies $ \frac 12 \equal{} \frac {2a_t}{(1 \plus{} a_t)^2}$ and so $ a_t \equal{} 1$ $ \forall t > \frac k2$ and so $ a_n \equal{} 1$ $ \forall n > 1$ and the sequence is $ \{a,1,1,1,...\}$
In this case, question (b) is obvious and the best $ \lambda$ is $ \max(a,\frac 1a)$

Consider now that $ a_n\neq 1$ $ \forall n$ and let $ b_n$ be the sequence $ b_n \equal{} \frac {1 \plus{} a_n}{1 \minus{} a_n}$

We get $ a_n \equal{} \frac {b_n \minus{} 1}{b_n \plus{} 1}$ and the equation becomes $ b_nb_m \equal{} b_pb_q$ $ \forall m,n,pq\in\mathbb N$ such that $ m \plus{} n \equal{} p \plus{} q$

It is then easy to solve this equation and to get $ b_n \equal{} b_1^{2 \minus{} n}b_2^{n \minus{} 1}$ and so $ a_n \equal{} \frac {b_1^{2 \minus{} n}b_2^{n \minus{} 1} \minus{} 1}{b_1^{2 \minus{} n}b_2^{n \minus{} 1} \plus{} 1}$ $ \equal{} \frac {b_2^{n \minus{} 1} \minus{} b_1^{n \minus{} 2}}{b_2^{n \minus{} 1} \plus{} b_1^{n \minus{} 2}}$ and :

[u]General expression for $ a_n$[/u] :
If $ a \equal{} 1$, then $ b \equal{} 1$ and $ a_n \equal{} 1$ $ \forall n$
If $ a\neq 1$ and $ b \equal{} 1$, then $ a_n \equal{} 1$ $ \forall n > 1$
If $ a\neq 1$ and $ b\neq 1$, then $ a_n \equal{} \frac {\left(\frac {1 \plus{} b}{1 \minus{} b}\right)^{n \minus{} 1} \minus{} \left(\frac {1 \plus{} a}{1 \minus{} a}\right)^{n \minus{} 2}}{\left(\frac {1 \plus{} b}{1 \minus{} b}\right)^{n \minus{} 1} \plus{} \left(\frac {1 \plus{} a}{1 \minus{} a}\right)^{n \minus{} 2}}$

Then :
[u]Question (a)[/u] : $ \frac {1 \plus{} a}{1 \minus{} a} \equal{} 3$ and $ \frac {1 \plus{} b}{1 \minus{} b} \equal{} 9$ and so $ a_n \equal{} \frac {9^{n \minus{} 1} \minus{} 3^{n \minus{} 2}}{9^{n \minus{} 1} \plus{} 3^{n \minus{} 2}}$ $ \equal{} \frac {3^n \minus{} 1}{3^n \plus{} 1}$

[u]Question (b)[/u] :
Let $ u \equal{} \frac {1 \plus{} b}{1 \minus{} b}$ and $ v \equal{} \frac {1 \plus{} a}{1 \minus{} a}$. In order to have $ a_n > 0$ $ \forall n$, we need to have $ |u|\geq |v| > 1$ and we can write $ a_n \equal{} 1 \minus{} \frac {2}{u\left(\frac uv\right)^{n \minus{} 2} \plus{} 1}$.

It's then rather easy to show, using that $ |v|\equal{}\frac{1\plus{}\min(a,\frac 1a)}{1\minus{}\min(a,\frac 1a)}$ $ \equal{}\minus{}\frac{1\plus{}\max(a,\frac 1a)}{1\minus{}\max(a,\frac 1a)}$,  that
$ \min(a,\frac 1a)\leq a_n\leq\max(a,\frac 1a)$ and so $ \boxed{\lambda \equal{} \max(a,\frac 1a)}$

I was going to write my solution when I found the nice pco's. I must however make the following comments.

By taking $ (m,n,p,q) \equal{} (1,k\plus{}1,2,k)$ one gets the recurrence relation $ a_{k\plus{}1} \equal{} \frac {(1\minus{}ab)a_k \plus{} (b\minus{}a)} {(b\minus{}a)a_k \plus{} (1\minus{}ab)}$ for the sequence $ \{ a_k \}$, with $ (1\minus{}ab)^2 \minus{} (b\minus{}a)^2 \equal{} (1\minus{}a^2)(1\minus{}b^2)$. Take $ \alpha \equal{} 1\minus{}ab$, $ \beta \equal{} b\minus{}a$, and $ f(x) \equal{} \frac {\alpha x \plus{} \beta} {\beta x \plus{} \alpha}$. Then $ a_{k\plus{}1} \equal{} f(a_k)$, so $ a_k \equal{} f^k(a)$.
But $ f$ is a homographic function (also known as Mobius transformation, or other names...), and the theory of its iterates is widely known, related to the eigenvalues of the associated matrix $ A(f) \equal{} \genfrac {(} {)} {0pt} {} {\alpha \ \beta} {\beta \ \alpha}$.

See for example [url]http://www.mathlinks.ro/viewtopic.php?t=4948[/url] in this very forum.

Thus the general form, and the ensuing discussions, are, from this point of view, a direct application of these classical techniques.


inequalities calculus derivative algebra polynomial AMC

I don't know if it is correct.

[hide]
Since $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)[(a+b)^{2}+(b+c)^{2}+(c+a)^{2}]$, $3abc=a^{3}+b^{3}+c^{3}$. Then

$f(a,b,c)=a^{4}+b^{4}+c^{4}-3abc$

$=a^{4}+b^{4}+c^{4}-a^{3}-b^{3}-c^{3}$

$=a^{3}(a-1)+b^{3}(b-1)+c^{3}(c-1)$

We can just put $a=1$ since we want the minimum and then $b+c=-1$. Then

$f(1,b,c)=b^{3}(b-1)+(b+1)^{3}(b+2)$

But I don't feel like taking the derivative of that and I suck at inequalities,  so I give up. :P
[/hide]

EDIT: The numbers didn't add up to 0 so I had to edit my solution. . .

An actual solution:

[hide]Applying AM-GM, $\frac{a^{4}+b^{4}+c^{4}-3abc}{4}\ge \sqrt[4]{-3a^{5}b^{5}c^{5}}$.
Simplifying, we have $a^{4}+b^{4}+c^{4}-3abc \ge 4abc \sqrt[4]{-3abc}$.
Therefore, if we know the value of maximum value of abc, we will  cover all the cases and will have found the minimum value.  Obviously, the maximum value of abc is 0.  Thus, $a^{4}+b^{4}+c^{4}-3abc \ge 0$.

And to think I was about to generate a polynomial!   :rotfl:   Good thing I didn't.[/hide]

[quote="falconwing64"]What isn't positive?
And I thought AM-GM worked for all nonnegative integers...[/quote]

then you're already assuming $a,b,c=0$

you can't use AM-GM on negative integers. therefore, by using it, you're assuming $a,b,c=0$ since $a+b+c=0$

[quote="falconwing64"]My God!  I didn't assume a,b,c = 0!  I [b]proved[/b] it.  Read the proof.[/quote]

Your proof is completely wrong.

[quote="falconwing64"]What's positive in my AM-GM Inequality? $a^{4}, b^{4}, c^{4}$ are all positive, and $-3abc$, which I am treating as a single term, must also be positive.[/quote]

If $-3abc$ is positive, then $a^{4}+b^{4}+c^{4}-3abc>0$

$-3abc$ can be negative or positive depending on the specific values of $a,b,c$.

oh.......ok...i'm wrong...i see what my mistake was...thx for pointing it out.  Sorry for being so rude  :blush:

AM-GM states that the arithmetic mean of any finite number of positive reals is greater than or equal to the geometric mean of the same set of numbers, with equality if and only if the numbers are all equal.

If $a=b=c$, the minimum is $-\frac{81}{256}$

what???

a=b=c can't be true, since $a\ge1$

[quote="falconwing64"]It's a very useful inequality.  Goto:

[url]http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means[/url][/quote]
Isn't AM-GM obvious?

That's a strange question, but I wouldn't say that AM-GM is "obvious" in the way that the trivial inequality is obvious (they don't call it the trivial inequality for nothing :P ). The proof is quite involved.

[hide="how about..."]
$a+b+c = 0 \implies a^{3}+b^{3}+c^{3}= 3abc$
since $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$. In particular, unless all variables are equal this is basically biconditional. Anyways,

$a^{3}(a-1)+b^{3}(b-1)+(-a-b)^{3}(-a-b-1) = a^{3}(a-1)+b^{3}(b-1)+(a+b)^{3}(a+b+1)$

and the goal is basically to minimize this sum for $a \geq 1$. This implies that $a^{3}(a-1) \geq 0$, so

$a^{3}(a-1)+b^{3}(b-1)+(a+b)^{3}(a+b+1) \geq b^{3}(b-1)+(a+b)^{3}(a+b+1)$

Finally, if $b$ is positive the above expression must also be positive. And if $b$ is negative, we can replace the original expression with an expression given as

$a^{3}(a-1)+b^{3}(b+1)+(a-b)^{3}(a-b+1) = b^{4}+b^{3}+(a-b)^{4}+(a-b)^{3}$

which is in turn equal to

$2b^{4}+2a^{4}+6a^{2}b^{2}+3ab^{2}-4a^{3}b-4ab^{3}-3a^{2}b$

so we need only show that for $a \geq 1$,

$2b^{4}+2a^{4}+6a^{2}b^{2}+3ab^{2}\geq 4a^{3}b+4ab^{3}+3a^{2}b$

By AM-GM, $2a^{4}+2a^{2}b^{2}\geq 4a^{3}b$.
Similarly, $2b^{4}+2a^{2}b^{2}\geq 4ab^{3}$. Then we simply need

$2a^{2}b+3ab \geq 3a^{2}$

$2b \geq 3 \iff b \geq \frac{2}{3}$, so we can assume $b < \frac{2}{3}$.

That's a start, at least...
[/hide]

AM-GM is pretty obvious for two numbers, but the proof of AM-GM for all sets of numbers is by no means obvious.


Let $ a,b,c>0$ such that $ abc\equal{}1$. Prove that
$ \frac{a}{{a^2  \plus{} k}} \plus{} \frac{b}{{b^2  \plus{} k}} \plus{} \frac{c}{{c^2  \plus{} k}} \le \frac{3}{{k \plus{} 1}}$

[quote="thanhak5"]Let $ a,b,c > 0$ such that $ abc \equal{} 1$. Prove that
$ \frac {a}{{a^2 \plus{} k}} \plus{} \frac {b}{{b^2 \plus{} k}} \plus{} \frac {c}{{c^2 \plus{} k}} \le \frac {3}{{k \plus{} 1}}$[/quote]
the condition of $ k$????  :maybe:

[quote="thanhak5"]Let $ a,b,c > 0$ such that $ abc \equal{} 1$. Prove that
$ \frac {a}{{a^2 \plus{} k}} \plus{} \frac {b}{{b^2 \plus{} k}} \plus{} \frac {c}{{c^2 \plus{} k}} \le \frac {3}{{k \plus{} 1}}$[/quote]
[quote="nguoivn"]Let $ a \equal{} \frac {x}{y}$...
It can rewrite:
$ \sum\ \frac {xy}{x^2 \plus{} 3y^2} \leq\ \frac {3}{4}$
We have: $ x^2 \plus{} y^2 \plus{} 2y^2 \geq\ 2\sqrt []{2}y\sqrt []{x^2 \plus{} y^2}$
So, $ \frac {xy}{x^2 \plus{} 3y^2} \leq\ \frac {1}{2\sqrt []{2}}\sqrt []{\frac {x^2}{x^2 \plus{} y^2}}$
 :)[/quote]
I think $ k \ge\ 3$
similar to nguoivn 's proof
We have: $ x^2 \plus{} y^2 \plus{} (k\minus{}1)y^2 \geq\ 2\sqrt []{k\minus{}1}y\sqrt []{x^2 \plus{} y^2}$


geometry
3D geometry
pyramid
analytic geometry
national olympiad

1.  Find all solutions in positive integers to:
\begin{eqnarray*} a + b + c = xyz \\ x + y + z = abc \end{eqnarray*}

2.  $F_n$ is the Fibonacci sequence $F_0 = F_1 = 1$, $F_{n+2} = F_{n+1} + F_n$. Find all pairs $m > k \geq 0$ such that the sequence $x_0, x_1, x_2, ...$ defined by $x_0 = \frac{F_k}{F_m}$ and $x_{n+1} = \frac{2x_n - 1}{1 - x_n}$ for $x_n \not = 1$, or $1$ if $x_n = 1$, contains the number $1$.

3.  $PABCDE$ is a pyramid with $ABCDE$ a convex pentagon. A plane meets the edges $PA, PB, PC, PD, PE$ in points $A', B', C', D', E'$ distinct from $A, B, C, D, E$ and $P$. For each of the quadrilaterals $ABB'A', BCC'B, CDD'C', DEE'D', EAA'E'$ take the intersection of the diagonals. Show that the five intersections are coplanar.

4.  Define the sequence $a_1, a_2, a_3, ...$ by $a_1 = 1$, $a_n = a_{n-1} + a_{[n/2]}$. Does the sequence contain infinitely many multiples of $7$?

5.  The points $D, E$ on the side $AB$ of the triangle $ABC$ are such that $\frac{AD}{DB}\frac{AE}{EB} = (\frac{AC}{CB})^2$. Show that $\angle ACD = \angle BCE$.

6.  $S$ is a board containing all unit squares in the $xy$ plane whose vertices have integer coordinates and which lie entirely inside the circle $x^2 + y^2 = 1998^2$. In each square of $S$ is written $+1$. An allowed move is to change the sign of every square in $S$ in a given row, column or diagonal. Can we end up with all $-1$s by a sequence of allowed moves?

1. http://www.mathlinks.ro/Forum/viewtopic.php?p=366273#p366273
2. http://www.mathlinks.ro/Forum/viewtopic.php?p=366275#p366275
3. http://www.mathlinks.ro/Forum/viewtopic.php?p=366276#p366276
4. http://www.mathlinks.ro/Forum/viewtopic.php?p=366277#p366277
5. http://www.mathlinks.ro/Forum/viewtopic.php?p=366280#p366280
6. http://www.mathlinks.ro/Forum/viewtopic.php?p=366282#p366282


geometry
3D geometry
prism
MATHCOUNTS
tetrahedron
puzzles

I have a very pricky fish. It only wants to go into a 50 gallon tank, but the problem is that I only have two 60 gallon tanks. If I have no measurements of any kind, how can I have a tank that is exactly 50 gallons?

Are the tanks in the shape of rectangular prisms?

ckck,

you meant a "very picky" fish, didn't you?

I say "Too bad fishy! You're going to the big tank!"

Anyhow, do you have anything to scoop the water from one tank to the other, or do you just dump it?

If your fish is being that picky you should just get rid of it and get something cool.............like a shark!

Yes tanks are in the shape of a retangular prism, and I meant picky fish. Sorry. :blush:

[hide]
Is there a solution without using two different color, immiscible liquids in the two containers? I can see no other way of measuring out one third...[/hide]

[hide]
If you are referring to my last hide tag, then a solution is along the lines of: measure out one half a container, measure out one third a container by:

putting equal amounts of the two immiscible multicolored liquids in one container such that the three volumes are equal

then the difference is one sixth a container
which is subtracted.

Note this isn't a real solution, which requires cunning, but a methd of the math behind the solution and what lengths are measurable[/hide]

EDIT: post 38^2 or for mathcounts people that 1444 :P

[quote="modeler"][hide]
Fill a 60 gallon tank, then put the fish into it.
Then scoop a little amount of water out, and ask the fish,
"Fishy, is this ok?"
If the answer is no, repeat the procedure.

Sorry for the nonsense, but the picky fish should give a sign about whether the amount of water is 50 gallon.
[/hide][/quote]
 :rotfl:  :rotfl: Good try.

[quote="SoxFan34"]Get a gallon-sized milk bottle and fill it up 50 times[/quote]
You have no kind of measurement at all.

[quote="probability1.01"]Fill up a 60 gallon tank and then pour some of it into the other tank such that the remaining liquid forms a tetrahedron sharing 3 edges with the tank. If you know what I mean.[/quote]

Clever... I assume you have found a way to stand the tank up on a vertex such that the plane containing the three adjacent vertices is parallel to the table... or make water go where you want it to go.

You mean a tank with 50 gallons of water in it, not a 50-gallon tank of water, correct?

Sixty gallons of water is really heavy IME.

A gallon is 3.79 (approximately) liters. Sixty gallons is 227.4 liters, which is 227.4 kg

Can [i]you[/i] lift 227.4 kg?

Not to mention the weight of the tank by itself you did not factor in.


Without replacement, what is the probability of drawing a 6 before a 9 when three cards are picked from a standard 52-card deck? (For example, 6 of diamonds, Queen of hearts, 9 of spades.)

[hide]  If there is a 6 in the first card, then the probability of drawing exactly one 9 in the other 2 is $\frac{4}{51}\times \frac{47}{50}+\frac{47}{51}\times \frac{4}{50}=\frac{188}{1275}$  The probability of drawing 2 9's is $\frac{4}{51}\times \frac{3}{50}=\frac{2}{425}$.  Since there is a $\frac{1}{13}$ chance of drawing a 6 on the first draw, this is a total probability of $\frac{188+6}{1275}\times \frac{1}{13}=\frac{194}{16575}$.

If the first card is a 9, the second a 6, and the third a 9, there is a probability of $\frac{1}{13}\times \frac{4}{51}\times \frac{3}{50}=\frac{2}{5525}$

If the first card is not a 6 or a 9, the second is a 6, and the 3rd is a 9, there is a probability of $\frac{11}{13}\times \frac{4}{51}\times \frac{2}{25}=\frac{88}{16575}$

This is a total probability of $\frac{194+6+88}{16575}=\frac{96}{5525}$.[/hide]

I'm not sure if this is right because it seems to be too ugly of an answer.

[hide]

 First of all, we need to find out the total number of possible combinations that you can get by drawing from a deck of 52
at three at a time.

 that would be $_{52}C_{3}= \frac{52 \bullet 51 \bullet 50}{3 \bullet 2 \bullet 1}= 22100$

 The number of possibilities that a $6$, a $9$ and another card (including drawing another $6$ or $9$) are drawn is $50(_{3}P_{1}) = 300$

The number of possibilities that a $6$, a $9$ and another card (excluding another $9$) are drawn is $47(_{3}P_{1}) = 282$

The probability that a $6$ is the first card drawn in those possibilities is $\frac{1}{3}$, so there are $100$ possibilities that a $6$ will be drawn first.

The second case is the probability of a $6$ being the second card drawn and the third card being $9$, provided that another $9$ isn't drawn as the first card which is $\frac{1}{6}$ so there are $47$ possibilities that a number other than $9$ will be drawn first, a $6$ drawn second and a $9$ drawn third.

The total possible ways of drawing three cards with a $6$ drawn before a $9$ is $100+47 = 147$

So the probability of drawing three cards where a $6$ is drawn before a $9$ in a deck of 52 cards is $\frac{147}{22100}$

[/hide]

remember

if you see a 69, 96, or 666 in a treething question you know you're either correct or very close


I try to insert a horizontal line into my pdf :
[code]\documentclass{article}
\usepackage[pdftex]{graphicx}
\usepackage{asymptote}
\begin{document}

Number   theory   maths

\begin{figure}[h]
\begin{asy}
................
\end{asy}
\end{figure}
Number   theory

\end{document}  [/code]

The length of the line is the  textwidth, what is the code ?

If the line is a dotted line , say, longdashed, then how to get that?

Many thanks!
[/code]

If the length of the line is not \textwidth , but half or one-third of \textwidth ,
and the line is on the middle of the paper, what is the code?

thangks!


If $a,b,c=(a^{2}+b^{2})/(ab+1)$ are integers, then prove that $c$ is a perfect square.

It's problème  6 IOM 1988
see
http://www.mathlinks.ro/Forum/viewtopic.php?p=352683#p352683

I've written another post to apologize for accidentally having posted it in here, instead of the Unsolved Problems, but it was cancelled, I guess.


algebra
polynomial
superior algebra
superior algebra solved

To which famous ring is

$\frac{\mathbb{Z}_{5}\left[X,Y\right]}{\left(Y-X^{2}, XY+Y+2\right)}$

isomorphic?

We can substitute any $y$ to $x^{2}$.
Thus we look for $\mathbb{Z}/(x^{3}+x^{2}+2)$.
If we would know that $\mathbb{Z}_{5}$ is the field with $5$ elements, we would see (cause of the polynomial clearly being irreducible) that the ring we get is $\mathbb{F}_{125}$, the field with $125$ elements.
If $\mathbb{Z}_{5}$ are the $5$-adic integers, we get an unramified extension of degree $3$.


You do realize that you can look under somebody's avatar to see their post "numbah", right?

[quote="pianoforte"][quote="1=2"]He's Treething. look at his sig, location, and avatar.[/quote]
huh
i don't think treething knows funcia, and takes pictures of her
otherwise, i would be worried for funcia
but tat's not the case

and junngi ist sezy[/quote]
eh i'd be worried too

asmallgoat is my sibling

treething->treething's profile

the converse isn't neccesarily true

[quote="Klebian"]so if i created a new user with your signature, location, and avatar, you would appreciate it if other people claimed i was 1=2?[/quote]

you mean like this?

Ego boost ++ for Treething.

If this ever gets out of hand, though, keep in mind that the administrators can easily read IP addresses from posts. I'm pretty sure that asmallgoat doesn't sneak into my house and post on my laptop when I'm out, so I should be fine.

[quote="Ignite168"][quote="bubka"]Too much off topic talk, that's the problem of this generation.
Mine is 274! A good deal more than yours.[/quote]

Wasn't yours like 3275? You posted -3000 times in 6 months?!?!?[/quote]

Ever here of the great depression? Well, the lesser depression happened, and all posts in the G&FF didn't count. Bubka fell from 1000 something to 3 something(disturbing...) and I fell from 1000 something to 200. Now that I hae solved a bunch of problems, I'm back on my feet, :)

[quote="miyomiyo"]Back off topic, isn't there also a "_funcia" with the same exact info as funcia?[/quote]

but that's funcia b/c my post number is 1337

and my don't asmallgoat and _funcia have the same ip almost all the time  :o

and doesn't asmallgoat post pictures of me all the time

that's annoying but i am not sure what i can do about it other than secretly finding
their pw one day...

[quote="ra5249"]mine is 999999999999999999999999999999999999999999999999999999999.53[/quote]

Even a double-precision floating point number cannot store such a large number with so many nines!

[quote="kyyuanmathcount"][quote="1=2"]He's Treething. look at his sig, location, and avatar.[/quote]

Drink less Evian water please.[/quote]

Or he could wait for his <plural body part> to drop before posting...


function
vector
algebra proposed
algebra

Does there exist two periodic function f(x),g(x),R->R,which causes to any x, has f(x)+g(x)=x?

That's a [url=http://cornellmath.wordpress.com/2008/01/22/periodic-functions-problem/]pretty famous[/url] question.

The answer is "yes", and it's fairly easy to do using a Hamel basis. This is a set $ H$ of real numbers which forms a vector-space basis for $ \mathbb{R}$ over $ \mathbb{Q}$. Thus every real number can be expressed as a linear combination of elements of $ H$ with rational coefficients, in a unique way.

It's now an easy exercise using the definitions of "basis" for a vector space to prove the following facts:

1. Every function $ f: H \to \mathbb{R}$ defines, in a unique way, a function $ \tilde{f}: \mathbb{R} \to \mathbb{R}$ such that $ \tilde{f}(h)\equal{}f(h)$ for all $ h \in H$.

2. If $ f: H \to \mathbb{R}$ satisfies $ f(h_0)\equal{}0$ for some $ h_0 \in H$, then $ \tilde{f}$ is periodic: $ \tilde{f}(x\plus{}h_0) \equal{} \tilde{f}(x)$ for all $ x \in \mathbb{R}$.

To solve the problem, all that remains now is to define two functions $ f,g: H \to \mathbb{R}$ such that $ f(h_0)\equal{}g(h_1)\equal{}0$ for some arbitrary basis elements $ h_0 \neq h_1 \in H$, in such a way that $ f(h)\plus{}g(h) \equal{} h$ for all $ h \in H$. This is easily done. Taking $ \tilde{f}$ and $ \tilde{g}$ yields two periodic functions which sum to the identity.


Hi!
is there anyone here who is or is thinking of studying maths at cambridge.
I live in germany but i am british and I am thinking of applying...

got some questions:

-how do I select the college?
-what exactly do I need to state in the UCAS application?
-what are the interviews like?
-how good do i need to be in maths to have a good chance of being accepted?
-is cambridge even a good place to study maths as a german student?

appreciate help...

Hi aaton,
I hope what follows is helpfull, although i'm only going to answer your questions partially.

1) How do you select a college ? Well, that's the last thing you should be concerned with :-) There are lots of excellent colleges e.g. trinity, johns, kings and many others. My advice is to check them out (try their websites for example or some relevant forums) and see which one would suit you most, also bare in mind some are more competitive to get in than others. You can also make an 'open' application in case you cannot decide.

2) What do you need to state in the UCAS app ? That's a bit special and the details change every year, so the best thing to try is to read carefully all the manuals, i'm sure they are designed to be accessible for everyone.

3) The interviews are very cool ! They give you the opportunity to show them how capable you are for doing maths there. They are very important for the admission ! They only ask math stuff (at least most of the colleges, i think) ! They don't assume extra knoledge and are designed to fit all the candidates. They don't want to test your knowledge, but your ability to think mathematically. Usually, you will be given a set of problems an hour before the interview and then you discuss your solutions with the interviewers.

4) How good do i need to be  ? Very good question ! You don't need to be a genious neither particularly clever. But you need to be able and enthousiastic, the later more !

5) From Germany ? Well, there are many students from Germany ! I think you should try to contact them (maybe you should write in the mathlinks-germany forum). Germany has excellent unis to do maths, but i know nothing more about them. There are brilliant students from Germany that decided to stay there ! I think whether you want to leave or not, it's a rather personal question. Something i can tell, is that Cambridge is also an excellent place to study maths, full a stimullating stuuf and students, very enthousiastic, it's really an amazing place.

I suggest you go to the website : http://www.maths.cam.ac.uk, and http://www.dpmms.cam.ac.uk (go to people-> Prof Koerner and check his website, it has many articles that would interest you), also check the student representatives website and the relative forums there e.g. the archimidean's forum. You can find more information there, and if you post there you're likely to get a better and more detailed answer.

Good luck with your decisions,
Stergios

thanks for your quick answere!

Still got some questions, hope you don't mind:
Do you think an open application will increase my chances of being accepted? I don't really care in which college I will end up in. Or are there big differences in the quality of tutoring?

Do you still have some problems from the interview? I would be really interested in seeing them...
So if I understood you correctly, extracurricular activities like sport or community work have no influence on the admission? I think I am fairly strong in that area and it would be great if they also look at those kind of things.
Did you have to take the STEP test or were you given an unconditional offer?

I also read somewhere that you can change to physics after studying maths for a year. Have you also heard of this?

I don't think that an open application will increase your chances of getting in. I know ppl who got in both ways. The quality of tutoring from college to college is roughly the same, there are a few differences tho. I think it's better to chose a college you like and apply to it. By doing an open app they "randomly" sent you to a college and then the college decides, so you might as well apply directly to one. (I'm not precise here but it works roughly like that)

As for the interview, i don't have any problems but i'm sure you can find a sample on the relevant websites.
Also, i suppose they do take in mind extracurricular activities but definitely don't know how much.
As far as STEP is concerned, there are cases were students are given unconditional offers (usually based on exceptional academic performances or cause they are coming from countries where it is not possible to take the test) Personally, i don't know anyone who was given a STEP-conditional offer and failed ! That doesn't mean STEP is so easy, it does require work and preparation but it's nothing to be afraid of and also it's the last thing you should bother with !

At last, you can indeed change from maths to physics. There is a math-physics choice the first year. But i encourage you to do straight math. There is a big choice of applied math courses and theoretical physics courses in the tripos so that you don't have to worry in case your interests lie in physics.

Good luck,
Stergios

In terms of cost, how does the maths and physics course differ from the maths (only) course?

Sorry to revive this old topic... I just wanted to know when the STEP is taken by the students... Is it like the SAT where people take it before they pass out of their school? Or, after one appears for the class 12 board exams?


Inequality
Sequence
algebra
IMO Shortlist

Let $ a_1 \geq a_2 \geq \ldots \geq a_n$ be real numbers such that for all integers $ k > 0,$

\[ a^k_1 \plus{} a^k_2 \plus{} \ldots \plus{} a^k_n \geq 0.\]

Let $ p \equal{}\max\{|a_1|, \ldots, |a_n|\}.$ Prove that $ p \equal{} a_1$ and that

\[ (x \minus{} a_1) \cdot (x \minus{} a_2) \cdots (x \minus{} a_n) \leq x^n \minus{} a^n_1\] for all $ x > a_1.$

Sorry to revive the post. I used induction instead.

[quote=WolfusA]$(x - a_1) \cdot (x - a_2) \cdots (x - a_n) = x^n - a^n_1\iff (n=1$ or $n=2\wedge a_2=-a_1)$[/quote]

sorry but what do you mean? :blush:

Condition for equality in the inequality is $$n=1\vee (n=2\wedge a_1=-a_2)$$

For the first part, it is easy to see that $k\to \infty$ for some $k,$ if the largest number $p$ is actually negative, then the first expression will not be true.

For the second part, could somebody help with me with the induction step. I managed to prove that if we assume that $n$ is true, then we just want to show that $a_{n+1}(x^n-a_1^n)+a_1^n(x-a_1)\geq 0$ is true, but I am not sure how to proceed from here. :(

very nice problem , Solved with [b]twbnrftw[/b]

[i]Solution:[/i]  1st part we observe that not all of the terms can be negative so at least one of the terms is positive, now since $a_1$ is the maximum of all $a_{i}'s$ we have $a_{1} \geqslant 0$, now we also have for each odd integer $k$ , $\sum_{i=1}^{n} a_{i}^{k} \leqslant na_{1}^{k}$ , now we claim that $\max{\{|a_{1}|, |a_{2}|, \cdots, |a_{n}|\}}=a_{1}$. This is equivalent to show that $\max{\{|a_{1}|, |a_{2}|, \cdots , |a_{n}|\}} \neq |a_{n}|$ , now FTSOC let us assume that is the case (which can only happen if $a_{n}<0$) , then we have for sufficiently large odd $k$, $a_{1}^{k}+a_{2}^{k}+\cdots+a_{n-1}^{k}+a_{n}^{k} \leqslant (n-1)a_{1}^{k}+a_{n}^{k} $ , now we show that for sufficiently large choosen $k$ we have $(n-1)a_{1}^{k}+a_{n}^{k} <0$, but that is actually true because if we had  $(n-1) \geqslant -\frac{a_{n}}{a_{1}}^{k}$ for large such odd $k$ we have $n-1 \rightarrow \infty$ which is not possible  since $n$ is fixed , now that implies for large enough odd $k$  we have  $a_{1}^{k} +a_{2}^{k}+ \cdots +a_{n}^{k} <0$ contradiction, hence  $\max{\{|a_{1}|, |a_{2}|, \cdots , |a_{n}|\}}=a_{1}$. $\square$
-----
Now we work on the second part clearly for for $x>a_{1}$ we have $x-a_{i}>0$ $\forall$ $i \in \{1,2, \cdots , n\}$ , hence by $\emph{AM} \geqslant \emph{GM}$ we have $(x-a_{2})(x-a_{3})\cdots (x-a_{n}) \leqslant  \left( x -\frac{\sum_{i=2}^{i=n} a_{i}}{n-1}\right)^{n-1}$ , also for $k=1$ we have $a_{1} \geqslant -(a_2+a_3+\cdots +a_n) \implies  \left( x -\frac{\sum_{i=2}^{i=n} a_{i}}{n-1}\right)^{n-1} \leqslant \left(x+\frac{a_{1}}{n-1}\right)^{n-1}$ , Now:\\

$\left(x+\frac{a_{1}}{n-1}\right)^{n-1} =x^{n-1}+ a_{1}\frac{\binom{n-1}{1}}{n-1}+a_{2}^{2}\frac{\binom{n-1}{2}}{(n-1)^2}+\cdots +a_{1}^{n-1}< x^{n-1}+x^{n-2}a_{1}+x^{n-3}a_{2}^{2}+\cdots +a_{1}^{n-1}=\frac{x-a_{1}^{n}}{x-a_{1}} \implies (x-a_{1})(x-a_{2})(x-a_{3})\cdots (x-a_{n}) \leqslant x^{n}-a_{1}^{n}$ , equality  holds at $n=1$ or $n=2$ with $a_{2}=-a_{1}$. $\square$


[quote="IP"]$x,y,z>0$ such as $xy+yz+xz=3$.
Pove that :         $\sqrt{x} + \sqrt{y} + \sqrt{z} \ge 3$[/quote]
This is not true.
Try $x=y\rightarrow\sqrt3,$ $z\rightarrow0.$ ;)

With the given condition the inequality $x^k+y^k+z^k\ge3$ is only valid for $k\ge2-\frac{\ln4}{\ln3}\approx.73814$.

[quote="spanferkel"]With the given condition the inequality $x^k+y^k+z^k\ge3$ is only valid for $k\ge\frac{\ln3}{\ln2}$.[/quote]
I think it is your mistake:  $x+y+z\geq3.$ ;)
The following problem is very interesting to me:
Let $x,y,z$ are non-negative real numbers and $xy+xz+yz=3.$
Prove that $\sqrt[4]{x^3}+\sqrt[4]{y^3}+\sqrt[4]{z^3}\geq3,$
but I do not manage to prove it. :(

Note to user IP that section "Inequalities Proposed & Own Problems" is here for "Problems that you have already solved and you are interested in second opinions or solutions."

If you don't know any solution to some problem and you don't know whether the statement of your problem is true or not, post it "Inequalities Open Questions".

PS.: Try first to get know with some rules, before you use the forum. :mad:

Nice day.

Not really a similar question but maybe useful:
http://www.mathlinks.ro/Forum/viewtopic.php?t=35839

[quote="conceive"]
If you don't know any solution to some problem and you don't know whether the statement of your problem is true or not, post it "Inequalities Open Questions".
[/quote]
Actually, in this case, it should rather go into the "Inequalities Unsolved Problems". ;)

[b]arqady[/b], your inequality is indeed interesting, I'll have a try at proving it! :)

[quote="mathmanman"]
Actually, in this case, it should rather go into the "Inequalities Unsolved Problems".
[/quote]
Thank you, mathmanman. I have already made it.

[quote="arqady"][quote="spanferkel"]With the given condition the inequality $x^k+y^k+z^k\ge3$ is only valid for $k\ge\frac{\ln3}{\ln2}$.[/quote]
I think it is your mistake:  $x+y+z\geq3.$ ;)
 :([/quote] right, I've edited  :blush:

The general problem had been posted here, but not complete solution.

http://www.mathlinks.ro/Forum/topic-57106.htmlhttp://www.mathlinks.ro/Forum/topic-57106.html

The solution's at http://diendantoanhoc.net/forum/index.php?showtopic=7982 but in VietNamese only.

Thank you hungkhtn.
Id est, this problem:
Let $x,y,z$ are non-negative real numbers and $xy+xz+yz=3.$
Prove that $\sqrt[4]{x^3}+\sqrt[4]{y^3}+\sqrt[4]{z^3}\geq3$
 is still unsolved. ;)

[quote="hungkhtn"]I don't think so, this has solution in the links I post. Even that It had solution in other famous problem.[/quote]
Yes you are right, hungkhtn.  I have understood it, at last.:D
Thank you!  :lol:


i downloaded the LaTeX months ago but i didnt actually use it

but i thought itd be useful for writing solutions for usamts so i started using it...

[code]\documentclass{article}
\begin{document}
Hello, world!
\end{document} [/code]

i put this in my TeXnicCenter, saved it, and pressed ctrl-f7, but nothing comes out in my output.

what is wrong?

What is in the log? How many pages did it say were created on the bottom line of the bottom window? You can see it in the window at the bottom of TexnicCenter and is also the file in the same place as your document; so if your file was called latex1.tex then the log is latex1.log. Is there a file called latex1.pdf?

even after i press ctrl+f7, there is nothing on the log...

Really? Are you saying there is no log or it is blank? Are any files created?


When private messaging, if you click on find a username, the screen seems to "overlap" making it impossible to type in the line to  input a username.  I've tried it with IE, and I use windows XP.

[url]http://www.artofproblemsolving.com/Forum/search.php?mode=searchuser[/url]

That was posted before [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99298]here[/url] but no one responded about it.

Whoops, I guess somebody delete this post or the other.


factorial
algebra
polynomial
number theory proposed
number theory

Suppose that $P(x)\in\mathbb Z[x]$ is a non-constant polynomial.
Prove that there is a natural number like $m$ such that $P(m!)$ is a composite number.

P.S. : I have a long solution, which uses some kind of asymptotics. Anyone has a short proof, or a proof without asymptotics?!

Nima Ahmadi Pour, could you please post your solution?
i would be very interested in a solution to this difficult problem.

Could you please post a soln? i dont mind if it is long and complicated, i think this problem is very difficult.


trigonometry
trig identities
Law of Cosines

Quadrilateral ABCD is inscribed in circle O with AB=3, BC=4, CD=5, and AD=6. Find, to the nearest hundredth, AC.

I know shamefully little about cyclic quadrilaterals, so please help and please explain.

[hide="Solution not using the above hint"]Since the quadrilateral is cyclic, angle $ ABC$ + angle $ CDA$ = $ 180$ degrees, so the sum of the cosines of the two angles is zero.

So we desire $ x$ such that $ 3^2 \plus{} 4^2 \minus{} 2\times3\times4x \equal{} 5^2 \plus{} 6^2 \minus{} 2\times5\times6(\minus{}x)$.

So $ 25 \minus{} 24x \equal{} 61 \plus{} 60x$. $ 84x \equal{} \minus{}36$. $ x \equal{}\minus{}\frac{3}{7}$. So $ \sqrt{25 \plus{} 72/7} \approx \boxed{5.94}$.[/hide]

That's brilliant (as in the Harry Potter British English "brilliant" :P), thanks.

no reason to be confused. There was an error in my post.

[quote="Jebus McAzn"]Now I'm confused, we have conflicting answers. O_O[/quote]

Actually, his answers matches mine.

He has $ x^2 \equal{} 25 \minus{} 24 \cos \theta$ and $ x^2 \equal{} 61 \plus{}60 \cos \theta$. The first equation is equivalent to $ 5x^2 \equal{} 125 \minus{} 120 \cos \theta$ and $ 2x^2 \equal{} 122 \plus{} 120 \cos \theta$. Add the two: $ 7x^2 \equal{} 247$. $ x \equal{} \sqrt{\frac{247}{7}}$, which is what I had.

[quote="pleurestique"]let $ AC \equal{} c$ and $ m\angle B \equal{} \theta$. then $ m\angle D \equal{} 180^{\circ} \minus{} \theta$ (this is a key property of cyclic quadrilaterals). by law of cosines in triangles $ ABC$ and $ ADC$, we have (since $ \cos(180^{\circ} \minus{} \theta) \equal{} \cos \theta$)
\[ x^2 \equal{} 3^2 \plus{} 4^2 \minus{} 2\cdot 3\cdot 4\cdot \cos \theta\]

\[ x^2 \equal{} 5^2 \plus{} 6^2 \plus{} 2\cdot 5\cdot 6\cdot \cos \theta\]
multiply the first by 5, the second by 2, and add...[/quote]

So the typo would be that $ \cos(180^{\circ} \minus{} \theta) \equal{} \minus{}\cos \theta$ instead of $ \cos \theta$?

$ \cos (180^{\circ} \minus{} \theta) \equal{} \cos 180^{\circ} \cos \theta \plus{} \sin 180^{\circ} \sin \theta \equal{} (\minus{}1) \cos \theta \plus{} (0) \sin \theta \equal{} \minus{} \cos \theta$.

yeah, that was a typo, but the equations I wrote are correct.


function
topology
complex analysis
complex analysis unsolved

Let $ G$ be a bounded open set in $ \bf C$ and suppose $ f$ is continuous function on $ \bar{G}$ which is analytic in $ G$. Then $ \max \{ |f(z)|: z \in \bar{G} \} \equal{} \max \{ |f(z)|: z \in \delta G \}$.

I need to prove this, using the previous theorem:

[quote]
If $ f$ is analytic in a region $ G$ and $ a$ is a point in $ G$ with $ |f(a)| \geq |f(z)|$ for all $ z$ in $ G$, then $ f$ must be constant.
[/quote]

Clearly we have not assumed connectedness in the theorem I need to prove. However, I can apply the previous theorem to the connected components and take the maximum. But what if there are infinitely many connected components. Isn't this a flaw in my proof?

No, you are still fine. Every point in $ G$ lies in one of the components and the boundary of each component is a subset of the boundary of the entire set. You may be required to prove this simple topological statement though. :)

Mmm, I still haven't figured this one out...

[quote]MMT (version 1) if $ f$ is analytic in a [b]region[/b] $ G$ and $ a$ is a point in $ G$ with $ |f(a)| \geq |f(z)|$ for all $ z \in G$ then $ f$ is constant.

Proof: let $ \Omega \equal{} f(G)$, put $ \alpha \equal{} f(a)$. If $ |\alpha| \geq |\zeta|$ for all $ \zeta \in \Omega$, then $ \alpha \in bound(\Omega) \cap \Omega$. The set $ \Omega$ can't be open. So the open mapping theorem yields that $ f$ must be constant.[/quote]

[quote]MMT (version 2)
Let $ G$ be a bounded open set in $ C$ and suppose $ f$ is continuous function on $ Cl(G)$ and is analytic in $ G$ . Then $ \max \{ |f(z)|: z \in \bar{G} \} \equal{} \max \{ |f(z)|: z \in \delta G \}$.

Proof: Since $ G$ is bounded there is a point $ a \in Cl(A)$ such that $ |f(a)|\geq|f(z)|$ for all $ z \in Cl(G)$. If $ f$ is constant we're done. If $ f$ is not constant the results follows from  MMT version 1.[/quote]

I don't understand the part

[quote]"If $ f$ is not constant the results follows from  MMT version 1"[/quote]

Where did  the connectedness go? How do we use MMT version 1?

What is the problem when you have what fedja said? Anyway, the proof you give is technically not enough, I guess. I suspect they wanted $ G$ to be connected as well.

For suppose that $ G \equal{} A \cup B$ with $ A$ and $ B$ sets that are both disjoint and open.  Then if we define $ f(z) \equal{} 1$ on $ A$ and $ f(z) \equal{} 0$ on $ B$ then $ f$ is analytic on the open set $ G$ but $ \mid f(z) \mid$ does not take its maximum on the boundary of $ G$.


if $ a,b,c \in \mathbb{R}^\plus{}$ , prove that :

$ (a^2\plus{}2)(b^2\plus{}2)(c^2\plus{}2) \geq 9(ab\plus{}ac\plus{}bc)$

 :D

[quote="Hidden Scofield"]if $ a,b,c \in \mathbb{R}^ \plus{}$ , prove that :

$ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2) \geq 9(ab \plus{} ac \plus{} bc)$

 :D[/quote]
The stronger one:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76508

[hide="Hint"]
Wlog $ (a\minus{}1)(b\minus{}1) \ge 0$ then $ (a^2\minus{}1)(b^2\minus{}1) \ge 0$ implying $ (a^2\plus{}2)(b^2\plus{}2) \ge 3(a^2\plus{}b^2\plus{}1)$ then we can use CS[/hide]

[quote="Hidden Scofield"]thank arqady about the link.

hey FOURRIER , Could you post a full proof.

 :D[/quote]

I did :P


AMC
AIME
geometry
3D geometry
email
geometric transformation
rotation

So here's part of an email I sent to the AMC people:

"I believe that question #5 on the 2008 AIME I can be interpreted in two valid ways which lead to different answers, and thus both answers should be accepted.

The question says that "the cone first returns to its original position on the table after making 17 complete rotations" but does not specify the axis/axes of rotation. The majority interpretation seems to be that this means 17 rotations around the cone's axis, which would make the answer 014.

However, if the cone makes n rotations around its own axis and returns to its original position, it also makes another complete rotation around an axis which is perpendicular to the table and goes through the vertex of the cone--that is, it sweeps out a circle on the table. Thus, "17 complete rotations" could mean that the cone makes 16 rotations around its own axis and 1 around this second axis, so that msqrt(n)=1*sqrt(255) and the answer is 1+255=256.

I would also argue that anyone who answers the problem assuming the second interpretation has already thought through the first interpretation and rejected it on the basis that it doesn't take the entirety of the cone's movement into account. They should not be punished for overthinking the problem. The second interpretation does not simplify the problem or change it in any substantial way."

In hindsight, I'm not sure what possessed me to interpret the problem this way despite the answer format and the fact that the AIME just doesn't use tricks like this...but at the time, I was pretty convinced that my answer was "more correct". I haven't found anything inherent in the problem that would eliminate this interpretation. So I guess my questions are 1) did anyone else do this/consider doing this, and 2) does my argument actually have any merit? Well, also 3) if my argument does have merit, is there anything else I should do with it to ensure that it's considered before grading starts?

I see where you're coming from
The question should have read "17 complete rotations about the cone's altitude" or something like that.

I actually did think of that interpretation, but I immediately discarded it as incorrect because it didn't seem to be going anywhere in my head. I didn't bother to justify why I discarded it, but looking back, that interpretation does seem valid.

I couldn't interpret that problem at all at first, but I eventually settled around it rotating about its own axis.

But here's the problem with your interpretation, bookaholic:

The earth [b]rotates[/b] about its own axis,
but it [b]revolves[/b] around the sun.

http://en.wikipedia.org/wiki/Rotation

"Mathematically, a rotation is, unlike a translation, a rigid body movement which keeps a point fixed. This definition applies to rotations within both two and three dimensions (in a plane and in space, respectively.) A rotation in three-dimensional space keeps an entire line fixed, i.e. a rotation in three-dimensional space is a rotation around an axis."

Both of the rotations that bookaholic referred to are contained by this definition.

The first keeps the central axis of the cone fixed relative to the rest of the cone.
The second keeps the vertex of of the cone fixed relative to the table.

[quote="distracted523"]But here's the problem with your interpretation, bookaholic:

The earth [b]rotates[/b] about its own axis,
but it [b]revolves[/b] around the sun.[/quote]
Your version of rotation of an object is (to the rest of the mathematical world) a rotation about some point in the object itself, and your definition of revolution of an object is a rotation about some point outside the object.  While such a distinction might be helpful in astronomy or something, in a mathematical context it is quite arbitrary.

Here is one serious objection to bookaholic's interpretation :  the cone was not engaged in simple rotation about this axis in the center of the tables.  If you follow one point on the cone as it moves, it will not trace a circular path about this axis.  Sure, topologically, the path is equivalent to a rotation, and in some sense the cone has rotated about this axis.  But the same is true for any of the infinitely many oblique axes that pass through the cone's vertex, so wouldn't you have infinitely many rotations whatever you did?  The only thing that distinguishes the axis perpendicular to the table is that the cone does indeed make a complete rotation about it if you identify all the cone's points with each other.  But in this case, rotation about the cone's central axis doesn't make very much sense at all, so I think that the mainstream interpretation is the only solid interpretation.

[quote]Here is one serious objection to bookaholic's interpretation : the cone was not engaged in simple rotation about this axis in the center of the tables. If you follow one point on the cone as it moves, it will not trace a circular path about this axis. Sure, topologically, the path is equivalent to a rotation, and in some sense the cone has rotated about this axis. But the same is true for any of the infinitely many oblique axes that pass through the cone's vertex, so wouldn't you have infinitely many rotations whatever you did? The only thing that distinguishes the axis perpendicular to the table is that the cone does indeed make a complete rotation about it if you identify all the cone's points with each other. But in this case, rotation about the cone's central axis doesn't make very much sense at all, so I think that the mainstream interpretation is the only solid interpretation.[/quote]
I'm not sure I see how I would be "identifying all the cone's points with each other". My argument is that from start to finish, each point on the cone moves exactly 360 degrees around the axis perpendicular to the table. The end result of this is the same as a simple rotation around the axis. The fact that the points do not trace a circular path is exactly accounted for by the 16 other rotations.

Also, the points on the cone's altitude do move in a circle. Since the problem doesn't specify an axis of rotation, I think it's problematic to insist that the points must stay on a single circle, because only the altitude points stay on a circular course without reference to a particular (moving) axis. That course goes around the axis perpendicular to the table. So the problem basically forces everyone to assume something about which axis/axes are being referred to--I just picked an unpopular assumption.

I think that answer is valid, but I don't think its the one that would be expected to be seen. (helping you bump too)

Thanks for the bump :D Yeah, I don't deny that I should have expected the obvious interpretation to be the correct one. But I don't think there's anything wrong with my interpretation outside of "AIME writers wouldn't do that", and I think relying on "AIME writers wouldn't do that" would set a bad precedent.

[quote="watermaximillion"]I understand your interpretation, but I think that is the least obvious of the three interpretations I came up with. I came up with the two that you came up with and a third. What if the 17 complete rotations is around the vertex of the cone and the original position means that the cone has to be in the exact same orientation aka the same part of the cone is touching the table? I am not sure whether or not that  gets a legitimate answer because I ended up just doing the mainstream one, but it could work because all it would be is that the number of rotations necessary for the cone to make it back to its original orientation, position and rotation of cone, could be 17 rotations if after one rotation a different part of the cone was touching the horizontal plane.[/quote]
After every integral number of rotations, the same part of the cone touches the table anyhow, so I'm pretty sure your new interpretation is the same as the mainstream one.

[quote="bookaholic"]I'm not sure I see how I would be "identifying all the cone's points with each other". My argument is that from start to finish, each point on the cone moves exactly 360 degrees around the axis perpendicular to the table.[/quote]
How can this be?  Degrees are a planar measure, and most of the cone's points do not follow a planar path around the perpendicular axis.

[quote]The end result of this is the same as a simple rotation around the axis.[/quote]
It is also the same as any integral number of rotations about this axis, as the cone returns to its original position.  We could also perform several translations such that each point follows a path topologically equivalent to a circle, and which return each point to its original position.  Although the end result is the same as a simple rotation about any axis we choose, I do not think anybody would call such a motion a rotation.

[quote]The fact that the points do not trace a circular path is exactly accounted for by the 16 other rotations.[/quote]
Could you please explain this sentence more thoroughly?

[quote]Also, the points on the cone's altitude do move in a circle. Since the problem doesn't specify an axis of rotation, I think it's problematic to insist that the points must stay on a single circle, because only the altitude points stay on a circular course without reference to a particular (moving) axis. That course goes around the axis perpendicular to the table. So the problem basically forces everyone to assume something about which axis/axes are being referred to--I just picked an unpopular assumption.[/quote]
The trouble is that as a point is rotated about a given axis, it traces a circle in a plane perpendicular to the axis of rotation.  No point on the cone (except points on the altitude, as you pointed out) traces a circlular path around your perpendicular axis.  On the other hand, every point on the cone traces a circlular path around the altitude as the cone---with the altitude---spins around the table.

Also, I still do not see what is special about the axis perpendicular to the table.  What distinguishes this axis from an arbitrary oblique axis passing through the cone's vertex?  Why can't we apply your arguments there?

What bookaholic meant by "The fact that the points do not trace a circular path is exactly accounted for by the 16 other rotations" is that, if you took out the other 16 rotations (aka the cone is somehow sliding around the table in a circle, with the same side always touching the table), each point on the cone would trace a circular path. Add in the other 16 rotations around the cone's axis, and you get what is happening in the problem.
You can't do that with just any axis passing through the vertex.

[quote="Alumen"]
You can't do that with just any axis passing through the vertex.[/quote]
The entirety of your argument rests on this point, but Boy Soprano's last two paragraphs prove it wrong.  [i]Any[/i] point which is rotating will trace a circle around the axis.

Okay, now I think I understand the argument, but I still don't think it's really valid, because you do not have unique representation of any motion as a sum of rotations.

For an example of this issue, consider a point in the plane rotating a full turn counterclockwise.  How many rotations is this?  You say it is one rotation, but I say it is three, because I am applying two counterclockwise rotations and one clockwise rotations at the same time; therefore, three rotations.

I don't know if this really helps your argument, but a girl at my school interpreted it the same way as you did and got it wrong (answer 256) because of it.

[quote="Boy Soprano II"]For an example of this issue, consider a point in the plane rotating a full turn counterclockwise.  How many rotations is this?  You say it is one rotation, but I say it is three, because I am applying two counterclockwise rotations and one clockwise rotations at the same time; therefore, three rotations.[/quote]
That's like saying the polynomial 2x+1 has three terms because you can write it as 3x-x+1.

[quote="bookaholic"]But there is no functional difference between one counterclockwise rotation and two counterclockwise rotations + one clockwise rotation, while there is a functional difference between the cone moving as it does in the problem and rotating in place, which would get rid of the "extra" rotation I'm arguing for. The additional rotation is necessary to fully represent the cone's movement.

P.S. Alumen, thanks for the great explanation![/quote]
Okay, now I have a better idea of what you are saying.  I think that your interpretation would be valid if the problem had said something like, "Seventeen is the smallest value of $ n$ such that the motion of the cone can be expressed as the composition of $ n$ rotations proceeding at the same rate."  However, I still think this is a bit of a stretch.


logarithms
real analysis
real analysis unsolved

Can anyone help me with this problem?

(x2-5x+5)x2-9x+20=1

I think you're supposed to start with logarithms, but I'm not sure. I'd appreciate any assistance.

see here : http://www.mathlinks.ro/Forum/viewtopic.php?t=59286


inequalities
function
inequalities unsolved

[quote="payman_pm"]if $a,b,c$ be positive find the minimum value of  $S$
$S=(\frac {a+b} {c})^2 +(\frac{a+c}{b})^2 +(\frac {b+c}{a})^2$[/quote]

Are you sure about your question? If it is this way round, then it's trivial:

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

$\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}=\left(\frac{b}{a}+\frac{c}{a}\right)+\left(\frac{c}{b}+\frac{a}{b}\right)+\left(\frac{a}{c}+\frac{b}{c}\right)$
$=\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)\geq 2+2+2=6$

(since $\frac{x}{y}+\frac{y}{x}\geq 2$ for any two positive x and y). Hence,

$\frac{\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}}{3}\geq \frac63=2$,

and by the AM-QM inequality,

$\frac{\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2+\left(\frac{a+b}{c}\right)^2}{3}\geq \left(\frac{\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}}{3}\right)^2\geq 2^2=4$,

so that

$\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2+\left(\frac{a+b}{c}\right)^2\geq 3\cdot 4=12$.

Thus, 12 is the required minimum (it is achieved for a = b = c).

  Darij

If someone want to complicate it then he should apply Jensen's inequality to the convex fucntion $\left(\frac{1-x}{x}\right)^2$.  :lol:

the solution is very simple!
we have
$S=\frac {a^2}{c}+\frac{b^2}{c}+\frac{2ab}{c}+\frac {a^2}{b}+\frac{c^2}{b}+\frac{2ac}{b}+\frac{b^2}{a}
+\frac {c^2}{a}+\frac{2bc}{a}$

and by $AM-GM$ inaquality we have
$S\ge 12$


Actually, this problem is from a book of Valentin Vornicu:
"Olimpiada de Matematica
De la provocare la experienta" at a Simulation test for IMO-BMO. Problem 1.
It's original formulation was:
" Prove that any triangle can be divided into n isosceles triangles, n>=4".
The solution is quite simple, but it doesn't check this following case, and I didn't find such a partition:

Can an equilateral triangle be partitioned into [i]5[/i] isosceles non- degenerate triangles?

Let $ABC$ be the equilateral triangle. Take $P$ inside the triangle s.t. $BPC$ is a right isosceles triangle ($\angle BPC=\frac {\pi}2$). Now take $M\in (AB),N\in (AC)$ s.t. $\angle MPB=\angle NPC=\frac {\pi}{12}=15^{\circ}$. The triangles we want are $BPC,PMB,PNC,PAM,PAN$.


Φέρνουμε $ EF\equal{}\parallel{}BD$
Τότε $ DBEF$ είναι ρόμβος και το $ EFC$ ισοσκελές με κορυφή το $ E$ άρα η γωνία $ BLF\equal{}90\minus{}A/2\minus{}C$είναι σταθερή και ο λόγος $ LF/FE\equal{}LF/FB\equal{}const$ αρα το $ F$ ανήκει στον απολλώνιο κύκλο των $ B,C$ με λόγο $ LF/FE$ και την ημιευθεία $ LF$ κλπ
Θέμα της ΕΜΕ 1972 την χρονιά που έδωσα και εγώ!

Suggnwmh alla mhpws mporeite na dieukrinhsete ti einai to L?
Epishs, poios einai o appolwnios kuklos??? :blush:

φίλε salonikie ο Απολλώνιος κύκλος είναι ο γεωμετρικός τόπος των σημείων Μ του επιπέδου που ο λόγος των αποστάσεων τους από 2 σταθερά σημεία Α και Β είναι σταθερός,βλέπε σελίδα 161 σχολικού γεωμετρίας Α ακι Β Λυκείου

r_boris και γω δεν κατάλαβα ποιό σημείο είναι το L
επίσης δεν κατάλαβα γιατί το ΕF=//BD μας εξασφαλίζει ότι το DBEF είναι ρομβος,
αυτό μας δίνει το παραλληλόγραμμο, εκτός αν ξέχασες κάτι να μας πεις ή είναι κάτι που δεν το έχω καταλάβει πως ισχύει!
όποτε βρεις χρόνο, πες μας λιγο πιο αναλυτικά τι εννοείς,
σε ευχαριστώ και πάλι!
καλό απόγευμα

[quote="spinos"]Συγνώμη είναι ρόμβος το BDEF, τώρα που τα ξαναείδα,
είχα πάρει το σημείο F πάνω από το Ε ενώ είναι κάτω απ΄το Ε στην ίδια ευθεία που είναι παράλληλη με το BD και μου έβγαινε εμένα η DE σαν διαγώνιο του παραλληλογράμμου BDEF
ενώ αν το πάρεις το F στην ίδια ευθεία κάτω από το Ε βγαίνει ρόμβος και η DE είναι πλευρά του[/quote]
ΣΥΓΝΩΜΗ ΠΑΙΔΙΑ ΟΠΟΥ ΒΛΕΠΕΤΕ $ L$ ΒΑΛΤΕ $ C$

[quote="salonikios"]Wraia, dhladh to F einai ston appolwnio kuklo twn B,C me logw CF/FE omws den katalavainw pws 8a ginei h kataskeuh thn stimgh pou den gnwrizoume pou vrisketai to shmeio F (afou h diereunush egine me vash ta E,D dedomena). Mporeite na to e3hghsete ligo parapanw?
Hlias[/quote]

Αφού η γωνία BCF είναι σταθερή η ημιευθεία CF είναι γνωστή. Αυτός είναι ο δεύτερος τόπος του F οπότε προσδιορίζεται

[quote="salonikios"]Tr to katalava auto..Alla mperdeuthka se ena allo shmeio. Ton logo CF/FE pws ton gnwrizoume? Afou den exoume prosdiorisei to shmeio F akoma...(Milaw gia thn kataskeuh)
Sorry an sas exw prh3ei!

Hlias[/quote]
από το γεγονός ότι το τρίγωνο ECF έχει σταθερές γωνίες  αρα από το θεώρημα ημιτόνων και σταθερό λόγο πλευρών

Wraia to katalava twra! Euxaristw polu!!

[quote="stelmarg7"]Κατασκευάστε με την μέθοδο της ομοιοθεσίας ένα ευθύγραμμο τμήμα DE, έτσι ώστε το σημείο D να βρίσκεται πάνω στην AB και το Ε πάνω στην AC σε δοθέν τρίγωνο ABC, έτσι ώστε BD=DE=EC.[/quote]
Να κατασκευαστεί τρίγωνο $ ABC$ από τα στοιχεία γωνία$ A$, πλευρές $ a\plus{}b , a\plus{}c$
Είναι σχεδόν ίδια άσκηση

[quote="r_boris"][quote="stelmarg7"]Κατασκευάστε με την μέθοδο της ομοιοθεσίας ένα ευθύγραμμο τμήμα DE, έτσι ώστε το σημείο D να βρίσκεται πάνω στην AB και το Ε πάνω στην AC σε δοθέν τρίγωνο ABC, έτσι ώστε BD=DE=EC.[/quote]
Να κατασκευαστεί τρίγωνο $ ABC$ από τα στοιχεία γωνία$ A$, πλευρές $ a \plus{} b , a \plus{} c$
Είναι σχεδόν ίδια άσκηση[/quote]
Αν πάρουμε γωνία $ xoy\equal{}A$ και στις πλευρές $ ox$ πάρουμε μήκος $ OB\equal{}a\plus{}b$ ενώ στην $ oy$ πάρουμε μήκος $ OC\equal{}a\plus{}c$ τότε η πλευρά $ DE\equal{}a$ προσδιορίζεται όπως στο προηγούμενο πρόβλημα


For integers $a,b,c,d > 1$, find the smallest $k$ such that it can be expressed as:

$k=a^{1}+1=b^{3}+3=c^{5}+5=d^{7}+7$

Can you generalize, for $n$ equivalent expressions? Will there always be a solution?

Well whatever $k$ is, it's something over $10^{25}$... :read:

Oh.  :rotfl:  How'd you find that?

(What I'm more interested in, tho, is how to go about solving this problem. I have basically no clue, and the only thing I can think of is mods... but I havent gotten that far with them. Any help?)

sam, problems that you just make up tend not to have elegant solutions...

:rotfl: Aww. Fine.  :P  Im still interested, though, on any ideas for this problem.


After the episode  with  [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=25734]mutton[/url]
It’s strict diet for [url=http://www.shopnbc.com/media/products/R/R10427_200.jpg]Button[/url]
After  five-sevenths of his [url=http://www.crofter.com/graphics/sorrel.jpg]sup'[/url]
He is left with five-sevenths  of a [url=http://www.drdandanner.com/Images/dog%20food.jpg]cup[/url].
How many cups of  food
Is he  gettin’ ?

[url=http://www.anvari.org/db/fun/Cute_Kids/Eating_Dog_Food.jpg]hint: Click here.[/url]

(You may like to click on  the highlighted text - eg  "mutton" etc above  for more clarity ) :)

[hide]
Well, x being the supper size, we have that he ate 5/7 x, and has 2/7 x left. Solving further, we get that 2/7 x = 5/7, and if we multiply each side by 7/2, we get 5/2, so Button only ate 2.5 cups of mutton today.
[/hide]

[hide]\frac{2}{7}d=5/7
d=35/14=5/2 cup to start with.[/hide]

Billy

[hide]d=dinner
2/7x=what he has left
2/7x=5/7 (cups)
x=5/2
>>2.5<<[/hide]


function
integration
real analysis
real analysis unsolved

$ f: [0,1]\minus{}>R$ with $ \int_0^1f(x)dx\equal{}0$ and let $ n$ be an odd number. Prove that there exists $ x$ s.t. $ f^n(x)\equal{}\int_0^xf(t)dt$.

I'll assume that $ f$ is continuous; otherwise the statement is false.

Sketch: Suppose that $ f^n(x)>\int_0^x f(t)\,dt$ for all $ x\in [0,1]$. Since $ f(0)>0$, the function $ \int_0^x f(t)\,dt$ attains a positive maximum at some point $ x_0\in (0,1)$. Then $ f(x_0)\equal{}0$, which contradicts our assumption that $ f^n(x_0)>\int_0^{x_0} f(t)\,dt$.