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Hello

I know that the following problem has already been posted before, and have seen the solutions, but I was wondering if someone could help clarify some issues that I have with the following solution that comes from Mathematical Olympiads.

Problem.
Show that for any positive integer $ n$, the polynomial $ f(x)=(x^{2}+x)^{2^{n}}+1$ cannot be written as the product of two nonconstant polynomials with integer coefficients.

Solution.
Note that $ f(x)=g(h(x))$, where $ h(x)=x^{2}+x$ and $ g(y)=y^{2^{n}}+1$.  Since $ g(y+1)=(y+1)^{2^{n}}+1=y^{2^{n}}+(\sum_{k=1}^{2^{n}-1} \binom{2^{n}}{k} y^{k})+2$, and $ \binom{2^{n}}{k}$ is even for $ 1 \le k \le 2^{n}-1$, $ g$ is irreducible by Eisenstein's criterion.

Now let $ p$ be a nonconstant factor of $ f$, and let $ r$ be a root of $ p$.  Then $ g(h(r))=f(r)=0$, so $ s=h(r)$ is a root of $ g$.  Since $ s=r^{2}+r \in \mathbb{Q}(r)$, we have $ \mathbb{Q}(s) \subset \mathbb{Q}(r)$, so $ \deg p \ge \deg (\frac{\mathbb{Q}(r)}{\mathbb{Q}}) \ge \deg (\frac{\mathbb{Q}(s)}{\mathbb{Q}})=\deg g=2^{n}$.

Thus every factor of $ f$ has degree at least $ 2^{n}$.  Therefore, if $ f$ is reducible, we can write $ f(x)=d(x)e(x)$ where $ d$ and $ e$ have degree $ 2^{n}$.

Next, observe $ f(x) \equiv (x^{2}+x)^{2^{n}}+1 \equiv x^{2^{n+1}}+x^{2^{n}} +1 \equiv (x^{2}+x+1)^{2^{n}} (mod 2)$; since $ x^{2}+x+1$ is irreducible in $ \mathbb{Z}_{2}[x]$, by unique factorization we must have $ d(x) \equiv e(x) \equiv (x^{2}+x+1)^{2^{n-1}} \equiv x^{2^{n}}+x^{2^{n-1}}+1 (mod 2)$.

Thus, if we write $ d(x)=a_{2^{n}} x^{2^{n}}+...+a_{0}$, $ e(x)=b_{2^{n}} x^{2^{n}}+...+b_{0}$, then $ a_{2^{n}}$, $ a_{2^{n-1}}$, $ a_{0}$, $ b_{2^{n}}$, $ b_{2^{n-1}}$, $ b_{0}$ are odd and all the other coefficients are even.  Since $ f$ is monic, we may assume WLOG that $ a_{2^{n}}=b_{2^{n}}=1$ also, $ a_{0} b_{0}=f(0)=1$, but $ a_{0} >0$, $ b_{0} >0$ as $ f$ has no real roots, so $ a_{0}=b_{0}=1$.  Therefore, $ ([x^{2^{n}+2^{n-1}}]+[x^{2^{n-1}}])(d(x)e(x))) \equiv (\sum_{i=2^{n-1}}^{2^{n}} a_{i} b_{2^{n}+2^{n-1}-i})+(\sum_{i=0}^{2^{n-1}} a_{i} b_{2^{n-1}-i}) \equiv a_{2^{n}} b_{2^{n-1}}+a_{2^{n-1}} b_{2^{n}}+a_{0} b_{2^{n-1}}+a_{2^{n-1}}b_{0} \equiv 2(a_{2^{n-1}}+b_{2^{n-1}}) \equiv 0 (\mod 4)$ as $ a_{2^{n-1}}+b_{2^{n-1}}$ is even.

But $ ([x^{2^{n}+2^{n-1}}]+[x^{2^{n-1}}])(f(x))=\binom{2^{n}}{2^{n-1}}=2\binom{2^{n}-1}{2^{n-1}-1}$, and $ \binom{2^{n}-1}{2^{n-1}-1}$ is odd by Lucas' Theorem, so $ ([x^{2^{n}+2^{n-1}}]+[x^{2^{n-1}}]) (f(x)) \equiv 2(mod 4)$, a contradiction.  Hence f is irreducible.

So, I have a good deal of questions from this soultion.

1) I know that $ \mathbb{Q}$ represents the rationals, but what exactly does $ (\frac{\mathbb{Q}(r)}{\mathbb{Q}})$ represent?

2) In connection with the previous question why is $ \deg p \ge \deg (\frac{\mathbb{Q}(r)}{\mathbb{Q}}) \ge \deg (\frac{\mathbb{Q}(s)}{\mathbb{Q}})=\deg g=2^{n}$?

3) Why are the coefficients $ a_{2^{n}}$, $ a_{2^{n-1}}$, $ a_{0}$, $ b_{2^{n}}$, $ b_{2^{n-1}}$, $ b_{0}$ odd and the others even?

That is all, thanks.

[quote="mathgeniuse^ln(x)"]
1) I know that $ \mathbb{Q}$ represents the rationals, but what exactly does $ (\frac {\mathbb{Q}(r)}{\mathbb{Q}})$ represent?

2) In connection with the previous question why is $ \deg p \ge \deg (\frac {\mathbb{Q}(r)}{\mathbb{Q}}) \ge \deg (\frac {\mathbb{Q}(s)}{\mathbb{Q}}) = \deg g = 2^{n}$?

3) Why are the coefficients $ a_{2^{n}}$, $ a_{2^{n - 1}}$, $ a_{0}$, $ b_{2^{n}}$, $ b_{2^{n - 1}}$, $ b_{0}$ odd and the others even?[/quote]
Let me try to explain these basic algebraic arguments. (This could take a while)
But first you have to look up the exact definitions of [url=http://en.wikipedia.org/wiki/Field_%28mathematics%29]field[/url] and [url=http://en.wikipedia.org/wiki/Vector_space]vector space[/url].

1)
Note that every vector space (defined over any field) has a basis, and any two bases
have the same number of elements, which is called the dimension of the vector space.

And now given two fields $ F\subseteq E$ (a so called [url=http://en.wikipedia.org/wiki/Field_extension]field extension[/url]),
then $ E$ is also a vector space over $ F$ simply by restricting the multiplication of $ E$ to $ F\times E\to E$.
But as such $ E$ has a dimension over $ F$, which is called the degree of the field extension, and is denoted by $ [E: F]$ or here $ \deg ({E\over F})$.
Also for any set $ S\subseteq E$ we denote $ F(S)$ the smallest subfield of $ E$ containing $ F$ and $ S$ (Meaning the intersection of all such subfields)
and for $ S=\{s\}$ we simply write $ F(\{s\})=F(s)$.

This should explain $ \deg (\frac {\mathbb{Q}(r)}{\mathbb{Q}})$.

2) Let's show $ m: =\deg p \ge \deg (\frac {\mathbb{Q}(r)}{\mathbb{Q}}) \ge \deg (\frac {\mathbb{Q}(s)}{\mathbb{Q}}) >= \deg g = 2^{n}$
First $ \deg (\frac {\mathbb{Q}(r)}{\mathbb{Q}}) \ge \deg (\frac {\mathbb{Q}(s)}{\mathbb{Q}})$ is clear, as $ \mathbb{Q}(r)\supseteq\mathbb{Q}(s)$.
(which follows by definition of $ \mathbb{Q}(s)$, since $ \mathbb{Q}(r)$ is a subfield of $ \mathbb{C}$ containing $ \mathbb{Q}$ and $ s=r^2+r$)

For $ \deg (\frac {\mathbb{Q}(s)}{\mathbb{Q}}) >= \deg g = 2^{n}$ we just need to show, that $ 1,s,s^2,\ldots ,s^{2^n-1}\in\mathbb{Q}(s)$ are linear independent over $ \mathbb{Q}$.
(and therfore extend to a basis with $ >=2^n$ elements)
Otherwise take $ \alpha_0,\ldots,\alpha_{2^n-1}\in\mathbb{Q}$ not all $ 0$ with $ \alpha_0+\alpha_1 s +\cdots +\alpha_{2^n-1} s^{2^n-1}=0$.
Then $ G(x): =\alpha_0+\alpha_1 x +\cdots +\alpha_{2^n-1} x^{2^n-1}\in\mathbb{Q}[x]\setminus\{0\}, \deg G<2^n=\deg g$ has root $ s$.
But $ g(x)$ is irreducible, and therefore coprime to $ G(x)$, and using the [url=http://en.wikipedia.org/wiki/Euclidean_algorithm]Euclidean Algorithm[/url] (repeated polynomial division)
we find $ u(x),v(x)\in\mathbb{Q}[x]$ with $ u(x)g(x)+v(x)G(x)=1$.
Then $ x=s$ gives the contradiction.

Lastly for $ m: =\deg p \ge \deg (\frac {\mathbb{Q}(r)}{\mathbb{Q}})$ we have to show, that the set $ \{1,r,r^2,\ldots,r^{m-1}\}$ spans the $ \mathbb{Q}$-vector space $ \mathbb{Q}(r)$,
meaning $ \mathbb{Q}(r)=\{\alpha_0 +\alpha_1 r+\cdots +\alpha_{m-1} r^{m-1}: \alpha_0,\ldots ,\alpha_{m-1}\in\mathbb{Q}\}$
or better $ \mathbb{Q}(r)=\{G(r): G(x)\in\mathbb{Q}[x],\deg G <m\}$.
For that we just need to show, that $ F: =\{G(r): G(x)\in\mathbb{Q}[x],\deg G <m\}$ is a subfield of $ \mathbb{C}$,
meaning closed under addition, subtraction, multiplication and division. ($ \mathbb{Q}(r)$ obviously contains $ F$)
But $ F$ is obviously closed under addition and subtraction, and for any $ G,H\in\mathbb{Q}[x],\deg G ,\deg H<m$
we can write $ G(x)H(x)=Q(x)p(x)+R(x)$ for some $ Q(x),R(x)\in\mathbb{Q}[x]$
with $ \deg R<\deg p=m$. (polynomial division)
Then $ x=r$ gives $ G(r)H(r)=R(r)\in F$, so $ F$ is also closed under multiplication.
To show $ {a\over b}\in F$ for all $ a,b\in F, b\not =0$, we take some basis $ \beta_1,\ldots ,\beta_k$ of the $ \mathbb{Q}$-vector space $ F$.
($ F$ is spanned by $ m$ elements $ 1,r,\ldots ,r^{m-1}$, so there is a finite basis)
Then $ b\beta_1,\ldots ,b\beta_k\in F$ are also linear independent over $ \mathbb{Q}$.
But then $ b\beta_1,\ldots ,b\beta_k$ is also a basis of $ F$ and we can find $ \alpha_1,\ldots ,\alpha_k\in \mathbb{Q}$ with
$ a=\alpha_1 \cdot b\beta_1+\cdots +\alpha_k\cdot b\beta_k$, and therfore $ {a\over b}=\alpha_1 \beta_1+\cdots +\alpha_k \beta_k\in F$.

3)
Now I have to take some issue with the proof.
If $ f(x)\in\mathbb{Q}[x]$ is reducible, we can write $ f(x) = d(x)e(x)$,
where $ d(x),e(x)\in\mathbb{Q}[x]$ have degree $ 2^{n}$ and leading coefficient $ 1$. That's ok.
But to assume $ d(x),e(x)\in\mathbb{Z}[x]$ to calculate $ \pmod {2}$ is not so trivial, and requires a lemma by Gauss.
Of course this wasn't part of the problem, and the proof is still correct like that, but still I wanted to make that clear.

Note that $ \mathbb{F}_2=\mathbb{Z}/2\mathbb{Z}$ is a field. ($ \mathbb{Z}_2$ is something else) So $ \mathbb{F}_2[x]$ is an [url=http://en.wikipedia.org/wiki/Euclidean_domain]Euclidean Domain[/url].
Also for all $ a,b\in\mathbb{F}_2[x]$ we have $ (a+b)^2=a^2+2ab+b^2=a^2+b^2$
and then by induction $ (a+b)^{2^n}=a^{2^n}+b^{2^n}$ for all $ n\in\mathbb{N}$.
This shows $ (x^{2} + x)^{2^{n}} + 1 =x^{2^{n + 1}} + x^{2^{n}} + 1 = (x^{2} + x + 1)^{2^{n}}$ in $ \mathbb{F}_2[x]$.
Now assume $ f(x) = d(x)e(x)$ with $ d(x),e(x)\in\mathbb{Z}[x]$ of degree $ 2^{n}$ and leading coefficient $ 1$.
Then $ \bar{d}(x),\bar{e}(x)\in\mathbb{F}_2[x]$ ($ d(x),e(x)$ taken modulo $ 2$) have also degree $ 2^{n}$ with $ \bar{d}(x)\bar{e}(x)=(x^{2} + x + 1)^{2^{n}}$.
Now write $ \bar{d}(x)=(x^{2} + x + 1)^{m}c(x)$ with $ c(x)\in\mathbb{F}_2[x]$ and $ m\in\mathbb{N}_0$ maximal.
Then $ x^2+x+1\nmid c(x)$, and since $ x^2+x+1\in\mathbb{F}_2[x]$ is irreducible (has no root), we have $ x^2+x+1$ and $ c(x)$ coprime.
Then again using the [url=http://en.wikipedia.org/wiki/Euclidean_algorithm]Euclidean Algorithm[/url] we get $ u(x)(x^2+x+1)+v(x)c(x)=1$ for some $ u(x),v(x)\in\mathbb{F}_2[x]$.
But $ c(x)\mid \bar{d}(x)\mid (x^{2} + x + 1)^{2^{n}}$, and therefore
$ 1=(u(x)(x^2+x+1)+v(x)c(x))^{2^n}$$ =u(x)^{2^n}(x^{2} + x + 1)^{2^{n}}+\sum_{k=1}^{2^n}({2^n\atop k})v(x)^k c(x)^k[u(x)(x^2+x+1)]^{2^n-k}$
is divisible by $ c(x)$. So $ 0\not =c(x)\in\mathbb{F}_2[x]$ is constant.
This gives $ c(x)=1$ and $ m=2^{n-1}$ and $ \bar{d}(x)=(x^{2} + x + 1)^{2^{n-1}}=x^{2^n}+x^{2^{n-1}}+1$.
So $ d(x)\equiv x^{2^n}+x^{2^{n-1}}+1\pmod{2}$, which means that all coefficients of $ d(x)$ are even, except $ a_{2^n}=1,a_{2^{n-1}}$ and $ a_{0}$.
Of course $ e(x)$ is analog.

Now please note, that I only used most basic algebraic arguments, but for [b]all fields[/b] and not just $ \mathbb{Q},\mathbb{R},\mathbb{C}$.
Also note, that everything I did is standard, which you'll find in every decent algebra book, and much more straight forward.
(I had to do some loops without the proper theorems)


linear algebra
matrix
modular arithmetic
combinatorics proposed
combinatorics

Double-faced coins are arranged with all the heads facing upward in the shape of $17\times17.$At one operation, you turn over $5$ consecutive coins in longitudinal or $5$ ones in transversal or $5$ ones in oblique at the same time.Now can you make all those reverses face upward when you repeat this operation?

I believe the answer is negative.

We fill a $5\times 5$ square with $0$ and $1$ as follows: $\begin{array}{ccccc}1&0&0&1&0\\1&1&1&1&0\\1&1&1&1&0\\1&0&0&1&0\\0&0&0&0&0\end{array}$. Now we fill a $20\times 20$ matrix with $16$ such $5\times 5$ squares, and take our $17\times 17$ matrix to be the one formed by the first $17$ lines and columns of the $20\times 20$ matrix. We regard an operation of turning around $5$ coins as adding the initial value to each of the $5$ places in the $17\times 17$ matrix (everything takes place modulo $2$). For instance, if we turn around the first $5$ coins in the first row, the $1\ 0\ 0\ 1\ 0$ will become $0\ 0\ 0\ 0\ 0$, and if we turn them around again it will become $1\ 0\ 0\ 1\ 0$ again.

If we assume we can turn around all the coins, this means that we can perform a sequence of moves s.t. each coin is turned around an odd number of times, so in this new setting, this would mean that we can perform a sequence of moves s.t. all the entries in the matrix become $0$ (everything is modulo $2$, remember). In particular, the sum of the entries in the matrix will be $0$ in the end. Now notice that the sum of every $5$ consecutive entries in a row, on a column, or on a diagonal is $0$, so the moves do not change the sum of the entries in the matrix. However, the matrix initially has sum $1$, so we have a contradiction (the sum is $1$ because the sum of every $5$ consecutive entries in a row or column is $0$ and $17\equiv 2\pmod 5$, so the sum is equal to the sum of the entris in the upper right $2\times 2$ corner, which is $\begin{array}{cc}1&0\\1&1\end{array}$).

I don't understand. If we flip an arrangement like $1\ 0\ 0\ 1\ 0$, shouldn't it become $0\ 1\ 1\ 0\ 1$ ?


induction
floor function
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algebra
functional equation

Let $f:\mathbb{N}\to \mathbb{N}$ , $f$ is strictly increasing : $f(f(n)) = 3n \quad  \forall n \in \mathbb{N }$. Find $f(2001)$.

Source: Tournament of Towns 1983, 7-8 grades

I remember this problem was at one of the second stages in the Polish MO
last years ...

Since $f(3n)=3f(n)$, it's enough to find $f(667)=f(\frac{2001}3)=\frac{f(2001)}3$.

It's easy to find $f(1)=2,f(2)=3,f(3)=6$. You can find a pattern in the sequence of values $f(n)$. If we list $f(1),f(2),\ldots$, we might try to show (by induction) that we obtain blocks of consecutive integers of the form $2\cdot 3^n,2\cdot 3^n+1,\ldots,3^{n+1}$, followed by blocks of the form $3^{n+1}+3,3^{n+1}+6,\ldots,2\cdot 3^{n+1}-3$. It's not hard from here.

What is tournament of towns , Myth ?????

It is well known international competition for school students (since 1980).
Main advantages of ToT is that you don't need to go to some place to participate in it.

[quote="grobber"]Since $f(3n)=3f(n)$[/quote]

grobber, I must be completely dense, but why is that?

Compute $f(f(f(n)))$ in two ways... Actually, when you have $f(f(n))=$something, applying $f$ in two different ways should be the first thing you consider :D.

d'oh.  really, I don't always need help with math  :blush:

Another [b]very[/b] nice similar problem:

Sequence of positive integers satisfies: $a_1=1$, $a_{n+1}=n+1-a_{a_n}$.
Find $a_{2004}$.

I don't understand what you mean , grobber . Pls, explain clearly .

It would be too tedious for me to write down all the details. Just try to find many values of $f$ and find a pattern.

No, it's 3816.
We can use a bulky idea that compute f(667). I have compute all f(n) with n from 1 to 667 and get the answer.( It cost 3 pages of paper)

[quote="grobber"]Is it $1239$?[/quote]
Yes, you are correct. And where is solution? ;)

@darkmaster. You misunderstood. Grobber answered to another question.
Why did you compute all f(n) for n=1..667 ??? :lol:

[quote="grobber"]I have the feeling (:)) that it can be shown by induction that $f(n)=\lfloor\alpha\cdot (n+1)\rfloor$, where $\alpha=\frac{\sqrt 5-1}2$.[/quote]
Yes, your feeling is right :) It is very nice, isn't it?


geometry
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geometry proposed

Consider in a triangle $ ABC$ the points $ M\in (BC)$ , $ N\in (CA)$ , $ P\in (AB)$ and $ Q\in BN\cap CP$ ,

$ R\in BN\cap MP$ , $ S\in CP\cap MN$ . Prove that $ \boxed {\ [MRQS] \equal{} [NQP]\ \Longleftrightarrow\ \frac {MB\cdot MC}{AN\cdot AP} \equal{} \frac {a^2}{bc}\ }$ .

By  Menelaus' theorem for $\triangle ABN , \triangle APC$ cut by transversals $\overline{PCQ} , \overline{BQN}$ we get

$ \frac {CQ}{PC} \cdot \frac{AP}{AB} \cdot \frac{BN}{QN}=1 \ , \ \frac {BQ}{BN} \cdot \frac{AN}{AC} \cdot \frac{PC}{PQ}=1$

$\Longrightarrow \ \frac {CQ \cdot BQ \cdot AP \cdot AN}{AB \cdot AC \cdot PQ \cdot QN} \equal{} 1$

$\frac {AP \cdot AN}{AB \cdot AC} \equal{} \frac {MB \cdot MC}{BC^2} \Longrightarrow \ \frac {CQ \cdot BQ}{PQ \cdot QN} \equal{} \frac {BC^2}{MB \cdot MC}$

$ \frac {[\triangle BQC]}{[ \triangle PQN]} \equal{} \frac {CQ \cdot BQ}{PQ \cdot QN}  \Longrightarrow \ \frac {[\triangle BQC]}{[\triangle PQN]} \equal{} \frac {BC^2}{MB \cdot MC}$

We must have:  $ \frac{[MRQS]}{[\triangle BQC]} \equal{} \frac {MB \cdot MC}{BC^2},$  which is not true for  arbitrary $M$ on $BC.$

Sorry again. Here is the correct enunciation. Now I am surely that it is right. Success !

[quote="Virgil Nicula"]Let $ \triangle ABC$ , the points $ M\in (BC)$ , $ N\in (CA)$ , $ P\in (AB)$ so that $ \underline {Q\in AM\cap BN\cap CP}$ and

$ R\in BN\cap MP$ , $ S\in CP\cap MN$ . Prove that $ \boxed {\ [MRQS] \equal{} [NQP]\ \Longleftrightarrow\ \frac {MB\cdot MC}{AN\cdot AP} \equal{} \frac {a^2}{bc}\ }$ .[/quote]

[u]Dedicated to [b]Luis González[/b][/u].

[quote="Virgil Nicula"] [color=darkred]Let $ \triangle ABC$ , the points $ M\in (BC)$ , $ N\in (CA)$ , $ P\in (AB)$ so that $ \underline {Q\in AM\cap BN\cap CP}$ and

$ R\in BN\cap MP$ , $ S\in CP\cap MN$ . Prove that $ \boxed {\ [MRQS] = [NQP]\ \Longleftrightarrow\ \frac {MB\cdot MC}{AN\cdot AP} = \frac {a^2}{bc}\ }$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ \boxed {\begin{array}{c} \frac {NA}{n} = \frac {NC}{1} = \frac {AC}{n + 1} \\
 \\
\frac {PA}{p} = \frac {PB}{1} = \frac {AB}{p + 1}\end{array}}$ . Apply [u][b]Ceva[/b]'s theorem[/u]  to $ Q/ ABC\ : \ \ \frac {MB}{MC}\cdot\frac {NC}{NA}\cdot\frac {PA}{PB} = 1$ , i.e. $ \boxed {\frac {MB}{n} = \frac {MC}{p} = \frac {BC}{n + p}}$ .

Apply [u][b]Van Aubel[/b]'s relation [/u] : $ \frac {QA}{QM} = \frac {PA}{PB} + \frac {NA}{NC}$ , i.e. $ \frac {QA}{QM} = n + p$ $ \implies$ $ \boxed {\frac {QA}{n + p} = \frac {QM}{1} = \frac {AM}{n + p + 1}}$ . Apply [u][b]Menelaus[/b]' theorem[/u]

to mentioned transversals/triangles : $ \left\|\begin{array}{c} \overline {BQN}/APC\ : \ \frac {BP}{BA}\cdot\frac {NA}{NC}\cdot\frac {QC}{QP} = 1\implies\frac {n}{p + 1}\cdot\frac {QC}{QP} = 1 \\
 \\
\overline {CQP}/ANB\ : \ \frac {CN}{CA}\cdot\frac {PA}{PB}\cdot\frac {QB}{QN} = 1\implies\frac {p}{n + 1}\cdot\frac {QB}{QN} = 1\end{array}\right\|$ $ \implies$ $ \boxed {\begin{array}{c} \frac {QC}{p + 1} = \frac {QP}{n} = \frac {CP}{n + p + 1} \\
 \\
\frac {QB}{n + 1} = \frac {QN}{p} = \frac {BN}{n + p + 1}\end{array}}$ .

Apply [u][b]Routh[/b]'s relation[/u] : $ \left\|\begin{array}{c} \overline{MRP}\ : \ \frac {RP}{RM} = \frac {BP}{BM}\cdot\frac {NA}{NC}\cdot\frac {BC}{BA}\implies\frac {RP}{RM} = \frac {n + p}{p + 1} \\
 \\
\overline{MSN}\ : \ \frac {SN}{SM} = \frac {CN}{CM}\cdot\frac {PA}{PB}\cdot\frac {CB}{CA}\implies\frac {SN}{SM} = \frac {n + p}{n + 1}\end{array}\right\|\implies$ $ \boxed {\begin{array}{c} \frac {RP}{n + p} = \frac {RM}{p + 1} = \frac {PM}{n + 2p + 1} \\
 \\
\frac {SN}{n + p} = \frac {SM}{n + 1} = \frac {NM}{n + p + 1}\end{array}}$ . Denote $ S = [ABC]$ .

$ \frac {[PQN]}{S} = \frac {[PQN]}{BQC]}\cdot\frac {[BQC]}{ABC]} =$ $ \frac {QP}{QC}\cdot\frac {QN}{QB}\cdot\frac {MQ}{MA} =$ $ \frac {n}{p + 1}\cdot \frac {p}{n + 1}\cdot \frac {1}{n + p + 1}\ \implies\ \boxed {[PQN] = \frac {np}{(n + 1)(p + 1)(n + p + 1)}\cdot S}$ .

${ \left\|\begin{array}{c} \frac {[RQM]}{[ACB]} = \frac {[RQM]}{[PQM]}\cdot\frac {[PQM]}{[PCM]}\cdot\frac {[PCM]}{[PCB]}\cdot\frac {[PCB]}{[ACB]} = \frac {MR}{MP}\cdot\frac {PQ}{PC}\cdot\frac {CM}{CB}\cdot\frac {BP}{BA} = \frac {np}{(n + p)(n + p + 1)(n + 2p + 1)} \\
 \\{
\frac {[SQM]}{[ABC]} = \frac {[SQM]}{[NQM]}\cdot\frac {[NQM]}{[NBM]}\cdot\frac {[NBM]}{[NBC]}\cdot\frac {[NBC]}{[ABC]} = \frac {MS}{MN}\cdot\frac {NQ}{NB}\cdot\frac {BM}{BC}\cdot\frac {CN}{CA} = \frac {np}{(n + p)(n + p + 1)(2n + p + 1)}}\end{array}\right\|$ $ \implies$ $ \boxed {\begin{array}{c} \frac {\left[ RQM\right]}{S} = \frac {np}{(n + p)(n + p + 1)(n + 2p + 1)} \\
 \\
\frac {\left[ SQM \right]}{S} = \frac {np}{(n + p)(n + p + 1)(2n + p + 1)}\end{array}}$ .

Thus, $ [MRQS] = [RQM] + [SQM] =$ $ \frac {np}{(n + p)(n + p + 1)}\cdot\left(\frac {1}{n + 2p + 1} + \frac {1}{2n + p + 1}\right)\cdot S\ \implies$

$ \boxed {[MRQS] = \frac {np[3(n + p) + 2]}{(n + p)(n + p + 1)(n + 2p + 1)(2n + p + 1)}\cdot S}$ . Denote $ \sigma = n + p\ ,\ \pi = np$ . Thus, $ \boxed {\ [MRQS] = [PQN]\ }$  $ \Longleftrightarrow$

$ \frac {3(n + p) + 2}{(n + p)(n + 2p + 1)(2n + p + 1)} =$ $ \frac {1}{(n + 1)(p + 1)}\ \Longleftrightarrow\ (3\sigma + 2)(\sigma + \pi + 1) =$ $ \sigma (2\sigma ^2 + 3\sigma + \pi + 1)\ \Longleftrightarrow$

$ \sigma ^3 - \sigma\pi - 2\sigma - \pi - 1 = 0\ \Longleftrightarrow (\sigma +$ $ 1)(\sigma ^2 - \sigma - \pi - 1) = 0\ \Longleftrightarrow\ \sigma ^2 =$ $ \sigma + \pi + 1\ \Longleftrightarrow\ \boxed {\ (n + p)^2 = (n + 1)(p + 1)\ }\ \Longleftrightarrow$

$ \frac {n}{n + p}\cdot\frac {p}{n + p} =$ $ \frac {n}{n + 1}\cdot\frac {p}{p + 1}\ \Longleftrightarrow\ \frac {MB}{BC}\cdot\frac {MC}{BC} = \frac {AN}{AC}\cdot\frac {AP}{AB}\ \Longleftrightarrow\ \boxed {\frac {MB\cdot MC}{AN\cdot AP} = \frac {a^2}{bc}}$ .

[b][u]Remark[/u].[/b] Show easily that $ [MRQS] = [PQN]\ \Longleftrightarrow\ \left(\frac {QA}{QM}\right)^2 = \frac {CA}{CN}\cdot\frac {BA}{BP}$ . Here are some particular cases :

 $ \blacktriangleright$ [b]IF[/b] $ Q: = I$ is [u]incenter[/u] of $ \triangle ABC$ [b]THEN[/b] $ \boxed {\ [MRQS] = [PQN]\ \Longleftrightarrow\ (b + c)^2 = (a + b)(a + c)\ \Longleftrightarrow\ \frac {RP\cdot SN}{RM\cdot SM} = 1\ }$ ([b]nice ![/b]).

$ \blacktriangleright$ [b]IF[/b] $ Q: = N$ is [u][b]Nagel[/b]'s point[/u] of $ \triangle ABC$ [b]THEN[/b] $ \boxed {\ [MRQS] = [PQN]\ \Longleftrightarrow\ bc = a^2\ }$ ([b]nice ![/b]).

$ \blacktriangleright$ [b]IF[/b] $ Q: = \Gamma$ is [u][b]Gergonne[/b]'s point[/u] of $ \triangle ABC$ [b]THEN[/b] $ \boxed {\ [MRQS] = [PQN]\ \Longleftrightarrow\ bc = a(p - a)\ \Longleftrightarrow\ (a - b)(a - c) + bc = 0\ }$ ([b]nice ![/b]).[/color]


calculus
integration
function
ratio
algebra
polynomial
algebra unsolved

hi evrybody
can one give me the primitve of $f(x)=ln(cos$ $x)$ ( if exist)

He means compute:
$F(x)$ such that $F'(x)=f(x)$

you first use integration by parts and you get
x*ln*(cos(x))-I(x*tg(x))dx
by I i mean primitive of the function x*tg(x)
Then you use the formula tg(x)=2tg(x/2)/(1-tg(x/2)^2)
you make a change of variable taking y=tg(x/2) and then you do one more integration by parts and you aer left with computing the primitive of a ratio of two polynomials which should be pretty easy i guess (i know there are some methods to do this, but i don't remember them).


geometry
circumcircle
analytic geometry
cyclic quadrilateral
geometry proposed

$ABCD$ is a cyclic quadrilateral, and $E$ is the intersection of $AD,BC$.
$M$ moves on the segment $AD$. $N$ is on the segment $BC$ and satisfies $AM/MD=BN/NC$.
Prove that the circumcircle of triangle $MNE$ passes through a fixed point.

Doesn't this follow from http://www.mathlinks.ro/Forum/viewtopic.php?t=68157 ?

  Darij

Really? In my opinion, the points M and N are related by a condition of the kind $BN=k\cdot AM$ for a fixed real k (in fact, $k=\frac{BC}{AD}$, since $\frac{BN}{NC}=\frac{AM}{MD}$ yields $\frac{BN}{BC}=\frac{AM}{AD}$ and thus $BN=\frac{BC}{AD}\cdot AM$), and thus, according to the thread I linked to, the circumcircle of triangle MNE passes through a fixed point... and the quadrilateral ABCD needs not be cyclic.

Have I misread something?

  darij

I think $A\in Ox$ and $B \in Oy$ implies $\angle AOB = \pi /2$.

:rotfl: Ox and Oy are not the axes of a coordinate system, just a bad notation for two lines through a point O. See my post #3 for a better formulation of the problem...

  Darij


I was given the equation w^2+3w<18 the answer is (-6,3)

Given the problem x^2-7x-30>/=0

I gave the answer 10,-3. This was marked wrong. I understand that to answer some inequalities you must write out the answer similar to this (3, infinity sign) U (-infinity,-6) - this is just an example. I was just wondering how you determine when you should write the answer just as (x,y) and when you should write it as (infinity,x) U (x,infinity).

Please help.

I hope this makes sense post if there's any questions.

[quote="pianoforte"][hide](-6,3)
means that the solutions for w are all values between -6 and 3, not including -6 and 3

[-6,3]
would mean all values between -6 and 3 including -6 and 3

( means it doesn't include that number [ means that is does

[x,y)
would be all values between x and y, including x, excluding y

(-infinity, -6) U (3, infinity sign) would mean all values from -infinity to -6, exclusive, [b]and[/b] all values 3 to infinity, exclusive.[/hide][/quote]

But how would I determine which is which?

Make one side zero, and then factor the other side into polynomials of degree 1 (disregard ones with complex roots). The sign of the side with the polynomial changes everytime you go by a root of the polynomial on the interval.

$w^2+3w<18 \rightarrow w^2 + 3w - 18 < 0$.  To solve this, imagine the graph of $w^2 + 3w - 18$.  We know, since the leading coefficient is 1, that it opens upwards.  So the right and left sides of the graph both approaches infinity.  So, the only possibility of the graph being < 0 is if the graph has zeroes, and if there are zeroes, then it will be < 0 between the zeroes.  Factoring, we get the zeroes to be $w = -6, 3$.  Therefore, $w^2 + 3w - 18 < 0 \text{ }\forall w \in (-6, 3)$


What is the smallest positive value of x such that x + 5678 results in a palindrome?

[hide]the next closest palindrome is 5775
so x is 97[hide][/hide][/hide]

[hide] The smallest palindrome after $ 5678$ is $ 5775$. Hence, $ x\equal{}5775\minus{}5678\equal{}\boxed{97}$.[/hide]


The diagram below shows a rectangle and a cicle with the center of the circle below the rectangle, one side of the rectangle tangent to the circle, and two vertices of the rectangle on the cicle. The longest side of the rectangle is $16$, and the radius of the circle is $10$. Find the area of the rectangle.

Basically the rectangle lies on the upper part of a circle with a region that is outside, in case.  :wink:

[hide]
The distance from the center of the circle to one vertex of the rectangle is 10. It makes a triangle with the side of the rectangle closest to it and the radius of the circle that touches the opposite side of the rectangle. Half the length of the rectangle is 8, so the triangle is 6-8-10. Since the distance from the center of the circle to the opposite side of the rectangle is 10 and the distance from the center of the circle to the side closest to it is 6, the width of the rectangle is 4. The area is $64$.[/hide]

what rectangel?
you forgot to put the diagram i think

Draw radius to vertices and tangent side
You get other side 16cm
So Area is$256cm^{2}$

either theres no diagram

OR

the fact that its 118 in the morning is getting to me...

did you forget the diagram?

btw, this is my 100th post, woo.  :D

ubemaya has it right....

well she got what the problem meant.  :)


geometry
trapezoid
inradius
geometry solved

Consider a quadrilateral $ABCD$. Let the incircle of $ABCD$ intersects $AC$ at $K,L$ and $BD$ at $M,N$ such that $M\in{BN}$ and $K\in{AL}$. It is given that $AK\cdot{LC}=16$ and $BM\cdot{ND}=\frac{9}{4}$. Find the radius of incircle.

You forgot to mention that the quadrilateral $ABCD$ is a trapezoid. See the thread  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28485]not hard (circumscribed trapezium...)[/url].

Hmm... Strange... My teacher didn't say me that it is a trapezoid. Maybe it is harder to solve without that condition but still solvable.What do you think?

Without the condition that the quadrilateral $ABCD$ is a trapezoid, you do not have enough information to calculate the inradius. Quadrilaterals with an arbitrary inradius do exist, for which the conditions $AK \cdot CL = 16$, $BM \cdot DN = \frac 9 4$ are satisfied. On the other hand, if the quadrilateral is a trapezoid, it is still not determined uniquely, but the inradius is fixed and can be calculated.


Prove Hardy-Littlewood-Polya's Inequality:

If f(x) is convex over an interval, a_k and b_k are in that interval, and the sequence a_1, ..., a_n majorizes b_1, ..., b_n, then

f(a_1) + ... + f(a_n) >= f(b_1) + ... + f(b_n)

Maybe some people need to know what majorizes means.

We say that a sequence {a1, ... ,an} majorizes {b1, ... ,bn} if:

a1 >= b1
a1+a2 >= b1+b2
...

a1+a2+...+an-1 >= b1+b2+...bn-1

and

Sum(ai) = Sum(bi)

Thanks. Yeah, I probably should have put that bit in.

is this a challenge problem, or are you just interested in seeing the proof?

I guess I'd be interested in seeing the proof, if its not too long. It could be a challenge too, I suppose.

Everything in the Advanced forum is a challenge :)

well, I've seen the proof before and I wouldn't want to ruin your fun, I'll wait a while and then post it

Ok, might as well start with the simple case and see where we can go from there. Maybe it will lead to some insight.

Let's just do it for two variables.

We wish to show

[tex]f(a_1) + f(a_2) \ge f(b_1) + f(b_2)\hspace*{0.5in} (*)[/tex]

Let S = a1 + a2 = b1 + b2.

[tex] \displaystyle a_1 \ge a_2 \Rightarrow a_1 \ge \frac{S}2 \Rightarrow a_2 \le \frac{S}2[/tex]

Adding (1) and (2) gives

[tex]\displaystyle b_1 \ge b_2 \Rightarrow b_1 \ge \frac{S}2 \Rightarrow b_2 \le \frac{S}2[/tex]

Therefore
[tex]a_2 \le b_1[/tex].

Since [tex]\displaystyle a_1 \ge b_1 \ge a_2, \exists t_1, t_2[/tex]
such that

[tex]\displaystyle t_1 + t_2 = 1 [/tex]

and

[tex]\displaystyle  t_1a_1 + t_2a_2 = b_1. [/tex]

By Jensen's inequality,

[tex]\begin{equation}
t_1f(a_1) + t_2f(a_2) \ge f(t_1a_1 + t_2a_2) = f(b_1)\hspace*{1.0in}
\end{equation}
[/tex]

Furthermore,
[tex]\displaystyle (1-t_1)  + (1-t_2) = 2 - (t_1+t_2) = 1,[/tex]
so also by jensen:

[tex]\displaystyle (1-t_1)f(a_1) + (1-t_2)f(a_2) \ge f(a_1 + a_2 - (t_1a_1 + t_2a_2) \\
\hspace*{0.4in} = f(a_1+a_2 - b_1) = f(S-b_1) = f(b_2) \hspace*{1.0in}(2)[/tex]

Adding (1) and (2) gives (*), as desired.

Now hopefully this can be extended to more variables than two.

Addendum:

I think that if we can show there exist nonnegative [tex]\displaystyle t_{i,j}, 1 \le i,j \le n[/tex] such that

[tex]\displaystyle  (i) \quad \sum_i t_{i,j}a_i = b_j [/tex]

[tex]\displaystyle  (ii) \quad \sum_j t_{i,j} = 1 [/tex].

[tex]\displaystyle (iii) \quad \sum_i t_{i,j} = 1 [/tex].

We should be in the clear, following an analogous procedure to the two-variable case...

It looks like it probably holds.


prove that for every a,b,c>0 and a 2 +b 2 +c 2 =1 we have
 \sum (abs(a-b)) \geq  \sum 2(c 2 +ab)/(a+b).
good luck.  :D

Arosisi,what will happen if a=b=c..... :?

sorry anhcuong, maybe my inequality was wrong but i will correct it.

ah, my problem must be
prove that for every a,b,c>0 and a  2  +b  2  +c  2 =1 we have
  \sum (abs(a-b))  \geq   2(c  2  - ab)/(a+b).
try again  :D


In the same vein as texas137's post about a Differential Equations text in the College forum (except that I'm not necessarily looking to test out of Abstract Algebra in college, though I would hope I could), what are your recommendations for an Abstract Algebra textbook?
I've been leaning towards Gallian's [i]Contemporary Abstract Algebra[/i] as it seems to be a well-liked and "fun" easy-to-self-study book, but it's apparently not very rigorous...? What books do you prefer?

I used Artin's Algebra, however for that it helps to already have a fairly strong background with proofs and abstract mathematics, and preferably some previous familiarity with linear algebra.  It might work well to do Gallian and then Artin, or to use Artin as a supplement to Gallian.

A lot of people like Dummit and Foote, but I haven't used it, so I'll have to wait for others to weigh in.

I haven't used Gallian, but Dummit and Foote is quite good, in my opinion. But if Gallian's book is anything like Gallian the person, then his is definitely the book to get.

Dummit and Foote is the way to go. Artin is too hard to start out with. It's really designed as a Graduate Text for people who already know some abstract algebra.

Gallian is okay, but it won't prepare you well enough to exempt out of a good college course. You might use it along with Dummit and Foote for an alternative view of things.

I learnd from I.N. Herstein, which used to be a very popular book at one time, but doesn't seem to be used as much anymore. I thought it was pretty good. But I would recommend Dummit and Foote over it.

Edit: You know what, looking at my Dummit and Foote book this book may be too hard too. You might want to go for the Gallian book first to get a feel for what is going on, and then move on to Herstein.

The problem with a lot of these abstract algebra books is that they try to include tooo many things so that they can appeal to both the undergraduate market and the graduate student market. Most of these books do not have answers to the exercises in them either - so you'll have to get someone to check them for you.

I have a rather old book entitled "Modern Algebra: A Survey" by Birkhoff and MacLane. It seems to be at a pretty introductory level, and I planned on reading it during the summer. Has anyone ever heard of or read this book?

I used Herstein as a supplement to Artin.  I inherited my copy from my father, who used it as an undergrad at Caltech, so I found it a little "old-fashioned", in notation and stuff.  However, there are newer editions, that should correct some of that.  Oh, and I love to tell my Herstein story, so I'll have to retell it here.

When I was maybe 10, my parents brought up a whole lot of their old math books from the basement.  I didn't have the mathematical background for almost all of them, but I noticed this cute slim black book, whose title seemed at least comprehensible.  "Topics in Algebra!", I thought, bright little kid that I was, "I know algebra!"  So I brought it back to my room, and found it was all about groups and fields and rings and all sorts of weird stuff.

Haha. Did you keep on reading? :)

To tetrahedr0n: That book is in my school library...That's all I know about it, though. ;) I meant to go take a look at it and the Artin book today, but the library was remodeling and couldn't.

I have a copy of Birkhoff and MacLane at home. When I get back (very soon), I'll take a look at it and tell you if I like it. I bought it from a used book store when I was in seventh grade.


algebra
polynomial
modular arithmetic
algebra unsolved

Find all the polynomials $f(x)$ with integer coefficients such that $f(p)$ is prime for every prime $p.$

You don't need $ f\in\mathbb{Z}[x]$!

Take $ f\in\mathbb{C}[x],n\equal{}\deg f$ with $ f(p)$ prime for all primes $ p$.
Then $ f(p_1),f(p_2),\ldots,f(p_{n\plus{}1})\in\mathbb{Z}$ for $ n\plus{}1$ different primes $ p_1,p_2,\ldots,p_{n\plus{}1}$
gives $ f\in\mathbb{Q}[x]$ using the [url=http://en.wikipedia.org/wiki/Lagrange_polynomial]Lagrange interpolation[/url]. So $ mf(x)\in\mathbb{Z}[x]$ for some $ m\in\mathbb{N}$.

If $ f(x)\not \equal{}x$, then $ f(p)\not \equal{}p$ for infinitely many primes $ p$. (otherwise $ f(x)\minus{}x$ has infinitely many roots).
If $ f(p)|m$ for all such $ p$, then for some prime divisor $ q$ of $ m$ we must have $ f(p)\equal{}q$ for infinitely many $ p$.
(since $ m$ has only finitely many prime divisors) And then $ f\equiv q$ prime.

Otherwise $ p\not \equal{}f(p)\nmid m$ for some prime $ p$.
But then for $ q\equal{}f(p)$ prime we have $ \gcd (p,q)\equal{}1$, and using the heavy [url=http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions]Dirichlet prime number theorem[/url]
we find infinitely many primes $ r\equiv p\pmod{q}$.
Then $ mf(r)\equiv mf(p)\equal{}mq\equiv 0\pmod{q}$ for all such $ r$ (using $ mf(x)\in\mathbb{Z}[x]$!)
and then $ f(r)\equal{}q$ for all such $ r$, as $ q\nmid m$ and $ f(r)$ prime.
Again this gives $ f\equiv q$ prime.

So either $ f(x)\equal{}x$ or $ f$ is constant prime.


trigonometry
geometry
rectangle
cyclic quadrilateral
geometric inequality

Suppose $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $AB \cdot BC \cdot CD \cdot DA \geq 4$, prove that $ABCD$ is a square.

Is $f(x)=\sin{x}$ concave on $[0, \pi]$? If so, then this problem can be proved by $l=2R\sin{\theta}$ then AM-GM then jensen.

Hm... I'd say that by Ptolemy we have \[ 4 \geq AC \cdot BD = AB \cdot CD + AD \cdot BC \geq 2 \sqrt{AB \cdot BC \cdot CD \cdot DA} \geq 4 \] (since $AC$ and $AB$ are not longer than a diameter of the circle).

So equality must hold everywhere. Hence $AC$ and $BD$ are diameters, $ABCD$ is a rectangle, and also $AB \cdot CD = AD \cdot BC$ which implies that $AB^2 = AD^2$, so $ABCD$ is a square, and we're done.

Oh...... I didn't think about using Ptolemy. Nice proof Arne.

Anyway, is my method correct?

Yeah, I think so...

Could you write a full solution? Then it will be easier to judge :)

OK.

Let $\angle{AOB}=2a$, $\angle{BOC}=2b$, $\angle{COD}=2c$, $\angle{DOA}=2d$. Then $a+b+c+d=180^o$.
Since we also have: $AB=2R \sin{a}$ and etc, Hence:

\[ AB \cdot BC \cdot CD \cdot DA=16R^2 \cdot \sin{a} \sin{b} \sin{c} \sin{d} \]
By AM-GM: $\sin{a} \sin{b} \sin{c} \sin{d} \leq (\dfrac{\sin{a}+\sin{b}+\sin{c}+\sin{d}}{4})^4$.

Then jensen: $\dfrac{\sin{a}+\sin{b}+\sin{c}+\sin{d}}{4} \leq \sin{(\dfrac{a+b+c+d}{4})}=\dfrac{\sqrt{2}}{2}$.

Thus $16R^2\cdot \sin{a} \sin{b} \sin{c} \sin{d} \leq 16 \cdot \frac14=4.$

using Ptolemy's theorem,we have $AB.DC+AD.BC=AC.BD\le4$ with equality iff $AC$ and $BD$ are diameters.
so , we have , by AM-GM ,$AB.BC.CD.AD\le (\frac{AB.DC+AD.BC}{2})^2 \le(4/2)^2=4$ with equality iff $AB.DC=AD.BC$
so , from the given condition in the problem , we must have , $AB.BC.CD.DA=4.$
so , from the equality conditions and simple geometry , it is obvious that ABCD is a square.

[quote=SRIDEV]Dear @shobber ,

You wrote AB = 2RSina
Thus AB.BC.CD.DA should = 16R^4. Sina.Sinb.Sinc.Sind

But you wrote
AB.BC.CD.DA = 16R^2. Sina.Sinb.Sinc.Sind

Please do clarify, how R^4 becomes R^2 ?[/quote]
Don't forget that $R=1$  ;)


I thought it would be fun to have a joke marathon.
One person posts a joke
the next person takes a guess
the the poster of the joke posts the punchline
the someone else gets to post a joke!

Does anyone want to start?

i'll start:  What dessert did the math teacher make?

What feature does Windows have that Linux doesn't?
Answer since i can't be here often[hide]crash.dll[/hide]

Man, I don't have anything to post, all of my jokes are either sexual or racist.  :(

what's brown and sticky?
[hide="answer"][hide="actually try to guess before you look"]a stick  :D [/hide][/hide]

in lieu of mathbomberii's joke that I didn't get...

So a guy walks into a bar.
[hide]ouch.[/hide]

[quote="math92"][quote="mathbomberII"]30 people ran into a building. you would think one of them could see   [hide]they all [u]ran[/u] into a building.[/hide][/quote]
i dont get it :?[/quote]
he means that the people are running into the wall of a building.

[quote="math92"]i don't get this "you would think one of them could see"[/quote]
if you can see, then you wouldn't run into a wall.

and here's another joke:
what's $\frac{\sin x}{n}$?
[hide="stupid answer"]6 as the the n cancels out in the numerator and denomenator which leaves six.  :D [/hide]

why doesn't the equation A=pi$r^2$ make sense?

[hide="my interpretation"]There are 5 people, as the speaker mentions. He was referring to 3 people, which couldn't be him, and a person who doesn't talk. Since the speaker is talking, he can't be the guy who doesn't talk. This means he has to be that other guy.... [/hide]

wow
that like totally killed the joke
I'm guessing none of you guys have ever seen The Simpsons

[quote="236factorial"][quote="dinoman"]I'm guessing none of you guys have ever seen The Simpsons[/quote]

Nope :noo:[/quote]

Oh wow...well, "I'd hate to be him" is just ironic because he is talking about himself there.

why did tigger look in the toilet?
[hide][hide]he was looking for pooh[/hide]
[/hide]

Where'd the bunny go when the bunny got hurt?

To the HOP-spital!  :lol:

Here's the real joke I was gonna post:

At Heathrow Airport today, an individual, later discovered to be a public school teacher, was arrested trying to board a flight while in possession of a compass, a protractor, and a graphical calculator. Authorities believe he is a member of the notorious Al-Gebra movement.

He is being charged with carrying weapons of math instruction!

Shouldn't that be a graphing calculator?  :huh:

How do you put a rhinoceros in the refrigerator?
[hide="Answer"]Open the door, put the rhinoceros in, and close the door.[/hide]
How do you put a giraffe in the refrigerator?
[hide="Answer"]Open the door, take the rhinoceros out, put the giraffe in, and close the door.[/hide]
There is a meeting in the jungle. All the animals go to the meeting except one. Which one?
[hide="Answer"]The giraffe. It's still in the refrigerator.[/hide]
How do you get past a river inhabited by many crocodiles, when you don't have a boat or other vehicle?
[hide="Answer"]Just wade through. All the crocodiles are at the meeting.[/hide]

[quote="Drunken_Math"]

Here's the real joke I was gonna post:

At Heathrow Airport today, an individual, later discovered to be a public school teacher, was arrested trying to board a flight while in possession of a compass, a protractor, and a graphical calculator. Authorities believe he is a member of the notorious Al-Gebra movement.

He is being charged with carrying weapons of math instruction![/quote]
 :lol:  :lol:  :lol:
nice one.

[quote="math92"]Why is 6 afraid of 7?
[hide="answer"]How should i know?[/hide]
 :rotfl:  :D  :lol:  :P   ;)  :)   :|  :([/quote]

because 7 8 9
that one is sooooo old...

EDIT NVM someone alreay said that....

ive heard the fridge one too.It's fun to tell people that one usually.

Three boys from different villages meet and start boasting
1st-my village is so cold,u have to wear 5 blankets and sleep
2nd-my village is so cold,u have to wear 10 blankets and sleep
3rd-my village is so cold that our cows give ice-cream instead of milk!!!

doesn't that signature come from calvin and hobbs?  The thing about f and uck and firetruck?  Correct me if i am wrong, but that was his exact words :lol:

What's the difference between a large pizza and a math PhD?

[hide]
A large pizza can feed a family
[/hide]

Three statisticians went duck hunting. One fired just in front of the duck, one fired just behind the duck, and the third said, "We got him!"

A statistician put his head in an oven and his toes in ice water and said, "On the average, I feel quite comfortable."

1)My sign

2)A taxi driver is driving very rashly and there's a lady in it.
Lady:Oh my,please be careful,I have 8 children at home and no husband.Who'll take care of them?

Driver:You have 8 children?

Lady:yes,so please be careful
Drievr:You have 8 children and you are telling me to be careful?

3)A guy in an electric chair starts getting hiccups.
the man who's going to *push the button* asks him for any last wish.

guy:Yes,could you please do something to scare me? :rotfl:


Be ABC a triangle. Draw the heights which will be called FB, CE and AD. Now you've a new triangle (FED) and the point "I" will be the orthocentre. Prove that the angles FEI=IED, FDA=ADE, EFB=BFD.

I must be the orthocentre of $\Delta ABC$
[hide="Hint"]$\widehat{FEC}=\widehat{DAC}=\frac{\pi}2-\widehat{C}$
The same for the others[/hide]


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_{\minus{}2}^2 x^4\sqrt{4\minus{}x^2}\ dx$.

We have:
$ I \equal{} \int_{\minus{}2}^2 x^4\sqrt{4\minus{}x^2}\ dx \equal{} 2\int_{0}^2 x^4\sqrt{4\minus{}x^2}\ dx \equal{} 2^7\int_{0}^{\frac{\pi}{2}} \sin^4{\theta}\cos^2{\theta}d\theta \equal{} 4\pi$.


[hide] The volume of a cylinder is $ \pi r^2 h$. Since the diameter increased by a factor of $ 1.3$, and the radius increases proportionally with the volume, the new radius will be $ (1.3)^2\equal{}1.69$ times the old radius. Thus, the height will increase by $ \frac{3}{1.69}\approx\boxed{77.5\%}$. Thus, the answer is A.[/hide]

[quote="sri340"]If the diameter of a cylindrical can is increased by 30 percent, by approximating what percentage should the height be increased to triple the volume of the can ?

(A)77.5 (B) 54.8 (C) 50.0 (D) 44.2 (E) 36.8[/quote]

Let's say the diameter of a can to be 20 and the height 10.  The volume of this can is $ 10^2\pi(10) \equal{} 1000 \pi$.  The triple of this volume is $ 3000\pi$, and the new diameter would be $ 20 \times 1.3 \equal{} 26$, making the radius 13.  We now plug in all the info we have into the formula to find the volume of a cylinder and solve for $ h$:

$ 3000\pi \equal{} 13^2 \pi h$
$ \frac{3000}{169} \equal{} h$

We now calculate the percent of change between the original height and the new one:

$ \frac{\frac{3000}{169} \minus{} 10}{10} \equal{} \boxed{\mathbf{(A) 77.5 \%}}$


function
algebra
polynomial
quadratics
combinatorics unsolved
combinatorics

I need help find a function for the following recurrence relation

h$ _{n}$=2*h$ _{n\minus{}1}$+2*h$ _{n\minus{}2}$

It's getting to big and messy for me to recognize any pattern.

Use the characteristic polynomial method...

Consider the polynomial $ x^2\minus{}2x\minus{}2$.  If $ r_1, r_2$ are the two roots of this quadratic, then the sequences $ h_n\equal{}r_1^n$ and $ h_n\equal{}r_2^n$ both satisfy the recursion.  Therefore, any linear combination of these two sequences satisfies the recursion.  Given two initial conditions (ex. $ h_r, h_s$, for some integers $ r, s$), we can determine $ a, b$ in the equation $ ar_1^n\plus{}br_2^n\equal{}h_n$, because we have two equations in two unknowns.


inequalities
function
three variable inequality

[quote="mateivld"]Since it is homogeneus we can assume $\sum a=1$.The inequality becomes $(a+b)(b+c)(c+a)\geq 24abc(1-2\sum ab)$,so we have $\sum ab-abc\geq 24abc-48abc\sum ab$,so $\displaystyle \sum ab\geq \frac{25abc}{48abc+1}$.From Schur we have $\displaystyle\sum ab\geq\frac{1+9abc}{4}$.So if we intercalate we have $1+27\cdot 16\cdot a^2b^2c^2\geq 43abc$,so $(1-27abc)(1-16abc)\geq 0$,wich is true from $27abc \leq 1$. :)[/quote]
Sorre,but You are wrong,[color=red]$\displaystyle\sum ab\geq\frac{1+9abc}{4}$?????????????????????? [/color]  :o  ,I can say you that $4(ab+bc+ac)\leq 1+9abc$ :D

Yes, Vlad, Tiks is right here. From Schur, we have $ab+bc+ca\leq\frac{1+9abc}{4}.$

We know that $a+b+c=1$=>$4(ab+bc+ac)=<1+9abc$ <=>$4(ab+bc+ac)(a+b+c)\leq (a+b+c)^3+9abc$<=>[color=blue]Chur's ineq.[/color](open brackets :D ) ;)

[quote="Soarer"]Haven't checked this, but it seems like we can prove the stronger inequality
$(a+b+c)^3(ab+bc+ca) \ge 27abc(a^2+b^2+c^2)$ by
$((\sum ab)(\sum a) - 9abc)(\sum (a-b)^2) + 6 \sum a^2(b+c)(b-c)^2 \ge 0$[/quote]
 :o  :o ........................Soearer I just now see that your stronger ineq. is WRONG!!! :P .
Take $a=1,b=\frac{1}{2},c=\frac{1}{2}$
So this problem still unsloved :P .

[quote="arqady"][quote="Tiks"][quote="arqady"]What did you mean, Tiks? Thank you.[/quote]
About what? :?[/quote]
That you meant, Tiks, when spoke: Very-very nice!!!!!!!!!?
What your proof of this inequality?  :roll:
Thx. :)[/quote]
I see,for first,I have proposed this problem long time ago(then this problem was be Very-very nice for me :D ),also I want ask you,can you advise me any method for make ineq.-s to this kind of forms$=\sum(b+c)(b+c-a)^2(a-b)(a-c)=(b+c)(b+c-a)^2(a-b)(a-c)+(b-c)((a-c)(a+b)(a+b-c)^2-(a-b)(a+c)(a+c-b)^2)\geq0,$ :? ?.
In my solution I take $a+b+c=1$ and used function methods for $f(ab)$ ;) .
Thank you.

[quote="arqady"][quote="Tiks"] I want ask you,can you advise me any method for make ineq.-s to this kind of forms$=\sum(b+c)(b+c-a)^2(a-b)(a-c)$ :? ?.
[/quote]
Dear Tiks!
To me has simply carried! I have noticed only, that if $b=c$ then
$(a+b)(b+c)(a+c)(a+b+c)^2- 24abc(a^2+b^2+c^2)=2b(a-2b)^2(a-b)^2.$ ;)[/quote]
O.K,thank you,I think I see... :)

[quote="Tiks"]Let  $ x,y,z$ be nonnegative numbers, prove that

$ (y \plus{} z)(z \plus{} x)(x \plus{} y)(x \plus{} y \plus{} z)^2\geq 24xyz(x^2 \plus{} y^2 \plus{} z^2),$

with equality if $ x \equal{} 2y \equal{} 2z$.[/quote]$ (y \plus{} z)(z \plus{} x)(x \plus{} y)(x \plus{} y \plus{} z)^2 \minus{} 24xyz\left(x^2 \plus{} y^2 \plus{} z^2\right)$

$ \equal{} \frac {y \plus{} z}{4}(x \minus{} y \minus{} z)^2(2x \minus{} y \minus{} z)^2$

$ \plus{} \frac {(y \minus{} z)^2}{16}\big[12(2x \minus{} y \minus{} z)^3 \plus{} 35(y \plus{} z)(2x \minus{} y \minus{} z)^2$

$ \plus{} 6\left(9z^2 \plus{} 2yz \plus{} 9y^2\right)(2x \minus{} y \minus{} z) \plus{} 3(y \plus{} z)\left(9y^2 \minus{} 14yz \plus{} 9z^2\right)\big]\geq 0$,

which is clearly true for $ x \equal{} \max\{x,y,z\}$.

$ (b\plus{}c)(c\plus{}a)(a\plus{}b)(a\plus{}b\plus{}c)^2\minus{}24abc\left(a^2 \plus{} b^2 \plus{} c^2\right)\\
\equal{}\left(b\minus{}a\right)^2\left({\left(7c\plus{}11b\minus{}2a\right)\left(c\minus{}b\right)^2\plus{}4\left(5c\minus{}3b\plus{}2a\right)\left(b\minus{}a\right)^2\plus{}a^2\left(15c\minus{}5b\minus{}8a\right)\plus{}32a\left(b\minus{}a\right)\left(c\minus{}b\right)}\right)\plus{}\\
\plus{}\left(\left(c\minus{}b\plus{}2a\right)\left(b\minus{}a\right)\plus{}2a\left(c\minus{}b\right)\right)\left(c\minus{}b\minus{}a\right)^2\left(c\minus{}b\right)\geq 0 \text{ for } a\leq b\leq c$
 :roll:

Because inequality is homogeneous, we can assume that $ a+b+c=3 $
We note $ ab+bc+ca=3-3t^2,0\le t<1 $, result $ a^2+b^2+c^2=3+6t^2 $ and
$ (1+t)^2(1-2t)\le abc\le (1-t)^2(1+2t) $
$ (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc $
Inequality returns to:
$ abc\le\frac{9(1-t^2)(1+2t^2)}{9+16t^2} $
It is sufficient that:
$ (1-t)^2(1+2t)\le\frac{9(1-t^2)(1+2t^2)}{9+16t^2}\Leftrightarrow $
$ 32t^4+2t^3+20t^2\ge 0 $ which is true!
________
Sandu Marin

Let $\ c = \max\{a, b, c\}$
$\ \ Ineq\Leftrightarrow \prod (a+b)\geq 24abc\frac{\sum a^2}{\left ( a+b+c \right )^2}\Leftrightarrow \prod (a+b)-8abc\geq 8abc\left [ \frac{3\sum a^2}{\left ( a+b+c \right )^2}-1 \right ]\left ( \bigstar  \right )$
We have:
$\ \prod \left ( a+b \right )-8abc=2c\left ( a-b \right )^2+\left ( a+b \right )\left ( a-c \right )\left ( b-c \right )$
And $\ \frac{3\sum a^2}{\left ( a+b+c \right )^{2}}-1=\frac{2\left ( a-b \right )^2+2\left ( a-c \right )\left ( b-c \right )}{\left ( a+b+c \right )^{2}}$
$\ \left ( \bigstar  \right )\Leftrightarrow S_1\left ( a-b \right )^{2}+S_2\left ( a-c \right )\left ( b-c \right )\geq 0$
And we can check $\  S_1=2c-\frac{16abc}{\left ( a+b+c \right )^{2}}>0$ and $\ S_2=a+b-\frac{16abc}{\left ( a+b+c \right )^{2}}>0$ with $\ c=\max\{a, b, c\}$ by $\ Am-Gm$.
:lol:


calculus
derivative
function
Support
integration
vector
real analysis

Hi,

let $\emptyset\neq E\subset\mathbb{R}^n$
 denote with $E_{\epsilon} =\{x: d(x,E)\leq \epsilon\}$

let $\rho$ be a function in ${C_{\infty}(E_n})$ with compact support $[\rho]=\bar{B}(0,1)$, with $\rho(x)$ only depending on $||x||$ , with integral 1, and positive

let $\rho_{\epsilon}(x)= \rho(\frac{x}{\epsilon}) \frac{1}{\epsilon^n}$

let $f=\chi_{E_{\epsilon}}$ thus a characteristic function
a very nice results is that $g=f*\rho_{\epsilon}$ will be 1 in E, and zero outside $E_{2 \epsilon}$

But how to see this?  For every positive integer $k$; there is $C_k$ independent of $\epsilon^$, such that

\[ sup_{x\in\mathbb{R}^n}| D^{\alpha} (f)|\leq \frac{C_k}{\epsilon^k}  \]

where $\alpha$ is een multi index vector, of length k

We can use this:

$\frac{\partial}{\partial x_j}(g*h)= (\frac{\partial g}{\partial x_j})*h=g*{\frac{\partial h}{\partial x_j}}$

From that we can get that

$D^{\alpha}(\rho_{\epsilon}*f)=(D^{\alpha}(\rho_{\epsilon}) )*f$

Now, $D^{\alpha}(\rho_{\epsilon})(x)=\frac1{\epsilon^{n+|\alpha|}} (D^{\alpha}\rho)(\frac x{\epsilon})$

It now follows by a change of variables that

$\|D^{\alpha}(\rho_{\epsilon})\|_1=\frac1{\epsilon^{|\alpha|}} \|D^{\alpha}\rho\|_1$

Then $\|D^{\alpha}(\rho_{\epsilon})\|_{\infty}\le \|D^{\alpha}(\rho_{\epsilon})\|_1\|f\|_{\infty} \le \frac{C_k}{\epsilon^{|\alpha|}}$

thanks!

small remark, could it be that in this case not both derivative formulas are correct : I mean you are convolving a non differentiable with a differentiable right?

Sure. The general principle is that you can put a derivative on either factor of a convolution that you wish to - provided that factor is differentiable. If it isn't, you can't.


Not sure if this one is correct.

[hide]
$x^5 - x^3 + x=x(x^4 - x^2 + 1)=x(x^2-\sqrt{x^2-1})(x^2+\sqrt{x^2-1})$[/hide]

[quote="jpjoe"]Can somebody please factor out: [code]x^5 - x^3 + x[/code]

please include the solution. and perhaps a brief explanation.. :) thanks![/quote]

First, factor out $x$:

$x(x^4 - x^2 +1)$
The second term is a quadratic in terms of $x^2$, so, using the quadratic formula:

$x(x^2-\frac{1+\sqrt{-3}}{2})(x^2-\frac{1-\sqrt{-3}}{2})$

$x(x^2-\frac{1+3i}{2})(x^2-{\frac{1-3i}{2})}$

$x(x+\frac{\sqrt{2+6i}}{2})(x-\frac{\sqrt{2+6i}}{2})(x+\frac{\sqrt{2-6i}}{2})(x-\frac{\sqrt{2-6i}}{2})$

You can stop at any step you'd like.

If you want it factored over the real numbers, then it's simply $x(x^4-x^2+1)$


Let $a,b,c \in [0,1]$
Find max $\sum \frac{a}{bc+1}$

[hide="solution"]Just use the convex nature of the function - the maximum must occur for $a = 0$ or $a = 1$, and similarly for the other variables.  Testing all four possibilities: $(0,0,0)$, $(1,0,0)$, $(1,1,0)$, and $(1,1,1)$, which covers all cases by symmetry, we find that

$\sum \frac{a}{bc+1} \leq 2$ with equality for $(a,b,c) = (1,1,0)$.[/hide]


function
real analysis
real analysis unsolved

Give an example of a continuous function $ f: R->R$ such that for each $ a \in R$ the equation $ f(x)=a$ has exactly three solutions.

First,we define $ f$ on $ [0,3)$ :
$ f(x)=x,x\in[0,1);f(x)=2-x,x\in[1,2);f(x)=x-2,x\in[2.3)$
Then we extend $ f$ on $ R$ by putting $ f(x+3)=f(x)+1,\forall x\in R$


geometry
inequalities
calculus
inequalities unsolved

Let a, b, c be the lengths of the sides of a triangle. Prove that
1) $(a+b+c)^3+9abc \ge 4(ab+bc+ca)(a+b+c)$
2) $(a+b-c)(b+c-a)+(c+a-b)(a+b-c)+(b+c-a)(c+a-b) \le \sqrt{abc} (\sqrt{a}+\sqrt{b}+\sqrt{c})$

The incircle substitution will help...

Bomb

inequality 2) was proposed by Mircea Becheanu at the first imo test selection in 2001. By calculus we
  reduce the inequality to 2(ab+bc+ca)-(a^2+b^2+c^2)<=asqrt(bc)+bsqrt(ca)+csqrt(ab) which is equivalent
  to Schur.

[quote="cezar lupu"]here's a better one: if a,b,c are the lenghts of triangle the following inequality holds:

               (b+c-a)(c+a-b)+(a+b-c)(c+a-b)+(b+c-a)(a+b-c)<=9abc/(a+b+c).[/quote]
it's also schur...

it seems very useful to substitute $a=x^2, b=y^2, c=z^2$. Then $x,y,z$ are still sides of a triangle and specifically of an acute-angled triangle.

[quote="ehku"]Prove them![/quote]
the first one is equivalent to schur and cezar lupu has already posted a better inequality for no.2 which is also schur. So they are proved.

well continuing my "solution" for 2) the ineq transforms into
$\Sigma_{cyclic}(x^2+y^2-z^2)(x^2+z^2-y^2)%Error. "leqxyz" is a bad command.
(x+y+z)$.
$LHS=16A^2$ where $A$ the area of the triangle with sides $x,y,z$. Now, due to the equalities $xyz=4AR$ and $x+y+z=2s$ it suffices to prove that $4A\leq2Rs$ which due to $A=rs$ is equivalent to $2r\leq R$ which is valid.


function
irrational number
algebra solved
algebra

Let $Q$ denote the set of rational numbers. Suppose that the functions $f, g: Q\to Q$ are strictly monotone increasing functions which attain every rational value. Is it necessarily true that the range of values of the function $f+g$ is also the whole set $Q$?

[i]Komal, problem A. 228 January 2000[/i]

[i]Edited by Myth[/i]

Since Q is dense in R we may speak of x->a for any real a. The question is equivalent to this: can there exist an irrational number a and 2 irrational numbers u and v s.t. u+v is rational and lim{x->a}f(x)=u and lim{x->a}g(x)=v?

f and g obviously have finite limits in every real point, so we may talk about those limits.

If such numbers exist then it's clear that (f+g)(x) can't take the value u+v for any rational x. In order for such numbers to exist we would need strictly increasing bijections between (a,00) \cap Q and (u,00) \cap Q, as well as (a,00) \cap Q and (v,00) \cap Q. Can such bijections exist? Remember that u+v is rational.

I hope that the following is a good counterexample:
  Take f the easiest one, f(x)=x for all x. Now we need for g to be something like -1/(x+a) on an interval. Let us take g(x)=-1/x for 1<=x<=2. We define g for the other points to be somewhat linear: g(x)=2x-3 if x<1 and g(x)=x-2,5 if x>2. Then g is strictly increasing, takes all the values, but we cannot find x such that f(x)+g(x)=1.

It looks like you're right! Here's what I was thinking:

There exist increasing bijections from (a,00) \cap Q to (b,00) \cap Q for all a and b in R (try to prove this; it's not hard). Then we do what I said in the previous post. But it looks like your solution is much better since it gives a clear example of f and g.


Evaluate $ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{(\sin ^3 \theta \minus{}\cos ^ 3 \theta \minus{}\cos ^ 2 \theta)(\sin \theta \plus{}\cos \theta \plus{}\cos ^ 2 \theta)^{2007}}{\sin ^ {2009}\theta \cos ^ {2009}\theta}\ d\theta$

[quote="kunny"]Evaluate $ \int_{\frac {\pi}{4}}^{\frac {\pi}{3}} \frac {(\sin ^3 \theta \minus{} \cos ^ 3 \theta \minus{} \cos ^ 2 \theta)(\sin \theta \plus{} \cos \theta \plus{} \cos ^ 2 \theta)^{2007}}{\sin ^ {2009}\theta \cos ^ {2009}\theta}\ d\theta$[/quote]

$ \textbf I \;\equal{}\;  \int_{\frac {\pi}{4}}^{\frac {\pi}{3}}  \dfrac { (\sin \theta \plus{} \cos \theta \plus{} \cos ^ 2 \theta)^{2007}} { \left(\sin\theta \cos \theta\right)^ {2007}}\cdot \dfrac { (\sin ^3 \theta \minus{} \cos ^ 3 \theta \minus{} \cos ^ 2 \theta)}{ \sin ^ {2}\theta \cos ^ {2}\theta} \; d\theta$

$ \equal{}\;  \int_{\frac {\pi}{4}}^{\frac {\pi}{3}}  \left( \dfrac {1} {\cos\theta} \plus{} \dfrac{1}{\sin\theta} \plus{} \dfrac { \cos\theta}{\sin\theta} \right) ^ {2007} \cdot \left( \dfrac { \sin\theta }{\cos ^ 2 \theta} \minus{} \dfrac {\cos \theta}{\sin ^2 \theta} \minus{} \dfrac {1}{ \sin ^2 \theta} \right)  \; d\theta$

$ \equal{}\;  \int_{\frac {\pi}{4}}^{\frac {\pi}{3}}  \left( \dfrac {1} {\cos\theta} \plus{} \dfrac{1}{\sin\theta} \plus{} \dfrac { \cos\theta}{\sin\theta} \right) ^ {2007} \cdot \left( \dfrac {1} {\cos\theta} \plus{} \dfrac{1}{\sin\theta} \plus{} \dfrac { \cos\theta}{\sin\theta} \right) '$

$ \equal{}\;  \dfrac{1}{2008}  \left( \dfrac {1} {\cos\theta} \plus{} \dfrac{1}{\sin\theta} \plus{} \dfrac { \cos\theta}{\sin\theta} \right) ^ {2008} \; \bigg| _{\frac {\pi}{4}}^{\frac {\pi}{3}}$

$ \equal{}\;  \dfrac{1}{2008} \left\{  \left( 2 \plus{} \dfrac { 2 } {\sqrt 3} \plus{} \dfrac {1 } {\sqrt 3}\right) ^ {2008} \minus{}  \left( \sqrt 2 \plus{}\sqrt 2 \plus{} 1 \right) ^ {2008} \right\}$

$ \equal{}\;  \dfrac{2^{2008}}{2008} \left\{  \left( 1 \plus{}  \dfrac  {\sqrt 3}{2} \right) ^ {2008} \minus{}  \left(  \dfrac {1}{2 }\plus{}\sqrt 2 \right) ^ {2008} \right\}$

i am not really sure if further simplification is possible here.  :maybe:

Thank you for nice problem and nice solution! :lol:


[b][1][/b]  Let $ n>m\ge 2 $ and $ A $  be a given $ n\times m .$
[b] How many matrices   $ X $ , of type $ m\times n , $ exist, such that  $  AX =I  $ ? [/b] ( $ I $ is the unit  $n\times n $ matrix).

[b][2][/b] If $ A $ is $ n\times m $ and $  B $ is a matrix of type $ m \times n $ , $m\ne n, $ [b] find $ \det(AB)\cdot\det(BA) .$[/b]

[quote="rgep"]You need to remind us of your definition of the determinant of a non-square matrix.[/quote]
Hi rgep, many thanks ! I reposted.

[quote="rgep"]Thanks, that makes more sense!  The answer to (2) is now zero, since is $A$ is $m \times n$ and $B$ is $n \times m$ then the rank of $AB$ and $BA$ each cannot exceed $\min\{m,n\}$.  But they are $m\times m$ and $n\times n$ and if $m \not= n$ then one of $AB$, $BA$ must be singular.[/quote]
Of course, thank for interest and nice reponse!

The answer to [1] is $0$, since
\[
rank(AX)\leq rank(A)\leq m<n=rank(I_n)\,.
\]

[quote="mchis68"]The answer to [1] is $0$, ... \][/quote] Indeed, thank you  for inteerst,Alex


First,I want to say I will be sorry,if someone feel uncomfortable when reading my post because of my poor English.I am trying to write more to improve my English skill.

To my certain matter,I am now learning ESP(English for specific purpose)and I had some difficulties when spelling in full some maths equations
For example :
1.$ \sqrt {n}, \sqrt [n]{a},x^2,x^3,..,x^n$
2.$ \frac {a^2 \minus{} b^n}{a_1.b_1}$
3.$ \frac {x^2}{a^2} \plus{} \frac {y^2}{b^2} \equal{} 1$
.....etc

Can someone help me by showing the way to spell them in full.
Thanks so much  :|  !

Is there some official way to say math expressions? I don't think so... and if there is one, it is only useful to test takers... This is the way I would write out the expressions and equations listed (and very rarely will you have to write out or verbally say anything like this...)

1. the square root of $ n$, the $ n$th root of $ a$, $ x$ squared, $ x$ cubed, [$ x^4$] $ x$ to the fourth power [or simply] $ x$ to the fourth [or] the fourth power of $ x$, the $ n$th power of $ x$ [or] $ x$ to the $ n$th power [or] $ x$ to the $ n$th

2. $ a$ squared minus $ b$ to the $ n$th, all over $ a$-one $ b$-one [or] the difference of $ a$ squared and $ b$ to the $ n$th power, divided by the product of $ a$-sub-one and $ b$-sub-one [commas indicate where one would pause if speaking]

3. $ x$ squared over $ a$ squared plus $ y$ squared over $ b$ squared equals one


function
limit
Asymptote
calculus
calculus computations

Find the function rule $ f$ for a graph with a maximum at $ (a,5)$ that also satisfies $ \lim_{x \to \infty} f(x) \equal{} 4$. Moreover, on the interval $ (a,\infty)$, $ f(x)$ must always remain $ > 4$.

[quote="Prime factorization"]the function rule[/quote]   :!: There's only one such function?   :wink:   More seriously, rational functions are a good thing to think of -- we need a horizontal asymptote at infinty, so the degree of the numerator should be no larger than the degree of the denominator.  We also want a maximum, so the denominator shouldn't have any real zeroes.  The simplest function that fits that description is $ f(x) \equal{} \frac{1}{1 \plus{} x^2}$.  Now ....

Never mind that question, I just need a function that has oscillatory behavior over a finite interval then converges without oscillation. Is there a field of study based purely on generating graphs? In general, I'm having a lot of trouble coming up with function rules that obey a pattern/curvature I'm aiming for when making asymptote pictures.


Let a,b,c  be positive real.Prove that:
$ \sqrt[3]{3[2(a^2\plus{}b^2\plus{}c^2)\minus{}(ab\plus{}bc\plus{}ca)]}$ :geq $ \sqrt[3]{a^2\plus{}b^2\minus{}ab}\plus{}\sqrt[3]{b^2\plus{}c^2\minus{}bc}\plus{}\sqrt[3]{c^2\plus{}a^2\minus{}ca} \geq \sqrt[3]{9(a^2\plus{}b^2\plus{}c^2)}$

[quote="ThanhPhuc_Huế"]Let a,b,c  be positive real.Prove that:
$ \sqrt [3]{9[2(a^2 \plus{} b^2 \plus{} c^2) \minus{} (ab \plus{} bc \plus{} ca)]} \geq\ \sqrt [3]{a^2 \plus{} b^2 \minus{} ab} \plus{} \sqrt [3]{b^2 \plus{} c^2 \minus{} bc} \plus{} \sqrt [3]{c^2 \plus{} a^2 \minus{} ca} \geq \sqrt [3]{9(a^2 \plus{} b^2 \plus{} c^2)}$[/quote]

\[ \sqrt [3]{3[2(a^2\plus{}b^2\plus{}c^2)\minus{}(ab\plus{}bc\plus{}ca)]}\ge\]
\[ \sqrt [3]{a^2\plus{}b^2\minus{}ab}\plus{}\sqrt[3]{b^2\plus{}c^2\minus{}bc}\plus{}\sqrt[3]{c^2\plus{}a^2\minus{}ca}\]

This does not always hold, try $ a\equal{}b\equal{}c$

:)

The correct problem statement is in the second post, I think.

Oh yes,can you prove that its?.I thinks it really difficult for me, and you?
PS: left-hand side is not difficult, but right-hand side is a problem


geometry
3D geometry
pyramid
calculus
integration
real analysis

We all know that the volume of any kind of pyramid is given by 1/3 b*h.  Don't use an integral, but use the riemann sum associated with the integral, to prove that this is the case for any pyramid.

(A pyramid is a shape in which the cross-sections are all similar, but decrease in size linearly -- a horizontal cross-section twice as far from the vertex is twice as large.)

Euclid proved it without doing calculus. Hehe in the Elements.

of course i can. i'm taking calculus right now its fresh  :D
let t=h let h/x=r and let r be a constant
so h/r=x
lets integrate

int((h/r)^2,h,0,t)
1/r^2*(h^3/3)
subsititue
(x^2/h^2)*(h^3/3)=x^2*h/3 which is base times height divide 3

i gueess that you the integral method is the same as riemann sums method.

Methinks you are not doing what the problem is asking.

oh i see you want riemann sums which are easier than integrals.
lets define area to be A(y)=x^2/h^2*y^2
sum(A(i*h/n)*h/n,i,1,n) which i will later say n goest to infinity
so that equlas h*x^2/n^3*summation(i)
equals h*x^2/n^3 * n(n+1)*(2n+1)/6
lets take the limit n->infinity
which gives us h*x^2/3
methinks that the quetion was really easy!!!!!!!!!!!! if you have taken calculus it should be in beginner calculus section in AOPS

uh goly gee are all pyrmaids 1/3*b*h even irregular ones?

Yes -- note that your x^2, which you used as area of the base, didn't change at all throughout.  It would have worked the same no matter what.


I have been reading and learning Linear Algebra from Hoffman and Kunze for some time now. However, there are only a limited number of problems in  this book (~ 15 problems maximum per section). I would like to know if anyone knew any good books for problems to supplement Hoffman and Kunze.

Thank you

I went through that book before and I didn't like it very much. What do you think about Halmos' book published by Dolciani?

http://www.amazon.com/Algebra-Problem-Dolciani-Mathematical-Expositions/dp/0883853221/ref=pd_rhf_p_t_2


Let's consider two squares: $ 1.7 \times 2.4$ and $ 1.7 \times 2.5$

now we want to use three squares to override the above squares

they are: square1: $ 7 \times 1$
              square2: $ 1.7 \times 0.6$
              square3: $ 3.3 \times 0.1$

you can divided the above three squares into small ones to done the work.

and how to divided will be convenience

(thanks sunken rock , I have made a mistake.....)

See that the total area of second group of 'squares' is $ 8.35$, while the first group is $ 8.50$!


I just have a question about the AIME Seminar...

What kind of things do we cover in the seminar??

I know that some past AIME pbls..and reviewing some stuff..

Is there anything other than that??

Thanks,

We'll talk a little bit about general strategies, but mostly we'll be working through past AIME problems of all difficulty levels.

As MCrawford said in another thread (I can't remember exactly where), when we say "working through past problems", we don't just mean presenting solutions, we mean that we'll try (as much as possible) to answer the question "How would I have thought of that?".

If we took the AIME Problem Series B, will there be much of the topics and problems (i.e. Is it worth taking?) ?

If you took the AIME B Problem Series -- think of the seminar as a very similar level of class using different problems.


probability Princeton college Harvard AMC USA(J)MO USAMO

I know the PSAT's not used for college exactly, just for the National Merit Scholarship competitions...but anyone want to talk about how they did and what kinds of strategies they used? :D

i got a 233... and i just went through the questions and answered them... well, ok... strategy, really? um... i go through and if a question looks boring or tedious, i skip it. there's ALWAYS time to go back if you are a quick reader. if you aren't, you are kind of already out of luck but for younger folks that have some time, i would suggest getting on the stick about reading for pleasure. it helps you out later if you read now, because you will be a faster reader (and comprehend more). also, only skip if you have NO idea what's going on. don't skip if you know half the words and are kinda sure but not really sure. probability is with you.

[quote="Iversonfan2005"]How have you guys already taken the PSAT, I thought the only dates to take it were October 12 and 15 this year?[/quote]

Yeah...but the test is offered every year :P

I took it last year as a sophomore and got 225... missed a math question due to my inability to add. But yeah, the writing section comes with lots of practice (or natural talent, I suppose; the curve is forgiving, though) and the reading... is not my thing. 225 was enough for NM semifinalist in WA, though, so hopefully I can make it (preferably 235+, aka definitely no math mistakes).

Practice drastically improves your SAT/PSAT score.  They test a lot of the same types of stuff so those who know what to do before hand have a distinct advantage.  Go and check out an SAT review book at your local library.  It is preferable to get one with a lot of practice SAT's.

[quote="furious"]i got a 206 last year. good enough for commended scholar. english isn't my forte and i'm would say i'm just ok at the writing section. and i missed 1 math question (due to not being able to read my own handwritting) and that dropped my score down 4 points. now how is that fair?[/quote]

Ditto on that. 65+7 = 62 cost me 4 PSAT points :mad:. Missing 3 questions on the writing section dropped the same amount. Why don't they make a better curve (make the questions harder)? Instead of punishing careless mistakes SO heavily.

Windslicer, I thought you were a sophomore?  :?:

[quote="WindSlicer"]Can someone explain to me the National Merit Scholar notification timeline? I still don't quite understand it.. does it take them a year to notify you...?

If I take the PSAT in October (yeah I understand that everyone else does too) when will I be notified of Semi-Finalist/Finalist/etc. if I made it.[/quote]

It's about a year later, yeah. You get notified at the beginning of your senior year I believe.

[quote="paladin8"]It's about a year later, yeah. You get notified at the beginning of your senior year I believe.[/quote]

Quite so.

About the math - they should make it hard. *mutters* Harder than RC or Writing thinks I. :D It's not our fault public education in America just kind of fails to teach problem solving of any non-trivial kind.

Does anyone know what kind of scores you need to make on the new SAT to continue to advance within National Merit? I had to take the new SAT (yesterday) for them, and I have this sick feeling I didn't do "great." Do I need to do "great" or will "decent" (2100+) suffice? I'm not even sure I made a 2100, the essay was AWFUL.

As far as I know, decent will suffice. Keep in mind - 15,000 of about 16,000 semifinalists advance, so you shouldn't be too worried.

General opinion is that the SAT score is the only real important factor, but the application has an essay, counselor rec. letter, etc... on it.

[quote="lowest_reynolds"][quote="rain"]Paladin, i heard of some really talented people who got in thier sophmore year so i was just wondering...kinda stupid question on my part[/quote]

Yeah...one sophomore got in last year.  She took Harvard Math 55 or something???  Apparently it's the hardest Math class ever...

No USAMO, no research; there are many ways to showcase your talent to the RSI committee.  One things for sure this at least the second year in a row Canadians won't be participating in RSI. :([/quote]

yea, i heard that half the kids at rsi dont know how they got in because they never did anything national recognized (usamo, imo, any olympiad, isef, research, etc)
kinda wierd, what do u guys think? are they taking like polymer chem or intro to quantum physics?

[quote="rain"][quote="paladin8"]Well someone from my school got in two years ago without anything nationally recognized. He was very talented at math and an experienced programmer, which probably looked pretty good to the RSI people. After RSI he did get Westinghouse semifinalist, though. Yes, it's definitely hard to get in. But it can't hurt to try, yeah? And I'd like to think I have a pretty decent chance. :D[/quote]

lol, im sure you do. when you apply for rsi, what is ur area of specialty?  do they already have tons of math/physics people?[/quote]

mines probably math/cs... im not sure how much they care about diversity of field, so cant say whether it'll help or take away from my chances.

Do they evaluate you in comparison to other people who chose your field?

Who knows?  I think RSI applications might be more mysterious than college applications.

213 (don't remember breakdown)

Very good score, except when you look at last year's cutoff - 214, it kinda makes me sad :(

Kinda stinks how a misbubble (111 instead of 1111) could cost me 10,000$\$$ but I guess that's how the test works :P

[quote="paladin8"]*sigh* We get them Friday... *anticipation builds up*.[/quote]

SAME HERE! along with our report cards...im so nervous about PSAT...report cards are a joke anywayz, my school too easy

Our scores just got back today. My breakdown is as follows:
Math: 74 :mad: (missed 2 easy questions due to not reading them properly)
Reading: 71 :) (big improvement from last year)
Writing: 80 ;)
Overall Score: 225 (enough for semifinalist status in Washington)

[quote="xxreddevilzxx"]i didnt do well at all.

217.[/quote]

i believe the cutoff fo FL last year was a 214 so u would have made the floridian cutoff
what grade are u in?

Finally.

Reading: 75
Math: 80
Writing: 78

Total: 233.

Was looking for the big 240, but guess I was being a little too hopeful :P.

wow you people make me feel so stupid :( i thought i did so well with my 219- was one of the highest in my school, but i guess this community is not quite a cross section of US population.
it's kind of interesting that AoPSers do well overall, since i used to assume that most peope getting 800 in math would get around 700 verbal at most (my friend actually got 800 verbal and 500 math on the old sat- go figure!)


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14059[/img]

 :(

HaTa is the sideline BC, the perpendicolar to HaTa from Ha is the A-altitude. Draw the line homotetic of HaTa with center G and radio -2, that meet the A-altitude on A. Call $ M: AG \cap HaTa$, that is the midpoint of AB, draw the perpendicolar line to HaTa on M that meet ATa on N. Then N is the midpoint of the arch BC that not contain A of the circumcircle of ABC. Then the perpendicolar axes of AN meet the perpendicolar line to HaTa on M on O that is the circumcenter. The circle with center O and radius OA meet HaTa on B and C.

"[b]Then the perpendicolar axes of AF[/b]"

where's F   ?

 :(

:blush:  yes i have edited i was said AN not AF  :lol:

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14069[/img]

 :thumbup:


ratio
arithmetic sequence
geometric sequence

若a1,a2,a3,a4,a5成等差數列,且a1,a2,a3,a4,a5成等比數列,
則a5可能為2000"試判斷此句話之真偽及其理由 .

[color=darkblue]Translate:[/color]
Suppose $ a_1,a_2,a_3,a_4,a_5$  is in $ A.P.$ , and also in $ G.P.$
Then the possible value of $ a_5$  is $ 2000$ ,how may it be justified ,if any, give resons.

[hide="solution"]If one of the numbers is negative, then the numbers must alternate between negative and positive, but this is impossible because the numbers are in arithmetic progression. If one of the numbers is 0, then all of the numbers are 0, but this is impossible since $ a_5\equal{}2000$. Thus they are all positive. From AM-GM, $ \frac {a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} a_5}{5}\geq \sqrt [5]{a_1a_2a_3a_4a_5}$, or $ a_3\geq a_3$, with equality when all numbers are equal. Thus $ a_1 \equal{} a_2 \equal{} a_3 \equal{} a_4 \equal{} a_5 \equal{} 2000$[/hide]

please correct me if i'm wrong but the problem is so weird.

for example if a1 = a2 = a3 = a4 = a5 = 1, then they are clearly in arithmetic progression ( difference of 0 ),  and they are clearly also in geometric progression ( ratio of 1 ). So a5 can be any real number.  Why does it have to be 2000 only??   So weird.   :maybe:


Let a,b,c>0.Prove that  $ \sqrt[3]{ \frac{a}{b\plus{}c}}\plus{}\sqrt[3]{ \frac{b}{c\plus{}a}}\plus{}\sqrt[3]{ \frac{c}{a\plus{}b}}>2$              :?:

You apply $ GM \geq HM$ for $ \frac{a}{b\plus{}c},1,1$ and others and we sum them;


It is quite simple to show that the best possible k is [i]not [/i] k=n^2 :( but k=4 for any n \ge 2. :?

If n=2 the ineq becomes (letting x :=x_1, y := x_2)
 (x+y)^2 . 2xy \ge 2k . x^2.y^2
or
(x-y)^2 \ge (k-4)xy
so kmax=4.
If n>2 choose x_3=...=x_n=0, ending up with the same ineq. as for n=2 (both terms with x_1x_2 in the cyclic sums are counted once here, whereas twice if n=2). So k cannot be >4.
It remains to prove the original ineq with putting k=4. Because of homogenity, we can assume WLOG \sum x_i = 1. So the ineq to prove becomes
\sum x_i x_(i+1) \ge 4.\sum x_i^2 x_(i+1)^2
or \sum x_i x_(i+1) (1 - 4 x_i x_(i+1) ) \ge 0.
In this last form, each term of the sum is obviously \ge 0. Because otherwise we would have for a certain i
 4 x_i x_(i+1) > 1 = (\sum x_j ) ^2 \ge x_i^2 + x_(i+1)^2,
i.e. 0 >  (x_i - x_(i+1))^2 contradiction ! :P


Show there exist polynomials   $p(x)$, $q(x)$ with integer coefficients, such that $p(x)(x + 1)^{2n} + q(x) (x^{2n} + 1) = k$, for some positive integer $k$. find the least   $k$ for given  $n$.

I think it is fair to say that if $n=2^tp$ then $k_{min}=2^p$

India International shortlist= Shortlist 96


How many 4-letter words can be formed from the word assisting?

[quote="SM4RT"][hide="hint"]Use multinomial coefficients[/hide][/quote]
What are multinomial coefficients?

Multinomial coefficients is basically the expansion of the binomial theorem..so

$(x_{1}+x_{2}+x_{3}\cdots+x_{m})^{n}=\sum_{k_{1},k_{2},\cdots,k_{m}}\binom{n}{k_{1},k_{2},\cdots,k_{m}}x_{1}^{k_{1}},x_{2}^{k_{2}},\cdots,x_{m}^{k_{m}}$ as $\sum_{i=1}^{m}k_{i}=n$

So in your case:
$n=4$ (since you want 4 letter words)
$m=6$ (since there are 6 distinct letters)

so let:

$k_{1}\implies\text{a}\implies 0\le k_{1}\le1$
$k_{2}\implies\text{g}\implies 0\le k_{2}\le1$
$k_{3}\implies\text{i}\implies 0\le k_{3}\le2$
$k_{4}\implies\text{n}\implies 0\le k_{4}\le1$
$k_{5}\implies\text{s}\implies 0\le k_{5}\le3$
$k_{6}\implies\text{t}\implies 0\le k_{6}\le1$

Basically the maximum value for $k_{i}$ is the number of times the letter appears in the given word

:) hope that helps

[quote="Leao"]ok thanks, I got it :)[/quote]

ya no problem..


linear algebra
matrix
linear algebra unsolved

If $ AX\equal{}0$ and $ BX\equal{}0$ has the same solutions

then can we get some relationship between matrix $ A$ and $ B$ ? :roll:

For the question to make sense, we do need to assume that $ A$ and $ B$ are the same shape - say, they are both $ m\times n$ matrices.

In order to have the same null spaces, $ A$ and $ B$ would have to be row equivalent. That is, $ A$ and $ B$ are row equivalent iff there exists an invertible matrix $ R$ such that $ RA\equal{}B.$

We can argue this through a canonical form. In this case that canonical form is reduced row echelon form.


function
geometry
number theory
number theory open

http://www.mathlinks.ro/Forum/viewtopic.php?t=77591

Well, just look here, there is a proposed proof for Riemann Hypothesis... however, this wasn't officially accepted I think, so it would be nothing... and it has only 7 pages, so, or it's marvelous, or it's a joke, or it is empty of logic... well just see it.

Surely it is a joke cos if it was true the author could have published it just anywhere.

[quote="nfrer"]Can you just take some time and find any flaws to reject it with reason. This will definitely help and since you are so sure it is nothing it won't take any time for you I assume..

Thnx Nick[/quote]
I agree that it is nothing serious, but it could even take some time to read and understand it;
either: the proof is meant serious (I doubt this!), then at least the author, who should have some knowledge about the subject, didn't see a flaw
or: it is a bad joke, in that case it isn't worth noticing (professional prooffakers put the flaw into a position where it isn't easy to find ;) ).
When the author would really think it works, he would not only propose it in that way...

I just took a look at the abstract, which doesn't even apply to the zeta function. The theorem discussed limits the zeros of a strictly decreasing, zero-converging function based on the LaPlace transform. Anyway, the zeta function most certainly does not converge to 0!!! Depending on what size infinities we are dealing with, the zeta function either converges to 1 or diverges.

Well I can assure you all that it is a serious attempt and not trying to trick with some logic flaw or something.  It is not published yet (that will take 4 months) and the first version of this was only released a couple of weeks ago.  I probably should just have waited for it to be published before releasing it on the web but I got over excited.
I agree that the Zeta function is not positive decreasing, but the function I use which is positive decreasing has as its two sided Laplace transform the Riemann Xi-Function(s+1/2) which has the same zeros as the Zeta(s+1/2) function in the critical strip (its formula is a bit too long to put in here).
Proof has been updated to improve some of the weak areas and personally I think its now indisputable (but I would).
I realise that its a lot to ask for someone to look through but the proof is fairly accessible and mainly comes from the first result which anyone with a basic knowledge of Laplace should be able to verify.  Amazingly I think this result is new and this is the root of the proof.  Anyone who has any feedback (good or bad) or questions is more than welcome to contact me.

http://www.aronpalmer.com/rproofv2.pdf

Post a link to your paper at the sci.math newsgroup; they are usually willing to look over such things there and will probably debunk what you wrote pretty quickly.

Its already been posted on sci.math.research (moderated) and got zero feedback, which is most likely because no-one could be bothered to go through it, which is my biggest problem at the moment (though optimistically you could say they couldn't find something wrong easily).  Doubt anyone on this board will go through it either, but you can't say you didn't have the chance.

Try sci.math instead of sci.math.research; it gets more activity.

I feel your pain, Aron; you should be able to get people to look at your proof seriously.  (Although I must say that a seven-page proof of the Riemann Hypothesis sounds unlikely to be legitimate from the onset.)  I posted something in sci.math for you, in case you haven't already, and I'll write back here if I hear anything.  Make sure to write back with any developments, Aron.

If anyone here has a chance, I think it'd be great for he or she to look through the proof, and give some feedback; I'm not really knowledgeable about Laplace transforms, so I'm not really that much of help.  Let's do our best to encourage another member of AoPS!  :)

Its up to nine pages now, clearly seven was not enough, but as you say, given that the website is called the art of problem solving I don't feel too guilty about it appearing here (not by me),  in an area called "Number Theory Open Questions".
Ta,

Aron.

The Riemann Hypothesis is just such an important conjecture that people are sometimes a little intimidated by it, I think, and just as with many other unsolved mathematics questions, there are a lot of people with very unsound "proofs" out there.  When someone says that they have a proof for the Riemann Hypothesis, it's easy to think that the proof must be automatically flawed if they aren't a very well-known mathematician.  But, of course, we shouldn't judge your proof from length alone; it is definitely possible that a relatively short proof exists for the Riemann Hypothesis, and you happened to be incredibly clever in figuring it out.

Keep us posted, and check out http://mathforum.org/kb/message.jspa?messageID=4535273&tstart=0 from time to time; that's where I posted.

[quote="aronpal"]... It is not published yet (that will take 4 months) and the first version of this was only released a couple of weeks ago.  I probably should just have waited for it to be published before releasing it on the web but I got over excited. ...[/quote]
You say it will take four months for it to be published.  Does that mean that it's been accepted for publication?

...[/quote]
You say it will take four months for it to be published.  Does that mean that it's been accepted for publication?[/quote]
It hasn't been accepted or rejected yet, in fact they haven't even found anyone to referee it yet, the wheels turn slowly.


superior algebra
superior algebra unsolved

Let F,E,K be fields with $ F \leq E \leq K$. Prove that if E is algebraic over F, and K is algebraic over E, then K is algebraic over F. You must [i]not[/i] assume the extensions are finite.

All we know is that for each k in K, $ \sum_{i \equal{} 0}^n e_i k^i \equal{} 0$ and that for each i $ \sum_{j \equal{} 0}^{m_i} f_j e_i^j \equal{} 0$. How does that help?

this cannot be proven directly, you have to use

$ a \in K$ algebraic over $ F$ iff $ F(a) / F$ is finite

To be more exact: it can be proven using this, but you will die on the way  :wink:
It's about $ 23$ times shorter to use the way -oo- suggested.

really? can it be done without repeating those finiteness-arguments?


81% of all scuba divers have experienced nitrogen narcosis at depths of 100 feet and deeper. Eighty divers are randomly selected asked if they had ever experienced nitrogen narcosis at depths of 100 feet and deeper. Given that at least 60 have experienced nitrogen narcosis, what is the probability that 70 to 75 divers inclusive have experienced nitrogen narcosis? Use the normal approximation or binomial probability to solve this problem .

Thanks.

[hide]Let $ A$ be the event that 70 to 75 of the 80 divers (inclusive) have experienced this nitrogen problem. Let $ B$ be the event that at least 60 of the 80 divers have experienced it. Let $ p \equal{} .81$
We know that $ P(A|B) \equal{} \frac {P(A \cap B)}{P(B)}$

$ P(A) \equal{} P(A \cap B)$ = binomialcdf(80,p,75) - binomialcdf(80,p,69) $ \approx .0852$
$ P(B)$ = 1 - binomialcdf(80,p,59) $ \approx .9305$
Therefore $ P(A|B) \approx \frac {.0852}{.9305} \approx .0916$ or $ 9.16 \%$[/hide]


Which AIME years were the easiest? hardest?

i found 2003 to be pretty easy but I would like to know other years like this
Thanks

[url=http://www.unl.edu/amc/e-exams/e7-aime/archiveaime.html]The AMC website[/url] contains statistics on a number of past AIME exams that you can look through, if you so desire.

I think 1988 AIME was one of the easiest since I got a 11 on it, not a score I can get on a normal AIME.


geometry
circumcircle
symmetry
ratio
geometric transformation
reflection
cyclic quadrilateral

in triangle $ABC$ draw a perpendicular line $\ell$ to $OH$ at $H$ , $D,E,F$ are intersection of tangents to circumcircle at point $A,B,C$ with $\ell$ .

Points $A',B',C'$ are symmetry of $D,E,F$ wrt $A,B,C$ respectivly prove that $A',B',C'$ are collinear

[quote="amir2"]draw a perpendicular line $\ell$ to $OH$[/quote]

In my opinion, $\ell$ should be the perpendicular to the line OH at the point H (and not just any arbitrary perpendicular to OH).

Beautiful problem, by the way! (Though quite easy after an additional observation about the line A'B'C'...)

  Darij

Darij is right, as always, in saying that this problem is quite easy.

 I agree, it is a pretty nice composition  of a simple property usage.

  In fact, this problem is so simple it has to be moved to High School section,

  only because there is no Pre-school forum here.  Isn't it a ripe time to open one?

  If the problem  starts to get a bit complicated and involved, you should know

  you are on the wrong track !

  Thank you.

  M.T.

$\text{armpist}$ are you joking? :?:

I found it difficult  and prove it hardly.

Would you send your solution clearly.

Let K, L, M be the feet of the A-, B-, C-altitudes and $t_A, t_B, t_C$ the circumcircle tangents at the vertices A, B, C. WLOG, assume that $\angle B < \angle A < \angle C$. Denote $\phi = \angle DHA > 0$. The following angles are right, the 1st three by definition, the 2nd three because circumcircle radii OA, OB, OC are perpendicular to tangents $t_A, t_B, t_C$:

$\angle OHD = \angle OHE = \angle OHF = 90^\circ$

$\angle OAD = \angle OBE = \angle OCF = 90^\circ$

It follows that the quadrilaterals OADH, OBEH, OFCH are all cyclic. From the isosceles triangle $\triangle A'OD$ and from the cyclic quadrilateral OADH,

$\angle AOA' = \angle DOA = \angle DHA = \phi$

Let t be a tangent from the point A' to the circumcircle (O) different from $t_A$ and T its tangency point. Then $\angle A'OT = \angle AOA' = \phi$ and $\widehat{(t_A, t)} = 2 \phi \mod 180^\circ$.  From the isosceles triangle $\triangle B'OE$ and from the cyclic quadrilateral OBEH,

$\angle B'OB = \angle BOE = \angle BHE = \angle BHK - \angle EHK = \angle C - \angle DHA = \angle C - \phi$

Let u be a tangent from the point B' to the circumcircle (O) different from $t_B$ and U its tangency point. Then $\angle UOB' = \angle B'OB = \angle C - \phi$ and $\widehat{(t_B, u)} = 2(\phi - \angle C) \mod 180^\circ$. But since $\widehat{(t_A, t_B)} = 2 \angle C \mod 180^\circ$, it follows that

$\widehat{(t, u)} = \widehat{(t, t_A)} + \widehat{(t_A, t_B)} + \widehat{(t_B, u)} = -2 \phi + 2 \angle C + 2(\phi - \angle C) = 0^\circ \mod 180^\circ$

This means that the circumcircle tangents $t \equiv A'T$ and $u \equiv B'U$ are identical and their tangency points $T \equiv U$ are also identical. Likewise, from the isosceles triangle $\triangle C'OF$ and from the cyclic quadrilateral OFCH,

$\angle COC' = \angle FOC = \angle FHC = \angle FHK + \angle KHC = \angle DHA + \angle B = \phi + \angle B$

Let v be a tangent from the point C' to the circumcircle (O) different from $t_C$ and V its tangency point. Then $\angle C'OV = \angle COC' = \angle B + \phi$ and $\widehat{(t_C, v)} = 2(\phi + \angle B) \mod 180^\circ$. But since $\widehat{(t_C, t_A)} = 2 \angle B \mod 180^\circ$, it follows that

$\widehat{(t, v)} = \widehat{(t, t_A)} + \widehat{(t_A, t_C)} + \widehat{(t_C, v)} = -2 \phi - 2 \angle B + 2(\phi + \angle B) = 0^\circ \mod 180^\circ$

and again, this means that the circumcircle tangents $t \equiv A'T$ and $v \equiv C'V$ are identical and their tangency points $T \equiv V$ are also identical. Consequently, the points A', B', C' are collinear, lying on the same tangent t to the circumcircle (O).

[quote="armpist"]In fact, this problem is so simple it has to be moved to High School section,

  only because there is no Pre-school forum here.  Isn't it a ripe time to open one?
[/quote]
Your rudeness is revolting.

[quote="mecrazywong"][quote="armpist"]
In fact, this problem is so simple it has to be moved to High School section,
only because there is no Pre-school forum here.  Isn't it a ripe time to open one?
[/quote]
Personally I don't agree with you-just have a glance at the other posts and you will see many much easier problems which can be solved by simple angle chasing or ratio computationh.[/quote]

  No offence was intended, as it always has been the case with my posts.

  Mecrazywong, if you glance at Yetti's solution, you will realize, no doubt, that

  it was nothing more than an angle chase.  Darij was the first one to call this problem

  'quite simple' , so 'gang-up' on him, if you like.

  Amir2 is the only one who possibly has the right to get offended, although

  I do not see a reason for such reaction: problem difficulty is and have always

  been very subjective issue.

  Naturally, I sent my take of this problem in a private message to amir2

  and I still think it is a beautiful rendition of a nice idea (maybe un-intentional

  on amir's part).   I would call this problem a fine sample of 'illusion' :

  solvers are made to believe something which is different from what

  really takes place (e.g.  tangency of that line is accidental).

  Compared to the 'beauty' level aspect, the complexity of the problem is nowhere

  near what one might expect at the first glance, and that was the essence of my

  'controversial' post.

  Thank you.

  M.T.

This problem can be easly than using complex numbers...

Sorry for being late with posting my solution. Anyway, I will prove more than the problem asks for:

[color=blue][b]Theorem 1.[/b] Let ABC be a triangle with circumcenter O and orthocenter H. Let $ \ell$ be the perpendicular to the line OH at the point H, and let the tangents to the circumcircle of triangle ABC at the points A, B, C intersect the line $ \ell$ at the points X, Y, Z, respectively. Denote by X', Y', Z' the reflections of the points X, Y, Z in the points A, B, C.

Denote by T the Anti-Steiner point of the line $ \ell$ with respect to triangle ABC. Then, the points X', Y', Z' lie on the tangent to the circumcircle of triangle ABC at the point T.[/color]

[i]Note.[/i] The Anti-Steiner point T of the line $ \ell$ with respect to triangle ABC is defined as the point where the reflections of the line $ \ell$ in the sidelines BC, CA, AB of triangle ABC concur. The fact that they actually concur is proven in [b][1][/b] (it follows from the fact that the line $ \ell$ passes through the orthocenter H of triangle ABC); what is also proven there is that the Anti-Steiner point T lies on the circumcircle of triangle ABC.

[i]Proof of Theorem 1.[/i] We will use directed angles modulo 180°.

According to [b][1][/b] Lemma 1, the reflection B' of the orthocenter H in the sideline CA lies on the circumcircle of triangle ABC. Now, since the line $ \ell$ passes through the point H, the reflection of this line $ \ell$ in the line CA must pass through the reflection of the point H in the line CA, i. e. through the point B'. Furthermore, this reflection passes through the point T, as we know. Hence, the reflection of the line $ \ell$ in the line CA is the line B'T. Thus, $ \measuredangle\left(B^{\prime}T;\;CA\right) \equal{} \minus{} \measuredangle\left(\ell;\;CA\right)$. Now, since H is the orthocenter of triangle ABC, we have $ BH\perp CA$, so that < (CA; BH) = 90°, and since we are working with directed angles modulo 180°, we have 90° = - 90°; thus,

$ \measuredangle\left(B^{\prime}T;\;BH\right) \equal{} \measuredangle\left(B^{\prime}T;\;CA\right) \plus{} \measuredangle\left(CA;\;BH\right) \equal{} \measuredangle\left(B^{\prime}T;\;CA\right) \plus{} 90^{\circ}$
$ \equal{} \minus{} \measuredangle\left(\ell;\;CA\right) \minus{} 90^{\circ} \equal{} \minus{} \measuredangle\left(\ell;\;CA\right) \minus{} \measuredangle\left(CA;\;BH\right) \equal{} \minus{} \measuredangle\left(\ell;\;BH\right) \equal{} \measuredangle\left(BH;\;\ell\right)$.

In other words, < TB'B = < BHY.

On the other hand, since the line BY is the tangent to the circumcircle of triangle ABC at the point B, while the point O is the center of this circumcircle, we have $ BY\perp OB$, so that < OBY = 90°. Since $ \ell\perp OH$, we have < OHY = 90°. Thus, the points B and H lie on the circle with diameter OY. Consequently, < BHY = < BOY.

In the right-angled triangle OBY, we have 90° - < BOY = < OYB.

Now, since the point Y' is the reflection of the point Y in the point B, the point B is the midpoint of the segment YY'; hence, from $ BY\perp OB$, it follows that the point O lies on the perpendicular bisector of the segment YY', and thus, by symmetry, < OYB = - < OY'B. In other words, < OYB = < BY'O.

Finally, since O is the center of the circumcircle of triangle ABC, which passes through the points B and T, the central angle theorem for directed angles modulo 180° yields < BTO = 90° - < TB'B. Thus,

  < BTO = 90° - < TB'B = 90° - < BHY = 90° - < BOY = < OYB = < BY'O.

Thus, the points B, O, T and Y' lie on one circle, so that < OTY' = < OBY'. But $ BY\perp OB$ leads to < OBY' = 90°, and thus < OTY' = 90°, so that $ TY^{\prime}\perp OT$. Since the point T lies on the circumcircle of triangle ABC and the point O is the center of this circumcircle, this shows that the line TY' is the tangent to the circumcircle at the point T. Thus, the point Y' lies on the tangent to the circumcircle of triangle ABC at the point T. Similarly, the same holds for the points Z' and X', and Theorem 1 is proven.

[color=red][b]References[/b][/color]

[b][1][/b] Darij Grinberg, [i]Anti-Steiner points with respect to a triangle[/i]. http://www.cip.ifi.lmu.de/~grinberg/

You were right, by the way, the problem is not that easy. I said it was easy since this was my first impression after solving the problem, as I am generally used to problems of this kind being very hard.

  Darij


geometry
circumcircle
combinatorics unsolved
combinatorics

Color the plane in 3 colors. Prove that there are two points of the same color 1 unit apart.

Say the colors are Red,Green,Blue. We'll prove it by contradiction, so suppose not. Then consider a regular hexagon of circumradius 1 with center (say Red) and verticies A1,A2,A3,A5,A5,A6, taken in counterclockwise order. Wlog, assume A1 to be Green(if its red then we are done) then A1,A3,A5 is green and A2,A4,A6 are all Blue. Now consider the point P such that the triangle PA1A3 is an isoceles triangle with PA1 = 1 and PA3 = A1A3 = sqrt(3). P cant obviously be colored Green, Thus it must be colored Red or Blue. Now take the unit segment QR such that QR bisects PA3 and vice Versa.It is trivial to check that no matter what Q and R are colored (as long as its Red, Green or Blue), that one of unit segaments PQ, PR, QR, A3Q,A3R, has monochromatic ends. Contradiction.


Hello, I am student from Latvia and am interested ih math, especially olympiad math. I know, that there is such book as "Mathematical olympiad challenges" by Titu Andreescu & Gelca.
Could you tell me where from I can download this book. :?:
Thank's.

You can't download it.

If you really want it, you can buy it off [url=http://www.amazon.com/exec/obidos/tg/detail/-/0817641556/qid=1111352382/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/002-8080026-5904045?v=glance&s=books&n=507846]amazon.com[/url]

May be you are talking about these books

[url=http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc96-97-01feb.pdf]Andreescu, Kedlaya, Mathematical Contests 1996-1997: Olympiad problems from around the world, with solutions, AMC 1998.[/url]

[url=http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.pdf]Andreescu, Kedlaya, Mathematical Contests 1997-1998: Olympiad problems from around the world, with solutions, AMC 1999.[/url]

Just in case, click on them, you can download them directly. Those books are two of a big collection (I am not sure, but I think they started in 1994 or 1995), but just these two are freely available online. Probably you may find the others, or the specific book you asked for in a P2P or IRC net, what is, let me advise you, ilegal in most of the countries, because it is copy righted material not freely waived by the author.

Any way, there are a bunch of good books that you can buy. I feel a little lazy today to find it, but I know that in this same site someone already did a good list of titles.

Best,

_sasha_, did you suceed to purchase the book via amazon.com?
I don't know about in Latvia, but in Lithuania I've managed to get it by ordering it in the bookshop.

Yes, I have a huge collection of math books, and I think I got this one at Barnes & Noble by special order.

It's not available for download, try ordering at your local book store or via amazon.com as suggested.
I enjoyed Mathematical Olympiad Treasures more than Challenges though.


Counting with Division Part 1

We learn how to use division to correct for overcounting.  We start with a couple basic problems in which we count pairs of items by counting all the items first, and then dividing by 2.  We also learn a little secret about counting problems in which the number of possibilities is of the form 1 + 2 + 3 + . . . + n for some number n.

to count n numbers,
the formula is (n(n-2))/2

To find the number of handshakes, he could have also done: $\binom{56}{2}$ .

However, Mr. Rusczyk used the formula $\frac{(n)(n-1)}{2}$, where $n$ is the number of people.  I would have used $\frac{n!}{(n-2)! \cdot 2!}$, where $n$ is also the number of people.  Both formulas would generate the same answers.

I will show that it is true.  We will write:

$\frac{n!}{(n-2)! \cdot 2!}$   =   $\frac{(n)(n-1)}{2}$.

This is true because if we cross multiply, we have:

$2 \cdot n! = (n-1) \cdot (n) \cdot (n-2)!\cdot2!$ ,
$n! = (n) \cdot (n-1) \cdot (n-2)!$ .

And so it is obviously correct.