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Here is a truly magnificient result: suppose $ p>3$ is a prime number and $ G$ is a subgroup of $ GL_2(F_p)$ such that for any polynomial $ f\equal{}X^2\plus{}aX\plus{}b$ with nonzero $ b$ there is a matrix in $ G$ whose characteristic polynomial is $ f$. Then $ G\equal{}GL_2(F_p)$. I'm wondering if there's a generalisation of such a result in higher dimensions.  :idea:

use the well-known classification of maximal subgroups

hutsiutsi,
You might be right, but which "well-known classification of maximal subgroups" are you referring to?

normalizer of (non-)split Cartan subgroup and Borel. used e.g. by Serre and Swinnerton-Dyer

Ok, I'm finally ready for this proof.
Let $ q=p^n>3$ be some prime power and $ G$ such a proper subgroup of $ GL_2(q)$.

Actually there is a classification of all subgroups of $ PSL_2(q)$ done by [url=http://en.wikipedia.org/wiki/Leonard_Eugene_Dickson]Dickson[/url],
but it's not "well-known" (at least not to me :P ) and very complex with many cases.
In fact all possible subgroups have the form
(1) cyclic groups of order dividing $ q\pm 1$
(2) dihedral groups of order dividing $ 2(q\pm 1)$
(3) $ P\rtimes \langle x\rangle$ with some elementary abelian $ p$-group $ P$ and some $ x$ of order dividing $ |P|-1$
(4) $ A_4,S_4$ or $ A_5$
(5) $ PSL_2(p^m)$ for $ m|n$ or $ PGL_2(p^m)$ for $ 2m|n$

In fact I had to look it up in [url=http://www.zentralblatt-math.org/zmath/en/search/?q=an:0217.07201]Huppert I[/url] to complete the following proof.

$ \mathbb{F}_{q^2}^\times=\langle \zeta\rangle$ is cyclic of order $ q^2-1$
with $ \mathbb{F}_{q}^\times=\langle \zeta^{q+1}\rangle$ the only subgroup of order $ q-1$.
The extension $ \mathbb{F}_{q^2}/\mathbb{F}_{q}$ is galois with only one nontrivial automorphism $ x\mapsto x^q$.
So we find $ A,B\in G$ with characteristic polynomials $ (X-\zeta)(X-\zeta^q),(X-1)(X-\zeta^{q+1})\in \mathbb{F}_{q}[X]$.
$ B$ has eigenvalues $ 1\not=\zeta^{q+1}$ in $ \mathbb{F}_{q}$, so it is diagonalizable and we find $ u\in GL_2(q)$ with $ u^{-1}Bu=\begin{pmatrix}1 &\\&\zeta^{q+1}\end{pmatrix}$.
Then after replacing $ G$ by the conjugate $ u^{-1}Gu$ we may assume $ B=\begin{pmatrix}1 &\\&\zeta^{q+1}\end{pmatrix}\in G$.
Analog $ A$ has eigenvalues $ \zeta\not=\zeta^q$ in $ \mathbb{F}_{q^2}$, so we find $ u\in GL_2(q^2)$ with $ u^{-1}Au=\begin{pmatrix}\zeta &\\&\zeta^q\end{pmatrix}$.
So $ \mathcal{A}_1: =\langle A\rangle$ has order $ q^2-1$ and $ A^{q+1}=u(\zeta^{q+1} I_2)u^{-1}=\zeta^{q+1}I_2$
gives the centre $ Z: =\mathbb{F}_q^\times I_2=\langle A^{q+1}\rangle\le\mathcal{A}_1\le G$.

If $ I_2\not=C\in \mathcal{A}_1\cap \langle B\rangle$, then $ C=A^n$ must have eigenvalue $ 1$ for some $ n\in \mathbb{Z}$.
But then $ \zeta^n=1$ or $ \zeta^{qn}=1$ gives the contradiction $ q^2-1|n$ and $ C=A^n=I_2$.
So $ \mathcal{A}_1\cap \langle B\rangle=1$ and $ |G|\ge|\mathcal{A}_1 \langle B\rangle|=|\mathcal{A}_1|\cdot | \langle B\rangle|=(q^2-1)(q-1)$.
Also $ G$ has the subgroup $ \mathcal{A}_2: =Z \langle B\rangle=\begin{pmatrix}* &\\&*\end{pmatrix}$ of order $ (q-1)^2$.

But $ \mathcal{A}_2$ gives all reducible characteristic polynomials $ f=X^2+aX+b\in\mathbb{F}_q[X]$ with $ b\not=0$,
and for $ f$ irreducible, it has conjugate roots $ \zeta^n,\zeta^{nq}\in\mathbb{F}_{q^2}$ for some $ n\in\mathbb{Z}$,
and $ f$ is characteristic polynomial of $ A^n$.
So $ \mathcal{A}_1\cup\mathcal{A}_2$ gives all possible characteristic polynomials over $ \mathbb{F}_q$
and we may assume $ G=\langle\mathcal{A}_1,\mathcal{A}_2\rangle=\langle A,B\rangle$.

And now we need to study the subgroups of $ GL_2(q)$ a little deeper.

[b][i]Lemma:[/i][/b]
1)
For any maximal abelian $ p'$-subgroup $ \mathcal{A}$ of $ G$ (i.e. $ p\nmid |\mathcal{A}|$) we have $ Z\lneq \mathcal{A}$ and $ |N_G(\mathcal{A}): \mathcal{A}|\le 2$.
2)
$ \mathcal{A}\cap\mathcal{B}=Z$ for any maximal abelian $ p'$-subgroups $ \mathcal{A}\not=\mathcal{B}$ of $ G$
3)
$ \mathcal{A}_1,\mathcal{A}_2$ are maximal abelian in $ G$.
4)
The conjugates $ g\mathcal{A}_1g^{-1},g\mathcal{A}_2g^{-1}$ for $ g\in G$ are all the maximal abelian $ p'$-subgroup of $ G$.
5)
$ G$ is a $ p'$-group
[b]Proof:[/b]
1)
$ \mathcal{A}\le Z\mathcal{A}$ is an abelian $ p'$-subgroup of $ G$, so $ Z\le Z\mathcal{A}=\mathcal{A}$.
Also $ Z\lneq \mathcal{A}_1$ is an abelian $ p'$-subgroup of $ G$, so $ Z\lneq \mathcal{A}$.

Take $ C\in \mathcal{A}\setminus Z$ with eigenvalues $ \alpha,\beta\in\mathbb{F}_{q^2}$.
If $ \alpha=\beta$, then $ C-\alpha I_2$ is nilpotent and $ 0=(C-\alpha I_2)^{p^k}=C^{p^k}-\alpha^{p^k}I_2$ for $ k\in\mathbb{N}$ large enough.
But $ C$ is $ p'$-element, so $ C\in\langle C^{p^k}\rangle \le Z$. Contradiction!
So $ \alpha\not=\beta$ and we find $ u\in GL_2(q^2)$ with $ u^{-1}Cu=\begin{pmatrix}\alpha &\\&\beta\end{pmatrix}$.
Then $ u^{-1}\mathcal{A}u$ lies in the centralizer $ \mathcal{C}: =C_{GL_2(q^2)}(\begin{pmatrix}\alpha &\\&\beta\end{pmatrix})=\begin{pmatrix}* &\\&*\end{pmatrix}$ abelian of order $ (q^2-1)^2$.
So $ \mathcal{A}\le G\cap u\mathcal{C}u^{-1}$ is an abelian $ p'$-subgroup of $ G$, which gives $ \mathcal{A}= G\cap u\mathcal{C}u^{-1}$.
Each $ g\in u^{-1}N_G(\mathcal{A})u$ normalizes $ u^{-1}\mathcal{A}u\le \mathcal{C}$, so $ g\begin{pmatrix}\alpha &\\&\beta\end{pmatrix}=\begin{pmatrix}\gamma &\\&\delta\end{pmatrix}g$ for some $ \gamma,\delta\in\mathbb{F}_{q^2}$,
which easily gives $ g\in \mathcal{N}: =\begin{pmatrix}* &\\&*\end{pmatrix}\cup \begin{pmatrix} &*\\ *&\end{pmatrix}=\mathcal{C}\rtimes \langle \begin{pmatrix} &1\\ 1&\end{pmatrix}\rangle$.
But $ \mathcal{N}$ normalizes $ \mathcal{C}$, so $ N_G(\mathcal{A})\le G\cap u\mathcal{N}u^{-1}$ normalizes $ G\cap u\mathcal{C}u^{-1}=\mathcal{A}$,
which gives $ N_G(\mathcal{A})= G\cap u\mathcal{N}u^{-1}$ and
$ |N_G(\mathcal{A}): \mathcal{A}|=|u^{-1}Gu\cap \mathcal{N}: u^{-1}Gu\cap \mathcal{C}|=|(u^{-1}Gu\cap \mathcal{N})\mathcal{C}: \mathcal{C}|\le |\mathcal{N}: \mathcal{C}|=2$.

2)
Take $ C\in \mathcal{A}\cap\mathcal{B}\setminus Z$ with eigenvalues $ \alpha,\beta\in\mathbb{F}_{q^2}$.
Like above we have $ \alpha\not=\beta$ and $ u^{-1}Cu=\begin{pmatrix}\alpha &\\&\beta\end{pmatrix}$ for some $ u\in GL_2(q^2)$.
Also like above $ u^{-1}\mathcal{A}u,u^{-1}\mathcal{B}u$ lie in the centralizer $ \mathcal{C}: =\begin{pmatrix}* &\\&*\end{pmatrix}$ of $ \begin{pmatrix}\alpha &\\&\beta\end{pmatrix}$ in $ GL_2(q^2)$.
And again like above we get the contradiction $ \mathcal{A}= G\cap u\mathcal{C}u^{-1}=\mathcal{B}$.

3)
$ \mathcal{A}_2=\begin{pmatrix}* &\\&*\end{pmatrix}$ is the centralizer of $ B=\begin{pmatrix}1 &\\&\zeta^{q+1}\end{pmatrix}$ in $ GL_2(q)$, so $ \mathcal{A}_2$ is clearly maximal abelian.
Assume therefore some $ C\in G\setminus \mathcal{A}_1$ commutes with $ A$.
If $ C$ has some eigenvalue $ \alpha\in \mathbb{F}_q$, then $ C\not\in Z\le \mathcal{A}_1$ gives the
$ 1$-dimensional $ A$-invariant $ \mathbb{F}_q$-subspace $ \ker (C-\alpha I_2)$ of $ \mathbb{F}_q^2$.
This gives $ \ker (C-\alpha I_2)=\mathbb{F}_qv$ and $ Av=\beta v$ for some $ 0\not=v\in\mathbb{F}_q^2$ and $ \beta\in\mathbb{F}_q$.
But then $ A$ has three different eigenvalues $ \beta,\zeta,\zeta^q$ in $ \mathbb{F}_{q^2}$. Contradiction!

So $ C$ must have conjugate eigenvalues in $ \mathbb{F}_{q^2}\setminus \mathbb{F}_q$.
But $ u^{-1}Cu$ lies in the centralizer $ \begin{pmatrix}* &\\&*\end{pmatrix}$ of $ u^{-1}Au=\begin{pmatrix}\zeta &\\&\zeta^q\end{pmatrix}$  for some $ u\in GL_2(q^2)$.
So $ u^{-1}Cu=\begin{pmatrix}\alpha &\\&\alpha^q\end{pmatrix}$ for some $ \alpha\in\mathbb{F}_{q^2}\setminus \mathbb{F}_q$ and then $ \alpha=\zeta^n$ for some $ n\in\mathbb{Z}$ and
$ C=u\begin{pmatrix}\zeta^n &\\&\zeta^{qn}\end{pmatrix}u^{-1}=A^n\in\mathcal{A}_1$. Contradiction!

4)
Now $ \mathcal{A}_1,\mathcal{A}_2$ are maximal abelian $ p'$-subgroups of $ G$ of orders $ q^2-1,(q-1)^2$.
If $ \mathcal{A}_3$ is another maximal abelian $ p'$-subgroup of $ G$ not conjugate to $ \mathcal{A}_1,\mathcal{A}_2$,
then by 2) the different conjugates $ g(\mathcal{A}_i\setminus Z)g^{-1}$ for $ g\in G,i=1,2,3$ are all pairwise disjoint.
Also by 1) $ Z\lneq\mathcal{A}_3$, so $ |\mathcal{A}_3: Z|\ge 2$.
This and 1) gives
$ |G|\ge\sum_{i=1,2,3}|G: N_G(\mathcal{A}_i)|(|\mathcal{A}_i|-|Z|)\ge {1\over 2}\sum_{i=1,2,3}|G: \mathcal{A}_i|(|\mathcal{A}_i|-|Z|)$
and
$ 1\le \sum_{i=1,2,3}{1\over |\mathcal{A}_i: Z|}\le {1\over q+1}+{1\over q-1}+{1\over 2}\le {1\over 6}+{1\over 4}+{1\over 2}<1$ for $ q\ge 5$,
and for $ q=4$ we must have $ |\mathcal{A}_3: Z|\ge 3$ odd and
$ 1\le {1\over q+1}+{1\over q-1}+{1\over 3}= {1\over 5}+{1\over 3}+{1\over 3}<1$.
Contradiction!

5)
Assume $ G$ has some Sylow $ p$-subgroup $ P\not=1$.
For $ I_2\not=C\in Z(P)$ of order $ p$ we have $ 0=C^p-I_2=(C-I_2)^p$.
So $ C-I_2\not=0$ is nilpotent, and using the jordan normal form we get $ u^{-1}Cu=\begin{pmatrix}1 &1\\&1\end{pmatrix}$ for some $ u\in GL_2(q)$.
Then $ u^{-1}Pu$ lies in the centralizer $ Z\begin{pmatrix}1 &*\\&1\end{pmatrix}$ of $ \begin{pmatrix}1 &1\\&1\end{pmatrix}$ in $ GL_2(q)$,
which has normal Sylow $ p$-subgroup $ Q: =\begin{pmatrix}1 &*\\&1\end{pmatrix}$ of order $ q$.
So $ u^{-1}Pu\le Q\cong \mathbb{F}_q^+$ elementary abelian.
Then $ P\le G\cap uQu^{-1}$ is a $ p$-subgroup of $ G$, so $ P=G\cap uQu^{-1}$ abelian.
All $ g\in u^{-1}N_G(P)u$ normalize $ u^{-1}Pu\le Q$, so we must have $ g\begin{pmatrix}1 &1\\&1\end{pmatrix}=\begin{pmatrix}1 & t\\&1\end{pmatrix}g$ for some $ t\in\mathbb{F}_q$.
This gives $ u^{-1}N_G(P)u\le \mathcal{N}: =\begin{pmatrix}* & *\\&*\end{pmatrix}=Q\rtimes \mathcal{A}_2$ of order $ q(q-1)^2$.

Now if $ N_G(P)\le PZ\le C_G(P)$, then by [url=http://www.matheplanet.com/default3.html?call=fav.php?op=view&fav_id=23296&uname=Gockel&select=ddddff&ref=http%3A%2F%2Fwww.google.com%2Fsearch%3Fie%3DUTF-8%26oe%3DUTF-8%26gfns%3D1%26sourceid%3Dnavclient%26rls%3Dcom.google%3Ade%3Aofficial%26q%3Dn_g%2528p%2529%253Dc_g%2528p%2529]Burnside's transfer theorem[/url] $ G$ must have a normal Hall $ p'$-subgroup $ H$.
But then $ H$ contains all $ p'$-elements of $ G$, so $ A,B\in H$ and $ G=\langle A,B\rangle\le H$ and $ P=1$. Contradiction!
So $ PZ\not\ge N_G(P)$, which is generated by the normal Sylow $ p$-subgroup $ P$ and all its $ p'$-elements.
This gives some $ p'$-element $ D\in N_G(P)\setminus Z$ and let $ \mathcal{A}$ be some maximal abelian $ p'$-subgroup of $ G$ with $ D\in \mathcal{A}$.

But now $ \mathcal{N}=Q\rtimes \mathcal{A}_2$ is solvable with Hall $ p'$-subgroup $ \mathcal{A}_2$ and $ p'$-element $ u^{-1}Du\in\mathcal{N}$.
Then by [url=http://en.wikipedia.org/wiki/Hall_subgroup]Hall's theorem[/url] we find some $ v\in \mathcal{N}$ with $ v^{-1}u^{-1}Duv\in \mathcal{A}_2$, and after replacing $ u$ by $ uv$
we can assume $ u^{-1}Du\in \mathcal{A}_2$ and still $ u^{-1}Pu\le Q$ and $ P=G\cap uQu^{-1}$ and $ u^{-1}N_G(P)u\le \mathcal{N}$.

Then $ D\not\in Z$ gives $ u^{-1}Du=\begin{pmatrix}\alpha &\\&\beta\end{pmatrix}$ for some $ \alpha,\beta\in\mathbb{F}_q,\alpha\not=\beta$,
and then $ u^{-1}\mathcal{A}u$ lies in the centralizer $ \mathcal{A}_2=\begin{pmatrix}* &\\&*\end{pmatrix}$ of $ \begin{pmatrix}\alpha &\\&\beta\end{pmatrix}$ in $ GL_2(q)$.
But 4) gives $ |\mathcal{A}|\ge (q-1)^2=|\mathcal{A}_2|$, so $ u^{-1}\mathcal{A}u=\mathcal{A}_2$ normalizes $ Q$
and $ \mathcal{A}= u\mathcal{A}_2u^{-1}$ normalizes $ G\cap uQu^{-1}=P$, so $ \mathcal{A}\le N_G(P)$.

Now $ \begin{pmatrix}t &\\&1\end{pmatrix}\begin{pmatrix}1 &1\\&1\end{pmatrix}\begin{pmatrix}t^{-1} &\\&1\end{pmatrix}=\begin{pmatrix}1 &t\\&1\end{pmatrix}$ for all $ t\in\mathbb{F}_q^\times$.
So $ u^{-1}\mathcal{A}u=\mathcal{A}_2$ operates transitiv by conjugation on $ Q\setminus\{I_2\}$ and normalizes $ u^{-1}Pu\le Q$.
This gives $ u^{-1}Pu= Q$ of order $ q$ and $ |N_G(P)|\ge |P\mathcal{A}|=|P|\cdot |\mathcal{A}|=q(q-1)^2$.

And now all elements of $ N_G(P)\le u\mathcal{N}u^{-1}=u\begin{pmatrix}* & *\\&*\end{pmatrix}u^{-1}$ have eigenvalues in $ \mathbb{F}_q^\times$
and all elements of $ \mathcal{A}_1\setminus Z$ have conjugate eigenvalues in $ \mathbb{F}_{q^2}\setminus\mathbb{F}_q$.
This gives $ N_G(P)\cap \mathcal{A}_1=Z$ and
$ |G|\ge |N_G(P)\mathcal{A}_1|={|N_G(P)|\cdot |\mathcal{A}_1|\over |Z|}\ge (q^2-1)(q^2-q)=|GL_2(q)|$. Contradiction!
[b]q.e.d.[/b]

And now for the final step.
Using 1) of the Lemma we get $ s_i: =|N_G(\mathcal{A}_i): \mathcal{A}_i|\le 2$ for $ i=1,2$
and using 2),4),5) we get $ G\setminus Z$ the disjoint union of the
different conjugates $ g(\mathcal{A}_i\setminus Z)g^{-1}$ for $ g\in G,i=1,2$ and like in 4) we get
$ |G|-|Z|=\sum_{i=1,2}|G: N_G(\mathcal{A}_i)|(|\mathcal{A}_i|-|Z|)=|G|\sum_{i=1,2}{1\over s_i}\left(1-{1\over |\mathcal{A}_i: Z|}\right)$.
Also $ |G|\ge (q^2-1)(q-1)$ and 5) gives $ q\ge |GL_2(q): G|$ divisible by $ q$.
So $ |G|= (q^2-1)(q-1)$ and

$ 1-{1\over q^2-1 }={|G|-|Z|\over |G|}={1\over s_1}\left(1-{1\over q+1}\right)+{1\over s_2}\left(1-{1\over q-1}\right)$.
Then for $ s_1=1$ or $ s_2=1$ this gives
$ 1>{1\over 2}\left(1-{1\over q+1}\right)+\left(1-{1\over q-1}\right)$
and
$ 1<{1\over q+1}+{2\over q-1}\le {1\over 5}+{2\over 3}<1$ for $ q\ge 4$. Contradiction!
And for $ s_1=s_2=2$ we get the contradiction
$ 1-{1\over q^2-1 }={1\over 2}\left(1-{1\over q+1}\right)+{1\over 2}\left(1-{1\over q-1}\right)=1-{q\over q^2-1 }$.

This completes the proof!
I know this is a lot to take in and I might have made some mistakes.
So if there are any problems or comments feel free to ask.

@harazi
Dickson's classification can in fact be generalized to higher dimensions
with more than exponentially increasing casework and possibilities. (I think $ n=3,4$ are done  :maybe: )
But I don't think there is a general result for this problem in all dimensions,
unless maybe if it uses the classification of all finite simple groups.

I didn't read olorin's proof, but this looks like it could be slightly shorter (although I'm sure many of the same ideas appear in both). The drawback is that it only works for the initial problem, where the field is prime, so it proves less than olorin's argument.

Given a generator $ g$ of the cyclic group $ \mathbb F_{p^2}^*$, multiplication by $ g$ gives a cyclic subgroup $ H$ of $ GL_2(\mathbb F_p)$ of order $ p^2 \minus{} 1$. Also, consider the group $ K$ consisting of matrices of the form $ \begin{pmatrix}t & 0 \\
0 & 1\end{pmatrix}$ with $ t\in\mathbb F_p$. By our condition on $ G$, there are conjugates of both $ H$ and $ K$ contained in $ G$. We may as well assume that $ G$ contains both $ H$ and $ K$. Because the eigenvalues of the elements of $ H$ are conjugate over $ \mathbb F_p$ and so can't be different elements of $ \mathbb F_p$, no conjugate of $ H$ meets a conjugate of $ K$ non-trivially. This implies that any double coset $ HxK$ (for $ x\in G$) has cardinality $ |H|\cdot|K| \equal{} (p^2 \minus{} 1)(p \minus{} 1)$. Since $ G$ is partitioned into such double cosets, we already have a lot: the cardinality of $ G$ is divisible by $ (p^2 \minus{} 1)(p \minus{} 1)$, so there are two cases - either $ G$ is the whole of $ GL_2$, or $ |G| \equal{} (p^2 \minus{} 1)(p \minus{} 1) \equal{} \frac {|GL_2|}p$ (this is where it was important that the field had a prime number of elements rather than a prime power). We have to prove that the latter can't be true.

Suppose $ |G| \equal{} (p^2 \minus{} 1)(p \minus{} 1)$. It follows then, with the same notations as above, that $ G \equal{} HK$. Being a product of two cyclic groups, $ G$ must then be metabelian (this was the "kicker" in the proof; a beautiful result of Ito, I think, states that a group which is the product of two abelian subgroups is metabelian). This means that $ G'$, the commutator subgroup, is abelian. It's also a subgroup of $ SL_2$ of order prime to $ p$, and it's easy to see that every such abelian group must be cyclic (it's simultaneously diagonalizable over the algebraic closure of $ \mathbb F_p$, or even over $ \mathbb F_{p^2}$, etc.). The elements of $ G'$ have pairs of eigenvalues of the form $ \lambda,\lambda^{ \minus{} 1}\in\mathbb F_{p^2}$, and given an element $ g\in G'$, the only other element sharing its set of eigenvalues is $ g^{ \minus{} 1}$. This implies that either $ G$ centralizes $ G'$, or that it acts on it by conjugation through a quotient $ \mathbb Z/2\mathbb Z$ sending $ g\in G'$ to $ g^{ \minus{} 1}$. Either way, because $ p > 3$, $ G'$ is centralized by a non-trivial element of $ K$. It follows that $ G'$ consists of diagonal matrices. Unless $ G'$ is central (i.e. either trivial or $ {\pm 1}$), some non-scalar element $ a$ of $ H$ commutes with a non-trivial element of $ G'$, which means that $ a$ is diagonal - absurd by the definition of $ H$. We've now shown that $ G'$ is either trivial or consists of $ \pm 1$ (here $ 1$ is the identity of $ \mathbb F_p^2$). This is clearly impossible: in either of the two cases, all matrices of the form $ \begin{pmatrix}t^2 & 0 \\
0 & 1\end{pmatrix},\ t\in\mathbb F_p$ must be central, hence all elements of $ G$ are diagonal - absurd.

P.S.

We really do need $ p > 3$. For $ p \equal{} 3$ there is a group $ G$ of order $ (p^2 \minus{} 1)(p \minus{} 1) \equal{} 16$ which is easily seen to satisfy the hypothesis. Still using the notation above, we can take $ H$ to be the group generated by $ \begin{pmatrix}1 & 2 \\
1 & 1\end{pmatrix}$. It's cyclic of order $ 8$, and the $ 2$-element group $ K$ generated by $ \begin{pmatrix}2 & 0 \\
0 & 1\end{pmatrix}$ acts on $ H$ by $ x\mapsto x^3$. The commutator $ G'$ is the $ 4$-element subgroup of $ H$.

You'll be surprised what happens when you fill in all the details, I was. :blush:
Like $ |G: C_G(G')|\le 2$ is just what I did in 1).
In fact 5) is the only redundant part of my proof for $ q \equal{} p$ prime.

Also Ito's result is pretty heavy, and I didn't want to use it,
especially since I haven't read its proof myself yet.
(could you tell me where to find it?)
But without it I was stuck at $ |G| \equal{} (p^2 \minus{} 1)(p \minus{} 1)$, just like you.
And that's when I went for Dickson's proof, and decided to go
with a counting argument using all maximal abelian subgroups.

I saw a proof some time ago, but I can't remember where, unfortunately. It must have been in a downloaded paper. I do remember that it was very short and simple (roughly half a page, and no "heavy" stuff), but I forgot it and it would probably take a while to reproduce it. Better just keep looking on the web; I'm sure you'll find it somewhere.

Ok, found it again in [url=http://www.zentralblatt-math.org/zmath/en/search/?q=an:0217.07201]Huppert I[/url]. :lol:
And you are right. It's a really short but tricky calculation with commutators.
But I wouldn't excactly use it as "well-known".  :maybe:
Though great of you to notice it.

Also with "heavy" I actually meant its generalisation to products of nilpotent groups,
which I heard about.


trigonometry
geometry
circumcircle
projective geometry

I think you made a mistake... Unless I am making a dumb mistake, P does not exist.

Please read the problem again, I think there is no mistake.
$EF$ and $GD$ are lines, not segments, otherwise I would have written $[EF]$ and $[GD]$.
I can even prove that $P$ always exists!  :wink:

$\left\{\begin{array}{c}R\in EP\\\\ CR\perp EP\end{array}\right\|$ $\Longrightarrow \left\{\begin{array}{c}ER=CE\cdot\sin\frac{A}{2}=(p-c)\sin\frac{A}{2}\\\\ EP=EG\cdot\cot\frac{C}{2}=FG\cdot\sin\frac{A}{2}\cdot \frac{p-c}{r}=2(p-c)\sin\frac{A}{2}\end{array}\right\|$ $\Longrightarrow$ $EP=2\cdot ER$ $\Longrightarrow$ $CD=CE=CP$.

Thanks for your interest and your solution Virgil Nicula.

Initially, my problem asked to prove that $PC \parallel AB$, but unfortunately, I changed the problem into proving that $PC=CE$.
Both statements are equivalent, but in hindsight, I think that the original formulation would perhaps have been more interesting. I'm still learning much about creating problems.
When I look at the problem now, I fear that I have made it easier instead of harder, although it's perfectly possible to prove that $PC \parallel AB$ without using the equal segments.

Anyway, I found two completely different synthetic solutions to this problem that are not too hard to come up with. Some people told me they wanted to give it a try, so I won't give it away yet. I'll post them soon. :wink:

[quote="April"][quote="Jan"][color=darkred]The inscribed circle of $\Delta ABC$ touches the sides $BC,\,AC,\,AB$ in the points $D,\,E,\,F$ respectively. $GF$ is a diameter of the inscribed circle.
Point $P$ is the intersection point of $EF$ and $GD$. Prove that $\left|CE\right|=\left|CP\right|$[/color].[/quote]
[color=red]It's not so hard problem. And here is another proof[/color]  :lol:

[color=indigo][b]Proof.[/b] We have: $\angle EGP=\angle EFD=90^\circ-\frac{\angle ACB}{2}$
$\Longrightarrow\angle EPD=90^\circ-\angle EGP=\frac{\angle ECD}{2}$
So $C$ is circumcenter of triangle $DPE$ and so $CP=CE$[/color][/quote]

[color=red]Congrats  April![/color]

 Extra nice solution !

 [u]Babis[/u]

[quote="Jan"][color=darkred]The inscribed circle of $\Delta ABC$ touches the sides $BC,AC,AB$ in the points $D,E,F$ respectively.
$GF$ is a diameter of the inscribed circle. Point $P$ is the intersection point of $EF$ and $GD$. Prove that $\left|CE\right|=\left|CP\right|$.[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] The property $C\in PR$ is well-known, where $R\in DF\cap EG$. Indeed, apply the Pascal's theorem to the degenerated cyclic hexagon $DDFEEG$ :

$\left\{\begin{array}{c}C\in DD\cap EE\\\ R\in DF\cap EG\\\ P\in FE\cap GD\end{array}\right\|$ $\Longrightarrow$ $C\in PR$. The points $D$, $E$ belong to the circle with the diameter $[PR]$ because $PD\perp FD$, $RE\perp FE$

and the point $C$ is the its center because $CD=CE=p-c$. In conclusion, the points $D$, $E$, $P$, $R$ belong to the circle $\mathrm w(C,p-c)$, where $2p=a+b+c$.[/color]

Here is my proof for the equivalent problem: "Prove that $PC \parallel AB$"

Call $S$ the intersection point of $ED$ and $GF$.
Clearly, $PC$ is the polar of $S$ wrt the circle. This polar is perpendicular to the diameter through $S$, i.e. $GF$.
Therefore, $PC \parallel AB$.

It now follows that $\left|PC\right|=\left|PE\right|$ because of the similarity of $\Delta AFE$ and $\Delta CPE$.

In fact this problem is too easy. Thanks all for your interest and the nice solutions and comments Virgil Nicula, Babis Stergiou and April!

After having shown $P-C-R$ collinear (very nice, [b]Virgil Nicula[/b]!), obviously $G$ is the orthocenter of $\Delta FPR$, hence $PC\parallel AB$ and $\Delta CEP\sim\Delta FEA$;  as $\Delta AEF$ is isosceles, the conclusion follows.

Best regards,
sunken rock


[quote="mailo"]Βρηκα σε ενα βοηθημα μια ασκηση και ζηταω την βοηθεια σας για να την λυσω  :maybe:

Αν $\lim_{x \to 0}\frac{ae^{x^{3}}+b-ax^{3}}{sin^{6}x}=1$  να υπολογιστουν τα $a,b\in R$[/quote]

[b]HINT[/b]

Το όριο του αριθμητή είναι ίσο με 0, οπότε β=-α.Αντικαθιστούμε το β και το όριο γίνεται:

 $L = a \lim_{x \to 0}\frac{e^{x^{3}}-1-x^{3}}{sin^{6}x}$

Πάρε de L'Hospital και μετά σπάσε το κλάσμα σε :

 $\frac{1}{2}\cdot$  $\frac{1}{\cos x}$ $\cdot (\frac{x}{\sin x})^{2}$ $\cdot \frac{ e^{x^{3}}-1}{\sin^{3}{x}}$

Στο τελευταίο ξανακάνεις τον κανόνα και το όριο αυτό είναι $1$. Αν δε μου ξέφυγε κάτι στην πληκτρολόγηση , αυτή είναι η πορεία.Βγαίνει α=2 , β = -2 ή όχι ;

 Μπάμπης

Ευχαριστω πολυ για την λυση, πανω κατω ειχα φτασει μεχρι εκει αλλα δεν σκεφτικα να εφαρμοσω $de L'Hospital$   :roll:

Και ναι βγαινει α=2 και β=-2

Ομως πως προκυπτει οτι το οριο του αριθμητη ειναι 0?
Δηλαδη πρεπει να πουμε οτι για να μας δινει το οριο αυτο πραγματικο αριθμο πρεπει να εχουμε στον αριθμητη 0 ωστε να εχουμε απροσδιοριστη μορφη $\frac{0}{0}$ ?

Skepsou to anapoda! Ean den itan to orio tou arithmiti iso me miden (alla diaforo) tote tha ipirxe to orio? kai an ipirxe tha itan pragmatikos ('h mono $+\infty$, $-\infty$ ??)

Alexandros

[quote="mailo"]Ναι, αν ηταν διαφορο του μηδενος τοτε θα ειχαμε $\pm\infty$ αρα πρεπει να ειναι $0$[/quote]

[b][color=red]Ή Διαφορετικά:[/color][/b]

 Θέτεις το κλάσμα $g(x)$ , λύνεις ως προς τον αριθμητή και παίρνεις όριο στο $0$. Αφού η $g$ έχει όριο 1 στο 0 , θα είναι τελικά και το όριο του αριθμητή μηδέν.
 ;Oμως και η απλή λύση - εξήγηση του Αλέξανδρου είναι πολύ καλή. Αυτή χρησιμποποιώ και εγώ συνήθως :) .

 [u]Μπάμπης[/u]


Prove or disprove that in any triangle we have: $\displaystyle 8(r_a+r_b+r_c)^{3} \geq 27(r_{a}+r_{b})(r_{a}+r_{c})(r_{b}+r_{c}).$

cheers! :D :D

I thought you would be able to apply AM-GM to the numbers $r_b+r_c$, $r_c+r_a$, $r_a+r_b$ on yourself, or is today a kind of April 1st in Romania? ;)

SCNR...

  Darij

I think it is a joke too..
anyway, may i know what is SCNR?


The Math Olympiad for Elementary Schools ([url]http://www.moems.org[/url]) is a contest for 5th and 6th graders.  The organizers publish a book (Math Olympiad Contest Problems for Elementary and Middle Schools) that contains 80 of their old contests.  I give the old contests to students in my daughter's school as practice for the real thing.  Before doing so, I tried all the problems in the book.  Of the 400 problems, I got 12 wrong.  I am listing 11 of them below as challenges for you young students.  (I omitted one problem because I didn't feel like drawing a 3-dimensional figure.)

My getting the problems wrong doesn't mean that they are difficult.  I may have been going too quickly.  In any case, if you can solve one or more of the problems, then you can truly say that you beat the grownup!  Remember to explain your work, and use the "hide" button so that other students can try too. Good luck!

[b]1.[/b]  A book has 500 pages numbered 1, 2, 3, and so on.  How many times does the digit 1 appear in the page numbers?

[b]2.[/b]  A boy has the following seven coins in his pocket: 2 pennies, 2 nickels, 2 dimes, and 1 quarter.  He takes out two coins, records the sum of their values, and then puts them back with the other coins.  He continues to take out two coins, record the sum of their values, and put them back.  How many different sums can he record at most?

[b]3.[/b]  Suppose five days before the day after tomorrow was Wednesday.  What day of the week was yesterday?

[b]4.[/b]  If a kindergarten teacher places her children 4 on each bench, there will be 3 children who will not have a place.  However, if 5 children are placed on each bench, there will be 2 empty places.  What is the smallest number of children the class could have?

[b]5.[/b]  The product of three natural numbers is 24.  How many different sets of 3 numbers have this property if the order of the 3 numbers in a set does not matter?

[b]6.[/b]  Let $N$ be a number that divides 171 with a remainder of 6.  List all the two-digit numbers that $N$ can be.

[b]7.[/b]  The sum of all digits in the numbers 34, 35, and 36 is 24 because $(3+4) + (3+5) + (3+6) = 24$.  Find the sum of all digits in the first twenty-five natural numbers: 1, 2, 3, 4, 5, ..., 23, 24, 25.

[b]8.[/b]  In the target of three concentric circles at right [diagram omitted here], ring A, ring B, and circle C have different point values.  The sum of the point values of A and B is 23, of B and C is 33, and of A and C is 30.  What is the sum of the point values of A, B, and C?

[b]9.[/b]  $(1, 1, 9)$ is a triple of natural numbers whose sum is 11.  We consider $(1, 1, 9)$, $(1, 9, 1)$, and $(9, 1, 1)$ to be the same triple because each triple has the same three numbers.  How many other triples of natural numbers have a sum of 11?

[b]10.[/b]  I have a drawer which contains 40 socks in the following numbers and colors: 12 tan, 9 brown, 11 gray, and 8 blue.  Suppose I am blindfolded.  What is the fewest number of socks I must pick from the drawer to be absolutely certain that I have two socks of the same color among those I have picked?

[b]11.[/b]  $ABC$ and $DEF$ are three-digit numbers, and neither of the lead digits $A$ and $D$ is zero.  Different letters represent different digits.  If $DEF$ is subtracted from $ABC$, what is the largest remainder that can occur?

1. A book has 500 pages numbered 1, 2, 3, and so on. How many times does the digit 1 appear in the page numbers?
[hide]1, 10 - 19, 21,31,41,51,61,71,81,91, 100-199,  (1, 10 - 19, 21,31,41,51,61,71,81,91) add 200, (1, 10 - 19, 21,31,41,51,61,71,81,91)add 300, (1, 10 - 19, 21,31,41,51,61,71,81,91)add 400. so 1, 10 - 19, 21, 31, 41, 51, 61, 71, 81, 91 have 19, that times 4 is 76. 76 + 100. so I think the answer is 176. But it's probably not right because I didn't bother to think through the whole thing.

[/hide]
2. A boy has the following seven coins in his pocket: 2 pennies, 2 nickels, 2 dimes, and 1 quarter. He takes out two coins, records the sum of their values, and then puts them back with the other coins. He continues to take out two coins, record the sum of their values, and put them back. How many different sums can he record at most?

[hide]um... this is an interesting question, I think it would be the combination of 4 coins chosen 2 at a time plus 3, so i'm guessing 9?[/hide]

3. Suppose five days before the day after tomorrow was Wednesday. What day of the week was yesterday?

[hide]Friday?[/hide]

4. If a kindergarten teacher places her children 4 on each bench, there will be 3 children who will not have a place. However, if 5 children are placed on each bench, there will be 2 empty places. What is the smallest number of children the class could have?

[hide]this is how I figured this one, if it's x/5 = Reminder 2, it ends in either 7 or 2, and only numbers ending with seven will have a reminder of 3 when divided by 4. so it's the LCM of 4 and 5 plus 7 ...... [b]27[/b]

[/hide]

I'll do the rest later

[quote="LuCy4EvA"]1. A book has 500 pages numbered 1, 2, 3, and so on. How many times does the digit 1 appear in the page numbers?
[hide]1, 10 - 19, 21,31,41,51,61,71,81,91, 100-199,  (1, 10 - 19, 21,31,41,51,61,71,81,91) add 200, (1, 10 - 19, 21,31,41,51,61,71,81,91)add 300, (1, 10 - 19, 21,31,41,51,61,71,81,91)add 400. so 1, 10 - 19, 21, 31, 41, 51, 61, 71, 81, 91 have 19, that times 4 is 76. 76 + 100. so I think the answer is 176. But it's probably not right because I didn't bother to think through the whole thing.
[/quote]

ooo, sorry, I forgot the ones in there's some 2 or even 3 ones in the one hundreds, okay, um... but I think it's 195

[quote="Ravi B"]Lucy, your answers to Questions 2 and 3 are fine.  On Question 4, your answer of 27 and my original answer of 7 both turn out to be wrong for the same reason -- we misread the problem. :(  Look at it again.

Question 1 has stumped you: 195 is incorrect.

Keep going...[/quote]

ooo, I get it now, forgot that 11 have two 1s.

and I get number 4 now.... wait, I'm gonna work it out now:
[hide]4. If a kindergarten teacher places her children 4 on each bench, there will be 3 children who will not have a place. However, if 5 children are placed on each bench, there will be 2 empty places. What is the smallest number of children the class could have?
so if 5 children on each bench have 2 EMPTY places, then it could end in 3 or 8, and if 4 on each bench will have a reminder of 3, it can end in 3, 7,  and 11, 15, and 19... and it goes on, and so 3 works for both, and so 23 doesn't work, neither does 33, and so 43 works. [/hide]

[quote="Ravi B"]The Math Olympiad for Elementary Schools ([url]http://www.moems.org[/url]) is a contest for 5th and 6th graders.  The organizers publish a book (Math Olympiad Contest Problems for Elementary and Middle Schools) that contains 80 of their old contests.  I give the old contests to students in my daughter's school as practice for the real thing.  Before doing so, I tried all the problems in the book.  Of the 400 problems, I got 12 wrong.  I am listing 11 of them below as challenges for you young students.  (I omitted one problem because I didn't feel like drawing a 3-dimensional figure.)
[/quote]

I was in it last year, our sponsor got the sheet without having to buy the book.  I can post good problems on here if that's not too easy. I've always thought that Math Olympaid for Elementary Schools was too easy for this forum because it's only for 5 and 6th grade.

Treething, nice work on Question 1.  An alternative method is to count the number of 1's in the units digits, the number of 1's in the tens digits, and the number of 1's in the hundreds digits, and finally add them up.

[quote="Ravi B"][b]4.[/b]  If a kindergarten teacher places her children 4 on each bench, there will be 3 children who will not have a place. However, if 5 children are placed on each bench, there will be 2 empty places. What is the smallest number of children the class could have?[/quote]LuCy, now that I look this problem over, I realize that its wording is not great.  (I copied it verbatim from the book.)  For example, is the number of benches meant to be the same in the two scenarios (4 per bench versus 5 per bench), or can it be different?  If the number of benches is meant to be the same, then the answer is 23.  (That's the answer the book gives, although their solution never says anything about the number of benches being equal -- they just ignore the case of 3 children.)  If the number of benches can differ, then I believe the answer is 3, as you say.

LuCy, go ahead and post some problems from MOEMS in a new topic.

k, I posted it in the classroom math section


I would like to apply for this program, but I am unsure of when (and how) to pay the 50$ for the application fee. Also, wouldI would like to know if I still have to pay in US funds even if I live in Canada. Thanks in advance.

I'm pretty sure that you have to pay in american currency, since if you pay in canadian, you pay less and that would probably not be acceptable.


geometry
trigonometry
AMC
AIME
AMC 10
vector
AMC 8

If there is a follow up book, then it would be a good idea to read the first book. This is because the follow up book will probably rely on concepts taught in the first book, and if you don't know these, it will be frustrating to have to learn them all over again.
In addition, a lot of the concepts taught in the first book (sigma, matrices) will appear on higher level contests, such as the AIME and USAMO. Unless the second book covers these topics, I suggest strongly against skipping the first book :)

(I might be wrong, but hopefully this helped (a little :D))

The best books to buy are made in Alphine, California.  You can probably guess who makes them. :D

[quote="akhil0422"]I have a 10th grade (Indian) textbook that has geometry, sets, algebra, trigonometry, etc. in it. However, I know half of the stuff in there (like similarity) and other stuff like disjunctions and conjunctions seem unimportant (I want to atleast qualify for AIME and try to get a perfect score on AMC10) for such competitions. Also, there are only answers, no solutions and the book is of poor quality. Some answers are wrong, some answers are forgotten, the questions sometimes are not understandable, etc. There are hard things that I have no idea of though like Sigma and Matrices. But once again, these seem unimportant. That's why I'm thinking to skip this book. The next book covers trigonometry, vectors, and Algebra. Do you all think I can skip this book, if it's such a good idea? I think I can brush up a little on the Internet if I don't get some of the concepts in the next book. Please help! I need to ATLEAST make a perfect score on AMC10 (this is really hard because I got 20/25 on AMC8 and now, since it's been 2 years, I'm probably worse)![/quote]

If the book is bad, then why would you get the next one? I would try to buy a new book from a totally different manufacturer.

Do you really need to get a 150 on the AMC 10?

The next book [i]is[/i] from a different manufacturer.


[solved]
it means Quod Erat Demonstrandum

I thought someone said that "quite elegantly done."

 :roll: I didn't say that... I read in this site so don't sue me. :D

So there are many meanings...We often use Valentin's one at the end of a problem, right?

[quote="sheepwarrior"][quote="Valentin Vornicu"]It's Latin for "proof is finished".[/quote]

i thought it meant "as was shown" or something along those lines...? or "as was proven"[/quote]

quod = what
est = is
demonstrandum = to demonstrate

So litterally it means that that was what we had to proof. (I knew the meaning I just didn't know the demonstrandum part :))

i've always wondered what that meant.  Thanks for telling me! :)

[quote="JBL"][quote="Peter VDD"]So litterally it means that that was what we had to proof.[/quote]

More elegantly: "Which was to be proven."[/quote]

That were the words I tried to find  :thumbup:

i remember the meaning as "which was to be demonstrated," and that sorta helps with remembering the acutal latin words...

[quote="ZetaX"][quote="Peter VDD"]
quod = what
est = is
demonstrandum = to demonstrate
[/quote]

it should be

quod = what
erat = was
demonstrandum = to demonstrate
 ;)[/quote]
that's the right one :lol:

[quote="Psycho"]and to be totally precise:

quod=(that) what
erat=was
demonstrandum=to [b]be[/b] demonstrated

demonstrandum is a gerundium or gerundivum form... i had my last latin exam today (ever :P ), and i can't even remember which of the two it is anymore :D[/quote]

Given the translation provided, demonstrandum appears to be the passive infinitive. But then, I'm not fluent in Latin ;) and have only studied it on my own. And shouldn't quod be quam? After all, quam is what. Oh, and I used to think that it was "quite elegantly done" :blush: . Apparently that's a common mistake.

no it shouldn't be 'quam'

'quod' is the sexless subject of 'quis' which means 'what'
for 'quam' you have two options: the female object of 'quis' or 'than'... however, none of the two work with 'erat'  :)

I see... I'll have to go back to my studying :) . But next year Latin will be offered, so I'm taking it. Besides, it doesn't matter as long as everything starts with q.

[quote="jb05"]But next year Latin will be offered, so I'm taking it. [/quote]
good luck :P i got it for six years, and at times, it can be fun (depends mostly on the teacher ;) )

[quote="jb05"]Besides, it doesn't matter as long as everything starts with q.[/quote]
there is a huge amount of common latin words which start with a "q"... they just all look alike, but have total different meaning  :wacko:

[quote="Psycho"]
there is a huge amount of common latin words which start with a "q"... they just all look alike, but have total different meaning  :wacko:[/quote]

Yes, the "q" words are probably the hardest part of Latin.  There are so many of them...I still don't know many of them, lol. :oops:

[quote="Psycho"]and to be totally precise:

quod=(that) what
erat=was
demonstrandum=to [b]be[/b] demonstrated

demonstrandum is a gerundium or gerundivum form... i had my last latin exam today (ever :P ), and i can't even remember which of the two it is anymore :D[/quote]
demonstrandum is gerundivum :roll:

It should be mentioned that the phrase comes from Euclid's Elements (or its translations), and became a piece of the common culture outside of any mathematical usage.
This goes back to Ithe old days when Euclid and Latin were part of what any learned European studied.

Quod
Erat
Demonstrandum

Quod from the latin relative pronoun Qui, quae, quod which means who, whom, which or whose in different cases

Erat is from the active imperfect translation of the verb "to be" sum, esse, fui, futurus, a, um so it is translated as "was" and this verb is modified by the pronoun which.

Demonstrandum, from the 4th principal part of the verb demonstro, demonstrare, demonstravi, demonstratus, a, um which means to demonstrate and is translated as the perfect passive participle which is translate as to be demonstrated

So after all the grammar we get

Quod Erat Demonstrandum
"[that] which was to be proved"


inequalities
search
Cauchy Inequality
inequalities proposed

if i'm not wrong, the condition is unnecessary... anyway, here's my solution

we'll prove that

$3(a^4+b^4+c^4)\geq 3abc(a+b+c) \geq 2abc(a+b+c)+\frac{3a^2b^2c^2(ab+bc+ca)}{a^2b^2+b^2c^2+c^2a^2}$.

the first inequality holds by muirhead and/or chebyschev, both of them deal with it.

for the second we'll have to prove that
$(a+b+c)(a^2b^2+b^2c^2+c^2a^2)\geq 3abc(ab+ac+bc)$, a little bit of multiplications yield this is equivalent to
$\sum_{sym}a^3b^2\geq \sum_{sym}a^2b^2c$, which follows by muirhead

that's my solution, hope there's no miscalculation

[quote="kunny"]Let $a,b,c$ be positive reals such that $a^2,b^2,c^2$ are the side lengths of a triangle.
Prove the following inequality.

\[ 3(a^4+b^4+c^4)\geq 2abc(a+b+c)+\frac{3a^2b^2c^2(ab+bc+ca)}{a^2b^2+b^2c^2+c^2a^2}.  \][/quote]

  Yeah i agree with Campos the conditions are not necessary in this problem...
And also this one will be stronger :
$(a^2+b^2+c^2)^2 \geq 2abc(a+b+c)+\frac{3a^2b^2c^2(ab+bc+ca)}{a^2b^2+b^2c^2+c^2a^2}$
My solution for both mine and kunny's is very simple using Cauchy inequality only.....

Kummer,Could you show your full solution?

If you show it, I will post my solution and show the back ground of this problem.

kunny

[quote="Kummer"]  Yeah i agree with Campos the conditions are not necessary in this problem...
And also this one will be stronger :
$(a^2+b^2+c^2)^2 \geq 2abc(a+b+c)+\frac{3a^2b^2c^2(ab+bc+ca)}{a^2b^2+b^2c^2+c^2a^2}$
My solution for both mine and kunny's is very simple using Cauchy inequality only.....[/quote]

  Ok here is my solution:
$2abc(a+b+c)= \sum a^2(2bc) \leq \sum a^2(b^2+c^2)=2(a^2b^2+b^2c^2+c^2a^2)$
But $2(a^2b^2+b^2c^2+c^2a^2) \leq \frac{2}{3}(a^2+b^2+c^2)^2$  (1)
Also $\frac{3a^2b^2c^2(ab+bc+ca)}{a^2b^2+b^2c^2+c^2a^2}= \frac{3(ab+bc+ca)}{\sum\frac{1}{a^2}}$   But $\frac{3(ab+bc+ca)}{\sum\frac{1}{a^2}} \leq \frac{3( \sum a^2)}{ \sum \frac{1}{a^2}} \leq \frac{3( \sum a^2)}{ \frac{9}{ \sum a^2}}= \frac{1}{3}(a^2+b^2+c^2)^2$ (2)
So finally we have:
(1) : $2abc( \sum a ) \leq \frac{2}{3}( \sum a^2)^2$
(2) : $\frac{3a^2b^2c^2( \sum ab ) }{ \sum a^2b^2} \leq \frac{1}{3}( \sum a^2 )^2$
Take sum of (1) and (2) we get:
$(a^2+b^2+c^2)^2 \geq 2abc(a+b+c)+\frac{3a^2b^2c^2(ab+bc+ca)}{a^2b^2+b^2c^2+c^2a^2}$ Proved
How was my solution Kunny? :)

[quote="campos"]nice proof, although $(a^2+b^2+c^2)^2\geq 3abc(a+b+c)$ :D[/quote]

  Ok Campos turn to ur problem...
Really i don't know what is Murhead?...So i will produce 1 proof by using Chebyshew inequality only...
If i do not mistake ur problem was:
$abc( \sum a ) \geq \frac{3a^2b^2c^2( \sum ab )}{ \sum a^2b^2 }$ which  is equivalent with:
$( \sum a^2b^2 ) \frac{a+b+c}{abc} \geq 3 \sum ab$
Or : $( \sum a^2b^2 )( \sum \frac{1}{ab} ) \geq 3 \sum ab$
Bcause in general a,b,c are equivalent so we can legally assume that $a \geq b \geq c$...which can lead to
$a^2b^2 \geq a^2c^2 \geq b^2c^2$ (1) $\frac{1}{ab} \leq \frac{1}{ac} \leq \frac{1}{bc}$ (2)
Applying Chebyshew inequality for two sequences (1) and (2)( remark that they have opposite direction ) we have:
$( \sum a^2b^2 )( \sum \frac{1}{ab} ) \geq 3\sum ab$ ( proved )
( Here i have to use the positive condition ).... :)

oh... nice! there are some little mistakes in your proof, cause you're missing the coefficient $3$ in the $LHS$ in
i) $( \sum a^2b^2 ) \frac{a+b+c}{abc} \geq \sum ab$
ii) $( \sum a^2b^2 )( \sum \frac{1}{ab} ) \geq \sum ab$, both times...

but the rest is nice!

concerning muirhead you may check [url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=muirhead+inequality&t=15896[/url], or search in the internet for some kyran kedlaya's or t. mildorf's notes on inequalities!

Thank u Campos ,i have edited my post and corrected the mistake in typing....And again thank u for Muirhead...


geometry
3D geometry
inequalities
inequalities unsolved

For every positive numbers $a,b,c,x,y,z$ prove the inequality

$\sqrt[3]{a(b+1)yz}+\sqrt[3]{b(c+1)zx}+\sqrt[3]{c(a+1)xy}\leq \sqrt[3]{(a+1)(b+1)(c+1)(x+1)(y+1)(z+1)}$

Have been post before by rogue. :)
See here:http://www.mathlinks.ro/Forum/viewtopic.php?t=89595


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ H$ be a Sylow $ p$-subgroup of $ G$ and $ K \equal{} N(H)$. Prove or disprove: $ K \equal{} N(K)$.

using the fact that $ K$ does not contain Sylow $ p$-subgroup of $ G$ other than $ H$. :)

Thanks. I haven't realized that $ H$ is the only one Sylow p-subgroup in $ K$.

Since all Sylow p-subgroups are conjugated, and $ H$ is normal in $ K$. Follows that $ H$ is the only Sylow p-subgroup.
We have $ K \subset N(K)$. Assuming that $ \exists x, x \in N(K), x \notin K$. Since $ xHx^{\minus{}1} \subset xKx^{\minus{}1} \equal{}  K$. And $ xHx^{\minus{}1}$ is a Sylow p-subgroup contained in $ K$. But $ H$ is the only Sylow p-subgroups. Hence $ xHx^{\minus{}1} \equal{} H \Leftrightarrow x \in N(H) \equal{} K$. Contradiction. Hence,  $ K \equal{} N(K)$.


Εστω f συνεχης πραγματικη συναρτηση τετοια ωστε

$ f(x)\equal{}f(x\plus{}1)\equal{}f(x\plus{}\sqrt{2})$

για καθε $ x$ πραγματικο.

Να αποδειχθει οτι η $ f$ ειναι σταθερη

Cute.

[hide]Το $ \mathbb{Z}[\sqrt{2}]$ είναι πυκνό στο $ \mathbb{R}$[/hide]


how come my Invisionfree forum doesn't support LaTeX? is there a way to turn it on?

Is this like a IB forum? Whats your website? Maybe I can help you out...

This has nothing to do with AoPS MathLinks. Please ask these questions at the support board for IB forums.


During football season, 25 teams are ranked by three reporters (Alice,Bob and Cecil). Each reporter assigned all 25 integers (1 through 25) when ranking the twenty-five teams. A team earns 25 points for each first-place ranking, 24 points for each second-place ranking, and so on, getting one point for a 25th place ranking. The Hedgehogs earned 27 total points from the three reporters. How many different ways could the three reporters have assigned their rankings for the Hedgehogs? One such way to be included is Alice - 14th place, Bob - 17th place and Cecil - 20th place.

This is standard Balls and Urns: the number of positive integer solutions to $ a \plus{} b \plus{} c \equal{} 27$ is $ \binom{27 \minus{} 1}{3 \minus{} 1} \equal{} \binom{26}2\equal{} \frac {26\cdot25}2 \equal{} 13\cdot25 \equal{} \boxed{425}$.

Edit: Haha, thanks. I might also note that all solutions work because we can't have $ a,b,c>25$, or else one must be nonpositive.

[quote="math154"]This is standard Balls and Urns: the number of positive integer solutions to $ a \plus{} b \plus{} c \equal{} 27$ is $ \binom{27 \minus{} 1}{3 \minus{} 1} \equal{} \binom{26}{2} \equal{} \frac {26\cdot25}2 \equal{} 13\cdot25 \equal{} \boxed{425}$.[/quote]

Basically fixed?

[quote="funsize"]i think you mean 325[/quote]
Yes it should be $325$ since $(13)(25)=\boxed{325}$.

I want to sue the person who wrote this question. I dont want to read it all. There was a question earlier thats scroll bar was the size of an atom. I gave up, and my opponent guessed b after 45 seconds and got it right. I was so triggered


I am new at Mathlinks and would like to join the contest.Can I do it with the next edition and when will it be announced?Also I wanna know more details about this WOOT program and the exact dates of the exams?

[quote="Makaveli"]Also I wanna know more details about this WOOT program and the exact dates of the exams?[/quote]You can get all the details by clicking the link in my signature and also [url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php]here[/url].

Shobber is right, the MathLinks contest, 6th edition, will begin sometime in fall 2005 (probably september something, but I am not sure yet :) ).

[quote="mathboy_cntHDVN"]i can not see  mathlinkscontest
can you give me the link ? :maybe:[/quote]
I think you are looking for this
[url]http://www.mathlinks.ro/Forum/contest.php[/url]
 :wink:

[quote="Peter VDD"]I think he means the nonexisting current versions.[/quote]
Then we are looking for it too :D .
Probably only Valentin knows when and where we can find it :lol: .


[quote="varun"]Prove that n^3 - n is divisible by 6
Also prove that n^5 - n is divisible by 30. :?:[/quote]

[hide] $n^3-n=(n-1)n(n+1)$ . Since this is a product of three consecutive number , which mean there would surely be a factor of $2$ and $3$ in it by Pigeon Hole .

similarly in $n^5-n=(n^2+1)(n-1)n(n+1)$ there already exist factor of $6$ from above result . We now need to proof it is divisible by $5$ . If the three consecutive number have $5$ as a factor , then it is true. If all the three consecutive number dont have $5$ as factor ,the only case is $n=5k-2$ or $5k-3$ for some integer $k$ . But pluging these into $n^2+1$ gives us $25k^2-20k+5$ and $25k^2-30k+10$ which are both mupltiple of $5$ . Hence we are done . [/hide]

The basic thing about pigeon hole is this --

If you have fewer pigeon holes than pigeons and you put every pigeon in a pigeon hole, then there must result at least one pigeon hole with more than one pigeon.

But if you think it out carefully , you'll realise that there is a lot you can prove with that simple statement above...

The Getting Started forum is not intended for proofs in general.  Please stick to the posting guidelines.  I'm moving this to Intermediate.

[quote="varun"]Prove that n^3 - n is divisible by 6
Also prove that n^5 - n is divisible by 30. :?:[/quote]

[hide="1"]
$n(n^2-1)=(n-1)(n)(n+1)$
Of course one of these must be even. If we can prove that one of these is $\equiv 0 \bmod {3}$ then we are done.

- If $n \equiv 0 \bmod {3}$ then we're done.
- If $n \equiv 1 \bmod {3}$ then $n-1 \equiv 0 \bmod {3}$ and we're done.
- If $n \equiv 2 \bmod {3}$ then $n+1 \equiv 0 \bmod {3}$ and we're done.
[/hide]

[hide="2"]$n(n^4-1)=(n)(n^2-1)(n^2+1)=(n-1)(n)(n+1)(n^2+1)$

We have to prove that either $n^2+1 \equiv 0 \mod {5}$, $n \equiv 0$, $n \equiv 1$, or $n \equiv 4$.

- If $n^2+1 \equiv 0$, then $n^2 \equiv 4$. Then, from the original equation, $n(n^4 - 1)$, $n^4 - 1 \equiv 0 \mod {5}$ so we are done.

- If $n \equiv 2$, then $n^2 \equiv 4$ and we are done.

- If $n \equiv 3$, then $n^2 \equiv 9 \equiv 4$ and we are done.

- If $n \equiv 1, 4, \text { or } 0$, then we are done.[/hide]

[quote="236factorial"]A usage of the Pigeon Hole:

Allison has sneezed exactly 1 million times in her life (a lot of sneezing). Because of her age however, there must have been at least one day where she sneezed 101 times. What is the oldest allison can be, in days?[/quote]

If there are $n+k$ pigeons in $n$ holes, then at least one hole must contain more than one pidgeon.

The holes are the days, the pidgeons are the sneezes. $n+k=10^6$. $k=100$, so $n=999900$?


function
algebra
polynomial
calculus
integration
rational function
Polynomials

For non-negative integers $n$ and $k$, let $P_{n, k}(x)$ denote the rational function \[\frac{(x^{n}-1)(x^{n}-x) \cdots (x^{n}-x^{k-1})}{(x^{k}-1)(x^{k}-x) \cdots (x^{k}-x^{k-1})}.\] Show that $P_{n, k}(x)$ is actually a polynomial for all $n, k \in \mathbb{N}$.

I use the fact that $\mathbb{Z}[x]$ is a UFD.
Lemma: if $HCF(a,b)=1$, $d(x)=HCF((x^{a}-1), (x^{b}-1))=x-1$.
Proof: Direct application of Bezout's Lemma. We know that $(x^{a}-1)|(x^{ka}-1)$ for integral $k$, so we pick an integer $s$ such that $as=bt+1$ for some integer $t$, we get that $d(x)|(x^{bt+1}-1)$, $d(x)|(x^{bt}-1)$. Setting $(x-1)e(x)=d(x)$, $e(x)|x^{bt}, e(x)|x^{bt}-1$, implying that $e(x)=1$, proving the lemma.

Cancelling the $x$s, it becomes \[P_{n,k}(x)=\frac{(x^{n}-1)(x^{n-1}-1)...(x^{n-k+1}-1)}{(x^{k}-1)(x^{k-1}-1)....(x-1)}\]

It follows from the fact that $\binom{n}{k}$ is an integer that $k!|n(n-1)...(n-k+1)$. Applying the lemma and considering unique factorisation into irreducibles over the field of polynomials, this directly implies that the above function is a polynomial.


topology
invariant
real analysis
real analysis unsolved

Show, that for any continuous map  $ f: S^{n} \to S^{n}$  degrees a zero, there  will be points  $ x,y \in S^{n},$  such that $ f(x) \equal{} x$ and $ f(y) \equal{} \minus{} y$

Degree is invariant under homotopy. Consider $ S^n \subset \mathbb{R}^{n\plus{}1}$. If there is no $ x$ such that $ f(x) \equal{} \minus{}x$, the maps $ g_t \equal{} \frac{tf(x) \plus{} (1\minus{}t)x}{| tf(x) \plus{} (1\minus{}t)f(x)|} : S^n \rightarrow S^n$ is well-defined and form a homotopy joining $ f(x)$ and $ x$ . But $ f$ has degree 0, contrary with $ x$ has degree 1 (Hence, the argument still holds if the degree of $ f$ is different from 1.

Similar argument holds for $ f(y) \equal{} y$.

.


The points of a circle are coloured by three colours. Prove that there exist infinitely many isosceles triangles with vertices on the circle and of the same colour.


what's the best specific area of physics that's best for research for high school students?

[quote]best for research for high school students[/quote]

Well, I didn't really follow that statement....Did you mean the best area to do some project or something like that in high school....? Or is it something else that you meant ??

thanks. i meant areas of physics that i can study, then do a research as a high school student. I'm planning to do a research in a physics related area for programs like siemens. But i want to konw beforehand which areas are more accessible to high school students.

well its up to you, the internet is a great source so look into areas that interest you.  You won't do well if you choose a subject area that you are not very interested in.  Use the research to spark further interest.  Good Luck


function
arithmetic sequence
algebra unsolved
algebra

Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that $|f(x)-f(y)| \leq |x-y|$, for all $x,y \in \mathbb{R}$.

Prove that if for any real $x$, the sequence $x,f(x),f(f(x)),\ldots$ is an arithmetic progression, then there is $a \in \mathbb{R}$ such that  $f(x)=x+a$, for all $x \in \mathbb R$.

Sometimes try to check visual appearance of your replies ;)

i really don't understand why it appears like that

Don't use <= or => in your code. Use $\leq$ or $\geq$ (\leq and \geq)


vector
calculus
integration
function
calculus computations

Determine the maximum value of this integral
\[\int_{C}\mathbf{F}\cdot d\mathbf{r}\]
where
\[\mathbf{F}(x,y,z)=xy^{2}\mathbf{i}+(3z-xy^{2})\mathbf{j}+(4y-x^{2}y)\mathbf{k}\]
and $C$ is simple closed curve in the plane $x+y+z=1$. oriented anticlockwise as seen fom points far up on the $z$-axis.

Also give the curve $C$ for which this maximum occurs.

Let $\vec n$ be the unit normal vector to the plane. Calculate $(\nabla \times \vec F)\cdot n$ and relate it to the integral in question.

I got $\rm{rot}\, \mathbf{F}=(1-x^{2},2xy,-4xy)$ since $\bar{n}=(1,1,1)$ their scalar product is
$(1-x^{2}-2xy)$ and usig Stokes thm we can relate this to the given integral
\[\int_{C=\partial D}F\cdot dr=\iint_{D}(1-x^{2}-2xy)dxdy \]
So I have to maximize the last integral but how ?

Well, I got $\nabla\times\vec F=\langle 1-x^{2},2xy,-y^{2}-2xy\rangle$, hence the integral is $\iint_{D}(1-x^{2}-y^{2})\,dx\,dy$, where $D$ is the [b]projection[/b] of the part of the plane bounded by $C$ onto the $xy$-plane. If we want to choose $D$ so that the integral is maximal possible, we should include all positive values of $(1-x^{2}-y^{2})$ and none of the negative. Thus, $D$ should be the unit disk.

You should recheck that calculation. I get that the curl of $\mathbf{F}$ (U.S. textbooks say "curl" instead of "rot" but it's the same thing) is $\nabla\times\mathbf{F}=\langle 1-x^{2},2xy,-y^{2}-2xy\rangle.$ With $d\mathbf{r}=\langle 1,1,1\rangle\,dx\,dy,$ we get that

$(\nabla\times\mathbf{F})\cdot d\mathbf{r}=(1-x^{2}-y^{2})\,dx\,dy$

Now: that function, $1-x^{2}-y^{2},$ is positive in some places and negative in some places - and note the the set on which it is positive is a bounded set. Every portion of a region on which the function is positive that you include increases the integral; every portion of a region on which the function is negative that you include decreases the integral. Can you see how to proceed from there?

---

I'm giving a final exam on Monday, in precisely this course. Should I include this question on my final?

... Nah, don't think so. I do want people to actually pass the exam.

Sorry for my mistake, thank you all for your help.

[b]Mr. Merryfield:[/b] So this is considered a bit hard?

Depends on whether or not you have seen a problem of this kind before. If you have not, it would be hard to come up with a solution at once.

[quote]I do want people to actually pass the exam.[/quote]
You are too kind to them. :diablo:


geometry
circumcircle
trigonometry
geometry proposed

Let ABC be a triangle, and M and N respectively be the points where incircle touches AB and AC respectively.
Let P be the point where the C - excircle touches BC, and Q the point where the B - excircle touches BC.
Prove MNPQ is cyclic iff <BAC=90.

If  $PMNQ$ is cyclic then obviously its center will be a point $K$- the midpoint of arc(BC) of the circumcircle if $\triangle{ABC}$.
Now just apply cosine theorem in $\triangle{PBK}$ and $\triangle{AMK}$, Ptolomeu’s theorem to express $AK$ in terms of sides and $BK$, and use $PK=MK$ to find the relation:
$(\cos{\frac{A}{2}})^2=\frac{1}{2}$ and we are done.
The converse works in a similar way.


Columbia
ARML
Duke
college
Princeton
geometry
number theory

The Columbia Math Circle is indeed a go this year!! Patrick and I will be organizing the monthly meetings which will be held at Hammond School in Columbia, SC. Some of the activities that we hope to have include interesting mathematical talks, guest speakers, problem solving sessions (AMC-10&12/AIME/USAMO), sponsoring teams for the ARML Power Contest and possibly the Duke and Princeton Math Contests, and possibly helping students find topics and mentors for research projects. The first meeting of the Math Circle will be Saturday, August 30th from 1pm to 5pm.

The talks and activities will be geared towards advanced high school students with a strong interest in mathematics. They may also be accessible to very strong middle school students. There will be a fee of \$30 per student for each session to help cover the cost of materials, speakers, activities, and our time organizing.

e-mails will go out to teachers and students whose addresses we have, and something may appear in the State Newspaper. But help spread the word!!! We hope to get a good level of participation and make this something exciting for South Carolina students.

Marcus Neal
Hammond School mathematics teacher

I'll bring along my sister and this other kid. Both are in 7th grade, and I think my sister can handle it if it's not too hard combinatorics (not certain completely about the other kid). We'll see how they do...

Just a reminder that the Columbia Area Math Circle will meet this Saturday, August 30th from 1pm-5pm at Hammond School. Unfortunately we scheduled it on Memorial Day weekend, but if you can come we hope you will.

Another reminder that the Columbia Area Math Circle will have its next meeting on Saturday, September 27th from 1-5pm at Hammond School. I will be giving the talk this week on a topic blending algebra, number theory, complex numbers, and a little bit of philosophy. I hope to see you there.

This is a reminder that the next meeting of the Columbia Area Math Circle will be this Saturday, Nov. 8th at Hammond School from 1pm to 5pm. Hope you all can make it!

Marcus


I have known this problem for a while but still do not know how to solve it. Here it is:

There are 2 non parellel lines and a point X between these two lines. Find the line through X s.t. the triangle formed by this line and the 2 original lines has

a)the minimum area
b)the minimum perimeter.

The first part is relatively easy but the second part is more difficult(in my opinion, that is).

(a) See [url]http://www.mathlinks.ro/viewtopic.php?t=175235[/url].
(b) See [url]http://www.mathlinks.ro/viewtopic.php?t=188859[/url].


inequalities
function
inequalities proposed

Let $ a,b,c,d > 0,abcd \equal{} 1$. Prove that:
$ \sum \frac {a}{(abc \plus{} ab \plus{} a \plus{} 1)^2} \ge \frac {1}{a \plus{} b \plus{} c \plus{} d}$
And extended result for n numbers is also true:
Let $ , a_1,a_2,...,a_n > 0;a_1a_2...a_n \equal{} 1$. Prove that:
$ \sum_{k \equal{} 1}^{n} \frac {a_k}{\left(\prod\limits_{i \equal{} 1}^{k}a_i \plus{} \prod\limits_{i \equal{} 2}^{k} a_i \plus{} ... \plus{} a_k \plus{} 1\right)^2} \ge \frac {1}{\sum\limits_{j \equal{} 1}^{n} a_j}$

[quote="pvthuan"]Nice stuff, Dung. I think you got inspired from an inequality that appeared last year in the entrance exam to Hanoi University of Science, High school for gifted learners.[/quote]
Yes, Mr pvthuan! :blush:  I think HUS's inequality very nice! The inequality:
Let $ a,b,c>0,abc\equal{}1$. Prove that:
$ \sum \frac{a}{(ab\plus{}a\plus{}1)^2} \ge \frac{1}{a\plus{}b\plus{}c}$. :lol:

By standard substitution, $ \sum_{cyc} \frac {a}{abc + ab + a + 1} = 1$. Also we have by applying weighted Jensen on the convex function $ f(x) = x^2$:
\begin{align*} \sum_{cyc} \frac {a}{(abc + ab + a + 1)^2} & = \sum_{cyc}af\left(\frac {1}{abc + ab + a + 1}\right) \\
& \ge (a + b + c + d)f\left(\frac {\sum_{cyc} \frac {a}{abc + ab + a + 1}}{a + b + c + d}\right) \\
& = \frac {a + b + c + d}{(a + b + c + d)^2} \\
& = \frac {1}{a + b + c + d} \end{align*}

Nice problem! :)
(Also, this generalization worked out better than the one in Pre-Olympiad :D)

General case can be proven in the same way, I think.


search
combinatorics theorems
combinatorics

I think there is a problem with this file. I download it but I couldn't see it. acrbot show errors continuously!

It works with Foxit Reader. I suggest updating to the latest version.

Um sorry I can't see why the devil can't just trap the angel from above and below using the strategy described in the article for trapping it from above and mirroring it for below. (using two moves per turn and increasing the angel's power two-fold) :wacko:


The problems from Argentina IMO TST 2009:

Day 1 - May 14, 2009
Problem 1: http://www.mathlinks.ro/viewtopic.php?t=276869
Problem 2: http://www.mathlinks.ro/viewtopic.php?t=276862
Problem 3: http://www.mathlinks.ro/viewtopic.php?t=276858

Day 2 - May 15, 2009
Problem 4: http://www.mathlinks.ro/viewtopic.php?t=276955
Problem 5: http://www.mathlinks.ro/viewtopic.php?t=277113
Problem 6: http://www.mathlinks.ro/viewtopic.php?t=277116

Day 2, Problem 2 is linked incorrectly in the resources section.

Okay, that was a copy&paste error of mine. I hope it is okay now.


inequalities
function
induction
algebra proposed
algebra

do you know Jensen's proof for Jensen's inequality for convex functions?

What do you mean? His original proof?
Are you sure that Jensen's inequality is really due to Jensen ;)  ?
In another hand, it is just an induction on the number of variables.

Pierre.


Two distinct integers are chosen from $ S \equal{}\{\minus{}1,0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,11\}$ The probability that their sum is 10 would be increased if an element $ n$ was removed from $ S$ . What is the value of $ n$ ?

I guessed 5 (and I got it right!), but I don't really understand the solution. Can someone clarify it?

It is 5 because u take the ends which is -1+11=10
then work your way in the last number without a pair is 5


induction
algebra
polynomial
Ring Theory
superior algebra
superior algebra unsolved

Assuming that the polynomials have coefficients in a commutative unitary ring $ R$ and $ n > 0$, expand the product to get
\[ \sum_{0\le k\le n,0\le l\le m, d \equal{} k \plus{} l}a_kb_l \equal{} 0\]
for $ 0 < d\le m \plus{} n$. For the induction step $ r \minus{} 1\to r$ (with $ 1\le r\le m$) multiply the sum  above evaluated for $ d \equal{} m \plus{} n \minus{} r$ with $ a_n^r$ and analyse what remains using the induction hypotheses for all $ r'\le r \minus{} 1$.

Thinking a bit more about the problem, here another solution where the arithmetic in the polynomial ring does all the work: The case $ r \equal{} 0$ is trivial by looking at the coefficient of highest degree of the product. For the induction step $ r \minus{} 1\to r$ we get that
\[ a_n^r\sum_{k \equal{} 0}^mb_kx^k \equal{} a_n^r\sum_{k \equal{} 0}^{m \minus{} r}b_kx^k\]
is a polynomial of degree at most $ m \minus{} r$. Multiplying with the other polynomial yields the claim by looking at the highest degree of the resulting product.


analytic geometry
graphing lines
slope
LaTeX
calculus
derivative
calculus computations

The height of a certain hill (in feet) is given by:

h(x;y) = 10(2XY -3X2-4Y2-18X+28Y+12)

*side note X2 and Y2, the 2 is a subscript*

were Y is the distance (in miles) north and x is the distance (in miles) east of the base camp (the origin)

a. Where is the top of the hill located?
b. How high is the hill?
c. How steep is the slope (in feet per mile) at a point 1 miles north and 1 mile east of Base camp? In what direction is the slope steepest at that point?

[quote="Ace123"]*side note X2 and Y2, the 2 is a subscript*[/quote]  A subscript would be like $ X_2$ (in which case your hill would be five-dimensional) -- you mean a superscript $ X^2$ (for the exponent).  See [[LaTeX]] and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 to learn how to format things nicely on the forum.

Anyhow, this is a pretty standard exercise -- what have you tried, and what parts are giving you trouble?

[quote="Ace123"]Just the last part. The first two parts are just partial derivatives. The point that I got was (-2, 3) those are in miles. I don't know how to find part c.[/quote]

since you had get the partial derivatives, then part C is that compute the partial derivatives of (1,1), assume (a,b) is the   partial derivatives of (1,1), then the steepest direction is (-a,-b)

[quote="JBL"][quote="Jolie"](a,b) is the partial derivatives of (1,1)[/quote]  For "the partial derivatives," one normally says "the gradient."[/quote]

yes!


I don't know if this is the nicest proof, but I believe this works okay:

Note that the inequality is equivalent to proving that $ a^{5}+b^{5}+c^{5}\ge abc(3)^{\frac{1}{3}}$.  Assume w.l.o.g. that $ 0<a\le b\le c$.  Because $ (a^{3}+b^{3}+c^{3})(a^{2}+b^{2}+c^{2})=a^{5}+b^{5}+c^{5}+\sum_{cyc}a^{3}(b^{2}+c^{2})$, we have that

$ a^{5}+b^{5}+c^{5}=a^{2}+b^{2}+c^{2}-\sum_{cyc}a^{3}(b^{2}+c^{2})$.

However, by the Rearrangement Inequality, we know that $ a^{5}+b^{5}+c^{5}\ge a^{3}b^{2}+b^{3}c^{2}+c^{3}a^{2}$ and $ a^{5}+b^{5}+c^{5}\ge a^{3}c^{2}+b^{3}a^{2}+c^{3}b^{2}$ meaning that $ \sum_{cyc}a^{3}(b^{2}+c^{2})\le 2(a^{5}+b^{5}+c^{5})$.  Then using the inequality above, we have that

$ a^{5}+b^{5}+c^{5}\ge a^{2}+b^{2}+c^{2}-2(a^{5}+b^{5}+c^{5})\Rightarrow a^{5}+b^{5}+c^{5}\ge \frac{a^{2}+b^{2}+c^{2}}{3}$.

Now by the AM-GM, we have that $ \frac{a^{2}+b^{2}+c^{2}}{3}\ge (a^{2}b^{2}c^{2})^{\frac{1}{3}}=abc\cdot (abc)^{\frac{-1}{3}}$.

But since $ 1=a^{3}+b^{3}+c^{3}\ge 3abc$ by either the power means inequality or use of the Rearrangement inequality again, we have that $ (abc)^{\frac{-1}{3}}\ge 3^{\frac{1}{3}}$.  Hence we have the desired result, with equality iff $ a=b=c=3^{\frac{-1}{3}}$.


Hey, you guys need to know about this:
It's a new show on CBS called "NUMB3RS" and its about this guy who is a mathematician (YAY!) and his brother is a private investigator(?)... anyways, the show is about how he uses his mad math skills and his brother uses his investigative skills and together, they solve crime problems! HOW COOL IS THAT? OK, well I'm not giving a very good synopsis of this but you all need to watch it... all you need to know about is apparently, in the fourth episode, the mathematician dude is like, FREAKING out about P vs. NP.... *laughs*... should be fun to watch.

Another interesting note: the guy they got to play the mathematician failed algebra. TWICE. ([size=59]so he must be a good actor, huh?[/size]).

It's on tonight right after Pats-Colts. Maybe not on the west coast though.

I might watch it if I'm too lazy to get up after the game...

The series premiere was last night, but i missed it due to a soccer game. It's supposed to be on Friday nights from now on, so I hope i can watch it then.

I took it that she was playing soccer.  I don't think there are too many soccer games on TV right now...but then again, I would be the last person to know about that  :D

[quote="Valentin Vornicu"][quote="MathFiend"] [...] But i missed it due to a soccer game. [/quote]American Girl watching soccer ... unbelivable :)[/quote]

correction...american girl playing soccer...

And yea, my friend said it's a realli good show. I will most DEFINITELY obtain a copy of the DVDs when they become available.

[quote="Valentin Vornicu"][quote="MathFiend"] [...] But i missed it due to a soccer game. [/quote]American Girl watching soccer ... unbelivable :)[/quote]

joml88 and miths are right. i was playing soccer, not watching it. i did hear all about the show tho from the MC coach at the middle school where i assistant coach. he wouldn't be quiet about it. :)

That's more unbelivable. Although soccer is the main sport in Romania, girls are hardly ever convinced to watch or to play this sport (ther are exceptions though) so you might understand my wonder :)

Remember, the US has a much more established and respected women's soccer tradition than many countries which have vastly more history and skill at men's soccer.  (I think especially of Brazil, but it is true of most European countries, too.)  In 4 Women's World Cups, we have never finished lower than 3rd, have won twice, and have had some of the best individual players in the world.  Only Norway and Sweden have even made it into the top 3 twice.  So, all in all, it's not so surprising to find American girls playing soccer, although unfortunately WUSA has shut down so they won't be able to do it professionally.

(I also should add that in the 2000 presidential election here, there was a group of voters who everyone kept referring to as "Soccer Moms."  So, it's pretty prevalent.)

Soccer is very big with children in the US. I would say it ranks right up there with Youth Basketball and Little League. The popularity goes right up until high school or college. In the fall, I always see little kids playing soccer on a soccer field whenever I pass one.

JBL beat me to it ..  ( boasting  about US   womens soccer etc.. .. Mia Hamm  has better records than any soccer player in the world (male or female etc.)

Last but not least - Don't underestimate Soccer Moms  ... They   are the one who  have  determined all of the  last few elections in USA.. :)

As for the  TV show, sorry  but   I don't think writer has  ever met any mathematician (or a FBI detective for that matter) in real  life .. Show is worse than  I could imagine...

The worst part about the show was how they said 'THE EQUATION' all the time.

Otherwise, it wasn't bad.

finally got to watch this (a la TIVO). very much like CSI with a mathematician. interesting but... the blood and nasty kind of was a little much. i dont do human bodies. however, the guy was COOL. i like him a lot. question, though: does anyone believe that it is possible, or even probable, that a serial rapist/killer would behave like a sprinkler system? That is, do you think you could trace a man to his place of residency by finding the point of origin from the scattered points of where his victims were dumped?.... and anyways, how do you do that? im not exactly experienced with that level of mathematics- should i go to the problem forums for this? and what level, exactly?

My science teacher showed us the episode of NUMB3RS with the relation to the sprinkler system thing.
He was telling us how to explain more about the graphs.

So tell me, how exactly does one use pure math to calculate the location of a murderer based on his victims?

I guess just average the locations.  Like find the center of all of them.  I guess.  I still want to see it, but I couldn't steal the TV from parental units, so I've missed it.  I'll try to watch it next week.

i watched, i think, all but the first episode and i think it's pretty interesting
but the math part, like on the 2nd episode with bank robbers, is that even possible?  does any one know?  i think he predicts where and when they strike


Given $ a,b,c \ge 0$.Prove that
$ \sqrt{\frac{a^2\plus{}bc}{b^2\minus{}bc\plus{}c^2}}\plus{} \sqrt{\frac{b^2\plus{}ca}{c^2\minus{}ca\plus{}a^2}}\plus{} \sqrt{\frac{c^2\plus{}ab}{a^2\minus{}ab\plus{}b^2}} \plus{} \frac{2(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2} \ge 4$
With this problem,we can prove by AM-GM.:)

[quote="secrets"]Given $ a,b,c \ge 0$.Prove that
$ \sqrt {\frac {a^2 \plus{} bc}{b^2 \minus{} bc \plus{} c^2}} \plus{} \sqrt {\frac {b^2 \plus{} ca}{c^2 \minus{} ca \plus{} a^2}} \plus{} \sqrt {\frac {c^2 \plus{} ab}{a^2 \minus{} ab \plus{} b^2}} \plus{} \frac {2(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \ge 4$
With this problem,we can prove by AM-GM.:)[/quote]

Nice but not hard (I think)

[b]Solution:
[/b]
Given $ a\ge b \ge c$

$ bc \ge 0 \ ;\ c^2 \minus{} bc \le 0 \ ; \ ca \ge 0 \ ; \ c^2 \minus{} ca \le 0 \ ; \ c^2 \ge 0 \ ;$

$ \frac {2(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \minus{} \frac {2ab}{a^2 \plus{} b^2} \equal{} \frac {2c[(a \plus{} b)(a^2 \plus{} b^2) \minus{} abc]}{(a^2 \plus{} b^2 \plus{} c^2)(a^2 \plus{} b^2)} \ge 0$

$ \equal{} > LHS \ge \frac {a}{b} \plus{} \frac {b}{a} \plus{} \sqrt {\frac {ab}{a^2 \minus{} ab \plus{} b^2}} \plus{} \frac {2ab}{a^2 \plus{} b^2}$

$ \frac {a}{b} \plus{} \frac {b}{a} \plus{} \sqrt {\frac {ab}{a^2 \minus{} ab \plus{} b^2}} \plus{} \frac {2ab}{a^2 \plus{} b^2} \minus{} 4$

$ \equal{} (a \minus{} b)^2(\frac {1}{ab} \minus{} \frac {1}{a^2 \plus{} b^2} \minus{} \frac {1}{\sqrt {ab(a^2 \minus{} ab \plus{} b^2)} \plus{} (a^2 \minus{} ab \plus{} b^2)})$

$ \ge (a \minus{} b)^2(\frac {1}{ab} \minus{} \frac {1}{2ab} \minus{} \frac {1}{2ab})$

$ \equal{} 0$

Our proof completed. Equal $ < \equal{} > a \equal{} b ; c \equal{} 0$ ......


I played a major role in getting my school to administer the AMC 10 and 12 tests, and I scored fairly well.
Next year I want to make it to the next level, but my school is not interested in helping me prepare. :(
If anyone would be so kind as to help me out with any suggestions or things that worked for you I would appreciate it.

I've moved this topic to the AMC forum where it's likely to get more response.

buy both art of problem solving books and read them.

take lots of practice tests


linear algebra
matrix
abstract algebra
algebra
polynomial
linear algebra solved

fairly easy!
assume that $A,B$ are complex matrix and $AB-BA$ is linear combination of $A$ and $B$ prove that they have common eigenvector.

Suppose first that we have $AB-BA=tA$ for some $t\in\mathbb C$.

If $t=0$, it's well-known that $A,B$ have a common eigenvector, so suppose now that $t\ne 0$.

Let $x\ne 0$ be an eigenvector of $B$, s.t. $Bx=\lambda x$. Then $\lambda Ax-B(Ax)=tAx$, so either $Ax$ is an eigenvector of $B$, corresponding to the eigenvalue $\lambda-t$, or $Ax=0$. If $Ax\ne 0$, we repeat this with $A^2x$, and so on. Since $\lambda-nt,\ n\in\mathbb N$ can't all be eigenvalues of $B$, we will eventually reach a point where $A^kx\ne 0,A^{k+1}x=0$, and $A^kx$ is an eigenvector of $B$ with eigenvalue $\lambda-kt$. Then $A^kx$ is a common eigenvector of $A,B$.

Now suppose that $AB-BA=aA+bB$, with $b\ne 0$. This means that $A\left(B+\frac ab A\right)-\left(B+\frac ab A\right)A=b\left(B+\frac ab A\right)$, which means, from the paragraph above (with $B+\frac ab A$ instead of $A$), that $A$ and $B+\frac ab A$ have a common eigenvector, and this will also be an eigenvector of $B$, as can easily be seen.

$A$ and $B$ generate a two dimensional Lie algebra which is either isomorphic to the abelian Lie algebra or to the Lie algebra given by $[x,y]=x$. Such algebras are easily seen to be solvable, so by Lie's Theorem, they have a comon eigenvector.

Nice proof, huh?

In the argument given above by grobber can we actually have $ Ax \neq 0$? For, if $ Ax \neq 0$ then

$ \lambda Ax \minus{} B(Ax) \equal{} tAx < \equal{} > (B \plus{} tI)(Ax) \equal{} \lambda (Ax)$ (1)

which means $ \lambda$ is also an eigenvalue of $ B \plus{} tI$. But the eigenvalues of this matrix are obtained from the eigenvalues of $ B$ by adding $ t$. Indeed, the characteristic polynomial of $ B$ is of the form

$ \phi_B(\lambda) \equal{} \det(I \lambda \minus{} B) \equal{} \prod_{i \equal{} 1}^s (\lambda \minus{} \lambda_i)^{\alpha_i}$.

Then,

$ \phi_{B \plus{} tI}(\lambda) \equal{} \det(I \lambda \minus{} B \minus{} tI) \equal{} \det[I(\lambda \minus{} t) \minus{} B] \equal{} \phi_B(\lambda \minus{} t) \equal{} \prod_{i \equal{} 1}^s [\lambda \minus{} (\lambda_i \plus{} t)]^{\alpha_i}$.

So, although equality (1) can be observed for some eigenvalue, it can't be true for all eigenvalues (and we can write such equality for every disctint eigenvalue of $ B$), so this demands $ Ax \equal{} 0$, which means $ x$ is also an eigenvector of $ A$.

Am I missing something?


geometry
3D geometry
combinatorics unsolved
combinatorics

A large cube consisting of $n^3$ small cubes sits on the table. We wish to remove small cubes one by one until no small cubes are left, but we can only remove cubes whose upper face is visible. Find the smallest $n$ such that we can do this in more than $10^{18}$ ways.

Label the $n^2$ stacks you can see from above by $a_1,a_2,...,a_{n^2}$. The number you are looking for is then the number of words in these letters having every letter $n$ times.
Thus the desired number is $\frac{(n^3)!}{n!^{n^2}}$.
Setting in $n=3$ gives a number that is big enough.

This also solves the (at the first sight) harder problem for that we are allowed to take any cube that is visible: the number of ways to take is higher than here, but still smaller than the desired number when $n=2$.


this is a fairly long question that i was trying to solve while training for a national competition in my country, i think i am missing something. is there a elegant solution to this? i cannot find it.

Here are two examples of a number triangle. There are four consecutive positive integers, 10,11,12 and 13, on the bottom line, in some order. Starting from the left, each number in the bottom line is added to the number next to it and the answer is written above the space between them. This is repeated for each line, until there is a single number ontop.

Number triangles with different numbers in the bottom line, or with the same numbers but in a different order, are regarded as different. The two examples shown are, therefore, different.

...........94
.......45...49
....21...24....25
10...11...13...12

and

...........94
.......49....45
...25...24....21
12...13...11...10

a) By changing the arrangment of the numbers 10, 11, 12 and 13 in the bottom line, what top numbers are possible.

(now this is were it gets harder)

b) Another number triangle has six consecutive integers in some order in the bottom line and the top number 2006.

i) Give two examples of this: one with 60, 61, 62, 63, 64, and 65, in some order, in the bottom line, and the other with 61, 62, 63, 64, 65 and 66 in some order, in the bottom line.

ii) Show that these are the only possible sets of six consecutive numbers for the bottom line of a number triangle with top number 2006. [ NO IDEA how to even start this ]

iii) How many different number triangles are there with six consecutive numbers in the bottom line, in some order, and with top number 2006? (Recall that two number triangles are different if their bottom line numbers are different or their bottom line orders are different)

if possible a full solution to each problem, or whatever problems you can do would be greatly appreciated.

this is in the australian maths challenge, but im not sure if its right to post solutions but i dont mind.

yep, as i said in the other thread, i am not trying to cheat as i think that is a very lowly thing to do and that it is immoral.

i should have put it in my post that it was from a contest that was held this year, anyway, my school has already finished with it, solutions have been submitted although i am the only person from my year group that wanted to do it... and ironically my entry was too late.

im still having trouble with the problem, i cant really do any of the questions except the first

As sapphyre pointed out, if the numbers on the bottom of the triangle are $x_{1},x_{2},x_{3}\cdots x_{6}$, then the number on the top of the triangle can be expressed as $x_{1}+5x_{2}+10x_{3}+10x_{4}+5x_{5}+x_{6}=2006$.  Our variables are $60,61,62\cdots 65$ in some order.

To make the problem simpler, we can let $x_{i}=y_{i}+60$ for each $i$.  Then the equation is $y_{1}+60+5(y_{2}+60)+10(y_{3}+60)\cdots =2006$.  The new variables are $0,1,2\cdots 5$.  You can simplify the equation further, and then you'll be able to efficiently guess and check values for the variables.

ohhh, yea i understand the hint now, but i still dont know how to solve b) ii and iii, can someone help me out?

Let the six consecutive integers be $n,n+1,n+2,\cdots,n+5$.  We want to show that $n=60,61$.

Consider $n=59$.  Try to maximize $x_{1}+5x_{2}+10x_{3}+10x_{4}+5x_{5}+x_{6}$.  Where should we put each of the six consecutive integers to make the expression as large as possible?  Show that the expression can't be as great as $2006$.

thats brilliant... i wouldnt have seen that.

iv been trying to do iii but no luck, im trying to apply what was posted in this thread earlier but im having alot of trouble, the question is difficult :( any more hints/solutions?

By part ii, the numbers are $60,61,62\cdots 65$ or $61,62,63\cdots 66$.  Start with $60,61,62\cdots 65$.  Use the same method you used in part i to find a possible arrangement - only this time you want to find all possible arrangements.

Since your equation is $x_{1}+5x_{2}+10x_{3}+10x_{4}+5x_{5}+x_{6}=2006$, once you find one solution, you can find many more just by switching $x_{1}$ and $x_{6}$, $x_{2}$ and $x_{3}$.  To be exact, you can find $7$ more solutions just by switching pairs of variables from one solution.  (Do you see why?)

This problem is much more difficult that part i.  In part i, you only had to find one solution.  In this problem, you have to find ALL solutions and then show that there can't be any more.  (Even if you weren't required to prove your answer, you should still be able to show that no more solutions exist.)  Good luck!  :)


function
number theory
least common multiple
trigonometry
floor function
algebra unsolved
algebra

If $m=\text{lcm }(a,b,c)$, then of course $f+g+h$ will have period $m$, since $f(x+m)=f(x),\ g(x+m)=g(x),\ h(x+m)=h(x)$. The lcm might not be the smallest period, though: consider the functions $4\cos^{3}x$ and $-3\cos x$. They both have period $2\pi$, but their sum happens to equal $\cos 3x$, which has period $\frac{2\pi}{3}$.

[quote="pardesi"]is this necessarily true
we have say three functions
$f(x)$,$g(x)$,$l(x)$ having periods $a,b,c$(either all rational or all purely irrational) then the function
$F=f+g+l$ must have a period which say can be found easily that is may be in some sense the "LCM" of $a,b,c$

for ex consider the function
$f(x)= \sin x+\tan^{3}x+\cot x$

it has period $2 \pi$ but is this fact true in all cases[/quote]

It is easy to see that if $a=\frac{p}{q},b=\frac{r}{s},c=\frac{t}{u},$ are  rational, then $F$ is periodic (take for example $T=p*r*t$).
It is rather easy too to see this not necessarily the case when $a,b,c$ are all irrational :

For exemple, take :
$f(x)=gx)=x\frac{\sqrt{2}}{2}-\frac{1}{2}\left\lfloor x\sqrt{2}\right\rfloor$ periodic with period $\frac{1}{\sqrt{2}}$
$l(x)=x\sqrt{3}-\lfloor x\sqrt{3}\rfloor$ periodic with period $\frac{1}{\sqrt{3}}$
Then $F(x)=f(x)+g(x)+l(x)=x\sqrt{2}-\lfloor x\sqrt{2}\rfloor+x\sqrt{3}-\lfloor x\sqrt{3}\rfloor$ is not periodic.

yes i was almost sure but wanted to double check with some examples i didn't thnk of
thanks :)


Consider a regular monopoly dice. Assume this die is "fair". On average, how many rolls of the dice will it take in order for all six faces to appear.

(It's for my Stats class ,so i'm not sure, it might a a trick problem like since its "fair" it would get all 6 faces after 6 rolls.)

still try to solve it!

[hide="Hint"] Recursion. Find the expected value to get one face, two faces, etc. :D [/hide]

Two things:

1) If a die has an expected value of 6 to get all 6 faces at least once it is certainly [i]not[/i] fair! For example, the first four rolls are 1,2,3,4. Does it mean that there is a 1/2 chance of getting a 5 and a 1/2 chance of getting a 6, and 0 chance of getting 1,2,3,or 4 on the next roll? Doesn't sound fair to me!

2) "Dice" is actually the plural form of "die". Since we are talking about one die in this problem, you teacher correctly uses the word "die".

[hide="hint"]You can add expected values. Recursion, as lotrgreengrapes7926 said.
[hide="more"]The expected number of rolls to get a unique value on the first roll is 1, because all of the rolls will produce a unique value. What about the 2nd unique value? Remember the probability that this happens on the [i]next[/i] roll is 5/6, and the number of rolls that it takes to achieve this "success" is a geometric distribution. Recur!
[/hide][/hide]

[hide="answer"]Let us examine the expected number of rolls it takes to get the n-th unique roll (by unique roll we mean a roll that has not yet appeared in our sequence of rolls). The first unique roll has an expected value of 1 (indeed, it is always 1).

The second unique roll has a probability of 5/6 to occur. Let us keep rolling until we get the second unique roll. This is a geometric distribution, and the expected value of geometric distributions is 1/p. So it takes an average of 6/5 rolls after the first unique number to get the second unique number.

In general, the n-th unique roll has a probability of (6-n+1)/6 to occur, and has an expected value of 6/(6-n+1) rolls after the (n-1)th roll to occur. Once you have gotten 5 of the numbers, it will take an average of six more rolls to get the sixth number! Adding all of these expected values we get:

$\frac{6}{6}+\frac{6}{5}+\frac{6}{4}+\frac{6}{3}+\frac{6}{2}+\frac{6}{1}$
which equals a surprisingly high [b]14.7 rolls[/b] in order to get all of the numbers 1-6.

(Next time you see a cereal box with a collectable toy inside and the command, "Collect all 6!", you'll know how many boxes you'll actually need to get (on average) in order to do this)
[/hide]


Exam Proctor: Okay, you may all now open your test booklets and begin.

Three minutes later: Fire alarm sounds.

O....M....F....G.

R...O....F....L...

what's the ending of the story?

OR..

Sabotage from another school (:O)

dun dun dun...

hahah thats hilarious.
so what happened? did u continue with it afterward? wouldn't that be cheating :P

seriously, i would just take my calc, run outside, and sit on a bench in the cold working on my amc

wouldn't that be hilarious, especially if in the ghetto.

picture this:

A bunch of math nerds sitting outside in the cold in the ghetto doing a math test.  :rotfl:  :rotfl:

We think we might have a chance of letting it happen since it was like the first 2 minutes.  I mean if it were in the middle, I could just be working on a hard problem and be thinking it over outside during the thing.  Instead we were all on like number 5.


logarithms
ARML
number theory
prime numbers

I've always been fascinated by prime numbers.  I figured out long ago that all prime numbers (other than 2) must end in a 1,3,7 or 9, and that when testing to see if a number is prime, you must only test for divisibility by the other primes up to the square root of the number.  I recently read in Fermat's Enigma that all prime numbers (other than 2) take the form 4n-1 or 4n+1 for an integer n.  This is rather fascinating, so I was wondering if there are any other rules/restrictions, specifically restrictions on n for the equations I just gave.
For example, n=1 works for both 4n-1 and 4n+1, n=2 works for 4n-1 but not 4n+1, n=3 works for both equations, n=4 only works for 4n+1,  and I haven't bothered to take it any farther than that yet.  Is there a pattern? n which do not work (I assume there are)?  Anyway...thoughts?

I figured out long ago that all prime numbers (other than 2) must end in a 1,3,7 or 9, [quote][/quote]

Don't forget 5.  :D

And as for basic prime information, perhaps you should read up about it in Hardy and Wright's [u]Introduction to Number Theory[/u]. A pdf version can be found on Mindspa's site: http://hem.bredband.net/lukhor/index.html .

Don't forget 5 and 0! :)
I don't know how to prove Fermat's Enigma, but 4n and 4n:pm:2 is both composite. Think about it...(correct me if I'm leading him the wrong way :) )

P.S. Fiertycoon's post came in after I started writing, so there are parts that double...

[quote]Don't forget 5 and 0!  [/quote]
Oh boy, if that's serious, I'm very sad.  If it's a joke, I'm just an idiot for not catching it.  Either way, one of us is stupid  :roll:

Anyway, thanks for the links... I'll go read those now.

Well one of us lacks a sense of humor, that's all. :) (0 isn't a prime since the definition of prime states "An integer larger than or equal to 1" or something of the equivilant at the beginning)

Well, also that 0 has quite a few factors, not just 2 :), infinitely many factors actually..
And not only is a prime 4n+1 or 4n-1, but apart from the trivial 2 and 3, every prime is 6n+1 or 6n-1.. thats real easy to prove (just think 6a, 6a+1, 6a+2, 6a+3, 6a+4, 6a+5, and quite a few of those factorize).

Yeah....solving the congruences for 1 mod 2 and 1 or 2 mod 3 give 1 or 5 mod 6. If you throw in a 1/2/3/4 mod 5 in there, you'll get all the possible thingies mod 30, etc.

[color=cyan]As for patterns, there aren't any.  I mean, there is simply no nice method for finding "the next" prime number or anything.  Only general, large-scale trends that don't tell you very much about specific primes or anything.  Sorry.

Unsolved questions: Are there infinitely many pairs of primes which differ by 2?  (Twin Prime Conjecture says yes).
Can every composite even positive integer be written as the sum of two (not necessarily different) prime numbers?  (Goldbach Conjecture says yes)

Things which are not unsolved:
There are roughly ln(n) primes less than n, where ln(x) is the logarithm of x base e.  There are more exact expressions for this.  This one is nice and simple, though.
There is always a prime between n and 2n, inclusive, for all positive integers n.  I think this may be an approachable problem, but I'm not positive.[/color]

[color=cyan]Oh, okay, so nix my last sentance then.  I just remember that I learned it from an ARML power.[/color]

I think you meant that the number of primes less than n is approximately n / ln(n).

[quote="ComplexZeta"]Bertrand's Postulate is actually quite tough to prove. The simplest proof is probably Erdos's proof, which involves estimating 2n choose n and proving a bunch of rather esoteric theorems about the upper bound of the kth prime.[/quote]

An even more nice proof is due to Ramanujan. Everytime I try to look at it I start crying because of it's beauty.

can you post that proof or give a link to it?

A good popular site http://www.utm.edu/research/primes/


997 is the largest 3-digit prime number.

Find the arithmetic mean of the two largest 3-digit prime divisors of:
$\prod^{\lfloor \frac{997}{2}\rfloor}_{n=1}(2n)-\prod^{\lfloor \frac{997}{2}\rfloor}_{n=1}(2n-1)$

[hide=" :) "]
[hide="Solution"]
This is really just $(2)(4)(6)(8)\dots (994)(996)-(1)(3)(5)(7)\dots (993)(995)$.

Note that all primes under $997$ are a factor of the second product, so a prime under $997$ is a factor of the difference iff. it is a factor of the first product.

The first product contains $2(1)$ through $2(498)$, so the only prime factors of the first product are primes less than $498$. So, primes between $498$ and $997$ exclusive are not factors of the difference.

What about $997$ itself? It doesn't divide either of the products. However,
$(2)(4)(6)(8)\dots (994)(996)-(1)(3)(5)(7)\dots (993)(995)$
$=(2)(4)(6)(8)\dots (994)(996)-(997-996)(997-994)\dots (997-4)(997-2)$
$\underline{=}(2)(4)(6)(8)\dots (994)(996)-(-996)(-994)(-992)\dots (-4)(-2)$
$\underline{=}(2)(4)(6)(8)\dots (994)(996)-(-1)^{448}(996)(994)(992)\dots (4)(2)$
$\underline{=}(2)(4)(6)(8)\dots (994)(996)-(996)(994)(992)\dots (4)(2)$
$\underline{=}0 (\text{mod }997)$

So $997$ is a divisor of the difference, and is the largest prime divisor. The second largest prime divisor is the largest prime under $498$. Quick checking reveals that $497$, $495$, and $493$ are not prime and that $491$ is prime.

So, the two largest prime divisors are $491$ and $997$, and their arithmetic mean is $\boxed{744}$
[/hide][/hide]

Well, I've been working on this for a while and gotten it down to:

$(2n)!!-\frac{(2n!)}{(2n)!!}$

where in this case n=498.

Edit:  Actually from there it can be simplified to:

$(x)!!-(x-1)!!$, where x is 2n, so x is 996.

Click the smiley face in the original problem post to reveal the solution  :)

BTW: The first instance of "all primes under 997" should say "all 3-digit primes under 997" because 2 is even

Whoa!  Where did that smiley come from?  :huh:

Anyway, that's a good solution.  I was getting too complicated, trying to find what the result would be and then factoring from there...


modular arithmetic
number theory unsolved
number theory

I found this property.
$\sum_{p/2 < u < p}u^{-1}\equiv \frac{2^{p}-2}{p}\pmod p$
(p : odd prime)
I can't prove it.
Please, help me.

Sorry, I don't good at English.

I don't know if this helps... \[\sum_{\frac{p}{2}<u<p}u^{-1}\equiv \sum_{0 < u < \frac{p}{2}}[-u]^{-1}\equiv (-1)^{\frac{p-1}{2}}\sum_{0 < u < \frac{p}{2}}u^{-1}\] So then \[\left(\sum_{\frac{p}{2}<u<p}u^{-1}\right)^{2}\equiv (-1)^{\frac{p-1}{2}}(p-1)! \equiv (-1)^{\frac{p+1}{2}}\] But then, now that I look at it more closely, did you check $\boxed{p=7}$?   :)

The fact is that the left side of your equation is a square root of $1$ if $p \equiv 3\mod{4}$, and the only such values for it are thus $\pm 1$.  But the right side is not necessarily $\pm 1\mod{p}$.

I don't understand.. why $(LHS)^{2}\equiv (-1)^{\frac{p-1}{2}}(p-1)!$ ?
and, p=7 : LHS=11, RHS=18 $11 \not\equiv \pm 1\pmod 7$ !

Oh I see...my bad.  For some reason, I read the sum as a product...

I don't see how you got the last couple steps...

because

${p-1 \choose i}\equiv (-1)^{i}\pmod p$

( ${p-1\choose i}+{p-1\choose i+1}={p\choose i+1}\equiv 0 \pmod p$)

see also http://www.mathlinks.ro/Forum/viewtopic.php?t=38958 .

  darij

what does $mod p$ mean(for rational number)?

They are not rational in the way you are probably thinking about it. When he says $u^{-1}$ mod p he means the inverse of u mod p. For instance, the inverse of 3 mod 5 is simply 2 because 3 * 2 = 1 mod 5. Given this, then $\frac{a}{b}$ mod b equals $a \cdot b^{-1}$ mod p.


trigonometry
geometry
geometric transformation
reflection
circumcircle

In $\triangle{ABC}$, $AB=AC$, $\angle{A}=80^o$. $O$ is inside the triangle. $\angle{OBC}=10^o$, $\angle{OCB}=30^o$. Find $\angle{OAB}$.

[quote="Iversonfan2005"]can you please show how you did it? :?[/quote]
There are many many ways to get it. You can use Trig Ceva, or some other constructions.

[hide="One of the construction methods"]
Reflect point $O$ over $BC$ and call it $D$.
Since $\angle BDC=140^\circ =180^\circ-\frac{1}{2}\angle A$, $A$ is the circumcenter of $\triangle BDC$.
Thus $AB=AD=AC$.

Then $\triangle ABD$ is an equilateral triangle ($\because AB=AD$ and $\angle ABD=60^\circ$), so $AB=BD$.
We know that $BO=BD$, so $BA=BO$.

Therefore, $\angle OAB=\frac{1}{2}(180^\circ -40^\circ)=\boxed{70^\circ}$.
[/hide]

[quote="frt"][hide="One of the construction methods"]
Reflect point $O$ over $BC$ and call it $E$.
[/hide][/quote]

Do you mean $D$?

Yeah, sorry for the typo :oops:
(It's fixed now)