geometry
geometric transformation
reflection
incenter
homothety
modular arithmetic
circumcircle

[quote="mathVNpro"]Let (I) be the incircle of triangle ABC. Let M, N be the reflections of I through AB, AC respectively. IM, IN intersect (I,IM) by the points M', N' respectively. Prove that BN' intersects CM' at a point wich lies on the circle (I,IM) :D[/quote]
I have a solution by complex numbers.
The main idea in the proof:
Let $ I$ be the center of the complex plane, and let $ m'',n'',p''$ be the points where $ (I)$ touches $ AB,AC,BC$ respectively, and assume wlog $ |n''| \equal{} |m''| \equal{} |p''| \equal{} 1$. Then notice that $ m' \equal{} \minus{} 2m''$ and $ n' \equal{} \minus{} 2n''$ and that the circle $ (I,IM)$ has radius $ 2$. We can also get $ a,b,c$ by $ a \equal{} \frac {2n''m''}{n'' \plus{} m''}$, and similar with $ b,c$. Then let $ z$ be the intersection point of $ BN'$ and $ CM'$. Then we have $ (b \minus{} n') \times (z \minus{} n') \equal{} 0$ and $ (c \minus{} m') \times (z \minus{} m') \equal{} 0$. And after some computations we get $ z \equal{} 2\frac {n''m'' \plus{} m''p'' \plus{} p''n''}{n'' \plus{} m'' \plus{} p''}$. So $ |z| \equal{} 2 \iff |n''m'' \plus{} m''p'' \plus{} p''n''| \equal{} |n'' \plus{} m'' \plus{} p''|$, but that is obvious since $ |n''m'' \plus{} m''p'' \plus{} p''n''| \equal{} |\overline{n''m'' \plus{} m''p'' \plus{} p''n''}| \equal{} \left | \frac {n'' \plus{} m'' \plus{} p''}{n''m''p''} \right |$ $ \equal{} |n'' \plus{} m'' \plus{} p''| \iff |z| \equal{} 2 \iff z \in (I,IM)$. QED :)

[quote="mathVNpro"]Huh?? I have never thought that this problem can be solved in complex number.... I used inversion to solve this. Is there anyone who can solve this beautiful geometry problem just using pured plane geometry ( transformation....., etc.). The problem still wait for you. :D[/quote]
I would like to see the solution by inversion.

Still no solution for this problem, my dear mathlinkers???? :D

[quote="nsato"]$ BN'$ and $ CM'$ intersect at the reflection of the incentre through the Feuerbach point.  It follows from this result:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221027[/url][/quote]
Can you make yourself clearer, in detail. :)  :lol:

OK, there is a good-looking solution now. And I was extremely surprised since this one was came up with by a friend of mine who is studying specialized in Literature class of my school...

Let $ (I)$ tangents to $ BC,CA,AB$ at $ A_0,B_0,C_0$ respectively. Consider the homothety with centre $ I$ which turns $ B\mapsto B_2,C\mapsto C_2,M'\mapsto B_1,N'\mapsto B_2$, thus $ B_0B_1,C_0C_1$ are two diameters of $ (I)$. We are to prove that $ B_1B_2$ and $ C_1C_2$ intersect each other at a point $ F\in (I)$.

We have $ A_0B_1\bot A_0B_0, IC\bot A_0B_0$ then $ A_0B_1//IC$ and so $ A_0(B_1C_2IC) = - 1$. But $ A_0I\bot A_0C$ so they are inner and outer angle bisector of $ \widehat{B_1A_0C_2}$. Similarly, $ A_0I$ and $ A_0C$ are inner and outer angle bisector of $ \widehat{B_2A_0C_1}$. This implies that $ (A_0B_1,A_0C_1)\equiv (A_0B_2,A_0C_2) \pmod \pi$.

We also have $ (C_1B_1,C_1A_0)\equiv (B_0B_1,B_0A_0)\equiv (CI,CA_0)\equiv (C_2B_2,C_2A_0)\pmod \pi$ so $ \triangle A_0B_1C_1 %Error. "sims" is a bad command.
\triangle A_0B_2C_2$ and we have $ \frac {A_0B_1}{A_0B_2} = \frac {A_0C_1}{A_0C_2}$, which implies $ \triangle A_0B_1B_2 %Error. "sims" is a bad command.
A_0C_1C_2$ and since $ (B_1B_2,B_1A_0)\equiv (C_1C_2,C_1A_0)\pmod \pi$, which means $ B_1B_2$ and $ C_1C_2$ intersect $ (I)$ at the same point. That's what we have to prove.

Look like you have mistyped something, Mashimaru, can you check it again???

I have edited the solution posted above. :blush:

[b]Lemma:[/b] $\triangle ABC$ has circumcircle $(O)$ and $M,N$ denote the midpoints of $AC,AB.$ Let $ M'$ and $ N'$ be points on rays $ MO$ and $NO,$ such that $OM' \equal{} 2OM$ and $ ON' \equal{} 2ON.$ Then circles $\odot( OBM')$ and $\odot( OCN')$ meet at a point lying on $(O).$

[b]Proof:[/b] Note that the distance from the orthocenter $H$ to $B$ is twice the distance from $O$ to $AC$ and similarly for $A,C.$ Hence, $ BM'OH$ and $ CN'OH$ are parallelograms $\Longrightarrow$ $ \angle BPO \equal{} 180^{\circ} \minus{} \angle BHO$ and $ \angle CPO \equal{} 180^{\circ} \minus{} \angle CHO$ $\Longrightarrow$ $ \angle BPC \equal{} 360^{\circ} \minus{} \angle BHO \plus{} \angle CHO \equal{}180^{\circ} \minus{} \angle A$ $\Longrightarrow$  $ P \in (O).$
______________________________

[b]Problem:[/b] Incircle $(I)$ of $\triangle ABC$ is tangent to $ AB,$ $AC,$ $BC$ through $ X,Y,Z.$ $X'$ and $ Y'$ are the antipodes of $X,Y$ WRT $(I)$ and $ M$ and $ N$ denote the midpoints of $BI$ and $ CI.$ Show that $MY'$ and $NX'$ meet at a point $P$ lying on $ (I).$

Let $ V$ and $ W$ be the midpoints of $YZ$ and $XZ.$ Negative inversion WRT $ (I)$ takes $X'$ into $X,$  $ Y'$ into $Y$ and $M,N$ into the points $ D,E$ on rays $WI$ and $VI,$ such that $ ID \equal{} 2IW$ and $ IE \equal{} 2IV,$ since this inversion is the composition of a direct inversion and a central symmetry about $I.$ Hence, lines $ MY'$ and $ NX'$ are taken into the circles $\odot( YID)$ and $ \odot(XIE).$ According to the previous lemma, the latter circles intersect a point on $ (I),$ hence their primitive lines intersect on $(I)$ as well.

P.S. Unfortunately, I'm not able to describe the intersection with this reasoning.

Your solution is so like mine. But I ignore the lemma and consider it as a "destination" I have to come up with while I am solving this problem. This problem is pretty nice, isn't it, huh?? :D

[quote="mathVNpro"]Let (I) be the incircle of triangle ABC. Let M, N be the reflections of I through AB, AC respectively. IM, IN intersect (I,IM) by the points M', N' respectively. Prove that BN' intersects CM' at a point wich lies on the circle (I,IM) :D[/quote]
Nice problem!!
My teacher  solved the problem , but the other form.
He proofed.
Lema .- Let $ ABC$ , and $ I$ be the incenter of $ ABC$, and the incircle touch to $ AB$,$ BC$ and $ CA$ at $ C'$,$ A'$ and $ B'$ , respectively.
Let $ A_0$ in $ IA'$such that $ IA_0 \equal{} k.A'A_0$.
$ B_0$,$ C_0$ , similarly.
Then $ AA_0$,$ BB_0$,$ CC_0$ are concurrent at $ X$

the problem is a case of lemma.

If $ E$ the point of tangency of incircle with the nine circle of $ ABC$ . Then $ I$,$ E$ and $ X$ are collinear and $ IE \equal{} EX$.

PD: my teacher sayed "the lemma is the theorem karilla ".

[quote="nsato"]$ BN'$ and $ CM'$ intersect at the reflection of the incentre through the Feuerbach point.  It follows from this result:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221027[/url][/quote]
Could you please show us how to apply the previous result?I think your solution is very interesting :)
I have made two figures for you :)

Oh,I see,Thank you for your nice proof :lol:

[quote="nsato"]Take this result in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221027[/url], and then apply a homothety with centre $ I$ and dilation factor 2.

Also, note the result in post #5.[/quote]
To nsato,do you know any synthetic proof for it?(without using the property of $ X_{80}$) :)
To luigeometria,can you come up with a solution for point $ P$ in your post is Feuerbach's point :)
And I have a proof using a bit long calculation,though I think it looks ugly,but I like it very much.

After one night thought,I came up with a solution using complex number just like in post 3 (without much calculation)
I pesonally think complex number is a good choice :)

[quote="hqthao"]dear Mathias_DK, I don't think your computation is right, I have the coordinate of z is more complicated.  :)[/quote]
I will show you all of my computations:

We can prove that if $ Z$ is the intersection of $ AB$ and $ CD$ then:
$ z \equal{} \frac{\frac{\overline{a}b\minus{}a\overline{b}}{a\minus{}b} \minus{} \frac{\overline{c}d\minus{}c\overline{d}}{c\minus{}d}}{\frac{\overline{a\minus{}b}}{a\minus{}b}\minus{}\frac{\overline{c\minus{}d}}{c\minus{}d}}$.

(In the case where $ |a|\equal{}|b|\equal{}|c|\equal{}|d|\equal{}1$ this gives the wellknown $ z \equal{} \frac{ab(c\plus{}d) \minus{} cd(a\plus{}b)}{ab\minus{}cd}$ )

In this case $ z$ is the intersection of $ BN'$ and $ CM'$.
$ b \equal{} \frac{2m''p''}{m''\plus{}p''}, c \equal{} \frac{2n''p''}{n''\plus{}p''}, n' \equal{} \minus{}2n'', m' \equal{} \minus{}2m''$
$ z \equal{} \frac{\frac{\overline{b}n'\minus{}b\overline{n'}}{b\minus{}n'} \minus{} \frac{\overline{c}m'\minus{}c\overline{m'}}{c\minus{}m'}}{\frac{\overline{b\minus{}n'}}{b\minus{}n'}\minus{}\frac{\overline{c\minus{}m'}}{c\minus{}m'}}$.

$ b\minus{}n' \equal{} 2\frac{m''p''\plus{}n''m''\plus{}n''p''}{m''\plus{}p''}$
$ c\minus{}m' \equal{} 2\frac{m''p''\plus{}n''m''\plus{}n''p''}{n''\plus{}p''}$

So:
$ \frac{\overline{b\minus{}n'}}{b\minus{}n'} \equal{} \frac{m''\plus{}p''\plus{}n''}{n''(m''p''\plus{}p''n''\plus{}n''m'')}$

$ \frac{\overline{c\minus{}m'}}{c\minus{}m'} \equal{} \frac{m''\plus{}p''\plus{}n''}{m''(m''p''\plus{}p''n''\plus{}n''m'')}$

$ \overline{b}n'\minus{}b\overline{n'} \equal{} \frac{4}{(m''\plus{}p'')n''} (m''p''\minus{}n''^2)$

$ \overline{c}m'\minus{}c\overline{m'} \equal{} \frac{4}{(n''\plus{}p'')m''} (n''p''\minus{}m''^2)$

$ \frac{\overline{b}n'\minus{}b\overline{n'}}{b\minus{}n'} \equal{} \frac{2(m''p''\minus{}n''^2)}{n''(m''p''\plus{}p''n''\plus{}n''m'')}$

$ \frac{\overline{c}m'\minus{}c\overline{m'}}{c\minus{}m'} \equal{} \frac{2(n''p''\minus{}m''^2)}{m''(m''p''\plus{}p''n''\plus{}n''m'')}$

$ \frac{\overline{b}n'\minus{}b\overline{n'}}{b\minus{}n'} \minus{} \frac{\overline{c}m'\minus{}c\overline{m'}}{c\minus{}m'} \equal{} \frac{2(m''\minus{}n'')}{n''m''}$

$ \frac{\overline{b\minus{}n'}}{b\minus{}n'} \minus{}\frac{\overline{c\minus{}m'}}{c\minus{}m'} \equal{} \frac{m''\minus{}n''}{m''n''}\frac{m''\plus{}p''\plus{}n''}{(m''p''\plus{}p''n''\plus{}n''m'')}$

So: $ z \equal{} 2 \frac{m''p''\plus{}p''n''\plus{}n''m''}{m''\plus{}n''\plus{}p''}$

Do you agree now ? :)

See the generalization at here:
[url]http://www.mathlinks.ro/viewtopic.php?t=325489[/url]

[quote="hqthao"]:blush: oh, sorry Mathias_DK, I had count again and you right. I think that time I had used some wrong formula (because I had checked so carefully).
@ Mathias_DK: can you tell me the relative of $ K$ and $ \overline K$ when $ K$ is a point belong to a line thought $ M$ and perpendicular/paralel to a line contain $ A$ and $ B$ ( I had read in TiTu book, and doesn't know so much  :blush:  and the way he talk about the line in complex plane not like in IMO compedium)[/quote]
If $ K(k)$ lies on a line which includes the point $ M(m)$ and has direction $ \alpha$, we can say the following about $ K$:
$ \frac {k \minus{} m}{\overline{k \minus{} m}} \equal{} \frac {\alpha}{\overline{\alpha}}$.
This is because $ k \minus{} m$ can be written on the form $ k \cdot e^{iv}$ for some $ k \in \mathbb{R}, v \in [0;2\pi]$. And so $ v$ denotes the direction of the line from $ M$ to $ K$. Then $ \frac {k \minus{} m}{\overline{k \minus{} m}} \equal{} e^{i2v}$, and if $ \alpha \equal{} k_2 \cdot e^{iu}$ we have $ \frac {\alpha}{\overline{\alpha}} \equal{} e^{i2u}$. So $ \frac {k \minus{} m}{\overline{k \minus{} m}} \equal{} \frac {\alpha}{\overline{\alpha}}$ iff $ v \equiv u \bmod \pi$. That is $ k \minus{} m$ and $ \alpha$ has the same direction.
In Titu's book he writes that $ a$ is parallel to $ b$ iff $ a \times b \equal{} 0 \iff \frac {1}{2} (\overline{a}b \minus{} a\overline{b} ) \equal{} 0$. This is of course equivalent to $ \frac {a}{\overline{a}} \equal{} \frac {b}{\overline{b}}$ if we assume $ a,b \neq 0$. (When i say that $ A(a)$ and $ B(b)$ are parallel I mean that $ \vec{OA}$ and $ \vec{OB}$ are parallel)
If you want to know if to $ a$ is perpendicular to $ b$ you just check if $ a \times (ib) \equal{} 0 \iff \frac {1}{2}i(\overline{a}b \plus{} a\overline{b}) \equal{} 0$ which is equivalent to $ a \cdot b \equal{} 0 \iff \frac {1}{2} (\overline{a}b \plus{} a\overline{b}) \equal{} 0$ or $ \frac {a}{\overline{a}} \equal{} \minus{} \frac {b}{\overline{b}}$.

I'm not quite sure what you asked exactly but I hope this answers it.

[quote="hqthao"]oh, thanks Mathias_DK. but May I have a question for you, direction $ \alpha$ ? how can we dertermine it ? (for example, we have it when the point go through two point or something else....) thanks so much :blush:[/quote]
In the case were it should be parallel to $ AB$ we would put $ \alpha \equal{} a\minus{}b$ and in the case where it should be perpendicular $ \alpha \equal{} e^{i\frac{\pi}{2}}(a\minus{}b) \equal{} i(a\minus{}b)$

oh, I know. thanks very much. I must learn a lot about complex number in geometry :blush:


Gauss
number theory
relatively prime
number theory unsolved

suppose this equation: x 2 +y 2 +z 2 =w 2   . show that the solution of this equation ( if w,z have same parity) are in this form:
x=2d(XZ-YW), y=2d(XW+YZ),z=d(X 2 +Y 2 -Z 2 -W 2 ),w=d(X 2 +Y 2 +Z 2 +W 2 )

i think i have a proof of this one using conditions gcd(x,y,z,w)=1
it's pretty long and i have to check it.it uses the fact that every prime of the form 4k+1 can be expressed as the sum of 2 squares . and then also uses factorization in the gauss number ring z(i)
i'll post it later.(in case it turns out to be correct)

This one? Hard!!!!! I obtained the formula for w and z, but I have no idea how to obtain those for x and y. Iranian olympiad represents for me together with the russian one the ultimate challenge. These problems are simply impossible.

so we have x^2+y^2+z^2=w^2, and z,w have the same parity.
firstly, if w and z are both even, we know that x^2+y^2 = 0 mod 4, or x and y are both even, in which case we can divide through by 4.  So WLOG, w and z are odd.

lemma 1: there exists integers u,v s.t. u+v = w and u-v = z with (u,v) = (w,z)
proof:  since w,z have the same parity we know that u=(w+z)/2 and v = (w-z)/2 are both integers.  Now, suppose (u,v) = e and (w,z) = d first, then since u+v = w and u-v = z, e|w and e|z, so e \leq d.  But also, 2u = w+z, so d|2u, but since w and z are odd, d is odd so d|u, and by similar means d|v then d \leq e => d = e.

Now, using the result of lemma 1, let w = u+v, z = u-v, and subbing in we see that x^2+y^2 = 4uv. we know from lemma 1 that (u,v) = d, so let u=d*u', v=d*v' where (u',v')=1, so then x^2+y^2 = (2d)^2*u'*v'.  Now, the LHS is the sum of two squares, so the RHS must have an even number of prime factors = 3 mod 4.  Now, (2d)^2 is a square, so it already has an even number of prime factors = 3 mod 4, and u' is relatively prime to v' so they must each have an even number of prime factors = 3 mod 4 (otherwise the product couldn't be written as the sum of two squares).  Therefore, u' and v' can be written as the sum of two squares.  Then, let u' = X^2+Y^2, v' = Z^2 + W^2, subbing in and using the identity (a^2+b^2)*(c^2+d^2) = (ac - bd)^2 + (ad + bc)^2
we get that: x^2 = (2d)^2(XZ-YW)^2, y^2 = (2d)^2(XW+YZ)^2
=> x = 2d(XZ-YW) and y = 2d(XW+YZ)
and we know that z = d*(u'-v') = d*(X^2+Y^2-W^2-Z^2)
                          w = d*(u'+v') = d*(X^2+Y^2+W^2+Z^2).

hey zscool, I was just wondering... is your solution similar to solving a problem from a Waring's variant? Just a thought though ;)

Dear zscool I think that in your solution exist very little mistake because I solved this as well in examination three month ago but Igot 0 inthis question :(  :D

I think the mistake is due to the fact that a^2+b^2=c^2+d^2 does not neccesarily imply a= \pm c and b= \pm d.

my bad, maybe somebody can finish it up along those lines?

I also tried with that idea and led nowhere.

Harazi, how did you obtain the formulae for w and z? Here's what I did:

Let's assume (x,y,z,w)=1. Then we have [(w-z)/2]*[(w+z)/2]=x'2+y'2 (x=2x' and y=2y', since it's easy to show that both x and y must be even). [(w-z)/2]*[(w+z)/2] can be written as sum of 2 squares, so every prime of the form 4k+3 has an even exponent in [(w-z)/2]*[(w+z)/2].

Let's assume that (w-z)/2 can't be written as sum of 2 squares. Then there is a prime p=4k+3 which has an odd exponent in (w-z)/2, so it has an odd exponent in (w+z)/2, so p|w and p|z. At the same time, p|x2+y2, and from the fact that p=4k+3 we get p|x and p|y (this is well-known), so p|(x,y,z,w)=1, which is false, so the assumption that (w-z)/2 can't be written as sum of 2 squares is wrong. We also get that (w+z)/2 can be written as sum of 2 squares, so (w-z)/2=X2+Y2 and (w+z)/2=W2+Z2.

I'm sure we can continue with some arithmetics in Z[i], but I'm not used to that and I think I'm missing something.

I'm not used to apply Z[i] too ,so who can finish this problem.SOS :?

[i]My friend tell me  he doesn't remember very well that there exist the following lemma or not  :maybe: :

  [u]Lemma[/u] : If $ m, n,a, b, c, d$ are positive integers such that :

  $ m^2\plus{}n^2\equal{}(a^2\plus{}b^2)(c^2\plus{}d^2)$ then there exists positive integers $ a_1,b_1,c_1,d_1$ such that :

   $ a^2\plus{}b^2\equal{}a_1^2\plus{}b_1^2, c^2\plus{}d^2\equal{}c_1^2\plus{}d_1^2, m\equal{}a_1b_1\plus{}c_1d_1, n\equal{}a_1d_1\minus{}b_1c_1$

   I have checked that if that lemma is right , I will have a complete solution for that nice one

  But I failed to prove the lemma :oops:

   Can anybody help me to prove or disapprove it ? [/i]

[quote="Waclaw Sierpinski "]
All the solutions of the equation
x²+y²+z²=t² (37)
in natural numbers x, y, z, t, with even y, z, are obtained from the formulae
$ x\equal{}\frac{l^2\plus{}m^2\minus{}n^2}{n},y\equal{}2l,z\equal{}2m,t\equal{}\frac{l^2\plus{}m^2\plus{}n^2}{n}$
l, m being arbitrary natural numbers, and n being the divisors of  l²+m² less than $ \sqrt{l^2\plus{}m^2}$.Every solution is obtained exactly once in this way.
...
It is worth-while to notice that, as has been proved by R. D.
Carmichael [4], pp. 39-43, all the solutions of equation (37) in natural
numbers can be obtained from the identity
(2d(XZ-YW))²+(2d(XW+YZ))²+(d(X² +Y² -Z² -W² ))²=(d(X² +Y² +Z² +W² ))²
[/quote]
CARMICHAEL, R.D.,
[4] Diophantine Analysis (New York 1915, reprint New York 1959).


geometry
geometric transformation
rotation
reflection
abstract algebra
group theory
superior algebra

i would like to find all nontrivial homomorphisms of  $\phi:D_{4}\rightarrow Z_{4}$, i know the operations must be preserved $\phi(xy)=\phi(x)+\phi(y)$ , but im a little confused going from a dihedral group to an integer group

If one of the rotations of order 4 maps to an element of order 4, we're in trouble; the image of each reflection must have order dividing 2, and the product of a reflection and a rotation is a reflection. This reduces the problem to the question of homomorphisms to $\mathbb{Z}_2$. Since the rotation of order 2 is a square, it must map to the identity. This leaves two independent choices: does a rotation of order 4 map to the identity or an element of order 2? does the generating reflection map to the identity or an element of order 2?

If a dihedral group maps to an abelian group, the image is isomorphic to a subgroup of $\mathbb{Z}_2^2$

so if $\rho$ mapped to 2 and $\tau $ mapped to 2  then
$\phi(\rho \tau)=2+2=0$ and 0 has order 1 ,
or $\rho$ maps to 0 and $\tau$ maps to 2 then $\phi(\rho \tau)=0+2=2$ and 2 has order 2 which divides the order of $\rho \tau$,   :?


linear algebra
matrix
vector
limit
trigonometry

Lets $A$ be a square matrix of order $n$. $A^{n}=0, A^{n-1}\neq 0$. Prove that, there exists a square matrix $P$ such that: $P^{-1}AP=B$ with $B: = \left[ \begin{array}{cccc}0 & ... & ... &0 \\ 1 & 0 & ...&0 \\ ... & ... & ...&0 \\ 0 & ... & 1&0 \end{array}\right]$.
Help me! Thanks!

[quote="anhminh"]Lets $A$ be a square matrix of order $n$. $A^{n}=0, A^{n-1}%Error. "nequa" is a bad command.
0$. Prove that, there exists a square matrix $P$ such that: $P^{-1}AP=B$ with $B: = \left[ \begin{array}{cccc}0 & ... & ... &0 \\ 1 & 0 & ...&0 \\ ... & ... & ...&0 \\ 0 & ... & 1&0 \end{array}\right]$.
Help me! Thanks![/quote]

The question consists answer.

There is Jordan decomposition of any matrix A=TJT^-1, case J=B.
http://mathworld.wolfram.com/JordanMatrixDecomposition.html

From A=TBT^-1 follows T^-1AT=B. So matrix P is T.  It exists, naturally, due there is Jordan decomposition of any matrix A=TJT^-1. Proved.

Details. Column matrix T has whole rank and consists of one eigen vector T1, AT1=0, and Jordan chain of vectors AT2=T1, AT3=T2, and so on T=[T1 T2 ... ], see link. J is diagonal matrix of simple eigen values or matrix B, there is mixed case (B presented on diagonal partially).

A^2=A*A=TJT^-1TJT^-1=TJ^2T^-1. So A^n=TJ^nT^-1.

If J is diagonal matrix of eigen values, then from A^n=0 follows J^n=0
(by RankT=n), and for diagonal matrices it could be only if J=0 (all eigen values are simple, equal 0). In mixed cases part of eigen values are simple and J^(n-1)=0. But if J=B, it's easily seen B^n=0 and B^(n-1)<>0. We have it: from A=TBT^-1 follows T^-1AT=B. So matrix P is T.

Choosing the right basis is the key. Let $x$ be a vector with $A^{n-1}x\neq0$, and let $v_{0}=x,v_{1}=Ax,v_{2}=A^{2}x,\dots,v_{n-1}=A^{n-1}x$.
If $v_{0},\dots,v_{n-1}$ are a basis, the matrix of $A$ with respect to this basis is the desired companion matrix.
It's pretty easy to verify linear independence; $x_{k}$ cannot be a linear combination of the later $x_{i}$, since $A^{n-k-1}x_{k}=x_{n-1}\neq 0$, and $A^{n-k-1}v=0$ whenever $v$ is a combination of the $x_{i}$ with $i>k$. Since there are $n$ of these vectors, they are a basis.

[quote="Kent Merryfield"]Quoting the Jordan canonical form in order to claim this is true seems a little backwards to me (in the same way that using L'Hôpital to prove that $\lim_{x\to0}\frac{\sin x}x=1$ is backwards): shouldn't we prove this first, and then use it to prove the existence of the Jordan form in general?[/quote]

To prove existance of the Jordan form after Jordan seems to me  unsense. In any case u manipulate this or that well known fact of matrix algebra. If it's your  college courses, u can give facts, in consecuences, how u want only, and, naturally, use your righ keys how u want. Is the question of taste. Is outside of matematics, sorry. Is methodology. I hope, i'll not very mix this process :)


geometry
trapezoid
search
combinatorics unsolved
combinatorics

Each point of a circle is painted in one of the N color (N>1).Prove that there exists an inscribed trapezoid such that all its vertices are painted the same color.

Have a look here (post #13) :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1950400325&t=164125

Pierre.


[hide]$-0+0-4+1-2+0-n+3=n$ and n must be divisible by 11(divisibility rule of 11) so n=0 :idea: our answer[/hide]

In the "Art of Problem Solving" Volume 1, it explains the divisibility rule for 11.  Where did you get the rule, jhollenbeck?

[quote="236factorial"]For what value of n will the number 200,412,0n3 be divisible by 11?[/quote]
[hide="answer"]use the add every other and subtract rule
6-6-n=0 n can only be 0
n=0[/hide]

I've heard the divisibility by 11 rule as the difference between the sums of every other integer. (In other words, $3766565$ is divisible by 11 because $3+6+5+5=7+6+6$, and it also works for every difference of a multiple of 11.) Therefore, this will also work by doing the following: $3-7+6-6+5-6+5=0$ (or any multiple of 11).


superior algebra
superior algebra unsolved

[b]Problem. [/b]  Let $G$ be a finite group with the following property. For every $g \in G$, there exists a positive integer $k \geq 2$ such that $(k-1,|G|) = 1$ and $g^k \sim g$ are conjugated. Prove that $G = \langle 1 \rangle$.

Enjoy!  :)

--Vesselin

Is this your own creation, vess? It's really cool :). Makes you wonder why it's still unsolved :).

Suppose the contrary and take some $g\in G$ whose order is the smallest prime $p$ which divides $|G|$. We have $aga^{-1}=g^k\ (*)$ for some $a\in G$.

By successively multiplying $(*)$ to the left with $a$ and to the right with $a^{-1}$ ($|G|-1$ times) we get $g=g^{k^{|G|}}$, or, in other words, $g^{k^{|G|}-1}=1$. This means that $p|k^{|G|}-1$. Since we also have $p|k^{p-1}-1$, we get $p|(k^{|G|}-1,k^{p-1}-1)=k^{(|G|,p-1)}-1=k-1$, which contradicts the hypothesis $(k-1,|G|)=1$.

Thanks!  :D I made this problem some time ago; glad you liked it!

Nice solution! Mine was similar (following a standard idea), but not that short and cute. Here's an outline. Of course, the starting point again is to choose the smallest prime divisor $p$ of $|G| > 1$, and to take an element $g$ of order $p$. Then, $g = h^{-1}g^kh$ for some $h \in G$ and $k$ with $(k-1,|G|)=1$. Since $p \nmid k-1$, we have $[g,h] \neq 1$: $g$ and $h$ cannot commute. Hence, $h \in N_G(\langle g \rangle) \setminus C_G(\langle g \rangle)$, which in particular implies that the quotient $Q:= N_G(\langle g \rangle)/C_G (\langle g \rangle)$ is non-trivial. But $Q$ can be embedded in an obvious way into $\mathrm{Aut} \, (\langle  g \rangle) \cong \mathbb{Z}_{p-1}$, hence its order $|Q| > 1$ divides both $p-1$ and $|G|$, which of course is a contradiction, as the minimality of $p$ gives $(|G|,p-1) = 1$.

This, of course, is a very well-known idea, so I don't really consider it 'my' problem.  :)


Maybe you could do:
p+10d+5n=25
that's just a guess :?

hi xenon, if you are sure that you want it to equal 25 cents, it's quite easy:

as said, write the equation

p + 5n + 25q = 25, where (p, n, q) is an ordered triplet representing the numbers of pennies (p), nickels (n), and quarters (q) respectively.  to reword the problem then, we want to find the number of ordered triplets (p,n,q) that satisfy that equation.  an important thing to realize is that we don't need to know the triplets, just how MANY there are.

becuase its 25 on the right side, you can either have a quarter or you dont.  for the case where you don't, you can either have 0, 1, 2, 3, 4, or 5 nickels (6 possibilities), and for each choice of nickels there is some number of pennies that will get you up to 25 (you dont need to know what that number is).  thus there's the one case with a quarter, and 6 possibilities without a quarter.

1+6=7.

[quote="xenon_nightmare152"]this may sound stupid...but i don't know any dramatically easy way to do this problem. is there a formula?

How many combinations of pennies, nickels and/or dimes are
there with a total value of 25¢?

i never have been good @ counting, prob. standard deviation, normal  distribution etc  :blush:[/quote]

There is no stupid question...dont be afraid to ask anything, asking will only help you and possibly other people too.

I don't believe that there's really a formula.
What I would do for these short problems is to make an organized list, but make sure it's systematic!

I think I've seen the formula, but it is very complicated, so I suggest a table and list the possibilites

i did a systematic list...but i just wish there was a faster method for competition.

anyway, thanks


can someone explain how applying Ptolemy to ACDE yields $ x\cdot z\plus{}81^2\equal{}y^2$?
[url]http://www.artofproblemsolving.com/Wiki/index.php/1991_AIME_Problems/Problem_14#Solution[/url]

It should probably be on ACDF instead...


Max has a spinner that lands on 1 with a probability of $ \frac{1}{2}$, lands on 2 with a probability of $ \frac{1}{4}$, lands on 3 with a probability of $ \frac{1}{6}$, and lands on 4 with a probability of $ \frac{1}{12}$. If Max spins the spinner, and then Zack spins the spinner, then what is the probability that Max gets a larger number than Zack does?

I'm somewhat certain there is a better solution (a shortcut), but I don't see it.

First get common denominators:
1 with prob. 6/12
2                3/12
3               2/12
4                1/12

Now casework:
If Max gets 1, there is 6/12=1/2 chance Zack gets larger number.
If he gets 2, chance is 3/12=1/4.
If he gets 3, chnace is 1/12.
If he gets 4, chance is 0/12=0.

These cases are weighted, so we must take the average after weighting the possibilities.
$ [(6*1/2)\plus{}(3*1/4)\plus{}(2*1/12)\plus{}1(0)]/12]$

Simplifying,
$ (3\plus{}3/4\plus{}1/6\plus{}0)/12$
$ (47/12)/12$
$ 47/144$

Let's say that they are playing a game.
Who gets the largest number wins.

By the symmetry, we just need to take out the "draw" cases and divide by 2.

(1-(1/2)²-(1/4)²-(1/6)²-(1/12)²)/2

[quote=birdmath]There is 1/2 chance that Max gets a 1 and a 1/2 chance that  Zack gets a larger number.
1/2 * 1/2 = 1/4
There is 1/4 chance that Max gets a 2 and 1/4 chance that Zack gets a larger number.
1/4 * 1/4 = 1/16
There is 1/6 chance that Max gets a 3 and 1/12 chance that Zack gets a 4.
1/72
There is 1/12 chance that Max gets a 4 and no chance that Zack gets a larger number.

1/4 + 1/16 + 1/72  =  36/144 + 9/144 + 2/144  =  47/144[/quote]

I did this. And then I divided by 6.  :wallbash_red:

[quote=birdmath]There is 1/2 chance that Max gets a 1 and a 1/2 chance that  Zack gets a larger number.
1/2 * 1/2 = 1/4
There is 1/4 chance that Max gets a 2 and 1/4 chance that Zack gets a larger number.
1/4 * 1/4 = 1/16
There is 1/6 chance that Max gets a 3 and 1/12 chance that Zack gets a 4.
1/72
There is 1/12 chance that Max gets a 4 and no chance that Zack gets a larger number.

1/4 + 1/16 + 1/72  =  36/144 + 9/144 + 2/144  =  47/144[/quote]

(why do we care about the chance that max get 4, wouldn't it not matter because it would be a draw sometimes?)


Let $ ABC$ be a triangle and$ E$ be Euler centre of $ ABC$ let $ E_a,E_b,E_c$ be Euler centres of $ EAB ,EBC,ECA$ respectively .Provethat $ AE_a,BE_b,CE_c$ are concurent.

See this post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=18424[/url]


find all positive integer numbers a1a2..an such that (a1a2...an) ^2=(b1b2...bm) and (an..a1)^2=(bm.....b1), m is a positive integer.
for example: 13^2=169, 31^2=961
                    12^2=144,21^2=441.

there are infinitely many of them, all integers 10...02 and 20...01 which have the same number of "0" 's fit for an example.


Find all positive integers $ x, y$   such  that   equation  $ \sqrt{x}\plus{}\sqrt{y} \equal{} \sqrt{468}$ .
Find all positive integers $ x, y, z$   such  that   equation  $ \sqrt{x}\plus{}\sqrt{y} \plus{}\sqrt{z}\equal{} \sqrt{468}$ .
Find all positive integers $ x_{1}, x_{2}, \cdot \cdot \cdot , x_{l} , a$   such  that   equation  $ \sqrt{x_{1}}\plus{}\sqrt{x_{2}} \plus{}\cdot \cdot \cdot \plus{}  \sqrt{x_{l}}\equal{} \sqrt{a}$ .

[hide]Consider that $ \sqrt {468} \equal{} 6\sqrt {13}$. Hence $ \sqrt {x} \equal{} a\sqrt {13},\sqrt {y} \equal{} b\sqrt {13},\sqrt {z} \equal{} c\sqrt {13}$.

Then, we want
a) all ordered pairs of positive integers $ (a,b)$ such that $ a \plus{} b \equal{} 6$
b) all ordered triples of positive integers $ (a,b,c)$ such that $ a \plus{} b \plus{} c \equal{} 6$

The answers follow from here.[/hide]


MATHCOUNTS probability geometry rectangle AMC AIME

This problem was first posted in the MATHCOUNTS forum but no one could solve it, so I'm moving it here:
[quote][size=200]Problem[/size]
Each unit square in a 4 by 4 square can be colored yellow, or white.
The probability of a square being colored yellow or white is equal.
What is the probability that when colored, the 4 by 4 square will contain at least one 2 by 2 block of white?[/quote]

Hopefully someone here can do it...

[hide="Solution"]Think about it this way...in how many ways can we select a 2 by 2 block of white from the 4 by 4 square? There are $ 9$ such blocks that are possible.
Then, there are $ 12$ squares left to be filled. Each square can be colored either yellow or white, so there are $ 9 \cdot 2^{12}$ ways to have at least one 2 by 2 block of white in a 4 by 4 square.
The total ways we can color the square without regard to the 2 by 2 block of white is $ 2^{16}$.
Hence, the probability is $ \frac {9 \cdot 2^{12}}{2^{16}} \equal{} \boxed{\frac {9}{16}}$[/hide]

No: you've just counted the all-white square 9 different times in your numerator.  Inclusion-exclusion will do the job, but it's ugly.  I'm thinking about a nicer way, maybe focusing on which of the four middle squares are yellow and which are white?  That'll lead to a decent amount of casework, but it's maybe better than PIE.  Also, a recursive solution is possible (count the number of ways to tile a $ 4 \times n$ rectangle so there is no $ 2 \times 2$ white block), but the casework there isn't much better than just doing it directly, if at all.

Ooops...I forgot to take into account that there can be more than one.

Maybe $ 1\minus{}P\left(\text{no blocks of white}\right)$?

Is the answer $ \frac {86}{2^9}$ ?  :maybe:
EDIT: and yes, the casework is ugly, which is why i shall post my logic after someone confirms the answer   :wink:

PM [b]ZhangPeijing[/b] about the answer. He's the one that posted the question. I just moved it.

I brute-forced the problem with a computer, and I got the following for an answer; also, I think I recall seeing this problem in a past AIME for which the solution is in the Wiki, but I don't remember which one.

[hide="Answer."]
$ \frac{365}{1024}$.[/hide]

[hide="SOLUTION"]I basically used a lot of the ideas already in this forum.
I count the number of ways that there is no 2 x 2 white block, based on the colorings of the 4 squares in the center. Here's my breakdown:
4 yellow tiles: $ 2^{12}$
3 yellow tiles: 4 cases, $ 7 \times 2^9$
2 adjacent yellow: 4 cases, $ 5\times 2^9$
2 diagonal yellow: 2 cases, $ 49\times 2^6$
1 yellow: 4cases, $ 113\times 2^4$
and summing these you get $ 659 \times 2^6$, so the prob. that there is no white 2x2 box is $ \frac{659}{1024}$, and hence the answer is $ \frac{365}{1024}$.[/hide]

I just wrote a program like Begoner, and I got the same number.
EDIT: good solution, SCHNOGGINVOGGINSCHWITZERW

oops i didnt see the "atleast one" part of it   :oops:
but if it's [i]exactly[/i] one 2*2 square, then i guess my answer is right  :|


Determine whether each ordered pair is a solution of the given equation.

y=2xsquared (1,2);(3,18) answer:yes;yes

I don't know what to do when I get to the y and I have to 2=2xsquared.

How do I unsquare the x or do I need to do that?

Please show work. Thank You

PS does the same rule apply when the x is tripled or quadruple times?

You just plug in the $ x$ and $ y$ coordinate values.

$ y \equal{} 2x^2$

$ 2 \equal{} 2 \cdot 1^2$

$ 2 \equal{} 2 \cdot 1$

$ 2 \equal{} 2$

This is a solution.

$ y \equal{} 2x^2$

$ 18 \equal{} 2 \cdot 3^2$

$ 18 \equal{} 2 \cdot 9$

$ 18 \equal{} 18$

This is also a solution.

I was wondering what you do when you plug in the y number and it looks like,

2=2xsquared ? Is there anything special that I do with the x squared?

Well, define "special".

You would square the $ x$ value, and then multiply it by $ 2$.

For any equation in terms of $ y \equal{} mx^a \plus{} b$, you would take $ x^a$ (you understand powers, right?), and then multiply it by $ m$ for the $ y$ value.

Hope this helps.

I guess I'm asking if the answer could be 2=x squared.
Do I have to unsquare the x somehow? That's what I don't get.

To "unsquare" something, take the square root of both sides. Like this:

$ 2 \equal{} x^2$

$ \sqrt{2} \equal{} \sqrt{x^2}$

$ \sqrt{2} \equal{} x$

Make sure you see why $ \sqrt{x^2} \equal{} x$.

note that when taking a square root, you must be sure to include the negative solution
for example
$ 2\equal{}x^2$
$ \sqrt{2}\equal{}\boxed{\pm}\sqrt{x^2}$
because you must remember a number, n, squared equals its opposite, -n, squared

But also note that, in this question, actually having to take the square root and solve is unnecessary because you can plug in values. (And solving it would be impossible anyways, because you are given one equation, two variables.)

I get it now. I was only plugging in one at a time instead of both of them at once. Boy, I feel silly. Thanks for all the help. With all this great help I might actually make a good grade on my math test tomorrow.


inequalities
function
calculus
derivative
inequalities unsolved

Put u=1/p^3-1 etc.
Then u+v+w=0.
Now the inequality writes 1/(u+1)^(5/3)-(1/u+1)^2+..>=0
You can now use Jensen I think

little nice mathematica helped me to derive the function and the second derivative of $\frac{1}{(u+1)^{\frac{5}{3}}}-\frac{1}{(u+1)^2}$ is $-\frac{6}{(1+u)^4}+\frac{40}{9(1+u)^{\frac{11}{3}}}$, so we get some little problems with jensen(u could be negative). but i don't see any other argument right now...

Peter

[Edit: derivative now correct]

at least it is wrong the other way round.
proof: $p^6\frac{1}{p}+q^6\frac{1}{q}+r^6\frac{1}{r}\leq (p^9+q^9+r^9)^(2/3)(\frac{1}{p^3}+\frac{1}{q^3}+\frac{1}{r^3})^(1/3)\leq p^6+q^6+r^6$,
where the first one is hlder inequality with one exponent 1/3, the other 2/3, the second one is powermean inequality with exponents 2/3 and 1.

Peter

vasc, can you explain me your use of muirhead more explicitly? i don't see it...

Peter

The homogeneous inequality has the form

$(a^6+b^6+c^6)^3(a^{-3}+b^{-3}+c^{-3})\geq(a^5+b^5+c^5)^3(a^0+b^0+c^0)$

Now is it clear, Peter ?

ok(why didn't i see?)...but using muirhead is somewhat brutal :)

Peter

Yes my Jemsen proof doesn't work..
But can't we mix 2 of u,v into their arithmetic mean?
The inequality will become in 2 variables and much easier

maybe iura, but i'm too tired to do these computations this evening. sorry :(
look at my proof above, or vascs proof.

Peter

Peter,
The right inequality (power mean) is not correct.
Correct is $\geq$, not $\leq$. ;)

oh, that's right. i compared "$2/3\leq 1$" and took the same sign, but that's nonsense. however, muirhead works.

Peter

[quote="Vasc"]The homogeneous inequality has the form
$(a^6+b^6+c^6)^3(a^{-3}+b^{-3}+c^{-3})\geq(a^5+b^5+c^5)^3(a^0+b^0+c^0)$
[/quote]

sorry for my stupidity, I still don't see it  :blush: . the only version of muirhead i know is http://planetmath.org/encyclopedia/MuirheadsInequality.html.

and i think that's the only form that exists. but just multiply everything, the terms $a^{15}+b^{15}+c^{15}$ cancel on both sides and on the left side, we still have $\sum a^6b^6c^{-3}$, on the right $\sum a^5b^5$, so you can use muirhead now.

Peter

[Edit: what i wrote is nonsense, but expanding out helps, i'm too lazy to actually do that]

oh ok  :blush:  :blush: . im such an idiot.

i cant start a new topic for some reason :? :? it says "please submit source" and there is no source field. :(. ill just post the following inequality here ( i think its difficult)

prove for all reals a,b,c:
a^2*b^2*c^2 ( 1/a + 1/b + 1/c ) + (a^3*b^2 + b^3*c^2 + c^3*a^2)
<= 2(a^3*c^2 + a^2*b^3 + b^2*c^3)

This inequality is not true for a=3, b=1, c=0.
 :)

I definitely agree that Muirhead's Theorem is somewhat brutal (expanding out that nice expression! :( ).  Anyway, here's a (less whacking) solution:

It suffices to prove the case when p, q, r are positive.

From the Power Mean inequality, we obtain
\[3(p^6+q^6+r^6)^5 \geq (p^5+q^5+r^5)^6.\]

Hence it suffices to prove that $p^6+q^6+r^6 \geq 3$.  However from the QM-AM and AM-HM inequalities we have

\[\sqrt{\frac{p^6+q^6+r^6}{3}} \geq \frac{p^3+q^3+r^3}{3} \geq \frac{3}{1/p^3+1/q^3+1/r^3} = 1,\]
and thus we have the desired result.

I may be wrong, but isn't this problem trivial? I mean, take $ a=\frac{1}{p^3} $ and so on. Then reduce it to $\frac{1}{a^2}+...\geq (\frac{1}{a})^{\frac{5}{3}}+... $ . But this follows immediately from the Power Mean Inequality ( god knows what is the correct name of it) and the fact that $\frac{1}{a}+...\geq 3 $.


LaTeX
calculus
inequalities unsolved
inequalities

[quote="anthony19891019"]x,y,z>0
(1). if xy is a ''fixed value'', please calculate the minimum value of
        1/[(1+x)^3] + 1/[(1+y)^3]

(2) if xyz=1. Prove sym{1/[(1+x)^3]}>= 3/8[/quote]
Please write in $\LaTeX$. It should be
[quote="anthony19891019"]$x,y,z>0$
(1). if $xy$ is a ''fixed value'', please calculate the minimum value of \[\frac{1}{(1+x)^{3}}+\frac{1}{(1+y)^{3}}.\] (2) if $xyz=1$. Prove $\sum_{sym}\frac{1}{(1+x)^{3}}\geq \frac{3}{8}$. [/quote]

[quote="Xevarion"]I think they are both just AM-GM. Thus: minimized at $x=y(=z)$. This gives $\frac{1}{4}$ for the first one.[/quote]

How do you proceed it? Can you post a solution please?

2 times square root of the product

Did you mean ''2sqrt(xy)"?

[quote="Rzeszut"][quote="Xevarion"]Well in the first one it is just $\frac{1}{(1+x)^{3}}+\frac{1}{(1+y)^{3}}\ge$ 2 times square root of the product. This is minimized when the two terms on the LHS are equal, so $x=y$. As $xy = 1$, it is $\frac{1}{8}+\frac{1}{8}= \frac{1}{4}$.
The second problem is similar...
( :| am I missing something important? )[/quote]

No, he meant $2\sqrt{\frac{1}{(1+x)^{3}}\cdot \frac{1}{(1+y)^{3}}}$.

To Xevarion: for what constraint the product $ab$ is minimized for $a=b$?[/quote]

But how can we continue this step?

[quote="Xevarion"][quote="Rzeszut"]To Xevarion: for what constraint the product $ab$ is minimized for $a=b$?[/quote]
[quote="anthony19891019"]xy is a ''fixed value''[/quote]

I assumed it means $xy=k$, from which we may as well assume $xy=1$ (and just change back at the end). So really the min should be
$\frac{2}{(1+\sqrt{k})^{3}}$
and I think there's nothing more to say, unless anthony# meant something else.[/quote]

Yes. xy=k=A constant.

I know the minimum is $\frac{2}{(1+\sqrt{k})^{3}}$.. But I don't know how to approach it..

It I use AM-GM (Like your above mentions), then I should consider the
minimum value of 1/[(1+a)^3 * (1+b)^3], hence we have to consider the maximum value of (1+a)(1+b), which equals to ab+a+b+1. Then we have to consider the maximum value of a+b. But we can't continue it because a+b doesn't have maximum value.

[quote="Xevarion"]AM-GM has equality at $a=b$... That minimizes the sum, as we want... I'm sorry if I'm an idiot and missing something important...[/quote]
But $\text{anthony19891019}$ wanted to [b]maximize[/b] the sum. And for $xy=1$ in fact $x+y$ doesn't have maximum.
Writing it with details: \[\frac{1}{(1+x)^{3}}+\frac{1}{(1+y)^{3}}\geq 2\sqrt{\frac{1}{(1+x)^{3}}\cdot \frac{1}{(1+y)^{3}}}=\]  \[=2\cdot \left(\frac{1}{(1+x)(1+y)}\right)^{\frac{3}{2}}= 2\cdot \left(\frac{1}{2+x+y}\right)^{\frac{3}{2}},\] so if you want to minimize it, $\text{Xevarion}$, you have to maximize $x+y$ for $xy=1$. It's impossible.

[quote="Xevarion"][quote="Rzeszut"]To Xevarion: for what constraint the product $ab$ is minimized for $a=b$?[/quote]
[quote="anthony19891019"]xy is a ''fixed value''[/quote]

I assumed it means $xy=k$, from which we may as well assume $xy=1$ (and just change back at the end). So really the min should be
$\frac{2}{(1+\sqrt{k})^{3}}$
and I think there's nothing more to say, unless anthony# meant something else.[/quote]
[quote="Xevarion"]OK I understand now, thanks. Here is another method.

$xy = 1$ gives $y = \frac{1}{x}$
$\frac{1}{(1+\frac{1}{x})^{3}}= \frac{x^{3}}{(1+x)^{3}}$
so we want to minimize
$\frac{1+x^{3}}{(1+x)^{3}}$
But this is just
$1-3\frac{x^{2}+x}{(1+x)^{3}}= 1-3\frac{x}{(1+x)^{2}}$
So we want to maximize $\frac{x}{(1+x)^{2}}$
Note $\frac{x}{(1+x)^{2}}= \frac{1}{1+x}-\frac{1}{(1+x)^{2}}$
We are maximizing $u-u^{2}$ which I hope no one will complain about when I say it obviously occurs at $u = \frac{1}{2}$.
$\frac{1}{1+x}= \frac{1}{2}\Rightarrow x = 1$

Is it ok?
(PS it can also be done with calculus but that would be much uglier ;))[/quote]
Yes, you method is right and nice. Thank you. But you can't solve (2) using it. I tried also $\mathrm{HM-GM}$, it sufficies to maximize $\sum{x}+\sum{xy}$ (impossible).

How many times second problem was posted?
http://www.mathlinks.ro/Forum/viewtopic.php?t=36567


geometry
trapezoid
geometric transformation
reflection
parallelogram

Is given  an isoscele trapezoid . Its diagonals are perpendicular. Prove that the line which connect the middle points of the congruent sides is equal with the trapezoid's altitude

[hide="Solution."]
Reflect the trapezoid $ ABCD$ as below, forming parallelogram $ AEFD$. Now the altitude from $ C$ to $ AE$ is also a median, since $ \triangle ACE$ is right, and so $ CX\equal{}\frac{1}{2}AE\equal{}\frac{1}{2}MO\equal{}MN$, where $ MN$ is the median of $ ABCD$, as desired.
[/hide]


The sum of two numbers is twice the sum of the reciprocals of the same two numbers. What is the product of the two numbers?

Let the numbers be a and b. a+b=2((1/a)+(1/b)). Therefore, a+b=2((b+a)/ab). Dividing by a+b leaves 1=(2/ab), so ab=2.

[hide="Explanation in Latex"]
Given that:

$ a+b=2(\frac{1}{a}+\frac{1}{b})$

$ a+b=2(\frac{b+a}{ab})$

Divide by $(a+b)$

$\Longrightarrow 1=\frac{2}{ab}$

[b]Multiply ab to both sides
[/b]
$\Longrightarrow ab=\boxed{\boxed{\boxed{\boxed{2}}}}$

[/hide]

a+b could equal 0. Then the answer 2 is not completely right. ab can be any negative number.

No because the denominator cannot equal 0


MATHCOUNTS probability geometry rectangle ratio inequalities

1.
Darts are thrown randomly at a board showing three concentric circles of radii 2, 3, and 4.  If a dart hits inside one of the circles, what is the probability, expressed as a common fraction, that it is in the interior of the circle of radius 3 but not in the interior of the circle of radius 2?

2.
Three tennis balls are tightly sealed in a right cylindrical can, so that the balls cannot move inside the can.  If the radius of each ball is 4 cm, find the number of cubic centimeters in the volume of the space within the cylinder, not taken up by the three tennis balls.  Express your answer in terms if $\pi$.

3.
A custom made window is a single pane of glass consisting of a semicircular sheet of glass on top of the narrow end of a rectangular sheet of glass.  The rectangle has dimensions of 4 feet by 3.5 feet.  How many feet of weather-stripping will be needed to go around the glass?  Express your answer to the nearest integer.

4.
The numbers from 1 through 8 are separated into two sets, A and B.  The numbers in A are multiplied together to get $a$.  The numbers in B are multiplied together to get $b$.  The larger of the two numbers $a$ and $b$ is written down.  What is the smallest possible number that can be written down using this procedure?

5.
Eric's Florist Shop sells roses for 20 dollars per dozen, daises for 5 dollars per dozen, and chrysanthemums for 10 dollars per dozen.  At these rates how much would an arrangement of 9 roses, 6 daises, and 18 chrysanthemums cost in dollars?

6.
Given the equation $\displaystyle \frac{1}{m} + \frac{1}{n} = \frac{2}{15}$, where $m$ and $n$ are positive integers, find the value of $m+n$ such that $m>n$ and $m$ is not a multiple of $n$.

7.
Use the pattern below to express $100^2$ in the form $a^2+b^2-c^2$.  Then find $a+b+c$.
\begin{eqnarray*}
12^2 = 8^2 + 9^2 - 1^2\\
14^2 = 10^2 + 10^2 - 2^2\\
16^2 = 12^2 + 11^2 - 3^2\\
18^2 = 14^2 + 12^2 - 4^2
\end{eqnarray*}

8.
The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length.  If the width is 2, how many units are in the length of the diagonal?

9.
Suppose a number $x$ satisfies $x^2-4x-1=0$.  How much greater is $x$ than its reciprocal?

10.
What is the difference between the largest and the smallest values of $x$ satisfying the following system of inequalities?
Express your answer as a mixed number.
\begin{eqnarray*}
2x-6 \le 6\\
4x+3 \ge 13
\end{eqnarray*}

ok, i don't have any scratch paper on me, but most of these are pretty straight forward.  There's a good way to do number 9, so i'll just show that one
9. [hide]Ok... we're essentially told to find X-(1/x).  If we combine these two, we get (X^2-1)/X= 4X/X= 4. There's ur answer[/hide]

[hide="10"]$ 2x\minus{}6\le6$
$ 4x\plus{}3\ge13$
$ x\le6$
$ x\ge2.5$

Answer is thus $ 3.5$[/hide]
[hide="8"]
Let the length of the diagonal be $ d$.

Then the length of the rectangle is $ \sqrt{d^2\minus{}4}$

$ \frac{\sqrt{d^2\minus{}4}}{2}\equal{}\frac{d}{\sqrt{d^2\minus{}4}}$
$ d^2\minus{}4\equal{}2d$
$ d^2\minus{}2d\minus{}4\equal{}0$
$ d\equal{}\frac{2\pm2\sqrt{5}}{2}$
$ d>0$, $ d\equal{}\sqrt{5}\plus{}1$[/hide]
[hide="7"]
As we can see, the first term on the RHS is always two less than the one on the left hand side.
Second term increases by one every time, as does the third term.

$ 100^2\equal{}96^2\plus{}53^2\minus{}45^2$
Or you could just notice that for all multiples of $ 4$ on the left hand side, the last two terms on the RHS must be odd. So ya XD. I'm lazy today
$ a\plus{}b\plus{}c\equal{}194$[/hide]


logarithms
geometry
3D geometry
algebra
factorization
difference of cubes
special factorizations

$a,b \in \mathbb{R^{*}}$ and $(x_{n})_{n \ge 1}$ , $x_{1}=a$ , $x_{n}=b \sqrt[3]{x_{n-1}}$
find a and b so as $x_{n}$ converges

Assume $a, b > 0$. Let $y_{n}= \log x_{n}$, $A = \log a$ and $B = \log b$, then we have

$y_{1}= A$
$y_{n+1}= B+\frac{1}{3}y_{n}$

Then you can check that $y_{n}$ always converges. So $x_{n}$ always converges to some positive number $x$.

Not let's consider more general cases. If $ab = 0$, then $x_{n}$ converges to 0, so we only check the case $ab \neq 0$. Since $|x_{n}|$ always converges to some positive number as we have seen, we only have to consider the convergence of $\text{sgn}(x_{n})$. If $x_{n}$ alternately changes its sign, then $x_{n}$ cannot converge. By simple observation, we know this can happen iff $b < 0$.

Therefore, $x_{n}$ converges if and only if $a = 0$ or $b \geq 0$.

(Actually, the limit is $\text{sgn}(a) b^{3/2}$)

[hide]
If $a=0$, then $x_{n}$ is 0 everywhere and trivially converges. If $a<0$, then negate $x_{n}$ to get a sequence starting at $-a$ and converging if and only if the original sequence converges. Therefore, it suffices to check the case $a>0$.

If $a>0$ and $b<0$, then the sequence can't converge because if it converged to $L$, then $L=b\sqrt[3]{L}$ which requires $b$ to be nonnegative. If $b\geq 0$, then the only way the sequence can converge would be if it converged to $b^{3/2}$. So we consider the sequence
\[d_{n}=|b^{3/2}-x_{n}|\]
So $d_{1}=|b^{3/2}-a|$ and, with the help of difference of cubes,
\[d_{n}=d_{n-1}\cdot \frac{b}{b+\sqrt{b}\sqrt[3]{x_{n-1}}+x_{n-1}^{2/3}}\]
Now let $m=\min \{ a,b^{3/2}\}$. We can show $x_{n}$ bounded below by $m$ (easily verified via induction). Therefore,
\[d_{n}\leq |b^{3/2}-a|\left( \frac{b}{b+\sqrt{b}\sqrt[3]{m}+m^{2/3}}\right)^{n-1}\]
and the upper bound on $d_{n}$ tends to 0, so $d_{n}$ does as well. Therefore, $x_{n}$ converges to $b^{3/2}$.

Overall, we have that $x_{n}$ converges if and only if $a=0$ or $b\geq 0$.
[/hide]


probability
floor function
LaTeX
symmetry
number theory
divisibility tests

A special box is designed. It has ten sides. The numbers from 1 to 10 are written on the ten sides of the box in one fixed order.( Just like how the numbers on a dice are in one configuration only).
Now the box is rolled six times. The number on the top face  of the box when it is rolled is the number we choose. What is the probability that the numbers we choose (i.e. numbers on top face) on each of 2nd , 3rd,4th,5th and 6th tries is at least 1 greater than the number we choose on the previous try. That is the number on 2nd try should be greater than number on 1st try, number on 3rd try should be greater than  the number on the 2nd try and so on till the last try.

[quote="befuddlers"]A special box is designed. It has ten sides. The numbers from 1 to 10 are written on the ten sides of the box in one fixed order.( Just like how the numbers on a dice are in one configuration only).
Now the box is rolled five times. The number on the top face  of the box when it is rolled is the number we choose. What is the probability that the numbers we choose (i.e. numbers on top face) on each of 2nd , 3rd,4th,5th and 6th tries is at least 1 greater than the number we choose on the previous try. That is the number on 2nd try should be greater than number on 1st try, number on 3rd try should be greater than  the number on the 2nd try and so on till the last try.[/quote]

Um, is it just $ \frac{\binom{10}{6}}{6^{10}}$?

Tram tickets have six-digit numbers(from 000000 to 999999). A ticket is called lucky if the sum of its first three digits is equal to that of its last three digits. A ticket is called medium id the sum of all its digits is 27. Let A and B denote the number of lucky tickets and medium tickets, respectively. Find A - B.

The first part requires the condition that each face has an equal probability of being the top face.  Perhaps a more interesting question is to actually exhibit an example of such a box.  (You can do [i]twenty[/i] faces - that's an icosahedron.)

[quote="1076207"]That first part doesn't quite work. You're overcounting for something like 164835, where it goes to 11, not 0. I think.[/quote]
Yes that is right. Alternating differences of $ 11$ and $ 22$ work for divisibility by $ 11$ too.
For example consider the number $ 616000$. This number is divisible by 11 but no permutation of the digits will satisfy the condition that the sum of the first $ 3$ digits equals the sum of the last $ 3$ digits.

[hide]Let $ P_n$ be the number of ways we can have a 3 digit number that sums to $ n$.

Then, $ A \equal{} {P_{0}}^{2} \plus{}{P_{1}}^{2} \plus{} ... \plus{} {P_{27}}^{2}$ (sorry I've forgotten how to LaTeX summation signs)
and $ B \equal{} P_{0}P_{27}\plus{}P_{1}P_{26}\plus{}...\plus{}P_{27}P_{0}$

But since $ P_n$ is the coefficient of $ x^n$ in $ {(1\plus{}x\plus{}x^2 \plus{}...\plus{}x^9 )}^3$, we have by symmetry $ P_n \equal{} P_{27\minus{}n}$

Therefore, $ A\minus{}B\equal{}0$[/hide]

Befuddlers, are you using [i]A path to combinatorics for Undergraduates[/i]....... :D        :rotfl:  [/i]

omg, it's a fusion of Arvind Narayan and my extremely annoying ex-Algebra 2 classmate Nakkul Sreenivas!

Either that or its just some person from India.

[quote="iniquitus"][hide]Let $ P_n$ be the number of ways we can have a 3 digit number that sums to $ n$.

Then, $ A \equal{} {P_{0}}^{2} \plus{} {P_{1}}^{2} \plus{} ... \plus{} {P_{27}}^{2}$ (sorry I've forgotten how to LaTeX summation signs)
and $ B \equal{} P_{0}P_{27} \plus{} P_{1}P_{26} \plus{} ... \plus{} P_{27}P_{0}$

But since $ P_n$ is the coefficient of $ x^n$ in $ {(1 \plus{} x \plus{} x^2 \plus{} ... \plus{} x^9 )}^3$, we have by symmetry $ P_n \equal{} P_{27 \minus{} n}$

Therefore, $ A \minus{} B \equal{} 0$[/hide][/quote]

Nice solution. Alas, I thought I found a nice bijection...

BTW, the latex for summation is \sum_{i=0}^{n}{f(i)} to produce $ \sum_{i\equal{}0}^{n}{f(i)}$.

Dude. U stink Sitan. I will shoot you when i come down there again.

But let's revive it anyway! So here's my counting/probability question that I thought of before, but have absolutely no idea how to do mostly because it's extremely ugly. Good luck to anyone who tries, and keep in mind that you don't have to (and probably shouldn't) simplify.

You are playing the game Eskimo Rat Screw with two other people. A $ slap$ is when there is a pair (when two cards of the same number are played on top of each other) or a sandwich (when two cards of the same number are played with one card in between the two) is played. At the start of the game, the 52-card deck is split as evenly as possible and you go first, with the three of you taking turns. After the whole game is played (all cards have been played), what is the probability that there is at least one $ slap$?

And ... what about false slaps? Do we assume ya'll play perfectly?

I think we do. Unless you want to try putting in false slaps....
There are no royal rules. It's basically the whole deck is just put in random order and played out. Also, I suppose it should say "all cards have been played exactly once."

I thought it was Egyptian rat screw...o.0

In cordial company, we say that S stands for slap. [...I can't think of any time I've used s-w as a bad word, even like that. And I'm certainly not going to torture rats like that.]

[quote="Arvind_sn"]<MEAN STUFF ABOUT SITAN IN HIS SIGNATURE>[/quote]\

Ooooooooooooohhhhhhhhhhh....  So that's what it means.................................

Wait, hey!

made have you some calculation mistakes in [quote=timwu][hide]Note there is a bijection between the lucky integers and multiples of 11, so $ A \equal{} \lfloor 999999/11 \rfloor \plus{} 1 \equal{} 90910$.

Note there is a bijection between the medium integers and integer 6-tuples satisfying
$ x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \plus{} x_5 \plus{} x_6 \equal{} 27 \text{ } (0\leq x_i\leq 9)$
By principle of inclusion and exclusion, $ B \equal{} \binom{32}{5} \minus{} \binom{6}{1}\binom{22}{5} \plus{} \binom{6}{2}\binom{12}{5} \equal{} 55252$.

Thus, $ A \minus{} B \equal{} 35658$.
[/hide][/quote]

B


I'm prety sure it all revolves around the fact that if we put $x_n=a_n-n$ we get $x_{m+n}+x_{m-n}-1=\frac{x_{2m}+x_{2n}}2$.

I didn't do anything else yet because I didn't have any other obvious ideas :).

Come to think of it, after putting $n=1$, a very simple induction shows us that $x_n=n^2+1$, so $a_n=n^2+n+1$.

I guess the answer is $4018021$. That's not that nice, is it? :)

We don't really need $x_n$, but the expression for $a_n$ would have been harder to guess from the beginning.

Let $y_n = a_n - n - 1$.  Then the given recurrence becomes $y_{m+n} + y_{m-n} = \frac12 (y_{2m} + y_{2n})$.
Setting $m=n$: $y_{2n} + y_0 = y_{2n}$ so $y_0 = 0$.  Further, the above substitution implies that $y_1 = 1$.
Setting $(m, n) = (1, 0)$: $2y_1 = \frac12 (y_2+y_0)$ implying $y_2=4y_1=4$.  Finally, setting $m = n+2$ and $z_n = y_{2n}$ gives, $y_{2n+2} + 4 = \frac12 (y_{2n+4} + y_{2n})$ or $z_{n+1} + 4 = \frac12 (z_{n+2} + z_n)$.

We prove by induction that $z_n = 4n^2$.  $z_0 = y_0 = 4(0)^2$ and $z_1 = y_2 = 4 = 4(1)^2$, providing our base case.  Now assume that $z_n = 4n^2$ and $z_{n+1} = 4(n+1)^2$ for some $n \geq 0$.  Then $z_{n+2} = 2z_{n+1} + 8 - z_n = 8(n+1)^2 + 8 - 4n^2 = 4n^2 + 16n + 16 = 4(n+2)^2$ completing the induction.

Therefore, $a_{2n} = y_{2n} + 2n + 1 = z_n + 2n + 1 = 4n^2 + 2n + 1$.  Setting $n = 1002$ gives the required result, namely $a_{2004} = 4(1002)^2 + 2005$.

[quote="Valentin Vornicu"]The sequence $\{a_n\}_{n\geq 0}$ of real numbers satisfies the relation:
\[ a_{m+n} + a_{m-n} - m + n -1 = \frac12 (a_{2m} + a_{2n})  \]
for all non-negative integers $m$ and $n$, $m \ge n$. If $a_1 = 3$ find $a_{2004}$.[/quote]

This problem has appeared in many other Olympiad and Training camps

by taking m+n=x, m-n=y you obtein a relation for all x and y st x>=y and x and y are with same parity! after that put b_n=a_2n!
i got the same expression that found michael
bye

Problem1: m=0, n=0 then  a_0=1      ,      n=0 then a_2m=4a_m - 2m - 3

                m=1         then  a_2=7      ,      m=2 then a_4=21

                m=2, n=1         then  a_3=13

Let's prove that a_n=n^2 + n + 1 by induction...

1) a_0=1 and a_3 are known.
2) Suppose k <= m
3) Using a_2m = 4a_m - 2m-3 show a_2m=4m^2 + 2m +1
4) Using the substitution n=1 prove that a_(m+1)=(m+1)^2 + m + 2

                                        Therefore a_2004=2004^2 + 2005
fatih sahin

Hmm... I turned this problem into an F.E:
Let f(n) = $a_n$.
$f(m+n) + f(m-n) - m + n - 1 = \frac{f(2m) + f(2n)}{2} $, same conditions apply for m and n.

Let $P(m,n) = f(m+n) + f(m-n) -m + n - 1$.

P(m,m) $ \Rightarrow $ f(2m) + f(0) -1 = f(2m) $\iff$ f(0) = 1.
P(m,0) $ \Rightarrow f(m) + f(0) - m - 1 = \frac{f(2m) + f(0)}{2} \Rightarrow f(2m) = 4f(m) - 2m - 3$
From this we get that $f(2) = 7$.
In the hopes of a linear recurrence, we plug in P(m,1) to get:
$f(m+1) + f(m-1) - m = \frac{f(2m) + 7}{2} = \frac{f(2m)}{2} + \frac{7}{2}$
Plugging in our result from P(m,0) gives:
$f(m+1) = 2f(m) - f(m-1) + 2 \iff f(m+2) = 3f(m+1) -3f(m) +f(m-1)$

So we use the characteristic equation combined with the fact that f(0) = 1, f(1) = 3, f(2) = 7 to get the answer.

Let $m=n=0$ so $a_0=1.$  Now, let $n=0$ so $a_{2m}=4a_m-2m-3.$  Let $n=m$ in the past equation to get $a_{2n}=4a_n-2n-3.$  Substituting these in the original equation gives $$a_{m+n}+a_{m-n}+2n+2=2a_m+2a_n.$$  Let $n=1$ so $a_{m+1}=2a_m-a_{m-1}+2.$  We can prove, by strong induction, that $a_n=n^2+n+1.$  The base case $n=0,1$ are already computed.  Assume that for any nonnegative integer $k$, $a_k=k^2+k+1$ and $a_{k-1}=(k-1)^2+(k-1)+1.$  Hence, $$a_{k+1}=2a_k-a_{k-1}+2=2(k^2+k+1)-(k^2-k+1)+2=k^2+3k+3=(k+1)^2+(k+1)+1,$$ and by strong induction, it holds for all nonnegative integers.  In particular, $a_{2004}=\boxed{2004^2+2004+1=4018021}.$

Making stuffs unnecessarily compilcated,
[hide=lawl]\begin{align*}
&a_{m+n} + a_{m-n} -m +n -1 = \frac{1}{2} (a_{2m} + a_{2n})\\
\implies & a_m + a_n = 2a_{\frac{m+n}{2}} + 2a_{\frac{m-n}{2}} - m+ n -2
\end{align*}
Now it is easy to find that $a_0 = 1.$ Finding some other values, we find a pattern that $a_n = n^2 + n + 1.$ We will prove this with strong induction.
	[b]base case[/b]	Trivial.
\begin{align*}
	&a_n + a_{n+2} =2a_{n+1} + 2a_1 - n -2 +n -2 \\
	\implies & n^2+n+1+a_{n+2} = 2a_{n+1} +2 \\
\implies &n^2+n+1+a_{n+1} = 2a_n + 2 \\
	\implies &a_{n+1} = n^2+n+3 = (n+1)^2+(n+1)+1 \\
\end{align*}
Hence, we are done getting $a_{2004} = \boxed{4018021}.$[/hide]


trigonometry
function
logarithms
inequalities
inequalities unsolved

Let $ a,b,c>1$ be such that $ a\plus{}b\plus{}c\equal{}abc$.

Prove that $ (a^2\minus{}1)(b^2\minus{}1)(c^2\minus{}1)\leq 8$.

We can make the substitution $ a \equal{} \tan x_{1}, b \equal{} \tan x_{2}, c \equal{} \tan x_{3}, x_{1} \plus{} x_{2} \plus{} x_{3} \equal{} \pi$ (because $ \tan(\pi \minus{} x \minus{} y) \equal{} \frac {\tan x \plus{} \tan y}{\tan x \tan y \minus{} 1}$)

define the function $ f(x) \equal{} \ln({\tan{x}^2 \minus{} 1})$ (it's well defined because a,b,c>1).
$ f''(x) < 0$ (for $ \tan x > 1$), hence f is concave (I -did- calculate the derivative...) for $ \frac {\pi}{2} > x > \frac {\pi}{4}$. by Jensen:

$ \sum_{i \equal{} 1}^{3} \ln({\tan{x_{i}}^2 \minus{} 1}) \le 3f(\frac {pi}{3}) \equal{} \ln{8}$
if we use the monotonicity of $ e^x$, we get: $ \prod_{i \equal{} 1}^{3} (\tan{x_{i}}^2 \minus{} 1) \le 8$
or $ (a^2 \minus{} 1)(b^2 \minus{} 1)(c^2 \minus{} 1)\leq 8$

equality case: $ a \equal{} b \equal{} c \equal{} \sqrt {3}$

[quote="discredit"]Let $ a,b,c > 1$ be such that $ a \plus{} b \plus{} c \equal{} abc$.

Prove that $ (a^2 \minus{} 1)(b^2 \minus{} 1)(c^2 \minus{} 1)\leq 8$.[/quote]
This inequality is true for all positive $ a,$ $ b$ and $ c$ such that $ a \plus{} b \plus{} c \equal{} abc.$ :wink:
By the way, a maximal $ k$ for which the inequality
\[ (a^2 \minus{} k)(b^2 \minus{} k)(c^2 \minus{} k)\leq (3\minus{}k)^3\]
holds for all positive $ a,$ $ b$ and $ c$ such that $ a\plus{}b\plus{}c\equal{}abc$ is $ k\equal{}1.$

But bambaman said the function is concave only in that range. That's why I asked...

I have a nice proof which is used pqr,there let $ \sum\ ab$ =$ \frac {p^{2}\minus{}q^ {2}}{3}$ and $ p\equal{}r$,the proof is not hard

[quote="Allnames"]I have a nice proof which is used pqr,there let $ \sum\ ab$ =$ \frac {p^{2} \minus{} q^ {2}}{3}$ and $ p \equal{} r$,the proof is not hard[/quote]

It seems strange. Can you show it?

because i have a short time so i cant post it now,to prove it you can write this ineq in the form with pqr i have just said,
then you can simplely find $ r\leq\ (p \minus{} 1)^{2}(p \plus{} 2q)$,the end is easy and you can finish by yourself,i believe that

[quote="discredit"][quote="bambaman"]hence f is concave (I -did- calculate the derivative...) for $ \frac {\pi}{2} > x > \frac {\pi}{4}$.[/quote]

Then what if $ x$ is not in this range?[/quote]

It is concave in every interval of the form $ (\frac {\pi}{4} \plus{} \pi k,\frac {\pi}{2} \plus{} \pi k)$ or $ ( \minus{} \frac {\pi}{2} \plus{} \pi k, \minus{} \frac {\pi}{4} \plus{} \pi k)$. I chose k=0 (it doesn't matter, tan is periodic), and I wrote $ x > \frac {\pi}{4}$ because $ a,b,c > 1$, and we want $ \ln(\tan x^{2} \minus{} 1)$ to be defined, i.e $ tan x^{2} > 1$  (we can choose $ x_{1},x_{2},x_{3}$ to be in that interval)

$ \ln(\tan^{2}(x) \minus{} 1)' \equal{} \frac {2 \tan x \frac {1}{\cos^{2}(x)}}{\tan^{2}(x) \minus{} 1} \equal{} \frac {2\tan x}{ \minus{} \cos 2x}$
$ \frac {2\tan x}{ \minus{} \cos 2x}' \equal{} \frac {\frac { \minus{} 2\cos 2x}{\cos^{2}(x)} \minus{} 4\tan x sin(2x)}{\cos^{2}(2x)} < 0$ iff the numerator is negative, iff $ \sin^{2}(2x) \plus{} \cos(2x) > 0$ or $ a \plus{} 1 \minus{} a^2 > 0, a \equal{} \cos(2x)$. Oh no. It doesn't hold... I will check my computations again :o


When I try artofproblemsolving.com I get the following:

[quote]
log.enquisite.com

This server collects data. There is really nothing to see here.

If you'd like more information, please visit http://www.enquisite.com.
[/quote]

When I use:

http://www.artofproblemsolving.com/index.php

it's fine.

You're misspelling the site's name I guess.

No. I may have typed it wrong in the report, but the bookmark is the one I've been using for months.

I've also cleared my cache in case it was on my side.

Weird, I cleared my cache again and now it seems fine. Sorry for the bother.

Tom

Don't think so. I'm pretty aggressive about running anit-spyware programs. The other interesting thing is that it seems to be fine today.


algebra
polynomial
linear algebra
matrix
number theory proposed
number theory

Hello world!
I'm trying to find a bijection between Lyndon words ot length $ n$ on an alphabet $ A\equal{}\{1,\cdots,q\}$ (lyndon words can be seen as aperiodic necklaces of length $ n$ with $ q$ colours) and the set of irreducible polynomials of degree $ n$ in $ F_{q}[x]$...

How can i transform a lyndon word in a polynomial?

Thanks in advance.

The bijection is as follows.  Fix an irreducible polynomial $ f$ of degree $ n$, so that $ \mathbb{F}_q[x]/(f(x)) \simeq \mathbb{F}_{q^n}$.  Then any element has a unique representation as

$ a_0 \alpha \plus{} a_1 \alpha^p \plus{} ... \plus{} a_{n \minus{} 1} \alpha^{p^{n \minus{} 1}}$

for unique $ a_0, ... a_{n \minus{} 1}$ where $ \alpha, \alpha^p, ...$ are the roots of $ f$.  The Frobenius map cyclically permutes the coefficients $ a_i$, and an element of $ \mathbb{F}_{q^n}$ is the root of an irreducible polynomial of degree $ n$ if and only if it has order $ n$ under the Frobenius map, hence if and only if $ a_0, ... a_{n \minus{} 1}$ is aperiodic, in which case one can identify the corresponding Lyndon word with the minimal polynomial.

Thanks for your reply.
Maybe, i could understand better ( :blush: ) your post by means an example.
Let $ n\equal{}4$ be the length of the words on an alphabet with $ q\equal{}2$ symbols.
I know that the lyndon words are:
$ L_1\equal{}0001$
$ L_2\equal{}0011$
$ L_3\equal{}0111$
Let $ f\equal{}x^4\plus{}x\plus{}1$ be an irreducible polynomial of degree $ n\equal{}4$.
Now, how should I continue?

Any element of $ \mathbb{F}_2[x]/(f(x))$ is uniquely a linear combination $ a_0 x \plus{} a_1 x^2 \plus{} a_2 x^4 \plus{} a_3 x^8$ where $ a_i \in \mathbb{F}_2$.  The bijection I described sends $ L_1$ to $ x^8$, $ L_2$ to $ x^4 \plus{} x^8$, and $ L_3$ to $ x^2 \plus{} x^4 \plus{} x^8$.  I claim that the minimal polynomials of each of these are precisely the irreducible polynomials of degree $ 4$.

Be patient, please! :blush:
I do not understand the method for transforming the three lyndon words into the three irreducible polynomials of degree four...
I know that the three lyndon words are:
$ L_1\equal{}0001$
$ L_2\equal{}0011$
$ L_3\equal{}0111$

Furthermore, the three irreducible polynomials are:
$ p_1(x)\equal{}x^4\plus{}x\plus{}1$
$ p_2(x)\equal{}x^4\plus{}x^3\plus{}1$
$ p_3(x)\equal{}x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1$

O.K.
Now, which is the method to transform $ L_i$ in $ p_j(x)$?

Thanks a lot!

Do you know how to compute minimal polynomials?  Here one is lucky because the minimal and characteristic polynomials coincide, so you can write down the matrix of left multiplication by any of the three numbers I wrote down and compute a determinant.

Note that this bijection is not canonical; it depends on which polynomial we start with.

:(
Which is the matrix?
Could you explain me the procedure?
Maybe, it is better (easy) in the case $ n\equal{}3$?
Thanks very much.

How much linear algebra do you know?  If you can't write down minimal polynomials perhaps you should take a closer look at field theory before thinking about this question.

I can write down the minimal polynomials... I don't understand which is the matrix associated at this problem...
 :wallbash:

Left multiplication $ a \mapsto xa$ acts on $ \mathbb{F}_2[x]/(f(x))$ as a linear transformation.  In the basis $ \{ 1, x, x^2, x^3 \}$ the matrix of this linear transformation is particularly easy to describe (it's called the [url=http://en.wikipedia.org/wiki/Companion_matrix]companion matrix[/url] of $ f(x)$).  This matrix contains all of the information about $ x$; the determinant of this matrix is the norm of $ x$, its trace is the trace of $ x$, and its roots are the conjugates of $ x$.  The characteristic polynomial of this matrix is the characteristic polynomial of $ x$, and for the numbers I wrote down the characteristic and minimal polynomials coincide, so one can compute minimal polynomials by finding characteristic polynomials of certain matrices.

Thanks for your patience!

A easier example for me...
$ n \equal{} 3$,$ q \equal{} 2$

$ L_1 \equal{} 001 \mapsto x^4$ that is $ a_0x^0 \plus{} a_1x^1 \plus{} a_2x^2 \plus{} a_3x^3 \plus{} a_4x^4$ that is $ (0,0,0,0,1)$
$ L_2 \equal{} 011 \mapsto x^4 \plus{} x^2$ that is $ (0,0,1,0,1)$

Now, the companion matrix is:
$ \left[\begin{matrix}0 & 0 & 0 & \minus{} a_0 \\
1 & 0 & 0 & \minus{} a_1 \\
0 & 1 & 0 & \minus{} a_2 \\
0 & 0 & 1 & \minus{} a_3\end{matrix}\right]$

For $ x^4$ is:
$ \left[\begin{matrix}0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0\end{matrix}\right]$.

For $ x^4 \plus{} x^2$ is:
$ \left[\begin{matrix}0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & \minus{} 1 \\
0 & 0 & 1 & 0\end{matrix}\right]$.

Now, i must compute the minimal polynomial... Are the matrices correct? :oops:

What irreducible polynomial did you pick?

So confused.
I fix $ f\equal{}x^3\plus{}x\plus{}1$.
Dividing by $ f$, I obtain:
$ x^4 \equiv x\plus{}x^2$
$ x^2\plus{}x^4 \equiv x$

Now, should I compute the associated companion matrices for $ x\plus{}x^2$ and $ x$?

Hence, we have two $ 2\times2$ matrices... :(

The basis is $ \{ 1, x, x^2 \}$, so you should get $ 3 \times 3$ matrices.

Hence
for $ x$:
$ \left[\begin{matrix}0 & 0 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0\end{matrix}\right]$

and for $ x\plus{}x^2$:
$ \left[\begin{matrix}0 & 0 & 0\\ 1 & 0 & 1\\ 0 & 1 & 1\end{matrix}\right]$

Is it correct?
Now I should compute the minimal polynomial.

For
$ \left[\begin{matrix}0 & 0 & 0 \\
1 & 0 & 1 \\
0 & 1 & 1\end{matrix}\right]$
the minimal polynomial seems to be $ \alpha^3 \plus{} \alpha$, but this is not an irreducible polynomial... Where is the error? :cursing:  :help:


Let $m\geq n$, $B_{m}={1, ..., m}$, $B_{n}={1,..., n}$. The mappings from $B_{m}$ onto $B_{n}$ are called surjections. Find the number $S(m,n)$ of surjections from $B_{m}$ onton$B_{n}$

see http://en.wikipedia.org/wiki/Stirling_number  :wink:


calculus
integration
trigonometry
calculus computations

Calculate $ \int_0^1 dx \int_0^x dy \int_0^y \frac{\sin{z}}{(1\minus{}z)^2} dz$

I would change the order of integration.  Maybe integrating with respect to $ y$ first.


Let $a,b,c,d,x,y,z,t$ be positive real numbers such that $ax+by+cz+dt=xyzt$. Prove that
$x+y+z+t > \frac{4}{3}\left(\sqrt{1+3\sqrt{a+b}+3\sqrt{a+c}+3\sqrt{b+c}+3\sqrt{b+d}+3\sqrt{c+d}}-1\right)$

[quote="The soul of rock"]A same (easier) problem
Let $a,b,c,x,y,z$ be positive real numbers such that $ax+by+cz=xyz$. Prove that
$x+y+z>\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1899

So how about the first inequality? No one interested in?

No idea?
P/s: Sorry for spam but I don't want it be forgot!


geometry
calculus
counting
distinguishability
HMMT
probability
linear algebra

Um... situation

I start middle school in 7th grade. So I skipped Pre-Algebra and took Algebra 1 regular (not honors, schedule conflict)

In 8th grade (now), I am assigned to Geometry Honors. But I wanted to take Precalc. But the school didn't let me skip Geo Honors. So I had to take Geometry Honors, even though I am doing Alg3/Interm C&P/ Precalculus outside of school. So geometry is really boring right now.

Well... looks like I have to wait to 10th grade to take Calculus, even though I know lots of people who are in the same phase of AoPS and are taking Precalc or Calc right now.

Is there a place to put this in a college application? (better written, of course :lol:)

[quote="qntty"]well if you take AOPS calculus next year and take the AP calculus exam, you'll be able to take more advanced courses (linear algebra and calc III) at a community college in 10th grade and it wont make a whole lot of difference. Although there is space on many applications to explain special circumstances, I wouldn't worry about it. To a college, the stuff you did in middle school will make little or no difference. Even if you had limited options in high school, it wouldn't be a huge deal. An admissions officer cares about how you took advantage of the opportunities that you had, not that you took class X in grade Y. Compared to the other people in your precalculus/Alg3/interm C&P classes, your situations are almost indistinguishable as long as you both get good grades in the classes.[/quote]

So, you're basically saying that it doesn't matter as long as I get good grades in my classes and do good on contests?

Thanks :)

[u]A few comments, in no particular order:[/u]

0. I don't think I would call that a "limited math opportunity". I think it's very, very normal for a public school to be skeptical about a freshman taking calculus. Nor does anyone expect you to take it freshman year.

1. If you were to complain anywhere on your college admissions about your school not allowing you to take calculus freshman year, it will probably be negatively received. If you don't understand why, you will after reading points (2) and (4).

2. Law of diminishing marginal returns: admissions will not be any more impressed by you taking calculus in freshman year than in sophomore year.

3. Take the good out of your situation. A good grade will help stabilize your GPA. An easy class will take off stress and give you a light start, allowing you more time to explore extracurricular options and to get comfortable with your high school experience. It's a great chance to expand your horizons and become aware of opportunities available for smart high school students.

4. If you know the material, you can still apply it out of class to projects (such as science/engineering fairs), math competitions (HMMT calc round, applied math competitions such as HiMCM and Moody's), or just dabbling with other subjects whose equations are written in the language of calculus.

5. Calculus isn't the holy grail. Consider exposing yourself to linear algebra and probability theory.

Interesting position you're in since I happen to be a freshmen taking calculus.

@TZF

AT -3: I would argue that learning more material and expanding one's knowledge outweighs the impact of a couple of percentage points on a report card. To me it's not really that stressful because of the scope of the class. I personally find it more enriching when I go through the content more deeply by using an alternative text such as Spivak's Calculus.

AT- 4: This is certainly true but being able to complete coursework at your own pace opens more opportunities for the interested student. I even feel that being exposed to more mathematics at a younger age gives me more information to make a decision on which career I want to go into. Even a cursory glance will prove to be useful in the future.

Of course none of this should be a compromise for depth, but sometimes an inquisitive student may want to cover more material.

My two cents: If you are interested and your school offers such an opportunity, then take it. If not then you can always study mathematics independently at your own inclination.

In terms of your current situation, I'm going to go along with mathemagician and recommend using an alternate text for geometry - go through Geometry Revisited, for example.

[b]@ mathemagician1729:[/b] I agree with what you've said. My main point (which addresses the OP's question) is that including a complaint about this 'lack of opportunity' will likely hurt the OP's application directly, and will definitely hurt indirectly due to space wasted.

The application should focus on positive things. This means focusing on opportunities taken rather than opportunities lost. A clause expressing that the OP's education was somehow limited by the system is certainly not along those lines, and will probably give the reader a negative impression of the OP's character. In my previous post, I gave an example of some possible opportunities to take that are probably a better use of the OP's time than a high-school level class on calculus.

My other main point is that it's not at all important to have taken calculus [i]from the high school on paper by freshman year[/i] -- sophomore year is just as good. And if there's any time or stress saved as a result, it can be put to good use.

In any case, someone who is advanced enough to be ready for calculus in freshman year is probably going to learn much more from exploration and inquisition than from the class itself, so the educational value-added of taking it at school is zero, and the total value-added is close to zero.


HMMT Mafia ARML quadratics trigonometry inequalities algebra

So who's going? I know it's early, but hey! It's a great tournament (probably becuz it has pizza  :D ). Also it's my 2nd favorite (after JV State, duh.)

I like it. I want to stop Harrison from winning Varsity 4 straight yrs (as if that is possible), but Science Olympiad is on the same day.  :(

Harrison won't win this year. And neither will Billy or I. The three of us will be at HMMT. This will be my third year of high school not being able to go to GA Tech. :(

im going, although i will do terribly.  my math tutor gave me some copies of previous jv tests, they were so hard.
 :(

And like half of my school just went to 11th grade, so I'm one of the few on JV  :D

Is Claus (haha i spelled it right this time) coming? Also, is he on AoPS? If so what's his username?

And I'm assuming Miles is coming? Or is he also going to HMMT? Did he go last year and stuff?

NICE! no harrison? people have a chance now....

Yes, Claus will be there as will Slok. Henry will be there on varsity. And Howard, it's really funny that you are cheering Harrison being gone!  :D Maybe we should take up a collection to send YOU to HMMT!

My school isn't signed up for this but it's one I really want to make it to. If worst comes to worst I'm pretty sure I can just sign up as a walk-on individual.

Though I'm not sure on the rules regarding this, but if any other schools need a substitute or something and I'm allowed  to fill it I'd be more than happy to - just make sure to let me know. It's always more fun to have a group. Of course if there's a rule that you have to represent your own school (even if they're not participating) then it can't be helped.

i'm pretty sure you actually have to attend a school to be a member of their team. i'm not sure about the deadline for registration, but i think it may have already passed.

Hmm, looking forward to meeting Slok, i don't think i've seen him before...and yay harrison's gone that makes me one place up!!! yeah!!! that's 1/200 (or less) more chance of me winning!!! XD

But Miles is going (right?) so there's no chance for first. But wait, he's varsity, right? OK, I have a chance... :D

How much do you need in JV Proof to place?

Oh My Gods...

Dude I forgot about Science Olympiad being on the same day!!

Gasp I think I will go weep in a corner now for not being able to go to Georgia Tech.

I won last year in the non-Harrison and non-senior division  :D but that doesn't really count for anything  :P

Okay time to go and weep  :(

Hm, it's a shame about having to be a student at the school you'd represent, but expected and not surprising. And yeah, registration is closed - but they do offer some walk-on spots, so I'll be there early to get in.

[quote="captcha000"]blegh...so karth wont be there either? i got second in the non-harrison, non-karth, non-senior division =P[/quote]

I got 1st in that division.  :)

I tied for first among those who weren't there (unfortunately, I didn't get a trophy, presumably because it was a more than six billion person tie).

[quote="captcha000"]i almost got a perfect on the written. although i guess in the end karthik's protest doesn't really change anything in terms of the final placings[/quote]

Haha, we tied then... 19/20?

My thoughts:

1. Easy... the answer's no
2. What the crap... I got the polynomial and then didn't solve for $ a$ because I thought that $ a$ had some random nonsensical value.
3. The one with $ n!!$ was ridiculously easy... induction? I'm really interested in seeing a combinatorial proof for this.
4. One-liner... Euler's formula.
5. Kinda confusing, but got it at the last second.

Dang it if I'd gotten #2, my score would've been higher. Sob.

Heh, I did a combinatorial/algebraic proof for number 3. It wasn't very elegant or pretty, but it was kind of cool. I can post it if you'd like, though being that I don't know TeX it will not be the most beautiful thing to look at.

Edit: And actually I'd need the problem again because I don't remember it in entirety.

When do u think the GA tech site will be updated w/ all of the stuff (tests, pics, etc.)?

[quote="Benjamin Hu"]I feel [b]really[/b] bad about getting two places lower on [b]JV[/b] than last year.  :(  Even though I got a perfect score on written (like Sitan), I got beaten by Yubo, which makes me feel like an idiot. :(
(sigh)There's always next year (in varsity, of course)... I hope.[/quote]

I don't know what i should do next year, varsity or jv. probably jv...anyway there's still jv state :) If Claus is coming, and if Yubo is too, four way battle of the eighth graders!!! XD

Actually, technically, we'll be 9th graders by that time.

No, like battle of eighth graders this year at jv state.

btw the phrase "battle of eighth graders" was coined by Mr Fulton... :ninja:

Well, I may or may not have coined the phrase, but it's reasonable obvious to me that we have a very talented group of 8th graders (and younger!) I'm still looking forward to Dickerson vs. River Trail at MATHCOUNTS!

Yeah, me too...I never got to meet Slok though. Is he a middle schooler?

[quote="gtf"]Yep, another 8th grader. I can't wait to see that ARML team when all of you guys are seniors![/quote]

Haha, personally, I wouldn't be surprised if we see an IMO member on our team, or at least a few black MOPpers.

OK, extend Battle to 5 participants...yay.

Yeah, it's gonna be pretty fun. Unless it's still at UGA, in which case it will be a one-day tournament and it will feel like UGA. Except with more people. And a longer acronym. :ninja:

How did Slok do at GT? He went, right?

No, i didn't go to tech because of Science Olympiad :(

... But I got two medals there (Science Olympiad)!  :)  :)

There's science olympiad in middle school?  Hmm...

Yep. Dickerson also qualified to go to state Science Olympiad, but it happens to be on 3/29/08, which (if I'm not mistaken) is the same day as JV State....

I think you see my predicament.

Dude...go to JV state so i can finally meet you...XD


Find all polynomials $ P\in \mathbb {R}[x]$ such that \[ P(x^2\plus{}2x\plus{}1)\equal{}(P(x))^2\plus{}1\] for all real $x$.

[quote="mathVNpro"]Find all polynomial $ P\in \mathbb {R}[x]$ such that: $ P(x^2 \plus{} 2x \plus{} 1) \equal{} (P(x))^2 \plus{} 1$[/quote]

Hello MathVNpro !

First, we can see that we have no constant solution and that the only degree $ 1$ solution is $ P(x) \equal{} x \plus{} 1$

Then let $ Q(x) \equal{} P(x) \minus{} (x \plus{} 1)$ : $ Q((x \plus{} 1)^2) \equal{} P((x \plus{} 1)^2) \minus{} (x \plus{} 1)^2 \minus{} 1$ $ \equal{} P(x)^2 \minus{} (x \plus{} 1)^2$ $ \equal{} Q(x)(P(x) \plus{} x \plus{} 1)$
So, if $ r$ is a real root of $ Q(x)$, $ (r \plus{} 1)^2$ is too a root of $ Q(x)$ and so $ Q(x)$ as infinitely many real roots.
So either $ Q(x) \equal{} 0$, either $ Q(x)$ is a polynomial whose degree is even with no real roots.
So, solutions $ P(x)$ different from $ x \plus{} 1$ have an even degree.

$ P((x \plus{} 1)^2) \equal{} P(x)^2 \plus{} 1$ $ \implies$ $ P((( \minus{} x \minus{} 2) \plus{} 1)^2) \equal{} P( \minus{} x \minus{} 2)^2 \plus{} 1$ $ \implies$ $ P(x)^2 \equal{} P( \minus{} x \minus{} 2)^2$ and so $ P(x) \equal{} P( \minus{} x \minus{} 2)$ since degree of $ P(x)$ is even.
So $ P(x) \equal{} H((x \plus{} 1)^2)$ for some polynomial $ H(x)$ and we have $ H(((x \plus{} 1)^2 \plus{} 1)^2) \equal{} (H((x \plus{} 1)^2))^2 \plus{} 1$ $ \forall x$
$ \implies$ $ H((x \plus{} 1)^2) \equal{} (H(x))^2 \plus{} 1$ $ \forall x\geq 0$
$ \implies$ $ H((x \plus{} 1)^2) \equal{} (H(x))^2 \plus{} 1$ $ \forall x$
And so $ H(x)$ is also a solution.
At the end of this descent processus, we get the unique degree $ 1$ solution $ x \plus{} 1$

And so the solutions are the polynomials $ P_n(x)$ defined as :
$ P_1(x) \equal{} x \plus{} 1$
$ P_{n \plus{} 1}(x) \equal{} P_n((x \plus{} 1)^2)$


inequalities
algebra
polynomial
geometry
pigeonhole principle
induction
number theory
\/closed

Two courses are now available starting this September.

We are re-offering the Intro to Number Theory class for those of you who couldn't join us this summer.  We are also offering an Intermediate Counting class that will pick up where Introdution to Counting left off.

The Intro to Number Theory class is suitable for those of you just getting started in problem solving, particularly those of you who are not familiar with modular arithmetic.

The Intermediate Counting class is more advanced, and is primarily appropriate for those who have already mastered the basics of counting such as combinations and permutations but have not yet progressed to inclusion-exclusion, more advanced combinatorics, or generating functions.

Suitability guides are available in the Online Classes section of our pages, or post here if you have any questions about the courses.

are the lesson parts of the couses held in the VIRTUAL CLASSROOM gonna be held the same as the math jams and such. along with all the other perks of the class?

The actual 18 hours of instruction will be similar to the Math Jams.  The main differences are

(1) The classes are longer - 90 minutes instead of ~60.
(2) The material is more 'cohesive'.  By this, I mean that the classes are designed to teach you a set of concepts, techniques, and methods of thinking in a much more rigorous and thorough manner than our Math Jams.  Our Math Jams are typically a collection of various problems from a single source - one problem may or may not lead to the next.  The courses are tightly organized and more focused (and more thorough) than the Math Jams.

As for the other differences, these courses have problem sets with solutions, and a message board strictly for the class for additional instruction.

When are more advanced classes going to be offered?  I have done well on various math tests, and I am looking more for things like classes on such harder subjects as group theory, or differental equations.

Can you help?

Possibly - we still haven't set our courses for the spring.  At some point we'll be teaching an inequalities class that will be pretty advanced, as well as a number theory course & advanced problem solving.  It may be a while before we get to those, though.  It's unlikely we'll be doing diffEQ's, as that's a subject that's pretty well covered elsewhere (and, frankly, not that terribly interesting from a problem solving perspective until a whole lot of basic stuff has been covered - or until the tools are applied to, say, physics).

You may want to take a look at the intermediate counting course - while it's not diffEQ, quite a bit of it will be a good deal harder than differential equations are.

can someone explain what is the difference between auditing a class and evaluating a class?

"Students opting for an 'Evaluated' enrollment can submit their work on the two challenge sets for detailed feedback from instructors. We also offer a less expensive 'Audited' enrollment for students who are cost conscious and are eager to use the class message board to seek feedback on their work. Feedback from the instructors on the work on the challenges is the only difference between the two options."

If you choose to evaluate, you get feedback from the instructors. If you audit the class, you simply get access to the message board. That's what I think.

That's basically right.  If you opt for Evaluated, you send in your work on the two Challenge sets to the instructor for thorough feedback.  If you opt for Audited, you don't.  We do provide full solutions for all the problem sets and you have access to a message board in this forum that's just for the given course where you can ask questions under either enrollment.

Other than the feedback on the two Challenge sets, the two enrollment options are entirely the same.

Aside from Counting and Number Theory, what other general fields will classes be offered for?

Over time we hope to hit tons of fields.  In the short term, we will have an algebra class in the fall (polynomials and other stuff).  Next year we'll likely roll out geometry and a couple other topics (maybe inequalities, or complex numbers, or another algebra class).  As the school grows, we'll probably branch into some more 'off the beaten path' topics like game theory or group theory, but it will likely be some time before we get to topics like those.

We also expect next year to develop a course on broad problem solving strategies, such as pigeonhole, invariance, induction, etc.


modular arithmetic
number theory unsolved
number theory

Let $p$ be a prime number that is greater than $3$. Show that there exist some integers $a_{1}, a_{2}, \cdots a_{k}$ that satisfy: \[-\frac{p}{2}< a_{1}< a_{2}< \cdots <a_{k}< \frac{p}{2}\] making the product: \[\frac{p-a_{1}}{|a_{1}|}\cdot \frac{p-a_{2}}{|a_{2}|}\cdots \frac{p-a_{k}}{|a_{k}|}\] equals to $3^{m}$ where $m$ is a positive integer.

Very nice problem!

We will show that taking all $a_{i}$ between $-\frac{p}{2}$ and $\frac{p}{2}$ congruent to $p \pmod 3$ makes the product a power of 3.

For example, for $p = 11$, we take $\{-4,-1,2,5\}$ as the $a_{i}$ so the product becomes \[\frac{15}{4}\cdot \frac{12}{1}\cdot \frac{9}{2}\cdot \frac{6}{5}= 3^{5}.\] Note that the denominator, the product of the $|a_{i}|$, is the product of all positive integers less than $\frac{p}{2}$ that are not multiples of 3. The numerator is the product of all the multiples of 3 from $\frac{p}{2}$ to $\frac{3p}{2}$. In other words, it is a power of 3 times all the positive integers from $\frac{p}{6}$ to $\frac{p}{2}$. However, this is exactly the numerator -- for every multiple of 3 between $\frac{p}{6}$ and $\frac{p}{2}$, we can divide out all powers of 3 to get a positive integer less than $\frac{p}{6}$, and conversely, every positive integer $\frac{p}{6}$ can be matched with a multiple of 3 between $\frac{p}{6}$ and $\frac{p}{2}$. Thus, we are left with a power of 3, as desired...

[i]A different solution:[/i]
[rule]
We will construct $a_i$ recursively. Define $b_0 = 1$. Now having defined $b_i$, choose a $a_{i+1} \in \left( \frac{-p}{2} , \frac{p}{2} \right)$ such that $\frac{p-a_{i+1}}{b_0}$ is a power of $3$ (it is easy to see such a $a_{i+1}$ exists, by looking at $b_i,3\cdot b_i,3^2 \cdot b_i,\ldots$) and define $b_{i+1} = a_{i+1}$. By induction we have that
$$\prod_{j=1}^{i+1} \frac{p-a_j}{|a_j|} = \frac{\text{power of }3}{b_{i+1}}$$
Now by PHP, we will have $b_m = b_n$ for some $m < n$. But then
$$\prod_{j=m+1}^n \frac{p-a_j}{|a_j|} = \text{power of }3 $$
So we are done. $\blacksquare$


How do you write long division in LaTeX?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=226670]Search[/url].  I assumed you wanted polynomial long division, as that's the topic of the Algebra 3 class today.


geometry
area of a triangle
Heron\u0027s formula

In triangle ABC, the three altitudes are $24, \frac{408}{25}$ and $\frac{204}{13}$. So, what are the lengths of AB, BC and CA?

[hide]
I think Stewart will do...
$h_a= \frac{2}{a}\cdot\sqrt{p(p-a)(p-b)(p-c)}$, where $p=\frac{a+b+c}{2}$.

Then solve the equations.[/hide]

[quote="luimichael"]a , b and c are unknowns. One equation is not enough to solve all of them.[/quote]
you can get another two equations...

$h_b=\frac{2}{b} \sqrt{p(p-a)(p-b)(p-c)}$
$h_c=\frac{2}{c} \sqrt{p(p-a)(p-b)(p-c)}$

Yes, we got three complicated equations.

My method of solution:
[hide] Apply Heron's Formula,the area of a triangle can be expressed in terms of the 3 altitudes.

Let $\Delta$ be the area of triangle, while $h_a, h_b$ and $h_c$ be the three altitudes.
Denote $heron(a,b,c)=\frac{\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}}{4}$ as the Heron's formula
for a triangle with sides a, b and c.
Note that the above expression is homogeneous of degree two in a,b and c.

$\Delta = heron(a,b,c)$
$\Delta = heron(\frac{2{\Delta}}{h_a},\frac{2{\Delta}}{h_b},\frac{2{\Delta}}{h_c})$
By the homogenity of the heron's formula,
$\Delta = {{(2\Delta)}^2}heron(\frac{1}{h_a},\frac{1}{h_b},\frac{1}{h_c})$
Hence,$\Delta = \frac{1}{4{heron({h_a}^{-1}},{h_b}^{-1},{h_c}^{-1})}$
$=\sqrt{({h^{-2}_a}+{h^{-2}_b}+{h^{-2}_c})^2-2({h^{-4}_a}+{h^{-4}_b}+{h^{-4}_c})}$

Knowing the area, it is then easy to determine the three sides from the altitudes.

[/hide]


group theory
abstract algebra
number theory
prime factorization
superior algebra
superior algebra unsolved

How do I show that $ \mathbb{Q}$ cannot be generated by a finite set of elements? I can show that it is not cyclic, but I can't seem to generalize my argument.

[hide="Hint"]What about the quotient $ \frac{1}{p}$ where $ p$ is a prime number which does [u]not[/u] occur in every of the prime factorization of the elements in a finite generating system?[/hide]

every finitely-generated subgroup of Q is cyclic (proof: directly or structure theorem, rank(Q)=1). thus it already suffices to show that Q is not cyclic.


How do I add a note to the each end of a page?

[quote="stevem"]Have a look at [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuidePack.php#fancy]fancyhdr[/url][/quote]
Thanks a lot for your help!

Edit: Here is another problem. I use the package this way:

\usepackage{fancyhdr}

\pagestyle{fancy}

and then:

\lhead{}
\chead{}
\rhead{}
\lfoot{}
\cfoot{}
\rfoot{The Text}

But after that, there a horizontal line at the top of every page. How can I get rid of the line?

[quote="leibniz"]You can handle the thickness of this line by using the command \headrulewidth, so to rid it :

[code]\renewcommand{\headrulewidth}{0pt}[/code][/quote]
Thank you very much! Now the problem is solved!  :D


function
induction
algebra proposed
algebra

Find all $ f: \mathbb R\to\mathbb R$ such that \[f(x)+f(y)=f(f(x)f(y))\] for all $x,y\in\mathbb R$.

[quote="MJ GEO"]Find all $ f(x)$ from $ \mathbb R\to\mathbb R$ such that $ f(x)\plus{}f(y)\equal{}f(f(x)f(y))$ $ \forall x,y\in\mathbb R$[/quote]

The equation may be written : Assertion $ P(x,y)$ : $ f(xy)\equal{}x\plus{}y$ $ \forall x,y\in f(\mathbb R)$

So, $ x,y\in f(\mathbb R)$ $ \implies$ $ x\plus{}y\in f(\mathbb R)$ and an immediate induction shows that $ nx\in f(\mathbb R)$ $ \forall x\in f(\mathbb R)$, $ \forall n\in\mathbb N$ (set of positive integers)

Then $ P(px,qy)$ $ \implies$ $ f(pqxy)\equal{}px\plus{}qy$
And $ P(pqx,y)$ $ \implies$ $ f(pqxy)\equal{}pqx\plus{}y$

And so $ pqx\plus{}y\equal{}px\plus{}qy$ $ \forall x,y\in f(\mathbb R)$, $ \forall p,q\in\mathbb N$ and so $ x\equal{}y\equal{}0$

So $ f(\mathbb R)\equal{}\{0\}$ and so the unique solution :

$ f(x)\equal{}0$ $ \forall x$ (which is indeed a solution)