geometry
ratio
circumcircle
geometric transformation
reflection
rectangle
analytic geometry

Let be $ \triangle ABC$.A line $ p$ intersects $ AB,BC,CA$ at poits $ C_{0},A_{0},B_{0}$ (respect) .The lines which through
$ C_{0},B_{0},A_{0}$ perpendicular to $ AB,AC,BC$ intersect at poits $ D,E,F$. $ H_{1},H_{2}$ are orthocenters of triangles $ ABC,DEF$ .Denote $ f(p)$ be the ratio which $ p$ intersect $ [H_{1}H_{2}]$. Prove that $ f(p)$ is a constant number for all line p in the plane.

If the ratio, in which $ p$ cuts $ H_1H_2,$ is constant, it has to be $\tfrac{1}{2}$. If the normals to $ BC,\ CA,\ AB$ at $ A_0,\ B_0,\ C_0$ happen to concur at a point, the $ \triangle DEF,$ including its orthocenter $ H_2,$ degenerates to the point $ H_2,$ which is then pole of the Simson line $ p$ of the $ \triangle ABC.$ Any Simson line cuts the segment $ H_1H_2$ between its pole $ H_2$ and the triangle orthocenter $ H_1$ in half. If the normals do not concur and if $ \mathcal T$ is the defined transformation $ \triangle DEF \equal{} \mathcal T(\triangle ABC,\ p),$ then $ \triangle ABC \equal{} \mathcal T(\triangle DEF,\ p),$ so that $ \mathcal T$ is an involution (its own inverse) and again, if the ratio is constant, it has to be 1/2. Changing notation:

[color=blue]Let an arbitrary line $ k$ intersect the triangle sidelines $ BC,\ CA,\ AB$ at points $ X,\ Y,\ Z.$ Let normals $ x,\ y,\ z$ to $ BC,\ CA,\ AB$ at the points $ X,\ Y,\ Z$ meet at $ A' \equiv y \cap z,\ B' \equiv z \cap x,\ C' \equiv x \cap y.$ The line $ k$ cuts the segment $ HH'$ between the orthocenters $ H,\ H'$ of the $ \triangle ABC,\ \triangle A'B'C'$ in half.[/color]

The $ \triangle ABC \sim \triangle A'B'C'$ are similar, having perpendicular their corresponding sides $ BC \perp x \equiv B'C',$ $ CA \perp y \equiv C'A',$ $ AB \perp z \equiv A'B'.$ Let $ (O),\ (O')$ be the respective circumcircles of these two triangles. Let $ P \in (O)$ be pole of the Simson line $ p$ of the original $ \triangle ABC$ parallel to the line $ k \equiv XYZ.$ Let $ F_a,\ F_b,\ F_c$ be feet of normals from $ P$ to $ BC,\ CA,\ AB,$ forming the Simson line $ p \equiv F_aF_bF_c$ of the $ \triangle ABC$ with the pole $ P.$ The quadrilaterals $ AF_bPF_c \sim AYA'Z$ are centrally similar with similarity center $ A,$ having parallel sides $ AF_b \equiv AY,\ AF_c \equiv AZ,$ $ PF_b \parallel A'Y$ (both $ \perp CA$), $ PF_c \parallel A'Z$ (both $ \perp AB$) and diagonals $ F_bF_c \equiv p \parallel k \equiv YZ.$ Consequently, $ PA'$ passes through the similarity center $ A,$ $ A' \in PA,$ and similarly, $ B' \in PB,\ C' \in PC.$ Let $ Q \in (O)$ be the diametrically opposite point of $ P,$ the pole of the Simson line $ q$ of the $ \triangle ABC$ perpendicular to the line $ k \equiv XYZ.$ Since $ PQ$ is a diameter of $ (O),$ $ A'P \equiv AP \perp AQ,$ $ B'P \equiv BP \perp BQ$ and since the $ \triangle ABC \sim \triangle A'B'C'$ have the corresponding sides perpendicular, $ A'B' \perp AB.$ As a result, the $ \triangle A'PB' \sim \triangle AQB$ are similar, which means that $ P \equiv Q' \in (O')$ is also pole of the Simson line $ q' \perp q$ of the $ \triangle A'B'C',$ hence $ q' \parallel p \parallel k.$

Let $ F_a',\ F_b',\ F_c'$ be feet of normals from $ P$ to $ B'C',\ C'A',\ A'B',$ forming the Simson line $ q' \equiv F_a'F_b'F_c'$ of the $ \triangle A'B'C'$ with the pole $ P.$ Furthermore, let $ P_a,\ P_b,\ P_c$ be reflections of $ P$ in $ BC,\ CA,\ AB$ and $ P_a',\ P_b',\ P_c'$ reflections of $ P$ in $ B'C',\ C'A',\ A'B',$ forming Steiner lines $ s \equiv P_aP_bP_c \parallel F_aF_bF_c \equiv p,$ $ s' \equiv P_a'P_b'P_c' \parallel F_a'F_b'F_c' \equiv q'$ of the $ \triangle ABC,\ \triangle A'B'C',$ respectively, $ s \parallel s' \parallel k \equiv XYZ.$ The Simson lines $ p,\ q'$ cut in half the segments $ PH,\ PH'$ from their common pole $ P$ to the orthocenters $ H,\ H',$ hence $ H \in s,\ H' \in s'.$ The quadrilateral $ PF_cZF_c'$ is a rectangle, because $ PF_c \parallel ZF_c' \equiv A'B'$ (both $ \perp AB \equiv ZF_c$), $ PF_c' \parallel ZF_c \equiv AB$ (both $ \perp A'B'$, hence $ ZF_c \equal{} PF_c'.$ But $ F_c'$ is the midpoint of $ PP_c',$ so that $ ZF_c \equal{} F_c'P_c'.$ Since $ Z \in k,\ F_c \in p,\ F_c' \in q',\ P_c' \in s',$ the distance between the parallels $ k \parallel p$ is equal to the distance between the parallels $ q' \parallel s'.$ It follows that the distance between the parallels $ k \parallel s'$ (the given line and the Steiner line of the $ \triangle A'B'C'$) is equal to the distance between the parallels $ p \parallel q'$ (the Simson lines of the $ \triangle ABC,\ \triangle A'B'C'$). In exactly the same way, we can show that the distance between the parallels $ k \parallel s$ (the given line and the Steiner line of the $ \triangle ABC$) is equal to the distance between the Simson lines $ p \parallel q'$ of the two triangles. Even simpler, the distance between the Steiner lines $ s \parallel s'$ is obviously twice the distance between the Simson lines $ p \parallel q'$ of the two triangles. Thus $ k$ is the midparallel of the parallels $ s \parallel s',$ cutting in half the segment $ HH',$ where $ H \in s,\ H' \in s'.$

See [url=http://mathworld.wolfram.com/ParalogicTriangles.html]Paralogic Triangles[/url], [url=http://mathworld.wolfram.com/SondatsTheorem.html]Sondat's Theorem[/url].

Let us define the rectangular coordinates $ A:(0,a) \ , \ B:(b,0) \ , \ C:(c,0)$ and $\tau$ is an arbitrary line with equation $ \tau: px \plus{} qy \plus{} r \equal{} 0.$ Under this reference the orthocenter $ H(x_0,y_0)$ of $\triangle ABC$ and the sidelines $AB,AC$  are

$ H \left (0 \ , \ \minus{} \frac { bc}{a} \right)  \ , \  AB \equiv  ax \plus{} by \minus{} ab \equal{} 0  \ , \  AC \equiv ax \plus{} cy \minus{} ac \equal{} 0.$

Line $\tau$ intersects $ AB,AC,BC$ at points $B_0,C_0,A_0$ with coordinates:

$ B_0 \left (b \cdot \frac {r \plus{} aq}{aq \minus{} bp}  , \ a \cdot \frac {r \plus{} pb}{bp \minus{} aq} \right) \ , \ C_0 \left (c \cdot \frac {r \plus{} aq}{aq \minus{} cp} , \ a \cdot \frac {r \plus{} pc}{cp \minus{} aq} \right)  , \ A_0 \left (\minus{} \frac{ r}{p} \ , \ 0 \right )$

Equations of the perpendiculars $\beta,\gamma$ to $AB,AC$ through $B_0,C_0$ are

$\beta \equiv  y \minus{} a \cdot \frac {r \plus{} pb}{bp \minus{} aq} \minus{} \frac {b}{a}\left (x \minus{} b \cdot \frac {r \plus{} aq}{aq \minus{} bp} \right) \equal{} 0$

$\gamma \equiv  y \minus{} a \cdot \frac {r \plus{} pc}{cp \minus{} aq} \minus{} \frac {c}{a} \left (x \minus{} c \cdot \frac {r \plus{} aq}{aq \minus{} cp}\right ) \equal{} 0$

$ E \left  [\minus{} \frac { r}{p}\ , \ \beta \left (\minus{} \frac { r}{p} \right) \right]  \Longrightarrow \ E \left[\minus{} \frac {r}{p} \  , \ \frac {(r \plus{} pb)(bq \plus{} ap)}{p(bp \minus{} aq)} \right]$

$ D \left [\minus{} \frac { r}{p} \ , \ \gamma \left (\minus{} \frac {r}{p} \right) \right] \Longrightarrow \  D \left [ \minus{} \frac { r}{p} \ , \ \frac {(r \plus{} pc)(cq \plus{} ap)}{p(cp \minus{} aq)}\right]$

Parallels through $D,E$ to $AB,AC$ meet at the orthocenter $H:(x_1,y_1)$ of $\triangle DEF.$

$ H_1 \equiv  \left  (\frac {bc}{a(c \minus{} b)} \cdot (n \minus{} m) \minus{} \frac {r}{p} \ , \ \frac {cm \minus{} bn}{c \minus{} b} \right )$

$ m \equal{} \frac {(r \plus{} pb)(bq \plus{} ap)}{p(bp \minus{} aq)} \ \ , \ \ n \equal{} \frac {(r \plus{} pc)(cq \plus{} ap)}{p(cp \minus{} aq)}$

$\Longrightarrow \ |px_1 \plus{} qy_1 \plus{} r| \equal{} \left |  \frac { bcq}{a} \minus{} r \right | \ \ , \  \ |px_0 \plus{} qy_0 \plus{} r| \equal{} \left | \frac {bcq}{a} \minus{} r \right |$

$\Longrightarrow \ | px_1 \plus{} qy_1 \plus{} r| \equal{} | px_0 \plus{} qy_0 \plus{} r|$ $\Longrightarrow$ $H,H_1$ are equidistant from $\tau.$

Dear Mathlinkers,
for me it seems that this problem come from Sondat?
Any reference?
Sincerely
Jean-Louis

Links to references are at the end of the 1st reply ?

Dear Mathlinkers,
an article concerning another Sondat's theorem can be found on
http://perso.orange.fr/jl.ayme/  vol. 5
Sincerely
Jean-Louis

I'll prove the midpoint of $ H_1H_2 $ lie on $ \overline{A_0B_0C_0} $ .

My solution:

Let $ X, Y, Z $ be the midpoint of $ AD, BE, CF $, respectively .
Let $ M $ be the Miquel point of complete quadrilateral $ AC_0BCA_0B_0 $ .
Let $ H $ be the orthocenter of $ \triangle XYZ $ .

Easy to see $ \triangle ABC \sim \triangle DEF $ .

Since $ \angle C_0DB_0=\angle C_0AB_0 $ ,
so we get $ D \in (AB_0C_0) $ . ie. $ A, D, B_0, C_0, M $ are concyclic
Similarly, we can prove $ E \in (BC_0A_0M) $ and $ F \in (CA_0B_0M) $ ,
so $ M $ is also the Miquel point of complete quadrilateral $ DC_0B_0A_0EF $ ,
hence $ M $ lie on the circumcircle of $ \triangle DEF $ .

From Peterson-Schoute theorem
we get $ \triangle ABC \sim \triangle XYZ \sim \triangle DEF $ and $ H $ is the midpoint of $ H_1H_2 $ .
Since $ M $ is the spiral similar center of $ \triangle ABC $ and $ \triangle DEF $ ,
so $ M $ also lie on the circumcircle of $ \triangle XYZ $ .
Since $ X, Y, Z $ is the center of $ (MB_0C_0), (MC_0A_0), (MA_0B_0) $, respectively ,
so we get $ YZ, ZX, XY $ is the perpendicular bisector of $ MA_0, MB_0, MC_0 $, respectively ,
hence $ \overline{A_0B_0C_0} $ is the Steiner line of $ M $ WRT $ \triangle XYZ $ .
ie. the midpoint $ H $ of $ H_1H_2 $ lie on $ \overline{A_0B_0C_0} $ .

Q.E.D


Anybody familiar with this? Used for finding the ratios of lines in a triangle. For example, in triangle ACD, B and E are on sides AC and AD respectfully. The ratio of AB to BC is 2:3 and the ratio of DE to AE is 2:3. F is where EC and BD are concurrent. What is the ratio of EF to FC?

Using mass point, the problem can be solved easily. I can only use mass point to that degree, though. Can anyone possibly post exactly how mass point works?

yeah, I asked for help with more complex mass points a couple days ago, but it got bumped off the page with no replies.

You understand the law of equilibrium, right?  Knowing that can help with many applications, but maybe you're having the same trouble that I am.

As far as this one goes.
[hide]Apply 6 units of pressure to point 6 and 4 to C, totaling 10 on B.  Then putting 9 on D will balance the scales with a total of 19 at F.  Since C has 4 and E has 15, law of equilibrium says EF:FC = 4:15[/hide]

Here is my problem, although I wasn't planning on getting direct help on it  :(  indirect got no replies tho, so...  In triangle ABC, D is the midpoint of BC.  E, F, and G divide AC into 4 congruent segments (in the order).  AD meets BE, BF, and BG at H, I, and J respectively.  If AH:HI:IJ:JD = w : x : y : z, find (w+y)/(x+z).

Now I'm not [i]sure[/i] that this is a mass points problem, but having looked at it and noticing that I is the centroid and you're trying to find the sum of ratios (like force at vertices), I figured it might be.  Any ideas?

I learned a lot from [url=http://mathcircle.berkeley.edu/mpgeo.pdf]Tom Rike's article[/url] on mass point geometry.

well, here is another solution
[hide]
Let the intersection of $AF$ and $CD$ be $G$, then, by using Ceva, we get:
\[\frac{CG}{GD}=\frac94\]
Let $S_{\triangle{ADE}}$ be $4$ units then just calculate other pieces of squares and then get:$\frac{15}{4}$
[/hide]

btw, i [b]don't[/b] understand the law of equilibrium, so can you introduce something about that??

New here. Never heard of mass point geometry before. And I thought I was good at math... are there answers to the pdf? Or other easy-to-learn-from documents?

This is my first math forum. Lotsa stuff I haven't done, and stuff I have done. Looking forward to becoming part of this community.

I think of it as a torque problem. In order for the triangle to be balanced a a chosen point, say the intersection of two lines, the mass you "place" on a point on one side times its distance from the balancing point must be equal to the mass you "place" on a second point times its distance from the balancing point (provided that the three points are collinear).
In physics, it's conservation of angular momentum.
In math, I think it's called Barycentric coordinates. You pick three points where you put equal masses onto and name other points with the masses required at that point to make it so that if that point is the fulcrum, the plane would balance out. I most likely remembered it wrong, so someone else clarify what I said.

Actually, I get what it's about; I just want more problems.

(Love how there's math symbols on the left to use...)


calculus integration limit logarithms factorial function ratio

Find $\lim_{n\to\infty} \left(\frac{_{3n}C_n}{_{2n}C_n}\right)^{\frac{1}{n}}$

where $_iC_j$ is a binominal coefficient which means $\frac{i\cdot (i-1)\cdots(i-j+1)}{j\cdot (j-1)\cdots 2\cdot 1}$.

i still don't do it, but
[hide]
maybe taking logarithm and consider it as Rienmann Sum.
[/hide][/hide]

[hide]writing them as factorials,

(3n n)/(2n n) = [(3n)!/(2n)!n!] / [(2n)!/2(n!)]

= (3n)!n! / 2(2n)!
= [(3n)(3n-1)(3n-2)...(2n+2)(2n+1)] / [(2n)(2n-1)(2n-2)...(n+2)(n+1)]

clearly, each term has n elements, and these can be grouped in the following way:

= [(3n)/(2n)][(3n-1)/(2n-1)][(3n-2)/(2n-2)]...[(2n+2)/(n+2)][(2n+1)/(n+1)]

each term here is in the form [A(n) + n] / [A(n)], with "a" a linear & increasing function of n; the the rth factor, A(r,n) = (2n-r)

so we want to find the product from r=0 to r=(n-1) of [A(n) + n] / [A(n)] = [(2n - r) + n]/(2n - r) = (3n - r)/(2n - r), and the limit of the nth root of this as n tends to infinity. I wish It was as simple as saying r is finite and so the limit is simple 3/2, but I don't think this is rigorous or even correct as r varies from 0 to (n-1), and so it is not finite...[/hide]

Ill leave this here, because I'm not educated enough about limits or even products (I dont know if they work in the same way as sums but Id guess there is some way of evaluating them and analysing them).. interesting question though.

Im tempted to just have a stab in the dark and say 1.

I can assure you that the answer is not 1.

[hide="Hint."]What is the radius of convergence of the series $\sum_{n=0}^{\infty}\frac{{{3n}\choose n}}{{{2n}\choose n}}x^n$ ?

Use both the ratio test and the root test.[/hide]

[hide="Beginners answer"]$\frac{27}{16}$[/hide] is correct.

The purpose of my hint in #4 was to elicit the computation that Moubinool did in #7.


calculus
integration
function
logarithms
calculus computations

Study the convergence of

(i)$\int_{0}^{1}\frac{(-1)^{[1/x]}}{x}dx$

(ii)$\int_{1}^{\infty}(x-[x])^{x^{2}}dx$

Good fun!

(ii) The integral (which is of a nonnegative function) equals

$\sum_{n=1}^{\infty}\int_{n}^{n+1}(x-n)^{x^{2}}\,dx= \sum_{n=1}^{\infty}\int_{0}^{1}u^{(n+u)^{2}}\,du$

$\le \sum_{n=1}^{\infty}\int_{0}^{1}u^{n^{2}}\,du= \sum_{n=1}^{\infty}\frac{1}{n^{2}+1},$ which converges.

The integral in (i) should also converge, but not absolutely.


let$ c\equal{} min (a,b,c )$  we have : $ c \leq 1/3$
  hence :  $ f(a,b,c)  \geq f(t,t,c)$*   where$ t \equal{}\frac{a\plus{}b}{2}$
because  (*)$ \to  (a\minus{}b)^2\frac{1\minus{}3c}{2} \geq 0$
  $ minf\equal{} minf(1/2,1/2,0) \equal{}\frac{1}{2}$

the problem is equivalent to a^3+b^3+c^3+6abc>=a^2b+a^2c +b^2a+b^2c +c^2a+c^2b
 which is clear by Schur  :)

[quote="greentreeroad"]Silly question, what is SMV?[/quote]
It's is ''strong mixing variables''.I have forgotten a link about this :D

[quote="greentreeroad"]Silly question, what is SMV?[/quote]
see [url=http://www.mathlinks.ro/viewtopic.php?t=113248]here[/url] :)

Thank you :D

Also, when I read that post and is induced to other posts. I see something like ABC sequence and AC method. Anyone know what they are?

[quote="M.A"]let a+b+c=1 then find the minimum of (a^2 +b^2 +c^2 +6abc )
 (its good to use SMV theorem) :)[/quote]
We have :
$ (a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)\plus{}6abc \geq \frac{1}{2}\equal{}\frac{(a\plus{}b\plus{}c)^3}{2}$
$ \Longleftrightarrow a^3\plus{}b^3\plus{}c^3\plus{}6abc \geq a(b^2\plus{}c^2)\plus{}b(c^2\plus{}a^2)\plus{}c(a^2\plus{}b^2)$
It is right because Schur inequality


Can someone get to MOP studying just the 2 volumes of AoPS? I'm so bad at this type of problem solving...I mean I took calc bc last year and got a 5, but I'm in Danny Zhu's grade and he's won math contests I've never even heard of...it's really depressing considering I could have had SOMETHING in relation with him if I did 10 points better on the Math SAT in the 8th grade talent search....

[quote="OVB"]10 points better on the Math SAT in the 8th grade talent search....[/quote] :rotfl: I did that too.

I doubt the 2 volumes of AoPS alone will get you to MOP. Lots and lots of problems will help. :D You could also get Problem Solving Strategies and other Olympiad books.


abstract algebra
calculus
integration
algebra
function
domain
Ring Theory

I look for a $ A$-module $ M$ with the following properties:
- there are $ f_1,...,f_n \in A$ with $ f_1 \plus{} ... \plus{} f_n \equal{} 1$, such that $ M_{f_i}$ is free of rank $ 1$ over $ A_{f_i}$ (i.e. $ \widetilde{M}$ is a locally free module sheaf of rank $ 1$).
- $ M$ is not free over $ A$.

any ideas? if we could find an integral domain $ A$, which is no field, and $ f_i$ as above, so that for every $ i$, every element of $ A$ divides some power of $ f_i$, then we may take $ M \equal{} Q(A)$.

Take M = nonprincipal ideal of a Dedekind domain.

Because the localisations are discrete valuation domains, in particular PIDs. M is projective and, over a local ring, projective = free.

$ A_p$ is a PID for prime ideals $ p$, but we're talkin' about $ A_f$ where $ f$ is an element of $ A$. so which $ f_i$ do you take?

$ f_i$ for two different principal prime ideals.

You can take e.g. $ (2, 1 \plus{} \sqrt{\minus{}5})$

sorry I still don't understand. so you take $ A \equal{} \mathbb{Z}[\sqrt { \minus{} 5}]$ and the ideal above, but what are the $ f_i$ ...

thank you :).

if we consider $ M\equal{}(2,1\plus{}\sqrt{\minus{}5})$ as a module over $ A\equal{}\mathbb{Z}[\sqrt{\minus{}5}]$, $ M$ is not free, but $ M_2$ is free over $ A_2$ with generator $ 2$ and $ M_3$ is free over $ A_3$ with generator $ 1\plus{}\sqrt{\minus{}5}$.


abstract algebra
superior algebra
superior algebra unsolved

Hi,

let $ M$ be a finitely generated $ R$-module where $ R$ is a commutative ring with unity.
Let $ S$ be a multiplicatively closed subset of $ R$. Let
$ \text{Ann}(M) \equal{} \{r\in R | \text{For all } m\in M: r\cdot m \equal{} 0 \}$.

I read a proof which in one step assumed that $ S$ contained
an element of $ \text{Ann}(M)$, and then concluded that the
localization $ S^{\minus{}1} M$ equalled $ \{0\}$ $ \emph{because}$ $ M$
was finitely generated. I do not see why this is essential for the
localization being zero in this case. Am I correct that it is zero
even when $ M$ is not finitely generated?

Thanks.


function
LaTeX
algebra
domain
algebra unsolved

Find all functions $ \ f$ satisfy:
$ \ f(t(1\minus{}t)x\plus{}ty)\ <\equal{}min({f(x),f(y)})$
for all 0<t<1.

"Sorry for my bad Latex"

What are the domain and range of $ f$? Did you mean $ (1 \minus{} t)x \plus{} ty$ or is $ (1 \minus{} t)tx$ correct? Is it minimum or maximum? (case it is minimum, $ f(a) \le f(b)$ for any $ a, b$...)


[quote="Vasc"]If $a,b,c$ are positive numbers, then
$\sum\frac 1{b+c} \geq \sum\frac{2a}{3a^2+bc}$.[/quote]
Any solution :? .

[quote="toetoe"][quote]
$\sum\frac 1{b+c} \geq \sum\frac{2a}{3a^2+bc}$
[/quote]

It is false for $a=b=c=1$  :D .[/quote]
This ia bad joke.

[quote="Soarer"]any non-SOS solution?
sum them up then we get
$c\sum (a-b)^2 \frac{2a^2+2b^2+2c(a+b)-4ab}{(3a^2+bc)(b+c)(3b^2+ac)(a+c)} \ge 0$[/quote]

Shouldn't it be
$\sum c(a-b)^2 \frac{3a^2+3b^2+5c(a+b)-3ab-c^2}{(3a^2+bc)(b+c)(3b^2+ac)(a+c)}$ ?

Put $x=\sqrt{\frac{bc}{a}}$, $y=\sqrt{\frac{ca}{b}}$ and $z=\sqrt{\frac{ab}{c}}$. The given inequality becomes
$\sum_{cycl}\frac{1}{x(y+z)}\ge\sum_{cycl}\frac{2}{3yz+x^2}$
We have
$S=\sum_{cycl}\left(\frac{1}{x(y+z)}-\frac{2}{x^2+3yz}\right) =\sum_{cycl}\frac{(x-y)(x-z)+2yz-xy-xz}{x(y+z)(x^2+3yz)}$
Note that
$x(y+z)(x^2+3yz)-y(z+x)(y^2+3zx)=(x-y)[z(x-y)^2+xy(x+y)]$ (*)
So (by Vornicu-Schur lemma)
$S\ge \sum_{cycl}\frac{2yz-xy-xz}{x(y+z)(x^2+3yz)} =\sum_{cycl}\frac{z(y-x)+y(z-x)}{x(y+z)(x^2+3yz)}$
$=\sum_{cycl}(x-y)\left( \frac{z}{y(z+x)(y^2+3zx)}-\frac{z}{x(y+z)(x^2+3yz)} \right)$
$=\sum_{cycl}z(x-y)^2\frac{z(x-y)^2+xy(x+y)}{xy(x+z)(y+z)(x^2+3yz)(y^2+3zx)} \ge 0$
where we again use (*).


What type of value does a regression need to be near to be a linear line?   What do negative regressions mean?

What does it mean if the slope of a graph is postive but the coorelation coefficient is negative?

[quote="namstap"]hm i think you're talking about r, the correlaton coefficient??

in that case... if |r| is close to 1, its very much linear....

a perfect linear set of data has r = 1

negative coefficients mean the slope of the data is negative.. so r = -1 is a perfect linear relation also.[/quote]

hm, i didn't remember that happening.... its been a year since i've had any statistics.... maybe i'm just crazy lol, i thought i remembered the sign of r was the same as the sign on the slope though


[quote="akai"]If $ a,b,c$ are positive numbers and $ abc \equal{} 1$ then $ 2(ab \plus{} bc \plus{} ca)^2\ge3(ab \plus{} bc \plus{} ca \plus{} ab^2 \plus{} bc^2 \plus{} ca^2)$[/quote]
I think, it's wrong. After homogenization try $ a\rightarrow\plus{}\infty.$ :wink:


[quote="mr. math"]The number n factorial has 12 terminal zeroes when multiplied out. what is the largest possible integer for n?

how would you do this problem?[/quote]

[hide]Basically, you have to find $n!$ so that there are 12 factors of 5 in it.

By guess and check, I got $\boxed{n=54}$[/hide]


geometry
3D geometry
sphere
geometric transformation
rotation

$Question 1$
There is a container that has a finite volume. You can therefore
fill it with paint. However, if you want to paint the inside surface of the container you will need an infinite quantity of paint. Sounds paradoxical isn't it? However, it is very easy to visualise such a container and to describe the shape geometrically. Try it!

$Question 2$
What are the last three digits of the decimal number $7^{9999}$ ?

[hide="#1"]If we disregard fractal objects, one example could be a string of spheres with radii $r_{n}={1\over n^{2/5}}$. Then the surface area is proportional to $\sum_{n=1}^{\infty}{1\over n^{4/5}}$, which is divergent, and the volume is proportional to $\sum_{n=1}^{\infty}{1\over n^{6/5}}$, which is convergent.[/hide]

[quote="junggi"][hide="#2"]
From Euler's totient thrm,
$7^{40}=1\mod100.$
So we reduce this to finding $7^{39}\mod100$
Since 7 is coprime to 100, we find find a unique multiplicative inverse of 7 in mod 100. We can quickly find that inverse to be 43(this means that $7\cdot43=1\mod100$).
So,
$7^{39}=7^{39}\cdot1=7^{39}\cdot7\cdot43=7^{40}\cdot43=\boxed{43}\mod100$
from Euler's totient thrm again.
[/hide][/quote]

[hide]how did you find $7^{9999}$ from what you have shown?[/hide]

[quote="junggi"][hide="#2"]
From Euler's totient thrm,
$7^{40}=1\mod100.$
So we reduce this to finding $7^{39}\mod100$
Since 7 is coprime to 100, we find find a unique multiplicative inverse of 7 in mod 100. We can quickly find that inverse to be 43(this means that $7\cdot43=1\mod100$).
So,
$7^{39}=7^{39}\cdot1=7^{39}\cdot7\cdot43=7^{40}\cdot43=\boxed{43}\mod100$
from Euler's totient thrm again.
[/hide][/quote]
oh sorry I misread the problem :blush:

[quote="t0rajir0u"][quote="xxxyyyy"]how did you find $7^{9999}$ from what you have shown?[/quote]

Well, he only found the last two digits, but...
[hide="The totient theorem says..."] $gcd(a, m) = 1 \implies a^{\varphi(m)}\equiv 1 \bmod m$ [/hide] [hide="Which implies..."] $7^{\varphi(100) = 40}\equiv 1 \bmod 100$ [/hide] [hide="So that..."] $7^{9999}= 7^{249(40)+39}\equiv 7^{39}\bmod 100$ [/hide]

[hide="1"] Many solids of rotation satisfy this requirement.  For example, rotate the graph of $y = \frac{1}{x}$ around the $x$-axis. [/hide]
[hide="2"] $a^{ 400}\equiv 1 \bmod 1000$ by Euler.  So

$7^{9999}\equiv 7^{-1}\bmod 1000$

Now, we know that $7 \cdot 11 \cdot 13 = 1001$, so

$7^{-1}\equiv \boxed{ 143}\bmod 1000$ [/hide][/quote]

Yours was an excellent explanation.. I also arrived at the same building upon what was mentioned earlier. Thanks.


[b]Calculate:[/b] $S=1\cdot 2\cdot 3+3\cdot 4\cdot 7+7\cdot 8\cdot 15+\cdots+\left(2^{k}-1\right)2^{k}\left(2^{k+1}-1\right)$ :wink:


Let $a,b,c >0$.Prove that the following inequality holds:

      $ (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 3(1+{(\frac{(a+b)(b+c)(c+a)}{abc})}^{\frac{1}{3}}) $ ;)

Let $x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$.  Upon expanding the left side and the right side, we find that the inequality is equivalent to:
$3+x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq{3+3((x+\frac{1}{y})(y+\frac{1}{z})(z+\frac{1}{x}))^{\frac{1}{3}}}$
Subtracting 3 from each side and rearranging the left side to $(x+\frac{1}{y})+(y+\frac{1}{z})+(z+\frac{1}{x})$ the result is immediate by AM-GM.

By AM-GM,
$ (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=3+\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}$
$\geq 3+3\sqrt[3]{\frac{(b+c)(c+a)(a+b)}{abc}} =3(1+{(\frac{(a+b)(b+c)(c+a)}{abc})}^{\frac{1}{3}}). $

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=56938&hilit=inequality
Let $a, b, c >0.$ Prove the following inequality:
$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\ge\frac{1}{3} (\sqrt [3] {\frac{a}{b}}+\sqrt [3] {\frac{b}{c}}+\sqrt [3] {\frac{c}{a}})^3.$

Because $ (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=3+\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c} $
inequality returns to $ \sum_{cyc}\frac{b+c}{a}\ge 3\sqrt[3]{\frac{b+c}{a}\frac{c+a}{b}\frac{a+b}{c}} $
which is true by Gm_Am
_____
Marin Sandu


OMG WE JUST HAD A TORNADO WARNING THAT ENDED ABOUT AN HOUR AND FOURTY-FIVE MINUTES AGO OMG OMG

We had a flood warning yesterday, and a nearby county had a tornado warning.

You guys are lucky. A few days ago we had a [i]terrorist attack[/i] warning!!

Omg a few years ago we got a Surgeon's General Warning for smoking.

Wow! Did you have to freak out in the middle of the night ?!  It is stil bad if you didn't but I find it is  the worst when you do... :wacko:

By the way this just struck me: Did the warning say that a tornado has ended one hour and forty five minutes ago?

[quote="nutz_for2.718281828"]boy, I just had a MAFIA warning!!(in the game. :dry: )

Well, I didn't hear the earthquake warning, and I slept right through it, while it woke up the rest of the house. :huh:[/quote]

That has also happened to me. I have even slept right through earthquakes and tsunamis!


Determine the minimum of $|z|$, if $z$ satisfies: $|z-3\imath |+|z-4|=5$.

Suppose $A(0,3)$ and $B(4,0)$. Point $P$ satisfies $|AP|+|BP|=5=|AB|$ thus $P$ is on segment $AB$. hence $\min |OP|$ is obtained when $OP$ is orthogonal to $AB$. $\min |z| = \min |OP| = 12/5$.

Thx liyi.
I was trying it algebraic and didn't know why  can't I handle it.


modular arithmetic
induction
number theory proposed
number theory

Let $ a,n\geq 2$ be two integers, which have the following property: there exists an integer $ k\geq 2,$ such that $ n$ divides $ (a\minus{}1)^k.$ Prove that $ n$ also divides $ a^{n\minus{}1}\plus{}a^{n\minus{}2}\plus{}\dots\plus{}a\plus{}1.$

[quote="Maxim Bogdan"]Let $ a,n\geq 2$ be two integers, which have the following property: there exists an integer $ k\geq 2,$ such that $ n$ divides $ (a - 1)^k.$ Prove that $ n$ also divides $ a^{n - 1} + a^{n - 2} + \dots + a + 1.$[/quote]

$ n$ divides $ (a-1)^k$ only means that prime divisors of $ n$ are prime divisors of $ a-1$, with any order (up to $ k$).

So the problem is only to show that $ a=1\pmod{p^m}$ (with $ p$ prime) $ \implies$ $ \sum_{i=0}^{\alpha p^m -1}a^i=0\pmod{p^m}$, which is easy to show with induction on $ m$ :

$ m=1$ : $ a=1\pmod{p}$ $ \implies$ $ \sum_{i=0}^{\alpha p -1}a^i$ $ =\sum_{i=0}^{\alpha p -1}1$ $ =\alpha p=0\pmod{p}$

$ m>1$ : $ \sum_{i=0}^{\alpha p^m -1}a^i$ $ =(\sum_{i=0}^{p -1}a^i)(\sum_{i=0}^{\alpha p^{m-1} -1}a^{pi})$

First factor is obviously $ 0\pmod p$ and second factor may be written ${ \sum_{i=0}^{\alpha p^{m-1} -1}b^i}$ with $ b=a^p$ and is divisible by $ p^{m-1}$ thru induction hypothesis (since $ b=1\pmod p$)

Hence the result

[quote="TTsphn"][quote="Maxim Bogdan"]Let $ a,n\geq 2$ be two integers, which have the following property: there exists an integer $ k\geq 2,$ such that $ n$ divides $ (a \minus{} 1)^k.$ Prove that $ n$ also divides $ a^{n \minus{} 1} \plus{} a^{n \minus{} 2} \plus{} \dots \plus{} a \plus{} 1.$[/quote]
The condition is equivalent to $ rad(n)|a \minus{} 1$ (all prime divisors of n are divisors of $ a \minus{} 1$) .The rest follows from lifting exponent lemma :
Have two cases here :
i) If $ p$ is odd then by the lifting exponent lemma we have $ v_p(a^n \minus{} 1) \equal{} v_p(a \minus{} 1) \plus{} v_p(n)$ ,it gives $ v_p(\frac {a^n \minus{} 1}{a \minus{} 1}) \equal{} v_p(n)$ .
ii) If $ p \equal{} 2$ ,we can prove that $ v_p(a^n \minus{} 1)\geq v_p(a \minus{} 1) \plus{} v_2(n)$ by induction .
Hence ,in all case we have $ v_p(\frac {a^n \minus{} 1}{a \minus{} 1})\geq v_p(n)$ ,therefore $ n|\frac {a^n \minus{} 1}{a \minus{} 1}$ . That ends the proof .[/quote]
It's also my proof  :)


calculus
integration
trigonometry
calculus computations

1) $ \int \frac {sin(2x) \minus{} sin(2k)}{sinx \minus{} sink \plus{} cosx \minus{} cosk}dx$

2)$ \int \frac {dx}{\sqrt {cos^3(x \plus{} a)sin(x \plus{} b)}}$

3$ \int \sqrt {\frac {cos(x \minus{} a)}{cos(x \plus{} a)}} dx$

4)$ \int \frac {cotx \minus{} 3cotx}{3tan3x \minus{} tanx} dx$

If possible plz explain your approach.I mean tell me from where to start and what to eliminate and like that.

i did the last $ 3$ but am too lazy to type them out right now, i am going to go to sleep. c ya.  :)

i could do the second and fourth

For second one,
pull out $ \cos^4(x \plus{} a)$ and when it goes to the numerator it becomes $ \sec^2 (x\plus{}a)$

Now put x+a = t and simplify the denominator by using $ \sin A \plus{} B$ formula....set denominator equal to p, then Nr becomes dp....

For the last one convert everything to tan and then use formual for tan 3x......next partial fractions.......  :sleeping:  :sleeping:

For  No1, just  note  that   sin2x-sin2k =$ (sinx+cosx)^{2} -(sink+cosk)^{2}$

No3.$ \frac{\sqrt{cos(x-a)}}_{\sqrt{cos(x+a)}}$ =$ \frac{cos(x-a)}_{\sqrt{cos(x+a)cos(x-a)}$ =$ \frac{cosxcosa}{\sqrt{cos^2x -sin^2a}}$ +$ \frac{sinxsina}{\sqrt{cos^2x-sin^2a}$ =$ \frac{cosxcosa}{\sqrt{cos^2a-sin^2x}}$ +$ \frac{sinxsina}{\sqrt{cos^2x-sin^2a}}$ which is reduces to the sum of two standard forms of integrand after substituting u=sinx in the first member and v=cosx in the second.

No4.There is a typing mistake in the numerator

Thank You all.
@Michael: The question is straight from book but it maybe a little wrong but the main thing is I wanted to know the logic.


trigonometry
geometry
algebra unsolved
algebra

let$a,b,c$ be three numbers positive real  such that:
 $a^2+ab+b^2 \leq18$ and  $b^2 +bc+c^2 \leq 6$
 prove that :
      $ab+bc+ca \leq 12$

I don't know why U always post the problem in diendantoanhoc.net there.. And consider it was your own problem ??

Consider a point $O$ and three rays $OA,OB,OC$ such that $\angle{AOB}=\angle{BOC}=\angle{COA}=120^{\circ}$ and $OA=a,OB=b,OC=c$. The cosine law in triangles $AOB,BOC$ gives $AB\le \sqrt{18},BC\le \sqrt{6}$, hence the area $S$ of $ABC$ satisfies $S\le \frac12 \cdot \sqrt{18}\cdot \sqrt{6}=3\sqrt{3}$.
On the other hand, we have $S=\frac12 \cdot (ab+bc+ca)\frac{\sqrt{3}}{2}$. The conclusion follows  easily.


Consider equation I : x + y + z = 46 where x, y, and z are positive integers,
and the equation II : x + y + z + w = 46 where x, y, z, and w are positive
integers. Then
(A) I can be solved in consecutive integers.
(B) I can be solved in consecutive even integers.
(C) II can be solved in consecutive integers.
(D) II can be solved in consecutive even integers.
(E) II can be solved in consecutive odd integers.

[hide="hint"](A) I can be solved in consecutive integers.
[b]a+a+1+a+2=3a+3=46, impossible[/b]
(B) I can be solved in consecutive even integers.
(C) II can be solved in consecutive integers.
[b]a+a+1+a+2+a+3=4a+6=46, a=10. [/b]
(D) II can be solved in consecutive even integers.
(E) II can be solved in consecutive odd integers.

so C is the answer[/hide]


quadratics
algorithm
algebra
polynomial
MATHCOUNTS
quadratic formula

What's the cubic formula? And what's the proof?

It is long: And the proof is really cool, I forgot how to get it.

It's actually not that great (math from the European Dark Ages—what do you expect?).  And it's definitely not worth writing out.  But here it is anyway.  It's in AoPS, by the way.

First you "depress" the cubic by making a substitution that makes your $x^{2}$ term go away.  (To be explicit, you use the substitution $y-\frac{b}{3}=x$, when your cubic is $x^{3}+bx^{2}+cx+d$.)  So then you have something that looks like $y^{3}+ey+f=0$ (I'm assuming you divided out the constant term right away).  And then you realize that maybe $y= \sqrt[3]{u}+\sqrt[3]{v}$.  So then you realize that would look like $u+v+(3\sqrt[3]{uv}+e)(\sqrt[3]{u}+\sqrt[3]{v})+f = 0$.  And then you notice that that will work if we have $uv =-e^{3}/27$ and $u+v =-f$.  But then you remember that we can just set up a quadratic equation to solve that, and so you do it if you really want to at this point.  Then you have found one of your roots, which you can then factor out of the original equation so it turns into a quadratic and you can find the other two roots less laboriously.

As you can tell, few sane humans would elect to do this on a regular basis.  (The quartic algorithm is more elegant, anyhow.)  It would be useful if you (a) wanted to program a computer/calculator to solve cubics for you, or (b) were trying to find a general method of finding a root of a general polynomial, which could have been a pretty smooth method of proving the Fundamental Theorem of Algebra.  But we now know that (b) is impossible.  But people back in the day in Europe didn't, so this doubtless attracted quite a bit of attention when it was first discovered.

Source: [url=http://www.math.vanderbilt.edu/~schectex/courses/cubic]The Cubic Formula[/url]

$ax^{3}+bx^{2}+cx+d=0$

The cubic formula is:

$x=\sqrt[3]{p+\sqrt{p^{2}+q^{3}}}+\sqrt[3]{p-\sqrt{p^{2}+q^{3}}}-\frac{b}{3a}$

Where:

$p=\left(\frac{-b^{3}}{27a^{3}}+\frac{bc}{6a^{2}}+\frac{-d}{2a}\right)$

$q=\left(\frac{c}{3a}-\frac{b^{2}}{9a^{2}}\right)$

I also suggest not memorizing this.

Some one at national mathcounts wanted to trade the cubic formula for several GA pins. Fortunately, I was smart enough to decline.

Dude, it's so much longer than the quadratic formula, but it is quite interesting to derive... I saw a site that had the formulas for the three distinct solutions...
The URL for the site is <http://planetmath.org/encyclopedia/CubicFormula.html>.


linear algebra
matrix
probability
probability and stats

Let {X_n} be a Markov chain taking values in S with transition matrix P=[p_(i,j)], i,j in S and let {π_i : i in S} be a probability on S. Define a new Markov chain {Y_i} with transition matrix Q given by:
Q_(i,j)=min(p_(ij),(π_j/π_i)*p_(ji))
For i not equal to j, and
Q_(ii)=1-sumQ_(ij) (sum over j in S, j not equal to i).
Prove that {Y_(n)} is reversible and from there that it has a stationary distribution π.

I am having some fun with this. I am not sure how to define Q_j,i for starters.... I have an idea but no justification.


Solve equation system
$ x^3\plus{}y^2\minus{}2\equal{}0 \\ x^2\plus{}y^2\plus{}xy\minus{}y\equal{}0$

hello, solving the first equation for $ y^2$ we have $ y^2\equal{}2\minus{}x^3$, plugging this in the second equaton and solving for $ y$ we get $ y\equal{}\frac{x^3\minus{}x^2\minus{}2}{x\minus{}1}$, inserting this in the second equation and simplifying we have to solve the following polynomial $ x^6\minus{}x^5\minus{}x^4\minus{}3x^3\plus{}2x^2\plus{}4x\plus{}2\equal{}0$.
This polynomial has only complex solutions.
Sonnhard.


Find every function $ f: R \longrightarrow R$
$ f( f(x) \minus{} f(y) ) \equal{} (x\minus{}y)^2 f(x\plus{}y)$

[quote="navidj"]Find every function $ f: R \longrightarrow R$
$ f( f(x) - f(y) ) = (x - y)^2 f(x + y)$[/quote]
obvously $ f(x)=f(-x),f(0)=0$, wa take $ X=x+y,Y=x-y$
we have $ Y^2f(X)=f( f(x) - f(y) ) =f( f(x) - f(-y) ) =X^2f(Y)$ gives $ f(X)/X^2=f(Y)/Y^2=a$ for some $ a$ constante
$ f(x)=ax^2$ and we must have $ a^3=a$ gives ${ a\in\{\pm 1,0}$

[quote="aviateurpilot"]
obvously $ f(x) \equal{} f( \minus{} x)$[/quote]

Why ?   :?:

if $ f\equiv 0$ then $ f(x)\equal{}f(\minus{}x)\equal{}0$
$ f(f(x)\minus{}f(y))\equal{}f(f(y)\minus{}f(x))$ so we have just to prove that $ \{f(x)\minus{}f(y): \ x,y\in R\}\equal{}\mathbb{R}$ and we get $ f(x)\equal{}f(\minus{}x)$
if $ f\not\equiv 0$ we take $ b$ such that $ f(b)\neq 0$
we have $ \forall x\in\mathbb{N}: (2x\minus{}b)^2f(b)\equal{}f(f(x)\minus{}f(b\minus{}x))\in Im(f)\equal{}\{f(s): \ s\in\mathbb{R}\}$
we suppose for exemple that $ f(b)>0$ then gives $ \mathbb{R}^{\plus{}}\subset Im(f)$ so $ \{f(x)\minus{}f(y): \ x,y\in R\}\equal{}\mathbb{R}$


I installed GeoGebra but  :blush: how can I get what I draw in the forum  :?:

Thank you I knew it now. But is there a way to work on the program and just paste it in the forum or something?

You can save the file and then open Geogebra on the forum, and choose Open from the menu.


In section 7 (inequalities), problem #2b of Engel's [u]Problem Solving Strategies[/u], I am a bit confused. The question reads as follows:

#####################################
Prove that, for $ a$, $ b$, $ c > 0$,

$ \frac {a^3 \plus{} b^3 \plus{} c^3}{a^2 \plus{} b^2 \plus{} c^2} \ge \frac {a \plus{} b \plus{} c}{3}$.

#####################################

The solution says that it is Chebyshev's inequality, and I understand that much:

$ 3(a^3 \plus{} b^3 \plus{} c^3) \ge (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)$

$ \frac {a^3 \plus{} b^3 \plus{} c^3}{3} \ge \frac {a \plus{} b \plus{} c}{3}\cdot \frac {a^2 \plus{} b^2 \plus{} c^2}{3}$

Which is true by Chebyshev's inequality. However, the definition of Chebyshev's inequality that is stated earlier in the book says that this inequality is for "similarly sorted sequences" (meaning they are either both increasing or decreasing).

But nowhere in problem 2b does it say that $ a < b < c$ or that $ a > b > c$.

Am I misunderstanding something?

You may assume $ a\geq b\geq c$ without loss of generality, because the equation is "symmetric" with respect to the variables.

EDIT: I realize I didn't read the above post correctly.  Ignore this.

Similarly sorted does not refer to the variables in the problem, but rather the terms in each of the two sequences. For example, if you want to apply Chebyshev's inequality to the two sequences \[ \{a_1,a_2,a_3,a_4\},\qquad\{b_1,b_2,b_3,b_4\}\] and the first sequence is sorted \[ a_3<a_2<a_1<a_4,\] then for the two sequences to be similarly sorted, the following must hold: \[ b_3<b_2<b_1<b_4.\]As you can see, similarly sorted does not necessarily mean monotonically sorted.

In this case, the solution indicates that $ \{a,b,c\}$ and $ \{a^2, b^2, c^2\}$ are similarily sorted.

OH! I was assuming monotonicity!

Thanks for your replies. And since $ a,b,c>0$, we know that they are similarity sorted.


Maybe it might be a good idea to make it so post count doesn't count in the individual class forums, or maybe just for the intro ones, etc. I've been noticing that some new members have been posting rather spam - ish posts in the forums (not gonna say name, but one person made a topic called something along the lines of "look inside, really cool" and then inside it just says "haha, made you look") to increase post count. I'm not sure, because a lot of the posts are also for math problems, but yeah...

The number of such users is very limited, and topics get deleted in the end, so their post count will drop once that happens.

Okay, thanks. I was just wondering. Should this topic be deleted?


What is the radius of the smallest circle that can contain $1000$ circles of radius $1$?

[hide="easy way out"]
umm...it's 1. the 1000 circles are all stacked on top of each other[/hide]
[hide="otherwise"]
if the circles don't overlap, our answer will be $2n+1$ where the $n^{th}$ hexagonal number exceeds 1000. 1,7,19,37,... are the hexagonal number. the nth hexagonal number in $1+3(n)(n-1)$, so the 19th hexagonal number is greater than 1000. our radius is 39.[/hide]

I disagree with the answer given concerning the 1000 nonintersecting circles, each of which has radius of 1.

In one style of packing, there is a "giant" hexagonal shape made, with six wide
sides to it allowing more circles to be placed against each side's center, but underneath the circle's circumference at their respective regions.  However, in the "giant" hex shape at each of the six corners the same number of circles are absent so that the expected diameter is not achieved as was referred to in the above post.  The packing allows the circle to come in closer to the main shape, but still place more circles connected to each of the six sides until they touch or get as close as possible to the circumference.

There ARE other ways, but I was trying to point out a variation on the hexagonal numbers mentioned, AND from the sheer numbers of circles, noticeable gaps develop for additional circles, as well as removal (repositioning of circles) of certain ones to bring down the size of the large radius being sought after.

I just didn't feel like thinking - youre probably right.

[quote="sapphyre571"]umm...it's 1. the 1000 circles are all stacked on top of each other[/quote] :roll: I meant if no circles could overlap.

And you're including that the 1000 circles and the larger circle to be all in the same plane.  Because I COULD, but I know it is NOT the case here, picture these 1000 circles of radius 1 being placed othogonally or skew-like fashion, with the larger circle going around them like a belt.


Say we have a product of symmetric terms (a simple example would be (x+y)(y+z)(z+x)). If we expanded out this product, would the resulting expression necessarily be a symmetric sum?

Thanks.

Yes, as long as you provide complete symmetry.

(i.e. (x+y)(y+z) does not make a symmetric sum, but you might argue that each term itself is symmetric)

Yes, if the whole thing is symmetric with respect to every variable.  In fact, any symmetric expression in n variables is a symmetric sum of itself divided by n!.


$\left\{\begin{array}{c}z\in \mathrm C\ ,\ |z|\le 1\\\ |z^{2}-2z+2|\le 1\\\ |z^{2}-3z+3|\le 1\end{array}\right\|$ $\Longleftrightarrow z=1\ .$

Vezi reply - ul meu de la topicul http://www.mathlinks.ro/Forum/viewtopic.php?t=136521


Let $ A(a_{1},a_{2},a_{3}),B(b_{1},b_{2},b_{3}),C(c_{1},c_{2},c_{3})$ be three non-collinear points in the space. Find the area of triangle $ ABC$

Distance formula + Heron's? :D

Or... cross-product, / 2

[quote="not_trig"]
Or... cross-product, / 2[/quote]

Oh yeah, how could I forget this. Stupid me!

Or more generally, for the area of a polygon formed by planar points in space, you can transform the plane into the Cartesian one, using a transform that keeps area constant, and then apply the [url=http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf]Shoelace Formula[/url]. This is actually basically the same thing as half the cross-product, though in practice it has a lot of extra useless steps in this particular case.

Full process:

- Find the cross product of any two vectors between any 3 points. (it will always get the same direction if the points are indeed planar, though some might point the opposite way)
- You get a result $ <a,b,c>$. Divide each component by the magnitude $ \sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}$ to get a new vector with magnitude 1. This is the normal vector to the plane; call it $ \vec{n}$.
- Take any vector between two of the points, and multiply it by the inverse of its magnitude (so that the result has magnitude 1). Call this $ \vec{u}$. Now define $ \vec{v}\equal{}\vec{u}\times\vec{n}$; $ \vec{v}$ will also have magnitude 1.
- Pick any point on the plane to be your origin, say $ P_{0}$. Then every point on our plane is $ P_{0}\plus{}a\vec{u}\plus{}b\vec{v}$ for some reals $ a,b$. Find the corresponding $ (a,b)$ for every point in the polygon.
- Apply Shoelace Formula. Yay.


topology
LaTeX
superior algebra
superior algebra unsolved

Repost it here on the forum, using $\LaTeX$!


I got this ?equation? in wich i must isolate Q.

It goes like this

y^2 = Q * x - Q * a + k

How would i go about isolating Q. THe closest i can come to is this.

Q - Q = (x * a + k) / y^2

Is that correct ??

And my second problem is this one.
I have to find out what x is.
And what do you call this kind of equation.

(x + 1)(x - 9) > 0

This is how i solved it.

(x + 1)(x - 9) <=> x^2 - 9x + x - 9 <=>  x^2 -8x - 9

D= -8^2 - 4 * 1 * -9 = 100

(-8 +/- Squareroot of 100) / 2 * 1

x = 9 or x = -1

Can you verify if this is correct

[quote="Cless"]I got this ?equation? in wich i must isolate Q.
It goes like this
y^2 = Q * x - Q * a + k
How would i go about isolating Q. THe closest i can come to is this.
Q - Q = (x * a + k) / y^2
Is that correct ??[/quote]
From what I see I can conclude that $Q=\frac{y^2-k}{x-a}$. :?

[quote]And my second problem is this one.
I have to find out what x is.
And what do you call this kind of equation.
(x + 1)(x - 9) > 0
This is how i solved it.
(x + 1)(x - 9) <=> x^2 - 9x + x - 9 <=>  x^2 -8x - 9
D= -8^2 - 4 * 1 * -9 = 100
(-8 +/- Squareroot of 100) / 2 * 1
x = 9 or x = -1
Can you verify if this is correct[/quote]
Sorry, but what actually are you trying to solve?
I guess it is the inequality $(x + 1)(x - 9) > 0$. In this case, answer is $x\in (-\infty,-1]\cup[9,+\infty)$.

BTW, it looks like Getting Started. Choose right section more wisely next time :)

[quote]
From what I see I can conclude that $Q=\frac{y^2-k}{x-a}$. :?

---------------------

Sorry, but what actually are you trying to solve?
I guess it is the inequality $(x + 1)(x - 9) > 0$. In this case, answer is $x\in (-\infty,-1]\cup[9,+\infty)$.

BTW, it looks like Getting Started. Choose right section more wisely next time :)[/quote]

I forgot to write the last part of my second problem.

This is and should be the final result.
L = ]-in ; -1] U ] 9 ; in[
[/quote]

im not too sure on what im supposed to prove though : And its not any easier when i can't explain it well in english since i don't know any mathematical terms in english.

Thanks on the "Isolation of Q" part. It seems right.

And im sorry on not posting this stuff in the right forum.
I wasn't to sure on where it was supposed to be.

EDIT: CAn you show me the process of how you isolated Q if its not to much of a hazzle??

Well. I will move it.

About $Q$:
$y^2=x\cdot Q-Q\cdot a+k$, so $y^2-k=Q(x-a)$.
So what kind of explanations do you need else?

I switched on 2 parts.

y^2 = Q * x - Q * a + k

Is the answer still the same.

Also, can anyone help me with this equation.

I don't know where i should begin with it???

x^3 - 4x = 0

EDIT: <<Well. I will move it. >>

That sounded ironic/sarcastic, could i have moved the thread myself??

If so then please forgive me.

Edit 2: ? y^2 = x * Q - Q * a + k ?  Why do the q and x switch places. When it was  y^2 = Q * x - Q * a + k to start with

[quote="Cless"]I switched on 2 parts.

y^2 = Q * x - Q * a + k
[/quote]
That doesn't change anything. Because of the commutative property of multiplication which says:

$ab=ba$

it doesn't matter if anything's switched since $Q \cdot x=x \cdot Q$.

[quote="Cless"]Also, can anyone help me with this equation.

I don't know where i should begin with it???

x^3 - 4x = 0
[/quote]
Well, I'll give you a hint to point you in the right direction:
[hide="Hint"]Same thing that Myth did in the isolating Q question--reverse distribute[/hide]

By the way, it looks like this post was moved for you.

Thanks for all of your help guys.

I seem to have understood the Q thing almost perfectly.
One thing i don't understand is how Q - Q becomes Q

y^2 = Q * x - Q * a + k <=> y^2 - k = Q * x - Q * a + k - k <=> y^2 - k = Q * x - Q *a

y^2 - k = Q * x - Q *a

At this point, shouldn't the Qs go out with eachother if its  +Q - Q

for the

x^3 - 4x = 0

x^3 - 4x = 0 <=> x^3 - 4x + 4x = 4x <=> x^3 = 4x

divide with x on both sides and we get x^2 = 4 ; x = Squareroot of 4

Is that the correct way to approach it.

[quote="Cless"]One thing i don't understand is how Q - Q becomes Q

y^2 = Q * x - Q * a + k <=> y^2 - k = Q * x - Q * a + k - k <=> y^2 - k = Q * x - Q *a

y^2 - k = Q * x - Q *a

At this point, shouldn't the Qs go out with eachother if its +Q - Q [/quote]
No. The Qs do not cancel each other out because we do not know that $x$ and $a$ are equal. If let's say $x=4$ and $a=3$, then we end up with $4Q-3Q=Q$ which does not cancel. But $x$ and $a$ can be anything, so the only real way we can do this is reverse distribute and get:

$y^2-k=(x-a)Q$

Another way to think of this is using the $x=4, a=3$ example. We end up having $4Q-3Q=(4-3)Q$. Substitute 4 and 3 with x and a respectively and you'll see what I mean.

[quote="Cless"]x^3 - 4x = 0 <=> x^3 - 4x + 4x = 4x <=> x^3 = 4x [/quote]

Not quite. Because it is a cubic equation, we know that there are 3 solutions. The way you did it divided by x and transformed it into a quadratic, which only has 2 solutions. Therefore, your way only finds 2 solutions out of 3 (remember $x=\pm\sqrt{4}$).

The way I would do it is reverse distribute like with the Q. If you cannot get it still, read this: (but [i]do[/i] try first please :D)
[hide="Solution"]
$x^3-4x=0 \implies x(x^2-4)=0$

Since $(x^2-4)$ is a difference of two squares, and we know that $a^2-b^2=(a+b)(a-b)$, we end up with:

$x^3-4x=0 \implies x(x^2-4)=0 \implies x(x+2)(x-2)=0$

In order for this to be true, either $x, (x+2)$ or $(x-2)$ has to be 0. This means $x=0$ is a solution. When $x+2=0$, $x=-2$, and when $x-2=0$, $x=2$. This means that the solutions are ${0,\pm2}$.

Notice that when you did it your way, you had two of the three possible answers.
[/hide]

[quote="Zyloch"][quote="Cless"]One thing i don't understand is how Q - Q becomes Q

y^2 = Q * x - Q * a + k <=> y^2 - k = Q * x - Q * a + k - k <=> y^2 - k = Q * x - Q *a

y^2 - k = Q * x - Q *a

At this point, shouldn't the Qs go out with eachother if its +Q - Q [/quote]
No. The Qs do not cancel each other out because we do not know that $x$ and $a$ are equal. If let's say $x=4$ and $a=3$, then we end up with $4Q-3Q=Q$ which does not cancel. But $x$ and $a$ can be anything, so the only real way we can do this is reverse distribute and get:

$y^2-k=(x-a)Q$

Another way to think of this is using the $x=4, a=3$ example. We end up having $4Q-3Q=(4-3)Q$. Substitute 4 and 3 with x and a respectively and you'll see what I mean.

[quote="Cless"]x^3 - 4x = 0 <=> x^3 - 4x + 4x = 4x <=> x^3 = 4x [/quote]

Not quite. Because it is a cubic equation, we know that there are 3 solutions. The way you did it divided by x and transformed it into a quadratic, which only has 2 solutions. Therefore, your way only finds 2 solutions out of 3 (remember $x=\pm\sqrt{4}$).

The way I would do it is reverse distribute like with the Q. If you cannot get it still, read this: (but [i]do[/i] try first please :D)
[hide=Solution]
$x^3-4x=0 \implies x(x^2-4)=0$

Since $(x^2-4)$ is a difference of two squares, and we know that $a^2-b^2=(a+b)(a-b)$, we end up with:

$x^3-4x=0 \implies x(x^2-4)=0 \implies x(x+2)(x-2)=0$

In order for this to be true, either $x, (x+2)$ or $(x-2)$ has to be 0. This means $x=0$ is a solution. When $x+2=0$, $x=-2$, and when $x-2=0$, $x=2$. This means that the solutions are ${0,\pm2}$.

Notice that when you did it your way, you had two of the three possible answers.
[/hide][/quote]

Thanks for the brilliant explanation on isolation Q, Zyloch, you have my utmost gratitude for that.
Completly understand the process now.

And for the other one. Im sorry but i couldn't resist looking.
My best bet was the 2 out of 3 answer.

I had no idea you could solve it in your way.

What annoys me the most is that i for some reason try to find the easiest way of solving a problem.
This means i exclude all other ways of solving problems, its gotten to a level where i get really sucky grades because of this problem.

Does it just pop into your head that it should be that solution?

Heavens no lol, it doesn't just pop in my head (I wish it did, then I'd be God *sigh*)

I'm having trouble even doing some of the problems in the Intermediate section. Basically, if you do enough problems, you'll get to used to some of the things that they always employ and by looking at a problem, you'll get an idea at what you need for them.

For instance, when you're isolating, you have to realize that if you have a square root, you'll always want to get rid of that first. Then, you just try to move things around until you have what you need on one side. It's not always clean, but you'll have your answer.

There are many questions on these forums and it will really help to try some of them (make sure you don't look at the answers :P) and it will increase your problem solving skills.

As for searching for the easiest way to solve problems, there's nothing wrong with that. If I solved a problem that was difficult for me, I probably won't really even want to [i]look[/i] at it again lest try another path to the solution. However, you have to really make sure that you have the correct (or in this case, all) of the answers.

From the look of things, it looks more that you lack knowledge (like me lol) not that you lack problem solving skills. For instance, you should [i]know[/i] that a cubic has 3 solutions and not just 2. I recommend getting the AoPS books Volume 1 and 2--they teach some stuff that are pretty useful.


Find all primes $ p$ such that $ 4p^2\plus{}1$ and $ 6p^2\plus{}1$ are both primes.

When $ p\equal{}5$, $ 4p^2\plus{}1\equal{}101$ and $ 6p^2\plus{}1\equal{}151$ are both primes.
When $ p\equal{}5k \pm 1$, $ 4p^2\plus{}1\equal{}4(25k^2 \pm 10k \plus{}1)\plus{}1\equal{}5(20k^2 \pm 8k \plus{}1)$ is composite excepth when k=0, but then , p is not prime.
When $ p\equal{}5k \pm 2$, $ 6p^2\plus{}1\equal{}6(25k^2 \pm 20k \plus{}4)\plus{}1\equal{}5(30k^2 \pm 24k \plus{}5)$ is composite.

So, $ p\equal{}5$ is the only solution.


Diophantine equation
number theory unsolved
number theory

Given any positive integer n, show

( sqrt(2) - 1 ) ^n = sqrt(N) - sqrt(N-1)

for some positive integer N

for $ n$ odd, note that $ (\sqrt {2} \minus{} 1)^n \equal{} a_{n}\sqrt {2} \minus{} b_{n}$, where $ a_n, b_n > 0$ and $ (b_n,a_n)$ is a solution to the Pell equation $ x^2 \minus{} 2y^2 \equal{} \minus{} 1$. so we have $ a_{n}\sqrt {2} \minus{} b_{n} \equal{} \sqrt {2a_n^2} \minus{} \sqrt {b_n^2} \equal{} \sqrt {b_n^2 \plus{} 1} \minus{} \sqrt {b_n^2}$. for $ n$ even, $ (\sqrt {2} \minus{} 1)^n \equal{} a_{n} \minus{} b_{n}\sqrt {2}$, where $ a_n,b_n > 0$ and $ (a_n,b_n)$ is a solution to the Pell equation $ x^2 \minus{} 2y^2 \equal{} 1$, so we have $ a_n \minus{} b_n\sqrt {2} \equal{} \sqrt {a_n^2} \minus{} \sqrt {2b_n^2} \equal{} \sqrt {2b_n^2 \plus{} 1} \minus{} \sqrt {2b_n^2}$.


hey I am new does anyone have any tips for learning calculator?

Kate u need to specify the clalculator type some people use different calculators than we do remember this is more of a math counts site (oh and dont mention my name please my dad will get mad)

yes I mean how to use it I have a TI 83 plus and I can't figure it out.

do you have a manual?

they're pretty straightfoward.

Its sort of kind of hard to teach that over the internet be more specific on what u want to know

[quote="BHorseMath"]I have no idea how to do anything past turning it on and some really stuff like adding and subtracting.  I am tecnologicaly challenged.[/quote]

Here you are on the Web, so you must know how to get help in using a Web browser. So here are links that may help with your other questions.

Calculator Links

General Advice on Getting Calculators

http://www.purplemath.com/modules/calc.htm

Calculators Approved for Various Tests

http://www.collegeboard.com/student/testing/psat/about/calculator.html

http://www.collegeboard.com/student/testing/sat/testday/calc.html

http://www.act.org/aap/taking/calculator.html

http://www.collegeboard.com/ap/students/calculus/exam007.html

http://mathcounts.org/Program/rules.html

Calculators Are Needlessly Bad

http://www.uclic.ucl.ac.uk/harold/srf/hucalc.pdf

Graphing Calculator Instructions Sites

http://www.kean.edu/~rnarasim/general/gcguides.html

http://college.hmco.com/mathematics/latorre/guides.htm

http://www.prenhall.com/divisions/esm/app/calculator/

http://www.esc4.net/math/Calculator/intro.htm

http://webpages.marshall.edu/~mitchelk/calculator.html

http://www.abc.se/~m9847/matre/calcula.html

Calculators in Mathematics Instruction

http://www.atcminc.com/mPublications/EP/EPATCM98/ATCMP015/paper.pdf

TI Calculator Programming

http://guide.ticalc.org/index.html

http://tifaq.web1000.com/tifaq.htm?index.html

http://www.angelfire.com/realm/ti_tiplist/

Calculator Bugs and Errors

http://www.math-atlas.org/99/calc_errors

http://www.vcalc.net/ti-accry.htm

http://triton.towson.edu/users/bbhatt1/ti/bugs.pdf

Hope this helps!

What else do you need to do on it? It came with a manual, by the way. Read it.

yes it came with a manual confuted but I'll be darned if I can understand the stuff it's telling me.  The only reason I am trying to learn is because I joined the Math Team at my school and I do well in the type of math thats on the calculator tests.

Rep-I can do the test just not on the calculator I compete with aalinsey100 but not in calculator.  I have only gone 1 and I need to figure out how to do the calculator so that I can help my team.

What does that mean? I couldn't read that post.

I am having trouble with calculator even though I can do calculator problems on paper I can not do them on the calculator.

[quote="Rep123max"]btw a 83+ isnt gonna be good for the calculator plus. you want an hp or casio[/quote]No, you don't. 'specially not a casio.

Yea I compete in the competitions hes talking about u deffinately want to use the hp but I dont know I wish I could do other contests oh well. My school doesnt have any other thing besides that program and whats sad is that is basically the only extra ciricular activity there besides chess club and hadit and teens for christ but Im only on the math team the otherstuff just seemed sort of cheesey to me. Im not going to complain though because Im just gratreful that we even have a math program there so I will just be thankful that we have one and shut my mouth.

Oh, I guess I misunderstood you. As long as you agree that TI makes better calculators than Casio (and HP imho), it's all good.

Yes we do and I sort of agree with the pledge but I will not voice my opinion and it isnt illegal because it is extrciricular at the highschool they have a pagans club and they have teens for christ also I didnt like the idea of it though because I knew it would offend some of those whom arent believers of christianity yet none of the people at my school are jewish we just have athiasts and juhovas witness. I have told the principle what I think thats part of the reason he doesnt like me but I told him what I thought and now he knows of at least one person who is concerned. I dont write for the school newspaper I write for the PTO newsletter though. But u cant voice ur opinion in there. I went to one teens for christ meeting and they dont even sxcept the catholic religion there was a big protest at my school for it so the principle allowed it. There is no onw who agrees with me that its worth it to fight it so I just ignore it now

Well thats true but the law is that there is aloud to be things like that if administered and set up by a student (which this is not by an adult the principle just gave consent to those who wanted to start it because there was a school petition signed and last year they asked me to sign it and I refused to.) Anyways, pagans dont worship our God but they do worship a God or Gods so still that is a religious thing because thats what they believe in. When we are saying the pledge does it occur to u that some people might think of it as their God and not our God? Maybe the say Gods instead of God. Think, if everyone was allowed to express in the pledge who their God was then would u say the pledge? Just a question for u to ponder on there for a sec.

[quote="Rep123max"]so your saying you go to a public school and they have teens for christ? that is 100% illegal.[/quote]

There have been court cases about that, and a Congressional statute about that which was reviewed and approved by the United States Supreme Court in the case [url=http://caselaw.lp.findlaw.com/scripts/getcase.pl?navby=search&court=US&case=/us/496/226.html]Westside Community Board of Education v. Mergens[/url], 496 U.S. 226 (1990). IF (and we could say, a la mathematics, "if and only if") the club is organized by students, it is perfectly legal. The term "separation of church and state" doesn't occur in the United States Constitution or any of its amendments--look it up if you don't believe me. The term "free exercise of religion" does occur in the First Amendment to the Constitution, the first article of the Bill of Rights, and that is what is being put into practice by such student clubs. If the teachers in your school teach you otherwise, they are giving you wrong information at taxpayer expense.

The phrase you mention comes from a personal letter written by Thomas Jefferson (who, by the way, had no role in drafting the United States Constitution, being United States representative to France at the time). Let's hear what Thomas Jefferson said, by the way, about compulsory school attendance:

"It [should not] be proposed to take ordinary branches [of education] out of the hands of private enterprise, which manages so much better all the concerns to which it is equal." -- Thomas Jefferson, sixth annual message to Congress (1806), reprinted in The Writings of Thomas Jefferson (memorial ed. 1907), vol. 14, p. 384

"But if it is believed that these elementary schools will be better managed by the Governor and Council, the commissioners of the literary fund, or any other general authority of the government, than by the parents within each ward, it is a belief against all experience. Try the principle one step further and amend the bill so as to commit to the Governor and Council the management of all our farms, our mills, and merchants' stores." Letter of Thomas Jefferson to Joseph Cabell, Feb. 2, 1816, reprinted in Political Writings of Thomas Jefferson (1955), p.98 and The Writings of Thomas Jefferson (mem. ed. 1904), v. 14, pp. 420-21

"Is it a right or a duty in society to take care of their infant members in opposition to the will of the parent? . . . It is better to tolerate the rare instance of a parent refusing to let his child be educated, than to shock the common feelings and ideas by the forcible asportation and education of the infant against the will of the father. What is proposed here is to remove the objection of expense, by offering education gratis . . ." Letter of Thomas Jefferson to Joseph Cabell, September 9, 1817, reprinted in Writings of Thomas Jefferson (mem. ed. 1904) v. 17, p. 423

Hmmm, this is a pretty bad case of topic drift now, and this part of the thread really belongs in the Round Table forum, but anyway let me recommend a link, sponsored by a BROAD variety of concerned organizations, that explains the law on this subject.

http://www.firstamendmentcenter.org/rel_liberty/publicschools/topic.aspx?topic=religious_clubs

I mean like if we just said our Gods name like if u were muslim u would say I pledge.... one nation under allah and wed say one nation under God. Like let people express in the pledge the God they believe in to diversify the US's religion and let other countries know that we have full spiritual diversity and religious freedom here in the US. Sure we all wouldnt be saying the same thing but it would satisfy those who dont believe in our God.

The 'seperation of church and state' isn't the free exercize clause, it's this one:  "Congress shall make no law respecting an establishment of religion."  In this case what that means is that there can be no preference for one religious group over another -- so the school couldn't prevent a Jewish student group from meeting while allowing a Protestant group to meet.  It isn't unconstitutional for religious groups to use school property, however.  If the school paid someone in to teach them Christianity, it would be problematic.

well I dont know about that I guess u might say the presidents name if that came to be the case (sarcasim.) I have another question for u. Does ur hate for Bush influence ur decision in any way just because he Republican? I happen to think hes doing a good job with this. Let me ask u do u know everything thats going on right now with the world and the country and did u vote (well that has an obvious answer we are to young to vote?) Well  then how can u judge the president? what information do u have to convict him of doing anything wrong? Do u realize that humans despite how hard we try are not perfect and never will be?

[quote="Rep123max"]i havent finished what you read tokenadult but i didnt want to forget to comment. you are correct about separation of church of state not appearing in the constitution at all (i didnt look it up, i know most of the key phrases of the constitution by heart, i really enjoy us history) [b]but[/b] it does say "congress shall make no law respecting an establishment of religion" which means endorsing a religion.[/quote]

Actually, what the "Congress shall make no law respecting an establishment of religion" (note the indefinite article, and I DID just look this up to be absolutely sure my memory is correct, as it has been for years about the wording of the Establishment Clause) phrasing means, in historical context is

BOTH

1) Congress is not permitted to establish a national religion, a policy I fully agree with,

AND

2) Congress is not permitted to disestablish any religion established by a state (as Congregationalism was in Massachusetts at that time, or some other denominations were in some other states at that time).

Years after the amendment to the federal Constitution became law, some states that still had established religions quietly disestablished them by legislative action in each state. Many more years after that, about a whole lifetime after the drafting of the Constitution, school attendance first became compulsory. I think I am fully agreed with you that it would be an outrage to have any compulsory religious observance in a school, enforceable by law (the majority of my ancestors left Europe to come to America precisely with regard to such issues). But I hope you can understand that it is also an outrage to compel attendance in what some people attempt to make religion-free zones, namely the current government-operated school system. Privately organized activities to exercise religious convictions ought to maintain the freedom the framers of the Constitution marked out for them. The United States is enormously diverse in the religious heritages that have gone into making up the current United States populace. (I have specifically studied and observed the religious diversity of the United States in work I had as a Chinese-English interpreter for official visitors and as a constitutional lawyer.)

Just as the government isn't allowed, and shouldn't be allowed, to prefer one religion over all others, it also isn't allowed, and shouldn't be allowed, to carve out whole areas of human life--for example, educating young people--as areas in which religiously motivated conduct is disparaged or discouraged. The phrase in the Constitution "free exercise [of religion]" was carefully chosen to describe a condition of general freedom--a condition of freedom that was a powerful inducement to many people around the world to become Americans, both before and after the drafting of the Constitution. It's not your business to tell other people what they should or shouldn't discuss in their club meetings, unless their club is a criminal conspiracy or something like that.

Whew! This is BIG TIME better suited for the Round Table forum now than for this Getting Started forum.

I haven finished reading ur post tolkenadult but this is in reference to Reps statements and questions. The answer to the allies question none I totally agree that we are in a mess right now and well Ill email u my thing I wrote about it later Rep but I dont know I will not judge by what I see. Like a famous person said u cant judge a book by its cover and thats what we are doing is judgeing the situation that we are in on how it appears to look and Im totally agianst that.

Well I dont know why we get sent to the principals office for not saying the pledge but I do know why he edits the paper well come to think of it our paper doesnt pass through an adult before it is printed hmmm.... u lost me on that one. What I do know is that we are supposed to be a nation, a country and what are we doing upsetting it. Why arent there any simple solutions anymore now days? Why do we have to worry about the rest of the world? Why do we have to argue over someones opnion? It seems like thats the only reason a war ever starts is a conflict of intrest. I mean why is my question? We are arguing with other countries we are arguing within our own. How long will it take for people to realize that everyone has their own opinion and its not going to stop just because we argue over it? How come?

So, instead of disagreeing with policies that we don't believe in, we should just let them happen?  That's totally opposed to the idea of democracy -- the whole point of democracy is that everyone has a voice in the political process, everyone can complain if something is going wrong.  What you're advocating -- "Just stop complaining, everybody" -- is the principle that dictatorships work on.  If you'll just be quiet and let Dear Leader run the country, everything will be fine.

As for disagreeing with Bush -- I don't think all Republicans are bad.  The governor of my state (New York) has quite reasonable politics in many respects.  I can provide you with a long, long list of reasons on the domestic front that I don't like him as president.  I think you're too willing to assume that just because you haven't heard about anything particularly bad nothing particularly bad has been done by this administration.

okay I can't let this happen.  My parents were in the military and I understand the setup that bush is in.  He really did not have much chioce of what to do.  He could have left it alone but that would probally have been worse they had weapons of mass desttruction and they could have used them against us and many others. As for allies if he did not make that move they could have left us saying we were afraid to go to war if nessasary.  Don't forget they could have gotten attacked too.  Bush may not get realected but I hope he does I fully support him but somethings that he could not prevent happened in his term.

Just because I said that we need to realize that we need to stop arguing doesnt mean I think we should sit by and watch are country be torn to peices by one person. I think u should speak out but I dont think we should argue over the little things and make them seem big. I believe in standing up for what u believe in if I didnt I wouldnt be having this conversation right now Im an optomist. I like Bush for some reasons but I dont for others. I just dont think u can judge a person unless u know all the facts which we dont and I know there are problems  right now and I think Bush had something to do with getting us into them but I dont think it was all his fault and alot of the problems that appear are not the real problems. Media makes it look as if so but its really not. Why dont we just fight our own war we have in our own country we are a country at eachothers necks and we dont do anything about it.

I totally agree with u there we shouldnt of done anything but what if he had had something? It was a possibility we were trying to save the world which was sort of stupid to do because we need to save our country before we can save anyone elses.

rep my parents were in the military I knowwhat happens.  The solders do not die at 2 a day.  one of the raesons that we are in Iraq is because they hate us and if they have weapons and they still could the military is not sure if they did or do or what, they would probally use them against us if the U.S. had not steped in!  My parents were intellingence officers and they have told me exacly what is going on.

I know but has texas ever had a giid governer?

I know but has texas ever had a good governer?

Bush was a good govener and now he is a good president!

Yes I know about the reconstruction but I didnt know that we had a governer named ann richards. I dont normally pay attention to governers I need to start though.

We aren't doing anything about North Korea right now true but we did not know that Iraq did not have WMDs.  Besides look at the repercusions of what might have happened if we did not step in.  Iraq agreed years ago about the weapons.  Bush also tried to avert war but Iraq did not listen did they?

BHorseMath, I may be misinformed about this, but I was under the impression that Bush did not "try to avert war." He made some rather unreasonable demands (Iraq must allow inspectors into every suspicious area of the nation, Iraq must immediately cease all "weapons production" within a tiny window of time). It seems more likely that Bush tried to goad Iraq into declaring war first, but of course they didn't -- he then tried to find some excuse to get it started.

Oh sorry Ive only been into politics for about 3 years so I had no clue who the governer before Bush was Im embarassed to say.

[quote="Syntax Error"]can you like move this topic to round table or something[/quote]
I agree, I have absolutely no idea how a discussion about a calculator turned into [b]this[/b] :lol:

Reference: [url=http://www.albinoblacksheep.com/video/bushbush.ram]George "Dubya" before Presidency[/url]
Lesson to be learned: Never trust a politician


linear algebra
matrix
algebra
polynomial

So I got most of the problem below, but am unsure how to complete the remainder of it.  Namely, the last sentence is what is getting me.  I've posted what I've figured out thusfar.

$ \underline{\bold{\text{Problem: }}}$ Let $ T$ be a linear operator on $ \mathbb{R}^3$ which is represented in the standard ordered basis by the matrix:
\[ \left[ \begin{array}{ccc} 6 & \minus{} 3 & \minus{} 2 \\
4 & \minus{} 1 & \minus{} 2 \\
10 & \minus{} 5 & \minus{} 3 \end{array} \right]
\]
Express the minimal polynomial of $ T$, $ M_T(x)$, in the form $ p_1 p_2$, where $ p_1$ and $ p_2$ are monic and irreducible over the field of real numbers.  Let $ W_i$ be the null space of $ p_i(T)$, $ W_i \equal{} N[p_i(T)]$.  Find bases $ \mathcal{B}_i$ for the spaces $ W_1$ and $ W_2$.  If $ T_i$ is the operator induced on $ W_i$ by $ T$, find the matrix of $ T_i$ int he basis $ \mathcal{B}_i$ (above).

$ \underline{\bold{\text{Solution: }}}$ We have $ V \equal{} \mathbb{R}^3$, $ T \in L(V,V)$ such that $ T \equal{} T_{\mathcal{A}} \equal{} A$.

Then $ C_A(x) \equal{} M_A(x) \equal{} (x \minus{} 2)(x^2 \plus{} 1)$ where $ C_A(x)$ is the characteristic polynomial.

Then $ V_1 \equal{} N[A \minus{} 2I_3] \equal{} \text{col. space of } (A \minus{} 4I_3)^2$

$ \implies V_1 \equal{} \text{span} \left\{\left( \begin{array}{ccc}5 \\
0 \\
10 \end{array} \right) \right\}$ with basis $ \mathcal{B}_1 \equal{} \left\{\left( \begin{array}{ccc}1 \\
0 \\
2 \end{array} \right) \right\}$

And $ V_2 \equal{} N[(A \minus{} 4I_3)^2] \equal{} \text{col. space of } A \minus{} 2I_3$

$ \implies V_2 \equal{} \text{span} \left\{\left( \begin{array}{ccc}4 \\
4 \\
10 \end{array} \right) , \left( \begin{array}{ccc} \minus{} 3 \\
\minus{} 3 \\
\minus{} 5 \end{array} \right) \right\}$ with basis $ \mathcal{B}_2 \equal{} \left\{\left( \begin{array}{ccc}1 \\
1 \\
0 \end{array} \right) , \left( \begin{array}{ccc}0 \\
0 \\
1 \end{array} \right) \right\}$

I don't know how to do this last part.

This means
$ T_1 : W_1 \rightarrow W_1$ and $ T_2 : W_2 \rightarrow W_2$.
$ T_1 : v \mapsto Av$, $ T_2 : v \mapsto Av$.

You now have 2 smaller mapping induced from the mapping $ T$. And you want to find a matrix representing it with respect to your basis. (When talking about matrix representation of a linear map, we need to specify the basis.) These representations are exactly their companion matrices.

$ [T_1]_{\mathcal{B}_1} \equal{} (2)$

and $ [T_2]_{\mathcal{B}_2} \equal{} \begin{pmatrix} 0 & \minus{} 1 \\
1 & 0 \end{pmatrix}$

By the way, I think that you 've made a mistake.

$ W_1 \equal{} span \{ \begin{pmatrix} \minus{}1 \\ 1 \\ 0 \end{pmatrix} \}$

and

$ W_2 \equal{} span \{ \begin{pmatrix} \minus{}2 \\ 0 \\ 1 \end{pmatrix} ; \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\}$

How did you precisely attain those companion matrices?  My professor went over this problem in class today but failed to demonstrate how he did this.  There is a language barrier so I'm asking it here.

Also, he got the same solution as I for $ W_1$ and $ W_2$.  I also checked it using an alternative method and found the same result.


Well, I know the answer, but I don't know how to prove it. :?

lol :lol:

[hide="hint"]
[hide="manipulate the sum..."]
[hide="divide by..."]
3! :D
[/hide]
[/hide]
[/hide]

[quote="mathnerd314"][quote="danthemanvt"]What is Mu Alpha Theta?    :?[/quote]

Notice that they are three Greek letters that make the sounds m, short a (as in cat), and th (as in bath). And it spells out "m-a-th".[/quote]

:lol: Lol, I never knew that, but here's their actual site: http://www.mualphatheta.org/

[hide]
The sum would be $S=\frac{2}{9}+\frac{4}{27}+\frac{6}{81}+.....$ Notice that we can't use the infinite series formula just yet, because the numerator is all screwy. Thus, we want to make the numerator unscrewy by making the numerator all the same number, 2. So, $\frac{S}{3}=\frac{2}{27}+\frac{4}{81}+\frac{6}{243}+.....$.

$S - \frac{S}{3}=$
$\frac{2}{9}+\frac{2}{27}+\frac{2}{81}+....$

Summing this using $\frac{a}{1-r}$, we get

$\frac{\frac{2}{9}}{1-\frac{1}{3}}=$
$\frac{1}{3}$
[/hide]

Nice...finish your solution though! :) (pay attention to what you did; you got a sum, but did you get the answer yet??)

Oh, darn it  :oops: , the solution would be
[hide]
$\frac{2S}{3}=\frac{1}{3}$
$S=\frac{1}{2}$[/hide]

[quote="danthemanvt"]What is Mu Alpha Theta?    :?[/quote]check out the website [url=http://mualphatheta.org]mualphatheta.org[/url]


Let three non intersecting circles and draw their six common internal tangents. Consider the hexagone that these six lines define. Prove that its main diagonals concur.

It has already been posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=140096

Kostas vittas.


Here's solution:

$2006=\sum x_i\geq n\cdot \sqrt[n]{\prod x_i}$

sorry , $x_{i} \in Z^{+}$   hence we can't use AM-GM


topology
geometry
calculus
derivative
Intel STS

I am a high school student in India and am planning to apply for US Admission for the Class of 2010.

Since I started thinking about my college entrnace tests and applications, I've been hearing about the research high school students in America. I've heard about the Intel STS , ISEF, Seiemens Competition and RSI etc... My question is regarding what exactly do the students  do?? I'm not asking about the camps, competitons, etc,... but just what research do students do?? Is the best one sort of pathbreaking!? Or is it just a sort of (in Maths and Science) proving or verification of corrolaries/conjectures etc..? Can you please give some examples og the best high school research? I've heard many universities asking for a research paper(optional) ?? Does it have to be published in a Journal???

[b]If You didn't have the time to go through the poorly written para above, my basic question is:
1) Is the best 'research' in American High school system in Math and Science(say at RSI, INTEL,) pathbreaking and pure genius? Or is it sort of verification and maybe a different, inegenious approavh/proof of theorems, previous results, generlaisationsetc.? SAy, eg, the research on manifolds in INTEL, ... was it really 'senior research' level? ( I hope you understand!! I can clarify if required)

2) PLease give examples of the best High School research in Math and Science(Physics).[/b]
Thanks!!!

go to the websites for the various contests. They will have at least abstracts, and sometimes links to entire papers, of previous winners.

Can you plz give an example? Also, what do I compare those papers with? How do I know whether that work is 'original' and 'pathbreaking' (no offence) in their respective fields???

In my high school, students wrote letters to potential mentors, who were usually like research scientists in universities. This is during sophomore year, I think. If someone responds, which is highly probable because our school has a bit of a reputation, the student would meet with the mentor occasionally to help him/her do his/her research. This research is original, not like the school science labs. You learn a lot about the subject and can branch away from the mentor's experiments to do your own research, but you will still have access to the mentor, as a source of guidance and resources.

[url=http://www.people.fas.harvard.edu/~gcarroll/math/intel.pdf]Here's[/url]
 an example of a project.

regarding the research project posted, since I havent much knowledge of the subject it is on, can you plx tell me whether such work has been done before or not???

I don't know it either, but note the "Thanks to ... for confirming the originality of the main results".

It's dangerous to take anything Gabriel Carroll did as an example- that was a winner, and no ordinary person could duplicate it.

The technical background for that paper is really graduate-level math.

I also had a paper published when I was still in high school- since it wasn't very long and the timing wasn't right for the contests (this result grew out of discussions with the professor of an analysis course I was taking), we went for publication in a journal, with my name one of three coauthors. That was an original result too, on an obscure problem that had only been studied for a few years. It wasn't very technical, but it was a confusing argument.
It's still a powerful bragging point (my Erdos number is 2!)- any original research is exceptional.

Sorry for confusing anyone with that example - obviously I have no idea what it's about. I just wanted to point out one of the most succesful examples of a research project.

[quote="Rushil"]Do you think a project on the geometrical analysis of the general Fermat point be considered original???[/quote]

Only if it's actually original.  This is impossible to answer without you having done anything or even offering some hints as to what kind of results you hope to establish.  If you're doing research, it's your job to find out what is already known.

How hard is it to find something truly original at a highschool level? (I'm talking something more like EE material, not Intel, blahblahblah)

[quote="WindSlicer"]How hard is it to find something truly original at a highschool level? [/quote]
It's very hard. It's still hard at the college level, and most of the accessible problems that are not already done require postgraduate material to understand and attack.
You need quite a bit of luck/focused depth to do it, and there are no guarantees of success. It's essentially impossible without knowing a significant amount of college material in the particular area you're focused on, but narrow knowledge might be enough.

By the way, what is EE?

Extended Essay, for the IB (International Baccalaureate) Program, which is something like a 3000 word essay on any subject (i'm not quite sure on the guidelines for math). I think blahblahblah might know more details, but I can't specifically remember if he did his EE in mathematics or not. Either way, the content doesn't have to be like that of the Intel example above, but I think it has to be at least semi-original. Or maybe I'm wrong, do you know blahblahblah?

I did my EE in math - it was essentially a survey paper on the basics of calculus of variations and some applications to lagrangian physics.  The word limit, plus the lack of depth of the IB curriculum (I had to spend a large amount of my word count explaining things like partial differentiation, etc) meant that my essay covered roughly the first five weeks of a 400-level course at my university.  In particular, it certainly wasn't original - I drew most of my material from a pair of dover books, added a couple of basic examples of my own devising, read a bit of a russian text that was too advanced for me at the time, and wrote a bit of Maple code.  Now, I make it sound trivial in retrospect, but I still probably put more than 50 hours into it.

blahblahblah,

Did you get good grade for your EE? One student (really good with math) in the IB program was warned by the IS teacher that the highest grade students from that IB program ever got was C for their EE with math topics.

Yes, I got an A - I was one point from a perfect score, and you can lose about 5 points and get an A.  Another person from my school did his on the history of the Riemann hypothesis (I helped him out a great deal  :D ); he also got an A.

As long as the student makes sure that he follows the IB criteria (word limits, documentation, use of 'technology', etc...) and makes sure that he doesn't confuse the markers, he should do fine.  If you want, I can send you my EE as a point of reference.


A bug crawled along a 45 degree arc of length 5 cm. How many centimeters long is the circle's radius? Express your answer as a common fraction in terms of $ \pi$.
[asy]draw((-1,0)..(0,1)..(1,0)..(0,-1)..cycle);
draw((12/13,-5/13)--(0,0)--(12/13,5/13));
label("5 cm",(1,0),E);[/asy]

The arc is $ \frac {45}{360} \equal{} \frac {1}{8}$ of the entire circle, so the circle's circumference is $ 5 \cdot 8 \equal{} 40$ cm. Thus, the radius is $ 40 \div 2\pi \equal{} \frac{20}{\pi}$.

EDIT: Typo, thanks jxl. :)

Izzy, don't you mean $ 40\div(2\pi)\equal{}\boxed{\frac{20}{\pi}}$?


All of the triangles in the figure and the central hexagon are equilateral. Given that $ \overline{AC}$ is 3 units long, how many square units, expressed in simplest radical form, are in the area of the entire star?
[asy]import olympiad; import geometry_dev; import graph; size(150); defaultpen(linewidth(0.8));
pair[] vertices;
for(int i = 0; i < 6; ++i){
  vertices[i] = dir(30 + 60*i);
}
draw(vertices[0]--vertices[2]--vertices[4]--cycle);
draw(vertices[1]--vertices[3]--vertices[5]--cycle);
label("$D$",vertices[0],NE); label("$C$",vertices[1],N); label("$B$",vertices[2],NW); label("$A$",vertices[3],SW);
label("$F$",vertices[4],S); label("$E$",vertices[5],SE);[/asy]

[hide="Solution"]
[hide="Formula for area of equilateral triangle"]An equilateral triangle with side length $ s$ has area $ \frac{s^2\sqrt{3}}{4}.$[/hide]
[hide="Insight-ish"]We see the whole figure is composed of $ 12$ equilateral triangles with side length 1.[/hide]
[hide="Area"]The area of each triangle is $ \frac{1^2\sqrt{3}}{4}\equal{}\frac{\sqrt{3}}{4}.$ Since there are 12, the total area is $ 12\left(\frac{\sqrt{3}}{4}\right)\equal{}\boxed{3\sqrt{3}}.$[/hide][/hide]


limit
probability
inequalities
probability and stats

suppose that $ X_i$ are independant random variables such that $ \sum \frac{Var(X_k)}{k^2} < \infty$. Prove that if $ \lim \frac{\sum_1^n E(X_k)}{n} \equal{} L <\infty$, then $ \lim \frac{\sum_1^n X_k}{n} \equal{} L$ almost everywhere

This is known as Kolmogorov's strong law of large numbers :) The proof can be found in many (but not all) probability textbooks. It is short enough to be posted here but not completely trivial. In a nutshell, what you can do is to assume that all expectations are $ 0$, split the sum into dyadic pieces, and use Kolmogorov's favorite inequality $ P\Bigl\{\max_{1\le k\le n}\sum_{j\equal{}1}^k X_j\ge t\Bigr\}\le t^{\minus{}2}\sum_{j\equal{}1}^n V_j$ for each piece, after which Borel-Cantelly yields the result in almost no time. Interestingly enough, the most often used strong law for the identically distributed $ X_j$ with finite expectation is a simple corollary of this result. Can you see why?

thanks for your proof, it is very neat.
Since I did not know this Kolmogorov inequality, let me give a proof of it:
Define $ M_n \equal{} \sum_1^n X_k$. This is a martingale, and $ X_n^2$ is a submartingale. Introduce the stopping time $ T\equal{}\inf\{n \geq 1: M_n > t \}$ and notice that $ E(M_{\min(T,n)}^2) \leq E(M_n^2)$ since $ \min(T,n) \leq n$ and $ M_n^2$ is a submartingale. Because $ t^2 1_{T \leq n} \plus{} M_n^2 1_{T \geq n\plus{}1} \leq M_{\min(T,n)}^2$, the conclusion easily follows: $ t^2 P(T \leq n) \leq E(M_n^2 1_{T \leq n}) \leq E(M_n^2) \equal{} \sum_1^n V_k$

And this is surprising that the strong law of large number follows from this !! I think that one can do as follows:
take $ m_i$ such that $ P(|X_i|>m_i)$ satifies $ \sum_i P(|X_i|>m_i) < \infty$ and $ E(X_i 1_{|X_i|>m_i}) \to 0$. So, by Borel-Cantelli, one knows that almost surely $ X_n(\omega) \equal{} Y_n(\omega)$ for $ n$ large enough, where $ Y_n \equal{} X_n 1_{|X_n| \leq m_n}$. So (almost surely) $ \frac{\sum_1^n X_n(\omega)}{n}$ converges towards $ \lim_{n \to \infty} \frac{\sum_1^n E(Y_k)}{n}$. And because one can suppose that $ E(X_n)\equal{}0$, then $ E(Y_n) \to 0$ since one assumed that $ E(X_n 1_{|X_i|>m_n}) \to 0$, so  $ \lim_{n \to \infty} \frac{\sum_1^n E(Y_k)}{n}\equal{}0$.

(I am surprised because the only proofs I have seen before of the strong law of large number were way more complicated, and I did not use the fact that the random variables had the same law, just that a first moment exists and independance ? there is a flaw somewhere ..)

You forgot to check that the variance condition is satisfied :( but otherwise you are on the right track: the cutoff for $ X_i$ should be done at level $ i$. Then you'll have two conditions to check: one for Borel-Cantelli and one for variances (plus, of course, there will be a stupid remark about slightly shifted expectations) :)

Thanks, I got it  :)  Now, it remains to give a counter example to the false law of large number where the $ X_i$ are independant, have zero expectation, but do not have the same distribution (wich is easy now that one knows where was the flaw in my 'proof') :lol:

PS: for another proof of the initial problem, one can notice that $ M_n \equal{} \sum_1^n \frac {X_k}{k}$ is a martingale bounded in $ L^2$, thus converges almost everywhere if one suppose $ E(X_n) \equal{} 0$. Hence $ \sum_1^{\infty} \frac {X_k(\omega)}{k}$ exists almost everywhere, which shows that $ \frac {1}{n} \sum_1^n X_k(\omega)$ converges towards to $ 0$ almost everywhere.


Let $f(x)= 2x^6- 3x^4+x^2$. Prove that $f(sin \alpha)+f(cos \alpha)=0.$

EDIT: OOPS! This isn't up to Pre-Olymp level, it is pretty easy. Can someone move to intermediate?

[hide]
\[ f(x) = x^2(2x^4 - 3x^2 + 1) = x^2(1-x^2)(1-2x^2) \]

so when we substitute we have $\cos^2{\alpha}\sin^2{\alpha} * -cos^2{2\alpha} + \sin^2{\alpha}\cos^2{\alpha}*\cos^2{2\alpha}=0$[/hide]


What were the results at state? I know that Albert Chu and Derryl Wu Got first and second, and that the team winners were most likely Lakeside, Odle, and Washington, but I would like all of the results.

I'm not from WA but I did know that Kathleen Zhou from Odle MS got 3rd, and John Zhang got 4th.

Also Darryl Wu got 1st in written with a 46 (perfect!) and Albert Chu started out 10th in written, before the countdown round, and beat everyone else to get 1st.


Let $S=\{0,1,2,\dots,1994\}$. Let $a$ and $b$ be two positive numbers in $S$ which are relatively prime. Prove that the elements of $S$ can be arranged into a sequence $s_{1},s_{2},s_{3},\dots,s_{1995}$ such that $s_{i+1}-s_{i} \equiv \pm{a} \ \mbox{or} \pm{b}$ $(\mod 1995)$ for $i = 1,2,\dots,1994$.

[i]Lemma.[/i]
For $p,q\in N$ so that $gcd(p,q)=1$ all of the members of the sequence
$0, p, 2p, 3p, \dots , (q-1)p$ are different mod $q$.
[i]Proof.[/i]
Otherwise there would have to be a member $k p$ ($q>k>0$) such that $q | k p$, i.e. $q | k$.
Back to the main proof, let $m=gcd(1995,a)$, $n=gcd(1995,b)$ and $k=\frac{1995}{m}$. I define members of the rearranged sequence as follows (everything mod 1995):
$s_1=0$; $s_i=s_{i-1}+a$ for $i=2, 3, \dots, k$
$s_{k+1}=s_k+b$; $s_i=s_{i-1}+a$ for $i=k+2, k+3, \dots, 2k$
$\dots$
$s_{1995-k+1}=s_{1995-k}+b$; $s_i=s_{i-1}+a$ for $i=1995-k+2,1995-k+3, \dots, 1995$
From the lemma it follows that all the numbers $s_i$ are different, because $gcd(a,k)=1$ and $gcd(b,m)=1$, and from the definition the sequence satisfies the necessary property.