$|x+2|+|y-3|=1$ is an equation of a square. How many units are in the lengths of the diagonals of the square?

[quote="Thales418"]2

this is the same as

$|x|+|y|=1$

because of simetry and the slopt

$|0|+|y|=1$   $y=1$ or $y=-1$
$|0|+|0|=1$   $x=1$ or $x=-1$

the total distance is 2...

its easy, just see the graphic...[/quote]

What graphic?

Based on your information, the diagonal would be $\boxed{2\sqrt{2}}$.

[quote="Thales418"]this is the same as

$|x|+|y|=1$[/quote]

How?

What is an equation of a square?

My bad, the answer should be $\boxed{2\sqrt{2}}$.

The diagonal of a square is $\sqrt{2}$ times the side length.

no

by symetry  the line $y=x$  have 45º

the square is rotate 45 degres...

the vertices is on $(-1,0)$  $(1,0)$  $(0,1)$ $(0,-1)$

a proof of this is just solving the equation.........

blargh

Exactly! Therefore the side length of the square is $2$, and the diagonal would be $2\sqrt{2}$.


[quote="amirhtlusa"][hide="this seems to work"]
let $x=b+c-a,y=a+c-b,z=a+b-c$; x,y,z>0
the LHS turns into:
$\frac{1}{2}(\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{x}{z}+\frac{z}{x})$
done.[/hide][/quote]

so, how are you supposed to come up with that substitution? :wink:

It is a natural substitution if $a,b,c$ are the lenghts of triangle sides.

[quote="e^pi"]given $a,b,c$ are sides of a triangle; prove:

$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\geq 3$[/quote]
Another way:
Let $a+b+c=3.$ Then
 $\sum_{cyc}\frac{a}{b+c-a}-3=\sum_{cyc}\left(\frac{a}{3-2a}-1\right)=$
$=\sum_{cyc}\left(\frac{a}{3-2a}-1-3(a-1)\right)=\sum_{cyc}\frac{6(a-1)^{2}}{3-2a}\geq0.$ :)

Arqady, you might want to explain your final step a little more, this is the beginners forum... it isn't really so obvious...

[hide="my solution"]
\[\iff \frac{a^{2}}{ab+ac-a^{2}}+\frac{b^{2}}{ab+bc-b^{2}}+\frac{c^{2}}{cb+ac-c^{2}}\ge 3\]
We can apply Cauchy since we are given that $a+b-c$ etc. is positive, thus
\[LHS \ \ge \frac{(a+b+c)^{2}}{2\sum ab-\sum a^{2}}\]

Now \[(a+b+c)^{2}=\sum a^{2}+2\sum ab\geq \sum a^{2}+2\sum ab+(4\sum ab-4\sum a^{2})=\]\[=3(2\sum ab-\sum a^{2})\]

It follows that \[\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\ge \frac{(a+b+c)^{2}}{2\sum ab-\sum a^{2}}\ge 3 \][/hide]


calculus
integration
function
calculus computations

If K(x) is a continuous function and $\int^{8}_{0}K(x)dx=5$, find $\int^{2}_{0}3xK(2x^{2})dx$.

I have been out of school for a while and forget some calculus concepts.  Any help would be appreciated.

Let $2x^{2}=t,$ we have that your integral can be written as $\int_{0}^{8}3\cdot \frac{1}{4}K(t)dt=\frac{3}{4}\cdot 5=\frac{15}{4}.$


calculus
integration
analytic geometry
trigonometry

[quote="kstan013"]a) why are you posting in classroom math
b) why are you posting in a middle school forum
c) a mod should move this
d) this isn't the only forum, you know :wink:[/quote]

yeah this should be in pre olympiad or something...

[quote="Josef"]Could anyone help me with this?
[img]http://www.astro-grpr.wz.cz/img20005.jpg[/img][/quote]
WTF????? What does that sign mean? Why is this in classroom math? Why am I shouting? I don't even know the answer to my own question.

[quote="Josef"]Could anyone help me with this?
[img]http://www.astro-grpr.wz.cz/img20005.jpg[/img][/quote]

Jesus christ! Is this a joke?

[quote="bpms"]This isn't a joke, it's only integration, this guy is obviously new. So this should be sent to the Calculus forum.[/quote]

Thanks.

Since no one has solved it yet...

$\int^{1}_{0}\sqrt{x(2-x)}= \int^{1}_{0}(x(2-x))^{1/2}= \int^{1}_{0}(2x-x^{2})^{1/2}$

and then apply the Power Rule

then subract $\text{f(1)-f(0)}$

i'm not sure if the last part will work to ur advantage my calc book says to unsimplify a lot of things

at least i think thats right...i just learned basic integration this morning  :lol:

Ahhhh.... Making no sense. Can a mod please move this?

[quote="Josef"]It's pi/4 :) well there's simple solution... if you release that y^2 = 2rx-x^2 is equation of a circle.[/quote]

i don't think thats right...nothing in the problem that has anything to do with $\pi$ or circle equations

In Cartesian coordinates, the equation of a circle of radius r centered on (r,0) is
(x-r)^2+y^2=r^2 => y=sqrt(2rx-x^2). r=1 => y=sqrt(2x-x^2).

on another note here is a question about solving integrals

when you have a question where the lower limit i think it's called on an integral is 0, can't you just solve for $\text{f(n)}$ where $n$ is the upper limit?

[quote="Necrowarrio0"][quote="Josef"]Could anyone help me with this?
[img]http://www.astro-grpr.wz.cz/img20005.jpg[/img][/quote]

Either your 'Classroom' math is insanely difficult, or you posted this in the WRONG forum.[/quote]

please do not go and revive old threads...the last post was almost a month ago

To solution is i*Pi/4

the solution is imaginary

please, read a calc book for more information

Wow That difference between the sections on this site is vast..on this one no one even puts dx on their integrals, and on others people are solving Complex Contour integrals ...

Here is the proper solution to the problem.

$\int \sqrt{x(2-x)}dx = \int^{1}_{0}\sqrt{x}\sqrt{2-x}dx$

Now make the substitution x=2 sin^2 u. Differentiate, and we get dx = 4 sin u cos u du.  Putting those into the equation and use the pythagorean identity sin^2 u + cos^2 u =1.

$\int \sqrt{2\sin^{2}u}\sqrt{2-2\sin^{2}u}4 \sin u \cos u du = \int \sqrt{2}\sin u \sqrt{2}\cos u 4 \sin u \cos u du$

Now simplify the constants and write it like this : $2\int (2\sin u \cos u)(2\sin u \cos u) du$

Now recognise that sin 2u=2 sin u cos u to simplify it to $\int 2\sin^{2}(2u) du$

Make the substitution y=2u, dy = 2 du, the integral becomes

$\int sin^{2}y dy$

Now use the cos double angle formula to simplify sin^2 y and finish it off.

EDIT: O btw the solution is pi/4.

grr...like I said, don't revive posts

hmm is it mandetory to put dx after the integral and f(x)? sometimes my book doesnt bother but most times they do

Sorry i was not trying to revive this thread, its just I didn't see anyone post a correct solution with working.

and yes, the dx is completely essential! it tells you that you want to integrate WITH RESPECT TO X. Now I may know the sometimes eg in a question like this threads, it is obvious what they want you do to. However, we really need the dx to tell us that we want to integrate with respect to x.

If I didnt have a dx, eg in my solution, i wouldn't have replaced the dx with 4 sin u cos u du and I would have gotten to wrong solution.

If you didn't have the dx, i could have said let $u=\sqrt{x(2-x)}$ and it would have become $\int u$. Now its not so obvious what to do huh? If you saw that, by the "obvious" way, you would say u^2/2 + C, but that is not correct.

You apparently know your way around calculus  :D . And frankly, I believe you should be interested in the High school section no matter what grade you are in.  :) . Oh yes I forgot u needed dx after [i]integration[/i] but sometimes in word format for derivitives it is not neccesary.


Compare the numbers:
$A= \sqrt{x}+ \sqrt{x+4}$ ; $B= \sqrt{x+2} + \sqrt{x+3}$, where $x>0$.

cheers! :D :D :cool:

We have B>A. Indeed, squaring we obtain (x+2)(x+3)>x(x+4), which is true.


[size=150]ABCD is a cyclic quadrilateral. The tangent at A meets the ray CB at K, and the tangent at B meets the ray DA at M, so that BK = BC and AM = AD. Show that the quadrilateral has two sides parallel. [/size]

We have AM/MD = (AB^2)/(BD^2) or AB√2 = BD
(BM  exsemidian in triangle ABD)
Analogously AB√2 = AC=BD.                 (e=f)
 Using second theorem of Ptolemy - in inscribed quadrilateral
    e/f = (ad+bc)/(ab+cd)
Hence  ad+bc = ab+cd  or (a-c)(b-d) = 0
As a result a=c or b=d. Using that e=f it is easy to obtain that
our quadrilateral  trapez with two opposite equal sides.

$AK$ is the $a$-exsymmedian in triangle $ABC$, so $\frac 12=\frac{KB}{KC}=\frac{AB^2}{AC^2}\ (1)$. In the same manner we deduce $\frac 12=\frac{BA^2}{BD^2}\ (2)$. $(1)$ and $(2)$ imply $AC=BD$, and that's pretty much it.


integration
trigonometry
calculus
derivative
geometry
area of a triangle
calculus computations

Help me find the taylor series of arcsinx, thanks.

Using that $ \arcsin x \equal{} \int_0^x \frac {1}{\sqrt {1 \minus{} x^2}}dx$, the problem is reduced to finding the Taylor series of $ (1 \minus{} x^2)^{ \minus{} 1/2}$. Just tell if you have any problem with the latter.

Hi.

Differentiating $ \sin^{\minus{}1}$, we get $ \frac{1}{\sqrt{1\minus{}x^{2}}}$.

Differentiating again, we get $ \frac{x}{(1\minus{}x^{2})^{\frac{3}{2}}}$.

Differentiating again, we get $ \frac{2x^{2}\plus{}1}{(1\minus{}x^{2})^{\frac{5}{2}}}$.

Differentiating again, we get $ \frac{3x(2x^{2} \plus{}3)}{(1\minus{}x^{2})^{\frac{7}{2}}}$.

Differentiating again, we get $ \frac{24x^{4} \plus{}12x^{2} \plus{}9}{(1\minus{}x^{2})^{\frac{9}{2}}}$.

This gives $ \sin^{\minus{}1} x \equal{} x \plus{} \frac{x^{3}}{6} \plus{} \frac{3x^{5}}{40} \plus{} ...$.

I don't know how many terms you want.

In fact, the series can be written explicitly, because yes, we do know the Taylor expansion of $ (1\plus{}t)^a$ for a real $ a$.

Using what zhoraster just said, which is to say the binomial series for arbitrary exponent, was one of Isaac Newton's best tools. Using that series was a signature method for him.

And Newton was the first to derive the power series for the arcsine. But he didn't to it by integrating $ (1\minus{}x^2)^{\minus{}1/2}.$ He didn't do that because he didn't yet know what the derivative of the arcsine (or, for that matter, the sine) was.

Draw a picture of the arc of the unit circle in the first quadrant. Draw a raw which makes an angle $ \theta$ with the $ y$ axis, and draw the vertical line from where that ray crosses the circle. Call that $ x,$ so that $ \theta\equal{}\arcsin x.$

The area of the wedge between that ray and the vertical axis is $ \frac12\theta.$ We can compute this by finding the area under a curve and then subtracting the area of a triangle:

$ \frac12\theta\equal{}\frac12\arcsin x\equal{}\int_0^x\sqrt{1\minus{}t^2}\,dt\minus{}\frac12x\sqrt{1\minus{}x^2}.$

Now use the binomial series for $ \sqrt{1\minus{}x^2},$ twice, integrate a series to get another series, and subtract.

And now the real work begins: [b]invert[/b] that series to solve for $ x$ in terms of $ \theta,$ hence obtaining the power series for $ \sin \theta.$ Yes, Newton did that, too.

Help me to check the solution:

$ *\ \ {{\left( \arccos x \right)}^{'}}\equal{}\frac{\minus{}1}{\sqrt{1\minus{}{{x}^{2}}}}\equal{}\minus{}{{\left( 1\minus{}{{x}^{2}} \right)}^{\minus{}1/2}}\equal{}\minus{}\sum\limits_{i\equal{}1}^{n}{\frac{{{x}^{2i}}}{i!{{.2}^{i}}}.\prod\limits_{k\equal{}1}^{i}{\left( 2k\minus{}1 \right)}}\minus{}1\text{,   where }\left| x \right|<1$

$ \Rightarrow \arccos x\equal{}\minus{}\int{\left( \sum\limits_{i\equal{}1}^{n}{\frac{{{x}^{2i}}}{i!{{.2}^{i}}}.\prod\limits_{k\equal{}1}^{i}{\left( 2k\minus{}1 \right)}}\plus{}1 \right).dx}\equal{}\minus{}\sum\limits_{i\equal{}1}^{n}{\frac{{{x}^{2i\plus{}1}}}{i!{{.2}^{i}}\left( 2i\plus{}1 \right)}.\prod\limits_{k\equal{}1}^{i}{\left( 2k\minus{}1 \right)}}\minus{}x$

$ \Rightarrow \arccos x<0\text{  }\forall x\text{ ?}$

Help me to explain this
$ \Rightarrow\arccos x < 0\text{ }\forall x\text{ ?}$
$ \arccos 0\equal{}0\text{  }?$


Let $a,b,c>0$ Prove the inequality:
$\frac{a+b^2+c^3}{ab+c^2}+\frac{b+c^2+a^3}{bc+a^2}+\frac{c+a^2+b^3}{ca+b^2}\geq \frac{9}{2}$

No one interested?  :(
There is a very nice solution to this inequality :) .

Ok,nice one.here goes my not nice enough solution:
Solution:Rewrite \[ \frac{a+b^2+c^3}{ab+c^2}+\frac{b+c^2+a^3}{bc+a^2}+\frac{c+a^2+b^3}{ca+b^2}  \]\[ =\sum \frac{a+b^2+c^3}{ab+c^2}=\sum \frac{a+c^3}{ab+c^2}+\frac{b^2}{ab+c^2}  \]
Following,I will prove:
\[ \sum \frac{a+c^3}{ab+c^2} \ge 3  \]
and \[ \sum \frac{b^2}{ab+c^2} \ge \frac 32  \]
For the first one:by Cauchy-Schwarz's we get:
\[ \sum \frac{a}{ab+c^2}= \sum \frac{a^2}{a^2b+ac^2}=\ge \frac{(a+b+c)^2}{\sum_{sym} a^2b}  \]
and \[ \sum \frac{c^3}{ab+c^2}= \sum \frac{c^4}{abc+c^3}\ge \frac{(a^2+b^2+c^2)^2}{3abc+\sum a^3}  \]
thus we have:\[ \sum \frac{a+c^3}{ab+c^2} \ge \frac{(a+b+c)^2}{\sum_{sym}a^2b}+\frac{(a^2+b^2+c^2)^2}{3abc+\sum a^3}  \]
\[ \ge 2\frac{(a^2+b^2+c^2)(a+b+c)}{\sqrt{(3abc+\sum a^3)(\sum_{sym}a^2b)}} \ge 4\frac{(a^2+b^2+c^2)(a+b+c)}{3abc+\sum a^3+\sum_{sym}a^2b} \ge 3  \]
For the second one:by Cauchy-Schwarz's we get:
 \[ \sum \frac{b^2}{ab+c^2} = \sum \frac{b^4}{ab^3+b^2c^2} \ge \frac{(a^2+b^2+c^2)^2}{\sum ab^3+\sum b^2c^2} \ge \frac 32  \]
(Here I used the Vasc's famous inequality:$(a^2+b^2+c^2)^2 \ge 3(ab^3+bc^3+ca^3)$.
So we are done.


How important are Newton's Sums (as referred to in AoPS v2)? I tried searching for more information on them on google, but came up with almost no matches. Does it go by some other name as well?

Fierytycoon

What do you mean by important? There was once a math contest in ARML style with a power round dedicated to them. (My team was quite fortunate and guessed that this would be the theme and studied the AoPS chapter in advance.) I think you can prove it fairly easily by expanding (r1+r2+r3+...+rn)^1, ^2, ^3, ..., ^k and equating coefficients.

Hmm, after more extensive searching, it seems that another, more popular term for the concept is the newton-girard identities. my use of the term 'important' was sorta linked on with how i could not (previously) find anything on newton's sums on the internet

Fierytycoon


how to subtract $ 1100\minus{}1001$ in binary form

$ ..1100$
$ \minus{}1001$
----------

for the right column i.e. 0-1 where will the 0 borrow a 1? will it be from the 1 (third from the right)

similar qn
---------------

$ 11010\minus{}10111$

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

do as usual subtraction.dont worry about binary.
borrow 1.
add 1 to third 0 in bottom row.
subtract 0-1.
again borrow 1
add it to 2nd 0 in bottom row.
do 1-1.
no borrow...

what you mean dont worry about binary?? :|

what is final answer?

It means, use the same algorithm that you use for decimal subtraction:

11010
10111

Consider the one's digits. You must borrow:

11002
10111

Consider the two's digits. You must borrow (\ = -1):
11\22
10111

Consider the four's digits. You must borrow:
10122
10111

Consider the eight's digits. You don't need to borrow.
Consider the sixteen's digits. You don't need to borrow.

Final form:
10122
10111

Now, just subtract column by column:
00011 = 11 (or, 3 in base 10)

Alternatively $ 1100_2\equal{}12_{10}$ and $ 1001_2\equal{}9_{10}$, $ 12\minus{}9\equal{}3\equal{}\boxed{11_2}$.

However, this is more work than is necessary and is inefficient when dealing with large numbers, so it's better to do the subtraction directly.


Find the number of units in the interior diagonal $ AD$ for the rectangular prism.
[asy]pair A,a,b,c,D;
a = (1,0); b = (0,4); c = (1.5,1); D = a+b+c;
draw(D--b+c--b--A--a--a+c--D--a+b--a); draw(b--a+b); draw(A--D,dashed);
label("$A$",A,W); label("$D$",D,E); label("3",D--D-a,N); label("4",a--a+c,dir(-45)); label("12",A--b,W);[/asy]

There's a formula
root(a^2+b^2+c^2)
Plugging into this formula you get
root(9+16+144)=root(169)=[b]13[/b]

Explaining this formula, note that this line is the hypotenuse of the right triangle with lengths

$ c, \sqrt{a^2\plus{}b^2}$. Thus, the diagonal is $ \sqrt{a^2\plus{}b^2\plus{}c^2}$

[geogebra]ec6f5545ce788b0f5f8eae73af57d90a26e4bd71[/geogebra]

Funny, you would think that the diagonal would bisect the rectangular prism

Another way of looking at it is that one side is a diagonal of another right triangle, a 3-4-5, and that this larger right triangle would then be a 5-12-[b]13[/b] triangle.


Let $ x, y, z$ be positive real numbers satisfying

$ \frac{1}{3} \leq xy\plus{}yz\plus{}zx \leq3$.

Determine the range of values for (i) $ xyz$, and (ii) $ x\plus{}y\plus{}z$.

hello, for (i) we get $ 0<xyz\le 1$.
Sonnhard.

[quote="AndrewTom"]Let $ x, y, z$ be positive real numbers satisfying

$ \frac {1}{3} \leq xy \plus{} yz \plus{} zx \leq3$.

Determine the range of values for (i) $ xyz$, and (ii) $ x \plus{} y \plus{} z$.[/quote]

$ (x \plus{} y \plus{} z)^2 \ge 3(xy \plus{} yz \plus{} zx) \ge 1$,So $ x \plus{} y \plus{} z \ge 1$.

For $ x \equal{} y \equal{} z \equal{} 1/3$,We get $ x \plus{} y \plus{} z \equal{} 1$.

Let $ x \equal{} y \to 0$,$ z \to \plus{} \infty$, So the range of $ x \plus{} y \plus{} z$ is $ [1, \plus{}\infty)$.

[quote="AndrewTom"]Thanks Sonnhard and frankvista.

Sonnhard, how did you get your result?[/quote]

For AM-GM inequality,we have $ 3 \ge xy \plus{} yz \plus{} zx \ge 3(xyz)^{\frac{2}{3}}$.

So $ xyz \le 1$.

Similiarly,Suppose $ xy \plus{} yz \plus{} zx \equal{} 1$.

Let $ x \equal{} y \to 0$, $ z \equal{} \frac{1\minus{}xy}{x\plus{}y} \equal{} \frac{1\minus{}x^2}{2x}$.

$ xyz \equal{} \frac{x(1\minus{}x^2)}{2} \to 0$.

hello, for (ii) we get
$ 1\le x\plus{}y\plus{}z<\infty$
Sonnhard.

Hi Sonnhard. Would you show us your workings? Thanks.


inequalities
graph theory
combinatorics proposed
combinatorics

Let G be a graph which has k points.G has exactly n values of distances made from two of points in it.Prove that:k<=$n^2+2$

u can prove that $n\geq \frac{1}{\sqrt{2}}(\sqrt{k-\frac 34}-\frac 12)$.

Yes Sam, and you can even delete the $ \frac 1 { \sqrt {2}}$, which leads to the first result on this type of problem (and due to Erds), but it is still slightly worse than desired.

A trivial remark, but the graph-theoretic statement of the problem is not judicious, because the word 'distance' in a graph has a specific sense, which is obviously not the one that was meant by the proposer (otherwise just choose a complete graph with $k$ vertices to prove the inequality wrong)

Pierre.


logarithms
MATHCOUNTS
LaTeX
geometry
3D geometry

At how many points do the graphs of the equations y =x to the power of 3 and y = x to the power of 2+ 4x intersect?

Do you mean the graphs \[ y\equal{}x^3\] and \[ y\equal{}x^{4x\plus{}2}?\] if so...
Setting the two expressions equal gives \[ x^3\equal{}x^{4x\plus{}2}.\] Taking the base $ x$ logarithm of both sides gives \[ 3\equal{}4x\plus{}2\implies4x\equal{}1\implies x\equal{}\frac{1}{4}\]
this shows that there is only $ \boxed{1}$ solution

[quote="prezcoin"]Wouldn't it be 3?  Because x=0,1/4,or 1 (I think???)

Just wondering..[/quote]

Hey yeah! It would! Cause $ 0^3\equal{}0^2$ and $ 1^3\equal{}1^6$. Good one there prezcoin.  :)

didn't he mean $ y\equal{}x^3$ and $ y\equal{}x^2\plus{}4x$?

that's what i thot when i read the prob...

[quote="RoFlLoLcOpT"]\]
Taking the base $ x$ logarithm of both sides gives
\[ 3 \equal{} 4x \plus{} 2\implies4x \equal{} 1\implies x \equal{} \frac {1}{4}
\]
this shows that there is only $ \boxed{1}$ solution[/quote]

When you take the base logarithm, you always need to consider base = 0, 1 (which is always true)
So there are 3 solutions.

Now that I think about it, there could be $ x^{2\plus{}4x}$ or $ x^2\plus{}4x$...  :(

Except the second interpretation was a 2008 MathCounts chapter problem. :)

okay, fine. if it is $ x^3\equal{}x^2\plus{}4x$, then:

$ x^3\equal{}x^2\plus{}4x$

$ x^2\equal{}x\plus{}4$

$ x^2\minus{}x\minus{}4\equal{}0$

$ \frac{1\pm\sqrt{(\minus{}1)^2\minus{}4(1)(\minus{}4)}}2$

$ \frac{1\pm\sqrt{17}}2$

Clearly, there are $ \boxed2$ solutions.

[quote="nikeballa96"]okay, fine. if it is $ x^3 \equal{} x^2 \plus{} 4x$, then:

$ x^3 \equal{} x^2 \plus{} 4x$

$ x^2 \equal{} x \plus{} 4$

$ x^2 \minus{} x \minus{} 4 \equal{} 0$

$ \frac {1\pm\sqrt {( \minus{} 1)^2 \minus{} 4(1)( \minus{} 4)}}2$

$ \frac {1\pm\sqrt {17}}2$

Clearly, there are $ \boxed2$ solutions.[/quote]
What about $ x\equal{}0$?

When you divided by $ x$ you lost a root, namely $ x\equal{}0$

[quote="nikeballa96"]:wallbash_red: FAIL FAIL FAIL.

yeah. $ \boxed3$.

Blah.[/quote]
Don't feel bad... u didn't do terribly... anyway, k00lperson needs to learn how to use LaTex, that way there wouldn't be the confusion. anyway, nikeballa96 didn't even need to solve the equation. he could just check to see if there were double roots. because it is a cubic, there always be 3 roots unless it is the cube or factor of a square of a binomial. I know this sounds complicated, but this way is really easy for me. Tell me if this doesn't work or anything...


Prove that $ 2\cos{A} \plus{} 3\cos{B} \plus{} 4\cos{C} > 1$ holds for all $ \triangle{ABC}.$


When I try to compile my round 2 USAMTS solutions, a window pops up that says I need to install that package fancyhdr. I could not figure out how to do this, so I completely reinstalled LaTex, thinking that the configuration wizard would allow me to reconfigure TeXnic Center for use with MikTex. However, the configuration wizard did not appear when I started TeXnic Center after reinstalling it, and now I cannot compile anything. What should I do? Please help!

Try [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuidePack.php#fancy]clicking here[/url] and following the directions you find.


AMC
AIME
geometry
analytic geometry
USA(J)MO
USAMO
number theory

Would anyone be interested in doing one?

(I'll co-write it; I know Mildorf wants to write most of it, right?)

I would, but don't count on getting my answers.

[quote="Fermatprime"]... I know Mildorf wants to write most of it, right?)[/quote]

I will write whatever part you don't write. ;)

For me, coming up with good problems is a very random process - I can't just will myself to think of a nice Number Theory problem and then a nice Combinatorics problem... right now I have about 5 problems, 1 is N.T. and 4 are Geometry. As was just reinforced by the Mock AIME, which was 32% impossible :lol:, I have a tendency to write difficult problems.

I am not sure what is going to happen with a prospective Mock USAMO...

I'm always willing to take these tests... I can only get better, right?

I'll take it!  Now that my internet is back on.   :maybe:

hey why not.

maybe i can miss cutoff by a point and get somewhere by finding another typo.

I'll take it. My computer kinda froze on me during the first test.

I'm in - I love your questions Tom. The more geometry, the better, btw ;)

I'll be there, whereever and whenever it is.

Messing around on John Scholes's site I came up with three more ideas (Can you guess which three problems I drew inspiration from? :lol: ) I'm already halfway there :lol:... Mystic Terminator will be happy to know that 5/8 of my ideas are Geometry problems. ;)

Please don't make it all geometry.  At least throw some other stuff in there so I have at least a little chance of getting a decent(for me) score.

i'll have to take it - not that i will do that good. hopefully my score will get better.  :D

I'll give it a try but don't count on getting any answers from me as I won't send them in unless I am sure they make at least a little sense.

[quote="white_horse_king88"]Can there be less difficult combinatorics problems - the past few on the Mock AIME A were very difficult in my opinion. Geometry isn't as bad as combinatorics.[/quote]

No way! Writing hard combinatorics problems is FUN!!!! :D  :D

Yeah, so judging by the 11 questions that Fermatprime and I have so far, it looks like it will be harder than Mock AIME A :? .  Better petition Fermatprime to send me a bunch of easy problems :lol:.

The real point of this post, I think we can run this contest basically the same way. I will be sending a link to the problems via PMs to those who can't make the appointed date and time... which is yet to be decided.  It will have to wait a few days since the contest isn't done yet and the answers/solutions will need to be verified.  Uhh... how about this Sunday - August 8, from 2:00-5:00PM EST?

ooh, that's bad for me if it's on sunday...but it's ok if you pm me the ?'s before then though...

Sunday should be fine.  Don't make it too hard.  The last Mock AIME was good, maybe a little hard but good. Try to keep it around that level maybe a little harder if you want to but not too much harder.

[quote="Mildorf"]I will be sending a link to the problems via PMs to those who can't make the appointed date and time... which is yet to be decided.  It will have to wait a few days since the contest isn't done yet and the answers/solutions will need to be verified.  Uhh... how about this Sunday - August 8, from 2:00-5:00PM EST?[/quote]

Do you think the test will be ready by Wednesday? I'd like to take it then, if possible.

Also, Mildorf, if you want to coordinate the Mock USAMO, I can start bugging some advanced users and ask them to send you questions for the Mock USAMO.

[quote="JSRosen3"]Also, Mildorf, if you want to coordinate the Mock USAMO, I can start bugging some advanced users and ask them to send you questions for the Mock USAMO.[/quote]

* I guess you could do that. :D I would be willing to coordinate / run the Mock USAMO at any time since this really doesn't require me to sit and actively watch over it for 4 and a half hours.  However, it seems to me that it won't happen for a while.  Hence, I don't think I'll be able grade it as school leaves bereft of free time.  I don't know how the grading will happen but I suppose that the AMC takes at least a few days to grade anyway so we could work something out.

*As soon as the Mock AIME B problems are finished I will send you a link but of course you won't get your score back for a while. ;)

[quote="genius1601"]sunday sounds good. nothing better than taking a 3 hour math test the day before school starts :lol:[/quote]

Of course! And school will be a cake walk compared to this particular test, eh?

[quote="Mildorf"][quote="JSRosen3"]Also, Mildorf, if you want to coordinate the Mock USAMO, I can start bugging some advanced users and ask them to send you questions for the Mock USAMO.[/quote]

* I guess you could do that. :D I would be willing to coordinate / run the Mock USAMO at any time since this really doesn't require me to sit and actively watch over it for 4 and a half hours.  However, it seems to me that it won't happen for a while.  Hence, I don't think I'll be able grade it as school leaves bereft of free time.  I don't know how the grading will happen but I suppose that the AMC takes at least a few days to grade anyway so we could work something out.[/quote]

When does school start for you? I'm on vacation Aug. 6-16, but I'd be able to take the test any time between the 17th and the second week of September.

[quote="Mildorf"]*As soon as the Mock AIME B problems are finished I will send you a link but of course you won't get your score back for a while. ;)[/quote]

Thanks!  ;)  I'm a little bit afraid of this test, though...I only got a 4 on the last one!  :(

can you take it if you missed the mock amc's? if you can, count me in, ill just not submit my answers

[quote="Mildorf"][quote="takuanitromars36"][quote="Mildorf"]I will be sending a link to the problems via PMs to those who can't make the appointed date and time... which is yet to be decided.  It will have to wait a few days since the contest isn't done yet and the answers/solutions will need to be verified.  Uhh... how about this Sunday - August 8, from 2:00-5:00PM EST?[/quote]

If we don't take it 2-5 P.M. EST and instead we do it beforehand, when do the answers have to be submitted?[/quote]

* I used the honor system last time; I don't think there were any "problems" so I will do the same this time. You give yourself three hours and then submit your answers. I don't really care when you take it, so long as it's before the main AoPS contest time has ended.

* I just finished writing the contest, although I need to solve one of the problems to determine whether I need to add "remainder when divided by 1000..."  If you really can't make the 2:00-5:00PM EST time on Sunday, August 8 then you can take it independently ahead of time.  PM me and I will direct you to the problems.  This time, however, I will not be able to tell you your score and send a link to solutions as I have not written the solutions yet!

* Like the other Mock AIME, this is "official" only for those who qualified through Mock AMCs, although everyone is invited to take it and see how they do.  I would say that the difficulty is [i]about[/i] the same as Mock AIME 1.[/quote]

If you're done writing it, I can look over it and maybe write some of the solutions.

This sounds great. How may one get a copy of the test (when it comes out of course)?

The contest has already happened, however it can be found at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14899[/url]. (The first Mock AIME can be found at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14682[/url].) I have plans for a third Mock AIME to be on some weekend in September.

can we make that weekend NOT the weekend after the AHSIMC competitions?

i think i'll be gone then.


Prove that if $\prod_{j=1}^{n} x_j \geq (n^2-1)^n$, then \[\sum_{j=1}^{n} \frac {1}{\sqrt{1+x_j}} \geq \frac {n}{\sqrt{1+{\sqrt[n]{\prod_{j=1}^{n} x_j}}}}\].

EDIT: Corrected source since it's now confirmed.

Ok, I greatly suspect Vasc to be the author of this magnificient problem. Don't you find any similarity with the topic " A beautiful inequality" ? Come on, it must be Vasc!


[quote="BOGTRO"][quote="GameBot"]Kat has a collection of porcelain dolls that have heights, in millimeters, of 24, 32, 24, 27, 40, 36 and 34.  What is the average height of the dolls in millimeters?[/quote]

$ 24,24,27,32,34,36,40$

Adding and dividing by 2 gives 31.[/quote]

This is correct, except you should divide by 7 instead of 2.


In a small town of 100 men, 85 are married, 70 have a telephone, 75 own a car, and 80 own their own home. On this basis, what is the smallest possible number of men who are married, have their own telephone, own their own car, and own their own home?

I have no clue..blah..although I saw this problem many times..
[hide="Answer"]10 men [/hide]

[quote="distracted523"][hide]
We subtract all these numbers from 100. 15 are single, 30 don't have a phone, 25 don't own a car, and 20 don't have their own home. Since one negation would exclude a member from the union, 15+30+25+20 people, at maximum, be [i]not[/i] in the group. Thus the answer is $ \fbox{10}$[/hide][/quote]
Oh! I get it now. Thanks.

I'm not sure why this works, but it does. :rotfl:  One thing i like doing is adding all the x numbers, and subracting (x-1)(100 in this case;100 men...). So here using my method you  would get 85+70+75+80-3(100)=10

this is my first post ever lolz so im new to this. :starwars:  :maybe:

i like the smileys.....


[quote="bigbang195"]with
$ 2$  monotonous $ (a,b,c) \minus{} \minus{} (x,y,z)$ .Prove that
$ ax(a \minus{} b)(a \minus{} c) \plus{} by(b \minus{} c)(b \minus{} a) \plus{} cz(c \minus{} a)(c \minus{} b)\ge 0$[/quote]

I think the variables need positive?

$ ax(a \minus{} b)(a \minus{} c) \plus{} by(b \minus{} c)(b \minus{} a) \plus{} cz(c \minus{} a)(c \minus{} b)$

$ \equal{}ax(a\minus{}b)^2\plus{}(ax\minus{}by\plus{}cz)(a\minus{}b)(b\minus{}c)\plus{}cz(b\minus{}c)^2$


Putnam
geometry
conics
hyperbola
analytic geometry
inequalities
rectangle

Find the least possible area of a convex set in the plane that intersects both branches of the hyperbola $ xy\equal{}1$ and both branches of the hyperbola $ xy\equal{}\minus{}1.$ (A set $ S$ in the plane is called [i]convex[/i] if for any two points in $ S$ the line segment connecting them is contained in $ S.$)

The key here is the formula for the area of a polygon based only on the coordinates of its vertices. If the polygon has vertices $ (x_1,y_1),(x_2,y_2),\dots,(x_n,y_n)$ traced out as a simple closed counterclockwise curve, then the area is

$ \frac12[(x_1y_2 \minus{} x_2y_1) \plus{} (x_2y_3 \minus{} x_3y_2) \plus{} \cdots(x_ny_1 \minus{} x_1y_n)].$

The convex set we're looking for much contain at least one point on each of the four curves. The smallest convex set containing a specific set of points is the convex hull of those points, and in this case that will be a quadrilateral. Let me write the four points, one on each branch, as

$ (x_1,y_1),( \minus{} x_2,y_2),( \minus{} x_3 \minus{} y_3),$ and $ (x_4, \minus{} y_4)$

such that all of $ x_1,\dots,x_4,y_1\dots,y_4$ are positive numbers and $ x_1y_1 \equal{} x_2y_2 \equal{} x_3y_3 \equal{} x_4y_4 \equal{} 1.$

The area is

$ A \equal{} \frac12[(x_1y_2 \plus{} x_2y_1) \plus{} (x_2y_3 \plus{} x_3y_2) \plus{} (x_3y_4 \plus{} x_4y_3) \plus{} (x_4y_1 \plus{} x_1y_4)]$

Now replace $ y_i$ by $ \frac1{x_i}$ for $ i \equal{} 1,2,3,4$ to get

$ A \equal{} \frac12\left[\frac {x_1}{x_2} \plus{} \frac {x_2}{x_1} \plus{} \frac {x_2}{x_3} \plus{} \frac {x_3}{x_2} \plus{} \frac {x_3}{x_4} \plus{} \frac {x_4}{x_3} \plus{} \frac {x_4}{x_1} \plus{} \frac {x_1}{x_4}\right].$

But since for $ u > 0,$ we have $ u \plus{} \frac1u\ge 2$ (think of it as a very specific case of the AM-GM inequality), we get that

$ A\ge\frac12[2 \plus{} 2 \plus{} 2 \plus{} 2] \equal{} 4$

The criterion for equality would be that $ x_1 \equal{} x_2 \equal{} x_3 \equal{} x_4,$ which is to say that the quadrilateral is a rectangle with sides parallel to the axes.

What I did was basically the same but didn't require to recall that formula as all you need to do is show that the origin has to be in S and then $ (OAB)\geq 1$ as an area where A and B are points in two consecutive branches. i.e. $ \frac{x_1^2\plus{}x_2^2}{2x_1 x_2}\geq 1$ QED.

I didn't know the general formula for the area of a polygon given the coordinates, so I just applied the triangle version twice. This made it somewhat more tedious, but it's essentially the same solution.

A cool idea suggested by one of my friends was to note that if you fix the two points that are on opposite branches of the same hyperbola, then the best you can do is choose the other two points to be at the tangents to the other hyperbola which have the same slope as the line connecting the first two points. Doing this to both pairs of points gives you a rectangle, and the rest is easy. It seems difficult to make this argument precise, though.

[quote="Xevarion"]Doing this to both pairs of points gives you a rectangle, and the rest is easy.[/quote]

How to justify this..?

I don't know. This guy figures things out in some idiosyncratic way that I can never understand. This is an obstacle for him on contests like this, of course. He didn't give me a full justification of his approach, just a sketch.

[quote="Albanian Eagle"]It seems like the "mixing points" version of the mixing variables in the inequalities section, that some members here are really familiar and could justify it. :D[/quote]

The reason I asked was actually because I had the same idea: I didn't know it can go down to rectangle (I'm actually quite skeptical about that), but I sort of think it would reduce the area with some other things to go down to force it become a rectangle, but I didn't get anywhere. Eventually I used Prof Kent Merryfield's method to solve it.

If it becomes a rectangle it would be really nice and solve the problem, so I want to know how to show that.

I originally tried this one geometrically. It was easy to show each diagonal was at least $ 2\sqrt{2}$, but then there was the sine of the angle between the diagonals to deal with. The solution page says that the right transformation can deal with that, but maybe there's a way to do it without transformation (i.e. get a bound on the diagonal length in terms of the angle between the diagonal and $ y\equal{}\pm x$, and then relate that to the angle between the diagonals).

In the end, though, I split into four triangles with vertex and origin and used determinant, it works perfectly fine.

[quote="Albanian Eagle"]The reason I asked was actually because I had the same idea: I didn't know it can go down to rectangle (I'm actually quite skeptical about that), but I sort of think it would reduce the area with some other things to go down to force it become a rectangle, but I didn't get anywhere. Eventually I used Prof Kent Merryfield's method to solve it.

If it becomes a rectangle it would be really nice and solve the problem, so I want to know how to show that.[/quote]

Maybe not a rectangle, but possibly a rhombus, and from there you can minimize the area almost as easily. Someone I know did it that way, I think.

[quote="Kent Merryfield"] The smallest convex set containing a specific set of points is the convex hull of those points, and in this case that will be a quadrilateral.[/quote]

My solution was almost exactly the same as yours, but now I'm not sure about something. In the case where three of the four points that must be on the branches are on the same line, wouldn't the convex hull be a triangle?

The relevant area formula doesn't require the quadrilateral to be convex, or for adjacent sides not to be parallel. It's possible that the convex hull is larger than the quadrilateral, and that doesn't hurt. It only requires the vertices to be in the right order, which we get by showing that the origin is in the convex hull.


Prove or disprove this statement :

If $a^m-b^n=a-b$ for $a,b,m,n\in\mathbb{N}$ ;$a\neq b$ . Then for any given $a,b$ ,  the only solution for the equation is $m=n=1$ .

Oh , that is true . But what about other than $2^3-3^2=1$ . Is the statement true other than this special case  :?

[quote="zhaoli"]How about a = b? Or more specially, a= b = 1?[/quote]

Ok , i editted so that $a\neq b$  ;)

That shouldn't be so complicated to investigate. For example assume a > b and m > n. For example m = n+1. Write a = b + r, then you get
(b+r)^(n+1) - b^n = a - b = r. If r is a positive integer greater than one then one clearly sees this cannot hold so for even larger m that surely cannot hold. So like that you could prove the other cases.

this is false because 16^2 - 3^5 = 16 - 3. There are likely an infinite number of counterexamples.

So I guess that statement is wrong afterall  .... so does anyone have any idea what conditions should we add for $a,b$ in order for $m,n$ to have the only solution which is $m=n=1$ ?


In a triangle the circumcircle's radium is $ 2$. The lengths of the heights are integers. Prove that the triangle is equilateral.

Let $ AB \equal{} c$ and $ BC \equal{} a$ and $ CA \equal{} b$ whe have : $ \text{sinA} \equal{} \frac {h_1}{b}$ and $ \frac {a}{\text{sinA}} \equal{} 4 \implies ab \equal{} 4h_1$

On the other hand we have:
$ 16S^2 \equal{} (a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) \Leftrightarrow a^2b^2c^2 \equal{} 4(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)$
And using The flowing inequality $ abc\ge (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)$ we obtain

$ 4(a \plus{} b \plus{} c)\ge abc \Leftrightarrow \frac {1}{h_1} \plus{} \frac {1}{h_2} \plus{} \frac {1}{h_3} \ge 1$

Since The heights are an integers we can easly see that $ h_1 \equal{} h_2 \equal{} h_3 \equal{} 3$ are the possible values of $ h_i$ with $ h_i>2$

And it follows that $ a \equal{} b \equal{} c$ which ends our problem


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hi all!

Let $G$ be a finite group, $f\in \mathrm{Aut}(G)$, and $p$ be a prime. Further, let $p$ be a divisor of $\vert G\vert$ and $f^{p^k}=\mathrm{Id}_G$ for some nonnegative integer $k$. Show that there's an $x\in G$ for which we have $f(x)=x$ and $ x\neq e$.

Hanno

Let the subgroup $\langle f\rangle$ of $\mbox{Aut}\ G$ act on $G$ in the obvious manner: $f$ takes $x$ to $f(x)$. Now write the class equation for this action: $|G|=|S|+\sum |\mbox{Orb}(x)|$, where $S$ is the set of elements of $G$ which have trivial orbits, and the second sum is taken after all non-trivial orbits of the action.

Since $|\mbox{Orb}(x)|$ is a multiple of $p$ for all $x$ which have non-trivial orbits (since $|\mbox{Orb}(x)|$ is the index of the stabilizer of $x$ in $\langle f\rangle$, and this stabilizer is a proper subgroup of $\langle f\rangle$ if the orbit is non-trivial), and so is $|G|$, so $|S|$ must also be divisible by $p$. Since $S$ contains the identity of $G$, it contains at least one other element, and this is the $x$ we're looking for.

A point of language: you don't want to call an automorphism of anything nilpotent. A nilpotent endomorphism would eventually reduce everything to the trivial object.
Calling this an automorphism of prime power order would suffice.


Is it possible sum of n (n>1) odd numbers to be a prime number? (Why?)

If n is even, the it isn't because that sum is even.
If n is odd, of course it is, for $ n\equal{}3$, $ 1\plus{}3\plus{}7\equal{}11$.

Bye

I asked this question, because I found this problem in Russian in the Net with answer: "No".

Clearly it is possible for $ n\equal{}3$, but is it possible for all odd $ n$ ($ n>3$)?

I think it is possible for some numbers greater than 3 - for example: 1+3+5+7+13=29 ... also nowhere is not said that some of the numbers may not be equal ... In my opinion the answer in the original source is wrong, but I'm not sure at 100 % ... there is no solution to this problem.


Find all functions $ f: \mathbb R\to \mathbb R$ satisfying $ xf\left(x\right) \minus{} yf\left(y\right) \equal{} \left(x \minus{} y\right)f\left(x \plus{} y\right)$ for all reals $ x$ and $ y$.

let $ y\equal{}\minus{}x \minus{}> f(x)\equal{}\minus{}f(\minus{}x)$
replace y by -y $ \minus{}> \frac{f(x)}{x} \equal{}const$
esay to continue! :wink:

[quote="Allnames"][quote="Math pro"]let $ y \equal{} \minus{} x \minus{} > f(x) \equal{} \minus{} f( \minus{} x)$
replace y by -y $ \minus{} > \frac {f(x)}{x} \equal{} const$
[/quote]
Why,my friend????????( I see you found $ f(0)$ can you post it[/quote]

With the Math pro method you obtain $ (x \plus{} y)f(x \minus{} y) \equal{} (x \minus{} y)f(x \plus{} y)$ . You continue :  $ x \equal{} \frac {t \plus{} 1}2,y \equal{} \frac {t \minus{} 1}2$ and so on .. :cool:

but his result is wrong,I guess $ f(x)\equal{}ax\plus{}b$ :(

Sorry, actually you have to use $ g(x)\equal{}f(x)\minus{}f(0)$ .  :D
$ xf(x) \minus{} y(f(y) \equal{} (x \minus{} y)f(x \plus{} y) \Longrightarrow$$ \ x(f(x) \minus{} f(0)) \minus{} y(f(y) \minus{} f(0)) \equal{} (x \minus{} y)(f(x \plus{} y) \minus{} f(0))$ . Let's denote $ g(x) \equal{} f(x) \minus{} f(0)$ .
Now we obtain $ g(x) \equal{} \minus{} g( \minus{} x)$ and  $ (x \plus{} y).g(x \minus{} y) \equal{} (x \minus{} y).g(x \plus{} y)$, so $ g(x) \equal{} x.g(1)$. Finally $ f(x) \equal{} x.(f(1) \minus{} f(0)) \plus{} f(0) \equal{} ax \plus{} b$ .
Do you agree ?  :cool:

sorry all bacause my proof is too spare!
I explained all names in our class:
Case1:$ f(0)=0$->done!
Case2:$ f(0)\#0,let g(x)=f(x)-f(0)$->done!
My solution is similar to diogene! :wink:

For $ x\neq0$, we have
$ P(x,\minus{}x): xf(x)\plus{}xf(\minus{}x)\equal{}2xf(0)\Rightarrow f(x)\plus{}f(\minus{}x)\equal{}2f(0)$.
Define $ g(x)\equal{}f(x)\minus{}f(0)$. We see $ g(x)$ is an odd function such that
$ xg(x)\minus{}yg(y)\equal{}(x\minus{}y)g(x\plus{}y)$
Replace $ y$ with $ \minus{}y$ in the above formula, we have
$ xg(x)\minus{}yg(y)\equal{}(x\plus{}y)g(x\minus{}y)$.
Compare the two formula above, we have
$ \frac{g(x\minus{}y)}{x\minus{}y}\equal{}\frac{g(x\plus{}y)}{x\plus{}y}$.
For any distinct nonzero $ u$ and $ v$, we plug in $ x\equal{}\frac{u\plus{}v}2$ and $ y\equal{}\frac{u\minus{}v}2$ and will get
$ \frac{g(u)}u\equal{}\frac{g(v)}v$.
Hence we have $ g(x)\equal{}cx$ for some constant $ c$ and $ x\neq0$. Since $ g(0)\equal{}0$, $ g(x)\equal{}cx$ for any $ x$.

Conclusion: $ f(x)\equal{}cx\plus{}f(0)$.


Gauss
trigonometry
quadratics
LaTeX
number theory
superior algebra
superior algebra unsolved

Hello,

I was wondering what was the technic used by Gauss in order to compute $cos(\frac{2\pi}{17})$. I will not copy the value here (many roots, this is an algebraic number of degree $2^{3}$), but I would be interested in the method.

we did it in class
he used a general thing in cyclotomics extension called periods
call $w=e^{\frac{2 \pi i}{17}}$

then he considered

$a=w+w^4+w^9+w^{16}+....$ basically all exponents occur that are squares mod 17
$b=w+w^{13}+w^{16}+w^3$  here we used the fourth powers mod 17
$c=w+w^{16}$which is twice the cosine you seek

now it turns out that you can always calculate the b from a by a quadratic extension, the c from the b etc

come to think of it, when you solve $x^4+x^3+x^2+x+1=0$ to find the formula for the pentagon you use the variable $x+\frac{1}{x}$ which is actually doing the exact same technique with periods in that cyclotomic extension to split $x^5-1$ instead of $x^{17}-1$

I don't understand this method  :blush:  How do you do those computations? Sorry if my question seems really dumb

if you can wait sixty hours, i will be out of exams then, i can even make a maple file about it

right now i have to number theory, coincidentally for the very same professor who taught me how to construct the 17-gon! :ninja:

Still there?

  :oops:

I must be crazy as I am just out of exams, but ... let us try again (I agree my original explanation was very vague)

the end result is at the bottom

$w=e^{\frac{2 \pi i}{17}}$

now this implies $w^{17}=1$ and also $w^{16}+w^{15}+\ldots+w=-1$

it also means $w^k=w^{k-17}$ so you work mod 17

also note that $w^1+w^{16}=w+w^{-1}=2 cos(\frac{2\pi}{17})$ and this of course very interesting

our goal is to work the other way around, find nice equations that you can solve by taking just square roots

call

$a=w+w^{13}+w^{16}+w^4+w^9+w^{15}+w^8+w^2$
and $b=w^3+w^{10}+w^5+w^{11}+w^{14}+w^7+w^{12}+w^6$

one easily checks that $a+b=-1$ due to the above
and $(a-b)^2=17$ this one is less fun, either you work it out completely, or you just believe me, or you can ask me for a nice reason for this

now we know the sum of a and b  and their difference up to sign, we get two possible values for a :
$\frac{-1+\sqrt{17}}{2}$ and $\frac{-1-\sqrt{17}}{2}$

now this is shady, depends on how picky you are, but believe me a is the first and b the second (your computer will confirm it numerically)

now decompose again :

$c=w+w^{13}+w^{16}+w^4$
$d=w^9+w^{15}+w^8+w^2$

note $c+d=a$

$e=w^5+w^{12}+w^3+w^{14}$
$f=w^6+w^7+w^{11}+w^{10}$

so $e+f=b$

when you write out all terms, $cd=ef=-1$

this means c satisfies the equation   $x^2-ax-1=0$
again one will find two expressions and you will have to check numerically which is the one we seek and which is d
so you find c and d

completely the same goes for e and f, they satisfy $x^2-bx-1=0$ again check numerically which is which (we are still gonna need them)

now we have $c=w+w^{13}+w^{16}+w^4$

call $g=w+w^{16}$ and $h=w^{13}+w^4$
we know $g+h=c$ and when you work it out $g h = w^{14}+w^5+w^{12}+w^3=e$

this again means that g and h satisfy $x^2-c x+e=0$
again we find two expressions
we check again numerically which is g

half of that g is exactly the cosine you requested

if you need to know more, just ask it rarely happens i can help here

this is what maple gives me as expression, i am sorry but i don't know how to quickly convert maple to latex :

-1/16+1/16*17^(1/2)+1/16*(34-2*17^(1/2))^(1/2)+1/16*(68+12*17^(1/2)-2*(34-2*17^(1/2))^(1/2)+2*17^(1/2)*(34-2*17^(1/2))^(1/2)-16*(34+2*17^(1/2))^(1/2))^(1/2)

Thanks for the reply. The method is very nice, although the computations do not fall in the same category.

Also, write $w^{11}$, not $w^11$.

well apparently gauss did this at 16 ;)
The idea is the same as I said for a 5-gon

now $w+w^4=w+w^{-1}=2 cos(\frac{2\pi}{5})$

now you must calculate $((w+w^4)-(w^2+w^3))^ 2=5$

this leads to the cosine being $\frac{\sqrt{5}-1}{4}$

as i said that first step with the square being 17 or 5, there is more to it than that

For $p=7$ or $11$, we get $(a-b)^2=-7$ or $-11$.

So $(a-b)^2 = \pm p$ depending on whether $p \equiv 1$ or $3 \left( \bmod 4 ~ \right)$. Right?

How do we prove it?

Yes that is the property.

For a prove, one needs to know a thing or two about working mod p

first of all $w^p=1$
secondly $w^1+\ldots+w^{p-1}=-1$

also if we call $\mathbb{F}_{p}$ the field op p elements (thus working mod p)
exactly half of the nonzero elements are square, the other is not, the product of two squares is a square, the product of a square with a nonsquare is a nonsquare
and the product of two nonsquare is square again

whether or not something is square in
$\mathbb{F}_{p}^{\times}$ is expressed with the legendre symbol

$(\frac{i}{p}) =1$ if i is square, -1 if i is not

note that this symbol is multiplicative

it can be proven (in quite elementary ways) that -1 is square mod p( p an odd prime) if
and only if p=1 mod 4

now the proof : $a-b=\sum_{i} (\frac{i }{p})w^i$  (so i runs through all nonzero elements)
the square is then
$\sum_{i,j}(\frac{i,j}{p}) w^{i+j}$

now write $j=ti$ to see that this is

$\sum_{t} (\frac{t}{p}) \sum_{i} w^{i(1+t)}= \sum_{t\neq 1} (\frac{t}{p})\sum_{i}w^{(1+t)i} + (\frac{-1}{p})(p-1)= \sum_{t\neq 1} (\frac{t}{p})\sum_{i}w^{i} + (\frac{-1}{p})(p-1)= \sum_{t\neq 1} (\frac{t}{p})(-1) + (\frac{-1}{p})(p-1) = \sum_{t} (\frac{t}{p})(-1) + (\frac{-1}{p})(p) = (\frac{-1}{p})(p)$

the last equation is true because exactly half of the nonzero elements is square, so the sum of the plus and minus ones is zero

and thus , we get +p if p is 1 mod 4 and -1 if it is 3 mod 4


linear algebra
matrix
vector
floor function
pigeonhole principle
combinatorics proposed
combinatorics

Let $S_n=${$1,2,...,n$},$1\le k\le n$.$A_1,A_2,...,A_m\subset S_n$such that:$|A_i|=k\forall 1\le i\le m$.Find $m_{max}$if:
$a)$    $|A_i\cap A_j|=1\forall 1\le i<j\le m$
$b)$    $|A_i\cap A_j|\le 1\forall 1\le i<j\le m$ :)

Ignore this... I got confused... :(

Tom

We must have $k\leq n/2$ for the problem to have any meaning.  The answer for part (a) is $\begin{pmatrix} n-1 \\ k-1 \\ \end{pmatrix}$. It follows from the Erdos-Ko-Rado theorem:

Let $\{A_1,\dots,A_m\}$ be a collection of $m$ distinct $k$-element subsets of $\{1,\dots,n\}$ where $k\leq n/2$ with the property that any two subsets have a nonempty intersection.  Then $m\leq\begin{pmatrix} n-1 \\ k-1 \\ \end{pmatrix}$.

Tom

I think you were wrong, because ${n-1\choose k-1}$ is too large.

Now, there are exactly ${n\choose 2}$  2-subset of $S_n$. But for each $2-$subet, there is at most one member in $\{A_1;A_2;...;A_m\}$ which contains it.

hence, $m{k\choose 2}\le {n\choose 2}$$\Rightarrow m\le \frac{n(n-1)}{k(k-1)}$.

But I haven't  finished yet. So I guess $m=[\frac{n(n-1)}{k(k-1)}]$ for part (b).

note:in part (b),if $k=3$,we have a well-known problem.

Deleted: I got confused and solved another problem.

Sorry  :) ,the answer of $a$is
$f(k,n)=max(k^2-k+1,[\frac{n-1}{k-1}])$if$n\ge k^2-k+1$
with$n<k^2-k+1$ my answer is not simple :? so I won't post it.

Actually, I got confused and solved a completely different problem: I proved that $n$ is an upper bound for the number of subsets of $S_n$ such that any two have $k$ elements in common :). Sorry about that. I've erased my message.

With part $(b)$,my answer is not simple ,too.(use part $(a)$) :)

[quote="Aryabhatta"]The answer for a) is $m \leq n$ and answer for b) is $m \leq {n \choose 2}$

Both of these can be done using Linear Algebra.

For part a)

Let $v_i$ be the bit vector which corresponds to $A_i$. Then we have that $v_i \cdot v_j = 1 \forall i \neq j and v_i\cdot v_i = k$.
(the $r^{th}$ bit of $v_j$ is 1 iff $r \in A_j$)

We can easily show that the $v_i's$ are linearly independent hence $m \leq n$

Part b) is more complicated, but still can be done using Linear Algebra. I think part b) follows immediately from Ray-Chaudhary-Wilson theorem.

Both bounds are valid, as can be seen by taking $n=3$ and $k=2$[/quote]
But it isn't the best answer :)

For part (b), I believe we have
\[ m \leq \left\lfloor \frac{n}{k} \left\lfloor \frac{n-1}{k-1} \right\rfloor \right\rfloor .  \]
We want to show $mk\leq n\left\lfloor \frac{n-1}{k-1} \right\rfloor$, so suppose not, i.e., $mk > n\left\lfloor \frac{n-1}{k-1} \right\rfloor$.  Since the total number of elements in $A_1,\dots,A_m$ is $mk$ (counting multiplicities),  by the pigeonhole principle, there is some element, say 1, which is contained in more than $\left\lfloor \frac{n-1}{k-1} \right\rfloor$ of the sets $A_i$, say specifically the element 1 is contained in $A_1,\dots,A_r$ with $r=\left\lfloor \frac{n-1}{k-1} \right\rfloor +1$.   Becuse $A_1\setminus\{1\},\dots,A_r\setminus\{1\}$ are disjoint, we have $r(k-1)\leq n-1$ hence
\[ \left\lfloor\frac{n-1}{k-1}\right\rfloor + 1 = r \leq \frac{n-1}{k-1}  \]
a contradiction.

Tom


Consider a circle of radius 1, and let  [tex]A_1,A_2,\ldots,A_n[/tex] be the vertices of a regular n-gon inscribed in the circle. Join [tex]A_1[/tex] to  [tex]A_2,A_3,\ldots,A_n[/tex] by segments of a straight line. You obtain  segments of lengths [tex]l_2,l_3,\ldots,l_n[/tex]. Show that [tex]\displaystyle{\prod_{i=2}^n l_i = n}[/tex].

Using complex numbers :

[hide]

Ai -> zi= e^(2*i*pi/n), A0, ..., A(n-1) (just a shift because its more convenient).

Product (z-zi) = z^n - 1 for all z.

Then l0 * l1 * .. * l(n-1)  = | product (z0 - zi) | (for i != 0) = 1^0+1^2 +..+1^(n-1) = n

Hope I managed at last to make something correct ![/hide]


Euler
modular arithmetic
abstract algebra
number theory
least common multiple
function

so,i have a big number like this one

\[N=2003^{(2005^{2007})}\]

and i need to find its last 5 digits.
now my approach is kind of lame but i really had to think on it because
i'm not that good at maths.
The problem is that i've made 2 programs in C++ calculate this and they
give different results and when i calculate this using calc.exe from windows
i get another result.

so let's start with the beggining.

i thought like this:
last 5 digits of N are actually \[N mod 10^6\]
but i know that euler theorem states that:
\[2003^{\Phi(10^6)} = 1 \pmod {10^6}\]
\[\Phi(10^6)=10^6-6\cdot 8=999952\]

i will continue the article later

One of the easy way to do it is to break it down in two parts:
Since $100000 = 2^5 \times  5^5$
We find the value modulo $32$, and  $5^5 $
First we know for any  odd $a$
 $a^8   \equiv  1 \bmod 32$   so all we have to find is $2005^{2007}  \bmod 8$

But again for any odd number $a^2   \equiv 1 \bmod 8 $ so   $2005^{2007} \equiv 5 \bmod 8 $

[size=150]So  $2003 ^ {2005^{2007}} \equiv  19^5  \equiv  3 \bmod 32 $[/size]

Now do the same thing  for mod $5^5$   or $3125$

For simplicity let us call $n = 2005^{2007} $
and we have to find  $2003^n  \bmod 3125  $

since  $\phi(5^5) =  4 \times 5^4 $

So we have to find  $ n  \bmod  2500 $

but   again   $2500 = 4 \times 5^4$   and thus  $ a^{ 500}  \equiv 1 \bmod 2500 $ etc ..

so we have to only find  $2005^7 $   which is   $625$  mod  2500

And you have to calculate $(2000+3)^{625} \equiv  3^{625} \bmod  5^5 $
Which you have to calculate by repeated squaring..

Hope this helps...

(Answer will, when all said and done,  last 5 digits of  $3^{625}$  or$85443 $)

[quote="spx2"]so,i have a big number like this one

$ N=2003^{(2005^{2007})}$

and i need to find its last 5 digits.
last 5 digits of N are actually \[N mod 10^6\]
but i know that euler theorem states that:
\[2003^{\Phi(10^6)} = 1 \pmod {10^6}\]
\[\Phi(10^6)=10^6-6\cdot 8=999952\]
[/quote]

BTW I am not very clear what you are trying to do; but  ...

First - to  find last 5 digits, you have to consider mod (10^5) and not (10^6)
Also  phi(10^5)    can be calculated by using phi(2^5) mutipled by phi(5^5) ....

etc..$ /phi(10^6)  =  2^5   \times 4\times 5^5  or 400000 $

[quote="spx2"]Find the last 5 digits of the number
\[N=2003^{(2005^{2007})}\]
[/quote]
So, u need to do module $10^5$ as someone purpposed before, $\varphi(10^5)=2^6\cdot 5^4=40.000$, then by
\[
(2005)^{2007}=(2000+5)^{2007}=5^{2006}*2007*2000+5^{2007} \ mod \ 40.000=-10.000+5^{2007} \ mod \ 40.000= -10.000+5^{999} \ mod \ 40.000=30.000-3\cdot 5^4 \ mod \ 40.000,
\]
then you only need to calculate $2003^{5^{999}+30.000} \ mod \ 10^5=3^{5^{999}+30.000} \ mod \ 10^5$ then by newton expansion and using modules again, we get that this number is  module $10^5$
\[
3^{-3\cdot 5^4}=3\left(1- \right)=13 \ mod \ 32=125 \ mod \ 125,
\]
then using Chinesse remainder theorem we get the result we get easily that the last digits are $35443$.

Of course you can use a comp. but I needed 53 mins to do it  :blush:  :blush:  :blush:  sorry I hope you do faster.

Best Regards.

RX TCM

Few comments: Hope they are helpful:
[quote]
 So, u need to do module $10^5$ as someone purpposed before, $\varphi(10^5)=2^6\cdot 5^4=40.000$, then by [/quote]

Actaully you need a number  smaller just  5000 instead  of 40,000 (generalized phi function). Instead of multiplying
$\phi(2^5) $ and $\phi(5^5)$ you take LCM , and  throw in a factor of half because $a^{(\phi(n)/2)} = 1 \bmod n$ if  $n = 2^k$

This makes it a little easier..

[quote]Of course you can use a comp. but I needed 53 mins to do it  :blush:  :blush:  :blush:  sorry I hope you do faster.
[/quote]

I used rexx   just to check the answer,  (I wrote myself a simple calculator type program), just typed it in  two steps $2005^{2007}$ took last 5 digits  (It gave all 6000+ digits in less than a second..) and then type $2003^n$ using last 5 digits (knowing that   5000 is a factor of 100000) and got all the digits.. all under a second. :)

(I don't have any fancy new mathcad  or calc type programs - just   20/30+ years old  tools - glad that some still run  under windows...)

i want to thank you because you helped me allot.
but the problem is i don't yet know chineese reminder theorem
also there is alot of stuff that you typed here that i don't know of
yet but i really want to ask you some things.
i read in a book that $\phi(m\cdot n)= \phi (m) \cdot \phi (n)$

i really think that trying to find last five digits and using mod 10^6 is not wrong
you actually find a digit more.
but i can't really find something wrong in this.

$\phi(10^6)=4^6=4096$

so i know that $2003^{4096}=1\pmod{10^6}$

so now i want to find a number K with the property that it is verry close
with $2005^{2007}$ and is divisible with 4096 because that number will have the property:
$2003^K=1\pmod{ 10^6}$
so actually i want to find $2005^{2007} \pmod{4096}$ wich is 3357
so now i just need to find $2003^{3357}=1\pmod{10^6}$
and using newton's binome i can make a program to calculate this
in 0.01 seconds but the result it gives about 910463 or something like this
if i use calc.exe from windows it gives :66563
so i ask you where am i wrong because you people beeing much better
than me at mathemathics are surely right.

[quote="spx2"]i
i read in a book that $\phi(m\cdot n)= \phi (m) \cdot \phi (n)$
[/quote]

Only true  if  $m$, and $n$ do not have any common factors. .. And $\phi(10^6)$ is definitly not $4^6$

Take  a small number, say  $12$, and you can see,  {1,5,7,11}  and \$phi(12) = 4 = \phi(4) \times \phi(3) $ but
you can not factor  $12 = 6 \times 2$  and expect $\phi(6) \times \phi(2) = 2 \times 1 $ to give correct answer.
For example  $\phi(100) $ is  $40$ and  NOT $16$. similary $\phi(10^6) = 4000000$
(That's why you may like to stay at 5 digits, as extra digit may make your work a little more time consuming
Hope that helps....
BTW if you  are curious .. last 7 digits here are $1985443$  .. if you want to try and check for 6 or 7 digits too? :)

why is this ?
could you tell me a site to understand and to learn all of
this.
and where am i wrong here ?
could you gimme a yahoo messenger id to talk to you please ?

thanks gyan for the reply
i understand now.
i really want to be better at number theory
what should i do ?
i don't want to be stupid but really i want to make a career on this actually
i know it sounds stupid.

could you please answer to my last questions please!!!!

\\
\\
Euler theorem states that:
If GCD(a,n)=1,\\ $a>1 , n>1$ then $a^{\varphi(n)}\pmod{n}=1$

We try to do this with number classes.\\
We are working in the class $Z_{10^5}$ to be able to find the last 5 digits
of the number N.
\\
\\
So we can reformulate Euler's theorem here in this class
of numbers $\LARGE {Z_{10^5}}$ like so:\\
\[a^{\varphi(n)}=\hat{1}\]\\
We must calculate $\varphi(n)$.It is the function that calculates\\
the number of numbers lower than n coprime with n.\\
And we have another definition for it:\\
\\
\\
If we write n like this:\[n=p_1^{\alpha_1}\cdot p_2^{\alpha_2} \cdot  p_3^{\alpha_3} \ldots p_n^{\alpha_n} \]
where $p_1,p_2,p_3,\ldots p_n$ are the prime numbers in the decomposition of n and $\alpha_1,\alpha_2,\alpha_3 \ldots \alpha_n$
are the powers associated to those numbers then :\[\varphi(n)=n\cdot(1-\frac{1}{p_1})\cdot(1-\frac{1}{p_2})\cdot(1-\frac{1}{p_3})\ldots (1-\frac{1}{p_n}) \]
\Large
So we must find $\varphi(10^5)=10^5\cdot \frac {4}{10}=4\cdot 10^4=40.000$

Now we know that in $Z_{10^5}$ we have $2003^{40.000}=\hat{1}$.Also we know for sure that $2005^{2007}\gg 40.000$
So we will have to do $2005^{2007} \bmod 40.000$ .
We can do a program in C++ altough it would be rather strange,or we could\\
apply newton's binom. Or more rapid and more lame at the same time we could use
calc.exe from windows. We do 2005 y 2007 percent 40.000 and we get 28125
this meaning that we had \[N=2003^{2005^{2007}}=2003^{M(40.000)+ 28215}=2003^{28215}\]
and now we use calc.exe again and we have\\ further helping ourselvs with newton's binom:

\[2003^{28125} mod 10^5=\]
\[32240721080810243^{5625} mod 10^5=\]
\[10243^{5625} mod 10^5=\]
\[112755014093508659443^{1125} mod 10^5=\]
\[59443^{1125} mod 10^5=\]
\[742170345549770876105443^{225} mod 10^5=\]
\[5443^{225} mod 10^5=\]
\[4777400730055835443^{45} mod 10^5=\]
\[35443^{45} mod 10^5=\]
\[55930972884964555985443^{9} mod 10^5=\]
\[85443^{9} mod 10^5=\]
\[623777155433307^{3} mod 10^5=\]
\[33307^{3} mod 10^5=\]
\[36949328585443 mod 10^5=85443\]
\\
So 85443 is the result.\\
I tried to make a C++ program that would calculate this\\
but i noticed that it has pretty big problems \\
when it makes this square $74721^2$ on unsigned longs.\\
For C++ this number is :$1288260545$.\\
Very weird because in reality it is:\\
5583227841.
This stupid error has unabled me to make a serious\\
program to make me $N\bmod 10^5$ in C++.


induction
number theory unsolved
number theory

Define a sequence of fractions by $ c_1 \equal{} 1,c_2 \equal{} \frac {8}{5},c_{n \plus{} 1} \equal{} \frac {2c_n^3}{c_n^2 \plus{} c_{n \minus{} 1}^2},$ then numerator of  the irreducible fraction $ c_n$ is $ 2^{2^{n \plus{} 1} \minus{} n \minus{} 3}.$

[quote="Ji Chen"]Define a sequence of fractions by $ c_1 = 1,c_2 = \frac {8}{5},c_{n + 1} = \frac {2c_n^3}{c_n^2 + c_{n - 1}^2},$ then numerator of  the irreducible fraction $ c_n$ is $ 2^{2^{n + 1} - n - 3}.$[/quote]

Let $ c_n=\dfrac{a_n}{b_n}$, where $ a_n$, $ b_n$ are integers and $ (a_n, b_n)=1$.
We will prove using induction that $ a_n=2^{2^{n + 1} - n - 3}$ and $ b_n\vdots b_{n-1}^2$. We know that $ b_n$ is not divisible by $ 2$.
Base for $ n=1, n=2$ is trivial.
Now suppose that $ a_{n-1}=2^{2^{n} - n - 3}, a_n=2^{2^{n + 1} - n - 3}, b_n=k\cdot b_{n-1}^2$ Substituting in the initial relation we get:
$ \dfrac{a_{n+1}}{b_{n+1}}=\dfrac{2^{2^{n+2}-n-2}}{b^2_n(\cdot2^{2^{n+1}}+k)}$. The down part of the second fraction is not divisible by $ 2$, due to that the second fraction is also irreducible. We get that $ a_{n+1}=2^{2^{n+2}-n-2}$ and $ b_{n+1}=b_n^2(\cdot2^{2^{n+1}}+k)$, which completes the induction.


algorithm
limit
Support
calculus
calculus computations

Consider the interactive process

X(k+1) = 0.5 { X(k) + a/X(k)} where a >0 . Assuming that the process converges. To what does it converge ? What is the order of convergence?

A truly ancient algorithm! Let's start by assuming that $X_{1}>0.$

We have $X_{k+1}=\frac12\left(X_{k}+\frac{a}{X_{k}}\right)$

Suppose $\lim_{k\to\infty}X_{k}=X$ and $X>0$ We'll support that later.

$X=\frac12\left(X-\frac{a}{X}\right)$

simplifies to $X^{2}=a,$ and, since $X>0,$ we have $X=\sqrt{a}.$

If no one else does it, I'll be back in an hour or so to talk about the order of convergence (it's second order.)

$\underline{\mathrm{Heron}}\mathrm{\ : \ \ }$ $\left\{\begin{array}{c}a>0\ ,\ n\in N^{*}\ ,\ x_{1}>0\\\\ x_{n+1}=\frac{1}{2}\cdot\left(x_{n}+\frac{a}{x_{n}}\right)\end{array}\right\|$ $\Longrightarrow$ $\lim_{n\to\infty}x_{n}=\sqrt a\ .$

$\underline{\mathrm{Newton}}\mathrm{\ : \ \ }$ $\left\{\begin{array}{c}a>0\ ,\ \{n\ ,\ p\}\subset N^{*}\ ,\ x_{1}>0\\\\ x_{n+1}=\frac{1}{p+1}\cdot\left(px_{n}+\frac{a}{x^{p}_{n}}\right)\end{array}\right\|$ $\Longrightarrow$ $\lim_{n\to\infty}x_{n}=\sqrt [p+1]a\ .$

Write $X_{k+1}=\frac{X_{k}^{2}+a}{2X_{k}}$

Now let $u_{k}=X_{k}-\sqrt{a},$ so that $X_{k}=\sqrt{a}+u_{k}$

$u_{k+1}=\frac{a+2u_{k}\sqrt{a}+u_{k}^{2}+a}{2(\sqrt{a}+u_{k})}-\sqrt{a}= \frac{2a+2u_{k}\sqrt{a}+u_{k}^{2}-2a-2u_{k}\sqrt{a}}{2(\sqrt{a}+u_{k})}$

$=\frac{u_{k}^{2}}{2(\sqrt{a}+u_{k})}$

All of this is based on the assumption that $X_{1}>0,$ hence $u_{1}>-\sqrt{a}.$ Note that we immediately get that $u_{k}>0$ for $k\ge 2$ (That's $X_{k}>\sqrt{a}.$)

We have that $|u_{k+1}|\le\frac{u_{k}^{2}}{2\sqrt{a}}.$

That's second order convergence.


topology
abstract algebra
group theory
superior algebra
superior algebra unsolved

I dont remember that i post this one but it is really nice :
Let $G$ be an infinite compact topological group with a hausdorff topology .prove that $G$ contains an element $g\neq 1$ such that the set of all powers of $g$ is either everywhere dense in $G$ or nowhere dense in $G$.

This is going to be rather long.. :)

The compact abelian groups which contain no element $g$ such that $\langle g\rangle$ ($\langle\cdot\rangle$ detones the subgroup generated by whatever is inside the brackets) is nowhere dense can be determined, and, unless I made some mistakes (which is possible since I'm not very familiar with all the results I'm using here), they turn out to be $(\mathbb Z_{p},+)$, the additive groups of the rings of $p$-adic integers $\mathbb Z_{p}$ (here, $\mathbb Z_{p}$ is not the ring $\mathbb Z/p\mathbb Z$ of residues modulo $p$, but the ring of $p$-adic integers).

Suppose $G$ is as above (i.e. all closed subgroups $\overline{\langle g\rangle}$ for $g\ne 1$ have non-empty interior). This simply means that all closed subgroups $H\le G,\ H\ne 1$ have non-empty interior. Now consider the Haar measure $\mu$ on $G$. It's finite, and $\mu(H)>0,\ \forall H=\overline H\le G$. Since the left cosets of $H$ have the same measure as $H$, there cannot be infinitely many of them. We have thus obtained the result that all closed non-trivial subgroups $H$ of $G$ have finite index.

In order to prove that $G$ is abelian, I'll now imitate the proof [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=cyclic&t=43091]here[/url] (notice how that problem seems to be a discrete analogue of this one).

Take an element $g\ne 1$ of $G$, and let $x_{1}=1,x_{2},\ldots,x_{n}$ be a complete system of representatives of the left cosets of $\overline{\langle g\rangle}$. For every $k\in\overline{2,n}$ there is an integer $m_{k}$ such that $g^{m_{k}}\in\overline{\langle x_{k}\rangle}$ (because $g^{m},\ m\in\mathbb Z$ can't all be distinct modulo $\overline{\langle x_{k}\rangle}$), so $g^{m_{2}\ldots m_{n}}$ is a non-trivial central element of $G$. This means that the center of $G$ is a non-trivial closed subgroup of $G$, and hence has finite index. By the problem cited in that other thread, it follows that the derived subgroup $G'$ is finite, and in the present situation this implies that it's trivial, i.e. $G$ is abelian.

Now we're free to use whatever [url=http://en.wikipedia.org/wiki/Pontryagin_duality]duality[/url] results we please.

Because all non-trivial closed subgroups of $G$ have finite index, it follows that all proper subgroups of the dual group $G^{*}$ (which is discrete because $G$ is compact) are finite. From this it follows that $G^{*}$ is some $\mathbb Z_{p^\infty}$ (for a prime $p$), i.e. the group of all roots of $1$ whose order is a power of $p$. I found this as a problem in Rotman's "An Introduction to the Theory of Groups", and I remember I used quite a few of the results that were developed in the main text of the book. Posting a proof would be too tedious, but I am sure, however, that the result is true.

Now we've determined $G$: it's the dual of $\mathbb Z_{p^\infty}$, and it's easy enough to see that this is precisely $(\mathbb Z_{p},+)$. As for what we initially wanted to prove, we immediately find that $\mathbb Z=\langle 1\rangle\le\mathbb Z_{p}$ is dense. Note that in the first paragraph I made a stronger assertion, namely that conversely, all groups $(\mathbb Z_{p},+)$ have the property that every non-trivial closed subgroup has non-empty interior. It shouldn't be too hard to prove that if $\nu_{p}(x)=k\ge 0$ for some $x\in\mathbb Z_{p}$ ($\nu_{p}(x)$ i the $p$-adic valuation of $x$; it equals the greatest exponent $k\in\mathbb N$ such that $p^{k}|x$), then the subgroup generated by $x$ is dense in $p^{k}\mathbb Z_{p}$.


The AMC-12 A 2003 #18

Let $n$ be a 5-digit number, and let $q$ and $r$ be the quotient and remainder, respectively, when $n$ is divided by 100. For how many values of $n$ is $q+r$ divisible by 11?

A) 8180
B) 8181
C) 8182
D) 9000
E) 9090

It's possible I miscounted by one, but I think my answer is correct.

[hide="Answer"]Obviously, we will have $9$ for every hundred numbers for one hundred out of every eleven, in which we have $10$. Therefore, for every eleven hundreds, we have $100$ numbers that work. Therefore, up through $99099$ we have $8100$ such numbers. Over the next $900$ we have another $81$, so our answer is $8181\Rightarrow \boxed{B}$.[/hide]


Ever thought of making an IRC channel for AoPS?

More importantly, what would be the point? IRC is not a suitable place for math - it's worse than AIM, even.


geometry
3D geometry
tetrahedron
octahedron
dodecahedron
icosahedron
probability

Starting next month (in other words, starting tomorrow), I will post a puzzle each day. Each of the puzzles will have something to do with logical thinking. The difficulty of the puzzles will range from ridiculously easy to brain-frying merciless, in other words, there's a bit of everything and every sort. So if you're a fan of logic puzzles, you'll (hopefully) love this thread. Remember to come check every day for the new logic puzzle.

For each puzzle, I will initially present:
-The puzzle
-Number of points it is worth
-Whether you can PM me for a hint

If you think you have the answer to a puzzle, send your answer, along with a possible solution, to me [b]by Private Message[/b]. That means, [b]do NOT post your solution on the thread!!![/b]

For some problems, I will award partial credit for certain answers. If you PM me for a hint, you get half the number of points you would have gotten if you gave the same answer without the hint. I will keep a scoreboard. As of now, I will post answers on the thread; you may PM me to ask for a solution if you reach a certain score in this, have at least 250 counted posts on AoPS, are a member of AAMAFTW, go to my school, or are special for some other reason. Feel free to PM me suggestions for this thread.

[quote="r15s11z55y89w21"]As of now, I will post answers on the thread; you may PM me to ask for a solution if you reach a certain score in this, have at least 250 counted posts on AoPS, are a member of AAMAFTW, go to my school, or are special for some other reason. Feel free to PM me suggestions for this thread.[/quote]

That's biased. Why can't everyone ask for solutions? And what does AAMAFTW have to do with it?

Some people deserve a special right; also, I am trying to positively contribute to this site, so I am "rewarding" people that are very helpful, like the ones with many posts and the members that made AAMAFTW. Of course, I also like neighbors, like you, so you're also in the special group. These people are like VIP for solutions. I feel that people who spam or perform incorrect activities here do not deserve to be able to take advantage of as many things as those who are positively contributing.

EDIT: 850th post!

So I have to fulfill at least one of the clauses to be VIP? OK, I have almost 1500 posts :P

[size=200]Puzzle #1[/size]

Out of Alberto, Bill, Chris, Danny, and Ernesto, Ernesto is the tallest. Chris is shorter than Danny, and Bill's height is both the median and the mode. Alberto's height is the mean of Danny and Chris's heights. Order the five boys from shortest to tallest.

This question is worth 5 points. You may NOT ask for a hint.

[hide="Submissions"]4[/hide]

[hide="Answer"]Chris, Alberto=Bill, Danny, Ernesto[/hide]

[hide="Points Leaderboard"]mathforlife 4
vahalla 4
rd5493 4
Bugi 3[/hide]

[hide="Perfect Answers Leaderboard"][/hide]

[hide="Submissions Leaderboard"]Bugi 1
mathforlife 1
vahalla 1
rd5493 1[/hide]

[size=200]Puzzle #2[/size]

Out of Kay, Kaila, and Kylan, one owns a cat, one owns a dog, one owns a donkey, one owns a llama, and one owns a turtle. None of the three owns more than two of the above, and none of them own an animal that shares two or more letters with their name. Kylan does not own a dog, but owns a cat. The person who owns the llama does not own any other animals. Who is the owner for each of the five animals?

This question is worth 5 points. You may ask for a hint.

[hide="Submissions"]6[/hide]

[hide="Answer"]Kay owns a llama, Kaila owns a dog and a donkey, Kylan owns a cat and a turtle[/hide]

[hide="Points Leaderboard"]mathforlife 9
vahalla 9
rd5493 9
lotsofmath 5
cyberspace 5
1=2 5
Bugi 3[/hide]

[hide="Perfect Answers Leaderboard"]lotsofmath 1
cyberspace 1
1=2 1
mathforlife 1
vahalla 1
rd5493 1[/hide]

[hide="Submissions Leaderboard"]mathforlife 2
vahalla 2
rd5493 2
Bugi 1
1=2 1
cyberspace 1
lotsofmath 1[/hide]

[size=200]Puzzle #3[/size]

The inhabitants of the planet of Jyrub use six coins in their currency system: the Barj, the Berj, the Birj, the Borj, the Burj, and the Byrj. A Borj is worth $ 8 \frac {1}{2}$ Berjes, a Birj is the equivalent of $ \frac {1}{6}$ of a Barj. A Byrj is worth a Birj less than a Burj, but is also worth a Berj more than a Borj. The mean of the values of the Borj and the Burj is $ \frac {1}{3}$ of a Birj less than the value of a Byrj. A Barj is worth less than a Borj, but more than a Berj. Order the six coins from greatest to least value.

This question is worth 6 points. You may ask for a hint.

[hide="Submissions"]3[/hide]

[hide="Answer"]Burj, Byrj, Borj, Barj, Berj, Birj[/hide]

[hide="Points Leaderboard"]mathforlife 15
cyberspace 11
vahalla 9
rd5493 9
SuperMathBoy 6
lotsofmath 5
1=2 5
Bugi 3[/hide]

[hide="Perfect Answers Leaderboard"]mathforlife 2
cyberspace 2
lotsofmath 1
1=2 1
vahalla 1
rd5493 1
SuperMathBoy 1[/hide]

[hide="Submissions Leaderboard"]mathforlife 3
vahalla 2
rd5493 2
cyberspace 2
SuperMathBoy 1
Bugi 1
1=2 1
lotsofmath 1[/hide]

[size=200]Puzzle #4[/size]

Find the three terms represented by question marks: Y, Y, H, L, Y, E, Y, T, R, ?, ?, ?

This question is worth 3 points. You may NOT ask for a hint.

[hide="Submissions"]6[/hide]

[hide="Answer"]R,R,R[/hide]

[hide="Points Leaderboard"]mathforlife 18
vahalla 12
cyberspace 11
rd5493 9
SuperMathBoy 9
lotsofmath 8
1=2 5
Bugi 3
andersonw 3
james4l 3[/hide]

[hide="Perfect Answers Leaderboard"]mathforlife 3
cyberspace 2
lotsofmath 2
vahalla 2
SuperMathBoy 2
andersonw 1
james4l 1
1=2 1
rd5493 1[/hide]

[hide="Submissions Leaderboard"]mathforlife 4
vahalla 3
rd5493 2
lotsofmath 2
SuperMathBoy 2
cyberspace 2
Bugi 1
1=2 1
andersonw 1
james4l 1[/hide]

[size=200]Puzzle #5[/size]

Jack likes 24 but not 34. He likes 104 but not 116. He likes 1568 but not 92. He likes 4 but not 64. He likes 44 but not 4444. Does he like 2004? Explain why or why not.

This question is worth 4 points. You may ask for a hint.

[hide="Submissions"]2[/hide]

[hide="Answer"]Yes, he likes numbers in the form $ p\plus{}1$, where $ p$ represents a prime number.[/hide]

[hide="Points Leaderboard"]mathforlife 18
vahalla 12
cyberspace 11
rd5493 9
SuperMathBoy 9
lotsofmath 8
1=2 5
Bugi 3
andersonw 3
james4l 3[/hide]

[hide="Perfect Answers Leaderboard"]mathforlife 3
cyberspace 2
lotsofmath 2
vahalla 2
SuperMathBoy 2
andersonw 1
james4l 1
1=2 1
rd5493 1[/hide]

[hide="Submissions Leaderboard"]mathforlife 5
vahalla 4
rd5493 2
lotsofmath 2
SuperMathBoy 2
cyberspace 2
Bugi 1
1=2 1
andersonw 1
james4l 1[/hide]

[size=200]Puzzle #6[/size]

Gert, Gil, Gina, Ginny, and Gwynette are sisters. Gina is not the shortest. Gert is shorter than Gwynette. Ginny is taller than both Gert and Gwynette. Gert and Gwynette are neither tallest nor shortest. Gil is shorter than Gwynette. Ginny is not the tallest. Order the five sisters from shortest to tallest.

This question is worth 5 points. You may NOT ask for a hint.

[quote="r15s11z55y89w21"]If anyone has any questions, feel free to PM me before July 21. :)[/quote]
What happens on July 21? I would volunteer and continue this, but I already have little time to do the PoTD and Tip of the Week.

Hahahaha, I have stopped playing this a long time ago, then came to 75, solved it, came back on leaderboard with a score of [b]42[/b]  :rotfl:

Congratulations, Bugi. As a sidenote: on the same puzzle, vahalla reached 42 correct answers. Interesting, huh? :)

Hi, I'd like to thank the creator of this competition; you helped my friend and I really refine our own competition (see my signature). (Hopefully I didn't plagiarize!  :o )

Wait, WTF at r15's signature. Did he leave AoPS?

[quote="isabella2296"]Yes, and on Second Pi Day, too.[/quote]

Well, that sucks. (And I'm guessing that the 355th day of the year (December 21 or 20, depending on whether it's a leap year or not) at 1:53 AM (the 113rd minute of the day) is Third Pi Day?)

Anyway, seeing as he's gone, let's post puzzles for the heck of it? No scoreboard, obviously.

[size=200]Puzzle #75.01[/size]

Michael and Ajid are playing a game. They place a counter on the top-left corner of a 5-by-5 grid. Each person takes a turn to move the counter either to the right or down, with Michael moving first. They can only do one of these at a time, and must move at least one square.

The person to move onto the bottom-right square wins.

1. Who has the winning strategy?
2. Who has the winning strategy if the person to move onto the bottom-right square [i]loses[/i]?
3. Who has the winning strategy on an $ n$-by-$ n$ grid?
4. On an $ m$ by $ n$ grid?

um

[quote="r15s11z55y89w21"]No, I would wish this thread to stay as is.[/quote]

[quote="r15s11z55y89w21"]I bequeath (verb choice?) LTM to Dojo (er?), all my HMC claims to dragon96 and james4l, [b]and DLP to james4l and/or AIME15, if he (they?) wish(es) to continue it (I don't think so, but I'll just give permission for the heck of it)[/b], and I donate anything of mine that could have been helpful to AoPS (was there any at all?) to the welfare of the AoPS community.[/quote]

[quote="r15s11z55y89w21"]I bequeath (verb choice?) LTM to Dojo (er?), all my HMC claims to dragon96 and james4l, [b]and DLP to james4l and/or AIME15, if he (they?) wish(es) to continue it (I don't think so, but I'll just give permission for the heck of it)[/b], and I donate anything of mine that could have been helpful to AoPS (was there any at all?) to the welfare of the AoPS community.[/quote]

Damn it. Well, then, I'll just ignore this thread for the time being then...

[quote="r15s11z55y89w21"]Please! Keep this thread clean of derogatory language![/quote]
I'm sorry...  :blush:
So you're back! I'm guessing that you won't continue this thread?

It may be restarted after I reconstruct everything else I destroyed.

Is it going to restart? We've waited a month.
I guess this thread is officially dead then.


Mom bought x containers of frozen juice concentrate. Each container of juice concentrate makes a gallon of juice. My little brother drinks 3/4 of each gallon of juice. If he drank 9 gallons of juice from these x containers, calculate $ x^2$.

[hide]Well, if he drank 9 gallons or 3/4 of [i]x[/i] gallons, [i]x[/i]=12, and [i]x[/i]²=[b]144[/b][/hide]

If the brother drank 9 gallons of juice, then 9 has to be 3/4 of the total number of gallons. 9 is 3/4 of 12, so there are 12 gallons. Each container has 1 gallon of juice, so there are 12 containers. Finally, if there are 12 containers, then x=12, and so x^2 is the answer, which is [b]144[/b].

P.S. Please tell me if I am wrong.


For a postive integer $\ x$ let $\ x'$ be the number obtained by reversing the order of its digits.Find the smallest
integer $\ a$ for which the equation $\frac{x'+1}{x}=a$ has infiniteley many solutions.

I think the answer should be 2.
We transfrom the equation into $ax=x'+1$
First, for a=2 we have solution 37, 397, 3997, ... , 39...97 whose number is obviously infinitely many.
Then for a=1 we find$x=x'+1$, the equation has no solution modulo 9.


geometry probability trigonometry AMC USA(J)MO USAMO MATHCOUNTS

without calculators it seemed a little bit harder, but the test itself was pretty straight-forword.

I didnt take it but looking over the content strictly, Ill say it was easier than the previous AMC 10s.

considerably easier than last year.

-_-

I actually got around 20 points less this year...
Maybe just because I was really tired at the end of the day.

I went from 126 last year to at most 118.5, an embarrassing drop of 7.5

uhh well factoring the no calculator rules, ill say this is about even or maybe slightly harder than last years...

but just based on content, this year's test was easier than last years (there were tons of geometry).

i hope the B has this much geometry because i really want to get a high score on it and geometry is the easiest one for me (compared to algebra, number theory, and probability)

I hated the geometry part. I can't do geometry for my life :(
Most problems were pretty easy besides #25 (it kinda took me a whiel to see that triangle).

Eaiser than the last few.

I hate geometry too! :|

I didn't take the 10, but I looked it over - I thought it looked harder. oh, wait a sec, I took the 10B last year, I think...

Dude, the 10B was way too hard for me last year. Too many geometry problems there.

Eh I hated the geo too, but I'm okay when it comes to AMC/MC type geo, where its a pretty obvious pythagorean theorem application or something like that.

i like geometry but i suck at it. can't do trig to save my life, and most AMC 10/12 hard geometry problems can just be trig bashed.

how can i do that? 2 consecutive posts in under 10 minutes. SKILLZ. SORRY MODS DON'T TAKE IT AS SPAMMING IT WAS TOTALLY BY ACCIDENT.

Wow, how'd you post 3 times in the same minute?  Anyways, I felt like the questions on this year's 10A were worded a little trickier, but the math was about the same as last years, making the test a little harder.

Oh jeez. I saw to many geo problems and my brain fried. But I solved everything else.

I thought it was a bit harder than past years.

lets just say, its VERY easy to make careless mistakes on geometry. but the concepts were trivial this year.

and yes, i thought the questions were purposely worded in a confusing way this year also. i took last years mathcounts state sprint the period before i was going to take the test and i noticed that the wording was somewhat similar (more than previous years, at least). i think the AMC people were kind of going for that this year on the 10.

[quote="serialk11r"][quote="cognos599"]how can i do that? 2 consecutive posts in under 10 minutes. SKILLZ. SORRY MODS DON'T TAKE IT AS SPAMMING IT WAS TOTALLY BY ACCIDENT[...]i have no idea. but anyway getting back to the real topic...[/quote]
That right there is totally spamming. Its easy to tell that you accidentally double posted. You spammed to point out that you were in fact not trying to spam. And don't go off topic like you did on other threads too, personal stuff is done by PM ;)
[quote="mathcrazed"]Everyone hates geo except me? Geometry is the decisive subject!! xD The only proofs or USAMOs I can do that aren't number 1 are geometry xD[/quote]
Wow...I really love geometry, but I'm completely horrible at it. This year if I make usamo then I'll pretty much be relying on the number 1 and 4s being non-geometry problems :([/quote]

actually i triple clicked and realized that i couldn't delete 2 of them so i decided that i would edit the last one to explain my impudence instead of posting another post as you thought i had done only with 2 posts in your theory, that would have been spamming and even this post is spamming.


integration
calculus
real analysis
real analysis unsolved

[hide="Suggestion"]$ S = \int^{1}_{0}\frac{1+x^{2}+x^{4}+x^{6}}{1+x^{8}}\,dx = \frac{1}{4}\int^{+\infty}_{-\infty}\frac{1+x^{2}+x^{4}+x^{6}}{1+x^{8}}\,dx$, and this integral can be calculated using complex analysis. However, there is certainly an easier way.
[/hide]


Given that $ a,b,c$ are non-negative numbers, prove that:

\[ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b} \geq \frac{3(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)}{4(ab\plus{}bc\plus{}ca)^2}\plus{}\frac{5}{4}\]

Notice that there are more than one equality points.

[quote="shaaam"]Given that $ a,b,c$ are non-negative numbers, prove that:
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \geq \frac {3(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}{4(ab \plus{} bc \plus{} ca)^2} \plus{} \frac {5}{4}\]
Notice that there are more than one equality points.[/quote]
Since $ a^2b^2\plus{}b^2c^2\plus{}c^2a^2\equal{}(ab\plus{}bc\plus{}ca)^2\minus{}2abc(a\plus{}b\plus{}c),$ the inequality can be written as
$ \sum \frac{a}{b\plus{}c}\plus{}\frac{3abc(a\plus{}b\plus{}c)}{2(ab\plus{}bc\plus{}ca)^2} \ge 2.$
Now, by using the known inequalities $ ab\plus{}bc\plus{}ca \le \frac{1}{3}(a\plus{}b\plus{}c)^2$ and
$ (ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c) \le \frac{9}{8}(a\plus{}b)(b\plus{}c)(c\plus{}a),$
we get
$ \frac{3(a\plus{}b\plus{}c)}{2(ab\plus{}bc\plus{}ca)^2} \ge \frac{9}{2(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c) } \ge \frac{4}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}.$
Therefore, it suffices to prove that
$ \sum \frac{a}{b\plus{}c} \plus{}\frac{4abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \ge 2,$
which is true because it is the third degree Schur's Inequality in fractional form.

[quote="can_hang2007"][quote="shaaam"]Given that $ a,b,c$ are non-negative numbers, prove that:
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \geq \frac {3(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}{4(ab \plus{} bc \plus{} ca)^2} \plus{} \frac {5}{4}\]
Notice that there are more than one equality points.[/quote]
Since $ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \equal{} (ab \plus{} bc \plus{} ca)^2 \minus{} 2abc(a \plus{} b \plus{} c),$ the inequality can be written as
$ \sum \frac {a}{b \plus{} c} \plus{} \frac {3abc(a \plus{} b \plus{} c)}{2(ab \plus{} bc \plus{} ca)^2} \ge 2.$
Now, by using the known inequalities $ ab \plus{} bc \plus{} ca \le \frac {1}{3}(a \plus{} b \plus{} c)^2$ and
$ (ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c) \le \frac {9}{8}(a \plus{} b)(b \plus{} c)(c \plus{} a),$
we get
$ \frac {3(a \plus{} b \plus{} c)}{2(ab \plus{} bc \plus{} ca)^2} \ge \frac {9}{2(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c) } \ge \frac {4}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}.$
Therefore, it suffices to prove that
$ \sum \frac {a}{b \plus{} c} \plus{} \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \ge 2,$
which is true because it is the third degree Schur's Inequality in fractional form.[/quote]

WOW, you are so quick! I can't believe how someone can find solution in couple of minutes. You amaze me ~
Now, let me absorb the solution  :lol:

[quote="shaaam"]Given that $ a,b,c$ are non-negative numbers, prove that:
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \geq \frac {3(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}{4(ab \plus{} bc \plus{} ca)^2} \plus{} \frac {5}{4}\]
Notice that there are more than one equality points.[/quote]

from the  inequality  assume $ a\geq b\geq c$

the inequality is  equivalent  to

$ f(a,b,c) \equal{} ((4bc \plus{} 2c^2 \plus{} 2b^2)a^3 \plus{} (7b^2c \plus{} 7c^2b)a^2 \plus{} (cb^3 \plus{} c^3b \plus{} 8b^2c^2 \minus{} 2c^4 \minus{} 2b^4)a)(a \minus{} b)(a \minus{} c) \plus{} (6(b \plus{} c)^2bca \plus{} 2b^2c^2(b \plus{} c))(b \minus{} c)^2\geq 0$

$ f(a,b,c)\geq((4bc \plus{} 2c^2 \plus{} 2b^2)a^3 \plus{} (7b^2c \plus{} 7c^2b)a^2 \plus{} (cb^3 \plus{} c^3b \plus{} 8b^2c^2 \minus{} 2c^4 \minus{} 2b^4)a)(a \minus{} b)(a \minus{} c)$
$ \geq((2c^2 \plus{} 2b^2)a^3 \plus{} ( \minus{} 2c^4 \minus{} 2b^4)a)(a \minus{} b)(a \minus{} c)$
$ \geq ((2c^4a \plus{} 2b^4a \plus{} ( \minus{} 2c^4 \minus{} 2b^4)a)(a \minus{} b)(a \minus{} c)\geq 0$

so..............


A person 2 meters tall walks directly away from a streetlight that is 8 meters above the ground. If the person is walking at a constant rate and the person's shadow is lengthening at the rate of $ \frac{4}{9}$ meters per second, at what rate, in meters per second, is the person walking?

let the shadow length be l, and the distance of the person to the light be x
then l/(l+x)=2/8, l/x=1/3
then l=x/3
then 3dl/dt=dx/dt
and dl/dt=4/9, then you got the answer


function
calculus
derivative
trigonometry
calculus computations

hi

it is well known that if in a 'ball' around a point, a function in two variables exists and has continous partial derivatives , it is differentiable

however, is the converse true?  are there functions that are defined in a ball around a point, that is differentiable in that point, and with partial derivatives in that ball that are not continuous in that point?

i'm sure there are, so there must be elementary examples?

thc

A simple example, based on bad things that can happen for functions of one variable:
$f(x,y)=(x^2+y^2)\sin\left(\frac1{x^4+y^4}\right)$ (and $f(0,0)=0$). That function is differentiable everywhere, but its derivative and the partial derivatives are not continuous at the origin.


integration
logarithms
calculus
calculus computations

Find a primitive on $(0,1)$ of \[f(x) = \frac1{x \left( 1-x^{4}\right)^{n}}, \, n \in \mathbb N .\]


Find the smallest positive integer which can be expressed as a difference of two different perfect squares in 8 different ways.

I think it would be [hide]$3^{3}*5*7$, right?  The different of squares is restricted in that it is the product of two factors which are of the same parity...so throwing in twos would screw things up, because you'd have large even numbers having to multiply with small odd ones to get the same number.  So the smallest such number is odd.  Then it's easy: we find the smallest odd number with $16$ factors (or we can have $15$ factors if we allow nonzero squares): that number is $3^{3}*5*7$.[/hide]

This seems more like intermediate/pre-olympiad...unles I screwed up horribly there


Circle with center $I$ is inscribed in $\triangle ABC$, and touches $AB$ at $D$, $AC$ at $E$ and $BC$ at $N$. $BI$ intersects $DE$ at $M$, $CI$ intersects $DE$ and $K$. Prove that $I$ is the incenter of $\triangle KMN$.

As the point $M,$ lies on the bisector of the angle $\angle ABC$ and $BD = BN,$ we conclude that the triangles $\bigtriangleup BDM,$ $\bigtriangleup BNM,$ are congruent and so, we have $\angle BMD = \angle BMN.$

That is the segment line $BI,$ is also the bisector of $\angle DMN.$

Similarly the segment line $CI,$ is the bisector of $\angle EKN$ and the proof is completed.

Kostas Vittas.


According to [url]http://en.wikipedia.org/wiki/American_Mathematics_Contest[/url], they're going to start giving 1.5 points instead of 2.5 point for omitted questions on the AMC, starting in 2007.  This is supossed to stop people from leaving a whole bunch of questions blank to qualify for the AIME.

What are your thought on this?

NOOOOOOOOOOOO!...

 :P jk lol...Oh well. I usually leave it blank when I have a bad chance of guessing right.

Well that would be nice. I hate leaving stuff blank.

Well, it is Wikipedia. Can we really trust it?

[quote="sam3.14"]So does this mean that you will have to get more questions right in order to qualify for the AIME?[/quote]

This is the precise reason for the change.  The AMC would like to see students have an incentive to try the harder problems.  The old system led to way too many students not even looking at the last 10 problems, which defeats the purpose of the test to some degree.

At least it is not next year......

That'll really make it tough to now get above a 130 now!

on the bright side, now there is absolutely no question that if you get 24 right, you HAVE to guess on the 25th one. no leaving it blank. you have to go all out

i think it's a change for the best. the current system does make it rather easy to qualify. i am happy that the change does not affect me tho (graduating 2006  :) )

Oh well, I go for the gusto anyway. :lol:  :lol:

Well, I only left one blank on the 10 last year, and that was because I realized the answer was wrong and was trying to fix it at the last minute, but didn't manage to. :P I also got two wrong, but whatever. ;)

Meh I just answer every single on on the 10 anyways.  Dunno about the 12, will have to see.

In two AMC-10s, I haven't left a question blank. In three AMC-12s, I left 6 blank. In my last AMC-12, I left 5 blank. That was my best score :). So I'm all for this adjustment.

[quote="chess64"]Well if you leave $x$ questions blank and get the rest of them right, then $2.5x+6(25-x) \geq 120 \Rightarrow x \leq 8...$ So you can skip 8 of them and still qualify (on AMC 10)

In the new system, $1.5x+6(25-x) \geq 120 \Rightarrow x \leq 6...$ So you can only skip 6 and still qualify. Not much difference, though.[/quote]
For AMC 12, the change would be like this:

Current: $2.5x+6(25-x)\ge 100\Longrightarrow x\le 14$
New: $1.5x+6(25-x)\ge 100\Longrightarrow x\le 11$

You can still leave 11 questions blank to qualify.

Raargh there should be 0 points for leaving blank.  I always guess anyway because leaving blank is just not cool.

watch this system get changed yet again later on.

people will complain that this system encourages guessing.

Pah, I just do all of them.

Guessing is for lamers, just like leaving it blank. I just... do all of them cuz I know how to.

This is good, i usually never left any blank on the AMC 10s

I think the new system will be better.

The new system forces people who try to get about 100 or 120 points (depending on which AMC they take) to try harder, which is good. Unfortunately, for those of us who try to solve all of them (and can't), it slightly hurts our score for the problems we leave blank.

I think it's a good change.  On test day, I hear way too many people asking, "OK, how many do I need to answer to advance?"  Unfortunately, these people (as far as I know) usually just do the 11 easiest questions and then stop, figuring they have nothing to gain by answering more.  Now they will need to answer 14, and I'm guessing many of these students are capable of getting 14 right.  Yes, it will make it harder to make AIME--and some of the students I speak of won't make it--but I don't see a problem with that.  After all, it should still be easier than on the old AHSMEs.

By the way, 1234567890, you can trust the info on wikipedia because (a) it originally came from the 2005 results book, and (b) I'm the one who added that particular piece of info   :)

Hehe, I was one of those that tried to figure out the "strategy" to get into AIME, as in the minimum I need to get right.

Last year I got around 18 on AMC 12, so this shouldn't affect me that much...

From the book ("Message from the AMC 12 chair", p. 15):

"Beginning in 2002 the number of points awarded for a blank response on the AMC 12 was increased from 2 to 2.5.  As a result, students could qualify for the AIME by solving as few as 11 problems.  Analysis of student performances since then reveals that the average number of blank responses by AIME qualifiers has increased from about 8 to 11.  This trend, suggesting that many bright students are simply ignoring problems of even moderate difficulty, runds counter to our goals.  For that reason, beginning in 2007, the reward for a blank response will decrease from 2.5 to 1.5 points.  It should be emphasized that in making this change, we are not attempting to reduce the number of AIME qualifiers.  Because we are simultaneously working to improve the overall success rate on the first ten problems, we expect that the number of students achieving the minimum qualifying score of 100 will remain roughly constant."

Pretty well said, I think.  Come to think of it, it doesn't specifically say that the change will apply to the AMC10, but I would assume it will.

The fact that the first ten will be lowered in difficulty will keep the same number of kids going to AIME, but will probably lwer the USAMO index a tad ...

JB

[quote="IntrepidMath"]What if they (like the canadian competicions) had the last..7 or so problems worth more than the first 18? That would make more people look at the last ones.[/quote]

I've seen many competitions like that (even standardized tests). That would change everything though.

Erm, I think 1.5 is a little too low. I mean, some people like the fact that they are able to not be 100% sure and not get punished. It's a bad idea to do that. I don't mind that much, but 1.5 means people will have to do much better - and 4 problems is much better for alot of people. Or is it 3? I'm not 100% sure.

Well the penalty for random guessing should be balanced so that its not "worth it" to guess. Or, if you randomly guess, you have a 1/5th chance of getting it right, so leaving it blank should give 6/5 pts. Not that I really want a lowered guessing penalty

Well if you think of this beyond the AMC, this could weed out some people that otherwise would have made it to the AIME, so the floor for that could either be [b]very high[/b] due to the high % of Good AMCers or [b]very low[/b] to allow about the same amount of ppl as usual into the USAMO (around 250). :o Of course AMC might change the floor. Idk, just a thought.

The thing is, now, you could leave the test blank and still get 62.5  :roll:

That's a good way to look at it... :roll:


geometry
analytic geometry
graphing lines
slope
vector

[quote="Interval"]When two straight lines meet each other perpendicularly, the product
of their slopes is -1. However, x and y axes meet at 90 degrees, but
the product of their slopes is not -1

because slope of x axis  =  0 and slope of y axis  =  infinity.
Why?[/quote]
$\infty=\frac{-1}{0}$
pwnage

Because the original assumption is flawed.

[quote="Interval"]How do you know it's flawed?[/quote]It's flawed because you need to consider the conditions for that proposition. Below are things to consider for all of the above posts:

1) The condition is that the two perpendicular straight lines are not both parallel to the axes.

2) There is a more general theorem about perpendicular lines in Cartesian coordinates using vectors and matrices.

3) Infinity is not a number, it's a concept.

4) Dividing by zero is undefined.

[quote="Interval"]How do you know it's flawed?[/quote]
Isn't the definition of the word flawed practically anything that doesn't work and gives wacky results?


Does anyone know a website where I can get solutions to the free response sections? I'm specifically looking for the years 94 and 99. So, anyone else cramming for AP chemistry?

hells yeah dude...three days left!!! wooo i can't wait to get it on wednesday!! good luck!!