geometry
circumcircle
trapezoid
AMC
USA(J)MO
USAMO
similar triangles

For a triangle ABC , let tangent lines of the circumcircle of ABC at B and C meet at D. Prove that AD is a symmedian of ABC.
Please help me. I will be happy for your help. Thank you.

Let be the point $ P\equiv (O)\cap AD$ and we denote as $ E,\ F,$ the orthogonal projection s of $ P,$ on $ AB,\ AC,$ rerspectively.

From the similar right triangles $\bigtriangleup EBP,\ \bigtriangleup FCP,$ we have that $ \frac {PE}{PF} \equal{} \frac {PB}{PC}$ $ ,(1)$

From the pairs of similar triangles $\bigtriangleup BAD,\ \bigtriangleup PBD,$ and $ \bigtriangleup CAD,\ \bigtriangleup PCD,$ we have that $ \frac {PB}{AB} \equal{} \frac {BD}{AD} \equal{} \frac {CD}{AD} \equal{} \frac {PC}{AC}$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac {PE}{PF} \equal{} \frac {AB}{AC}$ and the proof is completed.

Kostas Vittas

PS. Change please, the points $E,\ F$ in the attached figure, in to each other.

Sorry for late thank :blush: enormous chores  :(

Nice solution, [b]vittasko[/b]!

I will present two more:

[hide="Solution 1"]
Consider $ X$ on $ AB$, $ Y$ on $ AC$, and $ Z$ so that $ ACDX$ is cyclic, $ ABDY$ is cyclic, and $ Z$ is the intersection between $ XC$ and $ YB$.  Furthermore, let $ M$ be the midpoint of $ BC$.  Now, $ \angle DCY \equal{} 180 \minus{} \angle DCB \minus{} \angle ACB \equal{} 180 \minus{} \angle BAC \minus{} \angle ACB \equal{} \angle ABC$ since $ DC$ is tangent to the circumcircle of $ \triangle ABC$.  Similarly, $ \angle DBX \equal{} \angle ACB$.  Yet, since $ ACDX$ is cyclic, we have that $ \angle ABC \equal{} \angle DCY \equal{} \angle AXD$, so $ DX\parallel BC$.  Similarly, $ DY\parallel BC$, so $ X$, $ D$, and $ Y$ are on a line parallel to $ BC$.  This means that $ BCYX$ is a trapezoid, and so it is well-known that $ AZ$ meets $ BC$ and $ XY$ at their midpoints, so $ A$, $ M$ and $ Z$ are collinear.  Since $ ACDX$ and $ ABDY$ are cyclic, we see that\[ 180 \minus{} \angle BZC \equal{} 180 \minus{} \angle XZY \equal{} \angle ZXY \plus{} \angle XYZ \equal{} \angle DAC \plus{} \angle DAB \equal{} \angle BAC\]so $ ABZC$ is cyclic.  This implies that\[ \angle MAC \equal{} \angle ZAC \equal{} \angle ZBC \equal{} \angle ZYX \equal{} \angle DAB\]so $ AD$ is a symmedian, as desired. [/hide]

[hide="Solution 2"]
Invert around point $ D$ with radius $ DB$ that maps any point $ X$ to $ X'$.  Let the center of the circle be $ O$ and let $ MD$ meet the circumcircle of $ \triangle ABC$ at $ Y$.  Then, the circumcircle of $ \triangle ABC$ will invert to itself, but for any point $ P$ on the circle, $ P'$ will be the intersection between $ DP$ and the circle.  This is since if $ Q$ is the intersection between $ DP$ and the circle, then $ DP\cdot DQ\equal{}DB^2\equal{}DP\cdot DP'\implies DP'\equal{}DQ$, so $ Q\equal{}P'$.  Thus, $ A'$ is the intersection between the circumcircle of $ \triangle ABC$ and $ DA$.  Now, since $ DB\equal{}DC\implies MD\perp BC$, so $ M'$ is where $ M'B\perp DB$ and $ M'C\perp CD$.  Yet, $ OB\perp BD$ and $ OC\perp CD$, so $ M'\equal{}O$.  Now, notice that since $ Y$ is the intersection between the perpindicular bisector of $ BC$ and the circumcircle of $ \triangle ABC$, we have that $ Y$ is the midpoint of arc $ BC$.  Therefore, $ \angle BAY\equal{}\angle CAY$.  Yet, $ \angle BAY\equal{}\angle BAA'\plus{}\angle A'AY$ and $ \angle CAY\equal{}\angle CAA'\minus{}\angle YAA'$, so $ \angle BAA'\plus{}2\angle A'AY\equal{}\angle CAA'$, so $ \angle BAD\plus{}\angle A'OY\equal{}\angle DAC$.  Yet, the inversion maps $ \angle A'OY\equal{}\angle A'OM\equal{}\angle A'AM$, so $ \angle BAM\equal{}\angle BAA'\plus{}\angle MAA'\equal{}\angle CAA'\equal{}\angle CAD$, so $ AD$ is a symmedian of $ \triangle ABC$, as desired. [/hide]

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1187408#1187408[/url] for a proof. I think it's the same as vittasko's.

EDIT: no, it is slightly different  :)

I used this lemma on USAMO 08 #2.

Denote $ N \equal{} AD\cap (O)$ and $ M$ the midpoint of $ BC$. We have $ \triangle DBN\sim\triangle DAB$ and $ \triangle DCN\sim\triangle DAC$, so $ \frac {DB}{DA} \equal{} \frac {BN}{AB}$ and $ \frac {DC}{DA} \equal{} \frac {CN}{AC}$ $ \Longrightarrow \frac {BN}{AB} \equal{} \frac {CN}{AC}$, i.e. $ AC\cdot BN \equal{} AB\cdot CN$. Applying Ptolemy theorem to the cyclic quadrilateral $ ABNC$, we conclude that $ AN\cdot CM \equal{} AC\cdot BN$. Therefore $ \triangle ACM\sim\triangle ANB$, hence $ \measuredangle CAM \equal{} \measuredangle NAB$, as desired.

Dear April

I can't help but thank you  :)

[quote="77ant"] [color=darkred] Denote the tangents $ BB$ , $ CC$ to the circumcircle of $ \triangle ABC$ at $ B$ and $ C$

respectively and the its intersection $ D$ . Prove that $ AD$ is a $ A$ - symmedian of $ ABC$ .[/color] [/quote]

[quote][color=darkred][b][u]Lemma 1 (simply !).[/u][/b] Let $ w$ be the circumcircle of the cyclical convex quadrilateral $ ABCD$ . Denote $ I\in AC\cap BC$

and $ AB\equal{}a$ , $ BC\equal{}b$ , $ CD\equal{}c$ , $ DA\equal{}d$ , $ AC\equal{}e$ , $ BD\equal{}f$ . Then there is the chain of equal ratios :

$ \frac {IA}{da}\equal{}\frac {IB}{ab}\equal{}\frac {IC}{bc}\equal{}\frac {ID}{cd}\equal{}\frac {e}{da\plus{}bc}\equal{}\frac {f}{ab\plus{}cd}\equal{}\sqrt {\frac {\minus{}p_w(I)}{abcd}}$ , where $ p_w(I)$ is the power of $ I$ w.r.t. $ w$ .[/color][/quote]

[quote][color=darkred][b][u]Lemma 2.[/u][/b] Let $ ABC$ be a triangle. Denote the intersection $ T$ between the tangent $ AA$

in $ A$ to the circumcircle $ w$ of $ \triangle ABC$ with the sideline $ BC$ . Then $ \frac {TB}{TC}\equal{}\left(\frac {AB}{AC}\right)^2$ . [b]Ind.[/b] : $ \triangle TAB\sim\triangle TCA$ a.s.o.[/color] [/quote]

[color=darkblue][b][u]Proof of the proposed problem.[/b][/u] Denote $ S\in AP\cap BC$ and the second intersection $ P$ of the line $ AD$ with the circle $ w$ .

Apply the  [b]lemma 2[/b] to the triangles $ ABP$ , $ ACP$ , i.e. $ \frac {DP}{DA}\equal{}\left(\frac {BP}{BA}\right)^2$ and $ \frac {DP}{DA}\equal{}\left(\frac {CP}{CA}\right)^2$ . Thus, $ \frac {PB}{PC}\equal{}\frac {AB}{AC}\ (*)$ .

Apply the [b]lemma1[/b] to the cyclical quadrilateral $ ABPC$ , i.e. $ \frac {SB}{BP\cdot BA}\equal{}\frac {SC}{CP\cdot CA}$ . Thus, $ \frac {SB}{SC}\equal{}\frac {PB}{PC}\cdot\frac {AB}{AC}$ .

Using the relation $ (*)$ obtain $ \frac {SB}{SC}\equal{}\left(\frac {AB}{AC}\right)^2$ , i.e. the line $ \overline {ASPD}$ is the $ A$ - symmedian in the triangle $ ABC$ .[/color]

I have discovered this nice (?) picture at some mathematician's home page right now.
completely direct proof ... :(

Thanks for nice solutions , Virgil


limit
logarithms
real analysis
real analysis unsolved

For each $ n \in \mathbb{N}^*$ denote $ x_n$ the real solution of the equation $ 2^x+x=n$. Show that the limit exists and it is finite, and evaluate:

\[ \lim_{n \to \infty} n \left( \frac{x_n}{\log_2{n}}-1 \right).\]

Let $ f(x) \equal{} x \plus{} 2^x$. Then it is easy to see that $ \lim_{x\to\infty} f(x) \left( \frac{ x}{\log_2 f(x)} \minus{} 1 \right) \equal{} \minus{}\frac{1}{\log 2}$.

Put $ y \equal{} f(x)$. Since $ x \to \infty$ as $ y \to \infty$ and $ x_n \equal{} f^{\minus{}1}(n)$,  we have

$ \lim_{n \to \infty} n \left( \frac{ x_n}{\log_2 n} \minus{} 1 \right) \equal{} \lim_{y \to \infty} y \left( \frac{ f^{\minus{}1}(y)}{\log_2 y} \minus{} 1 \right) \equal{} \minus{}\frac{1}{\log 2}$.


ratio
geometry
rectangle
linear algebra
linear algebra unsolved

Let $C$ be an arbitrary circle of a plane.
find all affine transformations $f$ such that: $f(C) = C$

Our transformation turns two distinct diameters into two segments with endpoints on the circle, intersecting in their common midpoint, which means that they are also diameters, and the transformation fixes the center.

Since it preserves ratios of areas, it turns squares into squares, because it turns rectangles inscribed in the circle into rectangles, and squares have the largest areas. Using the fact that it preserves ratios of areas again, we find that it turns rectangles into similar rectangles. This means that the trace of the transformation is orthogonal, because it preserves scalar products. Conversely, it's clear that every orthogonal affine transformation which fixes the center of $C$ is is good.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $q=p^n$. Prove that for $p=2$ all the elements of the field $F_q$ are squares and for $p>2$ all the squares of the group $F^*_q$ form in it the subgroup $F^{*2}_q$ of index $2$ and $F^{*2}_q=Ker(t \to t^{\frac{q-1}{2}})$

Note that the multiplicative group is cyclic with order $p^n-1$, thus isomorphic to $\mathbb{Z}/ (p^{n}-1) \mathbb{Z}$.
Now $p^n-1$ is divisible by two for odd $p$, the mapping gets surjective onto the 'even' residue classes, giving the result.
Now for $p=2$, the mapping $a \mapsto 2a$ is bijective in $\mathbb{Z}/ (p^{n}-1) \mathbb{Z}$, so all numbers are squares.

ZetaX, why the order of the miltiplicative subgroup is $\phi(q)=(p-1)p^{n-1}$- by $F_q$ i meant the finite field of order $q$ ?

I don't know what I thought when writing that stupid nonsense  :blush:  :blush:
It should be correct now.


Does anybody know where can I find egyptian symbols?
Thanks

Go to http://tug.ctan.org/find.html and in the Search the Catalogue For a Keyword search look for egypt. You will get links to 3 packages:

[url=http://tug.ctan.org/tex-archive/help/Catalogue/entries/archaic.html]archaic[/url] A collection of archaic fonts.
[url=http://tug.ctan.org/tex-archive/help/Catalogue/entries/hieroglf.html]hieroglf[/url] The "poor man's" Egyptian Hieroglyphic font.
[url=http://tug.ctan.org/tex-archive/help/Catalogue/entries/hieroglyph.html]hieroglyph[/url] Hieroglyph fonts and other support.

I have found it but does it support and numbers? This is what I realy need. Have you tried it?

I haven't tried any of the packages but if you look at the pdfs at [url=http://tug.ctan.org/tex-archive/help/Catalogue/entries/hieroglf.html]hieroglf[/url] you will see all the font characters. It does say these are only a fraction of the ones at [url=http://tug.ctan.org/tex-archive/help/Catalogue/entries/hieroglyph.html]hieroglyph[/url]

Now I have problem with installation. Please, help me. I have WinXP.

That's a bit difficult unless you give details of your LaTeX system,  the problems you are having or which package you are working with.

If you are using MiKTeX use its Package Manager which will install packages automatically and refresh the database for you. Then read the package instructions for how to use it and what else may need installing (eg hieroglf requires oands).


geometry
linear algebra
matrix
topology
advanced fields
advanced fields unsolved

An infintesimal Lorentz transformation (of Minkowski space-time) and its inverse can be written as

$ {x'}^{\alpha} \equal{} (g^{\alpha\beta} \plus{} {\epsilon}^{\alpha\beta})x_{\beta}$

$ x^{\alpha} \equal{} (g^{\alpha\beta} \plus{} {{\epsilon}'}^{\alpha\beta}){x_{\beta}}'$

where $ {\epsilon}^{\alpha\beta}$ and $ {{\epsilon}'}^{\alpha\beta}$ are infinitesimal. For those of you that are not familiar with the geometry of Minkowski space-time, it is just a metric space described by the metric tensor $ g^{\alpha\beta}$ which can be written explicitly as a 4 by 4 diagonal matrix with signature (1,-1,-1,-1), and Lorentz transforms leave the norm invariant.

Show that:

(a) $ {{\epsilon}'}^{\alpha\beta} \equal{} \minus{} {\epsilon}^{\alpha\beta}$

(b) $ {\epsilon}^{\alpha\beta} \equal{} \minus{} {\epsilon}^{\beta\alpha}$

I'll write the transformation as $ x^{\mu} \rightarrow x' ^{\mu} = \Lambda_{\nu}^{\mu} x^{\nu}$. So if we were to take an infinitesimal transformation (away from the identity) we can write \[ \Lambda_{\nu}^{\mu} = \delta _{\nu}^{\mu} + \epsilon_{\nu}^{\mu}\] Now our space-time interval invariance requires \[ g_{\rho \sigma} = g_{\mu \nu} \Lambda_{\rho}^{\mu} \Lambda_{\sigma}^{\nu}\] Plugging in gives \[ g_{\rho \sigma} = g_{\mu \nu} (\delta_{\rho}^{\mu} + \epsilon_{\rho}^{\mu}) (\delta_{\sigma}^{\nu} + \epsilon_{\sigma}^{\nu}) \\ = g_{\mu \nu} (\delta_{\rho}^{\mu} \delta_{\sigma}^{\nu}  + \delta_{\rho}^{\mu} \epsilon_{\sigma}^{\nu} + \delta_{\sigma}^{\nu} \epsilon_{\rho}^{\mu} + O(\epsilon^2) ) \\ = g_{\rho \sigma}+ g_{\rho \nu} \epsilon_{\sigma}^{\nu} + g_{\mu \sigma} \epsilon_{\rho}^{\mu} \\ = g_{\rho \sigma} + \epsilon_{\rho \sigma} + \epsilon_{\sigma \rho}\]
So \[ \epsilon_{\rho \sigma} + \epsilon_{\sigma \rho} = 0\] The inverse is the same because your transformations form a group.


calculus
derivative
vector
trigonometry
inequalities proposed
inequalities

Let $ x,y,z \in [-1,1] $ and $ x+y+z = a $ . Find the maximum of the sum : $ x^2+y^2+z^2 $

proof by Lagrange method:
$f(x,y,z)=x^2+y^2+z^2+\lambda(x+y+z-a)$
$f'_x=2x+\lambda=0$
$f'_y=2y+\lambda=0$
$f'_z=2z+\lambda=0$

so $x=y=z$
but we have to check $x=-1$ and $x=1$

some calculation

and we are done

Can you expand your thought ? I'm new to Lagrange method though I'd like to learn it eventually - but I just can't get the result you give. With $\lambda$  I get $x^2+y^2+z^2=\frac{a^2}{3}$ - is it correct ? How to check boundary points ?

I'll be really thankful if someone would explain it all to me

Can you please tell me what Lagrange method is ?

This discussion is moot- $x=y=z$ is a minimum. Look at the second derivative; $x^2+y^2+z^2$ is convex, so its maximum will come on the boundary. Since the set is a polygon, we can even show that the maximum must come at a vertex.
For $1\le a\le3$, the maximum occurs when $x=y=1$ and $z=a-2$, with value $2+(a-2)^2.$
For $-1\le a\le1$, the maximum occurs when $x=1,y=-1,z=a$, with value $2+a^2.$
For $-3\le a\le-1$, the maximum occurs when $x=y=-1$ and $z=a+2$, with value $2+(a+2)^2.$

Lagrange multipliers are calculus anyway- that isn't really necessary for a problem like this, especially since high school olympiads don't include calculus.

Just idea.

Consider the scalar product of two vectors $(1,1,1)$ and $(x,y,z)\in |x|\leqq 1,|y|\leqq 1,|z|\leqq 1$.

Let $\theta $ be the angle forming these vectors, $a=(1,1,1)\cdot (x,y,z)=\sqrt{3}\sqrt{x^2+y^2+z^2}\cos \theta$


algebra
polynomial
function
superior algebra
superior algebra solved

Find a field $K$ and an irreducible polynomial $f \in K[X]$ having a multiple root.

I can't find ANY such polynomial :huh:

Look at the rational functions $\mathbb{F}(x)$ over some finite field of characteristic $p$.
Then the polynomial $y^p-x$ is irreducible, but has only one root.

Silly question: why is it irreducible? (You know, the field is somewhat... weird...)

Note that $x$ is prime in $\mathbb{F}[x]$ and that the last one is a unique factorisation domain. So the polynomial $y^p-x$ is irreducible by Eisensteins criterion there, so also in the quotient field.

I'm sorry, but I can't follow that example. What is $y$, and how do you conclude that it has a multiple root?

$y$ is the variable of the (irreducible) polynomial (thus $y=X$).
If $a$ is a root of $y^{p}-x$, then $y^{p}-x=y^{p}-a^{p}=(y-a)^{p}$, thus $a$ is a multiple (and the only) root.


let $ n \ge 2$; $ n \in \mathbb{N}$ and $ a_1,a_2,...a_n \in \mathbb{R}$
such that  $ \sum_{i\equal{}1}^{n}a_i \equal{}0$
prove that

$ \sum_{i<j} |a_i\minus{}a_j| \ge \frac{n}{2}\sum_{i\equal{}1}^{n}| a_i|$

For $ i,j\equal{}1,2,...,n$ we get

$ \left.\sum^n_{i\equal{}1}\left|a_j\minus{}a_i\right|\geq \left|\sum^n_{i\equal{}1}(a_j\minus{}a_i)\right|\equal{}\left|na_j\minus{}\sum^n_{i\equal{}1}a_i\right|\equal{}n\left|a_j\right|\ \Rightarrow\ \sum^n_{i,j\equal{}1}\left|a_j\minus{}a_i\right|\equal{}2\sum^{ }_{1\leq i<j\leq n}|a_j\minus{}a_i| \geq n\sum^n_{j\equal{}1}\left|a_j\right|\right.$

Q.E.D.


[hide]
Well, if y can be zero (i.e. they don't intersect at all), the answer is clearly 0.
If y can be one (i.e. it only intersects one circle or it only touches the point of tangency), the answer is clearly 1.

However, if the parallelogram has to intersect both circles at least twice, I get x=9 and y=4, so the answer is (reaches for calculator) 262144...

I think.
[/hide]

Edit: Oh, wow, my 888th post.

[quote="mathnerd314"][hide]
Well, if y can be zero (i.e. they don't intersect at all), the answer is clearly 0.
If y can be one (i.e. it only intersects one circle or it only touches the point of tangency), the answer is clearly 1.

However, if the parallelogram has to intersect both circles at least twice, I get x=9 and y=4, so the answer is (reaches for calculator) 262144...

I think.
[/hide][/quote]

One of those is correct, and it definitely isn't the weird-looking one.

They don't have to intersect so $y=0$ and $y^{x}=0$

[quote="Treething"]A parallelogram can intersect two internally tangent circles at most x times and at least y times. Compute $y^{x}$.[/quote]

[hide]what a wierd question... i shouldve known, its by treething  :D

anyways, it doesnt have to intersect at all so y=0 whichmakes the answer 0[/hide]


calculus
derivative
logarithms
calculus computations

For |x| < 1, the derivative of y = ln * sqrt(1-x^2) is?

should have said $ ln(\sqrt {1 \minus{} x^2})$

[quote="ProblemSets"]For |x| < 1, the derivative of y = ln * sqrt(1-x^2) is?[/quote]

Do you mean $ ln(\sqrt{1\minus{}x^2})$ or $ ln(x) \cdot \sqrt{1\minus{}x^2}$?

the first one-- my mistake

$ ln(\sqrt {1 \minus{} x^2})$

$ \ln\left(\sqrt{1\minus{}x^2}\right)\equal{}\frac12\ln(1\plus{}x)(1\minus{}x)\equal{}\frac12(\ln(1\plus{}x)\plus{}\ln(1\minus{}x))$

Now it's easy to take the derivative.

[quote="Hetidemek"]$ \ln\left(\sqrt {1 \minus{} x^2}\right) \equal{} \frac12\ln(1 \plus{} x)(1 \minus{} x) \equal{} \frac12(\ln(1 \plus{} x) \plus{} \ln(1 \minus{} x))$

Now it's easy to take the derivative.[/quote]

You need to use absolute value "|    |".


Let $1<a_{1}<a_{2}<\ldots <a_{n}$ be an infinite sequence of integers. Prove that \[\sum_{n=1}^{\infty}\frac{2^{a_{n}}}{a_{n}!}\] is irrational.

I think it's standard application of Dirichlet theorem:
$r$ is irrational iff there are exists integer numbers $p_{n}$ and $q_{n}$ such that $r-p_{n}/q_{n}= o(1/q_{n})$ and $q_{n}$ is increasing sequences.
Isn't it?


Prove that if p is a prime and n :ge: p is a positive integer, then
C(n, p) - [n/p]
is divisible by p.

Here C(n, p) denotes the binomial coefficient n!/(p!(n - p)!).
[x] denotes the greatest integer :le: x.

Nice problem...solution in spoiler.

[hide]

I apologize for some of the notation...

We wish to show that C(n,p):equiv:[n/p] (mod p).  To do this we will consider C(n, p) MOD p.  Let C(n, p) MOD p = x. Then:

C(n, p):equiv:x (mod p)

n!/(p!(n-p)!):equiv:x (mod p)

(n*(n-1)*...*(n-p+1)) / (p!) :equiv:x (mod p)   (Here we just simplified by dividing (n-p)! into n!.)

Now in the set {n-p+1,...,n} there is exactly one number that is a multiple of p, since these are p consecutive integers.  Let this number be kp; it is the largest multiple of p that is :le:n.  Then we can simplify more:

k*(n*...*(n-p+1) (excluding the number which was divisible by p)) / (p-1)! :equiv: x (mod p)

multiply both sides by (p-1)!:

K*(product of the p-1 terms not divisible by p) :equiv: x*(p-1)! (mod p)

Now, observe that those p-1 terms on the LHS all have different remainders modulo p (because they came from a set of p consecutive integers), and that none of them are divisible by p.  Hence, in some order, they are congruent to 1,2,...,p-1 modulo p.  We thus have:

k*(p-1)!:equiv:x*(p-1)! (mod p)

and using the fact that we can "divide out" since p is prime, or, more comfortably, using Wilson's Theorem, we have:

k:equiv:x (mod p).

But what was k?  k was defined so that kp was the largest multiple of p that is :le:n.  Hence n=kp+r, where 0:le:r<p, and so [n/p] = k.   So x-[n/p]:equiv:0 (mod p), and thus C(n, p)-[n/p]:equiv:0 (mod p), which is what we wanted.[/hide]


MATHCOUNTS blogs probability rotation geometry

how come i haven't seen many new middle-schoolers on this site? (i tried to tell some of mua friends about this site...but when they found out that it was a math site........even though i tried to tell them it was more than that and that it was really kinda intersting...)

I kindof agree with you

I tried to convince many of my friends to join this website, but I guess they hate math :(

You can lead a horse to water, but you can't make it drink.

It's also true that the more adventurous mind will discover greater adventures.  Though most people fail to see the adventure in mathematics, those who do typically find ample reward.

Fantastic!

As more middle schoolers join the site (and there are already scores of MATHCOUNTS participants here), it should become more and more enjoyable for them.  Middle schoolers will eventually learn about some of the fun features of the site such as blogs and the Games & Fun Factory forum.

[quote="KBabe"]wow! i just broke my record!!! my whole class wants to come to this site right now...(becuz i just decided to whow them one of the fun probs....) isn't that alsome???except i doubt that many of them are actually gonna come, but i know  some will.... YAY![/quote]

Which problem? I haven't been able to find anything that interests my classmates yet.

Different people will find different ideas and problems interesting of course.  Some people will simply never allow themselves to find ideas interesting at all.  I think the key is to find what interests a particular person.

[quote]You can lead a horse to water, but you can't make it drink.

It's also true that the more adventurous mind will discover greater adventures. Though most people fail to see the adventure in mathematics, those who do typically find ample reward[/quote]

This sounds so amazing...

I appreciate KBabe for her effort to bring more kids to this site..

I tell my nieghbors about "Albatross" problems. That helped a lot.

I always like that name

Albatross....

what's the other one?

well ,I also appreciate A+MATH 's effort

[quote="KBabe"]i think what interests a lotta my friends is like: stuff that makes them go "WOW!! How did u do that? Teach me!".
Stuff like the prob about how u start with one number and then add.. subtract...and do whatever and then end up with the same number and stuff..[/quote]

If you enjoy interesting math relationships, there is a book called Mathematical Circles that you might really enjoy.

How about the Monty problem? That one's always interesting.

Or you could do some other problem which appears to be hard, but has a really neat and elegant solution.

i'm really lost.... with the names and all...what's Albatross? and Monty??

Albatross is the name made up by A+MATH ..( Treething uses it all the time )

and the Monty problems : I'm pretty sure that it's the name of a book

Proof: Math is NOT pure boredom.

Math is better than nothing (Given by someone)
Nothing is better than...... [something very good]
Therefore math is better than [something very good]

Assuming (trivial) that [something very good] is not boring, math is not boring either. QED

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

What are the criteria to become a moderator? OOSC (out of sheer curiosity)

;) i don't think you need to prove that to anyone here :D

My whole MathCounts group thinks that this site is awesome, and one even decided to take a class with me! :) (Junior) Yay! I am homeschooled so i can't tell about this site to anyone but friends and brothers.  One of my brothers decided he wants to do it, and the other hates math :(

[quote=Treething]Math [b]is[/b] pure boredom.[/quote]

So I had to prove that. ;)

Then PM everyone on AoPS to join AoPS!!! Wait, that wouldn't work... um... If your friends aren't all homeschooled too, they can tell people.

[quote="jensen"]i think this site is very useful for someone who like math and like solving problem and enjoy it.but a lot of these people who likes math dont come because they think that they can do it alone and they think it wll be more useful.what should we do with these students?(Even i think this site is very useful for teachers an anybody on this site will learn alot of thins) :lol:[/quote]

These kids aren't (or are) rare.

Anyways, I think that you should try to convince them to take a look at this site. Maybe they'll like it. Maybe they won't. Just use persuation. :D

[quote="Sinistra"]...

I'm new... What do you guys do here?

Someone please give me a maths problem?

Thanks :huh:[/quote]

Hello Sinistra,

When you visit the main forum page, you see that people post plenty of math problems. Click on a link to see one.

It's always nice to have new members on this forum. Welcome to AoPS!  :)

[quote="math92"][quote="KBabe"]hey!!!! I'm back!!!! soo many new ppl!!!! so excited!!! hey everybody!!!
-kristy[/quote]
welcome back, i haven't seen you post in a long time.[/quote]

thx for the welcome part! and well... I don't really remember anybody I've seen on this site before, cuz it's really been a long time since I was here......but really glad to be back! :lol:

haiz.. in singapore, i guest all are too lazy! i THINK that the no. of singaporeans on this website, i can count wif my FINGERS??? :lol:

[quote="ashpotter"]i'm only in 6th grade![/quote]

what's wrong with sixth grade? hey, I was in sixth grade last year!

My friend showed me the site and my mom had me sign up for one of the classes.  Then I never posted on the forum and now my parents make me get off (they consider it play).


Every year, since 2003, the Bergen County Academies, located in Hackensack, New Jersey, holds a Mathematics Copmetition for middle schoolers. The math team worked hard, since last spring, to write and edit problems, get proctors, graders, volunteers, and our school to host it. It will take place on Sunday, October 18th, 2009 at the Bergen County Academies.  Our address is

200 Hackensack Avenue
Hackensack, NJ 07601

http://sites.bergen.org/mathcompetition/index.asp is the main website.
You can register at http://sites.bergen.org/mathcompetition/register2.asp
and previous years problems can be found here: http://sites.bergen.org/mathcompetition/exams.asp.

Good luck all middle schoolers!

[quote="CEM_College_Champ"]How would you compare this to other contests for this age bracket?[/quote]

You can see the previous contests to get an idea of what the problems are like. The 7th grade competition (what I took last year), was somewhere between states and nationals MATHCOUNTS.

From what I see:
4th grade:  REALLY easy MOEMS
5th grade:  School Mathcounts Level
6th grade:  easy Chapter Mathcounts
7th grade: State-ish Mathcounts level
8th grade:  hard state/national Mathcounts(plus trig)

4th grade: Not very hard, most people here would get 40+
5th grade: Maybe School Mathcounts at best
6th grade: Between School and Chapter Mathcounts
7th grade: At best state mathcounts
8th grade: not quite sure.

[quote="isabella2296"]Out of curiosity, any AoPSers taking it? I know I am.[/quote]

You know I am. And Epicfailure.

This is not good  :noo: BOGTRO and Epicfailure are both in my grade (7th, correct me if I am wrong).

Anyways I'm taking it.

Heh, the top score for 7th grade was 36 :| But apparently the hardest 7th grade test ever had a high score of 20-ish.

Really want to see the 8th grade test

[quote="BOGTRO"]7th grade: At best state mathcounts [/quote]

Not really ;)

[quote="limac"]8th grade this year I thought was by far the hardest ever...What do you guys think?[/quote]

If I remember correctly, 2003 was harder, but I took that as a practice a long time ago. This year was definitely more difficult than last year's. I'm not sure how well I "really" did, because I apparently can not fill out an answer sheet correctly.

Lol, nobody got my problem (7th grade #50)...

That includes you, BOGTRO. You did put $ 2\plus{}\sqrt{2}$ :P

RoFlLoLcOpT, did you place?

aha I put $ 2\plus{}\sqrt2$ also :P

Yeah.

It was actually $ 2\sqrt{2}$...

What place did you get?

The toughest year was actually 2005.  That year, on the 8th grade test, the top score was 29 (Ian Osborn), with 2nd place getting 17.  I didn't record the top 7th grade score, but it was also low.  The top 6th grade score that year was 33.  The top 5th grade score was also 33.  The top 4th grade score was 27, with 2nd place getting 21.

Edit: Actually, I do have the 7th grade scores from 2005.  The top score was 15.

I attempted 31 questions and know got a couple wrong, so my score is like in the high 20s. So maybe #1 got like low 30s? This year was definitely harder than all the years (their highs were probably in the 40s), except for maybe 2005 (a little easier that that maybe).

FantasyLover, were you in the grading committee thingy? What was the highest 8th grade score?

High 30s. Dunno the exact score though...

[quote="isabella2296"]Out of curiosity, any AoPSers taking it? I know I am.[/quote]

i am too
it already happened
 on sunday.

umm...
Are the results out yet?
It said they should be out in 72 hrs.... its way past 72..

i don't think so
i hope they will be out soon


trigonometry
inequalities
geometry
incenter
circumcircle
LaTeX
inequalities unsolved

Denote incentre of the triangle ABC by I, and the orthocentre by P.
R is circumradius,r is the inradius.Assume triangle is acute.

(IP)^2 =2(r^2) - 4(R^2).CosA.CosB.CosC     (1)

r = 4R.SinA/2.SinB/2.SinC/2                          (2)

(1) implies the inequality CosA.CosB.CosC. <=( r^2)/2.(R)^2, the right side of which  becomes 8.[SinA/2.SinB/2.SinC/2]^2 * using the square of equation (2), and this* becomes (1-cosA)(1-CosB )(1-cosC) using identity (1-CosX) =2.[SinX/2]^2.

This inequality  leads to the  nice result that  AI.BI.CI >=AP.BP.CP

I hope this is not too hard to read.There is no facility for Latex on public computers

To have latex appear in your posts you only have to include dollar signs around your work. The rest is handled by the forum~.

[quote="Centy"]Show that if $A + B + C = 180^{\circ}$ then \[ (1 - \cos A)(1 - \cos B)(1 - \cos C) \geq \cos A \cos B \cos C  \][/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=4972 :)


Prove[b](without derivates!)[/b] that the function \[ f: \left[0,\ 1\right]\longrightarrow\mathbb{R}\;,  f\left(x\right)\equal{}\frac{x}{x^{2}\plus{}x\plus{}1}\] is concave

[hide="Solution"]A function is concave iff $ f(\frac {x \plus{} y}{2})\ge\frac {f(x) \plus{} f(y)}{2}$.

Let $ x \equal{} 0$ and $ y \equal{} 1$.

We have to prove that $ f(0.5)\ge\frac {f(0) \plus{} f(2)}{2}$.

Since $ f(0) \equal{} 0, f(0.5) \equal{} \frac {2}{7}, f(1) \equal{} \frac {1}{3}$, we now have to prove that $ \frac {2}{7}\ge\frac {1}{6}$, which is true.

Q.E.D.[/hide]

EDIT: Meh, fail :(

[quote="FantasyLover"][hide="Solution"]A function is concave iff $ f(\frac {x \plus{} y}{2})\ge\frac {f(x) \plus{} f(y)}{2}$.

Let $ x \equal{} 0$ and $ y \equal{} 1$.

We have to prove that $ f(0.5)\ge\frac {f(0) \plus{} f(2)}{2}$.

Since $ f(0) \equal{} 0, f(0.5) \equal{} \frac {2}{7}, f(1) \equal{} \frac {1}{3}$, we now have to prove that $ \frac {2}{7}\ge\frac {1}{6}$, which is true.

Q.E.D.[/hide][/quote]

The statement is that inequality holds of all x,y. But you just proved it for two particular values of x,y...


September 1752 - Just 19 days ??

Have u ever seen the calendar for September 1752 ???

Look at the following picture carefully:

[img]http://www.freeimagehosting.net/uploads/86e40ce59f.png[/img]
A month with whole of eleven days missing.

This was the time England shifted from Roman Julian Calendar to the Gregorian Calendar, and the king of England ordered those 11 days to be wiped off the face of the month of September of 1752. (What couldn't a King do in those days?!) And yes, the workers worked for 11 days less, but got paid for the entire 30 days.

This shud have been in the believe it or not thread  :spam:

I such a thing is possible in India too (by a generous amount of bribes ofc  :D )

[hide]
At a rally in Mumbai, Raj Thackeray showed two ration cards in name of Osama and Kasab to show corruption in congress gov :rotfl:
[/hide]

[quote="Sagar_Vaze"]This shud have been in the believe it or not thread  :spam: [/quote]

Lol, that's no more a 'believe it or not' thread.. has turned into election campaign :rotfl:  :rotfl:  :rotfl:

[quote="Sybarite"]Lol, that's no more a 'believe it or not' thread.. has turned into election campaign[/quote]

then it shud b renamed to 'believe it not' thread  :rotfl:

[quote="Sybarite"]Lol, that's no more a 'believe it or not' thread.. has turned into election campaign :rotfl:  :rotfl:  :rotfl:[/quote]
You must be a really impractical to 'believe' an 'election campaign'!  :rotfl:  :rotfl:

[quote="Rajiv"]Somebody teach me how to manipulate EVMs.[/quote]

Electronic Voting Machine?! :huh:

[quote="Sagar_Vaze"][hide="EVMs"][b]E[/b]ndo[b]V[/b]*****l[b]M[/b]achines[/hide]  :huuh:[/quote]

Wtf are those *s?  :ninja:

[quote="Sybarite"]
Wtf are that *s?  :ninja:
[/quote]
[hide="*****"]
EndoVag*nalMachines  :rotfl:
[/hide]

Btw, to return to the topic, that is really true if someone's wondering abt it.. I couldn't swallow it at first, but then searched the net for the answer!

The above posts make it clear that we require a moderator  [hide="...  "]or do we  :rotfl: [/hide]

[quote="Sagar_Vaze"]itz gud that there is no mod  :gathering:
our post count keeps ticking   :spam:[/quote]
Vote for me, and it will SOAR!  :rotfl:

[quote="Sagar_Vaze"]u'll b post_count_alchemist  :rotfl:[/quote]
Great, I have my first political pseudonym!  :rotfl:

[quote="Rajiv"][quote="Gaurav Shukla in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1530760#1530760]#2[/url]"]
...CHEM SUCKS...

[/quote]

You have my vote.[/quote]
Thank you :D


knowing that $\frac{a}{b}*b=a$, anything $*0=0$, and $\frac{1}{0}$=error, what is $\frac{1}{0}*0$?

well the known rule that $\frac{a}{b}\cdot b=a$ is defined  for all numbers except when $b=0$

So $\frac{1}{0}\cdot0 = und$

Whenever division by 0 occurs in an equation like $\frac{1}{0} \cdot 0$ the equation remains undefined. Even though we might want to think that the multiplication by 0 somehow makes the equation equal to 0, it stays undefined.

You're talking about fundamental axioms of the number system, which isn't really the type of problems posted here.

Anyway, one axiom is that there is a number $1$ such that $a\cdot1=1\cdot{a}=a$ for all $a$, and whenever $b\ne0$, there is a number $b^{-1}$ such that $b\cdot{b}^{-1}=1$. Also, $\frac{a}b$ can be defined as $a\cdot{b}^{-1}$, so we have for all $a$ and all $b\ne0$, $\frac{a}b\cdot{b}=(a\cdot{b}^{-1})\cdot{b}=a\cdot(b\cdot{b}^{-1})=a\cdot1=a,$ since there is an axiom stating that for all $a,b,c$ we have $a\cdot(b\cdot{c})=(a\cdot{b})\cdot{c}$.

To prove that $0\cdot{x}=0$ for all $x$, we use the distributive property (another axiom) and $0+0=0$:

$0\cdot{x}=(0+0)\cdot{x}=0\cdot{x}+0\cdot{x}$, and if $a=a+b$ for some $a,b$ then $b=0$.

The symbol $\frac10=1\cdot0^{-1}$ is undefined ("error" is a computer term, not math) since $0$ doesn't have an inverse. Thus $\frac10\cdot0$ is also undefined.

[hide]
1/0 is undefined. We cannot use that expression at all. Thus, 1/0 * 0 is undefined.

[/hide]


Hey,

I guess this problem is well-known, but anyway:

find all (n,p) s.t. p is prime, n<=2*p and n^(p-1) | (p-1)^n + 1.

Does anyone know any solutions apart from the one at kalva?

thanx in advance.

Find all [tex]( n, p )[/tex] where [tex]n \leqslant 2 p[/tex] and [tex]p \in P[/tex], such that [tex]n^{p -
1} | ( p - 1 )^n + 1[/tex]

Case 1: [tex]( n, p ) = 1[/tex]
By Fermat's little theorem we have [tex]n^{p - 1} = 1 ( \mbox{mod } p )[/tex].
If [tex]n[/tex] is odd we obtain [tex]( p - 1 )^n + 1 = 0 ( \mbox{mod } p )[/tex] and if [tex]n[/tex] is even we get [tex]( p - 1 )^n + 1 = 2 ( \mbox{mod } p )[/tex].
So for any [tex]n > 1[/tex] there is no solution. Thus we must have [tex]n = 1[/tex], then [tex]n^{p - 1} = 1[/tex] and for any [tex]p[/tex] the couple [tex]( 1, p )[/tex] is a solution

Case 2: [tex]( n, p ) > 1 \Rightarrow p|n \Rightarrow n = pk[/tex], where [tex]k \in N[/tex]. But because [tex]n \leqslant 2 p[/tex] we have [tex]n = p \vee n = 2 p[/tex].
If [tex]n = p[/tex] then note that for any [tex]n \geqslant 4[/tex] we have (Edit: SORRY: Here was made a mistake ! I'll try to correct it as fast as I can). So enumarating all possibilites we remark that [tex]p = 2[/tex] and [tex]p = 3[/tex] satisfy the condition.
If [tex]n = 2 p[/tex] then [tex]( 2 p )^{p - 1} = 0 ( \mbox{mod } p )[/tex] but [tex]( p - 1 )^{2 p} + 1 = ( - 1 )^{2 p} + 1 = 2 ( \mbox{mod } p )[/tex], so there is no solution.

WARN : It hasn't been carefully checked  ;)

i'm sorry, i might be missing something very obvious but i dont quite see how Case 1 works.
How do you get that 'thus for n>1 there is no solution'?


Let $ n\ge 3$ be a natural number. A set of real numbers $ \{x_1,x_2,\ldots,x_n\}$ is called [i]summable[/i] if $ \sum_{i\equal{}1}^n \frac{1}{x_i}\equal{}1$. Prove that for every $ n\ge 3$ there always exists a [i]summable[/i] set which consists of $ n$ elements such that the biggest element is:
a) bigger than $ 2^{2n\minus{}2}$
b) smaller than $ n^2$

[quote="Bugi"]Let $ n\ge 3$ be a natural number. A set of real numbers $ \{x_1,x_2,\ldots,x_n\}$ is called [i]summable[/i] if $ \sum_{i \equal{} 1}^n \frac {1}{x_i} \equal{} 1$. Prove that for every $ n\ge 3$ there always exists a [i]summable[/i] set which consists of $ n$ elements such that the biggest element is:
a) bigger than $ 2^{2n \minus{} 2}$
b) smaller than $ n^2$[/quote]
What if $ 2^{2n\minus{}2} > n^2$?

The cases are 2 different, you must prove that each one of them holds, not both of them in one.

[quote="Bugi"]Let $ n\ge 3$ be a natural number. A set of real numbers $ \{x_1,x_2,\ldots,x_n\}$ is called [i]summable[/i] if $ \sum_{i \equal{} 1}^n \frac {1}{x_i} \equal{} 1$. Prove that for every $ n\ge 3$ there always exists a [i]summable[/i] set which consists of $ n$ elements such that the biggest element is:
a) bigger than $ 2^{2n \minus{} 2}$
b) smaller than $ n^2$[/quote]

a) Induction on $ n$ but i think it should be bigger than $ 2^{2(n\minus{}2)}$
For the case of $ n\equal{}3$, $ S_3\equal{}\{2,3,6\}$ works
Notice for the inductive step that if $ S_i$ is [i]summable[/i] then $ \{2\} \cup 2 \cdot S_i$ is also [i]summable[/i] .

[quote="mszew"][quote="Bugi"]Let $ n\ge 3$ be a natural number. A set of real numbers $ \{x_1,x_2,\ldots,x_n\}$ is called [i]summable[/i] if $ \sum_{i \equal{} 1}^n \frac {1}{x_i} \equal{} 1$. Prove that for every $ n\ge 3$ there always exists a [i]summable[/i] set which consists of $ n$ elements such that the biggest element is:
a) bigger than $ 2^{2n \minus{} 2}$
b) smaller than $ n^2$[/quote]

a) Induction on $ n$ but i think it should be bigger than $ 2^{2(n \minus{} 2)}$
For the case of $ n \equal{} 3$, $ S_3 \equal{} \{2,3,6\}$ works
Notice for the inductive step that if $ S_i$ is [i]summable[/i] then $ \{2\} \cup 2 \cdot S_i$ is also [i]summable[/i] .[/quote]
Then $ S_4 \equal{} \{2,4,6,12\}$ But $ 12 < 2^{2(4 \minus{} 2)} \equal{} 2^4 \equal{} 16$.

Bugi: If $ x_1,...,x_n$ is real numbers, then it is trivial. Shouldn't it be $ x_1,..,x_n$ are whole numbers?

For $ b$ we just set $ x_1 \equal{} x_2 \equal{} ... \equal{} x_n \equal{} n < n^2$.

Sorry for the mistake, I believe that all $ x_i$ should be different.

Notice that $ \frac {1}{n} \equal{} \frac {1}{n \plus{} 1} \plus{} \frac {1}{n(n \plus{} 1)}$
then the inductive step will be $ \{x_1,\cdots,x_{n \minus{} 1},x_n\} \rightarrow \{x_1,\cdots,x_{n \minus{} 1},x_n \plus{} 1,x_n \cdot (x_n \plus{} 1)\}$
For example $ \{2,3,6\} \rightarrow \{2,3,7,42\} \rightarrow \{2,3,7,43,1806\}$

I don't know, I just posted this:
http://www.mathlinks.ro/viewtopic.php?t=212129
Look at that topic

I've got the shortlist form orl and there is this problem a bit different:

    [b]Problem A3.[/b]  Let  $ n \geq 3$ be an integer. A set of positive integers $ x_1,x_2,...,x_n$ is called [i]summable[/i] if $ \frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} ... \plus{} \frac {1}{x_n} \equal{} 1$.
    (i) Prove that for every $ n \geq 3$ there exists a summable set of $ n$ integers having the value of the largest element greater than $ 2^{2^{n \minus{} 2}}$;
    (ii) Prove that for every $ n\geq 3$ there exists a summable set of $ n$ distinct integers having the value of the largest element less than $ n^2$.

what's the original problem?!it's wrong for $n=3,4$!

I have an approach through inequality.
We define the set {x1,x2,.....xn} such that the sequence is strictly increasing. Thua, without loss of generality,suppose xn is greatest. Now we have to chose xn twice, once such that it is bigger than 2^(2n-2) and when it is smaller than n^2.
Now, Working Backward, suppose the set is summable. Then, x1+x2+x3+.....xn >= n^2. [AM>= HM]. Then we can chose all of x_i to be less than n^2 but greater than n, and we see thus for such a selection, the inequality holds. Thus, the set is trivially summable for such a selection.(2nd part proved)
       Again, we are done with the first part if we can select xn> 2^(2n-2). Suppose all x_i greater than 2^(2n-2). Then also, we see that the inequality reduces to n. 2^(2n-2) > n^2 , which is obvious for n>=3.


R'nt I the best hing this forum needs to become popular?

I find this utter bull*****y self blasphemy and I request Pardesi to shut this topic, and if possible delete it too..
Who said the forum wasnt illustrious before?? It was probably even better..
Pardesi, delete this topic, else you will see abusive words in this topic.. Even if I post abusive words, you can understand, but if Shruthi does, will we be able to even take clues from her English??? :D

wats the meaning of this topic....i can't seem to get it :maybe:

The simple meaning is verizon is the person who is the most valuable addition to our AOPS indian forum he claims and it seems his joining here is going to make us popular...:D:D

verizon do tell properly what u meant,..... :ninja:

he meant i was the best thing that happened to this world Capsice?  :P   :rotfl:   :D


vector
trigonometry
integration
geometry
3D geometry
sphere
geometric transformation

From Tipler and Mosca:

Find the magnetic moment of a spherical shell of radius R and surface charge density sigma rotating about its diameter with angular velocity omega.

Let $x$ be the an axis passing vertically through the center of the
shell. Let $\alpha$ be the angle between the $x$ axis and the
vector $\vec{R}$ with the origin in the center of the shell and
the ending on a point on the surface.

So, the solution as follows:

$dS=2{\pi}{R^2}{\sin(\alpha)}^2{d\alpha}$

$dQ=dS\sigma={\sigma}2{\pi}{R^2}{\sin(\alpha)}{d\alpha}$

$dI=\frac{dQ}{T}=\frac{\omega}{2\pi}{\sigma}2{\pi}{R^2}{\sin(\alpha)}{d\alpha}$

$dN=SdI=\pi{R}^2{\sin(\alpha)}^2dI=\omega\pi\sigma{R}^3\sin(\alpha)^3d\alpha$

And so, for the final value of N we should get $N=\int\omega\pi\sigma{R}^3\sin(\alpha)^3d\alpha$

or

$N=\frac{4}{3}\omega\pi\sigma{R}^3$

 I hope there have been no maths mistakes.

Just one mistake--you dropped a factor of $R$. The correct answer is $N=\frac{4}{3}\omega\pi\sigma{R}^4$
Here's another way of solving it that I figured out after seeing that the answer contains that factor $\frac{4}{3}$.
We can integrate over the sphere by taking slices of height $dh$ along the axis of rotation. It turns out that the current going around every slice is the same.
This is because the velocity of the surface charge is multiplied by a factor of $\sin \theta$ (where $\theta$ is the angle between that radius and the axis of rotation) but the thickness of the slice is divided by $\sin \theta$.
Therefore, we can restrict our attention to the equator. It has a line charge density of $\sigma dh$ and the charges move at a velocity $\omega R$, so $I = \sigma \omega R dh$ and $m=A \sigma \omega R dh$ (where $A$ is area.)
Finally,
$N = \int \sigma \omega R A dh = \sigma \omega R \int dV = \sigma \omega R V = \frac{4}{3}\omega\pi\sigma{R}^4$

[quote="schulmannerism"]Just one mistake--you dropped a factor of $R$. [/quote]

Oops!


8. Cleo used 19 cents stamps and 29 stamps to mail a package. She spent exactly $ \$ 5.88$ (there is no tax on stamps) How many 29 cents stamps did Cleo buy?

letting $ x$ be number of 19 cents stamps and $ y$ be the 29 cents stamps,
I got to the point where you mod out 29, then I created equation $ y\equal{} 8\minus{}19x$
What should I do from here?

Or do you have a better solution?

[hide]Okay, so you see that 588 (mod 19) = 18 (mod 19); 29y =  10y (mod 19) and so you have to find a value of y that satisfies:

18 (mod 19) - 10y (mod 19) = 0 (mod 19)

or 10y (mod 19) = 18 (mod 19)
and so from here you see the pattern between y={1,2,3...} and the corresponding value (mod 19) = {10,1,11,2,12,3,13...} and count ahead in the pattern to find

y = 17

So he bought 17 stamps.[/hide]

[quote="shentang"][hide]Well, I just said 19x + 29y= 588
put that in mod 19,     10y=18(mod 19)
multiple both sides by 2,        20y=36(mod 19)
Simplify,           y=17(mod 19)
Answer,              y=17  [/hide][/quote]

How do you go from 20y to y?

EDIT:
Oh, got it.

[quote="shentang"][hide]Well, I just said 19x + 29y= 588
put that in mod 19,     10y=18(mod 19)
multiple both sides by 2,        20y=36(mod 19)
Simplify,           y=17(mod 19)
Answer,              y=17  [/hide][/quote]

Hmm.. so you rather mod out 19. Also, can you perhaps explain how you simplified 20y=36(mod 19) to y=17(mod 19). Sorry, I am kind of slow.

In shentang's solution...

$ 10y\equiv18\mod19$.

However, we can rewrite this to $ 10y\equiv\minus{}1\mod19$ this implies that $ 19x\minus{}1\equal{}10y$ for some $ x$.

Since 19x must end in a $ 1$, we have that $ x\equal{}9$ which yields $ y\equal{}17$


o problema foarte frumoasa, adaptata dupa problema nr 5 de la un concurs international din kazahstan
eu sper ca are solutie
se partitioneaza $(0,\infty)$ in trei multimi $A,B,C$, si se considera un $\alpha \in (0,\pi)$ fix.
a) sa se arate ca exista un triunghi de laturi $a,b,c$ cu $a\in A,b\in B,c\in C$.
b) este obligatoriu sa existe un triunghi care sa aiba un unghi egal $\alpha$ silaturile $a,b,c$ cu proprietatea de mai sus ?

nicio parere ? :(

in concurs problema s`a dat pentru $\alpha = 90^{\circ}$.
nu stiu care e raspunsul pentru $\alpha$ oarecare, insa cred ca este acelasi, doar ca rationamentul este mai complicat.
mie problema mi s`a parut foarte draguta


Hello, Robert0201!

[quote]Can someone refresh me on the formula for the sum of the n powers of 2 and give a proof?[/quote]
Are you referring to: $1 + 2 + 4 + 8 + 16 + \cdots + 2^n$ ?

This is a geometric series with first term $a = 1$ and common ratio $r = 2.$

The formula for the sum of the first $n$ terms is: $S_n \;= \;a\frac{1 - r^n}{1 - r} $

For this series: $S_n \;= \;\frac{1 - 2^n}{1 - 2} \;= \;2^n - 1$


Is it common for beginners to think that you have the right solution, and explain it to others, and realize that you haven't accounted for one tiny thing or made one logical misstep that ended up screwing you over?

I don't have problems with careless mistakes, I don't have problems with finding creative "solutions," but it's hard for me to know if my logic is correct, ESPECIALLY in counting problems.

It's all about practice, right?
=)

Believe me, it is. Those things usually get less frequent as you have more experience, though. Last year I had 4 stupid mistakes, this year I had 2, and I'm aiming for 0 next year.  :P

Logical missteps will kill you on the USAMO, so better brush up on it.

In math class the other day the teacher had let me do a solution to a hard problem. Every time I went through the math with the class, I found another mistake and had to redo all the work after that. Eventually I got an answer which was totally nonsensical intuitively, but when we checked it with the computer it was right!

I did what worthawholebean was talking about for AIME #5... somehow. I still don't know what I did...  :huh:


Prove  that every multiple of 6 can be written as the sum of four third powers of integers

Dutch MO 1998

Did you use some method to get the answer or did you just "see" it?

I seeked for such cubes that gave 6, I seeked some to get 12 and I noticed they each differed 1. I filled in for one more and -wow- got 18, so then I got to the general case :)


probability
number theory unsolved
number theory

thank you for your answer, i tought it was true because i saw it in my book that we have to consider this situation to proove that probability of mistake in Miller-Rabin is $\frac{1}{4}$, but it seems more difficult. Do you know where i can find this proof??


SO....

the AMCs are about 3 months away,

the AIMEs about 4 1/2,

[b]and the 2009 USAMO 5 1/2 away. [/b]

Well....I'm in South Korea here. (Duh)

Was qualified for 2008 USAMO, took it (at 1:30~6 in the morning - I HATE time zones), and thought

"Geez, next year, I must do some proper preparation."

[b]Is there any way I can prepare for the USAMO?[/b] (South Korean education facilities

all focus on the Korean Mathematical Olympiads - and they are almost TOTALLY different from the USAMO  :( )

Through AoPS, or through any kind of materials, books, etc...?

Awaiting some good advice  :lol:

Best way to practice: take practice tests at 1:30-6:00 AM.

All of the USAMO's from the time of the current format (1996 or so onward) are in the 'Contests' section toward the top of your screen. It's pretty self-explanatory how to get there from that link. IMO's are also easily found there, of which there are even more tests (but keep in mind that the further back you go, the easier the problems get).

As for the time issue, I remember reading that if you are in a time zone that makes it unreasonable for you to take the USAMO at that time, you can arrange with the AMC office to take it at a more reasonable time. Look around the AMC website, or just e-mail the AMC office directly.

Some Olympiad practice suggestion: 1.APMO-though full of "Asian flavour" as Zuming mentioned, I think the problems are in a lot ways similar to USAMO. Moreover, able to work on 5 problems within 4 hours will make you comfortable in USAMO.
2.IMO shortlist-the shortlist contains the best part of problems in the year(and definitely hard) except for..
3.Russian MO-I'm surprised at how this can advance one's technique(especially in combinatorial) Just a personal experience, problem number 1 from 2006 ARMO ninth grade gave me all I need for last USAMO's problem 3. Also, it's even worth to go back to Russia(Soviet) problems 20-30 years ago.

Try not to focus on China MO and alikes. Though hard, they doesn't relate to USAMO that much.

[quote="1=2"]The main thing is to learn math, for one. AoPS volume 2 is all you need.

Then you need to realize that the only way that you're gonna do well on the USAMO is to try things. Don't just stare at a problem and pray for an epiphany. It's not gonna happen. Extend a line, square the expression, do ANYTHING![/quote]

volume 2 has helped me, but not that much. You will probably need something more advanced than volume 2.

lol what? listen to greentreeroad - he knows what he's talking about. jb05 is joking (maybe); catalyst is right too; 1=2 is not completely right - you need to know more math than AoPS 2 has, and more importantly, solving olympiad problems comes from experience!

are the problems in the USSR problem book olympiad questions from RMO's or are they just practice problems?


For $ a,b,c$ positive reals prove that

$ \sum \frac{\sqrt{a\plus{}b}}{c} \geq 9\sqrt{\frac{\sum a}{2\sum ab}}$

Try $ a \equal{} b \equal{} c\equal{}1$  :(

[quote="manlio"]For $ a,b,c$ positive reals prove that

$ \sum \frac {\sqrt {a \plus{} b}}{c} \geq 9\sqrt {\frac {\sum a}{2\sum ab}}$[/quote]
I think you meant this:
$ \sqrt{\frac{b\plus{}c}{a^2}}\plus{}\sqrt{\frac{c\plus{}a}{b^2}}\plus{}\sqrt{\frac{a\plus{}b}{c^2}} \ge 3\sqrt{\frac{2(a\plus{}b\plus{}c)}{ab\plus{}bc\plus{}ca}}.$
See here: http://canhang2007.wordpress.com/2009/11/09/inequality-60-unknown-author/


calculus
integration
trigonometry
function
limit
calculus computations

Evaluate $\int_{0}^{2 \pi}\frac{dx}{4+\cos x}$.

Guess what? Residues work just fine here! Let's explain the general phenomenon a little bit. :)

If we have to calculate an integral on the form \[I=\int_{0}^{2\pi}R(\cos x, \sin x)dx\] we will proceed as it follows: \[I=\int_{0}^{2\pi}R(\cos x, \sin x)dx=\int_{0}^{2\pi}R\left(\frac{e^{ix}+e^{-ix}}{2}, \frac{e^{ix}-e^{-ix}}{2i}\right)dx\] Further we have \[I=\int_{d}R\left(\frac{z^{2}+1}{2z}, \frac{z^{2}-1}{2iz}\right)\frac{dz}{iz},\] where $d: [0,2\pi]\to\mathbb{C}$, $d(x)=e^{ix}$.

 Now, we denote by \[f(z)=R\left(\frac{z^{2}+1}{2z}, \frac{z^{2}-1}{2iz}\right)\] and \[S=\{z\in\mathbb{C}/ z pole \}.\] Now, by Residure theorem we get \[I=\int_{0}^{2\pi}R(\cos x, \sin x)dx=2\pi i\sum_{z\in S, |z|<1}Res(f, z).\] Application:

Calculate the following integral \[\int_{0}^{2\pi}\frac{1}{a+b\cos x}dx.\] where $a, b>0.$

 Solution.

  Let us consider the function \[f(z)=\frac{2}{i(bz^{2}+2az+b)}\] Now, since \[z_{1}=\frac{-a-\sqrt{a^{2}-b^{2}}}{b}\] and \[z_{2}=\frac{-a+\sqrt{a^{2}-b^{2}}}{b}\] are poles and
  $|z_{1}|>1, |z_{2}|<1$, by residue theorem we get \[Res(f,z_{2})=\lim_{z\to z_{2}}(z-z_{2})f(z)=\frac{2}{ib(z_{2}-z_{1})}=\frac{1}{i\sqrt{a^{2}-b^{2}}}.\] Now, it is clear that \[\int_{0}^{2\pi}\frac{1}{a+b\cos x}dx=\frac{2\pi}{\sqrt{a^{2}-b^{2}}}.\]

I was sure that a solution using residues existed, but I didn't perform the computations.

Anyway, here's an alternative way:
Split $\int_{0}^{2 \pi}$ into $\int_{0}^{\pi}\, \,+\, \, \int_{\pi}^{2 \pi}$. The two integrals are equal, so it suffices to compute the first. We have that $\cos x = \frac{1-\tan^{2}\frac{x}2}{1+\tan^{2}\frac{x}2}$, so make the substitution $y = \tan \frac{x}2$. We get an integral of the form $\int_{0}^\infty \ldots \, dy$, which is pretty easy to compute.
Now just double the result.

Another...

We set $x= 2w$ and use the identity $\cos 2w = 2\cos^{2}w-1$ .

Then $\frac{1}{4+\cos 2w }= \frac{1}{3+2\cos^{2}w }= \frac{1}{3 \sin^{2}w+5 \cos^{2}w }$ .

Now we divide by $5 \cos^{2}w$ and we get  $\frac{1}{4+\cos 2w }= \frac{\frac{1}{5}\cdot \sec^{2}w }{1+\frac{3}{5}\cdot \tan^{2}w }$.

This is  $\left[ \frac{1}{ \sqrt{15}}\cdot \arctan \left( \sqrt{ \frac{3}{5}}\cdot \tan w \right ) \right]'$ and the rest is easy...

 :)


If $ a$, $ b$, $ c$ and $ d$ are positive real numbers such that $ a \plus{} b \plus{} c \plus{} d \equal{} 4$ prove that
$ \frac {a}{1 \plus{} b^{2}c} \plus{} \frac{b}{1 \plus{} c^{2}d} \plus{} \frac {c}{1 \plus{} d^{2}a} \plus{} \frac {d}{1 \plus{} a^{2}b} \geq 2$

From Cauchy,
\[ \sum\frac{a}{1+b^2c}\ge\frac{(a+b+c+d)^2}{a+b+c+d+\sum ab^2c}\]
Hence it remains to show that $ ab^2c+bc^2d+cd^2a+da^2b\le 4$.

We have
\begin{align*}&ab^2c+bc^2d+cd^2a+da^2b=(ab+cd)(ad+bc)\le\left(\frac{ab+bc+cd+da}{2}\right)^2\\
&=\left(\frac{(a+c)(b+d)}{2}\right)^2\le\frac 14\left(\frac{a+b+c+d}{2}\right)^4=4\end{align*}

Another way to reduce it to $ ab^2c \plus{} bc^2d \plus{} cd^2a \plus{} da^2b \le 4$ is to use Jensen's on $ f(x) \equal{} 1/x$, it follows pretty quickly. Actually, I think it might be the same thing indirectly.

nayel, what about the following inequality:
Let $ a,$ $ b,$ $ c$ and $ d$ are non-negative numbers such that $ a\plus{}b\plus{}c\plus{}d\equal{}4.$ Prove that $ a^2bc\plus{}b^2cd\plus{}c^2da\plus{}d^2ab\leq4.$ :wink:

[quote="canada"]Very very nice and simple Thank you for this beautiful solution.[/quote]

Thank you. :blush:

[quote="arqady"]nayel, what about the following inequality:
Let $ a,$ $ b,$ $ c$ and $ d$ are non-negative numbers such that $ a + b + c + d = 4.$ Prove that $ a^2bc + b^2cd + c^2da + d^2ab\leq4.$ :wink:[/quote]

It is equivalent to $ ac(ab + cd) + bd(ad + bc)\le 4$.

Case 1: $ ab + cd\le ad + bc$

Then
\begin{align*}ac(ab + cd) + bd(ad + bc)\le (ac + bd)(ad + bc)\le\left(\frac {ac + bd + ad + bc}{2}\right)^2
= \left(\frac {(a + b)(c + d)}{2}\right)^2\le \frac 14\left(\frac {a + b + c + d}{2}\right)^4 = 4\end{align*}

Case 2: $ ad + bc\le ab + cd$

Then
\begin{align*}ac(ab + cd) + bd(ad + bc)\le (ac + bd)(ab + cd)\le\left(\frac {ac + bd + ab + cd}{2}\right)^2
= \left(\frac {(a + d)(b + c)}{2}\right)^2\le \frac 14\left(\frac {a + b + c + d}{2}\right)^4 = 4\end{align*}

Hope it is correct.  :)

it is a nice proof,but we can solve it by AM-GM ineq

Can anyone give the proof of the former inequality using arqady's technique, I mean Jensen's Inequality? I am fade up trying it


Here it is the very first run of this new [b]feature[/b] of the forum:

[b][color=red]Problem 02/09.[/color][/b] Suppose that $ \{(1,0,1), (0,1,2), (2,1,12)\} \in \mathbf{Z}_{8} \oplus \mathbf{Z}_{2} \oplus \mathbf{Z}_{16}$.

[b]Which[/b] group is the quotient $ \displaystyle \frac{\langle (1,0,1),(0,1,2) \rangle}{\langle(2,1,12)\rangle}$ isomorphic to?

Look forward to reading your solutions quite soon.  :wink:

P.S. Hope you all like this proposal of mine.

The [b][color=red]end[/color][/b] of the month approaches... Won't somebody post his/her solution to the [b][color=red]February[/color][/b] proposal?  :(

[quote="Riquelme"]I was about to reply to [b][color=purple]vict85[/color][/b]'s solution proposal... Wonder what's happened to it.[/quote]

There was an error and I have deleted it...

This is the right one, I hope:

$ a \equal{} (1,0,1)$
$ b \equal{} (0,1,2)$
$ G \equal{} \mathbf{Z}_{8} \oplus \mathbf{Z}_{2} \oplus \mathbf{Z}_{16}$

$ A_1 \equal{} \langle a, b \rangle$
$ \langle a \rangle \triangleleft A_1$
$ \langle b \rangle \triangleleft A_1$
$ A_1 \equal{} \langle a \rangle \plus{} \langle b \rangle$

$ \langle a \rangle \cong \mathbf{Z}_{16}$
$ \langle b \rangle \cong \mathbf{Z}_{8}$

And $ \langle a \rangle \cap \langle b \rangle \equal{} \{(0,0,0), (0,0, 8) \}$

$ |A_1| \equal{} \frac{|\langle a \rangle| \cdot |\langle a \rangle|}{|\langle a \rangle \cap \langle b \rangle|} \equal{} \frac{16\cdot 8}{2} \equal{} 2^{4\plus{}3\minus{}1} \equal{} 2^6$

The element $ a \in G$ is an element of maximal order in $ G$ and so it's an elements of maximal order in $ A_1$. Then $ A_1 \cong \langle a \rangle \oplus C$ for some subgroup $ C$ of $ A_1$. With $ |C|\equal{}4$ and $ C \ne \langle b \rangle$.

$ A_2 \equal{} \langle(2,1,12)\rangle \equal{} \langle(2,0,2) \plus{} (0,1,10)\rangle \equal{} \langle 2a \plus{} 5b \rangle$ and so $ A_2 \triangleleft A_1$. $ A_2 \cong \mathbf{Z}_{4}$

$ A_2 \cap \langle a \rangle \equal{} \{(0,0,0)\}$ and so $ C\equal{}A_2$ and $ A_1 \equal{} \langle a \rangle \oplus A_2$

So $ A_1/A_2 \cong \frac {\langle a \rangle \oplus A_2}{A_2} \cong \langle a \rangle \cong \mathbf{Z}_{16}$

Your answer is correct, dear [b][color=red]vict85[/color][/b]...

Be on the lookout for more [i]challenges[/i] of this kind in the near future.  :wink:

Regards,
José.

P.S. We're looking forward to reading the proposals of the rest of you, [i]guys[/i]! Don't hesitate to post your own solution-approaches.


[hide]Define the value of a point as the number of ways to reach that point.  The value of any point X can be calculated by adding the values of the points north and west of it.
For every way to get to the points north and west of X, there is a way to get to X (It's kinda hard to explain without a good diagram).  Set the value of point A as one, and add a lot.  You get
[hide]
160, I think.
[/hide]
as the value of B.  If that's unclear, tell me. It probably is, lol.
[/hide]

EDIT: Good answer kohjhsd, that's much better than mine. :oops:

[quote="Math Geek"]Could anyone help me with this 2001 National Mathcounts Sprint Roung question?  It reads:

Moving only south and east along the line segments, how many paths are there from A to B?

Note: The figure's attached. :roll:[/quote]

[hide]Two segments were removed in the picture. So we need to subtract the number of paths that pass through these segments from the total. Since the two erased segments are on the same row, one cannot pass through both segments in one path. Therefore, passing through each of the two is exclusive. Then use combinations:
\[\binom{12}{3}-\binom{5}{1}\cdot \binom{6}{1}-\binom{6}{1}\cdot \binom{5}{1}= 220-60=\boxed{160}. \]
[/hide]

Could someone explain the combinatorics please?

I'll try:

[hide]If all the lines were there, then you would need 9 east moves "E's" and 3 south moves "S's".  For example, you could have the string of moves EEEEEEEEESSS.  There are $\binom{12}{3}$ ways to do this.  But there are 2 removed line segments, so we have to subtract the paths using those.  There are $\binom{5}{1}$ ways to get from A to the top of the 1st removed segment and $\binom{6}{1}$ ways to get from the bottom of that segment to B.  The other removed line segment follows the same logic.  So it's $\binom{12}{3}-\binom{5}{1}\cdot\binom{6}{1}-\binom{6}{1}\cdot\binom{5}{1}=160.$[/hide]

EDIT: Whoops, forgot to hide it! :blush:

[hide]My plan is to label each node (intersection) of the map with the number of ways of arriving at that node.  The problem says we start at point A.  Begin by labeling point A with "1", since there is just one way to get there.  For each of the remaining nodes, we can arrive at the node from the left or from above.  Therefore, the number of ways of getting to a node equals the sum of the number of ways of getting to the node above plus the number of ways of getting to the node to the left.  In several cases, there is no path from the left or above.  Following is my labeled diagram, which I completed by working from the upper left to the lower right.  The completed diagram shows the total number of ways of arriving at point B is 160, which is the answer.  [Note the resemblance of the upper left corner of our graph with Pascal's Triangle.][/hide]


[hide]Multiply both sides of the equation by $ 2^{4009}$ to get

$ \sum_{k\equal{}0}^{2005}2^{2005\minus{}k}\binom{k\plus{}2004}{k}\plus{}\sum_{k\equal{}0}^{2004}2^{2004\minus{}k}\binom{k\plus{}2005}{k}\equal{}2^{4010}$
$ \sum_{k\equal{}0}^{2005}2^{2005\minus{}k}\binom{k\plus{}2004}{2004}\plus{}\sum_{k\equal{}0}^{2004}2^{2004\minus{}k}\binom{k\plus{}2005}{2005}\equal{}2^{4010}$

Now we use counting arguments.  We claim that the first sum is the number of subsets of $ \{1,\ldots ,4010\}$ with at least 2005 elements.  Let $ k\plus{}2005$ be the 2005th element.  Then there are $ \binom{k\plus{}2004}{2004}$ ways to choose the first 2004 elements, and $ 2^{2005\minus{}k}$ ways to choose the rest of the elements.  As $ k$ ranges from 0 to 2005, we get all subsets with at least 2005 elements.

Similarly, the second sum is the number of subsets of $ \{1,\ldots ,4010\}$ with at least 2006 elements.  Let $ k\plus{}2006$ be the 2006th element.  Then there are $ \binom{k\plus{}2005}{2005}$ ways to choose the first 2005 elements, and $ 2^{2004\minus{}k}$ ways to choose the rest of the elements.  This time, as $ k$ ranges from 0 to 2004, we get all subsets with at least 2006 elements.  By symmetry, the sum is also equal to the number of subsets with at most 2004 elements.

Therefore, the two sums together count all subsets of $ \{1,\ldots ,4010\}$, so the left side of the equation is equal to $ 2^{4010}$ as desired.[/hide]

[i] My solution is very elementary

  I think Darij and Scorpius119 can solve this problem with nice proof :lol: [/i]


vector
advanced fields
advanced fields unsolved

Is there a proof that doesn't exist an order relation on $ \mathbb R^n$, for $ n>1$, that preserves the usual structure of vector space (of $ \mathbb R)$?

Please specify your requirements more precisely (in particular those which are not satisfied by the lexicographical order). owk

Sorry, I forgot the keyword "total". I need a proof that any total order on $ \mathbb R^n$, $ n>1$, (e.g. the lexicographic order) isn't compatible with the vector space structure.

First of all, what does it mean that a total order relation is compatible with the vector space structure?

[url=http://en.wikipedia.org/wiki/Ordered_vector_space]Wikipedia's definition[/url] allows the lexicographical order, therefore I was asking you what [u]your[/u] definition is. owk

So the lexicographical order is compatible with the vector space structure? I thought that it was impossible to define [b]any[/b] compatible total order compatible with the vector space structure.

The order I had in mind was $ x \le y$ iff $ x_i \le y_i$, for $ i\equal{}1,\ldots,n$.

If we also ask for the supremum property, besides the properties of the vector space structure, we have to discard the lexicographic order..

What I need is a proof that no [b]total[/b] order that preserves the vector space structure and the supremum property can be defined on $ \mathbb R^n, n > 1$.


calculus
integration
trigonometry
calculus computations

show :D

$ \int_0^\infty \; \frac{(x\;\cos\; x \minus{} \sin \; x)^2}{x^6} \; dx \equal{} \boxed{\frac{\pi}{15} }$


I added my goals for 2009 as my signature, and it saved successfully, but it doesn't appear at the bottom of my posts.

The search function is a wonderful thing ;)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=246137


Find a 4-digit number x that when raised to the 2003rd power, has its last four digits 2003.

I can't find a $4$-digit number.  I found a $3$-digit number and a $5$-digit number.  :(

Or a six or a seven or an eight digit number :)
In fact any number which ends in ...0587 will do, and no other number will work :)
[hide="hint"]In fact all  you have  to check that  3rd power ends in 2003  (becuase it will have the same last four digits as 2003rd power)[/hide]

[quote="Gyan"][hide="hint"]In fact all  you have  to check that  3rd power ends in 2003  (becuase it will have the same last four digits as 2003rd power)[/hide][/quote]

I managed to prove this, but it didn't give me a $4$-digit answer so I gave up :(

How do you solve
[hide]

[quote]In fact all you have to check that 3rd power ends in 2003 (becuase it will have the same last four digits as 2003rd power)[/quote]

[/hide]
that, even if it leads to a 3 digit number?

Ignore - If you  already know  ....

[hide]Can you prove $a^3$ and $a^{2003}$ will have the same last four digit?
(In fact   this is true for  $a^{503}$   or $a^{1003}$ )

(where $a$ is any natural number, and for simlicity, assume that $a$  is not divisible by 2 or 5)

Also even if you did have a calculator, how do you get trhe number,   short of trying every four digit number[/hide]

Gian, how do you comput that large interger? or do you get it from some other source?

I'm curious :roll:

You can use Mathematica or some other math program.

Yes Maple too will do it, there are few BASIC versons, and REXX  which can work with 100's or 1000's of digits...

[hide]
[b]Lemma.[/b] For all integers $a$, with $a\perp 10000$, the last four digits of $a^{2003}$ are the same as the last four digits of $a^3$.

[i]Proof.[/i] By Euler's theorem, $a^{\phi(10000)} \equiv 1 \pmod {10000}$, and since $\phi(10000)=10000(1/2)(4/5)=4000$, we have $a^{4000} \equiv 1 \pmod {10000}$. Taking the square root of both sides, we get $a^{2000} \equiv \pm 1 \pmod {10000}$. Since a perfect square cannot end in $9999$, which is congruent to 3 modulo 4, we can only have $a^{2000}\equiv 1 \pmod {10000}$, and multiplying by $a^3$ gives us $a^{2003}\equiv a^3 \pmod {10000}$, as desired. $\boxed {}$

Now we need $x$ s.t. $x^3 \equiv 2003 \pmod {10000}$.
[/hide]

have to go... :(

[quote="chess64"]
[b]Lemma.[/b] For all integers $a$, with $a\perp 10000$, the last four digits of $a^{2003}$ are the same as the last four digits of $a^3$.

[i]Proof.[/i] By Euler's theorem, $a^{\phi(10000)} \equiv 1 \pmod {10000}$, and since $\phi(10000)=10000(1/2)(4/5)=4000$, we have $a^{4000} \equiv 1 \pmod {10000}$. Taking the square root of both sides, we get $a^{2000} \equiv \pm 1 \pmod {10000}$. Since a perfect square cannot end in $9999$, which is congruent to 3 modulo 4, we can only have $a^{2000}\equiv 1 \pmod {10000}$, and multiplying by $a^3$ gives us $a^{2003}\equiv a^3 \pmod {10000}$, as desired. $\boxed {}$
[/quote]

Actually you can get a little better number  than  $\phi(10000)=10000(1/2)(4/5)=4000$
Since $\phi(5^4)= (5^4)(4/5)=500$
$a^{500} \equiv 1 \pmod {625}$   and $a^4 \equiv 1 \pmod {16}$  Implies $a^{500} \equiv 1 \pmod {10000}$
(we assume $a$ is prime to 2 and 5 in above)
Hope that helps.

(In other words,  $a^4 \equiv 1 \pmod {10} , a^{20} \equiv 1 \pmod{100}, a^{100} \equiv 1 \pmod {1000}$ and for all larger numbers,
$a^{(5 \cdot 10^{n-2}) } \equiv 1 \pmod {10^n} , \ when \ n>3$ )
[quote]All right, but let´s focus only in finding the number.[/quote]
[hide]The number is $0587$ (as people have pointed out )[/hide]


Let $-1 < x < 1.$ Show that \[ \sum^{6}_{k=0} \frac{1 - x^2}{1 - 2 \cdot x \cdot \cos \left( \frac{2 \cdot \pi \cdot k }{7} \right) + x^2} = \frac{7 \cdot \left( 1 + x^7 \right)}{\left( 1 - x^7 \right)}. \] Deduce that \[ \csc^2\left( x + \frac{\pi}{7} \right) + \csc^2\left(2 \cdot x + \frac{\pi}{7} \right) + \csc^2\left(3 \cdot x + \frac{\pi}{7} \right) = 8. \]

How to obtained the relationship between cosecant of the above relation, please?

It actually is simply wrong (just plug in $x=0$ for instance). It also isn't quite what's written in my edition of the IMO compendium where it instead says
\[\csc^2 \frac{\pi}{7}+\csc^2 \frac{2\pi}{7}+\csc^2 \frac{3\pi}{7}=8\]
(and as a minor aside it says that this is problem 49 and not 44, just for all those who had trouble finding it...)
I'm not yet sure how to prove this with the preceding identity but it certainly does not seem impossible.
However, I find that there is a more conceptual approach by using the [url=https://artofproblemsolving.com/community/c6h1422994]identity[/url]
\[\sum_{k=0}^{n-1} \frac{1}{\sin\left(x+\frac{2k\pi}{n}\right)^2}=\frac{n^2}{\sin(nx)^2}\]
This gives that the sum in question is just
\[\frac{1}{2} \lim_{x \to 0} \left(\frac{49}{\sin(7x)^2}-\frac{1}{\sin(x)^2}\right)=\frac{49-1}{6}=8\]


Melinda will roll two standard six-sided dice and make a two-digit number with the two numbers she rolls. For example, if she rolls a 6 and 3, she can either form 36 or 63. What is the probability that she will be able to make an integer between 10 and 20? Express your answer as a common fraction.

[quote="$LaTeX$"]The only restriction is that she must roll a $ 1$. So our answer is $ 1\minus{}\frac{5}{6}^2\equal{}\boxed{\frac{11}{36}}$. I think...[/quote]
Yes, 11/36 is correct, alternatively we can approach this directly and see all the cases which make a number with tens digit 1.

11 - 1 Way
12 - 2 Ways
13 - 2 Ways
14 - 2 Ways
15 - 2 Ways
16 - 2 Ways
17 - 2 Ways
18 - 2 Ways
19 - 2 Ways

Total: 11 Ways

The total number of cases us $6\times6=36$ so the probability is $\boxed{\frac{11}{36}}$.

But if she is using standard dice, then wouldn't the numbers only go up to 6... 17,18, and 19 wouldn't be possible.

Complementary counting

There must be at least one 1, so just find the probability of no 1s: (5/6)^2 = 25/36

1 - 25/36 = 11/36


geometry
rectangle
symmetry
circumcircle
perpendicular bisector
geometry solved

Let $ABCD$ be a rectangle. Determine the set of all points $P$ from the region between the parallel lines $AB$ and $CD$ such that $\angle APB=\angle CPD$.

[quote="shobber"][quote="Rushil"]Just one point the point of intersection of diagnols?????!!![/quote]
I think it should be all points on the perpendicular bisector of AD and BC[/quote]

Indeed, just draw those p.b. to divide ABCD into four regions, and assuming any other solution, prove that it leads to a contradiction in every one of those 4 cases (depending on where P lies in the 4 smaller rectangles)

Yes.
Assume $P$ is not on the line, then $PA<PB$ and $PD<PC$. Then $\angle{PAD}<\angle{PBC}$ and $\angle{PDA}<\angle{PCB}$. Hence $\angle{DPA}>\angle{CPB}$. Cheers!

err...i think it's rather:

By symmetry, we only have to consider one of the four cases. Assume P lies to the left of the p.b. of AB, and below the p.b. of AD, then AP > DP, then <PDC < <PAB. Similarly, BP > CP, so <PCD < <ABP. So overall, <DPC > <APB. Done. :)

[b]Generalization.[/b] Let ABCD be a parallelogram.
Ascertain the geometrical locus of the point P for which  $\widehat{APB}\equiv\widehat{CPD}.$

[i]Solution[/i] (own). I note $I\in AC\cap BD$. For the any point P, from the plane of the $\triangle ABC$, with the property $\widehat{APB}\equiv\widehat{CPD}$, the circumcircles $C_1(O_1,R_1),C_2(O_2,R_2)$ of the triangles APB, CPD have the same  radius $R_1=R_2,\ O_1 O_2\parallel AD,\ IO_1=IO_2$ and the point I belongs to their common cord which is perpendicularly on AD. Thus, the geometrical locus of the point P is the line which pass through the point I and is perpendicularly on the line AD.

[b]Remark.[/b] Come on to extend at a convex quadrilateral ABCD !


function algebra polynomial calculus integration search trigonometry

random problem: Solve cos(cos(cos(cos(x))))=sin(sin(sin(sin(x)))) over the reals.

engel problems:

1. If a>b>0, then prove (a-b)^2/8a<(a+b)/2-sqrt(ab)<(a-b)^2/8b.
2. If f(n) is a function defined on the nonnegative integers by f(0)=0 and f(n)=n-f(f(n-1)), find a closed-form formula for f(n).
3. Given a polynomial with integral coefficients, let a_n be the digital sum in the decimal representation of f(n). Show that there is a number that occurs infinitely often in the sequence a_n, as n takes on the values of the positive integers.
4. Let f(x) and g(x) be nonzero polynomials, with f(x^2+x+1)=f(x)g(x). Show that f(x) has even degree.
5. A polynomial f(x)=x^4+[]x^3+[]x^2+[]x+1 has three undertermined coefficients denoted by []. If two players A and B move alternately (A goes first), and they move by replacing one [] with a real coefficient, show that B can always win if the condition for B winning is that at least one zero of f(x) is real (A wins if all zeroes are complex).
6. It is known of a polynomial over the integers that p(n)>n for every natural n. Consider x_1=1, x_(n+1)=p(x_n). We know that for any natural N, there exists a term of this sequence divisible by N. Prove that p(x)=x+1.
7. Find all polynomials P in two variables with the following properties: (i) For a positive integer n and all real t, x, y, P(tx,ty)=t^n*P(x,y), (ii) for all real a, b, c, P(b+c,a)+P(a+c,b)+P(a+b,c)=0, and (iii) P(1,0)=1.
8. Find all polynomials f over the complex numbers satisfying f(x)f(-x)=f(x^2).

As always, post solutions, i.e. don't just say (for example): 2. f(n)=[(1+sqrt(2))n] (;))

Don't you think these should be in advanced?

[quote="MysticTerminator"]
Solve cos(cos(cos(cos(x))))=sin(sin(sin(sin(x)))) over the reals.
7. Find all polynomials P in two variables with the following properties: (i) For a positive integer n and all real t, x, y, P(tx,ty)=t^n*P(x,y), (ii) for all real a, b, c, P(b+c,a)+P(a+c,b)+P(a+b,c)=0, and (iii) P(1,0)=1.
8. Find all polynomials f over the complex numbers satisfying f(x)f(-x)=f(x^2). [/quote]

Not solutions, just posting my thoughts for later when it's not after 1 AM.

Solve cos(cos(cos(cos(x))))=sin(sin(sin(sin(x)))) over the reals.
Lets do this case by case.  We can restrict our search to - :pi: < x :le: :pi:.
On (- :pi:, 0), sin(x) is in (-1, 0) which is a subset of (- :pi:, 0).  Thus, the RHS of the equation will also be in (-1, 0).  On (- :pi:, 0), cos(x) is in (-1, 1), a subset of (- :pi:/2, :pi:/2).  Thus, cos(cos(x)) on our interval gives us a subset of (0, 1).  Taking two more cosines will still give us some part of that interval.  Thus we get a LHS in (0, 1).  Thus there are no solutions in (- :pi:, 0).
So, onto [0, :pi:].  If there is a solution in ( :pi:/2, :pi:], it will have a corresponding solution ( :pi: - x) in [0, :pi:/2).  Thus, we need only examine this final quadrant to determine if there are any solutions.   (We can see by checking that :pi:/2 is not a solution.)
Now, sin is increacing over [0, :pi:/2), and its range is a subset of [0, :pi:/2) over that domain, so that means the RHS is an increasing function of x.  In particular, this means the largest value it achieves is sin(sin(sin(1))) :approx: 0.67843.
Now, for cosine.  Cosine is decreasing on [0, :pi:/2), and also maps into [0, :pi:/2).  This means that cos(cos(x)) is increasing, cos(cos(cos(x))) is decreasing, and cos(cos(cos(cos(x)))) is increasing.  Thus, the minimum value of the function over this interval will be cos(cos(cos(1))) :approx: 0.65429.  Bother.  So we're close, but not quite there yet.  I could always do at this point more numerical calculations: find the inverse of sinsinsinsin at 0.65429, which would give me some value up near :pi:/2, and then show that from that point onwards coscoscoscos > max(sinsinsinsin), which would work.  I don't know if there's a less-calculator-intensive method, though.

7. First condition gives us the polynomial P(x, y) = a0*xn + a1*x(n - 1)y + ... + an*yn.  Third condition gives us a0 = 1.

Looking at small cases, n = 1:
P(x, y) = x + a1*y
2(a + b + c) + a1*(a + b + c) = 0, a1 = -2, P(x, y) = x - 2y
n = 2:
P(x, y) = x2 + a1*xy + a2*y2
2(a2 + b2 + c2 + ab + bc + ca) + 2a1*(ab + bc + ca) + a2*(a2 + b2 + c2), a1 = -1, a2 = -2, P(x, y) = x2 - xy - 2y2 = (x - 2y)(x + y)

Actually, we know (x - 2y) is a factor of every Pn by letting a = b = c.
Letting mathematica solve the third case for me, we get P3 = (x - 2y)(x + y) :^2:, leading to the guess that Pn(x, y) = (x - 2y)(x + y)n - 1

Now, we can check without any trouble that such a polynomial works, which is good.  Now, for uniqueness, I'm not too sure ... I'll keep thinking about it.  It definitely is unique, though.

8. f(x)f(-x) = f(x2)
f(x) = (x - z1)(x - z2)...(x - zn)
f(-x) = :pm:(x + z1)(x + z2)...(x + zn)
f(x2) = (x2 - z1)(x2 - z2)...(x2 - zn), which has roots :pm: :sqrt:zi for all i.  What this means is that given the roots of f(x), they must also include all of their own squareroots.  (The f(-x) takes care of the negative square roots, or at least the other half of them.)  That is, if z1 is a root of f(x), then :pm: :sqrt:z1 must be a root of f(x) (only one of the two), and then the square root of that must be a root, and so on.  The only ways this is possible is if f has no roots (is a constant function) or if every root of f is either 1 or 0.  Otherwise, we will get an infinite sequence of roots of f.  (Which I think also means that f could only end up being the analytic function 0, but I don't remember this exactly.  Because the sequence would converge to 0 or 1.)  Anyhow, my rememberings of complex analysis aside, we then claim that the only possibly polynomial will have only zeros and ones as roots.  I also claim that it must be of even degree, and if it's of odd degree, we will get solutions to f(x)f(-x) + f(x2) = 0.

So, it looks like this: f(x) = xn(x - 1)m, f(-x) = (-1)nxn*(-1)m(x + 1)m = (-1)m + nxn(x + 1)m
f(x)*f(-x) = (-1)m + nx2n(x2 - 1)m = (-1)m + nf(x2).  So only if (m + n), the degree of f, is even, does it work.  But in that case it's an identity.  All done.

I think I found that there were no solutions in a mathchat. I might have messed up though. Basically, cos(cos(cos(cos(x))))=sin(sin(sin(sin(x)))) => cos(cos(cos(x)))+sin(sin(sin(x)))=pi/2. I think, by graphing it, they never intersected, if I remember correctly. In any case, I don't think calculators were allowed. Yeah, maybe these should go in advanced. but i can't move the thread.

What subject do you need to know to solve these?

(im wondering whether i should try to solve them or not)

I graphed it first before I tried my proof, and there aren't any solutions.  My method works if you have a scientific calculator handy.  I'm not sure how to improve on that to make a non-scientific calculator sufficient.  I was thinking for a while about using calculus (they have fun-looking derivatives), but I don't think it'll work.

Scrambled:  what you need to do these would be algebra. They do, after all, deal with polynomials.  But they are far more difficult than anything you'd see in an Algebra 1 course in high school (or middle school, as the case may be).  Mostly, they require ingenuity and at least a passing experience with questions of a similar type.

[quote="Mystic"]5. A polynomial f(x)=x^4+[]x^3+[]x^2+[]x+1 has three undertermined coefficients denoted by []. If two players A and B move alternately (A goes first), and they move by replacing one [] with a real coefficient, show that B can always win if the condition for B winning is that at least one zero of f(x) is real (A wins if all zeroes are complex). [/quote]

My earlier solutions are now much more finished -- I've done number 8 and am close to number 7 on the previous page of this thread.

Now then:
Lets assume that A changes either the third or first degree coefficient and that B changes the other of the two.  A has to pick the second degree coefficient so that the function is always positive.  Since it's positive at zero, A can just choose some astronomically large value for the second degree coefficient and B is doomed.  Thus, we know if A alters the third or first degree coefficients first, B much pick a good 2nd degree coefficient.

Let's have A pick either the 1st or 3rd degree coefficient.  Then we want B to pick some (large, negative) coefficient for the 2nd degree term, so that if the as-yet unlabelled coefficient were zero, the function would assume negative values for both positive and negative values of x.  Then no matter what A picks for the final coefficient, on one side or the other of zero, the function must still assume negative values (since the amount that A adds will be positive on one side of zero and negative on the other, no matter what.)  Since the function assumes positive values for very large x and is continuous, it must assume the value 0 somewhere in between, and B wins.

Thus, for A to have any chance, she must alter the 2nd degree coefficient.  So after A changes the coefficient, we have (x4 + nx2 + 1).  Now, we know if n :le: -2 that B can pick zero for either coefficient and A will be stuck in the case 1 scenario.  Thus, n > -2.  Let B pick a linear coefficient of -(2n + 6), converting our expression to (x4 + nx2 - (2n + 6)x + 1).  At x = 1, this function has a value of (1 + n - (2n + 6) + 1) = (-4 - n) < 0.  Thus, A must pick a coefficient of at least (4 + n) for the x3 coefficient.  So the final function is (x4 + mx3 + nx2 - (2n + 6)x + 1), where m > (4 + n).  Let us consider the value of this expression at x = -2:
(16 - 8m + 4n + (4n + 12) + 1) = (8n + 29) - 8m < (8n + 29) - (8n + 32) = -3 < 0.  Thus A must lose at either 1 or -2.  So B wins no matter what.

[quote="MysticTerminator"]1. If a > b > 0, then prove (a - b)2/(8a) < (a + b)/2 - :sqrt:(ab) < (a - b)2/(8b).[/quote]

Just the left side for now:  we wish to show that (a - b)2/(8a) < (a + b)/2 - :sqrt:(ab), or (rearranging) that
:sqrt:(ab) < (a + b)/2 - (a - b)2/(8a), or (rewriting and combining) that
:sqrt:(ab) < (3a2 + 6ab - b2)/(8a).  We can let b = ac, 0 < c < 1, to get
a :sqrt:c < (3a2 + 6a2c - a2c2)/(8a), or
:sqrt:c < (3 + 6c - c2)/8.

Will finish later -- I don't think it's too hard from here.

So, I was right, it's not too hard from there.  We square out and combine terms on the right to get
0 < c4 - 12c3 + 30c2 - 28c + 9.

Now, we note that if a = b, we have an equality condition in the original expression.  This occurs when c = 1.  So this thing better have a root of 1.  And it does.  In fact, if we take out that first root of 1 and then keep going, we see it has a triple root at one.  If factors as
0 < (c - 1)3(c - 9), which is always positive on 0 < c < 1.  Everything we did is reversible.  So the left side inequality holds nicely.  Now, no promises, but the right side should work exactly the same way, I think.

Anyhow, BUMP.

Sorry for reviving this thread - I am commenting on problem 8.:

[quote="JBL"]8. f(x)f(-x) = f(x2)
f(x) = (x - z1)(x - z2)...(x - zn)
f(-x) = :pm:(x + z1)(x + z2)...(x + zn)
f(x2) = (x2 - z1)(x2 - z2)...(x2 - zn), which has roots :pm: :sqrt:zi for all i.  What this means is that given the roots of f(x), they must also include all of their own squareroots.  (The f(-x) takes care of the negative square roots, or at least the other half of them.)  That is, if z1 is a root of f(x), then :pm: :sqrt:z1 must be a root of f(x) (only one of the two), and then the square root of that must be a root, and so on.  The only ways this is possible is if f has no roots (is a constant function) or if every root of f is either 1 or 0.  Otherwise, we will get an infinite sequence of roots of f.[/quote]

This argument is wrong: For instance, $f\left(x\right)=1+x+x^2$ is a solution of the problem, too; the roots of this polynomial are the nontrivial third roots of unity.

Unfortunately, the solutions that Arthur Engel gives in his book are not complete either. The problem occurs in Arthur Engel's book twice: the first time as problem 70 in chapter 10, and the second time as problem 36 in chapter 11. Not only are the solutions in these two chapters different from each other and even end up with different answers, but it also looks like both solutions are just incorrect. Engel claims to prove that all roots of the polynomial must be 0, 1 or third roots of unity, but (2n+1)-th roots of unity can work as well - for instance, $f\left(x\right)=1+x+x^2+x^3+x^4$ is a solution of the problem which clearly doesn't fit into Engel's general form $f\left(x\right)=\left(-x\right)^p\left(1-x\right)^q\left(x^2+x+1\right)^r$, which he claims to cover all solutions in chapter 11, whereas in chapter 10, he asks the (rhetorical?) question: "Are these all polynomial solutions?" - Alas, not.

It's interesting whether there exists a general form for all polynomials f(x) satisfying the functional equation $f\left(x\right)f\left(-x\right)=f\left(x^2\right)$. Clearly, all such polynomials are products of constants, polynomials of the kind $x^k$, and powers of cyclotomic polynomials, but not every such product satisfies $f\left(x\right)f\left(-x\right)=f\left(x^2\right)$ ...

  darij

The first one is easy.  There are no solutions.  You're taking the cos and sin of a very small number, which means that the first will be close to 1 while the other will be close to 0.  This can easy be formalized, but it's the easiest and most intuitive approach.

The first question, about sines and cosines, is not that easy.  I and rcv posted solutions in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=7344]Trig Problem[/url].


MATHCOUNTS geometry trigonometry AMC USA(J)MO USAMO AIME \/closed

My first question is: I am going to national mathcounts this year and will be going to high school next year. I have taken algebra 1 and 2, geometry, and trig (as of the end of this year). Geometry is my weak point and I intend to make the USAMO next year (I was one AIME point away this year). Olympiad geometry is far above my level but introductory geometry appears too easy. Is there a course I could take to improve my understanding of geometry?

My second question is: Is the course for trig and complex numbers just an introductory to the two topics (I know a lot about them already), or does it go more in depth? I really want to learn how to do problems like the complex number one on the AIME2 (which I took this year) or just AIME-level complex number problems in general. Would this course be suitable, or should I look into something else?

Thanks for any help.

There are piles of AIME level and higher problems in the trig/complex numbers class. It's an introduction in the sense that it starts from the basic definitions, but it moves fairly quickly. If you have trouble with the complex number problems on the AIME, then you can learn a lot from this class.

[quote="jb05"]My first question is: I am going to national mathcounts this year and will be going to high school next year. I have taken algebra 1 and 2, geometry, and trig (as of the end of this year). Geometry is my weak point and I intend to make the USAMO next year (I was one AIME point away this year). Olympiad geometry is far above my level but introductory geometry appears too easy. Is there a course I could take to improve my understanding of geometry?[/quote]
At some point we will offer an Intermediate Geometry course, but the earliest that it might be offered is late this year or early next year.

Given that you solve enough AIME problems to put yourself on the verge of qualifying for the USAMO, you might consider taking an AIME Problem Series.  That's been the most popular class at AoPS so far and about a third of the class pertains to geometry.  You might even improve your algebraic, combinatoric, or number theory problem solving skills in the process.

If you are interested in a more intense, guided curriculum with personal sessions with AoPS instructors, you might consider one of the Intermediate Independent Studies.  The Intermediate A covers exactly the level of geometry I would expect a study who gets half of all AIME problems but is weak in geometry would want to cover.

[quote="jb05"]My second question is: Is the course for trig and complex numbers just an introductory to the two topics (I know a lot about them already), or does it go more in depth? I really want to learn how to do problems like the complex number one on the AIME2 (which I took this year) or just AIME-level complex number problems in general. Would this course be suitable, or should I look into something else?
[/quote]
To add to what Simon said, the Intermediate Complex Numbers/Trigonometry class is currently our hardest Intermediate class.  It takes students up to what might be called "Pre-Olympiad" problem solving.  Unless you can solve nearly any complex numbers/trigonometry problems that would show up as one of the last 3 problems on an AIME, and easier USAMO problems of those varieties, this class will contain plenty of material that you will find valuable.

Thanks. I'll take this into account if I decide to take a course this year.


Two pulleys, one with radius 4 inches and the other with radius 10 inches are connected by a belt. If 4-inch pulley is caused to rotate at 3 revolutions per minute, determine the revolutions per minute of the 10-inch pulley.

I'm not sure...

[hide="answer?"]6/5?[/hide]

If I'm right, here's the solution:

[hide="solution"]Since the 4 inch pulley rotates 3 rev per min, the belt moves 24pi per min, and since the circumference of the 10 inch pulley is 20pi, 24pi/20pi=6/5.[/hide]

I think your solution is correct...  :lol:

my solution:
r1=4
r2=10
t1=3

r1/r2=t2/t1

4/10=t2/3
12=10*t2
t2=6/5

I get the same answer...


There are $ b$ boys and $ g$ girls at present at a party, where $ b$ and $ g$ are positive integers satisfying $ g\geq 2b \minus{} 1$. Each boy invites a girl for a dance (of course, two different boys must always invite two different girls). Prove that this can be done in such a way that every boy is either dancing with a girl he knows or all the girls he knows are not dancing.

[hide="Additional question which can spoil a solution of the above"]
[i]Additional question:[/i] (asked by me, and I don't know an answer) Is there a solution without using the Hall theorem?[/hide]

  darij

I have a solution using Hall. We can avoid Hall by imitating its proof, but its unnatural.
If this graph satisfies Hall condition, then each boy can choose girl than he knows.
Now we will suppose that there are k boys which together know less than k girls.
Choose maximal k and let boys b1,...,bk know girls g1,...gl with k>l.
Delete b1,..,bk and g1,...,gl. Remaining graph satisfies hall condition (because k is maximal),
so here every boy can take girl which he knows. There are at least
k girls left (because g>=2b-1) and boys b1,...,bk can take these girls.
Hope its clear.


Ring Theory
calculus
integration
algebra
polynomial
function
domain

(i) Show that (2) +(x) ={ f2 + gx: f, g in Z[x]} is not a principal ideal. So Z[x] is not a PID even though Z is a PID

(ii) Let r in R* where R is an integral domain. Show that (x) + (r) is a principal ideal <==> r in U(R). Hence show that R[x] is a PID <==> R is a field.

This is my attempt for part (i)

Let (a) = (2) + (x) for some a belongs to Z[x]
Clearly, (a) is the least principal ideal which contains both (2) and (x). Hence (2,x) exists and (2,x) = (a)
Since (a) = (2)+(x) we have (2) is the subset of (a) and (x) is a subset of (a). So, a|2 and a|x.
But also, a is in (a) = (2) +(x) and so a = 2f + gx.
Up to here, I dont know how to show it leads to contradiction so that (a) is not a principal ideal

Thank you in advanced for helping me

From $ a \mid 2$ follows that $ a \equal{} \pm 1$ or $ a \equal{} \pm 2$. But $ \pm 2 \nmid x$, so $ a \equal{} \pm 1$. However, this is a contradiction, since if we assume $ \pm 1 \equal{} 2f \plus{} xg$, then it is also true in $ x\equal{}0$, hence $ \pm 1 \equal{} 2f(0)$ in $ \mathbb{Z}$, absurd.

[quote]Another way: $ \mathbb{Z}[x]/(2, x)$ is finite, but $ \mathbb{Z}[x]/(a)$ can't be.[/quote]

... unless $ a \equal{} \pm 1$. But then $ | \mathbb{Z}[x]/(2, x) | \equal{} 2$, while $ | \mathbb{Z}[x]/(\pm 1) | \equal{} 1$.

Sorry; I meant for $ (a)$ not equal to the unit ideal.

The ideal is equivalent toeverything of the form $ 2a \plus{} x(p(x))$, where $ a \in \mathbb{Z}$ after simplifying any polynomial addition. $ 2$ is prime (do you know that if $ p$ is prime then $ p\mathbb{Z}$ is prime and maximal?) the constant coefficient's only extension is $ 1a$ for constant terms. But thats then the whole ring $ \mathbb{Z}[x]$.

Thank you very much guys.
How about part (ii), I dont know how to do it. Can you show me please?
I am really appreciated

For part (ii): Assume that $ r$ is non-zero in the integral domain $ R$ such that $ (X,R)$ is a principal ideal in $ R[X]$. So there is $ f(X)\in R[X]$ with $ (f(X)) \equal{} (X,r)$. Then $ r \equal{} f(X)g(X)$ for some $ g(X)\in R[X]$, hence $ 0\neq r \equal{} f(0)g(0)$, thus $ f(0)\neq 0$. Futhermore $ X \equal{} f(X)h(X)$ (*) for some $ h(X)\in R[X]$ which implies $ 0 \equal{} f(0)h(0)$, hence $ h(0) \equal{} 0$ or $ h(x) \equal{} Xh_1(X)$ with $ h_1(X)\in R[X]$. Since $ X$ is a not a zero-divisor in $ R[X]$, (*) gives $ 1 \equal{} f(X)h_1(X)\in (r,X)$. Hence $ 1 \equal{} f(X)a(X) \plus{} rb(X)$ with $ a(X),b(X)\in R[X]$ which finally gives $ 1 \equal{} rb(0)$ or $ r$ is a unit in $ R$. This other implication is trivial.

One implication of the second assertion of (b) follows from the above. The reverse implication can be derived by the fact that $ K[X]$ is a PID as Euclidean ring.

SIMPLER recall that PIDs have dimension $ \le 1$, i.e. nonzero prime ideals are maximal. Therefore
domain $ R \equal{} R[x]/(x) \Rightarrow (x)$ prime in PID $ R[x] \Rightarrow (x)$ maximal in $ R[x] \Rightarrow$ field $ R[x]/(x) \equal{} R$.
The prior "element-ary" proof is essentially a special case of this "structural" proof.


algebra
polynomial
calculus
integration
Galois Theory
number theory unsolved
number theory

Let p(x) be a polynomial of degree at least 1, with integer coefficients.

Suppose that for all positive integer n, p(x) ≡ 0 (mod n) always has a solution in x (of course, x is an integer here).

Is it true that there exists an integer c such that p(c) = 0?

OK, so the statement is not true for all polynomials.

How about if I fix a particular polynomial and ask if it is true?  Is there any general method to determine its validity?

See example below.

Fix integers a, b > 1, suppose that for any positive integer n > 1, the equation x^b - a ≡ 0 (mod n) always has an integral solution x, prove that there exists an integer c such that c^b = a.

Now it should be clear what I mean.

I don't understand.  If you pick a number for $ b$ then pick $ a$ that is not a power of $ b$ ...

Then by those choices, $ b^{c}\neq{a}$

So you have to pick $ a$ such that it is a power of $ b$.  If you do that then automatically $ b^{c} \equal{} a$

Sorry, I mistyped the statement.  It should be a polynomial in x.  I have made a correction in the above post.

Basically, I am asking, if a polynomial P(x) ∈ Z[x] has a zero in Z_n for all n, i.e., P(x) ≡ 0 (mod n) has a solution for all n, does it imply that P(x) has a zero in Z.

There are examples that the above statement is not true.  But there are also examples that the above statement is true (my example should be the polynomial x^b - a).

How do I in general determine when the statement is true or false by just looking at P(x)?

If I understood ZetaX's post correctly, it means your statement is true for irreducible P(x).  Otherwise, factor P(x) and analyze splitting fields to determine.

The question is subtler than you think.

The polynomial (x^2 - 2)(x^2 - 17)(x^34) is an example, which is irreducible in Z[x] but not in Z_n[x] for all positive integer n.

But what I am trying to say is different:

We know that if a polynomial is reducible in Z[x], then it is reducible in Z_n[x] for all n. And the converse is false for some polynomial (by the above example).  But when exactly does the converse hold then?  My example tells you one case.  Can you tell me, what kind of general conditions we need, in order that the converse also holds?  That's what I need.  Is it clear now?

JoeCrump is right, Copestone's polynomial isn't irreducible (by the way, the same question with "having a root" replaced by "being irreducible" is interesting as well; same if If replace "having a root $ \mod n$ for all $ n$" by "having a root $ \mod p$ for all primes $ p$" or additionally "...except for finitely many $ p$").

Copestone's example is http://www.mathlinks.ro/viewtopic.php?t=214712 .

In general, you ask when you can show equivalence of local (excluding reals) and global roots. Some cases are handled in the link (those where some element of the Galois group has no fixed point, including irreducible ones and the example if $ 2 \nmid b$). But there are others, like the example shows. I doubt there will be an easy characterisation, the only possibility I can even think of being perhaps possible is expressing it in cohomological terms...