geometry trigonometry AMC USA(J)MO USAMO AIME inequalities

Hey folks,

I decided that we needed to start a thread about what math books should be read. (Other than the obvious AOPS books :-) ) The criteria for this should be something along the lines of:

1. It's really a math book.
2. It has math in in (as opposed to just a history. There's nothing wrong with a good math history book. I just believe that the best ones include the actual mathematics.)
3. It's useful for improving problem-solving skills. This is AOPS after all.

I'm not trying to exclude any particular book (although Shakespeare might not be appropriate), just trying to keep the thread reasonably focused.

I'll start with one of my favorites.

[u]Prime Obsession[/u] by John Derbyshire. Perhaps not so great for problem-solving, but chocked full of math. Nice description of the Riemann Hypothesis. I like the way Derbyshire alternates chapters with math and math history. Not for the faint of heart. You should probably be pretty conversant with sequences and series before tackling it.

Fulton

How about "Problem Solving Strategies" by Arthur Engel. Perhaps the best math book I've ever come across...

Well we can definitely say that Fermat's Enigma, and The Music of The Primes should be read but I'm not quite sure of how they'd help you in contests(but if Prime Obsession can be included)....
I find A Primer of Abstract Mathematics(by Robert Ash) to be quite useful...however I can't read it straight through. :(

Another one that actually meets all of the criteria is:

[u]Geometry Revisited[/u] by H.S.M Coxeter. This is just an outstanding geometry book that is very readable by anyone that finished high school geometry and maybe a little trig. The proof of the extended Law of Sines is the very first thing covered, followed immediately by Ceva's Theorem. Morley's Theorem, cyclic quadrilaterals, and Ptolemy's Theorem are also included. Great, great book for those looking to learn a little more about geometry.

When you list a good book, how about telling a little bit about it. Why did you like it? What does it cover? etc.

Take care,
Fulton

I'll discuss some highlights of my (and my parents') mathematical library.  Of course, I have not read all of these (if any) cover-to-cover, but I have spent some time with all of them.

[b]Problem Books[/b]

USAMO & IMO Olympiad booklets.  Excellent.  Perhaps the best collections of solutions and problems that I have seen.

IMO Shortlists, with solutions.  Excellent.

Engel, [i]Problem-Solving Strategies[/i].  A good introduction to olympiad problem-solving.

Zeitz, [i]Art and Craft of Problem Solving[/i].  I'm not a big fan—especially since he added the chapter called "Geometry for Americans." :mad:

Andreescu and Andrica, [i]360 Problems for Mathematical Contests[/i].  Any of Titu's books is bound to be good.  This one seems to be an accessible transition from AIME to olympiad problems, and also includes a chapter on real analysis, which is not often covered in US high-school contests.  Chapters are Algebra, Number Theory, Geometry, Trigonometry, Mathematical Analysis, and Comprehensive Problems.

Djukić, Janković, Matrić, Petrović, [i]The IMO Compendium[/i].  This is useful as an archive of IMO problems; however, many of the solutions are lacking—they are crunched and often not the most elegant.  Alternate solutions are rarely presented.  Furthermore, the original wording of the problems is often changed, which I find irritating.

[b]Geometry[/b]

Kedlaya, [url=http://www-math.mit.edu/~kedlaya/geometryunbound/][i]Geometry Unbound[/i][/url].  Excellent.  I took the time (and paper) to print it off the internet.  The only book of these three which is aimed at olympiad geometry.

Coxeter, [i]Introduction to Geometry[/i].  Very nice.  Not for the faint of heart.

Hilbert, [i]Foundations of Geometry[/i].  Excellent for anybody interested in the axiomatic foundations of geometry.  This is the book that made Euclid's system of axioms obsolete.

[b]Inequalities[/b]

Andreescu, Cîrtoaje, Dospinescu, Lascu, [i]Old and New Inequalities[/i].  An excellent problem book of olympiad inequalities.  Problems range from mildly challenging to very difficult.

Steele, [i]The Cauchy-Schwarz Master Class[/i].  I just got this book yesterday, but it is already good.  In the first chapter, for instance, the Cauchy-Schwarz that we already knew and loved is generalized to inner products.  Some calculus is a prerequisite.

Hardy, Littlewood, Pólya, [i]Inequalities[/i].  The classic text on the subject.

[b]Analysis and Calculus[/b]

Thompson, [i]Calculus Made Easy[/i].  Despite its title, this book is an excellent introduction to calculus.  Its superior approach gives the elegant motivation behind calculus which is obscured so often by the cumbersome limit approach which the typical modern textbooks use.  It is also [i]small[/i] and [i]cheap[/i], unlike many of the twenty-pound monsters we see haunting modern bookshelves.  Unfortunately, I think that the recent additions somewhat detract from these latter qualities.

Phillips Exeter Academy Math Department, [url=http://math.exeter.edu/dept/materials/]Math Texts[/url].  Excellent, problem-solving approach; single-variable calculus is Math 4, which is less than 300 pages (note earlier comment about 20-pound monsters).

Rudin, [i]Principles of Mathematical Analysis[/i].  What can I say?  Excellent.  The classic introduction to mathematical analysis, as far as I know.  It is something beyond what you would find in a typical calculus class :  it is not calculation-obsessed.  Having had calculus before will probably make it easier for you, but is not a prerequisite.

This makes me realize that I have no books on number theory.  How odd.  I am not surprised by my lack of combinatorics books; after all, there are not very many theorems we use in contests, and these are more of a loose collection than a logical progression suitable for a course.

If you are interested in buying any of these, you should note that [i]360 Problems[/i] and [i]Old and New Inequalities[/i] are pulished by GIL, a Romanian publishing house, and are probably difficult to obtain in the US.  If you wish to buy them, I would suggest writing either to GIL, or to Titu Andreescu.


linear algebra
matrix
symmetry
algebra
polynomial
function
linear algebra solved

here is a hard one i think:

let p a prime number and E the set of n*n matrices whose coefficients are in Z/pZ and such that M^2=In. Find the cardinal of E.

Just first ideas

$M$ is diagonalisable since it is a symetry in $M_n(Z/pZ)$
its eigenvalue satisfy $k_j^2=1$
M is similar to $diag(1,..,1,-1,...,-1)$
where the number of 1 is $dim(Ker(M-I_n))$
the number of -1 is $dim(ker(M+I_))$

Hi,
   I have "encore" many stupidities to tell, so please forgive me for the non-senses that might appear in the following attempt of solution. I will treat just the case p odd.
   As Moubi pointed out, any such matrix is a symmetry and thus it is similar to a matrix of the form $ A_k=diag(I_k, -I_{n-k})$ for a certain k between 0 and n. The reciproque is also true, thus we need to count the matrices similar to one of the matrices $A_k$. Let $X_k$ be the set of matrices similar to $A_k$. Since no two $A_k$ have the same characteristic polynomial, we observe that no two $X_k$ intersect, thus they form a partition of the set of solutions. For k=0,n we find immediately that $X_k$ has 1 element. Suppose now that $k$ is between 1 and n-1 and define the function f from $GL_n(Z_p)$ such that $ f(P)= PA_kP^{-1}$. We will prove that each value of this function is taken the same number of times. Suppose that $ PA_kP^{-1}$ is such a value and consider $Q$ invertible such that $ f(P)=f(Q)$. This condition is equivalent to the fact that $P^{-1}Q$ commutes with $A_k$. This in turn implies that $P^{-1}Q$  is of the form $ diag(U,V)$ where U is kxk and V is (n-k)x(n-k). Moreover, the fact that Q is invertible is equivalent to U,V invertibles. Thus the value $ f(P)$ is taken $ card (GL_k(Z_p))\cdot card(GL_{n-k}(Z_p))$ times. It follos that the image of f, which is in fact $card X_k$ is $\frac{card(GL_n(Z_p))}{card (GL_k(Z_p))\cdot card(GL_{n-k}(Z_p))}$. Thus the number of such matrices should be, in my opinion, $2+\sum_{k=1,n-1}{\frac{card(GL_n(Z_p))}{card (GL_k(Z_p))\cdot card(GL_{n-k}(Z_p))}}$, where we know what a nice number is $card(GL_r(Z_p))=(p^r-1)(p^r-P)...(p^r-p^{r-1})$. I won't dare to find a nice closed formula for that simply because there must be something wrong in the solution.

It was Oral Examination ENS 1994, you solve it :)  harazi

U la la, this is I think the first problem from ENS oral algebra that I could solve. I promise this won't happen again.


geometry
rectangle
circumcircle
incenter
geometric transformation
reflection
trapezoid

Here is something interesting. When i was solving one problem i discovered something but i don`t know if it is true.

Let ABCD be a convex quadrilateral in a circle k with centre N and $O_1,O_2,O_3,O_4$ be the centres of the incircles of the triangles ABC, BCD, CDA, DAB. It is known that $O_1O_2O_3O_4$ is a rectangle (that is easy to prove). Let $N=AC\cap BD$ and T be the centre of $O_1O_2O_3O_4$.
[i]Prove or disprove[/i] that M,N and T are colinar.  :flex:

There is a typo, the point M is undefined and the point N os defined twice. Fixing this, assume that M is the circumcenter of the cyclic quadrilateral ABCD. Unfortunately, the proposition is false.

Let $\triangle ABC$ be an arbitrary triangle, M its circumcenter and D an arbitrary point on the arc CA of the circumcircle (M). As the point D moves on the arc CA of (M), the locus of the center (diagonal intersection) T of the rectangle $O_1O_2O_3O_4$ is a circle with the diameter C'A', where C', A' are midpoints of the arcs AB, BC of the circumcircle (M) (see the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=56605]Nice (circle pass through a constant point)[/url]). As the point D approaches the vertex A, the diagonal intersection $N \equiv AC \cap BD$ of the cyclic quadrilateral ABCD also approaches the vertex A. The incenter $O_1$ of the triangle $\triangle ABC$ remains fixed, the incenter $O_2$ of the triangle $\triangle BCD$ approaches the incenter $O_1$ and the incenters $O_3, O_4$ of the triangles $\triangle CDA, \triangle DAB$ approach the vertex A as well. Thus the diagonal intersection T of the rectangle $O_1O_2O_3O_4$ approaches the midpoint of the segment $AO_1$, which means that the line NT approaches the bisector of the angle $\angle CAB$, while the line NM approaches the line AM, which is the reflection of the A-altitude of the triangle $\triangle ABC$ in the angle bisector $AO_1$. Clearly, the angle bisector $AO_1$ is different from the isogonal conjugate AM of the A-altitude, unless the triangle $\triangle ABC$ is isosceles. Thus the lines NT, NM have different limits, which means they are generally different and the points N, M, T are generally not collinear.

The points N, M, T are collinear, if the quadrilateral ABCD is an isosceles trapezoid (just by symmetry) or if the quadrilateral ABCD is bicentric with an incenter I. Then the center T of the rectangle $O_1O_2O_3O_4$ is identical with the circumcenter of the quadrilateral $I_1I_2I_3I_4$, where $I_1, I_2, I_3, I_4$ are incenters of the triangles $\triangle ABN, \triangle BCN, \triangle CDN, \triangle DAN$. The cyclic quadrilateral $I_1I_2I_3I_4$ is centrally similar to the contact quadrilateral EFGH of the bicentric quadrilateral ABCD with the similarity center at the diagonal intersection N (see the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23192]Bicentric quadrilateral[/url]), which means that the points N, T, I are collinear. But the diagonal intersection N, the incenter I and the circumcenter M of the bicentric quadrilateral ABCD are collinear, hence all 4 points N, T, I, M are collinear.


quadratics
number theory proposed
number theory

a) There are infinitely many positive integers which are not the sum of a square and a prime.
b) There are infinitely many positive integers, which are not of the form $p+a^{2k}$ with $p$ a prime and $a, k$ positive integers.

[hide="Solution to part A"]
[b]A natural number $n$ cannot simultaneously be represented in the forms $a^{2}+p$ and $24k+3$ where $a, k \in \mathbb{Z}^{+}$ and $p$ is a prime greater than $3.$[/b]

It is trivial to see that all primes are congruent to $1, 5, 7, 11 \bmod{24}.$ In addition, the quadratic residues of $24$ are $0, 1, 4, 9, 12, 16.$ Irrespective of the different values we take and add them together, $3$ can never be formed. $\blacksquare$

Now, since there are infinitely many numbers which are congruent to $3 \bmod{24},$ we are done by the lemma.
[/hide]

[hide="Solution to part B"]
This follows from part a), as this number is the sum of a prime and a perfect square.
[/hide]

[quote="Karth"]It is trivial to see that all primes are congruent to $1, 5, 7, 11 \bmod{24}.$ [/quote]

What about $13, 17, 19$, and $23$?

Edit:  Anyway, you probably want to look at squares $k^{2}$ with the property that $2k-1$ isn't prime for part a.


Please suggest which one of the two (tmh and dasgupta) is better in terms of sufficient amount of theory,
solved examples and exercises, i.e. quality and quantity of content.

tmh - tata mac-graw hill
dasgupta - Asit Dasgupta
sound familiar ?


function
algebra
polynomial
limit
real analysis
real analysis unsolved

Show that a continuous real valued function on (0,1) is the uniform limit of a sequence of polynomial on (0,1) if and only if f can be extended to a continuous function on [0,1]

call our continuous real valued function on $ (0,1)$ $ f$.  if $ f$ can be extended to a continuous function on $ [0,1]$ (which we also denote by $ f$), then we can find polynomials $ p_n$ on $ [0,1]$ such that $ p_n\to f$ in $ L^\infty([0,1])$ by stone-weierstrass, and the result follows immediately.

for the other direction, since polynomials are bounded on $ [0,1]$, $ f$ is also.  say that $ f$ is the uniform limit of the polynomials $ q_n$ on $ (0,1)$ and let $ L \triangleq \limsup_{t\downarrow 0} g(t) < \infty$.  If $ n$ is chosen large enough that $ |q_n(t) \minus{} f(t)| < \varepsilon$ for all $ t\in (0,1)$, then $ |q_n(0) \minus{} L| < 2\varepsilon$ (why?).  Letting $ n\to\infty$ shows that $ \lim_{n\to\infty} q_n(0)$ exists and is equal to $ L$.  Setting $ l \triangleq \liminf_{t\downarrow 0} g(t) > \minus{} \infty$, the same argument shows that $ \lim_{n\to\infty} q_n(0)$ is $ l$.  So $ l \equal{} L$, but this means that $ \lim_{t\downarrow 0} g(t)$ exists and is finite, so we can extend $ f$ to a continuous function on $ [0,1)$ in the obvious way.  repeating this argument with the obvious modifications for $ 1$ instead of $ 0$, we extend $ f$ to a continuous function on $ [0,1]$.


The polynomial $4x^{3}+ax^{2}+bx+c$ is divisible by $x^{2}-1$, and when divided by $x+2$, the remainder is 3. Find a,b and c.

[hide="Alternatively"]
$f(x) = (4x+d)(x^{2}-1)$
By the Remainder Theorem
$f(-2) = 3(-8+d) = 3 \implies d = 9$
$\therefore f(x) = 4x^{3}+9x^{2}-4x-9$
[/hide]


Now I know.

[hide="the game"]You try to close the box, and then the lyrics show up. Hold down the enter key if you want to get out of there fast.[/hide]

That was the strangest thing I have seen in my life :huuh: . It scared the 3@%@ out of me. Dont do it again

[quote="alkjash"]Wtf!
I had to restart my computer.[/quote]

lolz really?

Annoying. The browser bounced around 3 or 4 full revolutions, and then came the dialogs... I prefer Space for no reason whatsoever.

Not going there anymore.

It said "Internet is serious business" for some reason... :huh:

holy cow, I hate that!

please no one else go there.

You are not curious!
I repeat.
[size=150]You are not curious!!![/size]

If people are complaining about that site, they must be getting viruses of some sort or something...

Because my computer is perfectly fine!!!!!!!

It doesn't do anything to your computer, it's just terribly irritating.

This is why SOME of us responsible people use Anti-script programs and Ad-block programs.
Others get hit with this same thing every time.
Anyways, I've seen how that site works on IE7. You just click close. And it says stuff like "are you sure?"
Just keep on pressing return and it'll close in awhile.


Find the value of c for which the roots of $x^3-6x^2-24x+c=0$ forma an arithmetic progression.

[quote="Iversonfan2005"][hide="hint"]Vietas[/hide][/quote]

[hide] What's vietas??[/hide]

it simply states what is the sum of the roots, the product of the roots and the combinations of products and roots which are needed to solve this problem :)

http://planetmath.org/encyclopedia/VietasFormula.html
http://mathworld.wolfram.com/VietasFormulas.html

:)

let the three terms be $a-d,a,a+d$ .

From Vieta , the coefficient of $x^2$ is the sum of roots which is $(a-d)+a+(a+d)=6$ . so $a=2$

The coefficient of $x$ is the sum of two-roots so $a(a-d)+a(a+d)+(a-d)(a+d)=-24$ solve to get $d=6$

So the three terms are $-4,2,8$

Hence $c=-(-4)(2)(8)=64$


$\boxed {\ \triangle ABC\Longrightarrow p^2\le (2R-r)(4R+r)\ }\ \ (2p=a+b+c)$

Hint to a nice solution:

[hide]write out everything in $r_a$, $r_b$, $r_c$...
[hide="more precisely"]$p^2=\frac{r_ar_br_c}{r}$; $4R=r_a+r_b+r_c-r$, $\frac{1}{r}=\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}$.[/hide][/hide]

  Darij

$3(r_ar_b+r_br_c+r_cr_a)\le (r_a+r_b+r_c)^2\Longrightarrow \boxed {\ p\sqrt 3\le 4R+r\ }\ \ (1)$
$\frac{p}{\sqrt 3}\le \frac{3R}{2}\ ,\ r\le \frac R2\ \Longrightarrow \frac{p}{\sqrt 3}+r\le 2R\Longrightarrow \boxed {\ \frac{p}{\sqrt 3}\le 2R-r\ }\ \ (2)$

The product of the relations $(1)$ and $(2)\Longrightarrow \boxed {\ p^2\le (2R-r)(4R+r)\ }\ .$


The incircle $w=C(I,r)$ of the triangle $ABC$ touches the its sides in the points

$D\in BC$, $E\in CA$, $F\in AB$. I note the middlepoint $M$ of the side $[BC]$ and

the intersection $P\in EF\cap DI$.

Prove that $\frac{PF}{PE}=\frac bc$ and ${P\in AM}$.

[u]Remark 1.[/u] This assertion is true and for the exincircles of the triangle $ABC$:

The exincircle $w_a=C(I_a,r_a)$ of the triangle $ABC$ touches the its sides in the

points $D_a\in BC$, $E_a\in CA$, $F_a\in AB$. I note the middlepoint $M$ of the side

$[BC]$ and the intersection $P_a\in E_aF_a\cap D_aI_a$.

Prove that $\frac{P_aF_a}{P_aE_a}=\frac bc$ and $P_a\in AM$.

[u]Remark 2.[/u] If we note the points $R\in EF,\ DR\perp EF$ and the middlepoints

$S$, $N$ of the segments $AR$, $EF$ respectively, then $M\in SN$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=61395

with simple angle chasing we get $\angle{FIP} = B$ and $\angle{EIP} = C$.

so, we have $EP/FP = (IE/IF) . (sinB/sinC) = sinB/sinC = b/c$.

now, let ${AM}\cap{EF} = X$.

we have $EX/FX = (AE/AF) . (sin(CAM)/sin(BAM))$

$= sin(CAM)/sin(BAM) = (b/c) . (BM/CM) = b/c$,

that implies $X = P$.


When Jane is one year younger than betty will be when jane is half as old as betty will be when jane is twice as old as betty is now, betty will be three times as old as jane was when betty was as old as jane is now.

One is in her teens and ages are in completed years. how old are they?

[hide]Jane is 19, Betty is 25 but it is very confusing so I might have made an error...

Let x be the age of Jane now and y be the age of Betty now.

When betty is as old as jane will be when she is twice the age that betty is now, she will be y + (2y - x) = 3y - x.  When Jane is half of this, she will be (3y-x)/2.  When Jane is (3y-x)/2, Betty will be y + (3y-x)/2 - x = (5y - 3x)/2.  When Jane is one year younger that, she is (5y-3x)/2 - 1.

When Betty was Jane's age now, Jane was 2x - y.  When betty is 3 times as old as this, she will be 6x - 3y.  So the entire sentence boils down to

when Jane is (5y-3x)/2 - 1, Betty will be 6x - 3y.  (5y-3x)/2 -1 + (y-x) = 6x-3y.  Simplifying, 13y - 2 = 17x.  The only integer solution with one of x and y in the teens is x = 19, y =25.  [/hide]


If $ \overrightarrow{x} \equal{} (1,\ 1,\ 1),\ \overrightarrow{y} \equal{} (a,\ b,\ c),\ \overrightarrow{z} \equal{} (P,\ Q,\ R)$, and $ \overrightarrow{x}\cdot \overrightarrow{y} \equal{} \overrightarrow{x}\cdot \overrightarrow{z} \equal{} \overrightarrow{y}\cdot \overrightarrow{z} \equal{} 0$,

Find the value of : $ \frac {a^2}{|\overrightarrow{y}|^2} \plus{} \frac {P^2}{|\overrightarrow{z}|^2}$

Let $ \overrightarrow{x} \equal{} (1,\ 1,\ 1),\ \overrightarrow{y} \equal{} (a,\ b,\ c),\ \overrightarrow{z} \equal{} (P,\ Q,\ R)$, the condition can be rewritten as $ \overrightarrow{x}\cdot \overrightarrow{y} \equal{} \overrightarrow{x}\cdot \overrightarrow{z} \equal{} \overrightarrow{y}\cdot \overrightarrow{z} \equal{} 0$.

[quote="kunny"]Let $ \overrightarrow{x} \equal{} (1,\ 1,\ 1),\ \overrightarrow{y} \equal{} (a,\ b,\ c),\ \overrightarrow{z} \equal{} (P,\ Q,\ R)$, the condition can be rewritten $ \overrightarrow{x}\cdot \overrightarrow{y} \equal{} \overrightarrow{x}\cdot \overrightarrow{z} \equal{} \overrightarrow{y}\cdot \overrightarrow{z} \equal{} 0$.[/quote]
So, what is the use of that?

[hide="Hint"]Note that $ \vec{z} \equal{} \alpha (\vec{x} \times \vec{y} )$, and wlog we can let $ \alpha \equal{} 1$, so $ P,Q,R$ are already out of the equation.[/hide]


In the preamble, use

[color=blue]\usepackage[bookmarks=false,colorlinks=true,linkcolor={blue},pdfstartview={XYZ null null 1.22}]{hyperref}[/color]

The [color=red]pdfstartview={XYZ null null 1.22}[/color] defines the % on the PDF, in this case we have the 122%

--

\href{post here the full URL}{name you like to set it}

\usepackage{url}
then
\url{put your link here}

That's for use in papers and can cope with syntax that might get messed up normally in LaTeX. But for a hyperlink in pdfs like this http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167767 which you can click on, you will need the [url=http://www.tug.org/applications/hyperref/manual.html]hyperref package[/url].
[code]\usepackage{hyperref}
\href{http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167767}{How to insert URL link in LaTeX? }[/code]


This code works perfectly well on my home PC but fails on the site. Any idea why?
[code]
import graph3;
currentprojection=orthographic((1,1,1));
size(200);
triple f(pair u)
{
real r=2+sin(u.y);
return (r*cos(u.x),r*sin(u.x),u.y);
}
draw(surface(f,(0,-7),(6.29,5),50),blue);
[/code]
[asy]import graph3;
currentprojection=orthographic((1,1,1));
size(200);
triple f(pair u)
{
real r=2+sin(u.y);
return (r*cos(u.x),r*sin(u.x),u.y);
}
draw(surface(f,(0,-7),(6.29,5),50),blue);[/asy]

It's possible that the site uses an older version. This happens to me sometimes.

Nope. I suspect that it is because of the new rendering of 3D images in the latest versions. Here is what I think (Valentin, please correct me if I'm wrong on some account).

The way Asymptote is handled by the AoPS server is like
1) extracting the code from the post;
2) parsing, checking for dangerous things, etc.;
3) creating an asy file with the code and some automatically generated name, say 12345.asy;
4) running Asymptote on that file
5) getting the file 12345.eps
6) converting it to some format suitable for displaying online
7) adding the resulting image to the page html code

I strongly suspect that it is the stage 5 that fails. With the current defaults, Asymptote creates no .eps file for the 3D images but opens the interactive viewer instead. To suppress this behavior, you need to add

access settings;
settings.render=0;

to the code. This will send Asymptote to its usual route bypassing the interactive picture mode. Unfortunately, user playing with settings is prohibited, so I cannot check this theory myself, but admins should be able to check it quite easily. :)

It seems like the trouble is with converting the .eps files to .png ones. If I knew what converter AoPS uses and had it, I would try to play with the settings a bit myself to see where the cause of the trouble lies. As it is, we'll have to wait for Valentin to figure it out. Anyway, the full power of Asymptote 3D graphing tools is rarely needed for the purposes of this site, so the issue can wait (but it is still somewhat irritating to construct the 2D images representing 3D objects by hand).


You know, I practice martial arts, and my instructor has a very simple but logic and efficient philosophy about success in basically every aspect of life,
[list]
[b]Work Hard[/b]: There is no substitute for hard work. There is no other way to become good doing something than doing it once and again and again. That is why IMO participants do something they call "training", that is why any sport team practice all the time, even when it is not their season. It can be frustrating at the begining, but eventually it will become more natural.
[b]Focus[/b]: Have a goal, and concentrate about what you have to do. If you put all your energy in something, probably you will do it faster than if not (anyway, some times you need to stand up, drink a coffee, or whatever you want, and afterwards go and do what you have to do)
[b]Think Positively[/b]: My friend, if you think that you can't do it, you can't! Don't start 'knowing' that you can't do it, start 'knowing' that it can be difficult, but you will succeed!
[/list]
On the other hand, understand other proofs before you try to do them by yourself. If you have no idea of how to speak spanish, I can't start talking you in spanish and spect you to follow the conversation. Spend your time, it is not something that you will be able to do tomorrow if you started today.

Best,

I faced a similar struggle to yourself, when I took my first number-theory course. It was the first course wherein I had to find creative solutions to problems

There are certain proof abilities vital: setting up induction (for proving stuff for all n) and setting a contradiction.
Also, for some if then statements, you might experiment with how the problem is phrased; maybe the contrapositive is easier to prove.

In general, work with what you are given in order to solve a problem. It helps to glance at proofs professionals have constructed in order to see their method.

[quote="msbaa672"]I am taking my first college proofs class. My abilities are fairly solid with regards to applied math, but to be honest, I am pretty bad with theory. Perhaps that says something about my true math ability. But regardless, I am very frustrated and discouraged with the class I am taking. It seems that no matter how much effort I invest, it just doesn't show. Can someone please recommend a course of action? What's the best way to improve in a theory-based class? How do I strengthen my proofs skills? Any suggestions would be appreciated. In particular, it may be helpful to hear from people who have had similar struggles.[/quote]

It depends partially on what kind of math course it is.  I consider analysis proofs different from number theory proofs, which are different from combinatorial proofs, etc.  What are you taking?

[quote="blahblahblah"]It depends partially on what kind of math course it is.  I consider analysis proofs different from number theory proofs, which are different from combinatorial proofs, etc.  What are you taking?[/quote]

number theory

Read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11946]this thread[/url] from the College Forum.

[quote="djimenez"]I have an article called "Tacticas Desesperadas" or "Desperate Tactiques", but it is in spanish. If you read it, I believe I can send it to you, otherwise, I believe I can translate it when I have the time (I am not sure if it is going to be soon).

It talks about some proving techniques, and how to improve them, and how to recognice which you should apply. A little basic, but very well written.

Best,[/quote]

I think I could read some... could you send it to me? filletwho@gmail.com
I could always ask my Spanish teacher for help. It would improve my spanish too.

Well,

For those who want to try, here you have another article. It is a draft. The original I can't find it, but it said pretty much the same than this one.

It is in spanish, and it is not finished yet! Any suggestion is very welcome. It is part of an slightly more extence draft that a couple of friends wrote some years ago.

Best,


Can someone clearly explain how the informal definition of a limit ($ L$ = the value that $ f(x)$ approaches as $ x$ approaches $ c$) translates to the formal definition (for every $ \epsilon > 0$ there exists $ \delta > 0$ such that $ |f(x) \minus{} L| < \epsilon$ whenever $ 0 < |x \minus{} c| < \delta$)?

I feel that not understanding this completely is making it harder to understand why you prove limit problems in a certain way.

Think of it a bit as a conversation:

Say I claim that $ f(x)$ approaches $ L$ when $ x$ approaches $ c$. I mean that $ f(x)$ gets "close"
to $ L$ whenever $ x$ is "close" enough to $ c$. I challenge you to disprove me. Perhaps you ask
me to show that $ f(x)$ gets within .5 of $ L$ as $ x$ gets close to $ c$. Then I show that for all $ x$ no more than, say, 2.5 away from $ c$, that $ f(x)$ is within $ .5$ of L. Here $ \epsilon \equal{} .5$ and $ \delta \equal{} 2.5$. And, if no matter which $ \epsilon$ you pick, I  can provide such a $ \delta$, you
must agree that $ f(x)$ gets "close" to $ L$ whenever $ x$ is sufficiently close to $ c$. You choose the
definition of "close" by picking $ \epsilon$. I respond by choosing my definition of close, by picking
$ \delta$.

Does this help?

"For any $ \varepsilon$ you pick, I can find a value $ \delta(\varepsilon)$ such that as long as $ x$ is within $ \delta$ of $ c$, $ f(x)$ will be within $ \varepsilon$ of $ L$."  We're playing a game: if the limit weren't $ L$, then you could pick an $ \varepsilon$ such that I can't force $ |f \minus{} L| < \varepsilon$.  If the limit is $ L$, then no matter what $ \varepsilon$ you pick, I can always find the right $ \delta$.  "$ f(x)$ approaches $ L$ means the same as "you can't keep $ f(x)$ away from $ L$."

Edit: redundant.

In some ways, I have an easier time finding a paraphrase of intermediate formality for the definition of the limit of a sequence.

$ \lim_{n\to\infty}a_n\equal{}L$ if and only if $ a_n$ eventually becomes arbitrarily close to $ L.$

"Arbitrarily close" translates to "for all $ \epsilon>0\ \dots\ |a_n\minus{}L|<\epsilon.$" and "eventually" translates to "there exists $ N$ such that for all $ n>N,\dots$."

The limit at a point is similar except for a somewhat less intuitive replacement for "eventually."

Thanks guys, a combination of your explanations has helped me understand it very well. I had to read it over many times before it sunk in though.


arithmetic sequence
algebra unsolved
algebra

Let $n\ge2$ is given integer, let $a_0,a_1,\cdots,a_n$ is arithmetic sequence in that order. Compute the sum:

$S=\sum_{k=0}^n(-1)^k\binom{n}{k}a_k$

[hide="Hint"]If we write the arithmetic sequence $a_k=a_0+kd$, then
$S=\sum_{k=0}^n(-1)^k\binom{n}{k}a_k=a_0\sum_{k=0}^n(-1)^k\binom{n}{k}+d\sum_{k=0}^n(-1)^k\binom{n}{k}k$.
These are easier sums (Newton, derivation,...).[/hide]

Anybody?
Well, taking $a_k=a_0+kd$,
$S=\sum_{k=0}^n (-1)^k \binom{n}{k}a_k = a_0 \sum_{k=0}^n (-1)^k \binom{n}{k} + d \sum_{k=0}^n (-1)^k \binom{n}{k} k =0$.
Let's prove both sums equal $0$.

First, we'll use Newton's identity,
$(1+x)^n=\sum_{k=0}^n \binom{n}{k} x^k$,
and then we'll substitute by $x=-1$:
$0=\sum_{k=0}^n \binom{n}{k} (-1)^k$.

Then, we can use derivation to get
$n(1+x)^{n-1}=\sum_{k=0}^n \binom{n}{k} kx^{k-1}$,
with a change of sign,
$-n(1+x)^{n-1}=\sum_{k=0}^n \binom{n}{k} kx^k$,
and then we'll substitute by $x=-1$:
$0=\sum_{k=0}^n \binom{n}{k} k(-1)^k$.

If $n$ is even, there's a way to avoid derivation.
With Newton's identity, we've proved that the sum is not dependent on $a_0$.
Well, let's choose another $a_0'$ that suits us best.  :)
Just try with $a_0'=-nd/2$, and - voilà :lol:, all is cancelled (the $k$-th term cancels the $(n-k)$-th).
If only it worked with $n$ odd.  :(


In rectangle ABCD, H is the midpoint of $ \overline{BC}$, E lies on $ \overline{AD}$, and F lies $ \overline{AB}$. In rectangle CEFG, H lies on $ \overline{FG}$ and $ HG \equal{} 3$ in. $ m\angle DEC \equal{} 45^o$. What is the number of square inches in the positive difference between the areas of the two rectangles?

Basically we use the $ 45^\circ\minus{}45^\circ\minus{}90^\circ$ triangles and the fact that $ BH\equal{}HC$ to get all the lengths, so the answer is clearly $ \left|6\sqrt2\cdot\frac{9\sqrt2}2\minus{}3\cdot9\right|\equal{}|54\minus{}27|\equal{}\boxed{27}$.


Some problems from OU Summer Math Institute daily Problem Solving Contest (fun, fun, fun :) ). I'll try to update this daily on weekdays starting on Tuesday (I'm going on vacation until then). Perhaps filletwho and tetrahedr0n may also choose to update this, but I'll attempt to update it myself. Here are problems from the last three days. So far, I've only solved the last two.

Day 1. Six individual pieces of string are unraveled and held parallel to each other. You hold the the centers of the strings with your hand, so you have six pieces on a side, six more on the other. On each side, you tie one string to only one other. What is the probability when you unravel the string, you will have a loop?

Day 2. On a trip from City A to City B, you travel these distances: uphill for some distance at 56 mph, downhill for some distance at 72 mph, and go on flat ground for some distance at 63 mph. You travel back the same route when going from City B to City A. It takes you 4 hours to get from city A to B, and 4 hours, 40 minutes to get from city B to A. What is the the total distance between the cities?

Day 3. What is the largest whole number that cannot be expressed as  a sum of 6a+ 9b + 20c?

Have fun!

1... not sure if i fully understood, but
[hide]$\frac{1}{15}$.  You know that you'll have a loop if and only if you tie the exact same two pieces of string together.  The probability of that is $\frac{1}{\binom{6}{2}}$.[/hide]

But can't you have all six in a loop, or five of them in a loop, etc.?


We are given a triangle $ABC$, and we have that $M$ is the midpoint of $AB$. There is a line $XY$ with $X$ on $CA$and $Y$ on $CB$ such that $XY$ pass trough $M$ and $CYX$ is isoceles in $C$. Let $O$ be the circumcenter of $CYX$, and let $D$ be the feet of the altitude from $C$ to $AB$. Prove that $OD=OM$

Indeed very nice. See http://www.mathlinks.ro/Forum/viewtopic.php?t=25424 .

  darij

Let $C=C(O,R)$ be a circumcircle of the $\triangle ABC$ and the middle point $M$ of the side [AB].

The proposed problem is in fact a speculation from the application of the Simson's lines

of the points $X,Y\in C(O,R)$ for which XY=OM !


geometry perimeter search circumcircle rotation trapezoid parallelogram

Given a circle $ (O) $ . $ P $ is inside $ (O) $ . P is fixed . 2 chords $ AB, CD $ through $ P $ satisfy : $ AB $ is perpendicular to $ CD $ . Find the maximum, the minimum of the perimeter of the quadrilateral $ ACBD $.

Have to use the property of a quad with perpendicular diagonals
   inscribed in a circle:

   the perpendicular to a side at it's  end point is equal to the opposite side
   of the quad.

    Maj.  Pestich

I remember posting this a very long time ago (either here or in the "geometry proposed problems" section). It did receive some answers. You might want to try searching for it, nttu.

Dear Grobber, I will be very thankful if you search it for me  :)

Let me denote the perpendicular chords $AC \perp BD$, the diagonals of a cyclic quadrilateral $ABCD$ with the fixed circumcircle $(O)$ and the fixed diagonal intersection $P$. As the lines $AC, BD$ rotate around the point $P$, the fixed line $OP$ alternately coincides with one of these diagonals or bisects the right angle between them. When one of the diagonals coincides with the line $OP$, the cyclic quadrilateral $ABCD$ is a right angle kite and when the line $OP$ bisects the right angle between them, the cyclic quadrilateral $ABCD$ is an isosceles trapezoid. We will prove that the kite perimeter is minimum and the trapezoid perimeter is maximum.

Let $R = OA$ be the circumcircle radius and $p = OP$ the distance of the point $P$ from the circumcenter $O$. Let $K, L, M, N$ be the midpoints of the chords $AB, BC, CD, DA$, forming a parallelogram with the opposite sides $KL \parallel MN$ and $LM \parallel NK$ parallel to the perpendicular diagonals $AC \perp BD$, i.e., a rectangle $KLMN$. The intersection $Q \equiv KM \cap LN$ of the rectangle diagonals is the centroid of the cyclic quadrilateral $ABCD$, hence, the midpoint between its circumcenter and anticenter. The anticenter of a cyclic quadrilateral with perpendicular diagonals is identical with its anticenter (the 2nd Brahmagupta's theorem). Thus the point $Q$ is the midpoint of the segment $OP$ and the quadrilateral $KOMP, LONP$ are parallelograms, because ther diagonals $KM, OP$ and $LN, OP$ cut each other in half. According to the solution of the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34217]Concurrency problem; related to Poncelet's theorem[/url], the side  midpoints $K, L, M, N $ of all considered cyclic quadrilaterals $ABCD$ lie on a circle $(Q, r)$ with the center $Q$ and radius

$r = QK = \sqrt{\frac{R^2}{2} - QP^2} = \frac{\sqrt{2R^2 - p^2}}{2}$

The segments $PK, PL, PM, PN$ are medians of the right angles triangles $\triangle APB, \triangle BPC, \triangle CPD, \triangle DPA$ from their common vertex $P$, hence, they are equal to half the sides $AB, BC, CD, DA$. As a result, the semiperimeter $s$ of the cyclic quadrilateral $ABCD$ is equal to

$s = \frac{AB + BC + CD + DA}{2} = PK + PL + PM + PN$

When the line $OP$ coincides with the diagonal $AC$, the diagonal intersection $P$ is the midpoint of the other diagonal $BD$ and the cyclic quadrilateral $ABCD$ is a kite with the right angles $\angle B = \angle C = 90^o$. From the right angle triangles $\triangle APB, \triangle BPC$

$DA = AB = \sqrt{AP^2 + BP^2} = \sqrt{AP^2 + AP \cdot CP} =$

$= \sqrt{(R + p)^2 + (R + p)(R - p)} = \sqrt{2R(R + p)}$

$CD = BC = \sqrt{CP^2 + BP^2} = \sqrt{CP^2 + AP \cdot CP} =$

$= \sqrt{(R - p)^2 + (R + p)(R - p)} = \sqrt{2R(R - p)}$

and the kite semiperimeter $s_k$ is

$s_k = AB + BC = \sqrt{2R(R + p)} + \sqrt{2R(R - p)}$

When the line $OP$ bisects the right angle $\angle APB$, the cyclic quadrilateral $ABCD$ is an isosceles trapezoid with the midline $LN$. The diagonals $KM \perp LN$ of the rectangle $KLMN$ are perpendicular to each other, hence, it is a square. Both the diagonal intersection $P$ and the cirrcumcenter $O$ of the isosceles trapezoid $ABCD$ lie on the square diagonal $KM$. Hence, the quadrilateral $LONP$ has perpendicular diagonals and it is a rhombus.  From the right angle triangle $\triangle LPQ$

$BC = 2LP = 2\sqrt{LQ^2 + PQ^2} = \sqrt{4r^2 + p^2} = \sqrt{2R^2 - p^2 + p^2} = R \sqrt 2$

$\frac{AB + CD}{2} = LN = 2r = \sqrt{2R^2 - p^2}$

and the trapezoid semiperimeter $s_t$ is

$s_t = \frac{AB + CD}{2} + BC = R \sqrt 2 + \sqrt{2R^2 - p^2}$

We will show that the trapezoid semiperimeter is greater than the kite semiperimeter, $s_t > s_k$:

$(?)\ \ R \sqrt 2 + \sqrt{2R^2 - p^2} > \sqrt{2R(R + p)} + \sqrt{2R(R - p)}$

Squaring this inequality (with both sides being positive), we get

$(?)\ \ 2R^2 + 2R \sqrt 2 \sqrt{2R^2 - p^2} + 2R^2 - p^2 > 2R(R + p) + 4R \sqrt{R^2 - p^2} + 2R(R - p)$

$(?)\ \ 2R\sqrt 2 \sqrt{2R^2 - p^2} > 4R \sqrt{R^2 - p^2} + p^2$

Squaring this inequality again (with both sides being positive), we get

$(?)\ \ 8R^2 (2R^2 - p^2) > 16R^2 (R^2 - p^2) + 8Rp^2 \sqrt{R^2 - p^2} + p^4$

$(?)\ \ 8R^2 - p^2 > 8R \sqrt{R^2 - p^2}$

Squaring this inequality again (with both sides being positive), we get

$(?)\ \ 64R^4 - 16R^2p^2 + p^4 > 64R^2 (R^2 - p^2)$

$(?)\ \ 48R^2 + p^2 > 0$

which is most certainly true.

Let $U, V$ be the midpoints of the diagonals $AC, BD$ and $0 \le \phi = \angle APO \le 45^o$ be the angle between the line $OP$ and the diagonal $AC$. The lines $OU, OV$ are the perpendicular bisectors of the diagonals $AC, BD$. From the isosceles triangles $\triangle AOC, \triangle BOD$, the diagonal lengths $AC, BD$ are

$AC = 2\sqrt{OA^2 - OU^2} = 2\sqrt{OA^2 - OP^2 \sin^2{\phi}} = 2\sqrt{R^2 - p^2 \sin^2{\phi}}$

$BD = 2\sqrt{OB^2 - OV^2} = 2\sqrt{OB^2 - OP^2 \cos^2{\phi}} = 2\sqrt{R^2 - p^2 \cos^2{\phi}}$

The angle $2\theta = \angle NQK$ formed by the rectangle diagonals $KM, LN$ is given by

$\tan \theta = \frac{NK}{KL} = \frac{BD}{AC} = \sqrt{\frac{R^2 - p^2 \cos^2{\phi}}{R^2 - p^2 \sin^2{\phi}}}$

The area $S_{ABCD}$ of the cyclic quadrilateral $ABCD$ is an increasing function of the angle $\phi \in (0, 45^o)$, simply because it is equal to twice the area of the rectangle $KLMN$. The angle $2\theta = \angle AQB$ is continuously increasing from $\tan \theta = \frac{\sqrt{R^2 - p^2}}{R} < 1, 2\theta < 90^o$ at $\phi = 0$ to $\tan \theta = 1, 2\theta = 90^o$ at $\phi = 45^o$, when the rectangle diagonals become perpendicular to each other and the rectangle becomes a square, having the maximum area. Formally,

$\frac{\text d\theta}{\text d\phi} = \frac{\text d}{\text d\phi} (\arctan{\sqrt{\frac{R^2 - p^2 \cos^2{\phi}}{R^2 - p^2 \sin^2{\phi}}}}) =$

$\frac{R^2 - p^2 \sin^2{\phi}}{2R^2 - p^2} \cdot \frac{2p^2 \sin \phi \cos \phi (R^2 - p^2 \sin^2{\phi}) + 2p^2 \sin \phi \cos \phi (R^2 - p^2 \cos^2{\phi})}{(R^2 - p^2 \sin^2{\phi})^2} =$

$\frac{2p^2 \sin \phi \cos \phi}{R^2 - p^2 \sin^2{\phi}} > 0$ [color=white].[/color] in [color=white].[/color] $(0, 45^o)$

because $\sin \phi > 0, \cos \phi > 0$ and $R > p$. The semiperimeter $s$ is an increasing function iff its square $s^2$ is an increasing function.

$s^2 = (PK + PL + PM + PN)^2 = PK^2 + PL^2 + PM^2 + PN^2\ +$

$+ 2(PK \cdot PL + PK \cdot PM + PK \cdot PN + PL \cdot PM + PL \cdot PN +  PM \cdot PN)$

Using the parallelogram rule for the parallelograms $KOMP, LONP$, we have

$PK^2 + PM^2 = \frac{KM^2 + PO^2}{2} = \frac{4r^2 + p^2}{2}$

$PL^2 + PN^2 = \frac{LN^2 + PO^2}{2} = \frac{4r^2 + p^2}{2}$

$PK^2 + PM^2 = PL^2 + PN^2 = \frac{2R^2 - p^2 + p^2}{2} = R^2$

Using Ptolemy's theorem for the cyclic quadrilateral $ABCD$ and the fact that it has poerpendicular diagonals,

$PK \cdot PM + PL \cdot PN = \frac{AB \cdot CD + BC \cdot DA }{4} = \frac{AC \cdot BD}{4} = \frac{S_{ABCD}}{2}$

$s^2 = 2R^2 + S_{ABCD} + 2(PK \cdot PL + PK \cdot PN + PL \cdot PM + PM \cdot PN)$

Since the circumradius $R$ is constant and the area $S_{ABCD}$ is increasing, it is sufficient to show that the remaining terms are also increasing:

$PK \cdot PL + PK \cdot PN + PL \cdot PM + PM \cdot PN = (PK + PM)(PL + PN)$

or that their square is increasing:

$(PK + PM)^2 (PL + PN)^2 =$

$= (PK^2 + 2PK \cdot PM + PM^2) (PL^2 + 2PL \cdot PN + PN^2) =$

$= (R^2 + 2PK \cdot PM) (R^2 + 2PL \cdot PN) =$

$= R^4 + 2R^2 (PK \cdot PM + 2PL \cdot PN) + 4PK \cdot PL \cdot PM \cdot PN =$

$= R^2 (R^2 + S_{ABCD}) + \frac{AB \cdot BC \cdot CD \cdot DA}{4}$

Again, it is sufficient to show that the remaining term is increasing (or constant). But this is exactly the case for any set of cyclic quadrilaterals with fixed circumcircle $(O)$, fixed diagonal intersection $P$ and fixed angle between the diagonals $AC, BD$, not necessarily with perpendicular diagonals !!! :cool: :cool: :cool: We need only the case of perpendicular diagonals. Denote

$\alpha_1 = \angle PAD = \angle PBC,\ \ \alpha_2 = \angle PAB = \angle PDC$

$\gamma_1 = \angle PCB = \angle PDA = 90^o - \alpha_1,\ \ \gamma_2 = \angle PCD = \angle PBA = 90^o - \alpha_2$

From the right angle triangles $\triangle APB, \triangle BPC, \triangle CPD, \triangle DPA$,

$AB = \frac{PA}{\sin{\gamma_2}} = \frac{PB}{\sin{\alpha_2}},\ \ \ BC = \frac{PB}{\sin{\gamma_1}} = \frac{PC}{\sin{\alpha_1}}$

$CD = \frac{PC}{\sin{\alpha_2}} = \frac{PD}{\sin{\gamma_2}},\ \ \ DA = \frac{PD}{\sin{\alpha_1}} = \frac{PA}{\sin{\gamma_1}}$

Using the power of the point $P$ to the circle $(O)$, i.e., $PA \cdot PC = PB \cdot PD = (R + p)(R - p) = R^2 - p^2$, we have

$AB \cdot BC \cdot CD \cdot DA = \frac{PA^2 \cdot PC^2}{\sin{\alpha_1} \sin{\alpha_2} \sin{\gamma_1} \sin{\gamma_2}} =$

$= \frac{4(R^2 - p^2)^2}{\sin{\alpha_1} \sin{\alpha_2} \cos{\alpha_1} \cos{\alpha_2}} = \frac{4(R^2 - p^2)^2}{\sin{2\alpha_1} \sin{2\alpha_2}}$

The numerator is constant, hence, we only have to show that the denominator is also constant.

${\sin{2\alpha_1} \sin{2\alpha_2}} = \frac 1 2 [\cos{2(\alpha_1 - \alpha_2)} - \cos{2(\alpha_1 + \alpha_2)}]$

Because of the rotation of the diagonal $AC$ counter-clockwise from the position coincident with the line $OP$ by an angle $0 < \phi < 45^o$, we have $\alpha_1 > \alpha_2$. Obviously,

$\alpha_1 + \alpha_2 = \angle PAD + \angle PAD = \angle A$

$\alpha_1 - \alpha_2 = \angle PBC - (90^o - \angle PBA) = \angle B - 90^o$

${\sin{2\alpha_1} \sin{2\alpha_2}} = \frac 1 2 [\cos{(2\widehat B - 180^o)} - \cos{2\widehat A}] = -\frac 1 2 (\cos{2\widehat B} + \cos{2\widehat A}) =$

$= -\frac 1 2 (\cos^2{\widehat A} - \sin^2{\widehat A} + \cos^2{\widehat B} - \sin^2{\widehat B}) = \sin^2{\widehat A} + \sin^2{\widehat B} - 1 =$

$= \frac{BD^2}{4R^2} + \frac{AC^2}{4R^2} - 1 = \frac{4(R^2 - p^2 \cos^2{\phi})}{4R^2} + \frac{4(R^2 - p^2 \sin^2{\phi})}{4R^2} - 1 =$

$= \frac{2R^2 - p^2 - R^2}{R^2} = \frac{R^2 -p^2}{R^2}$

$AB \cdot BC \cdot CD \cdot DA = \frac{4R^2(R^2 - p^2)^2}{R^2 - p^2} = 4R^2(R^2 - p^2)$

where we used the extended sine theorem for the sides $AC, BD$ of the triangles $\triangle ACD, \triangle BDA$ with the common circumcircle $(O)$ and the formulas for the diagonals $AC, BD$ as functions of the rotation angle $\phi$ obtained previously. The expression is obviously constant and this concludes the proof of the problem proposition. Was not easy.
:spider:
[b]Note:[/b] The proof that the product $AB \cdot BC \cdot CD \cdot DA$ is invariant under the rotation of the diagonals forming a fixed angle $\psi = \angle APB \neq 90^o$ is virtually the same as above, only it is necessary to use the sine theorem for the triangles $\triangle APB, \triangle BPC, \triangle CPD, \triangle DPA$ and the relation $\alpha_1 - \alpha_2 = \angle B - (180^o - \psi)$. The result is

$AB \cdot BC \cdot CD \cdot DA =  4R^2(R^2 - p^2) \sin^2{\psi}$

Congratulate ! A nice proof (although it's long and complex )  , Yetti !
  I also have a solution . I'll post it soon  :)

i have sloved this problem but with a long way thatr i don't posted
i'll happy nttu if you give a sort solution ;)


The upper portion of an inclined plane of inclination $ \alpha$ is smooth and the lower portion is rough. A particle slides down from rest from the top and just comes to rest at the foot. If the ratio of smooth length to rough length is $ m: n$, find the coefficient of friction.

You might as well assume the smooth portion has length $ m$ and rough length $ n$, so the incline has length $ m\plus{}n$.  Now you can figure out the height of the incline and you can figure out how fast the particle is traveling when it comes to the rough part.  Then you can figure out how high $ \mu$ must be in order to send the velocity to 0.

Let the lengths be $ mx$ and $ nx$.

By conservation of energy:

$ Mgh \equal{} nx\mu Mg \cos\alpha \\
 \\
\mu\equal{}\frac {(m \plus{} n)\not{x}\sin\alpha}{n\not{x}\cos\alpha} \equal{} \left(1\plus{}\frac{m}{n}\right)\tan\alpha$

Or we can go by JRav's wish!

$ v^2 \equal{} 2gmx\sin\alpha \\
 \\
a \equal{} \mu g \cos\alpha \minus{} g\sin\alpha \\
 \\
v^2 \equal{} 2as \\
 \\
\therefore 2gmx\sin\alpha \equal{} 2(\mu g \cos\alpha \minus{} g\sin\alpha)nx \\
 \\
\therefore \mu \equal{} \left(\frac {m \plus{} n}{n}\right)\tan\alpha$


function
trigonometry
integration
limit
real analysis
real analysis unsolved

Prove that a function \[ f(x)\equal{}\sum_{n\equal{}1}^{\infty}\frac{\sin(nx)}{\sqrt{n}}\]
is continuous at $ x\equal{}0\,.$
This problem is interesting because the series doesnt converge uniformly on
any interval containing zero.

[quote="shasha"]Prove that a function
\[ f(x) \equal{} \sum_{n \equal{} 1}^{\infty}\frac {\sin(nx)}{\sqrt {n}}
\]
is continuous at $ x \equal{} 0\,.$
This problem is interesting because the series doesnt converge uniformly on
any interval containing zero.[/quote]
yes this problem is really interesing. I will post later my solution. I think it's true if we wrote :
$ \sum \frac{\sin nx}{(n)^{\alpha}}$
where $ 0< \alpha <1$

It is just too good to be true: if the function were continuous at $ 0$, it would be continuous everywhere (the uniform convergence near every point except $ 2\pi k$ can be easly derived from the Abel-Dirichlet test). But then the series in the question would be its Fourier series. On the other hand, the sum of the squares of the coefficients is infinite, which makes it impossible. There is a weak point in this argument, of course: not every trigonometric series converging everywhere is a Fourier series of its sum. But here we are lucky to have an integrable majorant $ 1/\sqrt{|x|}$ near $ 0$, which puts everything in place. So, the statement is false.

What if we consider $ \sum_{n \equal{} 1}^{\infty}\frac {\sin nx}{n^{\alpha}}$ with $ \frac12<\alpha<1$?

I did some numeric computations for the original series (mostly because I'm learning MATLAB now) and they were not convincing one way or the other. This is the partial sum up to $ N \equal{} 10^7$ on the interval $ [10^{ \minus{} 8},2\cdot 10^{ \minus{} 6}]$ (the horizontal axis is rescaled). The maximum is of order $ N^{ \minus{} 1/2}$.  :|
[img]http://www.math.tamu.edu/~lkovalev/files/n1e7_dx1e-8.jpg[/img]

This is kind of ugly... Let $ f(x) \equal{} \sum_{n \equal{} 1}^{\infty}\frac {\sin nx}{n^{\alpha}}$, where $ 0 < \alpha < 1$. Then \[ F(x): \equal{} \int_0^x f(t)dt \equal{} \sum_{n \equal{} 1}^{\infty}\frac {1 \minus{} \cos nx}{n^{1 \plus{} \alpha}}\] by the dominated convergence theorem (domination by $ 1/x^{\alpha}$). The difference
\[ \sum_{n \equal{} 1}^{\infty}\frac {1 \minus{} \cos nx}{n^{1 \plus{} \alpha}}
\minus{}\int_{1}^{\infty}\frac {1 \minus{} \cos tx}{t^{1 \plus{} \alpha}}dt
\equal{}\sum_{n \equal{} 1}^{\infty}\left(\frac {1 \minus{} \cos nx}{n^{1 \plus{} \alpha}}
\minus{}\int_{n}^{n\plus{}1}\frac {1 \minus{} \cos tx}{t^{1 \plus{} \alpha}}dt\right)\]
is $ O(x)$ as $ x\to 0$, because
\[ \left|\frac {\partial}{\partial t}\left(\frac {1 \minus{} \cos tx}{t^{1 \plus{} \alpha}}\right)\right|\equal{}
\left|\frac {x\sin tx}{t^{1 \plus{} \alpha}}\minus{}(1\plus{}\alpha)\frac {1 \minus{} \cos tx}{t^{2 \plus{} \alpha}}\right| \le C\frac{x}{t^{1 \plus{} \alpha}}\]
With a change of variables we find
\[ \lim_{x\to 0}\frac {1}{x^{\alpha}}\int_{1}^{\infty}\frac {1 \minus{} \cos tx}{t^{1 \plus{} \alpha}}dt \equal{} \int_0^{\infty}\frac {1 \minus{} \cos u}{u^{1 \plus{} \alpha}}du \equal{} : C(\alpha)
\]
where $ C(\alpha) > 0$. Therefore, $ F(x) \equal{} C(\alpha)x^{\alpha} \plus{} O(x)$ as $ x\to 0$. In particular, $ F\not\in C^1$.

Does this actually work?  :maybe:

[quote="shasha"]Extremal, please, can you post your solution?
Thanks![/quote]
I'll post solution for $ \alpha \equal{}\frac{1}{2}$ and try to consider case when $ 0< \alpha < \frac{1}{2}$ after my exams, because now I preparing to my  exams.

fedja is right. Who think that serie is diferentiable please read again fedja's post. it.s not diferentiable for all $ 0<\alpha \leq 1$


calculus
integration
trigonometry
calculus computations

[hide="Hint"]If $ x$ is a real number, then $ \cos (e^{ix}) \equal{} \cos (\cos x) \cosh (\sin x) \minus{} i \sin (\cos x) \sinh (\sin x)$.[/hide]


Six distinct segments are drawn from one vertex of a triangle to the opposite side. What is the total number of triangles contained in the resulting figure?
[asy]draw((0,0)--(7,0)); for(int i = 0; i <8; ++i){draw((2.5,3.5)--(i,0));}[/asy]

Any two distinct points on the base will determine a unique triangle, so our answer is $ \dbinom82\equal{}\boxed{28}$.

Couldnt we also notice that there are 7 sie "1" triangles, and then 6 made of 2 size 1 triangles and so on so 7+....+2+1=8x7/2=28.

This can be considered one of those think about it problems. You know that the base of the big triangle is going to be the base of the smaller triangles also, so you only need to worry about the other two sides. The other two sides can be chosen from the eight lines. So the number of triangles is the number of ways you can choose two of those eight lines. math154 calculated it already.


Prove that there does not exist a non-decreasing function $ f$ from $ \Re\rightarrow\Re$ (ie $ x\le y\Rightarrow f(x)\le f(y)$) such that $ f(\Re) \equal{} \Re\setminus\{0\}$, where $ f(\Re) \equal{} \{f(x): x\in\Re\}$.

Can someone please help. Thank you very much. :P

Define $ W\equal{}\{w\in\mathbb{R}\,|\,f(w)<0\}$ and $ U\equal{}\{u\in\mathbb{R}\,|\,f(u)>0\}$, then we have:

(1) $ U\cup W\equal{}\mathbb{R}$
(2) $ U\cap W\equal{}\emptyset$
(3) $ u\in U, w\in W\Longrightarrow w<u$

(3) yields $ \sup W\leq \inf U$ and (1),(2) together with $ f(\mathbb{R})\equal{}\mathbb{R}^*$ then harden this to $ z: \equal{}\sup W\equal{} \inf U$. Using (1) we can assume wlog $ z\in W\setminus U$, i.e. $ f(z)<0$ and for all $ w\in W,u\in U$ we have $ f(w)\leq f(z)<0<f(u)$ (remark: the case $ z\in U\setminus W$ is done similarly)

Applying this we get the contradiction $ ]f(z),0[\not\subseteq f(\mathbb{R})\equal{}\mathbb{R}^*$


summer program
PROMYS
Ross Mathematics Program
number theory

I've never been to PROMYS, though I was at Ross for two summers, so I will simply comment about the Ross experience.

1. There is air conditioning. As far as I know, rooms are two room doubles, with desks in one room and beds in the other. Rooms have a refrigerator and a microwave, as well as a bathroom adjacent (that is not cleaned for you!). Dorms have lounges (good for talking about math!) and study rooms (also good for math!) and laundry rooms inside (could be good for math).
2. All problem sets are paper-and-pencil (no calculators necessary). I don't think computers are typically allowed for first-year students.
3. People definitely brought musical instruments to Ross. I believe the music building is open for practice in, or else people just practiced in their rooms or common areas. The focus of the program is to do math though, and you will not get as much out of it if you go there to pursue music.
4. The focus of the program tends to be problem sets, and you'll probably find that they take up a lot of your time. Many students have not seen number theory before going to Ross though. I suppose you could work on your number theory project in your own time, and I'm pretty confident that at least some of the counselors or teachers would help you on that if you had questions.
5. No clue. Learning structure at Ross is based on the problem sets, with an hour of lecture a day and sections 3 days a week. The emphasis of the teaching will probably be different from anything else you have seen before. I think the focus is to teach people how to be mathematicians, not just to teach people how to solve problems (though that is one aspect).
6. Ross is better :P . Honestly, I don't know what the differences are, other than Ross is longer by 2 weeks. PROMYS was started by Ross alumni, so there are probably some similarities.

[quote="buzzer11"]I am deciding between these two programs for this summer, and I have a few questions:

1. How are the living conditions in the programs? (A/C, sharing rooms, etc.)
2. Are the programs "paper-and-pencil" oriented, or "computer-usage" oriented?
3. I play the violin, and it would be nice to have time and rooms in which I can practice... I am very avid (my music pursuits surpass my mathematical pursuits), so this is very important to me. Which program would be more accommodating for this?
4. I also want to do some work on a number-theory-based research project I have been working on. At which camp do we have more opportunity to work on projects from home?
5. How are the camps different in what they teach, the learning structure, etc.?
6. What are some other important differences?

Thanks a lot in advance, I know theres a lot of questions up there... :)[/quote]
do you already know enough NT to do research?

Yes, I guess I do know enough to do research... But I'm not sure how much knowledge is "enough". :/


Three points are chosen randomly and independently on a circle. What is the probability that the triangle with vertices at these three points contains in its interior the center of the circle?

All acute triangles on the circle satisfy this (do you see why?).

Let the angles of the triangle be $ x,y,180\minus{}x\minus{}y$ such that $ 0<x,y$, $ x\plus{}y<180$ (requirements of a triangle), and the area formed be the total possible region.

Then the "successful region" is the area inside the aforementioned region that satisfies $ x,y<90$ and $ 180\minus{}x\minus{}y<90\implies x\plus{}y>90$ as well.

We quickly see that the probability is $ \boxed{\frac14}$.


limit
number theory unsolved
number theory

For any positive integer $ k$, let $ S(k)$ be the sum of its digits. Prove that

$ \lim_{n\rightarrow\infty}S(2^n)\equal{}\infty$

If it is not, then exist $ x_0,x_1,x_2,..,x_k,0<x_i<10$ suth that there are infinetely many n:
$ 2^n\equal{}x_0*1\plus{}x_1*10^{n_1}\plus{}...x_k*10^{n_k},0<n_1<...<n_k.$
We get $ 2^{n_1}|x_0,...,2^{n_i}|x_0\plus{}...\plus{}x_{i\minus{}1}*10^{n_{i\minus{}1}}$, therefore $ n_i<4n_{i\minus{}1}$ - contradition.

[quote="hollandman"][quote="Rust"] therefore $ n_i < 4n_{i \minus{} 1}$ - contradition.[/quote]

Why is this?  :)[/quote]
$ 2^{n_i}|x_i*10^{n_i}\plus{}...\plus{}x_k*10^{n_k}\to 2^{n_i}|y_i\equal{}x_0\plus{}...\plus{}x_{i\minus{}1}*10^{n_{i\minus{}1}}\to y_i\ge 2^{n_i}\to n_i<4n_{i\minus{}1}\plus{}4$.


function
absolute value
topology
real analysis
real analysis unsolved

Prove that there is no distance $d$ on the set of real functions defined on $R$ such that for any sequence of functions $f_n$, this sequence converges pointwise to a function $f$ if and only if $d(f_n,f)$ converges to 0.

In order to prove that a topological space is not metrizable, it suffices to show that there is a set $A$ and a point $x\in\overline A$ in the closure of that set, such that no sequence $(x_n)_n\subset A$ in that set converges to $x$.

Take $x=0_{\mathbb R}$, and take $A$ to be the set of functions which take the value $0$ in finitely many places, and the value $1$ everywhere else. Clearly, $x\in\overline A$. Now, if $(x_n)_n\subset A$, then choose a real $\alpha$ such that $x_n(\alpha)=1,\ \forall n$, and consider the neighborhood of $x$ consisting of all those functions which have absolute value $<1$ in $\alpha$. This open set contains none of the points $x_n$, so the sequence $(x_n)$ cannot converge to $x$.


A^3+B^3=C^3+D^3

A,B,C, and D are all distinct, positive integers.

Minimize the product ABCD.

I know the numbers, just cause there's a good story behind this problem.  They are [hide]1 and 12, 10 and 9[/hide].  But I don't know how to get them.

That'd be "minimize."  I don't think that there is a nice procedure for solving that particular diophantie equation in general, although I might be mistaken.

Well, the ordered quadruple that is needed contains fairly low positive integers anyways, so a few attempts at trial-and-error should yield the answer..I think.

[quote="Fierytycoon"][quote="blahblahblah"]Does the story involve Ramanujan?[/quote]

And Hardy and a taxi plate, I believe.[/quote]

Its the kind of story that screams apocryphal, but with him, I'm guessing it's not.


$x_1=0, x_n=1$
$x_1<x_2<x_3<...<x_n;$
$\forall i\in [2;n-1] \exists j,k:$
$0<j<i<k\leq n$ and $x_i = \frac{x_j+x_k}2$
is it possible to $\exists x_i \in R\backslash Q$ ???

No, it's not :).

[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=fantastic+algebra&t=4845[/url]

Post #6 contains a solution.


Sorry, that sounds off-topic,

What's TST? I know the word stands for Team Selection Test, but I have never seen TST of Japan Mathematical Olympiad. Why?

Japanese Communities Moderator

kunny

Each country has a way to select their teams.
In Brazil, for example, we take 4 tests and 4 or 3 training lists.

Thank you for your reply.

Maybe Japanese MO's TST is Board Training in Spring for 4 days.

How do you get TST problem in your country?

In each state, there is a coordinator who applies the tests.


calculus
integration
function
calculus computations

Use triple integral to find the volume of the given solid
17) The solid bounded by the cylinder $x=y^2$ and the planes $z=0$ and $x+z=1$.

I came up with $0\leq z\leq$ and then $y^2\leq x\leq1$ and therefore $-1\leq y\leq1$
Therefore: $\int_{-1}^{1}\int_{y^2}^{1}\int_{0}^{1-x}dzdxdy=-\frac{7}{15}$
The book gives: $\int_{0}^{1}\int_{-\sqrt{x}}^{\sqrt{x}}\int_{0}^{1-x}dzdydx=\frac{8}{15}$

What did I do wrong?

Those are the same regions, but somehow you've integrated a positive function over a positively oriented region and gotten something negative. Your mistake is in the calculation, not the setup.

Okay, I was so confident I had the right setup. Glad to know I did. Will go over my calculations to find the error. Thanks!


function
number theory proposed
number theory

Find the least amount of elements of set $ A $ satisfying there exists a function
$ f:N \to A $ where $ f(i) \neq f(j) $ if $ |i-j| \in P $
  $ i)P $ is the prime set.
  $ ii) N = \lbrace {1,2,... }\rbrace $

4 suffices: f(a) = a mod 4.
And 4 is necessary, as f(1), f(3), f(6) and f(8) are all distinct.


conics
hyperbola
calculus
calculus computations

one particle is moving at constant speed and another at constant acceleration
prove that the line joining them always touches a fixed hyperbola

actually the particles are moving parallel to each other


show that the closed interval $[0,1]$ cannot be written as a countably infinite disjoint union of closed sub-intervals of [0,1]. sorry if it has been posed already.

@zhaobin- it has been corrected. thanks for pointing out

$[0,1]=[0,0.5)\cup[0.5,1]$
I think you may forget some condition :wink:

Edit:
kouboy,But I am sorry :D  I think it is still wrong even if you add "infinite"
Let $A_{n}=(\frac 1{2(n+1)}+\frac{1}{2},\frac{1}{n}+\frac{1}{2}]$
Then $[0,1]=(\cup_{n=1}^{\infty}A_{n})\cup [0,\frac{1}{2}]$

Does the every interval should be closed?

yes, each sub-interval must be closed. so your thing doesnt hold.

One of your subintervals must contain $0$, say $I=[0,a_{1}]$, where $a_{1}<1$. Similarly, one of them must contain $1$, say $J=[a_{2},1]$, where $a_{2}>a_{1}$, since the intervals shall not overlap. Now take a point $x \in ]a_{1},a_{2}[$. $x$ must be contained in a certain interval $K=[a_{3},a_{4}]$, where $a_{1}<a_{3}\leq a_{4}<a_{2}$. Clearly enough, the set of those intervals created by repeating indefinitely the above process in-between the previously created intervals is idempotent to $\mathbb{R}$.

There's a proof in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=62703]this thread[/url] from last year. The Baire category theorem is necessary to make it rigorous.

here is my proof :) :
For the contrary:$[0,1]=\cup_{n=1}^{\infty}[a_{n},b_{n}]$
Consider the set $A={x: x=a_{n},or,x=b_{n},but,x\neq 0,x\neq 1}$
First prove tthat every point in $A$ is a limit point of $A$.Then we have $A$ is a $perfect$ set.
Second,obviously $A$ is nonempty,then by the well-known proposition that every perfect is uncountable,we get a contradiction.


[size=150][b]A is a given point and P is any point on a given straight line. If AQ=AP and AQ makes a constant angle with AP, find the locus of Q.[/b][/size]

Looks to me like the specifications describe a semi-circle, where the constant angle is a right angle.  Q must thus lie on this specific straight line...

Isn't it just the line rotated the constant angle about A?


The problem is :
Find all natural numbers a , b , c such that  a 2 +b 2 =abc-1
Thanks.

The problem is equivalent to :

Find all pairs of positive integers (a, b) such that

(a + b + 1)/(ab)

is a positive integer.

This problem was posted and solved somewhere on this forum.
I remember that the only possibility for c is c = 3 ...

Lets try to see the problem in another way:

a) Prove that if there exist a triplet of natural numbers (a,b,c) with the property that a^2+b^2=abc-1 then z=3.
b) compute all the triplets with that property ( a^2+b^2=abc-1 ).

[b]Solution:[/b]

[b]a)[/b] Let (a,b,c) be a solution with c<>3. Then a<>b, because otherwise we would get a^2(c-2)=1 impossible because (c-2)<>1. So we get that: 0=a^2+b^2+1-abc = (a-bc)^2+b^2+1+abc-b^2c^2 = (bc-a)^2+b^2+1- (bc-a)bc. So (bc-a,b,c) is also a solution in natural numbers, because a(bc-a)=abc-a^2=b^2+1>0 which gives us bc-a>0. We see that if a>b then a^2>b^2+1=a(bc-a). So a>bc-a, which shows us that the solution is smaller than the original one, because a+b>(bc-a)+b. So, using that a<>b we can continue doing this which will lead us to an infine sequence which is not possible according to the descendece of Fermat. So the equation has no solutions if c<>3.

[b]b)[/b] We can see that (1,1) is a solution of the equation a^2+b^2=3ab-1 (because if c<>3 we don't have a solution )

let (a1,b1) , a1>b1 be another solution. Then b1^2+(3b1^2-a1)^2+1=3b1(3b1-a) so also (b1,3b1-a1) is also a solution of the equation. from (a1-b1)(a1-2b1)=a1^2-3a1*b1+2b1^2=b1^2-1>0 we get that a1>2b1 so 3b1-a1<b1. So the new solution is smaller than the original one(from the one we started). If the ineqs a1>b1, 3b1-a1<b1 would be strict, and doing the same thing we would get an infinet sequence of descending natural numbers, also a contradiction from Fermat.

So all the solutions are obtained from (x1,y1)=(1,1) and given by the recurences: (x_(n+1),y_(n+1))=(y_n, 3y_n-x_n). The sequences x_n and y_n verifie the following relation:
z_(n+1)=3z_n-z_(n-1), z1=1, z2=2. So this recurence has the solutions the odd numbers of the Fibonacci sequence!!! So: (x_n,y_n)=(F_(2n+1), F_(2n-1)). for all n natural.

SO the solutions of the equation are: (1,1,3); (F_(2n+1),F_(2n-1),3) and (F_(2n-1),F_(2n+1),3) for all n natural!!!!!!!!!!!!!!!!

cheers! :D :D :D :D :D


trigonometry
algebra
polynomial
algebra proposed

Prove that for $n\geq 2\in\mathbb{N}$

$\prod^{n-1}_{k=1}\sin\frac{\pi k}{n}=\frac{n}{2^{n-1}}$

[quote="Rust"]Let f(x)=$\frac{x^{n}-1}{x-1}=x^{n-1}+x^{n-2}+...+1$
We have \[n=f(1)=\prod_{k=1}^{n-1}(1-exp(\frac{2\pi iK}{n}))=2^{n-1}\prod_{k=1}^{n-1}(sin \frac{\pi ik}{n}-icos{\frac{\pi ik}{n})sin\frac{\pi k}{n}}\] .
It give \[\prod_{k=1}^{n-1}\sin\frac{\pi k}{n}=\frac{n}{2^{n-1}}\] f(-1) give multiple of cos.
.[/quote]

other than a small typo which should be $\sin \frac{\pi k}{n}-i\cos{\frac{\pi k}{n}}$ , everything seems to be very nice  :)

$|\sin \frac{\pi k}{n}-i\cos{\frac{\pi k}{n}|=1}$ , and $\prod_{k}\sin\frac{\pi k}{n}>0$, therefore $\prod_{k}\sin{\frac{\pi k}{n}}=\frac{n}{2^{n-1}}$.

[quote="me@home"]Sorry, I don't understand Rust's proof... where did he get the complex numbers from? Is he factoring the polynomial into roots?[/quote]

yes,$\omega^{n}=1$ is the root of unity then we can write

$f(x)=(x-\omega)(x-\omega^{2})\cdots (x-\omega^{n-1})$

Ohhh now I get it! thank you... this is an ingenious technique, I see...
$\omega^{n}=1$,
$f(x)=\frac{x^{n}-1}{x-1}=x^{n-1}+x^{n-2}+...+1$ by observation has roots $\omega^{n}(n \in{1,n-1})$
And degree of this is clearly n-1, and we have found n-1 roots so that is all;
therefore $f(x)=x^{n-1}+x^{n-2}+...+1=(x-\omega)(x-\omega^{2})\cdots (x-\omega^{n-1})$

Sorry I still don't follow Rust's solution. Could someone elaborate on the details?

It seems to me that Rust is using this is in his proof:
[quote="Rust"] \[n=f(1)=\prod_{k=1}^{n-1}(1- e^{\frac{2\pi ik}{n}})=2^{n-1}\prod_{k=1}^{n-1}[ (\sin \frac{\pi k}{n}-i\cos{\frac{\pi k}{n})\sin\frac{\pi k}{n}]}\] [/quote]
I.e: \[2(\sin \frac{\pi k}{n}-i\cos{\frac{\pi k}{n})\sin\frac{\pi k}{n}=1-e^{\frac{2\pi ik}{n}}}\]  \[\leftrightarrow 2\sin^{2}\frac{\pi k}{n}-1-2i \cos{\frac{\pi k}{n}\sin\frac{\pi k}{n}=-e^{\frac{2\pi ik}{n}}}\] which is trivial :)

[hide=" :D "]
Edit:
Oh yes, we also must show that \[1=\prod_{k=1}^{n-1}(\sin \frac{\pi k}{n}-i\cos{\frac{\pi k}{n})}\] Evaluate product backwards (i.e. substitue $k\mapsto n-k$) gives: \[\prod_{k=1}^{n-1}(\cos \frac{\pi k}{n}+i\sin{\frac{\pi k}{n})=\prod_{k=1}^{n-1}(\sin \frac{\pi k}{n}-i\cos{\frac{\pi k}{n})}}\] But this is easy; again consider $P(x)$ as defined above, now \[P(0)=1=\pm \prod_{k=1}^{n-1}(\cos \frac{\pi k}{n}+i\sin{\frac{\pi k}{n})}\] (considering products of roots and values in alternate ways as above)
then we use the follow up post of Rust's where he shows it must be the positive value (i.e. $>0$)
and we're done! $\blacksquare$[/hide]

Stop reviving old topics with useless posts!


OK. I found this problem in "Do You Need This?" for AMC Counting class: A playoff series between two teams proceeds one game at a time until one team has won 5 games. What is the probability that the series last 9 games if each team is equally likely to win each game?

Here's what I did: [hide]To find all possible combinations for 8 games series with each 4 wins and 4 loses. Suppose the P for a combination of 8 games is 1/2^8. Now it can be figured out the possible combinations by considering the first occurance of the lose:

1. WWWWL.... 3C0=1

2. WWWL.......  4C1=4

3. WWL...........5C2= 10

4. WL..............6C3= 20

(1+4+10+20)2=70

Now let 70/2^8=70/256=35/128[/hide]

Here's the problem: when I tried to verify this on the Probability tree, the result was completely different. Isn't the probability=the chance of occurance of the event/ all possibilities? Then why are there two different Ps that satisfy this? Am I misreading or something? Please help!

How did you set up your probability tree?

Yes, that calculation is definitely correct (although you made it much more complicated then you needed to -- you could have just done C(8, 4) = 70 for the  number of 8-game sequences with 4 wins.)  So there is something wrong (probably) with your probability tree.


[quote="4everwise"]An equilateral triangle is circumscribed about a circle with radius 9. Find the area of the triangle.[/quote]

[hide]
Draw the medians (also the angle bisector) from A, B, and C in the triangle ABC circumscribed about the circle.  This creates 6 congruent 30-60-90 triangles.  The length between the 90 and the 60 angle is 3, so the area of that is $\frac{3\sqrt{3}}{2}$.  Six times that is $\boxed{9\sqrt{3}}$.
[/hide]


geometry
circumcircle
conics
parabola
perpendicular bisector
geometry proposed

Let ABC be a triangle, C(O) is the circumcircle and the points M', M" are on the side BC. Let C'(O') be the circle tangent to AM',
BM' (at D') and intouches the circumcircle C (at E') simultanously; let C"(O") be the circle tangent to AM", CM" (at D") and intouches
the circumcircle C (at E") simultanously. Prove that the bisectors of the angles <D'O'E', <D"O"E" and <O'OO" are concurrent. :)

Denote the bisector of the angle E’O’D’ and O’OO’’ meet at K, the bisector of the angle E’’O’’D’’ and O’OO’’ meet at K’
∠E’O’D’=2∠1   ∠E’’O’’D’’=2∠2
Then ∠O’OO’’=2∠1+2∠2    ∠OKO’=180-∠1-(180-∠1-∠2)= ∠2
The same way we acquire ∠OK’O’’=∠1
But ∠OKO’’=∠O’KO’’-∠O’KO=∠1+∠2-∠2=∠1=∠OK’O’’
So K=K’

<O'KO"=<1+<2. It's true, but how you prove this?

I am sorry , there is something wrong with my previous post
I will post my solution later

Denote A’  the midpoint of arc BC not contain A
Let ∠D’O’G=∠1   ∠D’’O’’G’=∠2
1.E’ G D’  and  E’’ G’ D’’  are collinear respectively
 See http://forumgeom.fau.edu/FG2003volume3/FG200325.pdf
2.D’,E’,E’’,D’’ are concyclic because A’B^2=A’C^2=A’D’*A’E’=A’D’’*A’E’’
(conclusion from thebault’s problem)
KG and KG’ are perpendicular bisector of E’D’ and E’’D’’ respectively.  So we acquire K is the circumcenter of quadrangle D’E’E’’D’’
3.K’  is the midpoint of E’E’’   =>  KK’ ⊥ E’E’’  but  OK’ ⊥ E’E’’
So K,O,K’ are collinear
4.∠O’KO’’=360-∠KGA’-∠KG’A’-∠D’A’D’’=360-90-90-(90-∠1+90-∠2)= ∠1+∠2
  ∠O’OO’’=∠E’O’D’+∠E’’O’’D’’=2∠1+2∠2
  ∠KHH’=∠E’HG=90-∠HE’G=90-∠HE’D’=90-∠E’’D’’C=90-∠2
   => ∠HKK’= ∠2   => ∠KOO’=180-∠1-∠2   => ∠O’OK=∠1+∠2=1/2∠O’OO’’
5.KK' is the  bisector of ∠O’OO’’
done!

Nice solution! Thank you!
       In my solution I use the next two properties:
1. Let ABC be a triangle, C(O) is the circumcircle and the point M is on the side BC. Let C'(O') be the circle tangent to AM, BC
    and intouches the circumcircle C simultanously. Then the point O' is on the parabola P with focus O which passes through
    the points B, C (the vertex of the parabola P and the point A are on the same side of the line BC).
2. Let P be a parabola with focus F and the points X, Y lie on P. Then the tangents to the parabola P at the points X, Y and the
    bisector of the angle <XFY are concurrent.

Hi everyone,
I think it is rather simple.
After we prove that D',E',E",D" are concylic, easy to see that the bisectors of angles D'OE',D"O"E",O'OO" is the perpendicular bisectors of segment D'E', D"E", E'E" and they are are concurrent at the circumcenter (D'E'E"D").


abstract algebra
group theory
vector
invariant
linear algebra
superior algebra
superior algebra unsolved

1. If $ G$ is a finitely generated abelian group such that $ G/G_t$ ($ G_t$ is the torsion subgroup) has rank n, and $ H$ is a subgroup of G such that $ H/H_t$ has rank m, then $ m \leq n$ and $ (G/H)/(G/H)_t$ has rank n - m.

2. How many subgroups of order $ p^2$ does the abelian groups $ \mathbf{Z}_{p^3} \times \mathbf{Z}_{p^2}$ have?

3. Let $ G$ be a finite abelian group and x an element of maximal order. Show that $ <x>$ is a direct summand of G.

Attempted solutions:

1. So $ G/G_t$ is the number of infinite cyclics in any decomposition of $ G$. So, basically I need to show that no decomposition of a subgroup can have more infinite cyclics than the original group. Pretty intuitive, but...

2. Here are the subgroups I found: $ < (np,1) >$ where n=0,1,2,...,p^2-1 and $ < np,m >$ n = 1,2,...,p-1 and m = 0,2,...,p^2-1. I suspect that is all.

3. I'm not exactly sure what they mean "$ <x>$ is a direct summand of G." Does that  just mean that I can find a way to write $ G$ as an internal direct sum where $ <x>$ is one of the summands?

That smells like applying the [url=http://en.wikipedia.org/wiki/Finitely_generated_abelian_group]structure theory[/url] of finitely generated abelian groups is pretty fruitful here.  :P

[quote="FibonacciFan"]1. If $ G$ is a finitely generated abelian group such that $ G/G_t$ ($ G_t$ is the torsion subgroup) has rank n, and $ H$ is a subgroup of G such that $ H/H_t$ has rank m, then $ m \leq n$ and $ (G/H)/(G/H)_t$ has rank n - m.[/quote]Tensoring the exact sequence of $ \mathbb Z$-modules \[ 0\rightarrow H\rightarrow G\rightarrow G/H\rightarrow 0\] with $ \mathbb Q$ yields an exact sequence of $ \mathbb Q$-vector spaces ("localisation preserves exactness"). Then use $ rk(G)\equal{}\dim_{\mathbb Q}G\otimes_{\mathbb Z}\mathbb Q$.

[quote]2. How many subgroups of order $ p^2$ does the abelian groups $ \mathbf{Z}_{p^3} \times \mathbf{Z}_{p^2}$ have?[/quote]What do the subgroups look like? To localise then in one or in both components of the given product tensor with the field $ \mathbb Z_p\equal{}\mathbb Z/p\mathbb Z$.

[quote]3. Let $ G$ be a finite abelian group and x an element of maximal order. Show that $ < x >$ is a direct summand of G.[/quote]This is a direct consequence of the [url=http://en.wikipedia.org/wiki/Finitely_generated_abelian_group#Invariant_factor_decomposition]invariant factor decomposition[/url].

[quote="Third Edition"]

Tensoring the exact sequence of $ \mathbb Z$-modules
\[ 0\rightarrow H\rightarrow G\rightarrow G/H\rightarrow 0
\]
with $ \mathbb Q$ yields an exact sequence of $ \mathbb Q$-vector spaces ("localisation preserves exactness"). Then use $ rk(G) \equal{} \dim_{\mathbb Q}G\otimes_{\mathbb Z}\mathbb Q$.
[/quote]

[quote="Third Edition"]
What do the subgroups look like? To localise then in one or in both components of the given product tensor with the field $ \mathbb Z_p \equal{} \mathbb Z/p\mathbb Z$. [/quote]

You're using language that I haven't learned yet: "exact sequence", " tensoring", " localisation". There must be a more elementary way.

 [quote="Third Edition"]This is a direct consequence of the [url=http://en.wikipedia.org/wiki/Finitely_generated_abelian_group#Invariant_factor_decomposition]invariant factor decomposition[/url].[/quote]

That's true but this exercise is actually used to obtain an alternative proof of that theorem.

Hmm, never mind. Let's call what I said mambojumbo.  :P  I think one of the basic tools you do need here is that the rank of a (finitely generated) abelian group G can be identified as the dimension of a vector space over the rationals derived from G. This allows you to use results in linear algebra to prove 1.

I agree that the other two parts could be solved by direct inspection since the given groups of order $ p^2$ are pretty simple to analyse.


On planet Zork there are some cities. For every city there is a city at the diametrically opposite point. Certain roads join the cities on Zork. If there is a road between cities $P$ and $Q$, then there is also a road between the cities $P'$ and $Q'$ diametrically opposite to $P$ and $Q$. In plus, the roads do not cross each other and for any two cities $P$ and $Q$ it is possible to travel from $P$ to $Q$.

The prices of Kriptonita in Urghs (the planetary currency) in two towns connected by a road differ by at most 100. Prove that there exist two diametrically opposite cities in which the prices of Kriptonita differ by at most 100 Urghs.

I knew it looked familiar! :)

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=automorphism&t=24585[/url]

The automorphism of order $2$ that we use here is $P\mapsto P'$. Besides, there's even a proof for a generalization there, by Leva.

Nice  :) I've done it in similar way but without refering to fleas  :lol:


integration function logarithms calculus derivative trigonometry algorithm

What about $\int x^xdx$, $\int\frac{dx}{x^x}$, $\int x^{\frac1x}dx$ or $\int\left(\frac1x\right)^{\frac1x}dx$?

the primitives, I believe, are not elementary functions...

No ideas at all?
Not necessarily elementary functions. Anything that you can tell about the primitives.

Expansion of power serie of each function and use theorem of itegration
for permuting $\int$ and $\sum $

Can you find the power serie of [tex]f(x)=x^x[/tex]? I can't. I can't find the general formula of [tex]f^{(n)}(x)[/tex].

Just write it as $e^{x\ln x}$ and then use the expansion of $e^x$, but I don't know if it helps..

The function $x^x$ doesn't have a bounded derivative as $x\to 0^+$. For that reason, it can't be represented as a power series centered at 0. Of course, it would be possible - but painful - to find power series centered at other points, such as 1. If you do what grobber suggests, you will have a series - just not a power series. Since you can (if you work hard enough) find expressions for the antiderivatives of $x^n (\ln x)^n$, you could write out the integral of $x^x$ as a rather unusual series.

Define "elementary function" as any function that can be obtained from powers of x, the exponential and the logarithm, the usual trig functions and the usual inverse trig functions, by combining them by addition, subtraction, multiplcation, division and composition. Any elementary function has an elementary derivative. (Proof by algorithm; component parts of proof include product rule and chain rule.) The converse is badly false; there are many elementary functions without elementary antiderivatives. If you can't find an elementary antiderivative of a particular elementary function, there are two possibilities:

1. You weren't clever enough, and you overlooked a method that works, or
2. There is no elementary antiderivative.

If the problem appears on the first midterm exam in your Calculus II course, #1 may be more likely, but in the "real world", #2 is far, far more likely.

Of course, asking whether there's an elementary antiderivative isn't that same thing as asking whether the antiderivative exists. Every continuous function has an antiderivative, guaranteed by the Fundamental Theorem of Calculus.

Since I am in my 1st course I am bound to make mistakes.
Talking of integrals of such functions,those integrals can only be carried within the disc of convergence.Finding those are not easy, i mean using Hamadard's theorem.For definite integrals this can be done by using varous approximation methods available.

In this context I remembered something
Find the nth derivative of e^(-x  2 )

I can feel that there will be some reduction or recursive type of formula but I want the result to involve only n,e,x.

the antiderivative of e^(-x^2) is computabel by using taylor expansions but not by reguar antiderivatives.

The value of $\int_0^1 x^x \ dx$ can be expressed as $1 - \frac{1}{2^2} + \frac{1}{3^3} - \frac{1}{4^4} + \frac{1}{5^5} - \cdots$, which is quite neat.

Hi adidasty using Taylor series expansion entered my mind.But as I was talking about in the last post these infinite series cannot be differentiated or integrated at will.These can be done only within the disc/radius of convergence.

some one from the site pointed out to use Hermite polynomials.

nth derivative of (e^(-x 2 ))=(-1)^n Hn(x)

u can easily find the H0 & H1. Using them & induction find a recursion relation which you can solve by generating functions.
Good luck........

Yeah such is the problem with the "diffcult to integrate" Taylor expansions. Hermite polynomials would also work. Good thinking.

If that last question was a reference to billzhao's post #11, look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=354567]here[/url] for an explanation. (There's a sign change, but it's the same idea.)

I've seen your nice solution Dr. Kent.  :)
Refering to the post #7 in this topic. You said that the expansion of $e^{x \ln x}$ will give the series not power series, why we can use Maclaurin series for this case? if $f(x) = e^{x \ln x}$ then $f'(0)$ , $f\"(0)$, .... ,$f^{(n)}(0)$ is what??? (it is defined?)
 And in your solution, you interchanged the sum and integral, we can do this if the series converges uniformly. Is $x^k \ln^k x$ converges uniformly for $x \in (0,1]$ ?

thanks for your attention

Edit: Of course I meant  $\frac{x^k \ln^k x}{k!}$  ;)

That's not the Maclaurin series for $x^x.$ It's a series expansion that is not a power series. The individual terms are powers of $x\ln x$ and are not analytic at zero. Of course, we used the well-known Maclaurin series for $e^u$ in order to find it.

As for interchanging the limit and the integral: Let $g(x)=x\ln x$ for $x\ne 0$ and $g(0)=0.$ Then $g$ is continuous on $[0,1]$ and $-\frac1e\le g(x)\le 0$ on that interval.

So, $\sum_{k=0}^{\infty}\frac{(g(x))^k}{k!}$ converges uniformly by the Weierstrass M-test and we can safely interchange limit and integral.

You asked if $x^k\ln^k x$ converges uniformly; that's not the question, since the terms in our series are actually $\frac{x^k\ln^k x}{k!}.$

That's clear enough.
Now I think I can use the integration by part to find the other solution. Many thanks prof :)


[quote="hien"]Let $a\in\mathbb{R}$. Find all function $f(x)$ satifies conditions:
1) $f(x+5)=f(x), \forall x\in R$
2) $af(x)+f(x+3)+f(x+4)=x^{2}+x+1, \forall x\in R$[/quote]]
It give contadition
$x^{2}+x+1=af(x)+f(x+3)+f(x+4)=af(x+5)+f(x+8)+f(x+9)=(x+5)^{2}+x+5+1$,
that is $10x+30=0$ for all $x\in R$.


hey, i'm stuck on this chemistry question for an assignment can some give me a hand please! (yes, i have left it to the last minute and it is due 2morro, oops)

1 g of hydrogen will react with 19 g of fluorine to form HF. By contrast, 1 g of hydrogen will react with 80 g of bromine to form HBr.

a. How many grams of each product will form in each reaction?

b. Why are different amounts of each element required to react with the same
amount of hydrogen?

c. Which of 's four principles does this observation support, and how does
it support it?

cheers, skipper

a) Think, "$1\text{g}\ \text{H}_2$ and $19\text{g}\ \text{F}_2$ go in; how much total comes out?"

b) What's the difference between $\text{F}$ and $\text{Br}$?

c) I belive you're missing a name here...

hey, thanks 4 tryin but i still dont understand (i'm only in yr 10)  :|

c. which of [i]what[/i]'s principles... :)

sorry about question c, that's a mistake (it was Dalton's principles btw)

a) $1\text{g}+19\text{g}=?$

b) Look at $\text{F}$ and $\text{Br}$ on a periodic table

[quote] a) [b]1g+19g[/b]=[/quote]

so thats goning to equal 20?

ok bare with me because i'm alittle slow, but when Fluorine (-1) is put together with Hydrogen (+1) they equal each other out!

thanks 4 the help by the way!

The overall charge is zero, yes ($\text{H}^+\text{F}^-=\text{HF}^0$).  But the same happens with bromine ($\text{H}^+\text{Br}^-=\text{HBr}^0$).  If you were to look at $\text{F}$ and $\text{Br}$ on a periodic table, various things would differ in the data.  Since the problem is asking about [i]masses[/i], what do you think you should look at?

Mass number?

Hydrogen- 1.00796

Flourine- 18.998403

Bromine- 79.904

i thought i had it before, but now i'm just lost.

we haven't learnt mole ratios in class  :|

did you actually find and answer for this because im in the same dilemma now and i think i need some help and this is due tommorow too...


$ y = f(x)$ is a solution of $ y''+e^{x}y = 0$. Show that $ f(x)$ is bounded.

This is a special case of a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157949]problem[/url] which you posted three days ago, and which was since solved by [b]jmerry[/b]. Do you even read your own posts? It looks like you are flooding the forums by mindlessly copying problems from old problem books. Too much of a good thing.


logarithms
calculus
calculus computations

(a) Find $ \frac{d}{dx}(x^{x})$.

(b) Find $ \frac{d}{dx}(x^{x^{x^{x^...}}})$. For what values of $ x$ is this undefined?

a) Let $ y\equal{}x^x$. Then $ \ln y\equal{}x\ln x$, and so $ \frac 1y\cdot\frac{dy}{dx}\equal{}1\plus{}\ln x$, which implies, $ \frac{dy}{dx}\equal{}x^x(1\plus{}\ln x)$.

Thanks nayel. I've come across the first part a number of times.For the second part I can do ${ x^{x^{x}}}$ and I'm thinking of building the general case up bit by bit but I'm not sure about it.

This might be useful: let $ y\equal{}x^{x^{x^{\dots}}}$ and $ D_n\equal{}\frac{d}{dx}(x^{x^{x^{\dots}}})\equal{}\frac{dy}{dx}$ (there are $ n$ $ x$'s). Then \[ D_{n\plus{}1}\equal{}\frac{d}{dx}(x^y)\equal{}\frac yx\plus{}\frac{dy}{dx}\cdot\ln x\equal{}\frac{x^{x^{x^{\dots}}}}{x}\plus{}D_n\ln x.\]

Are there infinite x's in the exponent? Or do you want it for $ n$ x's?
If infinite:

$ y\equal{} x^{x^{x^{x...}}} \equal{} x^y$
$ \ln y \equal{} y \ln x \Rightarrow \frac{dy}{dx} \equal{} \frac{y^2}{x(1\minus{}y\ln x)}$, which is undefined for $ x\equal{}0$ and $ x\equal{}e^{1/e}$.

Thanks to all of you.

kunny, I can see how $ x^{x} \equal{} e^{x \ln x}$ gives the first part; you can we use it for the second part?

[quote="kcn2rivers"]Are there infinite x's in the exponent? Or do you want it for $ n$ x's?
If infinite:

$ y \equal{} x^{x^{x^{x...}}} \equal{} x^y$
$ \ln y \equal{} y \ln x \Rightarrow \frac {dy}{dx} \equal{} \frac {y^2}{x(1 \minus{} y\ln x)}$, which is undefined for $ x \equal{} 0$ and $ x \equal{} e^{1/e}$.[/quote]

Note that $ y\equal{}x^{x^{\ldots}}$ only* converges for $ x \in [1/e^e, \; e^{1/e}]$, so $ dy/dx$ does not exist outside of that interval. Within that interval, your analysis is valid, so $ dy/dx$ is additionally invalid for $ x\equal{} 0, e^{1/e}$.

Thus, the final interval over which $ dy/dx$ is meaningful is $ [1/e^e, e^{1/e})$

[i]* -- I'm actually not entirely sure if the left-side bracket should be open or closed here.[/i]

Hi TZF. How would you go about finding the convergence interval?

Is $ x\equal{}0$ not outside the convergence interval as well?

There is more discussion on the infinite tower $ y\equal{}x^{x^{x^{x^{x...}}}}$ [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77942]here[/url].

Thanks Patterns_34; that just what I wanted.


If I were to take a summer course at my college, would I still be able to participate in the AMC competitions (since you are supposed to be in a high school math class)?

I believe so.  If you mean AMC 8 then I am not sure.  But for 10 and 12 I believe all that matters is your grade level.

It really does not matter, for AMC 8,10, or 12.

You don't have to be in a high school math class, you have to be enrolled in high school (or below).  Lots of people take college-level courses and take the AMC exams.  They don't punish you for being advanced, unless you graduate early.

The key thing in the AMC eligibility rules is your overall grade placement in your regular academic program. For homeschoolers it is your actual age. For the IMO it is not having finished your secondary schooling and started your undergraduate degree program.

Lots of IMO participants have taken lots and lots of "college-level" math courses, but they have taken them while still being formally high school (or junior high :)  ) students. The main point is to keep the twenty-two-year-old Ph. D.-holding "ringers" out of contests that are designed for younger learners. That wouldn't be fair. But no one can stop you from self-educating at whatever pace you can maintain, so no one can stop you from taking a summer course and declaring it to be be part of your late elementary or early secondary education.

Hope this helps! Welcome to the competitions; I'm sure you'll raise the standard for some of the other competitors to meet.

From the 2005 AMC 10/12 Information and Registration Brochure, Section VI: Eligibility

"Who is eligible to participate in the AMC 12? -- Any student who is in a program leading to a high school diploma and under 19.5 years on the day of the contest.
Who is eligible to participate in the AMC 10?  -- Any student under 17.5 years of age onthe day of the contest and not enrolled in grades 11 or 12 or equivalent."

The same statements appear in the 2004 AMC 10/AMC 12 Teachers' Manual and will be in the 2005 Teachers' Manual.

The long and short of it is that you can be taking college classes e.g. in summer, through distance education, through high school release, but must still not yet be graduated from high school.

Steve Dunbar
Director, American Mathematics Competitions

wow, the upper age boundaries are really high... i won't turn 17 until senior year and will be <17.5 by amc time, which means if i stayed back two years i could still take the amc 10 :D