probability geometry MATHCOUNTS trigonometry inequalities USAMTS AMC

Only post books YOU OWN PERSONALLY

Great Books I've Bought:
The Art of Problem Solving Introduction to Counting & Probability
The Art of Problem Solving Introduction to Geometry
The Art of Problem Solving Introduction to Number Theory
The All-Time Greatest MATHCOUNTS Problems (even though i've retired)

I've found these books to be the best so far

Please post your great books

AoPS Volumes 1 and 2 are excellent. I'm a MAO (Mu Alpha Theta) competitor and they've opened my eyes up to methods I've never seen before and have so far (I'm only through a few chapters) have introduced excellent review problems and problems more advanced then any I've seen on any MAO Test (And I've done probably over 100 tests).

Zeitz: The Art and Craft of Problem Solving
Engel: Problem-Solving Strategies

Recommended for after AoPS II.

Cool number theory book:
Joseph H. Silverman: A Friendly Introduction to Number Theory.

geometry revisited, must have geometry book

Are these supposed to be just competition oriented books?
I liked the AoPS Counting and Probability book,
Andreescu and Feng's 101 Trig problems,
Hardy, Littlewood and Polya's Inequalities, etc.

Oops sorry about that, it was the algebra book that had 101, that's pretty good too.
I agree with rnwang2

[quote="rnwang2"][quote="probability1.01"][quote="rnwang2"]Anything by Titu Andreescu/Zuming Feng is good.[/quote]
Huh how would you know?[/quote]
Ahem...I own Mathematical Olympiad Challenges, 103 Trigonometry Problems, and 104 Number Theory Problems (which is not to say I understand everything in them :blush:).[/quote]

But have you READ those books?  I'd wager you've simply leafed through them.

Books can be helpful, but, with a probability of 1.01, the essence of problem-solving is thinking.  A formulaic, books-oriented approach to olympiad success will be nearly impossible, intensely time-consuming, and ultimately far less rewarding.

Before you tackle a book such as Mathematical Olympiad Challenges, you should be able to consistently solve problems at the USAMO level.  Otherwise, you will wasting your time reading it (and being helpless in the problems section), and wasting your money buying it.

Accumulating books is not the answer.  [b]There is no "Must Have" list of math books.  [/b]

[/b]

[quote="13375P34K43V312"]Anything that USAMTS gives away as prizes is good

ie

CPIG
CPIA
360 Problems

etc etc[/quote]

Personally I don't think CPIA is so great.

[quote="gighiuhui"]Books can be helpful, but, with a probability of 1.01, the essence of problem-solving is thinking.  A formulaic, books-oriented approach to olympiad success will be nearly impossible, intensely time-consuming, and ultimately far less rewarding.[/quote]
Thank you for emphasizing this. I realized this very recently, and have now changed my preparation methods.  :)

Linear Algebra and Its Applications, by Gilbert Strang.

Introduction to Applied Math, by Gilbert Strang.

Anything by Gilbert Strang.

In physics, anything by Griffiths.  (And of course, anything by Feynman.)

(These are all college level books.)

1)S. Lang, Algebraic Number Theory.
2) Books by Tittu and Zumming.

I must say gighiuhui has a bit of a point, but books aren't completely useless, for example once you try to solve or have solved the problems on your own, you can check for alternative techniques from books like those by Titu Andreescu et al. This way you open up to a whole arsenal of techniques to tackle problems.
As for non competition oriented books my all time favorite is Hardy's Divergent Series. The style is like a unification of Gauss and Euler!

[quote="K81o7"]I borrowed Mathematical Olympiad Challenges from my teacher a couple weeks before the USAMO, that's a GOOD book...[/quote]

ugh the teachers at my school have no clue about these things.. it seems like

Hmm...I have an experience to share that might be helpful.

Today, I have tried to tackle poles/polars, circumscribed quadrilaterals, and branchion's theorem. In the past few hours, I have learned a large number of results. This is equivalent to what a book will teach you. However, it is clear to me that I would not be able to use anything that I have learned in an actual olympiad problem. You have to find a way to make that leap.

[quote="mihail911"]which is more appropriate for AIME/USAMO/IMO preparation...Zeitz's Art and Craft of Problem Solving....or Engel's Problem Solving Strategies?[/quote]

You know there's quite a gap between AIME and USAMO/IMO...

Engel is harder but more rigorous. Zeitz is better for beginners, especially in terms of its explanations.

If you are trying to improve your AIME score (say to at [i]least[/i] 10 out of 15) and keep it consistant at that score, what book would be best?

I am currently reviewing the AoPS Vol. 2 for that purpose, but would anything else be essential?


A polynomial $ P(x)$ of degree $ n \ge 5$ with integer coefficients and $ n$ distinct integer roots is given. Find all integer roots of $ P(P(x))$ given that $ 0$ is a root of $ P(x)$.


Given $ a, b, c > 0$ and $ abc \equal{}1$. Prove that: $ \frac{a\plus{}b\plus{}c}{3} \geq\ \sqrt[10]{\frac{a^3\plus{}b^3\plus{}c^3}{3}}$
It's a hard problem and I remember that there wasn't any "nice proof" for it.
Now, Am-Gm solved it  :)

Will your way work for the following:
$ \sqrt{\frac{a\plus{}b\plus{}c}{3}} \ge \sqrt[19]{\frac{a^3\plus{}b^3\plus{}c^3}{3}}$
with the same condition? ;)

With $ k\equal{}10$, Am-Gm solved it easily. But indeed, with $ k \equal{} 9.5$, my way is not effect  :oops:
[quote="can_hang2007"]Will your way work for the following:
$ \sqrt {\frac {a \plus{} b \plus{} c}{3}} \ge \sqrt [19]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}$
with the same condition? ;)[/quote]

Thank you, nguoivn! I proved it by $ uvw,$ but my proof is very ugly.
I posted this inequality here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=230715

We have the following lemma:
[i][b]Lemma. [/b]If $ c \ge \max\{a,b\} \ge 0$ and $ t\equal{}\frac{a\plus{}b}{2},$ then we have
$ a^2b^2(a^3\plus{}b^3\plus{}c^3) \le t^4(2t^3\plus{}c^3).$[/i]
Using this lemma, we can easily solve the following problem:
[i][b]Problem. [/b]Let $ a,b,c$ be positive real numbers such that $ abc\equal{}1$. Determine the least constant $ k>0$ such that the following inequality holds:
$ \frac{a\plus{}b\plus{}c}{3} \ge \sqrt[k]{\frac{a^3\plus{}b^3\plus{}c^3}{3}}.$[/i]

In my opinion, the mixing variables method is the best way for this problem (Certainly, except EV of Vasc). I would like to dedidated the lemma to arqady as you're the author of this nice problem. :)

[quote="nguoivn"]Given $ a, b, c > 0$ and $ abc \equal{} 1$. Prove that: $ \frac {a \plus{} b \plus{} c}{3} \geq\ \sqrt [10]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}$
It's a hard problem and I remember that there wasn't any "nice proof" for it.
Now, Am-Gm solved it  :)[/quote]
@arqady: And here is my proof with $ k \equal{} 10$. I hope that you'll like it  :)


Prove that for any positive reals $a_i,0\le i \le 8$
$\displaystyle\sum_{i=1}^8\left(\frac{a_i}{a_i+a_{i+1}}\right)^3\ge 1,$   $a_9=a_1$

Put $x_i=\frac{a_{i+1}}{a_i}$ Then $\prod x_i=1$ and we obtain
$\sum\frac1{(1+x_i)^3}\ge1$ which is a special case of [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=202141#p202141]the first one of the nice ineqs given here.[/url]


inequalities
linear algebra
linear algebra unsolved

Let $A$ be a real matrix. Suppose that $|a_{ij}| \leq M$ for some M > 0.
Prove the inequality
\[ \det A \leq n^{\frac{n}{2}}M^n. \]

Show that this estimate cannot be improved for $n = 2^m$.

You just have to prove that if all the coefficients satisfy $|a_{i,j}|<1$ then $|det(A)| \leq n^{\frac{n}{2}}$. And this is easy (Hadamard inequality) if you see a determinant as a volume.

[quote="alexilic"]
Show that this estimate cannot be improved for $n = 2^m$.[/quote]Hadamard matrices

I think it is an open problem to know for wich $n$ this inequality can be improved.

Another proof for hadamard ineq is using Gram-Schmidt method of orthonormal basis in n dimensions.
[this is not the place to expand this...]
For the cases when equality holds it may be helpful the following

[url=http://mathworld.wolfram.com/HadamardsMaximumDeterminantProblem.html]link[/url]
[url=http://www.answers.com/main/ntquery;jsessionid=330owgv4aw4tu?method=4&dsid=2222&dekey=Hadamard+matrix&gwp=8&curtab=2222_1&sbid=lc08a&linktext=Hadamard%20matrix]link2[/url]


function
algebra
partial fractions
complex analysis

When you're creating a Laurent series, does the annulus have to be just around one pole, or could it be around multiple poles.

For example, let $ f(x) \equal{} \frac{1}{x(x^2\plus{}1)}$. This has simple poles at $ 0$, $ i$, and $ \minus{}i$. You can have a Laurent series in the region $ 0<|z|<1$, centered at $ 0$. Do you also get a Laurent series (a different one) for $ 1<|z|<\infty$?

Can you make a Laurent series centered at a point that is not itself a pole, but for which the annulus contains poles. For example, for the function above, could you make a Laurent series around $ 1$ for $ \sqrt{2}<|z|<\infty$?

I think I've figured out the first part of my question.

For $ 0 < |z| < 1$, we have
\[ f(z) \equal{} \frac {1}{z}\left(1 \minus{} z^2 \plus{} z^4 \minus{} z^6 \plus{} \cdots\right)
\]
For $ 1 < |z| < \infty$, the series looks like
\[ \frac {1}{z^3} \minus{} \frac {1}{z^5} \plus{} \frac {1}{z^7} \minus{} \frac {1}{z^9} \plus{} \cdots
\]
The first was easy. For the second, I used the Residue Theorem to compute the coefficients. Is this the typical way to do it?

Also, why are there infinitely many negative terms in the second series even though there are no essential singularities?

You get a Laurent series for each annulus between poles. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=681920#681920]here[/url] for an example and a method for rational functions. Each term in the partial fractions expansion has a series at zero and/or a series at $ \infty$; the various Laurent series combine these based on which series converge on each region.


Let $\large a;b;c$ are real numbers. Prove that:
$\large a(a+b)^5+b(b+c)^5+c(c+a)^5 \geq  0$     :lol:


Many of you may have noticed that you once again have access to past class materials.  This will only be available for a couple of days, so please be sure to print out and save materials today or tomorrow.  This is the only time we will make these past materials available.

Ummm...I DIDN'T!  Are they still here? And how can I access them?

Sing - this was for people who were in classes long ago.  You can access the files to classes you are in currently by clicking on My Classes.

Oh, haha, I guess it was wishful thinking to have FREE transcripts. Thanks

Check out the old Math Jams - plenty of free transcripts there!


I'm currently working on a Mock State (sorry, no team :().  PM me if you would like to take it, so I can get a better sense to see if I can turn this into a competition.  :)

Competitors:

aleph0
AIME15
dnkywin
gauss1181

And [b]PM me[/b], cough cough...

Competitors:

aleph0
AIME15
AIME_is_hard
BOGTRO
Bugi
dnkywin
FantasyLover
gauss1181
highflyer_10
mewto55555
shentang?

And [b]PM me[/b], cough cough (mewto and gauss)...

i thought this was to get an idea....so if it is then it would have to depend on school work for me. So I might do it if I have time.

Competitors:

007math
aleph0
AIME15
AIME_is_hard
BOGTRO
Bugi
dnkywin
FantasyLover
gauss1181
highflyer_10
mewto55555
shentang?

And [b]PM me[/b], cough cough (mewto and gauss)...

Ok, this will become a competition.

Is it just gonna be like a 1 round competition, or like elimination?

A 1-rounder.  There might be a countdown round if enough people take it.  And it is due to come out next week.

Competitors:

007math
aleph0
AIME15
AIME_is_hard
andrewliu
BOGTRO
Bugi
dnkywin
FantasyLover
gauss1181
highflyer_10
isabella2296
mewto55555
shentang?

Progress:

30 sprint questions, typed
8 target questions, need to be typed
30 sprint solutions, 20 need to be typed
2 target solutions

Hopefully, it will be out by Sunday.

Sprint: 30 questions, 45 minutes (usually 40, but I'm giving more, you'll see why)

Target: 4 pairs of questions, 6 minutes each

These are the two main individual rounds that form the basis of MathCounts.

007math
aleph0
AIME15
AIME_is_hard
andrewliu
BOGTRO
Bugi
dnkywin
FantasyLover
gauss1181
highflyer_10
isabella2296
mewto55555
shentang
Dojo

pm'd

[quote="ernie"][quote="5849206328x"]I borrowed a template from mr frost about a year or so ago, then modified it a bit[/quote]

May I have that template? :roll:[/quote]

Whathe aid--but to me. :roll:  :D

you probably should pm me if you want it, I prefer not to have that comment reduce this thread to spam...

How would you know?  You're on your honor. :P

I HAVE RECEIVED ONLY ONE SET OF ANSWERS SO FAR.  PLEASE SUBMIT.  THANK YOU, THAT IS ALL.

oops, caps :lol:

[quote="ernie"]Extending the deadline to January 7, 8 pm EST.[/quote]
Good idea. That way I have more winter break time to do USAMTS. :D

No problem!

I need answers from:

#5-1Mole
007math
aleph0
AIME15
AIME_is_hard
andrewliu
BOGTRO
Bugi
Discrete_Math
dnkywin
Dojo
FantasyLover
gauss1181
highflyer_10
isabella2296
mewto55555
shentang
theprodigy
xpmath

Deadline: January 10, 8 pm EST

ok, since this is failing, i will send out PM reminders

deadline (re-extended): jan 12, 8 pm est

I've now already submitted my answers. :)


quadratics
induction
algebra
polynomial
limit
quadratic formula

Let $ f(x) \equal{} x^2 \plus{} 2007x \plus{} 1$. Prove that for every positive integer $ n$, the equation $ \underbrace{f(f(\ldots(f}_{n\ {\rm times}}(x))\ldots)) \equal{} 0$ has at least one real solution.

Solution 1 :
[u] Result [/u]
Consider the equation :
$ f(x) = a$ where $ a\geq \frac { - 2007^2 + 4}{4}$ has solution $ x_1,x_2$ then
${ x_1 = \max\{x_1,x_2}\geq \frac { - 2007}{2}$
Proof : From $ x_1 + x_2 = - 2007$
We prove by induction $ f_{n}x = a$ has at least real solution if $ a > \frac { - 2007^2 + 4}{4}$
Suppose it is true for $ n$
We prove it is true for $ n + 1$
$ f_{n + 1}(x) = f_n(x)^2 + 2007f_{n}(x) + 1$
If $ f_n(x_n) = x_1$ then $ x_n$ is a root of $ f_{n + 1}(x) = 0$
But we have $ x_1 > \frac { - 2007^2 + 4}{4}$ so the equation has solution .Call it is $ x_{11}\geq \frac{-2007}{4}$
Contine consider equation $ f_{n-1}=x_{11}$ ..
It has solution.
Our induction claim.
Apply this result our problem was be claim.
I know that have other solution so please contine discuss it .

[quote="cyshine"]Let $ f(x) = x^2 + 2007x + 1$. Prove that for every positive integer $ n$, the equation $ \underbrace{f(f(\ldots(f}_{n\ {\rm times}}(x))\ldots)) = 0$ has at least one real solution.[/quote]

The equation $ f(x)=x$ has two solutions $ a<b<0$.

Then, if $ g(x)=\underbrace{f(f(\ldots(f}_{n\ {\rm times}}(x))\ldots))$, we have :

$ 1)$ $ g(b)=b < 0$
$ 2)$ $ g(x)$ is a polynomial whose top term is $ x^{2^n}$ and so ${ \lim_{x\rightarrow+\infty}g(x)=}\infty >0$

And so $ g(x)$ has at least one real root in $ (b,+\infty)$

Define $ a\equal{}\minus{}\frac{2007}{2}\,,\,b\equal{}\minus{}a^2\plus{}1$ and the intervals $ D\equal{}[a,\infty[\,,\,W\equal{}[b,\infty[$, then we have $ 0\in D\subset W$ and $ f(x)\equal{}(x\minus{}a)^2\plus{}b$, i.e. $ f(\mathbb{R})\equal{}f(D)\equal{}W$ and therefore $ f(\ldots f(\ldots (\mathbb{R})\ldots )\equal{}W$

$ f(x)\equal{}q$ has solution if $ q>1\minus{}\frac{2007^2}4$. One solution for $ f(x)\equal{}q$ is $ x_1\equal{}\frac{\sqrt s \minus{}2007}2>\minus{}2007>1\minus{}\frac{2007^2}4$. So $ f(x)\equal{}x_1$ has a root $ x_2$ such that $ f(x)\equal{}x_2$ has a root $ x_3$..., and if we assume $ q\equal{}0$, we can do this process indefinitely, which is what we had to prove.

Here is general problem (can solve with my method )
Find then number of solution of the equation : $ f_{n}(x)\equal{}0$

I think that this works.

[b]Solution[/b]

Let $Q(x)=\underbrace{f(f(\ldots(f}_{n\ {\rm times}}(x))\ldots))$.

We have that
\[x \rightarrow \infty \quad \Rightarrow \quad f(x) \rightarrow \infty\]
Therefore we also have that
\[x \rightarrow \infty \quad \Rightarrow \quad Q(x) \rightarrow \infty\]
Hence there exists arbitrarily large value $\alpha$ such that $Q(\alpha)>0$.

Let $\beta$ denote a root of the equation $f(x)=x$. The quadratic formula yields that $\beta < 0$. This yields that $Q(\beta)=\beta<0$.

By the intermediate value theorem, there exists $x$ between $\alpha$ and $\beta$ such that $Q(x)=0$.

i have done a solution
if anything is wrong, please rectify it
we have$f(x)=x(x+2007)+1$ a parabola
where $f'(x)>0$ when $x>-\frac{2007}{2}$, $f'(x)<0$ when $x<-\frac{2007}{2}$
so $f(x)$ is increasing in $[-\frac{2007}{2},\infty)$ and decreasing in  $(-\infty,-\frac{2007}{2}]$
now $f(0)=1, f(f(0))=f(1)>f(0)=1, f(f(f(0)))>f(1)>f(0)=1$
so $f^n(0)>1>0$ for all n in $N$
again
$f(-2006)=-2006+1=-2005, f(f(-2006))=f(-2005)<f(-2006)=-2005$
or $f(f(f(-2006)))<f(-2005)<f(-2006)=-2005$
so $f^n(-2006)<-2005<0$ for all n in $N$
as $f^n(x)$ is continuous (for being polynomial) for all n in $N$
so by IVP $f^n(x)=0$ always has a solution

maybe there is something wrong while dealing with decreasing portion of $f(x)$

Note that $f^n(1)$ is always $>0$. Next, let $\alpha$ be the negative root of $x^2+2006x+1=0$. Thus,
\[f^n(\alpha)=\alpha\]
for all $n$, and so since $f^n(\alpha)<0<f^n(1)$, and $f$ is a polynomial, by IVT there is a real solution.

[color=#f00]Claim: [/color] If $f^{n}(r) = 0,$ then the value of $x$ which satisfies $f(x) = r$ will also satisfy $f^{n+1}(x) = 0.$
[i]Proof. [/i] We have $f^{n+1}(x) = f^{n}(f(x)) = f^{n}(r) = 0.$
It now suffices to prove that there's always a real solution to $f(x) = r,$ which occurs iff the discriminant is
\begin{align*}
&2007^2 + 4(r-1) \ge 0\\
&\Leftrightarrow r \ge \frac{4 - 2007^2}{4}
\end{align*}
Now, note that one solution to $f(x) = x^2 + 2007x + 1 = 0$ is $x = \frac{- 2007 + \sqrt{2007^2-4}}{2} > -1 $
[color=#f00]Claim: [/color] We can now use induction to make the claim that
If $r > -1$ satisfies $f^{n}(r) = 0$ for some positive integer $n$, then there exists a real $x > -1$ such that $f(x) = x^2+2007x+1 = r.$
[i]Proof.[/i] One $x$ that satisfies $x^2 + 2007x + 1-r = 0$ is $x = \frac{-2007 + \sqrt{2007^2 +4r-4}}{2},$ which is obviously real and greater than -1 for any $r > -1.$
Hence if we use the first claim, this second claim shows that because one of the roots to $f(x)$ is greater than -1, there will always be a real root greater than -1 to $f^{n}(x)$ for all positive integers $n.$ Hence $r > -1 \ge \frac{4 - 2007^2}{4},$ which is what we wanted to show and we're done.


Triangle $ OAB$ has $ O \equal{} (0,0)$, $ B \equal{} (5,0)$, and $ A$ in the first quadrant. In addition, $ \angle{ABO} \equal{} 90^\circ$ and $ \angle{AOB} \equal{} 30^\circ$. Suppose that $ \overline{OA}$ is rotated $ 90^\circ$ counterclockwise about $ O$. What are the coordinates of the image of $ A$?

$ \textbf{(A)}\ \left( \minus{} \frac {10}{3}\sqrt {3},5\right) \qquad \textbf{(B)}\ \left( \minus{} \frac {5}{3}\sqrt {3},5\right) \qquad \textbf{(C)}\ \left(\sqrt {3},5\right) \qquad \textbf{(D)}\ \left(\frac {5}{3}\sqrt {3},5\right) \\ \textbf{(E)}\ \left(\frac {10}{3}\sqrt {3},5\right)$

Answer:
[hide]This forms a 30,60, 90 triangle....
just draw it out with the angles and you'll find that point A is at:
(5,5/rt(3)) or (5,5rt(3)/3)
By rotating it 90 degrees counterclockwise, you move point to a point with (-,+) coordinates..
so you already know the answer is A or B.
By rotating Point A around point O, you're basically moving Point A to a point that is as far from point O as Point A is from Point O...

by looking, you can tell that B is just as far from Point O as Point A is as from Point O

Ans:B[/hide]

[hide]call any point on the negative x-axis, $ D$. So when you rotate $ OA$ by $ 90$ degrees, the angle $ A^1 OD(\mbox{a the image of a it is meant to be a prime but well okay}) \equal{} 180 \minus{} 30 \minus{} 90 \equal{} 60$. and since $ OA \equal{} \frac {10}{\sqrt {3}}$, the x coordinate and y coordinate are part of the 30-60-90 triangle and hence the coordinates are $ (\frac { \minus{} 5\sqrt {3}}{3}, 5) \longrightarrow \fbox{\mbox{B}}$[/hide]

had some LaTeX issues sorry.


I was trying to look at the problem and solution at a book and i can't understand some lines on the solution.
Problem:

Show by $\epsilon-\delta$ definition such that $lim_{x\to2}x^{2}+5 = 9$.

We want $x^{2}+5 = 9$ to be less than $\epsilon$
$|x^{2}+5-9| = |x^{2}-4|$
$=|(x+2)(x-2)| < \delta|x+2|$

We therefore cannot choose $\frac{\delta}{|x+2|}$ since $\delta$ has to be constant

Note that $|x-2| < 1$ implies that $x \in (1,3)$
[b]QUESTION[/b]: Where did 1 came from in $|x-2| < 1$ . Do we always use 1?

Hence if $|x-2| < 1$, we must have $|x+2|<5$
So that $|x+2\parallel x-2|<\delta|x+2|<5\delta$

Proof:

Let $\epsilon > 0$
choose $\delta = \frac{\epsilon}{5}$
...
...

[quote]QUESTION: Where did 1 came from in $|x-2| < 1 .$ Do we always use 1?[/quote]
You have to do something like that, to estimate the less-critical $|x+2|$ factor. But using 1 is just arbitrary. It's a convenient positive number. Do we always use 1? Let's just ask this: if I asked you to name a convenient positive number, what would you say?

But can we use 1 all the time? I think there are cases that you cannot use 1

Of course there could be cases where you might need something else. This isn't a rule - it's a convenience. Keep your eyes open.

We therefore cannot choose $\frac{\delta}{|x+2|}$ since  has to be constant

I am just wondering, where did $\frac{\delta}{|x+2|}$ came from? Isn't that suppose to be $\frac{\epsilon}{|x+2|}$

Sorry I'm so dumb at this.


What is the greatest hole number that MUST be a factor of the sum of any four consecutive whole numbers? (number 5)

John, Mike, and Chantel will divide a pile of pennies amongst themselves using the following process:

The number of pennies is counted:
If the number of pennies is even, Mike will get half the pile.
If the number of pennies is odd, one penny is given to Chantel, and then John will get half the pennies remaining in the pile.

If the number of pennies in the pile is 2005, how many pennies will Mike end up with?
Number eight. ^^^

it said consecutive odd whole numbers for number five right?
guess and check and you get 8.

5. (the way hwenterprise stated it)
[hide]The sum of 4 consecutive whole numbers can be expressed as $x+(x+1)+(x+2)+(x+3)$. You can simplify this to get $4x+6$. The greatest number that $4x+6$ is always divisible by is $2$.[/hide]

Edit: The actual question on the test said "What is the greatest whole number that MUST be a factor of the sum of any four consecutive positive [b]odd[/b] numbers?

5. (the way it was stated on the MATHCOUNTS chapter test)
[hide]The sum of 4 consecutive odd numbers can be expressed as $x+(x+2)+(x+4)+(x+6)$, which can be simplified to $4x+12$. Note that x must always be odd, so $4x=4 (mod 8)$. $12$ is also $4 (mod 8)$. $4+4=0(mod8)$, so $4x+12$ must always be divisible by 8.[/hide]

number 8 is just a long process. Just keep doing it over and over and over again.

For #5, I plugged in values and got it right. (I got all 8 in chapter target)

For #8:
The question begins with 2005 pennies in the pile. This is odd, so Chantel gets 1 and John gets 1002. There are now 1002 pennies in the pile.

1002 is even, so Mike getrs 501 pennies. There are now 501 pennies in the pile.

501 is odd, so Chantel gets 1 penny (total 2) and John gets 250 pennies (total 1252). There are now 250 pennies in the pile.

250 is even, so Mike gets 125 pennies (total 626). The pile now has 125 pennies.

125 is odd, so Chantel gets 1 (3 total) and John gets 62. There are 62 left.

62 is even, so Mike gets 31 (total 657). There are 31 left.

31 is odd, so Chantel gets 1 (total 4) and John gets 15. There are 15 left.

15 is odd, so Chantel gets 1 and John gets 7. There are 7 left.

7 is odd, so Chantel gets 1 and John gets 3. There are 3 left.

3 is odd, so Chantel gets 1 and John gets 1. There is one penny left. This then goes to Chantel. Mike has [b]657[/b] pennies.


logarithms
function
linear algebra
matrix
linear algebra open

Let $A \in M_n(\mathbb{C})$. What conditions must we impose on A such that there exists $B \in M_n(\mathbb{C})$ with the property that exp(B)=A.

Recall that $exp(X)=\sum_{k=0}^{\infty}\frac{X^k}{k!}$ for $X\in M_n(\mathbb{C})$.

There is a theorem:"spectral mapping theorem" (applied in this case, because it is more generally)
let $ \sigma=\{a\in C|A-aI_n \mbox{is not invertible} \} $ and f a function C-analitique such that we can define f(A). Then
$ f(\sigma(A))= \sigma(f(A)). $

[quote="amfulger"]Let $A \in M_n(\mathbb{C})$. What conditions must we impose on A such that there exists $B \in M_n(\mathbb{C})$ with the property that exp(B)=A.
[/quote]

As caugusts did, let $\sigma (A)=\{ a\in\mathbb{C} |A-a\cdot I_n \ is\ not\ invertible\}$

I'll prove that a matrix B such as exp(B)=A exists iff $0\ not\ \in \sigma (A)$.
It is easy to see, as caugust said, that $\sigma (\exp(B))=\exp(\sigma (B))$.

$ \Rightarrow $ tIf a matrix B exists such as exp(B)=A, then $\sigma (A)=\exp(\sigma (B)) \ not \ \ni 0$

$ \Leftarrow $ What if $0\ not\  \in \sigma(A)$?

$\exp(U\cdot B\cdot U^{-1})=\sum_{k=0}^{\infty} \frac{(U\cdot B\cdot U^{-1})^k}{k!}=U\cdot \exp(B)\cdot U^{-1}=U\cdot A\cdot U^{-1}$
This goes to prove that it is enough to solve the problem for matrices A in Jordan canonical form.
It is also easy to see that it is enough to solve the problem for matrices A that are actually Jordan cells.

Denote by $E=(e_{i,j})$, $e_{i,j}=\{ \begin {array}{cc} 1,&if\ j=i+1 \\ 0,& if j\neq i+1\end{array}$.
It is easy to see that $E^k=(e_{i,j}^k)$, $e_{i,j}^k=\{ \begin {array}{cc} 1,&if\ j=i+k \\ 0,& if j\neq i+k\end{array}$, for $k\geq 0$. Also for $k\geq n,\ E^k=0_n$.

Since A is a Jordan cell and $0\ not \in \sigma(A)$ ,$A=e^x \cdot I_n+E$ for some complex number x. I'll prove that there exists a matrix B of the form $B=x\cdot I_n+\sum_{k=1}^{n-1} a_k\cdot E^k$ such that exp(B)=A, $a_k\in \mathbb{C},\ for\ 1\leq k \leq n-1$ .

Lets solve the equation $\exp(x\cdot I_n+\sum_{k=1}^{n-1} a_k\cdot E^k)=e^x\cdot I_n+E$.
It is easy to see that if XY=YX, then exp(X+Y)=exp(X)exp(Y).
Let $T=\sum_{k=1}^{n-1} a_k\cdot E^k$. T is superior triangular with all the elements of the diagonal 0, so $\sigma(T)=\{0\}$. So T is nilpotent, so $T^n=0_n$.
$\exp(B)=\exp(x\cdot I_n+T)=\exp(x\cdot I_n)\cdot \exp(T)=e^x\cdot I_n\cdot ( \sum_{k=0}^{n-1}\frac{T^k}{k!})$.
So we have to solve the equation $\sum_{k=1}^{n-1} \frac{( \sum_{i=1}^{n-1} a_i\cdot E^i)^k}{k!}=e^{-x}\cdot E$.
A quick inductive argument shows that this equation is always solvable.

Statement: If $0 \ not\in \sigma(A)$ then for all $m \in \mathbb{N},\ m\geq 1$ there exists a matrix $C_m$ such that $C_{m}^{m}=A$.

Proof: If $0 \ not\in \sigma(A)$ then there exists B such as exp(B)=A. Take $C_m=exp(\frac{B}{m})$. It is easy to see that $C_{m}^{m}=A$.
---------------------------------------------------------------------------------------------
Open questions: If for all $m \in \mathbb{N}^*$ there exists $C_m$ such as $C_{m}^{m}=A$, does it follow that 0 is not in $\sigma (A)$? Does it follow that there exists a matrix B such as exp(B)=A?


[quote="Hawk Tiger"]Quite easy.But we should let $(a,n)=1$.[/quote]
I don't think so... Not must be $(a,n)=1$
(Is it right grammer?  :blush: )

[quote="mathematica"]Prove that for every two natural numbers like $a$ and $n$ we have $a^{n}=a^{n-\phi (n)}(\mod n)$.[/quote]
if $n-\phi (n)=0$ ,$a=n$,it is not true.(Am I right?)

It's always true and doesn't require any conditions.

@Hawk Tiger: wh do not have $n-\varphi(n)=0$ very often.

$(a,n)=1$ is not needed, just prove that $n-\varphi(n)\geq \beta$ for $p^\beta \parallel n$ ($p$ prime) which is very obvious :)

Sorry.I consider it as $a^{k}=a^{k-\varphi (n)}$ :blush:
Thanks.


Let A,B,C,D be a convex quadrilateral. Points K,L,M,N are the midpoints of AB,BC,CD,DA respectively. KM intersects AC and BD at P and Q respectively. LN intersects AC and BD at R and S respectively. Prove that if AP x PC=BD x DQ then $ \frac {PR}{QS} \equal{} \frac{SD \minus{} BQ}{RC \minus{} AP}$

you have got a typo: $ AP \cdot PC \equal{} BQ \cdot QD$.


Last year, Laura rotated the tires on her car so that all 5, including the spare, received equal wear. Laura drove 20,000 miles last year. How many miles of wear did she put on each tire during the year?

[color=00FFFF][code]The answer is 16000[/code][/color][/b]

at any given time, there are 4 wheels on a car, so the total miles on the wheels she has must be $ 20,000\cdot4\equal{}80,000$. however, these $ 80,000$ miles are spread out over five wheels equally, giving us $ \frac{80,000}5\equal{}\boxed{16,000}$, as prakhar23 said before me.


Find a vector with a direction opposite to v = 2i - 5j with thalf the length of v.


Let $ a,b,c,d > 0$ . Prove that :

$ (\frac {a \plus{} b}{a \plus{} b \plus{} c})^{2} \plus{} (\frac {b \plus{} c}{b \plus{} c \plus{} d})^{2} \plus{} (\frac {c \plus{} d}{c \plus{} d \plus{} a})^{2} \plus{} (\frac {d \plus{} a}{d \plus{} a \plus{} b})^{2}\geq\frac {16}{9}$

the inequality is wrong.try a=b=c=2 ,d=4 and the inequality is reversed


function
algebra
domain
complex analysis
complex numbers
complex analysis unsolved

I have gotten through most part of the question but i am totally blocked at this part, hope someone can help...

Let f be a non-constant entire function such that |f(z)|=1 for all complex numbers z satisfying |z|=1. Show that for any complex number y from the open unit disc, there exist a complex number z (from the open unit disc) such that f(z)=y.

I have proven the following statements which was given in the lecturer's hint.

(i) The above function f maps the open unit disc to the open unit disc.
(ii) g(z)= (z-a)/(1-a_bar*z) where |a|<1, is an analytic function mapping the unit disc to the unit disc.
(iii) Let h be an analytic non-constant function in the bounded domain D and continuous up to and including the boundary of D. If |f(z)| is constant on the boundary of D then h must have at least one zero in D.

Remarks: I strongly believed that schwarz's lemma needs to be used in the question. I have been trying to show, but to no success, that there exist some non-zero z in the open unit disc such that |f(z)|=|z|, which will allow me to complete the question.

Hope someone can help :roll:

let $ w \in\mathbb{D}\equal{}\{ z: |z|<1\}$ and consider $ F\equal{}f(z)\minus{}w$  and then use the Rouche theorem

Cobmine iii and ii to get the hypothesis of Schwarz Lemma.

iii says that there is a zero of $ f$ inside the disk. You need to move the zero to zero by composing with the function in ii, and then check that new composition of functions hasn't messed up mapping the disk to the disk.

Suppose $ f(a)\equal{}0$. Consider $ h(z)\equal{}f\left(\frac{z\plus{}a}{1\plus{}\overline{a}z}\right)$. Then $ h(0)\equal{}f(a)\equal{}0$, and it still maps the disk to the disk, so Schwarz applies to $ h$.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=277820]this topic.[/url]

I didn't specifically use Schwarz there, but I used arguments from the same family as the proof of Schwarz. But then fedja simplified it down from there. Go read his proof on the link.


MATHCOUNTS ARML trigonometry HMMT Duke college geometry

I thought it would be nice to add a little organization to this forum.  And as it stands, there aren't too many posts in this forum that it would be that tough of a task to maintain :D  If you have any suggestions to make this post better, I'd love to hear them.

\[ \text{\Large Practice Problem Lists}  \]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21038#]Mathcounts Practice[/url] - MCrawford
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20985]ARML Practice Problems[/url] - MCrawford
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21104]ARML Practice[/url] - MCrawford
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=36545]Two Trigonometry Problems[/url] - gauss202
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46375]Problem Marathon[/url] - Iversonfan2005
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61308]Some HMMT Problems[/url] - Iversonfan2005
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61776]Some Problems from the Duke Math Meet[/url] - gauss202
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=69252]Some Problem Shobhit Made Up[/url] - Iversonfan2005
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35645]Collection of Geometry Problems[/url] - gauss202
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=74217]Random AMC Type Problems[/url] - Iversonfan2005
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=74354]ARML NT Practice Problems[/url] - joml88
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=432882#p432882]NCML[/url] Problems - prybarczyk
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75721]ARML Practice Problems[/url] - joml88

\[ \text{\Large Lessons/Reference Material}  \]
\[ \text{Number Theory}  \]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75061]Divisibility  Rules[/url] - joml88

\[ \text{Geometry}  \]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75469]Triangle Area[/url] - joml88
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=434617#p434617]Circles[/url] -Iversonfan2005
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75573]Reference for Geometry Formulas[/url] - Iversonfan2005

\[ \text{Algebra}  \]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75738]Factoring[/url] - joml88

sweet :coolspeak: , it's amazing how much activity this forum's gotten in the last 2 weeks :roll:

I have a question, how exactly do we make pdfs, because it would be nice if we can compile all the problems into a single pdf.. :) , if someone can tell me, i can do it

[quote="Iversonfan2005"]I have a question, how exactly do we make pdfs, because it would be nice if we can compile all the problems into a single pdf.. :) , if someone can tell me, i can do it[/quote]

Well, I can think of two options:

(A) You can use TeXnic Center and type up all of the questions separately (well, you can just copy, paste, and do a little bit of editing to make it come out a little neater).  This requires you to have TeXnice Center downloaded on your computer and hooked up and everything which is good thing to have anyways.

or

(B) You can download CutePDF or some other program that converts files to pdf's.  This would allow you to take regular documents go to File > Print, change the printer to CutePDF and instead of actually printing something out on paper, it will output a PDF file on your computer of the document.  So basically, you could copy and paste everything into word and use CutePDF.

Option A would tend to be more time consuming.  Particularly, downloading all the stuff to get TeXnic Center working on your computer can be a hassle but I think it's well worth it if you plan on $\text{\TeX}$ing up documents every once in a while.  The plus side is that you'll have TeXnic Center working :) and you'll have more control over what the document looks like.

Option B is more practical and quicker for this specific task.

[quote="joml88"]I thought it would be nice to add a little organization to this forum.  And as it stands, there aren't too many posts in this forum that it would be that tough of a task to maintain [/quote]

LOL :rotfl:  :rotfl: , I think there's been so much activity on this forum now that you should just title it $\text{Lessons and References}$ and just put links for the lessons and stuff and probably the document files of questions posted.


If $ b$ is positive, what is the value of $ b$ in the geometric sequence $ 9, a, 4, b$? Express your answer as a common fraction.


We all know that P ( a U b ) = P(a) + P(b) - P(a int b)

Prove the addition principle for n events..
It comes in the form of alternate positive and negative summations with 2,3, ... n terms at a time.

Usually this is called the "Principle of Inclusion-Exclusion," frequently abbreviated "PIE" on the forums.  What have you tried?

Also, [[LaTeX]] is equipped on the forums -- read http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 and the first link in this sentence for an introduction.

Yea i know tht its calld principle of incusion-exclusion..n its very often used in probability..

thts y its posted here..

You have to prove something for a general $ n$ case. What does that suggest to you?


I would like to meet some penpals who go to competitions and who are willing to chare their experiances and problems with me.
If someone knows of some free on-line preparation program, please let me know!
my e-mail: nada011@yubc.net
please write,
regards,
Tijana

[quote="tijana_kovacevic"]If someone knows of some free on-line preparation program[/quote]

Why don't you just try the many nice problems that Arne, JBL, grobber and others freely and nicely give us ??


Prove that the equation $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\plus{}\frac{1}{abc} \equal{} \frac {12}{a \plus{} b \plus{} c}$ has infinitely many solutions $ (a,b,c)$ in natural numbers.

This is a very strange problem. Clearly if $ (a,b,c)$ is a solution, then so is $ (ka,kb,kc)$ where $ k\in\mathbb{Z}^\plus{}$. Hence the problem reduces to finding one solution.

From expanding, we see that $ (b\plus{}c)|11abc$. If we scale the variables sufficiently, we would have a solution such that $ b\plus{}c|bc$. From here, we can write $ b\equal{}r(s\plus{}t)s$, and $ c\equal{}r(s\plus{}t)t$. Back in the original equation, this gives
$ a\equal{}\frac{r}{2}\cdot(\minus{} s^2 \plus{} 9  s t \minus{} rt^2 \pm \sqrt{s^4 \minus{} 22 s^3 t \plus{} 75 s^2 t^2 \minus{} 22 s t^3 \plus{} t^4})$

which is certainly promising...but we still restated the problem in the sense that we need that expression to be a square...

I don't have a nice way to do this. I have checked $ 1\le s\le 200$, $ 1\le t\le 2000$ and not gotten a solution. I wonder what the solution is...I have tried root flipping but it is a quartic...

I think that you can work as here

http://www.mathlinks.ro/viewtopic.php?p=118691#118691

I'm very sorry, Altheman, for the mistake in my post.the problem is OK now.


Let $x;y;z$ be positive real numbers.Prove that:
$x^{2}(y+z)+y^{2}(x+z)+z^{2}(x+y)\ge 6xyz$[/QUOTE]

[quote="chien than"]Let $x;y;z$ be positive real numbers.Prove that:
$x^{2}(y+z)+y^{2}(x+z)+z^{2}(x+y)\ge 6xyz$[/quote][/quote]
ok
solution S.O.S
$xy(x+y)+yz(y+z)+zx(z+x)\ge 6xyz$
with
x/y+y/x=(x-y)^2/xy[/quote]

auhongan_au, you do nothing with SOS?
did you just use Cauchy?

[quote="mathubu"]Can anyone tell me what the s.o.s method is ?[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127

we can writ it like this, $x(y^{2}+z^{2})+y(z^{2}+x^{2})+z(x^{2}+y^{2}) \geq 6xyz$ wich becom easy by the fact $x^{2}+y^{2}\geq 2xy$

Why do we discuss this kind of inequalities?
It's obvious for
$x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)\geqslant3\sqrt[3]{x^{2}\cdot2\sqrt{yz}\cdot y^{2}\cdot2\sqrt{zx}\cdot z^{2}\cdot2\sqrt{xy}}=6xyz$


articles
geometry
geometric transformation

I have the following problem - I enjoy browsing news from the world, reading about politics, social affairs, economy, science etc., but I can' find a decent news site on the net that would satisfy my expectations. Portals in my country (Poland) are poor, they tend to write too much about Polish politics (which is getting more and more indigestible) and anyway, they are full of tabloid-style crap. Sites like NYTimes.com etc. would make a good reading if they allowed filtering information - but they're too chaotic , messy and full of useless stuff (also, their layout makes it hard to concentrate and find valuable news among the informational litter). I enjoy the style of writing about world affairs that "The Economist" presents, so a news site designed in a similar style would be OK. Any suggestions?

You can read the Onion if you want, but keep in mind that it is satire, not actual news.  I am not sure if Treething was trying to make a commentary on something or not, but I just want you to be aware of what you are getting into.

[url]http://news.bbc.co.uk/[/url] is about the best site I know.

The problem -of course  - is....how many languages do you speak. My experience is that that can be a huge barrier (I often compare articles on the internet).

Well you can translate it, but the automatic translations aren't very good...still though you can get a good amount of the information from it.

[url]http://babelfish.altavista.com/[/url]

Timesonline is quite a good one:
[url]http://www.timesonline.co.uk/tol/news/[/url]


Let $ n$ be a positive integer. Prove that for all real $ x,y$,
$ (nx)^n\minus{}((n\minus{}1)x\plus{}y)^n\plus{}((n\minus{}2)x\plus{}2y)^n\minus{}\dots\plus{}(\minus{}1)^{n\minus{}1}(x\plus{}(n\minus{}1)y)^n\plus{}(\minus{}1)^n (ny)^n$
$ \equal{}n!(x\minus{}y)^n$

for $ (x,y)\equal{}(1,0)$, the equation becomes $ n!\equal{}n^n\minus{}(n\minus{}1)^n\plus{}(n\minus{}2)^n\minus{}\cdots$, which is false. the correct formula is $ n!\equal{}n^n\minus{}\binom{n}{1}(n\minus{}1)^n\plus{}\binom{n}{2}(n\minus{}2)^n\minus{}\cdots$. you're missing the necessary coefficients.

Sorry about that  :blush:

The identity i meant was
\[ \binom{n}{0}(nx)^{n}\minus{}\binom{n}{1}((n\minus{}1)x\plus{}y)^{n}\plus{}\binom{n}{2}((n\minus{}2)x\plus{}2y)^{n}\minus{}\dots\]
\[ \plus{}(\minus{}1)^{n\minus{}1}\binom{n}{n\minus{}1}(x\plus{}(n\minus{}1)y)^{n}\plus{}(\minus{}1)^{n}\binom{n}{n}(ny)^{n}\]
\[ \equal{}n!(x\minus{}y)^n\]

corresponding coefficients are equal by a straightforward application of PIE.


Let $ a,b,c \in (0,\infty)$ such that $ b^2<ac$ and $ c^2<ab$. Prove that: $ \frac{bc}{ac\minus{}b^2}\plus{}\frac{ac}{ab\minus{}c^2} \ge \frac{4ab}{a^2\minus{}bc}$.

$ \frac {bc}{ac \minus{} b^2} \plus{} \frac {ac}{ab \minus{} c^2} \minus{}\frac {4ab}{a^2 \minus{} bc} \equal{} \frac {(a^2c \plus{} c^2b \minus{} 2b^2a)^2}{(ac \minus{} b^2)(ab \minus{} c^2)(a^2 \minus{} bc)} \geq 0$


[quote="Beat"]1. It is known that the sum of four non-zero numbers and the sum of their
reciprocals are both 0. Prove that among these four numbers there are
two whose sum is 0.[/quote]

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}$
equivalent to
$(x+y)(y+z)(z+x)=0$

I think you didn't understand socrates:
You have $\frac 1a + \frac 1b + \frac 1c = \frac{-1}{d} = \frac{1}{a+b+c}$ and this is equivalent to $(a+b)(b+c)(c+a)=0$, try out.

[quote="Beat"]maybe that was what socrates meant[/quote]

exactly!!  :P

 :D Enjoy yourself!


Factor $x^{8}+x^{4}+1$
(a) over the rationals
(b) over the reals
(c) over the complex numbers.

vishalarul, your first and fourth factors are equivalent.
[hide] I'll just finish the factorization.

Substitute $a=x^{2}$ to get $a=\frac{1\pm i\sqrt3}{2}=cis \pm\frac{\pi}{3}$

$x=\sqrt{cis \pm\frac{\pi}{3}}=cis \pm\frac{\pi}{6}, cis\pi\pm\frac{\pi}{6}$

$(x-e^\frac{i\pi}{6})(x-e^\frac{i\pi}{3})(x-e^\frac{2i\pi}{3})(x-e^\frac{5i\pi}{6})(x-e^\frac{7i\pi}{6})(x-e^\frac{4i\pi}{3})(x-e^\frac{5i\pi}{3})(x-e^\frac{11i\pi}{6})$[/hide]

[quote="Phelpedo"]I am a little confused by your factorization at the end. It's $\cis(\pi \pm \frac{\pi}{6})$, so shouldn't it factor as $(x-e^{i\frac{\pi}{6}})(x-e^{i\frac{5\pi}{6}})(x-e^{i\frac{7\pi}{6}})(x-e^{i\frac{11\pi}{6}})$?[/quote]
Yes, my final factorization was for $x^{8}+x^{4}+1$, not $x^{4}-x^{2}+1$.


What is the remainder when $ 99^{36}$ is divided by 100?

hello, we have $ 99\equiv \minus{}1 \mod 100$ hence $ 99^{36}\equiv 1 \mod 100$.
Sonnhard.

To put Dr. Sonnhard's answer simply for us little guys, the answer is 1.


analytic geometry
geometry
MATHCOUNTS
AoPSwiki

Find the area of a wuadrilateral whose vertices are (-3,4), (5,6), (3, -2), (-1, -7).

thanks, but i just looked at the theorum and shouldn't the answer be 57, because you need to add that last line again?

can someone help me with this one?

Find the distance between the parallel lines y=4x and y=4x-3

I don't have time to post a full solution, but you basically find a line perpendicular to both lines and find the length of that perpendicular line.

right, i knew that but I don't know how to do that :(

[quote="mathsterchief"]Find the area of a quadrilateral whose vertices are (-3,4), (5,6), (3, -2), (-1, -7).[/quote]

Usually, I encounter these types of problems on Mathcounts tests. So, I usually use the Shoelace Therom. Here's an aopswiki link that has a great explanation about it! Its hard to explain on the forum itself.

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

-Hope that helps

Sorry, my mistake. I placed them in a wrong order. I switched (3,-2) and (-1,-7). Your answer is right, but in your solution you don't need to multiply (-3)(4) and (4)(-3) twice.
It should be
$ \frac {1}{2} |(( \minus{} 3)*6 \plus{} 5*( \minus{} 2) \plus{} ( \minus{} 7)*3 \plus{} ( \minus{} 1)*4) \minus{} (4*5 \plus{} 6*3 \plus{} ( \minus{} 1)( \minus{} 2) \plus{} ( \minus{} 7)*( \minus{} 3))|$

$ \frac {1}{2} |( \minus{} 18 \minus{} 10 \minus{} 21 \minus{} 4) \minus{} (20 \plus{} 18 \plus{} 2 \plus{} 21)|$

$ \frac {1}{2} | \minus{} 53 \minus{} 61|$

$ \frac {1}{2} | \minus{} 114|$

$ \boxed{57}$


Last night I found the following stronger inequality:
  For all a,b,cx,y,z>0 we have a(y+z)/(b+c)+b(z+x)/(a+c)+c(x+y)/(a+b)>=sqrt(3(xy+yz+zx).
  Prove this inequality. It is trivial to see that it is stronger than Crux problem. And I have a 3-4 lines solution to this one.

Do not try this one! I didn't posted in the correct place. It is extraordinary difficult.

Wonderful, fantastic and beautiful inequality. It is really one of the best I have ever seen. I like a lot inequalities like the FAMOUS CRUX INEQUALITY and this is as nice as it.
I hope you will post solution as soon as possible.

Thank you very much, Harazi.

I promised a short solution. I hope it will be at most 4 lines:
  Take x+y+z=1. From Cauchy we have sum ax/(b+c)+sqrt(3 sum xy)<=sqrt[(sum x^2)*(sum (a/(b+c))^2)]+sqrt( sum xy*3/4)+sqrt(sum xy*3/4)<=sqrt(( sum x^2+2 sum xy)*(3/2+sum (a/(b+c)^2))<=sum a/(b+c) since it reduces to sum ab/(c+a)(c+b)>=3/4, trivial. So, 4 lines.

Fantastic solution! You virtuously manage with Cauchy. Congratulations!

Namdung

I am really astonished.

This is a SUPER FANTASTIC solution!!!!!


$ (a \plus{} b \plus{} c)(b\plus{}a\minus{}c)(a\plus{}c\minus{}b)(b\plus{}c\minus{}a)(\frac {1}{a^4} \plus{} \frac {1}{  b^4} \plus{} \frac {1}{c^4} )\ge \frac{9}{4}$
\]
for all $ a$ , $ b$ , $ c$ are the  lenghts of a triangle

[quote="ABRORBEK"]$ (a \plus{} b \plus{} c)(b \plus{} a \minus{} c)(a \plus{} c \minus{} b)(b \plus{} c \minus{} a)(\frac {1}{a^4} \plus{} \frac {1}{ b^4} \plus{} \frac {1}{c^4} )\ge \frac {9}{4}$
\]
for all $ a$ , $ b$ , $ c$ are the  lenghts of a triangle[/quote]
Try $ a\equal{}1.99,b\equal{}c\equal{}1$.


a,b are distinct postive integers s.t. $ a^2\plus{}ab\plus{}b^2 | ab(a\plus{}b)$
Prove that $ |a\minus{}b| > {(ab)}^{1/3}$

can someone help me check my solution plz
i am not sure if it is right

thanks

[quote="lgd0612"]a^2 + ab + b^2 | ab(a+b)

(a-b)^3 = -6a * ab (mod a^2 +ab +b^2)

since |-6a| is bigger than 1

|a-b| < (ab)^(1/3)

sorry it is kind of messy, i dont know how to use latex :([/quote]

The transition here is incorrect, i.e. you cannot go from the equivalence to the inequality.

hmm
ok from this step

(a-b)^3 = -6a * ab (mod a^2 +ab +b^2)

i can write it in anther form
(a-b)^3 = K(a^2 +ab+ b^2) - 6a*ab   and K is a non negitive integer
since a,b are both positive,  K(a^2 +ab+ b^2) is also a non negitive integer

if K is positive then
(a-b)^3 > -6a*ab
(b-a)^3 > 6a*ab
b-a > (6a)^(1/3)*ab^(1/3)
b-a > ab^(1/3)
|a-b|> ab^(1/3)

the only case that i need to worry about is when K = 0, (a-b)^3 is at its minimum value: -6a*ab
so i have
(a-b)^3 = -6a*ab
a-b = (-6a)^(1/3)*ab^(1/3)
b-a = 6a^(1/3)*ab^(1/3)

now i know that 6a^(1/3) > 1 (a is a positive integer)
so
b-a > ab^(1/3)
|a-b| > ab^(1/3)

From $ \textrm{[Russia 2001]}$ and the solution was:
Set $ g\equal{}\gcd(a,b)$ and write $ a\equal{}xg$, and $ b\equal{}yg$ with $ \gcd(x,y)\equal{}1$.
Then:
$ \frac{ab(a\plus{}b)}{a^2\plus{}ab\plus{}b^2}\equal{}\frac{xy(x\plus{}y)g}{x^2\plus{}xy\plus{}y^2}$ is an integer.
note that $ \gcd(x^2 \plus{} xy\plus{} y^2,x) \equal{} \gcd (y^2 , x) \equal{} 1$.
Similary: $ \gcd(x^2\plus{}xy\plus{}y^2,y)\equal{}1$. Because $ \gcd(x\plus{}y,y)\equal{}1$, we have
$ \gcd (x^2 \plus{} xy \plus{} y^2 , x\plus{}y) \equal{} \gcd(y^2 , x\plus{}y) \equal{} 1$
So we have:
$ x^2 \plus{} xy \plus{} y^2|g$
implying that $ g \ge x^2 \plus{} xy \plus{} y^2$. Therefore,
$ |a\minus{}b|^3$
$ \equal{} |g(x\minus{}y)|^3 \equal{} g^2|x\minus{}y|^3 g$
$ \ge g^2 (1)(x^2 \plus{} xy\plus{} y^2)$
$ > g^2 xy$
$ \equal{}ab$


induction
number theory proposed
number theory

Is there a sequence $ a_1,a_2,...,a_{100}$ of positive integers such that $ a_{i\minus{}1}<a_i$ and
$ lcm(a_{i\minus{}1}.a_i)>lcm(a_i,a_{i\plus{}1})$ , for $ i\equal{}2$ to $ 99$ ?

Assume that it was true for a sequence of $ k$ integers $ a_1,a_2,...,a_k$ (the proposition). (When k=3 you can easily find an example)
Then we can choose a large enough $ m$ and a large enough prime $ p < ma_1$ which coprime with $ ma_1$
so that $ lcm(p,ma_1) > lcm(ma_1,ma_2)$ because it's equivalent to $ pa_1 > lcm(a_1,a_2)$ and obviously $ L.H.S.$ can be made artitrary large.

Now from our assumption, $ lcm(a_1,a_2) > lcm(a_2,a_3) > .... > lcm(a_{k \minus{} 1}, a_{k})$
$ \Leftrightarrow lcm(ma_1,ma_2) > lcm(ma_2,ma_3) > .... > lcm(ma_{k \minus{} 1}, ma_{k})$

So $ p,ma_1,ma_2,...., ma_k$ will be a sequence of $ k \plus{} 1$ integers that satisfies the condition.

By the principle of mathematical induction, the proposition is true for all $ k \ge 3$, so it's true when $ k \equal{} 100$, done.


Spinners A and B are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' is even?

[img]http://img142.imageshack.us/img142/3076/spinnersls3.gif[/img]

$\text{(A) }\frac{1}{4}\qquad\text{(B) }\frac{1}{3}\qquad\text{(C) }\frac{1}{2}\qquad\text{(D) }\frac{2}{3}\qquad\text{(E) }\frac{3}{4}$

[hide]There are 12 total possibilities.

If the first spinner is 2 or 4, the product will always be even. If it is 1 or 3, only multiplying by 2 gives an even number.

The probability is $\frac{8}{12},$ or $\frac{2}{3}.$ The answer is $\text{D}.$[/hide]

[hide="answer"]D-The product of two #s is even if both aren't odd. The odds of both being odd are 1/2*2/3=1/3. So, the probability of an even # is 1-1/3=2/3. [/hide]

[hide]casework (longer) way:

to have an even product...

case 1 both spinners give even numbers. 1/2 * 1/3 = 1/6
case 2 1st spinner even, 2nd odd. 1/2 * 2/3 = 1/3
case 3 1st spinner odd, 2nd even. 1/2 * 1/3 = 1/6

1/6 + 1/3 + 1/6 = 2/3. D[/hide]

[hide]
D=2/3

I just looked at the possibilities:12 (4 x 3), and figured half already had even numbers, so there were 6. For the other numbers on the spinner with 4 numbers, they each had 1 even possibility, making 2 extra. So, 2 + 6 = 8. But, we have to remember its asking the probability. So, there are 8 out of 12, making 8/12. We put that in lowest terms to get 2/3 (divide the numerator and denominator by 4).

[/hide]

-jorian

[hide="solution"]The probability that the product of the numbers is NOT even is 1/2*2/3=1/3. The probability that the product is even is 2/3. D[/hide]


vector
induction
superior algebra
superior algebra unsolved

How many ordered sequences of $ k$ vectors in $ \mathbb{F}_q^n$ span?  ($ \mathbb{F}_q$ denotes finite field of $ q$ elements).

Let $ V$ be a sub-space of dimension $ 0\leq m\leq n$ and for $ k\in\mathbb{N}$ let $ A_{k,m}(V)$ denote the set $ \left\{\left.(v_1,\ldots,v_k)\in(\mathbb{F}_q^n)^k\,\right|\,V+<v_1,\ldots,v_k> =\mathbb{F}_q^n\right\}$, i.e. the set of ordered vector sequences, which together with $ V$ will span $ \mathbb{F}_q^n$. Obviously, the number of elements of $ A_{k,m}(V)$ does not depend on $ V$. Therefore, we will write $ a_{k,m}$ for the number of such ordered sequences.

For $ k=1$ we get: $ a_{1,m}=\left\{\begin{array}{ll}0 & \mbox{if }m<n-1\\[0.2cm]
q^n-q^{n-1} & \mbox{if } m=n-1\\[0.2cm]
q^n & \mbox{if }m=n\end{array}\right.$

and for $ m=n$ we have $ a_{k,n}=q^{nk}$. Now let $ m<n$ and $ k>1$ then we observe the following construction principle.

$ \begin{array}{ll}(v_1,\ldots,v_k)\in A_{k,m}(V)\Longleftrightarrow & v_1\in V,(v_2,\ldots,v_k)\in A_{k-1,m}(V)\mbox{ or }\\[0.2cm] & v_1\not\in V,(v_2,\ldots,v_k)\in A_{k-1,m+1}(V\oplus <v_1>)\end{array}$

This implies the recursion: $ a_{k,m}=q^m\cdot a_{k-1,m}+(q^n-q^m)\cdot a_{k-1,m+1}$ and using straightforward induction by $ k$ we can verify:

$ a_{k,m}=\left\{\begin{array}{ll}q^{kn}\cdot\prod_{i=1}^{n-m}\left(1-q^{-(k+1-i)}\right) & \mbox{if } k+m\geq n\\[0.2cm]
0 & \mbox{otherwise}
\end{array}\right.$

applying the usual convention $ 1=\prod_{\emptyset}\ldots$

The problem asked for the number of ordered vector sequences of length $ k$, which span $ \mathbb{F}_q^n$. This number is given by $ a_{k,0}$.


function calculus derivative integration geometry

this is homework as a source but ive already done it, just thought it was quite nice.

A uniform rope of mass per unit length $\lambda$ is coiled on a table. One end is pulled straight up with constant velocity $v$. Find the force exerted on the rope as a function of the height, $y$.

Find the total work done on the rope in lifting the end to a height $y$.

Compare (and explain the result) the instantaneous power needed to lift the rope with the rate of change of total mechanical energy (ie. potential plus kinetic).

Hey, there's something wrong with this problem :), I can't get the right answer!

it is quite a nice problem I think. hint..
[hide]
consider the rate of change of momentum. [/hide]

my solution to the first part:

[hide]there will be two reasons a force is required; to change the momentum of the rope from zero to some constant value when lifting it, and to keep the rope already lifted from falling. The second part is quite simple..

the time can be expressed by t = y/v, simple distance/speed

after a time t, there will be a length y in the air, giving a total mass of by (im using b for $\lambda$ because its easier!). The weight of this is gby, and so this is the force needed to keep the rope moving with constant velocity.

the second force is the force needed to change the velocity of the rope from 0 to v. the force is the time derivative of the momentum.

after a time t,

the rate of change of momentum

after a time t, there will be y = tv of the rope in the air. now have a small look at time t+dt, and there will be now v(t+dt) in the air; so v dt extra rope will be in the air. the mass of this little bit of rope is bv dt, and so the momentum is bv^2 dt, so bv^2 is the instantaneous change on momentum=force. (a bit of messy calculus but I think its allowed!)

so the total force is F = bv^2 + gby[/hide]

That's exactly what I did. The problem is that this can't be right! If you integrate the force you'll get the energy all messed up - you'll do more work than the rope gets energy.

Ok the rest of the problem,

[hide]ok so work done is

$W = \int^{y}_{o} F dx$
$W = \int^{y}_{o} F bv^{2} + gby dx$
$W = [xbv^{2} + xgby]^{y}_{0}$
$W = ybv^{2} + bvy^{2}$

Power

$P = \frac{dW}{dt} = bv^{3} + gbtv^{2}$

And the mechanical energy is the potential energy plus the kinetic energy. The kinetic energy is given by:

$T = \frac{mv^{2}}{2}$
$T = \frac{byv^{2}}{2}$

And the potential energy can be found by modelling the entire length of rope, mass by, and length y, as a particle of mass m halfway up the rope. So this is:

$U = mgh$
$U = \frac{bgy^{2}}{2}$

So the total mechanical energy is the sum,

$E = U + T = \frac{bv^{2}t^{2}g + bv^{3}t}{2}$

using $y = vt$

So $E = \frac{Pt}{2}$... I guess this is like saying is you draw a graph of power against time, the energy will be the area underneath...[/hide]

Two things:
[hide="#1"]Dude, how can you integrate the work like that?! You wrote:

$W=\int_0^HF(y)\text{d}x$

but it should really be this:

$W=\int_0^HF(y)\text{d}y$ !!!

And that's different. The result should be:

$W=Hbv^2+\frac{1}{2}H^2bv$.[/hide]
[hide="#2"]You say that $E=\frac{Pt}{2}$. If you apply my correction from #1, it turns out a little different but anyway.... shouldn't be $E=Pt$??? What happens to the rest of the work done? Shouldn't $E=W$?[/hide]


group theory
abstract algebra
superior algebra
superior algebra solved

$R$ is a ring. The product of any two non-zero left ideals of $R$ is not zero. If $RxR=0$ for some $x$. Show that $x=0$.

Let $I=Rx$. Then $I$ is a left ideal and $II=0$, so $I=0$.
Let $J=<x>$, the additive subgroup generated by $x$. Then $RJ=0$, so $J$ is a left ideal.
$I=Rx=0 \Rightarrow x^2=0$, so $JJ=0$, hence $J=0$ and $x=0$.

[quote="amfulger"]Let $J=<x>$, the additive subgroup generated by $x$. [/quote]
Nice solution. I never thought about that.


calculus
LaTeX
real analysis
real analysis unsolved

This is a case study problem on my textbook Applied Calculus 4th ed. Waner/Costenoble ISBN: 9780495384281.

about reducing sulfer emission. If you have this book it's on page 285 and 2 page long problem.

Now i am not sure if topic like this is appropriate in here, please advise me if i am in a wrong place.

Anyway if you have this textbook i would appreciate any help on on exercise 3~8.

Thank you

[img]http://ecx.images-amazon.com/images/I/51h%2BiSzavGL._AA240_.jpg[/img]

It's rather unlikely that anyone is going to have the same textbook as you.  Your options are to either scan the page or to do your own homework.

Actually, you have another option: choose one of the problems (or a similar problem), type it into the forum (using [[LaTeX]] -- see that link or [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]this one[/url]), tell us what you've done and where you're stuck, and someone will probably come along and give you some hints.


[quote="arkhammedos"]find all function with reel numbers wish satisfay
$f(x-1)+f(x+1)=2f(x)+x$[/quote]
Let $f(x)=g(x)+\frac{x^{3}}{6}$, then $g(x+1)+g(x-1)=2g(x)$.
It give $g(x)=g(\{x\})+[x](g(1+\{x\})-g(\{x\}))$, were g(x) is any function in $[0,2)$.

[quote="aviateurpilot"]then $f(x)=g(x)+2([x]x-\frac{[x]([x]+1)}{2})$
and we can choose any $g: [0,1[\to \mathbb{R}$[/quote]
No, this solution is incorrect.
Why do you define $f(x)$ using $g(x)$ for all $x\in\mathbb{R}$, if $g$ is only defined on $[0,1)$?

Rust's solution however is correct, and quite obvious.
$f(x)=g(x)+\frac{x^{3}}{6}$, then $g(x+1)+g(x-1)=2g(x)$, then
$h(x+1)=h(x)$ for $h(x)=g(x)-g(x-1)$, and then $h(x+n)=h(x)$ for all $n\in\mathbb{Z}$.
Then $g(x+n)-g(x) = h(x+n)+...+h(x+1) =nh(x+1)= n(g(x+1)-g(x))$
for all $n\in\mathbb{N}$.(and also for $n\in\mathbb{Z}$)
And this gives $g(x)-g(\{ x\})\! =\! g(\{ x\}+[x])-g(\{ x\})=[x](g(\{x\}+1)-g(\{x\}))$
and $g: [0,2)\rightarrow \mathbb{R}$ can be taken arbitrary.


advanced fields
advanced fields unsolved

Let $X$ be a topological space. If $\{X_j \}$ is a non-empty class of compact subspaces of $X$, each of which is closed, and if $\bigcap _j X_j$ is non-empty, show that $\bigcap _j X_j$ is also a compact subspace of $X$.

Any ideas?

[quote="alekk"]the intersection is closed and is a subset of the compact $X_1$ ..[/quote]

...and a closed subset of a compact set is compact. Thanks a lot!


inequalities
function
integration
real analysis
real analysis unsolved

Prove $ 2^{1 \minus{} q}|\zeta \minus{} \eta|^q\leq|\zeta \cdot |\zeta|^{q \minus{} 1} \minus{} \eta \cdot |\eta|^{q \minus{} 1}|\leq q|\zeta \minus{} \eta|(|\zeta|^{q \minus{} 1} \plus{} |\eta|^{q \minus{} 1})$,
$ 1\leq q < \plus{} \infty,\zeta,\eta\in \mathbb{R}$.

Use Convex function and Clarkson's inequality

Please give me information about Clarkson's inequality!!!

I must admit book_worm, your answer was pretty lame. Can you say something about Clarkson's inequality, or prove the Mazur's inequality using is?

I found the following inequalities and I hope that it is usefull

Let $ p \geq 2$, then for all $ x_1, x_2 \in \mathbb R^N$,
\[ \left| {x_2 } \right|^p  \geqslant \left| {x_1 } \right|^p  \plus{} p\left| {x_1 } \right|^{p \minus{} 2} x_1 \left( {x_2  \minus{} x_1 } \right) \plus{} \frac{{\left| {x_2  \minus{} x_1 } \right|^p }}
{{2^{p \minus{} 1}  \minus{} 1}}.
\]
If $ 1<p<2$, then
\[ \left| {x_2 } \right|^p  \geqslant \left| {x_1 } \right|^p  \plus{} p\left| {x_1 } \right|^{p \minus{} 2} x_1 \left( {x_2  \minus{} x_1 } \right) \plus{} c\left( p \right)\frac{{\left| {x_2  \minus{} x_1 } \right|^p }}
{{\left( {\left| {x_1 } \right| \plus{} \left| {x_2 } \right|} \right)^{2 \minus{} p} }}.
\]

[b]Proof[/b].
Case $ 2 \leq p$. The strict convexity of $ x \mapsto \left| x \right|^p$ implies that for any $ x_1, x_2 \in \mathbb R^N$
\[ \left| {x_2 } \right|^p  \geqslant \left| {x_1 } \right|^p  \plus{} p\left| {x_1 } \right|^{p \minus{} 2} x_1 \left( {x_2  \minus{} x_1 } \right).
\]
Writing $ \frac{{x_2  \plus{} x_1 }}{2}$ instead of $ x_2$ we obtain
\[ \left| {\frac{{x_2  \plus{} x_1 }}
{2}} \right|^p  \geqslant \left| {x_1 } \right|^p  \plus{} p\left| {x_1 } \right|^{p \minus{} 2} x_1 \left( {\frac{{x_2  \plus{} x_1 }}
{2} \minus{} x_1 } \right)
\]
or
\[ \left| {\frac{{x_2  \plus{} x_1 }}
{2}} \right|^p  \geqslant \left| {x_1 } \right|^p  \plus{} \frac{1}
{2}p\left| {x_1 } \right|^{p \minus{} 2} x_1 \left( {x_2  \minus{} x_1 } \right).
\]
Using Clarkson inequality
\[ \left| {x_1 } \right|^p  \plus{} \left| {x_2 } \right|^p  \geqslant 2\left| {\frac{{x_2  \plus{} x_1 }}
{2}} \right|^p  \plus{} 2\left| {\frac{{x_1  \minus{} x_2 }}
{2}} \right|^p
\]
we arrive at
\[ \left| {x_2 } \right|^p  \geqslant \left| {x_1 } \right|^p  \plus{} p\left| {x_1 } \right|^{p \minus{} 2} x_1 \left( {x_2  \minus{} x_1 } \right) \plus{} 2\left| {\frac{{x_2  \minus{} x_1 }}
{2}} \right|^p
\]
or
\[ \left| {x_2 } \right|^p  \geqslant \left| {x_1 } \right|^p  \plus{} p\left| {x_1 } \right|^{p \minus{} 2} x_1 \left( {x_2  \minus{} x_1 } \right) \plus{} \frac{1}
{{2^{1 \minus{} p} }}\left| {x_2  \minus{} x_1 } \right|^p
\]
Repeating this procedure we ca replace the constant
\[ \frac{1}
{{2^{ p \minus{} 1} }}
\]
by
\[ \frac{1}
{{2^{p \minus{} 1} }} \plus{} \frac{1}
{{4^{p \minus{} 1} }}.
\]
By iteration one obtains the constant
\[ \frac{1}
{{2^{p \minus{} 1} }} \plus{} \frac{1}
{{4^{p \minus{} 1} }} \plus{} \frac{1}
{{8^{p \minus{} 1} }} \plus{} ... \equal{} \frac{1}
{{2^{p \minus{} 1}  \minus{} 1}}.
\]

Case $ 1<p<2$. Fix $ x_1, x_2$ and expand the real function
\[ f\left( t \right) \equal{} \left| {x_1  \plus{} t\left( {x_2  \minus{} x_1 } \right)} \right|^p .
\]
Using Taylor formula
\[ f\left( t \right) \equal{} \left| {x_1  \plus{} t\left( {x_2  \minus{} x_1 } \right)} \right|^p .
\]
Then, provided $ f(t) \ne 0$ for all $ 0 \leq t \leq 1$,
\[ \left| {x_2 } \right|^p  \equal{} \left| {x_1 } \right|^p  \plus{} p\left| {x_1 } \right|^{p \minus{} 2} x_1 \left( {x_2  \minus{} x_1 } \right) \plus{} \int\limits_0^1 {\left( {1 \minus{} t} \right)f''\left( t \right)dt} .
\]
 At the same time
\[ f''\left( t \right) \equal{} p\left( {p \minus{} 2} \right)\left| {x_1  \plus{} t\left( {x_2  \minus{} x_1 } \right)} \right|^{p \minus{} 4} \left( {\left( {x_1  \plus{} t\left( {x_2  \minus{} x_1 } \right)} \right)\left( {x_2  \minus{} x_1 } \right)} \right)^2  \plus{} p\left| {x_1  \plus{} t\left( {x_2  \minus{} x_1 } \right)} \right|^p \left| {x_2  \minus{} x_1 } \right|^2
\]
Schwartz inequality yields
\[ f''\left( t \right) \geqslant p\left( {p \minus{} 1} \right)\left| {x_1  \plus{} t\left( {x_2  \minus{} x_1 } \right)} \right|^{p \minus{} 2} \left| {x_2  \minus{} x_1 } \right|^2 .
\]
Wow,
\[ \int\limits_0^1 {\left( {1 \minus{} t} \right)f''\left( t \right)dt}  \geqslant \frac{3}
{4}\int\limits_0^{\frac{1}
{4}} {f''\left( t \right)dt}
\]
Last
\[ c\left( p \right) \equal{} \frac{3}
{{16}}p\left( {p \minus{} 1} \right).
\]

I found the following paper

Annales de l’Institut Henri Poincare (C) Non Linear Analysis, Volume 15, Issue 4, July-August 1998, Pages 493-516

In this paper, Lucio Damascelli proved in page 501 a generalized one  :lol: .


Let $ S$ be a graph over a disc of $ \mathbb{R}^2$ with radius $ r > 0$. If $ S$ has mean curvature $ H \ge a$ at each point for some real number $ a$, show that $ ar \leq 1$

If $ z\equal{}f(x,y)$ is the equation, then $ 2H$ is the divergence of the field $ \vec G: \equal{}\frac{\nabla f}{\sqrt{1\plus{}|\nabla f|^2}}$. If $ H\ge a$ everywhere, then by the divergence theorem the flux of $ \vec G$ across the boundary is at least $ 2a\pi r^2$. On the other hand, the flux must less than $ 2\pi r$ because $ |\vec G|<1$.
http://en.wikipedia.org/wiki/Mean_curvature


[quote="dangvh"]Lets say i send my online application and quiz solutions on April 27 but i have to send my recommendations by snail mail... will I still get "full consideration" if the envelope the recommendations are postmarked on the 27th?[/quote]

Your recommendations should be sent so that they are [i]received[/i] by the 27th.

[quote="mcalderbank"]Same thing for solutions? I dont understand the def of postmark for snailmail.[/quote]

Yes, same for solutions -- they should be received by April 27, not postmarked April 27.
(The post office stamps the date on each letter on the day they receive it.)  As an alternative, you can fax your solutions to (617) 812-6339.

I'm still finishing up on my solutions and putting the finals details on the application (essay, background, and other stuff as well).

Exactly how much less consideration do you give late applications? And are they eligible for merit scholarships and financial aid still?

[quote="MithsApprentice"]Exactly how much less consideration do you give late applications? And are they eligible for merit scholarships and financial aid still?[/quote]

However many new applicants we have slots for, we'll admit that many people from the on-time applications.  Late applications will be considered for the waitlist alongside the remaining applications.   Yes, late applications are eligible for merit scholarships and financial aid.

[quote="aurelie"]Hmm...I have sent my rec and quiz solutions a little while ago, and they definitely should have arrived by now.  However, the deadline is getting dangerously close, and the status is still set to "Awaiting receipt".  Is that just because the page hasn't been updated yet, or should I consider faxing my solutions?[/quote]

Most likely the page hasn't been updated yet.  (Mira, who receives the mail at the Cambridge, MA address, is away at the moment.)

[quote="dangvh"]My teachers finally got around to writing me my recommendations... but I was planning originally going to snail mail them... would it be ok if i just scanned them and posted the file on my application?

edit- now that i think of it i could scan them and email them to you if that's ok[/quote]

the problem with that idea is, your letters may be confidential and your teachers may be expecting you not to read them; so this would be OK only if your teachers write to us to let us know that you have their permission to do this.  as an alternative, your teachers could email us the letters (either the original file, or a scanned copy).

[quote="bubala"]When you talk about faxing in your qualifying quiz, how can you change your option on the online app so that Mathcamp knows you are submitting by fax.  The app only has mail and web.[/quote]

If you'll fax, select that you're submitting by mail.

[quote="dangvh"]umm i got my 2 required recommendations and they will send via email... but i still have one that is in an envelope... could i send that one too? It will arrive late but i just want to know that if another recommendation arrives later would my application be considered late.[/quote]

If 2 recs are in by the deadline, that's fine.

[quote="picounts"]What time in what time zone does it have to be submitted by? I was unable to find it on the Mathcamp website. I know I shouldn't be procrastinating long enough to have to ask this, but I have a couple things left to tweek.[/quote]

Your home time zone is fine, whatever that may be.  We're not [i]such[/i] sticklers. (-:

What should I do if one of my recommenders said she sent the recommendation online today, but it says "awaiting receipt" on the site.  I really need to have an on-time application.

[quote="picounts"]It still says that the recommendation that I sent in by mail was not received. Is that page up to date?[/quote]

It is now up to date, so if it's not marked as received then we don't have it yet.  But don't worry -- we're not evil -- there is always slack on rec letters arriving by mail when the rest is on time.

[quote="The Original Pi Guy"]My solutions arrived on the 28th. Does that mean my applications is considered late? Everything else was in on time.[/quote]

Because of the computer difficulties, we're counting everything that arrived on the 28th as on time.

[quote="lfm"][quote]Because of the computer difficulties, we're counting everything that arrived on the 28th as on time.[/quote]

so, what is the final count of applications then? I assume it went above 209?[/quote]

we've had a few more applications, but not many.  i think the total is 219 now, but a handful of those may now be late applications.


I already have AOPS vol. 2 and I am pretty sure I will qualify for the USAMO, but my only proof experience is with the in-book vol.2 proofs and the USAMTS.  My goal is to qualify for Red MOP, and I'm not in WOOT.

wait. you are confident that you can qualify for USAMO just after completing Aops vol 2? sweet!

after that, i guess you could do art and craft of problem solving by paul zeitz for higher level, stepping into olympiad level problems.
then problem-solving strategies by william engle
for specific practice in particular areas, do the ...for the training of the USA IMO books
they're really good. i have 104 NT and it is awesome!

You could also look through old USAMO tests to practice.

Just do a ton of #1s, #2s and #4s from past USAMOs until you feel pretty confident in your ability to solve the majority of them. If you can solve at least two problems, you should be able to easily make Red.

[quote="Temperal"][quote="t0rajir0u"]Engel and Zeitz are both good resources.  I would also recommend [i]Mathematical Olympiad Treasures[/i] by Titu Andreescu.[/quote]

Isn't it called [i]Mathematical Olympiad Challenges[/i]? Perhaps I'm thinking of a different book...[/quote]

I believe there are two different books, one called Treasures and one called Challenges. I haven't seen Challenges, but I do know that Treasures book is pretty good.

The material in "Mathematical Olympiad N", where N $ \in \{Treasures, Challenges\}$, is generally pretty specialized. While the problems are excellent and numerous, don't expect to find Pompeiu's theorem on every contest you take.

I think you can also order the 21st Century Contests CD through the AMC Store. There are past real competitions from USAMO that you can download, print out, and practice as many times as you want to. Or you can print out the USAMO Competition Booklet/Pamphlet, which has all of the actual problems and solutions.

There are also a lot of Olympiad problems in the [url=http://www.artofproblemsolving.com/Forum/resources.php]AoPS contests[/url] section


Hi, can anyone tell me how to make a case differentation like this with $\LaTeX$?

Thanks

Use the cases environment (which requires the AMS package)
$S(x,a,b): = \begin{cases} 0 & x \leq a \\ 2*\left(\frac{a-x}{b-a}\right)^2 & a \leq x \leq \frac{a+b}{2} \\ 1-2*\left(\frac{b-x}{b-a}\right)^2 & \frac{a+b}{2} \leq x \leq b \\ 1 & x \geq b \end{cases}$


ratio
geometric sequence
arithmetic sequence

Prove that this has an infinite number of solution
a^3+b^3=c^2
and
x^3+r^3=ghj
where g, h,j are in geometric progress
and
same problem as above but g, h,j are in arithmetic progression.

For number 1, $ (a,b,c) \equal{} (m(m^3 \plus{} n^3),n(m^3 \plus{} n^3),(m^3 \plus{} n^3)^2)$ works for any $ a,b,c$.

For number 2, let $ g,h,j$ be in geometric progression with common ratio $ \alpha$, so $ \alpha \equal{} j/h \equal{} h/g\implies gj \equal{} h^2\implies ghj \equal{} h^3$. Then this turns into $ x^3 \plus{} r^3 \equal{} h^3$, which by Fermat's Last Theorem has no nontrivial solutions, so either $ x \equal{} 0$ and $ r \equal{} h$, $ r \equal{} 0$ and $ x \equal{} h$, or $ h\equal{}0$ and $ x\equal{}\minus{}r$.

Here is a family for #1 (may be encompassed in math154's though): $ (a,b,c)\equal{}(x^{2}, 2x^{2}, 3x^{3})$.

thx but could anyone solve the third question

It would certainly help if you posted all the questions together, but if you want to be cheap you can just let the arithmetic progression be $ h,h,h$ which gives infinitely many solutions... (it becomes the same as number 2).

Otherwise you can let $ (r,x,g,h,j) \equal{} (r,3r,r,4r,7r)$ motivated by
\[ (x \plus{} r)((x \plus{} r)^2 \minus{} 3xr) \equal{} h(h^2 \minus{} d^2),\]where $ d$ is the common difference.

Prove: if they have no solutions or an infinite amount of solution
1. a^3+b^3=3d where d is a prime greater than 3
2. a^3+b^3=2^c
3. a^3+b^3=3^c
4. a^3+b^3=p^c where p is a prime greater than 3

This is the rest of the problem.

Oh yeah and a>B and all unknowns are postivie integers

@Math154.

I think we have to take your second option because if we take the first we get that:

$ x^{3}\plus{}r^{3}\equal{}h^{3}$

This is false by Fermat's Last Theorem.

I did not realize that it was positive integers at the time of that post...


function
algebra
partial fractions
geometric series

Does anyone have any general tips for dealing with problems involving functions defined recursively? I know this question is somewhat general, but I am not very good at these types of problems.

I always find guessing to be useful ...

actually, that isn't that far off the mark.  Recursive functions are very often write-outable, so just put down the first many terms and see if you can find a pattern.  Things I look for tend to be the differences between terms and other possibly identifying characteristics, like prime factorizations of terms.

Guessing is definitely a good technique. However, for recursions, you'll often end up with something that you'd never guess. Thus I feel that generating functions are the way to go. Provided that you can sum a variety of infinite series (not always an easy task!), you should be able to solve most recursions that come up on math contests. Having said that, try guessing or whatever else hits you first before you resort to techniques that are way too powerful for the problem you're solving.

If you have specific questions, advice is easier to give.

Thanks for the advice guys. Though Simon, what do you mean by summing infinite series using generating functions? I thought (simple) generating functions were only used for counting problems.

To tell you the truth, I used generating functions for recursions long before I started using them for counting problems. Here's the method in action for a very simple recursion because I'm just lazy.

Problem: f(0)=0, f(n+1)=2f(n)+1. Find f(n).

Solution: Let's multiply both sides of the recursion by xn to get f(n+1)xn=2f(n)xn+xn. Now sum from n=0 to infinity, and call  :Sigma: f(n)xn from n=0 to infinity F(x). Note that the left sides is (F(x)-f(0))/x=F(x)/x. The right side is 2F(x)+1/(1-x). Thus F(x)/x=2F(x)+1/(1-x), or F(x)((1-2x)/x)=1/(1-x), or F(x)=x/(1-x)(1-2x). Now we expand this out by partial fractions and (hopefully!) get F(x)= :Sigma: (2n-1)x^n from n=0 to infinity. Thus f(n)=2n-1.

Cool! I don't think I've ever seen that type of problem solved that way before.

Just one minor detail:

[quote="ComplexZeta"]Now we expand this out by partial fractions and (hopefully!) get F(x)= :Sigma: (2n-1)x^n from n=0 to infinity. Thus f(n)=2n-1.[/quote]

When we expand it out, we actually get your sum from n=1 to infinity (since, after some algebra, F(x) = Sum from n=0 to infinity x^(n+1) * (2^(n+1) - 1)). Shouldn't we need to take this into account and subtract off the expression when n=0?

Here's a handout entitled [url=http://compgeom.cs.uiuc.edu/~jeffe/teaching/373/notes/recurrences.pdf]Notes on Solving Recurrence Relations[/url] that I found useful.

--Dan

[quote]Ah, but f(0)=0, so it doesn't matter. (How do you like the way I wriggled out of that mess?)[/quote]

:D

By the way, the notes are interesting petra.

Hmm try this one:

Find an explicit formula for a[size=75]n[/size] if a[size=75]n+2[/size] = (a[size=75]n[/size] + a[size=75]n+1[/size])/2.

2a_n+2=a_n+a_n+1
characteristic equation-->2x^2-x-1-->2x^2-2x+x-1-->(2x+1)(x-1)
how bout you tell me a_0 or something?

Use a0=1 and a1=2. (I just made those up, but why not?) Also, try to do it with generating functions. It's much more universal than characteristic equations (i.e. you can solve way more recursions).

Okay, here goes nothing.  Our terms are all an.  Let
f(x) = a0x0 + a1x1 + a2x2 + a3x3 + ...

Using our identity, we get that
f(x) = a0x0 + a1x1 + (a0 + a1)x2/2 + (a1 + a2)x3/2 + (a2 + a3)x4/2 + ...

And, recombining terms, we get
f(x) = a0x0 + a1x1 + 1/2*(a0x2 + a1x3 + ...) + 1/2*(a1x2 + a2x3 + ...)

which we can express in the form

f(x) = a0x0 + a1x1 + x2f(x)/2 + x*(f(x) - a0)/2

and combining terms with f(x) on the left, we get

f(x)*(1 - x/2 - x2/2) = a0 + a1x - a0x/2 = x*(a1 - a0/2) - a0

or

f(x) * (x2 + x - 2) = (a0 - 2a1)x + 2a0

or

f(x) = ((a0 - 2a1)x + 2a0)/(x2 + x - 2)

Now, I'm stopping here, but to finish it off, you need to split that into two fractions (one over (x - 1), the other over (x + 2) which we get from factoring (x2 + x - 2)).  Each of these will have a form similar to that for the summation of a geometric series, so you simply need to express them as the geometric series that they are and then add them together to find out what ever coefficient is.

(By the way, you can do the characteristic equation method for general first and second terms, too.  It just gets messy, as this does.  This would be much prettier if I had done it for given values of a0 and a1.


trigonometry
complex numbers
geometry unsolved
geometry

For a complex number $ z\equal{}x\plus{}iy$ we denote by $ P(z)$ the corresponding point $ (x,y)$ in the plane. Suppose $ z_1,z_2,z_3,z_4,z_5,\alpha$ are nonzero complex numbers such that:
$ (i)$ $ P(z_1),...,P(z_5)$ are vertices of a complex pentagon $ Q$ containing the origin $ O$ in its interior, and
$ (ii)$ $ P(\alpha z_1),...,P(\alpha z_5)$ are all inside $ Q$.
If $ \alpha\equal{}p\plus{}iq$ $ (p,q \in \mathbb{R})$, prove that $ p^2\plus{}q^2 \le 1$ and $ p\plus{}q \tan \frac{\pi}{5} \le 1$.

I think I have the first part...but I'm not very good at complex numbers so if I made a mistake, please correct me!

Let $ r = \max(|z_1|,|z_2|,|z_3|,|z_4|,|z_5|)$ so that $ P(z_1), P(z_2),...,P(z_5)$(and therefore $ Q$) lie inside the region $ |z|\le r$.
Since we are given that $ P(\alpha z_1),P(\alpha z_2),...,P(\alpha z_5)$ lie inside $ Q$ these points must also lie inside the region $ |z| \le r$.
Hence for all $ 0\le i \le 5$,
$ |\alpha z_i|=|\alpha\parallel{}z_i| \le r \Rightarrow |\alpha|\le \min\left(\frac{r}{|z_1|},\frac{r}{|z_2|},\frac{r}{|z_3|},\frac{r}{|z_4|},\frac{r}{|z_5|}\right) = 1$.


real analysis
induction
real analysis unsolved

Good Day, Everyone.

[quote="W. Rudin"]Construct a Borel set $E\subset R^1$ such that $0 < m(E\cap I)<m(I)$ for every nonempty segment $I$. Is it possible to have $m(E)<\infty$ for such a set?
[/quote]
Here $m$ is the Lebesgue measure.

I belive that the Cantor's nested set is what we need.  The Cantor's nested set can have finite measure. So the answer to the second question should be "Yes, it is possible". But i still doubt, since the second question implicitly points out that the answer should be "No".

If i understand you right, then for $I=[1/2-\varepsilon;1/2+\varepsilon]$ we will have $0=\mu (\emptyset) = \mu (E\cap I)$  :(

It is true for the Cantor's set, but not for the Cantor's [i]nested[/i] set. May be this terminology is not so popular as i presumed. So i'll explain what i mean.

First, (common) Cantor's set  $C_0$ can be constructed by removing a countable set of intervals from the segmen $[0,1]$: \[ C_0=[0,1]\setminus\bigcup_{s=1}^{\infty} I_s, \]
where $I_s=(\alpha_s,\beta_s)$ are pairwise disjoint intervals. Choosing different $I_s$ we can create Cantor's set such that $m(C_0)=a$, for every $a\in[0,1)$.

Second, we can [i]nest[/i] another Cantor's set in each of the intervals $I_s$. Let $D_s$ be a (common) Cantor's set constructed in the segment $\overline I_s = [\alpha_s,\beta_s]$ using the procedure described above (with $[\alpha_s,\beta_s]$ in place of $[0,1]$). Let\[ C_1 = C_0\cup\bigcup_{s=1}^{\infty}D_s. \]The measure of $C_1$ can have any value from $[0,1)$.

Third, $C_1$ has "holes" just like $C_0$ has. And we may nest second level Cantor's sets in this "holes" to obtain $C_2$. By induction we construct the sequence of sets\[ C_0\subset C_1\subset C_2\subset\ldots \]Finally, put $C=\bigcup_{s=1}^{\infty}C_s$. This is the [i]Cantor's nested set[/i]. The measure of $C$ can be any number in $[0,1]$.

Let $C_{[n,n+1]}$, $n\in\mathbb Z$ be a Cantor's nested set in the segment $[n,n+1]$ and let $m(C_{[n,n+1]})=a_n\in[0,1]$. I belive that the set\[ E=\bigcup_{n=-\infty}^{\infty}C_{[n,n+1]} \]satisfies the conditions of the problem.

My doubts concern the measure $m(E)=\sum a_n$  which can be any number in $[0,\infty]$ (it can be finite in particular).

P.S. Sorry for my poor English.

Can anyone check whether the described construction of the Cantor's nested set is correct. May be i am wrong. It would be nice at least if someone would verify that this set gives a solution for the first question of the problem (see the first post).

I also admit that the presented text may be uncomprehendable because of either my English or bad explanation. In this case, please, ask me for a clarification. Any comments will be very much appreciated.


calculus
integration
algebra
function
domain
pigeonhole principle
Ring Theory

I think its real easy. But it is just a an unexpected result.

Define a "free integral domain" to be a ring (not necessarily commutative) which has NO zero-divisors.

Classify all finite free integral domains.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=26475]here[/url]. That is more than two years old, of course; it doesn't hurt for something like this to come up again after all that time.

Actually I had a simpler answer.
[hide]
Let $\{a_{1},a_{2},...,a_{n}\}$ be the set of all the elements in the free integral domain.
Let $a$ be a non-zero element (assuming it exists).
Consider the set $\{aa_{1},aa_{2},...,aa_{n}\}$ none of these elements are equal to each other.
Because if that was the cas, $aa_{1}=aa_{j}$ so $a(a_{i}-a_{j})=0$ so $a_{i}-a_{j}=0$ because $a\not = 0$. By Dirichlet's box principle $\{aa_{1},...,aa_{n}\}$ is the set $\{a_{1},...,a_{n}\}$ in some order. So we can find a $aa_{g}$ so that $aa_{g}=1$. Meaning every non-zero element has an inverse.
We can think of this finite free integral domain as a division ring. Which is finite. By Wedderburn's Theorem it must be a field. So all free integral domains are powers of primes.
[/hide]

[quote="jmerry"]That's not really simpler- it's the same. My first step was independent of the rest, and I didn't assume the existence of a multiplicative unit.[/quote]
Whatever, I just happen to like pigeonhole arguments.


Simplify 7700 $ \div$ 0.007 and express your answer in scientific notation.


Solving this equation with a,b are positive intergers
$ a^{3b}$=$ b^{2a^2}$
P/s: I think it's hard for me ....

At first try to prove that $ a,b$ are powers of the same positive integer.


geometry
3D geometry
trigonometry
articles

Suppose we have a gas of randomly oriented (and in random motion) electric dipoles. Obviously the dipoles do not behave as free particles. How do we describe it's properties? Can we define potential etc. ?

well if the volume is quite large then  i don't see in ant problem assuming that there is no effect of one on the other for then the intermolecular distance would be far greater then the inter atomic distance.then each dipole would behave as if there were a neutral particle instead
well coming to defining potential since the field is conservative u can..but not always things are defined because they can be...they must have some 'utility'..
would surely like to know ur opinion

Actually I am looking for a solution without the large volume approximation. Is it possible to get an expression for the mean kinetic energy etc. of a partical given its total energy?

may be the things could be made more ordered over here .i think u should specify what's given and what's not so that it becomes simpler

I have cone to this problem working backwards from the Van der Waal's equation of state. I did not like the treatment given in standard texts. I am looking for a more fundamental way of arriving at that equation.

may i know which book have u come across and what is the treatment

The Treatment:
[i]Pressure correction p is proportional to :
(a) the number of molecules present per unit volume in the interior (bulk) of the gas, which are attracting the striking molecules;
(b) the number of molecules per unit volume striking the wall.
Each of these in turn depend on the density d of the gas. Hence p proportional to (d square.) But density proportional to 1/ V, hence p=a/(V square)[/i] :!:

I don't know the standard books on the topic. The book where I got this is not exactly a standard text or one to which I would usually turn for reliable information but it was all that I could get.
IIT chemistry by O.P. Agarwal :stink:

I found in this [url=http://arxiv.org/ftp/physics/papers/0508/0508167.pdf]paper[/url](page no. 4)

[quote]dipole field $ E\equal{}\frac{\mu}{4\pi\epsilon_{0}r^{3}}$ ...
... $ \frac{mv_{w}^{2}(r)}{2}\equal{}kT\minus{}(\frac{\mu_{r}^{2}}{4\pi\epsilon_{0}})(\frac{1}{d^{3}}\minus{}\frac{1}{r^{3}})$[/quote]
Shouldn't there be a term to take care of direction, $ U\equal{}\frac{\mu_{r}^{2}\cos\alpha}{4\pi\epsilon_{0}}$?, $ \alpha$ is angle between field and axis

I didn't read the article, but can you describe the question in more detail?

We have a bunch of dipoles -- is there an external E field applied?

No. They are moving under mutual fields. My problem is that the author seems to have assumed all dipoles to have the same orientation.


Let a(n) be a  sequence of positiv real with:
    1)   a(n+1)-a(n)--->0
    2)   a(n)---> :   :inf:
   Prove that {b(n)} is dense in [0,1] , where b(n)= a(1)+a(2)+...+a(n).[/list][/quote][/b][/u]

This has already been discussed (quite a few times, I might add :)).

Actually,I think it should look like this:

Given that $x_n\to\infty$ and $x_{n+1}-x_n\to 0$,show that $\{x_n\}$ is dense in [$0,1$].

the real problem is to show thet {b(n)} is dense in [0,1]


calculus
integration
trigonometry
LaTeX
calculus computations

[quote="Javkhaa"]hey all could you solve this integral pls
\int(cos(x)/(5+4cos(x)))dx[/quote]

This type of integrals is typically solved with the substitution $ y\equal{}\tan\frac{x}{2}$ to transform it into the integral of a rational function.  Try it!

[quote="Javkhaa"]
$ \int \frac{\cos x}{5+4\cos x}dx$

[/quote]

Put $ \$$ signs around your $ \text{\LaTeX}$ code.

Substitute $ t=\tan \frac{x}{2}$.


Two circles intersect at points A and B. An arbitrary line through B intersects the first circle again at C and the second circle again at D. The tangents to the first circle at C and the second at D intersect at M . Through the intersection of AM and CD, there passes a line parallel to CM and intersecting AC at K. Prove that BK is tangent to the second circle.

Really, not very hard.
Let the intersection between AM and CD be E. Then the most important thing is to notice A,D,M,C and A,B,E,K are both concyclic. The remaining is just some simple angle chasing.

I've tried to prove that A,D,M,C and A,B,E,K are both concyclic and finish the solution but I couldn't. Could you describe the angle chasing?

Okay, here is a detailed solution. I will work with directed angles modulo 180.

Denoting by E the point of intersection of the lines AM and CD, we have EK || CM.

Since the line CM is the tangent to the circle through the points A, B and C at the point C, we have < (CM; BC) = < CAB by the tangent-chord angle theorem. In other words, < (CM; CD) = < (CA; AB). Similarly, < (DM; CD) = < (DA; AB). Thus,

  < CMD = < (CM; DM) = < (CM; CD) - < (DM; CD) = < (CA; AB) - < (DA; AB) = < (CA; DA) = < CAD,

and it follows that the points A, D, M, C are concyclic. This yields < ACM = < ADM. On the other hand, since EK || CM, we have < AKE = < ACM, and thus < AKE = < ADM. But since the line DM is the tangent to the circle through the points A, B and D at the point D, we have < (AD; DM) = < ABD after the tangent-chord angle theorem. In other words, < ADM = < ABD. Hence, < AKE = < ADM = < ABD = < ABE, and this yields that the points A, B, E, K are concyclic. This entails < ABK = < AEK. Since EK || CM, we have < AEK = < AMC. And since the points A, D, M, C are concyclic, we have < AMC = < ADC. Thus,

  < ABK = < AEK = < AMC = < ADC = < ADB.

In other words, < (AB; BK) = < ADB. By the converse of the tangent-chord angle theorem, this yields that the line BK is the tangent to the circle through the points A, B and D at the point B.

$\blacksquare$

  Darij


floor function
combinatorics unsolved
combinatorics

Written the numbers from 1 to 100 on the cards. If there doesn’t exist two consecutive numbers on any chosen 5-cards amoung them then these 5-cards will be called as “good” . Find how many “good” verifiying this condition can be found ?

Is the question, "In how many ways can one choose 5 numbers from $ \{1, 2, \ldots, 100\}$ such that no two consecutive numbers are chosen?"  It's a standard exercise and the answer is $ \binom{100 \minus{} 5 \plus{} 1}{5}$.

In general, for $ N\equal{}\{1,\dots,n\}$ and $ k\leq \lfloor \frac{n}{2} \rfloor$, the number of k-subset so that $ a_i\minus{}a_j\neq 1\quad \forall a_i,a_j\in N$ is given by $ n\plus{}k\minus{}1\choose k$

Proof: Arrange k elements in ascending order. There have to be atleast k-1 elements, one each to be kept between 1st and the 2nd,2nd and the 3rd and so on upto (k-1)th and the kth. Rest of the n-2k+1 can be kept in k possible places (now including the space left and right of 1 and k respectively). This can be done in $ n\plus{}2k\minus{}1\plus{}k\choose k$ ways.


[quote="ckck"]When .2006, with 006 repeating, is expressed as a fraction whose numerator and denominator are relatively prime, what is the sum of the numerator and denominator?

I made 10000x = 2006. 006006 etc. and 10x= 2.006006 etc. and I subtracted, and got:

9900x=2004, and then 167/825, which would give a sum of 825+167=992. What did I do wrong?[/quote]
$ 10000\minus{}10 \equal{} 9990$, not $ 9900$.
So, $ 9990x\equal{}2004 \rightarrow x\equal{}\dfrac {2004}{9990}\equal{}\dfrac {334}{1665}$, and your answer is $ \boxed{1999}$.


if electronegativity differnce among the atoms in a covalent bond increase it's bond strength increases

If the electronegativity difference of the atoms in a bond increase, then the bond caracter changes from covalent to ionic. So we are to compare covalent versus ionic bonds, perhaps by comparing bond dissociation enthalpies with lattice enthalpies.

well actually i am seraching for a explaination of the order of acididty in case of halogen acids

For the halogen acids HX, although bond polarity decreases down the halogen group, the strenght of the bond between hydrogen and the halogen also decreases, and so the acids are increasingly stronger. Bond strenght decreases because the bond is formed by superposition of an H 1s orbital with an sp3 halogen orbital: however, as the halogen size increases the orbital combination becomes increasingly poor due to a incresing size difference - for iodine for example, the orbitals in question are H 1s and I "5sp3" - note the big size difference between the orbitals. Is is so expected that HI is the strongest acid, which is indeed the case. On the other hand, although the electronegativity difference in HF is the greastest (and so it would be expected for this acid to be the strongest) its covalent bond is the strongest because orbital overlap in this case is very efficient (due to the very similar atoms sizes): because of that, HF is only an weak acid in water.

A somewhat similar argument can be used to explain the anomalous behaviour of halogens regarding reactivity and orientation in aromatic electrophilic substitutions.


geometry
incenter
ratio
Euler
geometry unsolved

In acute angled triangle $ABC$, $BC=a$,$CA=b$,$AB=c$, and $a>b>c$. $I,O,H$ are the incentre, circumcentre and orthocentre of $\triangle{ABC}$ respectively. Point $D \in BC$, $E \in CA$ and $AE=BD$, $CD+CE=AB$. Let the intersectionf of $BE$ and $AD$ be $K$. Prove that $KH \parallel IO$ and $KH = 2IO$.

Let $CK$ meet $AB$ at $F$ then from Seva teorem
$\frac{AE}{EC}\frac{CD}{DB}\frac{BF}{FA}=1 \Longrightarrow AE=BD \Longrightarrow\frac{CD}{EC}=\frac{FA}{FB}\Longrightarrow \frac{CD+CE}{EC}=\frac{AB}{FB}$ $\Longrightarrow EC=FB$ and $AF=CD$
So $AF=CD=p-b, BF=CE=p-a, BD=AE=p-c$
Let $A_1,B_1,C_1$ be the feet of altitudes from $A,B,C$, respectively.Let $K_1,K_2$ be feet of perpendiculars from $K$to the side $BC$ and to the altitude $AA_1$ and let $S,T$ and $R$ be the feet of perpendiculars from $I$ and $O$ to the side $BC$, and foot of altitude $O$ to $IS$, respectively.
From the Seva teorem we get $\frac{AD}{KD}=\frac{p}{p-a}=\frac{AA_1}{KK_1} \Longrightarrow KK_1=\frac{p-a}{p}AA_1$
$|HK_2|=|HA_1-KK_1|=|AA_1-AH-AA_1\frac{p-a}{p}|=|\frac{a}{p}AA_1-AH|=|\frac{2S}{p}-AH|=|2r-2OT|$

$\Longrightarrow HK_2=2IR$ $(1)$
$KK_2=A_1K_1=A_1D-K_1D=A_1D-A_1D \frac{KD}{AD}=A_1D-A_1D\frac{p-a}{p}=\frac{a}{p}A_1D=\frac{a}{p}(CA_1-CD)=\frac{a}{p}(\frac{a^2+b^2-c^2}{2a}-(p-b))=b-c$
$\Longrightarrow KK_2=2(p-c-\frac{a}{2})=2ST$
$\Longrightarrow KK_2=2OR$$(2)$
Using $(1)$,$(2)$ and $\angle KK_2H=\angle ORI=\frac{\pi}{2}$
we get $\bigtriangleup KK_2H \sim \bigtriangleup ORI$ with ratio $2$.
So we get $KH \Vert OI$ and $KH=2OI$

We have $(p-a)\overrightarrow{KA}+(p-b)\overrightarrow{KB}+(p-c)\overrightarrow{KC}=\overrightarrow{0}$,   $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OH}$,  $a\overrightarrow{IA}+b\overrightarrow{IB}+c\overrightarrow{IC}=\overrightarrow{0}$. Now, it is easy ! :D

It's very easy problem.We have $ K $ is Nagel point of the triangle $ ABC $ and by Nagel line theorem $ K,I,G $ are collinear and  $ GK=2IG $, on the other hand $ H,G,O $ are collinear ( Euler line ) and  $ HO=2OG $ ,so the triangle $ IGO $ and the triangle $ HGK $ similar, so $ IO $ parallel to $ HK $ and $ HK=2IO $ .

Where $ G $ is centroid of the trianle $ ABC $.


Let  $P$ be an interior point of equilateral triangle ABC such that $\angle{PAB}+\angle{PBC}+\angle{PCA}=90^{\circ}$ . Find all $P$.

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=54016]Locus (nice, nice, ...)[/url].


Let triangle $ ABC, A\minus{}$excircle tagent $ AB,AC$ at $ M,N$ respectively,$ B,C\minus{}$excircle tagent $ BC$ at $ P,Q$ respectively. Prove that $ PM,QN,AH$ concurent (here $ AH$ be altitude of triangle $ ABC$)

See [b]vittasko[/b]'s lemma: http://www.mathlinks.ro/Forum/viewtopic.php?t=164092


I'm not sure if this has been posted before, but I like this problem a lot.

You have just found out that you have a deadly disease. Fortunately, your doctor has the cure. He/she gives you two bottles of pills: one labled A and one labled B. The pills in each bottle are exactly identical. The doctor tells you to take one pill from each bottle for the next 15 days. If you do this successfully, you live. If you fail, you die.

The first day is fine. You take your pills as planned and all goes well. The second day though, you put one A pill into your palm and when you get a B pill, two pills fall from the bottle into you palm. You don't which pills are which. What do you do to make certain that you live?

Good Luck!

Seen this one before.

[hide]
Take a pill from A.

Lay all four of them so that they are not close to each other.

Cut each one exactly in half.

Take one from each of the 4 piles and you have 2 1/2s A and 2 1/2s B.  Remember you have to separate the pills before you cut them.  Then you have to eat the remainder tomorrow.

QED[/hide]

[hide="I think the solution is..."]You crush those three pills and another A pill together, and then eat half of the mixture that day, and the next half the next day.[/hide]

[hide="Or..."]Just throw those pills away and eat new ones[/hide]

You can't throw those out because you only have 15 of each and the doctor doesn't have any more of them. But the other solutions work. Good job!

Another solution I've seen:

[hide]Add another A pill and dissolve all four pills in water. Then drink half the mixture today and half tomorrow. Eat the other pills as planned. (This only works if the pills dissolve in water)[/hide]

I personally prefer mathgeniuse's solution, but anything that works is fine too.

Was this a MathPath problem of the day? It sounds really familiar.

Use the Banach-Tarski Paradox to duplicate yourself some more pills!!! :roll:


geometry
geometric transformation
reflection
geometry proposed

ABC is a triangle.And the circumcenference of ABC is circleO. H is the orthocenter.
Given point P in the plane.AP,BP,CP meets the Circle O at A1,B1,C1.
The reflection of A1 with respect to the midpoint of BC is A2.B2,C2 similarly defined.
Prove that H,A2,B2,C2 are concyclic.
Also if we define A3,B3,C3 as the reflection point of A1 with respect to BC.
We also have H,A3,B3,C3 concyclic.

Dear Mathlinkers,
see the discussion : http://www.mathlinks.ro/Forum/viewtopic.php?t=303466
Sincerely
Jean-Louis


is a good author. Last year, when I had more time than I do now, I used to read many of his books. One of his best is False Memory.

Oh yeah, Dean Koontz rocks. He is one of the few authors even close to being as good as Stephen King. Watchers and Phantoms are his best books.

Personally I have never read Dean Koontz, but Pennywise here is a good friend of mine in person, so I thought I'd post and get him to elaborate on why they are so "good".:p

Dean Koontz is awesome because a.) his books are just as much humorous as they are creepy; he makes fun of all the horror cliches b.) the only fictional character even close to being as cool as Pennywise is Einstien, the dog, or The Outsider, both of which are in Watchers; finally, c.) he writes with a unique style, and doesn't fall into the classic horror stereotypes.


For a positive integer $r$, the sequence $(a_n)$ is defined by $a_1 = 1$ and $a_{n+1} = \frac{n a_n + 2(n + 1)^{2r}}{n + 2}$ for $n \geq 1$.

Prove that each $a_n$ is a positive integer, and find the $n$’s for which $a_n$ is even.

[quote="M4RI0"] Prove that each $a_n$ is a positive integer, and find the $n$’s for which $a_n$ is even.[/quote]

um, what about $n=1$ and $r=2$? i'm getting $a_2=\frac{17}{3}$...

Hm, we have $a_2= \frac{1+8r}3$.  But this isn't an integer for all $r$.  Maybe you're missing some piece of information?  Or maybe I'm misunderstanding the question?

edit: Jason beat me to it :P