induction
algorithm
geometry
geometric transformation
number theory
Euclidean algorithm
relatively prime

Look at these fractions. At firs step we have $ \frac{0}{1}$ and $ \frac{1}{0}$, and at each step we write $ \frac{a\plus{}b}{c\plus{}d}$ between $ \frac{a}{b}$ and $ \frac{c}{d}$, and we do this forever
\[ \begin{array}{ccccccccccccccccccccccccc}\frac{0}{1}&&&&&&&&\frac{1}{0}\\ \frac{0}{1}&&&&\frac{1}{1}&&&&\frac{1}{0}\\ \frac{0}{1}&&\frac{1}{2}&&\frac{1}{1}&&\frac{2}{1}&&\frac{1}{0}\\ \frac{0}{1}&\frac{1}{3}&\frac{1}{2}&\frac{2}{3}&\frac{1}{1}&\frac{3}{2}&\frac{2}{1}&\frac{3}{1}&\frac{1}{0}\\ &&&&\dots\end{array}\]
a) Prove that each of these fractions is irreducible.
b) In the plane we have put infinitely many circles of diameter 1, over each integer on the real line, one circle. The inductively we put circles that each circle is tangent to two adjacent circles and real line, and we do this forever. Prove that points of tangency of these circles are exactly all the numbers in part a(except $ \frac{1}{0}$).
[img]http://i2.tinypic.com/4m8tmbq.png[/img]
c) Prove that in these two parts all of positive rational numbers appear.
If you don't understand the numbers, look at [url=http://upload.wikimedia.org/wikipedia/commons/2/21/Arabic_numerals-en.svg]here[/url].

Hint for a
              [hide] If $ \frac{a}{b}$ and $ \frac{c}{d}$ are consecutive and $ \frac{c}{d}>\frac{a}{b}$then $ bc\minus{}ad\equal{}1$[/hide]
very nice solution for b
 [hide]Draw a line parallel to X-axis and tangent to biggest circlres.notice that by an inversion whit $ radius\equal{}0.5$ this figure doesn't change.now use induction...

[/hide]
Hint for c
[hide]all of fractions like $ \frac{a}{b}$ that $ 0\leq a\leq b\leq n$ appear in $ n\minus{}th$ row...[/hide]

Official Solution:
For part (a) we use induction. If $ bc\minus{}ad\equal{}1$ then $ b(a\plus{}c)\minus{}a(b\plus{}d)\equal{}ba\plus{}bc\minus{}ab\minus{}ad\equal{}bc\minus{}ad\equal{}1$ and $ (b\plus{}d)c\minus{}(a\plus{}d)c\equal{}bc\plus{}dc\minus{}ad\minus{}dc\equal{}bc\minus{}ad\equal{}1$ which proves the induction step. It is obvious that it is true at the beginning of the process.
Now if $ bc\minus{}ad\equal{}1$ then $ (a,b)\equal{}1$. So every fraction appearing in this process is simplified.
We will show that if $ x$ is a number generated in one of the processes, then $ \frac 1x,x\plus{}1,x\minus{}1$ are also produced.
For the first operation: If the $ i^{th}$ number, counting from the beginning of the sequence, is $ \frac ab$, then the $ i^{th}$ number, counting from the end of the sequence, would be $ \frac ba$. This is true at the beginning. ($ \frac 10,\frac 01$) And using induction, it is no hard to prove it.
Now note that the sequence generated by $ a\plus{}1,b\plus{}1$ as the first and the last fraction is exactly the sequence generated by $ a,b$ as the first and the last fraction, plus $ 1$. (that is a $ 1$ is added to each term of the sequence) Again this is easy to prove using induction. At the beginning it is true. If $ \frac ab,\frac cd$ are generated somewhere in the second sequence, then $ \frac {a\plus{}b}b,\frac {c\plus{}d}d$ are generated in the first sequence. The next fraction appearing between them is $ \frac{a\plus{}c\plus{}b\plus{}d}{c\plus{}d}$ which is one unit more than $ \frac{a\plus{}c}{b\plus{}d}$ (the one appearing in the second sequence)
Now note that, in the $ 1^{st}$ step of the process, the number appearing just before $ \frac 10$ is $ \frac 11$. Therefor the numbers appearing between $ \frac 11,\frac 10$ are exactly one unit more than the numbers appearing between $ \frac 01,\frac 10$. (which are all the numbers appearing in the process) So if $ x$ appears in the process, $ x\plus{}1$ does, and if $ x\plus{}1$ does, then so does $ x$.
Now for the second process: It is obvious that if $ x$ appears, then so does $ x\plus{}1$, because the first circles are one unit apart from each other. A unit inversion from the $ 0$ of the x axis, maps circles to circles, and tangent ones to tangent ones. The circle above $ 0$ is mapped to the line $ y\equal{}1$. So the circles tangent to it, are mapped to the circles tangent to $ y\equal{}0$ and $ y\equal{}1$. Each two consecutive circles among these ones are mapped to tangent circles. Therefor these circles are mapped to circles above natural numbers, with diameter $ 1$. (the first circles of the process)
Since the inverse of inversion is itself, the circles above natural numbers are mapped to circles appearing in the process and tangent to the circle above $ 0$.
This inversion maps the point $ x$ of the real line to $ \frac 1x$. Therefor if $ x$ appears in the process, $ \frac 1x$ appears in the image of the process under the inversion which also appears in the first sequence itself.
So in both processes, if $ x$ appears, then so does $ \frac 1x$ and $ x\plus{}1$ and $ x\minus{}1$. (if it's not negative)
Now consider the Euclidean algorithm for a pair of relatively prime numbers $ (a,b)$. In each step if our pair is $ (a,b)$ consider the fraction $ \frac ab$. Then a step consisting of swapping the numbers is the same as replacing our fraction $ x$ by $ \frac 1x$. A step consisting of replacing $ (a,b)$ by $ (a\minus{}b,b)$ is the same as replacing $ x$ by $ x\minus{}1$. Therefor after some number of moves of the form $ x\rightarrow\frac 1x$ and $ x\rightarrow x\minus{}1$, $ x$ becomes $ \frac 01$. (because $ (a,b)$ were relatively prime)
Using the inverse operations, which are $ x\rightarrow\frac 1x$ and $ x\rightarrow x\plus{}1$, we see that we can reach each rational number from $ \frac 01$ which is present in both processes.
So all the rational numbers appear in both processes. What remains to prove, is to show that only rational numbers appear in the second process. But it is a straightforward induction. Assume that two tangent circles $ C_1,C_2$ appeared in the process and we know that their tangency points with the x axis are rational numbers.
Using some unit translations and inversion from $ 0$, we can map $ C_1$ to the circle above $ 0$. We have already shown that these operations and their inverse maps, map circles appearing in the process to circles appearing in the process. So now we have two circles appearing in the process, and one of them is the one above $ 0$. We know that the circles tangent to this one are the circles above $ \frac 1n$'s. So the circle between $ C_1,C_2$ is mapped to a circle above $ \frac 1n$ for some $ n$ which has a rational tangency point. Since our operations map rational numbers to rational numbers, the original tangency point should have been a rational number; which completes the proof.


Hey all,

I posted this problem http://www.mathlinks.ro/viewtopic.php?p=1740588#1740588 almost a year ago in geometry, but it is as much calculus as geometry. If anyone wants to take a crack at it, go ahead, as it is still relevent to my work.

fedja posted a partial solution a while ago, which I missed until today, but he did ask in his post for someone to check it, so have a look and see what you come up with.


[quote="Pain rinnegan"]Let $ a,b,c > 0$. Show that:
\[ \sqrt {a^2 \plus{} b^2c^2} \plus{} \sqrt {b^2 \plus{} c^2a^2} \plus{} \sqrt {c^2 \plus{} a^2b^2} \plus{} 3\sqrt {2abc}\le a \plus{} b \plus{} c \plus{} ab \plus{} bc \plus{} ca\]
[/quote]
Try $ a\equal{}b\equal{}c\equal{}1.$ :wink:

[quote="arqady"][quote="Pain rinnegan"]Let $ a,b,c > 0$. Show that:
\[ \sqrt {a^2 \plus{} b^2c^2} \plus{} \sqrt {b^2 \plus{} c^2a^2} \plus{} \sqrt {c^2 \plus{} a^2b^2} \plus{} 3\sqrt {2abc}\le a \plus{} b \plus{} c \plus{} ab \plus{} bc \plus{} ca\]
[/quote]
Try $ a \equal{} b \equal{} c \equal{} 1.$ :wink:[/quote]

Yes, sorry i forgot a factor. The correct inequality is:

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

\[ \sqrt {a^2 \plus{} b^2c^2} \plus{} \sqrt {b^2 \plus{} c^2a^2} \plus{} \sqrt {c^2 \plus{} a^2b^2} \plus{} 3\sqrt {2abc}\le \sqrt{2}(a \plus{} b \plus{} c \plus{} ab \plus{} bc \plus{} ca)\]


college
group theory
abstract algebra
superior algebra
superior algebra unsolved

Hi,I think here is a nice problem:
Let $G$ is a finite group,and $|G|=2n$,$n$ is an odd,prove:if $o(a),o(b)$ is odd,then $o(ab)$ is odd.

I am not sure for this problem.
And $o(g)$ means the order of the element $g$.

You may look here http://www.mathlinks.ro/Forum/viewtopic.php?t=4451&highlight=berkeley
From there you result follows immediately.

Thanks,
and by this we can prove:
$N=\{g\in G | g$has an odd order$\}$ is the subgroup of $G$ has number of elements n,quickly.


does anyone wanna participate with me?  :D

What city are you from, Poincare?
Oh yay, someone acctually posted in the FL forum!

Venice in Sarasota County please please please participate with me! you only need to be from the same state...

It would be best to make a team with people in your city, that way doing the problems and communicating is easier.


what kind of program/software do you all use?  i use Garageband

Many different ones... Sonar and Pro Tools are good... Reason is amazing as well and I'd highly recommend it to electronic producers.


Im not sure if this is not right or wrong, but this came from the test booklet, and it said "Do not redistribute these or their questions" so i dont know if you should do these questions.

Okay, what are you talking about??

If you are [b]posting inside a topic, click "Post Reply"[/b]

If you are [b]posting a new topic, "click "New Topic"[/b]

I seriously lol'ed for like 15 seconds.

You can probably do the questions... you have the book so why not? No one will check anyways


algebra
polynomial
modular arithmetic
trigonometry
algebra unsolved

Find all polynomials $P(x)$ with no real roots and $ 1 \le \deg P \le 5$ such that
\[ P(x) \cdot P(-x) = P(x^2).\]

[quote="caubetoanhoc94"]$ P(x)$ is a polynomial no has root and $ 1 \le degP(x) \le 5$.Find all $ P(x)$ such that: $ P(x).P( \minus{} x) \equal{} P(x^2)$
My solution very long  :( .I need find nice solution![/quote]
I suppose that "no has root" means "with no real root".
So "no real root" and degree$ \le 5$ means degree=$ 0,2$ or $ 4$

1) degree 0 : $ P(x) \equal{} a$ and $ a^2 \equal{} a$ hence two solutions :
$ P(x) \equal{} 0$
$ P(x) \equal{} 1$

2) degree 2 : clearly $ P(x)$ is a monic polynomial (just look at equality of highest degree terms)
If $ x$ is a root, then $ x^2$ is too, and so $ x^4,x^8, ...$ and so all roots are $ 0$ or complex with modulus $ 1$.
Then the roots are $ e^{i\theta}$ and $ e^{ \minus{} i\theta}$ and we need $ e^{2i\theta} \equal{} e^{ \minus{} i\theta}$ and so $ \theta \equal{} \frac {2\pi}3$

So the solution $ P(x) \equal{} x^2 \plus{} x \plus{} 1$

3) degree 4 : clearly $ P(x)$ is a monic polynomial (just look at equality of highest degree terms)
And, once again, roots are complex with modulus $ 1$ and so :
$ e^{i\theta_1}$, $ e^{ \minus{} i\theta_1}$, $ e^{i\theta_2}$ and $ e^{ \minus{} i\theta_2}$ and :

either $ 2\theta_1 \equal{} \minus{} \theta_1\pmod{2\pi}$ and the solution $ (x^2 \plus{} x \plus{} 1)^2$
either $ 2\theta_1 \equal{} \theta_2\pmod{2\pi}$ and $ 2\theta_2 \equal{} \theta_1\pmod{2\pi}$ and the same solution.
either $ 2\theta_1 \equal{} \theta_2\pmod{2\pi}$ and $ 2\theta_2 \equal{} \minus{} \theta_1\pmod{2\pi}$ and the solution $ (x^2 \minus{} 2\cos(\frac {2\pi}5)x \plus{} 1)(x^2 \minus{} 2\cos(\frac {4\pi}5)x \plus{} 1)$

Hence the *[u][b]edited[/b][/u]* [u]four [/u]solutions (no real roots and degree at most 5) :
$ P(x) \equal{} 0$ [u][b]edited [/b][/u]: is not a solution since it has infinitely many real roots
$ P(x) \equal{} 1$
$ P(x) \equal{} x^2 \plus{} x \plus{} 1$
$ P(x) \equal{} (x^2 \plus{} x \plus{} 1)^2$
$ P(x) \equal{} (x^2 \minus{} 2\cos(\frac {2\pi}5)x \plus{} 1)(x^2 \minus{} 2\cos(\frac {4\pi}5)x \plus{} 1)$

[color=blue][Useless quote deleted. There is no need to quote the entire post immediately preceding yours.][/color]

Why degree=$ 0,2$ or $ 4$
I think degree$ 0,1,2,3,4,5$

[quote="caubetoanhoc94"] Why degree=$ 0,2$ or $ 4$
I think degree$ 0,1,2,3,4,5$[/quote]

But if degree is $ 1$ or $ 3$, $ P(x)$ has at least one real root and the problem specified "no real root", so ... .


find all $n$ such that $n!$ can be shown to multiplication of $n-3$ adjacent numbers
$n!=(x+1)(x+2)...(x+n-3)$

The most obvious solution comes when $n-3 = 1 \implies n = \boxed{4}$.  Next:

$n! = 1 \times 2 \times 3 \times ... \times (n-1) \times n$

We wish to represent this as a product of $n-3$ consecutive integers.  We can get rid of the $1$, obviously, and represent as a product of $n-1$ consecutive integers.  Next, because the 'bulk' of the factors need to be the same, the only way to remove two more factors is to remove $2, 3, 4$ and add a factor at the front $\implies n+1 = 24 \implies n = \boxed{23}$.  For $n > 23$ we will be unable to keep the 'bulk' of the factors consecutive.

Edit:  $\boxed{6!} = 8 \times 9 \times 10$ :(  But such results disappear for $n > 23$ :)


All the members of AoPS-ML who are semifinalists in the Semiens Westinghouse Competition post your name and your category!

We can use this same topic for the Regional Finalists (out in oct 31 it think) too so they (or we  :? ) can meet in the regional competitions.

I was the only semifinalist from my state Puerto Rico!  :)
with my (obviously)-math paper...

I don't mean to speak for him but I believe that Zach Abel is a semi-finalist.


geometry
circumcircle
trapezoid
rectangle
geometry proposed

The circle $c_1$ and $c_2$ are tangent at the point $A.$ A straight line $l$ through $A$ intersects $c_1$ and $c_2$ at the points $C_1$ and $C_2$ respectively. A circle $c$ which contains $C_1$ and $C_2$ and meets $c_1$ and $c_2$ at the points $B_1$ and $B_2$, respectively. Let $\omega$ be the circumcircle of triangle $AB_1B_2.$ The circle $k$ tangent to $\omega$ at the point $A$ meets $c_1$ and $c_2$ at the points $D_1$ and $D_2$, respectively. Prove that:

a.) the points $C_1,C_2,D_1,D_2$ are concircular or collinear
b.) the points $B_1, B_2, D_1, D_2$ are concircular if and only if $AC_1$ and $AC_2$ are diameters of $c_1$ and $c_2$.

Invert wrt $A$. $c_1,c_2$ are turned into two parallel lines, and let, for a general figure $\mathcal F$, $\mathcal F'$ be its image through the inversion.

$D_1'D_2'$ is a line parallel to $B_1'B_2'$, and, since $B_1'C_1'B_2'C_2'$ is an isosceles trapezoid (because it's a cyclic trapezoid), so is the quadraliateral with vertices $C_1',D_1',C_2',D_2'$, so it's also cyclic, meaning that the image of the figure $C_1D_1C_2D_2$ is a circle, so those four points are either concyclic or collinear.

Since $B_1'B_2'\|D_1'D_2'$, it means that those four points are concyclic iff they form a rectangle (they always form a parallelogram). This can happen iff $C_1C_2$ is perpendicular to the lines $c_1',c_2'$, because $C_1'C_2'$ and $D_1'D_2'$ must be equally inclined wrt a line perpendicular to $c_i'$. In turn, this condition holds iff $C_1C_2$ passes through the centers of $c_1,c_2$.


the object of this game is to post a pic that owns the pic above you

ill go first:

[img]http://media.teamxbox.com/dailyposts/halo2bios/01mc.jpg[/img]

[img]http://www.nearingzero.net/screen_res/nz267.jpg[/img]
 :rotfl: Quantum Intepretations.

[img]http://www.breaktaker.com/albums/pictures/animals/AngryDog.jpg[/img]

Ultimate pwnage.

This has been done before....use the search function. and I wanted to post Master Chief!

well you can't resist:

[img]http://www.gamertagpics.com/users/s/SC/SC_Knuckles/Plasma%20Pistol.jpg[/img]

oh yea noob combo pwns!

[img]http://www.nataliedee.com/011606/nerds.jpg[/img]

Picture courtesy of http://www.nataliedee.com

[img]http://www.ticalc.org/images/calcs/92-big.gif[/img]
Pwns four-functions and abacuses any day.

[img]http://www.kmhk.kmu.edu.tw/medhome/Intra_med/med_o/search/images/n1/Nitroglycerin2.jpg[/img]

Shouldn't have drank that....


Joel copied the digits the numbers of the days n the calendar one by one from January 1 to December 31 and arrnaged them in a row. In addition, he wrote the word CALENDAR repeatedly in the row below. How many times is the letter C under a "1"?


I tried to run a program, and I got the message "Illegal Instruction". At the bottom of the screen, it says "busy", and it's stuck like that.  I tried taking the batteries out and putting them back in, but that didn't work, because when I put them back in it just came up with the same screen again.  What can I do?

Read your manual - theres a combination of keys that when pressed resets the system. I remember doing this myself a few weeks back.

Try pressing the "on" button.  In addition to turning the TI-89 on, it also tells it to stop what it's doing.

If that doesn't work, then you should look in the manual, like tetrahedr0n said.

Have you tried tacking the small backup battery out in addition to the regular batteries?

You can really reset a TI-89T by pressing all of the following buttons at once:
Left + Right + 2nd + On

What program were you trying to run?  Was it a seemingly-harmless BASIC program, or was it an assembly/flash program of some sort?  Perhaps a shell like PreOS?

For further reference, try searching for "Illegal Instruction" + TI-89 on Google, and see
http://www.ocf.berkeley.edu/~pad/faq/asm.html

I tried taking out the back-up battery before I read any of the posts, and it worked. Thanks for your responces!

left - right - 2nd - on did bring my ti-89 back to life, just like new.  However, I lost all of my programs  :( .  I think there is another way to exit from the busy state without resetting everything, I just don't know what it is.


[quote="4everwise"]It is given that x varies directly as y and inversely as the square of z, and that x=10 when y=4 and z=14. What is x when y=16 and z=7?

[hide="hint"]Do you know what the terms inversely and directly mean? [hide="If you do"]Find the constant c such that $x=c \frac{y}{z^2}$[/hide][/hide][/quote]
[hide]x=(y/z)k
if 10=(4/14)k
    k=35
then x=(16/7)35
        x=80[/hide]

[color=darkred][size=75][Hide your answers!][/size][/color]

Notice: inverse as the [b]square[/b] of z.  ;)

[hide]According to the problem, we know that $x=\frac{ky}{z^2}\rightarrow 10=\frac{4k}{196}=\frac{k}{49}\rightarrow k=490$.  As a result, if $y=16$ and $z=7$,
$x=\frac{490(16)}{49}=160$.[/hide]

EDIT: Forgot to hide

Masoud Zargar

[hide]the answer is $160$,and nothing else.[/hide]

jensen could u plz send the solution?

because the little guys want to know how did u solve it ;)


Line m passes through (2,5) and the midpoint of the line segment with endpoints (2,7) and (4,13). If the x and y intercepts of line m are (a,0) and (0,b), then find ab.

[hide]The midpoint of the segment with endpoints $(2,7)$ and $(4,13)$ is $(3,10)$.

The equation of the line is $5x-y=5.$

$a=1$

$b=-5$

$ab=\boxed{-5}$[/hide]

[quote="tridecalogism_lover"]isn't the midpoint formula this:

y2-y1/2   + x2-x1/2      ( couldn't get the exponents to work.)

just wanted to check.[/quote]

you mean

$\frac{y_{2}+y_{1}}{2},\frac{x_{2}+x_{1}}{2}$

The midpoint formula is actually this:
$\frac{x_{1}+x_{2}}{2}$,$\frac{y_{1}+y_{2}}{2}$.

[quote="tridecalogism_lover"]isn't the midpoint formula this:

y2-y1/2   + x2-x1/2      ( couldn't get the exponents to work.)

just wanted to check.[/quote]
The midpoint of a segment with endpoints $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is $\left(\frac{x_{1}+x_{2}}2,\frac{y_{1}+y_{2}}2\right)$.

[quote="bos1234"]What is a segment?

Could someone please draw a diagram for this?[/quote]

For a segment, just think about two points and a line drawn that connects them.


For every positive integer $ n$ prove that
$ \sqrt{4n\plus{}1}<\sqrt{n}\plus{}\sqrt{n\plus{}1}<\sqrt{4n\plus{}2}$

[hide="Hint"]$ \sqrt {4n \plus{} 1} < \sqrt {n} \plus{} \sqrt {n \plus{} 1} < \sqrt {4n \plus{} 2} \Leftrightarrow 2n < 2 \sqrt{n^2\plus{}n} < 2n \plus{}1$[/hide]


function
floor function
inequalities
algorithm
logarithms
search
algebra

Define a sequence $ T_n$ by,
 $ T_n\equal{}\max\left(T_{\lfloor{\frac n2}\rfloor}\plus{}T_{\lfloor{\frac n3}\rfloor}\plus{}T_{\lfloor{\frac n4}\rfloor},n\right)$.
Also $ T_0\equal{}0$.

What would be a good method to use to find $ T_n$? Even if its not some simple formula in terms of $ n$. Really I just want to know how you can work around the floor function.

Compute terms and try to find a pattern you can induct on.  Note that we have the inequality $ T_n \ge n$; you might want to consider the sequence $ S_n \equal{} T_n \minus{} n$. What's the source for this problem?

It's not from a math book or anything.  It just popped up while I was trying to solve this programming problem http://www.spoj.pl/problems/COINS/.
I figured that a better understanding of the pattern would help me write a better algorithm.

These kind of recurrences are well studied in algorithms.  Let $ c$ be the unique real number such that
\[ \frac {1}{2^c} \plus{} \frac {1}{3^c} \plus{} \frac {1}{4^c} \equal{} 1.
\]
The value of $ c$ is approximately $ 1.0821315$.  Then $ T_n$ will be bounded above and below by a constant times $ n^c$.  You can prove it by induction.

A good algorithms book should have a discussion of these types of recurrences.

[quote="t0rajir0u"]The floor function isn't a problem from a programmer's perspective; I don't see anything wrong with doing this computationally.  It does seem like looking at $ T_n \minus{} n$ would be helpful, though.[/quote]

It's only a problem because i need to be able to calculate up to $ T_{10^9}$ with strict time constraints. So doing it one at a time will take a while doing them 1 at a time. Also I looked at $ S_n$ but dont see how that helps. maybe I'll see something if I plot it on a graph.

To compute $ T_n$, you can use the recursion formula as is, with one important optimization: store the values of the previous $ T$ values that you have computed, and when it comes time to compute a new $ T$ value, check whether you have already computed it.  (That optimization is sometimes called memoization.)

That approach will work well: You should have at most order $ (\log n)^3$ operations.  That's because there are at most $ \log_2 n$ halvings, at most $ \log_3 n$ divisions by 3, and at most $ \log_4 n$ divisions by 4.  The floors are not a problem because $ \lfloor \frac{\lfloor n / 2 \rfloor}{3} \rfloor$ is equal to $ \lfloor \frac{n}{6} \rfloor$.  So you will get a lot of duplicated values: dividing by 2 and then dividing by 3 is the same as dividing by 3 and then dividing by 2.

In fact, the number of steps will be order $ (\log n)^2$, because $ 4$ is a power of $ 2$.

For example, when $ n$ is a billion, there will be at most 30 halvings, at most 19 divisions by 3, and at most 15 divisions by 4.  So the total number of steps is less than 9000.  Try it!

ok so let me see if i understand what u are saying.
Run the recursion but store every $ T_n$ i get in some sort of array. Then when I need to find a new $ T$ just look through the array for the right vales? Is that it or am i missing somthing?

That's basically it.  One complication is that you would need to initialize the array (to -1 or 0), and initializing a billion values might take too much time.  (Try it and see.)  As an alternative, you can store the previous values in a binary search tree or hash table.  (In C++, you can use a map or hash-map.)

Here's an iterative way to organize the computation (dynamic programming).  Given $ n$, define $ f$ by
\[ f(i, j) \equal{} T(\left\lfloor \frac {n}{2^i 3^j} \right\rfloor).
\]
Then $ f$ satisfies the recurrence
\[ f(i, j) \equal{} \max(f(i \plus{} 1, j) \plus{} f(i, j \plus{} 1) \plus{} f(i \plus{} 2, j), \left\lfloor \frac {n}{2^i 3^j} \right\rfloor).
\]
So you can compute the $ f$ values starting from $ i$ and $ j$ large all the way down to $ f(0, 0)$, which is the desired value.  If $ n$ is at most a billion, then you can restrict $ i$ to at most 30 and $ j$ to at most 19.  So you have a two-dimensional array with at most 570 entries.


calculus
derivative
inequalities unsolved
inequalities

1) Let $a,b,c$ be nonegative real numbers such that
\[a^{2}+b^{2}+c^{2}=a+b+c, \]
find the best exponent $r$ such that
\[a^{r}+b^{r}+c^{r}\leq3. \]

2) Let $a,b,c,d$ be nonegative real numbers such that
\[a^{2}+b^{2}+c^{2}+d^{2}=a+b+c+d, \]
find the best exponent $r$ such that
\[a^{r}+b^{r}+c^{r}+d^{r}\leq4. \]

I understand that this problem is possibly-solved (because many theorem such as mixing variable, EV or SMV) can help us affirm that: it's sufficient to consider the case $a=b$ (first problem) and $a=b=c$ (second problem). The remaining work is checking that case by derivative, but it seems to be not a itelligent work to try  :lol:  So I stop at that step and wellcome all who are so diligent.


induction
combinatorial geometry
combinatorics solved
combinatorics

Given 2n coplanar points, show that it is possible to draw n segments having their endpoints among the given points, such that no 2 of these n segments have a common point.

Did I miss somethng or the problem is easy?

The result is trivial for n = 1. Now suppose that it holds for a given n > 0.
- If the 2n points are collinear, say M_1,..,M_(2n+2) in that order on a line. Then, draw the segments M_(2i+1)M_(21i+2) for i = 0,...,n.
- If they are not collinear, let C be the convex hull of the 2n points. Thus, C is a convex k-gon and wlog, we may assume that M_(2n+1) and M_(2n+2) are consecutive vertices of C (that means these two points are consecutive on a side of C, because C may be degenerated, with many of the points on each side). Thn just draw line segments for the set of the 2n remaining points as ensured by the induction hypothesis, and add the segment M_(2n+1)M_(2n+2)

Pierre.

Let X={((a1,b1),...,(an,bn))| a1,...,an,b1,...,bn are the 2n points in the problem text}
Let L(X) be the set of the sums of the lengths of the segments (ai,bi) from an n-tuple ((a1,b1),...,(an,bn)).

L(X) has a minimal element because X has finitely many elements.
Let t=((a1,b1),...,(an,bn)) be one of the n-tuples corresponding to the minimal element of L(X).

Suppose 2 of the segments from the n-tuple intersect.
If ABCD is a convex quadrilateral then AB+CD<AC+BD. This can be extended to:
If AC and BD intersect, then AC+BD>AB+CD or AC+BD>AD+BC.

So if (ai,bi) and (aj,bj) intersect, then (ai,bj)+(aj,bi)<(ai,bi)+(ci,di) or (ai,aj)+(bi,bj)<(ai,bi)+(ci,di).
This shows that if we replace t by the n-tuple obtained by removing (ai,bi), (aj,bj) and adding (ai,bj),(aj,bi) or (ai,aj),(bi,bj) has a smaller correspondig sum, which cannot happen.


limit
integration
logarithms
calculus
geometry
rectangle
real analysis

for integers $b>a$

what is

$\lim_{n\rightarrow \infty}{(\frac{1}{an}+\frac{1}{an+1}+\frac{1}{an+2}+...+\frac{1}{bn})}$?

[hide="Hint."]$=\lim_{n\to\infty}\frac 1n\left(\frac1a+\frac1{a+\frac1n}+\frac1{a+\frac2n}+\cdots +\frac1b\right).$

Now: could that possibly be considered a Riemann sum? (Feel free to add or delete one term if necessary to make it fit - one term isn't going to matter.)[/hide]

[hide]
As $n \to \infty$, $\frac{1}{an+k} \to \int_{an+k}^{an+k+1} \frac{dx}{x}$. So

$\lim_{n\rightarrow \infty}{\left(\frac{1}{an}+\frac{1}{an+1}+\frac{1}{an+2}+...+\frac{1}{bn}\right) = \lim_{n \to \infty} \int_{an}^{bn+1} \frac{1}{x} dx = \lim_{n \to \infty} \ln{\left(\frac{bn+1}{an}\right)} = \ln{\left(\frac{b}{a}\right)}}$.

This is equivalent to evaluating it as a Reimann sum, right? [/hide]

$\lim_{n\to\infty}\frac 1n\left(\frac1a+\frac1{a+\frac1n}+\frac1{a+\frac2n}+\cdots +\frac1b\right)$

is a riemann sum of 1/x (except for one term) over the interval [a,b].

$\lim_{n\to\infty}\frac 1n\left(\frac1a+\frac1{a+\frac1n}+\frac1{a+\frac2n}+\cdots +\frac1b\right)$ = $\lim_{n\to\infty}\frac 1n\left(\frac1a+\frac1{a+\frac1n}+\frac1{a+\frac2n}+\cdots +\frac1{a+\frac{nb-1}{n}}\right)+\frac{1}{nb}= \displaystyle\int^b_a \frac{1}{x}\,dx=ln{\frac{b}{a}}$

notice how the second limit expression is a left-endpt. riemann sum, which becomes an integral in the limiting case.

paladin, i don't understand your first step, can you explain?

$\frac{1}{an}$ = area of rectangle with width $1$ and height $\frac{1}{an}$. As $n \to \infty$, this approximates the area under the curve $\displaystyle \frac{1}{x}$ from $\frac{1}{an}$ to $\frac{1}{an+1}$. I suppose the Riemann Sum way works more generally.


Consider the sequence of positive integers which satisfies $ a_n\equal{}a_{n\minus{}1}^2\plus{}a_{n\minus{}2}^2\plus{}a_{n\minus{}3}^2$ for all $ n\ge 3$.
Prove that if $ a_k\equal{}1997$ then $ k\le 3$

It is true if $ 2007$ or any number form $ 4^k(8k\plus{}7)$.

[quote="polskimisiek"]That is not true.
For example let's define:
$ a_{1} \equal{} 2 \ a_{2} \equal{} 12 \ a_{3} \equal{} 43$
Then $ a_{4} \equal{} 2^{2} \plus{} 12^{2} \plus{} 43^{2} \equal{} 1997$ :wink:[/quote]
I think you have some misconceptions in problem.
If $ a_3 \equal{} 43,a_2 \equal{} 12$ and $ a_1 \equal{} 2$ then
$ a_0^2 \equal{} 43 \minus{} 144 \minus{} 4 < 0$...
If I am not mistaken this problem is rather famous,I'm pretty sure that I've already solved it...
To [b]Rust[/b]:
It seems to me that this problem is more inequality than N.T :wink:

If $ a_k\equal{}1997\equal{}5\mod 8$, and $ k>3$ then $ 44\ge a_{k\minus{}1}\equal{}a_{k\minus{}4}^2\plus{}a_{k\minus{}3}^2\plus{}a_{k\minus{}2}^2$.
Therefore $ a_{k\minus{}4}\le 6, a_{k\minus{}3}\le 6, a_{k\minus{}2}\le 6$.
If $ a_{k\minus{}1}\equal{}44\equal{}4\mod 8$ must be $ a_{k\minus{}2},a_{k\minus{}3},a_{k\minus{}4}$ are even - contradition.
If $ a_{k\minus{}1}\le 43$, then $ 148\equal{}1997\minus{}43^2\le a_{k\minus{}2}^2\plus{}a_{k\minus{}3}^2\le 36\plus{}36\equal{}72$ - contradition.


geometry
logarithms
calculus
integration
function
calculus computations

Express the area of the region bounded by $ y \equal{} \ln x$ and $ y \equal{} x \minus{} 3$ using integrals (no calculation needed).

I can't seem to figure out what region is bounded by these two curves...

Thanks in advance!

the equation $ \ln x\equal{}x\minus{}3$ is transcendental; its solutions cannot be expressed using standard operations like $ \sqrt{}, \log,$ etc

[quote="pleurestique"]the equation $ \ln x \equal{} x \minus{} 3$ is transcendental; its solutions cannot be expressed using standard operations like $ \sqrt {}, \log,$ etc[/quote]

Um, we haven't learned about transcendental functions, so I'm pretty sure there's a way to do this problem with easier methods.  We can use graphing calculators...

Thanks

Right. In the region that is enclosed, $ x \minus{} 3 < \ln x$, so you'd be integrating $ \ln x \plus{} 3 \minus{} x$ between its two zeros. The problem is that it's not particularly to express what those zeros are. If you just want a numerical approximation, you can just zoom in to the graph of this function and approximate the zeros.

That said, you can make a pretty good approximation to the integral.

[hide="Method how to approximate the integral"]Note that the area does not change if you flip everything over the line $ x \equal{} y$, ie use the inverse functions. Then, you want to find the area enclosed between $ x \plus{} 3$ and $ e^x$. The roots of this satisfy $ e^x \equal{} x \plus{} 3$. Call the roots $ c,d$, where $ c < d$. We are integrating from $ c$ to $ d$.

Now we need to find a convenient indefinite integral in terms of $ c,d$:

$ I \equal{} \int_c^d x \plus{} 3 \minus{} e^x dx \equal{} \left( \frac {x^2}{2} \plus{} 3x \minus{} e^x \right) _c^d$.

But we know that $ c,d$ satisfy $ e^x \equal{} x \plus{} 3$, so:

$ I \equal{} \left( \frac {x^2}{2} \plus{} 2x \minus{} 3 \right)_c^d \equal{} \left( \frac {x^2}{2} \plus{} 2x \plus{} 2 \right)_c^d \equal{} \frac {1}{2}\left( x \equal{} 2 \right)_c^d \equal{} \boxed{\frac {1}{2} (c \plus{} d \minus{} 2)(d \minus{} c)}$

Note that $ x \equal{} \minus{} 3$ makes the RHS 0 and the LHS very close to 0 (1/e^3), and since the RHS is increasing fairly rapidly and the LHS is not, we can guess that $ c \equal{} \minus{} 3$ is roughly the lower root. If you want to take the time, you can use Newton's recursion to find a closer $ c$.

Then, if you notice that the positive root is between $ 1$ and $ 2$ (intermediate value theorem), you can use a Taylor approximation for $ e^x \approx 1 \plus{} x \plus{} x^2 / 2 \plus{} x^3 / 3$. Setting this equal to $ x \plus{} 3$ results in the equation $ x^3 \plus{} 3x^2 \equal{} 12$. Trying some numbers, $ d \approx \frac {3}{2}$ is a decent approximation (you get 99/8 = 12.375 on LHS). If you take the time you can guess a $ d$ that is closer to solving that cubic equation.

Plugging in my approximation for $ c,d$ I get $ \boxed{I \approx \frac {45}{8} \equal{} 5.625}$

This isn't too far off the actual value of $ 5.69445 \ldots$.[/hide]

So if I just wanted to write the integral, would this be correct:

$ \int_{0.052}^{4.505} (\ln x \minus{} x \plus{} 3) dx$

Thanks in advance!

TZF - Thanks for your post.  I got some of it, but I think that a lot of your solution is more advanced than what we have covered.

Consider the tangent line $ y\equal{}x\minus{}1$ and two tangents at $ x\equal{}\frac{c\plus{}1}{2},\ \frac{d\plus{}1}{2}$ where $ c,\ d$ are defined by $ TZF$,

We are to consider a tarpezoid.

[quote="kunny"]Consider the tangent line $ y \equal{} x \minus{} 1$ and two tangents at $ x \equal{} \frac {c \plus{} 1}{2},\ \frac {d \plus{} 1}{2}$ where $ c,\ d$ are defined by $ TZF$,

We are to consider a tarpezoid.[/quote]

I'm a bit confused... Was my expression right? $ \int_{0.052}^{4.505}(\ln x\minus{}x\plus{}3) dx$

Thanks!


analytic geometry
calculus
derivative
complex analysis
complex analysis unsolved

$ \frac {d}{dt} re^{i\theta} \equal{} r\frac {d}{dt}e^{i\theta} \plus{} \frac {dr}{dt}e^{i\theta}$

I'm trying to teach myself complex analysis and this looks like a simple product rule, but why are they taking r to be a variable? I realize that these are probably horrendously simple questions, and I would appreciate it if anyone knows where I could learn the basics of complex analysis, other then buying an expensive text book. Thanks!

I got it from here:

http://www.math.gatech.edu/~cain/winter99/ch2.pdf

The idea is that we're representing an arbitrary curve in polar form $ r(t)e^{i\theta(t)}$. You need two coordinates to represent a generic point, after all.
The two terms of the derivative are the tangential and radial components of the velocity.


[color=orange][b]CLASA A XI-A[/b][/color]
[url=http://www.mathlinks.ro/viewtopic.php?t=266115]Problema 1[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=266112]Problema 2[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1444113#1444113]Problema 3[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1444109#1444109]Problema 4[/url]

Nu ai cumva si subiectul de la clasa a 12 a ?  :lol:

subiectele de la celelalte clase nu le am, dar ma gandesc ca vor fi afisate pe site-ul liceului cat de curand

[color=orange][b]CLASA A XII-A[/b][/color]
[url=http://www.mathlinks.ro/viewtopic.php?t=266728]Problema 1[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=266729]Problema 2[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=266730]Problema 3[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=266731]Problema 4[/url]

Subiectele de la clasele $ VII\minus{}X$ sunt urmatoarele:

[color=orange][b]CLASA A VII-A[/b][/color]
[url=http://www.mathlinks.ro/viewtopic.php?t=266792]Problema 1[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1447183#1447183]Problema 2[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1447187#1447187]Problema 3[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1447194#1447194]Problema 4[/url]

[color=orange][b]CLASA A VIII-A[/b][/color]
[url=http://www.mathlinks.ro/viewtopic.php?p=1447204#1447204]Problema 1[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1447211#1447211]Problema 2[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1447215#1447215]Problema 3[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1447221#1447221]Problema 4[/url]

[color=orange][b]CLASA A IX-A[/b][/color]
[url=http://www.mathlinks.ro/viewtopic.php?p=1447241#1447241]Problema 1[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1447248#1447248]Problema 2[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1447253#1447253]Problema 3[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1447260#1447260]Problema 4[/url]

[color=orange][b]CLASA A X-A[/b][/color]
[url=http://www.mathlinks.ro/viewtopic.php?p=1447276#1447276]Problema 1[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1447287#1447287]Problema 2[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1447294#1447294]Problema 3[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1447300#1447300]Problema 4[/url]


function
parameterization
parametric equation
real analysis
real analysis unsolved

[quote="duke_argonath"]i'm not asking for a total solution, but i am trying to come up with a function that can model the locomotion of a sting ray. basically, it's a circular path on the x-y plane and along that path, the height in the z axis changes sinusoidally. could you give me some starter hints so i can achieve this?[/quote]First make a parametric curve for the path (imagine seeing the sting ray from above, so you can't see the up and down motion).

Then think about the z-motion separately, since it is independent of the x-y motion.

[quote="duke_argonath"]ok, so i have the parametric equation for the x- y plane. that one was easy. BUt for the z axis, do i simply substitute in a sinusoidal function like z=Asin(b+ct) ? would that be ok? or do i need to modify it because it traces a circular path and not a straight one?[/quote]That looks general enough. You might think about the location of the sea floor It can't be at z=0.

right right.. haha.. forgot about translating it, but thanks a bundle anyway


Sign up with your highest rating

First eight to sign up participate in the tournament.

Sorry, I will have to drop out due to an extenuating circumstance.  Good luck to those participating!

Remember people, if you're not very active on FTW, please don't join this tournament. I'm hoping to end this before September.

i join. 1415   its funny how my rating is now 1173.

Ok, we now have 8 people. Deadline is August 21st. No extensions whatsover. First round is best 2 out of 3

#1 isabella2296 vs. #8 Maybach

#4 AIME15USAMO vs. #5 The Archer

#3 MCmaniac vs. #6 cgyao15

#2 Latex vs. #7 Dillinger

AIME15USAMO wins (4-2)(1-4)(4-3)
Last one was intense  :o

#1 isabella2296 vs. #8 Maybach,  Winner vs. AIME15USAMO

#3 MCmaniac vs. #6 cgyao15

#2 Latex vs. #7 Dillinger

bleh izzy maybach, i don't really see a AIME15USAMO

what do u mean u don't really see a AIME15USAMO?

he posted u winning beside izzy and maybach!

No, he posted "[b]Winner [/b]vs. AIME15USAMO", meaning whoever wins the isabella2296 vs. maybach game will face AIME15USAMO.

#1 isabella2296 vs. #4 AIME15USAMO-my pick to win

Izzy defeats maybach 4-2, 4-2

#3 MCmaniac vs. #6 cgyao15

#2 Latex vs. #7 Dillinger

#1 isabella2296
vs Winner of bottom half

#3 MCmaniac vs. #6 cgyao15

#2 Latex vs. #7 Dillinger

#1 isabella2296
vs Winner of bottom half

#2 Latex vs. #7 Dillinger, Winner vs #3 McManiac

I PMed Dillinger 3 days ago, but still no response, and I haven't seen him on....  :(

Edit: He hasn't even read it!

Note: Dillinger is banned from any of my future tournies

I would like this whole tournament to be done on Thursday next week. So the tournament's deadline is next thursday. Latex and McManiac's game is best 3 out of 5. Izzy's and winners game is best 4 out of 7 for championship. I will be gone until next thursday.

#2 Latex vs. #3 McManiac. Winner vs #1 isabella2296

#3 MCmaniac vs #1 isabella2296.

Best 4 out of 7.

I went to New York and Washington D.C from saturday to wednesday.


I was given this problem (without 8a), which was supposed to help solve the problem) and I came up with a solution without 8a). Unfortunately, like most of my solutions, I always seem to skip over some fundamental argument which eventually leads me with a flaw I can't detect:

8a) If $T= x^{2}+\frac{1}{x^{2}}$, determine the values of $b$ and $c$ so that $x^{6}+\frac{1}{x^{6}}= T^{3}+bT+c$ for all non-zero real numbers $x$.

8b) If $x$ is a real number satisfying $x^{3}+\frac{1}{x^{3}}= 2\sqrt{5}$, determine the exact value of $x^{2}+\frac{1}{x^{2}}$

Not having the first half of the question to use as a launch pad, I proceeded as follows for 8b):

[hide]
From the fact that $x^{3}+y^{3}= (x+y)(x^{2}-xy+y^{2})$
$x^{3}+\frac{1}{x^{3}}= (x+\frac{1}{x})(x^{2}-1+\frac{1}{x^{2}}) = 2\sqrt{5}$
So either $x^{2}-1+\frac{1}{x^{2}}= 2$ or $\sqrt{5}$
$x^{2}+\frac{1}{x^{2}}= 3$ or $\sqrt{5}+1$
I checked the answer and noticed that 3 was the intended answer, but what makes $\sqrt{5}+1$ wrong? Similarly, if I were to factor $2\sqrt{5}$ as $\sqrt{2}\sqrt{10}$, what would make those answers wrong? (I haven't checked all of them, some of them may produce a complex $x$, but I know that some of them don't). What did I miss?
[/hide]

Oh, and whoever wants to, feel free to solve it  :P

[hide="8b"]$x^{3}+\frac{1}{x^{3}}= 2\sqrt{5}$  (Square both sides)

$x^{6}+2+\frac{1}{x^{6}}= 20$

$x^{6}+\frac{1}{x^{6}}= 18$

so from a) you know:

$x^{6}+\frac{1}{x^{6}}= T^{3}-3T$ where $T=x^{2}+\frac{1}{x^{2}}$

$\rightarrow T^{3}-3T=18$

$\rightarrow T^{3}-3T-18=0$

$T=3$ satisfies the equation, so $T-3$ is a factor

$\therefore \left(T-3\right)\left(T^{2}+3T+6\right)=0$

$T^{2}+3T+6$ has no real solutions (since its discriminant is less than 0)

$\boxed{\therefore \ T=3}$[/hide]
[hide="Problem with your solution"]In your solution if $x^{2}+\frac{1}{x^{2}}= \sqrt{5}+1$ then
$x+\frac{1}{x}= 2$ since they both multiply to $2\sqrt{5}$

so

$x^{2}+1 = 2x$
$x^{2}-2x+1=0$
$(x-1)^{2}=0$
$\implies x=1$

but if u substitue $x=1$ back into $x^{2}+\frac{1}{x^{2}}$ you dont get $\sqrt{5}+1$[/hide]

So basically it's just something you can check after the fact? As in, if I had taken that into account and discounted my second solution I would have been correct?

Well your solution is okay..i guess..but you kinda got lucky if you think about if lets say if the number was 24 instead of $2\sqrt{5}$ then you wouldnt kno which were the right factors...and you wouldve had to check a lot of the values...the trick with Euclid is most of the time parts a) and b) are connected..

Yeah, it's true that I fell on a bit of luck in my factoring. If I had been given 8 a) I wouldn't have proceeded that way  :P. Thanks for clearing that up for me.


It seems that all you get is raw XML when you open it.  :(

And hide tags do even worse than before.

What do you mean? I get a feed on IE7. I don't actually use other readers but the format is standard RSS 2.0

Hmm... it seems to work with my reader, but not with the web broswer.

As in, it doesn't display like the one for the main website.

It's based on the same class, information is different that's all. I have no problems with it on my browser (IE7).


rotation
geometry
geometric transformation
probability
combinatorics solved
combinatorics

part 1)
There is a round table divided into 4 equal quadrants, with one cup in each quadrant. The quadrants are labeled with letters (A, B, C, D) that do not move. Initially, each cup is randomly face-up or face-down. You are blindfolded and put in front of the table. On each turn of the game, you instruct a genie to flip the cups in whichever positions you choose (e.g., you may say "flip the cups in A and B"), possibly choosing no cups or possibly all four. The genie complies. At this point, if all the cups on the table are face-up, the genie will tell you that you have won the game and are free to go. If not, he rotates the cups randomly (possibly not rotating them) and you play another turn. Give a strategy to win this game in a finite number of moves (the solution is not unique).

Remark: the outcome of "rotation" the four cups is one of the four possible positions: the cups originally at (A,B,C,D) can be at (A,B,C,D), (B,C,D,A), (C,D,A,B), or (D,A,B,C). It is not an arbitrary permutation.

part 2
Suppose that instead of 4, there are n divisions and cups. For which n is it possible to guarantee a win? Prove your answer is correct.

Answer to part 1)
[color=white]Flip none,
flip all,
flip A and C,
flip all,
flip A and B,
flip all,
flip A and C,
flip all,
flip A,
flip all,
flip A and C,
flip all,
flip A and B,
flip all,
flip A and C,
flip all. [/color]

i got the answer to part one, can anyone find a solution to part two?

Thanks for your time

this is an open contest right now at http://www.research.ibm.com!
if you really prove it for n=4, then you got the answer for general n!  ;)

You probably shouldn't be posting answers or soliciting solutions until the window for this month's challenge has closed.  [The IBM "Ponder This" challenges are every calendar month.]

If history serves as a guide, then on August 1, IBM will post an elegant and complete solution to the problem.  Until then, please work on your own, and submit your solutions (and questions) to the IBM Webmaster.  [url]http://www.research.ibm.com/ponder/[/url].

As of this date (July 20), at least 23 people have submitted correct solutions and are acknowledged on IBM's Web page.

Cheers and good luck!

[quote="heartwork"]this is an open contest right now at http://www.research.ibm.com!
[/quote]i'm closing this until next month. someone please remind me then to un-lock it :)

okay as promised the topic has been unlocked :) have fun.

PS this message will self distruct in 5-4-3-2 ... who am I kiddin, I'm no James Bond :P

[quote="tom11784"]ok - do you view their solution as "elegant", because i can't even finish reading it without getting a headache[/quote]

Less elegant than most of their solutions.  Which part is troubling you.  Proving that you can't win if n is not a power of 2?  Or proving that you can win if n is a power of 2?

[quote="tom11784"]let's start with the non-working cases: n=r*2^s

[quote="ibm"]Let the two "special" cups 0 and 2^s start out with opposite
orientations. In the requested move, consider the positions with labels (i*2^s, i=0,1,...,r-1).  Since r is odd, the requests (move, non-move) among these r positions cannot strictly alternate: [i]there are two requests, separated by exactly 2^s, which agree[/i] (either both are to move, or both are not to move). [/quote]

why are there only 2 requests, and why are they seperated by 2^s?
what does it mean for these 2 requests to agree?[/quote]

This is a little different than my proof, but the idea is quite similar...  IBM is proving that the genie can keep these two cups in opposite orientations.  [If you can't get these two cups to the same orientation, you can never be released.]

By dividing out the powers of 2, you are left with a table containing an odd number of cups.  According to IBM, the genie has chosen two adjacent cups (labelled 0 and 1).  The genie initially sets these two cups in opposite orientations (one up, one down), and the genie will always keep these two cups in opposite orientation.

Since r is odd, your list of orders to the genie must contain an adjacent pair of orders that match each other (..., keep, keep, ...) or (..., flip, flip, ...).  The genie predicts your instructions, rotates the table so there is a keep/keep or a flip/flip pair of instructions over the two special cups, and then performs your instructions.  The two special cups remain in opposite orientation.  And you remain imprisoned.

IBM's proof generalizes to consider any table that contains an odd factor, r.  Before the game begins, the genie chooses any set of r cups that are spaced a multiple of 2^s apart from each other.  The genie chooses 2 adjacent cups from among this set, labels these "0" and "2^s", and sets them in opposite orientation.  Hereafter, the genie only considers the r cups that are in his chosen set, and the genie only considers your instructions that apply to these r cups.

so what if the genie messes up and you flip only 1 of them, so they're both up?

i'd hate to think a genie can outthink me like that, but he's prolly had thousands of years to think this through

[quote="tom11784"]so what if the genie messes up and you flip only 1 of them, so they're both up?[/quote]
I think the operative words were "finite" and "guarantee":
[quote="IBM"]Give a strategy to win this game in a finite number of moves.  ...  For which n is it possible to guarantee a win?[/quote]
If there is *any possibility* the genie can outguess you, there is some probability the game will last forever.  Thus you cannot guarantee a win in a finite number of moves.

So, permitting the genie to predict your move doesn't change the result, it just makes the result more obvious.

I submitted a quite different solution to this challenge. And we might expect a surprise for the future  ;)
Really!


real analysis
algebra
function
domain
calculus
integration
real analysis theorems

Which books do you recommend in fourier analysis.
Please mention good introductory books and others.

[hide="Some examples, but please mention others (and if possible comment these)"]
Anyone read Fourier Analysis and Its Applications (Graduate Texts in Mathematics)?
Anyone read Fourier Analysis by T. W. Körner ?

[/hide]

"Fourier Analysis" by Korner is an interesting book, but it's a bit on the recreational reading side. I don't know the GTM text. My recommendation: "An introduction to harmonic analysis" by Katznelson.

It's very introductory in a good way: rigorous, with careful treatment of details, and with hard exercises. Just what you need if you want to actually learn a subject, as opposed to reading about it.

If you are not quite ready for Katznelson, consider another Dover book: "Fourier series" by Tolstov. It has more computations and illustrations, and not as much theory.

Katznelson does presume an understanding of Lebesgue integration.

For the undergraduate course I teach, I use my own notes and no textbook, but the notes are partially modeled on a textbook by David Kammler - I'm not sure if that's in print any more. I used to use Tolstov, and that is a nice book - but it's strongly focused on the case of Fourier series. For the course I want to teach, I prefer to interweave the Fourier transform on four domains: the line, the circle, the integers, and the integers mod $N.$

There's also a book by Bracewell - it's an engineer's book, written by an engineer, but it you can get past that, there's some useful material in it.

i'm currently reading korner and i dont think its recreational. its a very very nice book. i think its THE book to read for fourier analysis if you arent exposed to lebesgue integration( like me). i think it comes with a seperate excersise book, you might want to need that too

I think the first book with Fourier series that I encountered was Churchill's  elementary PDE book - which is eminently forgettable. The next book was E.M. Stein's [i]Singular Integrals and Differentiability Properties of Functions.[/i] - which I then found, when I got to grad school, was the standard ticket for admission to the group I wanted to join. There are, or of course, books that lie in between those extremes.

I would recommend [url=http://www.amazon.com/Introduction-Transforms-Springer-Undergraduate-Mathematics/dp/1852330155/ref=sr_1_1/104-6955053-1979126?ie=UTF8&s=books&qid=1177901952&sr=8-1]this book[/url] if you want to learn the very very basic of Fourier series and Fourier transforms and their applications in PDE's. But Harmonic analysis is a much much more serious course. Im just suggesting that if you are totally knew to it begin with something real easy


trigonometry
quadratics
function
trig identities

The problem is:
1 + sin^2x and match it with one of the 5 following answers
The answer was csc^2x - cot^2x + sin^2x

What steps is there to get to that answer?
Thanks!

cos^2x + sin^2x = 1
cot^2x + 1 = csc^2x
1 = csc^2x - cot^2x
csc^2x - cot^2x = 1

Use $\cos{2x} = \cos^2 x - \sin^2 x, \sin{2x} = 2\cos x\sin x$

Thanks kueh, but I figured it out with using the trig identities we used so far. Can you help me with a couple others?

(cosx/sinx) + sinx/(1+cosX)

And

cotx + tanx = secx X cscx

I figured out most of the other problems, now I am into verifying  :(

Basic trick for this stuff if to convert everything into sines and cosines and then multiply/divide/add/subtract/factor to simplify.  After enough of that, the identity should be obvious.  Be sure to watch out for sin^2 + cos^2 = 1.  It is a favorite in deriving identities with :).

Can someone help me out with
(cosx/sinx) + (sin)/(1+cos)

I really need someones help, literally 6 people I talked to can not figure it out. Pleeease help!
Thank you so so much :lol:  :D

\begin{eqnarray*}
\frac{\cos x}{\sin x}+\frac{\sin x}{1+\cos x} &=& \frac{\cos x+\cos^2 x+\sin^2 x}{\sin x(1+\cos x)}\\
&=& \frac{\cos x+1}{\sin x(1+\cos x)}\\
&=& \frac{1}{\sin x}\\
&=& \csc x
\end{eqnarray*}

Oh my god, thank you soooo much. My friend and I were working on that problem for about 90 minutes, literrally. I will post any problems I have with tommorrow's homework. Thank you everyone for helping me, I am going to do so good on this test, as opposed to everyone else.
Thanks! :lol:

I await JBL's discovery of this topic.  :?

This one is somewhat more OK than the other few topics. I had half a mind of deleting the other one with the quadratic functions. I'm tempted to delete some of this one too.

[quote="seminolecc45"]Hey, can you help me with a couple trig equations?
The first one is: tanX+1=  :sqrt: 3 + :sqrt: 3cotX

And

tanX - cotX

Please help, thanks![/quote]
$cot{x}=\frac{1}{tanx}$

for the first one, factor out the sqrt(3) and isolate all the trig functions on one side. YOu should get something like...

(tanX+1)/(cotX+1)=sqrt(3)
now use the fact that tanX=sinX/cosX and cotX=cosX/sinX and sub that into the equation. You should get:

(sinX+cosX)/cosX * sinX/(sinX+cosX) = sqrt (3)
simplifying:

sinX/cosX=sqrt(3)
tanX=sqrt(3)
therefore X=60 or 240 where 0<x<2pi

Remember that when solving trig equations, it's always useful to convert everything to sines and cosines and go from there. Hope this helps.

Cheers!

[quote="seminolecc45"](cos^2 :theta: - sin^2 :theta: ) / (sin :theta: cos :theta: )

This idenity chapter is very hard. Can anyone help me out with this problem?
Thanks![/quote]

$\frac{\cos^2 \theta -\sin^2 \theta}{\tan \theta}$

$\frac{1-\sin^2 \theta -\sin^2 \theta}{\tan \theta}$

$\frac{1-2\sin^2 \theta}{\tan \theta}$

$\frac{1}{\tan \theta}-2\frac{\sin^2 \theta}{\tan \theta}$

$\cot \theta -2\sin \theta \cdot \cos \theta$

$\cot \theta \cos 2\theta$


Sorry if this was already asked, but what is alcumus?

Search->Alcumus gives

http://www.mathlinks.ro/viewtopic.php?t=184487

It's an online educational tool that we're developing.  If all goes as planned, we'll have our first version of it available to the public within a couple months.

Basically, AOPS is a division of the CIA and Alcumus is a secret project. The next person who posts in this thread will be hunted down by the government.


number theory
greatest common divisor
algorithm
number theory unsolved

[quote="Summerburn"]For $m,n,a,b\in N$, is it true that

$\gcd (m^{a}-n^{a},m^{b}-n^{b})= m^{\gcd (a,b)}-n^{\gcd (a,b)}$?
[/quote]

This part only stands when $(m,n)=1$, and if they are not coprime you can easily find some counter-example.
\[(m^{a}-n^{a},m^{b}-n^{b}) = (m^{a}-n^{a},m^{a}-n^{a}-m^{a-b}(m^{b}-n^{b})) = (m^{a}-n^{a},n^{b}(m^{a-b}-n^{a-b})) = (m^{a}-n^{a},m^{a-b}-n^{a-b}) \]
because $(m,n)=1$. Now, as you can see, just continue this like in Euclidian algorithm, so you will get willing result.

The real question was already asked before. The answer is similar, but needs some cases. It's easy to find out yourself if you already know the first one.


APMO has been posted on the  [url=http://www.kms.or.kr/competitions/apmo/]official website[/url], so the problems may now be discussed.

[hide="Old Post"]As I understand, Asian Pacific Mathematical Olympiad (APMO) will be held in the upcoming week.

Your exam adminstrator will likely instruct you to not disclose the Olympiad questions until they have been posted on the internet (probably [url=http://www.cms.math.ca/Competitions/APMO/]here[/url]). You will proably not be allowed to bring home the problem sheet.

We should comply with this regulation on AoPS-MathLinks.

Several years ago (I think in 1997 and 1998), APMO results had to be cancelled due to premature leak of the problems. We do not want this to happen again.

So, please, [b]NO DISCUSSION OF APMO 2005[/b] until the problems are officially posted on the internet.

And I would ask all moderators and adminstrators to comply with this request.

I posted a similar announcement for [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26017]Tournament of the Towns[/url], yet some people still managed to leak the problems right here on AoPS - MathLinks ...  :mad:[/hide]

Please follow this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30956]link[/url] for the discussion of the problems.


What number, when added to the numerator and to the
denominator of $ \frac{5}{8}$, results in a fraction whose value is 0.4?

The only way I know how to do this is by using an equation.  Call the number $ x$.  We know that $ 0.4\equal{}\frac{2}5$
\[ \frac{5\plus{}x}{8\plus{}x}\equal{}\frac{2}{5}\]
Now, we cross multiply:
\[ 25\plus{}5x\equal{}16\plus{}2x\]
Now we simply solve:
\[ 3x\equal{}\minus{}9\implies x\equal{}\boxed{\minus{}3}\]

Let $ x$ be the number.

$ \frac{5\plus{}x}{8\plus{}x}\equal{}\frac{2}{5}$

Cross multiplying gives

$ 16\plus{}2x\equal{}25\plus{}5x$
$ 3x\equal{}\minus{}9$
$ x\equal{}\boxed{\minus{}3}$

EDIT: Beaten, again, darn. :(

Notice that $ 0.4 \equal{} \frac{2}{5}$, and that the numerator and denominator is 3 less than than that of $ \frac{5}{8}$. Thus, our answer is $ \boxed{\minus{}3}$.

Alternatively, we could have used algebra, setting up the equation $ \frac{5 \plus{} x}{8 \plus{} x} \equal{} 0.4 \implies 5 \plus{} x \equal{} \frac{16}{5} \plus{} \frac{2x}{5}$ and then solving. However, this is tedious and takes a long time, considering this is sprint and any bit of extra time helps.

What I usually do in general for these questions is to mentally list the first few fractions that equal the decimal in question. For example, for 0.4, I would list 2/5, 4/10, 8/20, etc., and then compare the fractions to 5/8.

EDIT: Wow, beaten? But my solution was a little different.


$\text{ If }m_{1}\text{ and }m_{2}\text{ are the two smallest solutions to the equation: }$

$2^{n}+3^{n}+4^{n}+70=m^{2}$

$\text{ where }m, n \in \mathbb{Z^{+}}\text{, find }m_{2}-m_{1}.$

[hide="Well this was rather unsatisfying"] It turns out that

$2^{3}+3^{3}+4^{3}+70 = 169 = 13^{2}$

And that

$2^{5}+3^{5}+4^{5}+70 = 1369 = 37^{2}$

So the difference is just $24$.  (I was calculating the first few values to ascertain patterns, but I ended up getting a solution immediately  :P ) [/hide]

[b]Follow-up:[/b]  Prove or disprove that there exist only a finite number of solutions.


inequalities
algebra
polynomial
calculus
derivative
inequalities proposed

Prove that
\[ \sum\frac{(a+b-c)(2a+2b-c)+(a-b)^{2}}{a^{2}+ab+b^{2}}\ge3 \]
I don't have a beautiful solution of this problem ... I propose this problem because it is symmetric with three variable without square root (and homogenous, with equality when a=b=c) but yet still seems difficult. And at least it is almost impossible if one tries to solve it by brute force (clear the denominator and lots of computations, AM-GM / Muirhead ...).

[quote="arqady"][quote="Darkseer"] And at least it is almost impossible if one tries to solve it by brute force .[/quote]
I don't agree with you! :D[/quote]Yes  :)   I admit that I should not be so sure. I will really appreciate if there is such a solution. :D

oh :D  excellent. Thank you very much.

After clearing the denominator the form is somehow different than I thought, but no doubt yours is right. I must have made some stupid mistake. And I didn't think of the possibility of using the term $ (a-b)^{4}$ ... clearly I was wrong.

I checked my draft and I didn't make a mistake in computation. In fact I got the same thing as your first line (or second?), but just fail to see the key point.

But that's not important here :P. What I would like to ask is a problem which motivated me to this problem I created. I considered $ a^{3}+b^{3}+c^{3}-a^{2}b-b^{2}c-c^{2}a-ab^{2}-bc^{2}-ca^{2}+3abc\ge 0$ first, which is a direct consequence of Schur's. However I know of no method of proving this one by direct combination of AM-GM (Muirhead is considered as a direct consequence of AM-GM), still maybe somebody can correct me in this point.

Then I square the equation. It becomes (if I am not mistaked)
\[ \sum_{3}a^{6}-2\sum_{6}a^{5}b-\sum_{6}a^{4}b^{2}+8\sum_{3}a^{4}bc+4\sum_{3}a^{3}b^{3}-6\sum_{6}a^{3}b^{2}c+15a^{2}b^{2}c^{2}\ge 0 \]
where the number below sigma denote the number of terms, and the summation is always taken over all symmetric terms.

My question is, how can we solve stuff like this? some analysis may show that the above polynomial has four zeroes when $ a=b=c$, e.g. the 0th to 3rd derivatives of the polynomial are zero when $ a=b=c$. (usual ineq only have two zeroes). But we can just add something to pollute it, e.g. add $ \sum a^{3}b^{2}c-6abc$. (I tried to pollute it in this sense, but apparently I wasn't successful :P). Then how can we solve it, if not using (either explicitly or implicitly) the fact that it is the square of the equation above? or is there any way that may lead us to such observation without lots of trial and error?

Let me have another try on my idea  :D  Prove that for any $ a,b,c>0$ we have

\[ 3\sum_{cyc}\frac{a(b+c)}{b^{2}+bc+c^{2}}\le5+\sum_{cyc}\frac{a^{2}}{b^{2}+bc+c^{2}}\]

The inequality

$ a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}a^{2}b\minus{}b^{2}c\minus{}c^{2}a\minus{}ab^{2}\minus{}bc^{2}\minus{}ca^{2}\plus{}3abc\geq 0$

is equivalent to

$ abc\geq (\minus{}a\plus{}b\plus{}c)(a\minus{}b\plus{}c)(a\plus{}b\minus{}c)$.

If $ a,b,c$ are not the sides of a triangle it is obvious, because $ LHS$ is negative, and if they are the sides of a triangle, it is $ R\geq 2r$.

Yes, that's nice of course, but I know no method to generalize such result/method (at least to me, it seems that we are seldom so lucky).

Any idea to my second problem?


As you probably know, the Simpsons Movie is coming out while camp is in session. From what I understand, there is a movie theater in Waterville. Does anyone want to go see it as a mini-field trip?

Count me in, though I'm not sure mini-field trip is the right word, but whatever...

If you need a staff member to take you on that field trip, count with me.  I had already thought of it!

I'm in.  Only sixty or so hours left until Mathcamp....


logarithms
analytic geometry
algebra
function
domain

[quote="bane"]$\begin{cases}\log_{x-y}8(x+y)=-2 \\ (x+y)\log_{4}(x-y)=\frac{1}{2}\end{cases}$

How to solve it? What to start with?[/quote]
[hide="how to solve it!!"] For the first equation, you get $x-y^{-2}=8(x+y)$. Solving this, you get $\frac{1}{(x-y)(x-y)}=8(x+y)$. I let us try something like moving all the variables to one side by first multiplying to rid ourselves of those nasty fractions. $\frac{1}{x-y}=8(x^{2}-y^{2})$. Now, let us finish off the multiplication. $1=8(x^{3}-xy^{2}-x^{2}y+y^{3})$. We got that so now, lets move on to the bottom equation. We divide both sides by $x+y$ and turn it into a non$log$ equation. You get $\sqrt{4}=x^{2}-y^{2}=2$. Factor the results of the first equation and substitute the $x^{2}-y^{2}$ and get $1=16(x-y)$. Now you know that $\frac{1}{16}=x-y$. Divide the $x^{2}-y^{2}$ in the second equation to get $x+y=32$. you have two linear equations with degree of one so the rest is pretty basic. $\begin{cases}x+y=32\\x-y=\frac{1}{16}\end{cases}$. $2x=32\frac{1}{16}\\=\frac{513}{16}$. Then, you get $x=\frac{513}{32}$. Which leaves $y=\frac{511}{32}$. If you are supposed to represent it as a coordinate, it is $(\frac{513}{32},\frac{511}{32})$.[/hide]

bigboy82892, you made a grave mistake in handling the second equation - it should be $x-y=2^{1/(x+y)}$ and not $x^{2}-y^{2}=2$

[quote="bane"]$\begin{cases}\log_{x-y}8(x+y)=-2 \\ (x+y)\log_{4}(x-y)=\frac{1}{2}\end{cases}$

How to solve it? What to start with?[/quote]

[hide]There are no integer solutions for x and y.

Solving the first equation:

$8\left(x+y\right) = \left(x-y\right)^{-2}$

This implies that the second equation results in:

$\left(x-y\right)^{\frac{1}{8\left(x-y\right)^{2}}}= 4^{\frac{1}{2}}= 2$

$\left(x-y\right) = 2^{8\left(x-y\right)^{2}}$

If $x$ and $y$ are integers then $x-y$ is an integer.

Let $x-y = n$ then,

$n = 2^{8n^{2}}.$

There is no integer for which the above expression is true.[/hide]

Hi, guys! You should the condition for which the logarithm can be defined.

What do you mean, kunny? Is there something wrong with my solution?

[quote="Kurt Gödel"][hide="I think this is correct"]Let $p=8(x+y)$ and $q=x-y$. We get that $\log_{q}p =-2$ and $p\log_{4}q = 4$. Therefore, $p = q^{-2}$ and $q^{p}= 4^{4}$. So $q^{2}= p^{-1}$ and $4^{8}= \left(q^{p}\right)^{2}= \left(q^{2}\right)^{p}= p^{-p}$. It's easy to prove that $f: \mathbb{R}^{+}_{0}\to\mathbb{R}^{+}_{0}: x\mapsto x^{-x}$ reaches an absolute maximum at $\frac{1}{e}< 4^{8}$. Therefore, there are no solutions.[/hide][/quote]

Very clever, nicely done. Except for one small error nothing is wrong with it. The absolute maximum of $x^{-x}$ occurs when $x = e^{-1}$ and thus $max\lbrace x^{-x}\rbrace = e^{e^{-1}}.$

[quote="mathgeniuse^ln(x)"]Nothing is wrong with it...he just wants you guys to state the domain of the logs...[/quote]

You should remark another condition. :wink:


algebra
polynomial
number theory
prime numbers

If $ p$ and $ q$ are prime numbers such that $ p|q^{3}\minus{}1$ and $ q|p\minus{}1$ prove that $ p\equal{}q^{2}\plus{}q\plus{}1$

[hide]
We know that the special form for factoring $ x^3\minus{}y^3$ is $ (x\minus{}y)(x^2\plus{}xy\plus{}y^2)$.
So we know that $ p \mid (q\minus{}1)(q^2\plus{}q\plus{}1)$.
Because $ q \mid p\minus{}1$, $ q<p$.
Therefore, $ p \nmid (q\minus{}1)$, which is less than p.
We now can see that $ p \mid (q^2\plus{}q\plus{}1)$.
$ (q^2\plus{}q\plus{}1)$ cannot be factored, so it is prime.
Because $ (q^2\plus{}q\plus{}1)$ is prime, its only divisors are 1 and itself.
But we know that $ p \mid (q^2\plus{}q\plus{}1)$, so $ p\equal{}q^2\plus{}q\plus{}1$
[/hide]

Just because $ q^2 \plus{} q \plus{} 1$ is an irreducible polynomial in $ q$ does not mean it is prime for all $ q$. A counterexample in this case is $ q \equal{} 7$, when $ q^2 \plus{} q \plus{} 1 \equal{} 57 \equal{} 3 \cdot 19$.


calculus
integration
calculus computations

It's been discussed. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=113208]Here[/url] is one link, and there are some further links inside that topic.


Find all finite sets $ S$ of nonnegative integers with the property that for any $ m,n\in S$ we have $ |m\minus{}n\plus{}1|\in S$.

I think that all nontrivial (that is, nonempty) solutions are $ S \equal{} \{1,1\plus{}k,1\plus{}2k,\ldots\}$.

[hide="Proof:"]It's obvious that $ 1\in S$. Denote the smallest element which is greater than one by $ 1\plus{}a$ and suppose that there exists a $ 1\plus{}b\in S$ with $ a\not|b$. Then $ b>a$, and thus $ 1\plus{}b\minus{}a,1\plus{}b\minus{}2a,\ldots$ are also elements of $ S$. Since $ 1\plus{}a$ is the smallest element greater than one, this process terminates with 1, so $ b\equal{}ka$, which is a contradiction.

Now denote by $ 1\plus{}c$ the largest element of $ S$; then $ 1\plus{}c\minus{}a,1\plus{}c\minus{}2a,\ldots$ are all elements of S. This finishes the proof.[/hide]

How do you guarantee $ 1 \plus{} b \minus{} 2a\in S$? It should be $ |1 \plus{} b \minus{} 2a|$ and similarly for the following terms I think.

Also, what if $ k\equal{}1$? Then from your theory we have $ 1\in S, 2\in S\implies |1\minus{}2\plus{}1|\equal{}0\in S$. So for $ M\equal{}\max S$ we'll have $ |M\minus{}0\plus{}1|\equal{}M\plus{}1\in S$ which is a contradiction.

[quote="bertram"]I think that all nontrivial (that is, nonempty) solutions are $ S \equal{} \{1,1 \plus{} k,1 \plus{} 2k,\ldots\}$.

[hide="Proof:"]It's obvious that $ 1\in S$. Denote the smallest element which is greater than one by $ 1 \plus{} a$ and suppose that there exists a $ 1 \plus{} b\in S$ with $ a\not|b$. Then $ b > a$, and thus $ 1 \plus{} b \minus{} a,1 \plus{} b \minus{} 2a,\ldots$ are also elements of $ S$. Since $ 1 \plus{} a$ is the smallest element greater than one, this process terminates with 1, so $ b \equal{} ka$, which is a contradiction.

Now denote by $ 1 \plus{} c$ the largest element of $ S$; then $ 1 \plus{} c \minus{} a,1 \plus{} c \minus{} 2a,\ldots$ are all elements of S. This finishes the proof.[/hide][/quote]

? if we do $ |(k\plus{}1)\minus{}(2k\plus{}1)\plus{}1|\equal{}k\minus{}1$

[hide="Solution"]The case m=n implies that 1 is in the set.

Now suppose 2. Then, |1-2+1|=0 must be in the set, and with 0 in the set, now, for k in the set, |k-0+1|=k+1 is also in the set, and we have the set of integers. Thus, neither 2 nor 0 can be in the set.

Now suppose that n is in the set, where n>2.

|n-1+1|=n is already in the set, and |1-n+1|=n-2 is in the set, and thus n-2k is in the set. Since n-2k should never be 2, n must be odd, and the numbers {1, 3, 5, ... n} are in the set.

No more numbers are necessary, and the set is complete.

Thus, S may be any set of consecutive odd natural numbers starting from 1.[/hide]

Yeah, I disagree too.

[hide="Solution"]$ S$ being empty is a possible answer, I believe. In the case that there is an element $ s \in S$, take $ m \equal{} n \equal{} s$ to get that $ 1 \in S$.

Lemma: $ 0 \notin S$. Suppose it was, then if $ a \in S$, $ |a \minus{} 0 \plus{} 1| \in S$, so $ a\plus{}1 \in S$. By induction, $ S$ is the set of nonnegative integers, contradicting $ S$ fineite.

Suppose $ a \in S$ and $ a \geq 2$. Then $ |1\minus{}a\plus{}1| \equal{} |2\minus{}a| \equal{} a\minus{}2 \in S$. So if $ a$ is even, we can repeatedly get elements $ 2$ less until we have $ 0 \in S$, contradicting the lemma.

There is no problem if $ a$ is odd, but we must have all odd numbers preceding it. Therefore, the solutions are:

$ \{ \}, \{1\}, \{1,3\}, \{1,3,5\}, \ldots$[/hide]

Blah beaten.

Suppose that S has at least two elements...

If these are m,n with m>n, then m-n-1 is in S and m-n+1 is in S, so if we call m-n-1=x, then x and x+2 are in S

I think that you guys use this observation in your proofs, but i wanted to state it more explicitly...

I am not sure if this leads to a nicer solution...It would be nice to say that generally, if m is in the set, then so is m+2k where k is an integer.


Let $f_{1}(x)=\frac{1}{1-2x}$, $f_{2}(x)=\frac{1}{1-2x-\frac{4x^{2}}{1-2x}}$, $f_{3}(x)=\frac{1}{\frac{4x^{2}}{1-2x-\frac{4x^{2}}{1-2x}}}$,...
let $f(x)$ be the limit of this sequence as $n$ tends to infinity for $x$ in some set $S$.
Prove that if for some given real $x$, the continued fraction $f(x)$ converges then the numer $y=f(x)$ satisfies: $y(1-2x-y)=4x^{2}$

I don't understand your sequence... maybe it would be better if you gave a recursive definition.

its hard to type..
what part you dont understand?

Well I don't see any relationship between $f_{1}(x)$ and $f_{2}(x)$, while $f_{3}(x) = \frac{1}{4x^{2}f_{2}(x)}$.

oops! $f_{1}=\frac{1}{1-2x}$ sorry. now it should be clear


[quote="Ligouras"]If $ a, b, c > 0$, then

              $ (a^2 \plus{} b^2 \plus{} c^2)^3\geq(a^3b \plus{} b^3c \plus{} c^3a)(a \plus{} b \plus{} c)^2$[/quote]

It's Vasc's inequality multiplied with $ 3(a^2\plus{}b^2\plus{}c^2)\ge (a\plus{}b\plus{}c)^2$

[quote="Ligouras"][quote="Pain rinnegan"][quote="Ligouras"]If $ a, b, c > 0$, then

              $ (a^2 \plus{} b^2 \plus{} c^2)^3\geq(a^3b \plus{} b^3c \plus{} c^3a)(a \plus{} b \plus{} c)^2$[/quote]

It's Vasc's inequality multiplied with $ 3(a^2 \plus{} b^2 \plus{} c^2)\ge (a \plus{} b \plus{} c)^2$[/quote]

Interesting.... thanks   :)
Do you have a solution that harmonizes the two solutions in an only?
If you answer "Yes" you are a great mathematician... Sincerely[/quote]

No , i can't find one. Do you have one?

[quote="Pain rinnegan"][/quote][quote="Ligouras"][quote="Pain rinnegan"][quote="Ligouras"]If $ a, b, c > 0$, then

              $ (a^2 + b^2 + c^2)^3\geq(a^3b + b^3c + c^3a)(a + b + c)^2$[/quote]

It's Vasc's inequality multiplied with $ 3(a^2 + b^2 + c^2)\ge (a + b + c)^2$[/quote]

Interesting.... thanks   :)
Do you have a solution that harmonizes the two solutions in an only?
If you answer "Yes" you are a great mathematician... Sincerely[/quote][quote="Pain rinnegan"]

No , i can't find one. Do you have one?[/quote]

It belongs to my new book... you have to wait 2010 November. Sorry


Putnam
combinatorics unsolved
combinatorics

This is some of my older work, related to [url]http://www.artofproblemsolving.com/viewtopic.php?t=77706[/url]

The issue is, given positive integer $ n > 2$, and $ N$ integers $ a_i$, with $ | a_i | \leq n$, $ 1 \leq i \leq N$, and with their sum $ S \equal{} \sum_{i\equal{}1}^N a_i$, to find as large a value for $ | S |$, such that we may always select some with their sum zero. Prove that this property is true

1. For $ N \equal{} 2n \minus{} 3$, not even for $ | S | \equal{} 1$;

2. For $ N \equal{} 2n \minus{} 2$ and $ | S | \leq N/2\minus{}1$, but not $ | S | \equal{} N/2$;

3. For $ N \geq 2n \minus{} 1$ and $ | S | \leq N \minus{} 1$, but not $ | S | \equal{} N$.

Case 1) is trivial, while case 2) is the most difficult. An even more difficult variant is

For $ N \equal{} 2n$, if we [b]postulate[/b] that not all of the integers have the same sign, prove that we may prolong this result to $ | S | \leq 3/2(N \minus{} 2)$, but not $ | S | \equal{} 3/2(N \minus{} 2)\plus{}1$.

Finally, let us notice that then, if exactly $ n$ of the elements are positive (so the other $ n$ are negative), we need no constraint at all on $ S$, as a Putnam problem particular case implies the existence of such zero sum.


calculus
derivative
function
calculus computations

Prove that the derivative of an odd function is even and the derivative of an even function is odd using the definition of the derivative (without using the chain rule)

I think you mean an 'odd differentiable function' and 'even differentiable function' respectively

Say $f$ is even.  Then $\frac{f(x+h)-f(x)}{h} = -\frac{f(-x)-f(-x-h)}{h}$, so $f'(x)=-f'(-x)$, and hence $f'$ is odd....


In any triangle $ ABC$ show that
\[ \cot\frac B2 \plus{} \cot\frac C2 > 2
\]

[hide]Because we are in a triangle, $ \angle{B}\plus{}\angle{C}<180$
That means that the maximum of $ \angle{B}\plus{}\angle{C}$ is just a bit smaller than 180.  That means that, because both angles have to be positive, $ \frac{B}{2}\plus{}\frac{C}{2}<90$
cot is an decreasing function in the first quadrant.
First, we assume that $ \frac{B}{2}\plus{}\frac{C}{2}$ is the maximum, 90. This will give us a bound on the values of
$ cot \frac{B}{2} \plus{}cot \frac{C}{2}$  If we make the sum of $ \frac{B}{2}\plus{}\frac{C}{2}$ any smaller, the value of cot will increase because it is a decreasing function (try graphing it)
Through manipulation of some variables
1) Split the cot into sin and cos
2) common denominator, sum of sin formula
3) multiplying top and bottom by two, change $ cos \frac{C}{2}$=$ cos \frac {180\minus{}B}{2}\equal{}sin\frac{B}{2}$, use double angle formula on denominator
4) do step 3 with changing B
5)get $ \frac{2}{sin B}>2$ and $ \frac{2}{sin C}>2$
Sin is an increasing function in the first quadrant so it is between 0<sinx<1.  Therefore, it is greater than 2.

[/hide]
I'm don't think it's rigorous but it works, for the most part.

[hide]It is easy to see that $ f(x) \equal{} \cot\frac {x}{2}$ is concave up and decreasing on $ x \in (0,\pi)$. So, by Jensen's: $ \cot\frac {B}{2} \plus{} \cot\frac {C}{2} \ge 2\cot\frac {B \plus{} C}{4} > 2\cot\frac {180^\circ}{4} \equal{} 2$ as desired.[/hide]

Let I be the incenter, so $ \angle IBC\equal{}(1/2)\angle B$, $ \angle ICB\equal{}(1/2)\angle C$ the altitude from I to BC is K, and $ IK\equal{}r$. So $ cot(\frac{B}{2})\plus{}cot(\frac{C}{2})\equal{}\frac{BK}{r}\plus{}\frac{KC}{r}\equal{}\frac{BC}{r}$. Draw the incrcle and the diameter parallel to BC, since the whole circle is fully inscribed in the triangle this diameter=2r must be less than BC, which completes the proof.


This is a probability problem (which I am not good at) :

You throw a needle of length $ a$ onto a grid of squares, each square having side of length $ 2a$. What is the probability the needle does not touch or cross any grid lines?

Someone said the answer will be $ \left(\frac{4}{\pi}\minus{}1\right)$. Can somebody please show me the formal solution ?

This is Buffon and Laplace Needle Problem
http://mathworld.wolfram.com/Buffon-LaplaceNeedleProblem.html


induction
quadratics
algebra
polynomial
system of equations

Not really helping answer the question but, AoPS has a nice geometric way of deriving this with the animated proofs.  :)

[quote="Carcul"]Knowing that $ S_1 \equal{} \sum_{k \equal{} 1}^nk \equal{} \frac {n(n \plus{} 1)}{2}$, and defining $ S_2 \equal{} \sum_{k \equal{} 1}^nk^2$, then consider

$ \sum_{k \equal{} 1}^n[(k \plus{} 1)^3 \minus{} k^3] \equal{} (n \plus{} 1)^3 \minus{} 1$

and expand both sides. To get $ S_1$ just consider $ \sum_{k \equal{} 1}^n[(k \plus{} 1)^2 \minus{} k^2] \equal{} (n \plus{} 1)^2 \minus{} 1$.[/quote]
can you please explain the seconde step

Another method ( But trivial )
[hide]
Use Induction.
[/hide]

Induction is probably an easy way to prove it, but if you wanted to find the formula you could try to express i^2 using combinatorials and then use hockeystick..

If anyone actually wants to read the proof,  [hide="here you go"]Prove it true for one:

$ 1^{2} \equal{} \frac {1(2)(3)}{6} \equal{} 1$ Yup.

Assume it true for some n.

*Assuming in progress, please wait.*

Prove it true for n+1.

For the sake of making it easier to type, let's replace the series $ 1^{2} \plus{} 2^{2}... \plus{} n^{2}$ with $ S$

Twirling the problem around a bit yields:

$ S \plus{} n^{2} \plus{} 2n \plus{} 1 \equal{} \frac {2n^{3} \plus{} 9n^{2} \plus{} 13n \plus{} 6}{6}$
$ 6(S) \equal{} 2n^{3} \plus{} 9n^{2} \plus{} 13n \plus{} 6 \minus{} 6n^{2} \minus{} 12n \minus{} 6$
$ 6(S) \equal{} 2n^{3} \plus{} 3n^{2} \plus{} n$
$ 6(S) \equal{} n(2n^{2} \plus{} 3n \plus{} 1)$

Factoring the quadratic and then dividing the equation by 6 yields the original conjecture, therefore, the equation is true for any n+1 if it is true for n. Since it is true for 1, it's true for any number, blah balh blah and all that good induction stuff.[/hide]

This just is an easy way to prove it, you can't derive it like this unless you are extemely lucky and guess a random correct formula.

[quote]I highly doubt someone could guess a formula like that, but then again.. who knows?[/quote]
That was more of a (really bad) humorous remark than a serious statement.

AoPS's graphical proof (mentioned b AstroPhys) involves creating grids of n by n squares (where n steadily increases) and arranging them to from a sort of tower shape, then rearranging that.

[quote="mountain.dew"]I highly doubt someone could guess a formula like that, but then again.. who knows?[/quote]

Actually while I was writing the formula as a program on my calculator, I "guessed" it. However, I have already seen it and remembered the divided 6 part and n part. I think this formula doesn't need to be memorized just as long as you know some of the numbers, it is easily guessed.

[quote="mountain.dew"]Well, I meant if you hadn't seen it before..[/quote]

Someone, at the very beginning must have got it. There are ways to work these things out (without any prior knowledge). I think bertram made a very good post about it a while back, see if you can find it

[quote="SimonM"]
Someone, at the very beginning must have got it. There are ways to work these things out (without any prior knowledge). I think bertram made a very good post about it a while back, see if you can find it[/quote]

Um, there's a derivation in this very thread; the fifth post actually. Maybe you should consider reading the thread next time.

[hide="more straightforward"]Let $ s_n$ be the nth square, and $ t_n$ the nth triangular number. We verify that $ t_{n\minus{}1}\plus{}t_n\equal{}\frac{(n\minus{}1)n}{2}\plus{}\frac{n(n\plus{}1)}{2}\equal{}\frac{n(2n)}{2}\equal{}n^2\equal{}s_n$. Now, letting $ S_n$ and $ T_n$ be the sum of the first n squares and triangular numbers, respectively, we see that

$ S_n\equal{}s_1\plus{}s_2...\plus{}s_n\equal{}(t_0\plus{}t_1)\plus{}(t_1\plus{}t_2)...\plus{}(t_{n\minus{}1}\plus{}t_n)\equal{}(t_0\plus{}t_1...\plus{}t_{n\minus{}1})\plus{}(t_1\plus{}t_2...\plus{}t_n)\equal{}T_{n\minus{}1}\plus{}T_n$. Now, the combinatorical definition of $ t_n$ states that $ t_n\equal{}\frac{n(n\plus{}1)}{2}\equal{}\binom{n\plus{}1}{2}$. We know that $ t_0\equal{}0$. Now, by Hockeystick, $ T_{n\minus{}1}\equal{}t_1\plus{}t_2...\plus{}t_{n\minus{}1}\equal{}\binom{2}{2}\plus{}\binom{3}{2}...\plus{}\binom{n}{2}\equal{}\binom{n\plus{}1}{3}$, and $ T_n\equal{}\binom{n\plus{}2}{3}$. Thus, $ S_n\equal{}\binom{n\plus{}1}{3}\plus{}\binom{n\plus{}2}{3}\equal{}\frac{(n\plus{}1)n(n\minus{}1)\plus{}(n\plus{}2)(n\plus{}1)n}{6}\equal{}\frac{n(n\plus{}1)((n\plus{}2)\plus{}(n\minus{}1))}{6}\equal{}\frac{n(n\plus{}1)(2n\plus{}1)}{6}$. Yay.[/hide]

[quote="perfect628"][hide="more straightforward"]Let $ s_n$ be the nth square, and $ t_n$ the nth triangular number. We verify that $ t_{n \minus{} 1} \plus{} t_n \equal{} \frac {(n \minus{} 1)n}{2} \plus{} \frac {n(n \plus{} 1)}{2} \equal{} \frac {n(2n)}{2} \equal{} n^2 \equal{} s_n$. Now, letting $ S_n$ and $ T_n$ be the sum of the first n squares and triangular numbers, respectively, we see that

$ S_n \equal{} s_1 \plus{} s_2... \plus{} s_n \equal{} (t_0 \plus{} t_1) \plus{} (t_1 \plus{} t_2)... \plus{} (t_{n \minus{} 1} \plus{} t_n) \equal{} (t_0 \plus{} t_1... \plus{} t_{n \minus{} 1}) \plus{} (t_1 \plus{} t_2... \plus{} t_n) \equal{} T_{n \minus{} 1} \plus{} T_n$. Now, the combinatorical definition of $ t_n$ states that $ t_n \equal{} \frac {n(n \plus{} 1)}{2} \equal{} \binom{n \plus{} 1}{2}$. We know that $ t_0 \equal{} 0$. Now, by Hockeystick, $ T_{n \minus{} 1} \equal{} t_1 \plus{} t_2... \plus{} t_{n \minus{} 1} \equal{} \binom{2}{2} \plus{} \binom{3}{2}... \plus{} \binom{n}{2} \equal{} \binom{n \plus{} 1}{3}$, and $ T_n \equal{} \binom{n \plus{} 2}{3}$. Thus, $ S_n \equal{} \binom{n \plus{} 1}{3} \plus{} \binom{n \plus{} 2}{3} \equal{} \frac {(n \plus{} 1)n(n \minus{} 1) \plus{} (n \plus{} 2)(n \plus{} 1)n}{6} \equal{} \frac {n(n \plus{} 1)((n \plus{} 2) \plus{} (n \minus{} 1))}{6} \equal{} \frac {n(n \plus{} 1)(2n \plus{} 1)}{6}$. Yay.[/hide][/quote]

I was talking about in time rather than this thread (unless I completely misunderstand you

Here is one way to show this with recurrence relations and without using the fact $ \sum_{k\equal{}1}^nk\equal{}\frac{n(n\plus{}1)}{2}$.

Let $ x_n\equal{}\sum_{k\equal{}1}^nk^2$ where $ x_1\equal{}1$.
Then $ x_{n\plus{}1}\minus{}x_n\equal{}(n\plus{}1)^2\equal{}n^2\plus{}2n\plus{}1$.
The characteristic equation $ r^2\minus{}r\equal{}r(r\minus{}1)$ har roots $ 0$ and $ 1$. So the homogenous solution $ x_n^h\equal{}(1)^nK_1\plus{}0^nK_2\equal{}K_1$ where $ K_1,K_2$ are some constants.
For the inhomogenous solution we guess a polynomial of degree 3,
$ a(n\plus{}1)^3\plus{}b(n\plus{}1)^2\plus{}c(n\plus{}1)\plus{}d\minus{}an^3\minus{}bn^2\minus{}cn\minus{}d\equal{}n^2\plus{}2n\plus{}1$
$ \equal{}n^2(3a)\plus{}n(3a\plus{}2b)\plus{}(a\plus{}b\plus{}c)\equal{}n^2\plus{}2n\plus{}1$
giving the system of equations:
$ 3a\equal{}1$
$ 3a\plus{}2b\equal{}2$
$ a\plus{}b\plus{}c\equal{}1$
which has solutions $ a\equal{}\frac{1}{3}$. $ b\equal{}\frac{1}{2}$ and $ c\equal{}\frac{1}{6}$. So the inhomogenous solution $ x_n^p\equal{}\frac{n^3}{3}\plus{}\frac{n^2}{2}\plus{}\frac{n}{6}\plus{}D$.
Then $ x_n\equal{}x_n^h\plus{}x_n^p\equal{}\frac{n^3}{3}\plus{}\frac{n^2}{2}\plus{}\frac{n}{6}\plus{}D\plus{}K_1$ and $ x_1\equal{}1$ gives that $ D\plus{}K_1\equal{}0$ since $ \frac{1}{3}\plus{}\frac{1}{2}\plus{}\frac{1}{6}\equal{}1$.
Hence $ x_n\equal{}\sum_{k\equal{}1}^nk^2\equal{}\frac{n^3}{3}\plus{}\frac{n^2}{2}\plus{}\frac{n}{6}\equal{}\frac{2n^3\plus{}3n^2\plus{}n}{6}\equal{}\frac{n(n\plus{}1)(2n\plus{}1)}{6}$.


geometry
parallelogram
circumcircle
complex numbers
algebra unsolved
algebra

Please help me solve the problem:

\[ C\equal{}\mbox{ Set of Complex Number}\]
Given that \[ x,y,z\epsilon C\] and \[ |x\minus{}y|\equal{}|y\minus{}z|\equal{}|z\minus{}x|\]

Prove that \[ (x\minus{}y)^{2}\plus{}(y\minus{}z)^{2}\plus{}(z\minus{}x)^{2}\equal{}0\]

I tried it so hard but disappointedly I cant solve it.
Isn't The above problem is like that:

Given that \[ a,b\epsilon C\] and \[ |a|\equal{}|b|\equal{}|a\plus{}b|\]
Prove that \[ a^{2}\plus{}b^{2}\plus{}(a\plus{}b)^{2}\equal{}0\].

Isn't the second problem also true??

$ 1)$ we have $ |x\minus{}y| \equal{} |y\minus{}z| \equal{} |z\minus{}x|$
so $ \exists h\in C$ such that $ x\minus{}y \equal{} hj,y\minus{}z \equal{} hj^{2},z\minus{}x \equal{} h$ where $ j \equal{} e^{i\frac{2\pi}{3}}$ ($ 1\plus{}j\plus{}j^{2}\equal{} 0,j^{3}\equal{} 1$)
then $ (x\minus{}y)^{2}\plus{}(y\plus{}z)^{2}\plus{}(z\minus{}x)^{2}\equal{} h^{2}(j^{4}\plus{}j^{2}\plus{}1) \equal{} h^{2}(j^{2}\plus{}j\plus{}1) \equal{} 0$

[quote="Oni"]Please help me solve the problem:
\[ C\equal{}\mbox{ Set of Complex Number}\]
Given that
\[ x,y,z\epsilon C\]
and
\[ |x\minus{}y|\equal{}|y\minus{}z|\equal{}|z\minus{}x|\]
Prove that
\[ (x\minus{}y)^{2}\plus{}(y\minus{}z)^{2}\plus{}(z\minus{}x)^{2}\equal{}0\]
I tried it so hard but disappointedly I cant solve it.
Isn't The above problem is like that:

Given that
\[ a,b\epsilon C\]
and
\[ |a|\equal{}|b|\equal{}|a\plus{}b|\]
Prove that
\[ a^{2}\plus{}b^{2}\plus{}(a\plus{}b)^{2}\equal{}0\]
.

Isn't the second problem also true??[/quote]

$ 2)$ notice that $ |b|\equal{}|\minus{}b|$ and $ |\minus{}b\minus{}a|\equal{}|a\plus{}b|$
then consider $ x\equal{}a$, $ y\equal{}O$ and $ z\equal{}\minus{}b$ and using $ 1)$
$ |x\minus{}y|\equal{}|y\minus{}z|\equal{}|z\minus{}x|$
$ |a\minus{}O|\equal{}|O\minus{}(\minus{}b)|\equal{}|\minus{}b\minus{}a|$
$ |a|\equal{}|b|\equal{}|b\plus{}a|$

[hide="using the equivalent second version of the problem"]
Let
\[ k\equal{}|a|^{2}\equal{}|b|^{2}\equal{}|a\plus{}b|^{2}\equal{}a\overline{a}\equal{}b\overline{b}\equal{}(a\plus{}b)(\overline{a\plus{}b})\]
Then $ a\overline{b}\plus{}\overline{a}b\equal{}\minus{}k$ after expanding the right most expression.

Multiply by $ ab$ to get $ a^{2}k\plus{}b^{2}k\equal{}\minus{}abk$, or $ k(a^{2}\plus{}ab\plus{}b^{2})\equal{}0$.

Of course, if $ k\equal{}0$, we must have $ a\equal{}b\equal{}0$ and then $ a^{2}\plus{}ab\plus{}b^{2}\equal{}0$ anyway. Therefore, $ a^{2}\plus{}ab\plus{}b^{2}\equal{}0$, which is equivalent to the result.
[/hide]

[quote="Oni"][color=darkred]$ 1\blacktriangleright$ Prove that for any $ \{x,y,z\}\subset\mathbb C$, $ |x-y| = |y-z| = |z-x|$ $ \Longleftrightarrow$ $ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}= 0\ .$[/color][/quote]
[color=darkblue]For any $ \boxed x\in\mathbb C$ I"ll denote the point $ \boxed X$ which has the affix $ x$ and I"ll write $ X(x)$. It is well-known, but I use same letter, lower case and capital,

for a affix and a point respectively. For example, $ XY = |x-y|$ and the triangle $ XYZ$ is equilateral if and only if $ |x-y| = |y-z| = |z-x|$.[/color]

[color=darkred][b]Lemma I (it is obviously).[/b] For any $ \{x,y,z\}\subset\mathbb C$, the relations $ \|\begin{array}{cc}\sum\ (x-y)(x-z) = 0\\ \\ \boxed{x^{2}+y^{2}+z^{2}= xy+yz+zx}\\ \\ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}= 0\\ \\ (x-y)^{2}+(x-z)^{2}= (x-y)(x-z)\\ \\ (y-z)^{2}+(x-y)(x-z) = 0\end{array}\|$ are equivalently.[/color]

[quote][color=darkred][b]Lemma II.[/b] Let $ ABC$ be a triangle for which denote $ \rho =\frac{AB}{AC}$. Then

$ \boxed{m(\widehat{BAC}) =\phi\Longleftrightarrow (a-b)^{2}+\rho^{2}\cdot (a-c)^{2}= (a-b)(a-c)\cdot\rho(\omega+\overline{\omega})}$, where $ \omega =\cos\phi+i\cdot\sin\phi$.[/color][/quote]

[color=darkblue][b]Proof.[/b] $ m(\widehat{BAC}) =\phi$ $ \Longleftrightarrow$ $ (a-b) = (a-c)\rho\omega\ \ \vee\ \ (a-b) = (a-c)\rho\overline{\omega}$ $ \Longleftrightarrow$

$ [\rho\omega (a-c)-(a-b)]\cdot[\rho\overline{\omega}(a-c)-(a-b)] = 0$ $ \Longleftrightarrow$ $ (a-b)^{2}+\rho^{2}\cdot (a-c)^{2}= (a-b)(a-c)\cdot\rho(\omega+\overline{\omega})$.[/color]

[color=darkred][b]Particular cases.[/b]

$ 1\blacktriangleright\triangle ABC$ is $ A$- isosceles, i.e. $ \rho = 1$ $ \Longleftrightarrow$ $ (a-b)^{2}+(a-c)^{2}= (a-b)(a-c)\cdot (\omega+\overline{\omega})\ .$

$ 2\blacktriangleright\triangle ABC$ is $ A$- right and $ A$- isosceles, i.e. $ \|\begin{array}{c}\rho = 1\\ \\ \phi =\frac{\pi}{2}\\ \\ \omega = i\end{array}\|$ $ \Longleftrightarrow$ $ (a-b)^{2}+(a-c)^{2}= 0$ $ \Longleftrightarrow$ $ 2(a-b)(a-c)+(b-c)^{2}= 0\ .$

$ 3\blacktriangleright\triangle ABC$ is equilateral, i.e. $ \|\begin{array}{c}\rho = 1\\ \\ \phi =\frac{\pi}{3}\\ \\ |\begin{array}{c}\omega =\frac{1}{2}\cdot(1+i\cdot\sqrt 3)\\ \\ \omega^{3}=-1\ ;\ \omega^{2}-\omega+1 = 0\\ \\ \overline{\omega}=\frac{1}{\omega }= 1-\omega =-\omega^{2}\end{array}|\end{array}\|$ $ \Longleftrightarrow$ $ (a-b)^{2}+(a-c)^{2}= (a-b)(a-c)$.

From the [b]lemma I[/b] results $ |a-b| = |b-c| = |c-a|$ $ \Longleftrightarrow$ $ \triangle ABC$ is equilateral $ \Longleftrightarrow$ $ (a-b)^{2}+(b-c)^{2}+(c-a)^{2}= 0$,  [b]q.e.d.[/b][/color]

[quote="Oni"][color=darkred]$ 2\blacktriangleright$ Prove that for any $ \{a,b\}\subset\mathbb C$, $ |a| = |b| = |a+b|$ $ \Longleftrightarrow$ $ a^{2}+b^{2}+(a+b)^{2}= 0\ .$[/color][/quote]
[color=darkblue][b]Proof.[/b] I"ll use the well-known identity $ |x+1|^{2}+|x-1|^{2}= 2(|x|^{2}+1)$ ([b]the rule of the parallelogram[/b])

which is equivalently with $ (x+1)(\overline x+1)+(x-1)(\overline x-1) = 2(x\overline x+1)$, what is evidently.

This identity is equivalently with the identity (seemingly more generally) $ \boxed{|a+b|^{2}+|a-b|^{2}= 2(|a|^{2}+|b|^{2})}$.

Return to the second [b]Oni[/b]'s equivalence. Suppose w.l.o.g. $ b\ne 0$ and denote $ x =\frac{a}{b}$. Then this equivalence becomes

$ |x+1| = |x| = 1\ (\ \overline x =\frac{1}{x}\ )\Longleftrightarrow$ $ x^{2}+1+(x+1)^{2}= 0$ $ \Longleftrightarrow$ $ x^{2}+x+1 = 0$. Using the [b]upper rule[/b] obtain $ |x+1| = |x| = 1$ $ \Longleftrightarrow$

$ |x-1|^{2}= 3$ $ \Longleftrightarrow$ $ (x-1)(\overline x-1) = 3$ $ \Longleftrightarrow$ $ (x-1)(\frac{1}{x}-1) = 3$ $ \Longleftrightarrow$ $ (x-1)^{2}+3x = 0$ $ \Longleftrightarrow$ $ x^{2}+x+1 = 0$, [b]q.e.d.[/b]

[b]Remark.[/b] Let $ ABC$ be a triangle with the circumcircle $ w = C(O,r)$ and the centroid $ G$. Choose the origin in the circumcenter $ O$,

i.e. $ o = 0$, $ |a| = |b| = |c| = r$ and $ 3g = a+b+c$. Therefore, $ \triangle ABC$ is equilateral $ \Longleftrightarrow$ $ G\equiv O$ $ \Longleftrightarrow$ $ a+b+c = 0\ .$

$ 1\blacktriangleright$ I"ll prove prove that for any complex numbers $ a$, $ b$, $ c$, $ |a| = |b| = |c| = r\ne 0$, i.e. $ a\overline a = b\overline b = c\overline c = r^{2}$ exists the equivalence

$ a+b+c = 0\Longleftrightarrow ab+bc+ca = 0$. Indeed,  $ a+b+c = 0\Longleftrightarrow$ $ \overline{\sum a}= 0$ $ \Longleftrightarrow$ $ \sum\overline a = 0$ $ \Longleftrightarrow$ $ r^{2}\cdot\sum\frac{1}{a}= 0$ $ \Longleftrightarrow$ $ ab+bc+ca = 0$.

$ 2\blacktriangleright$ I"ll prove that for any complex numbers $ a$, $ b$, $ c$ exists the implication $ \{\begin{array}{c}a+b+c = 0\\ \ ab+bc+ca = 0\end{array}\|$ $ \implies$ $ |a| = |b| = |c|$. Indeed,
we can consider $ a,b,c$ as the roots of the equation $ x^{3}= k\in\mathcal C$ $ \implies$ $ |x|^{3}= |k|$ $ \implies$ $ |a| = |b| = |c| =\sqrt [3]{|k|}$.

More precisely :[/color][color=darkred] if two from the relations $ \{\begin{array}{c}a+b+c = 0\\ \\ |a| = |b| = |c|\\ \\ ab+bc+ca = 0\end{array}$ are truly, then and the another relation is truly.[/color]

[color=darkblue]Particularly, for $ \{\begin{array}{c}a: = y-z\\ \ b: = z-x\\ \ c: = x-y\end{array}$ results $ |y-z| = |z-x| = |x-y|\Longleftrightarrow$ $ (y-z)(z-x)+(z-x)(x-y)+(x-y)(y-z) = 0$ $ \Longleftrightarrow$

$ \sum (x-y)(x-z) = 0$ $ \Longleftrightarrow$ $ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}= 0$ using the upper [b]lemma I[/b].[/color]

[size=117][color=green]$ \boxed{\mathrm{\ OFF\ -\ TOPIC\ }}$ [b][u]Oni[/u], is it O.K. now ? I like billiards, tennis, soccer (football) and ... geometry (the plane and the space).[/b][/color][/size]


[quote="shobber"]There is a number sequence:
$1,1,2,2,3,3,4,4........n,n.............$
Find $a_n$ and the sum of the first $n$ terms.[/quote]

[hide]$a_n$ part:

if $n$ is even, divide by two to get $a_n$

if $n$ is odd,  add one and then divide by two

sum:

separate the duplicate numbers and the sum of the two separate equations are $\frac{n\cdot (n+1)}{2}$

then add it again and get $n(n+1)$ as the sum[/hide]

[hide]If you don't want to do case by case analysis, you can express a_n=[n/2], where [k] indicates the smallest integer greater than k (the 'ceiling' function). This doesn't get rid of the case by case analysis, because it is still inside the [k] function, but I find it a little bit easier to read. And most of the time (in the contests that I see), they will actually give the two (or more) cases for a function, so both ways are quite good[/hide]


vector
inequalities
Cauchy Inequality
algebra proposed
algebra

Find all the values of k for which the system

a+b+c+d+e=1
a+2b+3c+4d+5e=k
a+4b+9c+16d+25e=k^2

has solution  :D

the system always has a solution. Did you forget to add something?

Do you mean $ a,b,c,d,e,k$ are integers?  :)

Let $ A\equal{}a\plus{}b\plus{}c\plus{}d\plus{}e$ and $ B\equal{}a\plus{}4b\plus{}9c\plus{}16d\plus{}25e$ and $ C\equal{}a\plus{}2b\plus{}3c\plus{}4d\plus{}5e$ .By couchy we have:
                             $ 1.k^{2}\equal{}A.B\geq C^{2}\equal{}k^{2}$
So there exist $ \lambda$ in $ \mathbb{R}$ such that : $ \lambda(a,b,c,d,e)\equal{}(a,4b,9c,16d,25e)$
So the solutions are:
$ (a,b,c,d,e,k)\equal{}(1,0,0,0,0,1) or (0,1,0,0,0,2) or (0,0,1,0,0,3) 0r (0,0,0,1,0,4) or (0,0,0,0,1,5)$
Is it true?  :huh:

Oh yeah sorry I forgot to say that a,b,c,d,e are integers  :P

That's ok shapeshift  :lol:

Don't we need nonnegativity of $ a,b,c,d,e$ to use cauchy  :?:  :)

Cauchy works for all reals. Remember that it has use for studying vectors, which almost never have restrictions to positives or negatives.

Cauchy in the form $ \left(\sum a_{i}^{2}\right)\left(\sum b_{i}^{2}\right)\geq\left(\sum a_{i}b_{i}\right)^{2}$ works for all reals obviously as you said, but in order to say $ \left(\sum a_{i}\right)\left(\sum b_{i}\right)\geq\left(\sum\sqrt{a_{i}b_{i}}\right)^{2}$ we need $ a_{i},b_{i}\geq 0$  :wink:

BUT we have :
                  $ \sum\sqrt\left a_{i}b_{i}\right$ $ \geq $ $ \sum c_{i}\sqrt\left a_{i}b_{i}\right$for $ c_{i}=-1 or 1$
So by couchy we have:
                    $ (\sum a_{i}^{2})(\sum b_{i}^{2})\geq \sum |a_{i}b_{i}|\geq \sum a_{i}b_{i}$
for $ a_{i},b_{i}$ in $ \mathbb{R}$

But , if $ a$ or$ b$ or $ c$ or $ d$ or $ e<0$ , then the inequality $ AB\ge C$ doesn't hold because you need to express $ a \equal{} x^{2}, b \equal{} y^{2},...$ to use Cauchy inequality  :(

I have a method as follow
From (1) we has
$ a\equal{}1\minus{}(b\plus{}c\plus{}d\plus{}e)$
Let replace to (2),(3)
$ b\plus{}2c\plus{}3d\plus{}4e\equal{}k\minus{}1$(4)
$ 3b\plus{}8c\plus{}15d\plus{}24e\equal{}k^{2}\minus{}14$(5)
From (4)
$ b\equal{}k\minus{}1\minus{}(2c\plus{}3d\plus{}4e)$
Replace to (5)
Use Lemma
Let $ (a_{1},....,a_{n})\in Z$ then the equation
$ \sum_{i\equal{}1}^{n}a_{i}.x_{i}\equal{}c$ has solution if and only if  $ \gcd(a_{1},....,a_{n})|c$
Use this lemma.

You are right quangpbc,I made a mistake...


Factorise the following as products of two non-constant polynomials with real coefficients:

(i) $ a^{3}\plus{}b^{3} \plus{} c^{3}\minus{}3abc$,

(ii) $ a^{4}\plus{}b^{4}$,

(iii) $ (a\plus{}b\plus{}c)^{3} \minus{} (a^{3}\plus{}b^{3}\plus{}c^{3})$

[hide="1"]
$ (a\plus{}b\plus{}c)^3\equal{}a^3\plus{}b^3\plus{}c^3\plus{}3ab^2\plus{}3a^2b\plus{}3bc^2\plus{}3b^2c\plus{}3ac^2\plus{}3a^2c\plus{}6abc$

$ (a\plus{}b\plus{}c)^3\minus{}(a^3\plus{}b^3\plus{}c^3\minus{}3abc)\equal{}3ab^2\plus{}3a^2b\plus{}3bc^2\plus{}3b^2c\plus{}3ac^2\plus{}3a^2c\plus{}9abc$
$ \equal{}3(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ac)$

$ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}(a\plus{}b\plus{}c)^3\minus{}3(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ac)\equal{}(a\plus{}b\plus{}c)((a\plus{}b\plus{}c)^2\minus{}3(ab\plus{}bc\plus{}ac))$

You can expand the second factor out if you want.
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I don't think 2 can be factored.

I suppose 3 shouldn't be too hard after looking at the solution to 1.

Hi grn_trtle.

For (i) I multilplied

$ (a^{2} \plus{} \lambda ab \plus{} b^{2})(as^{2} \minus{} \lambda ab \plus{} b^{2})$

to get $ (a^{4} \plus{} b^{4}) \equal{} (a^{2} \plus{} \sqrt{2}ab \plus{} b^{2})(a^{2} \minus{} \sqrt{2}ab \plus{} b^{2})$.

[hide="Solution to 2"]
Since the coefficients only need to be real, we can use irrational numbers, so
\[ a^{4}\plus{}b^{4}\equal{}\]
\[ a^{4}\plus{}2a^{2}b^{2}\plus{}b^{4}\minus{}2a^{2}b^{2}\equal{}\]
\[ (a^{2}\plus{}b^{2})^{2}\minus{}2a^{2}b^{2}\equal{}\]
\[ (a^{2}\plus{}b^{2}\minus{}\sqrt{2}ab)(a^{2}\plus{}b^{2}\plus{}\sqrt{2}ab)\]
[/hide]