geometry
3D geometry
search
quadratics
Ring Theory
polynomial
probability

Is there a quick way to solve $ p \equal{} a^2 \plus{} ab \plus{} b^2$ for prime $ p$ and integers $ a$, $ b$ ?

Use the ring $ \mathbb{Z}[\omega]$, where $ \omega\equal{}\frac12\plus{}\frac{i\sqrt{3}}{2}$ is a cube root of unity. The square of the absolute value of $ a\minus{}b\omega$ is $ a^2\plus{}ab\plus{}b^2$.
Any solution will come from factoring $ p$ in this ring, and will generate all other solutions by multiplying by units and taking conjugates (the conjugate is the other factor). There are either no solutions or exactly twelve solutions, except in the case $ p\equal{}3$ (six solutions).

As for finding that first solution- it's a search, but at least there are only finitely many places to look. Working mod $ p$ might help.

Why, are you in a hurry? :)

It is a classic result that this equation has a solution only for $ p \equal{} 3$ and $ p\equiv 1\ (mod\ 3)$.  Here's the quickest path to that answer I can think of:

* Show that there is no solution if $ p$ is not of the form above, by looking at the equation $ (mod\ 3)$.

* Show that $ \minus{} 3$ is a square $ (mod\ p)$ if $ p$ is of the form above.  This can be done with quadratic reciprocity.

* Show that $ p$ can be written as a sum $ x^2 \plus{} 3y^2$.  Use the above to write $ kp \equal{} r^2 \plus{} 3s^2$ and then factor in $ \mathbb{Z}[\sqrt { \minus{} 3}]$.  Since this ring is a UFD, we can write $ p \equal{} x^2 \plus{} 3y^2$. (observe that if a prime in $ \mathbb{Z}[\sqrt { \minus{} 3}]$ divides $ p$, its conjugate must also divide $ p$)

* Let $ a \equal{} x \plus{} y$, $ b \equal{} \minus{} 2y$ to solve the original equation.

It is probably possible to write a more intuitive proof that uses the Eisenstein integers ($ \mathbb{Z}[\omega]$) and cubic reciprocity.

EDIT: Note that jmerry and I are answering two different facets of the same question, namely:  "Given $ p$, what are all possible values of $ a$ and $ b$?" and "For which $ p$ do there exist solutions?"  For a "complete" solution you need both parts.

Since $ x^2 \plus{} xy \plus{} y^2\geq \frac12(x^2 \plus{} y^2)$, there is an obvious $ O(\sqrt p)$ algorithm.  But we can do better than that:

Pick $ \zeta\in(0,p)$ at random.  With probability $ \frac {\varphi(p \minus{} 1)}{p \minus{} 1} > \frac{c}{\log\log p}$, this will be a primitive root mod p, and $ x \equal{} \zeta^{\frac {p \minus{} 1}3}$ a primitive cube root mod p, so

$ p|1 \plus{} x \plus{} x^2 \equal{} (x \minus{} \omega)(x \plus{} \omega)$.

Now we can use the Euclidean algorithm in $ \mathbb{Z}[\omega]$ to quickly compute $ (p,x \minus{} \omega) \equal{} (a \plus{} b\omega)$, which has norm $ p \equal{} a^2 \plus{} ab \plus{} b^2$.

This should all be polynomial in $ \log p$.


ratio quadratics inequalities function algebra polynomial logarithms \/closed

Our Introduction to Algebra text is now available.  Topics covered in the book include linear equations, ratios, quadratic equations, special factorizations, complex numbers, graphing linear and quadratic equations, linear and quadratic inequalities, functions, polynomials, exponents and logarithms, absolute value, sequences and series, and much more!

This book can serve as a complete Algebra I course, and also includes many concepts covered in Algebra II. Middle school students preparing for MATHCOUNTS, high school students preparing for the AMC, and other students seeking to master the fundamentals of algebra will find this book an instrumental part of their mathematics libraries.

The text is 656 pages and the solution manual is 312 pages.

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=11]Click here for more details, and to order your copy[/url].  Excerpts, the table of contents, and the index are available on this page, as well.

Cool
This will be a good review fro Mathcounts State Competition :)

c'mon guys.. lets start up the excitement train?

or is nobody a big fan of algebra? lol.

It looks pretty good for the young'uns.

Based from the description by Mr. Rusczyk, I'd say it would be good for people who finished Algebra I, taking Geometry, and getting ready for Algebra II. After a year in Geometry you might tend to forget some of the important factors needed in Algebra II.  :)

It would also be a good book for students just entering Algebra I, but who want a greater challenge than they get in the typical Algebra I book.

[quote="archimedes1"]Will there be an Intermediate Algebra book?[/quote]

We hope to have it available in the end of this summer.

Haha...
Well, I'm rather looking forward to the Intermediate Counting book. (I suck at counting; I'm better at algebra). That should be ready by the end of the summer too, right?

[quote="bpms"]Very nice :D[/quote]
ooh, now i need to win mathcounts so i can be in an AoPS book :)

btw, what's Andrew's account?

I'm not sure he'd appreciate you declaring it to the world. ...

[quote="solafidefarms"]Haha, enshrining AoPSers in AoPS books. What other AoPSers have made it into books?[/quote]

Simon's favorite factoring trick is there :D

[quote="rrusczyk"]Good point; if he wants to announce himself, he can do so.  I don't think he has anywhere, so I snipped it.[/quote]
sorry :blush:


The area of the region consisting of those points (x,y) for which $x^2+y^2\le2$ and $(x-2)^2+y^2\le2$ is

a) 2
b) $\pi-2$
c) 3/2
d) $\frac{\pi}{2}-1$
e) 1

You can see that two circles are symmetric for the line $x=1$, then find two fans and a square.


inequalities
function
inequalities unsolved

[i][u]Problem :[/u]  Let $ x;y;z$ be positive real numbers such that : $ xy \plus{} yz \plus{} zx \plus{} xyz \equal{} 4$ . Prove that :

 $ \frac {x \plus{} y \plus{} z}{xy \plus{} yz \plus{} zx} \ \leq \ 1 \plus{} \frac {5}{247} \left( (x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} (z \minus{} x)^2 \right)$[/i]

[i]  ( sorry for my mistake , Now , the problem was correct)[/i] :lol:

this inequality should be used by replacing variable solution: x+y+z=p;xy+yz+xz=q;xyz=r

Are $ x,\ y,\ z$ integers?

Substitution by triangular function seems to be effective.

$ xy \plus{} yz \plus{} zx \plus{} xyz \equal{} 4 \Rightarrow$ there exist reals $ a,b,c$ such that $ x \equal{} \frac {2a}{b \plus{} c},y \equal{} \frac {2b}{c \plus{} a},z \equal{} \frac {2c}{a \plus{} b}$

if $ x,y,z$ are positive integers, then $ xyz \equal{} \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\le1\Rightarrow a \equal{} b \equal{} c\Leftrightarrow x \equal{} y \equal{} z \equal{} 1$

Then the inequality becomes trivial, are you sure $ x,y,z$ are positive integers? :lol:


A rectangular birthday cake measures 12 inches by 24 inches and is 3 inches thick. A thin layer of frosting is placed on the top and the four sides.  What per cent of the frosting is on the top of the cake?   Round to the nearest per cent.

A. 50%	B. 57%	C.  60%	D. 62%	E.  67%

[hide][quote]3 inches thick[/quote]
3 inches high.....

First we will find the surface area.
12(24)(2)+(12)(3)(2) + 3(24)2 = 792
Then we will subtract the surface area of the bottom to get our total frosted surface area.
792-12(24)=504
Since the top = the bottom, the percent of the top out of the total icing is
288/504=.571=57.1%
rounds to 57%
Our answer is B) 57%[/hide]


rotation
analytic geometry
geometry
geometric transformation
reflection

[quote="rnwang2"]The point $P=(1,2,3)$ is reflected in the $xy$-plane, then its image $Q$ is rotated by $180^{\circ}$ about the $x$-axis to produce $R$, and finally, $R$ is translated by $5$ units in the positive-$y$ direction to produce $S$. What are the coordinates of $S$?[/quote]

[hide]First of all, I don't think we are allowed to have 3-Dimension Space, but anyways, I don't care about that.

Notice that, when you reflect in the $xy$-plane, you get the z-coordinate changing signs, so it becomes $(1,2,-3)$.

Next, when you rotate this around the x-axis, you get that the y and z change sign, so it becomes $(1,-2,3)$.

Then, you just add 5 to the y-coordinate to get

$S(1,3,3)$[/hide]


quadratics
function
geometry
geometric transformation
special factorizations

How do I figure out the A in a(x-h)^2+k, from a vertex at (8,1) and pass through the point (11,2)?

Also what does the A really mean in the graph? I know A>0, the graph is up, and when A<0 the graph is down, but what else? Is A like the number that compress or expand the graph?

The vertex occurs when the $(x-h)^{2}$ term is 0.  Then we know that h is 8.  Then k=1.  Then we have that $a(11-8)^{2}+1=2\rightarrow 9a=1\rightarrow a=\frac{1}{9}$

Since the graph of $y=a(x-h)^{2}+k$ can be obtained by translation of $y=ax^{2},$ considering the case that $y=ax^{2}$ passes through the point $(3,\ 1),$
the answer is $1=a\cdot 3^{2}\Longleftrightarrow a=\frac{1}{9}.$

[quote="Farcus"]another question

How do you figure out how to find the maximum value of the quadratic function, when the function is like y=(-2x+7)(8x-4).

I know you have the complete the square of the problem. Initally using FOIL, I get -16x^2+64x-28, then simplify to get 4x^2-16x+7. Now we complete the square which I got 4(x-2)^-9. So the max value of the function is (2,9)?[/quote]
Actually, in this problem, that would be the minimum value. (If the quadratic is $ax^{2}+bx+c$, if $a$ is positive, it's the minimum, and if $a$ is negative, it's the maximum).
Once you complete the square and find that, for example, $y=4(x-2)^{2}-9$, you can notice that the $4(x-2)^{2}$ part is always nonnegative (because it's a square), and so the minimum of it is $0$, when $x=2$. Plug in $x=2$ to find that $y=-9$, and this will be the minimum.
In general, the minimum/maximum value of a quadratic $ax^{2}+bx+c$ occurs when $x=-\frac{b}{2a}$ (which can be shown by completing the square of $ax^{2}+bx+c$).

EDIT: Changed it to $y=-9$.

[quote="Farcus"][quote="kunny"]Your answer is incorrect, Farcus. :(

$y=(-2x+7)(8x-4)=-16x^{2}+64x-28$
$=-16(x-?)^{2}+?.$[/quote]

wait so I can't divide -4 for all of -16x^2+64x-28?[/quote]
You can only do that if you are trying to solve a quadratic [i]equation[/i], as in $-16x^{2}+64x-28=0$, but not if you wanted to do something like find the minimum value or factor it.

[quote="nebula42"] for example, $y=4(x-2)^{2}-9$, you can notice that the $4(x-2)^{2}$ part is always nonnegative (because it's a square), and so the minimum of it is $0$, when $x=2$. Plug in $x=2$ to find that $y=9$, and this will be the minimum. [/quote] Typo? $x=2, y=-9$

[quote="Farcus"]
Well for some reason I just did the -b/2a which is (-64)/2(-16)=2 so according to what you just said wouldn't it be just 2 and not 0 as you said?[/quote] When you divide by $-4$, the minimum x-value remained the same, but the y-value was divided by $-4$ also. So instead of $(2,-9)$, the minimum is $(2,36)$.

[quote="Farcus"][quote="nebula42"][quote="Farcus"]another question

How do you figure out how to find the maximum value of the quadratic function, when the function is like y=(-2x+7)(8x-4).

I know you have the complete the square of the problem. Initally using FOIL, I get -16x^2+64x-28, then simplify to get 4x^2-16x+7. Now we complete the square which I got 4(x-2)^-9. So the max value of the function is (2,9)?[/quote]
Actually, in this problem, that would be the minimum value. (If the quadratic is $ax^{2}+bx+c$, if $a$ is positive, it's the minimum, and if $a$ is negative, it's the maximum).
Once you complete the square and find that, for example, $y=4(x-2)^{2}-9$, you can notice that the $4(x-2)^{2}$ part is always nonnegative (because it's a square), and so the minimum of it is $0$, when $x=2$. Plug in $x=2$ to find that $y=9$, and this will be the minimum.
In general, the minimum/maximum value of a quadratic $ax^{2}+bx+c$ occurs when $x=-\frac{b}{2a}$ (which can be shown by completing the square of $ax^{2}+bx+c$).[/quote]

Well for some reason I just did the -b/2a which is (-64)/2(-16)=2 so according to what you just said wouldn't it be just 2 and not 0 as you said?[/quote]
No, actually the maximum would occur when $x=-\frac{b}{2a}$, or in this case, when $x=2$. If you want to find the maximum [i]value[/i], you would need to plug in $x=2$ and find the $y$ value for it, which would be $-16(2)^{2}+64(2)-28=36$.


search
function
number theory unsolved
number theory

Find all $a, b \in \mathbb N$ such that $a$ divide $b^2 + 1$ and $b$ divide $a^2 + 1$

Very easy when using search function ;)
http://www.mathlinks.ro/Forum/viewtopic.php?t=46767
http://www.mathlinks.ro/Forum/viewtopic.php?t=58763


inequalities
triangle inequality
calculus
calculus computations

turn to the attachment.[/youtube][/url][/hide][/list][/code][/quote][/u][/b][/i]

[quote="tancm"]
Let $ \xi _i,\ \theta _i\ (i = 1,\ 2,\ \cdots n)$ be complex number.

Prove the following inequality.

$ \sqrt {|\xi _1 + \theta _1| + |\xi _2 + \theta _1| + \cdots |\xi _n + \theta _n|}$

$ \leq \sqrt {|\xi _1|^2 + |\xi _2|^2 + \cdots + |\xi _n|^2} + \sqrt {|\theta _1|^2 + |\theta _2|^2 + \cdots + |\theta _n|^2}$[/quote]

That must be incorrectly stated. It's clearly false as written. Suppose that the left hand side equals $ A$ and the right hand side equals $ B.$ Now replace $ \xi_i$ and $ \theta_i$ by $ t\xi_i$ and $ t\theta_i$ for arbitrary positive real $ t.$ If this is a true inequality, we would have that $ \sqrt {t}A\le tB$ for all $ t > 0.$ This cannot hold for all $ t$ as $ t\to 0^ + .$

I assume that the statement was intended to be

$ \sqrt {|\xi _1 + \theta _1|^2 + |\xi _2 + \theta _1|^2 + \cdots |\xi _n + \theta _n|^2}$

$ \leq \sqrt {|\xi _1|^2 + |\xi _2|^2 + \cdots + |\xi _n|^2} + \sqrt {|\theta _1|^2 + |\theta _2|^2 + \cdots + |\theta _n|^2}$

Stated that way, it's the usual triangle inequality for the $ 2$-norm on $ \mathbb{C}^n.$ The usual path for the proof of that goes by way of the Cauchy-Schwarz inequality (or, to use the specific name for finite sums, Cauchy's inequality). You can find this proof in many places.

[kunny does appear to have faithfully transcribed the original post; hence this comment is directed to the original poster, not kunny.]

Oh, I should have noticed that, anyway thank you for the correction, Kent Merryfield.


a factor of polynomial is 3ac + 6ab +  4c + 8b  is

a) 3a + 2  b) 3a + 4   c) 2c   d) 2a+3 e) noneof these


[b][size=150]The 14th International Scientific Olympiad in Mathematics
for University Students
15-17 July 2009
Tehran, Iran[/size][/b]

I try to open the file it does not work, and I don't know what is djvu.

Could you put the problems in PDF, please ?

Thank you

Sorry for Delay!

Attached please find the PDF file.

Please note also, that the statement of the Linear Algebra problem No 1 is wrong.

Here is the counterexample for $ n\equal{}2$ due to Vardan Voskanyan:

Take \[ A\equal{}\left(\begin{array}{cc} 1 & 0\\ 0 & \minus{}1 \end{array}\right)\]
and $ B\equal{}I$. Then $ A^2\equal{}I$, $ BAB^{\minus{}1}\equal{}A\equal{}A^{\minus{}1}$, but every matrix of the form
\[ X\equal{}\left(\begin{array}{cc} \alpha & 0\\ 0 & \beta \end{array}\right)\]
with $ \alpha\not \equal{} \beta$ commutes with $ A$ and $ B$.

Actually, Vardan constructed a counterexample for any $ n\ge 2$.

Thank you Beginner and micael for the problems in PDF

For linear algebra 2

Here a hint
[hide]Proved $ tr(C^{k}) \equal{} 0$ for any k>0 integer
[/hide]

In fact, there is no $ f$ satisfying to the condition of Analysis problem 1.

Here is the solution:

[hide="Solution"]
Since $ f'(x)\equal{}f(f(x))>0$, then $ f$ increases  on $ R$. From the same identity it follows that
$ f'$ increases on $ R$, so $ f$ is convex.

As $ f(x)>0$ and $ f$ increases, then there exists the limit $ f(x)\to A\ge 0$ as $ x\to\minus{}\infty$.
The same reasoning shows, that there exists the limit $ f'(x)\to B\ge 0$ as $ x\to\minus{}\infty$. But the case $ B>0$ is impossible. This can be proved as follows:

If $ f'(x)\to B>0$,  as $ x\to \minus{}\infty$, then there exists a number $ E$ s.t. $ f'(x)>\frac{B}{2}>0$ for all $ x\le E$. Then
\[ f(x)\minus{}f(E)\equal{}f'(c_x) \cdot (x\minus{}E)< \frac{B}{2} (x\minus{}E)\quad for\quad x< E.\]
Hence
\[ 0<f(x)<f(E)\plus{}\frac{B}{2} (x\minus{}E) \to \minus{}\infty \quad as\quad x\to \minus{}\infty.\]

This proves that $ f'(x)\to 0$ as $ x\to \minus{}\infty$. Passing to the limit $ x\to \minus{}\infty$ in
$ f'(x)\equal{}f(f(x))$, we obtain $ 0\equal{}f(A)$, which is impossible.

[/hide]

For mathematical analysis problem 1

For complex solution

$ f(x) \equal{} ax^{b}$ for suitable $ a,b$

take let $ b$ complex number  such that $ b^2\minus{}b\plus{}1\equal{}0$ and $ a$ complex number such that $ b\equal{}a^{b}$

Numerical analysis problem 3

0/ If x(0)=0 the sequence is identically zero

1/ For the convergence of $ X(k)$ to a limit invertible

$ N(X_{0}.A^{t}.A \minus{} I_{n}) < 1$
 The limit    $ (A^{t}.A)^{ \minus{} 1}$

It is time for a café crème and a croissant.

[quote]Micael,

what about the case $ B \equal{} \plus{} \infty$ in your solution?

I mean, your proof of the impossibility of $ B > 0$ assumes
that $ B < \plus{}\infty$.
[/quote]

Well,  the case $ B\equal{}\plus{}\infty$ is also impossible:

If we assume that $ f'(x)\to\plus{}\infty$, then for every $ B_1>0$ there exits a $ E$ such that
$ f'(x)>B_1>0$ for all $ x\le E$. Then the rest applies with $ B_1$ in place of $ \frac{B}{2}$.


It is not true.
try
$a=b=\frac{1}{2},c=\frac{\sqrt 2}{2}$

Zhaobin the inequality is good even in this case.You should look for a solution. :D

sorry...I am wrong :(
and do you have a solution?
if you don't have,I think you should post in unsloved section :P

[quote="mateivld"]I have and it is very nice.  :)[/quote]
Yes! The vasc's hint: $9a^2b^2c^2\leq abc(a+b+c)$ kill it. ;)

[quote="cezar lupu"]Arqady, Vasc's hint is trivial because it reduces to $(a^{2}+b^{2}+c^{2})(a+b+c)\geq 9abc$, which follows from AM-GM. ;)[/quote]
I think Arqady know it ;)

Mateivld, why this ineq. is sharper than vasc's ?
They like each other as form, but are of different type.
For example, in your inequality, equality doesn'n hold for $a=b=c$.

[quote="black_puma"]please post full solution. thanks !!!!!!!!!   :huh:[/quote]
Sorry. I mistaked.

well it is sharper since $\displaystyle\frac{1}{5-6ab}=\frac{1}{2-3ab+2-3ab+1}\leq \frac{1}{9}\cdot\left(\frac{2}{2-3ab}+1\right)$.And equality holds for a=b=c.

[quote="Vasc"]Indeed, I was wrong, because equality holds for $a=b=c$.
Therefore, the both inequalities are of the same type. :lol:
But your inequality is not true.
Try, for example, $b=ka$ and $c=(k+1)a$ for any $k \geq 7$  ;)[/quote]
ok,and my suppose is right.though my counterexample is wrong ;)
and can you find the best form?

The inequality holds even in this case.Please try to think at a solution.

[quote="mateivld"]Prove that if $\sum a^2=1,a,b,c\geq 0$ then $\displaystyle\sum \frac{1}{2-3bc}\leq 3$.[/quote]
I found for you a nice contrexample: $a=b=\frac{12}{17},c=\frac{1}{17}$ ;)

Zhaobin, the best $k$ for the inequality
 $\displaystyle\sum \frac{1}{1-kbc}\leq \frac{9}{3-k}$
is $1.4558445..$. For this $k$, equality holds again when $a=0.1691...$ and $b=c=0.69692...$.
 :)
Therefore, a sharper than vasc's one is the following:
$\displaystyle\sum \frac{1}{3-4bc}\leq \frac{9}{5}$.  ;)

[quote="mateivld"]ok i will specify another condition for the problem $2-3bc\geq 0$ and the similar ones.Now the ineq works.At first I didn't see this.I will look today if one of this terms is negative,if the inequality is preserved.Sorry![/quote]
maybe I am blind.
but $2bc \le b^2+c^2 \le 1$
then $3bc \le \frac 3 2 <2$
is always true ;)

[quote="mateivld"]The ineq is good.So  nobody wants to find a solution,just looking for counterexamples. :([/quote]

Why can't you trust others..

Take arqady's example, then $\sum \frac{1}{2-3bc} = 3.0458727 > 3$ according to my calculator

I agree with the counterexample $a=b=12/17$, $c=1/17$.

[quote="Vasc"]Therefore, a sharper than vasc's one is the following:
$\displaystyle\sum \frac{1}{3-4bc}\leq \frac{9}{5}$.  ;)[/quote]

This one is equivalent to

\[ \sum_{sym} \frac{27}{2}a^6-51a^5b+81a^4b^2-51a^3b^3 + \frac{37}{2}a^4bc-14a^3b^2c+3a^2b^2c^2 \ge 0  \]

Which is rather sharp. It is also one of the first symmetric sixth degree inequalities I have seen that does not fall quickly to dumbassing. Vasc, do you have a solution?

[quote="Mildorf"]
\[ \sum_{sym} \frac{27}{2}a^6-51a^5b+81a^4b^2-51a^3b^3 + \frac{37}{2}a^4bc-14a^3b^2c+3a^2b^2c^2 \ge 0  \]
 It is also one of the first symmetric sixth degree inequalities I have seen that does not fall quickly to dumbassing.[/quote]
But it is fall! :)

[quote="Mildorf"][quote="Vasc"]Therefore, a sharper than vasc's one is the following:
$\displaystyle\sum \frac{1}{3-4bc}\leq \frac{9}{5}$.  ;)[/quote]
 Vasc, do you have a solution?[/quote]
I didn't try to solve it.  ;)
But I am a solution to my posted one (which is weaker, but nicer):
$\displaystyle\sum \frac{1}{5-6bc}\leq 1$.  ;)


The number of ounces of water needed to reduce $ 9$ ounces of shaving lotion containing $ 50\%$ alcohol to a lotion containing $ 30\%$ alcohol is:
$ \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 7$

If it is currently $ 50\%$ alcohol, there must be $ 0.5(9)\equal{}4.5$ ounces of alcohol in the lotion. To make $ 4.5$ ounces be $ 30\%$ of the solution, we need $ \frac{10}{3}(4.5)\equal{}15$ ounces of total solution (we use $ \frac{10}{3}$ because we are reverse-processing the percent, so we use the reciprocal of $ \frac{3}{10}$, the value of the percent). Subtracting the original $ 9$ ounces from $ 15$, we get $ 15\minus{}9\equal{}6$ ounces, which is choice $ \fbox{D}$.


If $ a,b,c\in\mathbb{R}$ such that

$ ab\minus{}5\equal{}3(a\plus{}b)$;
$ bc\plus{}3\equal{}3(b\plus{}c)$;
$ ca\minus{}12\equal{}3(c\plus{}a)$;

Find all solutions $ (a,b,c)$.

[hide="hint"]SFFT=Simon's Favorite Factoring Trick[/hide]


vector
combinatorics proposed
combinatorics

[b]A student wanted to copy a convex polygon M and draw it inside a circle $C(O,1)$.
At first, he draw the first side,then the second,then third and so on...
After he finished,he saw that the new <<polygon>> was not closed-surprisingly, the first and the last vertex were situated at distance $d$.It is known that, the student draw the angles just like the angles of M and the <<relative error>> at the length of each side is not more than $k$ ($k\in R$).[/b]
[size=150]
[b]Show that $d\leq 4k$[/b][/size]

[b][i][u]Nick Rapanos[/u][/i][/b][i][/i]

Okay i may give a hint!!

[hide]In my proof i use vectors![/hide]

Come on!Is it so difficult for everyone?????

What do you mean by <<relative error>>?
For example, if the original side has length 1, then the new will be between 1-k and 1+k?

not exactly... i mean that the new one will be 1+t where ItI<=k


ok so in book 4 in the priori incantatem james comes out before lily, meaning lily died first.

UH OH SO WHAT VOLDEMORT SAID HAPPENED IS NOT TRUE

or what possibly happened was that some james-impostor died in place of the real james before voldemort went over to (try to) kill harry, then voldemort killed lily. then, some time later, before bertha jorkins was killed, someone used voldemort's wand to kill the real james, or possibly ANOTHER impostor (snape?)

uh, that was a misprint, and it was fixed in later versions

Yeah, in my version Lily came out before James and said something like "your father is coming" or something.

Hmmm, I must have the misprinted one... James comes out first in mine :huh:

im not really sure about snape  :maybe:

of course, killing dumbledore wasnt the most innocent thing to do  :dry:

Actually you know how Dumbledore and Snape can prenetrate minds and interpret feelings. Maybe that's what they were doing on the tower! Maybe Dumbledore wanted Snape to kill him so it looks like Snape is on the bad side.

yeah, i know.

i love how jk rowling is actually unpredictable.

Yeah but she also said something about that Dumbledore had to die because something big is happening. Also we will find out more about Dumbledore and Wormtail's family I think. And apparently we will find out something extremely important about Lily and the fact that every says that Harry has her eyes is important.

Maybe he thought eyes meant color, but actually she had eyes that could see through things if she said the magic words, and that's what everyone meant

Am I the only one who has been overcome by an intense feeling of sadness upon hearing that yet another Harry Potter book would appear?

Most books are like this. The first one is really good and a bestseller and not meant to have sequals. The sequel gives a hint about how many books there will be. The rest of the books before the last book are pretty good and ok but leave you hanging. The last one has a happy ending but is pretty bad. Does anyone agree.

agreed, and also, i agree with BanishedTraitor. Maybe Dumbledore was asking for Snape to kill him when he was saying "please....Snape" or whatever...

Klebian that's an excellent idea :!:  never thought of that myself, maybe it's not the colour. Oh yeah, don't forget that Snape made the Unbreakable Vow and maybe that would come in play at some point. I strongly think that Harry would need to sacrifice himself for whatever purpose because that just seems to be quite fitting.

I have a theory.
Based on Rowling's unpredictability, it seems that Dumbledore is evil one, meanwhile Voldemort is the good one.
Note that I have said this theory, if my theory come true in the &th book. :)

[quote="sapphyre571"][hide="spoiler"]
I'm calling that Ginny dies. I was shocked when Rowling put them in a relationship together, and I've got a little feeling that they'll keep dating. Now, we should have noticed a pattern by now: Voldy (directly or indirectly) kills all the people whom Harry, for lack of a better word, loves. I mean his parents, Sirius, and Dumbledore. Ron and Hermione are exceptions because every now and then, Harry hates them. To make a long story short, Ginny sacrifices herself so Harry can continue to fight. It would be completely hilarious if Voldy walked in on them.[/hide][/quote]

Where did you get this? I wouldn't like HP7 if everyone died... :(

Dude no I think the fact that Harry broke up with her will come back and bite him in the ass because he is no longer loving anyone and therefore no longer exercising his power that the Dark Lord knows not as extensively.

[quote="life=tennis+math"]oook so dude i was watching harry potter 5 today right and i remembered something weird

alright so harry sees the thestrals for the first time in the fifth book right. and you're only supposed to be able to see the thestrals after you've seen death, that is you've seen someone die

but like, harry's parents are dead and he obviously saw them die because from the dementors he relives their death in a memory. if he wasn't awake, there's no way he would have the memory, so he was obviously awake, hence he has seen death. also he saw cedric diggory die in the fourth book but he did not see the thestrals then either.

so has j.k. rowling made some kind of explanation (that is has the question been asked) or have i just owned her[/quote]
uh when were threstrals even mentioned even previous books

as for snape i believe that he isn't bad i mean he was talking to dumbledore about how he doesn't want to do "something anymore" and dumbledore was taking "something" fer granted, so basically, dumbledore was taking his life for granted and that snape doesn't want to spy on the death eaters also

also, remember that weird part of the 4th book when there was a faint glint in dumbledore's eyes when he heard that voldy was back? well if you read very carefully there was a faint glint in snape's eyes when dumbledore sends on a mission to rejoin the death eaters. another time there is a faint glint in his eyes is when he tells harry that its his job to find out what the dark lord is telling death eaters......plus theres the time when his hand twitches and scrap like that.....hes probably not bad

[quote="life=tennis+math"]oook so dude i was watching harry potter 5 today right and i remembered something weird

alright so harry sees the thestrals for the first time in the fifth book right. and you're only supposed to be able to see the thestrals after you've seen death, that is you've seen someone die

but like, harry's parents are dead and he obviously saw them die because from the dementors he relives their death in a memory. if he wasn't awake, there's no way he would have the memory, so he was obviously awake, hence he has seen death. also he saw cedric diggory die in the fourth book but he did not see the thestrals then either.

so has j.k. rowling made some kind of explanation (that is has the question been asked) or have i just owned her[/quote]
I thought about Harry being able to see them a long time ago, but maybe there's some clever explanation? If not, I owned her first. :P

Maybe Professor Trewalney is going to make another prophecy in the 7th book.

[quote="sonny"]
uh when were threstrals even mentioned even previous books[/quote]
THAT'S EXACTLY MY POINT.
see cause harry mentions the "horseless carriages" prior to the 5th book and sees in the 5th book that the "horseless carriages" are in fact pulled by thestrals. what i'm saying is that he should have been able to see the thestrals even since the first book and certainly at the end of the 4th book, unless rowling has some explanation.

whats the deal with aunt petunia by the way? on a guess i'd say she's a squib or something because she mentions that her parents were proud to have a witch, her sister. while its also possible that she is a witch, i somewhat doubt it due to her repulsion of harry. if she was, by some chance, a witch, i'd say she could have been a death eater, since that would be an interesting turn of events. that could also explain the fear she felt when she heard of voldemort's return.

as a final point (the glints that sonny mentioned reminded me), wouldnt it be really cool if dumbledore turned out to actually be evil

the prophecy in the department of mysteries that said "at the solstice will come a new...and none will come after" is talking about dudley. pretty sure about it.
dudley was born on june 22nd or 23rd.

yes, i already have said this in my other thread. but just want to repeat it :D

[quote="bookaholic"]Do you guys seriously think something like that would have gone unnoticed until now? I believe JKR has an explanation for it somewhere on her site, probably in the FAQs or something. I think it involves Harry having been too young to understand/absorb what was happening on the night of his parents' death, and similarly not having had enough time to absorb Cedric's death by the end of GoF, but I don't feel like looking up the specifics right now. Of course, as always, the actual explanation is "the plot needed it to be that way".

P.S. I TRUST SEVERUS SNAPE![/quote]
GOOD FOR YOU

Harry didn't necessarily "see" his parents die because remember when Hagrid found him, he was fast asleep. Maybe his eyes were closed and he was about to fall asleep when his mum got killed and that woke him up (but he did not see her die) and that's why all he can remember was seeing heaps of green light.

[quote="7h3.D3m0n.117"][quote="Nerd_of_the_Ages"]i love how jk rowling is actually unpredictable.[/quote]

Agreed. Most good authors would be scared to kill off a popular character but JK Rowling doesn't mind. For all you know, in HP7 Harry loses and everyone good dies (Hermione, Ron, etc. etc.) and the magic world is succumed to Voldemort's wrath.

In an interview, JK Rowling says that Dumbledore IS dead and will not be coming back (not sure about the exact wording) in HP7. Although I would like to see him back  :lol: , I certainly don't have the power to make him alive again.[/quote]
But the thing is dumbledore [b]had[/b] to die. It's just like star wars where yoda dies, and every other epic, at the end, the hero [b]has [/b]to stand alone. In the end, Harry is the one who's going to have to win, not some old dude.

what if the reason Dumbledore defeated grindelwald is because grindelwald killed his parents? in the cave, dumbeldore seems to relive a very guilty and torturous part of his life.

In none of the books does it talk about dumbledore's family. "he had great teachers but was mostly self-taught." Never does it say anything about his family. perhaps, this explains the understanding of Dumbledore to Harry's current stage with his desire to defeat voldemort. dumbledore himself was caught in the same position.

Aberforth(the barman in the Hog's Head) was descrived as very odd. Could this be because of his parents were tortured and killed by grindelwald?

this makes perfect sense. The year dumbledore defeated grindelwald(1945) is the same year that tom riddle made his first horcrux when he killed his father and his grandparents. perhaps Tom Riddle searched for grinderwald to learn how to make a horcrux.

(yes i've been thinking about this a lot)

[quote="GoColts0626"]i think the bloody baron is grindelwald.

voldemort learned how to make horcruxes from grindelwald, which implies that grindelwald himself made a horcrux.
but grindelwald would only make a horcrux if he was afraid of death.

if he was afraid of death, he might have decided to become a ghost.

and the bloody baron fits, slytherin and peeves is scared of him.[/quote]

where did this grindelwald guy even become involved in this??? I dont remember reading about him anywhere ever....

[quote="indianamath"]book 1 when dumbledore was described in the chocolage frog card.  It said that dumbledore defeated the dark wizard grindelwald in 1945.

obviously, grindelwald has to affect book 7 right? jk rowling wouldn've put him in the story wihtout a purpose.[/quote]

lol ok i have a bad memory

[quote="sonny"]also, remember that weird part of the 4th book when there was a faint glint in dumbledore's eyes when he heard that voldy was back? [/quote]

the glint was when harry told him that voldy insisted on using harry's blood. I think I know why:

when dumbledore went tot he potters house, he did not plan to kill harry with the avada kedavra. He intended to replace his soul into harry's body to get rid of the threat stated in the prophecy.  But, Lily sacrificed herself and deflected (deflected, not absorbed) the spell and the horcrux spell bounced around the house and it finally hit harry. thus, harry got his scar. but voldy's soul did not destroy the current soul in harry's body. both souls lived inside one body. But the spell coontinued to bounce around the house, thus destroying the house. THE CURSE COULD NOT HAVE BEEN THE AVADA KEDAVRA BECAUSE THE AVADA KEDAVRA COULD NOT HAVE DESTROYED THE HOUSE. IT IS INTENDED TO LEAVE NO TRACE OF DAMAGE AT ALL. TOM RIDDLE SR. AND HIS PARENTS WERE PROOF.
     So, the horcrux curse hit voldy himself and destroyed his body. He would have been dead if not for his other horcruxes.

Now, up to the 4th book, it would have been impossible for harry to destroy all the horcruxes before he defeated voldemort. Why? it is because a piece of voldemort's soul was inside harry's own body from the night his parents died. But, Voldemort beleived in the prophecy too much. he desperately wanted to kill harry. Thus, he demanded harry potter's blood, instead of any ordinary wizard. By taking harry's blood, the part of his soul was returned to his regenerated body. If hvoldemort did not take harry's blood, voldy's soul would still be in harry's body, it would be impossible for voldemort to kill him because voldemort would be killing himself. however, he took harry potter's blood and his soul was returned to his regenerated body. THUS, HARRY COULD NOW DESTROY ALL THE HORCRUXES [b]AND[/b] DEFEAT VOLDEMORT. that would explain the gleam in Dumbledore's eyes.

lol i've been thinking about this for a long time.

p.s. some people might argue that voldy himself said "he set out to kill him(harry)" but there are many ways of killing somebody, or destroying their entire soul. Voldie intended to destroy harry's soul and replace it with his own.

I think I read somewhere that Rowling stated that the events in the 5th Harry Potter movie foreshadow the events in book 7. I think this specifically relates to the part when Voldemort tells Harry that "he will lose everything."

[quote="indianamath"] THE CURSE COULD NOT HAVE BEEN THE AVADA KEDAVRA BECAUSE THE AVADA KEDAVRA COULD NOT HAVE DESTROYED THE HOUSE. IT IS INTENDED TO LEAVE NO TRACE OF DAMAGE AT ALL. TOM RIDDLE SR. AND HIS PARENTS WERE PROOF. [/quote]

Well didn't Moody say that Harry was the only one who survived the Avada Kedavra Curse? I am pretty sure that Avada Kedevra can destroy a house I mean during that fight between Dumbledore and Voldemort in the 5th Book didn't Dumbledore make some golden statue come alive to protect Harry? When the curse hit them they deflected and destroyed a desk didn't it? Sorry I can't find my copy so I am not entirely sure if my memory is accurate.

the desk was lit with a small fire and stopped burning after a few seconds.


Prove that the equation \[1+2x+3x^2+...+nx^{n-1}=n^2\]has a rational root between $1$ and $2$ for all positive integers $n>1$

We have:
$P\left( x \right) = 1 + 2x + 3x^2  +  \cdots  + nx^{n - 1}  = \frac{{nx^{n + 1}  - \left( {n + 1} \right)x^n  + 1}}{{\left( {1 - x} \right)^2 }}$

and
$Q\left( x \right) = P\left( x \right) - n^2 $

then:
$nx^{n - 1}  +  \cdots  + 3x^2  + 2x + 1 = n^2  \Leftrightarrow Q\left( x \right) = 0$

Now prove that:  $Q\left( {\frac{{n + 1}}{n}} \right) = 0$


Let $ABC$ is a triangle with altitude AH. O is the midpoint of AH. The circle with center O and radius OA cut $AB$ and $AC$ at $E$ and $F$. $AM$ is the median of triangle $ABC$. Prove that two tangents at E and F of (O) intercept at point lie on $AM$.
[i]May be you cant understand my problem because my English is bad!!![/i]

Your English is quite OK. The problem was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=28075 .

  darij

Oh thanks very much!
But can you let me know the source of this problem (which contest...???)


Calculate $ \sum_{k\equal{}0}^{n} (k\plus{}1) \binom{n}{k}$.

[hide]
Write as $ \sum k\binom{n}{n\minus{}k}\plus{}\sum \binom{n}{k}$. The latter is just $ 2^n$, the former can be interpreted as picking an element of a n-element subset and then partitioning the remaining elements into two sets, so it's $ n2^{n\minus{}1}$. Summing, the answer is $ 2^n\plus{}n2^{n\minus{}1}\equal{}2^{n\minus{}1}(2\plus{}n)$. [/hide]

[hide]Let's do this by counting in two ways.

Suppose we have to choose a group from a set of $ n$ people and we have the option of picking a leader for the group. (the group may be empty also)

We can pick a group of size $ k$ (running from $ 0$ to $ n$). Then we have $ k \plus{} 1$ ways to solve the leader issue --- either no leader, or pick one from the $ k$ guys.

So, no. of ways = $ \sum_{k \equal{} 0}^{n} (k \plus{} 1) \binom{n}{k}$.

Counting in another way:

We split the counting into 2 cases here.

If no leader, $ 2^n$ ways to pick a group. (each of the $ n$ guys can either be selected, or not)

Else pick a leader ($ n$ choices) and then pick group from the remaining $ n \minus{} 1$ guys ($ 2^{n \minus{} 1}$ ways)

So, no. of ways = $ 2^n \plus{} n \cdot 2^{n \minus{} 1}$

Therefore, $ \sum_{k \equal{} 0}^{n} (k \plus{} 1) \binom{n}{k} \equal{} 2^n \plus{} n \cdot 2^{n \minus{} 1}$[/hide]

$ \sum (k\plus{}1) \binom{n}{k} \equal{} \sum (n\minus{}k\plus{}1) \binom{n}{n\minus{}k} \equal{} \sum (n\minus{}k\plus{}1) \binom{n}{k}$

Hence if you denote the required sum by S, we have $ 2S \equal{} \sum (k\plus{}1) \binom{n}{k} \plus{} \sum (n\minus{}k\plus{}1) \binom{n}{k} \equal{} (n\plus{}2) \sum \binom{n}{k} \equal{} (n\plus{}2)2^n$

Hence $ S \equal{} (n\plus{}2) 2^{n\minus{}1}$

We know from newtons formula,
$ (a\plus{}1)^n\equal{}\sum \binom{n}{k}a^k(1)$

so,

$ n(a\plus{}1)^{n\minus{}1}\equal{} \sum k \binom{n}{k} a^{k\minus{}1}(2)$

Setting a=1 and adding we get,

$ \sum (k\plus{}1)\binom{n}{k} \equal{} 2^n\plus{}n2^{n\minus{}1}$, done


I never noticed that the class transcripts were so long (35 pgs. for AIME A class). I remembered that AoPS used to have a printer-friendly option, but now it is gone. I also noticed that this forum uses phpBB. Doing a little Googling I found this forum post on the phpBB forum:

[url]http://www.phpbb.com/community/viewtopic.php?f=15&t=141443&st=0&sk=t&sd=a[/url]

Any chance of implementing this script on AoPS? It would be extremely helpful for those printing out transcripts, problem sets, etc.

[quote="1/(ln x)"]I also noticed that this forum uses phpBB. [/quote]The code for this site may have started with phpBB but now it has very little in common with it. We will revive that option but there are more pressing matters right now.


inequalities
search
induction
inequalities unsolved

Hi ! Let $ n$ a natural number.Prove that

$ {n}^{\frac{1}{n}} \plus{} {n}^{\frac{1}{n\plus{}1}} \plus{} {n}^{\frac{1}{n\plus{}2}} \plus{}  ... \plus{} {n}^{\frac{1}{2n\minus{}1}} \geq n ^{\sqrt[n]{2}}$

(I couldn`t find it by search :oops: )

It seems hard  :P  but I know it`s easy and AM-GM and CS kill it !! I`m waiting for your nice solution .
Thanks in advance .

they are natural number.
then induction must also help.
[b]mahanmath[/b] post your solution...


[color=blue][i]A retired mathematics teacher asked me to find a proof for this problem( he has not found a solution till now) :[/i][/color]

[b]PROBLEM[/b]
 The centers of the Euler circles of the four tringles in complete quadrilateral are concyclic !

 Is this a theorem ? I have never seen it before !

 Babis

According to my sketch the proposition is false.

I does not seem right. Suppose lines AB, CD intersect at X, and lines BC, DA intersect at Y. 9-point circle centers of triangles YAB, YCD, XBC, XDA are not concyclic. Which four triangles of this complete quadrilateral do you have in mind ?

Well. first thanks for your answers .Second, I wrote to Bil to send me the source of the problem !
 The only theorem  I have found is that if the quadrilateral is cyclic , then the Euler centers of  4 triangles ( I will write tomorrow which triangles are these, but surely they are not those of the complete one) are concyclic.

 Babis


limit
function
real analysis
real analysis solved

$f$ is continous and bounded on $(a,\infty)$. prove that $\forall T$, $\exists x_n\to\infty$ such that $\lim\limits_{n\to\infty} (f(x_n+T)-f(x_n)) = 0$

Let us fix $n $ and prove that we can find $ x=x(n)\geq n $ such that $|f(x+T)-f(x)|<\frac{1}{n} $. Suppose not, then define the function $ g(x)=f(x+T)-f(x) $ for $ x\geq n $. Then $ g$ is continuous and has no zeroes, so it must have constant sign. Suppose $ g(x)>0 $ for $ x\geq n $. Then $ f(x+T)-f(x)>\frac{1}{n} $ for all $ x\geq n $, which fails since $ f$ is bounded. Indeed, just take $ x=kT $ with k sufficiently large and add all inequalities. Thus, the conclusion. Moreover, we see that it suffices for $ f$ to have Darboux property.

Sorry for my ignorance, but
what is the Darboux proberty?

If $\lambda$ is between $f(a),f(b)$, then there is a $c$ between $a,b$ s.t. $f(c)=\lambda$. If $f$ has this property, then we say that it has the Darboux property. I think you might know it as the intermediate value property.


Could someone help on this problem:

a ^2+b^2+c^2=2004.

I found the solutions (4 of them) by writing a computer program, and I've been able to further narrow down the problem (by evaluating lhs and rhs mod 8 and the square roots mod 4) by finding that a has the form 4x+2, b has the form 4y and c has the form 4z. However, i can't seem to go on from here. The solutions by the way are (a,b,c)=(32,28,14), (34,28,8), (40,20,2), and (44,8,2). Thanks.

oops, the solutions are somehow in the wrong order following the fact a=4x+2, b=4y and c=4z

it should be (a,b,c)=(14,28,32), (34,28,8), (2,20,40), and (2,8,44).

in spoiler (highlight to read)

[hide]

Well its not too hard after you've done that. Plug in a, b, c and you get x^2 + x + y^2 + z^2 = 125. Now we just check some cases for x, and if you know the fact that a number is the sum of two squares if and only if it has no factors that are 3 mod 4..

Substitute in x from 0 thru 10 and check whether 125 - (x^2 + x) has a factor of the form 4k+3 in each case. Only possibilities are x=8, 3, or 0, and now we just make that result the sum of two squares, the first two options providing one solution, and the 0 producing the other two.

[/hide]

Thanks, i didn't know the fact that the sum of squares are not congruent 3 mod 4--then again, i should've seen that. That really helps a lot.

Well that just comes from the fact that squares are 0 or 1 mod 4. Its slightly harder to prove what I said - that it can be written as such if it doesn't have any factors 3 mod 4.. its very nice though.


Alex's uncle paid Alex $ \$5$ a week for doing a few minor chores.  However, if Alex failed to do all the chores in any week, he did not get paid for that week and he had to give his uncle $ \$7.50$ for that week.  During the first $ 52$ weeks, Alex failed to do the required chores during two weeks.  He started the year with $ \$89$, spent $ \$143$ on books, and saved the rest of the money he earned by doing chores.  How much money did he have at the end of the $ 52$-week period?

Alex got paid $ 5\times50$-$ 7.5\times2$ that year, or $ 235$ dollars. Thus he had $ 89$+$ 235$-$ 143$ at the end, or $ 181$.


cho tam giac ABC.CMR
$ \frac{1}{(a\plus{}b\plus{}c)^2}\plus{}\frac{1}{(ab\plus{}bc\plus{}ca)^2}\geq \frac{1}{3(ab^2\plus{}bc^2\plus{}ca^2)}\plus{}\frac{1}{3(a^2b\plus{}b^2c\plus{}c^2a)}$

[quote="ngocvantu"]cho tam giac ABC.CMR
$ \frac {1}{(a \plus{} b \plus{} c)^2} \plus{} \frac {1}{(ab \plus{} bc \plus{} ca)^2}\geq \frac {1}{3(ab^2 \plus{} bc^2 \plus{} ca^2)} \plus{} \frac {1}{3(a^2b \plus{} b^2c \plus{} c^2a)}$[/quote]
Y' cua? ban. la` j` khi viet ABC.CMR

zay ma cung hoi.a,b,c la 3 canh cua 1 tam giac day ban

Khong phai dau, y no hoi "abc.cmr" la ji? chac la khong bit "cmr" la chu viet tat cua "chung minh rang"(=prove) ay


algebra
function
domain
complex analysis

Hi could someone explain to me in details how to find laurent expansions for 1/(z-i) about the point z_0=1
What are the domains on which the series are convergent to the function?

Thank you much[/url][/code][/quote][/i]

Your starting point is the expansion $\frac{1}{1-w}=\sum_{n=0}^{\infty}w^{n}$, which is valid for $|w|<1$. There are two ways to fit your function into this formula:

(1) $\frac{1}{z-i}=\frac{1}{z-1+(1-i)}=\frac{1}{1-i}\frac{1}{1-(1-z)/(1-i)}$. Now use the above expansion with $w=(1-z)/(1-i)$:
$\frac{1}{z-i}=\frac{1}{1-i}\sum_{n=0}^{\infty}((1-z)/(1-i))^{n}=\sum_{n=0}^{\infty}\frac{(-1)^{n}(z-1)^{n}}{(1-i)^{n+1}}$. This expansion is valid when $|w|<1$, that is, for $|z-1|<\sqrt{2}$.

(2) $\frac{1}{z-i}=\frac{1}{z-1+(1-i)}=\frac{1}{z-1}\frac{1}{1-(i-1)/(z-1)}$.  Now proceed as in (1), but with $w=(i-1)/(z-1)$.

HI Mlok,

Thank you much for your reply
How do I proceed to get the domain of convergence to 1/(z-i)

Thank you


Given that it takes four miles of fence to enclose a square 640-acre field, how many acres are in a square field enclosed by two miles of fence?

Since area is squared, the answer is simply $ \frac{640}{2^2}\equal{}\fbox{160}$

In the initial problem, $1$ mile to enclose each side. But with 2 miles of fence $\frac{1}{2}$ to encode each side. So the area will be $\frac{1}{2}$ squared as much, giving $160.$


group theory
abstract algebra
induction
Galois Theory
superior algebra
superior algebra unsolved

Let $X$ and $Y$ be distinct subgroups of a finite group $G$. We say that $X$ and $Y$ are a weird pair if $|X| = |Y |$ and if no subgroup of $G$ other than $X$ and $Y$ has this same order.

(a) If $G$ is a group having a weird pair of subgroups, show that some subgroup of $G$ has a weird pair of normal subgroups.
(b) Show that a solvable group cannot contain a weird pair of subgroups.

Cute one :).

[b](a)[/b]

If $X,Y$ are not conjugated to one another, they're both normal, and we're done with (a). If, on the other hand, we can find $g\in G$ s.t. $Y=gXg^{-1}$, there are two cases: either $N_G(X)=N_G(Y)$ (the normalizers of the two subgroups coincide), in which case $X,Y$ form a pair of weird subgroups of $N_G(X)\le G$, or $N_G(X)\ne N_G(Y)$. We treat this latter case now. Since $X,Y$ each has precisely two conjugates in $G$, we have $[G: N_G(X)]=[G: N_G(Y)]=2$, so $N_G(X),N_G(Y)$ are distinct normal subgroups of $G$ with index $2$. This means that $G=N_G(Y)N_G(X)$, so we can write $g$ as $a_ya_x,\ a_x\in N_G(X),a_y\in N_G(Y)$. We get $Y=a_ya_xXa_x^{-1}a_y^{-1}\Rightarrow Y=a_y^{-1}Ya_y=a_xXa_x^{-1}=X$, which contradicts $X\ne Y$.

[b](b)[/b]

Thanks to (a), it suffices to prove (b) in the following form: a soluble group $G$ cannot have a weird pair of normal subgroups $X,Y$. We prove this by induction on $|G|$. Suppose it’s done for groups of cardinal smaller than $|G|$.

If $X\cap Y\ne\{1\}$, then $G/X\cap Y$ is a soluble group with cardinal $<|G|$, and if $X,Y$ form a werid pair in $G$, then $X/X\cap Y,Y/X\cap Y$ form a weird pair in $G/X\cap Y$, so we get a contradiction because of our induction hypothesis.

If $X\cap Y=\{1\}$, then $H=XY$ is actually the direct product of $X$ and $Y$, as can easily be seen. Since $H\ne G$ would contradict our induction hypothesis, we can assume that $G\cong X\times Y$. By a theorem of Hall (sorry I'm using this; I don't see a simple solution right now, although I'm sure there must be one :)), given a soluble group $K$ and naturals $a,b$ s.t. $(a,b)=1,ab=|K|$, the group $K$ has a subgroup of order $a$ (and thus one of order $b$ as well). We apply this to our groups $X,Y$. If $|X|$ is a prime $p$, then we easily find subgroups of $G=\mathbb Z_p^2$ of cardinal $p$, other than $X,Y$. If $|X|$ is a prime power $p^t,\ t\ge 2$, then $X$ has a subgroup $\tilde X$ of cardinal $p^{t-1}$, and $Y$ has a subgroup $\tilde Y$ of cardinal $p$, and we'll take the third group of order $p^t=|X|$ to be $\tilde X\tilde Y$. Finally, if $|X|$ is not a prime power, then we find $a,b>1,\ ab=|X|,\ (a,b)=1$ and apply Hall's Theorem cited above to find a subgroup $\tilde X$ of $X$ of order $a$ and a subgroup $\tilde Y$ of $Y$ of order $b$, and take our third group of order $|X|$ to be $\tilde X\tilde Y$ again.

P.S.

[url=http://en.wikipedia.org/wiki/Hall_subgroup#Definitions]Here[/url]'s that theorem I cited:


Hi,

I've put on the [url=http://mersenneforum.org/showthread.php?t=10670]GIMPS/Mersenne forum[/url] the description of 3 conjectures that are waiting for a proof. I've already done half the proof for one of them (the easy part...).
I've provided PARI/gp code that exercises the 3 conjectures.

I'll give [color=red][b][size=150]100Euro[/size][/b][/color] for the guy/lady who will provide one proof out of the three.

These conjectures deal with properties of a Digraph under x^2-2 modulo a Mersenne, a Wagstaff or a Fermat prime, and they use cycles of length q-1 (Mersenne and Wagstaff) or 2^n-1 (Fermat).

The final idea/dream is to use cycles of length (q-1)/n for proving that such numbers are NOT prime, in order to speed up the search and proof.

The conjecture about Wagstaff numbers is already implemented as a "Vrba-Reix PRP" test in the LLR tool by Jean Penné, based on Prime95 library, which is usually used for looking at primes of the form: k*2^n+/-1 .

Come and have a look, have fun, and compete !

(And come look how big are the two new Prime Mersenne numbers we found at [url=http://www.mersenne.org/]GIMPS[/url], and that I've verified.)

Tony

I've summarized the information in this [url=http://tony.reix.free.fr/Mersenne/SummaryOfThe3Conjectures.pdf]paper[/url].
And here is a link to a [url=http://www.cs.uwaterloo.ca/~tmjvasig/papers/newvasiga.pdf].pdf version of Shallit&Vasiga pape[/url]r.
Tony

Hi,

[b][size=84]Mister Robert Gerbicz has provided a proof for the first (the easier, but not so easy !) part of Conjectures 2 and 3. See [color=red][url=http://robert.gerbicz.googlepages.com/WagstaffAndFermat.pdf]his paper[/url][/color].[/size][/b]

Though this paper needs some cleaning and clarifications, 2 persons at least have read it. I've found no problem. Quite easy to understand. But guessing how he got the ideas. You are welcome to read it and make comments if you feel that some part is unclear or wrong.

This means that there is now a fast PRP test for Wagstaff numbers !
{Fast here means that this PRP test is about as fast as Prime95 is for Mersenne numbers.}

A first step forward.
Now, the converses (the hard part...).

Tony

Hello,
I've produced a [url=http://tony.reix.free.fr/Mersenne/SummaryOfThe3Conjectures.pdf]new version of the conjectures[/url] (thanks to "Dodo"), with more explanations, with new seeds, and with new PARI/gp code, and with information about the status of proof of the sufficiency, and who did what.
Hope it helps !
Only the necessity is now required to be proven !
(note that Lucas never provided a proof for the necessity of his theorems ! Then Lehmer came.)
Tony


Let A(n) denote the set of all 2n-digit numbers whose decimal representation conisists of exactly n 1's and n 0's. (no leading 0 is permitted.)
Show that there must be at least one number in the set A(n)that is evenly divisible by n.

Say, for n = 3, the set A(3) has 5!/(3!*2!) = 10 numbers.
A(3) = {100011, 100110, 101100, 111000, 110100, 110010, 110001, 101010, 101001, 100101} of which at least one should be divisible by 3. Here all of them are!!! That's a different story.

Prove it for any postitive number n irrespective of how big or small n is.

Regards,
Shamik Banerjee

This is Puzzles and Brainteasers not actual prove-it mathematical stuff. If a mod could move this that would be great.


function
algebra
domain
calculus
calculus computations

how many critical values does $ y\equal{}ln(\sin(x))$ have over (0, 4$ \pi$ )?

How many zeros does $ \cot(x)$ have in that interval?

I don't like the question, because there are whole intervals on which the function is not defined. "Over $ (0,4\pi)$ " doesn't make a lot of sense when we can't use $ (0,4\pi)$ as a domain for the function.


http://en.wikipedia.org/wiki/Magic_squares
is the whole article
 the formula:
$M(n) = \frac{n^{3}+n}{2}$
I don't get how to derive! Would someone tell me how I would?
Here is another link to it:
http://mathworld.wolfram.com/MagicSquare.html

Thanks,
Funcia

It's not hard.  A normal magic square contains the numbers $1, 2, ... n^{2}$.  The sum of all these numbers is $\frac{n^{2}(n^{2}+1)}{2}$.

The sum of all of the numbers can be viewed as the sum of each column sum, so the individual sum of all of the numbers in a column is that quantity over $n$, which is $\frac{n(n^{2}+1)}{2}= \frac{n^{3}+n}{2}$.

Since the row, column, and diagonal sums are all the same, they must be this sum.


I'm trying to increase the spacing between my lines in the {align} environment.  I've done it once before, but am not getting what I want this time.  Here is my code:

[code]
\renewcommand{\arraystretch}{2}
\begin{align}
\frac 68 &= \frac {2\cdot 3}{2\cdot 4} \\
\frac {2\cdot 3}{2\cdot 4} &= \frac 22 \cdot \frac 34 \\
\frac 22 \cdot \frac 34 &= 1 \cdot \frac 34\\
1 \cdot \frac 34 &= \frac 34
\end{align}
[/code]

It looks too crowded.  Any suggestions?

     -- Pawel

You could use for example \\[2ex] instead of \\


function
superior algebra
superior algebra unsolved

Let k, m, and n be positive integers such that k|mn. Show that the function F: [b]Z[/b]n -> [b]Z[/b]k defined by f([x]n)=[mx]k is a one to one correspondence if and only if k=n and (m,n)=1.

Can someone PLEASE help me! I don't even know where to start with this proof! There is too much information and I don't understand how to sort it out!!

Take a deep breath and walk through what you need to prove.

* An "if and only if" problem is really two problems.  Don't try to do both at once, focus on one direction at a time.  It may help to write one of the statements at the top of a blank sheet of paper.

* Showing that something is a 1-1 correspondence is also two problems: showing that it is injective and surjective.  Again, do not try to tackle both at the same time, but treat them as separate problems.

* One way to do the $ \Rightarrow$ direction is using the contrapositive.  That is, assume $ (m,n)\neq 1$ and try to show that $ f$ is not a 1-1 correspondence.  And then, (a separate problem!) assume that $ k\neq n$ and show that $ f$ is not a 1-1 correspondence.

* How can you show that something is not a 1-1 correspondence?  Show that it is not surjective, or show that it is not injective.  Which one do you think you should try to show?  Plug in some small values for $ m$ and $ n$ to see what happens and it may help you figure out where to start.

* There are many people willing to help with problems like this, but if you don't say what work you've done and where you've gotten stuck, people may assume that you are lazy and just asking us to do your homework.


$ u_{n}\equal{}\frac{cos1\minus{}cosncos(n\plus{}1)}{cosn.cos(n\plus{}1)}$
$ a_{n}$ is number of positve real in $ u_{n}$
$ b_{n}$is the number of negative real in $ u_{n}$
Find lim  $ \frac{a_{n}}{b_{n}}$

$ u_{n}\equal{}\frac{cos1\minus{}cos(2n\plus{}1)}{cos1\plus{}cos(2n\plus{}1)}$, therefore $ u_{n}< 0$ if and only if $ \minus{}\frac{1}{2\pi}<\{\frac{2n\plus{}1}{2\pi}\}<\frac{1}{2\pi}$, therefore $ lim_{n\to\infty}\frac{a_{n}}{b_{n}}\equal{}\frac{\pi\minus{}1}{\pi}$.


Hi,

Here you can find the rules of my new game as attachment. I called it "cleanumber".

The game is very hard (hard to program) because it involves billions of possibilities.
The goal of the game is to clean all the board with the best score. And the best score is the minimum of points to clean all the board.
In my example I gave only the first step (or move).
It is easy to imagine what are the next steps.
If you have any question post it.
I will try to answer.
Thank you.

The game at the step 6
Here is the picture as attachment.
At the right the steps with the values of a,b,c,d and S.
The digits removed are coloured in green.
The priority is to remove the initial 0's.
No digit isolated is another pririty unless we can not clean all the board.
I do it manually without program.
Any question or comment?

I did it! 30 points!!! It is not a good performance
But a program can reach less maybe 23 or 22 points.
Here the solutions and the steps as attachment

[quote="AIME15"]i still don't get what you do in this game...[/quote]

The goal of the game is to REMOVE ALL THE DIGITS on the board 10x10.
You have to combine 10 digits (0,1,2,3,4,5,6,7,8,9) to create 4 numbers.
If the sum S of the 4 numbers (a+b+c+d) is IDENTICAL to one of the numbers on the board you remove all the identical numbers.

Example :
On the board if you read it you have 3 times the number 77
By forming 77 as a sum of  7+20+14+36 (no digit is repeated twice) you remove all the 77 (green on the board).
And you continue doing so until cleaning all the board.
At each move or step you score one point.
I scored 30 to clean all the board.
You have to do less to win.
22 or 23 is reachable by programming.
Do it manually first because some problems appears :
- initial zeros (you have to remove the zeros first)
- isolated digit (if you have an isolated digit you can not solve the problem the minimum value of S is 10=1+2+3+4)

Good luck!

Ahh I get it....if anyone is good at programming, maybe post the game online?  so we don't hafta redraw every time...

My best score now is 26 points.
I'm trying to do less than 26 points.
Soon I will post if I found 25 points.
The game cleanumber is very hard to solve even with a program because it involves many cases.
The programming can only help to do better not to find the optimal solution.
Just try to solve it manually to understand what are the challenges involved.


geometry
AMC
USA(J)MO
USAMO
MATHCOUNTS
AMC 10
AIME
\/closed

Can someone illustrate when it is useful to get/use/learn-fromt the new geo book, and when alot of it will be review. Becuase I was reading through the exerpts, and I see it more as a source of problems, than 'much' of a teaching manual. aops vol 1 geo is enough? What do you all think?

Sure.
Chaps. 1-3 are good for USAMO training for the geometry stuffs.
Chap. 4 is good for Mathcounts and all of the AMCs.
Chaps. 5-7 are good for highschool tests.
Chap. 8-10 is more highschool stuff.
Chaps. 11-13 are good for AMC(10,12), AIME, USAMO.
Chaps. 14-15 is just more prep for all contests.
16-17, highschool.

Hope that helped!

[quote="mysmartmouth"]I am finding that i am very weak in (especially 3-d) contest geometry.  However, I can't read through the book, the stuff about area of a triangle and similarity postulates bores me to tears.  I've already taken Geometry, though I didn't get as much as I could have out of it.  I have the book, any suggestions as to how to go about reading it (which sections when)?[/quote]

Here's the rub with geometry -- you're really running up against a wall if you don't read through the stuff that isn't that exciting.  At some point, you'll have to put in those 50? hours of reading in order to provide yourself with the depth to get into the really cool stuff (it gets more exciting at the next levels).

If you're having trouble wading through some of the reading, [i]start with the problems[/i]!  Skim the text to give you a basic idea as to what the material's about if necessary, but go straight to the review/challenge problems.  This is probably an especially good strategy for students who have already taken geometry, but don't yet know it to this level (nearly every student who took geometry?).  Along the way, you'll have to go back and read the most important parts to solve problems or understand solutions, but that's perfect.  You get most of that reading done in a way that isn't quite as tedious.  That might even motivate you to complete those other few hours of reading.  There is nothing more motivating than success.

[quote="mysmartmouth"]I am finding that i am very weak in (especially 3-d) contest geometry.  However, I can't read through the book, the stuff about area of a triangle and similarity postulates bores me to tears.  I've already taken Geometry, though I didn't get as much as I could have out of it.  I have the book, any suggestions as to how to go about reading it (which sections when)?[/quote]

To echo Mathew's point, if you're just plowing ahead reading the book, you're taking the wrong approach.  There's a reason we put the problems at the start of each section.

(Also, I'll note that the chapter description given earlier isn't entirely accurate - we go to much more depth and more challenging problems than the typical geometry curriculum, so it's more than just high school stuff.)

[quote="rzsolt"]How many problems are there in the whole book? Also,  which subjects are developed more fully/deeper, than aops v1 / v2 ?[/quote]

Part of the point of our writing these new books is that they do cover topics more completely, with more depth, and include topics not included or undercovered in the old AoPS books (since clearly, math leading up to USAMO level math cannot be bound in two volumes).

As to the number of problems in the books -- I don't know those numbers off the top of my head, but there are hundreds.  Students reading the AoPS books might find only one or two problems at the end of a chapter testing each concept, but the new books should provide a dozen or more testing most concepts.


Hey guys,

We just increased the number of possible private messages you can store from 50 to 100.

Enjoy more private talk on AoPS - MathLinks :P :)

Thank you!

My private box was filling with replys for my contest and I've been tired of deleting them after I read them (so I can receive other's answers).

Thank you Valentin (and other people who worked on this)! :)

[quote="Valentin Vornicu"][quote="Silverfalcon"]Thank you Valentin (and other people who worked on this)! :)[/quote]Well it was a quick fix by me, but Harazi pushed me to it :P So  you can thank him :D[/quote]

Thank you harazi then! :)

Thank you harazi.  I have 45 right now, and if harazi hadn't upgraded it i would only be allowed 5 more!

[quote="236factorial"]Maybe we should also increase the amount of messages in the sent box, it's always getting filled in my account. I know that's not easy though. Maybe just 80 in the inbox?  :)[/quote]I am sorry, not going to happen anytime soon :)

How do you make a smileey in the main post in the forum like this one?

YOu mean this  :D ?
What do you mean?  :?

Yes, for exaple, in teh area where the icons  :arrow: , [img]http://www.artofproblemsolving.com/Forum/images/icons/userquestion.gif[/img], [img]http://www.artofproblemsolving.com/Forum/images/icons/icon3.gif[/img], and [img]http://www.artofproblemsolving.com/Forum/images/icons/icon1.gif[/img] are in the main forum, and there is a  :D in the place of that. How do i do it in one of my posts?

Hi Valentin:
    What about more attachments?

[quote="shobber"]What about more attachments?[/quote]

I think the attachments are more than enough. wHat do you need them for? (.gif files are only 2-5 kb each). PDF files?

then just host them on anotehr place and link there, or host a website wiht the files on your own computer(leave it on at the nights), and you just link like http://your ip address/your image file.gif or your pdf files.pdf.


limit
logarithms
calculus
calculus computations

find this limit: :D

$\lim_{k\to\infty}\;\left[ \left( \sum_{j=1}^{k}\; (-1)^{j}\cdot\ln{\left( 1+\frac{1}{j}\right) }\right)-\left( \sum_{j=1}^{k}\; (-1)^{j}\cdot\frac{1}{j}\right) \right]$

answer is hidden below.

[hide]
$answer: \;\;\;\quad \boxed{2\ln 2-\ln \pi}$
[/hide]

$\lim_{k\to\infty}-\left(\sum_{j=1}^{k}\frac{(-1)^{j}}{j}\right) = \lim_{k\to\infty}\left(\sum_{j=1}^{k}\frac{(-1)^{j+1}}{j}\right) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}= \ln 2$.

$\lim_{k\to\infty}\left[\sum_{j=1}^{k}(-1)^{j}\ln\left(\frac{j+1}{j}\right)\right] =-\sum_{n=1}^{\infty}(-1)^{n+1}\ln\left(\frac{n+1}{n}\right) =-S$.

For this last series, let $A_{2n-1}$ be the sum of the first 2n-1 terms. Carefull sum show that

$A_{2n-1}= \ln \frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot 6 \cdot 6 \cdot ... \cdot (2n)}{1 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot ... \cdot (2n-1) \cdot (2n-1)}$

$= \ln\left[\left[\frac{2 \cdot 4 \cdot 6 \cdot ... \cdot (2n)}{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1)}\right]^{2}\cdot \frac{1}{(\sqrt{2n})^{2}}\right]$

$= 2 \ln\left[\frac{1}{\sqrt{2}}\cdot \frac{2 \cdot 4 \cdot ... \cdot (2n)}{1 \cdot 3 \cdot ... \cdot (2n-1)}\cdot \frac{1}{\sqrt{n}}\right]$.

So:

$S = \lim_{n\to\infty}A_{2n-1}= 2 \ln\left[\frac{1}{\sqrt{2}}\cdot \lim_{n\to\infty}\left(\frac{2 \cdot 4 \cdot ... \cdot (2n)}{1 \cdot 3 \cdot ... \cdot (2n-1)}\cdot \frac{1}{\sqrt{n}}\right)\right]$

$\Rightarrow  S = 2 \ln\left(\frac{\sqrt{\pi}}{\sqrt{2}}\right) = \ln \pi-\ln 2$.

(To be more correct, we must find the expression for $A_{2n}$ and then show that lim of $A_{2n}$ is also this one).

Finally, your limit equals $-(\ln \pi-\ln 2)+\ln 2 = 2 \ln 2-\ln \pi$.


topology
algebra
function
domain
abstract algebra
advanced fields
advanced fields unsolved

Hi,

my definition of a differentiable n-manifold : a set M along with a set of maps A
such that every map in the set A ( achart) is defined on a subset of M, is injective, and has as image an open subset in $\mathbb{R}^n$

M has to be the union of all of those domains of maps in A

when two maps in A have a nonempty intersection of domains, and their respective images are U and V, the map from U to V (where possible) through M has to be in the class $C^{\infty}$ and vice versa

a topology is defined in this way on M : a set B is open if and only if for every map f with domain W,
$f(W\cap B)\subseteq \mathbb{R}^n$ is open

my definition of a topological n manifold :

a Haussdorf space
second countable : there is a countable basis of open subsets
locally euclidean : every point has a neighbourhood homeomorphic with an open subset of $\mathbb{R}^n$

Finally my question : are all differentiable manifolds topological manifolds?

They should be- if they're not, we have to change the definitions. The charts are homeomorphisms, so the space is locally Euclidean.

Hausdorff? Let's check a standard counterexample- the "line with two origins".
As a set, this is $\mathbb{R}$ together with an additional point $0'$. Our two charts are the identity and the map which sends $x$ to $x$ when $x\neq0$, and sends $0'$ to $0$. This satisfies the properties you listed, so we're in trouble; the space is not Hausdorff.

Second countable? This can go wrong in silly ways. Take an uncountable discrete set, with charts to $\{0\}$ that have single-point domains.

Both of these properties fail, so we don't have a very good definition. My textbook defines a smooth manifold $M$ as a topological manifold with a maximal smooth atlas; a smooth atlas is a set $A$ of charts to $\mathbb{R}^n$ such that:
Each chart $\phi\in A$ is a homeomorphism from an open set of $M$ to an open set of $\mathbb{R}^n$
For any two charts $\phi,\psi\in A$, if the domain of $\phi$ and the domain of $\psi$ intersect in $X$, $\phi\circ\psi^{-1}$ is a diffeomorphism from $\psi(X)$ to $\phi(X)$.
$M$ is the union of the domains of the charts in $A$.
A maximal smooth atlas is simply a smooth atlas not contained in any strictly larger smooth atlas- every smoothly compatible chart is already there.

[quote]such that every map in the set A ( a chart) is defined on a subset of M, is injective, and has as image an open subset in $\mathbb{R}^n$[/quote]

Does your second counterexample for the countability satisfy the condition that the image of the chart is open?

So what is your point now, your first line seems to suggest that 'my' differentiable manifolds are topological, but then you prove they aren't?

Sorry I don't understand, please elaborate.

Those are maps to $\mathbb{R}^0$. We could use uncountably many separated copies of $\mathbb{R}^n$ if we felt like it.

Yes, I'm saying that your definition allows differentiable manifolds to not be topological manifolds. That doesn't seem right, so we should probably use a different definition.

Yes I see now, my definition doesn't demand second countable or Haussdorf. It is locally euclidean though right???

A friend of mine takes a lie algebra course and his definition looks more like yours.  Just one question, when you define smoothness, do you also demand that $\psi(X)$ and $\phi(X)$ are open subsets of $\mathbb{R}^n$?


"I  hear  some  youngsters  playing  in  the  backyard,"  said
           Jones,  a  mathematics  student.  "Are  they  all  yours?"

           "No,"  explained  Professor  Smith,  "My  children  are  play-
           ing  with  friends  from  three  other  families  in  the  neighbor-
           hood,  although  our  family  happens  to  be  the  largest.  The
           Browns have a smaller number of children, the Greens have
           a still smaller number,  and the Blacks have the smallest of
           all."

           "How  many  children  are  there  all  together?"  asked  Jones.

           "Let me put it this way," said Professor Smith.  "There are
           fewer  than  18  children,  and  the  product  of  the  numbers  in
           the  four  families  happens  to  be  my  house  number,  which
           you  saw  when  you  arrived."

           Jones  took  a  notebook  and  pencil  from  his  pocket  and
           started  scribbling.  A  moment  later  he  looked  up  and  said,
           "I  need  more  information.  Is  there  more  than  one  child  in
           the  Black  family?"

           As  soon  as  Professor  Smith  replied,  Jones  smiled  and  cor-
           rectly  stated  the  number  of  children  in  each  family.

           Knowing  the  house  number  and  whether  the  Blacks  had
           more  than  one  child,  Jones  found  the  problem  trivial.   It
           is  a  remarkable  fact,  however,  that  the  number  of  children
           in each family can be determined solely on the basis of the
           information  given  above!

           Determine  the  number  of  children  in  each  family

Oh, god.  Wasn't this a Ciphering question at MAO last year?


geometry
inradius
incenter
geometric transformation
homothety
circumcircle
ratio

Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$, $\omega_{B}$ to $BC$ and $BA$, $\omega_{C}$ to $CA$ and $CB$, and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$. If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Hmmm slightly easy for #15.

First, apply Heron's formula (or anything else, w/e) to find that the area is 84. Also the semiperimeter is 21. So the inradius is 4. OK. Now consider the incenter I. Let the radius of one of the small circles be r. Let the centres of the three little circles tangent to the sides of ABC be X, Y, and Z. Let the centre of the circle tangent to those three circles be P. A homothety centred at I takes XYZ to ABC with factor (4 - r)/4. The same homothety takes P to the circumcentre of ABC, so PX/R = 2r/R = (4 - r)/4, where R is the circumradius of ABC. The circumradius of ABC can be easily computed by R = a/2sinA, so doing that reveals R = 65/8. Then 2r/(65/8) = (4-r)/4 ==> 16r/65 = 1 - r/4 ==> 129r/260 = 1 ==> r = 260/129, so the answer is 389. WOOHOO go me for not using any LaTeX.

I love this problem. Homothety is so fun.

I've actually seen this setup before. The goal was to prove that the circumcenter of the middle triangle was collinear with the incenter and circumcenter of the original triangle. I actually used areas to find the answer, and got the same answer as Alex.

BTW, I'm probably just not seeing this but how do you know that the ratio for the homothety is (4-r)/4?

Suppose $x=$ the radius we want...$w_{A}w_{B}w_{C}$ is similar to $ABC$ since the sides are parallel...hence, the inradius ratio=the circumradius ratio...$\frac{r}{r-x}=\frac{R}{2x}\implies x=(r^{-1}+2*R^{-1})^{-1}$...$R=\frac{65}{8}$ and $r=4$...$m+n=260+129=\boxed{389}$

hrmm...it is a nice problem...but how did this get to be #15...its like ok similar triangles...but at least there is not room for error here...

[quote="SamE"]
BTW, I'm probably just not seeing this but how do you know that the ratio for the homothety is (4-r)/4?[/quote]

[b]HINT[/b]

 The ratio ,is the ratio of the altitudes of the two similar triangles. But the distance between the parallel sides is just r and the inradius is 4, so the small altitude is 4-r.

[u] Babis[/u]

I know I'm reviving an old thread, but I'm not seeing why the homethety takes $ \omega$ to the circumcenter of $ \triangle ABC$.  Could someone please explain that to me?

I'm wondering the same thing. How did you know that the homothety brings the circle to the circumcenter?

$ P$ is the circumcenter of $ XYZ$, because $ PX \equal{} PY \equal{} PZ \equal{} 2r$ where $ r$ is the radius of $ \omega$.

Well is it possible for the triangle to be 13,14,15 in those conditions? And also the other given conditions?

[quote="supercomputer"]Hello I was doing this problem and I was wondering: couldn't w=w_a=w_b=w_c= the incircle of ABC? They didn't specify that the circles are distinct so wouldn't this satisfy all constraints?[/quote]
[quote="pkothari13"]and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$.[/quote]

[quote="rcv"][hide]Consider a 13-14-15 triangle.  $A=84.$  [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]

The inradius is $r=\frac{A}{s}=\frac{84}{21}=4$, where $s$ is the semiperimeter.  Scale the triangle with the inradius by a linear scale factor, $u.$

The circumradius is $R=\frac{abc}{4rs}=\frac{13\cdot 14\cdot 15}{4\cdot 4\cdot 21}=\frac{65}{8},$ where $a,$ $b,$ and $c$ are the side-lengths.  Scale the triangle with the circumradius by a linear scale factor, $v.$

Cut and combine the triangles, as shown.  Then solve for 4u:
[hide]$\frac{65}{8}v=8u$
$v=\frac{64}{65}u$

$u+v=1$
$u+\frac{64}{65}u=1$
$\frac{129}{65}u=1$
$4u=\frac{260}{129}$
$260+129=\boxed{389}$[/hide][/hide][/quote]

Wait why does 65v/8=8u?

I believe he's setting the radii of the incircle and circumcircle equal to each other after a suitable scaling.

lol when you don't notice the homothety

[hide=Sketch]
Diagram (from AoPS wiki): http://wiki-images.artofproblemsolving.com/2/23/2007_AIME_II-15.png

Let $AB=13, BC=14, CA=15$. First notice that from LoC and half-angle formula we find $\left\{\tan{\frac{A}{2}}, \tan{\frac{B}{2}}, \tan{\frac{C}{2}}\right\}=\left\{\frac{4}{7}, \frac{1}{2}, \frac{2}{3}\right\}$. By dropping the radii of $\omega_A, \omega_B, \omega_C$ to the sides, we can calculate the lengths of the tangents and in turn find the sidelengths of $\triangle{O_AO_BO_C}$. We find that $\triangle{O_AO_BO_C}\sim \triangle{ABC}$ with ratio $\frac{4-r}{4}$. The circumradius of $\triangle O_AO_BO_C$ is just $2r$, so from extended LoS we have \[\frac{O_AO_C}{\sin{O_AO_BO_C}}=\frac{15\left(\frac{4-r}{4}\right)}{\sin{O_AO_BO_C}}=\frac{15\left(\frac{4-r}{4}\right)}{\sin{ABC}}=\frac{15\left(\frac{4-r}{4}\right)}{\frac{12}{13}}=2\cdot 2r\implies r=\frac{260}{129}\longrightarrow \boxed{389}.\]
[/hide]

[hide=Solution]
Let $AB=13, BC=14, CA=15$, and let $X,Y,Z$ be the centers of $\omega_A, \omega_B, \omega_C$ respectively, and let $r$ be the radii of the four circles. Note that $X$, $Y$, and $Z$ are all $r$ away from the sides their respective circles are tangent to. So, $\overline{XY} \parallel \overline{AB}$, $\overline{YZ} \parallel \overline{BC}$, $\overline{XZ} \parallel \overline{AC}$. So $\triangle XYZ \sim \triangle ABC .$ Now all is needed is to find the similarity ratio. Note that $X,Y,Z$ are equidistant from the sides the circles are tangent to so they each lie on an angle bisector. Using this, we can compute (using the half angles from LoC) $XY = 13 - \tfrac{r}{\tan \tfrac{A}{2}} - \tfrac{r}{\tan \tfrac{B}{2}} = 13 - \tfrac{7r}{4} - \tfrac{3r}{2} = 13 - \tfrac{13r}{4}$ (or any other side similarly). So the similarity ratio is $1 - \tfrac{r}{4} .$ Then, connect $X,Y,Z,$ to the center of $\omega$ and notice that the circumradius of $\triangle{XYZ} $ is $2r$. Using the fact that the circumradius of a 13-14-15 is $\tfrac{65}{8}$ (can be found using areas and $\tfrac{abc}{4R}$) we have $2r = \tfrac{65}{8} ( 1 - \tfrac{r}{4})$ which gives $r = \dfrac{260}{129} \implies 389$.

[/hide]

Let $r$ be the desired radius, and $O, O_A, O_B, O_C$ be the centers of the circles. Obviously $O$ is the circumcenter of $\triangle O_AO_BO_C$. Furthermore, this triangle is similar to $\triangle ABC$ and has inradius $r$ less than that of $\triangle ABC$, or $4-r$. Then the similarity ratio is $\frac{4-r}4$, so $$2r = \frac{4-r}4 \cdot \frac{65}8,$$ which yields $r = \frac{260}{129}$.

Call the centers of $\omega_a$, $\omega_b$, and $\omega_c$ $x$, $y$, and $z$ respectively. Obvisously we have that $\triangle XYZ \sim \triangle ABC$. Let's write $XY=13x$, $YZ=14x$, and $XZ=15x$. The circumcenter of this triangle is the center of $\omega$ since we can connect the radii of $\omega_a$, $\omega_b$, and $\omega_c$ to the radii of $\omega$ to have $3$ equal segments (or a circumcenter) Since the area of this triangle is $84x^2$ we get $\frac{13x \cdot 14x \cdot 15x}{4R}=84x^2$ Solving we get $R=\frac{65x}{8}$. Or the radii are $\frac{65x}{16}$.

We can now draw the radii of the circles $\omega_a$, $\omega_b$, and $\omega_c$ which are tangent to the sides of $\triangle ABC$. Call the equal segments from $A$, $B$, and $C$ to the circles (the tangent segments) $p$, $q$, and $r$ respectively. We find that $q=6-6x$. Since $Bx$ is the bisector of $\angle ABC$ we know that $\frac{65x}{96-96x}= \tan\left(\frac{\angle ABC}{2}\right)$. We can use the half angle formula and simplify to get $x=\frac{64}{129}$. We multiply by $\frac{65}{16}$ to get our answer of $\frac{64 \cdot65}{129 \cdot 16}=\boxed{\frac{260}{129}}$. $\blacksquare$

A really synthetic AIME problem :) The homothety took some inspiration from a WOOT potd.

One can calculate that the circumradius $R$ of $\triangle{} ABC$ is $\frac{65}{8},$ and the inradius $r$ is equal to $4.$

Let the centers of $\omega_A, \omega_B, \omega_C, \omega$ be $A',B',C',O.$ Two things:
1. $\triangle{} ABC$ and $\triangle A'B'C'$ are homothetic with center $I,$ the incenter of $\triangle{} ABC.$ Let the scale of this homothety be $k;$ then the radius of $\omega_A, \omega_B, \omega_C, \omega$ is $r(1-k).$
2. $O$ is the circumcenter of $\triangle A'B'C',$ hence the distance from $O$ to $A'$ is just $kR.$
However, this radius is also equal to $2r(1-k)$ from the given condition, we get $$kR = 2r(1-k).$$ Plugging in values, we get $$\frac{65k}{8} = 8-8k.$$ Solving for $k$ gives $k = \frac{64}{129},$ hence the radius is $$r(1-k) = 4 \cdot \frac{65}{129} = \frac{260}{129}.$$ Our answer is $389.$

I trigbashed, but I think the result in the remark is interesting enough to merit a post.

Label the centers of the four circles as shown, and let $r$ be the common radius of the four circles. First note that $\triangle I_A I_B I_C \sim \triangle ABC$ and $I$ is the center of the circle $(I_A I_B I_C)$, which has radius $2r$.
[center][img width="50"]https://media.discordapp.net/attachments/1178875828008923176/1198117597443543170/07aimeii15.png[/img][/center]

Notice that the length $BC = 14$ is split into three parts. Since $\angle I_B I I_C = 2 \angle I_B I_A I_C = 2 \angle A$ we have \begin{align*} 14 = BC &= r \cot \tfrac12 B + 4r \sin A + r \cot \tfrac12 C \\ &= r \left(\frac{1+\cos B}{\sin B} + \frac{1+\cos C}{\sin C} + 4\sin A\right). \tag{$\star$} \end{align*} Since the circumradius $R = \tfrac{65}{8}$ of $\triangle ABC$, by Law of Sines (or by dropping the altitude from $A$) we have \begin{align*} \sin A &= \tfrac{56}{65}, & \sin B &= \tfrac{12}{13}, & \sin C &= \tfrac{4}{5}, \\ \cos A &= \tfrac{33}{65}, & \cos B &= \tfrac{5}{13}, & \cos C &= \tfrac{3}{5}. \end{align*} Substituting these in, \[ 14 = r \left(\frac32 + 2 + \frac{224}{65}\right) = \frac{903}{130} r \implies r = \frac{260}{129}. \quad \square \]

[b][color=red]Remark:[/color][/b] Notice that ($\star$) rewrites as \begin{align*} 2R \sin A &= r \bigl(\cot \tfrac B2 + \cot \tfrac C2 + 4 \sin A\bigr), \\ \frac{2R}{r} - 4 &= \bigl(\cot \tfrac B2 + \cot \tfrac C2\bigr) \csc A, \end{align*} and the right side of the last expression is not symmetric in $A$, $B$, $C$. This is interesting because it means this value is invariant over cycling $A$, $B$, and $C$.
Actually, it is not too hard to show this result algebraically: set $(\alpha, \beta, \gamma) = \bigl(\tfrac A2, \tfrac B2, \tfrac C2\bigr)$, and then use trig formulas, we have \begin{align*} \frac{2R}{r}-4 &= \frac{\cot \beta + \cot \gamma} {\sin\bigl(180^{\circ} - 2(\beta+\gamma)\bigr)} \\ &= \frac{\frac{\cos\beta}{\sin\beta} + \frac{\cos\gamma}{\sin\gamma}} {2\sin(\beta+\gamma)\cos(\beta+\gamma)} \\ &= \frac{\csc\beta\csc\gamma}{2\cos(\beta+\gamma)} \\ &= \frac{1}{2 \sin\alpha \sin\beta \sin\gamma}, \end{align*} where we used $\alpha + \beta + \gamma = 90^{\circ}$ in the last step.
This value is now symmetric in $\alpha$, $\beta$, and $\gamma$, so it is clearly invariant over cycling $A$, $B$, and $C$.


[quote="erdogankerem123"][hide]
$ \frac {(2000)(1999)}{2} \equal{} 1999000$
[/hide][/quote]
How did you get that?? Sum of n numbers in the sequence formula??

[quote="archimedes1"]Calculate $ 1999^2 \minus{} (1998^2 \minus{} (1997^2 \minus{} (1996^2 \minus{} \dots \minus{} (2^2 \minus{} 1^2))))$.[/quote]

[hide]
This seemingly annoying and time-wasting expression can be simplified to $ (1999^{2}\minus{}1998^{2})\plus{}(1997^{2}\minus{}1996^{2})\minus{}........$.  (and all that stuff)  Since the difference of 2 consecutive squares is the sum of their square roots, the whole thing simplifies to $ 1999\plus{}1998\plus{}1997\plus{}1996\plus{}....\plus{}2\plus{}1$.  We know this to be $ \frac{1999(2000)}{2}\equal{}1999000$ (hopefully).  And we're done.
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[quote="alanchou"][quote="archimedes1"]Calculate $ 1999^2 \minus{} (1998^2 \minus{} (1997^2 \minus{} (1996^2 \minus{} \dots \minus{} (2^2 \minus{} 1^2))))$.[/quote]

[hide]
This seemingly annoying and time-wasting expression can be simplified to $ (1999^{2} \minus{} 1998^{2}) \plus{} (1997^{2} \minus{} 1996^{2}) \minus{} ........$.  (and all that stuff)  Since the difference of 2 consecutive squares is the sum of their square roots, the whole thing simplifies to $ 1999 \plus{} 1998 \plus{} 1997 \plus{} 1996 \plus{} .... \plus{} 2 \plus{} 1$.  We know this to be $ \frac {1999(2000)}{2} \equal{} 1999000$ (hopefully).  And we're done.
[/hide][/quote]

Oh.. I think I get it now..


Whenever I try to program on my TI-89 titanium, if I am using a loop that goes through many times, the calculator terminates the program and says that it is out of memory... does anyone know how to solve this problem? could it have to do with using local variables?

Thank you!

If you have many loops in loops (generally considered good programming style), or anything that requires a lot of backtracking (just imagine you were carrying out the computations that the calculator is to figure this out), then you may encounter this problem. Yes, it's local variables. There is no way to increase this memory; the only way to fix this, as far as I know, is to resort to less elegant programming.

The vagueness and most likely un-usefulness of my post is due to me not knowing exactly what you are talking about. Can you say specifically what you are trying to make?

I have a TI84+ and I had the same problem, the problem is you must end every loop, sometimes one'll write:
for
...
goto xx
end
but this end doesn't really ends the loop because the calculator has already gone to xx.

[quote="buzzer11"]For lack of a better program:

[code]:xyz()
:Prgm
:rand(10)->x
:Prompt y
:If x=y Then
:Disp "same"
:Else
:Disp "different"
:EndIf
:t+1->t
:xyz()
:EndPrgm[/code]

The program crashes after around 52 iterations.[/quote]

Recursion fails. You want to use loops.

[quote="buzzer11"]really? I used recursion all the time on my TI83+ just because it was easier than For...End For... etc.
Maybe the 89 is more picky?[/quote]

What are you talking about?
It's
For(var,start,end[,inc]
STUFF HERE
End
You don't use End For
Recursion is slow. Try doing recursion on something like fibonacci numbers.

Yeah, I meant End. And I know recursion is slow and "improper" but I was just confused why the 83 allowed it and the 89 doesn't.


number theory
relatively prime
number theory unsolved

It may appear that RSA decryption will not work if you are unlucky enough to choose a message $ a$ that is not relatively prime to $ m$. Of course, if $ m\equal{}pq$ and $ p$ and $ q$ are large, this is very unlikely to occur.

Show that in fact RSA decryption does work for all messages $ a$, regardless of whether or not they have a factor in common with $ m$.

This is a special case of http://www.mathlinks.ro/viewtopic.php?t=192050 .


How many of the bracelts do you have? (by braclets i mean the rubber ones that have something written on it and example would be a "Livestrong Band")  I saw this kid who had 20 bands on.  It looked very uncomfortable.  I have two bands.  one is yellow and the other is orange.  I'm just wondering how many bands do you have.

lol none.  I could never figure out where everyone was getting them.

they were on websites, and most stores had them.(i got mine from Discovery store)

live strong and polio, though I don't know where they are anymore...

I have a Livestrong, Tsunami Relief, and a Duke Band.

None, because it's extremely pointless, why wear one of these bands, if you want to be aware of the different medical problems such as cancer or other diseases, why not just read about them and become aware, as well as honoring the survivors such as Lance Armstrong? The whole idea is completely pointless to me and almost hypocritical as well.

I've seen many people wear them, but I don't because I don't see why you should wear them.

I know that, but where do you get them?  :)

I got my LiveStrong at a gas station, I got my Tsunami Relief one at school, I got my Duke one at a duke game.

[quote="GoBraves"]Do you guys not understand that all the money that you use to purchase them go for cancer research and such?[/quote]

Yes, I understand that, but why wear them??? It's almost like showing off...

i dont wear them, and even if i did, what would be wrong with that. You bought a wrist band, and you are sopposed to wear wrist bands on your wrist. How is that showing off?

I don't want any.
 This topic is spam. especially since a very similiar one has already been posted.

[quote="Super Naterul"] a very similiar one has already been posted.[/quote]
it has...i didn't know that...

i have zero zilch NOODLE!!! (0)  my brother has 5 that have all been found lying around or given to him, he hasn't bought a single one.  A kid on my baseball team's sister came to practise and had 67 bands :ninja: it was crazy to spend that much money on bands!

i have a livestrong band that i cant find :P

1; it says South Carolina on it. I never wear it, but it makes a great mega-rubber band to shoot people with. :P

I have a livestrong, a Novi cares(for the tsunami), and this reading one from school.


logarithms
number theory unsolved
number theory

let write all the power of 2 like that :124816.....
Show that all integer can can be seen in this representation some where ( for exemple the number 248 ...)

It's enough to show that the powers of two can start with an arbitrary integer.

If $2^{n}$ begins with the integer $a$ then $\log_{10}\frac{2^{n}}{a}$ has small fractional part; but we know that we can make this fractional part arbitrarily small by some well-known theorem (I think Weyl's?).

but i dont know this theorem you are talking about , can you give more details , i didn't thought that powers of 2 can began with arbitrary integer , good idea man


My English is not well,so I don't speak english,ok.
given $ a,b,c \ge 0$.Prove that
$ ab^2\plus{}bc^2\plus{}ca^2$
I need a nice solution.

[quote="babyloverain"]My English is not well,so I don't speak english,ok.
given $ a,b,c \ge 0$.Prove that
$ ab^2 \plus{} bc^2 \plus{} ca^2$
I need a nice solution.[/quote]

where is the thesis ?  :huh:

Sorry.Given a,b,c$ \ge$0 and a+b+c=3
Provethat $ a^2b\plus{}b^2c\plus{}c^2a \leq 4$

[quote="babyloverain"]Sorry.Given a,b,c$ \ge$0 and a+b+c=3
Provethat $ a^2b \plus{} b^2c \plus{} c^2a \leq 4$[/quote]

Let x=a/3, y=b/3, z=c/3$ \Rightarrow x\plus{}y\plus{}z\equal{}1$
$ a^2b \plus{} b^2c \plus{} c^2a \leq 4\Leftrightarrow x^2y \plus{} y^2z \plus{} z^2x\le \frac {4}{27}$

It is Canada 1999.

http://www.mathlinks.ro/Forum/viewtopic.php?t=216


Find the real roots of the equation $ \sqrt[4]{16\minus{}x}\plus{}\sqrt[4]{1\plus{}x}\equal{}3$.

Conditions: $ \minus{}1\leq x\leq 16$

Let $ \sqrt[4]{16\minus{}x}\equal{}a, \ \sqrt[4]{1\plus{}x}\equal{}b$

Then,
$ \left\{\begin{array}{ll}a\plus{}b\equal{}3\\a^4\plus{}b^4\equal{}17\end{array}\right.$

Replacing $ b\equal{}3\minus{}a$ in the second equation we have

$ a^4\minus{}6a^3\plus{}27a^2\minus{}54a\plus{}32\equal{}0\Leftrightarrow(a\minus{}1)(a\minus{}2)(a^2\minus{}3a\plus{}16)\equal{}0$

$ \Rightarrow \left\{\begin{array}{ll}a\equal{}1\\b\equal{}2\end{array}\right., \ \left\{\begin{array}{ll}a\equal{}2\\b\equal{}1\end{array}\right.$

If $ a\equal{}1\Rightarrow x\equal{}15$

If $ a\equal{}2\Rightarrow x\equal{}0$


Let triangel ABC. Prove that
$ acosB\plus{}bcosC\plus{}ccosA\le \frac {3\sqrt{3}R}{2}$

I think it's same [url]http://www.mathlinks.ro/viewtopic.php?t=237441[/url]


geometry
perpendicular bisector
geometry proposed

Let $ABCD$ be a square and $a$ be the length of his edges. The segments $AE$ and $CF$ are perpendicular on the square's plane in the same half-space and they have the length $AE=a$, $CF=b$ where $a<b<a\sqrt 3$. If $K$ denoted the set of the interior points of the square $ABCD$ determine $\min_{M\in K} \left( \max ( EM, FM ) \right) $ and $\max_{M\in K} \left( \min (EM,FM) \right)$.

[i]Octavian Stanasila[/i]

Let the perpendicular bisector of $ EF $ meet the plane $ ABCD $ at line $ RS $ such that $ R%Error. "inCD" is a bad command.
$ and $ \s\in BC $.max{$ EM,FM $}in triangle $ RSC $is $ EM $ and in other places is $ FM $ so the required value is minimmum of this values. Min of $ EM $in triangle$ RSC $is$ EP $such that $ P $ is the midpoint of$ RS $.Min of$ FM $in other places is $ FP $so the minnimum of max{$ EM,FM $}is $ EP(=FP) $.We compute this value:let $ CP=x $then according to the Pythagurian equarion $ x^2+b^2=(a\sqrt{2}-x)^2+a^2 $ so $ x=\frac{3a^2-b^2}{2\sqrt{2}a} $ so
$ FP=\sqrt{b^2+x^2}=\sqrt{\frac{9a^4+b^4+2a^2b^2}{8a^2}} $.