[quote="miwalin"]Let $ 0\le a,b\le 1$, prove or disprove that

$ [(a^v \minus{} b^v) \plus{} (a^{1 \minus{} v} \minus{} b^{1 \minus{} v})]^2\le4[a \plus{} b \minus{} 2ab],$
where $ 0\le v\le 1$.

Remark. When $ v \equal{} 0, \frac {1}{2}, 1$, the inequality is valid.[/quote]

try $ a\equal{}\frac{1}{10000},b\equal{}\frac{1}{100},v\equal{}\frac{1}{10}$

Thank you, hedeng.  :)
Under the same condition, is it possible to find  $ f(v)$ defined on [0, 1]  such that $ [(a^{v}\minus{}b^{v})\plus{}(a^{1\minus{}v}\minus{}b^{1\minus{}v})]^{2}\le f(v)[a\minus{}b]^2$?


linear algebra
matrix
linear algebra unsolved

Let $A = \begin{bmatrix}2&2&1\\2&5&2\\1&2&2\\\end{bmatrix}$, $B = \begin{bmatrix}-x&-y&z\\0&y&2z\\x&-y&z\\\end{bmatrix}$ where $x,y,z$ are positive reals.
and Let $B^tAB = \begin{bmatrix}8&0&0\\0&27&0\\0&0&42\\\end{bmatrix}$.
Find $x,y,z$
(Note: can it be done without actually multiplying out $B^tAB$?)

OK, can it be done without actually multiplying $B^tAB$?


Let $ a,b,c,d \in \mathbb{R}$.

$ x_1, x_2 \in \mathbb{C}$ are the roots of  $ x^2\plus{}ax\plus{}b\equal{}0$

$ x_3, x_4 \in \mathbb{C}$ are the roots of  $ x^2\plus{}cx\plus{}d\equal{}0$

and $ |x_i|<1(i\equal{}1,2,3,4)$

$ x_5, x_6 \in \mathbb{C}$ are the roots of  $ x^2\plus{}\frac{a\plus{}c}{2}x\plus{}\frac{b\plus{}d}{2}\equal{}0$

Prove: $ |x_j|<1(j\equal{}5,6)$  :blush:

Let $ f(x)\equal{}x^2\plus{}ax\plus{}b,g(x)\equal{}x^2\plus{}cx\plus{}d$.
$ f$ admits 2 roots with modulus $ <1$ iff $ |b|<1,|a|<2,b>|a|\minus{}1$.
$ g$ admits 2 roots with modulus $ <1$ iff $ |d|<1,|c|<2,d>|c|\minus{}1$.
Then $ |(b\plus{}d)/2|<1,|(a\plus{}c)/2|<2,(b\plus{}d)/2>(|a|\plus{}|c|)/2\minus{}1\geq|a\plus{}c|/2\minus{}1$.   QED


Has anyone here played this game called Block Dude? It is a strategy based game in which you move blocks to get to a door.  I have it on my ti-83+ and I got to level 8.  Now i'm stuck on level 8. :D I was wondering if anyone could give me some help.  I think the ti-84 has block dude under puzzpack and the ti-89 definetly has it. So please help me!

Heres a link to the game if you don't have one of the calculators:
http://oos.moxiecode.com/examples/blockdude/index2.html

if you search on the internet it will give you how to do it in a video and instructions.

Im on level 10  :D Block dude is a good time waster

That game is really addicting. Very good for classes that are boring. I forget how far I got.

i got all the way through level 11!!!  there are only 11 levels.  the password for level 11 is [hide]wTF[/hide]

Level 11 is pretty fun. The iRC level is the most 'elegant' level, IMO.

What...
I haven't played for a couple years or something, but I ended on like level 8 or 9 and whenever I got stuck it was by one block or something.

block dude isnt meant to be searched for on the internet -.-"

[quote="dark_sunburst"]All hail Tetris![/quote]

I'll second that.
Last year, during AP Statistics, we only had to pay attention about half of the time to absorb the material, so we (the person sitting next to me and I) frittered away many happy hours playing tetris.

[quote="math92"]I have it on my ti-83+ and I got to level 8.  Now i'm stuck on level 8. :D I was wondering if anyone could give me some help. [/quote]

I'm also stuck on that level.

[quote="dark_sunburst"][quote="mysmartmouth"]Calculator games are such a waste of time.  Use that time to read a book, or do something productive.[/quote]
Hey, you can't insult Block Dude!

Or worse yet, TETRIS. :omighty: All hail Tetris!

Not that I don't like reading or anything. It's just that every time I go to the library, I get about fifteen books and end up wasting my entire weekend. The amount of time I waste on calculator games pales in comparison.[/quote]

That is exactly my opinion except for ZTRIS not TETRIS (same game, different graphics).

[quote="B-tone_Party"]
Last year, during AP Statistics, we only had to pay attention about half of the time to absorb the material, so we (the person sitting next to me and I) frittered away many happy hours playing tetris.[/quote]

That's what I'm going to do next year. The head of the math department at our school told me that AP Stat was for seniors only and said that we would fail that class if we took it. Fortunately, The magnet program director signed me up anyway.
AP Stat is probably the biggest slack class ever, and it gives a full quality point to your GPA. How sad.

I've gotten to lvl8. I'm stuck. please help me.

reviving topics=bad

anyways go to detachedsolutions.com


The trajectory of a missle translates itslelf by the equation y= 2/3x + 2.
An enemy plane is located at points (-2,5).
By how much did the missle miss its target if each unit is 10 metres?

Please help me out on this guys.. I'm really stuck.
I dont really know what to do.. or how to start.

Do you understand what the problem is asking you to do?  That's always the first step to solving word problems.

sry for not using latex

The problem is trying to make you find the shortest distance by which the missile missed, correct
(thats my understanding of the question...)

To find that we need to find the function that is perpendicular to the first one, and runs through the plane:

[hide]
y = -3/2x + 2
[/hide]

Then we have to find their intersection point:

[hide]
(0,2)
[/hide]

and then, find the distance from that point to the plane, if each unit is ten metres:

[hide]
20^2 + 30^2 = x^2
\sqrt {1300} = x
10 * \sqrt {13} = x
[/hide]

thats at least my understanding of the problem

I believe it asks for the distance between $ (\minus{}2, 5)$ and $ (\minus{}2, f(\minus{}2))$ (where
$ f(x) \equal{} \frac{2x}{3} \plus{} 2$), multiplied by $ 10$.


Fie triunghiul ABC dreptunghic in A.Demonstrati ca latura patratului inscris triunghiului este cel mult egala cu o treime din ipotenuza.

Fie $l$ lungimea laturii patratului si $h$ inaltimea. Fie patratul $EFGH$ cu $E$ apartine lui $[AB]$, $F$ lui $[AC]$, $G$ si $H$ lui $[BC]$.
Din Thales avem $\frac{AF}{AC}=\frac{l}{a}$ si $\frac{FC}{AC}=\frac{l}{h}$
Prin insumare avem $\frac{l}{a}+\frac{l}{c}=1$, de unde $l=\frac{1}{\frac{1}{a}+\frac{1}{h}}$, care trebuie sa fie mai mic sau egal cu $\frac{a}{3}$, de unde avem de demonstrat ca $a\geq\ 2h$, care e o chestie cunoscuta, usor de demostrat cu teorema inaltimii sau din faptul ca triunghiul e inchis intr-un cerc de diametru $[BC]$.


Write the standard form of the equation and the general form of the equation of the circle of radius r and center (h, k).

GIVEN:

r = 4; (h, k) = (2, -3)

Also: When (h, k) is given, does it mean that the circle has been moved from the origin (0, 0)?

I need basic steps in terms of writing the equation in standard form and the general form of the circle given the above data.

Thanks

[hide]The equation for the circle is $(x-h)^{2}+(y-k)^{2}=r^{2}$. Where $(h,k)$ is the center and $r$ is the radius. So the eqaution of the circle would be $(x-2)^{2}+(y+3)^{2}=16$.[/hide]

So, I simply plug and chug into the provided equations, right?

When you are giving the center of the circle, (h,k) that means that the center has been moved from the origin to the new point of (h,k)/


Take a = 1, b = x --> 0 c = 1/x then Sum --> 1/3.

(For example, a = 1, b = 1/10, c = 10 then Sum = 0.45...

Conjecture: Sum > 1/3.


modular arithmetic
number theory unsolved
number theory

Find all non-negative integers $ x,y,$ and $ z$ satisfying that $ 7^x \plus{}1 \equal{} 3^y \plus{} 5^z$

If one of $ x,y,z$ is $ 0$ or $ 1$,it's trivial and we have $ (x,y,z) \equal{} (0,0,0) (1,1,1)$,so let $ x,y,z\ge 2$.

we have $ \minus{} 1\equiv ( \minus{} 1)^z\pmod{3}$ so $ z$ is odd,so let $ z \equal{} 2Z \plus{} 1(Z\in \mathbb{N})$.
and we have $ ( \minus{} 1)^x\equiv ( \minus{} 1)^y\pmod{4}$,so $ x$ and $ y$ are both odd or both even,but if both even and we take $ \text{mod 8}$ yields contradiction,therefore $ x$ and $ y$ are both odd,so let $ x \equal{} 2X \plus{} 1,y \equal{} 2Y \plus{} 1(X,Y\in \mathbb{N})$,
the equation becomes

$ 7\cdot 49^X \plus{} 1 \equal{} 3\cdot 9^Y \plus{} 5\cdot 25^Z$.

take $ \text{mod 25}$ then we have
$ 7\cdot ( \minus{} 1)^X \plus{} 1\equiv 3\cdot 9^Y\pmod{25}$,
which is impossible by cheking all cases of $ (X,Y) \pmod{2}$,hence the solutions are $ (x,y,z) \equal{} (0,0,0)(1,1,1)$

[quote="Anavel_Gato"]If one of $ x,y,z$ is $ 0$ or $ 1$,it's trivial and we have $ (x,y,z) \equal{} (0,0,0) (1,1,1)$,so let $ x,y,z\ge 2$.
...
take $ \text{mod 25}$ then we have
$ 7\cdot ( \minus{} 1)^X \plus{} 1\equiv 3\cdot 9^Y\pmod{25}$,
which is impossible by cheking all cases of $ (X,Y) \pmod{2}$,hence the solutions are $ (x,y,z) \equal{} (0,0,0)(1,1,1)$[/quote]
Dear Anavel_Gato but if we check case X=even=2a Y=even=2b for some $ a,b\in N$
we will see that
 $ 7\cdot ( \minus{} 1)^{2a} \plus{} 1\equiv 3\cdot 9^{2b}\pmod{25}$
$ 3^{4b\plus{}1}\minus{}8\equiv 0 mod{25}$
but if we choose$ b\equal{}5k\plus{}2$ for some $ k\in N$ we will see that X,Y exist?

[quote="Rose_joker"]
Dear Anavel_Gato but if we check case X=even=2a Y=even=2b for some $ a,b\in N$
we will see that
 $ 7\cdot ( \minus{} 1)^{2a} \plus{} 1\equiv 3\cdot 9^{2b}\pmod{25}$
$ 3^{4b \plus{} 1} \minus{} 8\equiv 0 mod{25}$
but if we choose$ b \equal{} 5k \plus{} 2$ for some $ k\in N$ we will see that X,Y exist?[/quote]

You are right [b]Rose_joker[/b],its my bad :blush:
but I have no idea how to do from here.

[quote="apollo"]Find all non-negative integers $ x,y,$ and $ z$ satisfying that $ 7^x \plus{} 1 \equal{} 3^y \plus{} 5^z$[/quote]

If any of $ {x,y,z}\leq1$ we´ll have the solutions (0,0,0,) and (1,1,1).Suppose $ {x,y,z}\geq{2}$.
Obviously z is odd.Considering congruence $ mod 4$ we see that we must have $ {x}\equiv{y}(mod4)$.
We will divide the problem in 2 cases:
$ x$ and $ y$ even and $ x$ and $ y$ odd.
If they are even:
let $ x \equal{} 2a$ and $ y \equal{} 2b$ so we have $ (7^a \minus{} 3^b)(7^a \plus{} 3^b) \equal{} 4(5^{2a} \plus{} ... \plus{} 1)$.but $ (5^{2a} \plus{} ... \plus{} 1)$ is
odd, and $ 8$ divides $ LHS$,absurd.
Still missing the case $ x,y$ odd, but i´m working on it. :wink:


[quote="catcurio"]Let $A$, $M$, and $C$ be nonnegative integers such that $A+M+C=10$. What is the maximum value of $A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A$?

$\text{(A)}\ 49 \qquad\text{(B)}\ 59 \qquad\text{(C)}\ 69 \qquad\text{(D)}\ 79\qquad\text{(E)}\ 89$[/quote]

[hide]Since everything is multiplied with everything else, the max occurs when the numbers are closest together: (3,3,4). $A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A=(C)69$[/hide]

$(A + 1)(M + 1)(C + 1) = A + M + C + A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A + 1 = 11 + A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A$
Therefore, $(A + 1)(M + 1)(C + 1)$ must be maximum and ($A, M, C$) = ($4, 3, 3$).

haha funny answer :P
In general, max usually occurs when $A=M=C$ so use $4,3,3$.

[quote=Himandude101]When I was doing this I found that (4,4,2) resulted in 72 because 4x4x2+4x4+4x2+4x4=72. I chose 69 in the end because it was closest to 72.[/quote]

[hide](4*4*2)+(4*4)+(4*2)+([b]4*2[/b])[/hide]

[hide=Solution]We want to make $A,M,C$ as close as possible via AM-GM, so $A=3, M=3, C=4$ and the answer is $36+9+12+12 \implies 69 \implies \boxed{\text{(C)}}$.[/hide]

[hide=Solution]Since we want to maximize the sum of products, we want to make the product $A\cdot M\cdot C$ as high as possible, which happens when $A, M,$ and $C$ have values of $3, 3, 4$ in any order. Plugging these values into the original expression gives us:
$$3\cdot 3\cdot 4+4\cdot 3+3\cdot 4+3\cdot 3=36+12+12+9=69$$
Therefore, the correct answer is $\boxed{\text{(C)}\ 69 }$[/hide]


summer program
PROMYS
number theory
relatively prime
superior algebra
superior algebra unsolved

Let $\alpha = a+b\sqrt{10}$.  Prove that if $\alpha | 1$ then $a^{2}-10b^{2}= \pm 1$

So far I have the following:

Let $\beta = c+d\sqrt{10}$ such that $\alpha\beta=1$

Then:

$(ac+10bd)+(ad+bc)\sqrt{10}= 1$

So:

$ac+10bd = 1$
$ad+bc = 0$

Multiplying through the first by b and using the second equation to do a quick replace gives:

$a^{2}-10b^{2}= \frac{-b}{d}$

And similarly, multiplying through by a gives:

$a^{2}-10b^{2}= \frac{a}{c}$

So I'm getting close.  Or at least I think I am.  But I'm not sure what to do from here.

Let $N(a+b \sqrt{10}) = a^{2}-10b^{2}$.  You might find it helpful to prove the following lemmas:

[b]Lemma:[/b]  $N( \alpha \beta) = N(\alpha) N( \beta)$

[b]Lemma:[/b]  $\alpha | \beta \implies N(\alpha) | N(\beta)$

From $ac+10bd=1$, we have that $a$ and $b$ are relatively prime. Applied to the second equation, $a|c$ and $b|d$. Similarly, $c$ and $d$ are relatively prime, so $c|a$ and $b|d$. This gives $a=\pm c$ and $b=\pm d$. With a bit of algebra, we must have $c=a,d=-b$.

t0rajir0u, your second lemma is my next problem.  And I see how it can be directly used, but I have no idea how to prove it.

Oy... It can't be that easy.  Got it.

a | b, so b = ac for some c.  N(b) = N(ac), and so N(b) = N(a)N(c).

$\alpha | \beta \Leftrightarrow \exists \gamma \in \mathbb{Z}[ \sqrt{10}] \, : \, \beta = \alpha \gamma$ (PROMYS reflex!  :P )

Hence by the first lemma

$N (\beta) = N (\alpha) N( \gamma)$

Since $N(\gamma)$ is a nonzero integer it follows that $N(\alpha) | N(\beta)$.

Thanks for the help.  I edited my above post but did so after you posted.


[quote="tobeno_1"]let

   $ 10^{10^{10^{10}}}\equiv r(\mod{77})$

Find $ r$ and please tell me how :blush:[/quote]
A solution, which does not require you to now that $ 10^6 \equiv 1 \bmod 77$

$ 77 \equal{} 7 \cdot 11$, and it's easy to see that $ 10^{10^{10^{10}}} \equiv (\minus{}1)^{10^{10^{10}}} \equiv 1 \bmod{11}$
So we just have to find $ 10^{10^{10^{10}}} \equiv 3^{10^{10^{10}}} \bmod 7$
Generally $ a^{p\minus{}1} \equiv 1 \bmod p \forall (a,p) \equal{} 1$ so we will find $ 10^{10^{10}} \equiv 4^{10^{10}} \bmod 6$. Since $ 6 \equal{} 2 \cdot 3$ we easilly see that $ 4^{10^{10}} \equiv 0 \bmod 2$ and $ 4^{10^{10}} \equiv 1^{10^{10}} \equiv 1 \bmod 3$ so $ 10^{10^{10}} \equiv 4^{10^{10}} \equiv 4 \bmod 6$
Thus we know that $ 10^{10^{10^{10}}} \equiv 3^{10^{10^{10}}} \equiv 3^4 \equal{} 81 \equiv 4 \bmod 7$
and $ 10^{10^{10^{10}}} \equiv 1 \bmod 11$ which yields $ r \equiv 67 \bmod 77$


Let $ X$ be compact space and let $ f$ be closed(f(closed set) = closed set) continous function $ f: X\rightarrow Y$ (f is onto $ Y$). Prove that Y is Hausdorff's space

This can't be true, e.g. if $ Y$ is the disjoint union of $ X$ and a non-hausdorff space.

Can you explain it? I don't see any point in your post ...  :oops:

In formulas: Take $ Y \equal{} X \dot\cup Z$ where $ Z$ is not hausdorff and $ X$ is compact. Then the canonical embedding $ X \to Y$ is closed,but $ Y$ is not hausdorff.

[quote="zibi"]but f is continous...[/quote]
...so is the inclusion map in ZetaX's example. (My guess is that the $ f$ in the original problem should be assumed to be onto).

Yes, i forget to point it precisely, $ f$ is onto $ Y$. $ f$ in zetax post's is not onto... it's still open problem for me  :roll:

Then it depends on your definition of compact. If it includes hausdorff, then the statement is true.

My definition of compact includes fact that space is hausdorff. Can somebody solve it ?, because this problem starts to annoy me...  :wink:

here are some ideas about my problem:
We have to prove that for any to points $ x$,$ y \in Y$ there exists open sets $ U_{x}$ and $ U_{y}$ that have empty intersection.  Sets $ f^{ \minus{} 1}(x)$ and $ f^{ \minus{} 1}(y)$ are closed because $ X$ is hausdorff. Because of compactness of $ X$ sets $ f^{ \minus{} 1}(x)$ and $ f^{ \minus{} 1}(y)$
are compact. In hausdorff space for every compact subsets that don't intersect, we can choose open sets that contain those compact subsets and don't intersect. But how do we know that image of those open sets(under (f)) do not intersect ?


Let $ f(x)$ is differentiable on $ x=0$ and $ f'(0) \neq 0$. If for any $ %Error. "arbitrary" is a bad command.
x \neq 0$ there is exactly one $ \epsilon$ such that $ f(x) + f(0)=2f(\epsilon)$ then determine $ lim_{x \rightarrow 0} \frac{\epsilon}{x}$

The limit is 1/2.

Subtract 2f(0) from both sides of your equation and divide by x. Since the function is continuous at 0, e tends to 0 as x tends to 0.The left hand side approaches f'(0) as x approaches 0, while the right hand side approaches 2f'(0)e/x. Hence the limit is 1/2.


Source: 2005 AMC-10B #21

Forty slips are placed into a hat, each bearing a number $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, or $10$, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let $p$ be the probability that all four slips bear the same number. Let $q$ be the probability that two of the slips bear a number $a$ and the other two bear a number $b\ne a$. What is the value of $\frac{q}{p}$?

$\text{(A)}\ 162 \qquad\text{(B)}\ 180 \qquad\text{(C)}\ 324 \qquad\text{(D)}\ 360 \qquad\text{(E)}\ 720$

[quote="catcurio"]Forty slips are placed into a hat, each bearing a number $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, or $10$, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let $p$ be the probability that all four slips bear the same number. Let $q$ be the probability that two of the slips bear a number $a$ and the other two bear a number $b\ne a$. What is the value of $\frac{q}{p}$?

$\text{(A)}\ 162 \qquad\text{(B)}\ 180 \qquad\text{(C)}\ 324 \qquad\text{(D)}\ 360 \qquad\text{(E)}\ 720$[/quote][hide="Got it wrong, can someone tell me why?"]For all slips to bear the same number, the probability is $\frac{1}{39}\times\frac{1}{38}\times\frac{1}{37}$. For two to have the same number, there are a few different possibilities. Either the second, third, or fourth is the same as the first. If the second is the same as the first, then the probability is $\frac{1}{39}\times 1 \times \frac{1}{37}$. If the third has the same as the first, then the probability is $1 \times \frac{1}{38} \times \frac{1}{37}$. Lastly, if the fourth has the same as the first, $1 \times \frac{1}{38} \times \frac{1}{37}$ which is the same as the first = third case. Since these are equally likely to happen (I think, I could be very wrong here) then the probability total is the average of their sum, $\frac{1}{3}(\frac{1}{39\times37}+\frac{2}{37\times38})$. Putting this over the first gives $\frac{\frac{1}{3}(\frac{1}{39\times37}+\frac{2}{37\times38})}{\frac{1}{39}\times\frac{1}{38}\times\frac{1}{37}}$ which simplifies to $\frac{38+2\times39}{3}=\frac{116}{3}$ which is completely incorrect.[/hide]

[hide]There are a total of $\binom{40}{4}$ ways to pick four slips out of 40.

There are 10 ways to pick four which are the same number, so $p=\frac{10}{\binom{40}{4}}$.

In order to pick two of one number and two of another, there are $\binom{10}{2}$ ways to pick the two numbers. Then for each number, there are $\binom{4}{2}$ ways to pick two slips with that number. So there are $\binom{10}{2}\binom{4}{2}^2$ ways to pick two of one number and two of another. $q=\frac{\binom{10}{2}\binom{4}{2}^2}{\binom{40}{4}}$.

$\frac qp=\frac{\binom{10}{2}\binom{4}{2}^2}{10}=\frac{9*6^2}{2}=162\Rightarrow \boxed{A}$.[/hide]

Ok, I see my problem. I forgot to include the fact that there are 4 of each slip. Doh!  :roll:


Find all natural numbers n and m for which:

((n^3) + 1)/((m*n)-1) is natural number

 :blush:


number theory
greatest common divisor
number theory proposed

if $x,y \in N$ such that $3x^2+x=4y^2+y$  prove that
$x-y$ is perfect square.

Also posted here:

http://www.mathlinks.ro/Forum/viewtopic.php?t=30699


An ant is walking on a hexagon. At each step, he moves from the vertex on which he is standing to an adjacent vertex. After  steps, what is the probability that he is standing on the same vertex on which he began?

[quote="lisaz94"]An ant is walking on a hexagon. At each step, he moves from the vertex on which he is standing to an adjacent vertex. [b]After  steps[/b], what is the probability that he is standing on the same vertex on which he began?[/quote]

you need to tell us how many steps

but i think i remember doing this problem
if i remembered correctly, it was an odd number of steps

for the ant to get back to its starting point, it would have to go all the way around, or part of the way around and go back

for the second option, it is obviously it will travel an even number of steps (make sure you see why)

for the first option, we will take into consideration that the ant is walking on a hexagon
since it is a hexagon, it will take 6 moves to return to its starting point
this number of steps is also an even number

therefore, it is impossible for the ant to return to its starting position after an odd number of steps (if it is an odd number of steps)


inequalities
email
inequalities theorems

[quote="zweig"]Hi,

I'm interested in this book that's why I want to know how cost shipping for France in order to evaluate the total price. If someone from Gil could answer me, please.

Thanks,

Best regards.

(I didn't know where I could post this topic, so I posted here).[/quote]

Could you also give details about the price in the Channel Isles :)

Thanks

See http://www.gil.ro . The price is there 23 Euro

Virgil Nicula > Yes, I saw it. But I just want to know the shipping for France :).

[color=darkblue][b]Shiping[/b] ?! Yes, you can speak about it directly with the editor chief
[b]Mircea Lascu[/b] through the his Email [b]gil1993@zalau.astral.ro[/b] .[/color][color=darkred]
Happy new year ![/color]


For each positive real c, find a positive integer N such that for all naturals n>N-1, we have
   (1+c)^n>n.

Bomb

hello, it is $c>\sqrt[n]{n}-1$
Sonnhard.


Is this true: If every contraction on a metric space (X,d) has a fixed point, then (X,d) is complete metric space?

[url=http://www.ams.org/proc/2009-137-09/S0002-9939-09-09904-3/S0002-9939-09-09904-3.pdf]This[/url] paper seems to include a fair amount on converses to Banach's fixed point theorem.  Perhaps you should look at some of the references.


[quote="ProInCounter"]Let a,b,c  be positive real numbers,Prove that:
(a^3+b^3)^(1/3)+(b^3+c^3)^(1/3)+(c^3+a^3)^(1/3) <= (2^(1/3))(a+b+c) + ((a+b+c)^3)/27 :?:[/quote]
LaTeX'ed:

$ \sqrt [3]{2}(a \plus{} b \plus{} c) \plus{} \frac {(a \plus{} b \plus{} c)^3}{27} \ge \sqrt [3]{a^3 \plus{} b^3} \plus{} \sqrt [3]{b^3 \plus{} c^3} \plus{} \sqrt [3]{c^3 \plus{} a^3}$

This is false. Let $ a \equal{} 1k, b\equal{}2k, c\equal{}3k$, then $ \sqrt [3]{a^3 \plus{} b^3} \plus{} \sqrt [3]{b^3 \plus{} c^3} \plus{} \sqrt [3]{c^3 \plus{} a^3} > \sqrt [3]{2}(a \plus{} b \plus{} c)$. If you let $ k \to 0$ then  $ \frac {(a \plus{} b \plus{} c)^3}{27}$ will be much smaller than the others.


logarithms
algebra
polynomial
number theory
least common multiple
geometry
3D geometry

Try to do them without use of calculator.

1. At the beginning of the year, Maria invests 12,345 dollars in shares of Volatile stock. The stock's value increases by 25% the first month, decreases by 20% the second month, increases by 25% the third month, decrease by 20% the fouth month, etc. It continues montly to alternate 25% increases with 20% decreases through 12 months. At the end of 12 months, what is the value of Maria's investment?

2. To raise money, the Liberty High School debate team washes cars. Nine students can wash 30 cars in 1.2 hours. The nine students are equally efficient. How many cars one student wash in one hour? (P.S. for this question, I'm not sure but I got the answer in fraction. So, please post yours to see who's right :-D)

3. Compute:

$\sqrt {999^4+4 \cdot 999^3 + 6 \cdot 999^2 + 4 \cdot 999 + 1}$.

4. A patient, but slightly crazed, spider spins a web that begins at point (1,0) and then proceeds counterclockwise to (0, 1/2), (-1/4,0), (0, -1/8), and so forth forever, always ending a strand at the boundary between quadrants and each time cutting the distance to the origin by half. Assuming that the spider can finish this infinite process, what would be the sum of the lengths of all these strands?

5. Place these three numbers in order of increasing magnitude: $\frac {9}{2}, \sqrt {20}, 88^{\frac {1}{3}}$

6. Two hexagons with length one has a common side (the side that both hexagons share). Let's call the two points A and B, such that A and B are in different hexagons and that they're far away. What is the distance from A to B? (Direct distance. Not by crossing the sides of hexagon)

[code]
*/ \/ \
 |   |  |
  \ /\ /
         *[/code]

(P.S. This is a bad picture but you get the idea. *'s indicate point A and B)

7. A vehicle with wheels of radius 15 inches rolls forward at the rate of 10 feet per second. How rapidly are the wheels revolving on their axes (in radius per seconds)?

--Some of followings are posted before.---
8. In a circle of radius 2, what is the length of a chord that is the perpendicular bisector of a radius?

9. If $r,s,t$ are distinct positive integers for which $r = s^2 = t^3$, what is the smallest value of $r$?

10. Suppose that the positive integer $K$ has the property that:

$\log_{10} (\log_{10} (\log_{10} K)) = 0$. How many digits does $K$ have?

11. When the polynomial $p(x)$ is divided by $x^2-4$, the remainder is of the form $ax+b$. Find $a$ and $b$ if we know that $p(2) = 7$ and $p(-2) = 15$.

12. Three red dots and three blue dots are placed randomly in the plane, the only restriction being that no three of the six dots can be collinear. Nine line segments are then drawn, one from each red dot to each blue dot. What is the fewest possible number of intersections formed by these line segments, not counting intersections at the six dots?

13. We are given that $r(z)$ is a 2nd degree polynomial with the following properties:

$r(z) > 0$, if $z<3$ or $z>4$
$r(z) < 0$, if $3 < z < 4$,
$r(2) = 6$.

Find $r(0)$.

Enjoy! :)

1.
[hide]The stock's value increases by 25%, so 12,345 x 5/4 after the first month.
Then, the stock decreases by 20%, so 12,345 x 5/4 x 4/5 after the second month.
So on.. until after 12 months she still has 12,345 dollars. Unwise investment. [/hide]

2. I also got a fraction.
[hide]30 cars in 1.2 hours... 30/1.2 = 25 cars in 1 hour.
25 cars/9 people = 25/9 cars per person in an hour.[/hide]

3. [hide]That equals sqrt (1 + 999)^4 = (1 + 999)^2 = 1000^2 = 1,000,000.[/hide]

10.

[hide] Since log10(log10(log10(k))) = 0, log10(log10(k)) necessarily equals 1. Therefore, log10(k) necessarily equals 10. Finally,
k necessarily equals 10^10, which has 10 digits. [/hide]

11.

[hide] First of all, x^2 - 4 = (x + 2)(x - 2). The remainder theorem states that the remainder when a polynomial p(x) is divided by
x - r is equal to p(r). Therefore, the remainder when p(x) is divided by x - 2 is p(2) = 7, and the remainder when p(x) is divided by
x + 2 is p(-2) = 15. Don't really know where to go from here, but it can't be that far  ;)  [/hide]

[hide]
2. I am really bad at theses types of questions so if I get it wrong, don't be suprised.
The students work at a rate of $30$ cars in $72$ minutes.
$\displaystyle 9x=\frac {30}{72}=\frac 5{12} \Rightarrow x=\frac {\frac 5{12}}9$
So each can clean a car at a rate of $\frac 5{108}$ cars per minute.

5. Simplify into $\displaystyle 4.5, 2 \sqrt5, 2 \sqrt[3]11$
Arrange into $\displaystyle 88^\frac 13, \sqrt 20, \frac92$

9. Since $\displaystyle r^1=s^2=t^3$, we must find a common multiple with thoses three numbers and find the minimum value of $\displaystyle u^{lcm(2,3)}=u^6$ where $\displaystyle u\not=1$ since $\displaystyle s\not=r\not=t\not=1$. So the minimum value of $\displaystyle u$ is 2. Since $\displaystyle 2^6 = 64$ then $\displaystyle r=64$[/hide]

 :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:

To hiitsme,

Small mistake. $1.2 \text {hours} \neq 80 \text {min}$.

Thanks Silverfalcon :D Always love these problems since I'm not good enough for Intermediate yet, gotta practice :P

[hide="My Solutions"]
[quote="Silverfalcon"]1. At the beginning of the year, Maria invests 12,345 dollars in shares of Volatile stock. The stock's value increases by 25% the first month, decreases by 20% the second month, increases by 25% the third month, decrease by 20% the fouth month, etc. It continues montly to alternate 25% increases with 20% decreases through 12 months. At the end of 12 months, what is the value of Maria's investment?[/quote]
Answer: $12,345$

This is a very interesting problem. We can multiply out all 12 months to find that we have:

$(0.8)^6(1.25)^6(12345)=\frac{4^6}{5^6}\cdot\frac{5^6}{4^6}\cdot12345$ and the denominators and numerators cancel out to leave the final answer as $12,345$

[quote="Silverfalcon"]2. To raise money, the Liberty High School debate team washes cars. Nine students can wash 30 cars in 1.2 hours. The nine students are equally efficient. How many cars one student wash in one hour? (P.S. for this question, I'm not sure but I got the answer in fraction. So, please post yours to see who's right Very Happy[/quote]
Answer: $\frac{25}{9}\text{ cars/hour}$

We can begin with a proportion:

$\frac{30 \text{ cars}}{\frac{6}{5} \text{ hours}}=\frac{x}{1 \text{ hour}}$

Cross multiplication and rearranging gives us $x=25$ or 9 students can wash 25 cars in 1 hour. Therefore, each student can wash $\frac{25}{9} \text{ cars}$ each hour.[/quote]

[quote="Silverfalcon"]3. Compute:

$\sqrt {999^4+4 \cdot 999^3 + 6 \cdot 999^2 + 4 \cdot 999 + 1}$.[/quote]
Answer: $10^6$ (wow, what was I doing before? lol)

I had a really good answer for this question, and number 4 and 5, but somehow when I was typing it, it got erased. So I'll rush through this here because I hate doing repetetive work.

Notice that $(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$ Since we're looking for the square root of that, it's $(a+b)^2$.

Substituting $a=999$ and $b=1$, we get $1000^2=1,000,000=10^6$

[quote="Silverfalcon"]4. A patient, but slightly crazed, spider spins a web that begins at point (1,0) and then proceeds counterclockwise to (0, 1/2), (-1/4,0), (0, -1/8), and so forth forever, always ending a strand at the boundary between quadrants and each time cutting the distance to the origin by half. Assuming that the spider can finish this infinite process, what would be the sum of the lengths of all these strands?[/quote]
Answer: $\frac{\sqrt{5}}{2}$

Draw it out. We find that all the triangles formed are similar because of SAS Similarity (my first solution was a bit complete, but if you ask me to elaborate, I will). Therefore, hypotenuses are also proportional, which turns out to be $1:2$ going from strand to strand. Therefore, letting the hypotenuse of the triangle formed by the origin and the first two dots [(0,1/2) and (-1/4,0)], we have:

$x+\frac{1}{2}x+\frac{1}{4}x+\frac{1}{8}x+...=x(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8})$

The latter in parenthesis is equivalent to 2, so the answer is $2x$. From Pythagorean Theorum, we find that $x=\frac{\sqrt{5}}{4}$ and multiplied by 2 gives our answer as $\frac{\sqrt{5}}{2}$.

[quote="Silverfalcon"]5. Place these three numbers in order of increasing magnitude: $\frac {9}{2}, \sqrt {20}, 88^{\frac {1}{3}}$[/quote]
Answer: $88^{\frac{1}{3}}, \frac{9}{2}, \sqrt{20}$

Simplify $\sqrt{20}$ and $88^{\frac{1}{3}}$ to $2\sqrt{5}$ and $2\sqrt[3]{11}$ respectively. Divide everything by 2 and we are searching for:
$\frac{9}{4}\text{ ? }\sqrt{5}\text{ ? }\sqrt[3]{11}$
Raising everything to the sixth power (LCM of 1, 2, and 3), we have:
$\frac{9^6}{4^6}\text{ ? }5^3\text{ ? }11^2$
We can clearly see that $5^3>11^2$ so $\sqrt{5}>\sqrt[3]{11}$. Now, we first can compare $5^3$ with $\frac{9^6}{4^6}$. Taking the cube root of each side, we get:
$\frac{81}{16}\text{ ? }5$.
We see that $16\cdot5=90$ so $\frac{81}{16}<5$ and therefore, $\frac{9^6}{4^6}<5^3$.

Going over what we have so far:
$5^3>\frac{9^6}{4^6}\text{ ? }11^2$

We need to compare $\frac{9^6}{4^6}$ with $11^2$. Taking the square root of both sides leaves us with $\frac{9^3}{4^3}\text{ ? }11$. This is feasible by hand, so we do so and get $\frac{729}{64}\text{ ? }11$. We do $11\cdot64=704$ to realize $\frac{9^6}{4^6}>11^2$. Therefore, we arrive at our conclusion that $\sqrt{20}>\frac{9}{2}>88^{\frac{1}{3}}$. We need in increasing order, so order it backwards. I skipped a bit because it got boring, but you should be able to follow.

[quote="Silverfalcon"]6. Two hexagons with length one has a common side (the side that both hexagons share). Let's call the two points A and B, such that A and B are in different hexagons and that they're far away. What is the distance from A to B? (Direct distance. Not by crossing the sides of hexagon)[/quote]
Answer: $\sqrt{13}$

Assuming that A and B are on the vertexes of the hexagon and are as far away as possible, the attached image shows a drawing of it.

From the image, we see we can find the distance AB by using the Pythagorean Theorum. We know 1 side is 1 already. We need to find AC, the other side.

If we draw segment DE perpendicular to AC, since each angle of a regular hexagon is $120^\circ$, we have two separate 30-60-90 triangles (probably should've drawn it in diagram). That means $\frac{1}{4}$ of AC is the long side of the triangle whose hypotenuse is 1 and therefore, $AC=2\sqrt{3}$. Using Pythagorean, we find AB to be $\sqrt{13}$.

[quote="Silverfalcon"]7. A vehicle with wheels of radius 15 inches rolls forward at the rate of 10 feet per second. How rapidly are the wheels revolving on their axes (in radius per seconds)?[/quote]
Answer: $\frac{4}{\pi}$

I think you mean how many times the wheel turns in a second (if it's radius/sec, it just means how many 15 inches the wheel moves per second and isn't converting linear movement to angular movement and isn't much of a question).

The wheel has $30\pi$ circumference. Convert 10 feet to 120 inches. Every time the wheel rotates once, the distance it travels is $30\pi$. Each second, it travels 120. Therefore, the wheel rotates $\frac{4}{\pi}$ times each second or a bit over once.

[quote="Silverfalcon"]8. In a circle of radius 2, what is the length of a chord that is the perpendicular bisector of a radius?[/quote]
Answer: $2\sqrt{3}$

Since a radius that is perpendicular to a chord bisects it, we can form a right triangle with half the chord, half the radius that it bisects, and the radius. Recognizing a 30-60-90, we see half the chord is $\sqrt{3}$ meaning the chord is $2\sqrt{3}$.

[quote="Silverfalcon"]9. If $r,s,t$ are distinct positive integers for which $r = s^2 = t^3$, what is the smallest value of r?[/quote]
Answer: $64$

This means $r$ is the smallest integer that is both a cube and a square. Since all three numbers are distinct and positive, they can't be 0 or 1, the obvious answers. By guess and check, it's easy to see $s=8$ and $t=4$ and that $r=64$.

[quote="Silverfalcon"]10. Suppose that the positive integer K has the property that:

\log_{10} (\log_{10} (\log_{10} K)) = 0. How many digits does K have?[/quote]
Answer: $11$

We find that $\log_{10} (\log_{10} K) = 1$. Then, that means $\log_{10} K=10$. Finally, that means that $K=10^10$ and that means $K$ has 11 digits.

[quote="Silverfalcon"]11. When the polynomial $p(x)$ is divided by $x^2-4$, the remainder is of the form $ax+b$. Find $a$ and $b$ if we know that $p(2) = 7$ and $p(-2) = 15$.[/quote]
Answer: $\text{Er is this polynomial possible?}$

Does the polynomial $p(x)$ really exist? We know that $\frac{p(x)}{(x+2)(x-2)}$. We shouldn't be able to have $p(2)$ or $p(-2)$ since there would be division by zero...

[quote="Silverfalcon"]12. Three red dots and three blue dots are placed randomly in the plane, the only restriction being that no three of the six dots can be collinear. Nine line segments are then drawn, one from each red dot to each blue dot. What is the fewest possible number of intersections formed by these line segments, not counting intersections at the six dots?[/quote]
Answer: $0$

It is possible to have no intersections at all. If we draw six dots so that the conditions are fulfilled, the first dot can have 5 lines drawn from it with no intersections. For the remaining 4 segments, connect adjacent dots together so that no lines intersect.

Basically, this would be equivalent to a hexagon with all of its diagonals. It may even be easier thinking of a regular hexagon.

[quote="Silverfalcon"]13. We are given that $r(z)$ is a 2nd degree polynomial with the following properties:

$r(z) > 0$, if $z<3$ or $z>4$
$r(z) < 0$, if $3 < z < 4$,
$r(2) = 6$.

Find $r(0)$.[/quote]
Answer: $36$

Our aim here is to find $r(x)$. We know that $r(x)$ is an unique quadrilateral. Give the information, we also know that at $x=3$ and $x=4$, $y=0$. The x coordinate of the vertex is the average between these 3 and 4 and is $\frac{7}{2}$.

Because the standard form of the quadratic is $y=a(x-h)^2+k$, we can find $r(x)$. We know $h=\frac{7}{2}$ already. We also know when $x=3$, $r(x)=0$ and when $x=2$, $r(x)=6$. We aim to find a and k now.

We have two equations after substituting in $x=3$ and $x=2$:

$a+4k=0$
$9a+4k=24$

Solving, we find that $(a,k)=(3,\frac{-3}{4})$. Therefore, $r(x)=3(x-\frac{7}{2})^2-\frac{3}{4}$. Substituting $x=0$ in there, we find $r(0)=36$
[/hide]

Nice job.

I liked about your explanation on spider one. I thought it was talking about adding:

$1 + \frac {1}{2} + \frac {-1}{4} + \frac {-1}{8}...$

Stupid me :blush:

Yes, your answer is right. $\frac {\sqrt 5}{4}$ is the first term, so:

$\frac {\frac {\sqrt {5}}{4}}{1 - 1/2} \Rightarrow \frac {\sqrt 5}{2}$

Which came from infinite series sum.

I think you need to check your answer about square root one about 999 and 1. That doesn't seem correct answer.

7. you mean radians/sec, right? (it says radius)
[hide]Circumference of wheel = 2pi r = 30 pi inches = 2.5 pi feet/revolution.
Dividing 10 ft/s by the circumference gives us 4/pi revolutions per second.
Converting that into radians = 2pi radians/revolution x 4/pi revolutions/second = [b]8 [/b]rad/s.[/hide]

8. [hide]Drawing it out, we get: length of chord = 2 x sqrt (2^2 - 1) = [b]2 sqrt 3[/b].[/hide]

9. [hide]r = n^6 for some positive integer n.
Since r, s, and t are distinct integers, n can't be 1.. the smallest value of r = 2^6 = [b]64[/b].[/hide]

Thanks, Silverfalcon, I fixed it. Dunno [i]what[/i] I was thinking at the time.

If for #7, you're looking for radians/sec, then it should be what sky9073 has.

By the way, is #11 possible?

P.S: I also really liked the spider one. It's a very fun question that requires some thought :)

[quote="curl"]10.

[hide]  Finally,
k necessarily equals 10^10, which has 10 digits. [/hide]

[/quote]

10^10 has 11 digits

O.o! I missed that also! Corrected mine. Thanks :)

To Zyloch, why should 11 be impossible? If you are familiar with polynomials, this problem is straightforward

The thing is that I'm not familiar with polynomials or anything lol. For all the problems, I'm just moving stuff around and experimenting. I just needed confirmation that it's possible--I'll go try to figure it out  now :blush:

Let me give you a little hint, maybe this helps out your experimentation

[hide=""Hint""]$p(x)=(x^2-4)\cdot Q(x)+(ax+b)$, for some polynomial $Q(x)$[/hide]

So, is this helpful enough?

Ah. I see now.

[hide="Hidden"]You get two linear equations and from there it's easy to find that $a=-2$ and $b=11$.[/hide]

Yeah. The problem was that I was thinking something along the lines of that, but then I [i]convinced[/i] myself that I couldn't do it that way because for some reason, after I thought of it, I multiplied both sides by $x^2-4$ and figured that $(ax+b)$ remainder needed to be multiplied by $x^2-4$ also and pushed the original idea out of my head.

Oh well, it's all experience I suppose. Thanks  :lol:

For the ones who might not get it, would you mine showing your steps?

Silverfalcon :)

Sure :D

[hide="Polynomial Q"]
[quote="3X.lich"]$p(x)=(x^2-4)\cdot Q(x)+(ax+b)$, for some polynomial $Q(x)$[/quote]

Using what 3X.lich says, when $x=2$ or $x=-2$, $x^2-4=0$ because of difference of two squares, so $p(x)=ax+b$. Since we know $p(x)$ at 2 and -2, we substitute 2 and -2 for x to get two linear equations:

$p(2)=2a+b=7$
$p(-2)=-2a+b=15$

Solve and we get $(a,b)=(-2,11)$
[/hide]


There's that place for sticky messages up at the top, starts out pretty small. It is a grevious annoyance to me that the sticky-frame can only be expanded downawrd to about 2/3rds of my screen. When we're doing geometry, I can see either the problem and half a diagram, or the whole diagram; I'd much rather give up some more of the chat for the whole diagram. Is there a way to remove the frame-size limit please? :)

Hi, thanks for your suggestion. We have a request in to the creators of the classroom software for this and several other things.

I think the message character limit is too small -- I have had to split up a proof into two messages because of this.
Otherwise, I think I am fine with the new classroom :)

[quote="AIME15"]Also, when is the new classroom coming out for regular classroom use?[/quote]We used it started today for all our classes.

Yeah, I noticed in my geometry class today :P

they said that there's still a couple bugs, but it's good enough to use.


logarithms
calculus
integration
function
limit
real analysis
algorithm

:D today, a student asked me what could possibly be the slowest
diverging series ever! whilst the question seemed simple at
the time, it really made me think for a while before i could
give (or suggest) an answer.

first off, we know, for a series $ \mathbb S_{n}=\sum_{i=1}^{n}\;a_{i}$ to diverge, the sequence
of it's partial sums, i.e. $ \left\{ \;\mathbb S_{1},\;\mathbb S_{2},\;\mathbb S_{3},\;\mathbb S_{4},\;\mathbb S_{5},\;\mathbb S_{6},\;\cdots\;\right\}$
has to diverge.

here, of-course, i am going to overlook series like:
$ 1+1+1+1+1+1+\cdots+$
or
$ 1-1+1-1+1-1+\cdots+\;-$

instead, i am primarily interested in a sequence whose
terms [b]monotonically [/b] [i]decrease[/i] to zero, but the series [i]still [/i]diverges
(as does the harmonic series, but i know i can do better than that).
so, i came up with 2 flavors of series that i know will diverge
to infinity, but in this case, it will literally take them [b]forever[/b]
to do so (the divergence is just so, so, so, so, sooooooo very slow... !).

now which one of these 2 series will actually diverge [i]slower[/i],
that i do not know yet (also i don't know if there can be
an even slower series out there...)

here are the 2 series i came up with...  :|

$ \textbf{(i)}\;\;\;$ thinking along the lines that all of
$ \sum_{k=1}^\infty\;\frac{1}{k}$,
$ \sum_{k=2}^\infty\;\frac{1}{k\;\ln k}$,
$ \sum_{k=3}^\infty\;\frac{1}{k\;\ln k\;(\ln \ln k)}$,
$ \sum_{k=4}^\infty\;\frac{1}{k\;\ln k\;(\ln \ln k)\;(\ln \ln \ln k)}$,
$ \sum_{k=5}^\infty\;\frac{1}{k\;\ln k\;(\ln \ln k)\;(\ln \ln \ln k)\;(\ln \ln \ln \ln k)}$,
$ \sum_{k=6}^\infty\;\frac{1}{k\;\ln k\;(\ln \ln k)\;(\ln \ln \ln k)\;(\ln \ln \ln \ln k)\;(\ln \ln \ln \ln \ln k)}$,

etc... diverge, i then proposed
$ \boxed{\Omega_{1}{(s)}\;=\;\sum_{k=\aleph_{0}}^{\aleph_{1}}\;\frac{1}{\underbrace{k\;\ln k\;(\ln \ln k)\;(\ln \ln \ln k)\;\cdots\;(\underbrace{\ln \ln \ln \ln\cdots\ln}_{(\aleph_{0}-1)\;\;\;of\;\;\;them}k)}_{\aleph_{0}\;\;\;elements \;\;\;included \;\;\;within\;\;\; this\;\;\;product}}}$

will also diverge, but extremely, extremely [i]slowly[/i] (i have introduced
the $ \aleph_{0}\;\;\;and\;\;\;\aleph_{1}$ [b]infinite [/b]cardinals here). i believe the divergence
of this series can be proved by the aid of the integral test, but i don't
want to be the one showing it here... LOL! :P

$ \textbf{(ii)}\;\;\;$ and then, thinking along the lines of the riemann zeta function, i.e.
$ \zeta{(s)}\;=\;\sum_{k=1}^\infty\;\frac{1}{k^{s}}\;\;\;\quad$ (which diverges for $ 0<s\leq 1$), we have
all of the following divergent series:

$ \sum_{k=1}^\infty\;\frac{1}{k}$,
$ \sum_{k=1}^\infty\;\frac{1}{k^{0.1}}$,
$ \sum_{k=1}^\infty\;\frac{1}{k^{0.01}}$,
$ \sum_{k=1}^\infty\;\frac{1}{k^{0.001}}$,
$ \sum_{k=1}^\infty\;\frac{1}{k^{0.0001}}$,
$ \sum_{k=1}^\infty\;\frac{1}{k^{0.00001}}$,

etc. so, i proposed taking the limit $ s\to 0^{+}$, i.e. $ \lim_{s\to 0^{+}}\;\sum_{k=1}^\infty\;\frac{1}{k^{s}}$.
now seeing the numerator [i]stays[/i] at unity, and that the
denominator [i]goes[/i] to unity (since $ \lim_{k\to{+\infty}}k^{\frac{1}{k}}\;=\;1$), i suggested
to myself making the numerator smaller and smaller so we
eventually end up requiring a [b]googolplex [/b] number of terms for
the series to [i]ultimately [/i]diverge!

and so, the series i finally came up with (i.e. the 2nd one), looked like
$ \boxed{\Omega_{2}{(s)}\;=\;\lim_{s,t\to 0^{+}}\;\sum_{k=1}^{\aleph_{0}}\;\frac{t^{\frac{1}{t}}}{k^{s}}}$

well, this should [i]eventually [/i]diverge to infinity, don't you think so ?  :maybe:
also, the divergence of this series is (easily ??) provable,
by the integral test.

however, the question remains, which one of $ \Omega_{1}{(s)}$ and $ \Omega_{2}{(s)}$
actually diverges slower ? also, is there an even [b][u]slower[/u][/b] series possible ?

I can't help but pity your students.  :|

[quote="misan"]:D today, a student asked me what could possibly be the slowest
diverging series ever! whilst the question seemed simple at
the time, it really made me think for a while before i could
give (or suggest) an answer?[/quote]

Misan please, I am sure everybody wants to know, what class is this! Whatever it is, came I join it? I feel I can learn so much. And are you a professor?

And by the way I answered your sine/cosine integral using Complex Variables [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=158083]here[/url] exactly like you asked us to. (But you didnt respond "thank you" and I feel sad  :( ).

Taking natural logarithm $ \aleph_{0}-1$ times hardly makes any sense (for the variety of reasons). Also, $ \lim_{s\to 0}\sum_{k}$ is not exactly a series, so both of your suggestions do not make much sense. I think that you just missed a good opportunity to show your students that there is no such thing as the slowest divergent series, like there is no fastest growing sequence (except in the state of Kansas, where it is $ n!$, as one of my friends told me, but it is another long story :P) or the largest number. So, I'm inclined to second mlok's opinion :roll:.

Regarding this subject, all I know is what I remember from (baby) Rudin wherein he basically addresses this question and indicates that it cannot be done.

Having Rudin handy, he specifically says:

[quote]"One might thus be led to the conjecture that there is a limiting situation of some sort, a 'boundary' with all convergent series on one side, all divergent series on the other side -- at least as far as series with monotonic coefficients are concerned.  This notion of 'boundary' is of course quite vague.  The point we wish to make is this:  No matter how we make this notion precise, the conjecture is false.  Exercises 11(b) and 12(b) may serve as illustrations.  We do not wish to go any deeper into this aspect of convergence theory and refer the reader to Knopp's 'Theory and Application of Infinite Series' Chapter IX, particularly Sec. 41"[/quote]

Wow!  I have never seen Rudin to be so verbose  :D

For the enthused reader who does not have Rudin's book, I can provide these two exercises:

11. (b)  Suppose $ a_{n}>0$ and $ s_{n}=a_{1}+\cdots+a_{n}$ with $ \sum a_{n}$ diverging.  Prove that:  $ \frac{a_{N+1}}{s_{N+1}}+\cdots+\frac{a_{N+k}}{s_{N+k}}\geq 1-\frac{s_{N}}{s_{N+k}}$ and deduce that $ \sum \frac{a_{n}}{s_{n}}$ diverges.

12. (b)  Suppose $ a_{n}>0$ and $ \sum a_{n}$ converges.  Put $ r_{n}=\sum_{m=n}^{\infty}a_{m}$.  Prove that $ \frac{a_{n}}{\sqrt{ r_{n}}}< 2(\sqrt{r_{n}}-\sqrt{r_{n+1}})$ and deduce that $ \sum \frac{a_{n}}{\sqrt{ r_{n}}}$.

Enjoy!

P.S.  Subtracting the number one from an ordinal just makes no sense at all.

[quote="mlok"]I can't help but pity your students.  :|[/quote]

no, no... you don't understand, she (the student) was the one that started
it all.  :rotfl: i was merely trying to eat my turkey sandwich when she came
and interrupted me with the question (during lunch time) :D

[quote="fedja"]Taking natural logarithm $ \aleph_{0}-1$ times hardly makes any sense (for the variety of reasons). Also, $ \lim_{s\to 0}\sum_{k}$ is not exactly a series, so both of your suggestions do not make much sense. I think that you just missed a good opportunity to show your students that there is no such thing as the slowest divergent series...[/quote]

you are right, $ \lim_{s\to 0^{+}}$ is not exactly a series, but it was a suggestion
made to exemplify just how slow the divergence of certain series can be!
i wrote a quick scheme program last night to investigate the [b]rate [/b]at which the
2 series diverge, and i have reached a somewhat definite conclusion
as to which one of them actually diverges [i]slower[/i]...
i ran the program on my laptop for almost 6+ hrs and i only went as far
as $ k\approx 10^{10}$ terms, and both $ \sum_{k}$ still hovered at around $ \approx 10^{-50}$,  :(
such is the [i]sloooooooooooow[/i] divergence (as i had anticipated).
[ integral test, of-course, already claims they [u]both[/u] diverge ]

and yes, i agree there is no such thing as the slowest divergent series, but
i think the theme of the exercise was to find one that is [u]very slow[/u] (or just
get an idea and understand when and how a series diverges, without
delving into the rigors of real analysis), and i think the 2 series i came up
with come pretty close to achieving [u]extremely slow[/u] divergence...
of-course, the 2 series are not at all realistic (in fact, none of them will ever be),
but they do validate the point i was trying to make (in my opinion, that is), and
i am sure many can even construct or identify some even slower ones...

[quote="sylow_theory"]
Misan please, I am sure everybody wants to know, what class is this! [/quote]

sylow, i am not teaching the infinite series class. i suppose the student was
merely curious as to if i had investigated the divergence of series before
(using software programs that is), maybe she saw me referring to hardy's
"divergent series" and knopp's "infinite series" texts before... :|
[ also, i tend to meddle with scheme, C++ and python a lot... ]

[quote="edoo"]Regarding this subject, all I know is what I remember from (baby) Rudin wherein he basically addresses this question and indicates that it cannot be done.[/quote]

again, i agree, but that does not mean we cannot do our own investigation
using software programs (or otherwise) and come close to constructing them.
take for example, series representations for $ \pi$... one can use
$ \pi\;=\;4\left( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots \right)\;=\;-4\sum_{k=1}^\infty\;\frac{(-1)^{k}}{2k-1}$
which converges extremely slowly, or one can use one of ramanujan's
modular identity series
$ \frac{1}{12\pi}\;=\;\sum_{k=0}^\infty\;\frac{(-1)^{k}\;(6k)!\;(13591409+545140134\cdot k)}{(k!)^{3}\;(3k)!\;(640320)^{\frac{3(k+1)}{2}}}$
which converges so rapidly that it adds 14/15 digits for every value of $ k$ !
(let us forget [b]BBP[/b]-type formulas for now...)

following that note, i believe one can construct an even better and [i]faster [/i]
converging series (just like one can craft a [i]slower [/i]diverging series).
i suppose that, in itself, will always be a [u]neverending [/u]quest!

[quote="edoo"]
Wow!  I have never seen Rudin to be so verbose  :D [/quote]

 :rotfl:

[quote="JBL"][quote="misan"]$ \sum_{k=1}^\infty\;\frac{1}{k}$,
$ \sum_{k=2}^\infty\;\frac{1}{k\;\ln k}$,
$ \sum_{k=3}^\infty\;\frac{1}{k\;\ln k\;(\ln \ln k)}$,
$ \sum_{k=4}^\infty\;\frac{1}{k\;\ln k\;(\ln \ln k)\;(\ln \ln \ln k)}$,
$ \sum_{k=5}^\infty\;\frac{1}{k\;\ln k\;(\ln \ln k)\;(\ln \ln \ln k)\;(\ln \ln \ln \ln k)}$,
$ \sum_{k=6}^\infty\;\frac{1}{k\;\ln k\;(\ln \ln k)\;(\ln \ln \ln k)\;(\ln \ln \ln \ln k)\;(\ln \ln \ln \ln \ln k)}$[/quote]

Of these, only the first 3 satisfy your conditions of positive, monotonically decreasing terms.  $ \ln\ln\ln 4 \approx-1.119$, while $ \ln\ln\ln\ln 5$ is not a real number.  The fix is instead to consider
$ \sum_{k=1}^\infty\;\frac{1}{k}$,
$ \sum_{k=1}^\infty\;\frac{1}{k\;\ln (1+k)}$,
$ \sum_{k=1}^\infty\;\frac{1}{k\;\ln (1+k)\;\ln(1+\ln (1+k))}$,
$ \sum_{k=1}^\infty\;\frac{1}{k\;\ln (1+k)\;\ln(1+\ln(1+k))\;\ln(1+\ln(1+\ln1+k)))}$ and so on.

Incidentally, the fact that your student asked you a question ("started it") doesn't mean that you should give them a nonsensical answer.[/quote]

my series is perfectly fine, why do you assume i referred to taking logarithms
to the base $ 2.718281828\cdots$ ? just because i used the symbol $ \ln$ doesn't mean
the base will always be $ e$, for my software computation, i used a base of
$ \boxed{1.1}$ (so long as the [b]limit [/b]for the base $ \to 1^{+}$, the series
will always work in it's aim to diverge slowly).

FYI, in analysis of algorithms and of series, $ \ln$ or $ \log$ has no particular
meaning, it is [u]implicitly[/u] understood they just refer to "[i]a logarithm[/i]"
([i]without [/i]reference to any particular base, i.e. the base is completely irrelevant,
unless stated explicitly otherwise, e.g. $ \log_{e}$ and $ \log_{10}$).

and as for the index, i could very well have started at $ k=2$ (for all of them),
but i placed an [i]increasing [/i]order for the starting indices just for fun!

The symbol $ \ln$ [i]always[/i] refers to the natural logarithm; I have never seen it meaning anything else except in your post. If you want a logarithm with an unspecified base, use $ \log$ instead.

My favorite series with slow convergence/divergence is this one (Putnam 2002 A6):
Fix an integer $ b\ge 2$. Let $ f(1)=1,f(2)=2$, and for each $ n\ge 3$, define $ f(n)=nf(d)$, where $ d$ is the number of digits of base-b digits of $ n$.
For which values of $ b$ does $ \sum_{n=1}^{\infty}\frac1{f(n)}$ converge?

[quote="edoo"]

11. (b)  Suppose $ a_{n} > 0$ and $ s_{n} \equal{} a_{1} \plus{} \cdots \plus{} a_{n}$ with $ \sum a_{n}$ diverging.  Prove that:  $ \frac {a_{N \plus{} 1}}{s_{N \plus{} 1}} \plus{} \cdots \plus{} \frac {a_{N \plus{} k}}{s_{N \plus{} k}}\geq 1 \minus{} \frac {s_{N}}{s_{N \plus{} k}}$ and deduce that $ \sum \frac {a_{n}}{s_{n}}$ diverges.

[/quote]

How exactly do you use that inequality to get a contradiction if you assume that  $ \sum \frac {a_{n}}{s_{n}}$ is converging. To me it just seems that the inequality is of no help since $ 1 \minus{} \frac {s_{N}}{s_{N \plus{} k}}$ will go to $ 0$ as $ N$ gets bigger.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=305590]This recent problem[/url] seems pertinent.

[quote="misan"]

following that note, i believe one can construct an even better and [i]faster [/i]
converging series (just like one can craft a [i]slower [/i]diverging series).
i suppose that, in itself, will always be a [u]neverending [/u]quest!

[/quote]

Are you certain that this is true?! You say that for any $ a(n)$ such that $ L \equal{} \sum_n a(n)$ one can find such $ b(n)$ that $ L\equal{}\sum_n b(n)$ but it converges faster? How about $ a(n) \equal{} \{L, 0, 0, 0, ...\}$? There is no better series that this one - it converges in one step (has maximal convergence speed)! But when one wants a a series to diverge/converge slowly, smaller change is better - and the "smallest number" problem kicks in so the ultimate solution is unreachable (and that part of your opinion holds). Still, I might be (and probably am) totally mistaken..

[quote="milin"] Still, I might be (and probably am) totally mistaken..[/quote]
Only in one respect: if you look at the date of the post you quoted and at the date of the most recent post by misan, you'll realize that the chance that misan will read your objection is next to $ 0$. :roll: Mathematically you are completely correct.


I discovered the problems of the Dutch Mathematical Olympiad 2003 have been released today and I'll post them here.

The first one.

A pythagorean triangle is a triangle with one angle equal to 90 and sides of integer length. Find all pythagorean triangles for which the area equals twice the perimeter.

[hide]Let the three sides be a, b and c with a :le: b < c. (c is the hypotenuse)

we have: ab = 4 (a + b + c), and

a2 + b2 = c2

(a + b)2 - 2ab = c2

(a + b)2 - 8(a + b + c) = c2

- 8(a + b + c) = c2 - (a + b)2

- 8(a + b + c) = (c - a - b)(c + a + b)

- 8 = c - a - b

c = a + b - 8.

Sub this into either original equation, and we obtain

ab - 8a - 8b + 32 = 0

(a - 8)(b - 8) = 32

Valid possibilities: (keep in mind that a :le: b < c)

a - 8 = 1, b - 8 = 32 => a = 9, b = 40, c = 41.

a - 8 = 2, b - 8 = 16 => a = 10, b = 24, c = 26.

a - 8 = 4, b - 8 = 8 => a = 12, b = 16, c = 20.

[/hide]

You can also use the parametrized form of the pythagorean triples -- not any easier, just another method.

Osiris:  a challenge (because I'm too lazy to try):
do it where 2 is replaced by a general integer n.

i kno it doesn't apply here
but u should make a note of when the factors are both negative
that sometimes contains additional solutions

but, is it just me, or is this Dutch olympiad reallyl easy?

Well, I tried all of them. I listed only the valid ones, because when both factors are negative, exactly one of a and b is negative.

(Canadian Thanksgiving... I'll try the generalisation when I come back.)

Well ... The Dutch olympiad is easy indeed. But it is more difficult than last year :)
...
I'm was very surprised to see that the highest dutch scores were 34, 33, 31, 27 (/50) while I could do every problem in 5 minutes !
The Dutch people are our neighbours (I'm from Belgium) and I can tell you that the Belgian olympiads are harder ... well ... still not too difficult but more challenging. The Dutch niveau is decreasing last years ... at the IMO they got 30 points with the whole team while we got 70.


algebra
function
domain
Ring Theory
superior algebra
superior algebra unsolved

If $ R$ is a finite commutative ring, then every prime ideal is maximal.  Find a counterexample to the converse.  I'm pretty sure that maximal and prime are equivalent in a finite commutative ring, so is there a counterexample in the infinite or noncommutative case?

Also, if $ R$ is a local ring and $ J(R)$ denotes the set of nonunits, show that $ J(R)$ is the only maximal ideal in $ R$.

The ring $ \mathbb{Z}/(x)\cong \mathbb{Z}$ gives an example of prime ideal which is not maximal.

I don't think that's the converse of the statement, though.  Do there exist any rings which are maximal but not prime?

Surely in commutative rings every maximal ideal $ M \subset R$ is prime.
The quotient $ R / M$ is a field, hence certainly a domain, which means it is prime.

Okay, thanks.  Do you have any thoughts on the second question?

If you include any invertible element in an ideal, that ideal is the whole ring. Therefore, if the set of non-units in $ R$ is an ideal, it contains all ideals except $ R$ itself, and must be the unique maximal ideal.
Also, in general (for a commutative ring), the set of non-units is the union of all ideals other than $ R$; the principal ideals are enough to cover it already.

How this finishes depends on exactly what property you take as the definition of "local".

The book I'm using defines local to mean that the set of non-units is an ideal.  I'm aware of the alternative definition that a ring is local if it has a unique maximal ideal.


geometry
trapezoid
vector
geometry proposed

Let a quardilateral $ABCD$  with  $AB=AD$  and $\widehat B=\widehat D=90$.
At $CD$  we take point  $E$ and at  $BC$  we take point  $Z$  such that
$AE\bot DZ$.  Prove that  $AZ\bot BE$

Rename $Z$ as $F$. Now, consider the polar map with respect to the circle $\omega$ with center $A$ that passes through $B$ and $D$. We know that the pole of $E$ will pass through the points of tangency from $E$ to $\omega$ and will be perpendicular to $AE$ - note that $DF$ does indeed pass through one of these tangency points ($D$) and satisfies the orthogonality requirement; these two requirements uniquely determine the line, so $DF$ is the pole of $E$, i.e. $F$ lies on the pole of $E$ so $E$ must lie on the pole of $F$. As $B$ is a point of tangency from $F$ to $\omega$, the pole of $F$ is thus $BE$ and must be perpendicular to $AF$.

[b]A general enunciation:[/b]

Let $m,n$ be two lines (which can be parallels or concurrently) and a circle $w=C(I,r)$ which is tangent to the lines $m,n$ in the points $A\in m$, $B\in n$ respectively. Prove that for any points $M\in m$, $N\in n$ the following sentencies are equivalently:

$1^{\circ}.\ IM\perp AN\ ;\ \ \ \ 2^{\circ}.\ IN\perp BM\ ;\ \ \ \ 3^{\circ}.\ AM^2+BN^2=MN^2.$

[u]A particular cases.[/u]

$\blacksquare\ a).\ m\parallel n$. Let $AMNB$ be a trapezoid ($AM\parallel BN$) and the middlepoint $I$ of the side $[AB]$. Prove that $IM\perp AN\Longleftrightarrow IN\perp BM\Longleftrightarrow AM^2+BN^2=MN^2.$

$\blacksquare\ b).\ P\in m\cap n$. Let $ABC$ be a triangle. The its incircle $w=C(I,r)$ touches the its sides in the points $D\in BC$, $E\in CA$, $F\in AB$ and the its exincircle $w_a=C(I_a,r_a)$ touches the its sides in the points $D_1\in BC$, $E_1\in CA$. $F_1\in AB$. I note the intersections $L\in BC\cap EF$ and $L_1\in BC\cap E_1F_1$. Prove that: $IL\perp AD$, $LA^2=LD^2+AE^2$; $I_aL_1\perp AD_1$, $L_1A^2=L_1D_1^2+AE_1^2$. (A remarkable property !)

[hide="The my solution"]
[u]The my soluytion.[/u]

$\bullet\ \ (1^{\circ})\Longleftrightarrow IA^2-IN^2=$ $MA^2-MN^2\Longleftrightarrow$

$r^2-(r^2+BN^2)=MA^2-MN^2\Longleftrightarrow$

$AM^2+BN^2=MN^2\Longleftrightarrow (3^{\circ})$.

Thus, $(1^{\circ})\Longleftrightarrow (3^{\circ}).$

$\bullet\ \ (2^{\circ})\Longleftrightarrow IB^2-IM^2=$ $NB^2-NM^2\Longleftrightarrow$

$r^2-(r^2+AM^2)=NB^3-NM^2\Longleftrightarrow$

$AM^2+BN^2=MN^2\Longleftrightarrow (3^{\circ})$.

Thus, $(2^{\circ})\Longleftrightarrow (3^{\circ}).$

Therefore, $(1^{\circ})\Longleftrightarrow (2^{\circ})\Longleftrightarrow (3^{\circ}).$
[/hide]

Silouan,
  in my notes I see a vector solution(inner product), but I do not remember where I have found it. Do you have the same solution?

 Babis

[i]Stergiu[/i], read you the my extension of this problem (with three degrees of freedom) ?

@mr Stergiou : Yes my solution is as you said above. There exist another solution with elementary geom in your book for C' of Gymnasium at page 185.

Let $F\in AE\cap DZ, G\in BE\cap AZ$, From $\triangle ADE\implies AD^2=AF\cdot AE$, but $AB=AD$ ans so $AB$ is tangent to the circle $\odot BEF\implies\angle ABF=\angle AEB=\angle AEG$. $ABZF$ being cyclic, $\angle ABF=\angle AZF=\angle GZF$, consequently $\angle GZF=\angle GEF$ and $EFGZ$ is cyclic, hence $\angle EGZ=\angle EFZ=90^\circ$, done.

[color=#FF0000][b][size=120]Remark: $BF, DG, AC$ are concurrent, prove!
[/size][/b][/color]

My best wishes for a Merry Christmas and a Happy, and Prosperous New Year 2015 everybody!

Best regards,
sunken rock


Duke
college
geometry
probability
3D geometry
trigonometry
symmetry

Here are the Individual problems from the 2003 Meet. I think I have answers to all of these except for the last two. Give them a try, or if you have questions, post them here and maybe we can figure them out!

1. If Suzie has 6 coins worth 23 cents, how many nickels does she have?

2. Let a ¤ b = (a - b)/(a + b). If 8 ¤ (2 ¤ x) = 4/3, what is x?

3. How many digits does $x = 100000025^2 - 99999975^2$ have when written in decimal form?

4. A paperboy’s route covers 8 consecutive houses along a road. He does not necessarily deliver to all the houses every day, but he always traverses the road in the same direction, and he takes care never to skip over 2 consecutive houses. How many possible routes can he take?

5. A regular 12-gon is inscribed in a circle of radius 5. What is the sum of the squares of the distances from any one fixed vertex to all the others?

6. In triangle ABC, let D, E be points on AB, AC, respectively, and let BE and CD meet at point P. If the areas of triangles ADE, BPD, and CEP are 5, 8, and 3, respectively, find the area of triangle ABC.

7. Bob has 11 socks in his drawer: 3 different matched pairs, and 5 socks that don’t match with any others. Suppose he pulls socks from the drawer one at a time until he manages to get a matched pair. What is the probability he will need to draw exactly 9 socks?

8. Consider the unit cube ABCDEFGH. The triangle bound to A is the triangle formed by the 3 vertices of the cube adjacent to A (and similarly for the other vertices of the cube). Suppose we slice a knife through each of the 8 triangles bound to vertices of the cube. What is the volume of the remaining solid that contains the former center of the cube?

[hide="1"]1. 2 nickles.
there are obviously 3 pennies leaving 3 coins and 20 cents. the only possible way is with 2 nickles and a dime
[/hide]
[hide="2"]x=-30.  just an algebra equation with fun fractions. don't feel like showing all the steps.[/hide]
[hide="3"]3. 11digits
difference of 2 squares leaves 50 * 2.0 x $10^8$
this simplifies to 1.0 X $10^{10}$, which has 11 digits[/hide]

Does someone have an easy way of doing number $5$, or are you supposed to use law of cosines a lot of times?

i couldn't figure a way to $5$ quickly. however, a neat pattern emerged where the problem worked out nicely with the multiple law of cosines. i'm guessing the sum of the squares part of the problem is what would cause a simple solution. i'm interested in its solution. by the way, i got an answer of [hide="click"]600[/hide]

Using LoC on #5 [i]is[/i] fast...you just have to do it the right way.[hide]

Look at the picture...  Because of symmetry, we can just calculate the sum of the squares from $A$ to the vertices above the x-axis (as shown), multiple this by two (for those below the x-axis), and add $AG^2$ (since it's on the x-axis).

So we look at triangles $AOB, AOC, AOD, AOE,$ and $AOF$.  Notice that $m\angle AOB=\frac{\pi}6$, $m\angle AOC=\frac{\pi}3$, $m\angle AOD=\frac{\pi}2$, $m\angle AOE = \frac{2\pi}3$, and $m\angle AOF=\frac{5\pi}6$.  Using the Law of Cosines we wish to find (after rearranging some terms and using the fact that the radius is equal to 5)  \[ 250-(\cos \frac{\pi}6+\cos \frac{\pi}3+\cos \frac{\pi}2+\cos \frac{2\pi}3+\cos \frac{5\pi}6).  \]

Now, use the fact that $\cos(\pi-x)=-\cos x\Rightarrow \cos(\pi-z)+\cos x=0.$  That makes the whole sum of cosines in the parentheses equal to 0.  Thus, as an answer we have $2(250)+10^2=600.$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=391[/img][/hide]


If 2 cards from a standard deck are selected randomly, what is the probability that either two kings or at least 1 ace occurs? (There are 4 Aces, 4 kings and 52 total cards in a standard deck.)

alternative solution:

A = Chance of getting ace in first card
B = Chance of getting ace in second card
C = Chance of getting two kings = Chance of getting two aces.

If we calculate A+B, we the situation of getting two aces is covered twice. So, chance of getting at least one Ace is A+B-C.

Chance of getting at least one Ace or two Kings is A+B-C+C = A+B = 2A = 2*4/52=1/26      (B and C are mutually exclusive)

:)

I got a different solution:

We can split this into cases:

[b][u]Case #1[/u][/b]:[i]We draw two kings.[/i]
There are $\binom{4}{2}=6$ ways to choose which two kings that we can draw, and $\binom{52}{2}=1326$ ways to choose two cards.  Therefore, the probability that we draw two kings is $\frac{6}{1326}=\frac{1}{221}$.

[b][u]Case #2[/u][/b]:[i]We draw exactly one ace.[/i]
There are $\binom{4}{1}=4$ ways we can choose an ace and $\binom{48}{1}=48$ ways to choose the other card.  Additionally, there are $2$ ways we can draw exactly one ace, we can draw it first or second.  Therefore, the probability that we draw exactly one ace is $\frac{4\times48}{\binom{52}{2}}\times2=\frac{384}{1326}=\frac{64}{221}$.

[u][b]Case #3[/b][/u]:[i]We draw exactly two aces.[/i]
The probability that we draw exactly two aces is the same as the probability that we draw exactly two kings, which is $\frac{1}{221}$.

Last of all, we add up the probabilities from each case and get:

\[\frac{1}{221}+\frac{64}{221}+\frac{1}{221}=\boxed{\frac{4}{13}}\]

Yes, seanwang is right.

[hide="Solution"][hide="Case 1"]Two kings.  This is just $\frac{4}{52}*\frac{3}{51}=\frac{1}{13}*\frac{1}{17}=\frac{1}{221}$.[/hide]

[hide="Case 2"] At least one ace.  Use complementary probability to find the probability of picking no aces, and then subtract this from one.
            Probability of picking no aces: $\frac{48}{52}*\frac{47}{51}=\frac{12}{13}*\frac{47}{51}=\frac{188}{221}$.
             Probability of one ace: $1-\frac{188}{221}=\frac{33}{221}$.
[/hide]
[hide="Total"]Total: The total probability is $\frac{1}{221}+\frac{33}{221}=\frac{34}{221}=\frac{2}{13}$[/hide][/hide]

I came up with the alternate solution too. Here is a fixed version. The probability of getting 2 kings is 4/52 * 3/51 = 12/(52*51). The probability of getting an ace on the first card is 1/13. The probability of getting an ace on the second card is also 1/13. We have counted the chance of getting 2 aces twice. The probability of this is 4/52 * 3/51 = 12/(52 * 51). The answer is P(2 kings) + P(Ace on 1st) + P(Ace on 2nd) - P(2 aces), which is 12/(52 * 51) + 1/13 + 1/13 - 12/(52 * 51) = 1/13 + 1/13 = 2/13.

Just use complementary counting for probability of at least 1 ace. Note that there are 48 cards in a deck that aren't aces.
[hide= "Solution"]
$P = 1-(\dfrac{48}{52}\cdot \dfrac{47}{51}) = \dfrac{33}{221}$

Add this to $\dfrac{1}{221}$ and we get $\dfrac{2}{13}$.
[/hide]

I was off by 1. Can someone notify me of my error?

[hide] The total possible is $52 \cdot 51$ (without rt order). To get 2 kings, we have $4 \cdot 3=12$ choices for the suits. Now to compute the ways to get at least 1 ace. We use complementary counting. $52 \cdot 51$ ways to select the 2 cards, and $(52-4)(52-5)$ ways to select without any of the 4 aces. So our answer is $$\frac{12+(52 \cdot 51)-(52-4)(52-5)}{52 \cdot 51}=\frac{35}{221}.$$ [/hide]

Eugenis, I think you may have had a calculation error since the formula used was correct. Solving for (52*51)-(52-4)(52-5) we get 2652-(48)(47)=2652-2256=396. Plugging it in the main function we get (12+396)/2652=408/2652=102/663. This is when I think you made an error since 102/3 is 34 not 35. So using 34 we get 34/221=2/13. I hope this helps. :-D

[hide=solution]Divide the problem into three cases.
In the first case, the first card is an ace, so it satisfies the "win" situation no matter what the second card is. The "win" possibility of this case is $1/13$.
In the second case, the first card is a king. You would "win" when you draw either a king or an ace as the second card. The "win" possibility of this case is $1/13\cdot7/51=7/663$.
In the third case, the first card is neither king nor ace. You need to draw an ace in order to "win". The "win" possibility of this case is $(13-2)/13\cdot4/51=11/13\cdot4/51=44/663$.
Now add the "win" possibility of three cases together:
$7/663+1/13+44/663=102/663=2/13$[/hide]

no it doesnt matter @above
you can do it either way

umm

cases of two kings and at least one ace are both exclusive

soo we do complementary counting for the ace and just kinda add them together

i get alcumus is easy
but this should not be level 25

[quote=A1001][quote=smartninja2000][quote=Bestestsocks]This was easier than level 25?[/quote]

Alcumus covers a really basic range of problems; I would say that something that is final 5 on AMC 10 is at least level 23+[/quote]

except this is nowhere near final 5 amc10 lvl
i have a different [hide=sol]c1: 2 kings
4c2/52c2
c2: at least 1 ace
1-48c2/52c2 complementary counting
the answer is the sum which is 2/13, as desired[/hide][/quote]

Nowadays, I would predict this to be a problem 15 on amc10. :)


algebra
polynomial
abstract algebra
number theory
least common multiple

[hide="Proof"] Consider the polynomial $x^{p-1}-1 \bmod p$.  Fermat's Little Theorem tells us that $1, 2, 3, ... p-1$ are its roots.  In other words,

$x^{p-1}-1 \equiv (x-1)(x-2) ... (x-(p-1)) \bmod p$

Now, let $e$ be the least common multiple of all of the orders of $1, 2, 3, ... p-1 \bmod p$.  Then $1, 2, 3, ... p-1$ are all roots of

$x^{e}-1 \bmod p$

Which, by Lagrange's Theorem, can have at most $e$ roots; hence $e = p-1$.  It's a fairly easy lemma that if elements of order $x, y$ exist then an element of order $lcm(x, y)$ exists, hence an element of order $e$ exists.  QED. [/hide]


The sum of the squares of the first $ n$ positive integers is given by the expression $ \frac{n(n \plus{} c)(2n \plus{} k)}{6}$, if $ c$ and $ k$ are, respectively:

$ \textbf{(A)}\ {1}\text{ and }{2} \qquad
\textbf{(B)}\ {3}\text{ and }{5}\qquad
\textbf{(C)}\ {2}\text{ and }{2}\qquad
\textbf{(D)}\ {1}\text{ and }{1}\qquad
\textbf{(E)}\ {2}\text{ and }{1}$

Substituting 1 in, we see the answer is D

Solution, A little bit more explained:

Evaluate $ \dfrac{n(n\plus{}c)(2n\plus{}k)}{6}\equal{}1$, so that $ n(n\plus{}c)(2n\plus{}k)\equal{}6$.

Since in all the options $ c$ and $ k$ are positive, $ 2n\plus{}k>n\plus{}c>n$ such that each is a factor of $ 6$. The only triple that satisfies this is $ (1,2,3)$. We know that $ n\equal{}1$ so $ c\equal{}2\minus{}1\equal{}1$ and $ k\equal{}3\minus{}2(1)\equal{}1$, so the answer is $ \textbf{(D)}$.

[quote="varunrocks"]This is a relatively ez question if you no that n(n+1)(2n+1)/6 is the sum of the first n positive square numbers.[/quote]
... which is given in the problem.


inequalities
logarithms
inequalities proposed

Let $ m,n\in\mathbb{N}$ and for $ 1\leq i\leq m+n$ we have $ x_{i}\in\mathbb{R}^{+}$. Prove that:

$ \left(\frac{1}{m}\sum^{m}_{i=1}x_{i}\right)^{m}\left(\frac{1}{n}\sum^{m+n}_{i=m+1}x_{i}\right)^{n}\leq\left(\frac{1}{m+n}\sum^{m+n}_{i=1}x_{i}\right)^{m+n}$

Setting $ \sum_{i=1}^{m}x_{i}=a$ and $ \sum_{i=m+1}^{m+n}x_{i}=b$. Then, that inequality is equivalent to $ m\ln \frac{a}{m}+n\ln \frac{b}{n}\leq (m+n)\ln \frac{a+b}{m+n}$, this is true by Jensen's theorem! :D


let f(x)=$ {ax^2\plus{}bx\plus{}c}$ Satisfying $ {f(\minus{}1)\equal{}0}$ and for any $ {x \in R}$ we have $ {f(x)\minus{}x>\equal{}0}$  ; WHEN $ {x \in (0,2)}$ WE HAVE $ {f(x) \le (\frac{x\plus{}1}{2})^2}$;
find $ {f(1)}$;
when $ {x \in [\minus{}1,1]}$ function g(x)=f(x)-mx($ {m \in R}$) is only increasing or only reducing ,then find the range of m.
[Sorry for my bad English   :blush:  ]

Dear mathlinkers ,can you solve this problem ? :

This is my 1st post in the olympiad sections,so this answer is probably a 100% error,yet 1 can learn frm errors,so is the answer 1?(1st part)

Pocha,Well done ! Your answer is right . $ {f(x)\minus{}x>\equal{}0}$  $ {\Rightarrow}$ $ {f(1)  >\equal{}1}$  ;$ {f(x)\le (\frac{x\plus{}1}{2})^{2}}$ $ {\Rightarrow}$ $ {f(1) \le 1}$; so $ { f(1)\equal{}1 }$[ The first part is very easy ,but the difficulty is in the secong part ,Have a try dear mathlinkers! :) ]

[quote="CCMath1"]let f(x)=$ {ax^2 \plus{} bx \plus{} c}$ Satisfying $ {f( \minus{} 1) \equal{} 0}$ and for any $ {x \in R}$ we have $ {f(x) \minus{} x > \equal{} 0}$  ; WHEN $ {x \in (0,2)}$ WE HAVE $ {f(x) \le (\frac {x \plus{} 1}{2})^2}$;
find $ {f(1)}$;
when $ {x \in [ \minus{} 1,1]}$ function g(x)=f(x)-mx($ {m \in R}$) is only increasing or only reducing ,then find the range of m.
[Sorry for my bad English   :blush:  ][/quote]

$ f( \minus{} 1) \equal{} 0$ $ \implies$ $ a \minus{} b \plus{} c \equal{} 0$ and so $ c \equal{} b \minus{} a$
We get $ f(1) \minus{} 1\geq 0$ and $ f(1)\leq(\frac {1 \plus{} 1}2)^2 \equal{} 1$ and so $ f(1) \equal{} 1$ and so $ a \plus{} b \plus{} c \equal{} 1$ and so $ b \equal{} \frac 12$

So $ f(x) \minus{} x \equal{} ax^2 \minus{} \frac 12x \plus{} \frac 12 \minus{} a$ whose discriminant is $ \Delta \equal{} (2a \minus{} \frac 12)^2$ and so $ a \equal{} \frac 14$ else we could have $ f(x) < x$ for some $ x$ and so $ f(x) \equal{} \frac 14x^2 \plus{} \frac 12x \plus{} \frac 14 \equal{} (\frac {x \plus{} 1}2)^2$

Then $ g'(x) \equal{} \frac {x \plus{} 1}2 \minus{} m$ has a zero for $ x \equal{} 2m \minus{} 1$ and we want either $ 2m \minus{} 1\leq \minus{} 1$, either $ 2m \minus{} 1\geq 1$ and so $ m\in( \minus{} \infty, \minus{}0]\cup[\plus{}1, \plus{} \infty)$

perfect solution pco ! That's the right answer!


[quote="silouan"]Δίνεται το πολυώνυμο $P(x)$ το οποίο είναι 10ου βαθμού ώστε για κάθε x στο $R$  να ισχύει:
$P(x)+P'(x)+P''(x)+....+P^{(10)}(x)=x^{10}+2\cdot 9^9$

Νδο :$P(x)\geq 9^9$  για κάθε χ[/quote]

σιλουανε καλη!

$P(x)=x^{10}-10x^9+2\cdot 9^9$

και $P(x)\geq P(9)$ $\forall x \in \Re$

σωστος??

Nai Θανάση σωστά....Ωραία λύση


search
number theory unsolved
number theory

[quote="Beat"]Solve in prime numbers: $p^3-q^5=(p+q)^2$[/quote]
[b]Beat[/b] see this [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=p%5E3-q%5E5%3D%28p%2Bq%29%5E2&t=66828]www.mathlinks.ro/Forum/viewtopic.php?highlight=p%5E3-q%5E5%3D%28p%2Bq%29%5E2&t=66828[/url]
Ther are two solutions of this problem,my and [b]Jin[/b]'s  ;)

Thanks a lot Tiks. I tried to search it but no result..


calculus
derivative
trigonometry
calculus computations

I have no idea how to approach this partial derivative problem:

P(s,t) = R + r cos(t)

z(s,t) = r sin(t)
x(s,t) = P(s,t) cos(s)
y(s,t) = P(s,t) sin(s)

f(s,t) = [x(s,t) , y(s,t) , z(s,t)]

what is: df/ds , df/dt , and d^2f/dt^2 ?

Note: Some of the three answers may be impossibe!

____________________________________________

This was what I was thinking for the approach. Start with substitution.  Plug in P(s,t) into the z(s,t), y(s,t) and x(s,t).

Thus, I will get this:

z(s,t) = r sin(t)
x(s,t) = [R + r cos(t)][r cos(s)]
y(s,t) = [R + r cos(t)][r sin(s)]

Plug those into f(s,t), and we get:

f(s,t) = {[R + r cos(t)][r cos(s)], [R + r cos(t)][r sin(s)], r sin(t) }

Now, when it comes to finding the three derivatives. I am absolutely lost on what to do. Does anyone know how to approach this problem? Thanks!

You're having some serious difficulties with your plugging-ins.  While substituting $P$ into $x$ and $y$ you seem to have spontaneously generated some $r$s, changed an $s$ to a $t$, and interchanged a $\sin$ and a $\cos$.

After that, of course, the question is: do you understand how to take a partial derivative?  Because this particular problem is a straight-forward application, and if you are lost on it, it suggests that you aren't comfortable with the notion of partials and how they work.

I tried making the corrections, but it still looks like something is wrong here. Because P(s,t) = R + r cos(t) seems to be needed to be converted to something else.

Also, I have had experience with partial derivatives before, but this one is the most complex one I encountered.

However, I am not sure how to do the partial derivatives in this case. Does anyone have any ideas on how to get started?

I was thinking that we need to somehow convert P(s,t) = R + r cos(t) into terms of x, y, and z in order to solve it. But I am not sure how to do this, or if it is even legal.

Perhaps I am thinking too hard on this problem. You said it was straight forward. So, I was thinking that the answer would be something like this for f(s,t) = {[R + r cos(t)][r cos(s)], [R + r cos(t)][r sin(s)], r sin(t) }.

df/dt = (-r^2 cos[s] sin[t], -r^2 sin[s] sin[t], r cos[t] )

Is this correct? Thanks!


[hide="Solution"]

[hide="First"]
Get them to a common denominator: $ a^3b^3c^3$
So we have to prove: $ a^8c^3\plus{}b^8a^3\plus{}c^8b^3\ge a^7c^3b\plus{}b^7a^3c\plus{}c^7b^3a$.
[hide="Second"]
AM-GM?
[/hide]
[/hide]
[/hide]

[quote="Bugi"][hide="Solution"]

[hide="First"]
Get them to a common denominator: $ a^3b^3c^3$
So we have to prove: $ a^8c^3 \plus{} b^8a^3 \plus{} c^8b^3\ge a^7c^3b \plus{} b^7a^3c \plus{} c^7b^3a$.
[hide="Second"]
AM-GM?
[/hide]
[/hide]
[/hide][/quote]

yes AM-GM , I want a beautiful proof with only AM GM

[hide="Not AM-GM, but still solution..."]Using Cauchy-Schwartz Inequality, we have $ \left(\frac{a^3}{b}\plus{}\frac{b^3}{c}\plus{}\frac{c^3}{a}\right)\left(\frac{a^5}{b^3}\plus{}\frac{b^5}{c^3}\plus{}\frac{c^5}{a^3}\right)\ge \left(\frac{a^4}{b^2}\plus{}\frac{b^4}{c^2}\plus{}\frac{c^4}{a^2}\right)^2$.

Next, we wish to prove that $ \frac{a^4}{b^2}\plus{}\frac{b^4}{c^2}\plus{}\frac{c^4}{a^2}\ge\frac{a^3}{b}\plus{}\frac{b^3}{c}\plus{}\frac{c^3}{a}$.

Applying Cauchy-Schwartz Inequality again, we want to prove that $ \frac{a^3}{b}\plus{}\frac{b^3}{c}\plus{}\frac{c^3}{a}\ge \frac{a^3}{a}\plus{}\frac{b^3}{b}\plus{}\frac{c^3}{c}$.

Without loss of generality, assume that $ a\ge b\ge c$.

Since $ a^3\ge b^3\ge c^3$ and $ \frac{1}{c}\ge \frac{1}{b}\ge \frac{1}{a}$, the result follows by Rearrangement Inequality.[/hide]

Hopefully I didn't fail?

I think you cannot lose generality when assuming $ a \ge b \ge c$. The expression is cyclic, not symmetrical; you would also have to also consider $ a \ge c \ge b$. However, it doesn't matter, because $ \frac{a^3}a \plus{} \frac{b^3}b \plus{} \frac{c^3}c$ is always the "worst" pairing.

And, theoretically, your solution is by AM-GM.
CS can be proven by AM-GM, and the last step could be done by summing $ 3\frac{a^2}b \plus{} \frac{b^3}c \plus{} 9\frac{c^3}a \ge 13c^2$ cyclically.

Just maybe not that beautiful...

[quote="aauqen"]And, theoretically, your solution is by AM-GM.
CS can be proven by AM-GM.[/quote]
Just because CS can be proven by AM-GM, using CS doesn't mean using AM-GM...

You would say 'This inequality is true by Cauchy-Schwartz, and Cauchy-Schwartz is true by AM-GM (although you wouldn't need to say the second part)'

Plus, how would one prove that $ \frac{a^{3}}{b}\plus{}\frac{b^{3}}{c}\plus{}\frac{c^{3}}{a}\ge\frac{a^{3}}{a}\plus{}\frac{b^{3}}{b}\plus{}\frac{c^{3}}{c}$?

It looks considerably easy...

[quote="Evariste-Galois"]a,b,c>=0   Prove that :

$ \frac {a^5}{b^3} \plus{} \frac {b^5}{c^3} \plus{} \frac {c^5}{a^3}\geq \frac {a^4}{b^2} \plus{} \frac {b^4}{c^2} \plus{} \frac {c^4}{a^2}$

I want a nice solution with AM_GM !!!!! :D[/quote]
It's not nice, but it's Am-Gm..

$ \frac{41 \cdot \frac {a^5}{b^3} \plus{} 5 \cdot \frac {b^5}{c^3} \plus{} 3 \cdot \frac{c^5}{a^3}}{49} \ge \frac {a^4}{b^2}$ and cyclic..


I've tried to reproduce the expression on Problem 5b:
[code]\dfrac{\pi}{4} \left(1 + \cfrac{\sum_{i=1}^\infty \cfrac{1}{i^3}}{\sum_{i=1}^\infty \cfrac{1}{i^4}}\right)[/code]
\[ \dfrac{\pi}{4} \left(1 \plus{} \cfrac{\sum_{i\equal{}1}^\infty \cfrac{1}{i^3}}{\sum_{i\equal{}1}^\infty \cfrac{1}{i^4}}\right)\]

Unfortunately, the sum does not appear displaystyle. How do you make the sum appear correctly? (This is just helping with formatting.)

Use [code]\limits[/code] after the sum.

It looks like

[code]\dfrac{\pi}{4} \left(1 + \cfrac{\sum \limits_{i=1}^\infty \cfrac{1}{i^3}}{\sum \limits_{i=1}^\infty \cfrac{1}{i^4}}\right)[/code]

$ \dfrac{\pi}{4} \left(1 \plus{} \cfrac{\sum \limits_{i\equal{}1}^\infty \cfrac{1}{i^3}}{\sum \limits_{i\equal{}1}^\infty \cfrac{1}{i^4}}\right)$

[quote="electricj"]\frac{\pi}{4}( 1 +  \frac{\displaystyle\sum_{i=1}^{\infty} \frac{1}{i^3}}{\displaystyle\sum_{i=1}^{\infty} \frac{1}{i^4}})

This is my first time using Latex, so I have a basic "try it, and if it works leave it" way of using it, but this seemed to work just fine.[/quote]

\dfrac{\pi}{4}\left( 1 + \frac{\displaystyle\sum_{i=1}^{\infty} \frac{1}{i^3}}{\displaystyle\sum_{i=1}^{\infty} \frac{1}{i^4}} \right) works better; it has \left and \right, which make the parentheses larger, and the dfrac...well, I don't know the difference, but whatever.

dfrac means displayfraction, so the fraction will be in displaystyle.  It should be bigger and better looking in general.


Calculate the sum of the geometric series $ 1\plus{}\left(\frac{1}{5}\right)\plus{}\left(\frac{1}{5}\right)^2 \plus{} \left(\frac{1}{5}\right)^3 \plus{} \dots$. Express your answer as a common fraction.

hello, we have $ \sum_{i\equal{}0}^nq^i\equal{}\frac{q^{n\plus{}1}\minus{}1}{q\minus{}1}$ and for $ |q|<1$ exists the limit for $ n\minus{}>\infty$ and is equal to $ \frac{1}{1\minus{}q}$ after this formula we get the sum above as $ \frac{5}{4}$.
Sonnhard.


Barry is currently locked inside a room with a large metal door having a width of 5 metres. Right near the doorpost there are keyholes on both sides with corresponding keys in them. A wealthy gentleman, Mr. X (say) made a deal with Barry.

If Barry was successful in opening the door within two days starting right now at noon, Mr. X will give him an amount equivalent to $ 2,500,000. If not, Barry is free to go but without collecting the specified amount.

However, one of the conditions stipulated by Mr. X was that Barry must turn both keys at 1 o' clock in the afternoon in order to open the door.

There is a wall clock enabling Barry to keep track of time, and also an intercom so that Barry can communicate with Mr. X.

Are there any avenues whereby Barry can collect the amount under reference?

[quote=Are there any avenues whereby Barry can collect the amount under reference?[/quote]

Huh? i dont understand the question  :?

He's asking whether there's a way for Barry to win.  "Avenue" is being used in a non-literal way to mean a course of action, rather than a literal avenue.

Edit:  Though I must admit, the problem is trivial unless Barry's armspam is less than five meters... that is the point, I see.

Two possibilities:  1)  Barry turns the first key at one o'clock one day, and the second at one o'clock the other day.  2)  Barry uses the mechanism of the clock to turn one key, and turns the other key when he sees the first key turning (he would set it up so that the turning mechanism would turn the key at exactly one o'clock).  :P


group theory
abstract algebra
algebra
polynomial
induction
strong induction
superior algebra

Prove that for distinct integers $ m,n$ $ GL(n,\mathbb{R})$ and $ GL(m,\mathbb{R})$ are not isomorphic.

You could count the number of conjugacy classes of solutions to the equation $ x^2\equal{}1$.

I see how it works for $ GL(n,\mathbb{Z})$, but not for $ \mathbb{R}$.

There is one way to avoid the Jordan form (no, I don't mean the [url=http://en.wikipedia.org/wiki/Frobenius_normal_form]RCF[/url]  :P ).

At first, let me give a very rough sketch of the proof :
we let $ G$ be a finite subgroup of $ \mathbf{GL}_n(\mathbb{R})$ such that for any $ A \in G, \ \ A^2 \equal{} I_n,$ and we prove that $ G$ has at most $ 2^n$ elements. Next, we let $ f$ be an injective morphism from $ \mathbf{GL}_n(\mathbb{R})$ to $ \mathbf{GL}_m(\mathbb{R}),$ and show that $ n \le m.$ From here, the conclusion follows.

Firstly, it can easily be checked that such a group $ G$ is abelian.
Besides, the polynomial $ X^2 \minus{} 1 \equal{} (X \minus{} 1)(X \plus{} 1)$ annihilates any $ A \in G,$ so that $ A$ is diagonalizable over $ \mathbb{R}.$
Hence, the elements of $ G$ are in fact simultaneously diagonalizable (hint : strong induction over $ n$).
Thus, we may find $ P \in \mathbf{GL}_n(\mathbb{R})$ such that for any $ A, \ \ A \equal{} P \text{diag}(\pm 1, \ldots, \pm 1) P^{ \minus{} 1}$ and so $ |G| \le 2^n.$

Let $ G \equal{} \{\text{diag} (a_1, \ldots, a_n) \ / \ \forall i, \ a_i \in \{ \minus{} 1, 1\} \}$
Then $ |G| \equal{} 2^n$ and $ G$ is a subgroup of $ \mathbf{GL}_n(\mathbb{R})$ such that for any $ A \in G, \ A^2 \equal{} I_n.$
Hence $ f(G)$ is a subgroup of $ \mathbf{GL}_n(\mathbb{R})$ with $ 2^n$ elements since $ f$ is injective.
If $ B \in f(G), \ B \equal{} f(A), \ A \in G,$ so that $ B^2 \equal{} f^2(A) \equal{} f(A^2) \equal{} f(I_n) \equal{} I_m$ and so $ |f(G)| \le 2^m.$
Therefore $ n \le m.$

If you follow the proof, you'll see that $ \mathbb{R}$ is in fact not so relevant to the above proof, which works for any field whose characteristic is not 2, and gives us the conclusion :
\[ \mathbf{GL}_m(\mathbb{K}) \cong \mathbf{GL}_n(\mathbb{K}) \iff m \equal{} n.
\]


How do you edit a poll without having the old choices still remain in the poll? Do you just have to make a new post?

I'm sure you've all had bad experiences with this...

[quote="7h3.D3m0n.117"][quote="math92"]I was told to ask the moderator of the forum to do it for me.[/quote]

yeah same here

but why won't the admins let you edit your own?[/quote]
Perhaps because it could create odd results and people might have seemingly voted for something they didn't mean to etc.


Angle $ B$ of a triangle $ ABC$ is $ 120^\circ$. $ M$~--- midpoint of the side $ AC$. Points  $ K$ and $ L$ on the sides
$ AB$ and $ BC$ choosed that  $ KL\parallel AC$. Prove that $ MK\plus{}ML\geqslant MA$.
(S. Berlov)

If $ AK \le AB/2$, than $ MK \plus{} ML > KL \equal{} MA$.
If $ AK > AB/2$, than $ MK \plus{} ML > MK \plus{} AK > MA$.

[quote="csbonny"]why $ ML > AK$ ??[/quote]

อยู่แล้วววววววววว  :D

อสมการห่างได้ใจ  :roll:

[quote="csbonny"]Anybody Any idea[/quote]

Nobody & No idea


Does anyone know how to solve problem #16 on the 2006 AMC 10 A?
If so, would you kindly submit the solution?           :starwars:  :spider:

There's a reason that "Contests" link is there on the top.

http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43

You could probably also find it in the archives of this forum; just go to the bottom and click "2" or "3" or whatever.


quadratics
number theory proposed
number theory

Find all positive integers x,y,z,t such that
$x+y+t^2 = 4xyz $

I did some progress, but it started looking really awkward:

$t^2=4xyz-x-y\iff 4zt^2+1=(4xz-1)(4yz-1)$. We can see that it suffices to find $x,z,t$ s.t. $4xz-1|4zt^2+1\ (*)$, because then $y=\frac {\frac {4zt^2+1}{4xz-1}+1}{4z}$.

$(*)\iff 4xz-1|x+t^2$ (some easy-to-do computations). For quadratic residue arguments we see that $x,z$ must not be perfect squares. That's because for all primes $p|4xz-1$ we have $(\frac {-x}p)=1$, so if we choose $p$ s.t. $p=4s+3$ (there are such divisors of $4xz-1$) we get $(\frac xp)=-1$, so $x$ isn't a perfect square. On the other hand, $(\frac {xz}p)=1$, so $(\frac zp)=-1$, so $z$ can't be a perfect square either.

I know that such $t$ don't exist for any non-squares $x,z$ (take, for example, $x=z=2$). Is it possible that there's some mistake in the statement of the problem?

Here's how incredibly stupid I am: I went through all that trouble to get $4xz-1|x+t^2$, when from the relation we get $x+t^2=y(4xz-1)$. Sheesh!..

[quote="arthur"]that's indeed very comic, and it's not the first time for you, grobber.... :D[/quote]
I think you are true

we could make an entire topic entitled "grobber's mistakes". on the other hand, with 2200 posts, all can't be correct, and no one's perfect.

Thanks for the vote of confidence, although note that what I did above isn't exactly a mistake, just a long and useless way of achieving [b]nothing[/b] new :D:D.

Dont make this topic be a market or somewhere like that

Let x= $ A^2 p_1 p_2 ... p_k $, 4xz-1 = $ B^2 q_1 q_2 ... q_h $, where primes with different indices are different.
From our equation we get $ { x \choose q_i } = { -1 \choose q_i } $ ( the Legendre symbol),
So $\prod  { x \choose q_i } =  \prod { -1 \choose q_i } $
or  $\prod  { p_j \choose q_i } =  \prod { -1 \choose q_i } $ (*)
Set $ m= \sum \frac { p_j-1 } {2} $
      $  n=  \sum \frac { q_i-1 } {2} $
      $  l=  \prod q_i $
From (*) and the quadratic reciprocy law we get
$\prod  { q_i \choose p_j } = ( -1 ) ^{ m n + n}$.
On the other hand, $ { l \choose p_j } = { -1 \choose p_j} $
thus $\prod { q_i \choose p_j } = ( -1 ) ^{ m} $
Thus mn+m+n is even and hence m,n must be even.
Let r be the number of i such that $q_i$ = -1 mod 4, and s be the number of i such that $q_i$ = 1 mod 4.
Since n is even we get $\frac {( -r + s ) - ( r + s )} {2} $ is even, so r is even, but this is impossible since $\prod q_j $ = -1 mod 4.
Thus the original equation doesn't have positive integer roots.


is Canada/USA MATHCAMP better

or

AwesomeMath Summer Program(AMSP) at Texas better?

Also What are some other mathcamps that will improve my AMC/AIME/USAMO scores?

[quote="kimby_102"]I think they are about the same...[/quote]

From what I read, and from my experience (I went to Mathcamp), I get the opposite impression.  AwesomeMath seems very focused on problem solving and contest math.  I think its a brand new program, so its probably sort of hard to tell how good it will really be, but from the staff and description, it looked like the program would be rigorous and interesting.

Mathcamp, on the other hand, has a broader focus.  Problem solving at Mathcamp is significant, but there is also a whole range of other mathematical activities going on.  Classes are offered on a mileu of different interesting topics, many completely unrelated to contest problem solving.

In my personal opinion, the broader approach to learning math seems better, but if improving AMC/AIME/USAMO scores is your priority, AwesomeMath might be more helpful.

Yeah, I'm pretty sure AwesomeMath wins if you're looking to improve your AMC/AIME/USAMO scores. But as to whether that's what your goal should be, I can't say.

You already started a thread for this which was moved to Other US Competitions. PLEASE DO NOT MAKE ANOTHER ONE.