[quote="turboturtle"]There is a parallelogram ABCD.
Angle B is 82 degrees
There is also an altitude to side CD, which is a segment AE, with E being on CD.
There is also an altitude to side AD, which is a segment CF, with F on segment AD.
Find angle CFE.

The angles of the parallelogram are laid out as:
A                  B

D                   C
(just for clarification)
--I apologize if this turns out hard to follow, i really don't know how to do images.-- :blush:[/quote]
This angle is impossible to find out if we do not know the side lengths.

can you clarify? What do you mean by "side lengths"

he wants you to give some lengths like say $\overline{AB}=5$ or w/e..but @Quevvy I think it can be done w/o sidelengths like i was thinking parallel lines and something if u play around with it u might get it..btw is the answer

[hide]$\boxed{8^\circ}$[/hide]

I don't know. Can you post how you got your answer? I somehow got 64, my friend got 74, and his friend got 41. thus, its kind of strange.

I have attached a picture where everything follows the requirements. Obviously, $\angle{CFE}$ is different in both[/hide]

[quote="turboturtle"]Yes, I get the feeling that this problem is impossible to get a single value, and I also think that there is a range of values. can you post how you got your solution?[/quote]
Those were just examples that the angles can be different.
The range is $0^{\circ}$ to $82^{\circ}$ (Can easily be proved with triangles...)

ah. so its 0<x<82, by the fact that all triangles have 180 degrees.

[quote="Padmavathi"]Well..here's how I did it...seems pretty possible without knowing the length of the sides..[/quote]
So then what did you get?

[quote="Padmavathi"][hide="DUH!!"] 8 degrees[/hide]
Wasn't that obvious? :D[/quote]
1.) How'd you get that?
2.) What if the size of the parallelogram changed? (like in my picture)

Well...even if the parallelogram changes in size, the angles remain the same right? So you already know that in a triangle called DEF, the angles are 82, 90 and unknown. So Angles of a triangle add up to 180 degrees so find unknown.(See the figure)


A group of $ 75$ students in your school has just returned from a class trip to Washington, DC.
Twenty students visited the Washington Monument, thirty-five students visited the White House, and fifteen students visited the Kennedy Center. Ten students visited the White House and Kennedy Center, eight students visited the Washington Monument and White House, six students visited the Washington Monument and Kennedy Center, and four students visited all three sites. How many students visited none of the three sites?

$ 75 \equal{} 20\plus{}35\plus{}15\minus{}10\minus{}8\minus{}6\plus{}4\plus{}n$ where $ n$ is the students that visited none.
$ 75 \equal{} 50 \plus{} n$
$ n \equal{} \boxed{25students}$


A series circle contains a resistor with R=24 :cap: , an inductor with L=2H, a capacitor with C=0.005F, and a 12-V battery. The initial charge is Q=0.001C and the initial current is 0.

a.Find the charge and current at time t.
b.Graph the charge and current functions.

It's no real problem if it's a homework, however it's a problem to have it in size 18. Please use normal text in the future ;) :D

From the principle of conservation of energy:

\[
\begin{array}{l}
 U = \frac{1}{2}LI^2  + \frac{1}{2}\frac{{q^2 }}{C} \\
 \frac{{dU}}{{dt}} =  - I^2 R + \varepsilon V = LI\frac{{dI}}{{dt}} + \frac{q}{C}\frac{{dq}}{{dt}} \\

get \\ the \\ equation \\ of \\ circuit \\

 L\frac{{d^2 q}}{{dt^2 }} + R\frac{{dq}}{{dt}} + \frac{q}{C} = \varepsilon  \\
and \\  particular \\  solution \\
 q = \varepsilon C \\
If \\ for \\ the \\ homogeneous \\ equation \\
 u = \frac{{R \pm \sqrt {R^2  - \frac{{4L}}{C}} }}{{2L}} \\
then \\ general \\  solution \\
 q = Ae^{u_1 t}  + Be^{u_2 t}  + \varepsilon C \\
 \end{array}
\]
Put the number, find A and B with t = 0 , and will get the solution of the question


What is the probability that a randomly selected three-digit number
is divisible by 11? Express your answer as a common fraction.

The smallest three-digit number divisible by $ 11$ is $ 110$ and the largest is $ 990$.  Thus, the total number of them is $ 81$, so the desired probability is $ \frac{81}{900}\equal{}\frac{9}{100}$.


Prove that the number of 5-tuples a,b,c,d,e such that:

abcde = 5(abcd + bcde + cdea +deab + eabc) is odd.

Hello,
    Just see this link for solutions to recent INMO Papers.
http://www.isid.ac.in/~rbb/rmoqpapers.html
Sathej


inequalities
search
trigonometry
inequalities unsolved

Prove that :
 $ sin x + \frac{1}{2}sin 2x + \frac{1}{3}sin 3x + ... + \frac{1}{n}sin nx >0 $ with every $ x \in (0,\pi) $

 $ 1+ cos x + \frac{1}{2}cos 2x + \frac{1}{3}cos 3x + ... + \frac{1}{n}cos nx >0 $ with every $ x $

The sin inequality is called Jackson inequality and appeared at least once on forum - try to search for it

Exactly, I proved first one.
Though I have a bad presentiment that my post was lost. :?

Myth: it's not lost

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Jackson&t=20812

 :D

I am happy! :D

I am sure the same technique works for second inequality too.

[quote]I am sure the same technique works for second inequality too.[/quote]
There seems to be a little problem when we try to apply the same technique to the second inequality.

In [b]Myth[/b]'s proof of the first inequality (with sin), by appropriately choosing $x$, we arrive at

$\cos\frac{x}{2}<\cos\left(n+\frac{1}{2}\right)x$ and $\sin\frac{x}{2}\leq\sin\left(n+\frac{1}{2}\right)x$. (*)

By the assumption $x\in[0,\pi]$, we have $\cos\frac{x}{2}\geq 0$ and $\sin\frac{x}{2}\geq 0$. Therefore, all of the 4 numbers in (*) are $\geq 0$, so that we are allowed to square both inequalities, add them up to obtain a contradiction.

If we do the same with the inequality with cos, we will end up with

$\sin\frac{x}{2}>\sin\left(n+\frac{1}{2}\right)x$ and $\cos\frac{x}{2}\geq\cos\left(n+\frac{1}{2}\right)x$. (**)

Since the inequality signs are reversed, we are not sure that the two other numbers (in the right-hand sides) are $\geq 0$. There is actually no contradiction that can be derived from (**).

Could someone help me find a way to get around, please ? Thanks.

It really works for $cos$. If $S(x)<0$ for some $x\in(0,\pi)$ then there is $x\in(0,\pi)$ s.t. $S(x)=0$ and $S'(x)<0$.

I think it can be applied Abel's inequality.


To input a cube root in FTW (or anywhere else in $ \text{\LaTeX}$), put the 3 in brackets after the \sqrt operator. Same with higher-index roots. So here, this would be -3\sqrt[3]{154}: $ -3\sqrt[3]{154}$

Hmmm, the question changed to:

If $ \sqrt[3]{6}\cdot\sqrt[3]{-21}\cdot\sqrt[3]{33} $ is in simplified form $a\sqrt[3]{b}$, where $b$ is a positive integer, then what is $a+b$?

I think there should be different thread for the new question to avoid confusion.

[quote="gauss1181"]$ 6\equal{}2*3$
$ 21\equal{}3*7$
$ 33\equal{}3*11$

Product is simply $ \minus{}3\sqrt[3]{2*7*11}\equal{}\boxed{\minus{}3\sqrt[3]{154}}$.[/quote]
To answer JSGandora,
[hide]-3+154=151[/hide]


Does anybody else think this test was a lot easier than those in the past? :huh:

Lol, my brother got 6 to 8 questions wrong and guessed 2 right (the wedge one and the 42 one)

Anyone have the actual test. I cant remeber my answers.

It was much easier. They made it so because they made it so you could no longer use calculators.

Next year is gonna be liek an AMC 10...oh boy

BTW, the 1990 AJHSME was ridiculously easy, even easier than this one.

[quote="AIME_is_hard"]Easiest AMC 8 out of the three I've taken officially, in my opinion.

Number 24 was the hardest, but all it required was for you to list out all the possibilities.

Number 25 was very easy...answer was 42, that was funny. When I was checking, I thought I made a mistake by putting 42  :lol:[/quote]

Did you remember what the question#25 was?.. I did not remember :)

The winning art design thing with the shaded areas and all the circles with increasing radii by 2

I can post the entire question (to the best of my memory) if it is allowed.

Has anyone gotten their results?

I haven't got my test back and I took form b.

Yep, I officially got a 24.  Because, evidently, 8+9=24...

[quote="numbermaster"]Discussion for the AMC8-[/quote]
i am on the phail boat with a 16. holy crap i suck. crap i make mistakes in addition. crap i suck. crap.

I thought you can get more than 16/25 right on AMC tests...

i can but i made a crazy amount of stupid mistakes! crap. x-( my mom thinks that you can't do math without a tutor to help you

There is a copy of the 2008 AMC 8 on agmath.com if anyone's interested.

23.

Phail. :wallbash_red:

Phail. :wallbash_red:

Phail. :wallbash_red:

Yeah same athunder.  I thought that 8+9=24 for some stupid reason.

yeah it was pretty easy
i noticed that 1 - 15 were a bit harder than usual (which still makes them easy)
and that 16 - 25 were A LOT easier

i got 24 ... i missed 24 since i made a 10 * 10 grid ... and i caught part of my mistake so i got E

haha 24 on the 24th AMC 8 by missing number 24


Let $ a$ and $ b$ be nonnegative real numbers. Prove that
$ \sqrt {1 \plus{} a^2} \plus{} \sqrt {1 \plus{} b^2} \plus{} \sqrt {(1 \minus{} a)^2 \plus{} (1 \minus{} b)^2} \plus{} \left( 1 \plus{} \sqrt {2}\right) ab\ge 1 \plus{} \sqrt {5}.$
When does equality occur?

Remark. The following slightly weaker inequality was proved in book "Old and New Inequalities" written by Titu Andreescu, Vasile Cirtoaje (Vasc), Gabriel Dospinescu (Harazi) and Mircea Lascu, published by GIL, 2004

If $ a,b \in [0,1],$ then
$ \sqrt{1\plus{}a^2}\plus{}\sqrt{1\plus{}b^2}\plus{}\sqrt{(1\minus{}a)^2\plus{}(1\minus{}b)^2} \ge \left( 1\plus{}\sqrt{5}\right) (1\minus{}ab).$

[quote="can_hang2007"]If $ a,b \in [0,1],$ then
$ \sqrt {1 \plus{} a^2} \plus{} \sqrt {1 \plus{} b^2} \plus{} \sqrt {(1 \minus{} a)^2 \plus{} (1 \minus{} b)^2} \ge \left( 1 \plus{} \sqrt {5}\right) (1 \minus{} ab).$[/quote]
This inequality is [b]Problem 1.15[/b], page 28 of new book:

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://canhang2007.wordpress.com/2009/12/08/new-book-inequalities-with-beautiful-solutions-v-cirtoaje-v-q-b-can-t-q-anh/][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.


The sequence $a_0,a_1,a_2,...$ is defined as follows:
$a_0=2$, $a_{k+1}=2a_{k}^2-1$ for $k\geq 0$.
Prove that if an odd prime $p$ divides $a_n$ then $2^{n+3}$ divides $p^2-1$.
Actually, if $p$ had the form $12k+1 $ or $12k-1$, it becomes easy.
 Welcome a solution for this "pretty problem". ;)

it was posted here: http://www.mathlinks.ro/viewtopic.php?t=5174 (BTW it was also in some round of mathlinks contest, 3rd edition. there the problem was to find all n s.t. 2003|a(n), which becomes trivial with this result)

i think the following (stronger) result is true, but I cant prove it:
if p | A(n), and (3/p)L=-1, then 2^(n+2) | p + 1. (conversely, it's easy to prove that if (3/p)L=1, then 2^(n+2) | p - 1).

I don't feel pleasant about solution in that topic.Because it must have an elmentary solution.I am only take care of the case (3/p)=-1 but I don't understand in that topic!!! :(  However, your last idea is the same as me. ;)

i dont think it has an elementary solution ... the solution grobber gave is essentially the same as the official solution. and even if the result i thought about is true, i think it is much stronger than the required condition 2^(n+3) | p^2 - 1.


[b]Problem[/b] (Vu Thanh Tu). Let $ a,b,c$ be positive real numbers such that $ a\plus{}b\plus{}c\equal{}3$ and

$ 0 \le k \le \frac{3}{4}$.Prove that

$ k(a^2 b \plus{} b^2 c \plus{} c^2 a \plus{} abc) \plus{} \frac{{(a \minus{} b)^2 }}{c} \plus{} \frac{{(b \minus{} c)^2 }}{a} \plus{} \frac{{(c \minus{} a)^2 }}{b} \ge 4k$

We only need to prove your ineq with $ k \equal{} \frac {3}{4}$
But it's obviously trues by AM-GM and Cauchy Schwarts :)
PS: Notice that $ 4(a\plus{}b\plus{}c)^2 \ge (a\plus{}b)(4a\plus{}4b\plus{}c)$ with $ a \ge b \ge c$  :)

[quote="nguoivn"]We only need to prove your ineq with $ k \equal{} \frac {3}{4}$
But it's obviously trues by AM-GM and Cauchy Schwarts :)
PS: Notice that $ 4(a \plus{} b \plus{} c)^2 \ge (a \plus{} b)(4a \plus{} 4b \plus{} c)$ with $ a \ge b \ge c$  :)[/quote]
Dear nguoivn.Can you post your solution?  :)


I would want to ask.....the people who might know this.....what are the chances of a regular NMO participant after graduating the math section at the univesity in Bucharest to actually become a university peofessor there?and what would it take becoming one?...could anyone help me....cause I'm in a really big dilema of what I should do after graduating from high-school. :maybe:
Thank you!

Chances are pretty good, as all of the good people in maths decide to do something else in Romania, or just leave the country. So, you will probably be among the fewest students heading in that direction. :) Good luck!

I really wanna thank you for your encouragement cause I was really afraid I wouldn't stand a chance in front of other good math people....and I really wanna stay in the University after graduating it because if not...I really don't think I would like teaching at high-scool....
I'm not really convinced that I wanna make math after graduating high-school because money is little and...life is expensive...so I think it's between passion or money....tugh choice!damn!...
well anyway thank you!


Let $p > 2$ be a prime and let $a,b,c,d$ be integers not divisible
by $p$, such that
\[
\{ra/p\} + \{rb/p\} + \{rc/p\} + \{rd/p\} = 2
\]
for any integer $r$ not divisible by $p$.
Prove that at least two of the numbers $a+b$, $a+c$, $a+d$, $b+c$, $b+d$, $c+d$
are divisible by $p$.
(Note: $\{x\} = x - \lfloor x \rfloor$ denotes the fractional part of $x$.)

So far I have:
[hide]If a = x + yp with 0 < x < p and y integer, we have {ra/p} = {rx/p + ry} = {rx/p}, so WLOG let 0<a,b,c,d<p.  Since all of a,b,c,d are less than p, then we must prove that two of a+b,a+c,a+d,b+c,b+d,c+d are equal to p.  Suppose for some x and y, x + y = p.  {rx/p} + {r(p-x)/p} = {rx/p} + {r - rx/p} = {rx/p} + 1 - {rx/p} = 1.  So if a+b = p, then we have 1 + {rc/p} + {rd/p} = 2 =>(implies) {rc/p} + {rd/p} = 1.  So we only need to show that one of the numbers a+b,a+c,a+d,b+c,b+d,c+d are equal to p.[/hide]
And then I'm stuck.  Any hints?  I'm thinking a proof by contradiction would be best here, but I'm not quite sure how to proceed...

I fiddled with it for awhile, and then I recognized that it was USA 1999/A3.  This is commonly acknowledged as one of the two hardest USAMO problems ever.

Look at kalva, or here:

http://www.mathlinks.ro/Forum/viewtopic.php?t=29927


Hi,
  What are some computer science programs/internships/jobs for a graduating high school senior?

Thanks!

[url=http://public.ca.sandia.gov/casite/employment//students/index.php]Sandia Labs[/url] is one idea. Can vouch for them being a good employer.


Joe sold his dryer for $ \$60$ and made a profit of $ 25\%$ over the estimated value. He also sold his washer for $ \$60$ but had a loss of $ 20\%$ of its estimated value. What is the difference in dollars between the total estimated value for both items and the total sale price for both items?

Call the Estimated Value $ v \plus{} 1$
\[ v_1 \plus{} v_1/4 \equal{} 60
\]
$ v_1 \equal{} 48$.

Let the Estimated Value be $ v_2$
\[ v_2 \minus{} v_2/5 \equal{} 60
\]
$ v \equal{} 75$.

The Estimated Values are $ 123$.

The money sold for is $ 120$.

The difference is $ \boxed{ \minus{} 3}$.

However, when not stated, it is assumed that the second term of the "difference" phrase is subtracted from the first, and the answer is $ \boxed{\$3}$.


how many reel solution the equation :
$ E_{n}: \frac{1}{x}\plus{}\frac{2x}{x^2\minus{}1}\plus{}...\plus{}\frac{2x}{x^2\minus{}n^2}\equal{}a$ have? where n is positive integer and a reel number,
let $ x_{n}$ be the largest root of $ E_{n}$,prove that there exist a reel $ \alpha$ such that the sequence $ (x_{n}\minus{}n.\alpha)$ is convergent


logarithms
probability
number theory
prime numbers
number theory open

Are there infinitely many sums of cosecutive prime numbers (starting with 2) which are power of 10?

For example:

$2+3+5=10$
$2+3+5+7+11+13+17+19+23=100$

I haven't found any other even using computer...
Checked to $10^{15}$... Soon maybe more...

Mm, it seems interesting to me.
But, what is the program you use? Is it your friend who coded it or?

Where did you get this question from?  Somehow I doubt there is an elementary solution to this.

Yes, our friend (my and Bismarck's) has coded it. But, unfortunately we still don't have any idea concerning the problem.
JellyWorld: It was Bismarck who invented this problem. We didn't have any clue is this even possible to solve so he decided to post it here (that's way it is in the open questions). Actually I can second your doubts.

Is there a reason for picking 10?  Why not generalize this conjecture and ask if, for any $p$, there exists an infinite number of $k$ such that there exists some $n$ such that the first $n$ prime numbers sum to $p^k$.

Exactly [b]JellyWorld[/b]!
I was also wondering about that. :)

Maybe it's for coding reasons though...

[quote="jmerry"]Since the density of primes near $n$ is $\frac1{\ln n}$, the probability that $10^k$ is such a sum is $O(\frac1k)$. This indicates that there should be infinitely many such values, since $\sum_k \frac1k$ diverges. Unfortunately, this isn't a proof and it doesn't help in finding them.[/quote]I didn't understand your heuristic reasoning.  I estimated the "probability" that $10^k$ is such a sum as closer to $O(\frac{1}{5^k \sqrt{k}})$.


Find the smallest integer whose partiton in six different integers is such that the sum of any five of them is always a perfect square.

If we drop the "different" condition, the answer is of course 0.  If we take a partition into positive or negative integers, the answer is [hide]$11 = (-14) + (-5) + 2 + 7 + 10 + 11$[/hide]  If we require that they be positive, the answer is [hide]$911 = 11 + 70+127+182+235+286$[/hide]

The most general I can make the latter proposition, replacing 5 with a number $n$ and 6 with $n + 1$ is (I believe) to apply when $n$ is not divisible by 2 or 3, in which case the value we want is [hide]$\frac{6n^4+12n^3+8n^2+3n+1}{6}$[/hide]


Some probability questions about games ask for an "expected score" and they give probabilities for certain outcomes like winning vs. losing, each team equally likely to win a point, or every combination of wins and losses are equally likely in a best of 3 or best of 5 game, etc.

I'm not sure I'm wording this correctly, but what does a problem want you to say when they ask for an "expected score?" Is it the most probable outcome? Or is a average value of all possibilities - maybe a mean that may be impossible to actually score?

For example, if I was asked to find the "expected sum" of rolling two standard 6 sided dice, would my answer be the most likely sum of the two numbers I roll with the dice

OR

an average... taking into account every outcome and factoring them all in proportionally to their individual probabilities?

The most general definition of expected value is $ \sum p_{i}x_{i}$, where $ p_{i}$ is the probability of event $ i$ occurring, and $ x_{i}$ is the payoff for event $ i$. In english, you're basically taking all possible outcomes, multiplying the outcome by its probability of occurring, and then summing everything you get.

[quote="aufha"]an average... taking into account every outcome and factoring them all in proportionally to their individual probabilities?[/quote]

This is correct.  Another way to think of it is as the average of all of the outcomes you would get if you repeated a given random event arbitrarily many times.

i think they have an excerpt about this in AoPS vol 1


Maybe this was asked before ( in that case i apologize) but i'm interested does this book contain any olympiad topics?

Yes, it does contain hard problems and many ideas that you can use to solve Olympiad level problems (USAMO/IMO level). It's not designed specifically as a training book for that purpose, but you can definitely (and you probably should) use it as a material.


linear algebra
matrix
geometry
geometric transformation
homothety
invariant
linear algebra unsolved

(a) Let $A \in \mathcal M_{2}\left( \mathbb R \right)$ such that $\text{rank}\left( A+XY \right) = \text{rank}\left( A+YX \right)$, for all invertible matrices $X,Y \in \mathcal M_{2}\left( \mathbb R \right)$.
Prove that there is $a \in \mathbb R$ such that $A = a I_{2}$.
Can we generalize?

(b) Let $A \in \mathcal M_{n}\left( \mathbb R \right)$ ($n \geq 2$) be a matrix that isn't of the form $a I_{n}, \, a \in \mathbb R$.
Prove that there are $X,Y \in \mathcal M_{n}\left( \mathbb R \right)$ with $X$ invertible such that $\text{rank}\left( A+XY \right) < \text{rank}\left( A+YX \right)$.

[i]Marius Ghergu[/i]

[quote="perfect_radio"](b) Let $A \in \mathcal M_{n}\left( \mathbb R \right)$ ($n \geq 2$) be a matrix that isn't of the form $a I_{n}, \, a \in \mathbb R$.
Prove that there are $X,Y \in \mathcal M_{n}\left( \mathbb R \right)$ with $X$ invertible such that $\text{rank}\left( A+XY \right) = \text{rank}\left( A+YX \right)$.
[/quote]
Either you forgot an asumption, or it's obvious, just take $X=I_{n}$ and no need for "a matrix that isn't of the form $a I_{n}, \, a \in \mathbb R$."

i just wanted to notice that this is a very ugly problem :D .i managed to solve part a) and i have no time for b) now.but for a) the "big method" is just to be pacient enough to choose X and Y and calculate :(

For the a): if $A$ isn't a homothety then we may assume that \[A=\begin{pmatrix}0&a\\1&b\end{pmatrix}\].
We construct invertible $X,Y$ s.t. $det(A+XY)=0$ and $det(A+YX)\not=0$ .
case 1: $b\not=-1$. Let $\beta\notin\{0,-a\}$ and $\alpha=\frac{a+\beta}{2b+2}$. Choose:
\[X=\begin{pmatrix}\alpha&\beta\\0&1\end{pmatrix},Y=\begin{pmatrix}2&0\\0&-1\end{pmatrix}\]
case 2: $b=-1,a\not=0$. Choose:
\[X=\begin{pmatrix}1&-a\\0&1\end{pmatrix},Y=\begin{pmatrix}-1&0\\0&1\end{pmatrix}\]
case 3: $b=-1,a=0$. Choose:
\[X=\begin{pmatrix}1&1\\0&1\end{pmatrix},Y=\begin{pmatrix}1&0\\-1&1\end{pmatrix}\]
For the b): there exists a real $A$-invariant plane; moreover if $A$ isn't a homothety then we can choose this plane $\Pi$ s.t. $A|{\Pi}$ isn't a homothety.
 I don't know how to prove this last claim but I think that it's true.
Thus we may assume that $A$ is a block-matrix \[A=\begin{pmatrix}D&V\\0&W\end{pmatrix}\] where \[D=\begin{pmatrix}0&a\\1&b\end{pmatrix}\].
Let $\lambda$ a real s.t. $W+\lambda{I_{2}}$ is invertible.
Now choose \[X=\begin{pmatrix}X_{1}&0\\0&\lambda{I_{n-2}}\end{pmatrix},Y=\begin{pmatrix}Y_{1}&0\\0&{I_{2}}\end{pmatrix}\]
with $X_{1},Y_{1}$ as the $X,Y$ in part a).

1) In part b) of my last post change $I_{2}$ into $I_{n-2}$.
2) The plane $\Pi$ exists, I met it:
case 1: $A$ has no real eigenvalues.
$\lambda$ and $\bar{\lambda}$ are 2 non real $A$'s eigenvalues; let $v$ be an $\lambda$-eigenvector; then $\Pi=[Re(v),Im(v)]$ works.
case 2: $A$ has at least 2 different real eigenvalues.
It's easy.
case 3: $A$ has only 1 real eigenvalue.
Then we may assume that $A$ is a Jordan block of dimension $\ge2$; thus \[A=\begin{pmatrix}D&V\\0&W\end{pmatrix}\] where  \[D=\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}\] isn't a homothety.

[quote="loup blanc"]For the a): if $A$ isn't a homothety then we may assume that
\[A=\begin{pmatrix}0&a\\ 1&b\end{pmatrix}\]
).[/quote]
I wonder how can we assume
\[A=\begin{pmatrix}0&a\\ 1&b\end{pmatrix}\]
?
Thank you

Hi Zhaobin,
 if $A$ isn't a homothety then there exists $u\in\mathbb{R}^{2}$ s.t. {$u,Au$} is a free system. Choose {$u,Au$} as a new real basis.


calculus
integration
quadratics
algebra
number theory proposed
number theory

Find all possitive integer numbers (x,y,z) :
   $ (x+y)(1+xy) = 2^z $

[quote="Myth"]It was posted and solved some days ago, but I think all these post are lost due to DB crash.[/quote]
Yep,it was posted by pluricomplex(his name:Hung Ha Duy)but I don't know a person who solved it.Myth,please post the solution ;)

I want to give chance to other people. :)

We can set $x+y=2^l,1+xy=2^m\Longleftrightarrow x+y=2^l,xy=2^m-1$,thus $x,y$are the roots of quadratic equation for t,$t^2-2^lt+2^m-1=0$,the discriminant of this equation [b]D[/b] is $4^l-4(2^m-1)=4(4^{l-1}-2^{m+1})$ so $x\in\mathbb{N}$ and $y\in\mathbb{N}\Longrightarrow 4^{l-1}-2^{m+1}=4\cdot n^2$

Ha,I think that I had a solution.In fact, it is quite easy ;) .
Firstly,note that [tex] x,y,z \geq 0 [/tex] .I won't consider the trivial roots [tex] (0,2^a,a)... [/tex]
We have:
[tex] x+y=2^{a} [/tex] and [tex] 1+xy=2^{b} [/tex]. I will assume that [tex] b \geq a \geq 2 [/tex].
We have $ x+1+y+1=2^{a}+2 $ hence one of $ (x+1) $ and $ (y+1) $ has $ 4k+2 $ form. Just assume that is $ x+1 $.
We have: $ (x+1)(y+1)=2^{a}+2^{b}=2^{a}(2^{b-a}+1) $
So $ x+1=2m;y+1=2^{a-1}n $ where $ mn=2^{b-a}+1 $.
[tex] \Rightarrow x+y=2^{a-1}n+2(m-1) \Rightarrow n=1 [/tex]
and [tex] m=2^{b-a}+1 [/tex].
Use [tex] x+y=2^{a} [/tex] again to have [tex] b=2a-2 [/tex] and then [tex] x-y=2 [/tex].
So [tex] x=2^{a-1}+1;y=2^{a-1}-1 [/tex] and [tex] z=3a-2 [/tex] :)


Subiectele de la concursul A.Myller desfasurat la Iasi 12-13 martie . Subiecte frumoase (cel putin la a saptea si la a opta) :lol:

Se gasesc pe [url]http://www.epsilon.ro/[/url], de exemplu.

And who will translate it? Though I can understand about half of the text...

Ma scuzati am uitat sa atasez pdf.-ul :blush:

If someone will translate in english the problems and post on college playground
analysis or algebra proposed problem

Clasa a XI-a
problem 1
problem 2

Clasa a XII-a
problem 1
problem 2
problem 3
problem 4

Thank you

11th grade
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=187858#187858]Problem 1[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=187859#187859]Problem 2[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=187846#p187846]Problem 3[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30327]Problem 4[/url]

12th grade
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30329]Problem 1[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30330]Problem 2[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=187851#187851]Problem 3[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30334]Problem 4[/url]


geometry
circumcircle
parallelogram
trapezoid
incenter
cyclic quadrilateral
perpendicular bisector

Starting from a given cyclic quadrilateral $\mathcal{Q}_0$, a sequence of quadrilaterals is constructed so that $\mathcal{Q}_{k + 1}$ is the circumscribed quadrilateral of $\mathcal{Q}_k$ for $k = 0,1,\dots$. The sequence terminates when a quadrilateral is reached that is not cyclic. (The circumscribed quadrilateral of a cylic quadrilateral $ABCD$ has sides that are tangent to the circumcircle of $ABCD$ at $A$, $B$, $C$ and $D$.) Prove that the sequence always terminates, except when $\mathcal{Q}_0$ is a square.

Assume that the sequence does not terminate. Let $Q_0 \equiv A_0B_0C_0D_0$ be the initial cyclic quadrilateral with a circumcircle $\omega_0$. Let it be the contact quadrilateral of a tangential quadrilateral $Q_1 \equiv A_1B_1C_1D_1$. Since $Q_1$ is also cyclic with a circumcircle $\omega_1$, it is bicentric, which means that its contact quadrilateral $Q_0$ has perpendicular diagonals. Most simply, this follows from inversion of the tangential quadrilateral $Q_1$ in its incircle, identical the circuncircle $\omega_0$ of its contact quadrilateral $Q_0$. The vertices $A_1, B_1, C_1, D_1$ are carried into the Varignon parallelogram of the quadrilateral $Q_0$ inscribed in the inverted circumcircle $\omega_1'$ of the quadrilateral $Q_1$, i.e., into a rectangle. The diagonals of $Q_0$ are perpendicular to each other, because they are parallel to the pairs of the opposite sides of this rectangle.

Let the bicentric quadrilateral $Q_1$ be the contact quadrilateral of another tangential quadrilateral $Q_2 \equiv A_2B_2C_2D_2$. Assuming that $Q_2$ is also cyclic with a circumcircle $\omega_2$, i.e., bicentric, its contact quadrilateral $Q_1$ must have perpendicular diagonals. But the only type of a tangential quadrilateral with perpendicular diagonals is a kite. This follows from the fact that one diagonal of a tangential quadrilateral $Q_1$ passes through the pole of the other diagonal with respect to its incircle $\omega_0$ and the other way around. Since the kite $Q_1$ is cyclic, it has 2 opposite right angles. Its tangential quadrilateral $Q_2$ is an isosceles trapezoid, which is automatically also cyclic.

So, the quadrilateral $Q_2$ is a tangential isosceles trapezoid. Let $A_2B_2 \parallel C_2D_2$ be its parallel bases and $M_2, N_2$ the midpoints of its equal shoulder sides $B_2C_2 = D_2A_2$. Since its midline $M_2N_2 = \dfrac{A_2B_2 + C_2D_2}{2}$ is equal to the average of the bases and since the sum of the bases $A_2B_2 + C_2D_2$ is equal to the sum of the shoulder sides $B_2C_2 + D_2A_2 = 2B_2C_2 = 2D_2A_2$ ($Q_2$ being tangential), the midline $M_2N_2 = B_2C_2 = D_2A_2$ is equal to the shoulder sides. The incenter $O_1$ of its incircle $\omega_1$ is the midpoint of the midline $M_2N_2$. Consequently, the triangle $\triangle B_2O_1C_2$ has the median $O_1M_2 = \dfrac{M_2N_2}{2} = \dfrac{B_2C_2}{2}$ equal to half the opposite side, i.e., the angle $\angle B_2O_1C_2 = 90^\circ$ is right and similarly, the angle $\angle A_2O_1D_2 = 90^\circ$ is also right. The diagonals $A_2C_2, B_2D_2$ intersect at a point $X_2$ on the common perpendicular bisector of the bases $A_2B_2, C_2D_2$ and the midline $M_2N_2$. The circumcircles $(M_2), (N_2)$ of the right angle triangles $\triangle B_2O_1C_2 \cong \triangle A_2O_1D_2$ have this perpendicular bisector as their single common internal tangent, with the tangency point $O_1$. Either $A_2B_2 \neq C_2D_2$, the diagonal intersection $X_2$ lies outside of both the circles $(M_2), (N_2)$, which implies that the angle $\angle B_2X_2C_2 < \angle B_2O_2C_2 = 90^\circ$, or $A_2B_2 = C_2D_2 = M_2N_2 = B_2C_2 = D_2A_2$,  the diagonal intersection $X_2$ is identical with the incenter $O_1$, which implies $\angle B_2X_2C_2 = \angle B_2O_2C_2 = 90^\circ$ and $A_2B_2C_2D_2$ is a square.

Let $Q_2$ be the contact quadrilateral of another tangential quadrilateral $Q_3 \equiv A_3B_3C_3D_3$. Assuming that $Q_3$ is also cyclic with a circumcircle $\omega_3$, i.e., bicentric, its contact quadrilateral $Q_2$ must again have perpendicular diagonals. But we just demonstrated that this is possible only if $Q_2$ is a square. Then, of course, its contact quadrilateral $Q_1$ is also a square and the contact quadrilateral $Q_0$ of the square $Q_1$ is also a square.


Set up two equations in two unknowns:
$ 2x\plus{}2y\equal{}45$ and $ 7x\plus{}5y\equal{}97$
Express $ y\equal{}22.5\minus{}x$ and plug into the second equation to get $ 7x\plus{}112.5\minus{}5x\equal{}2x\plus{}112.5\equal{}97\Rightarrow 2x\equal{}\minus{}15.5\Rightarrow \boxed{3x\equal{}\minus{}23.25}$

But it says the answer is 18 for some reason...

huh i got -23.25 too...

but i remember the answer to this one was integer... (perhaps 18)

we should report this prob...

[quote="GameBot"]Given the following two equations, what is the value of the three pentagons?

[asy]for(int i = 0; i<5; ++i)
{
	draw(dir(18+72i)--dir(18+72(i+1)));
}
for(int i = 0; i<5; ++i)
{
	draw(((2,0)+dir(18+72i))--((2,0)+dir(18+72(i+1))));
}

label("+",(3.5,0));

draw(circle((5,0),1));
draw(circle((7.5,0),1));

label("$\LARGE{=45}$",(10,0));[/asy]

[asy]for(int i = 0; i<5; ++i)
{
	draw(dir(18+72i)--dir(18+72(i+1)));
}

for(int j = 0; j<6; ++j){
for(int i = 0; i<5; ++i)
{
	draw((2*dir(60j)+dir(18+72i))--((2*dir(60j)+dir(18+72(i+1)))));
}
}

label("+",(3.5,0));

draw(circle((7.5,0),1));

for(int j = 0; j<4; ++j){
	draw(circle((7.5,0)+2.5*dir(90j),1));
}

label("$\LARGE{=97}$",(12.5,0));[/asy][/quote]

I think I know why...

On FTW there are three circles for the equation, but here there are 2 circles for the first equation! :mad:  :o

Generalmotors is correct. If you solve the equation, 2x+3y=45, and 7x+5y=97. You get x(pentagon)=6, and y(circle)=11. Thus 3 pentagons is 18.


a,b,c,x,y,z are positifs numbers , prove that :
[img]http://www.gnux.be/latex/data/e1640c58a7494eb6608ec130968edccd.png[/img]

[quote="Altheman"]Then apply weighted jensen's on $ f(x) \equal{} \frac {1}{\sqrt {x}}$. Since $ f''(x) \equal{} \frac {3}{4*x^{2.5}} > 0$ so $ f$ is convex. With weights $ \frac {a}{a \plus{} b \plus{} c}$, etc to get:
\[ \sum \frac {a}{a \plus{} b \plus{} c}*\frac {1}{\sqrt {x}}\ge \frac {\sum \frac {a}{a \plus{} b \plus{} c}}{\sqrt {\frac {x \plus{} y \plus{} z}{3}}}
\]
[/quote]
Maybe
\[ \sum \frac {a}{a \plus{} b \plus{} c}*\frac {1}{\sqrt {x}}\ge \frac {1}{\sqrt {\sum\frac {xa}{ a\plus{}b\plus{}c}}}
\]
 :?:

[quote="Altheman"]Then apply weighted jensen's on $ f(x) \equal{} \frac {1}{\sqrt {x}}$. Since $ f''(x) \equal{} \frac {3}{4*x^{2.5}} > 0$ so $ f$ is convex. With weights $ \frac {a}{a \plus{} b \plus{} c}$, etc to get:
\[ \sum \frac {a}{a \plus{} b \plus{} c}*\frac {1}{\sqrt {x}}\ge \frac {\sum \frac {a}{a \plus{} b \plus{} c}}{\sqrt {\frac {x \plus{} y \plus{} z}{3}}} \equal{} \frac {\sqrt {3}}{\sqrt {x \plus{} y \plus{} z}}
\]

\[ \sum \frac {a}{\sqrt {3x}}\ge\frac {a \plus{} b \plus{} c}{\sqrt {x \plus{} y \plus{} z}}
\]
as desired.[/quote]
yes arqady
but Mr Atheman can you explain more the second line ?


So, how is that possible without knowing how many divisors 3n has?

I looked it up and 3n should have 30 positive divisors.  Here is the problem straight from the AHSME:

If n is a positive integer such that 2n has 28 positive divisors and 3n has 30 positive divisors, then how many positive divisors does 6n have?

I've done this one before so I ll leave the solution to someone who hasn't.

[hide]28=7*4
Therefore, 2n=2^6*3^3, so n=2^5*3^3

30=6*5
Therefore, 3n=2^5*3^4, so n=2^5*3^3
Both end up with the same n.  Therefore, 6n=2^6*3^4.  There are 7*5=35 total factors :D[/hide]

Oops.

The numbers must slipped off from the question because it's Christmas.

Edited! :D

You need to know that if there are x factors, the number can be

(m^(x/y - 1))(n^(y - 1)).

For example, a number that has 21 factors can be

(m^(7 - 1))(n^(3 - 1))
(m^6)(n^2)

Treething, that's not that good explanation.

[quote]If n is a positive integer such that 2n has 28 positive divisors and 3n has 30 positive divisors, then how many positive divisors does 6n have? [/quote]

The n is expressed as $2^{e_1}3^{e_2}5^{e_3}...$

For 2n, finding the factors way, would be like this: $2^{e_1 + 2}3^{e_2 + 1}5^{e_3+1}....$

For 3n, finding the factors way, would be like this: $2^{e_1+1}3^{e_2+2}5^{e_3+1}....$

Since 2 and 3 are important, we can set the rest of .... and 5^ as Q.

Then:

$(e_1+2)(e_2+1)Q = 28$
$(e_2+1)(e_1+2)Q = 30$

See if you can work from here.

Silverfalcon, don't you mean that

$2n = 2^{e_1+1}3^{e_2}5^{e_3}...$

with number of factors

$(e_1+2)(e_2+1)(e_3+1)...$?

Well, I was just expressing the n in prime factorization, not necessarily expressing the number of factors.

 :)

But thats my point exactly, 2n in prime factorization would be $\displaystyle 2n = 2^{e_1+1}3^{e_2}5^{e_3}... $


calculus
integration
geometry
3D geometry
Support
analytic geometry
number theory

This is a shaky argument which doesn't even convince me :D

Number the cubes $c_{i,j,k}$, with $(i,j,k)\in\overline{1,p}\times\overline{1,q}\times\overline{1,r}$. Since $p,q,r$ are pairwisely coprime, there are no lattice points on the planbe except for the three vertices of the parallelepiped defining the plane. This means that no two cubes $c_{i,j,k},c_{i+1,j+1,k+1}$ are cut by the plane at the same time.

Now, at each step, take a minimal cube $c_{i,j,k}$ of the set $P$, which initially consists of those cubes cut by the plane. By minimal cube $c_{i,j,k}$ I mean that $c_{u,v,t}$ with $u<i,v<j,t<k$ does not belong to $P$. Take such a minimal cube $c_{i,j,k}$ out of $P$, and, instead, put $c_{i+1,j+1,k+1}$ in $P$ (so $P$ has a constant number of cubes).

If we repeat this until it's no longer possible, I think it's pretty clear that the final set $P$ will consist of precisely those cubes $c_{i,j,k}$ for which $i=p$ or $j=q$ or $k=r$, and there are $pqr-(p-1)(q-1)(r-1)$ such cubes (which matches my previous guess).

Is it really easy and I'm just missing something? It's been frustrating me for about an hour now, because it seems so clear intuitively, and yet I couldn't find any rigorous and truly convincing arguments which don't get ugly (like using the equation of the plane and stuff like that, which I didn't want to do :))..

[quote="Myth"]I came to the same answer.[/quote]

Maybe I am wrong but I think the problem had a different answer.  :lol:

What was that answer, if you don't mind? My "solution" may very well be wrong, so I, for one, wouldn't be too surprised :).

[quote="orl"]A rectangular solid S of integral dimensions p x q x r, where p, q, and r are relatively prime in pairs, is cut into pqr unit cubes by planes parallel to its faces. How many of these unit cubes would be cut if a corner X of S were cut off by a plane going through three neighbouring vertices A, B, and C ? Determine the number in terms of p,q, and r.[/quote]
[quote="groober"]My instinct says pq+qr+rp-(p+q+r)+1, but, then again, that's just my instinct.. :D I have no proof to support this, so it might very well be wrong.[/quote]
[quote="Myth"]I came to the same answer.[/quote]
[quote="orl"]Maybe I am wrong but I think the problem had a different answer. :lol: [/quote]

Your mathematical instincts are right except that you forgot that you are dealing with only partial faces. The number is not the number of lattice points on the three faces cut by the tilted plane. The number is the number of lattice points on the triangular portions of those three faces that are on the same size of the tilted plane as vertex X. The p-by-q face has (p+1)(q+1) lattice points. Of them ((p+1)(q+1)-2)/2 = (pq+p+q-1)/2 points are on the proper side of the plane (two of them (p,0,0) and (0,q,0) are on the plane and we do not count them). The union of the three partial faces gives (pq+p+q-1)/2 + (pr+p+r-1)/2 + (qr+q+r-1)/2 - (p+q+r) + 1, which simplifies to (pq+pr+qr-1)/2.

I imagined a 2-by-3-by-5 solid and counted that the answer for it was 15 cubes. ((2)(3)+(2)(5)+(3)(5)-1)/2 = 15.

For a proof, let X=(0,0,0) and have the tilted plane P pass through (p,0,0), (0,q,0), and (0,0,r). Let L be the set of lattice points on the surface of solid S that are on the (0,0,0) side of plane P. As grobber said earlier, since p, q, and r are relatively prime, the only lattice points of solid S that are on P are (p,0,0), (0,q,0), and (0,0,r): do not count these three points as part of set L.

From any of those lattice points in L, create a line segment by proceeding through solid S in the (1,1,1) direction and ending at the opposite surface of S. That opposite surface is a face of S that is not cut by plane P. So the line segment has endpoints on opposite sides of plane P. Therefore, the line segment intercepts plane P. This point of interception is in a unit cube and plane P slices that unit cube. Of the line segments creates this way, only one such segment passes through the interior of any given cube. Thus, for each lattice point, we have a unit cube that is cut by plane P. This gives that |L| is less than or equal to the number of cut cubes.

Consider a unit cube in solid S that is cut by plane P. Take the corner of that cube that has the smallest coordinates and draw a line through it going in the (1,1,1) and (-1,-1,-1) directions. The intersection of that line and the solid S is a line segment. The line segment might pass through the interior of several unit cubes in S, but since that line segment intersects plane P only inside the chosen cube, only that cube is cut by plane P. The endpoints of that line segment are lattice points on the surface of solid S, one endpoint on each side of plane P. The number of endpoints on the (0,0,0) side of plane P cannot exceed |L|. So the number of cut cubes is less than or equal to |L|.

Therefore, the number of cut cubes is |L| = (pq + pr + qr - 1)/2.

Erin Schram

That formula doesn't work for a $1\times r\times q$ parallelpiped, where all the cubes are cut, for example... Maybe I am simply misunderstanding the problem.

[quote="grobber"]That formula doesn't work for a $1\times r\times q$ parallelpiped, where all the cubes are cut[/quote]
You are wrong.

Indeed, I forgot that we no need whole surfaces. I revised my calculations and obtained $\frac{pq+qr+rp-1}{2}$.

Hope we finally have right answer.

I think I'm beginning to see where I went wrong.. I can also modify my proof to make it correct (I have to move the cubes initially cut by the plane towards $(0,0,0)$ instead of upwards). Sorry for the stupid things which have been said :). This is a very nice proof, Erin.


calculus
integration
number theory proposed
number theory

Prove that the equation \[(x^{2}-2)^{2}+(y^{2}-2)^{2}= (z^{2}-2)^{2}+2\]
doesn't have any integral solutions.

[quote="Beginner"]Prove that the equation
\[(x^{2}-2)^{2}+(y^{2}-2)^{2}= (z^{2}-2)^{2}+2 \]
doesn't have any integral solutions.[/quote]

Mod 3, you have $(x^{2}-2)^{2}=1$, so LHS is $2$ mod $3$ and RHS is $0$ mod 3


algebra
polynomial
number theory theorems
number theory

[quote="Andrei"]Can anyone show me how to prove that for every prime $p$ there exist a number $q$ that has modulo order $p-1$? Or prove that for every prime $p$ there is a primitive root.[/quote]

The non-zero elements of the field $Z/pZ$ constitute a multiplicative group. Let $a \neq 0$ be an element of this group. The order of $a$ is a divisor of the order of the group, namely of $p-1$. Thus for every $k$ dividing this order, the polynomial $x^k-1=0$ has at most $k$ roots in the group. Thus, if you count the roots with various orders, you will see that there must be an $a*$ whose order is $p-1$.

Best regrds,
Murat Aygen


logarithms
calculus
calculus computations

Hi,

Could someone help me on this?

Determine whether the following converge or diverge:

1.) Summation of (k/(k+1))^(K^2) from k=1 to infinity.

2.) Summation of  1/((lnk)^lnk) from k=2 to inifinity.

Thank you so much!  :)

1: Try the root test.

2: Compare to a power of $k$; you may need to take logarithms to make the comparison.


topology
blogs
advanced fields
advanced fields unsolved

The compactness theorem in model theory states that a set $ S$ of first order sentences has a model if and only if every finite subset of $ S$ has a model.
This is a consequence of Godel's Completeness Thm.

The notion of compactness in logic is connected with the topological concept. I know there is a proof of the Compactness Theorem in Logic which usees Tychonoff's Thm of general topology (arbitrary products of compact spaces are compact), but I don't know it.
Can anyone post it, or try to figure it out?

Terry Tao wrote about this in his blog, [url=http://terrytao.wordpress.com/2009/04/10/the-completeness-and-compactness-theorems-of-first-order-logic/]here[/url]. I'm not sure if it provides a complete proof but it seems to get pretty close.

To inject some pseudo-mathematics: In some books Tychonoff's theorem is presented as if it were some kind of deep result, but in my opinion it is definitely not. Especially when stating compactness as the contrapositive of the usual definition it is clear that that the only ingredient in Tychonoff's theorem is Zorn's lemma. For all I care  Tychonoff's theorem could be an axiom that we use because we really really want it to be able to do analysis. Therefore, proving this logic result with Tychonoff might certainly look neat, but I really doubt it clarifies anything.


hi.

it seems that the keep unread flags feature takes a lot of memory, and it does not work that well.

since it was a request by someone (I don't remember who or where) I would like to see whether this feature is needed actually.

it may (but it's not sure if it will!) increase the speed of the site, so please vote.

for those of you who don't know what this feature does, it's simple: it remembers the topics you have not yet seen, even if you log out. it uses cookies to do that.

well I think the votes are going quite obviously: 0-4...


advanced fields
advanced fields unsolved

Let $ K$ be a number field contained in $ \mathbb{R}$ which is Galois over $ \mathbb{Q}$. Let $ L \equal{} K(i)$.

If $ U_K$, $ U_L$ denote the groups of units in $ \mathcal{O}_K$, $ \mathcal{O}_L$, show that $ U_K$ has finite index in $ U_L$.

Something with the Dirichlet theorem about units. But how?

(i) It is sufficient to prove that $ U_K, U_L$ have the same rank (e.g. tensor the short exact sequence $ 0\rightarrow U_K\rightarrow U_L\rightarrow U_L/U_K\rightarrow 0$ with $ \mathbb Q$ and use that the groups of units are finitely generated.)
(ii) The ranks are given by Dirichlet's unit theorem. Now $ K$ Galois and real => $ r_2(K)\equal{}0$, i.e. $ K\otimes_{\mathbb Q}\mathbb R\equal{}{\mathbb R}^{[K: \mathbb Q]}$. Tensor with $ {\mathbb Q}[i]$ over $ \mathbb Q$ to see that $ r_1(L)\equal{}0$.
(iii) Show $ rk(U_K)\equal{}rk(U_L)$.


Prove that the maximal number of edges in agraph of order n without an even cycle is $[\frac{3(n-1)}{2}]$.

We can assume WLOG that the graph is connected.

It's easy to see that each edge of the graph belongs to at most one cycle. Consider a spanning tree of our graph. It has $n-1$ edges. Each one of the other edges determines a cycle: it's the unique cycle we create by adding that edge to the spanning tree. For each such cycle, in addition to the edge we have just added, there are at least two other edges of the graph (of the spanning tree, actually) which belong to the cycle and thus cannot belong to any other cycles.

This means that we have a correspondence between the edges which do not belong to the spanning tree and some disjoint pairs of edges which do belong to the tree, which shows that there are at most $\frac{n-1}2$ edges which do not belong to the tree.

Hope it's correct.


Where do I find out my errors in LaTeX? Attached is an image showing what I see.

In two places I see:
[code]${p_b$}[/code]

I don't know what the other two are.

The window at the bottom of TexnicCenter is a bit small so drag it up to make it larger. It contains the log file (which is saved in the same directory as your tex file but is called [i]yourfilename[/i].log). Then press the ! buttons on the toolbar to show where each error is.

Click on the image below to see where I mean.


logarithms
inequalities
inequalities proposed

If a,b,c are positive real numbers with a+b+c=1 prove that

 2/3 \leq ln(a^5+b^5+c^5)/ln(a^7+b^7+c^7) \leq 5/7

and generalize it.

[quote="manlio"]If $a,b,c$ are positive real numbers with $a+b+c=1$ prove that

 $2/3 \leq \frac{ln(a^5+b^5+c^5)}{ln(a^7+b^7+c^7) }\leq 5/7$

and generalize it.[/quote]

$\frac{\ln(a^5+b^5+c^5)-\ln3}{\ln(a^7+b^7+c^7)-\ln3}\le\frac{5}{7}$ is a rearrangement of the power-mean inequality, so it is sufficient to show that $\frac{\ln(a^5+b^5+c^5)}{\ln(a^7+b^7+c^7)}\le\frac{\ln(a^5+b^5+c^5)-\ln3}{\ln(a^7+b^7+c^7)-\ln3}$, or $a^5+b^5+c^5\ge a^7+b^7+c^7$, which follows easily from homogenising.

This method is pretty easy to generalise as well.


I'd love to get a version with the Greek text, so I thought of the Fitzpatrick one

I'd like to know the difference between

http://www.amazon.com/Euclids-Elements-Richard-Fitzpatrick/dp/0615179843/ref=pd_bbs_sr_3?ie=UTF8&s=books&qid=1230974962&sr=8-3

and the 2

http://www.amazon.com/Euclids-Elements-Greek-Vol-Books/dp/1411626729/ref=pd_sim_b_7

http://www.amazon.com/Euclids-Elements-Greek-Vol-Books/dp/1411680871/ref=pd_sim_b_7

also can anyone compare Fitzpatrick's version with this one ?

http://www.amazon.com/Euclids-Elements-T-L-Heath-Translation/dp/1888009195/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1230974962&sr=8-1

thanks

If you can't decide which one to get, get them all!


induction
combinatorics proposed
combinatorics

Given two sets $A, B$ of positive real numbers such that: $|A| = |B| =n$; $A \neq B$ and $S(A)=S(B)$, where $|X|$ is the number of elements and $S(X)$ is the sum of all elements in set $X$. Prove that we can fill in each unit square of a $n\times n$ square with positive numbers and some zeros such that:

a) the set of the sum of all numbers in each row equals $A$;

b) the set of the sum of all numbers in each column equals $A$.

c) there are at least $(n-1)^{2}+k$ zero numbers in the $n\times n$ array with $k=|A \cap B|$.

Actually let me reformulate the problem to make it look better  :D

Given two sets $A,B$ (of positive numbers?) such that $n(A)=n(B)=n$, $A \neq B$ and $S(A)=S(B)$, where $S(X)$ is the sum of all elements in set $X$.

Prove that we can fill in each unit square of an $n$x$n$ table with positive numbers and some zeros such that:
(a) The sum of all numbers filled in each rows equals each element of $A$ .
(b) The sum of all numbers filled in each column equals each element of $A$.
(c) There are at least $(n-1)^{2}+k$ zero numbers in $n$x$n$ array , where $k= n( A \cap B)$

I think it is supposed to be that the row sums are $ A$, and the column sums are $ B$.  But yes, I can solve this.

Let $ A\equal{}\{a_1, a_2, \cdots, a_n\}$ so that $ a_1<a_2<\cdots<a_n$.  Similarly, let $ B\equal{}\{b_1, b_2, \cdots, b_n\}$ so that $ b_1<b_2<\cdots<b_n$.

Now put the numbers $ a_1$ to $ a_n$ on the left of the array going down, one assigned to each row.  Similarly, put the $ b_i$ along the top going right, one to each column.  We start in the top left square.  Place the largest number possible here so that the sum in the first column does not exceed $ b_1$ and the sum in the first row does not exceed $ a_1$.  Now, if $ a_1\equal{}b_1$, you go down to the second row, second column and do the same thing.  Otherwise, we have that we've either completed that row or that column.  Accordingly, either move right one square or down one square.  Repeat the process.  You have to end up at the bottom right square, because $ \sum\limits_{i\equal{}1}^{n}a_i\equal{}\sum\limits_{i\equal{}1}^{n}b_i$.  The maximum number of nonzero square will be $ k\plus{}2(n\minus{}k)\minus{}1\equal{}2n\minus{}k\minus{}1$, as desired.

This isn't completely rigorous, but you understand the argument, and it's not difficult to rigorize.

use of induction on $m+p$ that:if we have two disjoint Sets of positive reals $A,B$ such that $S(A)=S(B)$ and $\left | A \right |=m\geq 2$ ,$\left | B \right |=p\geq 2$ we can write exactly $m+p$ positive real in a $m\times p$ array such that the set of sum of rows is $A$ and the set of sum of columns is $B$.


LaTeX
logarithms
linear algebra
linear algebra unsolved

find $sup( \sum_{i \leq j} m_{i,j} , M \in O(n) )$ where $O(n)$ is the group of orthogonal matrices of $Mn(R)$

[quote="alekk"]I don't understand that Latex problem  :?[/quote]Do [b]not[/b] use <= inside a latex document. Use \leq ($\leq$) for <= or \geq ($\geq$) for >= instead. ;)

Alekk, did you find a solution for this one? I know a solution for the much easier problem when $i,j$ are independent, but this one is much more difficult, I think?

A PC* student said to me it is equivalent to $\frac{n\ln n}{\pi}$


function
real analysis
calculus
integration
counting
distinguishability
real analysis unsolved

A function f is Riemann integrable if and only if f is discontinuous on a set of measure zero. Still function g:[a,b]->[b]R[/b], g(x) = 1 if x is rational and g(x)= 0 otherwise, is known to be a non Riemann integrable function. How is this possible? Isn't [b]Q[/b] a set of measure zero and g is continuous elsewhere?

g is not continuous on irrational numbers hence it is not integrable.

The function $f(x)=\begin{cases}1,&x\in\mathbb{Q}\cap[0,1]\\0,&x\in[0,1]\setminus \mathbb{Q}\end{cases}$

is, as mentioned above, discontinuous at every point of $[0,1]$ and not Riemann integrable on $[0,1].$

The classic contrasting example is this:

$g(x)=\begin{cases}\frac 1q,&x=\frac pq\text{ in lowest terms}\\0,&x\in[0,1]\setminus \mathbb{Q}\end{cases}$

This function is discontinuous at all the rationals, continuous at all the irrationals, and (as can be quite explicitly shown even without the general theorem quoted above) Riemann integrable.

---

The Lebesgue integral attitude towards both of these functions is instructive. From the Lebesgue point of view, they're both integrable, but quite uninteresting, as they are both zero almost everywhere and thus indistinguishable from the function which is identically zero.


$\vec{a}= (\sqrt{3}, -1 )$ and $\vec{b}=(\frac{1}{2}, \frac{\sqrt{3}}{2})$.
(1). Prove $\vec{a} \perp \vec{b}$

(2). $\vec{x}=\vec{a}+(q^2-3) \vec{b}$, $\vec{y}= -p \vec{a}+ q \vec{b}$. If $\vec{x} \perp \vec{y}$, then express p with q in the form of $p=f(q)$. ( p, q $\neq 0$).

(3). Discuss the monotonicity of $f(q)$. (Do not use calculus.)

1)  Since $\vec{a} \bullet \vec{b}=0$
2)  $\vec{x}=(\sqrt{3}+\frac{q^2-3}{2},-1+\frac{\sqrt{3}(q^2-3)}{2})$
     $\vec{y}=(-\sqrt{3}p+\frac{q}{2},p+\frac{\sqrt{3}q}{2})$
     so $(\sqrt{3}+\frac{q^2-3}{2})(p+\frac{\sqrt{3}q}{2})+(-1+\frac{\sqrt{3}(q^2-3)}{2})(-\sqrt{3}p+\frac{q}{2})=0$
   we get $p=\frac{q(q^2-3)}{4}$
3)  How to...without calculus? If your teacher made you do so, :D  you may get the result then prove it in elementary way(Showing $f(x_1)-f(x_2)$)

It is because I used calculus, so I am interested to see a non-calculus solution.

What's monotonicity? Sorry about this question, but I have never come across such a term before.

A function f(x) is said to be monotone if

$x_1 \le x_2 $ implies $f(x_1) \le f(x_2)$

Pretty much, it's a function that preserves ordering.

So this function of f(x) is not considered monotone when x of power 2 and above?

What other rules apply for Monotonicity?

Here you go :) : http://en.wikipedia.org/wiki/Monotonic

Has anyone found a [b]neat[/b] solution for problem 3?


There are 9 people in a party . For every 4 of them , 2 of them know each other . Prove that there exist 3 people who know each other .

This should be an easy question but somehow my solution mess up  :(  . So I need some guide  :)

I'm confused by the problem. Is it If for every 4 people, exactly 2 know each other.

For example. If persons A and B are those people.

Every group of 4 involving A and B, would occur such that only A and B know each other.

Thus all interconnections between C, D... G, H, I would be impossible? For quadrilaterals A B _ _, and if that's true, then for other quadrilaterals, the people who don't know each other can't suddenly know each other. So I'm confused.

take 3 people, then do all the groups with those 3 and 1 of the other 6 people, each of those 6 people must know somebody from the initial 3, so by the pigeonhole, at least 1 of the initial 3 knew 2 people, if a person knew more than 3, that is done, but what if they know 2,2,2 people, i don't know where to go now

White_horse_king88 : Not exactly 2 people but at least 2 people.

Im not quite sure about the problem actually . Maybe it should be rewording like each person know only one language and every 4 of them , 2 can communicate in same language and we wish to prove there exist 3 people who can communicate in same language .  :?


Determine the range of values of $ a$ so that the equation $ x \plus{} \sqrt{1 \minus{} x^2} \equal{} a$, with $ 0 \leq x \leq 1$ admits real solutions.

[hide]$ 0 \leq x \leq 1$

$ 0 \leq x^2 \leq 1$

$ 0 \geq \minus{}x^2 \geq \minus{}1$

$ 1 \geq 1 \minus{} x^2 \geq 0$

$ 1 \geq \sqrt{1\minus{}x^2} \geq 0$

Adding this to the inequality $ 1 \geq x \geq 0$, we have that

$ 0 \leq \sqrt{1\minus{}x^2} \plus{} x \leq 2$

So $ \boxed{0 \leq a \leq 2}$.[/hide]

[hide="Solution"]By AM-GM, the maximum of $ x \plus{} \sqrt{1\minus{}x^2}$ is when $ x \equal{} \sqrt{1\minus{}x^2}$. This has solution $ x \equal{} \frac{1}{\sqrt{2}}$, which makes $ x \plus{} \sqrt{1\minus{}x^2} \equal{} \sqrt{2}$ a maximum. The minimum is $ 1$. Let $ x \equal{} \sin \theta$, so $ \sqrt{1 \minus{} x^2} \equal{} \cos \theta \equal{} y$. If $ x\plus{}y < 1$, then $ (x\plus{}y)^2 < 1$, contradiction, because $ \sin 2\theta$ is positive on the given range. So the range is $ \boxed{1 \le a \le \sqrt{2}}$.
[/hide]

[quote="p4wn3r"]Sorry, but your answer is wrong. I think you can't add the inequalities like that.[/quote]  You certainly can, and modularmarc's conclusion is correct, it just isn't [i]sharp[/i]: since we can't have both $ x \equal{} 1$ and $ \sqrt{1 \minus{} x^2} \equal{} 1$ at the same time, we can't ever get to the upper limit he computed.


inequalities
induction
number theory
prime numbers
number theory proposed

Let $ k\geq 2$ be a positive integer. A sequence $ p_1,p_2,...$ of prime numbers are such that for $ n\geq k \plus{} 1$, $ p_n$ is the greatest prime divisor of $ p_{n \minus{} 1} \plus{} p_{n \minus{} 2} \plus{} ... \plus{} p_{n \minus{} k} \plus{} 2008$. Prove that the sequence is bounded.

That's a really beautiful one, I don't know how to prove it, but I will show a proof of a very similar problem (the complete proof for that  requires I think only slight changes):

Prove that the following sequence is bounded:
$ p_{1},p_{2}\in \mathbb{P}$
$ p_{n}$
 is the greatest prime divisor of $ p_{n\minus{}1}\plus{}p_{n\minus{}2}\plus{}2008$

[u][b]Lemma:[/b][/u]
$ \forall_{n\geq 3} \ p_{n}\leq max(p_{n\minus{}1},p_{n\minus{}2})\plus{}2010$

Proof:
Case 1:
One of the numbers $ p_{n\minus{}1},p_{n\minus{}2}$ is equal to 2
Then from the assumption we have that that $ p_{n}$ divides $ 2\plus{}max(p_{n\minus{}1},p_{n\minus{}2})\plus{}2008$ so the inequality holds
Case 2:
Both of the numbers are not equal to 2
Then $ p_{n}$ must divide an even number $ p_{n\minus{}1}\plus{}p_{n\minus{}2}\plus{}2008$ so it must also divide
$ \frac{p_{n\minus{}1}\plus{}p_{n\minus{}2}\plus{}2008}{2}\equal{}\frac{p_{n\minus{}1}\plus{}p_{n\minus{}2}}{2}\plus{}1004\leq max(p_{n\minus{}1},p_{n\minus{}2})\plus{}2010$

Now let us prove by induction thet $ p_{n}$ is bounded by $ A\equal{}max(p_{1},p_{2})*2011!\plus{}2$
It' obvious that $ p_{1},p_{2}<A$
Assume $ p_{n\minus{}1},p_{n\minus{}2}<A$
Now, from out lemma we have an inequality:
$ p_{n}<A\plus{}2010$
Now observe that numbers $ A,A\plus{}1,A\plus{}2,...,A\plus{}2010$ are all composite, thus the equality cannot occur, hence we have a new boundary $ p_{n}<A$
Q.E.D. :wink:

And can it be extended to show for any $ k$?

That is what I wasn't able to do, but I'm quite sure that it is possible (I don't see any other method that could be aplied here) :wink:


A man, playing in a tough competition, loses.  In disgust, he shoots his opponent with the result that his opponent is never able to play again.  However, the police are not called to the scene, and the man escapes with only a fine and a warning.  How is this possible?

[hide]
One possibility is a chess game against his personal computer on some remote ranch in Texas but this wouldn't even result in a fine unless the neighbors who heard the shot called the police so this "solution" seems an overkill :?
[/hide]

That's the correct solution.
[hide]No, the problem doesn't have anything to do with Texas specifically.  (I think most places would only give you a fine for shooting a computer.)[/hide]


agar betavaan graphe $k_{n}$ ra be zir gerafhaaye $2$bakhshie $H_{1},H_{2},\dots, H_{m}$ efraaz kard sabet konid $m \geq n-1$

ba borhane kholf be sadegi sabet mishe , farz konid ke $m < n-1$
pas agar rase $i$ ro dar nazar begirid dar in soorat haddeaghal $2$ ras az ras haye motassel be rase $i$ too yeki az $H_{j}$ ha mian va chon graph $K_{n}$ hast pas be voozooh oon do ras ham hatman be ham vaslan . pas graphe 2 bakhshi dore be toole $3 ( fard )$ dare ke tanaghoze .:)


search
number theory unsolved
number theory

fibd positive integers a,b so that the number ab(a+b) isnt divided by 7 ,but the number -a^7 -b^7 +(a+b)^7 is diveded by 7.

im sorry but the there is smth wrong. the right one is that:
find all positive integers a,b so that the number ab(a+b) is not divided by 7 , but the number -a^7-b^7+ (a+b)^7 is divided by 7^7.

It is an IMO problem . Try to search on Forum.

$ (a\plus{}b)^{7}\minus{}a^{7}\minus{}b^{7}\equal{}...\equal{}7ab(a\plus{}b)(a^{2}\plus{}ab\plus{}b^{2})^{2}.$
Because the number ab(a+b) is not divided by 7 ,then we should find a,b so that the number $ a^{2}\plus{}ab\plus{}b^{2}$ is divided by $ 7^7$.
 But then $ a^{2}\plus{}ab\plus{}b^{2}\geq7^{7}$. So $ a\plus{}b)^{2}>a^{2}\plus{}ab\plus{}b^{2}\geq7^{3}\equal{}243$. That means that  $ a\plus{}b\geq19$.  With testing we find that a=18 and b=1.


How many solution be non-negative of an equation
$ x_1\plus{}x_2\plus{}...\plus{}x_{1000}\equal{}1000$

It is easy to find number of solutions
(1) $ x_1\plus{}x_2\plus{}...\plus{}x_k\equal{}n,x_i\ge 0.$
Consider
(2) $ 1\plus{}1\plus{}...\plus{}1\equal{}n$ (n times).
We can chose k-1 positions in $ n\plus{}k\minus{}1$ positions (2) and transform it to solution (1) by $ x_{i}\equal{}sum_{m_{i\minus{}1}}^{m_i} 1$.
For example n=2,k=3 give
$ ,,1\plus{}1\equal{}2\Longrightarrow x_1\equal{}0,x_2\equal{}0,x_3\equal{}2$,
$ ,1\plus{},1\equal{}2\Longrightarrow x_1\equal{}1,x_2\equal{}1$, et.c.
Therefore number of solutions is $ C_{n\plus{}k\minus{}1}^{k\minus{}1}$.


At what point should I be worried that my solutions haven't been received or are missing some sort of paperwork?

I have sent mine on the 26th of September and there are still no check marks next to the round.

thank you for your help

eh...i sent my solutions like a month ago and i haven't gotten anything yet

The solutions are not filed until after the deadline.

[quote=Rrusczyk on September 19]We'll probably start filing the papers in a week or so.[/quote]

Evidently, they are filed before the deadline. But anyway, we need to know if our papers got through.

I wouldn't be too worried. It takes a while. Sent mine in yesterday morning, so I'm not expecting anything for a while.  :wink:

We are still processing our received solutions.  We hope to have them all up to date sometime today or tomorrow.

If you sent your solutions before October 3, and do not have a checkmark showing your solutions as received in you USAMTS Profile/Scores, then post your USAMTS User ID#, the method you used to submit your solutions, and the date (or a guess of the date) for when you sent the solutions.

USAMTS ID: 7512
Method: email to solutions@usamts.org
Date: Monday, October 1

I am glad to announce that I have received the confirmation email  :D

USAMTS ID: 9332
Method: fax
Date: September 24 or 25.

I sent the entry form with this fax, and the My Profile page says that my entry form has already been submitted, so I would think that the fax went through.

I still haven't recieved a check mark.  I sent mine in through email on Oct. 5.  ID 7805.

Please do not post or otherwise contact us unless you sent in your solutions [b]before[/b] October 3 and have [b]not[/b] received a checkmark.  We are still in the process of filing solutions that came on or near the deadline.

Strange though- I sent on the deadline and it's confirmed that mine were recieved.

If I snail mailed mine on Tuesday, when should I expect to see check marks?

We are processing them as we speak.  We still have several hundred that are unprocessed.

If you mailed it on Tuesday, we might not even have it yet!  (We are all the way in the corner of the country and in a somewhat rural location.)

I will post an update towards the end of the day on Monday as to what our processing status is.  For now, once again, if you sent your solutions on or near the deadline, by any method, please don't panic if you don't have a checkmark.

[quote="123456789"]I dropped off my letter in a post box on Monday. My classmate mailed his out on Tuesday. He got a check mark on Friday. I haven't gotten mine and it's Saturday. We live in relatively the same area. Does that mean anything?[/quote]Not really.  We often mail two items at the same time [i]to the same address[/i] and they often arrive on different days.

Besides, last week Monday=Tuesday because Monday was a holiday.

[quote="123456789"]In the unfortunate case that you don't receive the solutions via mail, would there be any way to still have my solutions graded?[/quote]There's almost no chance of happening.  The Postal Service is very reliable.  We should know -- we are the largest customer in our zip code (which says more about the size of the town we're in than it does about us), and mail almost never just "disappears".

This is also why we ask you: please [b]do not send your papers via certified mail[/b].  All this means is that we have to wait in line at the post office to pick up your solutions.  (Plus it costs you more postage.)

Update, Monday 4 p.m. Pacific: To the best of our knowledge, all of the papers that we have received by mail or fax so far have been processed.  But, once again, if you sent your solutions by mail on Tuesday (10/9), there's still a reasonable chance that we don't have it yet (the postal service is reliable, but it is occasionally slow).  We are still processing email submissions; ironically, these take longer for us to process (since we have to detach all the files and print out the solutions for the graders), but we hope to have these finished in the next day or two.

[quote="DPatrick"][quote="dandan9002"]I sent my solution by fax on Fri. 10/05, but still didn't get a checkmark, should I be worried?[/quote]We think that we have all of the faxes filed, so if you sent yours by fax and we don't have it marked as received, there's a problem.

Send an email to usamts@usamts.org with all the details, including your USAMTS ID# and when you faxed your solution.[/quote]

I sent the email to usamts@usamts.org with my details.
What am I supposed to do now??

[quote="dandan9002"]I sent the email to usamts@usamts.org with my details.
What am I supposed to do now??[/quote]Wait for someone to reply.

I have to resend my solutions...
Can I still fax my solutions or should I resend by mail?

er, yeah, well I guess it can't tell that I sent a pdf. I keep getting emails asking me to resend. Should I be worried about this? I mean I didn't do so well, but at least want some points... :(

We've had a few people report bounced emails.  It turned out that it was because they tried to mail to usamts.com instead of usamts.org.  So please check carefully.

If you tried to mail to usamts.com and it got bounced, please resubmit it to the correct address, and include proof (e.g. the bounce message) that you tried to submit on time.

[quote="funcia"]well I guess it can't tell that I sent a pdf[/quote]I have no idea what you mean by this -- what is "it" and what exactly are you being told?

Are e-mail solutions still being processed? I'm getting nervous as I still don't have a checkmark...

[quote="funcia"]I don't know what "it" is either, but this is the message:
[quote="solutions@usamts.org"]
Greetings

We received your email attachment but were unable to print it. Please resend in a pdf or word document.

Thank you
[/quote]
 I [i]did[/i] send in a pdf. I'm pretty sure "it" happens to be some automatic mail processor or something, because when I emailed that my attachment was a pdf, sensing there was no attachment it sent me an email asking me to send a pdf.  I would hope any human who got an email like that would either not respond to, or acknowledge my message.[/quote]What we got was a .DAT file.  Once we figured out it was PDF, we were able to print it.

We have processed all of the solutions that we have received.  If your USAMTS ID# is among the following list:
8609 8629 10111 10115
then there are still outstanding issues with your submission.  (All of the above people have been contacted via email; if you have not received an email from us, please write usamts@usamts.org or call 619-445-0112.)

If you submitted Round 1 solutions and do not yet have a green checkmark when you log onto usamts.org and click "My USAMTS", then something bad has happened.  Please contact us at usamts@usamts.org or 619-445-0112 as soon as possible.  When contacting us, please include:
- your USAMTS username and ID#
- your submission method (fax, email, or mail)
- the approximate date (and time if fax or email) when you sent in your solutions
[b]Do not post in this thread or send a PM[/b] -- send email or call instead.

[b]Note:[/b] we have had a few issues with people emailing their solutions to usamts.com instead of usamts.org.  If you did this, your email should (eventually) bounce back to you.  Please resubmit to the correct address solutions@usamts.org, and include a copy of your email error message.


Yup, the formula being the first number plus the second number.

I find it easier to add even integers for some reason so I just do
2x-1. Right... but in any case it still is x^2-(x-1)^2, so either way it is completely acceptable.


I have the same problem lol.  If we're pdfing things, then it'd be a good idea to do that with problem sets.

[quote="chess64"]Would it be possible to post the class transcripts as .doc or .pdf files? That way it would be faster to save them to your computer. :)[/quote]

Just print to a PDF printer. This will allow you to save the files as PDF.  (Primo PDF does a good job.)

You could use OpenOffice.org Office  and export to PDF from there.

I have a PDF converter, but it takes a while to copy the transcript from the website into a word document. (Since they're so long.)


Prove that $\cos 40^{\circ}$ is irrational.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=80213 , bottom half of the topic, my approach works also for your problem.


AMC
AIME
quadratics
number theory
Diophantine equation

A nice faux-AIME problem:

Some examples of pythagorean triples of the for $ (a, a\plus{}1, a\plus{}k)$ are $ (3,4,5)$ where $ k\equal{}2$ and $ (20,21,29)$ where $ k\equal{}9$.  Find $ k$ in the next such triple.

There is a straightforward formula for any and all Pythagorean triples.  Just look on wikipedia.

Yes, but can you solve the problem?  I know of the formula, but you still have to use tricky methods to solve it.

[hide="Solution"] Clearly such a triple is primitive.  We therefore have the Diophantine equation

$ 2mn \minus{} (m^2 \minus{} n^2) \equal{} \pm 1 \Leftrightarrow$
$ (n \plus{} m)^2 \minus{} 2m^2 \equal{} \pm 1$

which is a Pell's equation.  For the positive sign the solutions are generated by $ (3 \plus{} 2 \sqrt{2})^n$.  $ n \equal{} 1$ gives $ (3, 4, 5)$.  $ n \equal{} 2$ gives $ m \equal{} 12, n \equal{} 5$ and the triple $ (119, 120, 169)$.

For the negative sign the solutions are generated by $ (1 \plus{} \sqrt{2})(3 \plus{} 2 \sqrt{2})^n$.  $ n \equal{} 1$ gives $ (20, 21, 29)$.  $ n \equal{} 2$ gives $ m \equal{} 29, n \equal{} 12$ and the triple $ (696, 697, 985)$.

The next value of $ k$ is therefore $ 169 \minus{} 119 \equal{} \boxed{50}$. [/hide]

I just found a quadratic in $ a$, and then applied the quadratic formula.  It follows easily from there.

[hide="Elaboration"]
Using some algebra, we get $ a^2 \plus{} 2a(1 \minus{} k) \minus{}(1\minus{}k^2) \equal{} 0$, and then we plug and chug. Unless there's a better way than plugging and chugging...?[/hide]

Fixed.

That depends on what you mean by plugging and chugging.  And you mean $ a^2 \plus{} 2a(1\minus{}k) \plus{} (1\minus{}k^2)$.  This approach also leads to a Pell's equation, but it characterizes the values of $ k$ such that this is true:  the $ (k\minus{}1)^{th}$ triangular number is a square.


find all pôlynomes p (x) such as:
[img]http://www.gnux.be/latex/data/c786f38b21f4af2bf91fc965bc9078bb.png[/img] [/img]

Solution
[hide]
Suppose $ P(x)$ - non constant.
If $ P(x) \equal{} a_n x^n \plus{} a_{n \minus{} 1} x^{n \minus{} 1} \plus{} ... \plus{} a_1 x \plus{} a_0$, then $ 2^{n}a_n \equal{} 8a_n$, so $ n \equal{} 3$.
We have that $ P(2) \equal{} 0$, $ P(8) \equal{} 0$, so $ P(x) \equal{} C(x \minus{} 2)(x \minus{} 8)(x \minus{} r)$, $ C\neq 0$:

$ C(x \minus{} 8)(2x \minus{} 2)(2x \minus{} 8)(2x \minus{} r) \equal{} 8C(x \minus{} 1)(x \minus{} 2)(x \minus{} 8)(x \minus{} r)$;

$ (x \minus{} 4)(2x \minus{} r) \equal{} 2(x \minus{} 2)(x \minus{} r)$.
$ r \equal{} 4$ is only one solution.

If $ P(x)$ - constant, then $ P(x) \equal{} 0$ is only one solution.

Answers:
$ P(x) \equal{} 0$;
$ P(x) \equal{} C(x \minus{} 2)(x \minus{} 4)(x \minus{} 8)$.[/hide]


#1 You are starting a business helping out neighbors who are going on vacation. You are going to offer three services: dog walking, plant watering, and taking in the mail/paper. You estimate that walking a dog will take 20 minutes per da, watering plants will take 15 minutes per day, and taking in the mail will take 5 minutes per day. You plan to charge 5/day for dog walking, 1/day taking out mail, and 2/day for watering. Your parents tell you cannot do more than 15 people's mail at one time due to storage issues. You plan to keep the number of watering jobs equal to the number of mail/paper jobs and the number of dig walking jobs will be no more than either of the two. You cannot work more than 8 hours a day. What should you schedule each week to max income?

Only need the equations

#2 A distributor of large screen plasma TV's has two warehouses W1 and W2. He currently has 80 TV's at W1 and 70 at W2. Two stores, BB and CC, order 35 and 60 TV's respectively shipping costs are in the chart. How many TV's should be shipped to each store from each warehouse so to minimize shipping costs?

graph http://img90.imageshack.us/img90/6585/mathgraphhc5.png

again only need equations.

Thanks for the help in advance thanks.

[quote="boo"]Let $ X_{ij}$ = # of TV shipped from warehouse i to store j
$ X_{11} \plus{} X_{21}\ge 35$
$ X_{21} \plus{} X_{22}\ge 60$[/hide][/quote]
Slight error here, should be $ \equal{}$ not $ \geq$.


analytic geometry
probability
calculus
geometry
conics
parabola
integration

Scores:
MysticTerminator: 4 Points
Everyone else: 0 points

Source: Past MMPC Exams

1998 2) Let N be a positive integer. Prove the following:
[tab]a) If N is divisible by 4, then N can be expressed as the sum of two or more consecutive
odd integers.
[tab]b) If N is a prime number, then N cannot be expressed as the sum of two or more
consecutive odd integers.
[tab]c) If N is twice some odd integer, then N cannot be expressed as the sum of two or more
consecutive odd integers.

1998 3) Let S = 1/(1^2) + 1/(2^2) + 1/(3^2) + 1/(4^2) + ....
[tab]a) Find, in terms of S, the value of 1/(2^2)+1/(4^2)+1/(6^2)+1/(8^2)+...
[tab]b) Find, in terms of S, the value of 1/(1^2)+1/(3^2)+1/(5^2)+1/(7^2)+...
[tab]c) Find, in terms of S, the value of 1/(1^2)-1/(2^2)+1/(3^2)-1/(4^2)+...

1997 1) A square 2 by 2 dartboard is situated in the xy plane with its center at the origin and its sides parallel to the coordinate axes. Darts that are thrown land randomly on the dartboard. Find the probability that a dart will land at a point closer to the point (0,1) than the bottom edge of the board.

[question removed]

Replacement:
1995 3) For any real number x, [x] denotes the greatest integer that does not exceed x. Given natural number N, denote as f(N) the following sum of N integers:
f(N) = [N/1] + [N/2] + [N/3] + ... + [N/N]
a) Evaluate f(7)-f(6)
b) Evaluate f(35)-f(34)
c) Evaluate (with explanation) f(1996)-f(1995)

1a) [hide]With N divisible by 4, we have two consecutive integers as (n-1)/2 and (n+1)/2.  There sum is N[/hide]

b) [hide]The sum of two odds is an even.  2 is the only even prime and the lowest prime.  Ergo, two consecutive odds are not prime when added together[/hide]

c) [hide]The odd is N/2.  With two consecutives, it would be (n-1)/2 and (n+1)/2, since N is even, the consecutives are not integers.  This leaves no possible value.[/hide]

Frodo:

1a)[hide]Incorrect[/hide]

 b)[hide]Two or more consecutive odd integers. Finish 'er off[/hide]

 c)[hide]If you correct part a, you'll likely see what's wrong with part c[/hide]

[quote="cats..."]the last one's just like the one in joel's contest with the oranges.[/quote]
Yeah, looks like it's rather similar. Don't do that one, I guess, and I'll put up a new one.

the sum

[hide]

first, its helpful to write it in sigma notation, then i think it becomes pretty obvious.

s=  :Sigma: 1/(n^2)

a)  this is just even ints, so :Sigma: 1/(2n)^2, which is 1/4*  :Sigma: 1/n^2, which is just s/4.

b)  this is all the odd ints, which is all numbers - even, or S-1/(2n)^2 = s-s/4 = 3s/4 ?

c) then this is just 3s/4-s/4 = s/2[/hide]

1. [hide]n=4k, n=(2k-1)+(2k+1); b) if n is the sum of m consecutive odd integers, m|n. as we are given that n is prime, we have m=n, so n is being expressed as the sum of n consecutive odd integers. In this case, all primes (except for 2) can be expressed as the sum of consecutive odd integers: 3=-1+1+3, 5=-3+-1+1+3+5, etc. (follow the pattern); c)the sum of an odd number of odd integers will be odd, thus it cannot :equiv:2 mod 4. So, we have that the number of odd integers is even, meaning that the mean falls right in between two terms of the sequence, i.e. it is even. However, the sum of this arithmetic series will be the number of terms times the mean, i.e. an even number of terms times an even mean will be :equiv:0 mod 4. contradiction[/hide]

2. [hide]S=:pi::^2:/6. yupyup. a)note how each term is 1/2:^2: times a corresponding term in S => i.e. if S converges (which is does, to :pi::^2:/6), this new sequence converges to S/4 (:pi::^2:/24); b)this is S minus the thing in part a, i.e. 3S/4=:pi::^2:/8; c)this is S minus twice the thing in part A, i.e. S/2=:pi::^2:/12[/hide]

For 1b:  [hide]The sum n consecutive postive integers is always composite by Gauss's Theorem: n*(a+z)/2.  Odd and even numbers apply to only natural numbers.[/hide]

3. [hide]edges of this dartboard will be (:pm:1,:pm:1). Okay, the boundary (where points are equally spaced between (0,1) and this bottom edge) is a parabola. This means we get to use calculus eventually. But first let's find the parabola. The focus is (0,1) and the directrix is y=-1. Busting out that analytical geometry, we find that the parabola is y=x:^2:/4. Taking the integral of this from x=-1 to 1 gives x:^3:/12 (-1 to 1) = 1/6. this is the area between the parabola and the x-axis. we want the area between the parabola and y=1, so we subtract this 1/6 from 2, and we get 11/6.[/hide]

4. grr hard problem grr. [hide]a) f(7)=7+3+2+1+1+1+1, f(6)=6+3+2+1+1+1, the difference is 2. b)the only time in which one of the terms in the expansion of f(35) will be diff. than in one of the terms of f(34) will be in the [N/1] term, and in any subsequent terms of the form [N/d], where d|35. To see why, observe that for any term of the form [N/k], k not dividing 35 and not equal to 1, 34 and 35 will have the same integer quotients upon division by k and thus that term will be the same. Ok, finally, note that for each term of the form [N/d], k|35, the difference between [35/d] and [34/d] will be 1, as [35/d] will have exactly one more factor of d. So essentially what the question is asking for is the number of divisors of 35. We write 35 in its PPF, 5*7, and note that it is (1+1)(1+1)=4. Using similar reasoning for part c), we factor 1996=2^2*499, and (2+1)(1+1)=6[/hide]

Mystic gets #1. Frodo, your clarification/correction didn't make sense.

Mystic also gets the fourth question. There will be more tomorrow.


calculus
geometry
ratio
limit
geometric series

***I'm new and dont know how to do "click here to reveal hidden content" so dont read ahead if you dont want my 2 cents***

ok, so the first term is 1/2, and every time after that we mutliply the term by 1/2

thus it will converge to (1/2)/(1- (1/2))
where the top term  is intended to be the first term, and the bottom is 1 - (what we mutliply every term by)

thus we get (1/2) / (1/2) = 1

not sure if this is the solution you're looking for, i just remember this formula from calculus and convergence

[quote="cezar lupu"]What's so hard about this problem. Just use the geometrical serie.Is this the Basics forum or not????[/quote]

hey man, i felt special cause i actually knew how to do a problem here! lol

and now you've killed it... :(

I think geometric series belong in getting started, even if they are simple ones. You've got to learn to formula one time or another.

The formula for a geometry series with a common ratio is $\frac{a}{1-r}$ Where $a=$the starting value and $r=$the ratio. We can see that this infinite series has the ratio $\frac{1}{2}$ because we multiply by $\frac{1}{2}$ to each term to get the next term. We can also see that the first value is $\frac{1}{2}$ So we plug the numbers in:

$\frac{\frac{1}{2}}{1-\frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = \fbox{1}$

If you have any other questions about infinite geometric sequences, feel free to ask here.

[quote="cezar lupu"]What's so hard about this problem. Just use the geometrical serie.Is this the Basics forum or not????[/quote]

The concept of infinity and geometric series are generally not taught until high school.  So this is the proper forum for this problem.

Rather than just stating the formula for the sum of an infinite geom. series, showing how to arrive at it is much more enlightening.

For our problem, call the sum we want $S=\frac 12 +\frac 14+\cdots+\frac{1}{2^n}+\cdots.$  Now notice that $\frac 12 S=\frac 14+\frac 18+\cdots.$  Thus:

\begin{eqnarray*}
S &=& \frac 12 + \frac 14 + \frac 18 +\frac{1}{16}+\cdots\\
-\frac 12 S &=& -\left(\frac 14 + \frac 18 + \frac 1{16}+\cdots\right) \\ \hline
\frac 12 S &=& \frac 12 \\
S &=& 1
\end{eqnarray*}

You can use joml88's specific example for that series and generalize to prove the formula for the sum of infinite series.

[quote="Drunken_Math"]Albatross wants you to Prove that $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}...=1$[/quote]

Just use the geometric series formula, so the first term a=1/2 and the ratio is 1/2, so $(1/2)/(1-1/2)=1$. It's pretty clear.

[quote="joml88"]

The concept of infinity and geometric series are generally not taught until high school.  So this is the proper forum for this problem.

Rather than just stating the formula for the sum of an infinite geom. series, showing how to arrive at it is much more enlightening.

For our problem, call the sum we want $S=\frac 12 +\frac 14+\cdots+\frac{1}{2^n}+\cdots.$  Now notice that $\frac 12 S=\frac 14+\frac 18+\cdots.$  Thus:

\begin{eqnarray*}
S &=& \frac 12 + \frac 14 + \frac 18 +\frac{1}{16}+\cdots\\
-\frac 12 S &=& -\left(\frac 14 + \frac 18 + \frac 1{16}+\cdots\right) \\ \hline
\frac 12 S &=& \frac 12 \\
S &=& 1
\end{eqnarray*}[/quote]

thats the same concept used to convert terminating decimals into fractions...
wow. that's cool.
thanks joml88

[hide]
Errm.. theres a nice way with the infinite series formula.

$\displaystyle \frac{a_1}{ |1-r| }=\frac{\frac{1}{2}}{|1-\frac{1}{2}|}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$

proved.
QED.
$W^5$(which is what we wanted)
[/hide]

Can anyone prove that $\frac{1}{2}+\frac{1}{4}+..............$=1 without using words (picture proof)?

Does it have to be like "proven"?

Or can we have like an animation (in theory) where there's a circle, and it starts out by shading a semicircle, and then half of the remaining semicircle, and then half of the remaining quarter circle, and on and on until the shading gets so small that it "looks" like 1.

That's exactly what I meant (except with a square) :)

well i think you can do it for any shape, b/c it'll eventually appear to be all shaded anyway

What about the race between the tortoise and the hare where the tortoise goes a certain number of units per second while the hare covers half the distance every second.  Who wins if the tortoise covers $\frac{1}{n}$ units per second and the hare still covers half the distance every second as $n \rightarrow \infty$  and as $t \rightarrow \infty$ where $t$ denotes time?

The sum never actually reaches of $1/2+1/4+1/8...$ never actually reaches $1$ but we can make it as close to $1$ as we want it to be, therefore it is just considered $1$.

[quote="JesusFreak197"][hide="3"]Simplify to the sum of three infinite series: if $1+\frac{1}{8}+...+\frac{1}{8^n}=x$, we have $x-\frac{1}{2} x -\frac{1}{4} x=\frac{1}{4} x=\frac{1}{4} \displaystyle \left(\frac{1}{1-\frac{1}{8}}\right)=\frac{1}{4}\left(\frac{8}{7}\right)=\frac{2}{7}$[/hide][/quote]

:yup: how about those other two?

I have a few questions. I'll make an addendum.

4) What is the sum of all 5-digit palindromes? Express your answer in scientific notation.

5)Write ${\displaystyle\{.3\overline{12}}_{5}$ as a base 10 fraction

6) Find the sum of all integers between 100 and 2000 inclusive that are not divisible by 3.

7)Evaluate the sum $\displaystyle\sum_{i=1}^\infty \frac{4n}{5^n}$

8)Find constants A and B such that $\displaystyle\ \frac{1}{x^2-9}=\frac{A}{x-3}+\frac{B}{x+3}$
and use this to evaluate $\displaystyle\sum_{n=4}^\infty \frac{1}{n^2-9}$

Source: MAO Denver Nationals '01 Theta Sequences and Series. I found it to be a rather good test. (Not arranged according to any sort of difficulty, although I'm unsure of how to do number 4...the solutions are messy and illegible)

[quote="pkothari13"][quote="Hessm"]
hey man, i felt special cause i actually knew how to do a problem here! lol

and now you've killed it... :([/quote]

haha nah, your moment is still alive for all i know.  to do the "click to reveal hidden content" just highlight what you want to hide, and then click "Hide" at the end of the buttons above the text box.

for any infinite series $a+ar+ar^2+...$ where $|r|<1$, then the sum  = $\frac{a}{1-r}$

try these problems:

1) Prove the above statement.

2) Find the infinite sum $\frac{7}{12} + \frac{25}{144} + \frac{91}{1728} + ... \frac{3^n + 4^n}{12^n}$

3) Find the sum of the infinite series $1-1/2 - 1/4 + 1/8 - 1/16-1/32 + 1/64-1/128-...$[/quote]

$\displaystyle\sum_{n=1}^\infty \frac{4^n+3^n}{12^n}$ Can be rewritten as two seperate sums. $\displaystyle\sum_{n=1}^\infty \frac{3^n}{12^n}$ and $\displaystyle\sum_{n=1}^\infty \frac{4^n}{12^n}$

Going from there we then factor $12^n$ into $4^n 3^n$ and therfore can cancel the $3^n$ and $4^n$ respectively to have a much more manageable series of $\displaystyle\sum_{n=1}^\infty \frac{1}{4^n}$ and $\displaystyle\sum_{n=1}^\infty \frac{1}{3^n}$ respectively. Now take the sum of the two different series $\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$ and$\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{\frac{1}{3}}{\frac{2}{3}}=\frac{1}{2}$ to get the final answer of $\frac{1}{3}+\frac{1}{2}=\frac{5}{6}$

[quote="DPopov"]Well the limit of the sum is one.[/quote]
$\lim_{n \to \infty}{S_n}=\lim_{n \to \infty}{\frac{1}{2} \cdot \frac{1-({\frac{1}{2}})^n}{1-\frac{1}{2}}}=1$

[quote="Pakman2012"]I have a few questions. I'll make an addendum.

4) What is the sum of all 5-digit palindromes? Express your answer in scientific notation.

8)Find constants A and B such that $\displaystyle\ \frac{1}{x^2-9}=\frac{A}{x-3}+\frac{B}{x-3}$
and use this to evaluate $\displaystyle\sum_{n=4}^\infty \frac{1}{n^2-9}$
[/quote]

4) Because the palindromes can be paired up, one can take the sum of the largest and smallest palindrome, 10001 and 99999, whose sum is 110000.  Now there are 900 5 digit palindromes (9*10*10*1*1), meaning there are 450 pairs.

Therefore, $110000(450) = 49500000 = 4.95*10^7$

8) I assume you meant $\displaystyle \frac{1}{x^2-9}=\frac{A}{x-3}+\frac{B}{x+3}$.  One can solve and see $A=\frac{1}{6}, B = -\frac{1}{6}$.

Therefore, you get $\displaystyle\sum_{n=4}^\infty \frac{1}{n^2-9} = \frac{1/6}{1} - \frac{1/6}{7} + \frac{1/6}{2} - \frac{1/6}{8} + \frac{1/6}{3} - \frac{1/6}{9} +...$

Everything cancels except $\frac{1/6}{1} + \frac{1/6}{2} +\frac{1/6}{3} + \frac{1/6}{4} +\frac{1/6}{5} + \frac{1/6}{6} = \frac{49}{120}$

i hope i did that correctly. i hope someone can check that for me.

Good job! That was a typo. Both correct.


A ferry boat that travels at 10mph has to travel repeatedly across a river, arriving at a port 4 miles downstream from the starting port.  If this trip takes 26 min, How many minutes will be saved by moving one of the ports directly opposite the river from the other?

[hide]$\frac{26}{60}*10=4.33...,$ and from Pythagoras, $\frac{13}{3}^{2}-4^{2}= \frac{5}{3}$ so from $t=\frac{d}{r}$ we know that it takes 5/30ths of an hour - or ten minutes to cross. Therefore it saves 26-10=16 minutes.[/hide]

EDITED: i was being stupid-  sorry

[hide="More Steps"]
Distance already being travelled $10 \cdot \frac{13}{30}= \frac{13}{3}$.
Then by the Pythagorean Theorem, $d = \sqrt{\frac{169}{9}-\frac{144}{9}}= \sqrt{\frac{25}{9}}= \frac{5}{3}$, the new distance $d$.
Then the time to travel that distance $d$ , $\frac{\frac{5}{3}}{10}= \frac{5}{30}= \frac{10}{60}\ \text{hours}$.
So, $26-10 = 16 \ \text{minutes}$
[/hide]


geometry
circumcircle
geometric transformation
reflection
rhombus

Consider triangle $ABC$ and its circumcenter $O$. Reflect $O$ with respect to $AB$, $BC$, and $AC$ to obtain points $O_{AB}, \ O_{BC}, O_{AC}$. Draw circles, with radius equal to the circumradius of $\triangle ABC$, with centers at each of these three points. Prove that these three circles intersect at a common point. Identify this point.

first note that $\angle BO_{BC}C=\angle BOC=2A$ so we get that $\angle AO_{AB}B+\angle BO_{BC}C+\angle CO_{CA}A=360$ now we will prove the followin generalization:

[b]for $\triangle ABC$ we construct triangles $\triangle ABC',\triangle BCA',\triangle CAB'$ s.t $\angle A'+\angle B'+\angle C'=360 or 180$ prove that the circumcircles of these three triangles are concurrent at a point?[/b]

[hide]name the circumcircles of $\triangle ABC',\triangle BCA',\triangle CAB'$ as $w_{1},w_{2},w_{3}$ respectively and
assume that $w_{1},w_{2}$ imtersect at a point $H$ (different from $B$) now we will prove that $w_{3}$ passes thruogh $H$:

first note that $H$ lies on $w_{1}$ so we get that $\angle AHB=180-\angle C'$ and also $H$ lies on $w_{2}$ so we get that
$\angle BHC=180-\angle A'$ so we get that:

$\Rightarrow \angle AHC=360-(360-\angle AHB-\angle BHC)=\angle C'+\angle A'=360-\angle B'$

so we get that $360-\angle AHC=\angle B'$ so we get that $AB'CH$ is cycelic which means that $w_{3}$ passes thruogh $H$
and the lemma is proved...

also like the method above we can prove that $w_{1},w_{2},w_{3}$ are concurrent if $\angle A'+\angle B'+\angle C'=360 or 180$
[/hide]

now according to this problem we get that the three desired circles are concurrent,now we claim that the concurrency point is
the orthocenter of $\triangle ABC$

[b]lemma:if three equal circles $w_{1},w_{2},w_{3}$ concur at a popint $H$ then $H$ is the orthocenter of $\triangle ABC$
where $A,B,C$ are the other intersection of the circles...[/b]

the proof of the above lemma is not hard:

[hide]

there are two ways to prove the lemma above:

$\boxed{\text{solution}1}:$

name the intersections of the circles different from $H$ as $w_{1}\cap w_{2}=C,w_{2}\cap w_{3}=A,w_{3}\cap w_{1}=B$
and also name the centers of the circles $w_{1},w_{2},w_{3}$ as $O_{1},O_{2},O_{3}$ respectively its sufficient to show
that $AH \bot BC$

now name the intersection of $AH,BC$ as $D$ so its sufficient to show that $\angle ADC=90$ or $\angle ADC=\angle ADB$
now its just some angle chasing:

in $w_{1}$ we get that $\angle HBC=\frac{\frown{HC}}{2}$

also in $w_{2}$ we know that $\angle HAC=\frac{\frown{HC}}{2}$

now we know that $w_{1}=w_{2}$ so we get that $\angle HBC=\angle HAC$

so according to $\triangle BHD,\triangle ADC$ its sufficient to show that $\angle BHD=\angle ACD$ which is equivalent to
$\angle BHA+\angle ACD=180$

but in $w_{3}$ we know that $\angle BHA=\frac{\frown{BA}}{2}$

and also we know that $\angle ACD=\angle HCB+\angle HCA=\frac{\frown{HB_{w_{1}}}}{2}+\frac{\frown{HA_{w_{2}}}}{2}=\frac{\frown{HB_{w_{3}}}}{2}+\frac{\frown{HA_{w_{3}}}}{2}$

so according to the last two equations we get the desired resault...

$\boxed{\text{solution}2}:$

denote $A',B',C'$ as $AH\cap O_{2}O_{3},BH\cap O_{3}O_{1},CH\cap O_{1}O_{2}$ repectively...

now the radius of the three circles are equal so we get that the quadrilaterals $AO_{2}HO_{3},BO_{3}HO_{1},CO_{1}HO_{2}$ are rhombus,so we get that points $A',B',C'$ are the midpoints of $HA,HB,HC$ and also the midpoints of $O_{2}O_{3},O_{3}O_{1},O_{1}O_{2}$ repectively and also we know that $HA \bot O_{2}O_{3},HB \bot O_{3}O_{1},HC \bot O_{1}O_{2}$

so we first get that:

$\frac{O_{1}C"}{C'O_{2}}=\frac{O_{1}B'}{B'O_{3}}=1\Rightarrow B'C' \parallel O_{2}O_{3}(I)$

$\frac{HB'}{B'B}=\frac{HC'}{C'C}=1\Rightarrow B'C' \parallel BC (II)$

now from $(I),(II)$ and the fact that $HA \bot O_{2}O_{3}$ we get that $HA \bot BC$ and samewise we get that $HB \bot AC,HC \bot AB$ so $H$ is the orthocenter of $\triangle ABC$ and the proof of the lemma is completed...[/hide]


factorial
calculus
integration
Ring Theory
superior algebra
superior algebra unsolved

Do there exist fields F such that R (the field of real numbers) is an algebraic field extension of F?

Take a transcendence base $ B$ of $ \mathbb R$ over $ \mathbb Q$. Then consider $ K\equal{}\mathbb Q(B)$: every element of $ \mathbb R$ is algebraic over $ K$ by definition of transcendence base. But if $ x \in B$, then $ \sqrt x \not\in K$, thus it's a nontrivial extension.

In fact we have also the following: if $ \mathbb R$ is an algebraic extension of some field $ K$, then it's degree/dimension is $ \infty$.

Thank you for your answer.

I've another related question: Does there exist a sub-ring of the real numbers, whose field of fractions equal the real numbers?

[quote="jmerry"]Of course there is: $ \mathbb{R}$ itself is a subring. You have to step carefully defining what you want, and I'm not sure there's a definition that makes sense with transcendental numbers out there.[/quote]

How about - a sub-ring of $ \mathbb{R}$, whose not a field, and whose field of fractions is  $ \mathbb{R}$?

Let $ B$ as before. Then take $ R \equal{} \mathbb Q[B]$: this is a factorial ring.
Now take it's integral closure $ \bar R$ in $ \mathbb R$. Then every element of $ \mathbb R$ is of type $ \frac rs$ with $ r \in \bar R$ and $ s \in R$, especially $ \mathbb R \equal{} \mathrm{Quot}(\bar R)$. And as $ R$ is factorial, we have $ \bar R \cap \mathrm{Quot}(R) \equal{} R$, thus $ \bar R$ being not a field (e.g. if $ x \in B$, then $ \frac 1x \not \in \bar R$).

Another construction:
Let $ \mathbb Q_p$ be the $ p$-adic numbers. Let $ \mathbb A_p = \bar{\mathbb Q}_p$ it's closure and ${ \mathbb C_p = \hat{\mathbb A}_p}$ the closure's completion. Then $ \mathbb C_p \cong \mathbb C$ as fields (in fact, all algebraic closed fields of same characteristic and uncountable cardinality are isomorphic).
We have a unique extension of the $ p$-adic valuation $ v_p$ to $ \mathbb C_p$. Now take $ \mathcal O_p = \{x \in \mathbb C_p |v_p(x) \geq 0 \}$: this is a subring as $ v_p$ is non-archimedian and it's quotient ring is $ \mathbb C_p$. We can restrict to $ \mathbb R$ now.

By the way, note that both "constructions" needed the Axiom of Choice.

ZetaX: i don't know exactly what you mean, but in $ \bar{\mathbb{Q}_p}$ there is an huge (uncountable) number of valuations that extend the $ p$-adic valuation of $ \mathbb{Q}_p$. In facts, every time that extending a ring of algebraic integers a prime $ P$ splits (for instance as $ Q_1Q_2$), the valuation over the corresponding field given by the prime $ P$ is induced by two valuations over the extended field, that correspond to the primes above $ P$.
But of course this does not mean that your example does not work! :)

No it's unique:
$ \mathbb Q_p$ is henselian (it is complete with it's valuation), thus it's valuation extends to any algebraic extension in a unique way.

The problem at your construction is: you get only different valuations in that rings of algebraic integers, not in the completions.
E.g. take $ \mathbb Q[i] | \mathbb Q$ and consider $ 5\equal{}(2\plus{}i)(2\minus{}i)$. But as $ i \in \mathbb Q_5$, we get that $ \mathbb Q_5[i] \equal{} \mathbb Q_5$ is the trivial extension (end clearly the valuation extends in a unique way).

Uh... you're right, thanks, i will have to study again these things... :)
I was not considering that even if you can extend the valuation in different ways in a bigger field, the bigger field can actually be embedded in different ways in $ \mathbb{Q}_p$, and different embedding can induce different extensions of the valuation :)


A circle has equation $x^{2}+y^{2}=8$
what are the equations of th tangents and normals where the circle crosses the coordinate axes?

Dude...

Radius is $2 \sqrt{2}$. Equations of tangents are $y= \pm 2 \sqrt{2}, \; x= \pm 2 \sqrt{2}$

This goes in HSB

sorry but i didnt get this, can somebody xplain this abit more, n what is the equation of the normals

What do you mean by normals???  The normals of the tangents?  Isn't that just...

$y = 0$ and $x=0$