Let $ x,y,z$ be the non-negative real number satisfying $ (x\plus{}y\plus{}z)^2\plus{}xy\plus{}yz\plus{}zx\equal{}2$.Prove that $ \frac{x\plus{}y}{\sqrt{z^2\plus{}xy\plus{}1}}\plus{}\frac{y\plus{}z}{\sqrt{x^2\plus{}yz\plus{}1}}\plus{}\frac{z\plus{}x}{\sqrt{y^2\plus{}zx\plus{}1}}$≥$ \frac{3\sqrt2}{2}$

I really like your inequality.
we can let:a=x+y,b=y+z,c=z+x
then we have ab+bc+ca=2
and we have to prove that:
$ \sum_{cyc}\frac{a}{\sqrt{a^2\plus{}3bc}} \geq \frac{3}{2}$(1)
but i can't solve (1) .
can you or somebody help me,pleasezzz>?
thanks
 :blush:

[quote="tuandokim"]I really like your inequality.
we can let:a=x+y,b=y+z,c=z+x
then we have ab+bc+ca=2
and we have to prove that:
$ \sum_{cyc}\frac {a}{\sqrt {a^2 \plus{} 3bc}} \geq \frac {3}{2}$(1)
but i can't solve (1) .
can you or somebody help me,pleasezzz>?
thanks
 :blush:[/quote]
You have a right idea.Your idea is same to mine.We have if a,b,c  are 3 sides of a triangle,then
$ \sum_{cyc}\frac {a}{\sqrt {a^2 \plus{} 3bc}} \geq \frac {3}{2}$(1)
This ineq is not hard to prove.
$ aGA^2\plus{}bGB^2\plus{}cGC^2\equal{}(a\plus{}b\plus{}c)IG^2\plus{}abc$
=>$ 2ab(a\plus{}b)\plus{}2bc(b\plus{}c)\plus{}2ca(c\plus{}a)$≥$ a^3\plus{}b^3\plus{}c^3\plus{}9abc$(**)
Now use Cauchy-Schwarzt ineq:
$ \sum_{cyc}\frac{a}{a^2\plus{}3bc}$≥$ \frac{(a\plus{}b\plus{}c)^2}{\sum_{cyc}a\sqrt{a^2\plus{}3bc}}\equal{}\frac{(a\plus{}b\plus{}c)^2}{\sum_{cyc}\sqrt{a}\sqrt{a^3\plus{}3abc}}$≥$ \frac{(a\plus{}b\plus{}c)^2}{\sqrt{a\plus{}b\plus{}c}\sqrt{a^3\plus{}b^3\plus{}c^3\plus{}9abc}}$
So it remains to prove that $ \frac{(a\plus{}b\plus{}c)^2}{\sqrt{a\plus{}b\plus{}c}\sqrt{a^3\plus{}b^3\plus{}c^3\plus{}9abc}}$≥$ \frac{3}{2}$
This is equivalent to $ 4(a\plus{}b\plus{}c)^3$≥$ 9(a^3\plus{}b^3\plus{}c^3\plus{}9abc)$
<=> $ 12\sum_{cyc}ab(a\plus{}b)$≥$ 5(a^3\plus{}b^3\plus{}c^3)\plus{}57abc$
From (**),we have
$ 12\sum_{cyc}ab(a\plus{}b)$≥$ 6(a^3\plus{}b^3\plus{}c^3)\plus{}54abc$≥$ 5(a^3\plus{}b^3\plus{}c^3)\plus{}57abc$
The ineq hold iff $ a\equal{}b\equal{}c$,this leads to $ x\equal{}y\equal{}z\equal{}\frac{\sqrt6}{6}$

[quote="tuandokim"]I really like your inequality.
we can let:a=x+y,b=y+z,c=z+x
then we have ab+bc+ca=2
and we have to prove that:
$ \sum_{cyc}\frac {a}{\sqrt {a^2 \plus{} 3bc}} \geq \frac {3}{2}$(1)
but i can't solve (1) .
can you or somebody help me,pleasezzz>?
thanks
 :blush:[/quote]
It can solve by Cauchy-schwarz and SOS :)

[quote="tuandokim"]
$ \sum_{cyc}\frac {a}{\sqrt {a^2 \plus{} 3bc}} \geq \frac {3}{2}$(1)
[/quote]
Just use Holder, let the LHS be S, then by Holder
$ S^2(\sum a(a^2 \plus{} 3bc)) \ge (a \plus{} b \plus{} c)^3$
So we have to show $ 4(a \plus{} b \plus{} c)^3 \ge 9(a^3 \plus{} b^3 \plus{} c^3 \plus{} 9abc)$ which is obvious by Muirhead

Whoops, I realized that right after you posted  :oops:

[quote="Altheman"]That last inequality that you want to show is false... b=c=0 gives 4a^3>=9a^3 which is obviously false...[/quote]
Why is that false? Isn't $ a,b,c$ sides of a triangle, since tuandokim has set $ a \equal{} x \plus{} y, b \equal{} y \plus{} z, c \equal{} z \plus{} x$ ? :)

I get $ 4(a \plus{} b \plus{} c)^3 \minus{} 9(a^3 \plus{} b^3 \plus{} c^3 \plus{} 9abc) \equal{}$ $ (7b \plus{} 7c \minus{} 5a)(a \minus{} b)(a \minus{} c) \plus{} (19a \minus{} 5b \minus{} 5c)(b \minus{} c)^2 \ge 0$, which is obvious if we assume $ a \ge b,c$ :)

[quote="Agr_94_Math"]quykhtn-qa1 are u telling about solving the original inequality by Cauchy Schwarz and SOS???
I would like to see it...........[/quote]
I will must prove that:
$ 2(ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ca(c\plus{}a)) \ge a^3\plus{}b^3\plus{}c^3\plus{}9abc$
$ <\equal{}> (3c\minus{}a\minus{}b)(a\minus{}b)^2\plus{}(3a\minus{}b\minus{}c)(b\minus{}c)^2\plus{}(3b\minus{}a\minus{}c)(c\minus{}a)^2 \ge 0$  (*)
Suppose $ a \ge b \ge \ge c$
We have: $ 3b\minus{}a\minus{}c \ge 3b\minus{}c\minus{}(b\plus{}c) \equal{}2(b\minus{}c) \ge 0 ,,(c\minus{}a)^2 \ge (a\minus{}b)^2\plus{}(b\minus{}c)^2$
then$ LHS \ge (3c\minus{}a\minus{}b\plus{}3b\minus{}a\minus{}c)(a\minus{}b)^2\plus{}(3b\minus{}a\minus{}c\plus{}3a\minus{}b\minus{}c)(b\minus{}c)^2  \ge 0$
We have Q.E.D
 :wink: :)


ok, so i think i can handle the math problems but the application is a no-no  :P ... how does mathematics fit in my special interests? how do u answer a question like that? "ummmm... i like math?"  :blush: lol... & describe ur experiences with mathematics... i guess i could list out competitions and math fairs and all that but thats not gonna be too long & its more of a list than an essay  :? ... *sigh*  i need help, any Ross veterans out there, please share ur thoughts!!!!!!!!  ;)

Well I've been going for the past two years, and am returning as a counselor...

I wouldn't worry too much about these questions, just write what comes to mind.

you should probably focus on why you enjoy mathematics, and what makes you want to pursue it.


integration
logarithms
calculus
calculus computations

Evaluate $ \int_{0}^{1}\frac {\log ( 1 \plus{} x ) }{x} dx .$

[b]log(1+x) = x-(x^2)/2+(x^3)/3...........
expansion of log(1+X) AND THEN INTEGRATE[/b]

But how do i integrate infinite terms . Can u please work it out ?? And i don't think we can do that.

If you do so, you would get an infinite series $ \sum_{k\equal{}1}^{\infty}\frac{(\minus{}1)^{k\plus{}1}}{k^{2}}$ which equals to $ \frac{\pi^{2}}{12}$.

Oh ya thats correct ,

Any other ideas without using the series expansions of $ log (1\plus{}x)$ ??

I think they aren't others, there's no way to get the result by not using series.

By the way to justify its convergence, it's enough to see that the integrand has limit as $ x\to0$ ?

Yes, another method can be: \[ J\left( \theta  \right) \equal{} \int_0^1 {\frac{{\log \left( {1 \plus{} \theta x} \right)}}
{x}} dx\underbrace  \Rightarrow _{Leibniz}J'\left( \theta  \right) \equal{} \int_0^1 {\frac{{dx}}
{{\left( {1 \plus{} \theta x} \right)}}}\]

And then integrate.


Can you please provide a brief discription of the math courses below?

(1) Differential Equations

(2) Applied Mathematics

(3) Mathematical Logic

(4) Axiomatic Geometry

(5) Game Theory and Linear Programming

Thanks

You can find information about your recent posts in Wikipedia or other websites.

Okay, I will do a research online.  I just wanted the insight of most tutors regarding other areas of math not common to people who did not major in math.

Thanks anyway.


geometry
geometric transformation
reflection
abstract algebra
superior algebra
superior algebra solved

Find all $n\in \mathbb{N}$ for which there is an epimorphism from $D_{2n}$ to $C_n$.
[i](in our notation, $D_{2n}$ has $2n$ elements)[/i]

Let such a epimorphism be given.
There are $n$ reflections in $D_{2n}$, all their images having order $1$ or $2$ in $C_n$.
Now any of the other elements can be written as product of two reflections and their images have, since $C_n$ is abelian,  also order $1$ or $2$. But thats just possible for $n=1$ or $2$, since $C_n$ is by definition cyclic.


modular arithmetic
number theory unsolved
number theory

Find the remainder of $ 5^{100}$ when divided by $ 1001$

$ 1001 \equal{} 7\cdot 11 \cdot 13$. find the remainders of $ 5^{100}$ modulo $ 7, 11, 13$, using fermat's little theorem. then use these to find the remainder modulo 1001...


During one season from May to September, a pair of mosquitoes can become parents and ancestors to $ 1.7\times 10^{21}$ mosquitoes. If a large swamp in Florida contains $ 10^{10}$ pairs of mosquitoes, how many mosquitoes can be produced in one season. Express your answer in scientific notation.

Because each pair can be parents and ancestors to 1.7 x 10^{21} mosquitoes, the answer is simply $ 1.7 * 10^{21} * 10^{10} \equal{} \boxed{1.7 * 10^{31}}$
Someone please check this...


Find all $(m,n) \in \mathbb{Z}^2$ that we can color each unit square of $m \times n$ with the colors black and white that for each unit square number of unit squares that have the same color with it and have at least one common vertex (including itself) is even.

so, is the problem to find all m,n for which an mxn rectangle's unit squares can be colored W/B, so that every black squares has vertices touching to 0,2 or 4 (a square number) of other black squares and vice versa for white?

But that cannot be, since a chessboard always fulfils the condition [always 0]... please clarify yourself.

Peter, I have no idea what you find hard to understand here.. :?

For a small square, call all the squares which share at least a vertex with it (including itself) its neighbours. We have to find those $m,n$ for which it's possible to color the squares of an $m\times n$ board in two colors, s.t. each square has an even number of neighbours colored in the same color as itself.

I believe the answer is: those $m,n$ for which $2|mn$ (i.e. at least one of $m,n$ is even). If $m$ is even, for example, we color the first two lines in white, the next two in black, and so on.

Conversely, let's assume we can find such a coloring for our $m\times n$ board. We look at the graph having the black squares as vertices and in which two different squares are connected iff they are neighbours. The hypothesis tells us that in this graph, each vertex has an odd degree, so it has an even number of vertices. The same goes for the graph formed by the white squares, so the board has an even number of squares, and this is what we wanted to show: $mn$ (the number of squares) is even.

For $m \times n$ table that $m$ is even, Divide the table in $\frac m2$ of $ n \times 2$ tables.
And color the tables black and white

That's what I said, but i guess my post looked too long to be worth reading :).

Is that long? :? I'm still waiting for Pierre's first post of that length :D

As for my misunderstanding: I skipped the last line and thus thought that the number of neigbours not including itself needed to be even... and then a chessboard pattern does the job :)


3(p^q+q^p)=n! p,q are prime numbers and n is natural number. Find these unknown values

[b]Case 1:[/b] $ p \neq q$ and $ p,q \neq 3$, then $ p,q>n$ or one of $ p,q |n!$, which is impossible. Now $ 3(p^q\plus{}q^p)>6n^n>n!$, contradiction.
[b]Case 2:[/b] one of $ p,q$ , say $ p\equal{}3$, then the equation becomes $ 3(3^q\plus{}q^3)\equal{}n!$. As $ 3 \nmid 3^q\plus{}q^3$ , the only possible values of $ n$ are $ 3,4,5$. After checking we see that they are impossible also.
[b]Case 3:[/b] $ p\equal{}q$, then the equation becomes $ 6(p^p)\equal{}n!$. For $ n \ge 4$, we have $ 5!|n!\equal{}6(p^p) \Rightarrow (2^2 \times 5|p^p$, which is impossible .So the only possible values of $ n$ are $ 3$ and $ 4$. After checking, $ n\equal{}4, p\equal{}q\equal{}2$ is the only solution.


The acute angles of a particular isosceles trapezoid each measure 40 degrees. What is the measure, in degrees, of each obtuse angle in the trapezoid?

[quote="math_explorer"][hide]There are two acute angles and two obtuse angles in an isosceles trapezoid; the two acute angles are congruent, and the two obtuse angles are also congruent. The angle sum of any quadrilateral is 360 degrees, so the two obtuse angles sum to $ 360 \minus{} 2 \times 40 \equal{} 280$ degrees, so one obtuse angle is 140 degrees.[/hide][/quote]

Nice explanation.  :idea:

All you need to do is that the consecutive angles are supplemantary. So the other angle must be 140 degrees.


Hi!!
I have a problem to solve by tomorrow, please help me :!:
You have a cube whose sides measure 1 cm and you section it with a plan which passes in the vertexs A, B,C
(look at the attachment). I need to find out the volume of the piramyd it originates from sectioning the cube.
thanks for your answers :)

[hide="EDIT Answer"]
$V= \frac{ah}{3}$
Area of the base
$a= \frac{\sqrt{2}^{2}}{2}\rightarrow a=1$
Height of Pyramid
$h=1^{2}-l^{2}$
$l= \frac{{\sqrt{2}}^{2}-\left( \frac{\sqrt{2}}{2}\right)^{2}}{2}\rightarrow l=\frac{3}{4}$
$h=\frac{7}{16}$
Volume
$V=\frac{1 \cdot \frac{7}{16}}{3}\rightarrow V= \frac{7}{48}$
[/hide]

:|  :blush: I am sorry I did it wrong.
I will post a correction.


Last month sue made $\$$200 by baby sitting for 12 hours and tutoring for 20 hours.  This month she worked the same number of hours at each job as she did last muchm but she increased her babysitting fee by $\$$2 per hour and her turtoring fee by $\$$1 per hour.  How much did se earn this month?

$244$

sorry idonthave time toprove it :( .

sorry I can't understand muchm or se
[hide]200 + 12 x 2 + 20 x 1
= 200 + 24 + 20
= [b]244[/b][/hide]

[hide]
244, 12*2=24 is the extra that she got for baby sitting and 20*1=20 for tutoring.
200+24+20=244[/hide]


invariant
modular arithmetic
combinatorics unsolved
combinatorics

1994 lamps which can be switched single threaded are arranged in a circle and initially switched on. At random the lamps' state is changed (on-> off, or off -> on) by the following two rules:

1.) four consecutive lamps,

2.) among five lamps all states are changed except the middle one.

Is it possible to reach a state where all lamps are switched out ?

please explain the rules again
u haven't said nothing about those 4
those five, how are they chosen, and what does middle mean.

I think it's pretty clear: the moves you can perform are the following:

a) You can choose any four lamps in consecutive positions on the circle and change switch all of them (turn a lamp on if it's off and vice-versa);
b) You can chose any five lamps in consecutive positions on the circle and switch the first two and the last two (the "middle" remains unchanged).

You choose these groups of lamps randomly (orl actually mentioned that).

Number the lamps starting from one of them and going around the circle. Now regard the indices $\pmod 4$. For a certain configuration of lamps, consider $s$ to be the sum of the numbers of the lamps which are on $\pmod 4$. We can see that $s$ is always odd, so we can't have $0$ lamps which are on.

One sees that an operation repeated twice cancels each other
Consider the following
xxxx
..xxxx
.xxxx=
x.xx.x
and
xx.xx
.xx. xx=
x.xx.x

So we see that two neighbour operations of the second type can be replaced by operations of the first type
Since two operations cancel each other we see
$O_iO_j=O_iO_{i+1} O_{i+1}O_{i+2} \cdots O_{j-1}O_{j}$.
$O_i$ is the operation of second type  applied with lamp #i
Hence every two operations of the second type can be replaced by operation of the first type.
Hence we must so\\look just at two cases:
1. All operations are of first type
2. All are of the first type execept one.

To solve the first case denote the opqrations $a_1\cdots a_n$ where $a_i$ is the
leftmost square changed at operation #i.
Since every lamp is changed odd number of times one gets
$a_i$ and $a_i-1$ are changed the same # of times which means $a_i-4$ also belong to $a_1, \cdots .. an$. Now we see that the operations cover $4k$ lamps which cannot be true since $1994=4k+2$.
Analogously one can handle the second case.

Here is a brief outline of my solution, it shares some resemblance with grobber's:

Label each lamp Ai (i=1,2,...,1994), now assign Ai=0 if lamp is on and Ai=1 if lamp is off. After each operation (either i) or ii)), calculate S=Sum (A_i after - A_i before)^2 over all 1994 Ai. Observe that S changes by 4 after each operation. Initially, S is, by definition, 0 (multiple of 4). So an easy contradiction argument follows because 1994 is not divisible by 4.

Next time, why not change 1994 to 2004 for some modernization, eh?  :D

But with the present conditions, that problem (with $2004$) would be trivial :).

Yes. In the way stated there it's a quite stupid problem. Either 4 $|$ n, then it's trivially possible, either 4 $\not|$ n, then it's trivially impossible

I see that al.M.V.'s solution is better than mine: for me it was crucial that $1994=2 \pmod 4$.

grobber, your solution is just as equally effective.  :)


Let $x,y,z>0$ such that:$x+y+z=1$.Prove that:
\[\frac{xy}{1+z}+\frac{yz}{1+x}+\frac{xz}{1+y}\leq \frac{1}{4}\]


I have 300 or so of these so if you want more help improving your logic, PM me and I'll type up some more problems.

On the show Quiz Kids, five contestants (Males: Caleb, Dean, and Paul; Females: Megan and Shirley) were questioned about pop music each from a different decade (40s, 50s, 60s, 70s, or 80s). In the next round, they each answered questions about a different 20th century figure. (Albert Einstein, Charles Lindbergh, Dwight Eisenhower, Pablo Picasso, and Sigmund Freud) Each contestant ended up with a different # of points and a different prize (Bomber Jacket, Cutlery, Dinner for Two, Snowboard, and a Vacation) Determine the decade and 20th century figure each contestant answered questions about, the number of points, and the prize they went home with.

Clues:
1. The three men are the ones who answered questions about Albert Einsten, the one who won the snowboard, and the one who scored exactly 5 points. The two women are the one who answered questions about 80s music and the one who won a vacation.
2. The three people who answered questions about Charles Lindbergh, Dwight Eisenhower, and Pablo Picasso are the one who discussed music of the 40s, the one who won a set of cutlery (who scored 30 points), and Shirley, in some order.
3. The sum created by adding 40 points to Paul's score is equal to either Megan's or Shirley's score. The difference created by subtracting Paul's score from 40 is equal to either Megan's or Shirley's score. Paul scored exactly 10 fewer points that the one who answered questions about 60s music.
4. Shirley didn't answer questions about 70s music. Caleb (who isn't the one who answered questions about Sigmund Freud) is not the one who won dinner for two at a swanky downtown restaurant.
5. The one who discussed 70s music (who didn't win a snowboard) ended the game with exactly 20 fewer points that the one who answered questions about Lindberg. The one who won the bomber jacket didn't answer questions about Picasso.

Suggestion: slap Hide tags around your clues.

Suggestion: the clues are necessary to solve the problem - without all of them, you can't solve it


ratio
geometry
trigonometry
quadratics
algebra
quadratic formula

Find $k$ if P,Q,R, and S are points on the sides of quadrilateral ABCD so that $\frac{AP}{PB}=\frac{BQ}{QC}=\frac{CR}{RD}=\frac{DS}{SA}=k$, and the area of quadrilateral PQRS is exactly 52% of the area of quadrilateral ABCD.

[hide="A (well-known, maybe) formula says that"]
\[\frac{ [PQRS] }{ [ABCD]}= \frac{k^{2}+1}{(k+1)^{2}}\]

[b]Try to prove this on your own.[/b] The solution is gives is $\boxed{k= \frac{2}{3}}$ or equivalently $\boxed{k= \frac{3}{2}}$. [/hide]

Never seen that formula before. :D Here's how I solved it.

[hide="Areas"]Let the area of the quadrilateral $ABCD$ be $x.$ Also, let $AB=a,$ $BC=b,$ $CD=c,$ and $DA=d.$ It follows that $AP=\frac{ka}{k+1},$ $PB=\frac{a}{k+1},$ and etc.

Note that\[ad\sin A+bc\sin C= 2x\]
and\[ab\sin B+bc\sin D=2x.\]

Additionally,
\begin{eqnarray*}2(48\%) &=& (AP)(AS)\sin A+(BP)(BQ)\sin B+(CQ)(CR)\sin C+(DR)(DS)\sin D \\ &=& \frac{ka}{k+1}\frac{d}{k+1}\sin A+\frac{kb}{k+1}\frac{a}{k+1}\sin B+\frac{kc}{k+1}\frac{b}{k+1}\sin C+\frac{kd}{k+1}\frac{c}{k+1}\sin D \\ &=& \frac{k}{(k+1)^{2}}(ad\sin A+ab\sin B+bc\sin C+cd\sin D)\\ &=&\frac{k}{(k+1)^{2}}(4x). \end{eqnarray*}
Hence, $\frac{96}{100}=\frac{k}{(k+1)^{2}}(4x)$ and simple quadratic formula yields $k=\frac{3}{2}, \frac{2}{3}.$

[/hide]

The way I solved this when I first saw out of the AoPS book used a little cheating... since it holds for any quadrilateral, it holds for a square.
Let the big square have area 25 so the little has area 13. Clear, the sides of the big are 5 by 5, and we find that AS to SD is 3 to 2 since 3^2+2^2=13.
The result follows :P

uhh...that formula on teh noah sheets was basically copying the problem here...so i suppose if you mean this is a well-known problem...which i think it is, then fine, but this is obviously a result that you would just prove on a contest

I think it's also pretty easy to use vector/cross products? You should get the 'well known formula' mentioned earlier.


I'm just wondering, do people at AOPs play trading card games like Magic, I played magic back in middle school with the same group of people I did math team with, but then we all kinda drifted away from it.  What about you people?

I have pokemon which I play with my sister and a couple friends.

Wow! I play Magic. Used to play Pokemon and Yu-gi-oh

What is the point of these pieces of paper that have pictures on them?

lol, what's the point of those large rectangular pieces of grass with lines painted on them?  or those strange boards with black and white squares on them and funny pieces on top. or for that matter --- does it matter if we can prove that if you have 6 points on a plane (no 4 coplanar) that you can connect 3 with one color and the other 3 with another color and one passes thru the interior of the other.

Seriously tho CTG's expand your mind in a way no other game does, besides the fact that they're SO much more fun then a limiting game like chess or a physical game like football. ;)

[quote="NAS"]lol, what's the point of those large rectangular pieces of grass with lines painted on them?  or those strange boards with black and white squares on them and funny pieces on top. or for that matter ---does it matter if we can prove that if you have 6 points on a plane (no 4 coplanar) that you can connect 3 with one color and the other 3 with another color and one passes thru the interior of the other.

Seriously tho CTG's expand your mind in a way no other game does, besides the fact that they're SO much more fun then a limiting game like chess or a physical game like football. ;)[/quote]

Then how come chess has survived so long. And so many books, strategies been developed.

I used to play Pokemon, switched over to Yu-gi-oh!

[i]Moderator note: Watch the language, Tare.[/i]

Realistically, CCGs are far more limiting than chess, as (in most cases), given two CCG players with roughly the same experience, the one with the more expensive deck will win. For example, assuming you're into magic... can you imagine a Storm deck with power 9 losing to a $5 stompy? Such a situation would never happen in chess, where money doesn't inevitably create an unfair playing ground. In general a good game of chess requires more thinking, as does finding a solution to a math proof.
I'm being quite hypocritical here since I've spent about $300 between magic in middle school and pokemon in fifth grade... But I was mostly undefeatable in both games against other players with years more experience who hadn't spent as much on their decks, which further proves my point.

I have to disagree, (I have to say I love chess, and math obviously) the key difference between chess and magic is that in magic half the fun (at least for me) is that building the decks.  The other part is the 'on your feet thinking' you draw seven cards, sure you have (or should have) a very good idea what kind of cards you're going to get but you never know.  You also (at least in the beginning) have no idea what kind of cards your opponent has.  In short, it's the before game creativity, and the in game randomness.  With chess, on the other hand, everything is out in the open and predictable aka in every chess game you play, you have exactly 20 opening moves.
Not to be too poetic, but the reason I like chess is the reason I like this site.  The reason I like Magic is the reason I like music.

[quote] can you imagine a Storm deck with power 9 losing to a $5 stompy [/quote]
Sure, it's certainly not very likely but that's the thing about Magic that I love(also the formats).  The Stompy deck has a better chance of winning that game by a person that's never played magic before then someone that's never played chess playing with Bobby Fischer.

Do you people who play magic have like really, really good cards? If you got something really good, please post it. My cards aren't very good.

Umm... probably the rarest card I have is an Alpha Sierra Angel and/or and alpha Nightmare

My bro's rarest card is planeshift Draco which isn't that rare but it's pretty good.

Alpha Serra Angel!!!!!!!!!!!!!!!!!!!!!!! WHOA!!!!!!!!!!!!!!!!!
Mine is Akroma, Angel of Wrath.

MTG rocks!!!

I have a counter control deck and a seismic assault deck which throws lands - how about you guys?

I agree with PI(up there aka 3.14159265358979 etc). Money is a major factor in Magic -  my friend spends lots on Magic and builds a great deck, but I don't and I have a mediocre deck. However, there is the point that the win loss is 50/50.

Everyone that plays magic w/me is also on my MC team - strange. It must take smart people then to build a good deck with little money.
With a few thousand different cards - there is always a good undiscovered strategy awaiting.

I used to play magic in the fifth grade....but its way to expensive so I dropped out in the sixth.  Chess limited! Pah! Magic is more limited because you actually need to spend money on it.  Its more like a "who has the most money to spend" game than chess where everyone is equal and takes lots of thinking to beat your opponents.

WoW!! Akroma is good!

Anybody have a dragon deck?

Uhh... I got an Elf, a Mercanary and a spectrum deck. They're pretty bad but oh well...

MAGIC rules!!!
I have several decks, and I luv arcbound creatures-DArksteel
Cheap combination-Broodstar, Mycosynth Lattice, and DARKSTEEL FORGE!!!!=
Superpowered indestructible, giant bashing BROODSTAR

I got 4 Darksteel Forges, 3 Memnarchs, a Phage the Untouchable, a Broodstar, 5 Mycosinth lattices, and a whole ton of other rares.
I build stomp decks that are semitribal with lots of support. AKA Darksteel Forge: Artifacts that you control are indestructible and Mycosynth Lattice:  All permanents become artifacts in addition to other types. All spells are colorless. Players may spend mana as though it were mana of any color and Broodstar: Power and toughness equal to the number of artifacts you control. Has flying too!
Arcbound creatures rule with Fangrens.

I have a pro-black (turns everything black, courtesy of Darkest hour, then gives me super pro black- Multiple "Sphere of Grace"s
, an Arcbound(modular) deck as mentioned above, and  black red nightmare burn deck.

Dragon decks are good if you use a lot of squirrel tokens and day of the dragons!

I had a beta/alpha (can't remember) Ancestral Recall..... I have a whole bunch of cards because I used to play it in 6th grade.  I guess the best deck I ever made never consisted of expensive cards, just cheap ones.

Horseshoe Crab + Hermetic Study

I believe trading card games are fun and I would love to play Magic if I had my own deck, but I'm not willing to spend my own money on them. Pokemon battling is fun too but doesn't require as much ingenuity and deck modifying skills as Magic.  Yet, since I have thousands of Pokemon in my closet, I would use them to battle if there were some Pokemaniacs out there who aren't afraid to admit Pokemadness.

Well the ancestral recall was in fair condition... I bought it for 20 dollars one day and I got 50 dollars the next! Ka ching!

pokemon OWNS! the game is pretty good, and its not "in fasion" so the cards are really really cheap and you can have a good deck without paying as much (those cost a dollar) as "in fasion" yugioh and magic decks. (those cost like 12 dollars).
sum1 already voted in my account.

whats wierd about tcgs are that popular tcgs are always denounced by some other group

i used to play magic, pretty seriously actually about 7 years ago, pokemon was either shortly before or shortly after (I think before)... I might still enjoy playing them, except I have more important things that consume my time these days.

I dont see the point in these games but I do know how to play Yu-gi-oh and pokemon for what its worth. I dont know any of the cards and Im not very good at it one of my best friends lil brothers taught me how to play. I still see no point why not just play regular cards there is so much more u can do with them. Or why not have baseball trading cards which totally rock. My brother collects football, baseball, and nascar trading cards why cant we just go back to those?

Does anyone who plays MTG have a Serra Avatar? Those things are like $30.

[quote=mikimoto12]14 year BOOOMP
Awesome! Another 3[/quote]

aaaaaaaarggggghhhhh


Let $\large a_0;a_1;. . . ;a_n$ be a sequence satisfying: $\large a_0=1; a_{n}=a_{[7n/9]}+a_{[n/9]}$ for all n=1;2;. . .
Prove that there exist a positive integer k with $\large a_k < \frac{k}{2001!}$

The expectation is that $a_n$ is within a small (e.g., bounded or logarithmic) factor of ${n^c}$ for some $c < 1$.   There is only one possible value of $c$, that is the one satisfying  $9^c = 7^c + 1$.  Proving it seems like a difficult problem about uniform distribution.


ceiling function
graph theory
combinatorics unsolved
combinatorics

Given a complete graph on $ k\ge 3$ vertices whose edges are colored in black and white s.t. every triangle has an even number of black edges (either $ 0$ or $ 2$). Prove that we can find a complete subgraph on $ \left\lceil\frac {k}{2}\right\rceil$ vertices all of whose edges are white .
it was in [url]http://www.mathlinks.ro/viewtopic.php?p=1652182#1652182[/url] but i couldn`t understand the proof .
(sorry for my bad English) :D

grobber wrote
[quote]It’s easy to see that the partial graph generated by all the vertices and the black edges is a complete bipartite graph. If $ A,B$ is its bipartition, the subgraphs induced on $ A,B$ are both complete graphs with white edges only, and at least one of $ |A|,|B|$ is $ \ge\left\lceil\frac k2\right\rceil$, so we’re done.[/quote]
Take two vertices $ x, y$, joined by a black edge, and put them in two classes $ A,B$. Any third vertex $ z$ is joined to exactly one of $ x,y$ by a black edge - put it in the opposite class. Now check each class cannot contain black edges, so they are complete subgraphs with white edges only.

By the way, you made a typo when copying. You wrote
[quote]even number of black edges (either $ 0$ or $ 1$)[/quote]
$ 1$ is odd. It should be (either $ 0$ or $ 2$), like in the original.

thank you very much !!! (I edited my typo)


calculus
integration
calculus computations

I just need help getting started mainly with problem # 1.  However if you find this problem to be a challenge go ahead and have at it.

Air resistance is a force that acts in the direction opposite to the motion and increases in magnitude as velocity increases, let us assume at least initially that air resistance r is proportional to the velocity: r = pv, where p is a negative constant. suppose a ball of mass m is thrown upward from the ground. The net force f on the ball is F = r – mg (the direction of the force r is downward (negative) when the ball is traveling upward and the direction of r is upward when the ball is traveling upward.)

1. Use the net force equation and the fact that F = ma to write a differential equation for the ball’s velocity.

2. Assume m = .5kg and p = .1. Make a direction field for the differential equation and sketch a solution of the initial value problem v(0) = 50 m/s.

3. Solve the initial value problem algebraically. Hint: be sure to take the constant of integration into account.

4. Find an equation of the height of the ball at time t.

5. When does the ball reach the apex of its trajectory? When does the ball land?

6. Does it take the ball longer to come up or come down?

From $ma=pv-mg$ it follows that $\frac{dv}{dt}=\frac{p}{m}v-g$.


Points $ B$ and $ C$ lie on $ \overline{AD}$. The length of $ \overline{AB}$ is $ 4$ times the length of $ \overline{BD}$, and the length of $ \overline{AC}$ is $ 9$ times the length of $ \overline{CD}$. The length of $ \overline{BC}$ is what fraction of the length of $ \overline{AD}$?

$ \textbf{(A)}\ \frac{1}{36} \qquad
\textbf{(B)}\ \frac{1}{13} \qquad
\textbf{(C)}\ \frac{1}{10} \qquad
\textbf{(D)}\ \frac{5}{36} \qquad
\textbf{(E)}\ \frac{1}{5}$

[hide]Since it doesn't matter the length since they are just ratios i took the length to be 10. and so BD = 2. and CD = 1 so BC = 1 and so BC/AD = $ \frac{1}{10}$ $ \fbox{\mbox{C}}$ sry i am too tired to do it in Latex now.[/hide]

[hide]Let BD = x and CD = y.

Thus,
AB = 4x and AC = 9y
AD = AB + BD = 4x + x = 5x
AD = AC + CD = 10y
AD = 10y = 5x
x=2y

BC = AC - AB = 9y - 4x = 9y - 8y = y
BC/AD = y/(10y) = 1/10[/hide]


calculus
integration
trigonometry
calculus computations


let  $\mu^{*}$ be an outer measure and $\mu^{-}$ be the  restriction
of $\mu^{*}$ on $\mu^{*}$ measurable sets. prove that $\mu^{-}$ is
saturated.

I don't know what a saturated measure is. Maybe you mean "complete" (that would be "$\mu^{-}(A)=0\implies$ all subsets of $A$ are $\mu^{-}$ measurable").

let$(X,\mathbf{B},\mu)$be a measure space
 we say  $E$ subset of $X$  is locally measurable
  if for each $B\in\mathbf{B}$ with $\mu(B)<+\infty$ we have  $E \cap B\in\mathbf{B}$

we say $\mu$ is saturated if each locally measurable set be measurable
 .


trigonometry
geometry
angle bisector
geometry unsolved

$ D$ is a point on the edge $ BC$ of triangle $ ABC$ such that $ AD\equal{}\frac{BD^2}{AB\plus{}AD}\equal{}\frac{CD^2}{AC\plus{}AD}$. $ E$ is a point such that $ D$ is on $ [AE]$ and $ CD\equal{}\frac{DE^2}{CD\plus{}CE}$. Prove that $ AE\equal{}AB\plus{}AC$.

Who know the solution of this problem?  :?:

[color=darkred][size=117][b]A nice and difficult (for who doesn't know the lower lemma) metrical problem ![/b][/size][/color]

[quote="anonymous1173"][color=darkred]$ D$ is a point on the edge $ BC$ of triangle $ ABC$ such that $ AD = \frac {BD^2}{AB + AD} = \frac {CD^2}{AC + AD}$. Denote

the point $ E$ is a point such that $ D\in [AE]$ and $ CD = \frac {DE^2}{CD + CE}$. Prove that $ AE = AB + AC$.[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] The relations from hypothesis suggest me to apply the well-known property :[/color]

[quote][color=darkred][b][u]Lemma.[/u][/b] Let $ ABC$ be a triangle. Denote the point $ D\in BC$ for which $ AD\perp BC$ .

Suppose $ D\in (BC)$ . Then $ b^2 = c(a + c)\Longleftrightarrow B = 2C\Longleftrightarrow DC = DB + BA$ .[/color]

[color=darkblue][b][u]Proof.[/u][/b] $ \boxed {\ b^2 = c(a + c)\ }\Longleftrightarrow b^2 - c^2 = ac\Longleftrightarrow$ $ DC^2 - DB^2 = ac\Longleftrightarrow$ $ \boxed {\ DC = DB + BA\ }\Longleftrightarrow$

$ b\cos C = c + c\cdot \cos B$ $ \Longleftrightarrow$ $ \sin B\cos C = \sin C(1 + \cos B)$ $ \Longleftrightarrow$ $ \tan\frac B2 = \tan C$ $ \Longleftrightarrow$ $ \boxed {\ B = 2C\ }$ .[/color][/quote]

[color=darkblue]Denote $ \|\begin{array}{c} m(\widehat {ABC}) = x \\
 \\
m(\widehat {ACB}) = y \\
 \\
m(\widehat {CED}) = z\end{array}$ . Apply the upper lemma to the triangles :

$ \|\begin{array}{cccc} \triangle\ ABD\ : & BD^2 = AD\cdot (AB + AD) & \implies & m(\widehat {BAD}) = 2\cdot\widehat {ABD}) = 2x \\
 \\
\triangle\ ACD\ : & CD^2 = AD\cdot (AC + AD) & \implies & m(\widehat {CAD}) = 2\cdot m(\widehat {ACD}) = 2y \\
 \\
\triangle\ CDE\ : & DE^2 = CD\cdot (CE + CD) & \implies & m(\widehat {DCE}) = 2\cdot m(\widehat {CED}) = 2z\end{array}$ .

Observe that $ A + B + C = 180^{\circ}$ $ \Longleftrightarrow$ $ (2x + 2y) + x + y = 180^{\circ}$ $ \Longleftrightarrow$ $ \boxed {\ x + y = 60^{\circ}\ }$ ,

$ m(\widehat {ABD}) + m(\widehat {BAD}) = m(\widehat {CED}) + m(\widehat {DCE})$ $ \Longleftrightarrow$ $ x + 2x = z + 2z$ $ \Longleftrightarrow$ $ \boxed {\ z = x\ }$ and

$ ABEC$ is cyclically. Thus, $ AE = AB + AC$ $ \Longleftrightarrow$ $ \sin \widehat {ACE} = \sin\widehat {ACB} + \sin\widehat {ABC}$ $ \Longleftrightarrow$

$ \sin (2x + y) = \sin y + \sin x$ $ \Longleftrightarrow$ $ \sin (2x + y) - \sin y = \sin x$ $ \Longleftrightarrow$

$ 2\sin x\cos (x + y) = \sin x$ $ \Longleftrightarrow$ $ \cos (x + y) = \frac 12$ $ \Longleftrightarrow$ $ x + y = 60^{\circ}$ , what is truly.[/color]

We have $ AD=\frac{BD^2}{AB+AD}\Rightarrow AB\cdot AD = BD^2 - AD^2 $.

Let the circle with center $D$ and radius $DA$ cut $BA$ at $X$.
The power of $B$ is $BD^2 - DA^2 = BA \cdot BX$.
So $BX = AD = XD \Rightarrow \angle BAD = 2\cdot \angle ABD = 2\alpha$.
Similarly, $\angle DAC = 2\cdot \angle ACD = 2\beta$.
We have $\alpha + \beta + 2\alpha + 2\beta = 180^\circ \Rightarrow \alpha + \beta = 60^\circ$.
Again similarly, and since $\angle EDC = \angle BDA$, $\angle DCE = 2\cdot\angle DEC = 2\alpha$.
Since, $\angle BCE = \angle BAE = 2\alpha$, $B,A,C,E$ are concyclic.

Let $H$ be foot of altitude from $B$ to $EC$.
Let $BH$ and $AE$ meet at $F$.
Since $\angle BAC = 120^\circ$, $\angle BEH = 60^\circ$.
Let $G$ be on $EC$ such that $\triangle EBG$ is equilateral.
Let $BG$ meet $AE$ at $I$.
Since $F$ is on the angle bisector of $\triangle EBG$, $EF=FG$ and $\angle IFG = 2\angle FEG = 2\alpha$. So $FG \parallel AB$.

By Sin Law, $BI/IG = AB/AC$.
If you want synthetic proof: $EI$ cuts $(BEG)$ at $J$. $\triangle JBG \sim \triangle ABC$. $JI$ is angle bisector of $\angle BJG$. So $AB/AC=BJ/JG=BI/IG$

$AB \parallel FG \Rightarrow AB/FG = BI/IG = AB/AC \Rightarrow FG = AC = EF$.
We have $\angle EFH = \angle BFA = 90^\circ - \angle AEC = 90^\circ - \alpha$ and also $\angle BAF = 2\alpha$. So $AB=AF$.
$AE=AF+FE=AB+AC$. $\blacksquare$


A collection S of integers is defined by the following three rules: (I) 2 is in S: (II)for every x in S, 3x and x+7 are also in S; (III) no integers except those defined by rules I, II are in S. What is the smallest integer greater than 2004 which is NOT in S?

Since the title of this topic is "Another Hard Question", I think I screwed this up...

[hide]Well first, you can notice that 2004 is one of the numbers that fall under the category of "x+7". (since $ 2\plus{}7\cdot286\equal{}2004$)
Then, 2005 Doesn't fall under the "x+7" category. Also, $ \dfrac{2005}{3}\equal{}\text{Integer}$. Hence, $ \boxed{2005}$ is the answer?[/hide]

If we look at this mod 7, we quickly see that all numbers of the form $ 2 \mod 7$ fit are in $ S$. Also if $ 2$ is in $ S$, then $ 6$ is in $ S$ and all numbers in the form $ 6 \mod 7$ are in $ S$ as well. Then all numbers in the form $ 4 \mod 7$ ($ 18 \equal{} 4 \pmod 7$) are in $ S$, and so are $ 5 \mod 7$, and $ 1 \mod 7$, and $ 3 \mod 7$, and back to $ 2 \mod 7$. So we see that all numbers are in $ S$ except some smaller numbers and eventually just the multiples of 7. So our answer is the smallest multiple of 7 greater than 2004, which is $ \boxed {2009}$.


inequalities
inequalities unsolved
acute

if a,b,c are the sides of a right triangle prove that $ c^n > a^n \plus{} b^n$.$ n>2$

I suppose $ c$ is the hypothenuse of the triangle. :maybe:

Is $ n>2$? For $ n\equal{}1$ or $ n\equal{}2$ for example your inequality does not hold...

[quote="aadil"]if a,b,c are the sides of a right triangle prove that $ c^n > a^n \plus{} b^n$[/quote]

first  we have $ p > 1$
 $ (x \plus{} y)^p > x^p \plus{} y^p$

$ x^p \plus{} y^p \equal{} (x \plus{} y)^p \minus{} (x \plus{} y)(x \plus{} y)^{p \minus{} 1} \plus{} x^p \plus{} y^p \equal{} (x \plus{} y)^p \minus{} x((x \plus{} y)^{p \minus{} 1} \minus{} x^{p \minus{} 1}) \minus{} y((x \plus{} y)^{p \minus{} 1} \minus{} y^{p \minus{} 1}) < (x \plus{} y)^p$

Therefore,

$ c^n \equal{} (c^2)^{\frac {n}{2}} \equal{} (a^2 \plus{} b^2)^{\frac {n}{2}} > a^n \plus{} b^n$

while $ n > 2$ and n is positive reals


How many positive integers $k$ leave a remainder of 2 when 101 is divided by $k$?

[hide]If the remainder when divided by $k$ is 2, then $k$ must be a factor of $101-2=99$.  The factors of 99 are: ${1,3,9,11,33,99}$.  However, 1 does not work, so the only values of $k$ which satisfy are ${3,9,11,33,99}.$

The number of values of $k$ which satisfy are $\boxed{5}$
[/hide]


algebra polynomial calculus derivative induction

How can we prove that a root a of a polynomial f(x) has multiplicity m iff f(a) = f'(a) = f''(a) = f'''(a) = ... = (m-1)th derivative of f(x) = 0 ?

One way is to use Taylor's theorem, which for polynomials goes like this.  Let f be a polynomial of degree n, and let a and x be numbers.  Then f(x) equals the sum (for k from 0 to n) of f(k)(a) (x - a)k / k!.

Do you see how Taylor's theorem implies what you want?

[quote]Do you see how Taylor's theorem implies what you want?[/quote]

Not really...but that might simply be because I am not too familiar with the Taylor Series and am missing something obvious.

[quote]Then f(x) equals the sum (for k from 0 to n) [/quote]

Isn't it from 0 to infinity (at least this is what my calculus textbook says)?

Hint:  Prove that if g'(x) has multplicty j of root k, and g(k)=0, then g has j+1 roots.

[quote]Well, you could think of the sum as an infinite sum, but for a polynomial of degree n, all the derivatives past the nth derivative are 0.[/quote]

Yes, but aren't we looking to show the opposite? We must show that f(a) = f'(a) = f''(a) = f'''(a) = ... = (m-1)th derivative of f(x) = 0, not that (m+1)th derivative of f(x) = (m+2)th derivative of f(x)  = ... = 0, which is what your statement suggested, no?

[quote]Prove that if g'(x) has multplicty j of root k, and g(k)=0, then g has j+1 roots.[/quote]

So induction, eh?

You've asked two separate questions.  One was your original post.  The second was why my statement of Taylor's theorem had a finite sum.  My last post, about the derivatives of a polynomial being 0 past a certain point, was meant to answer your second question, not your first.

Oh, I see, you were commenting on my question "Isn't it from 0 to infinity (at least this is what my calculus textbook says)?" Anyways, thanks for the responses.


I got one point off for Round 3 Problem 5 because the grader said that I did not prove the link between the Dirichlet series and the Phi Function. But in the solution, I proved it:

[img]http://img263.imageshack.us/img263/8661/prooffp3.jpg[/img]

Possibly the grader wanted you to prove that $ \sum_{d | h} \varphi(d) \equal{} h$.

Don't worry, you're not the only one who lost a point for citing wikipedia...

But I'm not just citing it...I'm showing the proof, which is completely valid.

Yeah, but I easily could have copy-pasted it from Wikipedia as well, since I'm assuming you didn't come up with that yourself.

I did the exact same thing as you and got 1 point off. In my case, I didn't use Wikipedia at all, but used knowledge about Dirichlet convolutions that I learned at Mathcamp last year. I figured that that would be common knowledge, but apparently it wasn't  :o

Did you remember cite Wikipedia? I copied the proof and cited the URL, and I got a 5.

I forgot to cite wikipedia...but the grader didn't cite that as the reason for taking points off...he said i didn't prove it.

The rubric specifically said that citing wikipedia should result in taking off a point.

I got a commended 5 and I didn't even go through all of those steps.  I jumped from the end of 123456789's line 1 to the end of line 3 by just saying that we sum along a different index.  I did show a simple proof of Gauss's lemma (that is, $ \sum_{d|k} \phi(d) \equal{} k$).  It's really quite easy if we consider reducing the fractions $ \frac {1}{k}, \frac {2}{k}, \cdots, \frac {k}{k}$.  But then again, I did come up with the "Dirichlet convolution" idea independently of an outside source.


[b]The Sagebrush student council has 6 boys and 6 girls as class representatives. Two committees, each consisting of 2 boys and 2 girls, are to be created. If no student can serve on both committees, how many different combinations of committees are possible?[/b]

I got a completely diff answer

[hide]

so I got

(6) *  (6)
(2)     (2)

so it is n!/(n-k)!k!

or 6!/(6-2)!2!

or 720/48

this simplifies to 15 and 15*15=

[225][/hide]

Edit: I realized that each committee had 2 boys and two girls, not one with 2 boys and one with 2 girls.

[quote="lokesh"][hide]$ \binom{6}{2}\cdot\binom{6}{2}\cdot\binom{4}{2}\cdot\binom{4}{2}$
This equals $ 15*15*6*6$ or $ 8100$. Now we divide this by $ 2!$ and get $ \boxed{4050}$. [/hide][/quote]

I agree with you on $ \binom{6}{2}\cdot\binom{6}{2}\cdot\binom{4}{2}\cdot\binom{4}{2}$.

But why do you divide it by $ 2!$?

[quote="alanchou"]In the original count, the committee group |AB|CD| was counted as a different committee than |CD|AB|.[/quote]
Oh, okay. Thanks.


As much as I hate to admit it, I found myself scrambling to solve my final unsolved problem late last night (solution had eluded me for a good week at that point).  The good news:  I solved it.  The bad news:  I had roughly 3 minutes until the deadline for submissions, and, being overly hasty, sent an e-mail titled simply "Round 3 Solutions," as opposed to "Round 3 - username - ID#."  Looking at the submission guidelines afterwards (wonderful idea, right?), I saw that my speed-titling was certainly a bad idea, but I also did not want to submit a second set of solutions, since that could potentially make the situation worse.  My question is simply... am I screwed (perhaps rightly so, for wasting someone's _free_ time on something like this), or is there something that I should do to correct this?  My assumption is that whoever opens the e-mail will see the lack of information in the title and check the attached PDF (which takes time, and thus, I should have done for them, obviously).  However, if a script runs through this stuff, then I'm in trouble.

Whatever happens, thanks for your time in advance.

My advice is to reply to your own e-mail, send it to the solutions address again, and attach the full title, explaining in the body your mistake.  I don't think you'll need to attach another set of solutions, as they'll see that your second e-mail is coming from the same e-mail address.  I've made this mistake before, and they've had no problem with that way of correcting it.

Humans go through it, not a script.  Don't worry about it unless you don't have a green checkmark next to Round 3 on the Scores page at the end of the week.

One thing you can always count upon in these forums are solid, quick responses to questions.  Thanks a bunch!

i always send with subject "Round # Solutions"

is that bad?  :blush:  :blush:

It's not 'bad', but we would like you to follow the instructions regarding including identifying information.  You can find them here:

http://www.usamts.org/Problems/U_ProblemsSoln.php

Not bad as in doomsday bad, as evidenced by the responses from Richard and Zanttrang, but I believe that it does make it easier on the one who runs through all of the e-mails if you have your ID number and name in the topic, as they ask for you to send them in that format on the submitting solutions page (http://www.usamts.org/Problems/U_ProblemsSoln.php).

ah ok.

i had my USAMTS ID in the body of the email, but I'll change next time :)


[quote="ckck"]Alfreda hiked half her planned trip the first day;tiring she only covered one-third the remaining distance the second day; rain slowed her to one-fourth the remaining distance the third day, and the fourth day she hiked three miles, which was one-fifth of what she had left. How many miles did she hike during the four days?[/quote]

[hide]On the first day she hiked 1/2, the second she hiked 1/6, and the third she hiked 1/12. This leaves 1/4 left.
3=x/20
x=60 total miles

So she hiked a total of 48 miles.[/hide]


For what value of $ x$ does $ \frac89 \equal{} \frac{x}{x\plus{}\frac{x}{x\plus{}x}}$?

[quote="GameBot"]For what value of $ x$ does $ \frac89 \equal{} \frac {x}{x \plus{} \frac {x}{x \plus{} x}}$?[/quote]

$ \frac{8}{9}\equal{}\frac{x}{x\plus{}\frac{x}{x\plus{}x}}\equal{}\frac{x}{x\plus{}\frac{1}{2}}$

$ 8x\plus{}4\equal{}9x$

$ x\equal{}4$.


Let a,b,c >0, a<1+b , b+c<1 . Prove
$ 1\ge 27abc(1\plus{}b\minus{}a)(1\plus{}b\minus{}c)(1\plus{}a\plus{}c)$

[quote="dragon1"]Let a,b,c >0, a<1+b , b+c<1 . Prove
$ 1\ge 27abc(1 \plus{} b \minus{} a)(1 \plus{} b \minus{} c)(1 \plus{} a \plus{} c)$[/quote]
It 's wrong .Try $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$ :wink:

[quote="quykhtn-qa1"][quote="dragon1"]Let a,b,c >0, a<1+b , b+c<1 . Prove
$ 1\ge 27abc(1 \plus{} b \minus{} a)(1 \plus{} b \minus{} c)(1 \plus{} a \plus{} c)$[/quote]
It 's wrong .Try $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$ :wink:[/quote]

Sorry

Let a,b,c >0, a<1+b , b+c<1 . Prove
$ 1\ge 27abc(1 \plus{} b \minus{} a)(1 \minus{} b \minus{} c)(1 \plus{} a \plus{} c)$


limit
integration
function
calculus
real analysis
real analysis unsolved

Assume $ 1\le p < \infty.$ Suppose $ f_k,f\in L^p$ such that $ f_k\to f$ a.e. and $ \|f_k\|_p\to\|f\|_p.$ Prove that $ \|f_k \minus{} f\|_p\to0.$

Here's a proof I have for $ p \equal{} 1$ that I don't see how to make work for $ p > 1.$

Decompose $ f$ and $ f_n$ into positive and negative parts (or positive and negative parts of real and imaginary parts). For the rest of the problem, assume that $ f\ge 0$ and $ f_k\ge 0.$

We know that $ \lim \int (f_k \minus{} f) \equal{} 0.$ Write $ \int (f_k \minus{} f) \equal{} \int (f_k \minus{} f)^ \plus{} \minus{} \int (f_k \minus{} f)^ \minus{} .$

Note that $ (f_k \minus{} f)^ \minus{} \le f.$ Therefore, we can use $ f$ as a dominating function and use the DCT to show that $ \lim\int(f_k \minus{} f)^ \minus{} \to0.$

(*) Since $ \int (f_k \minus{} f)^ \plus{} \equal{} \int (f_k \minus{} f) \plus{} \int (f_k \minus{} f)^ \minus{}$ and the limits both integrals on the right are zero, we have that $ \lim\int (f_k \minus{} f)^ \plus{} \equal{} 0.$

Therefore $ \lim \int|f_k \minus{} f| \equal{} \lim\int(f_k \minus{} f)^ \plus{}\plus{} \lim\int(f_k \minus{} f)^ \minus{} \equal{} 0.$

It's the step marked (*) that doesn't seem to extend to $ p > 1.$

Let $ h_{k} \equal{} (f_{k} \minus{} f)^{p}$. Then

$ |h_{k}|\leq \left(|f_{k}| \plus{} |f|\right)^{p}\leq 2^{p \minus{} 1}\left(|f_{k}|^{p} \plus{} |f|^{p}\right) \equal{} g_{k}$

On the other hand, $ g_{k}\rightarrow 2^{p}|f|^{p} \equal{} g$, which is an integrable function and $ \int g \equal{} \lim\int g_{k}$ (this follows based on the fact that $ \parallel{}f_{k}\parallel{}\rightarrow \parallel{}f\parallel{}$.)

Also $ h_{k}\rightarrow 0$.

Thus, $ \lim\int h_{k} \equal{} \int\limits \lim h_{k}\equal{}\int 0 \equal{} 0$.

Note that I have used the so called generalized Lebesque Convergence Theorem.

[quote="didilica"]Note that I have used the so called generalized Lebesque Convergence Theorem.[/quote]
What is the statement of this? Because I'm not seeing how the very last line follows.

Theorem: Let $ g_{n}$ be a sequence of integrable functions which converges a.e to an integrable function g.
Let $ f_{n}$ be a sequence of measurable functions such that $ |f_{n}|\leq g_{n}$ and $ f_{n}$ converges to $ f$ a.e.
If \[ \int g\equal{}\lim\int g_{n}\] then $ \int f\equal{}\lim\int f_{n}$.

You can apply Fatou's lemma to $ g_k \minus{} |h_k|$ and it falls right out

for the case $ 1<p<\infty$, we could replace the hypothesis "$ \|f_n\|_p\to \|f\|_p$" by "$ \limsup\limits_{n\to\infty}\|f_n\|_p\leq \|f\|_p$"


Σε ένα βιβλίο βρήκα το παρακάτω ενδιαφέρον λήμμα αλλά χωρίς απόδειξη, γι' αυτό θα παρακαλούσα όποιον μπορούσε να βάλει την απόδειξή του

[color=red][u]Λήμμα[/u][/color]

Σε τρίγωνο $ ABC$ έστω $ I$ το έγκεντρο, και $ a,b,c$ τα μήκη των πλευρών  $ BC,CA,AB$ αντίστοιχα. Τότε για κάθε σημείο $ M$ του επιπέδου ισχύει:

$ a*MA^2 \plus{} b*MB^2 \plus{} c*MC^2 \equal{} abc \plus{} (a \plus{} b \plus{} c)*MI^2$

Ίσως βγαίνει εύκολα με διανύσματα και μιγαδικούς. Το δοκίμασες έτσι;

Στέλιος

bilot4 λιγο ζορικο το λημμα!... κατα τυχη κατεβασα αυτο το pdf πανω σε ανισοτητες απο το ΑοPS  του Hojoo Lee που περιεχει τη λυση (σελ.9-10) :)


I have received 4 protests on 3 separate problems.  Each was received through the regular communication channels we publish on every piece of information we produce.  After consultation with the Contest Chairs, I have responded personally and in detail to each communication.

As for the 4 protests on 3 problems, obviously two protests were about one problem, but it was not the number 11 on the 2009 AMC 12B.

Steve Dunbar
MAA Director, American Mathematics Competitions

Perhaps I am mistaken, but I believe I filed my protest through the correct channel (amcinfo@maa.org), and I am currently looking at my Inbox that is without a response.  The only communication I have regarding my protest is: [quote]I will forward you concern onto the appropriate people.

Thank you[/quote]

[quote="AMCDirector"]I have responded personally and in detail to each communication[/quote]

Does that mean all protests have been rejected :D

[quote]perhaps I am mistaken, but I believe I filed my protest through the correct channel (amcinfo@maa.org), and I am currently looking at my Inbox that is without a response. The only communication I have regarding my protest is:[/quote]

amcinfo@maa.org received a message concerning this problem at:
Sun, 01 Mar 2009 23:53:20 -0500
The message was relayed to me, and I responded:
03/04/2009 11:54 AM
after consultation with the Contest Chairs.
The message was sent to an aol.com address.
One of the contest chairs informed me this weekend that he had seen the content of the message, so I know it was sent.
I have not received a bounce-back informing me the message was not deliverable.

Have you checked your spam filter?

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote]The message was sent to an aol.com address[/quote]

I assume this is the problem considering I sent the email from my one, and only, email address that is a Gmail account.  Would there be any way to have this resent?

[quote]is a Gmail account[/quote]

Send me a message via the Private Messages here on AoPS with your email address, I will resend.

Steve Dunbar
MAA Director, American Mathematics Competitions


I could not post in the problems thread, so I started this one.  In addition, the graphics are inversed and I can not seem to fix it.

What do you mean the graphics are inversed?

Also, what do you mean you couldn't post?  Try again - I've been messing around with the site all day as people identify little problems; you may have tried posting as one of the files was being replaced.

When I clicked on Post, which I did several times, it took me to a blank page, and never loaded anything.  By inverted colors, I mean that everything is yellow, green and gray instead of red yellow and blue.

As for the colors, I am seeing the same, although I think that such is intended (that the color scheme was changed).

yes the colors are lyk that for me too.... so i agree with fierytycoon. but im able to post normally

No duh, I'm shocked, I mean wow. </no offence, but that was kind of obvious :)>

Anyways the posting I have no problems with...well, I've had that once but as he says it was during when he was tweaking the phpBB so yeah, things happen, but they should be alright by the time everything's sorted out.


Prove that for all $ k \in  N$, there exist $ A$ is positive integer, p is a prime number such that $ ORD_p (A) \equal{}k$
$ N \equal{}{1,2,3....}$

WHO CAN ?

In other words, there exists a prime of the form $ nk \plus{} 1$.  In fact, there are infinitely many: see http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=155461 .


probability
number theory
relatively prime

When rolling a certain unfair six-sided die with faces numbered 1,
2, 3, 4, 5, and 6, the probability of obtaining face $ F$ is greater
than $ 1/6$, the probability of obtaining the face opposite face $ F$
is less than $ 1/6$, the probability of obtaining each of the other
faces is $ 1/6$, and the sum of the numbers on each pair of opposite
faces is 7. When two such dice are rolled, the probability of
obtaining a sum of 7 is 47/288. Given that the probability of
obtaining face $ F$ is $ m/n$, where $ m$ and $ n$ are relatively prime
positive integers, find $ m\plus{}n$.

Well. To get 7 there are 3 ways. Letting the faces be a,b,c,d each with probability 1/6, and F and f with probability 1/6+p and 1/6-p.

To get 7:

ab:1/18

cd:1/18

Ff:2(1/36-p^2)=1/18-2p^2

Thus 1/18+1/18+1/18-2p^2=1/6-2p^2=47/288

Therefore 2p^2=1/288
p^2=1/576

p=1/24

Thus the probability of rolling F is 1/6+1/24=5/24, so m+n=29


[quote="nickolas"]Let  [tex] a,b,c>0[/tex] .Prove that  [tex]\frac{ab}{ab+c^2}+\frac{bc}{bc+a^2}+\frac{ca}{ca+b^2}\geq\frac{3}{2}[/tex][/quote]
Will AM-HM help?

[quote="leepakhin"]Let $x=\frac{a^2}{bc}$ and so on, $xyz=1$
By AM-HM inequality,
$\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge 3(\frac{3}{1+x+1+y+1+z})$
It remains to prove $\frac{3}{3+x+y+z}\ge \frac{3}{2}$
$\Leftrightarrow x+y+z\le 3$
This is true by AM-GM inequality.[/quote]

Actually, it remains to prove $\frac{3}{3+x+y+z}\ge \frac{1}{2}$

and $x+y+z \le 3$ is not AMGM ineq, because $x + y + z \ge 3 \sqrt[3]{xyz} = 3$ is AMGM.

Bill:  a,b,c must be greater than zero. :)  So those do not work.

Of course they serve as counterexamples. Take anything close to it and try it out (like $(0.001, 1, 1)$).

It's just easier to see why the inequality it's wrong when we extend the domain to include 0.

Or if you want to be more formal, we can let $(a,b,c) = (\epsilon, 1, 1)$ and let $\epsilon \to 0$.

[quote="billzhao"][quote="Singular"]But $(1,1,0)$ majorizes $(4/3, 1/3, 1/3)$, so by muirhead we are done.[/quote]

It does?

Anyways, the inequality cannot be true either way. Letting $(a,b,c) \to (0, 1, 1)$ gives $LHS \to 1$ and $(a,b,c) \to (0, 0, 1)$ gives $LHS \to 2$.[/quote]

my bad, you are right.  i was rushing lol

Oops, my bad, I understand what you mean now (a limit).

Another way to realize that the inequality is wrong...  :P

$\sum \frac{ab}{ab+c^2} \geq \frac{3}{2} \iff \sum \frac{c^2}{ab+c^2} \leq \frac{3}{2}$

which after setting $x = c^2/ab$, etc or $p = ab/c^2$, etc becomes

$\sum \frac{1}{1+x} \geq \frac{3}{2} \iff \sum \frac{1}{1+p} \leq \frac{3}{2}$

under the conditions $xyz = pqr = 1$... prove one and the other's false.


Two different numbers are chosen from the sequence of consecutive odd integers 51, 53, 55, …,95, 97, 99. Which of the following could not be their sum?
a. 161 b. 168 c. 186 d. 196 e. NG

[hide]Well, all the numbers are odd.
An odd number plus an odd number is always an even number.
The only odd answer choice is [b](A) 161[/b][/hide]


geometry
rectangle
combinatorics unsolved
combinatorics

Let $n$ be a positive integer. Nancy is given a rectangular table in which each entry is a positive integer.  She is permitted to make either of the following two moves:

(1)  select a row and multiply each entry in this row by $n$;

(2)  select a column and subtract $n$ from each entry in this column.

Find all possible values of $n$ for which the following statement is true:

Given any rectangular table, it is possible for Nancy to perform a finite sequence of moves to create a table in which each entry is $0$.

May I clear up another old unsolved problem?

Clearly $n=1$ is not going to work if the entries in a column are not equal.

If $n>2$, then the rectangle consisting just of one column of a $2$ and a $1$ cannot be changed to $0$s- it is impossible to make the two values congruent $\mod (n-1)$.

To prove that it works for $n=2$:

Clearly if we can do it for one column, we can do it for any number of columns, because once a column is all zeros we can move on to another column. Also, we only need consider columns with two numbers in them because once we have made two numbers in a column equal, we can just treat them together and move on to the rest of the column.
So we only need consider $2 \times 1$ rectangles.
We can double both numbers in this $2 \times 1$ rectangle, and then subtract $2$ repeatedly, until one of the numbers is $2$. Then we double that to $4$, subtract $2$ again until both numbers are $2$. Then we can move onto the rest of the rectangle, if it is larger than a $2 \times 1$.

Hence $n=2$ is the only possibility.


pigeonhole principle
linear algebra
matrix
modular arithmetic
linear algebra unsolved

Let $n\geq 2$ and $A\in\mathcal M_n(Z)$ with $\det(A)=\pm 1$. Prove that in the sequence $A,A^2,A^3,...$ there exists at least one matrix with the property that each of it`s elements is congruent modulo $n$ with the element in the same postion in the corresponding inverse.

Yes, pigeonhole :). There is, however, something I don't quite get: that $n$ modulo which we have to work need not be the same $n$ as the dimension of the matrix. Was it a typo of yours, or did the problem really look like this? I'll replace the former with $m$.

Anyway, let $\widetilde A$ be the matrix reduced modulo $m$. Since $\mathcal M_n(\mathbb Z/m\mathbb Z)$ has only finitely many elements, we have $\widetilde A^k=\widetilde A^p$ for some $0\le k<p$. Since $\widetilde A$ is invertible, we have $\widetilde A^{p-k}=I\Rightarrow \left(\widetilde A^{p-k}\right)^2=I$, which means that $\widetilde A^{p-k}\equiv\left(\widetilde A^{p-k}\right)^{-1}\pmod m$, and this is exactly what we want.

In the original problem the $n$ with which we must modulate is the same as the dimension of the matrix, however since you generalised the problem, it does not matter any longer :) .( The problem is from a book of Gheorghe Andrei or C.Caragea about algebra.)


Today morning, for about 2 hours ET, the AoPS Forum gave a message:

Critical Error

Cannot connect to the database.

What happened? :maybe:

It happened for me for 4 hours (Some time around 10GMT to 3:30GMT). Bizarre, though I guess the admins knew about the problem (and fixed it)

i was wondering why i was the only one other than bots at 10 :ninja:

When I just went to http://artofproblemsolving.com or http://mathlinks.ro, it gave me:

This problem is temporary for 5 minutes. If the problem persists, email webmaster@artofproblemsolving.com or http://mathlinks.ro respectively

or something to that extent

Great that you did. This most likely appeared due to an automatic system update and some paths were screwed up :)

The same thing happened to me. I was checking the USAMTS website too, and I got the same message.

Aren't like the Mathlinks and USAMTS website on teh same server? :maybe:

I know this is totally not the place to report a new, distinct, problem, but whenever I try to go to the forum (by going to mathlinks.ro, or artofproblemsolving.com and clicking on forum, or by clicking a link to any subforum) I get an error message as such:

SQL requests not achieved

DEBUG MODE

SQL Error: 1030 Got error 134 from storage engine

SQL Request: SELECT COUNT(DISTINCT session_ip) AS count_session_ip FROM phpbb_sessions WHERE session_user_id = -1 AND session_time >= 1166987998

Line : 86
File : class_stats.php

However, all the header/background is still there, and I can click on "view my posts" and click on the topics I have posted in (hence this topic). Just about anywhere else I try to go in the forum, I get the same error. Is this a bug/temporary problem/only my problem? (Using IE7 on XP).

Again, sorry for reposting here but it is the only thing I can do at the moment.

It definitely is... emailed Mr. Vornicu again. He's probably working on it though...

I am getting this message:
[quote]
SQL requests not achieved

DEBUG MODE

SQL Error: 1030 Got error 134 from storage engine

SQL Request: SELECT COUNT(DISTINCT session_ip) AS count_session_ip FROM phpbb_sessions WHERE session_user_id = -1 AND session_time >= 1166991021

Line : 86
File : class_stats.php
[/quote]
For like the last two hours i got the other one

It appears one of our tables was messed up; I think it is fixed now.  Is anyone still having problems?

[quote="anirudh"]That message came since today morning for me.[/quote]

Is it still happening?

[quote="Valentin Vornicu"]Always on X-Mas something happens :)[/quote]

Yeah, I'm becoming a real Scrooge.

Yes. The Forums work. But the classroom doesn't work. :(

The classroom does not work for me either  :maybe:

It gives some kind of error.

Cannot connect to server

or something to that extent

Yes. the classroom works now.

Also, sometimes, latex just crashes in the classroom sometimes...

and only:

[code];$x$[/code]

works.

Nothing else works.

But it works fine now :)

Sometimes, though, when in discussion in the classroom, not in the class, it crashes such.

This is a little unrelated (and I guess I might be bumping this post... its been almost a month)

but after reading this, I remembered what I was going to mention - I use firefox, and I always have problems logging in. I will be working on a post and try to submit and it will say, "please log in" or something. Sometimes, every 15 minutes or so I get automatically logged out. Especially when I'm editing a post of mine or something.

Just wondering if someone could look into it r if other people have similar experiences.


Calculate $ \displaystyle \sum_{k\equal{}0}^m  \left( \begin{array} {c}  s\minus{}1 \\ m\minus{}k  \end{array} \right) \left( \begin{array} {c}  n\minus{}s \\ k\minus{}1  \end{array} \right)$, where $ n>s>m$.

The result is $ \binom{n}{m}$.
This is the number of ways to choose $ m$ balls from an urn which contains $ n$ balls numbered from $ 1$ to $ n$.
In another hand, for each $ k$, the $ k^{th}$ summand in the sum is the number of ways to choose $ m$ of balls from this urn such that the $ k^{th}$ greatest number of the chosen balls is $ s$.

Pierre.


I know that it's possible for an irrational number multiplied by an irrational number can equal a whole number, (i.e. take a 1 foot line bend it into a circle multiply the diameter times pi [both are irrational] and you should get 1 foot) but how can we apply this in real life? How can we tell when it will equal a whole number and how to do the problem?

This really shouldn't be here....

Well, we have $ \sqrt{2} \cdot \sqrt{2}\equal{}2$.

[quote="mathblitz"]How can we tell when it will equal a whole number[/quote]

Take $ \sqrt{a} \times \sqrt{b}$.  If $ ab$ is a square number, or if $ a\equal{}b$, or if $ a$ or $ b$ is $ 0$, it will equal a whole number.

[quote="mathblitz"]how can we apply this in real life? How can we tell when it will equal a whole number and how to do the problem?[/quote]
That depends on the problem.  The algebra of irrational numbers is exactly the same as the algebra of rational numbers, so nothing changes until you have to do numerical computations - is that what you're asking about?  Do you need help understanding roots or irrational numbers in general?  It would be helpful if you were more specific.


Players A and B play a (fair) dice game. "A" deposits one coin and they take turns rolling a single dice, "B" rolling first.

If "B" rolls an even number, he collects a coin from the pot. If he rolls an odd number, he put a coin (coins with same values always).  If "A" (plays and) rolls an even number, he collects a coin but if he rolls an odd number, he does NOT add a coin.

The game continues until the pot is exhausted.

Question: what is the probability that "A" wins this game (that is, exhaust the pot) ?

(Show solution, not just answer.)

You could solve this by getting ever-closer approximations by looking at the probability that the game ends after n rounds.  The probability that b wins after the first round is $\frac12$, which immediately sets an upper bound on the probability that a will win.

Of the remaining $\frac12$, there is a $\frac12$ probability that a will remove one of the two coins in the pot.  B might then also remove a coin, lowering the upper bound by $\frac38$.

There is also a $\frac18$ probability that a will not remove a coin, but b will.  Then a will have an immediate shot at winning, yielding a lower bound of $\frac1{16}$.

If, in the second round, b instead places a coin in the pot, there is no chance for a to win in the second round.

This is obviously not the most elegant solution to this problem, and really requires writing a program to get a decent approximation in a decent time interval, but it's the only thing I can think of.

roadnottaken:  Good approach!

I realized the A:B winning probabilities is NOT 1:1 using the very same (start of your) reasoning:
-------------
The probability that B wins after the first round is... sets an upper bound on the probability that A will win
-------------
and the begining of your argument is more-than-strong enough to garantee the probability asked is NOT 1, too...

But... I am still looking for an analytical-only solution and I will try a bit harder on the "total probability theorem" and related ideas.
Now it seems to me that my first difficulties are not silly, anyway. (I guess I had tried unecessary-hard paths, I mean.)

Let us wait other contributions. For myself, I will keep my word to come back to bring some options -- if not a solution, who knows? -- in the near future (that is, in some days)!

Best regards,
fskilnik.

P.S.: have you seen the "Dead Poets Society" movie? (A.Frost poem, beautiful!)

It said "A wins if the coins are exhausted".

If you want: B wins if B gets to exhaust the pile before A gets any, then it gets a litle more complicated.

Basically let R(i) be the probability of a random walk returning to 0 after2i steps. \[sum_{i=1}^{\inf}R(i)2^{1-2i}\]

Anyone know R(i)?

Sorry, Bill.

When I wrote

what is the probability that "A" wins this game (that is, exhaust the pot) ?

I should have written

what is the probability that "A" wins this game (that is, A exhaust the pot) ?

Not a realistic situation, because B could lose money and still win the game. Whatever.

The pot is 1 before A's turn:
A has a 1/2 chance. B has a 1/4 chance. There is a 1/4 chance of going up to 2.

The pot is 2 before A's turn:
B has a 1/4 chance. There is a 1/4 chance of it staying the same. There is a 1/4 chance of it going down to 1. There is a 1/4 chance of it going up to 3.

The pot is 3 before A's turn:
There is a 1/4 chance, respectively, of, 1, 2, 3, and 4

Therefore at the start $B$'s chance is $1/2+P(2)/2=2/3+P(1)/6+P(3)/6=2/3+1/24+P(2)/24+P(1)/18+P(2)/18+P(4)/18 =17/24+P(1)/18+7P(2)/72+P(4)/18 =17/24+1/72+P(2)/72+7/216+7P(3)/216+7P(1)/216+[lots]/18$

Minimum (for B): 163/216$,$53/216<3/8$Maximum (for B): 192/216=24/27=8/9$ $1/9>1/16$
My range is now $29/216=13.4%
$


Calculate $\left(2^{100}-2\right) \text{ mod } 100$.

[hide]mod 100 means finding the last two digits.

Okay then.....

$2^1=02$
$2^2=04$
$2^3=08$
$2^4=16$
$2^5=32$
$2^6=64$
$2^7=28$

.....

*feels bored, says K.W.M.A.N's solution is way better. Uses calculator because of sleepiness*

$2^{21}=52$
$2^{22}=04$
$2^{23}=08$

Hey a pattern, $2^2$, $2^22$, etc. have the same last digit. That means $2^{102}$ must also end with four, implying that $2^{100}$ ends with 76.

$2^{100}-2\equiv74\pmod{100}$. [/hide]


function
combinatorics unsolved
combinatorics

Find the number of lattice walks from $ (0, 0)$ to $ (n, n)$,
consisting of up and right steps of length one only,
such that paths do not cross the lines $ y \minus{} x \equal{} k$ and $ y \minus{} x \equal{} \minus{}m$.

If I'm not mistaken, [url=http://www.research.att.com/~njas/sequences/A080934]A080934[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url] gives the number of paths from $ (0, 0)$ to $ (n, n)$ lying above $ y \equal{} x$ and below $ y \equal{} x \plus{} k$ (encoded in some not-entirely-obvious form), a natural version of restricted Dyck paths.  One can write down recursions for these sequences and maybe generating functions for these, but they aren't overwhelmingly elegant.  Anything that will solve this case can probably be modified for $ y \equal{} x \minus{} m$ for $ m > 0$, as it's not all that different then the case where you're bounded between $ y \equal{} x$ and $ y \equal{} x \plus{} k \plus{} m$.


Order of operations
Multiplication comes first
4x3=12
12+6=18

easy, oreder of operations, 4 times 3 and then add 12 and six

hello, or so: $ 6\plus{}4\times3\equal{}2\times(3\plus{}2\times3)\equal{}6\times(1\plus{}2)\equal{}18$
Sonnhard.

ans is 6+4*3
12+6=18
so it is just using pemdas


geometry
combinatorics proposed
combinatorics

Let $ABC$ be a given triangle .
Prove that if we color the plane with 2 colors(Black and white) then
Either we have 2 white points with distance of unit or  we have 3  black points
such that they will form a triangle with equal area to $ABC$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=chromatic+number&t=26773]Here it is[/url]. In fact, that seems to be a stronger version (the triangles are congruent, not merely equivalent).


A claw is a graph consisting of 4 vertices and three edges, one vertex of degree three and three vertices of degree one (one vertex adjacent to other 3 that are pairwise non-adjacent). A graph is claw free if it does not cotained a induced claw as a subgraph.

Prove that if G is 4-connected and claw free then G is Hamiltonian.

[quote="manuel"]A graph is claw free if it does not cotained a induced claw as a subgraph.[/quote]
What does it mean? There are no vertices A,B, C, D s.t. A is connected to B, C, and D, and there are no edges between B, C and D. Right?

no.. my teacher is out of the country  :(

Manuel, i know you already explined me this, but what is k-conected? thanks!

Pascual here are some definitions of graph connectivity that maybe useful to you:
A [b]separating set [/b]of a graph G is a subset S of the set of vertices of G ( written V(G) ) such that G - S disconnectes the graph. The [b]connectivity[/b] of G, sometimes written k(G), is the minimum size of a separating set in G. A graph G is [b]k-connected[/b] if its connectivity is at least k. In oder words G is [b]k-connected [/b]iff when you remove any k-1 vertices the graph remains connected.
I also told you some theorems regarding this that if you are interested you can ask me... ;)


ok...im making a puzzle type thing where ill give you some clues and after you figure out one clue, it leads to another and another untill it stops (kind of like national treasurer)  The only thing you will need is you smarts and the AoPS website (i.e all the clues will be on the website...)  Too hard?  ask for a hint.  NO PMING OTHER PEOPLES

READ THIS PASSAGE:

There once was a man named bob.  he was a garbaje cllector.  he always wen on AoPS.  one day, he goot fired.  he had nothing do do so he went on the AoPS foru for hours.  his favorite was the communiti section.  Bobb friend was Molly.  one day moly died.  So bob started smamming in the getting started frum.  the moderators got really mad.  One moderator was Geroge.  geore was nice

THE END :D

the answer is someones username

pm me the answer when you are done

becuase i thought that if people already read it they wouldnt know i had edited it

jeez

its like notpron (in a way)

look for clues in the passage which will lead you to the next clue and so on

HINT: [hide]look for spelling errors[/hide]

HINT2: [hide]carreers in math...wondering...hmm[/hide]

HINT3: [hide]line-letter[/hide]

HINT4: [hide]porblem, getting started[/hide]

I understood what you are trying to do here but I can't figure it out. What's next after the spelling errors?

I'm working on it now :weightlift: Sorry if my brain is so slow :)

Does this username do all of the things you described?

i would like to congradulate 236! for being the first to solve the puzzle...using TONS of hints though

first...look for spelling errors which spell out"go to my blog"

next...i posted something called "careers in math" and said ii was WONDERING about careers in math.

go to the careers in math foruma and click wondering

all the pairs of numbers stand for a letter in the first post

it comes out as "go to mathcounts" or something

go to mathcounts and click on "problem"

BOB:lets go to the getting started forum...i like their MODERATORS

later on in the post after the problem was stated...

HINT:the answer is less than [b]10000[/b]

can you get the answer now?

[quote="jli"]There once was a man named bob.  he was a garbaje cllector.  he always wen on AoPS.  one day, he goot fired.  he had nothing [b]do[/b] do so he went on the AoPS foru for hours.  his favorite was the communiti section.  Bobb friend was Molly.  one day moly died.  So bob started [b]smamming[/b] in the getting started frum.  the moderators got really mad.  One moderator was Geroge.  geore was nice[/quote]
 :?
I get GOTODMYBLMOG.

sorry...its supposed to say go to my blog


calculus
integration
trigonometry
conics
ellipse
search
calculus computations

Please find (if its possible):

\[{ I=\int^{0}_{\frac{\pi}{2}}\sqrt{a^{2}\sin^{2}x+b^{2}\cos^{2}x}}\,dx\]

This calculation seems to be the arc length for the ellipse of a quarter,but  generally the calculation is impossible.

This is a classical example of something called [i]elliptic integrals[/i]. It turns out that finding the circumfurence of an ellipse isn't an easy problem. It leads to an non-elementary integral. Which is why a math reference book will write the circumfurence of an ellipse is $ \approx\pi(a\plus{}b)/2$, which is just an approximation. If you consider $ \bold{r}(t) \equal{} a\cos(t)\bold{i}\plus{}b\sin(t)\bold{j}$ for $ 0\leq t\leq 2\pi$ this is an ellipse. Its length is given by:
$ \int_{0}^{2\pi}\sqrt{a^{2}\sin^{2}(t)\plus{}b^{2}\cos^{2}(t)}dt$.
Which [b]isnt[/b] doable if $ a,b>0$ and $ a\not \equal{} b$.

Because of appearaces of this integrals in mathematical physics some work on tables and identities have been developed. You can find more [url=http://mathworld.wolfram.com/EllipticIntegral.html]here[/url] including your integral which is formula #108.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1082482064&t=55166]This topic[/url] discusses the situation and includes a discussion of various approximations.


inequalities
function
vector
induction
inequalities theorems

What  is the  general  form  of  Holder's  inequality??
Thanks  Harazi !

sum(i=1,n) p_i*a_i*b_i<=(sum(i=1,n) p_i*a_i^x)^(1/x)*(sum(i=1,n) p_i*b_i^y)^(1/y).

where 1/x+1/y=1 and x,y,pi,ai,bi>0 and n>2.

I hope this is what u wanted

I knew this one! I'm so glad. I thought he was asking about the most general form of his inequality and I don't know that. Anyway, open question-as far as I know- means a problem that was not solved by the one who thought about it and is proposed for the others to think about it. So, this question is not an open question.

This one should go to formulas and theorems corner! :D

Well  i  knew  that  one , but  i  was  wondering  if  there  is  a  more  general  form  like  schure's  inequality...Thanks

btw I have extracted the generalisation I gave for schur and put it into this folder if you want to take a look at it nickolas.

Let $a_{1,1},a_{1,2},...a_{1,n},a_{2,1},a_{2,2}...a_{2,n...}a_{k,n} > 0$ and  $\frac{1}{p_1}+\frac{1}{p_2}+...+\frac{1}{p_k}=\frac{1}{s}$. Than
$(a_{1,1}^{p_1}+a_{1,2}^{p_1}+...+a_{1,n}^{p_1})^{\frac{1}{p_1}} (a_{2,1}^{p_2}+a_{2,2}^{p_2}+...+a_{2,n}^{p_2})^{\frac{1}{p_2}}....(a_{k,1}^{p_k}+a_{k,2}^{p_k}+...+a_{k,n}^{p_k})^{\frac{1}{p_k}}\geqslant ((a_{1,1}a_{2,1}...a_{k,1})^s+(a_{1,2}a_{2,2}...a_{k,2})^s+...+(a_{1,n}a_{2,n}...a_{k,n})^s)^{\frac{1}{S}}$
generalization by jalowiec ( http://www.matematyka.org/forum/viewtopic.php?t=1893 - the site is in Polish)

proof 1 (by me)

let $\mathbb{R}_+^n \ni \vec{A_j} = [ \ a_{j,1} \  , \  a_{j,2} \  , \  a_{j,3} \  , \  ... \  , \  a_{j,n} \ ]$
let us define $\Vert \vec{A_j} \Vert_m = (a_{j,1}^m + a_{j,2}^m + ... + a_{j,n}^m)^{1/m}$
and a realtion - $\ell$-multiplication - denoted as: it's a function $\mathcal{M}_\ell : \mathbb{R}_+^{n \ell} \mapsto \mathbb{R}_+^n$ ($\ell$ vectors give one vector) and
$\mathcal{M}_\ell (\vec{A_1} , \vec{A_2} , ... , \vec{A_\ell}) = [ \  a_{1,1} a_{2,1} ... a_{\ell, 1} \  , \  a_{1,2} a_{2,2} ... a_{\ell , 2} \  , \  ... \  , \  a_{1,n} a_{2,n} ... a_{\ell , n} \  ]$ (analogically to usual vector addition, but with multiplication instead). of course $\ell$-multiplication is commutative, you can randomly change the order of the factors and the result is always the same due to commutativity of regular multiplication. $\ell$-multiplication has also this property:
$\mathcal{M}_2 (\mathcal{M}_r (\vec{A_1} , ... , \vec{A_r}) , \mathcal{M}_w (\vec{A_{r+1}} , ... , \vec{A_{r+w}}) ) = \mathcal{M}_{r+w} (\vec{A_1} , ... , \vec{A_{r+w}})$ - r-multiplication times w-multiplication is r+w-multiplication
first let us prove that $\Vert \vec{X} \Vert_t \Vert \vec{Y} \Vert_t \geqslant \Vert \mathcal{M}_2 (\vec{X} , \vec{Y} ) \Vert_t$ , ($\vec{X} = [ \  x_1 \  , \ x_2 \  , \  ... \  , \  x_n \  ] \  , \  \vec{Y} = [ \  y_1 \  , \  y_2 \  , \  ... \  , \  y_n \  ]$)
plotting it to our definition of norm we get
$(\sum_{i=1}^n x_i^t)^{1/t} (\sum_{i=1}^n y_i^t)^{1/t} \geqslant (\sum_{i=1}^n (x_i y_i)^t)^{1/t}$
it's visible that after putting both sides to the power $t$ and multiplicating the left side, we will get there products $x_i x_j$ with both $i=j$ and $i \neq j$ , while on the right side there will only be $i=j$.
above all this, the Holder inequality states that
${1 \over p} + {1 \over q} = 1 \  , \  a_i , b_i \geqslant 0 \Rightarrow (\sum_{i=1}^n a_i^p)^{1/p}(\sum_{i=1}^n b_i^q)^{1/q} \geqslant \sum_{i=1}^n a_i b_i$
setting ${p' \over s} = p$ i ${q' \over s} = q$ we get
$(\sum_{i=1}^n a_i^{p'/s})^{s/p'}(\sum_{i=1}^n b_i^{q'/s})^{s/q'} \geqslant \sum_{i=1}^n a_i b_i$
putting both sides to the power $1 \over s$ and substtuting $c_i = a_i^{1/s} , d_i = b_i^{1/s}$ we obtain
$(\sum_{i=1}^n c_i^{p'})^{1/p'}(\sum_{i=1}^n d_i^{q'})^{1/q'} \geqslant (\sum_{i=1}^n (c_i d_i)^s)^{1/s}$ and ${1 \over p'} + {1 \over q'} = {1 \over s}$
and this is equivalent to
$\Vert \vec{C} \Vert_{p'} \Vert \vec{D} \Vert_{q'} \geqslant \Vert \mathcal{M}_2 (\vec{C} , \vec{D}) \Vert_{(p'^{-1} + q'^{-1})^{-1}}$
we'll call this a "generalized Holder inequality for two sequences/vectors"
and now returning to the proper inequality:
we have to prove that:
$\Vert \mathcal{M}_k (\vec{A_1} , \vec{A_2} , ... , \vec{A_k}) \Vert_{(\sum_{i=1}^k p_i^{-1})^{-1}} \leqslant \Vert \vec{A_1} \Vert_{p_1} \Vert \vec{A_2} \Vert_{p_2} ... \Vert \vec{A_k} \Vert_{p_k}$
we'll do this with induction (variable - $k$). the latter has to imply in
$\Vert \mathcal{M}_{k+1} (\vec{A_1} , \vec{A_2} , ... , \vec{A_k} , \vec{A_{k+1}}) \Vert_{(\sum_{i=1}^k p_i^{-1} + p_{k+1}^{-1})^{-1}} \leqslant \Vert \vec{A_1} \Vert_{p_1} \Vert \vec{A_2} \Vert_{p_2} ... \Vert \vec{A_k} \Vert_{p_k} \Vert \vec{A_{k+1}} \Vert_{p_{k+1}}$
due to the second given by me property of $\ell$-multiplication we have:
$\Vert \mathcal{M}_2 ( \mathcal{M}_k (\vec{A_1} , \vec{A_2} , ... , \vec{A_k} ) , \vec{A_{k+1}}) \Vert_{(\sum_{i=1}^k p_i^{-1} + p_{k+1}^{-1})^{-1}} \leqslant \Vert \vec{A_1} \Vert_{p_1} \Vert \vec{A_2} \Vert_{p_2} ... \Vert \vec{A_k} \Vert_{p_k} \Vert \vec{A_{k+1}} \Vert_{p_{k+1}}$
let us substitute $\vec{M} = \mathcal{M}_k (\vec{A_1} , \vec{A_2} , ... , \vec{A_k} ) , \vec{\alpha} = \vec{A_{k+1}} , s^{-1} = \sum_{i=1}^k p_i^{-1}$ because i've had enough writing so far :)
hence we get
$\Vert \mathcal{M}_2 (\vec{M} , \vec{\alpha}) \Vert_{(s^{-1} + p_{k+1}^{-1})^{-1}} \leqslant \Vert \vec{A_1} \Vert_{p_1} \Vert \vec{A_2} \Vert_{p_2} ... \Vert \vec{A_k} \Vert_{p_k} \Vert \vec{\alpha} \Vert_{p_{k+1}}$
we can substitute $\Vert \vec{M} \Vert_s$ for $\Vert \vec{A_1} \Vert_{p_1} \Vert \vec{A_2}\Vert_{p_2} ... \Vert \vec{A_k} \Vert_{p_k}$ because due to inductive assumption the latter is larger, thus we make the right hand side smaller.
hence:
$\Vert \mathcal{M}_2 (\vec{M} , \vec{\alpha}) \Vert_{(s^{-1} + p_{k+1}^{-1})^{-1}} \leqslant \Vert \vec{M} \Vert_s \Vert \vec{\alpha} \Vert_{p_{k+1}}$
and this is true due to generalized Holder inequality for two sequences/vectors ($\vec{C} = \vec{M} , \vec{D} = \vec{\alpha} , p' = s , q' = p_{k+1}$)

proof 2 (by jalowiec)

let us denote $(a_{1,1}a_{2,1}...a_{k,1})^s+(a_{1,2}a_{2,2}...a_{k,2})^s+...+(a_{1,n}a_{2,n}...a_{k,n})^s=c$ and write the inequality this way:
$(\frac{(a_{1,1}a_{2,1}...a_{k,1})^s}{c}\cdot \frac{a_{1,1}^{p_1}}{(a_{1,1}a_{2,1}...a_{k,1})^s}+\frac{(a_{1,2}a_{2,2}...a_{k,2})^s}{c}\cdot \frac{a_{1,2}^{p_1}}{(a_{1,2}a_{2,2}...a_{k,2})^s}+...+\frac{(a_{1,n}a_{2,n}...a_{k,n})^s}{c}\cdot \frac{a_{1,n}^{p_1}}{(a_{1,n}a_{2,n}...a_{k,n})^s})^\frac{1}{p_1}\cdot$ $\cdot(\frac{(a_{1,1}a_{2,1}...a_{k,1})^s}{c}\cdot \frac{a_{2,1}^{p_2}}{(a_{1,1}a_{2,1}...a_{k,1})^s}+\frac{(a_{1,2}a_{2,2}...a_{k,2})^s}{c}\cdot \frac{a_{2,2}^{p_2}}{(a_{1,2}a_{2,2}...a_{k,2})^s}+...+\frac{(a_{1,n}a_{2,n}...a_{k,n})^s}{c}\cdot \frac{a_{2,n}^{p_2}}{(a_{1,n}a_{2,n}...a_{k,n})^s})^\frac{1}{p_2}\cdot...$ $...\cdot(\frac{(a_{1,1}a_{2,1}...a_{k,1})^s}{c}\cdot \frac{a_{k,1}^{p_k}}{(a_{1,1}a_{2,1}...a_{k,1})^s}+\frac{(a_{1,2}a_{2,2}...a_{k,2})^s}{c}\cdot \frac{a_{k,2}^{p_k}}{(a_{1,2}a_{2,2}...a_{k,2})^s}+...+\frac{(a_{1,n}a_{2,n}...a_{k,n})^s}{c}\cdot \frac{a_{k,n}^{p_k}}{(a_{1,n}a_{2,n}...a_{k,n})^s})^\frac{1}{p_k}\geq 1$
let us now use the inequality $q_1x_1+q_2x_2+...+q_nx_n\geq x_1^{q_1}x_2^{q_2}...x_n^{q_n}$ ($\sum_{j=1}^n q_j = 1$) for $q_j = \frac{(a_{1,j}a_{2,j}...a_{k,j})^s}{c}$ in each of the "wider" brackets. it substitutes each bracket with
$((\frac{a_{1,j}^{p_1}}{(a_{1,j}a_{2,j}...a_{k,j})^s})^\frac{1}{p_1}(\frac{a_{2,j}^{p_2}}{(a_{1,j}a_{2,j}...a_{k,j})^s})^\frac{1}{p_2}...(\frac{a_{k,j}^{p_k}}{(a_{1,j}a_{2,j}...a_{k,j})^s})^\frac{1}{p_k})^\frac{(a_{1,j}a_{2,j}...a_{k,j})^s}{c}$, which is equal to $1$
multiplying the latter from $j=1$ to $j=n$ we obtain the initial inequality.


Someone asked if there's a relationship between math and musical skill.  Let's see!

By the way, I play piano and viola.

[quote="Treething"]I'm a pianist.[/quote]
Same here.

[quote="Danbert"]Someone asked if there's a relationship between math and musical skill.  Let's see![/quote]

There's no "Don't play" choice there. So how do we say we don't play an instrument?

Piano since I was 5... so I've been playing for about 7 years.

played piano, hated it, gave it up after maybe 2 years....

[quote]There's no "Don't play" choice there. So how do we say we don't play an instrument?[/quote]

Oops.  Just check 'Other,' I guess.

[i]Are[/i] there any other instruments?

Kazoo.  :D

Seriously, piano since I was 7. Violin since I was 10, but seriously thinking about giving it up this year. Didn't join school orchestra, might just play by myself a little bit.

I'm an asian girl. Asian girls plays strings, woodwind and piano, it's like a law.

i have played the trumpet before (didn't work out so well) and i've played the piano for 10 years

I used to play sax but stopped and now I play mandoline :)

I play the piano, and I sing Indian music a bit. I can play a few songs on the guitar too! I have to admit, guitarists are very talented. :D

heh piano here... almost every asian guy at my school plays the piano or the violin lol

Vibraphone!!! (percussion)
Go Cadets of Bergen County

i play accordian.
BTW, which section does it belong to? Piano?

i used to play recorder, then i began playing guitar and now i play piano and drum... :)

I play both violin and piano. Piano for like 6 years and violin for like 1.

Alto Saxophone!

And learning the piano.

Piano since I was five...

Pecussionist

Trumpet Player

Harmonica (Hey this actually does take some skill)

Guitar...Kind of...  It a bit complicated.

 :P

Piano for ten years, singing in the school chorus for 3 years, trombone for school for 5 years.

Music is one of the single most important activities one can take part in. Music and sports. And...competetive math :)


A car-rental company charges $ \$50$ for the first 500 miles driven and $ \$0.25$ for each additional mile. Given that Meghan was charged $ \$239.25$ for total mileage, how many miles did she drive?

We first compute the number of miles she drove past $ 500$. Since she spent $ \$189.25$, on those miles, we have that $ \frac{189.25}{.25}\equal{}757$. Thus, the total number of miles is $ 757\plus{}500\equal{}\boxed{1257}$.


Ai biết đường thẳng Ơ-le cho tứ giác và n-giác không chỉ cho e với :?:

[quote="Kyoshiro"]Mọi người dịch hộ e cái này với
[url]http://forumgeom.fau.edu/FG2006volume6/FG200634.pdf[/url]
Dịch trang 2,3,4 thôi nha.
Cảm ơn nhìu! :)[/quote]
Không mở được em à!!! :(

E vẫn mở được mà.
Anh thử link này xem [url]http://209.85.175.104/search?q=cache:qCSjtWryIaUJ:forumgeom.fau.edu/FG2006volume6/FG200634.pdf+the+Euler+line+of+quadrilateral&hl=vi&ct=clnk&cd=6&gl=vn[/url]


algebra
polynomial
trigonometry
binomial theorem

please don't use binomial theorem as I have not learnt it yet

------------------------------

Use De Moivre's theorem to show that $ cos3\theta \equal{} 4cos^{3}\theta\minus{}3cos\theta.$ Hence solve $ 8x^{3}\minus{}6x\minus{}1 \equal{} 0.$Deduce that $ cos\frac{\pi}{9}\equal{} cos\frac{2\pi}{9}\plus{}cos\frac{4\pi}{9}$

I done the show $ cos3\theta\minus{}3cos\theta$ bit. So don't waste time on that.
How do I solve the eqn now?

$ cos{3\phi}= Re[(\cos\phi+i\sin\phi)^{3}]$
$ \Rightarrow\cos{3\phi}=\cos^{3}{\phi}-3\cos{\phi}\sin^{2}{\phi}$
$ \Rightarrow\cos{3\phi}=\cos^{3}{\phi}-3\cos{\phi}(1-\cos^{2}{\phi})$
$ \Rightarrow\cos{3\phi}=\cos^{3}{\phi}-3\cos{\phi}+3\cos^{3}{\phi})$
$ \Rightarrow\cos{3\phi}= 4\cos^{3}{\phi}-3\cos{\phi}$

Let $ x =\cos{\phi}$
$ 8x^{3}-6x-1 = 0$
$ \Rightarrow 2(4\cos^{3}{\phi}-3\cos{\phi})-1 = 0$
$ \Rightarrow\cos{3\phi}=\frac{1}{2}$
$ \Rightarrow 3\phi =\frac{\pi}{3},\frac{5\pi}{3},\frac{7\pi}{3}$
$ \Rightarrow\phi =\frac{\pi}{9},\frac{5\pi}{9},\frac{7\pi}{9}$
$ \Rightarrow x =\cos{\frac{\pi}{9},\cos{\frac{5\pi}{9},\cos{\frac{7\pi}{9}}}}$

oh I see it now.

thanks for your help. Appreciate it.

I have just figured out the last part of the question.

$ 8x^{3}-6x-1 = 0$
$ \Rightarrow x^{3}-\frac{3x}{4}-\frac{1}{8}= 0$

and we have:
$ (x-\cos{\frac{\pi}{9})(x-\cos{\frac{5\pi}{9})(x-\cos{\frac{7\pi}{9}) = 0}}}$
$ \Rightarrow (x-\cos{\frac{\pi}{9})(x+\cos{\frac{4\pi}{9})(x+\cos{\frac{2\pi}{9}) = 0}}}$
(since $ -\cos{\frac{5\pi}{9}=\cos{\frac{4\pi}{9}}}$ and $ -\cos{\frac{7\pi}{9}=\cos{\frac{2\pi}{9}}}$

Consider the coefficient of the term $ x^{2}$ of $ x^{3}-\frac{3x}{4}-\frac{1}{8}= 0$, then we have:
$ -\cos{\frac{\pi}{9}+\cos{\frac{2\pi}{9}+\cos{\frac{4\pi}{9}= 0}}}$
$ \Rightarrow\cos{\frac{\pi}{9}=\cos{\frac{2\pi}{9}+\cos{\frac{4\pi}{9}}}}$

[quote="p4fn2w"][quote="IrinaSharova"]Let $ x \equal{}\cos{\phi}$
$ 8x^{3}\minus{}6x\minus{}1 \equal{} 0$
$ \Rightarrow 2(4\cos^{3}{\phi}\minus{}3\cos{\phi})\minus{}1 \equal{} 0$
[/quote]

I have a small question about this substitution. Wouldn't this restrict $ x\in [\minus{}1, 1]$? So, hypothetically, if one were given another polynomial with a root that is $ <\minus{}1$ or $ > 1$, then this substitution would be ineffective, correct?

And also, how would one prove that all the roots in a polynomial is $ \in [\minus{}1, 1]$?[/quote]
Well, doesn't it increase/decrease without bound in both directions?

[quote="archimedes1"]
Well, doesn't it increase/decrease without bound in both directions?[/quote]

I'm not sure I follow... can you please elaborate more?

[quote="p4fn2w"][quote="IrinaSharova"]Let $ x \equal{}\cos{\phi}$
$ 8x^{3}\minus{}6x\minus{}1 \equal{} 0$
$ \Rightarrow 2(4\cos^{3}{\phi}\minus{}3\cos{\phi})\minus{}1 \equal{} 0$
[/quote]

I have a small question about this substitution. Wouldn't this restrict $ x\in [\minus{}1, 1]$?[/quote]

Actually, when $ \phi$ is allowed to be complex $ \cos$ can take on any value in $ \mathbb{C}$.  In particular, often it is desirable to [b]define[/b] the cosine via the complex exponential

$ \cos z \equal{}\frac{e^{iz}\plus{}e^{\minus{}iz}}{2}$

so that, for example, $ \cos i \equal{}\frac{e\plus{}\frac{1}{e}}{2}> 1$.


Adam grew rabbits. Adam grew lots of rabbits. Adam started with 229 rabbits that he bought from various stores in 1999. It is now 2006 and adam has 1727 rabbits. assuming that the rabbit population grew exponentially find,

a)The formula for the growth in (t) years

b) the year the rabbit population grew by 50%

[hide]Although a few more numbers would be nice, I think we can manage.

$y=a\cdot b^t$
$229=a \cdot b^0$
$229=a$

$1727=229 \cdot b^7$
$b\approx 1.33$

$y=229\cdot 1.33^t$

^Is assuming that the rabbits grew exponentially [b]per year[/b]

$\frac{229\cdot 1.33^{t+1}}{229\cdot 1.33^t} \neq 1.5$ (I think...)

Or, if you mean when the rabbits were 50% then the original level.. it's

$229\cdot 1.5=229\cdot 1.33^t$
$t\approx1.42$
So approximately 1.42 years after 1999.
[/hide]

surge got it right, if you round to 1.3346 you get a little more but that's O.K. otherwise, good job!

I was going to do that, but isn't the more general form of exponential growth:
\[ P = P_o * C ^{a*t + b} \]
or
\[ P = P_o * C ^{a*t + b} + d \]


geometry
perimeter
number theory
combinatorics unsolved
combinatorics

Good afternoon
Im having a trouble with this :(
Let $ n\geq 3$ be an integer. Suppose that a convex $ n$-gon is drawn in the plane, and then each pair of nonadjacent corner points is joined by a straight line through the interior. Suppose that no $ 3$ of these lines through the interior meet at a common point in the interior. Let $ f_n$ be the number of regions into which the interior of the $ n$-gon is divided. $ f_3 \equal{} 1, f_4 \equal{} 4$. Use Euler's Formula to find $ f_n$.

My attempt has been thus: Euler's formula is $ p \minus{} q \plus{} r \equal{} c \plus{} 1$, with $ p$ vertices, $ q$ edges, $ r$ faces and $ c$ components.
Since the graph is connected, $ c \equal{} 1$. Now the convex $ n$-gon has p vertices...plus the number of intersections of lines in the interior. I've figured that taking any $ 4$ points and drawing the diagonals we get $ 1$ intersection.
So the number of vertices we have is actually $ n \plus{} \binom{n}{4}$. I did this so that the graph becomes planar to use Euler's Formula...if I am on the right track.
So we have $ r \equal{} f_n \equal{} q \minus{} (n \plus{} \binom{n}{4}) \plus{} 1$. I don't remember why I substracted $ 1$ from the $ 2$. >_<
Anyways my problem is I cannot find a closed expression for the number of edges :(
Thanks for any help =)

Edit: I subtracted 1 because the outer region is not a region considered for the problem =)

Well, let's see... every interior point is an endpoint for exactly four edges, while every point on the perimeter is an endpoint for exactly $ n\minus{}1$.  So $ 4(p\minus{}n) \plus{} (n\minus{}1)n \equal{} 2q$.  Combining this with $ p\equal{}n \plus{} {n\choose 4}$ gives $ q \equal{} {n\choose 2} \plus{} 2{n\choose 4}$.

Finally, Euler's formula gives $ r \equal{} 1\minus{}p\plus{}q \equal{} 1 \minus{} n \plus{} {n\choose 2} \plus{} {n\choose 4}$.

Now try it for the [url=http://www-math.mit.edu/~poonen/papers/ngon.pdf]regular n-gon[/url] :)

Hmm yea I have read that paper before. Michael Rubenstein, one of the co-authors, is a professor at my school (Waterloo). He teachers most of the number theory courses (analytic, etc..)


When the measure of $ \angle A$ is added to the measure of the complement of $ \angle B$, their sum is $ 120^\circ$. $ \angle A$ measures $ 86^\circ$. How many degrees are in the measure of $ \angle B$?


geometry
incenter
geometric transformation
rotation
trapezoid
perpendicular bisector
angle bisector

On the internal bisector of the angle <ACB of the isosceles triangle ABC with <BAC = 100 degs take the point D so that CD = BC.
Find the angle <BAD.

Sorry to still bother with this kind of problems, but I just found two nice synthetical solutions...

Best regards,
sunken rock

Denote $ m(\widehat {BAD})\equal{}x$ , $ P\in AB\cap CD$ and $ R\in AC\cap BD$ . Thus, the point $ P$ is the incenter

of the triangle $ EBC$ and $ AEDP$ is cyclic. In conclusion, $ x\equal{}60^{\circ}\minus{}x$ , i.e. $ m(\widehat{BAD})\equal{}30^{\circ}$ .

Sometimes the proof you have in mind drives you to suppose the problem was more difficult as it actually is...
I was thinking the problem was college level.

Your E must be common to CA and BD.

I took as the easiest method the following:
A 20 degs rotation around C brings B to D and A to F, on (BA, hence DF = AC. Take now Q on (BF) so that FQ = FD; the triangles DFQ and ACF are congruent, thence DQ = AF and it's well known that the angle <ADF = 10 degs, which gives <BAD = 30 degs.

[i] Remark: In the triangle BDF (20-40-120 degs), if take A on (BF) s.t. AB = DF, then <ADF = 10 degs [/i]

A 2nd method:
The perpendicular bisector of AC intersects CD at M, then take N the symmetrical of M about the perpendicular bisector of BC, the triangle ANM is equilateral, the trapezoid BCMN isosceles with BM angle bisector of <NBC = 20 degs, hence BD = CM = AM and
<DBM = <AMB = 70 degs, thence DBMA is an isosceles trapezoid, with <(AB, DM) = 60 degs.

Best regards,
sunken rock


linear algebra
matrix
parameterization
graph theory
linear algebra unsolved

a strongly regular graph $srg(v,k,\lambda,\mu)$

is a graph with v vertices, each vertex has k neighbours, two neigbours have $\lambda$ common neighbours
two nonneighbours have $\mu$ common neighbours

You can prove that $(v-1-k)\mu =k(k-1-\lambda)$ (ask if you want this)

let us assume they are not disconnected or have a complement that is disconnected

Consider its adjacency matrix
One can prove that if $2 k\neq (v-1)(\mu-\lambda)$ , the eigenvalues are integers (just write out the eigenvalues)

Prove that all strongly regular graphs with noninteger eigenvalues are conference graphs :

$srg(2 k+1,k,\frac{k}{2}-1,\frac{k}{2})$

I find it weird that two equations on four parameters imply this.

Apologies if this is the wrong place.  I placed it here because of the eigenvalues.  What is in general the best place for graph theory?

Maybe a better place is the advanced forum?  :?
because i do realise it's more graph theory than matrices

i think i got it, for the interested :

 $2 k\neq (v-1)(\mu-\lambda)$

if the graph is not complete :$v-1>k$

this implies $\mu-\lambda<2$ but it has to divide $2 k$ and therefore

$\mu-\lambda=1$

this implies $v=2 k+1$

use the other equation i gave, you quickly find everything

apparently type I strongly regular graphs are also used for conference graphs


f is an even function, g and h are odd functions, all 3 are polynomials. given: f(1)=0, f(2)=1, f(3)=-5, g(1)=-1, g(-3)=2, g(5)=3, h(1)=3, h(3)=5, and h(5)=1.
Find f(g(h(1)))+g(h(f(3)))+h(f(g(-1)))

[hide="Question"]For the last one I simplified it down to h(0) and put unsolvable because in the previous given statements, h(0) was not given. However looking at the solution key, it stated(or assumed) that h(0)=0...can you assume such a statement in this case, or is that wrong?[/hide]

$ h$ is odd, so $ h(0) \equal{} h(\minus{}0) \equal{} \minus{}h(0)$, which is true if and only if $ h(0) \equal{} 0$.  This is true of any odd function.

[quote]An odd function means it's symmetric across the origin. An even is symmetric across the y-axis. For it to be symmetric across the origin, doesn't that mean it has to go through the origin? Which means for any odd function, f(0)=0.[/quote]

Consider $ f(x)\equal{}\frac{1}{x}$

Perhaps I should rephrase that. If an odd function is defined at x=0, then it goes through the origin. Nice catch grn_trtle!

But in this question, as they are polynomials, they are all defined at x=0.

I think that holds for all polynomials. Notice that if some we have some function $ p(x)\equal{}a_0x^1\pm a_1x^3\pm a_2x^5\pm \cdots\pm a_nx^{2n\plus{}1}$, clearly odd and continuous, then $ p(0)\equal{}0$. Basically, an odd polynomial cannot have a y-intercept not equal to 0 without losing its origin symmetry.

Or something like that.

"Odd function" implicitly means "odd function on $ \mathbb{R}$," and $ f(x) \equal{} \frac{1}{x}$ is not defined on all of $ \mathbb{R}$.  The only value to which it can be set at zero to make it odd is $ 0$.  The condition that the functions are polynomial is not necessary (and neither is the weaker condition that they are continuous).