pigeonhole principle
function
email
CEMC

Are we allowed to discuss now?:P

Oh, and if I sent my solutions as 14 image files (.JPEG)....
Is it ok?? :( I don't know how to use Latex..

I think we're allowed to discuss now. It's well into tuesday

Anyways, I think jpeg images are fine. I sent it as a pdf and the waterloo people said they're fine with it.

So let's start discussing from the end.  Anyone got a solution for question 10?

The two questions I had trouble with were 8 and 10.
I only got halfway through each question, because I ran out of time.

Should I post the questions up?

My answers for 8 was
a=3^(75+105t)
b=3^(60+84t)
c=3^(50+70t)
d=3^(43+60t)

10..I wrote like 5 paras..my main idea was that the sum of the balls never change, and there's only a finite number of distributions, and a move always goes one direction from a distribution to another..so it'll eventually form a loop.

For 8, I got (ak^105p, bk^84p, ck^70p, dk^60p) where a, b, c, d, p and k are all positive integers and where (a, b, c, d) is the lowest term that satisfies a^4 + b^5 + c^6 = d^7.

I couldn't find the lowest term.. :( so I just wrote my answer as above.

For 10, I used the pigeonhole theory...

EDIT
[quote]my main idea was that the sum of the balls never change, and there's only a finite number of distributions, and a move always goes one direction from a distribution to another..so it'll eventually form a loop.
[/quote]
Yea.. I got this, too... I said that after the total possible number of distributions, there must be a set or boxes that is repeated at least once, so the initial position must be achieved.

for #10, a stronger arguement is that there exists a distribution that occurs infinitely often. now you have to consider the box in which the move is made on for that particular configuration. since that particular configuration occurs infinitely often, there must also exist a starting box for that configuration that occurs infinitely often. from there, it is easy to see that this must be a loop and you are done.

yep, I got that too, after testing numbers with 3 boxes, I noticed that the starting box matters, and I defined a unique distribution with a unique startingbox as a "situation" and it's actually the situations that loop.
Also, I mentioned that for a sum of balls, there could be more than one loop, and kinda proved that loops don't overlap, or else, you could start from a situation but enter another loop that won't get you back..nvm, there's too many details, and I don't wanna post 5 paragraphs..
BTW, when do we know the results?

well, I'm expecting the results to be up around next friday...

These are my answers, and for the 10th question, so since all configurations lead to a unique configuration, all you have to do is prove that all configurations can only be arrived at from a unique previous configuration. Once you have this, there are a finite number of solutions for any given distribution and unique starting box, so you'll never be able to repeat a configuration unless you get through your starting configuration.

I forgot, 8 had two different answers that I'm aware of, one of which I used by grouping the numbers as powers of two, and they could also be done with powers of three (As I think was posted earlier in this thread).
[quote]I conjecture that (266+105n, 253+84n, 244+70n, 238+60n) for any non-negative integer n provides an infinite family of solutions as n varies[/quote]

I got everything except 8, which I used 3 as base, the same as you.
Ok, so I didn't make any big silly mistakes, as I always did before, this time...

[quote]This means that there is a unique configuration (due to 1) A that leads to a configuration B with a move, and thus B leads to A (and only A, because of 2) with an anti-move. As such, no other configuration C could lead to B with a move, because B would then have to lead to C with an anti-move, which is impossible.[/quote]

Umm, I don't think this is 100% true..
ex> (1, 1, 1, 1, 1, 1, 0, 1, 1) --> (1, 1, 1, 1, 1, 1, 0, 1, 1, 1)
also, (1, 1, 1, 1, 1, 2, 0, 0, 1) --> (1, 1, 1, 1, 1, 1, 0, 1, 1, 1)

I don't know if I understood you correctly, but aren't there many more cases like this as well???

true, but it'll end with different startingboxes, so they do have the same distribution, but differ by having different boxes to begin the next move.
And you can also consider the "anti-move", which tells us that there is only one way to get to a configuration.
"let a configuration be a distribution of golf balls, as well as a specific starting box"

you're right, and EastyMoryan's is right too. except that when you copy from his answers and paste it here, the power turns into multiplication..

I used superscript, so where it says (k-1)m it's actually a small m being used as a power, but you have (-1)^(m-1) which is wrong, I think.

Try substituting m=2 into your equation [the answer should be k(k-1)].[quote]f(m)=(k-1)^m +(k-1)*[(-1)^(m-1)]... [/quote]=(k-1)^2 +(k-1)*[(-1)^(2-1)]
=(k-1)^2 +(k-1)*[(-1)^(2-1)]
=(k-1)(k-1)+(k-1)*(-1)^1
=(k-1)(k-1)-(k-1)
=(k-1)(k-1)-(k-1)(1)
=(k-1)(k-1-1)
=(k-1)(k-2)
...which isn't what we're looking for. It should be (-1)^m

ugh,... so, when are we going to find out if we made CMO??

Did anyone receive any email fromthe CMS yet?/

:( I really hope i make CMO..

[quote="eddyliu"]Dear Student: Thank you for participating in the Canadian Open Math ChallengeRepechage. You did very well in the competition, but unfortunately youwere not among the top 25 students that will be invited to write theCanadian Math Olympiad. There were over 100 students participating andthe level of the competition was very high. We hope you enjoyed solving the problems and we look forward to seeingyou again in future competitions. Sincerely,Kalle Karu

-.- i got 40-50 and got rejected thats sucks[/quote]I also got 40-50 and got the same letter. =( Much sadness on my end :\

[quote="azn_rocker0322"]WAit, shouldn't this be fake??

I thought we would not receive any email if we don't make it...

and, I didn't receive any emails when I didn't make the CMO last year from the CMOR..

I'm pretty sure that this is a fake

EDIT also, why would this email be a reply and forwarded ??[/quote]

i dont think this is fake since everyone at my school who wrote repechage either got the invite or the other email you are talking about, btw, cuttoffs seem pretty high this year

Ugh got the same email today..
wow looks like so far nobody from the forum made it through repechage
oh well I guess I'll have to wait another year

Am I an exception?

Dear Student:

Many thanks for participating in the Canadian Mathematical Oplympiad Qualification Repacharge. Congratulations on getting an excellent result!

On behalf of the Canadian Mathematical Society I now wish to invite you to participate in the 2009 Canadian Mathematical Olympiad (CMO).

...

Yeah. But I don't believe that I'm the only one in this forum that made it.

lol i just found it funny that the cms people got the name of the contest wrong  :P
COMCR?? Not CMOR? lol   :|

The person that sent you guys the email, Kalle Karu, is new to CMS, I believe. And obviously not very professional.

Well by the sound of it, im just hoping that is this some like of practical joke. anyone going to say feburary fools?

hahaha i wish it was
unless jix and Heinrich planned this out together.. :ninja: probably not

the thing that made me mad was that the cemc person said that CMS would only contact me if I made CMO.. so I got all excited when I saw that mail in my inbox and then... kerplunk  :o

[quote]the thing that made me mad was that the cemc person said that CMS would only contact me if I made CMO.. so I got all excited when I saw that mail in my inbox and then... kerplunk [/quote]
[quote]You will hear directly from the Canadian Mathematical Society if (and only if) you are invited to the CMO.[/quote]

I know how you feel buddy :(

[quote]The CMS decided to contact the students who were not invited to write the CMO as well as those who were invited to write the CMO.  When we sent out the CMOQR results, we assumed (incorrectly) that they would only contact those who were invited.  Sorry for any confusion.
[/quote]

ha..ha....ha.....ha.......ha.........

o what a disappointment...*slashes wrist and trashes room*


A right circular cone has a volume of $ 12\pi$ cubic centimeters. The height of the cone is 4 cm. How many centimeters is the circumference of the base of the cone, in terms of $ \pi$?

$ V\equal{}\frac{\pi r^{2}h}{3}\equal{}12\pi\equal{}\frac{4\pi r^{2}}{3}\Rightarrow r\equal{}3\Rightarrow C\equal{}6\pi$


i need help with this problem: :rotfl:  :arrow:  :!:

what is the area of a parallelogram with sides measuring 4 cm and 14 cm and an interior angle measuring 120 degrees in cm?

Remember the formula
$A=ab\sin(\theta)$

Applying this to your parallelogram,
$A=4(14)(\sin120)=4(14)(\sqrt{3}/2)=28\sqrt{3}cm^{2}.$

If you would like me to prove the formula, please ask so in your reply.  :P

[hide="Proof"]Denote the height perpendicular to $b$ by $h$. Then, look at the triangle with height $h$, hypotenuse $a$, and the angle opposite $h$ $\theta$. Notice that:
$\sin(\theta) = \frac{h}{a}$
$a\sin(\theta) = h$
Since the area is $h$ times the other base ($b$):
$A = bh = b(a\sin(\theta)) = ab\sin(\theta)$

A simple geometry proof (with a little trig). [/hide]

[hide]Using the area of a triangle via sines, we get it to be

$4*14*\sin{120}=4*14*\frac{\sqrt{3}}{2}=28\sqrt{3}$[/hide]


Find then units digit in the decimal expansion of
\[ (15 \plus{} \sqrt {220})^{10} \plus{} (15 \plus{} \sqrt {220})^{82}
\]
Thanks in advance!

[quote="Yongyi781"][quote="314.math"]Find then units digit in the decimal expansion of
\[ (15 \plus{} \sqrt {200})^{10} \plus{} (15 \plus{} \sqrt {220})^{82}
\]
Thanks in advance![/quote]

Typo? I'm assuming both of them should be $ \sqrt {220}$, since the problem wouldn't leave $ \sqrt {200}$ as that and not simplify it to $ 10\sqrt 2$...[/quote]

Sorry, it should be $ \sqrt{220}$.  I'll edit that.  Thanks!

[hide="hint"]use the conjugate $ \left(15 \minus{} \sqrt {220}\right)$[/hide]
[hide="hint V2"]add $ \left(15 \minus{} \sqrt {220}\right)^{10} \plus{} \left(15 \minus{} \sqrt {220}\right)^{82}$[/hide]


geometry
geometric transformation
reflection

Suppose we have a quadrilateral $ ABCD$ inscribed in a circle. The tangents to this circle at the points $ A$ and $ C$ intersect at a point $ P$, such that this point $ P$ does not lie on the line $ BD$, and such that $ PA^{2} \equal{} PB\cdot PD$. Prove that the line $ BD$ passes through the midpoint of the segment $ AC$.

If $ PB$ meets the circle at $ D'$ and $ PD$ meets the circle at $ D'$, then by the power of point theorem, we have that $ PD\equal{}PD'$ and $ PB\equal{}PB'$. Hence we have a reflection $ R(OP)$ under which we have $ D \to D'$, $ B \to B'$ and $ A \to C$.

Therefore $ AC$, $ BD$ and $ B'D'$ are concurrent on some point on $ OP$. Since $ AC$ is a chord of the circle, so $ OP$ meets $ AC$ at its midpoint, which means $ BD$ passes through the midpoint of $ AC$.


graph theory
combinatorics unsolved
combinatorics

Let $ G$ be the complete graph with $ 2n$ vertices, and suppose that $ S_1,...,S_k$ are subgraphs of $ G$ with the following properties:

(a) Every $ S_i$ is a complete bipartite graph;

(b) Every edge of $ G$ is contained by an odd number of subgraphs $ S_i$.

Show that $ k\geq n$.

[b]EDIT:[/b] please discuss this in the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=271585

This is [url=http://www.komal.hu/verseny/feladat.cgi?a=honap&h=200903&t=mat&l=en]problem A. 477.[/url] from the KöMaL contest, the deadline is 15 April.

[b]Why did you post an active problem?[/b]


geometry
analytic geometry
Pythagorean Theorem
virat kohli
ind 21
rmo 2014 P6

A square has sides of length $2$. Set $S$ is the set of all line segments that have length $2$ and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set $S$ enclose a region whose area to the nearest hundredth is $k$. Find $100k$.

[hide]Consider the case where the two sides are the lower one and the one on the left (the other three will follow from symmetry).
If one endpoint of a segment on $S$ has coordinates $(2x,0)$ and the other endpoint has coordinates $(0,2y)$ then the midpoint will be $(x,y)$.
The length of the segment is 2, so by Pythagorean Theorem, $(2x)^2+(2y)^2=4\Rightarrow x^2+y^2=1$. Since $x$ and $y$ are greater than 0, the set of these midpoints are one a quarter-circle with radius 1.

The area enclosed in the midpoints will be inside the square but not inside any of the 4 quarter-circles. This area is given by $4-\pi\Rightarrow \boxed{086}$.[/hide]


How many permutations of n objects are there such that no object is in its original position?

This is probably well-known, but I thought I'd post it anyway.

Unless I'm mistaken, this is [i]very[/i] well-known. It's called a derangement.

http://mathworld.wolfram.com/Derangement.html

The contest-useful formula is $D(n)= \left[ \frac{n!}{e}\right]$, where $[x]$ is the closest integer function.

Ah! You're right.  I've seen that before, but didn't remember what it was.

I really posted this problem, because there was one in our math textbook that said "How many ways are there to arrange four letters such that exactly 1, (2, 3, etc...) are in their correct envelopes?"

And the answer book for it had one write out all permutations by hand and count, which I thought was sort of dumb...

I think for case n=6 the example is worked out in AoPS V2 but they don't go into derangements


What is the next term, and the way of forming the sequence

0, 1, 10, 2, 100, 11, 1000, 3, 20, 101, ....

[quote="bdpiang"]Looking at the terms in the prime positions:
2nd term=1
3rd term=10
5th term=100
7th term=1000
11th term =10000 (is this the solution?)[/quote]kind of related! But you are explaining only for the primes, what about the others!

[quote="ZetaX"]Well, it looks like prime factorisation
($n$-th digit gives multiplivativity of the $n$-th prime)[/quote]Yes, that is one way to look at it! The question would be, what is the term 1024 in the secuence  :D

[quote="ZetaX"]Hmm... $A$ ¿  ;)[/quote]jeje, then, in what basis are you working?, and if you are working in base k, what is the term $2^k$ in the sequence  ;)

There is a small way around, but basically, you have the answer.

[quote="ZetaX"]I work in basis $\infty$  :lol:  and see it more as sequence of sequences (of sequences of decimal digits  ;) )[/quote]Well, it works perfectly. Originally I had thought about $f(p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_n^{\alpha_n})=\alpha_1+10\alpha_2+\cdots+10^n\alpha_n$, and of course, if you want to vary the basis, you can just change 10 to your favorite basis. But of course, yours work too! :first:


I'd appreciate a hint on this one and not a solution :)

if $p$ and $p^2+2$ are primes, prove that $p^3+2$ is also a prime.

[hide="Hint"]Try to find all the numbers $p$ such that both $p$ and $p^2+2$ are prime.[/hide]

is this correct?

feel free to ridicule me if i made an error in my reasoning :)

[hide]Suppose we are looking at $p^2+2$ mod $3$. since we assume that $p^2+2$ is a prime and that the residues mod 3 of $p^2$ are either $0$ or $1$ and if it is $1$, then $p^2+2$ will be a multiple of 3 and hence not a prime, this means that $p^2$ must have residue $0$ in order for $p^2+2$ to be a prime.

This means that $p^2$ is a multiple of 3 and if $p^2$ is a multiple of 3 then p is a multiple of $3$. this leaves us the only option : $p = 3, p^2+2 = 11$ and$p^3+2=29$.[/hide]

Yes, this is correct. A bit shorter way is to employ the fact that any prime larger than $3$ can be written as $6k\pm 1$ (no other possible residue, since $2,4$ would make it even and $3$ would make it divisible by $3$). But then $(6k\pm 1)^2+2=36k^2\pm 12k+3$ is divisible by 3. Hence, $p$ can't be greater than $3$.


[quote="vipCD_A1"]Let $ a,b,c > 0; a^2 \plus{} b^2 \plus{} c^2 \equal{} 1; k\geq{1}$. Prove that:
$ \frac {a \plus{} b}{\sqrt {a^2 \plus{} 2ab \plus{} kb^2}} \plus{} \frac {b \plus{} c}{\sqrt {a^2 \plus{} 2bc \plus{} kc^2}} \plus{} \frac {a \plus{} c}{\sqrt {c^2 \plus{} 2ac \plus{} ka^2}}\geq\sqrt\frac {a \plus{} b \plus{} c}{\sqrt [3]{kabc} \plus{} 1}$

goodluck!  $ : maybe:$[/quote]
......................
to give up
ok
not solve

[quote="chess"]vipCD_A1 wrote: "VERY HARD FROM DIEU".
What is DIUE?  :huh:[/quote]
I'm Dieu
ok


group theory
abstract algebra
superior algebra
superior algebra theorems

Hello

let $G$ be a group, and $H$ a subgroup different from $G$

prove that you can always find a group $W$ and group morphisms $f_{1}$ and $f_{2}$ from $G$ to $W$ that are different, but have equal images for every element in $H$

it shouldn't be too hard but I can't find it.
For those familiar with category theory : this would then prove that every epic morphism in the category of groups has to be surjective

Take $W$ to be the amalgam $G*_HG$, and set $f_1$ to be the identity of $G$ onto the first copy of $G$ in $W$, and $f_2$ to be the identity from $G$ onto the second copy of $G$ in $W$.

By amalgam, you mean just the same as a pushout diagram?
Quite consufing in my opinio, you really have to 'tag' elements of the first copy of $G$ and the second in $W$,right, or you would just get $G$ again instead of $G* G$

Yes, if $f : K \to G,~ g : K \to H$ are homomorphisms of groups, the amalgated product $G*_K H$ is defined by a pushout-square

[img]http://www.xsploitz.org/hosted/helpar/martin/amalg.png[/img]

Like all pushouts, it can be constructed by coproducts and coequalizers, in this case as the quotient of the free product $G*H$ by the smallest normal subgroup containing all $f(k) g(k)^{-1}$.

In your case, one takes the amalgated product of the inclusions $H \hookrightarrow G,~ H \hookrightarrow G$.

Wow first reply since new layout!

Uhm, yes I have studied pushouts and free (abelian) groups quite briefly, but if you take both maps from $H$ to the amalgam, how are you so sure that they are different then? :huh:


trigonometry
limit
inequalities
real analysis
real analysis unsolved

Let $a , a \ge 2$  , be a given  integer and
\[ f_n(x)= \sin\left( a^n x\right) \; \; ,\; \; x\in {\mathbb R}. \]
[b]
Find all  $x$ for which the sequence $\left(f_n(x)\right)_{n=1}^{\infty}$ is   convergent.  In case of convergence, find the limit .[/b]

This is indeed a very hard problem. Even for $ a=2$ the problem is not at all simple. Anyway, is it an open question or does it have a solution? For $a=2$ I can solve it, but for the other values...

I found in one of my  ancient note-books, only  following informations regarding this problem :

[b]A)[/b] [i] Solution in case  $ a=$odd [/i] , $|a|>1\; .$
Consider the sets of numbers
\[(1)\; \; \; \;  M_1=\left\{ y=\frac{2m\pi}{a^b(a-1)}\; \; | \; \; m\in {\mathbb Z}\; ,\; b\in {\mathbb N}\right\}\]
\[(2)\; \; \; \;  M_2=\left\{ z=\frac{(2m+1)\pi}{a^b(a+1)}\; \; | \; \; m\in {\mathbb Z}\; ,\; b\in {\mathbb N}\right\}\; .\]
Then $\left(f_n(x)\right)_{n\ge 1} $ is convergent iff  :
[b]Case A.1 : [/b]  $\; \ ; x\equiv y \; \; \mbox{\rm (mod)}2\pi \; \; ,\; \; y\in M_1\; \; \; \mbox{\rm see}\;  (1) $,  in that case  $ \lim\limits_{n\to \infty}f_n(x)=\sin {\frac{2m\pi}{a-1}}\; . $

[b]Case A.2 : [/b]  $\; \ ; x\equiv z \; \; \mbox{\rm (mod)}2\pi \; \; ,\; \; z\in M_2\; \; \; \mbox{\rm see}\;  (2) $ . Then  $ \lim\limits_{n\to \infty}f_n(x)=\sin {\frac{(2m+1)\pi}{a+1}}\; . $

There is also specified that case $ a=7$ was a question raised by Ky Fan
(a famous mathematician which has deep results in  Tpopology, Convex sets,
Linear inequalities,... ) in Amer.Math.Monthly[-1949 ?].

Try to extend the solution when   a=2 to case [b]B)[/b] when  $a=$[i]even, arbitrary[/i].

It would be of interest. Regards,Alex


integration
calculus
calculus computations

The base of a certain solid is the circle $x^2+y^2=9$, and each cross section perpendicular to the x-axis is an equilateral triangle with one side across the base. What is the volume of the solid?


limit
vector
real analysis
real analysis unsolved

Let $E$ be a normed vectorial space. Suppose that $F$ is a closed vectorial subspace and $D$ is a line.
Prove that $F+D$ is closed.

Since $D$ is a shifted one-dimensional subspace, it suffices to show the required result for lines through the origin, i.e. to prove that if $F\le E$ is closed, then $F+<x>$ is also closed for $x\in E$ (we assume $x\not\in F$, otherwise it's clear).

Assume we have $a_n\in F,\ \lambda_n\in\mathbb R,\ u\in E\ a_n+\lambda_n\cdot x\to u$. If $(\lambda)_n$ is unbounded, then we can extract a subsequence $\lambda_{k_n}\to\infty$, and $\frac 1{\lambda_{k_n}}\cdot a_{k_n}\to-x$, which iis absurd. This means that $(\lambda_n)_n$ is bounded, and we can extract a convergent subsequence $(\lambda_{k_n})_n\to\ell$. If $\ell=0$, then it's clear that $u\in F$, and we're done. Otherwise, we divide $a_{k_n}+\lambda_{k_n}\cdot x\to u$ by $\lambda_{k_n}$, and we get $\frac 1{\ell}\cdot u\in F+<x>$, as desired.

Assume we have $f_n \in F$ and $a_n \in \mathbb{R}$ such that $\{f_n + a_nx\}$ is a Cauchy sequence converging to $c$. Then
\[\lim_{m,n\to \infty} \| (a_n -a_m)x - (f_n -f_m)x)\| = 0 \] hence \[(a_n - a_m)\text{d}(x,F) = \text{d}((a_n-a_m)x,F) \rightarrow 0\] That implies $\{a_n\}$ is Cauchy sequence, and then so is $\{a_nx\}$. Thus both $\{f_n\}$ and $ \{a_nx\}$ converges, to $f\in F$ and $ax $ respectively, so $c = f +ax \in F + <x>$, which shows that $F + <x>$ is closed.

I was looking through some older problems, and I found this. Just one comment: this holds for topological vector spaces $E$ (over $\mathbb R$ or $\mathbb C$) in general:

Suppose there is a [url=http://planetmath.org/encyclopedia/Net.html]net[/url] $(f_\alpha+t_\alpha d)_{\alpha\in A}\to x\in E$, where $(f_\alpha)_\alpha\subset F,\ d\in D$, and $t_\alpha$ are scalars.

Suppose first that you can find a bounded subnet $(t_\beta)_\beta$ of $(t_\alpha)_\alpha$ (this is true if there is some $M>0$ such that the set of indices $\beta$ for which $|t_\beta|\le M$ is cofinal in $A$). Then you can extract a convergent subnet $(t_\gamma)_\gamma\to t$. We thus have $t_\gamma d\to td\Rightarrow f_\gamma\to x-td\Rightarrow x-td\in F$, so $x\in F+ D$, as desired.

Now assume we have $|t_\beta|\to\infty\Rightarrow\frac 1{|t_\beta|}\to 0$. Since multiplication with scalars is continuous, we have $\frac 1{t_\beta}f_\beta+d\to 0$, so $d=0$ (otherwise a net in $\mathcal F$, which is closed, would converge to $d\not\in F$). We thus have $f_\alpha\to x\Rightarrow x\in F$. Again, we're done.


Al and Barb both tried to solve the same equation, of the form A(x^2) + bx+c=0. However, each made exactly one error when copying the equation. Al's solutions were -2 and -1.5, but he coped A wrong. Barb's solutions were 2 and 3, but her value of B was wrong. What are the solutions of the correct equation?

This is a really good question from my math summer packet  :)

[hide="solution"]From the problem we know what quadratics they made.

Al's: $ a(x \plus{} 2)(x \plus{} \frac{3}{2}) \equal{} ax^2 \plus{} \frac{7}{2}ax \plus{} 3a$

Barb: $ b(x \minus{} 2)(x \minus{} 3) \equal{} bx^2 \minus{} 5xb \plus{} 6b$

6b = 3a so 2b = a.

The right equation should be $ bx^2 \plus{} \frac{7}{2}ax \plus{} 6b$  Next we substitute a = 2b into the equation and find it's roots  $ bx^2 \plus{} 7bx \plus{} 6b \equal{} b(x^2 \plus{} 7x \plus{} 6) \equal{} b(x \plus{} 1)(x \plus{} 6)$

The roots are -1 and -6.[/hide]


Let $n$ be a positive integer. If the equation $2x + 2y + z = n$ has 28 solutions in positive integers $x$, $y$, and $z$, then n must be either:

(A) 14 or 15
(B) 15 or 16
(C) 16 or 17
(D) 17 or 18
(E) 18 or 19

[hide="Solution"]First, note that $28 = {8 \choose 2}$.

Rewrite the equation as $2(x+y) = n - z$.  This only has solutions when $n \equiv z \bmod 2$, so we divide into two cases:

Case 1:  $n \equiv 0 \bmod 2$.  We have $z = 2, 4, ... n - 4$ (it cannot equal $n - 2$ because then one of $x$ or $y$ must be 0, which is not a positive integer), giving us a total of $p = \frac{n-6}{2} + 1$ possibilities for the value of $z$, which translates to ${p+1 \choose 2}$ possibilities for the number of solutions.  This gives us $\frac{n-6}{2} + 2 = 8 \implies n = \boxed{18}$.

Case 2:  $n \equiv 1 \bmod 2$.  We have $z = 1, 3, ... n - 4$ for a total of $p = \frac{n-5}{2} + 1$ possibilities, and the same logic as above gives $n = \boxed{17}$.  So our answer is $\boxed{D}$.
[/hide]

[hide]

Using generating functions, you can find that the number of solutions to the equation is the coefficient of  in the expansion of . We can quickly see that  is the sum of the first 7 natural numbers, so it's the coefficients using the expansion of the first 7 terms in the second term of the generating function. We can see that either we use all odd powers to combine them, making the exponent 17, or all even powers to make it 18. Thus the answer is D.[/hide]

Elemennop:

Can you show the process for how you derived that generating function? I am sort of confused as to how you got the coefficients.

Sure.

$z$ can take on any positive integer value, and it can only take on any positive integer value in only one way (ie. it can only be $3$ if $z=3$). So since the coefficient of each term is the number of ways you can get the value in the exponent, it's generating function is $(x+x^2+x^3+x^4+x^5+x^6+\ldots)$, since it can only get $1$ one way, $2$ one way, and so on.

Instead of dealing with the $x$ term, we deal with the entire $2x$ term. $x$ can take on any positive integer value, but it is doubled so the term $2x$ can only take on even positive integer values, and it can only take it on one way. So we make it's generating function $(x^2+x^4+x^6+x^8+x^{10}+x^{12}+x^{14}+\ldots)$ in the same manner we made $z$'s generating function, but not including odd exponents because $2x$ can never be odd.

$2y$'s generating function is identical to $2x$'s.

To simplify the problem, we can multiply $2x$'s and $2y$'s generating functions; $(x^2+x^4+x^6+x^8+x^{10}+x^{12}+x^{14}+\ldots)\cdot(x^2+x^4+x^6+x^8+x^{10}+x^{12}+x^{14}+\ldots)$. Looking at what the product may be, we know that it will only have even exponents. We also note that as we go up, we get one more way to obtain the value. For $x^4$, we can only make it one way ($x^2\cdot x^2$). With $x^6$, we can either multiply the first $x^2$ and the second $x^4$, or vice versa, so there's two ways to get $x^6$. For $x^8$, we can either multiply $x^4$ and $x^4$, or $x^6$ and $x^2$ in two different ways, so there's three ways we can get $x^8$. We clearly see that each successive term's coefficient is one more than the previous, so we get the generating function between $2x$ and $2y$ to be $(x^4+2x^6+3x^8+4x^{10}+5x^{12}+6x^{14}+\ldots)$.

At this point we only have to multiply $z$'s generating function with the one we just derived, and we end up at the beginning of my earlier post.

[hide]Change the equation to $2(x+y)+z=n$.  For each possible value of $x+y$, there are $x+y-1$ ways to choose values for $x$ and $y$.  So for $x+y=2$ (the minimum), there is 1 solution, when $x+y=3$, there are 2 solutions, etc.  Since 28 is the sum of the first 7 natural numbers, the maximum of $x+y$ is 8.  The maximum of $x+y$ is 8 when $n=17,18$, so the answer is $\fbox{D}$[/hide]

How are there exactly $x+y-1$ ways to choose solutions? If $x+y=n$ wouldn't there be $(n+x+y-1)C(x+y-1)$ ways? so for intance $x+y=2$ would have $3$ solutions, and $x+y=3$ would have $4$ solutions, etc... Can someone explain that? Or is it because we can't have $0$ for either variable, so we subtract $2$ solutions from that?

All variables are positive integers, so if $x+y=5$, then $x$ can be any integer from 1 to 4 (not 5 b/c $y$ must be positive also)


Let $A_1,A_2,\ldots,A_n\in M_2(\mathbb{R})$. Prove that there is a choice of $\varepsilon_1,\varepsilon_2,\ldots,\varepsilon_n\in\{-1,1\}$ s.t.
$\det\left(\varepsilon_1A_1+\varepsilon_2A_2+\ldots+\varepsilon_nA_n\right) \ge \det A_1+\det A_2+\ldots+\det A_n$
and a choice for the reverse inequality.

This isn't hard at all if you use the formula $\det(A+B)+\det(A-B)=2(\det A+\det B)$ which holds in $\mathcal M_2(\mathbb C)$ (or any other field, for that matter), but I think there was this rule on the forum stating that we shouldn't post any problems which come from magazines less than one year old.

[quote]we shouldn't post any problems which come from magazines less than one year old.[/quote]
sorry... I forgot about that :(


trigonometry
function
inequalities proposed
inequalities

Prove that
\[ \frac {\sin{A}}{A} \plus{} \frac {\sin{B}}{B} \plus{} \frac {\sin{C}}{C}\leq \frac {9\sqrt {3}}{2\pi}
\]
where $ A,B,C$ are the angles of a triangle.

This is just Jensen.

$ f(x) \equal{} \frac {\sin x}{x}$ for $ x\in(0,\pi)$.
$ f'(x) \equal{} \frac {x\cos x \minus{} \sin x}{x^2}$
$ f''(x) \equal{} \frac {(\cos x \minus{} x\sin x \minus{} \cos x)x^2 \minus{} 2x(x\cos x \minus{} \sin x)}{x^4}$ $ \equal{} \minus{} \frac {x^2\sin x \plus{} 2x\cos x \minus{} \sin x}{x^3}$

$ g(x) \equal{} x^2\sin x \plus{} 2x\cos x \minus{} \sin x$
$ g'(x) \equal{} 2x\sin x \minus{} x^2\cos x \minus{} 2x \sin x \plus{} 2\cos x \minus{} \cos x \equal{} \cos x (1 \minus{} x^2)$.

$ \Rightarrow$ the function has local minimum at $ x \equal{} \frac {\pi}{2}$.

$ \Rightarrow g_{\min} \equal{} g\left(\frac {\pi}{2}\right) \equal{} \frac {\pi^2}{4} \minus{} 1 > 0$ or $ g_{\min}\equal{}g(0)\equal{}0$.

$ \Rightarrow f(x)$ is concave for every $ x\in (0,\pi)$

$ \Rightarrow \sum \frac {\sin \alpha}{\alpha}\le 3\frac {\sin \frac {\pi}{3}}{\frac {\pi}{3}}$.

And excepting this, most probably, the problem has been posted already.


(I [i]think[/i] this is the right forum to post these in. If not, sorry).

I'm trying these two questions and I'm completely floored on both of them. They're probably both really easy but I'm fed up of looking at them after an hour without having some flash of inspiration so I thought I might as well ask for some help, since you guys never cease to amaze me with your sheer mathematical ability.

The first is:
Find in $ \sqrt{3\minus{}2\sqrt{2}}$ the form $ p\plus{}q\sqrt{2}$ where $ p$ and $ q$ are integers. I know the answer is:

[hide]$ 1\minus{}\sqrt{2}$ and $ \minus{}1\plus{}\sqrt{2}$[/hide]

But I don't see how that answer is found.

And the second is solve the simultaneous equation:
$ 2^{x}\plus{}3^{y}\equal{} 5$
$ 2^{x\plus{}3}\minus{}3^{y\plus{}2}\equal{} 23$

I'm pretty sure this one involves logarithms of a sort, but I'm struggling to apply them without making it a mess.

[quote="kh-"]Find in $ \sqrt{3\minus{}2\sqrt{2}}$ the form $ p\plus{}q\sqrt{2}$ where $ p$ and $ q$ are integers. [/quote]

Questions of this type are common.  The method is to write $ 3\minus{}2\sqrt{2}\equal{} (p\plus{}q\sqrt{2})^{2}$ and calculate $ p, q$ by comparing coefficients.  :)

[quote="kh-"]And the second is solve the simultaneous equation:
$ 2^{x}\plus{}3^{y}\equal{} 5$
$ 2^{x\plus{}3}\minus{}3^{y\plus{}2}\equal{} 23$[/quote]

[hide="Hint 1"] $ 2^{x\plus{}3}\equal{} 8\cdot 2^{x}, 3^{y\plus{}2}\equal{} 9\cdot 3^{y}$ [/hide]
[hide="Hint 2"] Substitute $ a \equal{} 2^{x}, b \equal{} 3^{y}$. [/hide]

OK, the hints for the second question were very helpful, thanks. I feel a bit of a dunce for not seeing it before, but I'm not likely to forget how to do this again!

[hide="Q2 Solution"]
$ 2^{x}\plus{}3^{y}\equal{} 5$
$ 2^{x\plus{}3}\minus{}3^{y\plus{}2}\equal{} 23$

The second equation can be written as:

$ 8\cdot 2^{x}\minus{}9\cdot 3^{y}\equal{} 23$

Which allows us to substitute for $ a \equal{} 2^{x}$ and $ b \equal{} 3^{y}$, giving the simple enough simultaneous equations:

$ a\plus{}b \equal{} 5$
$ 8a\minus{}9b \equal{} 23$

So making $ a \equal{} 5\minus{}b$, we get for the second equation:

$ 8(5\minus{}b)\minus{}9b \equal{} 23$
$ 40\minus{}8b\minus{}9b \equal{} 23$
$ 40\minus{}17b \equal{} 23$
$ \minus{}17b \equal{}\minus{}17$
$ b \equal{} 1$

Which substituted back into $ b \equal{} 3^{y}$ means that $ y \equal{} 0$. We can then use the $ y$ value in the original equation, which is simplest in the first:

$ 2^{x}\plus{}3^{0}\equal{} 5$
$ 2^{x}\plus{}1 \equal{} 5$
So clearly $ x \equal{} 2$.
[/hide]
However, I do not understand your advice for the first question, and I'm unable to put it into practice.

The central concept behind problems such as #1 is that $ \sqrt{2}$ is irrational.  This means there do not exist integers $ p, q$ such that $ \sqrt{2}\equal{}\frac{p}{q}$.  This allows us to prove a useful lemma.

[b]Lemma:[/b]  If integers $ a, b, c, d$ exist such that $ a\plus{}b\sqrt{2}\equal{} c\plus{}d\sqrt{2}$, then $ a \equal{} c$ and $ b \equal{} d$.

Proof:  $ (a\minus{}c)\plus{}(b\minus{}d)\sqrt{2}\equal{} 0$.  Now, if $ a\minus{}c, b\minus{}d$ were not equal to zero then we could write $ \sqrt{2}\equal{}\frac{a\minus{}c}{d\minus{}b}$, but we know that $ \sqrt{2}$ is irrational, so this is a contradiction.  Hence $ a\minus{}c \equal{} b\minus{}d \equal{} 0$.

[hide="From here..."] $ 3\minus{}2\sqrt{2}\equal{} (p^{2}\plus{}2q^{2})\plus{}(2pq)\sqrt{2}$

From the lemma we now know that

$ 3 \equal{} p^{2}\plus{}2q^{2}$
$ \minus{}2 \equal{} 2pq\Leftrightarrow$
$ \minus{}1 \equal{} pq$

Because we are given that $ p, q$ are integers, the only solutions are $ (p, q) \equal{} (\minus{}1, 1), (1,\minus{}1)$, which give

$ p\plus{}q\sqrt{2}\equal{}\minus{}1\plus{}\sqrt{2}, 1\minus{}\sqrt{2}$ [/hide]

[quote="kh-"]
Find in $ \sqrt{3-2\sqrt{2}}$ [/quote]
\begin{eqnarray*}\sqrt{3-2\sqrt 2 }&=&\sqrt{\left({2-2\sqrt 2+1}\right)}\\ &=&\sqrt{\left({\sqrt 2-1}\right)^{2}}\\ &=&\sqrt 2-1\end{eqnarray*}

$ \sqrt{3-2\sqrt2}=1-\sqrt2$ it's not correct, 'cause square root of a positive real number it's a real nonnegative one.

[quote="t0rajir0u"]The central concept behind problems such as #1 is that $ \sqrt{2}$ is irrational.  This means there do not exist integers $ p, q$ such that $ \sqrt{2}\equal{}\frac{p}{q}$.  This allows us to prove a useful lemma. [i](snippy snip)[/i][/quote]

Thank you [i]very[/i] much for taking the time to go into such detail. I found the explanation of the lemma and the walk through of what was really going on very instructive, and understand everything that needs to be done in this question now: I think I could never have got this on my own. Your help was greatly appreciated!


[color=blue][b]A game of Jai Ali has eight other players and starts with players $ P_1$ and $ P_2$ on court and other players $ P_3,P_4,P_5,P_6,P_7,P_8$ waiting in a queue. After each point is played, the loser goes to the end of the queue, the winter adds $ 1$ point to his score and stays on the court, and he player at the head of the queue comes on to contest the next point. Play continues until someone has scored $ 7$ points. At the moment, we observe that a total of $ 37$ points have been scored by all eight players. Detrmine who has won and justify your answer.[/b][/color]

[quote="keyree10"]could you please post the solution to this problem?
I wasn't able to solve it.  :oops:[/quote]
[color=blue][b]Oh, I want to get a nice solution after that I will post my solution for more reference. So, be calm down...[/b][/color]

[hide="Solution"]Suppose $ P_{n}$ is the winning player.  Define an "entry" as the moment when $ P_{n}$ enters the court to play a point, and consider the number of points scored by other players between two consecutive entries.  After an entry, $ P_{n}$ wins some nonnegative number of points, then he loses a point, and then he waits in each of the six places in the queue for one point to be played.  Therefore, seven points are scored by other players between two entries.  Also, $ P_{n}$ must win the last point of the game, so no points are scored by other players after the last entry.  There are thirty points scored by other players, so there must be two points scored before the first entry.  Therefore, $ P_{4}$ is the winning player.[/hide]


1)If  $x+y=1$  and  $x^2-y^2=2$,  find $x^3+y^3$

2)$148x+231y=527$
   $231x+148y=610$

   Find $x$ and $y$

Since these are tooo easy,i would appreciate no hints given away by high schooler's for at least 2 days.

I don't think that #1 is possible. You can't find out what $xy$ is.

Yeah. Given $x+y$ and $x^2+y^2$, however, we can get $x^3+y^3$

Well without squaring or cubing any expressions that is.  But if we just substitute,

$x=\frac{3}{2}$ and $y=-\frac{1}{2}$

Oh oops.  Didn't see the footnote.

Number one can be solved using a system of equations method, and pianoforte even did it for us  :D

There is a nice solution for number 2.  :P

[hide="here is a little hint"]don't solve by substitution or addition (method so the equations can cancel). Try adding them without doing anything. [/hide]

[quote="pianoforte"]2.
[hide]
$148x+231y=527$
$231x+148y=610$

$379x+379y=1137$
$83x-83y=83$

$x+y=3$
$x-y=1$

$x=2$
$y=1$
[/hide][/quote]

I'm glad you didn't do it the hard way.

by the way, how are these problems "easy"?

[quote="shadysaysurspammed"]1)If  $x+y=1$  and  $x^2-y^2=2$,  find $x^3+y^3$
[/quote]
I know im in high school, but this problem has already been solved anyways, i just want to show a different approach.

$x+y=1$
$2x +2y=2$
$x^2-y^2=2$
$2x+2y=x^2-y^2$
$2(x+y)=(x+y)(x-y)$
$2=x-y$

   $x-y = 2$
$+$$x+y=1$
$-----------------$
$2x  = 3$
$x=\frac{3}{2}$
$y=-.5$

So, by plugging into $x^3 + y^3$, we get $3.25$

I guess they're easy because they aren't this-year-Nationals level.

[hide="hint for number 2"]Use Cramer's rule.[/hide]

[quote="nonie"][quote="GoBraves"]
$2x=3$
$x=3$
$y=-.5$

So, by plugging into $x^3 + y^3$, we get $3.25$[/quote]

i think you mean x= 3/2 right??(seems like a typo but I may be wrong)[/quote]

Yes 'twas a typo

The easiest method for the first one.

$x^2-y^2=2$
$(x+y)(x-y)=2$
$x-y=2$

$x+y+x-y=3$
$x=3/2$

If i may have the permission,i'll keep posting questions like these of different genres,like medium,brainy,etc....is it okay if i do so?


In this thread we post web sites, media, etc. that interest us, in the hopes that it interests our fellow AoPSers.

I'll start with

http://xkcd.com/  Geeky webcomic. "A webcomic of romance, sarcasm, math, and language." Updates Monday, Wednesday, and Friday. Obligatory.

http://turbulence.org/spotlight/thinking/chess.html A java applet that plays chess against you, except that it shows you where it's searching, which is cool.

http://www.fark.com/ "Fark.com, the Web site, is a news aggregator and an edited social networking news site." Basically people find strange, cool, and funny news articles, make up a snarky headline, and submit it to this site, where a collection of all such said news items are conveniently linked for you to read. Just keep in mind, 40% of all people who post in its discussion forums are trolls.

gentoo.org: The distro I trust on many of my machines. News thereof.

Slashdot.org: geek news

giantitp.com: A [i]funny[/i] comic, though some mildly bad language/references sometimes.

http://requiem.seraph-inn.com/ Everyone likes web comics. :rotfl:

Here(AoPS)

http://www.pathguy.com/chess/ChessVar.htm Chess variants

um.... That's it.

Megatokyo.com: Web comic of two guys stranded in Japan that find love. Tons of references to "leet" and PS games.

I'm going to try xkcd when I get home  :)

[quote="lingomaniac88"][quote="miyomiyo"]http://www.harveycartel.org/metanet/downloads.html[/quote]
I'm assuming you've played N?[/quote]Played N?  N is my life.  I had to turn off Caps Lock to right this because I was editing a map.

xkcd is sooooo awesome.. yet weird.. but still... awesome. :D

XKCD is so random, there's no plot...just random things.

He could use a couple lessons in drawing though...

[quote="bubka"][url]http://www.twcenter.net[/url]
[url]http://www.totalwar.org[/url]

the center and the org! guaranteed to keep you glued for the rest of your life...  :roll:

and...umm...what is N?[/quote]

http://en.wikipedia.org/wiki/N_%28game%29

N is really cool. I haven't played it in a while, though. I think there's a new version since the last time I played it.

Um... I don't go anywhere cool except here and xkcd... I wish I did...
And I have tried N, and I stink at it.

[quote="1=2"][quote="miyomiyo"][url=http://numa.notdot.net/browse?sort=rating&order=desc&q=author:miyoser]My N Maps[/url]

Back on topic now.[/quote]

How do we play them?[/quote]Open up N.
Press "~".
Copy the data from the text box in the map you want to play.
Paste it into the upper text box in N.
Hit "L" to load.
Hit "P" to play.
To restart, hit "~", "R", "P".

We're getting off topic here.

Here's a topic dedicated to N:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190081

To help get back on topic, I guess I'll list some websites:

http://xkcd.com/

http://www.yoyogames.com/

[b]Veoh:[/b] [url]http://www.veoh.com[/url] -- Lots of videos, including anime episodes and stuff. Anything before Nov 2007 is downloadable as the original file.

[b]Youtube:[/b] [url]http://www.youtube.com[/url] -- Everyone loves YouTube

[url]http://www.sodaplay.com[/url]
(cool website where you can make things that move)

[quote="7h3.D3m0n.117"]My tech teacher gave me that site 2 years ago but as a warning, it's very time consuming and tedious so you'll probably get frustrated.[/quote]
Na.

It's lots of fun!

I made a locomotive wheel that turns. ~Gasp!~

And a light with rays that extend when a little light switch goes up and down.  ~Incredible!~


Sorry but I dont know if this question belongs here.

If Area of A1 = Area A2, Find the area of the shaded part and the radius of the smaller circle. (See Drawing)

[quote="shadysaysurspammed"]I  cant  understans,what  is  A1?is it the area of  the  shaded prtion and  A2 is the area of the unshaded portion?
that means a1+a2=area of smaller circle?
please be clear :)[/quote]

A1 is the area of the shaded part and A2 is the area of the unshaded part of the smaller circle.

Yup, you're right. A1 + A2 = Area of the smaller circle.

[quote="Iversonfan2005"]Is the answer $\frac{10\sqrt{6}}{3}$? I formed a $45,45,90$ triangle in the bigger circle in which the legs of length $10$ were tangent to the smaller circle. And then using the radii of the smaller circle, I constructed two $30,60,90$ triangles which each having the bigger leg of length $5\sqrt{2}$, thus the hypotenuse or the radii of the smaller circle is $\frac{10\sqrt{6}}{3}$.[/quote]
shobhit, how dyou know that the legs of $10$ are tangent?

this question is some wrong i think.
we can put a circle with radius 10 and put a circle upon it.there is so much circles we can set the A1 and A2 areas are equal.

[quote="riddler"]yes something is missing. :P[/quote]

Maybe, it's a problem from a friend. I had a hard time figuring out how to solve it but I just cant. Maybe something is missing.


"Find all horizontal and vertical asymptotes of $ \frac {\sqrt {x^2 \minus{} 4}}{x \plus{} 2}$."

Obviously, the vertical asymptote is $ \minus{} 2$.  However, I'm not able to find the horizontal asymptote(s).

I do know the answer: $ \boxed{x \equal{} \minus{} 2, y \equal{} \minus{} 1, y \equal{} 1}$.

The horizontal asymptotes are, in simple terms, what value the function tends to as $ x$ goes to positive infinity and the value as $ x$ goes to negative infinity.

Rewriting the function as $ \sqrt{\frac{x\minus{}2}{x\plus{}2}}$ is helpful.

Er... I know how to find the horizontal asymptotes in general, but I've never had to do it for problems involving square roots.  Your insight on telling me to rewrite the equation is good, but what should I do from there?  Without the square roots, the horizontal asymptote would be the numerator's leading coefficient divided by the denominator's leading coefficient (since the leading terms of the numerator and the denominator are equal), giving an answer of $ 1$.  But that doesn't give me the $ \minus{}1$, or help me with the actual problem (which [i]does[/i] have the square roots).
A little more help please?

Let x go to infinity, then minus infinity. In MellowMelon's rewritten form which is the same as the function, but always positive, we see that both times we get $ \sqrt {1}$. Then look back at the original function, the numerator is always positive, but for big positive x the denominator is positive, for big negative x, the denominator is negative. So, the asymptotes are 1 as x goes to infinity, and -1 as x goes to minus infinity.

EDIT: The thing with square roots in functions is that when you move things in and out of the roots, you make sure that the sign of the new function is correct towards the value you're moving to.


For any different $ \{a,b,c\}\subset \mathbb R\ \implies\ \boxed {\ \sum\frac {b^2 \plus{} c^2 \minus{} a^2}{(a \minus{} b)(a\minus{}c)} \equal{} \minus{}2\ }$ .

This identity has many algebraic and geometrical applications.

[hide="Probably the first seen way"]

Multiply out. The denominators of the fractions 2 by 2 have 1 term common, so multiplying is natural. [/hide]


trigonometry
number theory proposed
number theory

Let $ p \equal{} 3k\plus{}2 \geq 11$ be a prime and let
\[ \lambda \equal{} \sum_{ k \geq 4 } { p^2 \minus{} 1 \choose 2k } \sum_{ j \equal{} 1, 3, 5 } \left( 2\cos \frac{j \pi}{6} \right)^{2k} \left( \cos \frac{ j k \pi}{3} \plus{} \cos \frac{ j (k\minus{}1) \pi}{3} \right) \]
Prove that $ \lambda$ is an integer and that $ 3p \; | \; \lambda \minus{} 24.$

[quote="bodan"]Let $ p \equal{} 3k \plus{} 2 \geq 11$ be a prime and let
\[ \lambda \equal{} \sum_{ k \geq 4 } { p^2 \minus{} 1 \choose 2k } \sum_{ j \equal{} 1, 3, 5 } \left( 2\cos \frac {j \pi}{6} \right)^{2k} \left( \cos \frac { j k \pi}{3} \plus{} \cos \frac { j (k \minus{} 1) \pi}{3} \right)
\]
Prove that $ \lambda$ is an integer and that $ 3p \; | \; \lambda \minus{} 24.$[/quote]
Note, that $ A_k\equal{}\sum_{j\equal{}1,3,5}(2cos\frac{j\pi}{6})^{2k}(cos\frac{jk\pi}{3}\plus{}cos\frac{j(k\minus{}1)\pi}{3})\equal{}2*3^k(cos\frac{k\pi}{3}\plus{}cos\frac{(k\minus{}1)\pi}{3})$ and $ 2cos\frac{k\pi}{3}\plus{}2cos\frac{(k\minus{}1)\pi}{3}\equal{}B_k$, $ B_k\equal{}0$ if $ 3|k\plus{}1$, $ B_k\equal{}3$ if $ k\equal{}0,1\mod 6$, $ B_k\equal{}\minus{}3$ if $ k\equal{}3,4\mod 6$. Therefore
\[ \lambda\equal{}\sum_{j\equal{}1}^{(p^2\minus{}7)/6]}\binom{p^2\minus{}1}{6j\plus{}2}3^{6j\plus{}3}(\minus{}1)^j\plus{}\sum_{j\equal{}2}^{(p^2\minus{}1)/6}\binom{p^2\minus{}1}{6j}3^{6j\plus{}1}(\minus{}1)^j.\]
But I don't undestand $ \lambda \equal{}\sum_{k\ge 4}\binom{p^2\minus{}1}{2k}A_k$ and $ p\equal{}3k\plus{}2$/ I think $ p\equal{}\minus{}1\mod 6, p>5$.


[quote="anirudh"][hide]$x=2$ and $y=2$. So, when you sixth that: $2^{6}+2^{6}=2(2^{6})=2^{7}=\boxed{128}$.[/hide][/quote]
this can't be, he said xy=2, not 4
-jorian

[quote="i_like_pie"][quote="7h3.D3m0n.117"]Find $x+y =4$ and $xy = 2$. Then find $x^{6}+y^{6}$.
Show work please.[/quote]
[hide]$x+y=4$

$x=4-y$

$xy=4y-y^{2}$

$2=4y-y^{2}$

$y^{2}-4y-2=0$

$y=\frac{4\pm\sqrt{16+8}}{2}$

$y=2\pm\sqrt{6}$

$x=2\pm\sqrt{6}$

One of them has to be $+$ and the other $-.$

$(2+\sqrt{6})^{6}+(2-\sqrt{6})^{6}=7760$[/hide][/quote]
While this method may be possible, I doubt you would be able to actually go through this step without a calculator:
[quote][hide]$(2+\sqrt{6})^{6}+(2-\sqrt{6})^{6}=7760$[/hide][/quote]
Try to find another method.
[hide="Hint"]You won't actually be finding $x$ and $y$.
By the way, kunny's hint is actually pretty good too (and probably better than mine  :P). [/hide]

[quote="haoye"][quote="i_like_pie"]
$2=4y-y^{2}$

$y^{2}-4y-2=0$
[/quote]

Check your algebra.[/quote]
Fixed. :blush:

[quote="kunny"]Ragnarock23 is right!

After finding the value of $x^{3}+y^{3},\ x^{6}+y^{6}=(x^{3}+y^{3})^{2}-......$[/quote]
that is not what he did. you can't distribute exponents over addition


[quote="silouan"]Let a triangle $ABC$  and  $a,b,c$  its angles  in  $rad$  so that  $a,b,c\in\left(0,\pi\right)$.

Prove that

$\frac{a^4+b^4}{(a+b)^3}+\frac{b^4+c^4}{(b+c)^3}+\frac{c^4+a^4}{(c+a)^3}\geq\frac{\pi}{4}$[/quote]
$\frac{a^4+b^4}{(a+b)^3}\geq\frac{a+b}{8}$
$\Leftrightarrow 7a^4+7b^4-4a^3b-4ab^3-6a^2b^2\geq 0$
This is true since $3a^4+3b^4\geq 6a^2b^2$ and $4a^4+4b^4\geq 4a^3b+4ab^3$ :D


physically(one cant judge the average of mental maturing)

Hehe, I don't think I've matured mentally yet.  :lol:
Actually, I've matured a lot the last couple of summers, mentally of course.

What's physically mature mean again?

how are "early", "average", and "late" defined

Though physical maturity on the basis of "early," "average," or "late" may be relative, I think he means parallel to your mental growth.  Maybe.

[quote="Scrambled"]I read somewhere that there is a theory that smarter people tend to mature earlier.  Im going to try to see if there is any credibility to this.(this is in comparison to the people who are around you)[/quote]

I think he meant early, average, and late as in relative to the people around you.

I think people who are tipically smart mature late mentally at least my advanced class is.  Theres still 8th graders in my class who eat glue...

[quote="BHorseMath"]Theres still 8th graders in my class who eat glue...[/quote] Hehe :D

Mentally... well I don't eat glue anymore :P (I never did actually) but I can still do crazy things like kids, some people call it being immature, others agree it's healthy not to age too fast ;-)

As for physically.. hmm around 11 yrs is that early enough? :)

Oops, I thought the original poster was talking about when you were born compared to when you were suppossed to, and put late instead of early.

I don't think maturity has a coorelation with intelligence.. There's people in my classes who "physically" matured around age 12, and words can't even describe their intelligence, or lack there-of. [This is only personal experience]

maturity is hard to try to rate.  like I said theres still kids in my class who eat glue and other rather discusting imature things.

Wait, are we talking about physical maturity or mental maturity?

define physical maturity

if physical maturity counts as outward signs, how do you see someone else's and "compare" ? (giggle)

Scrambled, you don't happen to be older than 18...do you?

peter wrote:i think scrambled is in 7th grade right now

um...[hide]perhaps[/hide]

no he's in 8th grade[hide]...right[/hide]?

...the fact is that Scrambled is not 18 yet...

Actually I know for afact that scrambled is in 7th grade.

[quote="Scrambled"]11
6th
nothing[/quote]

That post was made in Sun Feb 01, 2004 9:22 pm.

There ya go.  Scrambled is in 7th.  Settled.

[quote="gameworld7"]hmm...shouldn't there a poll for "not yet"? (i'm not saying that i haven't matured yet.)[/quote]
for those that are "not yet" don't vote...simplest solution i could think of

easiest? so they just wait until they do? that'll take awhile.

after reading all this... are u meaning HG&D "devolopment", or mental?

i dont think i've really "matured" in both ways...  :D

we were talking about phys.

but how do we compare???

the fact is that we cant compare unless we "see someone"

We're just looking at people's votes, I guess, and hope that they're being honest and not making themselves out to be like those girls in [i]Playboy[/i].

EEEEEEEWWWWWWWW

OMG!!!!

this is REAL innapropriate.....

*makes barfing sound*

obviously sunnyray didn't mature yet. right?


I wrote an English document in LaTeX where I started my document like this:

\documentclass[a4paper]{amsart}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{amssymb}
\begin{document}
\title{Title}
\author{My Name}
\begin{abstract}
Abstract
\end{abstract}
\maketitle

As I tried to compile my text LaTeX gave the following error:
! Package babel Error: You haven't defined the language finnish yet.

How can I tell LaTeX that I don't want any Finnish stuff here?

Try taking the babel package out of the code (if you can).


superior algebra
superior algebra solved

First of all , i wanna say its really nice to see a math forum like this  :)

Anyway , my question is the following

Let a and b be elements of a multiplicative group G . Show that there [i][u]exists[/u][/i] a [i][u]unique[/u][/i] element x $\in$G such that ax=b .

(Each step should be justified by an axiom or an elementary deduction)

x exists because $x=ba^{-1}$ and $a,b^{-1} \in G$ so $x \in G$ too.
ok
now if $ax=b$ with $a,x,b\in G$ and $ax'=b$ with $x' \in G$ and $x\neq x'$ (*)
we would have that $x=ba^{-1}$ and $x'=ba^{-1}$ but $x=x'$ contradiction
with (*) so $x$ is unique


logarithms
linear algebra
linear algebra open

Let $m,n \in N^*$ and $A \in M_n(\mathbb{C})$. What conditions must we impose on A such that there exists $B \in M_n(\mathbb{C})$ with the property $B^m=A$.

Let $Spec(A)=\sigma(A)=\{ a\in \mathbb{C}| A-aI_n\ is\ not\ invertible \}$.

I will prove that if 0 is not in Spec(A) then there exists B such as $B^m=A$.

$(U\cdot B\cdot U^{-1})^m=U\cdot B^m\cdot U^{-1}=U\cdot A\cdot U^{-1}$.
So it suffices to prove the problem for matrices A in Jordan canonical form and it is easy to see that is is actually enough to solve the problem for matrices A that are Jordan cells.

If 0 is not in Spec(A), then $A=x^mI_n+E$ for some $x \in \mathbb{C}^*$, where $E=(e_{i,j})$, where
$e_{i,j}=\{ \begin{array}{cc} 1, &\  j=i+1 \\ 0, &\  j\neq i+1 \end {array}$.
It is easy to see that $E^k=(e_{i.j}^k)$, where
$e_{i,j}^k=\{ \begin {array}{cc} 1, &\  j=i+k \\ 0, &\  j\neq i+k \end {array}$.
$E^k=0_n$ for $k\geq n$.

I will prove that there exists a B of the form $B=xI_n+\sum_{k=1}^{n-1}a_k\cdot E^k$. Let $T=\sum_{k=1}^{n-1}a_k\cdot E^k$.
We have to solve the equation $\sum_{k=0}^{m} C_m^k\cdot x^{m-k}\cdot T^k=x^mI_n+E$. $C_n^k$ is the binomial coefficient.
This is equivalent to solving $\sum_{k=1}^{m} C_m^k\cdot ( \sum_{i=1}^{n-1} a_i\cdot E^i )^k=\frac{E}{x^m}$.
I hope you realised by now that we look for the $a_k$s, not for x.

We get a system of the form
$m\cdot a_1=x^{-m}$
$m\cdot a_2+C_m^2a_1^2=0$
$m\cdot a_3+C_m^2a_1a_2+C_m^3a_1^3=0$
.......................................................................
It is easy to see that this system is solvable and has a unique solution for a fixed x.

Notice that in the first equation is vital to the solution that $x\neq 0$ .

So what do we do if $0 \in Spec(A)$?
The equation $B^m=E$ has no solution for m>1.
Proof: Suppose there is a B such that $B^m=E$.
If n>2 then r=[n/m]+1<n and $E^r=B^{mr}=0_n$ since mr>n and B is nilpotent. So $E^r=0_n$ and r<n. It is easy to see that this is contradiction.
If n=2. Then $B^m=E$ means that B is nilpotent wich means that $B^2=0_2$, so $B^m=E$ has no solution for m>1.

But the fact that the equation $B^m=E$ has no solution doesn't mean the the equation $B^m=A$ has no solution if $0 \in Spec(A)$.

[quote="amfulger"]Let $Spec(A)=\sigma(A)=\{ a\in \mathbb{C}| A-aI_n\ is\ not\ invertible \}$.
I will prove that if 0 is not in Spec(A) then there exists B such as $B^m=A$.
[/quote]

If you look in the topic matriceal logarithm, you will see another proof for this stament. I think that it is not an essentially different proof.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ G$ be a profinite group, $ N \leq G$ such that $ G/N$ is the largest quotient of $ G$ which is a pro--group.

 I want to prove that $ N \leq G^{(2)}$ with $ G^{(2)}\equal{}G^q[G,G], q\equal{}p^f$ and $ [G,G]$ the derived group (the set of all the commutators of $ G$)?
 Thanks in advance for some hint.

[quote="Bob75"]Let $ G$ be a profinite group, $ N \leq G$ such that $ G/N$ is the largest quotient of $ G$ which is a pro--group.[/quote]

I think something is missing.  I'll assume "pro-p-group".

[quote]I want to prove that $ N \leq G^{(2)}$ with $ G^{(2)} \equal{} G^q[G,G], q \equal{} p^f$ and $ [G,G]$ the derived group (the set of all the commutators of $ G$)?[/quote]

This is equivalent to proving that G/G^(2) is a pro-p-group, but G/[G,G] is an abelian group and G/G^(2) is an abelian group of exponent (dividing) p^f, so is a direct sum of cyclic p-groups.

Now I think such a direct sum is a profinite group if and only if it is finitely generated, BUT you have that G is profinite by assumption, so apparently everything works out.  That seems suspicious to me, so check it.

Effectively, I forgot the $ p$ of pro-$ p$-group. If, as you proved, $ G/G^{(2)}$ is a pro-$ p$-group, by the definition of  $ G/N$, effectively $ N \leq G/G^{(2)}$. Thanks you for your answer!


What is the probability that a random power of 2 starts with 1? (such as, 1, 16, 128, 1024, etc)

That only the case up to a certain point...
Notice that 3+3+4=10 and 2^10=1024...that is where this comes from.
My guess is that the probability would be
[hide]$\log_{10}{2}$[/hide]
Well, of course that would make sense, but i think that the possibilities are unlimited if you're considering exponents all the way up to $\infty$.

if you count negative powers (can you include those?), there would be an infinate number of fractions, thus making the probability approx. 0.

If you include fractions and are looking at the leftmost nonzero digit, it should still hold...

[hide="Quote from "Problem Solving Strategies""]

[quote]Let $a_n$ be the number of terms in the sequence $2^1,2^2,...,2^n$, which begin with digit 1. Prove that:

$\log2-\frac1n<\frac{a_n}n<\log2$

and, hence,

$p_1=\lim_{n\rightarrow +\infty}\frac{a_n}n=\log2\approx0.30103.$

One sometimes says that a randomly chosen power of two begins with 1 with probability $\log2\approx0.30103.$[/quote]

The proof isn't in the book however  :(  and i was unable to solve it myself too  :?
maybe if anyone of you feels like giving it a try :) [/hide]

The probability that 2^k(k>=0) starts with i (i=1,2,...,9) is lg(i+1)-lgi .

Shouldn't it be equal for any digit?  That would make sense...
For any digit, there is a set of numbers which the previous power of two could be to have the one iin question start with that digit.  For instance, if you're looking at 3, then the previous power of two would have to be anything from 15000... to 1999...
And if you're looking at 1, then the previous one would have to be anything in the range from 05000... to 0999... (I put the zero as a digit placeholder so that you can see that the range is the same)
Anyways, that's the logic that I would use...not totally sure that it works...probably doesn't...
Well, I suppose my logic kills itself, considering that if the chance is equal, then given a number, there is a 1/2 (or 5/9?  :huh: ) chance that the next number will start with 1 (if the first digit is something from 5-9)...
I can't do this...but how did you get your answer, lin?

[quote]he proof isn't in the book however Sad and i was unable to solve it myself too Confused [/quote]
One of the intutive and  very simple way to think about this is:

[hide="Hint One"]It really does not matter if you take powers of 2 or 3 or (1947 for that matter, as long as it is not a power of 10) the probablity of first digit being 1  (for  powers of any such number ) is  about $0.30103$

[/hide]
(And by the way this makes the answer almost onvious to anyone who used to use "log tables" in pre-calculator days! :)
[hide="Hint"]
First thing you noticeif    $a= \log_{10}2$  then  $a$  is  irrational.

Now consider $a , 2a,, 3a, 4a  \dots $   and  just take the fractional part .
The factional part can be anyting between $00000...$ and $.9999999..$ or IOW $0$ and $1$ with uniform distributional probablity. so that  it will be between $0$ and $\log_{10} 2 $ will be $\log_{10} 2 $  :)

[/hide]

And a lliitle more fun is;  what is the probablity the  $2^n$ starts with $2005$ ?

Added later: Sorry did not see all the posts above, the answer is already given...:)


[quote="quangdung_toank13"]Given $t_1,t_2,....,t_n\in{R}$ satisfy:$t_{k}=t_{n+1-k}$.Two sequence $(x_n),(y_n)$ defined the following:
$x_0=0,x_1=1,x_{k+1}=t_kx_k+x_{k-1}$
$y_0=1,y_1=0,y_{k+1}=t_ky_k+y_{k-1}$
Prove that $x_n=y_{n+1}$[/quote]
don't solve


[hide="hint"]
convert everything to $5^{x}$ and $6^{x}$

[/hide]
[hide=" solution"]
$5^{2x}= 5^{x}\cdot 6^{x}+6^{2x+1}$

Let $y = 6^{x}$ and $z=5^{x}$

$z^{2}= zy+6y^{2}\rightarrow z^{2}-zy-6y^{2}=0 \rightarrow (z-3y)(z+2y)=0$

So,
Case 1: $z=3y$
$5^{x}= 3 \cdot 6^{x}\rightarrow x = \frac{\ln(3)}{\ln(5)-\ln(6)}$

Case 2: $z=-2y$
$5^{x}=-2 \cdot 6^{x}$ yields no solution, assuming we are only dealing with real numbers :).
[/hide]

oooh, that's where i went wrong, I tried doing $2^{x},3^{x},5^{x}$...

if it's irreducible, doesn't it only imply that there's no real [b]integer[/b] solution?

you're right ... :D  :blush:

ooooo i haven't seen you do have a solution....

i really don't know why i can't get it

______________________________________

nice solution by the way

[quote="t0rajir0u"][quote="ElChapin"]but $6^{2x}-5^{2x}>0$[/quote]

False for $x < 0$.  Did you read Zellex's solution or not?[/quote]

yes, well, I've just read it and I still don't know why a get a contradiction

in that part I considerer $x>0$, look a the "transformation"

______________________________________________________

i think i know why, is the $5^{x-1}-6^{x-1}<0$ is only truth for $x-1>0$ so i guess solution is in $(0,1)$,(and beacuse of the "transformation" :oops:) is in $(-1,0)$ jiji how shameful... :rotfl:

(is there something to do with all this spam I've created, like erase it?)

[quote="ElChapin"]or $x<0$, let us rewrite equation as follows, with $x>0$

$\frac{1}{5^{2x}}=\frac{1}{5^{x}6^{x}}+\frac{1}{6^{2x+1}}$ then[/quote]

You applied the transformation $x \to-x$ incorrectly.  Specifically, the last exponent should be $2x-1$.


$ \sum\frac {a^2(b\plus{}c\minus{}a)}{b\plus{}c} < \sum\frac{ab}2$

 $ a,b,c \in R\plus{}$

I think you mean $ \le$ (try $ a \equal{} b \equal{} c \equal{} 1$)

[hide]$ \sum\frac {a^2(b \plus{} c \minus{} a)}{b \plus{} c}\le\sum\frac {ab}{2}$

$ \Leftrightarrow \sum a^2\le\sum\frac {ab}{2} \plus{} \sum\frac {a^3}{b \plus{} c}$

$ \Leftrightarrow \sum a^2\le \sum\left(\frac {ab}{4} \plus{} \frac {ac}{4} \plus{} \frac {a^3}{b \plus{} c}\right)$

$ \Leftrightarrow \sum a^2\le \sum a\left(\frac {b^2 \plus{} c^2 \plus{} 2bc \plus{} 4a^2}{4(b \plus{} c)}\right)$

$ \Leftrightarrow \sum a^2\le \sum \left(\frac {a(b \plus{} c)}{4} \plus{} \frac {a^3}{b \plus{} c}\right)$

$ \Leftrightarrow \sum a^2\le \sum a^2$

where the last step was by AM-GM.
[/hide]


[quote="msecco"]We are given 48 positive integers less than 70 with sum 140.Prove that there are some of that integers with sum 70.Find a counter-example for 47 positive integers.[/quote]
This puzzle is a special case of the following result (the easy proof by case distinctions can be found in  Applied Mathematics Letters 19, pages 1053-1056):

Consider a sequence of $ n$ positive integers that are less or equal to $ T$ and that add up to $ 2T$.
IF
(a) $ T\le n\minus{}1$, or
(b) $ n\le T\le 3(n\minus{}1)/2$ and $ T$ even, or
(c) $ 3(n\minus{}1)/2\le T\le 2n\minus{}4$ and $ T$ a multiple of 6,
THEN there exists some subsequence of these integers with sum T.

And if $ T$ and $ n$ do not satisfy any of the conditions (a),(b),(c), then there exist corresponding instances without any subsequence with sum T.

Savchev's problem falls under (b), since $ 48\le70\le70.5$ and $ 70$ even.
The counter-example for $ n\equal{}47$ and $ T\equal{}70$ consists of $ 46$ times the integer $ 3$ and once the integer $ 2$
(note that 70 is congruent 1 modulo 3).


Who is BTW?I've seen him name appear in many posts here.
Valentin,I still need that solution please.Maybe writing it down for me would be like remembering the old times. ;)
Another last thing.Claudiu Raicu is Albanian.I made a google search cus i've heard about him before as the winner opf a gold medal at the Balkan Olympiad and in the website of Lumina Institutions i got to know that he's Albanian. :lol:  :P
We Albanians are very inteligent. :D

Claudiu Raicu is Romanian.

BTW means "By the way" and is no person at all :D

Ahem, Claude is Romanian 100% :) he did study at the International High-School of Informatics in Bucharest, but, as blondu, he is still Romanian :)

BTW = by the way

[quote="Iris Aliaj"]Who is BTW?I've seen him name appear in many posts here.[/quote]
Iris, let me introduce my friends LOL and ROTFL!   :D

Is this a grammar and spelling website now? :maybe:
So how about everyone stops judging others for some pointless acronyms!  ;)

[quote="Ani"]So how about everyone stops judging others for some pointless acronyms!  ;)[/quote]It would be great if you'd start it! :D ;)

[quote="Iris Aliaj"]Oh what a keen humor.Really you have an exceptional sense of humor Mindflyer.What about socialising with my friends NMNF(non me ne fregga) and IMR(ignore mean remarks)
Wouldn't it be better if you stopped wasting precious time in making mean remarks about small pointless things.You could spenb that time in making useful things for yourself.It would be much better this way.

In goodwil Iris[/quote]

 ;)

PPS. I hope you will enjoy the following text. It was written by a 13 years old English student either as homework or on a test and it has really shocked the teacher, and it has caused a notable discussion about use and abuse of language, being written [i]completely[/i] in hieroglyphic SMS language. (I have found it on the website of Spiegel Online, a german newspaper, but now they have taken it from the web, hence I copy it below.)

Here comes the text. It may become quite difficult to you to decipher it! In any case, try to understand as much as you can. The solution is below under the spoiler.

[quote="Original Text"]My smmr hols wr CWOT. B4, we usd 2 go 2 NY 2C my bro, his GF & thr 3 :-@ kds FTF. ILNY, its gr8.

Bt my Ps wr so {:-/ BC o 9/11 tht they dcdd 2 stay in SCO & spnd 2wks up N.

Up N, WUCIWUG -- 0. I ws vvv brd in MON. 0 bt baas & ^^^^^.

AAR8, my Ps wr :-) -- they sd ICBW, & tht they wr ha-p 4 the pc&qt...IDTS!! I wntd 2 go hm ASAP, 2C my M8s again.

2day, I cam bk 2 skool. I feel v O:-) BC I hv dn all my hm wrk. Now its BAU ...[/quote]

[quote="and the same in old-school English"]My summer holidays were a complete waste of time. Before, we used to go to New York to see my brother, his girlfriend and their three screaming kids face to face. I love New York, it's a great place.

But my parents were so worried because of the terrorism attack on September 11 that they decided we would stay in Scotland and spend two weeks up north.

Up north, what you see is what you get - nothing. I was extremely bored in the middle of nowhere. Nothing but sheep and mountains.

At any rate, my parents were happy. They said that it could be worse, and that they were happy with the peace and quiet. I don't think so! I wanted to go home as soon as possible, to see my mates again.

Today I came back to school. I feel very saintly because I have done all my homework. Now it's business as usual...[/quote]

  dg

hey even i remember having read that thing (written by that english student) in a newspaper ,funny how this thing is becoming so common (acronyms and stuff)
so how many here think of it as deterioration (or maybe an insult ) of the beautiful english language :D  :D

c'mon guys break it up, mindflyer just made a joke. it may have been a stupid joke, but it WAS a joke.

you are taking all this too serious, so that is why I will use my SUPER-POWERS :P to split this topic in half *evil laughter* :D

Darij: ROTFL = Rolling On The Floor Laughing. It's stronger than LOL.

Iris: As Valentin said, mine was just a little joke. I was not going to criticize you or to judge you, the aim of my remark was very far from that! I found your question just funny, because you had mistaken BTW for a person, and that's all.   :)
I'm also aware that your English is much better than mine, so any criticism towards you would have been pointless. [i]Peace, my friend![/i]   :D

[quote="MindFlyer"]Darij: ROTFL = Rolling On The Floor Laughing. It's stronger than LOL.[/quote]

Aha, thank you. BTW, I well remember myself a year ago asking on another forum what LOL means...

  dg

[quote="darij grinberg"]PS. Ah, BTW, MindFlyer, I always wanted to know what ROTFL means... IMHO it's quite the same as LOL, but I'm not really sure... So PLS explain it to me ASAP.

SCNR. GRTZ, DG[/quote]

what does IMHO mean  :)  :) ?


algebra polynomial Vieta function domain quadratics

What is the product of the real roots of the equation \[x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}\,\,?\]

[hide="Solution"]If we expand this, we get a quartic polynomial, which obviously isn't very helpful.

Instead, we substitute $y$ for $x^2+18x+30$ and our equation becomes $y=2\sqrt{y+15}$.

Solving for $y$, we get $y=10$ or $y=-6$. The second solution gives us non-real roots, so we'll will go with the first. Substituting $x^2+18x+30$ back in for $y$,

$x^2+18x+30=10 \Rightarrow x^2+18x+20=0$. The product of our roots is therefore 20.[/hide]

[hide="Answer"]Squaring both sides, we have $x^4+36x^3+384x^2+1080x+900=4x^2+72x+180\Rightarrow$
$x^4+36x^3+380x^2+1008x+720=0$, which factors to $(x^2+18x+20)(x^2+18x+36)=0$. The second term gives extraneous roots due to squaring, so the two real roots multiply to $\boxed{20}$.[/hide]

[quote="4everwise"]What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$?[/quote]

[hide]Let us assume that $a=x^{2}+18x+30$.

This means that $a=2\sqrt{a+15}$.

So $a^{2}=4a+60$. We solve for a to get $a=10$ or $a=-6$.

So this means that $x^{2}+18x+30=10$, or $x^{2}+18x+30=-6$, or $x^{2}+18x+20=0$ and $x^{2}+18x+36=0$.  But we can show that the roots of the latter equation are not in the domain of $x^{2}+18x+45$, which should be nonnegative.  But the roots of the other quadratic are allowed, and we can show that that is true.

Hence the product of the roots is just the product of the first equation which is 20.

$20$[/hide]

[quote="srulikbd"]vieta's formuals
is it ok to revive a thread? sorry if not  :blush:[/quote]You can't use Vieta's formulas because you're only looking for the product of the real roots, and not all the roots are real.

How did you show that $ x^2 \plus{} 18x \plus{} 36$ does not work?

You don't even need to re-substitute $ a\equal{}\minus{}6$ into $ x^2\plus{}18x\plus{}36\equal{}0$. It per se is an extraneous solution of $ a\equal{}2\sqrt{a\plus{}15}$ since the principal square root cannot be negative.

let $ u \equal{} x^2 \plus{} 18x \plus{} 30 \implies u \equal{} 2\sqrt {u \plus{} 15}$ squaring both sides we have $ u^2 \equal{} 4u \plus{} 60 \implies u^2 \minus{} 4u \minus{} 60 \equal{} 0 \implies (u \minus{} 10 ) (u \plus{} 4) \equal{} 0 \implies u \equal{} 10$ so  $ x^2 \plus{} 18x \plus{} 20 \equal{} 0$ and by viet's the product of the real roots = [b]20[/b]

Note: u cannot equal -4 because then there are no real solutions  :D

[quote="JesusFreak197"][hide="Answer"]Squaring both sides, we have $ x^4 \plus{} 36x^3 \plus{} 384x^2 \plus{} 1080x \plus{} 900 \equal{} 4x^2 \plus{} 72x \plus{} 180\Rightarrow$
$ x^4 \plus{} 36x^3 \plus{} 380x^2 \plus{} 1008x \plus{} 720 \equal{} 0$, which factors to $ (x^2 \plus{} 18x \plus{} 20)(x^2 \plus{} 18x \plus{} 36) \equal{} 0$. The second term gives extraneous roots due to squaring, so the two real roots multiply to $ \boxed{20}$.[/hide][/quote]

How did you factor $ x^4 \plus{} 36x^3 \plus{} 384x^2 \plus{} 1080x \plus{} 900 \equal{} 4x^2 \plus{} 72x \plus{} 180\Rightarrow$
$ x^4 \plus{} 36x^3 \plus{} 380x^2 \plus{} 1008x \plus{} 720 \equal{} 0$ ?

[quote="4everwise"]What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$?[/quote]

[hide="Solution"]
Let $x^2 + 18x + 30 = a$. We get $a = 2\sqrt{a+15}$, or $a^2 = 4(a+15)$. Solving this, we get $a = 10, -6$, but $-6$ is extraneous.

Now, the quadratics $x^2 + 18x + 30 = 10$ has, by Vietas, product of (real) solutions $20$, so our answer is $\boxed{020}$.
[/hide]

Is the only way to check for extraneous roots to plug the roots back in to the original equation? Or is there an easier way?

All you have to do like @timwu said is realize that $a=-6$ is not even a solution to $a=2\sqrt{a+15}$ so it gets filtered out immediately.

[hide=Solution]Let $y=x^2+18x+30$. We have $$y=2\sqrt{y+15} \implies y=10.$$ Thus, $x^2+18x+30=10 \implies x^2+18x+20=0$, for which the product of the roots is $\boxed{20}$, given by the Vieta relations.[/hide]

[quote=HamstPan38825][b]Solution from Hamst's 60 AIME grind:[/b]

We wouldn't like to square to get a quartic - after all, the roots of the quartic might not even be all real. Instesad, we can notice that $x^2+18x+30$ occurs in both sides of the equation, so let $x^2+18x+30=a$. Then, $$a=2\sqrt{a+15},$$ from which we can finally square to get $a^2 = 4(a+15) \Rightarrow a^2-4a-60=0$, or $(a-6)(a+10)=0$.

Hence, either $x^2+18x+30=-6$ or $10$.

Before we go on, we check for extraneous solutions. $-6$ is obviously extraneous since the square root cannot be negative. Hence, $a=10$, so the two roots are roots of $x^2+18x+20=0$ and have product $\boxed{020}$.[/quote]

While I'm here, I just happened to notice this typo. It should factor as $(a+6)(a-10)$ instead.

Let $x^2+18x+30=k$.

We have $k=2\sqrt{k+15}$.

Squaring both sides gives $k^2=4k+60\implies k^2-4k-60=0\implies k\in \{10,-6\}$.

Now $k=-6$ is an extraneous solution, so $k=10$, which satisfies $k=2\sqrt{k+15}$.

This implies $x^2+18x+30=10\implies x^2+18x+20=0$. The product of the roots (which are both real) is $\boxed{20}$.

Let $y=x^2+18x+30$. Then, we have that $y=2\sqrt{y+15}\implies y^2=4y+60\implies (y-10)(y+6)=0$. We know that $y=-6$ doesn't satisfy the original equation (is an extraneous solution generated by squaring), so $x^2+18x+30=10$. As both of these roots are real which can be checked using the discriminant, we can use Vieta's Formulas to get $\boxed{20}$.

Let $u=x^2+18x+30$. We have $$u=2\sqrt{u+15}$$
$$u^2=4(u+15)$$
$$u^2-4u-60=0$$
$$(u-10)(u+6)=0$$

Thus, $u=10$ or $u=-6$. Plug in gives $x^2+18x+20=0$ or $x^2+18x+36=0$. Since they are all real by simply taking the discriminant, however, $-6$ is an extraneous solution. Thus, the product is equal to $\boxed{20}$.

[hide=Sol]We have

$$x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$$

Letting the equation be as it is leads to tricky quartic polynomials and such, so we decide to substitute: let $y = x^2 + 18x + 30$. Therefore

$$y = 2 \sqrt{y + 15}$$
$$y^2 = 4(y + 15)$$
$$y^2 - 4y - 60$$

Factoring gives us

$$(y + 6)(y - 10)$$

So our solutions are $y = -6$ and $y = 10$. Notice that our first solution is extraneous, even though calculating the discriminant in each case yields a positive number, since $\sqrt{y}$ refers to the principal square root only, so we are only left with one solution $y=10$. Plugging this back into our original expression gives

$$x^2 + 18x + 30 = 10$$
$$x^2 + 18x + 20 = 0$$

And hence the product of the real roots of this equation must be $\boxed{20}$.[/hide]

That is my solution too @s12d34.
I found this one the easiest on the test by one quadrillion

wow, this thread got revived from 2005...
That's amazing

[hide=sol]square both sides
$x^4+36x^3+384x^2+1080x+900=4x^2+72x+180$
$x^4+36x^3+380x^2+1008x+720=0$
factoring it, you get $(x^2+18x+20)(x^2+18x+36)=0$
so the roots are $-3\sqrt{5}-9, 3\sqrt{5}-9, -\sqrt{61}-9,$ and $\sqrt{61}-9$
but $x^2+18x+45 \ge 0$ and $x^2+18x+30 \ge 0$ to make sure that the signs are the same
plugging the roots back into the equation, only $-\sqrt{61}-9,$ and $\sqrt{61}-9$ work, which product to $\boxed{20}$

This good question is also included in the E. J. Barbeau Polynomials book page 126.

Let $y=x^2+18x+30$, so that the given equation rewrites as $y=2\sqrt{y+15}$. Squaring,
$$y^2=2^2(y+15)=4y+60\implies y^2-4y-60=(y+6)(y-10)=0,$$
which has solutions $y=-6$ and $y=10$. However, we require that $y\geq0$ in order for $y=2\sqrt{y+15}$, so we eliminate $y=-6$ to obtain $y=x^2+18x+30=10$, or $x^2+18x+20=0$. The product of the roots of this equation, both of which are real, are $\boxed{20}$ by Vieta's.

[hide=sol]Let $y = x^2+18x+30.$ So $y=2\sqrt{y+15}.$ solving for $y$ gives $y = -6 \text{ or } 10.$ However, $-6$ is an extraneous solution. Substituting $y=10$ in $y = x^2+18x+30$ (we can see that its roots are real from the discriminant) and applying Vieta’s gives $\boxed{20}.$[/hide]

Let $y=x^2+18x+30$, hence, $y=2\sqrt{y+15}$, now by solving this, we get $y=10$ or $y=-6$, we see that $y=-6$ gives us unreal roots, so if $y=10$ by Vieta's we have that $\boxed{20}.$

Let $k=x^2+18x+30$. We have $k=2\sqrt{k+15}$, from which we solve that $y=10$. Then, the product of the solutions is $\boxed{20}$ from Vieta's as the desired roots are indeed real.

[hide]We start off by substituting $a=x^2+18x+81$. This is for two reasons: for one, it allows us to better get a sense for the roots of the equation by making solving it easy (because vieta's seems unfeasible because we only want real roots), and also could catch any imaginary solutions if we get that $a$ is negative, because $x^2+18x+81$ must always be nonnegative by the trivial inequality. Doing this and squaring both sides, we get that $a^2-102a+2601=4a-144$. We can simplify to get $a^2-106a+2745=0$, and solve to get $x=\frac{106\pm 2\sqrt{53^2-2745}}{2}=45, 61$. These are both positive, so we will have $4$ real roots. However, we also need to take into consideration that $x^2+18x+30=a-51$ must be nonnegative as well because it is equal to a square root. Therefore, only the roots of $x^2+18x+81=61$ are valid. By vieta's their product is $20$. Main idea: substitute for quadratic to simplify things, make sure that after squaring, a square root is not negative.[/hide]


algebra
polynomial
geometry
3D geometry
quadratics
linear algebra
matrix

Let be $ A\in M_{4}(\mathbb{R})$ so that $ A^3\equal{}I_{4}$. Compute $ det(A\plus{}I_{4})$.

There are three possible values for this determinant: $ 1,4,$ or $ 16.$

Let $ p(x)\equal{}\det(A\minus{}xI)$ be the characteristic polynomial for $ A.$ The problem asks us to compute $ p(\minus{}1).$

We know these things about $ p(x).$

- It is a fourth degree real polynomial.
- All of its complex roots are cube roots of unity.

But since it is a real polynomial, any non-real roots must come in complex conjugate pairs. We can factor $ p$ into linear and quadratic factors. There are three possibilities:

$ \mathbf{(1): }\ p(x)\equal{}(x^2\plus{}x\plus{}1)^2.$ In this case, $ p(\minus{}1)\equal{}1.$

$ \mathbf{(2): }\ p(x)\equal{}(x^2\plus{}x\plus{}1)(x\minus{}1)^2.$ In this case, $ p(\minus{}1)\equal{}4.$

$ \mathbf{(3): }\ p(x)\equal{}(x\minus{}1)^4.$ In this case, $ p(\minus{}1)\equal{}16.$

We can easily produce example of all three cases. Let $ B\equal{}\begin{bmatrix}0&1\\\minus{}1&\minus{}1\end{bmatrix}$ and consider the following block-structured matrices:

$ A\equal{}\begin{pmatrix}B&0\\0&B\end{pmatrix},\ A\equal{}\begin{pmatrix}B&0\\0&I\end{pmatrix},\ A\equal{}\begin{pmatrix}I&0\\0&I\end{pmatrix}$

This funny.I tried everything I could think of and minutes after posting here I come up with a solution. I am not sure it is correct though.

Let $ \lambda$ be an eigenvalue of A. Since the polynomial $ X^3 \minus{} 1$  has A as a "root"( I know this is not expressed correctly but I don't know how to put it) then it also has its eigenvalue $ \lambda$ as a root. So 1 is an eigenvalue. Accordingly $ det(A \minus{} I_{n}) \equal{} 0$.

We have $ det(A \plus{} I_{4}) \plus{} det(A \minus{} I_{4}) \equal{} 2detA \plus{} 2detI_{4} \equal{} 4$ so $ det(A \plus{} I_{4}) \equal{} 4$.

Seeing the post above this one I wonder if the property I used(det(A+B)+det(A-B)=2det(A)+2det(B) holds for 4-dimensional matrices.(?) It does not.

[quote="bike_me"]... then it also has its eigenvalue $ \lambda$ as a root. So $ 1$ is an eigenvalue. Accordingly $ \det(A \minus{} I_{n}) \equal{} 0.$ [/quote]
That's not quite right. The eigenvalues must be among the roots of $ x^3 \minus{} 1 \equal{} 0$ but that does not mean that all roots of that polynomial are eigenvalues. So, while $ 1$ might be an eigenvalue, it is incorrect to insist that it must be an eigenvalue.

Look at the three matrices in the last line of my previous post. The first one does not have $ 1$ as an eigenvalue.

[quote]Seeing the post above this one I wonder if the property I used(det(A+B)+det(A-B)=2det(A)+2det(B) holds for 4-dimensional matrices.(?)[/quote]
Notice that you also missed the answer $ 16.$ Here's a test, inspired by that: let $ A \equal{} B \equal{} I_4.$ Then

$ \det(A \plus{} B) \plus{} \det(A \minus{} B) \equal{} \det(2I) \plus{} \det(0) \equal{} 16 \plus{} 0 \equal{} 16.$

$ 2\det(A) \plus{} 2\det(B) \equal{} 2\det(I) \plus{} 2\det(I) \equal{} 2 \plus{} 2 \equal{} 4.$

No, this alleged identity is not true. As a function, the determinant is [i]not[/i] linear, and except in a two-dimensional space, it is not quadratic either.


projective geometry
geometry unsolved
geometry

In the figure below:

[b]Problem[/b]

 The two small circles are tangent to the big one  at $ P,Q$ and meet each other at $ X,Y$.Line $ XY$ meets the circle at $ M,N$ and points $ A,B,C,D$ are as in the figure.Prove that $ AB\equal{}CD$.

[u]Babis[/u]

(Thanks to mr M. Miculita)- see http://www.mathlinks.ro/viewtopic.php?t=218883 )

Dear Mathlinkers,
let 0, 1, 2 the circles (PQMN), (PAXYD), (QBXYC) and Tm the tangent to 0 at M.
According to Monge's theorem, P, A, B and Q are cocyclic ; note 3 this circle.
According to Reim's theorem applied to 0 and 3, (AB) // Tm ; applied to 0 and 1, Tm // to the tangent at 1 in A ; then AB is the tangent to 1 at A.
Mutatis mutandis, AB is the tangent to 2 at B; in consequence,	AB is the common tangent to 1 and 2.
Mutatis mutandis, CD is the common tangent to 1 and 2.
…
Sincerely
Jean-Louis

just prove AB,CD are tangent to the circles
my solution is similar to jayme's

Dear Stergiu and Mathlinkers,
Thanks to Stergiu to remember us Igor Sharygin who died in 2004. He wrote a nice book intilted "Problemas de Geometria" Mir Editions.
Perhaps can some one precise the date of his birth and die?
Sincerely
Jean-Louis


Find all integer numbers $a,b,c$ such that $\frac{(a+b)(b+c)(c+a)}{2}+(a+b+c)^{3}=1-abc$.

[quote="pluricomplex"]It's obvious to see that the above equavalent to
$\prod_{sym}(2a+b+c)=2$
so we are done after few caculating.
Good luck to see why ... guy :D[/quote]

WOW, I will check later  :wink: Now, I go to teaches


Well, I'm actually reposting that same problem I posted under the name 'bands', because that post turned into something else when Arne posted another (nicer :D ) problem. Here it goes (again :D ):

Is there a set of points in the plane such that

a) every 2 points in the set have distance at least 1 between them
b) every band in the plane contains at least a point from the set in its interior   ?

[i]unknown source[/i]

Do you have a solution or is it an open problem Grobber?

Pierre.

I know the set of points of the form (m-n*31/2, m+n*31/2) is a solution (someone told me), but I haven't even tried to prove it (after that someone spoilt it for me, I wasn't interested anymore :D).

Aaargh...I thought about that one, but I had to miss something...

Pierre.

I'm sure there are other approaches. Another teacher tried to prove the existence of such a set without constructing an actual example, but I remember his proof was a bit faulty (I didn't understand much :)).

Hmmm... I do not think that your set of points does work :
Let b(e) be the band defined by the lines parallel to the line d with equation x+y+1 = 0 and which are e apart from d.
We want a point of the set to belong to this band.
But, the distance of any point (m + n \sqrt 3, m - n \sqrt 3) to the line d is |2m+1|/ \sqrt 2, so that none of them belong to the band if e is sufficientely close to 0.

Am I wrong or the problem is still open?

Pierre.


calculus
integration
function
algebra
floor function

Find the integral solutions to the equation $ [x][y] = x+y $.where $[ \ ] $ is the greatest integer fn.

Prove that all integral solutins lie on exactly two lines. Determine these lines in the cartesian coordiante system.

[hide]

well if ur looking for integral solutions to $(x,y)$, then i dun think the greatest integer function really applies

so you have $xy = x+y$

adding 1 to both sides and moving the $x+y$ over you have:

$xy - x-y +1 = 1 \Rightarrow (x-1)(y-1) = 1$

now you find integral solutions $(2,2), (0,0)$

These two points lie on the line $y=x$... but i'm not sure about the other line because i didnt get any other solutions

my first assumption might have been wrong but i dunno

[/hide]

I think that there are a few non integral solutions we have yet to find


[quote="calculuslover800"]Find the six-digit number is which the first digit is one less than the second, the third digit is half the second, the fourth digit is three times the third and the last two digits are the sum of the fourth and fith digit.  The sum of all the digits is 24.

Have fun.[/quote]
[hide]
563910

We have equations a=b-1, c=$\frac{b}{2}$, d=3c, [u]e[/u] [u]f[/u] = d+e, and a+b+c+d+e+f=24. then b-1+b+$\frac{b}{2}$+$\frac{3b}{2}$+e+f=24

4b+e+f=25 so since e has to be 1, f can't be 4 or above because then d+e can't sum up to [u]e[/u] [u]f[/u]. f=0.

4b=24, b=6, then following restrictions we have

[u]5[/u] [u]6[/u] [u]3[/u] [u]9[/u] [u]1[/u] [u]0[/u] [/hide][/u]


function
real analysis
real analysis unsolved

I can't remember the years for the problems I've posted today, that's why there's no year or date:

Let f:[a,b]->R be a function with finite limits in every point in [a,b]. Prove that f is Riemann integrable.

Take g(x)=lim(t->x)f(t) and show that the discontinuities of g-f form a countable set. :)


I am taking the AMC8. Is there any way I can also take the AMC-10 or maybe the AIME?

There was a topic like this somewhere. I think you can.

Yes, you may certainly take both the AMC 8 and the AMC 10 in this 2006 -2007 contest year.  The easiest way to do that is to have your school register for the AMC 8, given on Nov. 14, 2006, and to also have your school register for the AMC 10 given on Feb 6, 2007, or Feb 21, 2007.  If your school does not register for the Feb 21, 2007 date you may be able to take the AMC 10 B at a nearby college or university offering the contest through the college and university administrations.

If you score 120 points or more, or score in the top 1% of all scores on the AMC 10 then you will be invited to take the AIME.

Steve Dunbar
MAA Director of American Math Competitions

yes you can. i did that last year. in fact, you can take amc 10a and amc 10b as well.

Or even better, you could take the 10a and 12b  :D


Hello..?

Well.. I am a kind of new one who registered just right now..

And actually I wanna ask two questions about MATHCOUNTS Drills

If you get to the site that I wrote in Subject description,

You might get kind of website..

And there are sprint round problems, target round problems, team round and countdown round problems.

The first thing I wanna ask is,

Is the difficulty of those problems like state or national, or kind of for school round..?

Frankly, this is the first time that I ever joined mathcounts, and the first year that I stay in United States

Well, I am from Korea, so..

I will appreciate it if you can endure kinda grammar problems..

Well, the second question is,

I wonder if it is illegal or something...?

You know, if it is illegal, then it represents the possibility that I could be arrested or something because I involved in it..

Well...
.
.
.
Have a nice day.

It is not illegal.

Also, the difficulty on that website can range.  I'd say it's from about Chapter (Regional) to State, but there are a few National-level questions as well as School-level questions.

No, it's not illegal because the person who runs the site changes the wording of the problems, so that's why they're kind of confusing sometimes, and sometimes the solutions are... messed up. So if the solution is confusing/doesn't seem right, it probably isn't.

If you've used a site, it's most likely that hundreds of other MathCounts participants have too. Do you really think the police would hunt down a random person who used it?

Either way, if anyone got punished, it would be the site creator.

[quote="Elias Saab's Site"]
Most Problems are adapted From  MATHCOUNTS Competitions and Workouts.
CGI written By Elias Saab
Copyright © 1999-2001 Elias Saab
[/quote]

So no, it's not illegal.

Well..

Thank you for a lot of answers! :)  :)

As mathluver said, its not illegal because the "Elias Sabb" guy changes the numbers. Pretty smart. I wish I could do that.  :huh:


function
algebra
domain
calculus
derivative
analytic geometry
graphing lines

Let $ f(x) \equal{} x\sqrt{x\plus{}1}$.

Find the following information:
-domain
-range
-is it one to one?
- y intercept
- x intercept
- what symettry it has (y axis, x axis, origin, or none)

hello, the domain of definition is $ \minus{}1 \le x<\infty$ because of the square root.
Sonnhard.

[hide="Solution (except for the range)"]
Domain: [-1, Infinity)
X intercepts: 0, -1
Y intercepts: 0
No symmetry.

I'm not sure about the range, because at first glance it looks like [0, Infinity) because the lowest x can be is -1, and f(-1)=0. But when you think about it, a bit of the curve will dip below the X axis because when f(-.5), for example, is approximately -0.354.
[/hide]

[hide="Range"]I don't know if this can be done without calculus but..

Differentiate $ f(x)$ with respect to $ x$.  $ \frac{d}{dx}x\sqrt{x+1} = \sqrt{x+1}+ \frac{x}{2\sqrt{x+1}}$.
Set this equal to zero and solve for $ x$ to get $ x=-2/3$. ${ f(-2/3)=\frac{-2\sqrt{3}}{9}}$, so the range of $ f$ is $ \left[\frac{-2\sqrt{3}}{9}, +\infty\right)$.[/hide]

[quote="Poincare"]i know this solution sort of belongs to the calc forum but can you please explain why setting the derivative equal to zero will give you the range?

thanks[/quote]
Because if the derivative of a point in the graph of $ y\equal{}f(x)$ is zero, then the function stops increasing or decreasing (think of it as either the peak of a mountain or the lowest point of a valley). At these "critical points", the slope of the function will be zero (tangent line will be horizontal), and you test them to find the maxima/minima of a function.

Oh, that's smart. All my calculus is self-taught, so that didn't occur to me at all :)
Well, now I know.


When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to $0$ and is the same as that of getting heads exactly twice. Let $\frac ij$, in lowest terms, be the probability that the coin comes up heads in exactly $3$ out of $5$ flips. Find $i+j$.

I think you worded it wrong... you should say "out of 5 flips" there, since I don't think those two probabilities can be equal in normal circumstances...

Kalva says the same thing, although on kalva this is AIME 1989  :?

[hide]
Assume that the probability of getting heads on one flip is x. Since the probability of getting one head equals that of getting two heads,
(5C1)x(1-x)^4 = (5C2)x^2(1-x)^3
5x(1-x)^4 = 10x^2(1-x)^3
5x(1-x)^4-10x^2(1-x)^3 = 0
5x(1-x)^3(1-x-2x) = 0
x = 0,1,1/3
0 and 1 are not possible as probabilities. Therefore, the probability of getting a head when the coin is tossed is 1/3. From this, the probability of getting heads on 3 tosses out of 5 can be determined:
(5C3)(1/3)^3(1-1/3)^2 = 10(1/27)(4/9) = 40/243 = i/j
Therefore, i+j equals 40+243 = [b]283[/b][/hide]

[quote="agolsme"]This is not AIME 1993 Problem #5. I have the booklet - #5 is

Let $P_0(x) = x^3 + 313x^2 - 77x - 8.$ For integers $n \ge 1$, define $P_n(x) = P_{n-1}(x-n)$. What is the coefficient of $x$ in $P_{20}(x)$[/quote]Did I post this in the wrong year? Thanks for catching that.  :blush:

 :wallbash_red:

Yes it is from 1989 but it still says it's from 1993 also.  And I think you need to check the wording, I think you have to clarify in the first sentence that the two probabilities are equal when thrown 5 times.  It can be implied that the two probabilities are equal under all circumstances, which is not possible.

Let the probability of getting heads be $h$. Then $5h(1-h)^4 = 10h^2(1-h)^3$, or $1-h=2h \implies h = \frac{1}{3}$. Then the answer is $10 \cdot \frac{1}{27} \cdot \frac{4}{9} =\frac{40}{243} \implies \boxed{283}$.

[quote=mahaler][hide = One-liner]$5p(1-p)^4 = 10p^2(1-p)^3 \Rightarrow p = \frac{1}{3} \Rightarrow 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243} \Rightarrow \boxed{283}.$[/hide][/quote]

[quote=mathmax12]$5p(1-p)^4 = 10p^2(1-p)^3$, hence, $p=\frac{1}{3}$, so we get $10\cdot(\frac{1}{27})\cdot(\frac{4}{9})=\frac{40}{243}$, so we have $\boxed{283}.$[/quote]

I appreciate your great effort in secretly copying @2above


say you've got some chord BC of length 6 in a circle (with center O) of radius 5. and then you have this point on the circle A. and it's just happily walking around etc. etc. etc. and suddenly it finds that the circle with diameter AO is tangent to line BC. so then O wants to know, what's the sine of AOB, eh?

[hide]sine of AOB is 7/25:
OA=OB=5, The distance from the intersection of the perpendicular drawn from O to BC (call this D) is 3, so it is a 3-4-5 triangle, which makes sine BOD=3/5 and cos BOD = 4/5.  The circle with AO as diameter must have it's center 5/2 from BC, and dropping an altitude to DO from A gives another 3-4-5 triangle, so sin AOD = 4/5 and cos AOD = 3/5.  Angle AOB = AOD - BOD, so sine AOB = sine AOD * cos BOD - sin BOD * cos AOD = (4/5)^2 - (3/5)^2 = 7/25.[/hide]

[hide]Draw altitude OD, D on BC, creates BDO a 3-4-5 triangle.
Let the center of the circle with diameter OA be P, and where the perpendicular from P to BC be E, and the perpendicular from P to OD be F.
We know that OP=PE=2.5, so OF is 1.5 and PF is 2, so OPF is another 3-4-5 triangle.

Let $\sin B=\alpha$
So $\sin AOB=\sin(\alpha-(90-\alpha))=\sin(2\alpha-90)$
$=\sin^2\alpha\cos90-\cos2\alpha\sin90$
$=-\cos2\alpha=-(\cos^2\alpha-\sin^2\alpha)=\frac{7}{25}$ (after plugging in those values).[/hide]

Heh, a remnant of a time where MysticTerminator struggled with AIME problems.

I thought he poasted this after he qualified for the USAMO.  Twice.

Hm...it seems that you didn't phrase it properly.

You could have $\sin{\angle{AOB}}=\cos{(2\cos^{-1}{(\frac{4}{5})})}=\frac{7}{25}$, but you could have $A$ closer to $C$, which would make $\sin{\angle{AOB}}=1$

Where is the problem that this problem came from located?


Another method
Use Binet's Formula for the Fibonacci numbers:

$ F_{n}= \frac{1}{\sqrt{5}}\cdot\left(\frac{(1+\sqrt{5})^{n}-(1-\sqrt{5})^{n}}{2^{n}}\right)$

Then smack it with geometric series.

[quote="Lazarus"]I'm working off the assumption that $ F_{0}=F_{1}=1$ and $ F_{x+2}=F_{x+1}+F_{x}$.[/quote]

Then your thing would be wrong...because it would be 1/1+1/3+....

but still im working off of $ F_{1}=F_{2}=1$ and from the same logic it should be $ \frac{2}{3}$

[quote="Adam Shaffer"][quote="Lazarus"]I'm working off the assumption that $ F_{0}=F_{1}=1$ and $ F_{x+2}=F_{x+1}+F_{x}$.[/quote]

Then your thing would be wrong...because it would be 1/1+1/3+....[/quote]

Lol my thing is definitely right for $ F_{0}=F_{1}=1$ :X especially cuz it starts with $ n=1$, and $ 3^{1}\neq1$

uh lazurs, my bad I thought you summed it up from 0 to infinity so $ 3^{0}=1$ but 2/3 should be right if its f1=1 and f2=1 which is the general consensus.

or you can, more generally - use generating functions to express $ \sum F_{n}z^{n}$ as a rational function for $ |z|<\frac{\sqrt{5}-1}{2}$
and at the end just put $ z=\frac{1}{3}$


geometry
geometric transformation
rotation

A coin in put in movement under an axis, corresponding to a diameter, on a table. Due to the friction, it begins to stop, 'till it stops totally on the horizontal. The challenge is the find a mathematical expression that describes the form of how the frequency (of the sound we hear) varies.

Source(in Portuguese):
[hide]Source: [url=http://algol.fis.uc.pt/quark/viewtopic.php?t=313&postdays=0&postorder=asc&start=0]Quark[/url][/hide]

the problem sounds nice, but could you be a little bit more clear about it??

So do you put the coin into initial motion under an axis that is perpendicular to the plane of the table? That is initialy the coin will be in the vertical plane?

Let me see if I understand what you're asking about. When I refer to the 'central axis' of the coin, I mean the line perpendicular to the faces (obverse and reverse) of the coin and passing through the center. I'm modeling the coin as a ideal circular disk of uniform radius, R, and uniform thickness, W. The edges are perpendicular to the faces.

I think that you are interested in the case where, initially, a coin is oriented so that the axis is parallel to the table, with an edge in contact with the table surface. The initial axis of rotation is a line, perpendicular to the table top and passing through these two points: (1) center point of the segment of the edge touching the table top (2) the point in the geometric center of the coin (along the coin's central axis, halfway between the coin's faces.

In other words, the coin is spinning on its edge.  ;c)

If you are trying to figure out how long it takes for the coin to slow, wobble, tip over, and roll around its rim before coming to a stop, you [i]are[/i] presenting a difficult problem.

A few factors that you'll need to take into account are:
[list]Coefficient of friction between the coin and tabletop
Hardness of coin and table
Weight of the coin
Speed of rotation
Mass (for momentum) of the coin
Dimensions of the coin (R & W)
[/list]
There's at least one nonlinear transition in the process. The coin's rotation slows and, at the transition point, it tips over, but the remaining angular momentum keeps it wobbling.

I think that the coefficient of friction between the coin and the table is absolutely crucial (I'm neglecting other friction). If there's zero friction, the coin will spin forever. If the friction is [u]very[/u] small, but nonzero, the coin will gradually slow and eventually tip over. However, if the friction is small enough, I think the coin would almost immediately fall flat but continue to spin, flat on the table. If the friction is high enough, the coin will roll on its edge for longer, until enough angular momentum is used up for it to collapse and almost instantly drag to a stop.

Axis corresponds to the diameter.The uniform disk has I=(mr^2)/4.If the torque of friction is constant,then the coin will uniformly slow down with time,its frequency will decrease uniformly.I think the frequency of sound is directly related to the frequency of the coin.


2n points are on plane . prove that there exists a line with these properties:
1.it bisects the 2n points (half of the points on each side of the line)
2.the sum of the distances of points on one side from the line equals the sum of the distances of points of the other side from the line
(e.g for n=1 the perpendicular bisector of the points is the line with the two properties )

Aceasta problema era initial postata pe [b]subforumul in lb.rusa  [/b]([i]dar tot in engleza[/i]), asa ca i-am scris in [i]lb. rusa [/i]urmatoarele:
Daca $ G$ este centrul de greutate al sistemului de puncte {$ A_k; k \equal{} \overline{1,2n}$}, atunci orice dreapta $ d$ care trece prin punctul $ G$ are proprietatea [b]2).[/b]

[u][b]In english:[/b] [/u][i]  If[/i] $ G$ [i]is the centroid of the system of points [/i]{$ A_k; k \equal{} \overline{1,2n}$}, [i]then all the line [/i]$ d(G\in d)$ [i]have propertie [/i][b]2).[/b]
[b][i]Remark.[/i][/b] If $ n \equal{} 1$ and $ G$ midpoint(centroid) of $ A_1A_2$, then all line $ d(G\in d;d\not \equal{} A_1A_2)$ have properties: [b]1)[/b] and [b]2).[/b]