[color=blue][b]
Să se determine valoarea maximă a expresiei:
$E\left( {x,y} \right) = x^2 - y^2 + 2xy - 2x - 2y + 1$
[/b][/color]

Am o intrebare:
Daca derivez functia $E(x,y)$  odata in raport cu $x$ si odata in raport cu $y$ si pun conditia ca derivatele sa fie $0$ si daca obtin ca sistemul format din cele doua ecuatii are solutie atunci solutia reprezinta un punct de extrem?
Banuiesc ca nu este cea mai buna abordare a problemei dar este o curiozitatea de-a mea.

Referitor la problema :
$E(x,0)=(x-1)^2$ asa ca.....  :roll:

[b][color=red]Acum poate este mai dificilă ??[/color][/b]

[color=blue][b]
Să se determine valoarea maximă a expresiei: $E\left( {x,y} \right) = x^2 - y^2 + 2xy - 2x - 2y + 1$
când: $x^2 + y^2 \le 2x$
[/b][/color]

Asta este o problema care tine de optimizare neliniara, ramura a matematicii care nu se studiaza in liceu. Deasemenea, metodele de rezolvare sunt destul de laborioare: multe calcule, tabele, reguli etc.

In principiu, daca nu ar exista limitarea (inecuatia de jos), maximul s-ar afla rezolvand sistemul derivatelor partiale (a lui x, respectiv y).

[quote="void"]In principiu, daca nu ar exista limitarea (inecuatia de jos), maximul s-ar afla rezolvand sistemul derivatelor partiale (a lui x, respectiv y).[/quote]

E de ajuns?

[color=darkblue]Problemele de extrem conditionat, cu cel putin doua variabile, care necesita derivate partiale etc. nu se fac in liceu si nu vor aparea la bac sau admitere in facultati. Asadar, dupa parerea mea, pentru moment asemenea solutii nu sunt interesante, fiind mai greu de inteles.Cred ca Metru a pus problema pentru a i se oferi o solutie elementara, la nivelul programei scolare. Problema este frumoasa, se poate rezolva la nivel de liceu si oricand poate deveni subiect de examen la bac si/sau admitere.[/color][quote="metru"][color=darkred]Să se determine valoarea maximă a expresiei : $E({x,y}) = x^{2}-y^{2}+2xy-2x-2y+1$ pe multimea $\{\ (x,y)\in \mathbb R^{2}\ |\ x^{2}+y^{2}\le 2x\ \}\ .$[/color][/quote]
[color=darkblue][b]Metoda 1 (coordonate polare).[/b] $E(x,y)=(x-1)^{2}+2(x-1)y-y^{2}$ si domeniul $(x-1)^{2}+y^{2}\le 1$ este un disc $\mathbb D\ (C,1)$ cu centrul $C\ (1,0)$ si raza $1\ .$
Substitutia $z=x-1$ $\implies$ $\max_{y^{2}+z^{2}\le 1}(\ z^{2}+2yz-y^{2}\ )\ .$ Substitutia $y=r\cos \phi$ , $z=r\sin\phi\ ,$ unde $\phi\in [0,2\pi)$ si $r\in [0,1]$ $\implies$ $\max\ (\ r^{2}\sin^{2}\phi+2r^{2}\sin\phi\cos\phi-r^{2}\cos^{2}\phi\ )=\max\ [\ r^{2}(\sin 2\phi-\cos 2\phi )\ ]\ .$ Insa $r^{2}(\ \sin 2\phi-\cos 2\phi\ )\le \sqrt 2$ deoarece $0\le r\le 1$ si $|\sin 2\phi-\cos 2\phi |\le \sqrt 2$ $\implies$ [b]Maximul cautat este[/b] $\boxed{\ \sqrt 2\ }$ [b]si este atins, de exemplu, pentru[/b] $r=1$ [b]si[/b] $\phi =\frac{3\pi}{8}\ ,$ [b]adica[/b] $\boxed{\ x=1+\sin\frac{3\pi}{8}\ }$ [b]si[/b]  $\boxed{\ y=\cos\frac{3\pi}{8}\ }\ .$

[b]Metoda 2 (numere complexe).[/b] Se arata usor ca $u=\alpha+\beta \cdot i \Longrightarrow \alpha+\beta = \frac{1-i}{2}\cdot (u+i\cdot \overline u)\ .$ Se observa ca pentru $z=x+y\cdot i\in\mathbb C$ obtinem :

$\{\begin{array}{c}u=(z-1)^{2}=\alpha+\beta\cdot i\ , \mathrm{\ unde\ }\{\begin{array}{c}\alpha =(x-1)^{2}-y^{2}\ .\\\ \beta =2y(x-1)\ .\end{array}\\\\ |u|=|(z-1)^{2}|=|z-1|^{2}=(x-1)^{2}+y^{2}\le 1\ .\\\\ E(x,y)=\alpha+\beta =\frac{1-i}{2}\cdot (u+i\cdot\overline u)\ .\end{array}$ $\Longrightarrow$ $|E(x,y)|=|\frac{1-i}{2}\cdot (u+i\cdot\overline u)|=\frac{\sqrt 2}{2}|u+i\cdot\overline u|\le \sqrt 2\cdot |u|\le \sqrt 2\ .$ $\Longrightarrow$ $|E(x,y)|\le \sqrt 2$

[b]Observatie.[/b] Fara a restrange generalitatea, problema se putea formula mai simplu astfel : $\boxed{\ x^{2}+y^{2}\le 1\Longrightarrow |x^{2}+2xy-y^{2}|\le\sqrt 2\ }\ .$[/color]


Let $ B= \frac{\pi}{7}.$ Prove that
\[ \tan B \cdot \tan 2B +  \tan 2B \cdot \tan 4B + \tan 4B \cdot \tan B=-7.\]

[quote="vijaymenon"]if A = 2*(pie) / 7  prove that

[size=150][b](tan A*tan2A) + (tan 2A * tan 4A ) + (tan 4A*tan A) = -7[/b][/size][/quote]

Here is a very ugly solution.

[hide="solution"]Let $ B \equal{} \frac {\pi}{7}$ the equation becomes:

$ tan2Btan4B \plus{} (tan4B \plus{} tan2B)tanB \equal{} \minus{} 7$

As $ \frac {tan 2B \plus{} tan4B}{1 \minus{} tan2Btan4B} \equal{} \minus{} tanB$, we have the equation

$ \Leftrightarrow (\frac {tan2B \plus{} tan4B}{tanB} \plus{} 1) \plus{} (tan4B \plus{} tan2B)tanB \equal{} \minus{} 7$
$ \Leftrightarrow (\frac {tan2B \plus{} tan4B}{tanB})(1 \plus{} tan^2 B) \equal{} \minus{} 8$
$ \Leftrightarrow \frac {tan2B \plus{} tan4B}{sinBcosB} \equal{} \minus{} 8$
$ \Leftrightarrow \frac {1}{cos 2B} \plus{} \frac {2cos2B}{cos4B} \equal{} \minus{} 4$

$ \Leftrightarrow cos4B \plus{} 2cos^2 2B \equal{} \minus{} 4cos2Bcos4B$
$ \Leftrightarrow 2cos4B \plus{} 1 \equal{} \minus{} 2( \minus{} cosB \plus{} cos2B)$
$ \Leftrightarrow 2cos4Bsin3B \plus{} sin3B \equal{} 2sin3BcosB \minus{} 2sin3Bcos2B$
$ \Leftrightarrow (0 \minus{} sinB) \plus{} sin3B \equal{} (sin 4B \plus{} sin 2B) \minus{} (sin 5B \plus{} sin B)$
$ \Leftrightarrow sin3B \plus{} sin5B \equal{} sin4B \plus{} sin2B$, which is obviously true.[/hide]


How does one make a "not equivalent to" sign in $ \text{\LaTeX}$? Like $ \equiv$, but with a slash through it.

In general, \not puts a slash across whatever symbol follows. This can look silly ($ \not\alpha$), but it's generally pretty well spaced for relation symbols.


induction
inequalities
function
algebra solved
algebra

a1,a2,...,an>0 diferent  numbers.Prove that:
a1 2 +....+an 2   \geq  (2n+1) *(a1+...an)/3

I think you mean different positive integers?

Prolly i'm wrong, but:
We can suppose WLOG that a1<a2<a3<a4...<an.
Also we can use induction on N and also the thing that a(n)>a(i).
So it shud  be easy to prove that N^2 >= than (2*(n-1)+1/3)*sum(of all integer smaller than it = (n)*(n-1)/2) + 2/3*N.
For N sufficiently big, it's true.
BTW, when the olympic test in Romania were so easy?

firstly it's the first problem on that test ... secondly i'm not that sure that what you wrote there works (the inequality must be proven for all n-s) - anyway I don't even know who's N ... maybe a more detailed solution/idea ?

Lets consider the function f:R-->R, f(x)=x^2-(2x+1)/3 *x, for any x from R.

we can see that the function is strictly decresing on [0,(2x+1)/6] and strictly increasing on [(2x+1)/6, oo). The inequality can be rewritten like so f(a1)+...+f(an)>=0. we can verify wuite simple that f(1)+..+(n)=0. More if x>=n+1 then f(x)>f(k), k from 1 to n.
If A={a1,a2,..,an}={1,2..,n}=B the problem is solved! :D
if A<>B we consider then WLOG a1<..<an and then an>n. we will replace element an, with an element from B which cannot be found in A. We denote A1 that crowd (forgot the mathematical term :( ) and we can see that

sum (f(x))> sum (f(x)), in the 1st sum we have x from A and in the 2nd x from A1.
we continue doing this until  we get crowd B and then we have that
sum (f(x))>= sum (f(x)=0 1st sum we have x from A and in the 2nd x from B.

cheers! :D :D


$a,b,c,d>0$.

1.	Prove $\sum_\text{pairs}(a+b)(\frac ab + \frac ba) \geq\sum_\text{pairs}(a+b)(\frac cd + \frac dc)$.
2.	Prove $0\leq\sum_\text{pairs}(a+b)(\frac cd + \frac dc) - 6(a+b+c+d) \leq\sum_\text{pairs}(a-b)^2(\frac 1c +\frac 1d )$.
The sums are over the six possible pairs, as in the title.
I have a nice solution where the very last inequality comes as a by-product of the proof of the first.
I wonder if it is possible to prove the first inequality [b]directly [/b]by using a rearrangement argument?? At a first glance it is [seems obvious, doesn’t it? :)], at a second it isn’t [not at all :(], so now what?

I'd love to try the second one, since the first one does sound obvious.
but we can definitely use it for the second one.
$\ \sum_\text{pairs}(a+b)(\frac ab + \frac ba) \geq\sum_\text{pairs}(a+b)(\frac cd + \frac dc)$
Use this method might work
$\ \sum_\text{pairs}(a+b)(\frac cd + \frac dc) - 6(a+b+c+d) = \sum_\text{pairs}((a+b)(\frac ab + \frac ba) - 2(a+b)) = \sum_\text{pairs}(a+b) (\sqrt(\frac b a) - \sqrt(\frac a b))^2 = \sum_\text{pairs}(\frac{a+b}{ab})[( \sqrt(\frac b a) - \sqrt(\frac a b))^2 *ab] = \sum_\text{pairs}(\frac{1}{a}+\frac{1}{b})(b-a)^2$ $\ ??$
  Not True!!!  $\ \leq\sum_\text{pairs}(a-b)^2(\frac 1c +\frac 1d )$
And in order to prove $\ \sum_\text{pairs}(a+b)(\frac cd + \frac dc) - 6(a+b+c+d) \geq 0$ we can use this one which has already existed in the above stuffs:
$\ \sum_\text{pairs}(a+b)(\frac cd + \frac dc) - 6(a+b+c+d) = \sum_\text{pairs}(a+b) (\sqrt(\frac b a) - \sqrt(\frac a b))^2 \geq 0$

Frank Morgan

[quote="Kalimdor"]$\sum_\text{pairs}(\frac{1}{a}+\frac{1}{b})(b-a)^2 \leq\sum_\text{pairs}(a-b)^2(\frac 1c +\frac 1d )$

Frank Morgan[/quote]
hi,Frank Morgan.can you tell me why about it? :)

[quote="Kalimdor"]This is exactly the part I don't get, but sounds [b]correct[/b].

Frank[/quote]
but you can say the orgianal problem also  sounds [b]correct[/b]. :P

:oops:
(Based on the condition, if the original ineq is true)
In order to prove $\ \sum_\text{pairs}(\frac{1}{a}+\frac{1}{b})(b-a)^2 \leq\sum_\text{pairs}(a-b)^2(\frac 1c +\frac 1d )$
We can try it this way
gotta prove this:
$\ \sum (\frac{1}{a}+\frac{1}{b})(b-a)^2 \leq \sum(\frac{1}{a}+\frac{1}{b})(c-d)^2$
then $\ \sum(\frac{1}{a}+\frac{1}{b})(a-b+c-d)(a-b-c+d) \leq 0$
However, I can't foresee this one's result, true or false.
Frank

[quote="spanferkel"]$a,b,c,d>0$.

1.	Prove $\sum_\text{pairs}(a+b)(\frac ab + \frac ba) \geq\sum_\text{pairs}(a+b)(\frac cd + \frac dc)$.
2.	Prove $0\leq\sum_\text{pairs}(a+b)(\frac cd + \frac dc) - 6(a+b+c+d) \leq\sum_\text{pairs}(a-b)^2(\frac 1c +\frac 1d )$.
The sums are over the six possible pairs, as in the title.[/quote]

Sorry, but it seems to me that both inequality 1. and the right part of inequality 2. are wrong for a = 1, b = c = d = 2.  :(

  Darij

Oh, yeah, I must be drunk then.
but $\ 0\leq\sum_\text{pairs}(a+b)(\frac cd + \frac dc) - 6(a+b+c+d)$ should be true, indeed.:oops:

I'm really sorry, I missed a factor 2 in the middle of my 'proof'


I am attempting to read Shakespeare's Venus and Adonis. It is so hard with all those abbreviations and old vocabs... I am wondering if there are some anthologies out there that contain explanation of words and paragraphs or some modern translation? or like study guides?

I am ESL btw  :blush:

Thanks in advance.

i myself haven't read Venus and Adonis but think
http://en.wikipedia.org/wiki/Venus_and_Adonis_%28Shakespeare_poem%29#Literary_Background
would be good if you got at least some of it
these:
http://search.barnesandnoble.com/booksearch/results.asp?WRD=Venus+and+Adonis&z=y
might be useful

Venus and Adonis?  Are you reading it for school or for personal reasons?  If you have a choice, I'd suggest dropping it and reading Much Ado About Nothing, or another one of Shakespeare's works.  Venus and Adonis is just TOO MUSHY


Quadrilateral-zero is cyclic and has an in-circle. The four points of contact between Quadrilateral-zero and its in-circle form Quadrilateral-one. Quarilateral-two, -three etc are formed in the same way.
Quadrilateral-ten has whole number angles in degree measure. What are the degree measures of the angles in Quadrilateral-zero?


ratio
geometric series
geometric sequence

What is the sum of the infinite geometric series ${\sum_{-\infty}^{5}\{2^n}$?

A person decides to save money in savings account for retirement.  At the beginning of each year, $\$1,000$ is invested at 14% compounded annually.  How much is the retirement fund after $40 years$? (Remember that the end of $year 40$ is the beginning of $year 41$).

A ping-pong ball is dropped from a height of $16 ft$ and always rebounds $\frac{1}{4}$ of the distance of the previous fall.  What distance does it rebound the $6th$ time?  What is the total distance the ball has traveled after this time?

[quote="xxreddevilzxx"][hide]
$\frac{2^5}{1-\frac{1}{2}} = 64$
[/hide][/quote]

Can someone please explain the solution to this one please?

Thanx.

Well the summation means....

$2^5+2^4+2^3+....+2^{-\infty}$

Which is basically sum of infinite geometric sequence with the first term $32$ and the common ratio of $\frac{1}{2}$. Chug and plug in the formula and voila!


A list consists of $11$ distinct positive integers, each of which has distinct digits. The largest and smallest integers on the list are $85491$ and $6$, respectively. These $11$ integers are written using a total of $40$ digits, with each of $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ used $4$ times. When the leftmost digit of one of the $3$ digit integers in the list is erased, the sum of this new list of $11$ positive integers is $297, 388$. Compute the sum of the original list of integers.

You sure it's $297,388$? If every one of the eleven integers was equal to the maximum $8541$, they wouldn't sum to anywhere near that number.

Oops, sorry about that. The max value is supposed to be $85491$, not $8541$.


inequalities
logarithms
inequalities unsolved

$ (x,y,z>1)$       prove that
$ x^{x^2\plus{}2yz}y^{y^2\plus{}2xz}z^{z^2\plus{}2xy}\ge(xyz)^{xy\plus{}yz\plus{}zx}$

hello your inequality is equivalente to
$ Ln (x^{x^{2}\plus{}2yz}y^{y^{2}\plus{}2xz}z^{z^{2}\plus{}2xy})\geq Ln((xyz)^{xy\plus{}yz\plus{}zx})$
$ \Leftrightarrow Ln(x)(x\minus{}y)(x\minus{}z)\plus{}Ln(y)(y\minus{}z)(y\minus{}x)\plus{}Ln(z)(z\minus{}x)(z\minus{}y)\geq0$
and this is obvious if we suppose that $ x\geq y\geq z$.

[quote="sxgfly"]$ (x,y,z > 1)$       prove that
$ x^{x^2 \plus{} 2yz}y^{y^2 \plus{} 2xz}z^{z^2 \plus{} 2xy}\ge(xyz)^{xy \plus{} yz \plus{} zx}$[/quote]

as [b]saif[/b] has done, we take the logarithm of both sides (since $ f(x)\equal{}\log x$ is increasing).

This is equivalent to
$ \sum (x^2 \plus{}2yz) \log x \ge (xy\plus{}yz\plus{}zx)\sum \log x \Leftrightarrow \sum (x\minus{}y)(x\minus{}z)\log x \ge 0,$
which is a consequence of the generalized version of Schur. :D


Let we have a clock with the central second arrow. Speeds of all 3 arrows are constant.
What is the probability that second arrow is more close to minute arrow than to houre arrow?

[hide="Probably wrong"]
Say they start all at 12 o clock,
Distance between the 2nds and the minutes is=
$ 360*60-360=\frac{d}{t}$
$ 360*59t={d}$

Distance between the 2nds and the hours is=
$ 360*60-30=\frac{d}{t}$
$ 30*719t=d$

So actual distance is $ 59t mod360$
and
$ 11t mod360$
But we want the probability when $ 59t mod360 >-t mod360$
Is this the right way?
 and $ t\leq12$?
and so probability is $ \frac{1}{6}$?
[/hide]

[quote="riddler"]Probably wrong[/quote]
You are right - it is too far from the truth.

Damn! but how far am I off? Is this the method? A possible method?

Well this is not a very simple problem...
I've looked at him about 5 minutes and the first idea i have is that the answer may be around 50%, because if the arrows weren't always moving (ex: the minutes arrow passes directly from one minute to the next when the 2nds arrow completes 60 2nds) then the probabilty in each minute would be 50% because we could divide the clock with a diameter that is the bissector os the angle formed by hour's arrow and minute's arrow, and each half of the clock would be closer to the hour's arrow or the minute's arrow.
Not very clear? :lol:
Just an idea...

[quote="VascoMoreira"]Well this is not a very simple problem... [/quote]
You are right.
[hide="hint"]For the beginning try to consider three runners which start in the same direction from one point on circle distance with speeds $ \frac{1}{2}, \ 1, \ 2$[/hide]

Could someone please post a solution to this...?

Perhaps it is time to give another hint?

Here's what I got, but it's ugly. There may be some unnecessary stuff.
[hide]
Let's say our clock has a 1-meter radius, and thus a $ 2\pi$ meter diameter.

Also, instead of hands, let's have particles moving around the circumference to represent the hour, minute, and second hands, which move at $ \frac{\pi}{21600},\frac{\pi}{1800},\frac{\pi}{30}$ meters per second, respectively.

We are given that the second hand is between the hour and minute hand, so let's start with the case where the second and hour particles are in the same position, a distance of $ k\pi$ behind the minute hand. Obviously, once the second hand hits midway between the hour and minute hands, it will be closer to the minute hand, but it's closer to the hour particle until that point. Let our time from the hour particle until when the second particle reaches midway be $ t_{1}$, and from the hour particle to the minute particle be $ t_{2}$.

We have this:

$ \frac{t_{1}(\frac{\pi}{21600}\plus{}\frac{\pi}{1800})\plus{}k\pi}{2}\equal{}\frac{t_{1}\pi}{30}\rightarrow$

$ {t_{1}}\equal{}\frac{21600k}{1427}$

$ t_{2}\frac{\pi}{30}\equal{}t_{2}\frac{\pi}{1800}\plus{}k\pi\rightarrow$

$ t_{2}\equal{}\frac{1800k}{59}$

The fraction of the time where the second particle is closer to the minute particle is $ \frac{t_{2}\minus{}t_{1}}{t_{2}}\equal{}\frac{719}{1427}$

Now we have to do the case where the second particle starts with the minute particle and goes toward the hour particle. Let $ t_{3}$ be the time it takes for the second particle to reach midway, and let $ t_{4}$ be the total time the second particle takes to reach the hour particle. The distance between the minute and hour hands is $ k\pi$. We want $ \frac{t_{3}}{t_{4}}$

$ \frac{t_{3}(\frac{\pi}{1800}\plus{}\frac{\pi}{21600})\plus{}k\pi}{2}\equal{}\frac{t_{3}\pi}{30}\rightarrow$

$ {t_{3}}\equal{}\frac{21600k}{1427}$

$ t_{4}\frac{\pi}{30}\equal{}t_{4}\frac{\pi}{21600}\plus{}k\pi\rightarrow$

$ t_{4}\equal{}\frac{21600k}{719}$

$ \frac{t_{3}}{t_{4}}\equal{}\frac{719}{1427}$

If the second particle reaches the hour hand $ a$ times and the minute particle $ b$ times in a 12-hour period, the probability that the second hand is closer to the minute hand than hour hand is $ \frac{a\frac{719}{1427}\plus{}b\frac{719}{1427}}{a\plus{}b}\equal{}\frac{719}{1427}$

Wait...isn't the second hand always between the hour and minute hands?
[/hide]

Yeah it is haha.. but what does that have to do with the solution to this problem?

I suppose it's a time to give a slight but [hide="key hint"]Consider moments when the event "the second hand is more close to minutes hand than to hour hand" changes their logical value to their opposite value, i.e. changes truth to false or vice versa. [/hide]

Are we assuming that the second arrows and minute arrows move by the whole minute or second, or do they move along smoothly and able to be pinpointed at half or fractions of minutes and seconds?

[hide="Not sure, but sounds right"]We can introduce the fourth hand - let's call it bisector hand - which always bisects the angle between the hours and the minutes hand (hence also rotates at a uniform velocity) and is shaped as the [i]diameter[/i] of the clockface. Then the problem can be put thus: "Bisector hand divides the clockface into two halves. What's the probability that the seconds hand will be in the same half with the minutes hand?"

Since the space of desirable outcomes, at any given moment, is exactly twice smaller than the space of all the possible outcomes (semicircle against circle), the probability is $ {1\over 2}$.[/hide]

Agree, this sounds right. Not thinking deeply the same sought the author of the problem.  :lol:
You have shown the correct factor influencing on a sign of the event to be close to one arrow or to another one.
OK, let me formulate above [b]key hint [/b]in other words:

Are you sure this is the only factor?

[offtop]Amusing thing: I have answered earlier (Today, at 10:53 am) than you have written (Today, at 12:02 am)
What at us with hours? :lol: [/offtop]

To [b]jimhu[/b]: Of course movement of arrows are assumed continuous.


algebra
polynomial
linear algebra
matrix
Putnam
linear algebra unsolved

Let $G$ is finite set of $nxn$ matrices, which is group with the trivial $*$ and $+$ of matrices (e.g. 1)$A,B\in G$ =>$AB\in G$ 2)$E\in G$ 3)$A\in G$ =>$A^{-1}\in G$). prove that if $\sum_{M\in G}tr(M)=0$ then $\sum_{M\in G}M=0$.

[quote="alekk"]look at $P=\frac{1}{Card(G)}\sum_{g \in G}g$. What can you say about it ?[/quote]
what is Card(G)?

He means the cardinality of $G$ - that is, the number of elements. Many people would shorten the notation to $|G|.$ He's forming an average.

Note that if $M\in G,$ then $PM=MP=P,$ since multiplication by $M$ merely permutes the sum.

Then $P^{2}=\frac1{|G|}\sum_{m\in G}PM=\frac1{|G|}\sum_{m\in G}P=P.$

$P$ is an idempotent, meaning its minimal polynomial divides $x^{2}-x.$ This splits over any field and the only possible eigenvalues are $0$ and $1.$ But the sum of those eigenvalues is the trace of $P,$ and by the hypotheses of the problem, $\text{Tr}(P)=0.$ Hence, all of the eigenvalues of $P$ are zero, hence $P$ is nilpotent. The only matrix which is both nilpotent and idempotent is the zero matrix, so $P=0,$ as we wish.

This proof does seem to require that we be working over a field of characteristic zero.

You don't really need to divide by it. You could define $P=\sum_{M\in G}M.$ Then we get that $P^{2}=|G|P.$ From that the eigenvalues are all either $0$ or $|G|,$ but they sum to $0.$

To conclude from that that the eigenvalues are all zero we still need for the field to be of characteristic zero.

About that business of needing the characteristic of the field to be zero:

Consider matrices over $\mathbb{Z}/3\mathbb{Z}.$ Let

$G=\left\{\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}, \begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}, \begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}\right\}.$

This is a group.

$P=\sum_{M\in G}M=\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}.$

This matrix does have trace zero, and is nilpotent. In fact, $P^{2}=0.$ But $P$ is not the zero matrix.

Kent, your $G$ is not a group under addition of matrices. If I did not misunderstand this is a required condition in the hypothesis.
By the way, for any finite group $G$ we have $\sum_{g\in G}g=\sum_{g\in G, g+g=e}g$ where the summation is the group operation.
In our case we have that if the characteristic of our field is not 2 then we do not have an element of order 2 (with respect to addition) so the sum is 0 without any other assumption.

$S=\sum_{M\in G}M$
Pick $M\in G$ with $M$ inveritble. Then
$SM=S$
$S(-M)=S$
Adding the last relations $2S=0$. If char $K\neq 2$ then $S=0$.

Actually, it can't be a group under both addition and multiplication; we have to pick one, and multiplication is the interesting choice.

This is actually one of my favorite Putnam problems. I am pretty sure that this was B6 in 1985; check there if you want the exact wording of the original problem.


linear algebra
matrix
vector
linear algebra unsolved

find the maximum of k such that
we can find matrix n*k like B with element {+1,-1} such that for every vector A OF SIZE 1*n
with element{+1,-1} we have distinct AB.


function
algebra unsolved
algebra
Fixed point

I'm find a problem in an A-level book, but my knowledge still can't help me solve it:
         f(x)=5sinx-2x
a/ Show that the equation f(x)=0 has a root @ in the interval [2, 2.2] (in radian)
b/ Use the iteration formula:    x(0)=2.1;      x(n+1)= pi - arcsin(2/5x(n));
              to find @ correct to two decimal places.
             (x(n) means a element of a sequence)

Part (a): Intermediate value theorem. Evaluate the function at each endpoint of the interval. It's negative at one end and positive at the other.
Part (b): Get a spreadsheet or a calculator and iterate. It is easy to verify that the gixed point $y$ of $g$ is a zero of $f$ where $g(x)=\pi-\arcsin(.4x)$, and the iteration does converge to its fixed point (for all initial values between 1.5 and 2.5).
The convergence is very slow (first order with constant approximately -.75)- this is not a very good iterative method.

On something like this, there is no substitute for actual computation.

But in an A-level problem, can we use calculator to get sin(2pi) or sin(2.2pi). Usually we only get sin(pi), sin(pi/4), sin(pi/6),...

I have no idea what the rules are here, but this numerical exercise makes no sense without a decent calculating device. How else are you going to calculate $\arcsin(.84)$ or $\sin(2)$ or $\sin(2.2)$?

This question looks like an early exercise in a numerical analysis course.


I have a question, I've tried to find the answer but I'm not quite sure how to word it. Anyways, are ALL numbers that are in triples that work for the Pythagorean theorem always be Pythagorean triples?

I think you mean that if $ a^2\plus{}b^2\equal{}c^2$, is the triangle a right triangle. This isn't that hard to show.

Pythagorean triples are INTEGERS that satisfy the Pythagorean theorem.  As an example, if a is 2.5, b is 1.8, and c is 3.08058436; then the triangle is a right triangle.  Even though these numbers work in the Pythagorean theorem, they are NOT considered Pythagorean triples.  Here are some Pythagorean triples.

a=3, b=4, and c=5
a=5, b=12, and c=13

They satisfy the Pythagorean Theorem and are integers.

So not all numbers that fit the Pythagorean theorem work as Pythagorean triples?

Um... that is not ture. numbers satisfying the pythagorean theorem are pythagorean triples by definition.

Wait--so that means that pythagorean triplets are ALWAYS in the pythagorean theorum?

Well, obviously we can't include 0's or negatives, but since this is usually defined for positives, yes.

So that means this also includes square roots... right? So I could have 1,1, square root of 2... Just asking in case it doesn't necessarily include irrational numbers.

in the equation $ a^2\plus{}b^2\equal{}c^2$, $ a,b,$ or $ c$ can be any positive number, rational or irrational, integer or nonintegral, square root, anything. note the word POSITIVE, however, as we are talking about length here.

amd $ 1,1,\sqrt2$ is indeed a pythagorean triplet.

[quote="ZzZzZzZzZzZz"]No.  Pythagorean triples MUST be integers.  Look on Wolfram [url]http://mathworld.wolfram.com/PythagoreanTriple.html[/url][/quote]

This is true. My math teacher says so. :D

[quote="mathgeniusxx"]pythagorean triples are always integers.... yes sqrt3 sqrt4 sqrt5 work, but are they integers(all of them)? NO!
3^2(9), 4^2(16), and 5^2(25) are pythagorean triples because  they are all integers.[/quote]
But they have to fit the Pythagorean theorem, and 9, 16, 25 doesn't, so it isn't a pythagorean triple. .3, .4, .5 do satisfy the pythagorean theorem, but i don't think they would be considered as a triple since they're not integers. multiples of a pythagorean triple will ALWAYS fit the pythagorean theorem.


Let $ a,\ b,\ c$ be positive real numbers. Prove :

$ \frac{(b\plus{}c)^{18}}{(a\plus{}b)^{9}}\plus{}\frac{(c\plus{}a)^{18}}{(b\plus{}c)^9}\plus{}\frac{(a\plus{}b)^{18}}{(c\plus{}a)^{9}}\plus{}6\geq 72abc$.

[hide]Here's my approach:
Using AM-GM we have: LHS>=3(a+b)^3(b+c)^3(a+c)^3+6
We need to prove:[(a+b)(b+c)(c+a)]^3+2>=24abc, which is obviously true by AM-GM:
[(a+b)(b+c)(c+a)]^3+1+1>=3(a+b)(b+c)(c+a)>=3.8abc=24abc
Therefore, the original inequality is proved
The equality occurs when a=b=c=1/8(Q.E.D)
PS: It's nice but quite easy, kunny! :)
I think the following is also true(inspired by kunny's one :D ):
(b+c)^(k+9)/(a+b)^k +(c+a)^(k+9)/(b+c)^k + (a+b)^(k+9)/(c+a)^k +6>=72abc(k is real number and a,b,c,>0) :)[/hide]


a,b:Positive whole number
To solve an equation :
2.2007^3.a^3 + 1 = b^3

[hide]
b must be divisible by 2.
$b = 2c$
$2\cdot 2007^{3}\cdot a^{3}= 8c^{3}$
$2007^{3}\cdot a^{3}= 4c^{3}$
a must be divisible by 2
$a = 2d$
$2\cdot 2007^{3}\cdot d^{3}= c^{3}$
And we get no-where, they'll always have a different number of factors of 2, so there's no solution.
[/hide]

sorry JavaMan
Equation : 2.2007^3.a^3 + 1 = b^3

Is this the equation from HCMCity Round 2 (VietNam)?


a smooth rod is fixed vertically and infront of  rod of mass m and radius R is rolling without slipping on a rough horizontal surface.velocity of center of mass of disc is v.a rod of length L is connected  with disc at A , r/2 distance vertically up from center of disc with pin point (about which it can rotate freely) and other end with a small smooth ring which can freely in rod OP. at this instant rod makes an angle 30degrees with vertical.find  the angular velocity of rod at this instant.

Do not double post.
Ask a moderator or the administrator to delete the unnecessary post.


[quote="ajai"]If the square of Winslow's age is added to Abby's age, the sum is 209. If the square of Abby's age is added to Winslow's age, the sum is 183. What is the sum of their ages?[/quote]
[hide]There is a general way to solve these kinds of equations, but in this case, it is simpler to just use guess and check. We can see that Winslow's age must be about $14$, and that Abby's age must be about $13$. We can check these numbers to find that this is the actual solution, so our answer is $13+14=\boxed{27}$.[/hide]

[hide]Just use guess and check.
13+14=$27$[/hide]

[hide]Guess and check.  13+14=27.[/hide]

[quote="nebula42"][quote="lotrgreengrapes7926"]Anyone want to post an algebraic solution? :D[/quote]
Err... okay.
[hide="Algebraic Solution"]Let Winslow's age be $w$, and let Abby's age be $a$. Then we have
$w^{2}+a=209$ (1)
$a^{2}+w=183$ (2)
We can use equation (1) to get
$a=209-w^{2}$, and substituting this into (2), we get
$(209-w^{2})^{2}+w=183$, and then it gets pretty ugly...  :wink:  :|  [/hide][/quote]

...we then simplify and use synthetic division to find 14 as a solution. Substitute it into an original equation to find Abyy's age.

Just look at the sum 209 and see that $15^{2}$ is too large and Abby would be an old person if winslow was 13.

Then I just chose 14 for winslow and subtracted 196 from 209 to get 13, abby, and didnt bother to check because it has to be right.

well since abby and winslow are old people names i tried abby's age = 70 and winslow's age = 69 but that doesn't equal 209 or 183 when squared and now i'm stuck

[quote="kyyuanmathcount"]The guess and check won't always work.  In a national sprint, there was the EXACT same problem, except the people were old, and you had numbers in the thousands.[/quote]

What was that problem?

[quote="kyyuanmathcount"]The guess and check won't always work.  In a national sprint, there was the EXACT same problem, except the people were old, and you had numbers in the thousands.[/quote]
You can still approximate though, by trying to find positive integer squares about equal to the sums (for example, they were 209 and 183 in this problem). I'm pretty sure the answers were still integers (and I think the answer was what Treething randomly posted here  :P )


In a group of 12 people, among any 9, we can find 5 mutual friends (everyone knows everyone), prove that there are 6 mutual friends!

Pascual

in the group of 9, or in the group of 12?

In the group of 12........

I put this one was easy but i realised i made a mistake, so now, i dont think it is to easy.......

there must be some mistake in the question. Do you want to prove that there are 6 people such that each of them knows the other 5?

A counter example would be 3 groups of 4 people, each person in the 3 groups, knows only the other 3 persons in the group.

No matter how you take the 9 persons you will always find at least 6 (!) pairs of people that know each other ...

No, the mistake was maybe in the way I told the initial condition, I mean that in any group of 9 people, there were 5 such that each one know the other 4, then we have to prove there is a group of 6 (in the initial 12 guys) such that every guy knows the other 5.

Is it clear now?

And sorry abput my poor english...

I have a very long solution to the problem though I guess a shorter solution must be there.

First we prove a lemma : if among 9 people every eight contain five who all know each other, on the whole there are 6 who all know each other. This in itself is not very easy but it can be done.

Then take any 10 people. Suppose there is any one person A among them who is not part of a collection of 5 (among the 10) who all know each other. Remove any of the others. The remaining 9 contain a set of five who all know each other (a K5). But as A is not part of a K5 the remaining 8 form a K5. So of the 9, every 8 form a K5 and the 9 have a k6 and we are done.

So clearly each guy is part of a K5.

Take any 11 people. Consider any guy F knowing d people (d is the degree of the corr vertex in a corr graph). Then clearly if we remove 1 of these d people then in the remaining 10 people the guy knows at least 4 people so d >= 5. Now we want that whichever of the d people I remove there should be a K4 among the remaining d-1. But we also want no K5 because those five people along with F form a collection of 6 who all know each other.

It can again be seen that for d = 5, d = 6, d =7 it is not possible so clearly d >= 8.

Now among all the 12 people there is clearly at least one pair of people P, Qwho know each other. Remove Q. Then among the remaining 11, P knows at least 8. But then P knows at least 9 people.

But these 9 people contain a K5 that along with P gives a K6, or 6 people who all know each other.

I know the official (Russia '99) solution.
Basicly, it says that the graph doesn't contain an odd :) cycle, or it is biparite.
If it is bipartite, it's obvious, if not, then consider cases cycle of length 3,5,7,9,11.
Very funny, of course :D


search
combinatorics proposed
combinatorics

here is a question i recently came across:

There are three doors and behind one of the doors there is a prize and behind the rest there is nothing. There is a host running the show and he gives you the chance to guess where the prize is. You choose a door . Thereafter  the host opens one of the doors where there is no prize .He now asks you if you wanna change your mind. The question is, is it more profitable to change your mind or stick to your first choice?

 ;) [/b]

It has already been discussed here ;)
Search for : Monty Hall problem


[quote="Xaenir"]A positive integer is the product of n distinct primes.In how many ways can it be represented as the difference of two squares.[/quote]

We get $ p\equal{}x^2\minus{}y^2\equal{}(x\minus{}y)(x\plus{}y)$ and so $ x\equal{}\frac 12(d\plus{}\frac pd)$ and $ y\equal{}\frac 12|d\minus{}\frac pd|$ where $ d$ is any divisor of $ p$.

First, we need to have $ d$ and $ \frac pd$ both odd or both even and so $ p$ odd or $ 4|d$, which is impossible ($ p$ is product of $ n$ [b][u]distinct [/u][/b]primes)

Last, we have $ 2^n$ possibilities for $ d$, but $ d$ and $ \frac pd$ give the same difference $ x^2\minus{}y^2$

Hence the result :
$ p$ even $ \implies$ no solution
$ p$ odd $ \implies$ $ 2^{n\minus{}1}$ solutions

$ (a\plus{}b)(a\minus{}b)\equal{}p_1p_2p_3\ldots p_n$

If $ 2$ is the element of the set $ p_1,p_2,\ldots,p_n$ then because $ a\plus{}b$ and $ a\minus{}b$ have the same parity , so $ 2|a\plus{}b$ and $ 2|a\minus{}b$ , but $ 4$ doesn't divide $ p_1p_2\ldots p_n$, so it is impossible. We get the number of ways to represented the number is $ 0$

If not , since all of them are odd numbers and $ a\plus{}b$ and $ a\minus{}b$ have the same parity then $ a\plus{}b$ and $ a\minus{}b$ are the factors of $ p_1p_2p_3\ldots p_n$ , so the number of ways are $ 2^n$ (since $ a\plus{}b$ not always greater than $ a\minus{}b$)

[quote="matrix41"] ... so the number of ways are $ 2^n$ (since $ a \plus{} b$ not always greater than $ a \minus{} b$)[/quote]

I think you got twice too many.

Choose $ n\equal{}1$ and, for example $ p\equal{}3$. We obviously have $ 3\equal{}4\minus{}1$. Could you show us the second way you predicted ?


geometry
rhombus
vector
complex numbers
geometry proposed

Let $ABCD$ be a conves quadliteral such that $AC=BD$ and $AC\bot BD$.Contruct at the outside of the quadliteral the equilateral triangles $ABX,BCY,CDZ,DAT$.Prove that $XZ=YT$ and $XZ\bot YT$

[hide="Here is a solution using the complex numbers and there are some additional properties."]I shall use the complex numbers. Let $X(x)$ be the point $X$ with the affix $x$. Denote $A(a)$, $B(b)$, $C(c)$, $D(d)$, $X(x)$, $Y(y)$, $Z(z)$, $T(t)$  and $\omega =\frac 12+\frac{\sqrt 3}{2}\cdot i$. We observe $\omega ^3=-1$, $\omega ^2=\omega -1=-\overline {\omega}$.
[b]Lemma.[/b] $AC=BD\ \wedge\ AC\perp BD\Longleftrightarrow (d-b)=(c-a)\cdot i.$

[u]Proof of the proposed problem.[/u]
$x=a+\omega (b-a);\ y=b+\omega (c-b);\ z=c+\omega (d-c);\ t=d+\omega (a-d).$
$z-x=(c-a)+[(d-b)-(c-a)]\omega =(c-a)+[(c-a)i-(c-a)]\omega\Longrightarrow$
$\boxed {\ z-x=(c-a)[1-(1-i)\omega]\ }$
$t-y=(d-b)-[(c-a)+(d-b)]\omega =(c-a)i-[(c-a)+(c-a)i]\omega\Longrightarrow$
$\boxed {\ t-y=(c-a)[i-(1+i)\omega]\ }$
We observe that $(t-y)=(z-x)\cdot i$, i.e. (from the about lemma) $\boxed {\ XZ=YT\ \wedge\ XZ\perp YT\ }$

[b]Remark.[/b] Prove easily that: $XZ=\frac{\sqrt 6 -\sqrt 2}{2}\cdot AC$; if we denote the middlepoints $M$, $N$, $U$, $V$ of the segments $AC$, $BD$, $XZ$, $YT$, then the quadrilateral $MUNV$ is a rhombus with $m(\widehat {MUN})=60^{\circ}.$[/hide]

Also this problem have short solution with using vectors :) (but it is not more shorter to  Virgil's solution :) )


vector
linear algebra
linear algebra unsolved

Tr(AB)=Tr(BA), Det(AB)=Det(BA) hence, the sum and products of eigen values of AB are equal to those of BA.

Further, if c is an eigen value of AB, ABX=cX has a nontrivial solution X.
hence, multiplying by B, BABX=cBX, i.e., c is an eigen value of BA, if BX is not zero.

If BX =0, then, ABX=0, hence, c=0, and hence, det(AB)=0=det(BA); so 0 is an eigen value of BA also.

Now, my problem is, what do I do with multiplicity of eigen values? If a particular eigen value is a double root of the eigen equation of AB, the above argument doesnot show that it is a double of eigen equation of BA also. So, is it really true that the eigen values of AB and BA are the same?

Geometric multiplicities must not be the same as the example $ A\equal{}\left(\begin{array}{cc}0 & 0\\ 0 & 1\end{array}\right)\,,\,B\equal{}\left(\begin{array}{cc}0 & 1\\ 0 & 0\end{array}\right)$ shows. $ AB\equal{}0$ has two (Jordan) eigenblocks corresponding to the eigenvalue $ 0$ and $ BA\equal{}B$ has one eigenblock corresponding to the eigenvalue $ 0$. But the algebraic multiplicities will coincide, since $ \det(tI\minus{}AB)\equal{}\det(tI\minus{}BA)$ always holds.

P.S. For a proof of the above identity of determinants assume first that $ B$ is invertible, then $ \det(tI\minus{}BA)\equal{}\det(B^{\minus{}1}(tI\minus{}BA)B)\equal{}$ $ \equal{}\det(tI\minus{}AB)$. If $ B$ is not invertible, construct a series of invertible matrices $ B_k$ with $ B_k\rightarrow B$, e.g., choose $ B_k$ appropriately from the infinite set $ {\cal B}\equal{}\left\{B\plus{}\frac{1}{t}I\Big|t\in\mathbb{N},\det\left(B\plus{}\frac{1}{t}I\right)\not\equal{}0\right\}$. Then $ \det(tI\minus{}AB_k)\equal{}\det(tI\minus{}B_kA)$ and since $ \det$ is continuous, the claim has been verified.

Suppose $ \alpha \equal{} 0$ is an eigenvalue of AB: then det(AB - 0 I) = det(AB) = 0 = det (BA) = det(BA - 0 I), and $ \alpha \equal{} 0$ is also an eigenvalue of BA. If $ \alpha \neq 0$ is an eigenvalue of AB there is a non zero vector x such that $ ABx \equal{} \alpha x <\equal{}> BABx \equal{} \alpha Bx <\equal{}> BA(Bx) \equal{} \alpha (Bx)$, and $ \alpha$ is also an eigenvalue of BA (with associated eigenvector Bx). Therefore, AB and BA have the same eigenvalues.

[b]Carcul [/b], your observation was already shown in the initial post, the question was about multiplicities


number theory
greatest common divisor
algebra
polynomial
algorithm
relatively prime
Euclidean algorithm

For any integer $a$, show that $gcd(2a+1, 9a+4) = 1$

Prove that the product of four consecutive integers is 1 less than a perfect square.

Obtain integers $x$ and $y$ satisfying the following: $gcd(56,72) = 56x+72y$ (use the Euclidean algorithm).

#1:

[hide]
Special cases to be considered momentarily, but it is clear for almost all integers the magnitude of 9a+4 is greater than the magnitude of 2a+1. Thus:

$\frac{9a+4}{2a+1}=\frac{8a+4}{2a+1}+\frac{a}{2a+1}$

$=4+\frac{a}{2a+1}$

Clearly the second term is only an integer when a=-1, but then 2a+1=-1, so the gcd of the two numbers is still 1.

Now for special cases: There might be an interval where the magnitudes do not relate as previously stated. Investigating a=-1, 0, and 1, the initial assertion that 9a+4 has the greatest magnitude actually does hold, so the proof is complete.
[/hide]

[quote="brady00n"]How does your solution to #1 work? (I don't know).[/quote]

Any pair of integer solutions to the equation $ax+by=1$, where $a$ and $b$ are constants and $x$ and $y$ are the variables, have to be relatively prime I think. Because if they weren't, you could write $x=mk$ and $y=nk$ for some integer $k$, and the equation would become

\begin{eqnarray*}a(mk)+b(nk) &=& 1,\\ k(am+bn) &=& 1, \end{eqnarray*}

so $k$ would have to equal 1.

(In paladin's solution, $a=9, b=-2, x=2a+1, y=9a+4$)

#2: Prove that the product of four consecutive integers is 1 less than a perfect square.

Hint:
[hide]
Try some easy sets of 4 integers and look for a pattern between the integers and the square number involved.
[/hide]

Solution:
[hide]
I tried 1-4, 2-5, 3-6 and noticed that for $n(n+1)(n+2)(n+3)=k^{2}-1$, k seemed to equal $(n+1)(n+2)-1$. Polynomial multiplication shows this to be the case for all sequences of 4 consecutive integers (do the multiplication if you don't believe it).[/hide]

[quote="brady00n"]Hey...that gets me thinking.  I know that if two integers $a$ and $b$ are relatively prime, integers $x$ and $y$ exist such that $ax+by=1.$  But I never knew that $x$ and $y$ themselves were relatively prime, interesting.

Unfortunately, I didn't understand how the proof worked. (Fraxia)
I don't really see why $k=1$ from the two equations.[/quote]
The two equations are actually the same. The reason $k=1$ is that in the second equation, we see that 1 is divisible by $k$, which is ridiculous if $k>1$.

brady00n:

We know that $k(am+bn) = 1$. But since we are working only with integers, $k$ and $(am+bn)$ must be divisors of $1$. So they must both equal 1 (or -1; the proof still works either way).

[quote="brady00n"]How does your solution to #1 work? (I don't know).[/quote]
if you let $g=gcd(2a+1,9a+4)$
then $g|2a+1$ since $g|18a+9$
on the other hand $g|9a+4$ since $18a+8$
so $g|18a+9-18a-8=1$
so $g=1$

[quote="brady00n"]Hey...that gets me thinking.  I know that if two integers $a$ and $b$ are relatively prime, integers $x$ and $y$ exist such that $ax+by=1.$  But I never knew that $x$ and $y$ themselves were relatively prime, interesting.[/quote]

But if they weren't, then you would not be able to find integers $a$ and $b$ such that $xa+yb = 1$.  ;)

[quote="brady00n"]How does your solution to #1 work? (I don't know).[/quote]

I was wondering the same thing for a while, but I came to this conclusion:

If there was a number g such that it divides both $2a+1$ and $9a+4$, then it should also divide both $9(2a+1)$ and $2(9a+4)$, and since 9 and 2 are relatively prime, no new common factors are introduced.

If two numbers are both divisible by g, then their difference should be a multiple of g (ex. 40-25=15, which is divisible to once of their common factors: 5).
Their difference is 1 (note that the numbers 9 and 2 were chosen to eliminate the a's), which is a multiple of nothing but one. Thus g must be 1.


geometry
geometric transformation
homothety
ratio
perpendicular bisector

There is a circle having a radius of $1$ unit. A point, $P$, which is in the interior of the circle is also given.

$Q$ is an arbitrary point on the circumference of the circle.  Fold the paper in such a way so that $P$ and $Q$ becomes the same point. Note the straight line as the paper folds and draw a line along it.

Repeat this process for each point on the circumference.

$Question$
I guess these lines might enclose a region on of this region? Or it is not a well observed region (hotch-potch)?

Comments.

[hide="Observations"] Well, each line is just the perpendicular bisector of $PQ$... I think the region is enclosed by the midpoints of each segment $PQ$; and it looks like a circle w/ half the original radius no matter where you place the point $P$ (see the attached picture). If I have time, I'll get around to proving this... but in the meantime, anyone want to? [/hide]

well...a homothety ratio 1/2 maps the larger circle, with Q on it, to the smaller circle with the midpoint of PQ on it


[quote="Diddy"]Hi

I do not know if this is the right forum to put this thread in.

I have heard about a program called LaTeX which is very usefull so therefore can you please tell me where I can get it.

Thanks.[/quote]

Here in AoPS you can write with LaTeX... and just use search function or Google... Ah!!!! and in AoPS page you have a lot of information.

[quote="Diddy"]Can you recommend a program that I can to write math so that it looks properly.[/quote]

The're a lot, but start with LaTeX... it's preety!

I think [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]this[/url] is what you're looking for. It's a tutorial that tells you how to install the program and gives basic instructions.
You can incorporate LaTeX into your posts by putting dollar signs around your LaTeX code.

How to make a vector in LaTeX, a () you know a high one like divison just without the -.

Latex-related questions should go in the Latex forum.  I'm moving this thread there.


ok, i have the complex number 3-3i, and if I wanted to raise it to the 8th power, i could use demoivre's theorem, but for me to use that I need sines and cosines in the problem, which I don't have. also, since x needs to be equal for both sine and cosine, I can't get the polar form of this number!

[hide="hint"]
consider $ z \equal{} 3 \sqrt 2[\cos{( \minus{} \frac {\pi} 4)} \plus{} i \sin {(\minus{} \frac {\pi} 4 )}]$[/hide]

WOW, your geniusnous is way beyond my abilities

[quote="ph34rb0t"]but for me to use that I need sines and cosines in the problem, which I don't have.[/quote]

You can convert the complex number into it's trigonometric representation, namely $ z \equal{} x\plus{}iy \equal{} r \cos \theta \plus{} i r \sin \theta \equal{} \text{cis} \theta$.

The previous poster did convert it into cis, so you should be able to do DeMoivre's.

EDIT: Sarcasm won't get you very far, Milo. Do your homework yourself.


algebra
polynomial
induction
Ring Theory
superior algebra
superior algebra unsolved

If $ K$ has characteristic $ 0$ and $ p_1, p_2, ... , p_n$ are the newton sums in $ K[x_1,x_2,...,x_n]$ every symmetric polynomial can be expressed in terms of $ p_1,p_2,...,p_n$. Prove that this representation is unique.

I think this can be proven by induction using a partial ordering on monomials in $ x_1,\ldots, x_n$, say $ x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}\ge x_1^{b_1}x_2^{b_2}\cdots x_n^{b_n}$ for lexicographic $ (b_1,b_2,\ldots, b_n)\prec (a_1,a_2,\ldots, a_n)$.

Let $ e: K[x_1,\ldots,x_n]\to K[x_1,\ldots,x_n]$ be the map $ e: f(x_1,\ldots,x_{n-1},x_n)\mapsto f(x_1,\ldots,x_{n-1},0)$ and let $ f$ be a symmetric polynomial.
Note that $ e(p_n(x_1,\ldots,x_{n-1},x_n))=p_{n}(x_1,\ldots,x_{n-1})$.
By induction there exists a polynomial $ R\in K[X_1,\ldots,X_{n-1}]$ such that $ e(f(x_1,\ldots,x_n))=R(p_1(x_1,\ldots,x_{n-1}),\ldots,p_{n}(x_1,\ldots,x_{n-1}))$.
Define $ g(x_1,\ldots,x_n)=R(p_1(x_1,\ldots,x_n),\ldots, p_n(x_1,\ldots,x_n))$.
By construction $ e(f(x_1,\ldots,x_n)-g(x_1,\ldots,x_n))=0$.
Since $ K[x_1,\ldots,x_n]$ is an UFD, we must have that $ x_n$ divides $ f(x_1,\ldots,x_n)-g(x_1,\ldots,x_n)$, and since  $ f$ is symmetric $ s_n=x_1x_2\cdots x_n$ divides $ f-g$.

Now  $ \frac{f-g}{s_n}$ is a symmetric polynomial with total degree less than $ f$, so by induction on the total degree, $ \frac{f-g}{s_n}$ is expressible in $ p_1,\ldots,p_n$. Now all you have to do is express $ s_n$ in terms of $ p_1,\ldots,p_n$, which should not be difficult..

Thanks, with this occasion I learned what an UFD is.  :)
But, I think there are some typos and unclarities. Shouldn't the codomain of $ e$ be  $ K[x_1,\ldots,x_{n\minus{}1}]$ and  $ e(f(x_1,\ldots,x_n))\equal{}R(p_1(x_1,\ldots,x_{n\minus{}1}),\ldots,p_{n\minus{}1}(x_1,\ldots,x_{n\minus{}1}))$ ? Remember that you can only use $ p_1,\ldots,p_{n\minus{}1}$ for $ n\minus{}1$ variables.
So, if I understand correctly, you do induction on both the number of variables and the total degree, right?

But don't forget the question was
[quote] Prove that this representation is unique. [/quote]
and this is not yet clear. The fact that this representation exist is a direct consequence of the fundamental theorem of symmetric polynomials and the newton-type identities expressing elementary symmetric polynomials in terms of power sums.

[quote="Thor"]Thanks, with this occasion I learned what an UFD is.  :)
But, I think there are some typos and unclarities. Shouldn't the codomain of $ e$ be  $ K[x_1,\ldots,x_{n \minus{} 1}]$ [/quote]
This doesnt really matter, $ e$ was a priori just a selfmap of $ K[x_1,\ldots,x_{n }]$, but you are right, $ e$ goes surjectively on $ K[x_1,\ldots,x_{n \minus{} 1}]$,
[quote="Thor"]and  $ e(f(x_1,\ldots,x_n)) \equal{} R(p_1(x_1,\ldots,x_{n \minus{} 1}),\ldots,p_{n \minus{} 1}(x_1,\ldots,x_{n \minus{} 1}))$ ? Remember that you can only use $ p_1,\ldots,p_{n \minus{} 1}$ for $ n \minus{} 1$ variables.
[/quote]Yes you are right, it is neccesary to consider $ p_1,\ldots,p_{n \minus{} 1}$ for the induction hypothesis.
[quote="Thor"]
So, if I understand correctly, you do induction on both the number of variables and the total degree, right?
[/quote]Yes.

I think also the uniqueness part can be proven by similar induction.


Is it possible to select 102 17-element subsets of a 102-element set such that the intersection of any two of the setes has at most 3 elements?

I'm not sure about the case where the maximum allowed intersection is 3, but if the maximum allowed intersection is 2, it's not possible:

Let the 102-element set be S. Let A = sum of the magnitudes of the pairwise intersections of the 17 subsets. For an element x in S, define f(x) = the number of subsets which contain x. Then clearly, we have A = sum (f(x) C 2), where the sum is taken over all elements in S.

sum(f(x)) = 17*102, so
A = sum(f(x)*(f(x)-1)/2)
= 0.5*(sum(f(x)^2) - sum(f(x)))
>= 0.5*(102*17^2 - 102*17), by QM-AM
= 17*8*102

Now there are 102 C 2 pairwise intersections, so the average magnitude is
17*8*102/(102*101/2)
>2
By pigeonhole, at least one of the intersections contains 3 elements.

I think you can probably create an example where each has max 3 elements, but I'm not completely sure.
ps. sorry if this post isn't that easy to read - I need to learn how to use latex.

If I am not mistaken,this problem also appeared in St.Petersburg's city olympiad several years ago.And it seems to me that problem is too hard to be placed in [b]pre-olympiad[/b] section,isn't it?
But I can't checked the information above,as I don't have these books with me,so maybe someone will be so kind to check this information?Furthermore,I'm pretty sure that some of the ML members has this book at home. :wink:
I am also very interested in its solution...


It's taking 20 seconds to load any AoPS page! What's wrong? I'm on Bellsouth DSLand my connection is fine.

We have a small problem here with $\LaTeX$. Will be fixed soon.

OK, site will continue for a couple of days to be occasionally slow, because some of the $\LaTeX$ formulaes are being regenerated, but it's a (very) temporary thing, so don't worry about it :)


[quote="math92"]Find four consecutive integers such that twice the sum of the first and third is 11 greater than 3 times the second.[/quote]

[hide="click for solution"]
$2(2a+2)=11+3(a+1)\\ 4a+4=11+3a+3\\ a=10$

therefore the numbers are: $10,11,12,13$[/hide]

[hide][quote="math92"]Find four consecutive integers such that twice the sum of the first and third is 11 greater than 3 times the second.[/quote]

2(2x+2)=3(x+1)+11
4x+4=3x+14
x=10
11,12,13[/hide]


Is there a method to find the general term of any sequence, where the difference between the terms form their own sequence?

As an example, take this simple sequence:

$ 2,5,11,20,32,47,.....$

Well, yes, if the series of differences, $ b_n : \equal{} a_n \minus{} a_{n\minus{}1}$, is known, then you can really say:

$ a_n \equal{} a_0 \plus{} \sum_{j\equal{}1}^{n} b_n$

As for closed form, that's a different matter - that might not always be possible.

In your case it is possible - I'm assuming the differences go 3, 6, 9, ..., 3k, ...

$ a_0 \equal{} 2, \; b_k \equal{} 3k$
$ a_n \equal{} a_0 \plus{} \sum_{j\equal{}1}^{n} b_n \equal{} 2 \plus{} \sum_{j\equal{}1}^{n} 3j\equal{}2\plus{}\frac{3}{2}(n)(n\plus{}1)$

General methods for finding the general term of [b]any[/b] sequence don't exist.  Many different sequences are possible that follow many different possible rules.

We will deal with a particular subset of sequences, which are sequences that are [b]polynomial[/b] in the index; for example, sequences of the form

$ a_n \equal{} n^2$

or

$ a_n \equal{} 3n^3 \plus{} 2n^2 \plus{} n \plus{} 1$

Such sequences are easy to identify (given a sufficient number of terms) because they have the property that some sequence of differences [b]becomes constant.[/b]  For example, the first differences of your sequence are

$ 3, 6, 9, 12, 15, ...$

and the second differences are

$ 3, 3, 3, 3, ...$

Whenever a sequence behaves this way, we can quickly assign a polynomial to it using various summation rules such as

$ \sum_{k\equal{}1}^{n} k \equal{} \frac{n(n\minus{}1)}{2}$
$ \sum_{k\equal{}1}^{n} k^2 \equal{} \frac{n(n\minus{}1)(2n\minus{}1)}{6}$
$ \sum_{k\equal{}1}^{n} k^3 \equal{} \frac{n^2(n\minus{}1)^2}{4}$


3 different integers are chosen between 1 and 13 inclusive. What is the probability that the sum of the 3 integers is divisible by 4.

[quote="tim1234133"]I have a [b]long and ugly[/b] method that basicallly involves replacing the integers from 1 to 13 with their residue mod 4 and drawing a [b]huge[/b] tree diagram. It gives me:
[hide="Answer?"]$\frac{73}{286}$[/hide]
Is this the correct answer? I would be interested to see a shorter, cleaner solution if anyone has one.[/quote]
Yes thats the right answer. I was looking for a solution shorter than mine. Nevermind mine is short enough.
[hide="hint"]find the number of sets of 3 numbers that you can choose from. Then look at the ones divisble by 4 in mod 4. Then find the number of ways  they can be done in. And then, you should get your probability. [hide=" :) "]Alright! My solution is almost the same as the one provided.[/hide][/hide]

[quote="junggi"][hide="solution"]The only possible choices are
0,0,0 mod 4
1,1,2 mod 4
0,1,3 mod 4
0,2,2 mod 4 and
2,3,3 mod 4
1 way to choose the first case.
$\binom{4}{2}\binom{3}{1}=18$ ways to choose the second
$\binom{3}{1}\binom{4}{1}\binom{3}{1}=36$ ways to choose the third
$\binom{3}{1}\binom{3}{2}=9$ ways to choose the fourth
$\binom{3}{1}\binom{3}{2}=9$ ways to choose the last, for the total of 73.
Without restriction, we can choose $\binom{13}{3}=286$ so the probability is $\frac{73}{286}$[/hide][/quote]
yep, that's what i did. Thanks junggi :)


Friends, the following proof appears in Knuth and he as usual gives a sensational proof. However, the proof is not clear to me. The question alongwith the solution appears below---

There is a string which contains n-letters. U need to determine the number of [b]stack-derivable permutations[/b] of the string. A stack-derivable permutation is one which can be printed using a stack.

This printing is achieved in the following way. Every character of the string is pushed once and popped once at which time it is printed and this is how we print a string. We want to find the number of permuatations of string which can be printed using a stack. Hope the question is clear.

This is the solution according to Knuth---

Help me in understanding it....

Since every letter is pushed and popped (printed ) exactly once, we just need to focus on valid pushing and popping orders. Let us denote a push by S and a pop by X. Thus, the problem reduces to determining the number of permutations of a 2n-lettered string with n-S's and n-X's (n-pushes and n-pops for printing n-letters). in which no prefix substring contains more X's than S's.

The total number of permuattions of this kind of string W, without any restrictions is 2n[b]C[/b]n..

Sure enough this overcounts by including invalid strings which have a prefix substring with more X's than S's. To remedy this, consider any such string. Focus on the first position i of this string where X exceeds S for the first time. Next replace every X by S and vice-versa in the substring W[1].W[2]-------W[i]. Thus, you obtain a tring with n+1 S's and n-1 X's.

Thus answer is 2n[b]C[/b]n -- 2n[b]C[/b]n+1..

The above statement confuses me. Do we assume that there is a bijection b/w strings with a prefix substring with more X's than S's and the strings with n+1 S's ? Is it not possible that corresponding to one string with n+1 S's we may get 2 strings containing a prefix substring with more X's than S's???

Kindly do reply. Thanx in advance...

[quote="Akash Kumar"]
Sure enough this overcounts by including invalid strings which have a prefix substring with more X's than S's. To remedy this, consider any such string. Focus on the first position i of this string where X exceeds S for the first time. Next replace every X by S and vice-versa in the substring W[1].W[2]-------W[i]. Thus, you obtain a tring with n+1 S's and n-1 X's.

Thus answer is 2n[b]C[/b]n -- 2n[b]C[/b]n+1..

The above statement confuses me. Do we assume that there is a bijection b/w strings with a prefix substring with more X's than S's and the strings with n+1 S's ? Is it not possible that corresponding to one string with n+1 S's we may get 2 strings containing a prefix substring with more X's than S's???
[/quote]

Yes, there is such a bijection. You can apply exactly the same process in reverse: given a string with n+1 S's and n-1 X's, find the first position in the string where you have more S's than X's. This will be exactly the same position you found with the original string, so flipping S's and X's up to that point gives you the original string.


Describe all the primes in $ \frac {\mathbb{Z}}{m\mathbb{Z}}$.
In particular, say under what condition, there is only one prime, up to multiplication by a unit, in $ \frac {\mathbb{Z}}{m\mathbb{Z}}$.

What do the ideals of $ \mathbb{Z}/m\mathbb{Z}$ look like?  What about the prime ideals?


superior algebra
superior algebra unsolved

Find epimorphism $ \phi : S_4 \to S_3$ where $ S_n$ is a n permutation group.

$ V_4$ is normal in $ S_4$ of index 6 and since the commutator group $ (S_4/V_4)'\equal{}A_4/V_4$ cyclic of order 3 we have $ S_4/V_4\simeq S_3$


Find (and prove) a formula to the sum $ 1^p\plus{}2^p\plus{}...\plus{}n^p$, where $ p$ is a positive integer.

This is not elementary.  See http://mathworld.wolfram.com/FaulhabersFormula.html .

consider $ \sum_{i \equal{} 1}^n i^p \equal{} f(n)$ where $ f$ is a polynomialof degree $ p \plus{} 1$, i.e. $ f(n) \equal{} \sum_{i \equal{} 0}^{p \plus{} 1} a_i n^i$, and setting $ n \equal{} 0,...,p\plus{}1$ we get a system of equation which can be solved for the $ a_i$. (Actually, we can find a definition for the $ a_i$ which is related to the Bernoulli numbers). If you ever need the formula it's better to prove it for the value of $ p$ you need than citing the general proof.


parameterization calculus function vector logarithms derivative algebra

Suppose that $f(xy)=f(x)+f(y)$; determine an expression for $f$ involving one variable and two parameters ($x,y$ in positive Reals)
[hide="Note."]If $f(xy)=xf(y)$ is given, then to solve for $f$ with variable $x$ and one parameter would be: $f(x)=xf(1)$ for instance.[/hide]
[hide="Hint"][hide="Dont click until really sure!!!"]
I came up with this in calculus class today while doodling.
I am sure you can guess what the function is. Now what is the simple calculus property of the function?[/hide][/hide]

$f: \mathbb{R}\rightarrow\mathbb{R}$, then $f(x)+f(0)=f(x\cdot 0)=f(0)$ for all $x\in\mathbb{R}$ and $f\equiv 0$.
$f: \mathbb{R}_{+}\rightarrow\mathbb{R}$, then $g(x): =f(e^{x}),g: \mathbb{R}\rightarrow\mathbb{R}$ satisfies Cauchy's functional equation, and there are many many solutions.
(every map $B\rightarrow \mathbb{R}$ extends to one, where $B$ is some $\mathbb{Q}$-basis of the $\mathbb{Q}$-vector space $\mathbb{R}$)
And then $f(x)=g(\ln x)$ gives all solutions.

[quote="me@home"]Isn't anyone interested? 4 days until I post my solution[/quote]
Do not ask for more interest after 3 hours only! And the above post already tells you why the problem as stated is nonsense/wrong ;)

If you want functions that $f(xy)=f(x)+f(y)$ you can choose $ln(xy)=ln(x)+ln(y)$

 :wink:

Not exactly sure what ZetaX was trying to say...

[hide="but here was my idea:"]
$\frac{d}{dx}f(xy)=\frac{d}{dx}\left( f(x)+f(y) \right)\implies f'(xy)y=f'(x)$
taking partial derivative with $x$

Then, consider $g(x)=\frac{1}{f'(x)}$
So $g(xy)=y g(x)\implies g(x)=x g(1)=x f'(1)\implies \frac{1}{f'(x)}=xf'(1)\implies f'(x)=\frac{1}{xf'(1)}$

Now integrating yields $f(x)=\frac{\ln x}{f'(1)}+C$, substitute $x=1$ so that $C=f(1)$, thus $f(x)=\frac{\ln x}{f'(1)}+f(1)$

Which was kind of what I was looking for....

but today I realized a much more trivial proof, notice that:
$f(xy)=f(x)+f(y)\implies e^{f(xy)}=e^{f(x)f(y)}\implies e^{f(x)}=e^{f(1)}$ and the result follows like before...[/hide]

You never said the function has a derivative! Nor what the functions domain is!

Sorry...

I thought I kind of implied it, since I came up with it while doodling in high school calculus and I wasn't so formal about it...

[quote="me@home"]Not exactly sure what ZetaX was trying to say...

[hide="but here was my idea:"]

but today I realized a much more trivial proof, notice that:
$e^{f(xy)}=e^{f(x)f(y)}\implies e^{f(x)}=e^{f(1)}$ and the result follows like before...[/hide][/quote]

 I think you want to write something else , because this is wrong !
[u] Babis[/u]


geometry
3D geometry
AMC
AIME
similar triangles

How do shadows generally work?
How do you know what they will look like.
Can someone explain in these following problems
Thanks

1. A wooden cube, whose edges are one centimeter long, rests on a horizontal surface. Illuminated by a point source of light that is x  centimeters directly above an upper vertex, the cube casts a shadow on the horizontal surface. The area of the shadow, which does not inclued the area beneath the cube is 48 square centimeters. Find the greatest integer that does not exceed 1000x. (AIME #4, 1996)

2. (I hope my memory is correct) A six feet lamp post is directly 20 feet across the street from a 2 feet by 2 feet sign.  If the light on the top of the lamp post causes the sign to cast a shadow, what is the area of its shadow? (Mandelbrot, March 3-7, 08)

[quote="AstroPhys"][hide="HINT"]The "edges" of the shadow are made of similar triangles.[/hide][/quote]

but hwo do you know what shapes are made,

(for the shape:) Ok... so consider the light. It's a point. Draw the lines from the point to the vertices of the shape (because it's convex and 'polygonal'). Where these lines hit the plane that represents the ground, mark the points. Since the original shape is 'polygonal' (well, composed of straight lines), its vertices will also be the vertices of the shadow. Connect the images of the points, and you're done.

For the AIME problem, you can notice that drawing the point of shadow and the result of shadow are similar triangles. Apply similar triangles to get hte ansewr.


8. Sebuah lantai luasnya 3 meter persegi ditutupi lima buah karpet dengan ukuran masing-masing 1 meter persegi. Buktikan bahwa ada dua karpet yang tumpang tindih dengan luas tumpang tindih minimal 0,2 meter persegi.

A floor of a certain room has a $ 3 \ m^2$  area. Then the floor is covered by 5 rugs, each has an area of $ 1 \ m^2$. Prove that there exists 2 overlapping rugs, with at least $ 0.2 \ m^2$ covered by both rugs.

PIE being the Principle of Inclusion and Exclusion, not apple or pumpkin pie  :)

For $ R_i$ the rugs, and $ |R|$ the area of rug $ R$, an inequality derived from PIE implies

$ 3 \geq \left | \bigcup_{1\leq i \leq 5} R_i \right | \geq \sum_{1\leq i \leq 5} |R_i| \minus{} \sum_{1\leq i < j \leq 5} |R_i \cap R_j| \equal{} 5 \minus{} \sum_{1\leq i < j \leq 5} |R_i \cap R_j|$.
Therefore $ \sum_{1\leq i < j \leq 5} |R_i \cap R_j| \geq 2$, and as there are $ 10$ summands, there exists one of value at least $ 0.2$.

This is the best we can do: there is a model, with mutually disjoint $ 5$ patches $ P_i$ and $ 10$ patches $ P_{\{i,j\}}$, of area $ 0.2$ each (hence covering completely the floor), such that $ R_i \equal{} P_i \cup \left ( \bigcup_{j \neq i} P_{\{i,j\}} \right )$.


The problem is as follows:

[(x*2y^-1*z^-2)/(x*y^3*z^-4)]^-3

The first thing I did was flip the fraction and made the outer exponent positive. Then I cubed every term in the fraction to get the following:

(x^3*y^9*z^-12)/(x^3*8y^-3*z^-6)

Have I done everything right so far?

From there on out, I canceled out the x's and made the negative exponents positive:

(8y^12*z^6)/(z^12).

My answer was (8*y^12)/(z^6), but my teacher told me it was wrong. Please help me realize where I messed up. :blush:

:welcome: :welcome:

$\text{\LaTeX}$-ing (there's a LaTeX tutorial on this site by clicking the link LaTeX on the left):

This is what you said:

$(\frac{x \cdot 2y^{-1} \cdot z^{-2}}{x \cdot y^3 \cdot z^{-4}})^{-3}$

$\frac{x^3 \cdot y^9 \cdot z^{-12}}{x^3 \cdot 8y^{-3} \cdot z^{-6}}$

$\frac{8y^{12} \cdot z^6}{z^{12}}$.

$\frac{8y^{12} \cdot z^6}$

Look back over your cancellations carefully.

My answer was 8y^12/z^6.

I am now thinking that the answer is y^12/8z^6.  Am I right?

Mathnerd, you made a mistake with the 8:

$\frac{x^3 \cdot y^9 \cdot z^{-12}}{x^3 \cdot 8y^{-3} \cdot z^{-6}}$

$\frac{y^{12}\cdot z^6}{8z^{12}}$

$\frac{y^{12}}{8z^6}$

[hide]Don't worry about the parentheses first, simplify the inside expression (cancelling exponents) to get $(2y^{-4}z^2)^{-3}$.

$\frac{1}{8y^{-12}z^6}$

$\frac{y^{12}}{8z^6}$[/hide]

Somewhere else I saw "no more homework problems". This was a homework problem (because of "My teacher said..."). Therefore, I deliberately didn't give the right answer but merely reiterated what she said.

So yes, I did make a mistake. It was correct the first time but I edited my solution out.


In $ ABC$ we have $ AB \equal{} 25$, $ BC \equal{} 39$, and $ AC \equal{} 42$. Points $ D$ and $ E$ are on $ AB$ and $ AC$ respectively, with $ AD \equal{} 19$ and $ AE \equal{} 14$. What is the ratio of the area of triangle $ ADE$ to the area of quadrilateral $ BCED$?

$ \textbf{(A)}\ \frac{266}{1521}\qquad
\textbf{(B)}\ \frac{19}{75}\qquad
\textbf{(C)}\ \frac{1}{3}\qquad
\textbf{(D)}\ \frac{19}{56}\qquad
\textbf{(E)}\ 1$

Hm wow this must be the easiest AMC #25 ever (note: this is my opinion only).
[hide]
Since $ A\equal{}\frac 12 ab\sin\theta$, and we know that $ \theta$ is the same in both $ \triangle ABC$ and $ \triangle ADE$,

\[ \frac{[ADE]}{[ABC]}\equal{}\frac{19 \cdot 14}{25 \cdot 42} \equal{} \frac{19}{75}.\] On the other hand, \[ \frac{[BDEC]}{[ABC]}\equal{}1\minus{}\frac{[ADE]}{[ABC]}\equal{}\frac{56}{75}.\] Therefore,

\[ \frac{[ADE]}{[BDEC]}\equal{}\boxed{\frac{19}{56} (\textbf{D})}.\]
[/hide]

Or [hide="No Trig Solution"]
Notice AE=14 and EC=28. Then the ratio of $\frac{[ABE]}{[BEC]}=\frac{1}{2}.$ Also, BD=6 and DA=19, so $\frac{[BDE]}{[ADE]}=\frac{6}{19}.$ Therefore, if we let $[BDE]=6A$, then $[ADE]=19A,[BEC]=50A.$ Since we want $\frac{[ADE]}{[BDEC]}$, we get \[\frac{19A}{6A+50A}=\boxed{\frac{19}{56}}\implies\bold{D}\]
[/hide]

Can someone help me with the No Trig Solution by adding a diagram? I don't really get it...


Determine $ f: \mathbb{R} \to \mathbb{R}$ such that: $ f(f(x)) \equal{} 4x \plus{} 3$ and $ f$ has the degree equal with $ 1$

Here is my solution:

[hide]$ f(x)\equal{}ax\plus{}b \Rightarrow a^2x\plus{}ab\plus{}b\equal{}4x\plus{}3 \Rightarrow a^2\equal{}4$ and $ ab\plus{}b\equal{}3$

If $ a\equal{}2$ then $ b\equal{}1$

If $ a\equal{}\minus{}2$ then $ b\equal{}\minus{}3$

$ \Rightarrow f(x)\equal{}2x\plus{}1$ or $ f(x)\equal{}\minus{}2x\minus{}3$[/hide]

[quote="moldovan"]... has the degree equal with $ 1$[/quote]
Does that mean that it is like $ f(x)\equal{}ax\plus{}b$, or you just assumed?

$ f(x)\equal{}ax\plus{}b$ is a condition of this problem and not an assumption. Otherwise, the problem would be more difficult.


calculus
integration
algebra
polynomial
abstract algebra
superior algebra
superior algebra unsolved

Hello,
suppose $B$ is an $A$-algebra finitely generated and $G$ is a finite group of $A$-automorphisms. Prove that $B^{G}$ (the elements of $B$ fixed by every elements of $G$) is an $A$-algebra finitely generated.

$B$ is integral over $B^{G}$ but I don't if it helps.

Thanks

It's false. Consider $A=\mathbb{Z}$ and let $B$ be the ring of polynomials with coefficients in $\mathbb{Z}$ on two noncommuting variables $x,y$. Let $G$ be the group of two automorphisms; the trivial one and $x\mapsto y,y\mapsto x$.

Since $B^{G}$ decomposes as the sum of its homogeneous pieces, it is generated by finitely many homogeneous polynomials if it is finitely generated. If $p$ is homogeneous of degree greater than 1 and is fixed by $G$, $p(1,1)$ must be even. By the example of $x^{n}+y^{n}$, $p(1,1)$ can be any even number for any degree. On the other hand, every polynomial combination of homogenous elements of $B^{G}$ which is itself homogenous of degree $n$ evaluates to something divisible by 4 at $(1,1)$. We need generators of arbitrarily high degree to generate $B^{G}$, and therefore we need infinitely many generators.

Sorry I forget something. I read again the problem and $A$ is supposed to be noetherian !
There is a problem !

I suspected noetherian as a condition, actually. That doesn't break my example; $\mathbb{Z}$ is noetherian. I could also replace $\mathbb{Z}$ with something like $\mathbb{Z}/4\mathbb{Z}$.


how do you make stickies/announcements? do you have to be a mod?
how do you make animated avatars?

I think on any forum only mods/admins can sticky things, same with announcements.
Animated avatars, use an animated GIF file.

Someone else's topic can be stickied by another moderator.

huh i dont understand.
(not really important but)
check my work plz
1. discreetfouriertransform is not a mod.
2. mods can turn existing topics into sticky/announcement form.
3. a mod turned DFT's topic into an announcement.
4. this mod is a mod of another forum but came into our forum to do 3.
5. who is teh mod?

[quote="bowei"][quote="junggi"]i think you should ask one of the admins instead of the bay area forum which is pretty much composed of spammers(me, davids, i like pie, etc ) lurkers(ccy, bookaholic) and self proclaimed stupids(ubemaya)[/quote]
peculiar. the mods are lurkers.[/quote]lies

[quote="bowei"]hey i didnt write that junggi did
anyway whats a lurker[/quote]
A lurker on a forum would be best described as...someone who lurks, i.e. not posting but reading I guess you could say. I'm not too sure if that is an accurate description as the "lurkers" don't lurk. Rather, they just don't visit :P

[quote="bowei"]hey i didnt write that junggi did
anyway whats a lurker[/quote]

They are a Zerg unit.

I do too visit. Occasionally. Okay, I guess the fact that I only just responded to a post from Jan. 30 isn't helping my case.

Meh. /me doesn't like zerg...  :(  But lurkers are pretty sweet. :)

I wonder where I go in this contorted forum social structure, anyhow... junggi? you there?

[quote="bowei"]theres a soical structure?
if so where am i

i wanna be a mod
that would be ubercool
and id pay more attention than someone
*cough cough*

anyone know how?[/quote]
Uh, get an admin to like you/think you're pro or something?

Social structure? I=non-existent ghost who can't solve any problems. New class yay!


I asked my science teacher this and he told me to wait until next year. (Translation: He doesn't know.)

If the combined mass of two deuterium atoms is less than that of one helium atom (because of the energy released), why do stars expand when they reach the red giant stage? After all, there is less mass. And helium is denser than hydrogen--isn't it?

First of all, the difference in the mass is very small.

Second thing, I don't think that the density and hence the volume of gas depends much on this fraction of mass, mainly because the mass decrease happens in the nucleus, which is much, much smaller than the atom itself. It is true that in the stars where the temperature is in order of thousands or millions of kelvins, there are on atoms, but just a kind of plasma gas (I don't know how it is called exactly).

I think the reason why stars expand so much is quite complicated. It might be because the outer regions become thinner and they get blown off with the soler wind (i.e. flow of light and particles) to further distances. (But take this just as a speculation;)

[quote]They are believed to be stars of solar mass or higher which have exhausted the supply of hydrogen in their cores and started fusing hydrogen in a shell outside the core. Since the source of energy is closer to the surface, the star begins to expand. This makes the star more luminous (from 1,000 – 10,000 times brighter) but, counterintuitively, also reduces the effective temperature. This is because the radius (and hence surface area) of the star increases by a larger amount than the luminosity of the star. As a result, the star becomes larger, but cooler and redder—hence red giant.[/quote]

Given by Wikipedia

Kubus is correct, however.  The small difference in mass would have little difference towards the size of the star

This means that stars fuse from the inside out? Of course that would make sense. So it is possible to run out of fusable elements in the core before you run out of fusable elements farther from the core, if you have a big enough star. What would happen then?

It would then have a helium flash and become a red giant, fusing heavier elements in its core, and burn successively lighter nuclei in shells


Quantities $ r$ and $ s$ vary inversely. When $ r$ is 1200, s is 0.35. What is the value of $ s$ when $ r$ is 2400? Express your answer as a decimal to the nearest thousandths.

They are inversely proportional, so their product is constant.  When $ r$ is multiplied by $ 2$, $ s$ is divided by $ 2$.  $ \frac{0.35}{2}\equal{}\boxed{0.175}$.


Sorry for my english, but English is not my native language. My question is the next:
"Let $G$ be a gruop of order $p^{n}$ where $p$ is a prime. Show that any nontrivial subgroup $U$ normal in $g$ intersects the center of $G$ nontrivially"

Thanks!!

This would be more likely to receive an answer in one of the Superior Algebra forums.  Hopefully a moderator will move it there soon, but if not you should probably re-post it there.


geometry
parallelogram
geometry unsolved

In a regular pentagon, $4$ diagonals are parallels to the $4$ respective sides. Prove that the fifth diagonal is parallel to the respective side, too.

I think the regularity of the pentagon is unnecessary!!!

sorry, the pentagon is not regular. :blush:

Assume, that in a pentagon $ABCDE$:
$AC\parallel ED$
$BD\parallel AE$
$AD\parallel BC$
$BE\parallel CD$
I'll prove that $CE\parallel AB$.
Denote:
$F=BD \cap CE$
$G=CE \cap DA$
$H=DA \cap EB$
$I=EB \cap AC$
$J=AC \cap BD$
Obviously, $CDEI$ and $BCDH$ are parallelograms, therefore $EH=BI$. Similarly, $AI=CJ$. By the Thales theorem:

$\frac{DH}{HA}=\frac{CI}{AI}=\frac{AJ}{CJ}=\frac{DJ}{JB}$

hence $HJ\parallel AB$, so

$\frac{HI}{JI}=\frac{BI}{AI}=\frac{EH}{CJ}$

This gives

$\frac{AI}{BI}=\frac{CI}{EI}$ and finaly $AB\parallel CE$.


Solve:
Let P be a point inside triangle ABC such that:
angleAPB-angleC=angleAPC-angleB
let h feet of the perpendiculars from P to BC;CA and AB be X,Y and Z respectively.
Prove that YZ=PA.sinA

At first, you don't need any condition like angleAPB-angleC=angleAPC-angleB.

Actually the problem is more or less trivial. Are you asking for a solution or do you want others to try the problem?

  Darij


Hi,
This is probably easy but I need help with this task:

a) Find function $ f\equal{}(x_1\minus{}4)^2\plus{}(x_2\minus{}3)^2$ , where $ x_1, x_2\ge 0$ and restrictions are: $ \begin{cases}
2x_1\plus{}3x_2\ge 6 \\
3x_1\minus{}2x_2\le 18\\
\minus{}x_1\plus{}2x_2\le8
\end{cases}$ min and max value using [u]task geometric interpretation.[/u]

b) Find $ max f\equal{}3x_1\plus{}4x_2$ ,where $ x_1, x_2\ge 0$ and restrictions are: $ \begin{cases}
x_1^2\plus{}x_2^2\le 25\\
x_1^2*x_2^2\ge 4
\end{cases}$ using geometric interpretation.

hello, for a) we have $ (x_1\minus{}4)^2\plus{}(x_2\minus{}3)^2\geq0$ and the equal sign holds if $ x_1\equal{}4$ and $ x_2\equal{}3$.
Further we get $ (x_1\minus{}4)^2\plus{}(x_2\minus{}3)^2\le \frac{549}{4}$ and the equal sign holds for $ x_1\equal{}13$ and $ x_2\equal{}\frac{21}{2}$.
Sonnhard.

hello, for b) we have $ 3x_1 \plus{} 4x_2\le 25$ and the equal sign holds if $ x_1 \equal{} 3$ and $ x_2 \equal{} 4$.
Sonnhard.


geometry AMC AIME USA(J)MO USAMO ARML MATHCOUNTS

Here's a quick list of Bay Area math contests (without dates; I'm not as pro as Discreet)

CAML - 6 rounds in a year.
October thru March

Mandelbrot-5 rounds individual / 3 rounds team
October-March/ Jan-March

AMC's 10/12 Feb
AIME March
USAMO April

mathleague: 6 competition, 1 finalists competition
October-May

BAMO-february

SCU high school competition - November

SMT - march

SCVMA - march

ARML - may
tryouts through mathleague?

Julia Robinson - April?

Anything i missed?

more exact dates [i]possibly[/i] coming

And for little people like me, AMC 8, MATHCOUNTS, and the middle school Mathleague stuff.

[quote="Sly Si"]Actually, SMT will probably be in February.  Also, ARML tryouts, AFAIK, are definitely via Mathleague again this year.

A few of the planned Mathleague dates have been posted to the website: September 26th (next Saturday) in San Jose, January 16th at Stanford, February 13th at Berkeley, March 6th somewhere, and the regional meet on April 24th.  Also, the league championships (the tournament Amador went to last year) are scheduled for May 14th--don't forget those if you think you might want to go.[/quote]

hmm will smt conflict with hmmt again?

[quote="bowei"]That's a ridiculous distance to travel for just a math competition. How are you going to pay for the trip?[/quote]
Apparently AAST sent a bunch of people to SMT? (my source for this is serial, btw)
So I guess it's possible.

Although the idea of a Bay Area team sweeping down socal sounds scary. And cool. But mostly scary.

[quote="bowei"]That's a ridiculous distance to travel for just a math competition. How are you going to pay for the trip?[/quote]
If you weren't aware, Taylor did try to go to HMMT 2 years ago, but didn't make it since the planes were grounded by bad weather. Also ARML is even farther, and of course it's "just a math competition" as well. Don't some of the top high schools in China send kids over to HMMT as well?

Plane tickets are actually pretty cheap, I think you can fly down for less than a hundred bucks? My dad flys a crazy amount so I can probably get free mileage plus tickets. Carpool would be like 40 bucks for gas.

Yes AAST swims in money, they flew like 50 people to SMT or something. They also went to a math competition in China...judging by Alex Zhu's backpack.

I'll add that for the Southern California contest that I help run, namely CSULB Math Day at the Beach, we will only accept teams from Southern California. So don't plan to come for that.

For the Caltech/Harvey Mudd contest, you'll probably also have to add a night in a motel (or whatever) to the cost, since they'll start pretty early in the morning on Saturday.

If money is an issue, CHMMC might be able to help. Contact us if that's the case. Keep in mind that schools and local businesses sometimes give money for teams to go to these sorts of things, so you might want to ask them first.

I went to PUMaC and HMMT in my senior year of high school while living in North Carolina, and there were teams at the Duke Math Meet from Georgia, so people do indeed travel for math contests.

[quote="serialk11r"]Will SMT conflict with CHMMC? I'm thinking about going down there for fun, if others want to come.[/quote]

I just emailed Peter Smillie and he said that they will avoid February 20 when scheduling SMT. Assuming nothing goes horribly wrong, there shouldn't be a conflict.

fyi the mathleague.org dates have changed slightly; you can check them out at http://mathleague.org/hsevents.php..

i'll send more details out when i have a complete package of information for this school year..

FYI i just received an email from the organizer of the Santa Clara contest, and he indicated a plan to host that contest on November 14.  This of course conflicts with a mathleague.org contest i informed him of about a month ago when i asked what his date was going to be.  i am attempting to work with him to resolve this situation so that students can go to both contests, but if we are unable to reach a solution i am going to ask all of you to consider going to the mathleague.org contest because it was scheduled first and only after a concerted effort was made to avoid conflicts..

Of course, to be fair, i'm also going to encourage those of you who are eligible to go to the Miller Math Marathon on January 9 or the 12th grade contest on December 12 because they were scheduled before the mathleague.org high school contests on those dates.  i only scheduled contests on those dates because the majority of students who attend mathleague.org high school contests are ineligible to compete at either of these other events, but if you are eligible for one of those two contests you should definitely attend it instead..

One more note on a contest Bowei missed - there is an organization called JACL that runs a contest that usually occurs in January.  They have not yet gotten back in touch with me regarding their date this year or how to find out more information about it..

For those asking questions about SMT, I'm no longer the one to ask, since I'm not coordinator anymore.  Peter Smillie is SUMO president, and Patrick Thill is SMT coordinator.  They--along with problem coordinator Seva Tchernov--are your new contacts on SMT, and you can find their email addresses on the SUMO webpage.


[b]Assalam-o-Alaicom
I M doing Master in Pure Mathematics from Karachi University, Pakistan.
[color=darkred]I want to doing some short courses for my career, can anyone help me?[/color][/b] :?:

There is actually a whole sub-forum on this,
http://www.artofproblemsolving.com/Forum/index.php?f=338
You should go through it and see what you like. :)

P.S. As for short courses, keep in mind that Microsoft office and its aplications, and your typing speed. These are crutial to the workplace. Reasearch on this, and you might find something. I wish you luck, and welcome to AOPS math forum! :D


here's another one I am stuck on!!!

What is the intercepts form of the equation y=-3/5x+3

I end up with x=5 and y=3 but I am stuck on how to set up the rule for intercepts with my answer

What's intercepts form?

The closest thing I can think of is slope-intercept form---which its already in.

What do you mean by rule of intercepts? x-intercept is when y=0 and y intercept is where x=0 if that's what you mean...

x/E + y/F = 1 where E = x-intercept and F = y-intercept

y = -3/5x + 3

When x = 0, y = 3.
When y = 0, x = 5.

y-intercept = 3
x-intercept = 5

x/5 + y/3 = 1

Well, there's point slope form, which is also called point intercept form.
Could the question maybe be asking for you to graph the line?