Are there infinitely many squares $a^2_1, a^2_2, a^2_3, \ldots$ such that the sums $a^2_1 + a^2_2, a^2_1 + a^2_2 + a^2_3, \ldots$ are squares too ?

Yes.  This sequence can be derived from the common formula for Pythagorean Triples:

[tex]c^2 - (c-1)^2 = 2c-1[/tex]

So we start with [tex]a_1=3[/tex],[tex]a_2=4[/tex] which we know sum to [tex]5^2[/tex].

Now to determine what to make [tex]a_3[/tex] we use [tex]5^2=2c-1[/tex] giving [tex]c=12[/tex] and use this for [tex]a_3[/tex].

As this continues we use [tex]a_n=\frac{(\sum_{i=1}^{n-1}{a_i^2})-1}{2}[/tex] which always produces an even number (since our sums starting with [tex]5^2[/tex] are always odd and congruent to 1 (mod 4).

Therefore, we can create such a sequence, starting 3,4,12,84,3612,...


ARML
AMC
AIME
geometry
3D geometry
pyramid
trigonometry

Welcome to the Third ARML Practice Series, hosted by myself (Brut3Forc3) and pythag011! At the time of this post, it's right after the AMC B date, and AIME is coming up... but that means that ARML is coming in a couple of months! Time to practice with a problem a day!

[b]Rules/Procedures:[/b]

[u]Time Limit:[/u] I or pythag011 will post a problem each day. You will have until I or pythag011 posts the solution to the problem, amounting to about and at least 24 hours, to PM a solution/answer. Use the solution thread to check which problems have solutions posted already.

[u]Calculators:[/u] No calculators, computer programs, or any other tool to do calculations for you.

[u]PMing Solutions:[/u] Please PM your answer to the person who posts the problem. Also, when PMing an answer, please put at least the day number in the subject to make grading easier. Sending a solution is not required*, (i.e., only a numerical answer is needed), but we'd like to see how different people can tackle the same problem in different ways, so it would be appreciated if you did send in solutions. Also, we will post exemplary solutions in the solution thread. (Having your solution posted will net you a total of 1 point each.)
*Unless the problem is a proof question, of course. Should there be such questions, consider them as proof round practice. Or just general proof-writing practice. :P

[u]Scoring:[/u] Problems will be worth integer values from 1-10, weighted more heavily toward the middle of this range. There may be sporadic bonus point opportunities offered for no reason. If there is a proof involved, grading will be as unarbitrary as possible.

[u]Difficulty:[/u] Each week (starting on Saturday and ending on Friday), problems will progress from easy to hard, with a general increase in difficulty as the weeks go on. But clearly, difficulties will vary based on your strengths, as well as the ability of pythag011 and myself to judge the difficulty of these problems.

[u]Other:[/u] Please refrain from posting here, unless for a good reason, such as noting a typographical error. In particular, do not post solutions here (especially before the time limit!). That's what the solution thread is for.

Stupid 10 minute limit.
I'm starting it a bit earlier than I said I would, but you'll have over 24 hours (slightly busy tomorrow), so it should be okay.
Also, though the first post mentions problems being grouped in weeks, this week will be 8 days long (explaining the numbering below), so that the final problem of each week can be on Friday.
Now, let's begin!

[b][color=green][size=150]Day 0[/size][/color][/b]

Compute $ \frac {(1337 \minus{} 7)(1337 \minus{} 13)}{(37 \minus{} 7)(37 \minus{} 13)} \plus{} \frac {(1337 \minus{} 13)(1337 \minus{} 37)}{(7 \minus{} 13)(7 \minus{} 37)} \plus{} \frac {(1337 \minus{} 37)(1337 \minus{} 7)}{(13 \minus{} 37)(13 \minus{} 7)}.$
[i]Point Value: 4[/i]

Edit: Forgot to number it Day 0, since this "week" is eight days, and also forgot point value.

[b][size=150][color=Olive]Day 1[/color][/size][/b]

Note: This problem is a proof question. You MUST submit a proof to get full credit. Partial points will be given for only giving the answer.

We have a 2009x2009 board, and a and b are real numbers. Each square of the board contains a positive real number. Also, every two squares contain distinct positive real numbers. We define an operation O on the board that changes the numbers in the board. We choose two adjacent squares, and let the numbers in those two squares be x and y. Operation O changes x to ax + by and O changes y to ay + bx. We apply operation O a finite (but positive) amount of times, and we get the original board. Also, at all times, the board does not contain a negative number. Find all possible values of a and b, with proof.

[i]Point value: 5[/i]

Also, the first few problems won't be very hard...

EDIT: Brut3Forc3 has a better rewording.

We have a 2009x2009 board, and we place 2009^2 distinct positive reals into the board, one in each square. Let O be an operation that changes two adjacent numbers x,y into ax+by,ay+bx respectively, where a,b are positive reals. Suppose we apply O to the board a finite but positive amount of times, and that while we do this, the board never contains a negative number. Find all pairs (a,b) for which this is possible, [b]with proof[/b].

[quote="CatalystOfNostalgia"]Are you asking for all (a,b) such that for any 2009x2009 board we can operate in such a way, or just there exists a 2009x2009 board? because if it's the latter, you can just start with all 0's...[/quote]

Read the part about all the numbers in the board being distinct and positive.

Also, Brut3Forc3 has a better rewording:

We have a 2009x2009 board, and we place 2009^2 distinct positive reals into the board, one in each square. Let O be an operation that changes two adjacent numbers x,y into ax+by,ay+bx respectively, where a,b are positive reals. Suppose we apply O to the board a finite but positive amount of times, and that while we do this, the board never contains a negative number. Find all pairs (a,b) for which this is possible, [b]with proof[/b].

Alright fine, screw the all 0's thing, but I still don't understand - are we looking for (a,b) such that there exists an initial board where this is possible, or for all initial boards it's possible?

Hmm, sorry. For all initial boards, it must be possible.
So now let it be:
We have a 2009x2009 board, and we [i]arbitrarily[/i] place 2009^2 distinct positive reals into the board, one in each square. Let O be an operation that changes two adjacent numbers x,y into ax+by,ay+bx respectively, where a,b are reals. Suppose we apply O to the board a finite but positive amount of times, and that while we do this, the board never contains a negative number. Find all pairs (a,b) for which this is possible, [b]with proof[/b]

Edit: AGH I totally messed up pythag's problem with my revisions. ghjk: a,b are supposed to be any reals.

Since a,b>0 and x,y are also greater than 0, then we always have (ay+bx) and (ax+by)>0. How can we finally get a negative number no matter how many times we use this transformation?

[quote="jvenezuela716"]The wordings are still not consistent... do we have to get back the original board?  :([/quote]

Well, it has to be possible to get back.

If we operate an arbitrarily large (but finite) number of times, disregarding the fact that we get the original board, do we still have all nonnegative numbers in the board? Or is this true only before we get back to the original board?

Is it necessary to show that a pair $ (a,b)$ satisfying the conditions in Brut3Forc3's version of the problem allows a positive, finite amount of applications of operation $ O$ to revert the board back to its original position? Brut3Forc3's version does not seem to require this.

[I sent my answer in already] I'm pretty sure that that was just a typo in Brut3Forc3's statement; another statement of the problem can be:

Find, with proof, all pairs $ (a,b)$, such that given [u]any[/u] $ 2009 \times 2009$ grid of distinct positive real numbers, there exists an operation $ O$ on the grid that takes two adjacent squares $ x,y$ and replaces them with $ ax \plus{} by, ay \plus{} bx$, respectively; and that after a finite number of iterations of $ O$ on the grid, we [u]can[/u] return to the original grid, where at [u]any[/u] time (in this process) all numbers of the grid must be positive.

[quote="azjps"][I sent my answer in already] I'm pretty sure that that was just a typo in Brut3Forc3's statement; another statement of the problem can be:

Find, with proof, all pairs $ (a,b)$, such that given [u]any[/u] $ 2009 \times 2009$ grid of positive real numbers, there exists an operation $ O$ on the grid that takes two adjacent squares $ x,y$ and replaces them with $ ax \plus{} by, ay \plus{} bx$, respectively; and that after a finite number of iterations of $ O$ on the grid, we [u]can[/u] return to the original grid, where at [u]any[/u] time (in this process) all numbers of the grid must be positive.

(don't take my word though until confirmed)[/quote]

That restatement works...

Meh, sorry for being kinda late.

[b][size=150][color=red]Day 2[/color][/size][/b]
If the product of two sides of a triangle with lattice points is prime, and the area of the triangle is a prime, prove that the area must be 2. (1994 ARML Power Round)
[i]Point Value: 6[/i]

[b][size=150][color=#5183AC]Day 81[/color][/size][/b]
Compute the $ x$-coordinates of all points on the line $ y\equal{}6$ that form isosceles triangles with the points $ (\minus{}5,0)$ and $ (5,0)$.
[i]Point Value: 4[/i]

[b][size=150][color=#216E21]Day 83[/color][/size][/b]
Find the number of positive integer solutions to $ n^x\plus{}n^y\equal{}n^z$ with $ n^z<2009$.
[i]Point Value: 5[/i]

Sorry that I keep forgetting... >.<
I think I will have next Wednesday be the last day I post a problem, with Thursday being the last time I post solutions. Yay, ARML is soon!

[b][size=150][color=#9FA67A]Day 84[/color][/size][/b]
Let $ A,B,C,D,E,F,G,H,$ and $ I$ be distinct, non-zero digits. If $ \frac {A\cdot B\cdot C\cdot D\cdot E}{F\cdot G\cdot H\cdot I}$ can be written as a non-repeating decimal, determine its least possible value in decimal form.
[i]Point Value: 6[/i]

[b][size=150][color=#AF2769]Day 85[/color][/size][/b]
Let $ n$ be an integer. Of all fractions $ \frac {1}{n}$, the fractional part of $ \sqrt {123456789}$ is closest to one such fraction. Compute that value of $ n$.
[i]Point Value: 8[/i]

[b][size=150][color=#3B44A6]Day 86[/color][/size][/b]
The graph of $ f(x) \equal{} x^4 \plus{} 4x^3 \minus{} 16x^2 \plus{} 6x \minus{} 5$ has a common tangent line at $ x \equal{} p$ and $ x \equal{} q$. Compute the product $ pq$.
[i]Point Value: 9[/i]

[quote="Brut3Forc3"][quote="hurdler"]do you mean that the line is tangent to the graph at $ (p,f(p))$, and $ (q,f(q))$?[/quote]
Yes. One line, two points of tangency.[/quote]

I think the reason so many people missed this problem is that they were unable to understand it, not that it was difficult.

The question is ARML 2003 I-8, and that is the exact phrasing. The table in post #47 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=207240&start=40]here[/url] puts it as the third hardest individual question of the last six years. And yes, given that the solution isn't all that long-winded once you know what you're doing, it probably is the case that a significant part of the difficulty is in understanding what the question is asking for.

Yeah, I spent a while trying to decode the problem... but it's no excuse since I tried (and failed) a calculus bash. :P
Only 3 more days! :O (Whoa I somehow made it all the way through. (Minus the like 50 days I forgot.))

[b][size=150][color=#66CBE9]Day 87[/color][/size][/b]
A triangle has side lengths 18, 24, and 30. Find the area of the triangle whose vertices are the incenter, circumcenter, and centroid of the original triangle.
[i]Point Value: 4[/i]

EDIT: nvm I clicked the wrong link on my computer. Sorry about that.

Ack. Please try to ignore the post above if you see this post...
[color=green][I put that under "hide" for you. - K.M.][/color]
Here is the problem, printed again so you don't have to look up:

[quote="Brut3Forc3"][b][size=150][color=#66CBE9]Day 87[/color][/size][/b]
A triangle has side lengths 18, 24, and 30. Find the area of the triangle whose vertices are the incenter, circumcenter, and centroid of the original triangle.
[i]Point Value: 4[/i][/quote]

[b][size=150][color=#D61771]Day 88[/color][/size][/b]
If $ 0^{\circ}<x<180^{\circ}$, and $ \cos x \plus{} \sin x \equal{}\frac{1}{2}$, then $ \tan x \equal{}\minus{}\left(\frac{p\plus{}\sqrt{q}}{3}\right)$. Compute the ordered pair of integers $ (p,q)$.
[i]Point Value: 5[/i]

[hide="hint"]$ 1 \plus{} 2\times \cos{x} \times \sin{x} \equal{} .25 \implies \sin2x \equal{} \minus{} 3/4,$ then we compute $ \tan2x$ and finally $ \tan{x}$[/hide]

[b][size=150][color=#CA1774]Day 89[/color][/size][/b]
There are three pairs of real numbers $ (x_1,y_1), (x_2,y_2)$, and $ (x_3,y_3)$ that satisfy both $ x^3\minus{}3xy^2\equal{}2005$ and $ y^3\minus{}3x^2y\equal{}2004$. Compute $ \left(1\minus{}\frac{x_1}{y_1}\right)\left(1\minus{}\frac{x_2}{y_2}\right)\left(1\minus{}\frac{x_3}{y_3}\right)$.
[i]Point Value: 10[/i]


inequalities
inequalities proposed
China
BPSQ

What are your solutions please posts them.

Here is my solution to the second one
From simmetry let $a \geq b \geq c$
$b-c,a-c,a-b \leq 1$ So
$S=\underset{cyc}{\sum}\sqrt{a+\frac{(b-c)^{2}}{4}}\leq \sqrt{a+\frac{b-c}{4}}+\sqrt{b+\frac{a-c}{4}}+\sqrt{c+\frac{a-b}{4}}\leq \sqrt{a+\frac{b-c}{4}}+2\sqrt{\frac{b+\frac{a-c}{4}+c+\frac{a-b}{4}}{2}}=\sqrt{a+\frac{b-c}{4}}+\sqrt{a+\frac{3b+3c}{2}}$
$S \leq \sqrt{a+\frac{b-c}{4}}+\sqrt{a+\frac{3b+3c}{2}}\leq 2\sqrt{\frac{a+\frac{b-c}{4}+a+\frac{3b+3c}{2}}{2}}\leq 2$
${2a+\frac{7b}{4}+\frac{5c}{4}}\leq 2=2a+2b+2c$
$\Longrightarrow 0 \leq b+3c$
Equality holds iff $a=1,b=c=0$

I proved it going from equality case.Maybe we can solve the first one with this method.Do anyone know equality case of the first inequality? :maybe:

[b]@pvthuan[/b]:Where is my mistake? :maybe:

For the second, you can assume that $a \ge b \ge c \ge 0$, then we have the following Lemma:
\[\sqrt{b+\frac{(a-c)^{2}}{4}}+\sqrt{c+\frac{(a-b)^{2}}{4}}\le \sqrt{2(b+c)+\frac{(2a-b-c)^{2}}{4}}\]

Have fun! :)

Actually, the second can have a simpler solution. Can, the first looks interesting in the equality case. How do you solve?

MrThuan, both of 2 problems we can deduce to consider the cases $(a-b)(b-c)(c-a)=0$, or $abc=0$.

Can anybody say where is my mistake? :mad:
I used following identity:
$b-c \leq 1 \Longrightarrow \sqrt{a+\frac{(b-c)^{2}}{4}}\leq \sqrt{a+\frac{b-c}{4}}$
$a-c \leq 1 \Longrightarrow \sqrt{b+\frac{(a-c)^{2}}{4}}\leq \sqrt{b+\frac{a-c}{4}}$
$a-b \leq 1 \Longrightarrow \sqrt{c+\frac{(a-b)^{2}}{4}}\leq \sqrt{c+\frac{a-b}{4}}$
So
$S=\underset{cyc}{\sum}\sqrt{a+\frac{(b-c)^{2}}{4}}\leq \sqrt{a+\frac{b-c}{4}}+\sqrt{b+\frac{a-c}{4}}+\sqrt{c+\frac{a-b}{4}}$

[quote="cefer"]Can anybody say where is my mistake? :mad:
[/quote]

There is not mistake .Your solution is nice and true . :)
Thanks for sharing it with us .

The following solution is also simple:
Squaring, we have
$\sum\left( a+\frac{(b-c)^{2}}{4}\right)+2\sum\sqrt{\left( b+\frac{(c-a)^{2}}{4}\right)\left( c+\frac{(a-b)^{2}}{4}\right)}\le 4$
Note that
$2\sqrt{\left( b+\frac{(c-a)^{2}}{4}\right)\left( c+\frac{(a-b)^{2}}{4}\right)}\le b+c+\frac{(a-b)(a-c)}{2}$
(and similarly for other permutation). Thus it suffices to prove that
$3(a+b+c)+a^{2}+b^{2}+c^{2}-ab-bc-ca \le 4$ or
$a^{2}+b^{2}+c^{2}-ab-bc-ca \le 1 .$
But this obvious since $(a+b+c)^{2}=(a^{2}+b^{2}+c^{2}-ab-bc-ca)+3(ab+bc+ca).$

[quote="ductrung"]The following solution is also simple:

$2\sqrt{\left( b+\frac{(c-a)^{2}}{4}\right)\left( c+\frac{(a-b)^{2}}{4}\right)}\le b+c+\frac{(a-b)(a-c)}{2}$
[/quote]
Nice, ductrung. This is the same proof with my lemma. :)

[quote="toanhocmuonmau"]Let $a,b,c$ be nonnegative real numbers such that $a+b+c=1$. Prove that
\[\underset{cyc}{\sum}\sqrt{a+b^{2}}\leq \frac{11}{5}\]
[/quote]
No one like it. :(

Equality doesn't occur in (a).
We can prove it using the inequality

$\sqrt{a+b^{2}}+\sqrt{b+c^{2}}\leq \sqrt{c+b^{2}}+1-c$,

which is true if assume that $c$ is beetwin $a$ and $b$.

[quote=toanhocmuonmau]Let $a,b,c$ be nonnegative real numbers such that $a+b+c=1$. Prove that
\[\underset{cyc}{\sum}\sqrt{a+b^{2}}\leq \frac{11}{5}\]
\[\underset{cyc}{\sum}\sqrt{a+\frac{(b-c)^{2}}{4}}\leq 2 \][/quote]

[img]http://s9.sinaimg.cn/middle/006ptkjAzy76saY8p8Ab8&690[/img]

 For  $a,b,c\ge0 $ , and  $a+b+c=1 , 0<\lambda\le5$ . Prove that\[ \sqrt{\lambda a+(b-c) ^2}+ \sqrt{\lambda b+(c-a) ^2}+ \sqrt{\lambda c+(a-b) ^2}\le\sqrt{2(\lambda +4)}.\]
\[\sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\le \sqrt{3}.\]


geometry
incenter
ratio
trigonometry
geometry unsolved

Let ABC a triangle, I the incenter of ABC, A'B'C' the I-cevian triangle of ABC, A"B"C" the orthic triangle of ABC and R the point of intersection of the bissector of <AB"C" with AB.
Conclusion : B'R is parallel to CC".
Any reference ? Any synthetic proof ?
Sincerely
Jean-Louis Ayme

[hide]$ \triangle{ABC}$ and $ \triangle{AB''C''}$ are similar.  It then follows that $ \frac{AB''}{AB}\equal{}\frac{AR}{AB'}$.  The value of this ratio is $ \cos{A}$.  It then follows $ B'R \perp AB$, and we're done.[/hide]

Another proof :
triangles ABC and AB"C" are similar
the quadrilateral BB'B"R is circonscriptible, BB' being a diameter of the circle
it follows that B'R and AB are perpendicular, and we're done.
jean-Louis


calculus
integration
logarithms
function
real analysis
real analysis solved

[quote="kunny"]Is $\int_0^1 $ right, vandeville?[/quote] It's indeed that. Why?

Is the given definite integral improper integral?

[quote="vandeville"]Find three reals $a,b,c$ such as $\int_1^{e} {(\ln{(t)} - a -bt -ct^2)}^2 dt$ is minimal.[/quote] I have little change the problem ($\int_1^e$ instead of $\int_0^1$)  ;).

find the projection of $ln$ on $R_2[X]$ with the usual scalar product.

Another approach, consider the function$f(a,b,c)$ in three variable $a,b,c$ and use
multivariate calculus

Find the critical points $f'_{a}=0,f'_{b}=0,f'_{c}=0$ and sudy the sign of the Hessian


LaTeX
inequalities proposed
inequalities

prove this inquality for all $a,b,c>0$ and $a+b+c=3$
\[
\frac{a}{a+bc} + \frac{b}{b+ac} + \frac{c}{c+ab} \geq 1.5
\]

[i]Edited for $\LaTeX$[/i]


Let $ a,b,c \geq0$ such that$ a \plus{} b \plus{} c \equal{} 1$
prove $ (a\sqrt {a} \plus{} b\sqrt {b} \plus{} c\sqrt {c})^2 \plus{} 1\geq 6(a^2b \plus{} ab^2 \plus{} b^2c \plus{} bc^2 \plus{} c^2a \plus{} ca^2)$

I don't think it is hard  :wink:
The inequality is equivalent to:
\[ (a\plus{}b\plus{}c)^3\plus{}(\sum a\sqrt{a})^2\ge \sum 6ab(a\plus{}b)\]
S.O.S method will kill it easily  :wink:


This is a simple question. Either you do or you don't; there's nothing more to it.

[quote="LynnelleYe"][quote="ZennyK"]Either you do or you don't; there's nothing more to it.[/quote]
I can't have absolutely no clue and not really care?[/quote]
If you don't care, then don't enter a discussion about it.

[quote="ZennyK"][quote="djimenez"][quote="The American Heritage® Dictionary of the English Language"][b]Death[/b]:The act of dying; [b]termination[/b] of life.[/quote]
But on the other hand, there is life after death, the life of everyone else who hasn't died yet!  ;)[/quote]
lol
I mean life after death for the person who dies. After you die, do you get to live again somewhere/somehow?[/quote]I understood the question, or at least that is what I think, I was just trying to be a little sarcastic.

Any way, there are, as far as I know, three different theories about "life after death", beside the non-religious as the vampires and werewolves and such things. Those are the reencarnation, resurrection and "heaven". The first two basically say that your "soul" is going to be born again in another body, but in this same world. Resurrection says that eventually you are going to come back to life and live as the person you are and live in the same world, and heaven, that you go to other level of existence, say, another world. At least those are the three I know, and I know of at least a religion for each one of them that believe on them. Honestly, even when in a romantic way I would like to say that all the escence of what I am is not going to stop and is going to last forever, the rational part says that, no matter how great you are, after you died, that was it!

nah, but i have to admit it would be weird to not... exist. but then again you wouldnt be existing for you to think it was weird, so... nvm.

Something seems wrong about just flat-out saying "I don't like God"

As for an afterlife, I dunno so I won't start discussing it. As I said in the creation of the world topic, this is also something that if you think about too much, you'll just end up confusing yourself and getting a headache. So since I have no faith or unfaith to base it on, I'll go eat some jelly beans.

[quote="yif man12"]Something seems wrong about just flat-out saying "I don't like God"[/quote]

ditto :D

i believe in life after death, but i accidently pressed no. Sorry, god!

How 'bout "I don't believe in God." instead of "I don't like God."

Maybe after we die, if we're good, we get free Crest Original (the wintermint one) toothpaste :D !

[quote="Treething"]Maybe after we die, if we're good, we get free Crest Original (the wintermint one) toothpaste :D ![/quote]sad, but possible!

[quote="yif man12"]I'll go eat some jelly beans.[/quote]

Same here.  Jelly Beans rock, especially Jelly Bellies.

yeah.  I don't believe in God, I wouldn't say I don't like God.

no, i don't believe in life after death.  sorry

I like God...

But I am most certainly Athiest, and don't intend to change my mind any time soon. There would be nothing wrong with God if he existed :) .

To me it doesn't matter. There are two possibilities after life:
1. Existence ceases to exist and you really want option 2 to occur.
2. You live for eternity and get exceedingly bored and really want option 1 to occur.

Anyway, I never think deeply about this because it really depresses me...

[quote="jb05"]I like God...

But I am most certainly Athiest, and don't intend to change my mind any time soon. There would be nothing wrong with God if he existed :) .
[/quote]

To me this doesn't make sense. How can you like God if you don't believe he exists? You can't like or dislike God if you don't believe he exists.. and I believe in a heaven.

Haha good point. But you like the [i]idea [/i]of Santa Claus, which does exist, not the actual person. Personally, I'd be pretty scared if some fat guy with a red suit came up to me and said he was Santa Claus. And I guess people could like the idea of God.. but I don't know, if you don't believe that God exists.. then you can't actually like God, you might just like or dislike the idea of God.

It appears I suceeded in baffling you. I like the idea of a god, but there is no tangible evidence that God exists. Even some religious people admit that you have to blindly follow God and not question his existence...

[quote="math92"]whoa the votes are tied 13 to 13. :starwars:[/quote]

AWESOME, STAR WARS!!!

Yeah, I think the point of religion is that you can't prove it. I believe in God, but it takes faith, and that's like the main aspect of religion (in my opinion). I think it's good to question your beliefs, and not blindly follow your religion, but have a reason for it. But you can't prove that God exists or that he doesn't exist, so it takes faith for whatever beliefs you have.

[quote="polymorphic"]I like Santa Claus but he doesn't exist  :D[/quote]

NO!!!! YOU'RE WRONG!!! SANTA IS REAL!!!! AND HE DID EAT MY MILK AND COOKIES!!!!!!!

Seriously, I agree with ambierona. Almost every religion requires some faith to follow it. It's going to be darn near impossible to find a religion that one cannot question. Faith is a good thing, if used in a correct way. To me, even though I'm not very religious, religion answers some questions that man could not answer and it also gives people a model to live by. It is a very important part of society.

It definitely is an important part of society by holding us together, but it is also very important that there are people who do not believe in it, just as we need everybody to be different. I respect religion as a whole, but honestly, they need to come up with a better way then asking you to blindly follow a never-before-seen deity.

And for those of us with very analytical minds (like many of us here have), religion is a hard and nonsensical concept to grasp.

Going back to the main question, maybe we imagine that whatever we thought was going to happen happens. Maybe devout Christians imagine that they're up in heaven with God, and maybe devout Hindus or Buddhists or whatevers imagine that they are reincarnated. You have to admit, that is a logical assumption and may persuade you to follow Christianity (actually, I would be more persuaded to be a Buddhist).

[quote="jb05"]

And for those of us with very analytical minds (like many of us here have), religion is a hard and nonsensical concept to grasp.

[/quote]

I'm confused about why you think that everyone with an analytical mind has difficulty understanding religion.  Though many very intelligent people have been atheists, agnostics, whatever (if you want a list, djiminez made one in an earlier topic) there are also many historical intellectual giants who have been religious and have written very comprehensively not only on what, but why they believe what they believe.  I would say that J. R. R. Tolkien, T. S. Eliot, C. S. Lewis, Blaise Pascal, Rene Descartes, St. Augustine, Thomas Aquinas, Louis Pasteur, Hugh Ross, Francis Schaeffer, and Michael Faraday (and this is limited only to a small number of intellectual christians, i'm sure there are many for other religions) all had analytical minds.

By the way, those of you who claim to be "agnostic," I hope you realize what the word means in an etymological sense.  "Ag" and "nos" are greek roots meaning literally "without" and "knowledge."  Those of you who make the claim that you are "without knowledge" in a religious sense, have you actually sought the answers to the questions you have about religion, or are you sitting on them?  Try reading professional apologetic arguments sometime.  You might find that religion is a bit less nebulous than you think.

I believe in existence after death and a description of judgement and salvation as set forth by the bible (justification by faith in Christ), though I think we may be surprised by its nature.  For example, I think that the archetypal descriptions of heaven and hell (pearly gates, clouds, fire and brimstone, etc.), are largely attempts to use analogy to relate the nature of both places to something that we can understand (i. e. physical beauty and pain).  The only qualities of both places that scripture is definitive about is that one is eternal isolation from, and one eternal reconciliation with God.

[quote="jb05"]

Anyway, I never think deeply about this because it really depresses me...[/quote]

i'm an atheist and it only seems logical that when people die, that's the complete final end.  it is depressing though, so for people who believe in a religion or god, good for them.  blind faith becomes a little bit hard after olympiad type math questions though :D

I do not believe in life after death, and consider most religions phenomenally creative fabrications. I admire all the "founders" of all the popular religions. After all, the stories are creative enough to meet the high standards of the people in the world today. But now that science has come and challenged so many areas which used to be government by faith, I feel that the religions need to be updated a bit, since they are getting harder and harder to believe in.

Nevertheless, I do believe in some sort of impetus which drives the world on - it could be gravity, or perpetual motion, or the natural desire to persist. Whatever it is, it's something which is crutial to the success and continuation of the world, but most importantly the human race.

Again, I respect all religions as well as the followers of all religions, but I personally look at religion more admiringly than faithfully.

[quote="ambierona"]But you like the [i]idea [/i]of Santa Claus, which does exist, not the actual person... And I guess people could like the idea of God.. but I don't know, if you don't believe that God exists.. then you can't actually like God, you might just like or dislike the idea of God.[/quote]

Or one could like God if He existed, and not dislike the idea of God if He did not exist. It's kind of the sitting on the fence option, or "playing it safe" option.

I wonder how many people clicked "Yes" in the poll even though they personally believed "No". It's amazing how logic goes out the roof when there is fear involved.  ;)

[quote="txcowgirl09"][quote="MathFiend"]I believe in life after death in that if you have accepted Jesus Christ as your Lord and Savior, you will go to Heaven where you will spend the rest of eternity.[/quote]

i believe the same thing as you! i totaly love God.[/quote]

Me too  :yup: Even if i'm only 17 year old I can hardly wait to happen that :)


[img]file:///C:/Documents%20and%20Settings/HOME/Desktop/untitled.GIF[/img]

It shows a man of 60 kg standing on alight weighing machine kept in a box of mass 30kg. The box is hanging from the pulley fixed to the ceiling through a light rope, the other end of which is held by the man himself. If the man manages to keep the box at rest, what is weight shown by the machine ? What force should he exert on the rope to get his correct weight on the machine?

there is an option "Add an attachment" in the bottom  :D

Never seen a person standing on three legs :rotfl:

Well in the given problem we have to take into account the psuedo force applied by the person when he pulls the rope.
Total wt=900N
MA of the machine=2
Effort req=450N
the man applies the given force. in return the normal force(acting upon the weighing machine) decreases as he clings to the rope
Wt shown by the machine=600N-450N= 150N = 15kg
 :)

[quote="mittal"]

Total wt=900N
MA of the machine=2
Effort req=450N
the man applies the given force. in return the normal force(acting upon the weighing machine) decreases as he clings to the rope
[/quote]

wats MA? and how did you calculate? :oops:

Sorry for using abbreviations :ninja:

MA is Mechanical Advantage of the pulley system.
A simple defination to it is Load/effort

The load is 900  Newton, It is equally divided into 2 parts in the string.(one attached to the lift and other holded by the person) thus the pulley works as a force multiplier.
In other words the man has to apply less force(effort)
=load/MA

MA=2 since two strands of string support the same load.

P.S I know I dont sound convincing   :(   and their much better alternative methods to this questions. So please wait. Someone might give you a better method. :wink:

here i have a nicer one

Pre-Requisite Knowledge:
the weighing machine measures the normal force and nothing else

let tension= T
acceleration due to gravity=g
Normal force between block and man=N

now making the free body diagram we have

now since man is applying some force this is resulting in the tension force on the man by string

so FBD is

tension upwards, normal force upward, weight downwards

now FBD of block

tension upwards, weight downwards, normal force downwards

equating above two we have

T + N = 60g
N + 30g=T

From the above two we have N=15g=150N=15kg

now for the man to have his corrected weight

he should exert a force=120N

QED.................

 :)  :wink:

sorry there is an error in the last part :P

here is the correction

now the system will begin to move with some constant acceleration

now making the FBD's we have

of man

tension upward, weight downward, normal force upward and a net resultant acceleration upwards(say a)

Note:(here the frame is non inertial so a pseudo force will act downward having the magnitude equal to 60a)

we have T=120a+60g-N

Now since man is going upward so to maintain contact block will also move upwards with the same acceleration

now FBD of block

Tension upwards, weight downwards,normal force downwards, net resultant acceleration upwards

Note:(here the frame is non inertial so a pseudo force will act downward having the magnitude equal to 30a)

we have T=N+30g+60a

but since the machine is giving correct weight, therefore the value of N=60g

putting this in above 2 we have a=3g/2 and hence T=180g=1800N

qed

 :)  :P  :wink:  :lol:

btw mittal ur soln didn't contain the second part.....??????????

[quote="Jaswinder"]hey guys pls post a reply

and tell me whether my answer is correct or not

and if u found the solns useful or not?????????? :)  :roll:  :wink:[/quote]
The tension in the string is $ 2T$. And since the box is in equilibrium and so is the man, $ 2T \equal{} (60 \plus{} 30)g \equal{}  900N.$ Thus we get $ T\equal{} 450N$. The FBD of the man will reveal $ 600N$ downwards and $ 450N$ upwards. That gives you a balance of $ 150N$ which is provided by norman reaction force of the box (or the weighing machine for that sake). Hence the weighing machine reads $ 15kg$.
[hide=" :lol: "]Our friend anurags92 has found an innovative, lightning fast way of reducing your weight. I think he should have it patented, he'll make a lot of money :first: .!!!  :lol:  [/hide]

i guess rajiv u have answered 1st part

my last post includes the answer to the second part which no one else has answered................

[quote="Jaswinder"]hey guys pls post a reply

and tell me whether my answer is correct or not

and if u found the solns useful or not??????????[/quote]

Thanks for such an elaborate reply.......

@mittal & rajeev thanks to you both too :lol:  :lol:

[quote="Jaswinder"]

now for the man to have his corrected weight

he should exert a force=120N

QED.................

 :)  :wink:[/quote]

the answer is 1800N

anurag u have not seen my last post

where i have corrected my mistake

plz see previous posts beforing leaving any comment............ :mad:

[quote="aadil"]a similar type of problem is in hc verma's concepts of physics volume 1[/quote]
PLease read the topic sub-title. It says HCV (meaning H. C. Verma) 82pg.

[quote="Rajiv"][quote="anurags92"]oh sorry........ my fault...... :oops:[/quote]
Another fault-this question should belong to one of the Physics forums, for eg, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=247504[/quote]

dont go by the name maths olympiad...all questions of PCB by indians shud belong here :huh:  :huh:

Then for what have they created separate forums {Physics, Chem, Bio, etc. }. Please share your doubts with the people of other countries especially maybe they will post even better explanations which will really help . Finally just post the links to the discussions in a topic where u could post them one by one in the same topic itself.
THis will be better.

yes....this will be much better. :D

the older tradition used to be that The Question with thei no. would be posted in the "IIT JEE"sticky and only the solutions in subsequent topics.And involvement of P,C,B forums only on heavy discussion topic.... :lol: but i dont kno 4 how long u guys can follow it  :maybe:


complex analysis
function
abstract algebra
complex analysis unsolved

Prove that the set of holomorphic functions from $ U$ open in $ \mathbb{C}$ with values that have positive real parts forms a normal family!

The sequence $ \{f_n(z)\equal{}z\plus{}n\}$ on $ \{Re(z)>0\}$ seems to be a counterexample, since every subsequence is not compactly convergent. ;)

It is true if we say that a family is normal iff each sequence has a subsequence that converges compactly uniformly to an analytic function or the function that is identically $ \infty$. In fact, much more is true. It is a famous result due to Montel that a family of analytic functions on some region is normal if no function in the family takes either of the complex values $ a$ and $ b$. It is an immediate consequence of the Schottky theorem.

Here is the easiest argument I can currently think of in this case:
Suppose that $ (f_n)$ is a sequence in the family and let $ z_0 \in U$ with $ r$ such that the closed disc $ \overline{D(z_0, r)} \subset U$. It is enough to establish that $ (f_n)$ is normal at a neighborhood of $ z_0$.

Case 1: There is a subsequence, also written $ (f_n)$, such that $ |f_n(z_0)| \leq M < \infty$ for all $ n$. Write $ f_n \equal{} u_n \plus{} iv_n$. Since $ u_n$ is positive we have that $ \frac13 u_n(z_0) \leq u_n(z) \leq 3u_n(z_0)$ for $ z \in \overline{D(z_0, r/2)}$. To see this, write $ u(z)$ with the Poisson formula, estimate the kernel and use the mean value property. Hence the sequence $ \Re f_n$ is uniformly bounded on $ \overline{D(z_0, r/2)}$, and it follows that so is the sequence $ f_n$ on, say, $ \overline{D(z_0, r/4)}$.  So in this case, we have proven that the original sequence is normal at $ z_0$.

Case 2: We have that $ f_n(z_0) \to \infty$ as $ n \to \infty$. As in Case 1 we deduce that the sequence $ \left(\frac {1}{f_n(z) \plus{} 1}\right)_n$ has a subsequence $ \left(\frac {1}{f_\mu (z) \plus{} 1}\right)_\mu$ converging uniformly to some holomorphic function $ F$ on $ \overline{D(z_0, r/4)}$. If $ F$ is not identically zero, then it has no zero, since $ \frac {1}{f_\mu (z) \plus{} 1}$ doesn't. This is a contradiction to the case premise. Hence $ F$ is identically 0, and $ f_\mu$ converges uniformly to $ \infty$.

edit: clarified case 2. Also, since $ U$ is not necessarily connected we don't quite get the definition of normal family that I said.


Two pitchers of the same size contain pure orange juice.  One pitcher is 1/3 full and the other pitcher is 2/5 full.  Each pitcher is then filled with water.  The contents of both pitchers are poured into one large container.  What is the fraction of pure orange juice in the large container?

A. 1/8 	B. 1/3	C.  3/8	D. 11/30  E.  11/15

[hide]Lets say each pitcher was 150 oz.
Pitcher A has 50oz orange juice
Pitcher B has 60oz orange juice
Pitcher A would have an added 100oz of water
Pitcher B would have an added 90oz of water
In Container A it would be 110oz orange juice and 190oz water.
Therefore, the fraction of orange juice in Container A to the total is 110/300,
or 11/30.
So our answer is D) 11/30.[/hide]


polynomial
algebra
calculus
derivative
superior algebra
superior algebra unsolved

I just read [url=http://en.wikipedia.org/wiki/Eisenstein_criterion]Eisenstein's criterion from wikipedia[/url].
The polynomial $ X^2 \plus{} X \plus{} 2$ is mentioned in the article, and because $ X$ has coefficient 1 no prime will divide 1.
So the article suggests to make the change of variable $ X\to X \plus{} 3$ and the polynomial becomes $ X^2 \plus{} 7X \plus{} 14$ and now we can
apply Eisenstein.
But 3 doesn't seem to be chosen arbitrarily.

For example if we choose any other value up to 17 it doesn't work ([b]except for 10 and 17[/b]).
How does one choose this particular [b]shift value[/b] ?
Also if you know a detailed proof of Eisenstein criterion please let me know I'd like to read it.

Thank you for taking the time to read this

[quote="billwgd"]Only primes $ p$  dividing $ \rm{resultant}(f,f') \ne 0$ can work.
For such, simply check $ f(x \plus{} a)$ for $ a \equal{} 0,1,2, ..., p \minus{} 1$.
In the above example the resultant is 7. These results are
much clearer when viewed in terms of [b]Newton polygons[/b][/quote]

So according to wikipedia the resultant is "[b]the product of the differences of their roots[/b]" for monomials(which these polynomials are).
Thanks a lot for the answer! it will help a lot but I still have a few questions about what you wrote:
1) have you denoted by f' the polynomial after X->X+3 ? or is f' derivative of f ?
2) how did you calculate the resultant if the first polynomial is irreducible and has no roots in R ?
3) the $ p$ you mention, how do I compute that ?
4) a good read for Newton polygons ?

I thought a bit more.
The change of variable should be $ X\to X \plus{} n$
(let's denote with $ a_i$ coefficients of the polynomial after the variabla change, and the polynomial after variable change is $ g$)
where n is the smallest natural number for which $ gcd(a_0,a_1,...,a_{deg(g) \minus{} 1})\neq 1$ so that the criteria can be applied
(if p is the one in the wikipedia statement of Eisenstein and such a p exists then it will divide the above mentioned gcd).

[quote="billwgd"]1) Yes, $ f'$ denotes the derivative of $ f$
2) The resultant can be computed by elimination or extended GCD. Google it.
3) p denotes a prime dividing $ \rm{discriminant}(f) : \equal{} \rm{resultant}(f,f')$
4) Google Newton polygon[/quote]

1-3) thanks :)
4) Google gives a lot of results of which many have nothing to do with Eisenstein.
I was interested where you have read about Newton polygons and their relationship to Eisenstein's criterion.
It doesn't seem very obvious to me.

Perhaps you need to hone your searching skills. I find many of interesting results
searching on: Eisenstein "Newton polygon" via Google {Web, Scholar, Books}.

Your 2nd-last reply doesn't make any sense. See my initial post for a simple algorithm.

[quote="billwgd"]Only primes $ p$  dividing $ \rm{resultant}(f,f') \ne 0$ can work.
For such, simply check $ f(x \plus{} a)$ for $ a \equal{} 0,1,2, ..., p \minus{} 1$.
[/quote]
what if the resultant is 1 ? There's no prime dividing that, can we still shift ?

No, as I said, for the prime $ p$ to possibly succeed we  must have $ p\;|\;\rm{discriminant}(f) \ne  0$.


calculus
integration
trigonometry
LaTeX
calculus computations

$ \frac{5x^2}{x^2\plus{}x\plus{}1} \equal{} 5 \minus{} 5 \frac{x\plus{}1}{x^2\plus{}x\plus{}1} \equal{} 5 \minus{} \frac52 \frac{2x\plus{}1}{x^2\plus{}x\plus{}1} \minus{} \frac52 \frac{1}{\left(x \plus{} \frac12\right)^2 \plus{} \frac34}$

[quote="Carcul"]$ \frac {5x^2}{x^2 \plus{} x \plus{} 1} \equal{} 5 \minus{} 5 \frac {x \plus{} 1}{x^2 \plus{} x \plus{} 1} \equal{} 5 \minus{} \frac52 \frac {2x \plus{} 1}{x^2 \plus{} x \plus{} 1} \minus{} \frac52 \frac {1}{\left(x \plus{} \frac12\right)^2 \plus{} \frac34}$[/quote]
[hide]
$ \int\frac {5x^2}{x^2 \plus{} x \plus{} 1}dx \equal{}5\int dx \minus{} \frac52 \int\frac {2x \plus{} 1}{x^2 \plus{} x \plus{} 1}dx \minus{} \frac52 \int \frac {1}{\left(x \plus{} \frac12\right)^2 \plus{} \frac34}dx$

$ \int\frac {2x \plus{} 1}{x^2 \plus{} x \plus{} 1}dx \\
\\
\int\frac {d(x^2\plus{} x\plus{}1)}{x^2 \plus{} x \plus{} 1}\equal{}ln|x^2 \plus{} x \plus{} 1|$

$ \int \frac {1}{\left(x \plus{} \frac12\right)^2 \plus{} \frac34}dx\equal{}\frac{2}{\sqrt 3}\tan^{\minus{}1}\frac{2x\plus{}1}{\sqrt 3}$

by using $ \int \frac {1}{u^2 \plus{} a^2}du\equal{}\frac{1}{a}\tan^{\minus{}1}\frac{u}{a}$

$ \mathbb{ANS.}\equal{}5x\minus{}\frac52 ln|x^2 \plus{} x \plus{} 1|\minus{}\frac52.\frac{2}{\sqrt 3}\tan^{\minus{}1}\frac{2x\plus{}1}{\sqrt 3}$

$ \equal{}\boxed{5x\minus{}\frac52 ln|x^2 \plus{} x \plus{} 1|\minus{}\frac{5\sqrt3}{3}.\tan^{\minus{}1}\frac{\sqrt3 (2x\plus{}1)}{ 3}\plus{}C}$
  [/hide]
 :lol:  :D

[quote="polandbear193"]Integrate:
5x^2
---------  dx
x^2+x+1

What I tried doing:
Take out the 5, then do long division:
5*int(   1-(x+1)/(x^2+x+1)   dx)
Complete the Square for the denominator:
5*int(   1-(x+1)/[ (2x+1)^2/4 + 3/4]   dx)

Now what?[/quote]
Sorry,My guy . you should learn Latex to write your formulae ,your stated method is too guly to read.. :P

[quote="Carcul"]$ \frac {5x^2}{x^2 \plus{} x \plus{} 1} \equal{} 5 \minus{} 5 \frac {x \plus{} 1}{x^2 \plus{} x \plus{} 1} \equal{} 5 \minus{} \frac52 \frac {2x \plus{} 1}{x^2 \plus{} x \plus{} 1} \minus{} \frac52 \frac {1}{\left(x \plus{} \frac12\right)^2 \plus{} \frac34}$[/quote]
Hello
For
$ \int\frac{ax^2\plus{}bx\plus{}c}{px^2\plus{}qx\plus{}r}dx$
use this
$ ax^2\plus{}bx\plus{}c\equal{}\alpha(px^2\plus{}qx\plus{}r)\plus{}\beta(2px\plus{}q)\plus{}\gamma$

Find value of $ \alpha,\beta,\gamma$.
Thank u.


How many positive perfect squares less than 300 are multiples
of 9?

if the number squared is a multiple of three, it's a multiple of nine
3^2 = 3*3
6^2 = 3*3*2*2
9^2 = 3*3*3*3
and so on...
12^2
15^2
18^2 pasts 300 so there are 5

Since 3 squared is 9, any number divisible by 3 and has a perfect square under 300 will work. 3,6,9,12, and 15 all work, so there are 5.


integration
inequalities
inequalities proposed

$ t\ge0$ and $ x_n \equal{} \frac {1 \plus{} t \plus{} t^2 \plus{} ... \plus{} t^n}{n \plus{} 1}$

Prove that $ x_1\le\sqrt {x_2}\le \sqrt [3]{x_3}\le...\le\sqrt [n]{x_n}$

[quote="Arrigo-Sacchi"]The following proof is taken from page 430 of Andreescu-Dospinescu's "Problem from the book" (very nice book). $ x_n \equal{} {1\over t \minus{} 1} \int_1^t z^n dz \equal{} {1\over 1 \minus{} t} \int_t^1 z^n dz.$ The inequality becomes (use either of them according to $ t > 1$ or $ t < 1$)   a particular case of
$ \Bigl({1\over b \minus{} a} \int_a^b f^k(x) dx \Bigr)^{1/k} \le \Bigl({1\over b \minus{} a} \int_a^b f^{k \plus{} 1}(x) dx \Bigr)^{1/(k \plus{} 1)}$ with
$ k\ge1$ and $ f$ nonnegative but this is the Power Means Inequality for integrals.[/quote]

You're right!!!
I'm reading [b]Problem from the book[/b] right now :lol:


When I make parentheses around fractions, they turn out like this:

$ (\frac {1}{2})$

How can I make the parentheses "fit" better?

EDIT: Thanks, Dojo!

Try this:
$ \left(\dfrac{1}{2}\right)$
[code]$\left(\dfrac{1}{2}\right)$[/code]

More on this: \left and \right work with other symbols like [ ] | { } as well (you need \{ or \} for brackets). Each \left must be paired with a \right; use \left. or \right. to get an invisible version:
\left. \frac1x \right|_1^2 gives $ \left. \frac1x \right|_1^2$

or if you're too lazy to pair them, or you need to do something like this:

$ x^2\bigg|_1^2$

which won't look as nice (in my opinion) with \left and \right:

$ \left.x^2\right|_1^2$

you can use \big, \Big, \bigg, and \Bigg

$ \big|\Big|\bigg|\Bigg|$


there is a circle with radius 5096. There is a square inscribed in it, and a circle is inscribed in that square, and in that circle, another square is inscribed in it, and in that square, another one is inscribed in it. An isosceles right triangle is inscribed into that one. What is the area of the original circle that is outside of the isosceles right triangle?

[hide="Not sure of my answer"]Let's take it a step by step. So the diameter of the first circle will be the daigonal of the square. So the side length of the square will be $2548\sqrt{2}$. Thats also the diameter of the inscribed circle, and that will be the diagonal of the inscribed square. So $2548$. That will be the diameter of the circle, and that is the circle which we want to find the area of. So half and square it to get $1274^{2}=1623076$[/hide]


function
algebra
polynomial
number theory
prime numbers
Other problems

Hello guys!
It is a very fun problem. I am thinking on it when studying some Compilers-and-related books.

First, to the sake of completeness, I will define some things and so go to the problem. Click inside if you really don't know the subject.
[hide]
A Finite State Machine is a 5-tuple $ (\Sigma, Q, q, F, \delta)$, where
$ \Sigma$ is a set of symbols (here called letters of an alphabet)
$ Q$ is a set of another type of symbols (here called states)
$ q$ is a starting state (and of course it is one of $ Q$)
$ F$ is the ending states (and of course all of them are of $ Q$; the starting state can be a ending state, too; there is no a-priori restrictions on final states)
$ \delta$ is a function, $ \delta: Q \times \Sigma \rightarrow Q$.

Now we will define a valid string:
A string is just an array of letters. A finite state machine can process a string in the following way:
1 - Initially it chomps the first letter of string, and applies the function $ \delta$ to the starting state and the chomped symbol, and so compute the next state.
2 - And so on: it chomps the currently first symbol of string, and applies the function to the current state, and so changes the current state.
3 - It stops when ther is no more symbols to chomp.
4 - If the last state (after the automaton has stopped) is a final state, then the string is valid. Else, it is invalid.

As an example, I will build an automaton to recognize an even number in base 10 (left zero allowed, I am very sloth...).
$ \Sigma\equal{}\{0,1,2,3,4,5,6,7,8,9\}$
$ Q\equal{}\{e,o\}$
$ q\equal{}o$
and $ \delta$ is defined this way:
$ \delta(x,E) \equal{} e$
$ \delta(x,O) \equal{} o$
where $ E \in \{0,2,4,6,8\}$ and $ O\equal{}\{1,3,5,7,9\}$ and $ x$ is any state.

Well it is just a quick-and-dirty definition. See
[url=http://en.wikipedia.org/wiki/Finite_state_machine]this Wikipedia link[/url] for more (clean) info.

[/hide]
The question is:
We can build a finite state machine to validate the primes in base 10, and only them?

No.  Use the [url=http://en.wikipedia.org/wiki/Pumping_lemma_for_regular_languages]pumping lemma[/url].

[b]Edit:[/b]  Actually this argument isn't completely obvious, so I guess it's worth giving.  Suppose by contradiction that the language $ L$ consisting of all strings on the alphabet $ \{ 0, 1, ... 9 \}$ which are prime numbers is regular and let $ l$ be its pumping length.  Let $ p$ be a prime with more than $ l$ digits and consider its decomposition $ xyz$ into three substrings satisfying the conditions of the pumping lemma.  I claim that some string of the form $ xy^i z$ is a composite number, which provides the desired contradiction.

Suppose $ x$ has length $ a$, $ y$ has length $ b$, and $ z$ has length $ c$.  Then (the LHS is concatenation notation whereas the RHS is ordinary arithmetic)

$ s_i \equal{} xy^i z \equal{} x \cdot 10^{ib \plus{} c} \plus{} y \cdot (10^c \plus{} 10^{b \plus{} c} \plus{} ... \plus{} 10^{(i \minus{} 1)b \plus{} c} ) \plus{} z$.

Then there exist rational numbers $ u, v$ such that $ s_i \equal{} u \cdot 10^{ib} \plus{} v$.  By Fermat's little theorem, $ s_{s_1}$ is divisible by $ s_1$.

There is also a more conceptual, but more difficult, argument:  any language recognized by an automaton has exponential-polynomial asymptotics, which the sequence of primes of a certain length does not.  This is a trivial corollary of the PNT.

And if you were curious, a similar argument using the pumping lemma for context-free languages shows that the prime language $ L$ is also not context-free.


Diophantine equation
continued fraction
number theory unsolved
number theory

I read that the following problem (and generalizations of this problem) could be solved using continued fractions.  Does anyone know of how this is done or have a reference?

"Imagine that you are on a street with houses marked $ 1$ through $ n$. There is a house in between ($ x$) such that the sum of the house numbers to left of it equals the sum of the house numbers to its right. If $ n$ is between $ 50$ and $ 500$, what are $ n$ and $ x$?"

Thanks.

Basically, one needs to solve an equation:
\[ 1 \plus{} 2 \plus{} \dots \plus{} x \minus{} 1 \equal{} (x \plus{} 1) \plus{} (x \plus{} 2) \plus{} \dots \plus{} n
\]
or
\[ (2n \plus{} 1)^2 \minus{} 8x^2 \equal{} 1
\]
which is a Pell equation and can be solved using continued fraction of $ \sqrt {8}$.
See http://mathworld.wolfram.com/PellEquation.html for details.

Oh yeah, thanks for the reply.  But shouldn't that be $ \left(2n\plus{}1\right)^{2}\minus{}8x^{2}\equal{}1$?


algebra
polynomial
limit
integration
algebra proposed

Let $s$ be a real number such that $0 < s \leq 1$. Does there exist polynomials $ P(x),Q(x)\in \mathbb R[x]$ such that
\[\frac{P(n)}{Q(n)}=\sum_{i =1}^{n}\frac{1}{i^{s}},\]
holds for all $n\in \mathbb N$?

[hide="A hint"]$ \sum_{i\equal{}1}^{n}\frac{1}{i^{s}}\to\plus{}\infty$ when $ n\to\infty$.[/hide]

[quote="TTsphn"]$ 0 < s \equal{} < 1$
Dose there exist $ P(x),Q(x)\in R[x]$
$ \frac{P(n)}{Q(n)}\equal{}\sum_{i \equal{} 1}^{n}\frac{1}{i^{s}}$,for all $ n\in N$[/quote]

Let $ a_{p}x^{p}$ be the highest term of $ P(x)$ and $ b_{q}x^{q}$ be the highest term of $ Q(x)$ (with $ a_{p}\neq 0$ and $ b_{q}\neq 0$)

Then $ \lim_{n\to\plus{}\infty}n^{q\minus{}p}\frac{P(n)}{Q(n)}\equal{}\frac{a_{p}}{b_{q}}$

We have : $ \int_{k}^{k\plus{}1}\frac{1}{x^{s}}dx<\frac{1}{k^{s}}<\int_{k\minus{}1}^{k}\frac{1}{x^{s}}dx$ $ \forall k>1$

And so : $ 1\plus{}\int_{2}^{n\plus{}1}\frac{1}{x^{s}}dx<\sum_{i \equal{} 1}^{n}\frac{1}{i^{s}}<1\plus{}\int_{1}^{n}\frac{1}{x^{s}}dx$

If $ s\equal{}1$ : $ 1\plus{}\ln(n\plus{}1)\minus{}\ln(2)<\sum_{i \equal{} 1}^{n}\frac{1}{i^{s}}<1\plus{}\ln(n)$

Or, if $ 0<s<1$ : $ 1\plus{}\frac{(n\plus{}1)^{1\minus{}s}\minus{}2^{1\minus{}s}}{1\minus{}s}<\sum_{i \equal{} 1}^{n}\frac{1}{i^{s}}<1\plus{}\frac{n^{1\minus{}s}\minus{}1}{1\minus{}s}$

And so :
And, if $ s\equal{}1$ : $ n^{q\minus{}p}(1\plus{}\ln(n\plus{}1)\minus{}\ln(2))<n^{q\minus{}p}\sum_{i \equal{} 1}^{n}\frac{1}{i^{s}}<n^{q\minus{}p}(1\plus{}\ln(n))$

Or, if $ 0<s<1$ : $ n^{q\minus{}p}\frac{(n\plus{}1)^{1\minus{}s}\minus{}2^{1\minus{}s}\plus{}1\minus{}s}{1\minus{}s}<n^{q\minus{}p}\sum_{i \equal{} 1}^{n}\frac{1}{i^{s}}<n^{q\minus{}p}\frac{n^{1\minus{}s}\minus{}s}{1\minus{}s}$

And so, since $ 1\minus{}s\neq p\minus{}q$ is $ s<1$, $ \lim_{n\to\plus{}\infty}n^{q\minus{}p}\sum_{i \equal{} 1}^{n}$ can only be $ 0$ or $ \plus{}\infty$ and is always different from $ \frac{a_{p}}{b_{q}}\equal{}\lim_{n\to\plus{}\infty}n^{q\minus{}p}\frac{P(n)}{Q(n)}$

And so $ \frac{P(n)}{Q(n)}\equal{}\sum_{i \equal{} 1}^{n}\frac{1}{i^{s}}$,for all $ n\in N$ is impossible.

Other (Quite simple)
$ \sum_{i \equal{} 1}^{\plus{}\infty}\frac{1}{i^{s}}\to\ \plus{}\infty$
$ \frac{1}{n}(\sum_{i \equal{} 1}^{n}\frac{1}{i^{s}})\to 0$


geometry
trigonometry
cyclic quadrilateral
geometry proposed

problem 3
Suppose ABCD is a cyclic quadrilateral inscribed in a circle of center O and the diagonal is perpendicular
prove that a quadrilaterals AOCD and AOCB have the same aire

[quote="mathmanman"]It's french for "area". ;)[/quote]
OK, thanks.

Let X be the diagonal intersection and R the circumradius.

$\angle AOB = 2 \angle ACB \equiv 2 \angle XCB = 2 (90^\circ - \angle XBC) \equiv$

$\equiv 2(90^\circ - \angle DBC) = 180^\circ - \angle DOC$

The areas of the triangles $\triangle AOB, \triangle DOC$ are therefore equal:

$|\triangle AOB| = \frac{R^2}{2} \sin \angle AOB = \frac{R^2}{2} \sin \angle DOC = | \triangle DOC|$

Similarly, the areas of the triangles $\triangle BOC, \triangle AOD$ are also equal. Hence,

$|AOCD| = |\triangle AOD| + |\triangle DOC| = |\triangle BOC| + |\triangle AOB| = |AOCB|$

[quote="yetti"]Let X be the diagonal intersection and R the circumradius.

$ \angle AOB \equal{} 2 \angle ACB \equiv 2 \angle XCB \equal{} 2 (90^\circ \minus{} \angle XBC) \equiv$

$ \equiv 2(90^\circ \minus{} \angle DBC) \equal{} 180^\circ \minus{} \angle DOC$

The areas of the triangles $ \triangle AOB, \triangle DOC$ are therefore equal:

$ |\triangle AOB| \equal{} \frac {R^2}{2} \sin \angle AOB \equal{} \frac {R^2}{2} \sin \angle DOC \equal{} | \triangle DOC|$

Similarly, the areas of the triangles $ \triangle BOC, \triangle AOD$ are also equal. Hence,

$ |AOCD| \equal{} |\triangle AOD| \plus{} |\triangle DOC| \equal{} |\triangle BOC| \plus{} |\triangle AOB| \equal{} |AOCB|$[/quote]

Hi Yetti your proof is correct if O is inside te quadrilater , what happens if O is outside it , i think in that case , AOCD won't be convex anymore and we'll have rather :  [AOCD] = [APD] + [DOC] such that P is the point of intersection of (DC) and (AO) , can you tell how to deal with that .


Find all functions $f: R-{0} to R$ which satisfies
$xf(x+\frac{1}{y})+yf(y)+\frac{y}{x}=yf(y+\frac{1}{x})+xf(x)+\frac{x}{y}$ for all x,y in $R-{0}$.

hey!cant anyone solve this problem! :( or you just dont want to write it down!
I had 4,5 hours for 4 problems together with this! ;)


I just want to announce that I just wasted an hour of my life playing [url=http://www.kongregate.com/games/arawkins/dolphin-olympics-2]Dolphin Olympics 2[/url].

If you would like to waste one or more hours of your life playing a fun and highly addictive flash game, this might be a good choice!

[quote="cf249"]That game's pretty good, but I think Castle Wars is better.[/quote]
I haven't even played castle wars, but just by the name, all i have to say is:
you can't compare them.

I think oranges are better than chess (extreme example, but yeah).

It's ok I guess, my record was like 5,900,000 or something. I only got it once, i can't get it again. Maybe i just read my score wrong.

Just got 2 million. The game is too easy.

2 million isn't a particularly phenomenal score  :wink:

Have you seen the Diner?

If 2 million's not good, then I must be horrible.  I haven't even got 1 million yet.  :(

Haha I tried this out like last month but I never bothered to post. It's a fun game, and quite addicting.

Citing the relevent XKCD:

[img]http://imgs.xkcd.com/comics/flash_games.png[/img]

I don't get it...I suck at this.

I only got 40K points...

Haha, unfortunately, I only wasted half an hour playing this with a maximum of 800000 points or so. :P

I got over 12 million the other day!
How high in space have you gotten to?  My highest is 263 meters.


logarithms
function
algebra unsolved
algebra

Consider a positive real number $a$, with $a\ne 1$. How many solutions have the equation $log_a {x}=x$?

[quote="shobber"][quote="José"][quote="hydro"]it has exactly one solution (because both the functions are monotonic) for $0<a<1$, and it has no solutions for $a>1$[/quote]
got proof hydro???[/quote]
Draw their graphs.[/quote]

I see.... can anybody post them?

1. I am sorry, for 0<a<1 i put that $log_a (x)-x$ decrease, but put mistake signs near x=0 for this expression.
For case 0<a<1 one solution.

Indeed, for $0<a<1$ there is one solution. For $1<a<e^{\frac{1}{e}}$ there are two solutions, for $a=e^{\frac{1}{e}}$ there's only one solution, namely $x=e$, and for $a>e^{\frac{1}{e}}$ there are no solutions.


Fin $ x,y \in N$ such as $ \frac{1}{x}\plus{}\frac{1}{xy}\plus{}\frac{1}{y}\equal{}1$

Let $ x\ge y$. That means: $ \frac{1}{xy}\le\frac{1}{x}\le\frac{1}{y}$, so $ \frac{3}{y}\ge 1$.
It follows that $ y\le 3$.
Now it's easy.

$ (x,y)\equal{}{(3,2),(2,3)}$

Alternatively, multiply both sides by $ xy$ to get $ xy\minus{}x\minus{}y\equal{}1$.
Adding $ 1$ to both sides, we have $ (x\minus{}1)(y\minus{}1)\equal{}2$, and it is easy to see that $ (3,2);(2,3)$ are the only solutions.

[quote="balan razvan"]Now try this one $ \frac {1}{x^2} \plus{} \frac {1}{xy} \plus{} \frac {1}{y^2} \equal{} 1$[/quote]
[hide]I assume that x,y belongs to N*.
Assuming that x=<y, we have: 3/y^2>=1, so y^2=<3.Therefore, y=1.
From there we have: 1/x^2+1/x+1=1, which doesn't have solutions for x>0.
Therefore, this equation is inconsistent.
[/hide]

[hide="second"]Since $ 1<x,y$, $ max(\frac{1}{x^2}\plus{}\frac{1}{xy}\plus{}\frac{1}{y^2})\equal{}\frac{1}{2^2}\plus{}\frac{1}{2^2}\plus{}\frac{1}{2^2}\equal{}3/4<1$. Thus the equation has no solutions in positive integers.[/hide]


Let f be a linear function with the properties that f(1) greater then or equals f(2), f(3) less then or equals f(4) and f(5)=5. Which of the following statements r true?

a,   f(0) < 0
b,   f(0) = 0
c,    f(1) <  f(0) < f(-1)
d,   f(0) = 5
e,   f(0) > 5

[hide]
$f(1)\ge f(2)\\ f(3)\le f(4)$

Because this function is linear, it's either always increasing, always decreasing, or never changing. Assume $f(1)>f(2)$. Then, because of it's linearity, $f(3)>f(4)$, which cannot be true according to what was given. Therefore,

$f(1)=f(2)=f(3)=f(4)=f(5)=5$

So it follows that $f(x)=5$ is our function. So, $\boxed{D}$ is our answer.
[/hide]

-BLt


We can only discuss $ f$ defined on $ R^\plus{}$.

1) $ f(y)>y,\forall y>0$.
Suppose $ f(y)<y$. $ x\equal{}y\minus{}f(y)\Rightarrow f(x\plus{}y)\equal{}0$. Contradiction. If $ f(y)\equal{}y$, we have $ f(y)\equal{}0$. Contradiction!

Define $ g(y)\equal{}f(y)\minus{}y$. The equation becomes
$ g(x\plus{}g(y))\equal{}g(x)\plus{}y,\forall x>y>0$  (*)

2) $ g(g(y))\equal{}y$
We assume $ x$ is sufficiently large.
For $ a>b>0$,
$ g(x\plus{}g(a\plus{}g(b)))\equal{}g(x)\plus{}a\plus{}g(b)$,
$ g(x\plus{}g(a\plus{}g(b)))\equal{}g(x\plus{}g(a)\plus{}b)\equal{}g(x\plus{}b)\plus{}a$.
Hence, $ g(x\plus{}b)\equal{}g(x)\plus{}g(b)$.
Now we look back (*).
$ g(x)\plus{}y\equal{}g(x\plus{}g(y))\equal{}g(x)\plus{}g(g(y))\Rightarrow g(g(y))\equal{}y$.

3) $ g(x\plus{}y)\equal{}g(x)\plus{}g(y),\forall x,y>0$
By induction and (*), we have for $ a>b>0$,
$ g(a\plus{}ng(b))\equal{}g(a)\plus{}nb$, where $ n$ is natural.
Hence,
$ g(a\plus{}ng(b)\plus{}y)\equal{}g(a\plus{}y)\plus{}nb$ (**)

Since $ g$ is surjective from 2), there exists $ g(u)\equal{}y$. Again by 2), $ g(y)\equal{}u$.
We can always make $ a\plus{}ng(b)$ large by choosing large $ n$.
$ g(a\plus{}ng(b)\plus{}y)\equal{}g(a\plus{}ng(b)\plus{}g(u))\equal{}g(a\plus{}ng(b))\plus{}u\equal{}g(a)\plus{}nb\plus{}g(y)$. (***)
Compare (**) and (***),
$ g(a\plus{}y)\equal{}g(a)\plus{}g(y)$, where $ a,y>0$ are arbitrary

Cauchy equation and $ g(x)>0$ imply linearity.
$ f(x)\equal{}2x$.

I wish there is a shorter proof.


For some reason, the question is not parsing properly in this textbox. I've added a txt attachment. thanks!

It's the dollar signs -- they're used on the forum to create LaTeX codes.  (See [[LaTeX]] and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 for info.)  Get rid of the dollar signs and it will be fine.

[quote]The price for a bet on New York Yankees winning is USD79.97; the price for a Philadelphia Phillies winning is USD20.03 . Securities for the winning team pay off USD100.

(a) At this point, the Phillies will win only if they win both the next two games (games 6 and 7). Suppose you believe that there is a 60% chance of them winning game 6; if there game 7 is held, you believe there is a 50% chance of the Phillies winning it. Given these beliefs, what is the expected value of a Philadelphia Phillies bet (right now)?

(b) Given the same beliefs as in part (a), suppose that you bought one Phillies bet at the market price of USD20.03. What would your expected profit be?[/quote]

What part of the problem is giving you trouble?

both parts...I'm not sure how to proceed


I need to know how to use opening quotes in LaTeX. In my output pdf file they either don't show up or come up with a completely different symbol

I looked in TeXnic help and it said to use this

[code]$\"$[/code]

But that doesn't exactly work

Using ``Whatever text here'' hopefully does the trick.


hi all,

let $f:\mathbb{N}-\{0\}\to\mathbb{R}$ so that
(1) $f$ is increasing
(2) $f(1)=1$
(3) $f(mn)=f(m)f(n)$ if $m$ and $n$ are coprime

Let $p$ be a prime number and $x_0 >0$.
compute: \[ \inf_{x>x_0} \frac { f(p+x)}{f(x)}. \]

Hmmm...sounds strange.
A famous result of Erds (another one  :P ) states that if $f$ is a multiplicative and non-decreasing function from $N^*$ onto $R$ then $f$ is identically $0$ or  there exists a nonegative real number $\alpha$ such that $f(n) = n^{\alpha}$ for all $n$.

Since in our case we have $f(1)=1$, it follows that $f(n) = n^{\alpha}$ for some $\alpha \geq 0$.
Thus, $\frac {f(x+p)} {f(x)} = (1+ \frac p x )^{\alpha}$, and the inf is therefore equal to $1$. But, I do not see what the hypothesis $p$ is prime is here for...

Pierre.

i don't know but perhaps if we assume that p is prime it is not nescessary to use this result by Erdos ..

but what the proof of this result ? It's intersesting ?

Well, if you are trying to do it without that result, perhaps you could try to find a sequence of $x_n$ such that $\frac{f(x_n+p)}{f(x_n)}\to1$, since it's clear that the inf must be at least 1 (since the function is increasing).  Can you use the given properties to show, for example, that $\lim\frac{f(a_n+1)}{f(a_n)}=1$ for some sequence $a_n$?

Come on Alekk! Look at 'Mgamath', ex. An-13  :D

Pierre.

[quote="alekk"]let $f:\mathbb{N}-\{0\}\to\mathbb{R}$ so that
(1) $f$ is increasing
(2) $f(1)=1$
(3) $f(mn)=f(m)f(n)$ if $m$ and $n$ are coprime
Let $p$ be a prime number and $x_0 >0$.
compute: \[ \inf_{x>x_0} \frac { f(p+x)}{f(x)}. \][/quote]
A strange problem...
We know that $\inf \geq 1$. Suppose $\inf=a>1$. Then $\frac{f(x+pk)}{f(x)}\geq a^k$. Take now $x$ sufficietly large odd number, $pk<x$ and $a^k>f(2)$. We have $a^k\leq \frac{f(x+pk)}{(x)}<\frac{f(2x)}{f(x)}=f(2)$ -- contradiction.  :?


prove that $3^n+3^m+1$ is never a perfect spuare for integer $(n,m)$.

[quote="solafidefarms"][hide]We observe that in mod 8 all squares are 0,1,or 4. We observe that $3^n$ and $3^m$ are always 1 or 3 mod 8. There are 3 possibilities for the unordered set $(3^n,3^m)$: (1,1),(1,3), and (3,3).

In the first case the sum of the RHS is 3, which is not equal to 0, 1, or 4.

In the second case the sum of the RHS is 5, which again is not equal to the LHS.

In the last case the sum of the RHS is 7, again not possible.[/hide]

I remember this problem from Zeitz :)

Billy[/quote]
Grr, I was looking mod 4!!!

That's what I did too to start with, till I found it didn't work. Remember, mods 4 and 8 work well for squares. Can't remember what works for cubes... mod 5 maybe?

i think that mod 9 works well for cubes if i remember.
[hide="here is what i did"]
i first used parity (odd/even) to prove that they both could not be even or odd. i did this by looking at it in mod 4. then i assumed them to be one even and one odd and then viewed it in mod 8 (as suggested above).
it took me like 5min. before i tried mod 8. i was trying mod 4 for the longest time.  :rotfl: [/hide]

just a tip, but for any exponent $n$, $n+1$ or $2n+1$ work really well if either one of them is a prime; even if they aren't, $2n$ can work ok sometimes.

For squares just use as high a power of $2$ as you need for the mod.

For an exponent $n$, $\phi^{-1} (d)$ tends to work where $d | n$ :)

what is $\phi^{-1}(d)$?
i know what $\phi(d)$. are they related in anyway?

$\phi^{-1} (d)$ isn't really a function, as such, since it would give multiple answers (I mean the inverse function of $\phi(n)$), but generally I mean that if you are considering $n^{th}$ exponents then you want to take $\bmod m$ where $\phi(m) | n$ (or, I suppose, $n | \phi(m)$).

Edit:  For example, $\phi(7) = 6$ and one finds that $n^3 \bmod 7 = 0, 1, 6$ and $3 | 6$.  :)


[color=darkblue][b]These identities are well-known :[/b] $\{\begin{array}{c}\cos x=\frac{1-t^{2}}{1+t^{2}}\\\\ \sin x=\frac{2t}{1+t^{2}}\end{array}$, where $t=\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}=\frac{1-\cos x}{\sin x}$ (prove easily from the right side to the left side).[/color]

[quote="renatzu"]Recall that $\tan \Big(\frac{x}{2}\Big) = \sqrt{\frac{1-\cos x}{1+\cos x}}$

From that just plug in and simplify.[/quote]
Forthe value $x=\frac{4\pi}{3}$ this formulas is not verified ! Why, Tjhance, where is the mistake ?!

Actually, $\tan \Big(\frac{x}{2}\Big) = \pm\sqrt{\frac{1-\cos x}{1+\cos x}}$ depending on the quadrant of which the angle lies

[quote="Xaero"]Actually, $\tan \frac{x}{2}= \pm\sqrt{\frac{1-\cos x}{1+\cos x}}$ depending on the quadrant of which the angle lies[/quote]
O.K. ! Thus, correctly is $\left|\tan \frac{x}{2}\right| = \sqrt{\frac{1-\cos x}{1+\cos x}}$.


$ a> b> c> d$
$ a \plus{} b \plus{} c \plus{} d \equal{} 2048,$ and $ a^2 \minus{} b^2 \plus{} c^2 \minus{} d^2 \equal{} 2048$. What is the minimum value of $ a$?

Are $ a$, $ b$, $ c$, and $ d$ integers, or are they any reals?

@mathwizzardude: Does the problem have the restriction that a,b,c,d are positive real numbers??Without it, it's very tough(at least to me)

If a,b,c, and d are positive reals, then there is no minimum value; any value of a strictly greater than 32*sqrt(2) will work.

Assume positive integers. b^2 - c^2 + d^2 > 0, so

a^2 > 2048, so you immediately have a>=46. This is a very flexible quadratic form so chances are good that a minimal solution will exist.

If a = 46, then b^2 - c^2 + d^2 = 68.

c^2 - d^2 > 0, so b^2 > 68 and b >= 9.

Then see c^2 - d^2 = 13, and you have a solution at (a,b,c,d) = (46,9,7,6)


Can someone please tell me how to solve this:

22. The intersection of the graphic solutions of all three inequalities listed below form an enclosed region in the coordinate plane. What is the area of this region? Express your answer in terms of pi.
(x - 4)^2 + y^2 [color=blue]is less than or equal to[/color] 16
y [color=blue]is greater than or equal to[/color] x - 4
y [color=blue]is greater than or equal to[/color] -x/3

Thanks.
-12markkram34

[quote="12markkram34"]Can someone please tell me how to solve this:

22. The intersection of the graphic solutions of all three inequalities listed below form an enclosed region in the coordinate plane. What is the area of this region? Express your answer in terms of pi.
\begin{align} (x - 4)^2 + y^2 & \le 16 \\
y & \ge x - 4 \\
y & \ge - \frac x3 \end{align}

Thanks.
-12markkram34[/quote]
[hide]
Basically, you have a circle with radius $ 4$ and center $ (4, 0)$ by equation $ (1)$. This is cut in half by $ (2)$ because that line goes through the center of the circle. At this point, the desired region is the top-left half of the circle. Now, we add in equation $ (3)$ and the desired region is reduced further to a figure consisting of a triangle with vertices $ (0, 0), (4, 0), (x, y)$, where $ (x, y)$ is the point of intersection of $ (2)$ and $ (3)$, and a $ 135^\circ$ sector of the circle.

Now we will find the point of intersection of $ (2)$ and $ (3)$.

\begin{align*} x - 4 & = - \frac x3 \\
\frac {4x}3 & = 4 \\
x & = 3 \end{align*}

Then, $ y = - 1$ and the point of intersection is $ (3, - 1)$.

The area of the $ 135^\circ$ sector is $ \frac {135}{360}\pi \cdot 4^2 = 6\pi$. The area of the triangle is $ \frac 12 bh$.

The base is just $ 4$ and the height is just $ 1$. Then the area of the triangle is $ \frac 12 \cdot 4 = 2$.

Our final answer is the sum of the two areas, which is $ \boxed{2 + 6\pi}$, as desired.[/hide]


For $ A,B,C$ angles of a triangle

$ \sum sinA\geq 2\sqrt{3}(\sum sin\frac{A}{2}sin\frac{B}{2})$

[quote="manlio"]For $ A,B,C$ angles of a triangle

$ \sum sinA\geq 2\sqrt {3}(\sum sin\frac {A}{2}sin\frac {B}{2})$[/quote]

See here : http://www.artofproblemsolving.com/Forum/viewtopic.php?t=267890&sid=a5f26847dc2f5ab164e7c3921832cae4


In trapezoid ABCD, AD is parallel to BC, 3 x BC = 2 x AD, F is on segment BA, and DF bisects CA. Find the value of BF/BA.

[hide] Say that DF and AC intersect at O.  Draw OX parrelel to AD and BC such that O is on AB.  Say that AD=3x and BC=2x.  Since AO/AC=1/2 and Triangle AOX is similar to Triangle ACB, we have that OX=x.  Now, say that AD=4y.  Thus, we have that AX=2y.  Now, since Triangle FXO is similar to Triangle FAD, we have that OX/AD=FX/AF.  Now, OX/AD=x/3x=1/3.  Thus, FX/AF=FX/(FX+AX)=FX/(2y+FX)=1/3.  Solving, we have that FX=y.  Now, we can say that XB=4y-2y=2y.  Thus, we have that FB=2y-y=y, so FB/AB=y/4y=$\boxed{\frac{1}{4}}$. [/hide]


calculus
integration
trigonometry
calculus computations

Find the maximum and minimum values of $ \int_0^{\pi} (a\sin x\plus{}b\cos x)^3dx$ for $ |a|\leq 1,\ |b|\leq 1.$

We have $ f(a,b)\equal{}\int_{0}^{\pi}{(a\sin x \plus{} b\cos x)^3 dx}\equal{}\frac{4}{3}a^3 \plus{} 2ab^2\equal{}a(\frac{4}{3}a^2\plus{}2b^2)$
So, minimum will be $ f(\minus{}1,\minus{}1)\equal{}\minus{}\frac{10}{3}$ and maximum $ f(1,1)\equal{}\frac{10}{3}.$

Your answer is correct, Yuriy Solovyov. :)

[quote="kunny"][color=darkred]Ascertain $ f(a,b)\equiv\int_0^{\pi} (a\sin x \plus{} b\cos x)^3\ \mathrm {dx}$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] $ \int_0^{\pi} (a\sin x \plus{} b\cos x)^3\ \mathrm {dx} \equal{}$ $ \int_0^{\pi} (a\sin x \minus{} b\cos x)^3\ \mathrm {dx}$ $ \implies$

$ 2\cdot f(a,b) \equal{} \int_0^{\pi} \left[(a\sin x \plus{} b\cos x)^3 \plus{} (a\sin x \minus{} b\cos x)^3\right]\ \mathrm {dx}$ $ \implies$

$ f(a,b) \equal{} a\cdot\int_{\pi}^{0}\left[a^2\left(1 \minus{} \cos^2x\right) \plus{} 3b^2\cos^2x\right]\cdot (\cos x)'\ \mathrm {dx}$ $ \implies$

$ f(a,b) \equal{} a\cdot\int_{ \minus{} 1}^{1}\left[\left(3b^2 \minus{} a^2\right)t^2\plus{}a^2\right]\ \mathrm {dt} \equal{}$  $ 2a\cdot\int_0^1\left[\left(3b^2 \minus{} a^2\right)t^2\plus{}a^2\right]\ \mathrm {dt}$ .

Thus, $ f(a,b) \equal{} 2a\cdot\left(\frac {3b^2 \minus{} a^2}{3}\plus{}a^2\right)$ $ \implies$ $ \boxed {f(a,b) \equal{} \frac {2a(2a^2 \plus{} 3b^2)}{3}}$ .[/color]


Let a, b, c, d be positive integers with a > b > c > d such that

ac + bd = (b + d + a - c)(b + d - a + c)

Find the minimum number of prime factors (not necessarily distinct) of ab + cd, ac + bd, and ad + bc.

Well, you can simplify the equation to:
b :^2:+d:^2:-a:^2:-c:^2:+2*b*d-2*a*c = a*c+b*d
which simplifies to:
(b+d):^2:-b*d = (a+c):^2:+a*c

I have not figured out beyond that

[color=cyan]Wait, are we supposed to find the smallest number of factors [i]shared[/i] by those three numbers?  I definitely don't understand this problem.[/color]

Do you know the sixth problem of IMO 2001 ? Hint ...


Does anyone have useful information on the honors math programs at the Ohio State University or the University of Michigan (besides what is already online)?  Would you recommend one over the other?

Michigan all the way!

This should answer your question: http://www.theonion.com/content/from_print/ohio_state_uses_t_shirt

Well, I suppose someone should say this: I know nothing at all about Ohio State.  The Michigan honors program has a good reputation.  And Michigan is both more academically well-regarded than Ohio State overall and has a better graduate math program (though in institutions of that size, it's not clear how much difference that will make to an undergrad).


Show that the largest number of inversions of a permutation of {1,2,...,n} equals n(n-1)/2. Determine the unique permutation with this number of inversions. Determine all permutations with one fewer inversion.

I'm just a little confused at what this question is asking?

well it asks to show that the only permutation with $\frac{n(n-1)}{2}$ inversions
is n,n-1,n-2,....3,2,1 . and we can see that if we have
...a...b... and a<b then
...b...a... has more inversions.
So if a permutation has $\frac{n(n-1)}{2}-1$ inversions then it must be one change away from  n,n-1,n-2,....3,2,1 and it should be a change of two neighbour positions
For example n,n-1,....,k+2,k+1,[b]k-1,k[/b],k-2,k-3,....3,2,1. And we have n-1 such permutations :)


Did anyone else participate in either round of the AMATYCSML competition this year?

If you do not know what the AMATYCSML competition is and you attend a class or classes at a two-year college in the USA, I highly recommend that you visit this link for more information:

http://www.amatyc.org/SML/SML.html

Yes--got 40 in the fall, and we haven't found out scores yet for this spring.  I think some colleges are still giving it.  Unfortunately, I'm not actually a student at the college where I take it anymore, so it's just for practice; I can't compete officially anymore.

I believe today is the last day for administering round 2.

I'm part of the Bellevue Community College math team. We got 4th in the nation in round 1, which I think is our second-best ever. We have consistently placed in or around the top 10 nationally in past years. :) It's kind of weird taking the test because I'm an 11th grader and there are people like twice as old as me who take it, too :huh: That doesn't really matter, though, because the high-schoolers usually get the top scores. :lol:

I participated in AMATYC this year as part of the Harper College team. I think Harper placed first this year, which would be the first time it has done so. As with the other colleges, I think that most of the credit belongs to all those random high school students who take classes there.


calculus
integration
function
derivative
inequalities
real analysis
real analysis unsolved

Hi,

Let f:[0,1]->R a  continuous non decreasing function : x<y => f(x) < f(y) , do you think that $ \int_{[0,1]^2} \frac{f(x)\minus{}f(y)}{x\minus{}y}dxdy < \plus{}\infty$.

Thanks

Nothing is wrong with that example, but the question was whether this holds for all such $ f$ - and $ f(x)\equal{}x,$ or, for that matter, any $ f$ with a bounded derivative, is not the worst case example that holomorf is thinking of. Start thinking of things that don't have bounded derivatives that exist everywhere.

In fact, I was surprised but it's true.

For every increasing and continuous function f : [0,1]->R we have $ I\equal{}\int_{[0,1]^2}   \frac{f(x)\minus{}f(y)}{x\minus{}y}dxdy \leq 4  \sup\limits_{t\in [0,1]} |f(t)|$

Let denote $ K\equal{}\{ (x,y) \in [0,1]^2, x<y \}$, and we introduice

$ y\equal{}(1\minus{}t)u\plus{}t \qquad  x\equal{}(1\minus{}t)u \qquad dxdy\equal{}(1\minus{}t)dtdu$

When $ (u,t)$ goes in $ ]0,1[^2$, it comes the formula that I call "Formula 1"

$ I\equal{}2 \int_{K}        \frac{f(x)\minus{}f(y)}{x\minus{}y}dxdy   \equal{} 2\int_{]0,1[^2} \frac{f((1\minus{}t)u\plus{}t )\minus{}f((1\minus{}t)u)}{t} (1\minus{}t)dtdu$
Now let $ F(t)\equal{}\int_{0}^t f$, we integrate u

$ I\equal{}2 \int_{0}^1  \frac{F(1)\minus{}F(1\minus{}t)\minus{}F(t)\plus{}F(0)}{t} du      \leq 4     \sup\limits_{t\in [0,1]} |f(t)|$

In fact the change of variables $ (x,y)\rightarrow (u,t)$ is not natural!! But when I tried to study when f has continuous derivate, I wrote $ \frac{f(x)\minus{}f(y)}{x\minus{}y}\equal{}\int_{0}^1 f'((1\minus{}t)x\plus{}ty) dt$, and the Formula 1 came with the help of Fubini's theorem

Nice proof!
I accidentally found some [url=http://mathematicalremarks.wordpress.com/2009/07/24/an-interesting-inequality-due-perhaps-to-kashin/]related inequality[/url] in the internet a while ago, maybe you're interested.

Here is another proof with slightly more natural change of variable. Let $ y=x+t$. Consider the change of variable $ (x,y)\leftrightarrow(x,t)$. Integrating over $ x$ first and cancelling the common part, we get
\[ \int_{y>x}\frac{f(y)-f(x)}{y-x}\,dx\,dy=\int_{0<t<1}\frac{dt}{t}[\int_t^{1}f-\int_0^{1-t}f]=\int_{0<t<1}\frac{dt}{t}[\int_{1-t}^{1}f-\int_0^{t}f]
]\]
after which everything is obvious because the integral of $ f$ over an interval of length $ t$ does not exceed $ t\max|f|$. The second half ($ \int_{y<x}$) is the same.

thanks fedja, it's very short and natural!!


[hide]
1000cm/2cm=500cm, which means that the actual brick wall is 500*4.5cm=2250cm
So the answer is 2 and 1/4 meters[/hide]

[quote="math92"]Matthew is at a zoo. He takes a picture of a one-metre snake beside a brick wall. When he developed his pictures, the one-metre snake is 2cm long and the wall is 4.5cm high. What was the actual height of the brick wall in cm.[/quote]

[hide]The picture reduced the size of each of the objects by divided them by 50. So to change the size back, you have to multiply by 50.

So the height of the wall is $50*4.5=225$ cm. or 2.25 meters. [/hide]

[quote="math92"]Matthew is at a zoo. He takes a picture of a one-metre snake beside a brick wall. When he developed his pictures, the one-metre snake is 2cm long and the wall is 4.5cm high. What was the actual height of the brick wall in cm.[/quote]
[hide]
since the snake was 1 m, and in the picture it was 1/500m, the picture is 1/500 of the actual size. 500 x 4.5 = 2250 cm.
[/hide]


[quote="armpist"]Let R1, R2,R3 be 3 mutually tangent circles and r  the small Soddy circle
  touching all 3 externally.  Let a23 be the line through centers of R2 and R3
  with s and S1  being distances of centers  r   and R1  from line a23.

  Show that   s/r - S1/R1 = 2

  Any link to Euler's  V - E + F = 2 ???

  Thank you.

   M.T.[/quote]

You should familiarize yourself with the problem [url=http://www.mathlinks.ro/viewtopic.php?t=44942]Circle and Circles[/url]in the section [url=http://www.mathlinks.ro/index.php?f=150]High School » Intermediate Topics[/url], it is all there.

Nea.   Thanks for concern anyway.

  Too many irrationalities to deal with there. I've got a pretty bunch of my own
  and that's more than  enough for me.

  Friendly,
  M.T.


quadratics
number theory unsolved
number theory

Find all pairs of integers (x,y) satisfying the equality $y(x^{2}+36)+x(y^{2}-36)+y^{2}(y-12)=0$


abstract algebra
group theory
induction
inequalities
number theory
least common multiple
superior algebra

Use finite group theorems only and solve the following nice problem:(Do not use number theory.)

Let $G$ be a finite group and  let \[\forall n\in \mathbb{N},|\{x\in G: x^{n}=1\}|\leq n.\] Prove that $G$ is a cyclic group. Conclude that if $\forall n\in \mathbb{N},|\{x\in G: x^{n}\in Z(G)\}|\leq n$ then $G$ is the trivial group. ($Z(G)$ is the center of $G$.)

Prove or disprove the following:

If $G$ is a finite cyclic group then  $\forall n\in \mathbb{N},|\{x\in G: x^{n}=1\}|\leq n.$

[moderator: Problem 8 is not an appropriate title (and by some strong definition not even a source); so stop changing the title to that one!]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=75806 .
And what does "Do not use number theory" mean¿

[quote]Conclude that if $\forall n\in \mathbb{N},|\{x\in G: x^{n}\in Z(G)\}|\leq n$ then $G$ is abelian.[/quote]
Using that for $n=1$ gives that there is at most one element in the center...

And the last one is easy to do, just use the definitions and handle around.

[quote="ZetaX"]what does "Do not use number theory" mean¿[/quote]

See your second post in [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=75806[/url]

[quote="zhaobin"][quote="mofidy"]Use finite group theorems only and solve the following nice problem:(Do not use number theory.)

Let $G$ be a finite group and  let \[\forall n\in \mathbb{N},|\{x\in G: x^{n}=1\}|\leq n.\] Prove that $G$ is a cyclic group. Conclude that if $\forall n\in \mathbb{N},|\{x\in G: x^{n}\in Z(G)\}|\leq n$ then $G$ is abelian. ($Z(G)$ is the center of $G$.)[/quote]

$Lemma$,let $x$ and $y$ be elements of $G$ such that $o(x)=m$ and $o(y)=n$.Then that $G$ has an element $z$ such that $o(z)$ is the least common multiple of $m$ and $n$.

$Proof$ of lemma:If $gcd(m,n)=1$,then let $z=xy$,the lemma has been proved.
Then let $m=p^{k_{1}}_{1}p^{k_{2}}_{2}\cdots p^{k_{t}}_{t}$,and $m=p^{l_{1}}_{1}p^{l_{2}}_{2}\cdots p^{l_{t}}_{t}$
We may assume:$k_{1}\ge l_{1},k_{2}\ge l_{2},\cdots k_{i}\ge l_{i}$,$k_{i+1}\le l_{i+1},\cdots k_{t}\le l_{t}$
Then let $a=x^{p^{k_{i+1}}_{i+1}\cdots p^{k_{t}}_{t}},b=y^{p^{l_{1}}_{1}p^{l_{2}}_{2}\cdots p^{l_{i}}_{i}}$
Then we have $o(a)=p^{k_{1}}_{1}p^{k_{2}}_{2}\cdots p^{k_{i}}_{i},o(b)=p^{l_{i+1}}_{i+1}\cdots p^{l_{t}}_{t}$
Because $gcd(o(a),o(b))=1$
Let $z=ab$ we have $o(z)$ is the least common multiple of $m$ and $n$.
So we proved the $Lemma$.

For the prove,assume $g \in G$ have greatest order,if $o(g)=|G|$ then $g$ generate $G$.
If not,assume $o(g)=k<|G|$,then by the lemma we can show the order of any element of $G$ must divide by $o(g)$,(because $g$ has greatest odder.then  $a^{k}=e$ for all $a \in G$,contradiction,So we have proved that $G$ is cyclic. :)
[quote="mofidy"]
Prove or disprove the following:

If $G$ is a finite cyclic group then  $\forall n\in \mathbb{N},|\{x\in G: x^{n}=1\}|\leq n.$[/quote]
I think it is not hard.[/quote]
I remember ZetaX (or others, I can't really remember) said that if a is of order m, b is of order n, ab is not necessarily of order lcm(m,n)?

[quote="Soarer"]
I remember ZetaX (or others, I can't really remember) said that if a is of order m, b is of order n, ab is not necessarily of order lcm(m,n)?[/quote]
Here I assume that $G$ is abelian.
But in the problem,it didn't say that.

It is also wrong for abelian groups, take $1$ and $1$ in the group $\mathbb{Z}/4\mathbb{Z}$.
But it is true when they have coprime orders in the abelian case, and nothing else was used above.

I think the correct corollary is (at least it is correct in this form):
Conclude that if $\forall n\in \mathbb{N},|\{x\in G: x^{n}\in Z(G)\}|\leq n \cdot |Z(G)|$ then $G$ is abelian.

[quote="ZetaX"]It is also wrong for abelian groups, take $1$ and $1$ in the group $\mathbb{Z}/4\mathbb{Z}$.
But it is true when they have coprime orders in the abelian case, and nothing else was used above.
[/quote]
In my proof
I don't say that $ab$ have the order lcm(m,n).
I just proved there exist $z \in G$,which have the order  lcm(m,n).

Ok,can you prove the corollary you posted?Thank you.

Thanks for your discussion.

My Solution:

Let $G$ be a finite group and let $\forall n\in \mathbb{N},|\{x\in G: x^{n}=1\}|\leq n.$ I prove that $G$ is a cyclic group. It easy to see that by assumption if $a,b\in G$ and $|<a>|=|<b>|$ then <a>=<b>. So each cyclic subgroup is normal in $G$. Now we solve problem by induction on $|G|.$ By this, each maximal subgroup in $G$ is cyclic. If $G$ has  the unique maximal subgroup, $G$ is cyclic. So suppose $M_{1}$ and $M_{2}$ are maximal subgroup in $G$ and $M_{1}=M_{2}.$ In this case we have $G=M_{1}M_{2}$ because $M_{1}$ is the normal subgroup. Let $M_{1}= P_{1}P_{2}\cdots P_{k}$ and $M_{2}= Q_{1}Q_{2}\cdots Q_{t}$ where $P_{i},Q_{j}$ are Sylow subgroup. Then $G = P_{1}P_{2}\cdots P_{k}Q_{1}Q_{2}\cdots Q_{t}.$ Simplify this (for example if $P_{i}\leqslant Q_{j}$ delete $P_{i}$.) Now we can write $G=H_{1}H_{2}\cdots H_{s}$ such that $H_{i}$ is a normal p-group and $(|H_{i}|,|H_{j}|)=1 (i\neq j).$ So $G$ is cyclic.

Now let $\forall n\in \mathbb{N},|\{x\in G: x^{n}\in Z(G)\}|\leq n.$ Therefore $\frac{G}{{Z(G)}}$ is cyclic by the above problem. Then $G$ is abelian and so $Z(G)=G.$ Now $|G|=1,$ by assumption.

[quote="zhaobin"]In my proof
I don't say that $ab$ have the order lcm(m,n).
I just proved there exist $z \in G$,which have the order  lcm(m,n).[/quote]
In my post, I did say nothing else ;)
I was more answering to Soarer.

My proof for the corollary is more or less the above, but@mofidy: the corollary is wrong (or more exactly: senseless) as you stated it; but the proof is correct for the one I gave ;) (note that you have to use the condition for the quotient group, so also the inequality must contain the quotioent.

[quote="ZetaX"] the corollary is wrong (or more exactly: senseless) as you stated it; but the proof is correct for the one I gave ;) (note that you have to use the condition for the quotient group, so also the inequality must contain the quotioent.[/quote]

You are right. I corrected the corollary. In fact we have:

if $\forall n\in \mathbb{N},|\{x\in G: x^{n}\in Z(G)\}|\leq n$ then $|G|=1.$

Thanks


geometry
circumcircle
incenter
geometry unsolved

let $ o$ is circumcircle of ABC triangle.let K,L are on AB and BC arc respectively such that KL||AC.
consider $ I_1,I_2$ are incenters of ABK and BCL triangles.M is midpoint of ABC arc.
prove $ MI_1\equal{}MI_2$

Denote the midpoint of the arc $ AK$ and $ CL$ by $ T$ and $ Q$ then obviously $ TI_1\equal{}QI_2$ also $ MQ\equal{}MT$.
now it remains to prove that $ \angle I_2QM\equal{} \angle I_1TM$.
which can be proven easily: note that $ I_1,T,B$ are collinear similarly for $ Q,I_2,B$.
we r done!


You can purchase past contests from MOEMS.  See
[url]http://www.moems.org/PDBooks.htm[/url].

[quote="anirudh"]:mad:  :mad:  :mad: THOSE ARE THE BOOKS. READ MY 1ST POST[/quote]
Did you read the whole link?  You can purchase a year's worth of contest problems for 3 dollars.

I'm not sure why you are angry at someone who is trying to help you.  Next time I will think twice about assisting you.

oops. sorry. i didn't read that part :blush:  :blush:  :blush:

[quote="anirudh"]Does anyone have any materials for doing the MOEMS Level "M"? Past contest problems would be helpful. Thanks. I have already completed the Math Olympiads in the Elementary and Middle Schools and Creative Problem Solving by George Lenchner, so please don't recommend those. Once again, I am looking for [u]PAST CONTEST PROBLEMS.[/u][/quote]

MOEMS has booklets for each year starting with 1998-1999. Also the 1997-1998 problems were very tough. Go to the http://www.moems.org/ATSBooks.htm to access the order form.


[hide="Solution"]There are $ \binom{5}{2}$ ways to choose the elements for the intersection. Then, for the remaining $ 3$ elements, there are $ 2$ ways each to place them into each of two subsets. Since we have now counted each combination twice, we divide by 2:
\[ {\frac{\left(\binom{5}{2} \cdot 2^3\right)}{2} = 40 \Rightarrow \boxed{B}
}\]
[/hide]

[hide]Adding to 123456789, for the second part (how many ways can you subdivide 3 elements into two subsets) you could just represent that as 4 choose 1, since this situation is

@a@b@c@, where @ represents somewhere you can sort of place a "barrier" to divide the three terms (you know what a I mean?). So its (5 choose 2)(4 choose 1) or 10*4=40[/hide]

Drunken Math, your solution is incorrect.

First of all, you count $ \{d,e\} \cup \{a,b,c,d,e\}$ twice. Second of all, you don't allow for this possibility: $ \{d,e,b\} \cup \{d,e,a,c\}$

there are 5C2 ways to select the 2 elements that are in both subsets. Now for the remaining three terms:
There can be all 3 in 1 subset and none in the other, or 2 in 1 and one in the other. Because there are 3 terms there are 3 ways to choose the term that can be alone in the other subset. So the number of ways is:
 5C2 x (1+3)= 40

We can do case work.
Suppose A={a, b, c, d, e}. Then B should have only two elements. $\binom{5}{2}=10$ ways.
Now suppose A has 4 elements, say A={a, b, c, d}. Then B should look like {a, b, e}, {a, c ,e}, etc, which is   $\binom{4}{2}=6$ ways. There are 5 ways where A can have 4 elements.  Thus 5*6=30 ways. This consumes all the cases.
In all, there are 10+30=40 ways that satisfies the conditions of the problem.

[quote=worthawholebean]Two subsets of the set $ S\equal{}\{a,b,c,d,e\}$ are to be chosen so that their union is $ S$ and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?

$ \textbf{(A)}\ 20 \qquad
\textbf{(B)}\ 40 \qquad
\textbf{(C)}\ 60 \qquad
\textbf{(D)}\ 160 \qquad
\textbf{(E)}\ 320$[/quote]

By symmetry, the answer is clearly $$\binom{5}{2} \cdot \frac{2^{5 - 2}}{2} = \boxed{40.}$$

*when you realize this was a problem 23*

in all fairness 2021 amc 10a #23 isnt much harder than this

[hide=a question]
is the person who created this thread worth a whole bean?

[quote=awesomeguy856][quote=kante314]Wow, 10 years!

[hide=Solution]
WLOG let $a$ and $b$ be the union of the two sets. Now, we need to distribute $c,d,$ and $e.$ This can be done
$$\binom{3}{0}+\binom{3}{1}+\binom{3}{2}+\binom{3}{3}=2^3=8$$ ways. Counting for rearrangements and dividing by $2$ for double counting,
$$\frac{8 \cdot \binom{5}{2}}{2}=\boxed{40}$$[/hide][/quote]

why would you bump a 10 year old post and explicitly announce the fact it's 10 years old like its an achievement[/quote]

I didn't post it for the 10 years, I posted for the solution


The Art of Computer Programming - Donald Knuth 1.2.2 #25
Suppose that we have a binary computer and a number x, 1<=x<2
Show that the following algorithm ... may be used to calculate an approximation for [tex] y=log_bx[/tex]
L1 Set y<- 0, z<-x shifted right 1, k<-1
L2 If x=1, stop
L3 If x-z<1, set z <- z shifted right 1, k <- k+1, and repeat this step
L4 Set x<- x-z, z<-x shifted right k, y <- y+[tex]log_b \frac{2^k}{2^k-1}[/tex]
and find an error bound for this algorithm

also, can we maybe have physics/chem/CS sections in this board??

[quote="joml88"]I don't really think it is necessary to have those extra forums.  There are other science forums on the internet that focus more on these areas.  The board already has a lot of forums.[/quote]

I agree. This forum is based on math.

we should have at least an "Applied Math" section then... right now ppl with CS problems post randomly in different sections... some in "Other US Contest" some in "The Round Table", etc

[quote="bugzpodder"]some in "Other US Contest" some in "The Round Table", etc[/quote]

If it is a contest question, it belongs in other US contests, if not, it belongs in either. Is it that abnormal then? :)

Easiest solution would be renaming the first to Other US Math Contests, then they have only 1 choice left ;)

[quote]Should we have physics/chem/CS sections in this forum[/quote]

No!  The addition of MathLinks popped the number of forums from 10 to 200 or so.  (I don't understand why we need separate forums for solved, unsolved, proposed, and open problems.  And why are there 10 IMO-related forums?  Isn't one enough?  Or why all the undergraduate contest forums?  Why will one 'undergraduate-level' contest forum not suffice?  There are only ten contests at most, as compared to hundreds of U.S. Contests, which all share a single forum...argh.)