Da li neko zna BiH tim za Juniorsku Balkanijdu. I mozda zvanicnu listu i zadatke? :)

neam blage kako se zovu samo znam da su iz sarajeva,prijedora,banja luke,tuzle,zenice i nevesinja ( :D ) , nemam zvaničnu listu, zadatke i jedan prijedlog koji je umalo prošao znam...  :D

Zadaci objavljeni a i prijedlog na koji posebno treba obratiti paznju  :D

Ja sam bio samoprozvani observer kao Vedran na doigravanju za IMO :D

Evo rezultata

1. Ratko Darda; Prijedor (9. razred)
2. Osman Mehinagić; Zenica (1. razred)
2. Edin Husić; Tuzla (8.razred)
4. Vladimir Ivković; Nevesinje (9. razred)
5. Faruk _____; Sarajevo (8. razred)
6. Dejan _____; Banja Luka (9. razred)

Nemam rezultate kod sebe, samo sam ih vidio na kratko vrijeme, pa stoga "______" :D
Nijedan prošlogodišnji učesnik JBMO-a nije prošao ove godine. (bila su dva koja su imali šanse) :D

JBMO 2008 se održava u Albaniji, a slijedeće godine se JBMO 2009 može održati u Bosni i Hercegovini :D  :D  :D

[quote="hsiljak"]Obratili smo paznju na prijedlog (a i na one-line solution  :lol: )... Svidja mi se geografski raspored ekipe... moracu se malo raspitati za cijenjena imena ;)

Hajde da jos grupisem sve sto si pokacio, da se ljudi ne pate trazeci:

[url]http://www.mathlinks.ro/viewtopic.php?t=207592[/url]
[url]http://www.mathlinks.ro/viewtopic.php?t=207590[/url]
[url]http://www.mathlinks.ro/viewtopic.php?t=207595[/url]

i prijedlog... [url]http://www.mathlinks.ro/viewtopic.php?t=207597[/url][/quote]

 Moram naglasiti da onakvo one line solution uopće nije tačno rješenje tj. predstavlja samo loše objašnjeni dio jednog dijela zadatka, a cjelokupno rješenje zasad uopće nije dato ( tj. ono koje bi nosilo 10 odnosno 7 bodova), harune pozabavi se pa ti pokušaj  :D

[quote="Pink Floyd"]Ime člana ekipe iz Nevesinja :lol:  je Vladimir Ivković  (ništa ne može proći bez nekog Vlada).
Nadam se da će nam osvjetlati obraz, kao i prošle godine.

P.S. Vedrane, bolje sačuvaj svoje prijedloge za neka bolja vremena (recimo kroz dvije godine ), pošto sada ne nailaze na dobar prijem kod takmičara :P[/quote]
 ja ću samo reći da ste gotovi sljedeće godine  :rotfl:  :rotfl:

[quote="hsiljak"][quote="vedran6"] ja ću samo reći da ste gotovi sljedeće godine  :rotfl:  :rotfl:[/quote]

Predlazem da slijedece godine nastupimo udruzenim snagama, sa cijelim shortlistom (imam vec 2 NT i 2 A)... pa da bar 2 nasa zadatka udju medju 6 ;)[/quote]

   može...ali i ne mogu ući više nego 2  :D osim ako neki od nas se ne udruži s dr. Sušcem  :rotfl:  :rotfl:

[quote="hsiljak"]Suska se da cemo od slijedece godine napustiti sistem 2+2+2, nego da ce se problemi birati po kvalitetu, ljepoti, cemu vec...  :lol:[/quote]

  to se šuška otkad je uveden sistem  :D  :D  al je zasad sve ostalo na šuškanju  :D

evo ja sam bio na tom takmicenju
ali samo meni nije jasno u kojem gradu se odrzava takmicenje
mislim kojem gradu albanije nisam bas dobro cuo
a zadatke ne znam, kako da napisem

Hmm... Prema onom sto sam ja cuo, trebala bi biti ili u Tirani ili u Vloreu/Vlori (jako lijepom primorskom gradicu)...

Da li neko zna treba li za Albaniju viza? :(
Fali vam gore treći zadatak, bio je neki paralelogram, poslije ću ga napisati.
Ovaj Dejan se preziva Gvozdenac, a za Faruka ne znam.
PozZ

[quote="RaleD"][quote]Dobri ste  Vrlo dobri
Koji ste ekipno?[/quote]
Iza su Crna Gora, Kipar i Makedonija. :)
Albanija je bila bolja za koji bod...[/quote]
Jel Albanija imala jednu ili dvije ekipe?
Pa da vidim imamo li mi sanse za dvije ekipe slijedece godine :D

Imali su samo jednu...  :|
Ali to ništa ne mora da znači, Bugari su prošle godine imali dvije  :) .

Bugari lani nisu imali jednu, ni dvije nego tri ekipe: Bulgaria1, Bulgaria2 i Shumen-Varna...
A Moldavci su godinu ranije imali dvije ekipe
Pa da vidim hocemo li mi slijedece godine imati dvije ekipe, imamo li sanse za to... :D

Pa,Harune,kako mene da ne prepoznas!Bar sto sam Uljarevic,kad sam ne prepoznatljiv po matematici;)

Ja mislim da je nama i jedna ekipa dosta;)

***opet off-topic***

Pa pored toliko matematicara Hercegovaca, bilo je tesko sa sigurnoscu utvrditi ko se krije pod tvojim nickom... sreca pa se nesto dalo desifrovati iz email adrese  :lol:

***back on topic***

Naravno da je dosta jedna... ali eto, moze jos jedna revijalna... Nudim se da im budem leader :D

[quote="RaleD"]Pa je li ta treća ekipa išta uradila? Na onoj se listi ne vide.[/quote]

Ma samo idu revijalno i uzmu medalju al ne uticu na konacni poredak i dodjelu ostalih medalja...
Bugari su imali "kodove" BUL1, BUL2 i SHV

I ja sam zadovoljan s rezultatima ekipe, osim što se da primjetiti da im Doktor nije držao nikakve pripreme pa nema 6 desetki na prvom zadatku  :D  :D  :D

Samo se pitam ko ce nadmasiti onu slavnu ekipu BiH iz Grcke?!

Eeeeeeeeeeeeeee,bila su to sretna vremena :rotfl:

Ma moci su sjajni, ipak su ono bila druga vremena.
Ali se ipak vidi ko je spremao Ivkovica :D  :D
A jadna ona djeca iduce godine, nece imati sanse na Vecinim prijedlozima

Svaka cast za BiH team.

[quote]Samo se pitam ko ce nadmasiti onu slavnu ekipu BiH iz Grcke?!

Eeeeeeeeeeeeeee,bila su to sretna vremena Rotfl[/quote]

Hehe Dvije su olimpijade Ulja u Grckoj bile (uspjeh one ekipe sto je za srednje isla 2004. je nedostizan a ipak se zna koji je dosad najbolji BiH JBMO team kao i takmicar   :P )

[quote]I ja sam zadovoljan s rezultatima ekipe, osim što se da primjetiti da im Doktor nije držao nikakve pripreme pa nema 6 desetki na prvom zadatku Mr. Green Mr. Green Mr. Green[/quote]

Steta prava za taj 1. zadatak a teski sablon (doduse tip zadataka koji nije tako popularan kod nas u osnovnoj)

ima li iko rezultate bih za 2003. i 04 godinu? (na jbmo)

2003. nismo ucestvovali (znam, tad sam jos mogao ucestvovati-te je godine organzovana prva JBHMO, i nije formirana ekipa za Izmir).

Sto se 2004. tice-to Vedran sigurno ima ;)

2004- bronzane medalje : Marko Jovović (28), Ana Krkljić (28), Vedran Karahodžić (25), Salem Malikić (20) . Još su učestvovale Milena Milošević i Branka Kablar (obje oko 7-8 bodova).... Dobili smo Kipar, Albaniju i Makedoniju (iako za Makedoniju nisam 100 posto).... Leader: Vidan Govedarica; Deputy: Hamza Dobraća......


trigonometry
calculus
integration
calculus computations

Hi all, its me again.  :D

I have just started my Calculus 2 class and have been in there for a week, however i am noticing that we use a lot of Trig rules while doing integration. Can someone help me remember the formulas or rules that involve trig. Or is there a site i can go to with all the rule?

There's probably a table in your calculus textbook; in the book I have (Stewart), there's a table of basic antiderivatives in the middle of the techniques of integration chapter.


find the value of the following expression
$ \sum_{k\equal{} 0}^{[\frac {n}{2}]} \tbinom {n \minus{} k}{k}$
where
$ [x]$ is the largest integer less than or equal to $ x$

Let that expression be $ F(n)$
it is clear that $ F(0)\equal{}F(1)\equal{}1$ and that $ F(n\plus{}1)\equal{}F(n)\plus{}F(n\minus{}1)$ by $ \binom{n\plus{}1}{k}\equal{}\binom{n}{k}\plus{}\binom{n}{k\minus{}1}$
so $ F(n)\equal{}f_{n\plus{}1}$ (the $ n\plus{}1$'th Fibonacci number)
$ F(n)\equal{}\frac{1}{\sqrt{5}}\left(\left(\frac{1\plus{}\sqrt{5}}{2}\right)^{n\plus{}1}\minus{}\left(\frac{1\minus{}\sqrt{5}}{2}\right)^{n\plus{}1}\right)$

Scroll down to "excursion" [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=217361]here[/url].


function
analytic geometry
graphing lines
slope
geometry
geometric transformation
rotation

It merely means the slope approaches infinity as $x \to 0$.  Note that this is just a $90^{\circ}$ rotation of $f(x) = x^{3}$ which is perfectly nice.

[quote="mayhem"]Consider $f: R \to R , f(x)=x^{\frac{1}{3}}$ . This function is not differentiable  for $x=0$. But this function smooth and defined at $x=0$. So what is the problem?[/quote]

No the function is not smooth.

[b]Definition:[/b] A function $f$ that defined on $(a,b)$ that is differentiable [b]and[/b] $f'$ is continous is said to be "smooth"

[quote="sylow_theory"][quote="blahblahblah"]smooth usually refers to $C^{\infty}$, not $C^{1}$.[/quote]
Okay, just the way I studied it I always used it to refer to $C^{1}$ but since you were doing analysis longer than I have I would start using $C^{\infty}$ from now on. But anyway, my point is still correct, the original poster shouldnt have used "smooth" in his post.[/quote]

It's easy to be misled as there are textbooks which use smooth to refer to larger classes than $C^{\infty}$.  For example, my calculus III textbook called $C^{2}$ functions smooth.  But in almost every context smooth will mean $C^{\infty}$.

I am sure that the set of all papers and books that use "smooth" to mean something else than $C^{\infty}$ has positive measure. :) I can remember "smooth" being defined in the beginning of a book as "having all the derivatives that we need", i.e., as $C^{k}$, where $k$ varies from one theorem to another.  :huh: To avoid ambiguity, write $C^{1}$-smooth, $C^{2}$-smooth, etc.

It seems that [b]mayhem[/b] referred to the obvious smoothness of the graph of $x^{1/3}$. Indeed, the graph is a smooth curve in any sense (it's even analytic). But smoothness of a function is not the same as smoothness of its graph. Yeah, math language is confusing... what about those sets that are open and closed at the same time...  :roll:

[quote="mlok"]what about those sets that are open and closed at the same time...  :roll:[/quote]

clopen?

[quote="sylow_theory"][quote="blahblahblah"]
It's easy to be misled as there are textbooks which use smooth to refer to larger classes than $C^{\infty}$.  For example, my calculus III textbook called $C^{2}$ functions smooth.  But in almost every context smooth will mean $C^{\infty}$.[/quote]
I find it pretty funny that $C^{1}$ is bigger then $C^{\infty}$.  :rotfl:[/quote]

$C^{\infty}=\bigcap_{k\geq 1}C^{k}\implies C^{\infty}\subseteq C^{1}$  :?:

Some people are easily amused. An explanation: if $C$ were a number greater than $1$, then $C^{1}$ would be smaller than $C^{2}$, etc. However, with $0<C<1$ it's the opposite.


Prove  the spherical  concave mirror equation : $ \frac{1}{d}\plus{}\frac{1}{d_{1}}\equal{}\frac{1}{f}$

Well, I think these links should help:

http://www.tutorvista.com/content/physics/physics-ii/light-reflection/mirror-formula-concave.php
http://www.learnquebec.ca/en/content/curriculum/mst/opticks/chapter5/5_eqmircav.html
http://scienceworld.wolfram.com/physics/MirrorFormula.html

(it's all basically the same proof, just with varying level of details).


function
limit
algebra
domain
inequalities
triangle inequality
calculus

How would you prove that for any functions $f(x)$ and $g(x)$,
$\lim_{x \to a}\big(f(x)+g(x) \big) = \lim_{x \to a}f(x)+\lim_{x \to a}g(x)$
where $a$ is in the domain of $f(x)$ and $g(x)$

(There's an unstated condition: "provided both limits on the right exist.")

This is what the $\epsilon$-$\delta$ formalism is for - for proving theorems like this. This is one of several closely related theorems, and of that collection, this is the easiest one to prove, hence the one to start with.

Step 1: Our hypothesis is that $\lim_{x\to a}f(x)=L$ and $\lim_{x\to a}g(x)=M.$

Write out formally (this is the $\epsilon$-$\delta$ definition) what that hypothesis means.

[You don't need $a$ to be in the domains of $f$ and $g.$ You merely need $a$ to be a limit point of the intersection of the two domains.]

not at all rigorous but i think you can let $J(x)=f(x)+g(x)$.

[quote="Kent Merryfield"][You don't need $a$ to be in the domains of $f$ and $g.$ You merely need $a$ to be a limit point of the intersection of the two domains.][/quote]
Isn't that the same thing as $a$ being in the domain of $f$ and $g$?

[quote="Kent Merryfield"](There's an unstated condition: "provided both limits on the right exist.")

This is what the $\epsilon$-$\delta$ formalism is for - for proving theorems like this. This is one of several closely related theorems, and of that collection, this is the easiest one to prove, hence the one to start with.

Step 1: Our hypothesis is that $\lim_{x\to a}f(x)=L$ and $\lim_{x\to a}g(x)=M.$

Write out formally (this is the $\epsilon$-$\delta$ definition) what that hypothesis means.[/quote]
I am not too good at $\epsilon$-$\delta$ proofs but I'll give it a try. Here goes nothing.

So $\lim_{x\to a}f(x)=L$ means that for all $\epsilon > 0$, there exists a $\delta > 0$ suct that if $0 < |x-a| < \delta$, then $| f(x)-L| < \epsilon$.
Similarly, $\lim_{x\to a}g(x)=M$ means that for all $\epsilon > 0$, there exists a $\delta > 0$ suct that if $0 < |x-a| < \delta$, then $| g(x)-M| < \epsilon$.

Now given the following two information, I have to prove that $\lim_{x\to a}(f(x)+g(x))=L+M$ or, stated more precisely, for all $\epsilon > 0$, there exists a $\delta > 0$ suct that if $0 < |x-a| < \delta$, then $| f(x)+g(x)-L-M| < \epsilon$.

I am not quite sure how to proceed now. I am a bit confused. Am I suppose to use the same $\delta's$ and $\epsilon's$ or should I make seperate variables for seperate statements?

Let's start from the back.

We want $|f(x)+g(x)-L-M|<\epsilon.$

We have the triangle inequality on our side, so we can say

$|f(x)+g(x)-L-M|\le|f(x)-L|+|g(x)-M|.$

So if we can prove that the sum on the right hand side is less than $\epsilon,$ that will be good enough.

One (certainly not the only, but convenient) way to make a sum smaller than some number is to make each term smaller than half that number.

So now we want $|f(x)-L|<\frac{\epsilon}2$ and $|g(x)-M|<\frac{\epsilon}2$

Now: $\epsilon,\,\frac{\epsilon}2$ - they're all just arbitrary postive numbers. Either has the same status in the original definition.

Find $\delta_{1}$ such that $|f(x)-L|<\frac{\epsilon}2$ for $0<|x-a|<\delta_{1}.$

Find $\delta_{2}$ such that $|g(x)-M|<\frac{\epsilon}2$ for $0<|x-a|<\delta_{2}.$

These had to be $\delta_{1}$ and $\delta_{2}$ because we were making two separate and independent claims upon the definition of limit.

The next line starts "Let $\delta=\min(\delta_{1},\delta_{2})$." I'll let you figure out how it now fits together.

[quote]Isn't that the same thing as $a$ being in the domain of $f$ and $g$? [/quote]
No. Insisting that $a$ be in the domain is insisting that $f(a)$ and $g(a)$ be defined. We specifically do not want to insist on that. Note the use of $0<|x-a|<\delta,$ deliberately leaving $a$ out of things. We do need for $a$ to be "very near" or "next to" the domain, but that's not the same thing as in the domain.


trigonometry
analytic geometry
arithmetic sequence

I was wondering if anyone could help me out with these two problems :?

1.) An ant starts at the origin of the Cartesian plane and moves one unit at a time either left, right, up or down. Compute the number of ways for the ant to reach (4,7) in exactly 11 moves.

This is one is a bit more challenging:

2.) If a, b and c are the sides of the triangle ABC such that a 2, b 2, and c 2  form an arithmetic progression, and cot A+cot C = 3, compute cot B.

If anyone could show me how to solve these two problems, with the work shown, that would be excellent..thanks in advance!

1) Since ant must reach (4,7) from origin in 11 moves only allowed moves are up and right: 4 right moves and 7 up moves. And it is no matter how he mix these 4 right moves and 7 up moves, anyway he will reach (4,7). It means that each way from (0,0) to (4,7) can be encoded as binary sequence of 0 and 1, where 0 means right move and 1 means up move. For example: 00101101111, 00001111111, and so on. We are to calculate number of such sequences. In other words, we need to calculate the number of ways to point four positions where we will place 0 from 11 given positions. It is standart combinatoric question and answer is $C_{11}^4=\frac{11!}{4!\cdot 7!}$.

2) Since $a^2$, $b^2$ and $c^2$ form an arithmetic progression we conclude that $2b^2=a^2+c^2$. On the other hand, using cosine theorem we obtain $b^2=a^2+c^2-2ac\cos B$. These two equalities imply that
\[b^2=2ac\cos B.\qquad(1)\]
We have
\[3=\cot A+\cot C=\frac{\cos A}{\sin A}+\frac{\cos C}{\sin C}=\frac{\sin(A+C)}{\sin A\sin C}=\frac{\sin(\pi-B)}{\sin A\sin C}=\frac{\sin(B)}{\sin A\sin C}.\]
or $3\sin B=\frac{\sin B}{\sin A}\cdot \frac{\sin B}{\sin C}$. From sine theorem we know that $\frac{\sin B}{\sin A}=\frac{b}{a}$ and $\frac{\sin B}{\sin C}=\frac{b}{c}$. Hence
\[3\sin B=\frac{b^2}{ac}.\qquad(2)\]
Combining [b](1)[/b] and [b](2)[/b] we obtain $3\sin B=2\cos B$, i.e. $\cot B=\frac{3}{2}$.

:) Wow..thanks so much Myth!!! :)

However, I was wondering...what do you mean by "In other words, we need to calculate the number of ways to point four positions where we will place 0 from 11" :?

four positions where we will place 0 from 11 given positions.  Because each valid path has 4 moves right and 7 moves up, it's just the number of distinct arrangements of seven "1's" and 4 " 0's ".


Compute the sum  F[size=75]0[/size]/1 + F[size=75]1[/size]/2 + F[size=75]2[/size]/4 + ... + F[size=75]n[/size]/2^n + ...
where F[size=75]0[/size] = F[size=75]1[/size] = 1 and F[size=75]n + 1[/size] = F[size=75]n[/size] + F[size=75]n - 1[/size] for n :ge: 1.

Since JBL dosen't seem to appreciate generating functions  (he's wrong because it's a very powerful (and to my opinion elegant) tool) :

[hide]Ignoring problems of existence (convergence) of the sum :

If S = F0/1 + F1/2 + F2/4 + ... + Fn/2^n + ...

then

(1-1/2-1/4)S = 1 Hence S = 4[/hide]

Not that I don't appreciate generating functions -- I think they're very cool.  I just don't have any experience working with them.

This sum does converge -- there isn't any problem with it!  2>:phi:


I know that it will be hold at Albania . When it will be hold exactly ?  Also please post the problems here .

When they do the competition? Is there any official site of jbmo? Good luck to Serbian team.  :lol:

Thanks to Ahiles for the problems  :)

1st problem
http://www.mathlinks.ro/Forum/viewtopic.php?t=211771

2nd problem
http://www.mathlinks.ro/Forum/viewtopic.php?t=211772

3rd problem
http://www.mathlinks.ro/Forum/viewtopic.php?t=211774

4th problem
http://www.mathlinks.ro/Forum/viewtopic.php?t=211773

An easy JBMO . When the results will be announced I would like someone to post some here because it seems there is no official site  :(

Do you know official web-site of JBMO 2008? :D
I can't find it anywhere, like you silouan.

Same problem as you. THE WEB-SITE OF JBMO 2008!! I can't find it either.

Can somebody post results of JBMO here, especially I want to see Serbia results?

For Serbia it should be:

Teodor, Stevan 40,
Nevena 34,
Igor 32,
Rade 28,
Lola 25

They aim to get 2 gold(:lol:), 2 silver and 2 bronze...

[quote="raliicgomg"]Do you know something about the Bulgarian results?If you know the official site,please post a link.[/quote]

I know that Bulgaria in sum has 200points, but I think that this doesn't help you. :) I don't know official site...

What about the results of the romanian team? Thank you!

[quote="Bugi"]Same problem as you. THE WEB-SITE OF JBMO 2008!! I can't find it either.[/quote]

I think that the official website doesn't exist at all. :lol:  :D

[quote="AdrianDD"]OK. If there is no official site, I'm sure somebody has all the results.[/quote]

But who? :D

[color=darkblue]Here  are the the places...[/color]
[color=brown][u][b]1.[/b][/u] Romania (2 max scores and 4 silvers)...
[u][b]2.[/b] [/u]Turkey (2 max scores, 1 gold (38 points), 2 silvers and 1 bronze)...
[u][b]3.[/b][/u] Bulgaria (2 max scores and I don't know about others)
[u][b]4.[/b][/u] Serbia (2 max scores )
[u][b]5.[/b][/u] Moldova (3 silvers (one mine :) ) and 3 bronzes)
[u][b]6.[/b][/u] Kazakstan (they didn't recieve any medals)
[u][b]6.[/b][/u] Greece (1 silver, 4 bronzes)[/color]

[color=darkblue]Bronze was given from 13 points, silver from 30, and gold from 38.[/color]

[color=green]As silouan said the JBMO was easy. But it was too easy to take only 35 points. The 4th Combinatorics problem was quite hard...[/color]

[quote="Ahiles"][color=darkblue]Here  are the the places...[/color]
[color=brown][u][b]1.[/b][/u] Romania (2 max scores and 4 silvers)...
[u][b]2.[/b] [/u]Turkey (2 max scores, 1 gold (38 points), 2 silvers and 1 bronze)...
[u][b]3.[/b][/u] Bulgaria (2 max scores and I don't know about others)
[u][b]4.[/b][/u] Serbia (2 max scores )
[u][b]5.[/b][/u] Moldova (3 silvers (one mine :) ) and 3 bronzes)
[u][b]6.[/b][/u] Kazakstan (they didn't recieve any medals)
[u][b]6.[/b][/u] Greece (1 silver, 4 bronzes)[/color]

[color=darkblue]Bronze was given from 13 points, silver from 30, and gold from 38.[/color][/quote]

What about other countries? :lol:

Does anybody have complete results with points on each problem?

@Ahiles: It isn't supposed that all problems be easy it would be very stupid.

And beside that problem 4. is very nice (it is proposed by Serbia  :lol: ).

[color=darkblue]Proposers of the problems:
1. Bulgaria
2. Bulgaria
3. Turkey
4. Serbia[/color]

[quote="dule_00"]
@Ahiles: It isn't supposed that all problems be easy it would be very stupid.
[/quote]
I didn't say that all the problems must be easy. I completely agree with you. I think that problems could be much harder.

[quote="Bugi"]Who made that 4th problem?[/quote]
Dusan Milijancevic...who are you?

Lola, by the way, tell Dusan that his problem is very nice...

[quote="Bugi"]Well, look at my location...
If it doesn't help, tell me![/quote]
I know that you are from Novi Sad...but who are you?what is your name?

The bulgarian team has 2 gold medals(max rezults, one of them i mine, I am Zhivko Zhechev and have 2 gold and 1 silver medals from JBMO(2006,2007,2008) :wink: ), 3 silver medals(one of them with 37 points I think) and 1 bronze medal. The sum is 200. I think the problem were quite easy, just as the third problem on IMO this year  :lol:  :P .

here there is the ranking:
[url]http://www.mategl.com/Calude%20pentru%20site%202007%20PDF/REZ_JBMO_2008.pdf[/url]


three less than 11 times a # is the same as the # deacresed by 13. find the #

$ 11a \minus{} 3 \equal{} a \minus{} 13$

$ 10a \minus{} 3 \equal{} \minus{}13$

$ 10a \equal{} \minus{}10$

$ a \equal{} \boxed{\minus{}1}$


conics
parabola
analytic geometry
calculus
calculus computations

Draw a diagram to show that there are 2 lines tangent to both of the parabolas y = -x^2 and y = x^2 + 4. Find the coordinates of the four pts at which there tangents touch the parabolas.

Could i get some help on the question, step by step?

Let $P(s,\ s^{2}+4),\ Q(t,-t^{2}).$ You are to find the condition for which two tangent lines at $P,\ Q$ coincide.


...this is for creating a name for something else...please vote...

Uh I think the first and last choices are the same...just lettin you know

I realized...for last one I meant what I obviously meant...  :)

[quote="PowerOfPi"]I think the best one is the last one. $ leeticus$ $ factoriala$[/quote]

The last one should be leetica factorialis.


If $f(x) = x^{\left(x+1\right)}\times \left(x+2\right)^{\left(x+3\right)}$ then $f(0) + f(-1) + f(-2) + f(-3) =$

$\textbf{(A)}\ \frac{-8}{9} \qquad
\textbf{(B)}\ 0 \qquad
\textbf{(C)}\ \frac{8}{9} \qquad
\textbf{(D)}\ 1 \qquad
\textbf{(E)}\ \frac{10}{9}$

[hide]You basically plug all the number's in. You can automatically eliminate 0 and -2 because you see that there is a 0 as the base. YOu are left with -1 and -3 You plug in -1 where you get 1*1=1 and for -3 you get (1/9)*1=1/9 so you get 1/9+1=10/9 or 1 1/9. E [/hide]


[hide]she drove for 13 hours. to drive it in 12 hours, she needs to drive 54.166667 or $4\frac{1}{6}$ faster[/hide]

[quote="anirudh"][quote="aznmonkey1992"]gilda drove 650 miles at 50 mph. how much faster should she go to take 1 hour less?[/quote]

rt=d

50t=650

t=13

to take 12 hrs:

r12=650

r=54?[/quote]

They asked for how much faster, you're on the right track  :)

4 mph. i read the problem wrong again.  :wallbash_red:

[quote="aznmonkey1992"]cant there be someone else? moans[/quote]
Thers's me :lol:  :)   :D
[hide="myanswer"]650/50=13
650/13=54 and 1/6
she needs to go 4 and 1/6 mph faster [/hide]

[hide]
650/50 is 13 hours.  So she wants to drive it in 12 hours so 650/13=54 and 1/6. so the answer is 4 and 1/6 :) [/hide]

[quote="aznmonkey1992"]gilda drove 650 miles at 50 mph. how much faster should she go to take 1 hour less?[/quote]
[hide]
650/50 = 13 hours.
to get 12 hours total she shouldve drove an average of 650/12 = 54 and 1/6 mph.

so she needed to increase the average by 4 and 1/6.

[/hide]

uh oh. Do we have THINKING involved here? We will have no thinking in this fourm.
Moved to MOEMs. :D

[quote="aznmonkey1992"]gilda drove 650 miles at 50 mph. how much faster should she go to take 1 hour less?[/quote]
[hide]

She took 13 hours if she drove at 50 mph.
To get 12 hours, she would have to drive at 650/12 mph which is 54 and 1/6 mph
She needs to go $\boxed{4\frac{1}{6}}$ mph faster[/hide]


$ \Why\do\you%Error. "wantto" is a bad command.
%Error. "know" is a bad command.
\?$.

Lol.... I'll follow the pattern and say female.  :lol:

You know, I have a couple accounts named LilyGirl in some places so you can't assume anything on the internets.

I've got big jugulars and a hole between my legs so I must be a female. Oh wait, sorry, I forgot that I'm an ubék ambassador from the planot shnortlekzi.

Okay...so I'm curious to know...what was the purpose of this thread?  Was it the statement I made in the last such thread? (which is now Lazarus's sig)

So many of these threads and polls have been made, it really makes me wonder why.  We've seen so many of these male/female threads that you have little chance of getting truthful answers.  You can probably deduce about many people by the way they post, their blogs, their sigs, avatars, etc.

I'm going to go ahead and answer basketball.

[quote="Lazarus"]$ \text{Female}$.[/quote]

nice signature it works perfectly well with the topic

btw to break the chain i'd say male. No point of pretending to be the opposite gender.

[quote="i_like_pie"][quote="HangOut"]you a boy or girl? i mean i cant tell from ur online name unless its like lily10 or like john101010 or something like that[/quote]
This is at least the third time a topic like this has been created, and I'm getting tired of it.

[quote="K81o7"]I'm going to go ahead and answer basketball.[/quote]
Same here. :rotfl:[/quote]

LIAR, U R 11% MALE, 98% FEMALE, 1% PURE CRACK

Oh, I forgot to mention that I'm 12% violin, 23% orange and 57% enzyme

I think it's around 55% Mathematics, 13% Chemistry, 8% Physics, 10% Biology, 15% English, 9% Gender Confused.

Am I the only one that is 50% bisexual and 50% transexual?


Prove the following inequality for arbitrary positive real numbers a,b,c:

$ \sqrt {\frac {a}{8b \plus{} c}} \plus{} \sqrt {\frac {b}{8c \plus{} a}} \plus{} \sqrt {\frac {c}{8a \plus{} b}} \geq 1$

I've tried numerous methods, but there haven't been anything close to success..

[quote="qwerty414"]Prove the following inequality for arbitrary positive real numbers a,b,c:

$ \sqrt {\frac {a}{8b \plus{} c}} \plus{} \sqrt {\frac {b}{8c \plus{} a}} \plus{} \sqrt {\frac {c}{8a \plus{} b}} \geq 1$
[/quote]
The contradiction method works here:
Let $ \frac {a}{8b \plus{} c}\equal{}\frac{x^2}{9},$ $ \frac {b}{8c \plus{} a}\equal{}\frac{y^2}{9}$ and $ \frac {c}{8a \plus{} b}\equal{}\frac{z^2}{9},$ where $ x,$ $ y$ and $ z$ are positive numbers.
Then $ 57x^2y^2z^2\plus{}8(x^2y^2\plus{}x^2z^2\plus{}y^2z^2)\equal{}81$ and we need to prove that $ x\plus{}y\plus{}z\geq3.$
Let assume $ x\plus{}y\plus{}z<3,$ $ x\equal{}ku,$ $ y\equal{}kv,$ $ z\equal{}kw,$ where $ k>0$ and $ u\plus{}v\plus{}w\equal{}3.$
Then $ k<1$ and $ 57u^2v^2w^2\plus{}8(u^2v^2\plus{}u^2w^2\plus{}v^2w^2)>81,$ which is a contradiction.

[quote="arqady"][quote="qwerty414"]Prove the following inequality for arbitrary positive real numbers a,b,c:

$ \sqrt {\frac {a}{8b \plus{} c}} \plus{} \sqrt {\frac {b}{8c \plus{} a}} \plus{} \sqrt {\frac {c}{8a \plus{} b}} \geq 1$
[/quote]
The contradiction method works here:
Let $ \frac {a}{8b \plus{} c} \equal{} \frac {x^2}{9},$ $ \frac {b}{8c \plus{} a} \equal{} \frac {y^2}{9}$ and $ \frac {c}{8a \plus{} b} \equal{} \frac {z^2}{9},$ where $ x,$ $ y$ and $ z$ are positive numbers.
Then $ 57x^2y^2z^2 \plus{} 8(x^2y^2 \plus{} x^2z^2 \plus{} y^2z^2) \equal{} 81$ and we need to prove that $ x \plus{} y \plus{} z\geq3.$
Let assume $ x \plus{} y \plus{} z < 3,$ $ x \equal{} ku,$ $ y \equal{} kv,$ $ z \equal{} kw,$ where $ k > 0$ and $ u \plus{} v \plus{} w \equal{} 3.$
Then $ k < 1$ and $ 57u^2v^2w^2 \plus{} 8(u^2v^2 \plus{} u^2w^2 \plus{} v^2w^2) > 81,$ which is a contradiction.[/quote]
Can you give other examples?

[quote="Sunjee"]
Can you give other examples?[/quote]
See here ( my last post ):
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64122

is there any other solutions such as using MV method or Holder?? Thank you

[quote="arqady"][quote="qwerty414"]Prove the following inequality for arbitrary positive real numbers a,b,c:

Then $ 57x^2y^2z^2 \plus{} 8(x^2y^2 \plus{} x^2z^2 \plus{} y^2z^2) \equal{} 81$[/quote][/quote]

why? :maybe:

[quote="greentreeroad"][quote="arqady"][quote="qwerty414"]Prove the following inequality for arbitrary positive real numbers a,b,c:

Then $ 57x^2y^2z^2 \plus{} 8(x^2y^2 \plus{} x^2z^2 \plus{} y^2z^2) \equal{} 81$[/quote]

why? :maybe:[/quote][/quote]
Because $ 57x^2y^2z^2 \plus{} 8(x^2y^2 \plus{} x^2z^2 \plus{} y^2z^2)\equal{}\frac{57\cdot9^3abc\plus{}8\cdot9^2\sum ab(8a\plus{}b)}{(8a\plus{}b)(8b\plus{}c)(8c\plus{}a)}\equal{}81.$ :wink:

[quote="qwerty414"]Prove the following inequality for arbitrary positive real numbers $ a,b,c$:

$ \sqrt {\frac {a}{8b \plus{} c}} \plus{} \sqrt {\frac {b}{8c \plus{} a}} \plus{} \sqrt {\frac {c}{8a \plus{} b}} \geq 1.$[/quote][quote="arqady"]The contradiction method works here:

Let $ \frac {a}{8b \plus{} c} \equal{} \frac {x^2}{9},$ $ \frac {b}{8c \plus{} a} \equal{} \frac {y^2}{9}$ and $ \frac {c}{8a \plus{} b} \equal{} \frac {z^2}{9},$ where $ x,$ $ y$ and $ z$ are positive numbers.

Then $ 57x^2y^2z^2 \plus{} 8\left(y^2z^2 \plus{} z^2x^2 \plus{} x^2y^2\right)\equal{} 81$ and we need to prove that $ x \plus{} y \plus{} z\geq3.$

Let assume $ x \plus{} y \plus{} z < 3,$ Then[/quote]$ 57x^2y^2z^2 \plus{} 8(y^2z^2 \plus{} z^2x^2 \plus{} x^2y^2)$

$ < 57x^2y^2z^2\left(\frac {3}{x \plus{} y \plus{} z}\right)^6 \plus{} 8\left(y^2z^2 \plus{} z^2x^2 \plus{} x^2y^2\right)\left(\frac {3}{x \plus{} y \plus{} z}\right)^4$

$ \equal{} 81 \minus{} \frac {81}{2(x \plus{} y \plus{} z)^6}\sum{(y \minus{} z)^2\left[15x^4 \plus{} 6x^3(y \plus{} z) \plus{} 149x^2yz \plus{} 5x(y \plus{} z)(2y^2 \plus{} 11yz \plus{} 2z^2) \plus{} (y \plus{} z)^4\right]}$

$ \leq 81,$ which is a contradiction.


yo. I do. I suck at it. but it's nice to know a couple of other people that do.

I've heard of it before, but never actually got to see the cards up close.

@erabel:what types of decks do you have?

i have a white protection/lock deck
i have a green "stopmy" deck, where everything is less than 2 mana and i have alot of pumping cards->fast kill before turn 5
i have a blue/black deck that is a couter/destroy deck
i have a green snake deck
i  have an artifact deck(which is awesome, mycosynth golem ftw)
and i have more decks i'm too lazy to add...

I have a similar blue/black deck. I don't play anymore though.

I have a cousin who has designed a few of the cards, but I never play.

MTG FTW!! I play when I get the chance, but I had yet to find anyone else around here who actually played...sweet :lol:

I have a control deck, a block-constructed merfolk decking deck, and the usual burn spell deck.

My experiences with burn decks were bad. I usually lose before I burn them. Burn/Counter decks are better.

But burn spell decks need every spell they can get.  Add in a few cards like Browbeat or Harmonize to draw some cards, or Kavu Predator to prevent lifegain, and you have a good deck.

(my burn deck is Red/Green).

[quote="mathblitz"]what is mtg????????[/quote]

Magic the Gathering, as said in the subject. It's a card game.

Lol, I use an online version of Yugioh. Hehe.

Never played MTG, but I Yugioh, I have a burn deck, water deck (stil testing), and a toon deck (kinda sucks, still testing). Haven't been able to play recently, though.

[quote="cyberspace"]I prefer decks that the creatures have low mana cost so I can get them out really fast. It seems to work, strength in numbers I suppose.
Oh, and walls rock :lol:[/quote]

Walls do rock...when you animate them.  Or have Doran the Siege Tower.

[quote="cyberspace"]I prefer decks that the creatures have low mana cost so I can get them out really fast. It seems to work, strength in numbers I suppose.
Oh, and walls rock :lol:[/quote]

unless someone owns them with pro creatures and/or first/double strike (assuming your walls have first strike as some do)

[quote="azn6021023"]
Walls do rock...when you animate them.  Or have Doran the Siege Tower.[/quote]
What's Doran the Siege Tower do? I've never heard of that one... :huh:

new card.  Let's toughness decide combat damage instead of power.

the new cards are ruining the game.
that's probably a reasno why I quit.
it was ok in the 8th/9th editions

Yeah I haven't gotten any new cards for awhile, so all of mine are 8th edition.


[quote="color"]This was not that messy post but in the future use $\text{\LaTeX}$.

Your post would then read:
[hide="check out the code"]What is f(x) in this equation:f(x)+f(1-\frac{1}{x})=x([color=red]remember[/color] the dollar signs!)[/hide]
$\hline$
What is $f(x)$ in this equation:

$f(x)+f(1-\frac{1}{x})=x$[/quote]

Thanks man, I did as you told me.
But I still waiting for someone to help me :maybe:


function
complex numbers
number theory
relatively prime

A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a$ and $b$ are positive numbers.  This function has the property that the image of each point in the complex plane is equidistant from that point and the origin.  Given that $|a+bi|=8$ and that $b^2=m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$

[hide="Solution"]
We are given that $(a+bi)z$ is equidistant from the origin and $z.$  This translates to \begin{eqnarray*} |(a+bi)z-z| &=& |(a+bi)z|\\ |z(a-1)+bzi| &=& |az+bzi|\\ |z||(a-1)+bi| &=& |z||a+bi|\\ (a-1)^2+b^2 &=& a^2+b^2\\ &\Rightarrow& a=\frac 12 \end{eqnarray*}

Since $|a+bi|=8,$ $a^2+b^2=64.$  But $a=\frac 12,$ thus $b^2=\frac{255}4.$  So the answer is 259.[/hide]

I feel like the test makers could not have asked this question in a more ambiguous way if they tried.

PROBLEM
A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a_{}$ and $b_{}$ are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin. Given that $|a+bi|=8$ and that $b^2=m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n.$

SOLUTION
We know that $|f(z)|=|z-f(z)|$ because the image of each point in the complex plane is equidistant from that point and the origin, so
$$|f(z)|^2=|z-f(z)|^2 \implies$$$$|a+bi|^2 \cdot |z|^2 = |z|^2-z\overline{f(z)}-\overline{z}f(z) + |f(z)|^2 \implies$$$$64|z|^2=|z|^2-z\overline{z}(a-bi)-z\overline{z}f(z) + |a+bi|^2|z|^2 \implies$$$$64|z|^2=|z|^2-|z|^2(a-bi)-|z|^2(a+bi) + 64|z|^2 \implies$$$$64=1-(a-bi)-(a+bi) + 64 \implies$$$$a = \frac{1}{2} \implies b = \frac{255}{4}$$$255+4=\boxed{259}$

[hide=bash]Let $z=c+di$. We have that $f(z)=(a+bi)(c+di)=ac-bd+(bc+ad)i$. The distance from $f(z)$ and $z$ is the same as $f(z)$ and the origin, so we simply have $(ac-bd)^2+(bc+ad)^2=(ac-bd-c)^2+(bc-ad-d)^2$. After some [hide=expansion/simplification]$$(ac-bd)^2+(bc+ad)^2=(ac-bd-c)^2+(bc-ad-d)^2$$ $$\Longrightarrow a^2c^2+b^2d^2+b^2c^2+a^2d^2=a^2c^2+b^2c^2+a^2d^2+b^2d^2+c^2+d^2-2ac^2-2ad^2$$ $$\Longrightarrow c^2+d^2-2a(c^2+d^2)=0$$ $$\Longrightarrow (1-2a)(c^2+d^2)=0$$.[/hide] we are left with $(1-2a)(c^2+d^2)=0$. Since $c^2+d^2$ doesn't necessarily always equate to $0$, we have $1-2a=0 \Longrightarrow a=\frac{1}{2}$. Thus, from the fact that $|a+bi|=8$, we have $a^2+b^2=64 \Longrightarrow b^2=64-\frac{1}{4}=\frac{255}{4}$. Finally, $255+4=\boxed{259}$.[/hide]

Let $Z$ denote the point corresponding to $z$ and $Z_1$ denote the point corresponding to $f(z)$  in complex plane.Let $O$ be the origin. $|a+bi|=8 \implies OZ_1=|f(z)|=8|z|=8OZ$ . the problem condition gives $OZ_1Z$ isosceles triangle. Let $M$ be the midpoint of $OZ$. $arg(a+bi)=\arccos(\frac{OM}{OZ_1})=\frac{1}{16}$. So, $a+bi=8(\frac{1}{16}+\frac{\sqrt 255}{16}i) \implies b^2=\frac{255}{4}$. So, $m+n=259$

The condition that we are given can be written as $|(a+bi)z-z|=|(a+bi)z|$. We want to factor out $z$, so we can rewrite (using property that $|p||q|=|pq|$) our equation as $|z||(a+bi)-1| = |z||a+bi|$, so $|(a+bi)-1|=|a+bi|$. But this can only happen if $|a-1|=|a|$, so $a = \frac{1}{2}$. But we are given the absolute value condition, so $b^2 = \frac{256}{4} - \frac{1}{4} = \frac{255}{4}$. The answer is $\boxed{259}$.

The only thing you need to obtain $a=1/2$ is to plug in $z=1$, since now $f(z)$ must lie on the line $\mathrm{Re}(w)=1/2$, and the rest is easy. No actual computation is needed.


Let n>1 integer, p>2n prime, and \[S(p,n)=\sum_{i=1}^{n}(-1)^{i}\frac{{2n-3 \choose n-i}-{2n-3 \choose n-i-2}}{i}{ip\choose p}.\] 1. Prove that $p^{2n-1}|S(p,n)$.
2. Prove that exist infinetely many pairs (p,n), suth that $p^{2n}|S(p,n)$.
3. (I don't know solution) Prove that $p^{2n+1}\not |S(p,n)$.


function
group theory
number theory
superior algebra
superior algebra unsolved

Let us denote by $ E_n$ the set of integers $ r$ such that $ 1\leq r\leq n\minus{}1$, and $ (r,n)\equal{}1$. Now let $ G_n\equal{}\lbrace \hat{r} : r\in E_n \rbrace$. Then $ G_n$ is a group, and it is called the ``reduced set of residue classes''. It can easily be shown that the order of $ G_n$ is $ \phi(n)$.

My question:
Is it possible to find an action on $ G_n$, which, after using the orbit-stabilizer theorem, gives us some beautiful identity between arithmentic functions? I mean, the orbit-stbilizer theorem gives $ |orb|*|stab|\equal{}|G|$. Here, $ |G_n|\equal{}\phi(n)$, and thus the theorem becomes $ |orb|*|stab|\equal{}\phi(n)$. I think this formula shouts that there is something behind, joining group theory and number theory.

Is anyone aware of something?
Thanks.

This question is at once too vague and too specific.  There are lots of connections between group theory and number theory and many ways in which group actions are useful.  Some things you might want to look at:

http://qchu.wordpress.com/2009/06/16/gila-iii-the-orbit-counting-lemma-and-baby-polya/
http://qchu.wordpress.com/2009/06/21/gila-v-the-polya-enumeration-theorem-and-applications/
http://qchu.wordpress.com/2009/08/23/newtons-sums-necklace-congruences-and-zeta-functions/


Lets see, I see surds I think conjugates.. multiplying ksqrt(k+1) + (k+1)sqrt(k) by its conjugate gives us k(k+1).. so its equivalent to the sum of k(k+1)/(krt(k+1) - (k+1)rtk) which is the sum of rt(k+1) - rtk.
Summing this from k=1 to 4012008 should give us rt(4012009), which, oh look at that, comes out nicely to 2003.

Is this a challenge or a request for help?
[hint deleted]


Complementary Probability Part 1

We learn that sometimes it's easier to find the probability of what we don't want than what we do want.  We can then use the probability of what we don't want to quickly find the probability we want.


Find all $ m\in N$ such that

$ \frac{2^ma^m\minus{}(a\plus{}b)^m\minus{}(a\minus{}b)^m}{3a^2\plus{}b^2}$

is an integer for all nonzero integers $ a,b$.

If I"m not wrong, this problem is from Bulgarian competition  :wink:


$d$ is a positive integer and $f : [0,d] \rightarrow \mathbb{R}$ is a continuous function with $f(0) = f(d)$. Show that there exists $x \in [0,d-1]$ such that $f(x) = f(x+1)$.

It's easy to see that defining $g(x)=f(x+1)-f(x)$, $g$ is continuous and
\[ g(0)+g(1)+\ldots +g(d-1)=0 \]
so either $g(k)=0$ for some $k$, or in the sequence $g(0),g(1),\ldots ,g(d-1)$ there is a change of sign, so $g$ must be zero somewhere.
The problem is a version of the classical: let $f: [0,1]\to \mathbb{R}$ be continuous such that $f(0)=f(1)$. Prove that for any positive integer $n$ there exists $x$ such that $f\left( x+\frac{1}{n}\right)=f(x)$.
Very interesting is the converse (see Iaglom & Iaglom, "Non-elementary Problems With Elementary Solutions"): if we replace $\frac 1n$ with $d\ne \frac 1n$ for no $n$, the above is not true...

[quote="enescu"]if we replace $\frac{1}{n}$ with $d\ne \frac{1}{n}$ for no $n$, the above is not true...[/quote]
Can you post prove of this statement?

$d={3\over 2}$ and
$f(x)=\{\begin{array}{cl}x&\text{for }x\in [0,{1\over 2}]\\{3\over 2}-2x&\text{for }x\in ({1\over 2},1]\\ x-{3\over 2}&\text{for }x\in (1,{3\over 2}] \end{array}$
gives $f: [0,d]\to\mathbb{R}$ continuous with $f(0)=f(d)$ and $f(x)\not=f(x+1)$ for all $0\leq x\leq d-1$.

But with other $d$, In fact $d\not\in\{\frac{1}{n}|n\in\mathbb{N}\}$ ?

${1\over n+1}<d<{1\over n}$ for some $n\in\mathbb{N}$. Then $0<1-nd<d$ and
$f(x)=\{\begin{array}{cl}{nx-k\over 1-nd}&\text{for }x\in [kd,kd+1-nd],k=0,1,\ldots,n\\ &\\{(n+1)x-(k+1)\over 1-(n+1)d}&\text{for }x\in [kd+1-nd,(k+1)d],k=0,1,\ldots,n-1 \end{array}$

gives $f: [0,1]\to\mathbb{R}$ continuous with $f(0)=f(1)=0$ and $f(x)-f(x+d)=1$ for all $0\leq x\leq 1-d$.

The counterexample is very nice. Thanks all people!


Hey all,

I know this is a bit tooooo early for the UGA math tournament. But it is only two weeks away. Who is going?

We're coming (north FL school), but we only have 2 of our best...6+ people coming.  So doing well is not seeming too likely  :P

hm, it appears that the online website isn't working....anyone else having this problem? :-/

yeah, the cached version on google of the main website works, but none of the links work...>__>

[quote="timwu"][quote="Benjamin Hu"]Northview is sending a total of 14 people there, including me, Sitan, Arvind, Yeonji, Nathan, and Mohan and Krishna (of course, this is only a partial list).[/quote]

I remembered there was a limit of 12 people from one school?

Parkview will bring 4-5 people.

BTW, the website is back.[/quote]

The thing is, we originally were to send 12 people, but in the end, 3 people wanted to go as individuals. In total, that's 15 official people from Northview at UGA. :)

EDIT: I was a bit disappointed in my team's performance this year. If either the top scorer from the B team or the C team had replaced either of our two lowest scores, we could have won. :( Oh well...let bygones be bygones.
Also, I have reason to believe the machine graded my form with 2 blank instead of 4 (probably due to incomplete erasure on my part), but I'm not complaining about getting 4th place.

Congratulations to Oliver, Sean (whoever that is), Tim, and everyone else in the top 40, as well as the top 10 teams!

I'm Sean.  :P

Congrats to everyone who placed.  That was my first time at UGA but from what I've seen online I thought that was harder than previous years.

Hi folks,

Good tournament and congrats to all. It's always good to see competition from out of state.

Ben, you need to convince other folks from Northview to go to Furman in March. It's a great tournament (similar to UGA) and I'm pretty sure you would enjoy it.

Tom

Thanks, Mr. Fulton. I'll be sure to tell the team about it. :)


Does anyone know when the Toronto Math Circles starts?  :)

Also, is it at the Fields Institute again this year?  :)

[quote="eddyliu"][quote="rem"][quote="instantnoodles"]started awhile ago, its in the fields institute, same time (1-3pm)[/quote]
On Saturdays.[/quote]

Hey alex do you run that one with konstatine too?[/quote]
Yes we do, bi-weekly 11-1 on Sundays at University of Toronto.

[quote="rem"][/quote][quote="eddyliu"][quote="rem"][quote="instantnoodles"]started awhile ago, its in the fields institute, same time (1-3pm)[/quote]
On Saturdays.[/quote]

Hey alex do you run that one with konstatine too?[/quote][quote="rem"]
Yes we do, bi-weekly 11-1 on Sundays at University of Toronto.[/quote]
ahhh...sorry cant make those anymore...waterloo is a little too far. but maybe ill find time to drop by meh


geometric transformation
geometry solved
geometry

In a triangle $ABC$ circles $o_1,o_2,o_3$ are tangent respectively to $AB$ and $AC$, $BC$ and $AB$, $AC$ and $BC$. Circle $s$ is tangent externally to circles $o_1,o_2,o_3$ respectively in $K,L,M$. Show that $AK$, $BL$, $CM$ are concurrent.

[[b]Moderator edit:[/b] This is equivalent to the four circles theorem, discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=5364 .]

The triangles $ABC$ and $KLM$ are perspective with perspectrix the line throught the exsimilicenters of  $(O_1,O_2) ; (O_2,O_3); (O_1,O_3)$. Therefore $AK, BL, CM$ are concurrent.

Wow, I like your solution - I've learned many things from just reading these few words (and some more studying later  ;) ), thanks  :)


inequalities
search
triangle inequality
real analysis
real analysis unsolved

Could someone please help me show that the triangle inequality holds for Canberra metric ?
\[ d(x,y) \equal{} \displaystyle\sum_{i \equal{} 1}^{n}\frac {|x_i \minus{} y_i|}{x_i \plus{} y_i},
\]
x_i and y_i are nonnegative real numbers
Thanks!

It suffices to show the inequality for one dimension- adding up copies doesn't change anything.

WLOG, $ x>y>z$. We want to show $ \frac{x\minus{}y}{x\plus{}y}\plus{}\frac{y\minus{}z}{y\plus{}z}\ge \frac{x\minus{}z}{x\plus{}z}$.
$ \frac{x\minus{}y}{x\plus{}y}\plus{}\frac{y\minus{}z}{y\plus{}z}\minus{}\frac{x\minus{}z}{x\plus{}z}\equal{} \frac{(x\minus{}y)(y\plus{}z)(x\plus{}z)\plus{}(x\plus{}y)(y\minus{}z)(x\plus{}z)\plus{}(x\plus{}y)(y\plus{}z)(z\minus{}x)}{(x\plus{}y)(x\plus{}z)(y\plus{}z)}$
The numerator is
$ (x^2y\plus{}x^2z\minus{}y^2x\minus{}y^2z\plus{}z^2x\minus{}z^2y)\plus{} (x^2y\minus{}x^2z\plus{}y^2x\plus{}y^2z\minus{}z^2x\minus{}z^2y)\plus{} (\minus{}x^2y\minus{}x^2z\minus{}y^2x\plus{}y^2z\plus{}z^2x\plus{}z^2y)$
$ \equal{}x^2y\minus{}x^2z\minus{}y^2x\plus{}y^2z\plus{}z^2x\minus{}z^2y\equal{}(x\minus{}y)(x\minus{}z)(y\minus{}z)\ge 0$

Why "Canberra metric?"  (My search engine has failed me :(.)

I don't know why, but I know for sure that this is it's name...
Thanks!

may be [url=http://www.ucl.ac.uk/oncology/MicroCore/HTML_resource/Distances_detailed_popup.htm]this helps[/url]


How many positive integers < 500 has exact fifteen positive integers dividers.

A) 0
B) 1
C) 2
D) 3
E) 4

[hide="Solution"]

A number whose prime factorisation is $ p^{a}q^{b}$ has $ (a \plus{} 1)(b \plus{} 1)$ factors.

$ 15 \equal{} 1\times 15, 3\times 5, 5\times 3$ or $ 15\times 1$

$ (a \plus{} 1)(b \plus{} 1) \equal{} 15$ gives $ (a, b) \equal{} (0, 14), (2, 4), (4, 2)$ or $ (14, 0)$

Checking with the primes $ 2, 3, 5, 7$, the only ones which are smaller than $ 500$ are $ 324 \equal{} 2^{2}\times 3^{4}$, $ 144 \equal{} 2^{4}\times 3^{2}$ and $ 400 \equal{} 2^{4}\times 5^{2}$.

So the answer is D) 3.

[/hide]


[quote="espark52"]if b and 7 are solutions to the equation:$x^{2}-4x=a$, find the value of a+b[/quote]
[hide]So the equation is $x^{2}-4x-a=0$. So the factored result is $(x blank)(x blank)=0$. We need to fill up these blanks. But one of the roots is 7 so in the first blank is -7. Then whatever is in the second blank has to add up to -4. So the second blank is 3. So now we have $(x-7)(x+3)=0$. Foiling we get $x^{2}-4x-21=0$ or $x^{2}=4x=21$. So b=-3 and a=21, so a+b=18.[/hide]

[hide]Since the sum of the solutions is $-\frac{b}{a}$, $b+7=4$ and $b=-3$. The product of the solutions is $\frac{c}{a}$ so $-a=-21$ and $a=21$. $a+b=21-3=18$[/hide]

[hide=":maybe: "]yes, 18 is correct[/hide]<-- click me


algebra
polynomial
function
calculus
calculus computations

Prove:
If f is n times differentiable on an interval I and L is the degree at most n-1 polynomial which agrees with f at the points $ x_1 ,...,x_n$ in I then for each x in I there is an $ \xi$ in I for which
$ f(x) \minus{} L(x) \equal{} \frac{{(x \minus{} x_1 )...(x \minus{} x_n )}}{{n!}}f^{(n)} (\xi )$

done by cases:

case 1: $ x\equal{}x_k$ for some k=1,2.. n

then we are done, since both sides are zero

case 2: $ x\ne x_k$ for any k=1,2.. n

then create auxiliary function

$ g_x(y)\equal{}f(y)\minus{}L(y)\minus{}[f(x)\minus{}L(x)]\prod_{k\equal{}1}^n \frac{y\minus{}x_k}{x\minus{}x_k}$

then $ g_x$  has n+1 zeros, $ y\equal{}\{ x, x_1, ..., x_n\}$

using rolles thm n times we know that

$ \exists \xi \in (\min (x,x_1), \max(x,x_n) )$ s.t. $ g^{(n)}(\xi ) \equal{} 0$

so $ 0\equal{}g^{(n)}(\xi ) \equal{} f^{(n)}(\xi)\minus{}L^{(n)}(\xi)\minus{}[f(x)\minus{}L(x)]\left(\prod_{k\equal{}1}^n \frac{\xi\minus{}x_k}{x\minus{}x_k}\right)^{(n)}$

$ \equal{}f^{(n)}(\xi) \minus{} [f(x)\minus{}L(x)] (\xi^n \plus{} a_{n\minus{}1}\xi^{n\minus{}1} \plus{} ... \plus{}a_0)^{(n)}\prod_{k\equal{}1}^n \frac{1}{x\minus{}x_k}$
$ \equal{}f^{(n)}(\xi) \minus{} [f(x)\minus{}L(x)] \prod_{k\equal{}1}^n \frac{n!}{x\minus{}x_k} \to f(x)\minus{}L(x) \equal{} \frac{f^{(n)}(\xi)}{n!}\prod_{k\equal{}1}^n (x\minus{}x_k)$

thanks kenn4000, I knew either the Lagrange finite increment theorem (in the 'n times' way you used) or the Taylor theorem were necessary but I did not know what function to concoct. How did you figure out that auxiliary function?


[hide="hint"]
Find the intersection of the number of mulitples of even numbers up to 100 and is the number of mulitples of odd numbers up to 100. Then cancel out all the numbers in the intersection.
[/hide]

[quote="redsmartie"]On graduation day, 100 grade 8 students lined up outside the school.
As they entered the school, they passed their lockers. There are 100 lockers.
The first student opened all the locker doors.
The second student closed every second locker door.
The third student changed the position of every third locker.
The fourth student changed the position of every fourth locker.
This pattern continues.

Which doors are open after 100 students have entered the school?[/quote]
[hide="Solution"]Consider the factors of each locker number. All of them except perfect squares have an even number of factors, which means they will be closed after all 100 people go through. All the lockers with perfect square numbers will remain open because their position was switched an odd number of times.[/hide]

[quote="abacadaea"][hide="hint"]pigeon hole principle[/hide][/quote]
How does that help? My solution is perfectly fine without it. :huh:

sorry i didnt read it. but yeah your solution works too

The only ones that r left open r the ones with an odd number of factors, wat numbers have those?

Hint:

The first are 1,4,9,...

Quickster please stop reviving old topics.

[quote="besmart"]what if there are 1000 lockers?[/quote]

Then find the square number that is less than 100, which would be $ 31^2$

So there are 31 lockers open.

[quote="andrewliu"]the only way a number can have an odd number of factors if its a square.... or zero......[/quote]

Every positive integer is a divisor of $ 0$, so technically, $ 0$ has an infinite number of divisors.

I just used trial and error.  I tried the pattern on a sequence of 10 lockers and 10 students.

[hide][hide]I found that, after the first student... all lockers were open
after the second, all the even numbers were closed
after the third, numbers 1,5, and 7 were open.
after the fourth, numbers 1,4,5,6,7, and 8 were open.
after the 5th... numbers 1,4,6,7,8, and 10.
the 6th... 1,4,7,8,10.
7th... 1,4,7,8,10.
8th... 1,4,10
9th... 1,4,9,10.
10th... 1,4,9.

Then I saw how 1,4, and 9 are the first 3 perfect squares, and figured that, with 100 lockers and 100 students, only the perfect-square numbered lockers would be open.
[/hide]
Therefore, the open lockers are numbers: 1,4,9,16,25,36,49,64,81, and 100.[/hide]


calculus derivative geometry trigonometry function inequalities quadratics

Bah, you can. Although I didn't mean it was THAT ugly. :D

There are solutions without using analytic geometry if that's what you meant ;)

[quote="Peter VDD"]Be $R$ a positive real number. If $R, 1, R+\frac12$ are triangle sides, call $\theta$ the angle between $R$ and $R+\frac12$ (in rad).

Prove $2R\theta$ is between $1$ and $\pi$.[/quote]

I actually don't find it so ugly...

Here is my solution:

At first, clearly $0<\theta <\pi $, since $\theta$ is an angle of a triangle. We must prove $1<2R\theta <\pi $.

Since the numbers R, 1, $R+\frac12$ are the sides of a triangle, we must have $R+\left( R+\frac12\right) >1$, what simplifies to 4R > 1. We denote u = 4R and thus have u > 1.

Now, the semiperimeter of our triangle is

$s=\frac{1}{2}\left( R+1+\left( R+\frac{1}{2}\right) \right) =R+\frac{3}{4}=\frac{1}{4}u+\frac{3}{4}=\frac{1}{4}\left( u+3\right) $

(since $R=\frac{1}{4}u$, as u = 4R). After the half-angle formulas, we have

$\displaystyle \sin ^{2}\frac{\theta }{2}=\frac{\left( s-R\right) \left( s-\left( R+\frac{1}{2}\right) \right) }{R\left( R+\frac{1}{2}\right) }$.

This quickly simplifies to

$\displaystyle \sin ^{2}\frac{\theta }{2}=\frac{3}{u\left( u+2\right) }$.

Now we will show $1<2R\theta $. This is obviously equivalent to

$\displaystyle \frac{\theta }{2}>\frac{1}{4R}$,

i. e. to

$\displaystyle \frac{\theta }{2}>\frac{1}{u}$.

This, in turn, is equivalent to

$\displaystyle \sin ^{2}\frac{\theta }{2}>\sin ^{2}\frac{1}{u}$,

since the function $\sin ^2 x$ increases on $\displaystyle \left[ 0;\;\frac{\pi }{2}\right] $, and both numbers $\displaystyle \frac{\theta }{2}$ and $\displaystyle \frac{1}{u}$ lie in this interval (since $\theta <\pi $ implies $\displaystyle \frac{\theta }{2}<\frac{\pi }{2}$, and u > 1 implies $\displaystyle \frac{1}{u}<1<\frac{\pi }{2}$).

Now, in order to prove

$\displaystyle \sin ^{2}\frac{\theta }{2}>\sin ^{2}\frac{1}{u}$,

we will show

$\displaystyle \sin ^{2}\frac{\theta }{2}>\left( \frac{1}{u}\right) ^{2}$;

in fact, once this will be shown, from x > sin x for positive x we will immediately get $\displaystyle \sin ^{2}\frac{\theta }{2}>\left( \frac{1}{u}\right) ^{2}>\sin ^{2}\frac{1}{u}$.

Since $\displaystyle \sin ^{2}\frac{\theta }{2}=\frac{3}{u\left( u+2\right) }$, the inequality

$\displaystyle \sin ^{2}\frac{\theta }{2}>\left( \frac{1}{u}\right) ^{2}$

that we must prove simplifies to

$\displaystyle \frac{3}{u\left( u+2\right) }>\left( \frac{1}{u}\right) ^{2}$.

This one, after multiplication with $u^{2}\left( u+2\right) $, takes the form 3u > u + 2, or, in other words, 2u > 2, what is clearly true because u > 1. Thus we have proven the left bound $1<2R\theta $.

Remains to establish the right bound $2R\theta <\pi $. Again, this is equivalent to

$\displaystyle \frac{\theta }{2}<\frac{\pi }{4R}$,

i. e. to

$\displaystyle \frac{\theta }{2}<\frac{\pi }{u}$.

Now, this is trivial for $u\leq 2$ (since in this case we just have $\displaystyle \frac{\theta }{2}<\frac{\pi }{2}\leq \frac{\pi }{u}$); thus, in the following, we will only consider the case u > 2. Then, again, both numbers $\displaystyle \frac{\theta }{2}$ and $\displaystyle \frac{\pi }{u}$ lie in the interval $\displaystyle \left[ 0;\;\frac{\pi }{2}\right] $ (in fact, for $\displaystyle \frac{\theta }{2}$ we have already showed this, and for $\displaystyle \frac{\pi }{u}$ it's clear because $\displaystyle \frac{\pi }{u}<\frac{\pi }{2}$ follows from u > 2). Hence, instead of proving

$\displaystyle \frac{\theta }{2}<\frac{\pi }{u}$,

it is enough to show that

$\displaystyle \sin ^{2}\frac{\theta }{2}<\sin ^{2}\frac{\pi }{u}$,

i. e. to show that

$\displaystyle \frac{3}{u\left( u+2\right) }<\sin ^{2}\frac{\pi }{u}$

(here we have used our formula $\displaystyle \sin ^{2}\frac{\theta }{2}=\frac{3}{u\left( u+2\right) }$ again).

To prove the inequality $\displaystyle \frac{3}{u\left( u+2\right) }<\sin ^{2}\frac{\pi }{u}$, we take its reciprocal:

$\displaystyle \frac{u\left( u+2\right) }{3}>\csc ^{2}\frac{\pi }{u}$.

Now we subtract 1 on both sides:

$\displaystyle \frac{u\left( u+2\right) }{3}-1>\csc ^{2}\frac{\pi }{u}-1$.

But $\displaystyle \frac{u\left( u+2\right) }{3}-1=\frac{\left( u-1\right) \left( u+3\right) }{3}$ and $\displaystyle \csc ^{2}\frac{\pi }{u}-1=\cot ^{2}\frac{\pi }{u}$. Thus, we have to prove

$\displaystyle \frac{\left( u-1\right) \left( u+3\right) }{3}>\cot ^{2}\frac{\pi }{u}$.

Again, we take the reciprocal (we can do it since both sides of our inequality are positive, the left one since u > 1 and the right one since $\displaystyle \frac{\pi }{u}<\frac{\pi }{2}$):

$\displaystyle \frac{3}{\left( u-1\right) \left( u+3\right) }<\tan ^{2}\frac{\pi }{u}$.

In order to verify this, it is enough to show that

$\displaystyle \frac{3}{\left( u-1\right) \left( u+3\right) }<\left( \frac{3}{u}\right) ^{2}$;

in fact, once this will be shown, we will immediately get

$\displaystyle \frac{3}{\left( u-1\right) \left( u+3\right) }<\left( \frac{3}{u}\right) ^{2}<\left( \frac{\pi }{u}\right) ^{2}<\tan ^{2}\frac{\pi }{u}$,

where we have used $3<\pi $ and x < tan x for all $\displaystyle x\in \left[ 0;\;\frac{\pi }{2}\right] $. So we have to show that

$\displaystyle \frac{3}{\left( u-1\right) \left( u+3\right) }<\left( \frac{3}{u}\right) ^{2}$.

After multiplication with $u^{2}\left( u-1\right) \left( u+3\right) $, this takes the form $3u^{2}<9\left( u-1\right) \left( u+3\right) $. Dividing by 3 and simplifying, we can rewrite this inequality as $2u^2+6u-9>0$. But the roots of the quadratic trinomial $2u^2+6u-9$ are $\displaystyle -\frac{3}{2}+\frac{3}{2}\sqrt{3}$ and $\displaystyle -\frac{3}{2}-\frac{3}{2}\sqrt{3}$, both < 2. Since u > 2, we thus have $2u^2+6u-9>0$. And this completes the proof of the right bound.

This ends the solution of the problem. But what makes me a bit mistrustful is the wealth of traps and possibilities to make some dumb errors in this solution. I have fixed some before posting, but I am still not at all sure I have no mistakes left now...

  Darij

[quote="darij grinberg"]This ends the solution of the problem. But what makes me a bit mistrustful is the wealth of traps and possibilities to make some dumb errors in this solution. I have fixed some before posting, but I am still not at all sure I have no mistakes left now...
  Darij[/quote]

Hell yeah, such a solution! :D I'll let you know if I find a mistake, but it can take a while before I'm finished with such a long answer.

Anyway the official sols (there were 3 I guess) all were less than half a page ;)

[quote="darij grinberg"]This ends the solution of the problem. But what makes me a bit mistrustful is the wealth of traps and possibilities to make some dumb errors in this solution. I have fixed some before posting, but I am still not at all sure I have no mistakes left now...
  Darij[/quote]

  Darij,

  Time is ripe to fix your solution to this problem a bit.  It must be clear that

  when the difference between two sides of a triangle is fixed, then the vertex

  moves on hyperbola (in this case  focal distance 2c=1 and 2a=1/2).

  The upper bound is obvious, the lower can be worked out  rather quickly.

   I leave you the honors to clean up your own stuff, so that the gatekeeper

  of 'Posterity' will undoubtedly  induct you in there too.

   Thank you.

   M.T.


I want to add a horizontal line between two paragraphs in a document.  I tried

[code]
<text>
\[
\hline
\]
<text>
[/code]

but I get a slew of error messages.  Is there a simple way to add this horizontal line?  Thanks.

Try
[quote]<text> \\

<text>[/quote]
End first paragraph wit two backslashes, the leave a blank line in your text, then start new paragraph.

The reason that you get error messages is that \hline is for tables and \ [...\ ] is for maths mode ie mathematical expressions. Use
\rule[ [i]lift [/i]]{[i]width[/i]}{[i]height[/i]}
so for a line between two paragraphs you could write
[code]text \\
\rule{\linewidth}{0.1pt}
text [/code]
or if you want the line a little higher then use
[code]text\\
\rule[0.5ex]{\linewidth}{0.1pt}
text[/code]


logarithms
number theory unsolved
number theory

Find all pairs (a, b) of positive integers that satisfy (a^b)^2 = b^a.

I noticed that (0,0) and (1,1) would do fine. But I'm not able to proof this.

[quote="seamusoboyle"]What is $0^0$?

0 or 1? Or something else?[/quote]

it is mostly considered as 1, but actually it is not really defined (I think)

[quote="seamusoboyle"]So i dont see how $(0,0)$ is a solution.[/quote]

0^(2*0) = 0^0

The problem asks for POSITIVE integers...

see here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3845#p3845

[quote="seamusoboyle"]What is $0^0$?

0 or 1? Or something else?[/quote]

$0^0$ is not defined.  In practice, it will never be evaluated except as a limit, where you would use logarithms and then l'Hopital's.


[quote="gauss1181"][quote="tpulak"]heres the problem  " A fair coin is tossed four times, and at least one of the tosses results in heads. What is the probability that exactly two tosses result in heads? Express your answer as a common fraction."[/quote]
Of the possibilities, there can be either 1, 2, 3, or 4 heads. So there are a total of 15 possibilities. (2^4-1=15) this excludes the no-heads possibility. So there are 4 possible permutations for 1 head, 4 for 3 heads, 1 for 4 heads, and 6 for 2 heads. The probability is 6/15, which is simplified to 2/5.[/quote]
@gauss1181: Why do you subtract 1? I have done the same thing except subtracting 1 from $ 2^4$.. Can you explain?

[quote="tpulak"]heres the problem  " A fair coin is tossed four times, and at least one of the tosses results in heads. What is the probability that exactly two tosses result in heads? Express your answer as a common fraction."[/quote]
[hide="My Solution"]
Since at least one of the coin tosses results in a head, you are looking for the chances that only 1 of the 3 coin tosses results in a head.
So the chances that there is 1 head on the first toss and none on the last two is $ \frac{1}{8}$
Same thing for the second, and third tosses.
So $ (3)\frac{1}{8}\equal{}\frac{3}{8}$
Which is your answer.
[/hide]

I think the answer is [hide]6/15. As proof, these are the possible flips, and only 6 of them have exactly 2 heads.
(H,H,H,H), (H,H,H,T), (H,H,T,H), (H,H,T,T), (H,T,H,H), (H,T,H,T), (H,T,T,H), (H,T,T,T), (T,H,H,H), (T,H,H,T), (T,H,T,H), (T,H,T,T), (T,T,H,H), (T,T,H,T), (T,T,T,H)[/hide]

Yes, that's right, vamathletes, but it's just that the answer is 2/5 when expressed in lowest terms.

[quote="alkjash"]This is conditional probability:
It doesn't tell you which one comes up heads, it only tells you that one heads came up,
so you can actually only rule out one case, the one with all tails.[/quote]
Wait, so you're saying that the correct solution would be [hide="?"]$ \frac{3}{7}$?[/hide]

How did you get $ \frac{3}{7}$?!?

There are $ 2^4 \equal{} 16$ possible combos. You rule out the one that is all tails. There are 15.
6 have exactly two heads (number of ways to choose 2 out of four to be heads).
Thus, the probability is $ \frac{6}{15}\equal{}\boxed{\frac{2}{5}}$

[quote="alkjash"]How did you get $ \frac {3}{7}$?!?

There are $ 2^4 \equal{} 16$ possible combos. You rule out the one that is all tails. There are 15.
6 have exactly two heads (number of ways to choose 2 out of four to be heads).
Thus, the probability is $ \frac {6}{15} \equal{} \boxed{\frac {2}{5}}$[/quote]
 :oops:
Whoops.
I thought you were correcting the other people's work, so I simply did $ \frac{6}{15\minus{}1}$ to get $ \frac{3}{7}$

[quote="tpulak"]heres the problem  " A fair coin is tossed four times, and at least one of the tosses results in heads. What is the probability that exactly two tosses result in heads? Express your answer as a common fraction."[/quote]

Um... this requires conditional probability...

...in Australia you learn conditional probability in Year 11.

The formula for Pr(B|A) (B given A), is $ Pr(B|A) \equal{} \frac {Pr(B \cap A)}{Pr A}$

$ Pr(At least 1) \equal{} 1 \minus{} \left(\frac {1}{2}\right)^4 \equal{} \frac {15}{16}$

$ Pr(Exactly two heads) \equal{} {4 \choose 2} \left(\frac {1}{2}\right)^2 \left(\frac {1}{2}\right)^2 \equal{} \frac {3}{8}$

$ Pr(At least 1 \cap Exactly two heads) \equal{} \frac {3}{8}$

$ Pr(Exactly two heads | At least 1) \equal{} \frac {Pr(At least 1 \cap Exactly two heads)}{Pr(At least 1)}$$ \equal{} \frac {\frac {3}{8}}{\frac {15}{16}} \equal{} \frac {2}{5}$


I know, I know, week 2 answers still not up, yeah yeah yeah, I'll get around to it soon, I promise. In the meantime, discuss week 4 problems here. Yes, the sheets say week 3, sorry about that, I've been writing out my lessons without taking the practice sprints into acount. 9th graders, where were you? Not that I mind AT ALL that you guys didn't show up, but you said you would so why the change of plans?

Week 4 (says 3 on sheet) plane geometry, class set:
1. 40+20sqrt(5)
2. 36-9pi
3. 4+sqrt(2)
4. 3sqrt(2)-3 (You might have it in a different form. Rationalize please.)
5. 210
6. 25/2
7. 6sqrt(3)+11pi
8. 26
9. 5/3
10. 408

Homework set
1. c. 28
2. 63
3. 17/3
4. 3
5. 46
6. 5sqrt(5)
7. 4pi/3-2sqrt(3)
8. 150
9. 120/17
10. 672/5

I hope those are right.

lame... i don't come... cuz ccy is there

oh wait did I say I would come.....
sry kinda have a lot of hw today(history notebook due tomarrow and I still need to do 5 prompts aslkdhgalskdhgaskldghklashg+a test alskdghlkasdhgkladsgjlk)

ps.off topic, but I dont care. did you get [hide="ccy dont look"]
$2^{5}3^{2}$[/hide] for #4 w00t p set?

[quote="junggi"]ps.off topic, but I dont care. did you get [hide="ccy dont look"]
$2^{5}3^{2}$[/hide] for #4 w00t p set?[/quote]
Yes, I think so. Darn it, I still haven't even looked at any more of the problem set since those four problems. Still haven't done the message board problems either. Darn it. I should start.

woot? thats the math thing right? that this forum offers

I can't edit the first post anymore...darn it. Anyway, answers! Finally!
Week 4 (says 3 on sheet) plane geometry, class set:
1. 40+20sqrt(5)
2. 36-9pi
3. 4+sqrt(2)
4. 3sqrt(2)-3 (You might have it in a different form. Rationalize please.)
5. 210
6. 25/2
7. 6sqrt(3)+11pi
8. 26
9. 5/3
10. 408

Homework set
1. c. 28
2. 63
3. 17/3
4. 3
5. 46
6. 5sqrt(5)
7. 4pi/3-2sqrt(3)
8. 150
9. 120/17
10. 672/5

I hope those are right.


calculus
integration
logarithms
algebra
partial fractions
calculus computations

This is algorithmic ... decompose your rational fraction first ...

$\frac{A}{x}+\frac{B}{x-3}+\frac{C}{x+3}$
$A(x-3)+B(x)(x+3)+c(x-3)=2x^{2}-9x-9$
$Ax-3A+Bx^{2}+3B+Cx-3C$
$Bx^{2}+(A+C)x+(-3A-3C)$
B=2
A+C=-9
-3A-3C=-9

Is this what you meant?  I'm stuck!  I'm not even sure if I'm in the right direction.    :(

That's the general idea, but you need to check your algebraic details. For one thing, I get that $B=-1.$

To be specific, the second line of your post, just after you multiplied by the common denominator, is incorrect.

And are you familiar with shortcut ways of finding the partial fractions coefficients corresponding to simple ([i]i.e.[/i], non-repeated) linear factors that avoid systems of equations?


A basketball player makes her free throws $ 55\%$ of the time. What is the probability that she will make both of her next two free throws? Express your answer as a common fraction.


Does anybody here participate in the Science Olympiad competitions?

I'm new to Science Olympiad. I'm going to join this year though.  ;)

My high school is very new, this is its first year..my friend and I are trying to start a science olympiad team, but no one seems really interested.

I [i]was[/i] going to do it last year but somehow I forgot about it. :?

My middle school does it, and apparently idoes well. But I can't do it because it's quite hard to participate in mathcounts and science olympiad.

That reminds me...i need to also start some kind of math club at my school for AMC 10/12...my school is stupid

w00t for science olympiad! My schedule does not allow me to do it anymore, but I did div. C Chem Lab, Polymer Detectives, Mission Possible, Robot Ramble, Compute This, and Qualitative Analysis a couple years ago. Got 5th and 4th in state in Polymer Detectives and Chem Lab all by myself :)

What are your favorite events?

Hey, my sci oly team's shirt says "w00t" on it.  I do Physics Lab, Remote Sensing, Fermi Questions, Chem Lab, Forensics or whatever it's called and PDG.

I'm basically a mission possible maniac... I've been doing it for 4 years. I've placed in states a couple of times in MP.

[quote="ZennyK"][quote="frankthetank"]The MP guy at my school used to carry around his chemicals in Poland Spring bottles.  Once he left a bottle full of Pb(NO3)2 in his hotel room.  If someone got thirsty and decided to drink from the bottle, they would be in for a nasty surprise.  He also used to use a UPS so that he could solder stuff on the bus.[/quote]
LOL! :rotfl: I still have an Arrowhead water bottle full of sodium carbonate (not as toxic as lead nitrate, I'm afraid) sitting in my room! I guess I'm not the only one . . . how freaky is THAT? I guess we MP people are weird like that. :lol:[/quote]

That's hilarious.  :rotfl:

[quote="elementsofmathkid"]I was in science olympiad for like a week in middle school, but quit, because I wanted to do track and I couldn't stand it because the science olympiad had no organization (and still doesn't have any).

I want to join at my new school now, because it's a different coach.[/quote]
Yeah, I know from experience, too, what a nightmare SO can be if your team isn't organized!

[quote="Treething"]lol, science olympiad is so fun...

Third year this year, and I'm, well, upset that they took away Write it Do it and Picture This.

Incidentally, I'm from Communist Middle School, top placing Division B school in 2005. First place baby.[/quote]
Where is your school?

How come $NONE$ of you post in the physics forum??No,my school doesn't take part in olympiads..

I used to do it; My main event was Experimental Design, which I got medals at state and regionals for. I also did some of the other ones - mostly dealing with physics and chem.


Is there a connection between the surjectivity of a function onto a space V and the compactness of V?

This is too vague.  Without more information about the function and about the domain there's nothing to say.


$ \frac{1}{x^{2}+1}$ = $ \frac{Ai+B}{x+i}$ + $ \frac{Ci+D}{x-i}$ Find the real numbers A,B,C,D

[hide="solution"]$ R=\frac{(Ai+B)(x-i)+(Ci+D)(x+i)}{x^{2}+1}$, so it must be:
$ (Ai+B)(x-i)+(Ci+D)(x+i)\equiv 1$
$ x[i(A+C)+B+D]+A-C+i(D-B)\equiv 1$
$ \{\begin{array}{l}A+C=0 \\ B+D=0 \\ A-C=1 \\ D-B=0\end{array}$
And now solve it :P [/hide]


function calculus integration limit derivative algebra domain

Suppose D is an open subset of the complex field. Let f_n be an analytic function from D to the complex field for all naturals n.
Suppose that f_n converges to f locally uniformly on D. Show that f is also analytic.

Bomb

Since I haven't seen it elsewhere: locally uniformly=uniform on compact sets?

The Cauchy integral formula gets you the same convergence for the derivative. A standard real analysis theorem says that uniform convergence of both $ f_n$ and $ f_n'$ gets you that $ \lim f_n$ is differentiable, with derivative $ \lim f_n'$.

Local uniform convergence means that for each x in the domain, there exists an r > 0 such that f_n converges to f unformly on the open ball B(x,r). What is this standard real complex theorem that you refer to?
If its differentiable, then how is f analytic?

Since you can patch together any compact set from finitely many of those balls, it's equivalent to what I said. Compact sets are more natural for discussing uniform convergence everywhere.
Differentiable [b]is[/b] analytic. We are talking about the complex derivative here.

How did you deduce that f is differentiable from the Cauchy Integral formula?

I didn't, and I already said exactly where differentiability comes from: (locally) uniform convergence of the derivative.

Using the Cauchy Integral Formula with various circles gets you explicit estimates of the rate of convergence of the derivative inside those circles.

Use the Morera theorem, it takes only one or two lines.

What Arne points to (I think):

Being analytic can be expressed by terms of integrals only. By standard analysis, integrals are stable under uniform convergence. Then the result follows.


Without using integration,

what integer is this sum closest to?

(The numbers in the square roots increase in subsequent fractions by 1.)

1/sqrt(100.5) + . . . + 1/sqrt(9999.5)

Note:  For your possible help, there are 9900 terms here.

*** Everyone, as a change (because there has not been a seen attempt in this thread),

you MAY use integral calculus on this problem, BUT you don't need it.

"Lower level" math will do it also.

A question from the team round of a [url=http://www.csulb.edu/depts/math/mathday/index.htm]local contest[/url] I was involved in:
Find the integer part of $ \sum_{k\equal{}1}^{2007^{2}}\frac{1}{\sqrt{k}}$

This was intended as a calculus question; although many individual competitors didn't have calculus yet, every team had somebody in a calculus class.


Let $ X \subset \mathbb{R}^3$ be the union of n lines through the origin. Compute $ \pi_{1}\left(\mathbb{R}^3 \minus{} X\right)$.

$ S^2 \minus{} \{x_1,\ldots x_{2n}\}$ is a deformation retract of $ R^3 \minus{} X$
and $ S^2 \minus{} \{x_1,\ldots x_{2n}\}$ is homotopic to wedge sume of $ 2n \minus{} 1$ circles, so the fundamental group is isomorphic  to a free group with $ 2n \minus{} 1$ generators.


inequalities
geometry
circumcircle
inradius
incenter
inequalities unsolved

Let $ABC$ be a [b]acute[/b] triangle and let $C(I,r)$ be the its incircle.

Prove that the following inequality: $\boxed {IA+IB+IC\ge 3\sqrt{2Rr}}\ \ (1)$.

[hide="Remark."] From the [u]Erdos-Mordell's inequality[/u] results the inequality

$IA+IB+IC\ge 6r$ and $3\sqrt{2Rr}\ge 6r\ !$.

Therefore, the proposed inequality $(1)$ strengthens the Erdos-Mordell's

inequality in the particular case when the triangle $ABC$ is [b]acute[/b].[/hide]

Since $\displaystyle IA=\sqrt{\frac{bc(b+c-a)}{a+b+c}}$ we have to prove that $\displaystyle \sum\sqrt{\frac{b+c-a}{a}}\geq 3$.

And so on ? [u]Atention[/u]: in the [b]acute[/b] triangle !!

Let $i=C(I,r)$, $e=C(O,R)$ be the incircle and the circumcircle respectively for the [b]acute[/b] triangle $ABC$. I note the second intersections $A'$, $B'$, $C'$ with the circumcircle $e$ of the lines $AI$, $BI$, $CI$ respectively. Is it true the following inequality $IA'+IB'+IC'\ge IA+IB+IC\ ?$

It was posted but $\sum IA'\geq \sum IA$(see Gi1,Mircea Lascu)

Thanks, Vladut (Matei) How are you ? Write on private message.

The inequality to be proved is a special case of a (still unsolved) inequality submitted on the Olympiad Geometry Unsolved Problems Forum 24/5/2005 by Lomos Lupin under heading "Relations between distances of a point and triangle"

Take point P inside the triangle ABC.Denote the respective distances from P to the vertices A, B, C by x, y, z  and the respective distances from P to the sides BC, CA, AB by u, v, w.  R is the circumradius and r the inradius of the triangle. Then we have the inequality:

xy + yz + zx >= 2R/r.(uv + vw + wv)

Take P at  the incentre I and write AI =x,  BI =y,  CI =z. Then we will have u=v=w=r. And using the well known inequality,
 (x+y+z)^2 >=3(xy+yz+zx), the required result follows.

It is also true that AI+ BI +CI <= SQRT{ s^2 + 9(r^2)} where 2s =a+b+c. Can anyone prove this ?


Hey, does anyone know about the difficulty of the AMC 12B 2002 test? I took it as practice, and it seemed easier than the previous ones. Normally, I can answer the first 14 questions. But on this one, I answered the first 20 questions. So, for those of you who have done it, is it easier by comparison to the previous ones? Or have I improved over the last few weeks? Hopefully, this is not giving me false hope..thanks!

Also, how do you do #7?
Question: The product of 3 consecutive positive integers is 8 times the sum. What is the sum of the squares?
I set the equations equal to each other, but could not get anything that resembled the sum of the squares..help? Thanks again.

For #7:
[hide]Call the integers n-1, n and n+1
Then:
(n-1)(n)(n+1)=8(n-1 +n +n+1)
(n-1)(n)(n+1)=24n
 Cancel the n's:

(n-1)(n+1)=24
Expand:
n^2 -1=24
n^2 =25
n=5
The three numbers are 4,5, and 6,
the sum of whose squares is 16+25+36=[b]77[/b][/hide]

I've taken the 2002B, and to me it was only a TAD bit easy. I scored a 126.5 while my usual scores fall in the 115-127.5 range. So, I'd say that to improve 6 questions is quite an achievement! Great Job!

[quote="stanford1463"]^Thanks! That's good news. So yea, for #7, instead of naming the integers "n-1,n,n+1" I named them "n,n+1,n+2" which turned out to result in some complicated equation that I couldn't simplify.[/quote]
For reference, your way would also work, but you would want to keep the product of the integers factored:

$ n(n\plus{}1)(n\plus{}2)\equal{}8[n\plus{}(n\plus{}1)\plus{}(n\plus{}2)]$
$ n(n\plus{}1)(n\plus{}2)\equal{}8(3n\plus{}3)$
$ n(n\plus{}1)(n\plus{}2)\equal{}24(n\plus{}1)$
$ n(n\plus{}2)\equal{}24$
$ n^2 \plus{} 2n \minus{} 24 \equal{} 0$
$ (n\plus{}6)(n\minus{}4) \equal{} 0$
$ n\equal{}4$
$ 4^2 \plus{} 5^2 \plus{} 6^2 \equal{} 77$


Yes, the proof I saw was also a projective one.

[b][u]theorem[/u]:three points $ A,B,C$ lie on line $ \ell$ (the order isnt important),and points $ A',B',C'$ lie on $ \ell '$ now let $ X=(AB' \cap A'B),Y=(BC' \cap B'C),Z=(AC' \cap CA')$,then $ X,Y,Z$ are colinear.[/b]

in the following proof,by $ (AB,CD)$ we mean the cross ratio of points $ A,B,C,D$
[i]proof[/i]:
let $ BZ$ intersect $ \ell '$ at point $ D'$ and $ BZ'$ intersects line $ \ell$ at $ D$...
now we have that:
$ (AB,DC)=(C'D',B'A')=(A'B',D'C') \Rightarrow B'(ABDC)=B(A'B'D'C')$
now note that $ B'(ABDC)$ and $ B(A'B'D'C')$ have a common line $ BB'$,so the intersection of other three lines,respectively, are colinear,thus $ X,Y,Z$ are colinear...

[quote="BaBaK Ghalebi"]
now note that $ B'(ABDC)$ and $ B(A'B'D'C')$ have a common line $ BB'$,so the intersection of other three lines,respectively, are colinear,thus $ X,Y,Z$ are colinear...[/quote]

Can someone explain me this last part?

P.S. sorry for reviving this old thread.

[quote="aidan"]Can anybody explain what Pappus' theorem is?[/quote]
You can also prove Pappus' Theorem using Pascal's Theorem.
Since Pascal's Theorem holds true even when the plane undergoes a Projective Transformation, therefore, it holds true for every conic.
As a pair of straight lines is nothing but a degenerated conic, Pascal's must also hold true for it. But this application of Pascal's, is nothing but Pappus' Theorem. :)


probability
calculus
search
function
expected value
geometric series

[b]Problem:[/b] The probability that a certain coin will land heads when tossed is $ p$ (hence, it is not necessarily a "fair" coin).  This coin is continuously flipped until $ 5$ heads have been flipped.  What is the expected value $ E$ of the number of flips?

I think that this problem is equivalent to $ E = \limits\sum^\infty_{n = 5}n\binom{n - 1}{4}p^5(1 - p)^{n - 5}$.  I'm posting this in the Pre-Olympiad section because I don't want any calculus involved.  This is because I don't yet know enough calculus to be able to evaluate infinite series such as this using it, hence I won't understand a calculus solution to this problem and the solution will be meaningless to me.
I want to know how to do a problem like this using clever pre-calculus methods.  I don't know how to use calculus to find out if an infinite series converges, and since I want pre-calculus solutions, you can ASSUME CONVERGENCE in your solution.  I used Mathematica to evaluate this sum and I got a result, so I know that the series does indeed converge.

This is not a contest problem, at least I don't think it is.  So I'm not trying to cheat on some contest or anything; the fact of the matter is (as the title of this topic says), I need a solution quickly (within hours).  Feel free to post a solution even tomorrow and after, but know that it won't be of concrete use to me (i.e., it'll mathematically help me, but not in the real world in which I need to find the solution by tomorrow).  This isn't a problem from a test either, and I am not cheating in any way that I know of to post this here, so I'm allowed to find the solution with others' help.

After an hour of work or so, I've reduced this problem to $ E = \frac {5}{p}\limits\sum^\infty_{n = 5}\binom{n - 1}{4}p^5(1 - p)^{n - 5}$.  But I can't seem to evaluate $ E$ even with this (simpler) sum to calculate.

Help is much appreciated.

EDIT: I suppose that you can use SOME calculus in the solution, but nothing involving derivatives/integrals.  This pretty much rips you off of all the calculus tools you have EXCEPT for limits.  So the only calculus that one might use in the solution is probably limits as $ n\rightarrow\infty$.

Also, as for rigor, the rigor of your solution should be greater than or equal to that of the following proof:

[b]Claim:[/b] The sum of the terms of the infinite geometric series $ a,ar,ar^2,\ldots$ is $ \frac{a}{1-r}$, assuming the series converges.

[b]Proof:[/b] Call the sum S.  Then we have
$ S=a+ar+ar^2+\cdots$ and multiplying by $ r$ yields
$ rS=ar+ar^2+\cdots$.
Subtracting the second equation from the first gives
$ S(1-r)=a\Rightarrow S=\frac{a}{1-r}$, which completes our proof.

Now, I understand that the proof of my problem will be more complex and will require more steps than the above proof for the formula of the sum of the terms of an infinite geometric series.  Use your judgment to decide on the rigor... I don't expect a perfect pre-calculus solution, but not one with any major holes in it.

Consider the following:

$ E_5 \equal{} \frac {1}{2}(E_4 \plus{} 1) \plus{} \frac {1}{2}(E_5 \plus{} 1)$ because there is a $ \frac {1}{2}$ chance the coin flip was heads, in which case the expected value will be $ E_4 \plus{} 1$, and if the coin flip was tails, we will have $ E_5 \plus{} 1$.

Now, we can solve for $ E_5$, and now we have an expression for $ E_4$. Now we repeat the same idea for $ E_4$, etc. etc. until we get $ E_1$.

[quote="MathAndKnowledge"][quote="math154"]For $ p \equal{} \frac12$, refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2104964214&t=248714]this[/url]...the general result can be derived if you replace everything with $ p$ and $ 1 \minus{} p$...

So I get $ E_n \equal{} (1 \minus{} p)(1 \plus{} E_n) \plus{} p(1 \minus{} p)(2 \plus{} E_n) \plus{} \cdots \plus{} p^{n \minus{} 1}(1 \minus{} p)(n \plus{} E_n) \plus{} p^n\cdot n$.[/quote]

No, this problem is far more difficult (or at least more difficult).  When 5 heads have been FLIPPED is the condition, not when 5 heads have been FLIPPED IN A ROW.

So the example you gave me does not help solve the problem.[/quote]

My apologies for misreading the question. :oops:

But as rofler said, the relevant recursion is $ E_n\equal{}p(E_{n\minus{}1}\plus{}1)\plus{}(1\minus{}p)(E_n\plus{}1)\equal{}pE_{n\minus{}1}\plus{}p\plus{}(1\minus{}p)E_n\plus{}1\minus{}p$
$ \implies pE_n\equal{}pE_{n\minus{}1}\plus{}1\implies E_n\equal{}E_{n\minus{}1}\plus{}\frac1p\equal{}E_{n\minus{}2}\plus{}\frac2p\equal{}\cdots\equal{}E_0\plus{}\frac np\equal{}\frac np$.

So for this problem, $ E_5\equal{}\frac5p$, which should also be intuitively correct.

It is enough to compute the expected time until a single head, since expected values are linear.  The expected number of flips until five heads are flipped will be five times the expected number of flips to obtain a single head.

So, the question is, what is the expected number of flips required to obtain a single head?

You can condition on the event that you get a head for the first time on the $ n\minus{}$th flip.  This means you need $ (n\minus{}1)$ tails beforehand and a single head on the $ n\minus{}$th flip.  This has probability $ (1\minus{}p)^{n\minus{}1}p$

So the expected number is $ P \equal{} \sum^\infty_{n\equal{}0} np(1\minus{}p)^{n\minus{}1}$.  This can be done without calculus, but in practice no one ever does it that way (if you must, try to 'reorder' the sum).  But the calculus solution only requires that you know derivatives.

Consider the function $ f(x) \equal{} \sum^\infty_{n\equal{}0} (1\minus{}x)^n$.  This is a geometric series, and it has sum $ 1/(1\minus{}(1\minus{}x)) \equal{} 1/x$.  Differentiating both sides gives $ f'(x) \equal{} \sum^\infty_{n\equal{}0} \minus{}n(1\minus{}x)^{n\minus{}1} \equal{} \minus{}1/x^2$.  Cancel out the minus signs, plug in $ p$ in both sides and multiply through by $ p$ to get that $ \sum^\infty_{n\equal{}0} np(1\minus{}p)^{n\minus{}1} \equal{} p/p^2 \equal{} 1/p$.  So $ P \equal{} 1/p$.

Finally, here's another version of the recursive solution, which is basically the easiest one.  Condition on what happens after one flip.  Either you get a head, in which case you're done, or you get a tail, and you are in the same place as when you started.  So, calling the expected number of flips $ P$, you get the equation $ P \equal{} p \plus{} (1\minus{}p)P$.  Solving for $ P$, we get $ P \equal{} 1/p$.

In any event, the answer is $ 5/p$, as noted earlier.

[quote="math154"]Haha...I'm so stupid. One can simply note that, since the probability of getting heads is $ p$, it will take $ \frac1p$ flips on average to get a head, so $ E_n \equal{} \frac np$.[/quote]

that's not a rigorous argument (well, unless you know something about the [url=http://en.wikipedia.org/wiki/Geometric_distribution]geometric distribution[/url]!)


here's how you play the SLAP YOUR KEYBOARD GAME.

you slap your keyboard, first of all.

example: cvbnfm

make sure that the "slap word" as i'll call it does not contain any numbers of symbols.

the next person makes an acronym for the above person's slap word.
example for word above:

chickens victimize banana nerds for me.

don't choose your letters specificly either. it has to be random!!!! that's why you smack the keyboard!!!!

here's my entry:

dstrf

Dancing soap to racing fishies!

My turn!!!
grdfs

Hehehehehe >: )

>>Great robots destroy free sodas.

vmdos

A Solid Door Accidently Lets Seawater Drain

lolz don't ask :P

oijsfd

Joe never concocts heavy jello effortlessly

jboiosjgk

Just boring or insecure or secure just good k?

wteihjiawije

What the eggnog? I hit jelly in a walrus. It just evaporated.
xD
hjcdfxmj

What to expect if Harry jumps in accidentally without intending jumping, excitedly :?:

Never mind:
hjcdfxmj:
Harry jumped coz dexter freaked xaenir's mind jingling.

sdfghjkxsdgfbkivdi
( :wallbash: I slammed my face on the keyboard,ow it hurt!)

so dead for good hearts! just kidding! xylophones stink. dogs go! follow bob k? i vent delusional illusion..
lolz that wuz hard... idk

uifhvioarejsroifgeojr

Ulsa, if Frank has violent ill orangutans, ask Raymond. Extra juice sauce rinsing off its fuzzy gummy excellent orange-juice rain.

 :rotfl:  :rotfl: IDK!!!

asdfpqcxvbzxzxcvbnm

GL on this :D

I bet this game will be really unpopular among laptop users...

v

Hmm, maybe my slap was too firm?

Hoardy Jimmy Died Fishing So He's Jolly :huh: idk lol

bobdylan(oops...how did my random letters make a name? :P  :huh: )

ok
hm
My elephants hug unanimously grounding himself
mine is
alsdfl

Awesome loving sad destructive flammable loser

dfhkidj

Daisies fly heretically kissing insane daffodils jealously.

dsflkjf

Dallying faces hit karen in daren's jaw.
alksjdf

All Lonely Kangeroos Seek Jolly Dirt Faries

ajeslig

Ashes joyfully eat snakes loving it graciously

you forgot the whole point of the game: SLAP YOUR KEYBOARD!!! :|

I will slap for you:
iwpveno

I wish people vented ends not objectively.

grehunh

gods reply happily unknowingly not! Hah.

wegj

queer beer flow at your intestines :help:  :fool:

oabciw

Only acronyms bring cheer in winter.

iguhljsaf

I graciously utilized her loving jaw sitting about free
alskdjf

a loving son kicks daddy joe furiously

gdsndnsf

game's dead. save now? dumb necro stuff fails.

siodsfa


Find the sum of the first 19 terms in the following of the sequence:

$1, 4, 5, 9, 14 ,23, 37...$

if the 19-th term is 11,933 and the 20-th term is 19,308.

[quote="236factorial"]brute force
[hide]31237[/hide][/quote]

lol, my first impression was that brute force would in fact yield an answer within the average amount of time given for a target question (with a calculator of course) however, a non-brute force method exists.....

[hide]
Assuming that I see the sequence right (fibinocci), we can generalise the formula.

Start with $a_1$ and $a_2$.

The sum of 1 to 1 is $a_1$

The sum of 1 to 2 is $a_1+a_2$

the sum of 1 to 3 is $a_1 + a_2 + a_3 = 2a_1 + 2a_2$

1-4 = $2a_1 + 2a_2 + a_4 = 2a_1 + 3a_2 + a_3 = 3a_1 + 4a_2$

checking:

$3(1) + 4(4) = 3 + 16 = 19$
$1 + 4 + 5 + 9 = 19$

ok so it's good so far.

1-5 = $3a_1 + 4a_2 + a_5$
$3a_1 + 4a_2 + a_3 + a_4$
$4a_1 + 5a_2 + a_4$
$4a_1 + 6a_2 + a_3$
$5a_1 + 7a_2$

look at the coefficients of $a_1$:

1, 1, 2, 3, 5

You could intuitively guess that the next coefficient will be 8.

Look at the coefficients of $a_2$:
0, 1, 2, 4, 7. You can intuitively guess that the next coefficient will be 11. Let's check our intuitive guesses.

1-6
$5a_1 + 7a_2 + a_6$
$5a_1 + 7a_2 + a_4 + a_5$
$5a_1 + 8a_2 + a_3 + a_5$
$6a_1 + 9a_2 + a_5$
$6a_1 + 9a_2 + a_3 + a_4$
$7a_1 + 10a_2 + a_4$
$7a_1 + 11a_2 + a_3$
$8a_1 + 12a_2$

our second intuitive guess is wrong, but a new sequence pops up.

to get the next sum, add the previous $a_1$ coefficient to the current one, and add the current $a_1$ coefficient to the current $a_2$ one.

so we get:

1 0
1 1
2 2
3 4
5 7
8 12
13 20
21 33
34 54
55 88
89 143

and here we see a new sequence. To get the next $a_1$ term, add 1 to the current $a_2$ term.

144 232
233 376
377 609
610 986
987 1596
1597 2583
2584 4180
4181 6764 <--- This is #19.

So the sum of the first 19 terms should be:

$4181 + 4(6764) = \boxed{31237}$

That took a lot longer than the target round would allow >_<

[/hide]

I'm sure there's a better way (why else would they give you $a_{20}$), but I don't know it.

My initial solution the problem was something similar to Hamster1800's (well-over the given time for a target problem :) )  But the solution provided with the problem is as follows:

[b]Elias Saab's Solution[/b]

Let $s[n]$ be the nth term.  It is clear that

$s[1] + s[2] = s[3]$

$s[2] + s[3] = s[4]$

$s[3] + s[4] = s[5]$

$s[4] + s[5] = s[6]$

....

$s[n-2] + s[n-1] = s[n]$

$s[n-1] + s[n] = s[n+1]$

  Adding the above euqalities and simplifying, we get:

$s[1] + 2 s[2] + s[3] +... + s[n-1] = s[n+1]$

  Adding $s[n] - s[2]$ to both sides, we get:

$s[1] + s[2] + s[3] +... + s[n-1] + s[n] = s[n] + s[n+1] - s[2]$

  The answer is:

$s[1] + s[2] + s[3] +... + s[19] = s[19] + s[20] - s[2] =$

$11933 + 19308 - 4 = \boxed{31237}$

Ahh, nice solution. I was trying to find some way of simplifying it, but didn't see that.

[quote="solafidefarms"]It's a Fibonacci-like sequence; the sum of the first n terms of a Fibonacci sequence is the n+2th term -1. That doesn't work here; can you find a formula that does? (It makes it a lot faster than ya'll's methods!)[/quote]

It'll probably take the whole target round to [i]think[/i] of that formula  ;)

I don't really see how many people would think of that.

In this case, it's f_(n+2)-4 that is the sum; just try summing a few and adapting it to that formula...might take 30 seconds.

[quote="solafidefarms"]In this case, it's $f_(n+2)-4$ that is the sum; just try summing a few and adapting it to that formula...might take 30 seconds.[/quote]

i worked for 10 minutes and never thought of that, should've been pretty obvious, but I wasn't concentrating enough  :blush:

lol, I already knew the formula provided by Elias Saab: namely the sum of the first n terms of a Fibbonaci-like (I forget the name) sequence is the nth term plus the n+1th term...


If $ x,y,z\in \mathbb{R}_\plus{}$ such that $ x\plus{}y\plus{}z\equal{}5$ , find the maximum value of :

$ P\equal{}x^y\plus{}y^z\plus{}z^x$

hello, my computiations give me that it must be  $ x\equal{}y\equal{}z\equal{}\frac{5}{3}$ and we have
$ x^y\plus{}y^z\plus{}z^x \le P(5/3,5/3,5/3)$.
Sonnhard.


group theory
abstract algebra
complex numbers
superior algebra
superior algebra unsolved

Let $ \mathbb{C}^*$ be mutiplitive group of nonzero complex numbers, and $ H$ be a subgroup of finite index in $ \mathbb{C}^*$. Prove that $ H\equal{}\mathbb{C}^*$.

we may replace $ \mathbb{C}^*$ by an arbitrary divisible abelian group. and having this said, the proof is straight forward by considering the quotient.

More clearly, if the quotient has order $ n$, then every $ n$-th power lies in $ H$.  But every member of $ \mathbb{C}^{*}$ is an $ n$-th power.

If $ C^*/H$ has $ n$ elements, then the $ n\minus{}$th power of any element of $ C^*/H$ is the identity, which is $ H$. Now any element $ z$ of $ C^*$ must lie in some coset of $ H$, so $ z^n$ must lie in $ H$. But since any element of $ C^*$ can be written as $ z^n$ for some $ z$, we have that $ H\equal{}C^*$.


If $ \underset{x\rightarrow p}{\lim}f\left( x\right) \equal{} \infty$, then $ \underset{x\rightarrow p}{\lim}\frac {1}{f\left( x\right) } \equal{} 0$

If $ \underset{x\rightarrow\infty}{\lim}f\left( x\right) \equal{} \infty$ , then  $ \underset{x\rightarrow\infty}{\lim}\frac {1}{f\left( x\right) } \equal{} 0$

If $ \underset{x\rightarrow\infty}{\lim}f\left( x\right) \equal{} 0$ , and  $ f\left( x\right) \neq0$ ,then$ \underset{x\rightarrow\infty}{\lim}\left\vert \frac {1}{f\left( x\right) }\right\vert \equal{} \infty$

Are they all correct above?

If they couldn't, have any counterexample above?

They are all correct. The first two are pretty much the same thing.


Let $ X$ random variable with density function:

$ f_x(x) \equal{} 2xe^{ \minus{} x^2}$  when $ x > \equal{} 0$ and 0 otherwise.

Let $ Y \equal{} X^2$

Find the distribution and density functions of Y.

How can I do that with first finding the Distribution function of x? Integrating the density function that is.


We know that the sum of the angles is $ 540$, so $ P\equal{}540\minus{}111\minus{}113\minus{}92\minus{}128\equal{}540\minus{}(444)\equal{}\boxed{96}$.


What is the maximum number of points of intersection of the graphs of two different fourth degree polynomial functions $ y \equal{} p(x)$ and $ y \equal{} q(x)$, each with leading coefficient $ 1$?

$ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$

We must have $ p(x) \minus{} q(x) \equal{} 0$; $ \deg{p \minus{} q}$ is the number of solutions to this (at most). Since $ \deg p,\deg q \equal{} 4$, $ \deg{p \minus{} q}\geq4$. However, since both are monic, the $ x^4$ term will cancel, and $ \deg{p \minus{} q}\ne4\implies\deg{p \minus{} q}\geq3\implies\boxed{\textbf{(C)}\ 3}$.

I agree with the reasoning above, but why does the official answer key say that $D$ is the answer?

The answer key does say $\boxed{C}$
Link: [url]https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_12[/url]

I looked at the answer key here: https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Answer_Key


algebra
polynomial
logarithms
calculus
calculus computations

I'm having trouble finding the solution to the following DE problem using the power series method:

$3x^2y''+x(1+x)y'-y=0$ about the point x=0.

Any help is appreciated.

Power series about what point? Note that $x=0$ is a singular point - to be specific, a "regular singular point." Read what your textbook has to say about such points, including the "Method of Frobenius."

That's the only singular point; if the series is supposed to be around any other point, you would use techniques appropriate for ordinary points.

all the question states is use the power series method to solve the given DE about the point x=0.  I'm trying to solve it the way I have been solving these types of problems, but I can't seem to find the solution.

A too-brief introduction to the method of Frobenius:

You cannot assume that the solution has a power series in the usual sense around a regular singular point. You can, however, find a modified power series.

Assume that $y(x)=\sum_{n=0}^{\infty}a_nx^{n+r}$ where $a_0\ne0$ and $r$ is a number (not necessarily an integer) to be determined from the equation for the leading coefficient.

so i use the modified $y(x)$ to find $y'$ and $y''$ and then substitute those back into the origianal equation and solve that way?

Yes. As a practical matter, I would write $xy'$ and $x^2y'$ as completely separate terms - easier to keep track of what you're doing that way. I personally find that sigma notation gets in my way when I'm doing this - I'd rather spread everything out using lots of paper (or chalkboard). Somewhere out there, after a "...," write a generic term, aiming for the same power of $x$ in each row.

Your first equation should read $a_0(\text{polynomial in }r)x^r=0.$ Use the fact that $a_0\ne0$ to set the polynomial equal to zero and solve for $r.$ With a second order equation, the polynomial should be quadratic. Take each possible value of $r$ as a separate case.

Do look this up in a standard textbook - there should be more explanation there.

alright, thanx for all your help...i'll check it out in the text book and see if i can figure it out.  If I can't, then I'll be back.  Thanx again.

is it possible to get only one value for $r$ or should I have two values since its a second order DE.  i'm getting $r=0$ as the only value

The polynomial equation should be $3r(r-1)+r-1=3r^2-2r-1=0$ for roots of $r=-\frac13,\,r=1.$

It is possible to have a double root. Double roots and roots that differ by an integer complicate the system and force the introduction of logarithms in the second solution. Fortunately, that didn't happen here.

ok I got it.  I just made an algebra mistake.  Thanx for your help.


Let $ F$  be  a given CDF. By Skorohod construction we know how to construct an identically distributed sequence of random variables $ X_n$ on a prob. space with CDF $ F$. But how do we construct a probability space $ (\Omega, \mathcal{F}, \mathbb{P})$ and  a sequence of[b] independent[/b] and identically distributed random variables $ X_n$ on it with distribution $ F$?

For each random variable construct a different probability space carrying them.  Then take a direct product of all those spaces and define the value of the nth random variable only by the value of the nth coordinate.

By construction those rv are independent and by Skorohod construction you know that those identically distributed rv exist.