In triangle $ABC$, let the incircle touch side $BC$ at $P$. $R$ is an arbitrary point on side $BC$. The incircles of $ABR,ACR$ has centers $X,Y$ respectively. Show that $P$ lies on one of the internal tangent of the two circles.

It's sufficient to show that $\angle XPY=90$. To prove that we must show that $XRPY$ is cyclic. Let K,L be points of tangency of incinters of $ACR$, $ABR$.  By easy couting we can show that $RL=KP$.

[quote="wojto111"]It's sufficient to show that $\angle XPY=90$. [/quote]

Eh, why?

Take on $BC$ point $T$ such that lie on internal tanget to this circles. Mark points of tangency to the circles $ACR$,$ABR$ by $R$,$S$. Draw this and you will see :P.


Dakle... što vam se najčešće vrti u CD playeru? (ili gramofonu, kazetofonu, mp359 playeru ili čemu već)

Ja bih izdvojio Beatlese i Stonese, iako znam poslušati i balkanskije stvari tipa Rundek, Gibonni, Balašević, Azra, Indexi, Britney itd...

 :lol:

Meni ne treba klasična poezija jer, da citiram njemačkog matematičara Dehna, Теоријска математика је поезија логичних идеја. ("Pokradeno" s sajta Društva matematičara Srbije, hence the cyrillic.)

Uz neke iznimke dosta naginjem domaćim stvarima (naravno, ne folku  :noo:  :help:  :diablo: ), Balašević mi je for the win, a još povremeno slušam (puno rjeđe nego Balaševića, naravno  :D ) Zaklonišče prepeva, Die Toten Hosen, Hladno pivo, Zabranjeno pušenje, Edo Maajka, Slon in Sadež (dobro, i Rundek, Gibonni, Šerbedžija, Bijelo dugme, Van Gogh, Marčelo, Dubioza Kolektiv... u tragovima). E, da, Orthodox Celts su isto for the win, thanks Jelena.  :lol:

Eto, svašta, uglavnom.

Btw, baš razmišljam da napravim neku listu svih estradnih nadriumjetnika koje naprosto ne mogu podnijeti i onda, kad pričam s nekim, a on mi, recimo, ovlaš spomene kako mu je naš "najveći domoljubni pjesnik" (znate ga sigurno, fin jedan gospodin, prozvao se po automatskom oružju) srcu drag ili kako je upravo bio na nastupu godine nekog od najrenomiranijih turbofolk izvođača (u narodu poznatom kao "noć dugih noževa"), ja samo kažem "OK, ispričaj me sada, idem zaroniti glavu u akvarij... bit će mi ugodnije pričati s ribama." [size=75]No, dobro, ne valja biti isključiv, a i morao bih dosta često držati glavu u akvariju.[/size]

Nije fer... Da sad napišem šta slušam, izgledalo bi kao quote kolege Ornika...  :lol:

No, možda uspijem štogod dodati na listu: dakle, osim pobrojanih u postu iznad, u moj player zalutaju i slijedeći izvođači, pobrojani nasumičnim redom: [b]Miladin Šobić[/b] (umjesto pitanja: (t)ko je ovaj koje vam se upravo mota po glavi, navratite na dzuboks.de i skinite par pjesama  :lol:  ), [b]Letu štuke[/b], [b]Bajaga[/b], stare stvari od [b]Čorbe[/b], [b]Salvatore Adamo[/b], [b]Weird Al[/b], [b]Panic(!) at the disco[/b], [b]Nick Cave[/b], [b]Guns'n'Roses, Iron Maiden, Pink Floyd[/b]... razvuče se ovaj spisak, izgleda da slušam sve što mi dođe pod ruku i svidi mi se...

Ah da, zaboravio sam sve nabrojane ispod:

[quote="Bora Đorđević"]Na priredbi kod Frankfurta
peva Kurta, peva Murta
Sinan, Hasan, Jasar, Saban
Nino, Seki i jos neki i naravno Dzej

Mala Maja Marijana
Anci, Brzi i Dajana
Keba, Kemis, Rodja, Jami
svi zajedno u galami i naravno Dzej

Is’o sam kod oni stranci
da im otmem njini Franci
imam jednu zelju zarku
da im uzmem koju marku

Mica, Cica i Zorica
Zoki, Boki, Rodic Roki
Sladja, Kobra, Stela, Nela
slatka mala Karamela i naravno Dzej

Lepa Brena, Ekstra Nena
Lepa Lana i Cakana
Lepi Bora, Lepi Gagi
Lepi Mica, svi su dragi, i naravno Dzej

Era, Gedza, Bekic Beki
i naravno Super Sneki
i sa njima zmija Veki
Viki, Siki, Viki, Riki i naravno Dzej

Peva Sinko i Marinko
Mitar Miric, Skoric Mira
dok muzika grozno svira
peva hitic Mile Kitic i naravno Dzej
[/quote]

Ispričavam se zbog nepoznavanja srpske kulture ali tko bi bio gore više puta spomenuti gospodin Dzej? :D

Napisao je Čorba još puno dobrih stvari, kao ona [i]Neko mi je ukrao biciklo[/i] ili [i]Veče ne miriše na rakove i školjke[/i] (ili je to bio Bajaga? genijalna pjesma)

I ako netko voli punk-rock, preporučam album Marquee Moon benda Television koji je stvarno genijalan i ujedno zahvaljujem Bodanu što me upoznao s njim. :)

Džej Ramadanovski: Omaleni pjevač romskog podrijetla, s evidentnim manjkom kose, hiperaktivan na nastupima  :lol:

Što se pjesme "Kad hodaš (Veče ne miriše na rakove i šoljke)" tiče, da, napisao ju je Bajaga, tadašnji član Riblje Čorbe.

Ne mogu vjerovati da sam zaboravio Milana Manojlovića - Mancea (http://www.last.fm/music/mance - ima tamo i free download, pa poslušajte). Ima mnogo impresivnih pjesama, a svakako preporučam Studijski je zapis loš, Kafić bezveze i, recimo, Javorova grana. No, posebno je interesantna njegova pjesma Dva, čiji tekst ćete naći dolje:

[quote="Mance"]
Ima jedan broj
A djeljiv je sa pet
U nizu brojeva
Djeljivih sa šest

Svaki novi broj
Podjeljen sa sedam
Umanjen za tri
Pomnožen sa pet

To je dva, dva
dva je labud moj
dva, dva, dva je tajni broj

Pomnožen sa dvanaest
Umanjen sa tri
Ima jedan broj
Broj tramvajski

A to je dva, dva
Dva do Žitnjaka
Dva, dva, dva do neba sad

Ima mnogo brojeva
Djeljivih sa pet
Neki pak su djeljivi
I sa brojkom šest, sa šest

Ali dvaaaa, dvaaa
Dva do Žitnjaka
Dvaaaaaaaaaaaaa do neba

Dva je tajni broj
Dva je labud moj
Dvaaa, Dvaaa
Dva u dvanaest četiri
[/quote]

I ne mogu vjerovati, čitajući Harunov popis, da sam zaboravio navesti Bajagu i Weird Ala.  :(


1. 2004 can be expressed as the sum of distinct positive numbers  of the same digits. For example 725 + 752 + 527 = 2004 or 617 + 671 + 716 = 2004 or 509 + 590 + 905 = 2004.  2003 can also be expressed as the sum of positive numbers with the same digits. What are they?

2. A string is wrapped around the earth's equator (wow! that's a long string) and the two ends just touch.  Suppose that another string is tied to the original string so it becomes 30.48 m longer.  If this new string is placed around the equator and pulled tight so that it is suspended in air, how high would the string be above the ground?

please clarify what the radius of the earth is

i think it is 6.38*10^6 meters

The circumference of the earth is approximately 24, 900 miles, then how do you solve it?

[quote="Arne"]In the Flemish Mathematical Olympiad 2003, problem 3 was:

Let N be a positive integer which consists of 3 distinct digits, such that the sum of the other five positive integer which can be formed with the digits of N equals 2003.
Find all possible values of N.

[hide]127 + 172 + 271 + 712 + 721 = 2003[/hide][/quote]

  WOW, thats one of the easiest olympiad problems ive ever seen.......

I know. The hardest questions of the Flemish olympiad are even easy for the Pre-Olympiad section... It's sad. :)

2 is a mess of number and units, you know the circumfrence is changed by 30.48, how much must the diameter change?

http://www.google.com/search?hl=en&q=radius+of+the+earth

[hide]
$r=6 378.1 \text { km } = 6,378,100 \text { m.}$

I don't understand the question though :huuh:

EDIT: :blush: oh i see now
[/hide]

[hide]Using chess64's kind given measurements...  :D

Circumference = 20037.39 km.
New string = 20067.87 km.
Its circle would measure 6387.80,

6387.80-6387.1 = 0.70 km,
Divide that by two (see why?) to get 350 meters. [/hide]

[quote="mathnerd314"]It doesn't matter what the radius of the earth is. For any circle, when you increase the circumference by 30.48 m, the radius is increased by $\frac{30.48}{2\pi}$ which is approximately 4.85.[/quote]

Nice observation, so now I don't know how I got my answer....

[quote="guile"]
2. A string is wrapped around the earth's equator (wow! that's a long string) and the two ends just touch.  Suppose that another string is tied to the original string so it becomes 30.48 m longer.  If this new string is placed around the equator and pulled tight so that it is suspended in air, how high would the string be above the ground?[/quote]

OH this problem looks like a joke problem I have met before. It is also about a string is wrapped around the earth's equator, and the two ends just touch. If the string is 1m longer, there will be a hole. The question is if a normal cat will go through that hole  :rotfl: .


Given that $5r+4s+3t+6u =100$, where $r\ge s\ge t\ge u\ge 0.$ are real numbers,
find, with proof, the maximum and minimum possible values of $r+s+t+u$.

[hide=correct bound]we have $5r+4s+3t+6u=100$\\
notice $u-t\le 0$ and $-2t \le 0$ so we have \\
$5(r+s+t+u)+u-s-2t\le 5(r+s+t+u)$\\
which gives $r+s+t+u \ge \frac{100}{5}=20$\\
also we have $r-t \ge 0$ and $2u\ge 0$ so $r-t+2u \ge 0\implies 4(r+s+t+u)+r-t+2u\ge 4(r+s+t+u)\implies \frac{100}{4}=25 \ge (r+s+t+u)$\\
so we get $\boxed{20\le r+s+t+u\le 25}$ $\blacksquare$[/hide]


Points $ A$ and $ B$ are on the parabola $ y \equal{} 4x^2 \plus{} 7x \minus{} 1$, and the origin is the midpoint of $ \overline{AB}$. What is the length of $ \overline{AB}$?

$ \textbf{(A)}\ 2\sqrt5 \qquad
\textbf{(B)}\ 5\plus{}\frac{\sqrt2}{2} \qquad
\textbf{(C)}\ 5\plus{}\sqrt2 \qquad
\textbf{(D)}\ 7 \qquad
\textbf{(E)}\ 5\sqrt2$

Let the 2 points be $ (a,b)$ and $ (\minus{}a,\minus{}b)$ (since their midpoint is the origin), with $ a \ge 0$.

$ b\equal{}4a^2\plus{}7a\minus{}1$
$ \minus{}b\equal{}4a^2\minus{}7a\minus{}1$

Summing these 2, we have $ 8a^2\equal{}2$, so $ a\equal{}\frac{1}{2}$, so $ b\equal{}\frac{7}{2}$. Therefore, the 2 points are $ (\frac{1}{2},\frac{7}{2}$ and $ (\minus{}\frac{1}{2},\minus{}\frac{7}{2})$. The distance between the 2 by the point-point distance formula is $ \sqrt{1^2\plus{}7^2}\equal{}\sqrt{50}\equal{}2\sqrt{5}$

Therefore, the answer is A.

[quote="rofler"]$ \sqrt {50} \equal{} 2\sqrt {5}$

Therefore, the answer is A.[/quote]

A minor correction: $ \sqrt{50}\equal{}5\sqrt{2}$, making the answer E.  :)

[b]Solution:[/b] We know that the points have to be $(x, y)$ and $(-x, -y)$ so both of the equations: $y=4x^2 + 7x - 1$ and $y=-4x^2 + 7x + 1$ must be true. Adding this together gives us $2y = 14x \Rightarrow y = 7x$. Substituting this into the first equation, we get that $x = \frac{1}{2}$ and therefore $y = \frac{7}{2}$. This makes the distance between the points $\sqrt{7^2 + 1^2} = \sqrt{50} = 5\sqrt{2} = \boxed{E}$.


My four-digit code is 0235. Alice can't choose a code that is the same as mine in three or more of the four digit-positions, nor that is the same as mine except for switching the positions of two digits (so 3205 and 2035, for example, are forbidden, but 3025 is fine). Alice can otherwise choose any four-digit code. How many codes are available for Alice?

hmm this problem seems similar to a problem i did not so long ago  :rotfl:  :rotfl:  :rotfl:
Now there are four digits available, or 10000 different codes.
So which ones are not available?
Those with 3 digits same as 0235, or any set that is in the form
023x
02x5
0x35
x235
except that for the set being 0235, we have 4 values that overlap into one.
This gives us 10+10+10+10-3 or 37 different codes.
Moving on to the second condition, we have that there are 6 pairs of numbers (4 C 2) that can be switched around, or 6 different codes that are forbidden.

Now, we can sum the forbidden codes, 37+6=43;
Then subtracting this from our start point, we have
10000-43=[b]9957[/b] available codes!

To be more transparent in the second part:

To get a code the same in three places, we would pick any spot and exchange it for one of the other nine digits.  Total $ 9\plus{}9\plus{}9\plus{}9\equal{}36$


Find the shortace distance from the origin to the line segment containing (2,0) and (0,3)

[hide="a quickie spoiler DONT LOOK"]2 times area of triangle divided by hypotenuse :D
aww told you not to look :P[/hide]

oh so altitude is the shortest distance. oooh

i thought it was the median, no wonder i messed up  :lol:


limit
trigonometry
calculus
calculus computations

Calculate:

$ \lim_{x\rightarrow0}\frac {e^{x^2} \minus{} e^{3x}}{\sin(\frac {x^2}{2}) \minus{} \sin x}$

hello :) !!

i thing that $ lim \equal{} 3$

the prove !!!

$ e^{x^2} \sim_0 1\plus{}x^2$ and $ e^{3x}\sim_0 1\plus{}3x$ and $ sin(\frac{x^2}{2}) \sim_0 \frac{x^2}{2}$ and $ sin(x) \sim_0 x$

than:

$ f(x)\equal{}\frac{e^{x^2}\minus{} e^{3x}}{sin(\frac{x^2}{2})\minus{} sin(x)} \sim_0 \frac{x^2\minus{}3x}{\frac{x^2}{2} \minus{} x}$

than $ if  \ x\to 0$ than $ f(x) \to 3$

Plugging in, we get our favorite indeterminate form, $ \frac{0}{0}$, so we attack this thing with L'Hopital's rule:

$ \lim_{x\to 0}\frac {e^{x^2} \minus{} e^{3x}}{\sin(\frac {x^2}{2}) \minus{} \sin x}\equal{}\lim_{x\to 0}\frac{2xe^{x^2}\minus{}3e^x}{2x\cos\frac{x^2}{2}\minus{}\cos x}$.

Plug in 0 to get $ \frac{\minus{}3}{\minus{}1}\equal{}\boxed{3}$.

[b]mathema*[/b], what do you mean by $ \sin\frac{x^2}{2}\sim_0\frac{x^2}{2}$, and the others?

I believe [b]mathema*[/b] meant: $ a(x)\sim_0 b(x)$ is equivalent to "$ a(x)$ is asymptotic to $ b(x)$ when the variable (i.e. $ x$) tends to zero".


A cell phone plan costs $ \$50$ per calendar month. At this rate, how
many dollars does it cost for 10 years?

For every month, it's $50. 12 months in a year. do 50*12*10=6000.


I just realized that the quotes that were on the bottom of the page before are gone.
Why is the quotes gone?

Many people complained that they were hard to read, and they slowed down the site.  So we removed them.

If you enjoy mathematical quotes, there are many of them at http://www-groups.dcs.st-and.ac.uk/~history/Quotations/index.html


[quote="NC7"]3$\frac{ab+bc+ca}{a+b+c} \le \sum_{a,b,c}\frac{a^3}{b^2-bc+c^2}$
I think SOS is the best for solving this problem[/quote]
Are you mean to $a>0,$ $b>0,$ $c>0$ $?$
If so, I don't agree with you,since
$\sum_{cyc}\frac{a^3}{b^2-bc+c^2}\geq\sum_{cyc}a\geq\frac{3(ab+bc+ca)}{a+b+c}$ is easier. ;)

[quote]I don't agree with you,since
$ \sum_{cyc}\frac{a^{3}}{b^{2}\minus{}bc\plus{}c^{2}}\geq\sum_{cyc}a\geq\frac{3(ab\plus{}bc\plus{}ca)}{a\plus{}b\plus{}c}$ is easier. ;)[/quote]
So how can you prove this without SOS or Schur

See
http://www.mathlinks.ro/Forum/viewtopic.php?t=3221 ;
http://www.mathlinks.ro/Forum/viewtopic.php?t=6551 .

  darij

Actually, Arqady posed a good question: It seems that for any [i]reals[/i] $ a$, $ b$, $ c$, we have

$ \left(a\plus{}b\plus{}c\right)\left(\frac{a^{3}}{b^{2}\minus{}bc\plus{}c^{2}}\plus{}\frac{b^{3}}{c^{2}\minus{}ca\plus{}a^{2}}\plus{}\frac{c^{3}}{a^{2}\minus{}ab\plus{}b^{2}}\right)\geq 3\left(bc\plus{}ca\plus{}ab\right)$.

Any proof or counterexample?

  darij

[quote="Zubr"]Sorry, but what you called SOS, i can't catch meaning of this word.[/quote]

See http://www.mathlinks.ro/viewtopic.php?t=80127 and http://www.mathlinks.ro/viewtopic.php?t=65385 for different aspects of the method.

  Darij


(This is a continuation of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17222]another thread[/url], but it seemed appropriate to discuss this in a separate, considering all the flameage that had taken place there.)

[quote=white_horse_king88]Ahh... mathfanatic. You are in Exeter already? How's exeter - I will never make it, but I am curious![/quote]

Exeter is nice.  Admittedly, it's a bit different from what I expected and how it's probably been portrayed to you, but I'm glad to be here.

Classes are...well, classes.  Some of them are really good (I thoroughly enjoy my Physics and European History classes, which involve a lot more experimentation/thinking and discussion, respectively, than occurs in most schools).  Some of them are decent (my Spanish class is too slow, although that may be more of a placement issue than a teaching issue).  There's generally a fair amount of work, but it's not overwhelming.  Generally, I have 5-6 hours of classes per day and 2-3 hours of homework.

Outside of classes, there is generally a lot to do.  Sports are meant to take up some time (particularly if you're on a team...club sports are far less time-consuming) but Exeter sets aside about 2 hours each day for them.  There are over 100 clubs, from debate to drumming to philosophy to math team (or, as a friend calls it, "varsity math").  For math team, we usually work on a set of 12 problems/proofs on wednesdays, then discuss solutions on sundays.

Regarding the admissions process, I think many of you (judging by your work on AoPS) would make it.  I know plenty of not-very-qualified people who made it in; it's not as tough as you think.  If you're interested in Exeter, give it a shot and apply.

Hey mathfanatic, what were some of Exeter's Varsity Football team's recent scores?

Back on topic here, yeah, it's not too difficult to make it. I mean, I know this really lazy bum at my old school who almost never did his homework until the morning of classes, even flunked a test or two, and never knew when a quiz or test was, but he still got accepted to Exeter.

[hide]go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.go to choate.[/hide]

what's choate? and if your sig is true, how much are you like the "lazy bum"?

Choate is a utopia, a place of perfect harmony between everybody there. There is only total bliss and happiness, none of the evils that plague much of our world. At Choate, you'll find a place where there is no fighting or killing or name-calling. You will only find the pureness and soul-purging nature of your life.

As for that bum, let's just say we're so close, we're almost brothers.

[quote="yif man12"]Hey mathfanatic, what were some of Exeter's Varsity Football team's recent scores?  :D [/quote]

Meh.  Oooh, CRH is better than us at acting like gorillas, throwing irregularly-shaped balls over long distances, and making each other trip.

I'll believe the total-bliss-and-happiness line when I see it.

[quote="mathfanatic"][quote="yif man12"]Hey mathfanatic, what were some of Exeter's Varsity Football team's recent scores?  :D [/quote]

Meh.  Oooh, CRH is better than us at acting like gorillas, throwing irregularly-shaped balls over long distances, and making each other trip.[/quote]

Darn right we are!!!  :D

I also go to Choate. I really, really like it, and would encourage you all to apply. Choate is located in Wallingford, Connecticut, a fairly small town that is close to the city of New Haven and about half an hour from Hartford. The student body is a perfect size, 850, which (correct me if I am wrong, mathfanatic) is a few hundred less than Exeter. Also, it's a lot colder at Exeter...and you don't want that.

Our sports accomplishments speak for themselves (see yif man's above post). We've got a very successful math team as well as excellent debate and economics teams. There are many, many, many other clubs, including Indian Club, Chinese Club, Japanese Club, Korean Club, Russian Club, Chess Club, Asian Student Association, P.P.P.P.P.P.P.P.P.P.P.P.P.P. Club, Young Democrats Club, Young Republicans Club, Interact (Community Service), and far more, too many to name. Also, if you have are interested in starting a club that is not already in existence, you are encouraged to.

Classes are excellent...students are motivated, teachers are encouraging. Material is covered at a rapid pace, but if you are unsure of something you are encouraged to seek extra help from your teacher, who will always be willing to help you out.

All in all, Choate's a great school. I encourage everyone here who is considering private schools to apply!

Wow. My school sponsors all those clubs too, but we aren't nearly as successful. I would really care to learn more because my school is very slow paced. I often fall asleep in class - Figuratively speaking of course. Then again, private schools are meant to have higher standards.

You can find more information at [url=http://www.choate.edu]Choate's website[/url] and at [url=http://crh.choate.edu]Choate's student website[/url].

Correct me if I'm wrong, but Exeter is a highschool right?
And where is it located?

Okay I'm pretty sure it's a highschool.  I'm looking at Choate's website, and it seems like an awesome school to go to.  Way better than my public school...by the way, how much does a school like that cost?

Ignore that question too.  In fact, ignore them all.


What's the current edition of the AoPS books? (i.e., What edition will I get if I order them?) Neither the information pages on the books nor the ordering page seems to give this info. I hope you'll update that. (I've heard there's a 5th edition. If there's a new edition, [i][u]is the number of pages is the same[/i][/u]?)

We're on around the 6th edition now, but the books haven't materially changed since the 2nd edition (and there wasn't that great a change from the 1st to the 2nd).

(I saw 'around 6th edition' because some the of the books are now in their 5th, some in the 6th - as I noted, there haven't been major changes in many editions.)


$ a,b,c>0$,prove that $ \frac{a^3}{b^2}\plus{}\frac{b^3}{c^2}\plus{}\frac{c^3}{a^2}\ge a\plus{}b\plus{}c$.

[quote="math10"]very easy;only using AM-GM:
    $ \frac {a^3}{b^2} \plus{} b \plus{} b\geq3a$[/quote]

Thanks,it's a nice solution.

I didn't think of it!  :o

I use generalized Cauchy's-Schwarz here

but I think my solution is not so nice,

so I post the problem here to find some better solution.

[quote="kcn2rivers"]Does it not follow directly from the Rearragement inequality?[/quote]
You are right!

Holder;
$ \left(\frac{a^3}{b^2}\plus{}\frac{b^3}{c^2}\plus{}\frac{c^3}{a^2}\right)(b\plus{}c\plus{}a)^2\geq (a\plus{}b\plus{}c)^3$
then $ \frac{a^3}{b^2}\plus{}\frac{b^3}{c^2}\plus{}\frac{c^3}{a^2} \geq a\plus{}b\plus{}c$


A bag contains hundreds of glass marbles, each one coloured either red, orange, green or blue. There are more than $ 2$ marbles of each colour.
Marbles are drawn randomly from the bag, one at a time, and not replaced.
How many marbles must be drawn from the bag in order to ensure that at least three marbles of the same colour are drawn?

The worst case scenario is 2 red marbles, 2 orange, 2 green, and then 2 blue. Whatever marble is drawn afterwards will be a third, so we have 9.


limit
geometric series
calculus
calculus computations

Evaluate:

$ \lim_{n \to \infty}{\sum_{k\equal{}1}^n \frac{k}{2^k}}$.

$ \sum_{k\equal{}1}^{\infty} k p^k \equal{}  \frac{p}{(1\minus{}p)^2}$ for nice (what do I mean here?) $ p$.

But he actually asked for the partial sums.

This works pretty well on a "try it and see" basis. Explicitly compute $ \sum_{k\equal{}1}^n\frac k{2^k}$ for $ n\equal{}1,2,3,4,5,6.$ Is a pattern becoming apparent? Assume that the pattern you see is the correct formula, and try proving its validity by induction.

[hide]Let for $ |x| < 1$ we have: $ f \equal{} \sum_{k \equal{} 1}^{n}x^k \equal{} x\cdot\frac {1 \minus{} x^n}{1 \minus{} x}$

Then $ \sum_{k \equal{} 1}^{n}kx^{k} \equal{} x\cdot f_{x}'$

Easy to calculate that $ x\cdot f_{x}' \equal{} x\cdot\frac {1 \minus{} (n \plus{} 1)x^n \plus{} nx^{n \plus{} 1}}{(1 \minus{} x)^2}$ and we get [b]Tyl's[/b] result when $ n$ tends to infinity.[/hide]

uh, kunny --- since $ \frac12\plus{}\frac24\plus{}\frac38$ is already larger than $ 1,$ I don't think $ 1$ is the sum of the infinite series. Time to recheck the details.

Of course the next to last item on the line is correct: it essentially says that
\[ \sum_{k\equal{}1}^n\frac{k}{2^k}\equal{}2\minus{}\frac{n\plus{}2}{2^n}\]
Which can be gotten from what Yuriy Solovyov said, or could be found in the way I suggested earlier.

This is a series commonly known as the arithmetico-geometric series. Its closed form is fairly easy (though somewhat painstaking) to derive.

Simply substitute that closed form, then let $ n$ tend to infinity.

[quote="Kent Merryfield"]uh, kunny --- since $ \frac12 \plus{} \frac24 \plus{} \frac38$ is already larger than $ 1,$ I don't think $ 1$ is the sum of the infinite series. Time to recheck the details.

Of course the next to last item on the line is correct: it essentially says that
\[ \sum_{k \equal{} 1}^n\frac {k}{2^k} \equal{} 2 \minus{} \frac {n \plus{} 2}{2^n}
\]
Which can be gotten from what Yuriy Solovyov said, or could be found in the way I suggested earlier.[/quote]

You are right. :blush:

We have explained how to solve such kind of infinite series, haven't you?,  as for me at least I have answered 5 times, this time I will copy it.

Anyway, first of all we should teach beginners the way $ S_n \minus{} rS_n$. :wink:

To Japanese high school students, you need to prove $ \lim_{n\to\infty} \frac {n}{2^n} \equal{} 0$ in the university entrance examination. :P

[quote="grn_trtle"]This is a series commonly known as the arithmetico-geometric series. Its closed form is fairly easy (though somewhat painstaking) to derive.[/quote]
Really?  There are several straightforward ways depending on what you consider straightforward.

1.  Differentiation.  You just need to deal with a finite geometric series instead of an infinite one.
2.  Multiplying and taking differences, which amounts to summation by parts.
3.  The combinatorial proof [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=220351]here[/url].


logarithms
calculus
derivative
integration
function
search
projective geometry

[quote][b]Theorem:[/b] If $ G$ is either the whole plane $ \bf C$ or some open disk. then if $ u: G \to \bf{R}$ is harmonic, then $ u$ has a harmonic conjugate.
[/quote]

This happens to fail for $ u(z) \equal{} \log |z|$ on $ G \equal{} \bf{C} \minus{} \{0\}$

(because we can't define an analytic branch for the logarithm on $ G$)

I don't really see what goes wrong in the proof of the theorem though. They seem to use Leibniz' rule of advanced calculus (Differentiation under the integral sign) with $ u: (a,R) \times (b,R) \to \bf{R}$ continuous and assuming the open disk $ G$ is taken $ G \equal{} B(z,R)$ with $ z \equal{} a \plus{} bi$.

So Leibniz' rule is supposably to go wrong if we don't have a continuous function from a nice open interval or what is the main issue actually?

Leibniz rule is really a technicality and not the point here. Differentiating under the integral sign is supposed to feel natural.

The proof goes wrong because what is your well defined harmonic conjugate? The construction doesn't work when you've punched out a hole (because your lines of integration start intersecting the hole). Of course, locally around each point the construction works fine, so it has a harmonic conjugate locally, but this doesn't give a harmonic conjugate globally.

The construction should feel pretty natural, if you think about it; the construction should give the only possible harmonic conjugate up to an additive constant, if there is one at all. We simply must attempt to define our harmonic conjugate as we do.

The problem is closely connected to simple connectivity and Cauchy's theorem. Note that the things I said above could be said about Cauchy's theorem for a disc instead, but instead of a harmonic conjugate we're trying to define a primitive of an analytic function. Indeed, if you've done Cauchy's theorem on the course, this leads to a quick proof of this business:

Suppose there is a harmonic conjugate $ v$, so that $ f \equal{} u \plus{} iv$ is analytic in the disc. Then it follows that $ f' \equal{} \frac{\partial u}{\partial x} \minus{} i\frac{\partial u}{\partial y}$. So now, let us define $ h \equal{} \frac{\partial u}{\partial x} \minus{} i\frac{\partial u}{\partial y}$. Check that it is analytic. The function $ f$ we search for MUST be a primitive of $ h$. According to Cauchy's theorem such a primitive exists, call it $ f \equal{} u_1 \plus{} iv_1$. $ u_1$ has the derivatives as $ u$, so they only differ by a constant. It follows that $ v_1$ is a harmonic conjugate for $ u$

[quote]
 (because your lines of integration start intersecting the hole)
[/quote]
I"m not with you on this one....


calculus
integration
logarithms
calculus computations

Use the following formula:

$\frac{1}{2}\ln \frac{1+a}{1-a}=\sum_{n=0}^{\infty}\frac{a^{2n+1}}{2n+1}$ for $|a|<1$.

For $a=\frac{1}{2}$ we get that the value of your sum is $\frac{1}{2}\ln 3$.

Your answer is correct, didilica. :)

Can anyone solve in using integral? :lol:


Does anyone know any colleges that are good for computer science (i'm looking closer to the west coast, so colleges situated mostly on the west coast are most appreciated). Thanks. :)

You want to consider a whole range of selectivity, right? I frequently hear Stanford and University of California Berkeley mentioned as strong computer science colleges, and Caltech should be fine too. I know less about colleges that are perhaps easier to get into but that also are west coast colleges with strong programs in computer science.

What do the rest of you suggest?


What is the integer value of $ x$ in the arithmetic sequence $ 3^2, x, 3^4$?

Call the common difference $ d$. We then have,

$ 2d\equal{}3^4\minus{}3^2 \implies 2d\equal{}72 \implies d\equal{}36$.

We can then see that $ x\equal{}9\plus{}36\equal{}\boxed{45}$.


Hello,
How do I enable my blog or website to allow latex to appear as it does here? I will appreciate your help.

If you use an AoPS blog then it will already be able to use LaTeX. Otherwise, for WordPress blogs there is the LatexRender plugin at [url=http://sixthform.info/steve/wordpress/index.php]Using LaTeX in WordPress[/url] and for all other blogs and websites use [url=http://www.mayer.dial.pipex.com/tex.htm]LatexRender[/url].

LatexRender is for PHP programs and you will need LaTeX, GhostScript and ImageMagick already installed on the server. If that's not possible and you can run cgi programs then there are instructions for using [url=http://www.forkosh.com/mimetex.html]Mimetex[/url] in the download. Mimetex is not quite as flexible or give such good quality as LaTeX but it's an excellent alternative if you cannot run LaTeX.

hmm is it possible to get LaTeX on another forum? like invisionfree? i asked on their help boards but no one answered for like a month now

You can use LatexRender with any PHP program [i]provided[/i] LaTeX etc is installed on the server hosting the forum. You just to need to know where in the files to add a couple of lines.

[url=http://www.hsn.uk.net/forum/index.php]HSN.uk.net[/url] is an Invision Board which has LatexRender installed.


Suppose that $ f: \mathbb R\to\mathbb R$ is not bounded on $ [a,b]$. Prove that there exists $ c\in[a,b]$ such that $ f$ is not bounded on any open interval containing $ c$.

[quote="ifai"]Suppose that $ f: \mathbb R\to\mathbb R$ is not bounded on $ [a,b]$. Prove that there exists $ c\in[a,b]$ such that $ f$ is not bounded on any open interval in $ [a,b]$ containing $ c$.[/quote]

Apply compactness.

Assume the contrary that for any $ c\in[a,b]$, $ f$ is bounded on some open intervals containing $ c$. That is,

$ (\forall c\in[a,b])\,(\exists U(c)\subset\mathbb R)\,(\exists M_c\in\mathbb R)\,(\forall x\in U(c))\, [|f(x)|\le M_0],$

where $ U(c)$ is an open intervals containing $ c$. Clearly, we have

$ [a,b]\subset\bigcup_{c\in[a,b]} U(c).$

By Heine-Borel Theorem, there exist $ c_1, c_2,\cdots, c_n\in[a,b]$ such that

$ [a,b]\subset\bigcup_{i \equal{} 1}^n U(c_i).$

Hence, for any $ x\in[a,b]$, we have $ x\in U(c_k)$ for some $ k$ with $ 1\leq k\leq n$; and so

$ f(x)\leq M_{c_k}\leq\max_{1\leq i\leq n} M_{c_i}$ for all $ x\in[a,b]$.

However, it violates that $ f$ is not bounded on $ [a,b]$. The result follows.

How about a less-fancy version of compactness - a nested interval argument?

Let $ a_0\equal{}a$ and $ b_0\equal{}b.$ $ f$ must be unbounded on at least one of the intervals $ \left[a_0,\frac{a_0\plus{}b_0}2\right]$ or $ \left[\frac{a_0\plus{}b_0}2,b_0\right].$ Choose one of the sides for which $ f$ is unbounded and call that interval $ [a_1,b_1].$

Repeat and iterate. $ f$ is unbounded on at least one of the intervals $ \left[a_n,\frac{a_n\plus{}b_n}2\right]$ or $ \left[\frac{a_n\plus{}b_n}2,b_n\right].$ Chose a half on which $ f$ is unbounded and call that $ [a_{n\plus{}1},b_{n\plus{}1}].$

There is a unique point $ c$ such that $ c\in[a_n,b_n]$ for all $ n.$ Since any neighborhood of $ c$ must contain at least one of the intervals $ [a_n,b_n],$ then $ f$ is unbounded in any neighborhood of $ c.$

Or how about sequential compactness?

If $ f$ is not bounded (WLOG, above) on $ [a, b]$ then for every $ n$ there exists $ x_n$ such that $ f(x_n) > n$.  The sequence $ (x_n)$ takes values in a compact set, hence has a convergent subsequence that converges to $ c \in [a, b]$.  Any open interval about $ c$ contains infinitely many elements of $ (x_n)$, hence $ f$ is not bounded on such an interval.

So, like the proof of t0rajir0u, we can use the theorem of accumulation points (Bolzano-Weierstrass) to prove it.


calculus
analytic geometry
graphing lines
slope
derivative

Suppose f(x) = x2−3x+1: Find all values of c so that the tangent line to the graph of f(x) at the point (c; f(c)) intersects the point (−3; 0)

Elementary calculus gives a shorter solution, but it isn't necessary:

[hide="Hint"]Take the general equation of a line through $(-3,0)$ and put the condition that it should have exactly one intersection point with the function.[/hide]

[quote="M4RI0"]Is it $c=-3 \pm \sqrt{19}$ :?:[/quote]

I would say those are the slopes of the two tangents, but you still have to find the touching points.

[quote="lotrgreengrapes7926"][quote="wigianto"]Suppose $f(x) = x^{2}-3x+1$: Find all values of $c$ such that the tangent line to the graph of $f(x)$ at the point $(c; f(c))$ intersects the point $(-3; 0)$.[/quote]
[hide="Calculus Solution"] $f'(c)=2c-3$. The equation of a line through $(3,0)$ is $f(x)=mx+3m$. Plugging in $m=2c-3$, $x=c$ and $f(x)=c^{2}-3c+1$.

$c^{2}-3c+1=(2c-3)c+3(2c-3)$
$c^{2}-3c+1=2c^{2}-3c+6c-9$
$c^{2}+6c-10=0$
$c=\frac{-6\pm\sqrt{36+40}}{2}=\frac{-6\pm2\sqrt{19}}{2}=\boxed{-3\pm\sqrt{19}}$. [/hide]

M4RIO did have the correct answer. :D[/quote]

I stand corrected then. Haven't dealt with the problem in detail :)

[quote="Farenhajt"][quote="lotrgreengrapes7926"][hide="Non-Calculus Solution"] $f(x)=mx+b$
$0=-3m+b$
$b=3m$
$f(x)=mx+3m$
$f(x)=x^{2}-3x+1$
$x^{2}-3x+1=mx+3m$
$x^{2}+(-3-m)x+(1-3m)=0$

For $f(x)=mx+3m$ to be a tangent line, the derivative must be $0$.
$(-m-3)^{2}-4(1-3m)=0$
$m^{2}+18m+5=0$
$m=\frac{-18\pm\sqrt{324-20}}{2}=-9\pm\frac{\sqrt{304}}{2}=-9\pm2\sqrt{19}$.

$x^{2}+(-3-9\pm2\sqrt{19})x+(1-3(-9\pm2\sqrt{19}))=0$
Since the derivative is $0$, $x=\frac{3-9\pm2\sqrt{19}}{2}=\frac{-6\pm2\sqrt{19}}{2}=\boxed{-3\pm\sqrt{19}}$. [/hide][/quote]

I must say that using derivatives in a "non-calculus" solution is a controversial practice :)[/quote]

he meant discriminant

Of course, that much is obvious... but it still should be corrected :)


function
limit
calculus
calculus computations

Consider the sequence of function fn(x) = x^n.
Does it converges uniformly?

The question is meaningless as stated. Uniform convergence is tied to a set; the set is an inescapable part of the definition.

Does $ f_n(x)\equal{}x^n$ converge uniformly on $ [0,.9]?$ Yes, it does.

Does $ f_n(x)\equal{}x^n$ converge uniformly on $ [0,1)?$ No, it does not.

I forgot to mention that it is on the open interval (-1,1)

We have $ \lim_{x\to 1 \minus{} 0}x^n \equal{} 1$ for every $ n\in N$.

Hence, for every $ \epsilon > 0$ exists $ \delta > 0$ sutch that

$ x^n > 1 \minus{} \epsilon$ whenever $ 1 \minus{} x < \delta$ for every $ n\in N$.

That means we can get $ x^n$ as close to $ 1$ as we want on open interval $ ( \minus{} 1,1)$, no metter how large $ n$ is, hence $ x^n$ doesn't converge uniformly on $ ( \minus{} 1,1)$.


Is it possible to have a triangle with two 90 degree angles, and the
other two legs from the connected 90 degree angles meet?  Where would
you find such a triangle?  I thought maybe if the triangle is on a
sphere, but then the lines aren't straight.

In Euclidean Geometry, the sum of the angles of any triangle is $180^{\circ}$. If you have two $90^{\circ}$ angles, the third angle must be $0^{\circ}$, which is impossible.

On a sphere, I think it is possible to have a triangle with 3 $90^{\circ}$ angles.

right. consider a point on the north pole of the Earth. then draw two lines south perpendicular to each other to the equator. The north pole and the intersections at the equator make a triangle of 3 90 degree angles


Let $p$ be a fixed prime number. If $a(n)$ denotes the exponent of $p$ in the prime factorisation of $n$, determine the sum\[S(n)=a(1)+a(2)+...+a(p^n)\]

Exactly $p^{n-1}$ numbers will be divisible by $p$, $p^{n-2}$ numbers divisible by $p^2$, $p^{n-3}$ numbers divisible by $p^3$ and so forth.

So the sum will be $1+p+p^2+p^3+...+p^{n-1}$ or $\displaystyle\frac{p^n-1}{p-1}$.


geometry
3D geometry
tetrahedron
sphere
parallelogram
vector
perimeter

[b][size=150]IMO 1973 ShortListed Problems: (Posted in Forums)[/size][/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026247#p2026247]1. (BUL 6)[/url] Let a tetrahedron $ABCD$ be inscribed in a sphere $S$. Find the locus of points $P$ inside the sphere $S$ for which the equality
\[\frac{AP}{PA_1}+\frac{BP}{PB_1}+\frac{CP}{PC_1}+\frac{DP}{PD_1}=4\]
holds, where $A_1,B_1, C_1$, and $D_1$ are the intersection points of $S$ with the lines $AP,BP,CP$, and $DP$, respectively.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026254#p2026254]2. (CZS 1)[/url] Given a circle $K$, find the locus of vertices $A$ of parallelograms $ABCD$ with diagonals $AC \leq BD$, such that $BD$ is inside $K$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357923&#p357923]3. (CZS 6)[/url] [b]IMO1[/b] Prove that the sum of an odd number of vectors of length 1, of common origin $O$ and all situated in the same semi-plane determined by a straight line which goes through $O,$ is at least 1.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026257#p2026257]4. (GBR 1)[/url] Let $P$ be a set of $7$ different prime numbers and $C$ a set of $28$ different composite numbers each of which is a product of two (not necessarily different) numbers from $P$. The set $C$ is divided into $7$ disjoint four-element subsets such that each of the numbers in one set has a common prime divisor with at least two other numbers in that set. How many such partitions of $C$ are there ?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026258#p2026258]5. (FRA 2)[/url] A circle of radius 1 is located in a right-angled trihedron and touches all its faces. Find the locus of centers of such circles.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357905&#p357905]6. (POL 2)[/url][b]IMO2[/b] Establish if there exists a finite set $M$ of points in space, not all situated in the same plane, so that for any straight line $d$ which contains at least two points from M there exists another straight line $d'$, parallel with $d,$ but distinct from $d$, which also contains at least two points from $M$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026262#p2026262]7. (POL 3)[/url] Given a tetrahedron $ABCD$, let $x = AB \cdot CD$, $y = AC \cdot BD$, and $z = AD \cdot  BC$. Prove that there exists a triangle with edges $x, y, z.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026267#p2026267]8. (ROM 1)[/url] Prove that there are exactly $\binom{k}{[k/2]}$ arrays $a_1, a_2, \ldots , a_{k+1}$ of nonnegative integers such that $a_1 = 0$ and $|a_i-a_{i+1}| = 1$ for $i = 1, 2, \ldots , k.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026273#p2026273]9. (ROM 2)[/url] Let $Ox, Oy, Oz$ be three rays, and $G$ a point inside the trihedron $Oxyz$. Consider all planes passing through $G$ and cutting $Ox, Oy, Oz$ at points $A,B,C$, respectively. How is the plane to be placed in order to yield a tetrahedron $OABC$ with minimal perimeter ?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357934#p357934]10. (SWE 3)[/url] [b]IMO6[/b] Let $a_1, \ldots, a_n$ be $n$ positive numbers and $0 < q < 1.$ Determine $n$ positive numbers $b_1, \ldots, b_n$ so that:

[i]a.)[/i] $ a_{k} < b_{k}$ for all $k = 1, \ldots, n,$
[i]b.)[/i] $q < \frac{b_{k+1}}{b_{k}} < \frac{1}{q}$ for all $k = 1, \ldots, n-1,$
[i]c.)[/i] $\sum \limits^n_{k=1} b_k < \frac{1+q}{1-q} \cdot \sum \limits^n_{k=1} a_k.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357937&#p357937]11. (SWE 4)[/url] [b]IMO3[/b] Determine the minimum value of $a^{2} + b^{2}$ when $(a,b)$ traverses all the pairs of real numbers for which the equation \[ x^{4} + ax^{3} + bx^{2} + ax + 1 = 0 \] has at least one real root.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026275#p2026275]12. (SWE 6)[/url]Consider the two square matrices
\[A=\begin{bmatrix} +1 & +1 &+1& +1 &+1 \\+1 &+1 &+1&-1 &-1 \\ +1 &-1&-1 &+1& +1 \\ +1 & -1 & -1 & -1 & +1 \\ +1 &+1&-1 &+1&-1  \end{bmatrix} \quad \text{ and } \quad B=\begin{bmatrix} +1 & +1 &+1& +1 &+1 \\+1 &+1 &+1&-1 &-1 \\ +1 &+1&-1& +1&-1 \\ +1 &-1& -1& +1& +1 \\ +1 & -1& +1&-1 &+1 \end{bmatrix}\]

with entries $+1$ and $-1$. The following operations will be called elementary:

(1) Changing signs of all numbers in one row;

(2) Changing signs of all numbers in one column;

(3) Interchanging two rows (two rows exchange their positions);

(4) Interchanging two columns.

Prove that the matrix $B$ cannot be obtained from the matrix $A$ using these operations.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026289#p2026289]13. (YUG 4)[/url] Find the sphere of maximal radius that can be placed inside every tetrahedron that has all altitudes of length greater than or equal to $1.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357950&#p357950]14. (YUG 5)[/url] [b]IMO4[/b] A soldier needs to check if there are any mines in the interior or on the sides of an equilateral triangle $ABC.$ His detector can detect a mine at a maximum distance equal to half the height of the triangle. The soldier leaves from one of the vertices of the triangle. Which is the minimum distance that he needs to traverse so that at the end of it he is sure that he completed successfully his mission?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026280#p2026280]15. (CUB 1)[/url] Prove that for all $n \in  \mathbb N$ the following is true:
\[2^n \prod_{k=1}^n \sin \frac{k \pi}{2n+1} = \sqrt{2n+1}\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026281#p2026281]16. (CUB 2)[/url] Given $a, \theta \in \mathbb R, m \in \mathbb N$, and $P(x) = x^{2m}- 2|a|^mx^m \cos \theta +a^{2m}$, factorize $P(x)$ as a product of $m$ real quadratic polynomials.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357910&#p357910]17. (POL 1)[/url] [b]IMO5[/b] $G$ is a set of non-constant functions $f$. Each $f$ is defined on the real line and has the form $f(x)=ax+b$ for some real $a,b$. If $f$ and $g$ are in $G$, then so is $fg$, where $fg$ is defined by $fg(x)=f(g(x))$. If $f$ is in $G$, then so is the inverse $f^{-1}$. If $f(x)=ax+b$, then $f^{-1}(x)= \frac{x-b}{a}$. Every $f$ in $G$ has a fixed point (in other words we can find $x_f$ such that $f(x_f)=x_f$. Prove that all the functions in $G$ have a common fixed point.


[hide="Solution"]The equations imply $x,y,z>0$. From the first equation we get

\begin{eqnarray*}x-\sqrt{y^{2}-{1\over 16}}=\sqrt{z^{2}-{1\over 16}&\iff& x^{2}-2x\sqrt{y^{2}-{1\over 16}}+y^{2}-{1\over 16}=z^{2}-{1\over 16}\\ &\iff& 2x\sqrt{y^{2}-{1\over 16}}=x^{2}+y^{2}-z^{2}}\end{eqnarray*}

That gives

$2x\left(y+\sqrt{y^{2}-{1\over 16}}\right)=x^{2}+2xy+y^{2}-z^{2}=(x+y+z)(x+y-z)$
$2x\left(y-\sqrt{y^{2}-{1\over 16}}\right)=z^{2}-x^{2}+2xy-y^{2}=(-x+y+z)(x-y+z)$

Multiplying those two we get

$4x^{2}\cdot{1\over 16}=(x+y+z)(x+y-z)(x-y+z)(-x+y+z)\qquad(*)$

Similarly we find

$4y^{2}\cdot{1\over 25}=(x+y+z)(x+y-z)(x-y+z)(-x+y+z)$
$4z^{2}\cdot{1\over 36}=(x+y+z)(x+y-z)(x-y+z)(-x+y+z)$

Hence ${x^{2}\over 4}={4y^{2}\over 25}={z^{2}\over 9}=t^{2}$ for some real positive $t$, giving

$x=2t, y={5t\over 2}, z=3t$.

Plugging that into $(*)$ we get

\begin{eqnarray*}t^{2}&=&{15t\over 2}\cdot{3t\over 2}\cdot{5t\over 2}\cdot{7t\over 2}\\ &=& \frac{225\cdot 7t^{4}}{16}\end{eqnarray*}

which gives $t^{2}=\frac{16}{225\cdot 7}\implies t=\frac{4}{15\sqrt{7}}$

Now $x+y+z=2t+{5t\over 2}+3t={15t\over 2}={15\over 2}\cdot\frac{4}{15\sqrt{7}}={2\over\sqrt{7}}$

Hence $x+y+z={2\over\sqrt{7}}\implies m+n=9$[/hide]

[quote="D.P.L"][quote="Totally Zealous"]Given $x$, $y$, and $z$ are real numbers that satisfy
$x=\sqrt{y^{2}-\frac{1}{16}}+\sqrt{z^{2}-\frac{1}{16}}$
$y=\sqrt{z^{2}-\frac{1}{25}}+\sqrt{x^{2}-\frac{1}{25}}$
$z=\sqrt{x^{2}-\frac{1}{36}}+\sqrt{y^{2}-\frac{1}{36}}$
and that $x+y+z=\frac{m}{\sqrt{n}}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime, find $m+n$.[/quote]

interpreting it geometrically might give you a shorter solution :wink:[/quote]
Indeed, this is #15 of the 2006 AIME, and a solution is given in the math jam.

[quote="Farenhajt"]Then give a geometric interpretation or post a link to the mathjam solution.[/quote]
[url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=145#fifteen]Click this link for the solution[/url]


There are eight furlongs in a mile. There are two weeks in a
fortnight. The British cavalry traveled 2800 furlongs in a
fortnight. How many miles per day did the cavalry average?

If they go $ 2800$ furlongs in $ 14$ days, then in one day they go $ 200$ furlongs.

Since $ 200$ furlongs is $ 25$ miles, the armyu goes $ 25$ miles a day.

[quote="BOGTRO"]
Since $ 200$ furlongs is $ 25$ miles, the armyu goes $ 25$ miles a day.[/quote]

This is because $ \frac{200}{8} \equal{} 25$


Assuming that the top angle is a right angle, we can easily partition it to find an area of $ \frac12\cdot6\cdot8\plus{}\frac12(12\plus{}18)\cdot8\equal{}24\plus{}120\equal{}\boxed{144}$.

Think of this pentagon as a triangle on top of a trapezoid. The area of the pentagon would be the area of the two parts added together. The area of the triangle is $ \frac {1}{2}(6)(8) \equal{} 24$. The area of the trapezoid is $ \frac {1}{2}(12 \plus{} 18)(8) \equal{} 120$. Adding these two areas up, we find that the area of the pentagon is $ 24 \plus{} 120 \equal{} \fbox{144}$.

EDIT: Argggggh. Got beaten again.


Find all integer numbers $b$ and $d$ satisfying
\[ b^2 - d^2 +bd + 1 = 0 \]

[color=red]Edited by moderator[/color]

Crucial observation is probably that if (a,b) is solution then so is (a+b,a+2b) and so is (2a-b,b-a).

if (a,b) is solution with $a,b > 0$ then it'a easy to see that $a < b < 2a$, so starting with (a,b) solution, it's possible to build a strictly "smaller" solution, by infinite descent like argument, we end with (0,1).

Similar argument if $a < 0$.

So solutions are :
- $(x_n,y_n)$  with $x_0 = 1, y_1=1, x_{n+1} = x_n+y_n, y_{n+1} = x_n + 2 y_n$
- $(-x_{n+1},y_n)$
- $(x_{n+1},-y_n)$
- $(-x_n,-y_n)$

---------
High Algebra is an art of counting to 20 without taking off shoes or mittens  ;)


Hallo, kann mir bitte jmd. sagen, wie sich die Aufgabe 441144 lösen lässt? :blush:

Mit Q(n) sei die Quersumme der natürlichen Zahl n bezeichnet. Man beweise, dass dann Q(Q(Q(2005^2005)))=7 gilt.

(Ich habe hier auf Latex verzichtet, weil man 2005^2005 ja auch gerade noch so versteht  :wink:  )

Danke!

Hat das keiner gelesen oder will das keiner lösen?

Sorry, daß ich erst jetzt antworte - hatte beim ersten flüchtigen Lesen deines Posts nur das Wort "Quersumme" bemerkt und mich gleich weggeklickt...

Hab die Aufgabe selber mal in http://www.mathlinks.ro/Forum/viewtopic.php?t=36891 gepostet, die eigentliche Lösungsidee befindet sich auf http://www.mathlinks.ro/Forum/viewtopic.php?t=31409 .

  grüße,
  darij

[b]EDIT:[/b] 4444 Posts... wenn das nicht zu der Aufgabe passt :D

okay, danke.
(wieso klicktst du gleich weg, wenn du Quersumme liest)?

[quote="Blender"](wieso klicktst du gleich weg, wenn du Quersumme liest)?[/quote]

Sind meist die hässlichsten Aufgaben von Wettbewerben...
naja, Zahlentheorie in Wettbewerben ist eh meist hässlich.

  darij

Zahlentee in IMO: meist recht gut.
Zahlentee in QEDMO: no comment ;)
Zahlentee in DeMO: spam.
Zahlentee auf ML: je nach Author oft auch sehr nett.

Ich hab diesen Thread erst heute überhaupt bemerkt...

[quote="ZetaX"]Zahlentee in IMO: meist recht gut.[/quote]

Findest du daß die Zahlentee letztes Jahr gut war?

[quote="ZetaX"]Zahlentee in QEDMO: no comment ;)[/quote]

... ;)

[quote="ZetaX"]Zahlentee auf ML: je nach Author oft auch sehr nett.[/quote]

Bloß daß die netten Zahlenteeaufgaben auf ML meist über Wettbewerbsniveau angesiedelt sind.

  darij


The question is:

At the same time, two buses leave a depot and travel in opposite directions on a straight road. The first bus averages 5 mi/h more than the second. If they are 142.5 mi apart after 1 1/2 hours, how fast is each bus going?

Okay, I realize that the distance is the same, so the equations can have an equal sign in between.

Also one is going 5 mi more per hour than the other, so x+5. I'm not sure where the 1.5 hours comes in and do you need to have it on both sides of the equation. Since the buses travel in miles per HOUR, and it's an hour and a half, do you need to add 30 mins to get the answer?

I'm confused! Please help!  He needs this by Monday. I have others too.

Two Planes leave simultaneously from an airport, one flying East and the other West. They are 870 miles apart after 3/4 hour.  If the Eastbound plane averages 120 mi/h more than the Westbound lane, at what rate is each plane flying. I know we need to know the rate, and the distance is the same, so the equations should be equal to each other.  I am thrown off again by the 3/4 hour.  Help again!

Thanks so much!

You already solved it....

D=RT...plug in values for D, R, and T.  If it's unknown, use a variable.

Maybe I'm an idiot, but I still don't understand how the 1.5 hours and the 3/4 hours come into play.  Can you help me with a table or how to do it? My son needs to show his work with a table that has RxT=D on the top and I'm not sure when to set the distances as equal or when to combine numbers on one side... it's been 30 yrs since I did algebra. :)

Thanks!

Oh dang...tables.  I never learned to solve it that way...

I can help you with the 1 1/2 and the 3/4 hours part though.

If you are traveling at $ x$ miles per [b]hour[/b], then use hours in your equation.  So it would be D=x*3/4 or d=x*3/2.

I remember this from weighted averages. (The spacing hee is a little off, but you should be able to read it by lining up the signs.)

[u]D=r*t[/u]
1.5x=x*1.5 (bus 1)
142.5-1.5x=(x-5)*1.5 (bus 2)
-------------------------------
142.5=(2x-5)*1.5 (add equations)

142.5=3x-7.5
135=3x
45=x

So bus 1 is traveling at 45 mph, and since x-5=45-5=40, bus 2 is traveling at 40 mph.

Try doing the second one on your own. (And if the time throws you off, think of this. When you buy apples, and it says $ \$$3/lb, does that mean you have to buy exactly 1lb, or 2lb, etc? No! Just use a decimal instead. The same applies to time. Notice how I converted 1 1/2 hours to 1.5.)

[hide="#2"]
[hide="dont click here unless you tried to solve it"]
Okay.

[u]D=r*t[/u]
.75x=x*.75 (east)
870-.75x=(x-120)*.75 (west)
-------------------------
870=(2x-120)*.75 (adding)

870=1.5x-90
960=1.5x
1440=x

So the east plane is going at 1440 mph, and since x-120=1440-120=1320, the west one is going at 1320 mph.[/hide][/hide]

[quote="lifeinhim61"]Maybe I'm an idiot, but I still don't understand how the 1.5 hours and the 3/4 hours come into play.  Can you help me with a table or how to do it?[b] My son needs to show his work with a table that has RxT=D on the top and I'm not sure when to set the distances as equal or when to combine numbers on one side... it's been 30 yrs since I did algebra. :)[/b]

Thanks![/quote]

They need tables....

Tables aren't a good idea, in general.  They just waste time.

... "the kid has to use the tables, the teacher demands it". Sorry for your kid! Lots of teachers kill creativity!

[quote="lifeinhim61"]Thank you.... the kid has to use the tables, the teacher demands it, and he's frustrated and doesn't know where to put what... the way he learned doesn't make any sense to me... but I guess this is the way it is. I just can't do it without just reasoning it out in my head.  He has to have all this funky stuff going on. Thanks for your help. :)[/quote]
if its not a HUGE assignment just do it the equation way cause the tables are unnecessary work.

Well, I made "tables", but they're in plain text...

Essentially, just line up the signs, and then insert lines where they belong...(I don't really know how to do it here...)

Maybe like for the 2nd question:

\begin{tabular}{|c|c|}
Symbol &Meaning\\
\hline x&Speed of westbound plane\\
\hline P_e&Eastbound plane\\
\hline P_w&Westbound plane\\
\hline
\end{tabular}

\begin{tabular}{|c|c|c\parallel{}c|}\hline Time& P_e & P_w &Distance Apart\\
\hline 0 min& 0 & 0 & 0\\
\hline 15 min& (x+120)/4 & x/4 & (2x+120)/4 \\
\hline 30 min& (x+120)/4 +(x+120)/2 & x/4+x/2 & (6x+360)/4\\
\hline 45 min& (x+120)/4 +(x+120)/2+3(x+120)/4 & x/4+x/2+3x/4 & (12x+720)4\\
\hline \end{tabular}

$ \dfrac{12x+720}{4}=3x+180=870$
\begin{eqnarray*}370&=&3x+180\\290&=&3x\\\dfrac{290}{3}&=&x\end{eqnarray*}

I probably got that wrong but maybe something like that?

[quote="lifeinhim61"]

I'm confused! Please help!  He needs this by Monday. I have others too.

[/quote]

Guess what?  This was posted last Saturday...


geometry
calculus
3D geometry
probability
circumcircle

[quote="goldendomer"]These are some medium and challenging target round questions.

1. Let ABDE be a rectangle. Let C be a point on BD . If BC=11 cm, CD=9 cm and ACE is a right angle, what is the number of centimeters, to the nearest whole number, in the length of CE ?

2.  A school organization consists of 5 teachers, 7 parents and 8 students. A subcommittee of this group must be formed by choosing 2 teachers, 3 parents and 5 students. How many different subcommittees can be formed?

3. An equilateral triangle is inscribed in a circle with diameter 9 cm . How many square centimeters are in the area of the triangle? Express your answer as a decimal to the nearest tenth.

4. At the grading of the Advanced Placement Calculus exams, 390 math teachers graded 120000 tests in 50 hours. Each test had 6 problems. What is the average number of minutes it took one teahcer to grade 25 problems? Express your answer to the nearest minute.

5. Two number cubes (number cube=dice), each with the digits 1-6 on the six faces, are rolled. What is the probability that the product of the numbers on the top faces will be greater than 8 ? Express your answer as a common fraction in lowest terms.

Good luck! :)[/quote]

[hide="1"]$CE=\sqrt{180}$
AE=BD=20. AB=DE=x. BC=11, CD=9. AC=$\sqrt{x^2+121}$, CE=$\sqrt{x^2+81}$.
$x^2+121+x^2+81=20^2$
$2x^2+202=400$
$x^2+101=200$
$x=\sqrt{99}$
$CE=\sqrt{99+81}$
$CE=\sqrt{180}$
[/hide]

[hide="2"]$binom{5}{2}\cdot \binom{7}{3} \cdot \binom{8}{5}=10\cdot 35 \cdot56 = 19600$(?)[/hide]
[hide="3"]
since circumradius = $\frac{abc}{4A}$, $\frac{4a}{\sqrt{3}}=4.5$, $a=\frac{9\sqrt{3}}{8}$, area is $\frac{243\sqrt{3}}{256}$[/hide]

[hide="4"]total = 720000 problems, each teacher grades 480/13 problems in 1 hour, $\frac{25}{x}=\frac{\frac{480}{13}}{60}$, $x=40.625$, nearest minute = 41[/hide]
[hide="5"]
# less than  or = 8 are 16, greater than eight are 36-16=20, 20/36=$\frac{5}{9}$[/hide]


search
real analysis
real analysis unsolved

Consider the sequence of numbers (u_n) defined by $ u_n=n^{2^n} $ for all n=1,2,...
 Put $ x_n = \frac{1}{u_1}+\frac{1}{u_2} + ... + \frac{1}{u_n} $
 PRove that the sequence (x_n) has a limit when n tends to infinity and the limit is an irrational

If you put $t_1=u_1,t_n=\frac{u_n}{u_{n-1}},\ \forall n\ge 2$, then we see that $(t_n)_n$ is strictly increasing, and it's a well known fact that $\frac 1{t_1}+\frac 1{t_1t_2}+\ldots$ is irrational. The proof is pretty much the same as the usual proof that $e=1+\frac 1{1!}+\frac 1{2!}+\ldots$ is irrational.

[quote="grobber"]it's a well known fact that $\frac 1{t_1}+\frac 1{t_1t_2}+\ldots$ is irrational.[/quote]
Only if $t_j$ are integers, which is not the case here  ;)

Oh, sorry about that :). Now it looks to me like this number contains arbitrarily long runs of $0$'s in its decimal expansion, but I'm too lazy and/or tired to complete the estimations. Sorry if this is, again, wrong :).

I think this problem was solved before by Moubinool but I can't remember the topic

[quote="romano"]No solution ? :([/quote]

romano,

To ask for the solution of a problem three times in a row sometimes is considered spamming, and by the way, to be that insistent probably will have the oposite efect, that is, people just stop looking at the topic because it does not have much sence to see someone asking once and again the same question. Be patient, probably someone will solve it eventually.

The convergence of the series is very trivial (even when nobody has mention it). It is obvius that $2^n<n^{2^n}$ for $n>1$, then, by comparation test, or basically any test that you can imagine, it trivially convergence.

The irrationality does not seem to be easy at all. Probably you can prove (or attempt to) that if $x_n\rightarrow x$ then
$\left|x-\frac{p}{q}\right|<\frac{1}{q^\mu}$ have no solution for $p$ and $q$ integers if $\mu<2$, and by the irrationality measure theorem it turns to be irrational automatically.

By the way, about the Moubinool post that you mention, well, I couldn't find it, but there are search tools, just in case!

Best,

The following lemma should help.  Let $x$ be a real number, and let $p$ and $q$ be integers such that $p \ne qx$.  Then x is not a rational number with denominator at most $\frac{1}{|p - qx|}$.

For this problem, let $x$ be the limit.  Let $n$ be a large integer.  Let $p$ and $q$ be the numerator and denominator of $x_n$.  Then an estimate shows that $|p - qx|$ approaches 0.  By the lemma, $x$ must be irrational.

The basic (paradoxical) idea: To show that a number is irrational, show that it is very close (but not equal) to a rational number.  That works because two different rational numbers (with bounded denominators) can't be too close to each other.

[quote="djimenez"]

By the way, about the Moubinool post that you mention, well, I couldn't find it, but there are search tools, just in case!
[/quote]

Probably the post is in solved problem, I did not find it  :(


[quote="MCrawford"]The spindle speed of a CD player is 300 revolutions per minute.  How many revolutions does the spindle make while playing a song that lasts exactly 4 minutes?[/quote]
[hide="solution"]
300 x 4 = [b]1200[/b]
[/hide]

[hide]
300 revolutions per minute. 4 minutes.
300*4=1200[/hide]

[hide]1200
u can also use proportions
when it gets harder![/hide]

Well I think technically you are using proportions no matter how you solve it.

Yep, you just don't think that way because you think it's really easy.

[quote="MCrawford"]The spindle speed of a CD player is 300 revolutions per minute.  How many revolutions does the spindle make while playing a song that lasts exactly 4 minutes?[/quote]

By the way, you know that it is not true that the CDs always revolute to the same speed. Look at it. It goes faster at the beginning of the CD than at the end!  :D .

Well, now find the formula for that, then find the answer :lol:  :lol:  :D  :D  :D :lol: lol


inequalities
function
calculus
derivative
inequalities unsolved

[color=red][size=200]Let a,b,c non negatve numbers with a+b+c=3 and p>=1
prove that                            (1+a^p-a)(1+b^p-b)(1+c^p-c)>=1[/size][/color]

If I am not wrong, it belongs to Zhaobin and is unsolved in VASC's book.

I will try to solve this problem:
let us consider function:
$f(x)=1+x^{p}-x$ where $p\geq 1$ minimal valua hold's when $x=(\frac{1}{p})^{\frac{1}{p-1}}$ therefore we have
$g=f(x)f(y)f(z)$ where $x+y+z=3$ for wich we know when hol'd minimal for each function $f(x),f(y),f(z)$(here I must was more correct )  therefore (I think somewhere I have seen this statment, if you want I'll try to prove it, it's not hard) minimal hold's when $x=y=z$ when $p=1$ after is obviusly.

[quote="kunny"]Let $f(a,\ b,\ c,\ \lambda)=(1+a^{p}-a)(1+b^{p}-b)(1+c^{p}-c)+\lambda (a+b+c-3)$,

Using the method of Lagrange multipliers, taking partial derivative this for $a,\ b,\ c,\ \lambda$ and set those as 0, we obtain $a=b=c=1$.[/quote]
Try it, kunny! :wink:


floor function
inequalities unsolved
inequalities

[quote="long14893"]sr for asking but I searched the solution of this 4th problem(4th round) in this linkhttp://www.mathlinks.ro/Forum/resources.php?c=80&cid=101&year=1998 but have not found ... can anyone solve it >"< ?[/quote]
A lemma:
$ \frac{1}{n} > \frac{1}{10n\plus{}0} \plus{} \frac1{10n\plus{}1} \plus{} \frac{1}{10n\plus{}2} \plus{} ... \plus{} \frac1{10n\plus{}9}$ (Which is quite obvious...)

$ \sum_{i} \frac{1}{n_i}$ can reach $ 1 \plus{} \frac12 \plus{} .. \plus{} \frac19$ when $ n_i\equal{}i, i\equal{}1,2,3..,9$. Suppose another configuration $ m_i$ is such that $ \sum_{i} \frac{1}{m_i} > \sum_{i} \frac{1}{n_i}$ with the same constraints, and that the largest $ m_i$ contains no more than $ k \ge 2$ digits. Further more let $ m_i$ be that sequence with largest $ \sum_{i} \frac{1}{m_i}$. (If there are more than one then pick an abritary)

Let $ m_r$ be the largest number, and hence the number with the most digits.(Obviously $ m_r\ge10$) Let $ t \equal{} \lfloor \frac{m_r}{10} \rfloor$. Then there is at most $ 10$ numbers from $ m_i$(Including $ m_r$) on the form $ \overline{td}$, where $ d$ is a digit. The sum of the reciprocicals of these are strichtly less than $ \frac{1}{t}$. Hence there exist another sequence where all the numbers on the form $ \overline{td}$ is replaced with $ t$, which still contains no element with more than $ k$ digits, but whos sum of reciprocicals is strictly bigger than $ m_i$'s. Since there are only finitly many numbers with $ k$ digits - less than $ 10^{k\plus{}1}$ at least - we obtain a contradiction. Hence there doesn't exist such a sequence.

Now we have proved the problem when $ n_i$ has finitly many elements. How can we expand it to infinitly many elements?
Further more: The proof get's a bit clumsy.. Using $ \frac{1}{n} \ge \frac{1}{\overline{n0}}\plus{}\frac{1}{\overline{n2}}\plus{}...\plus{}\frac{1}{\overline{n9}}$ would do the job, but would be hard to argument with, if there were infinitly many elements..

[quote="Mathias_DK"][quote="long14893"]sr for asking but I searched the solution of this 4th problem(4th round) in this linkhttp://www.mathlinks.ro/Forum/resources.php?c=80&cid=101&year=1998 but have not found ... can anyone solve it >"< ?[/quote]
A lemma:
$ \frac {1}{n} > \frac {1}{10n \plus{} 0} \plus{} \frac1{10n \plus{} 1} \plus{} \frac {1}{10n \plus{} 2} \plus{} ... \plus{} \frac1{10n \plus{} 9}$ (Which is quite obvious...)

$ \sum_{i} \frac {1}{n_i}$ can reach $ 1 \plus{} \frac12 \plus{} .. \plus{} \frac19$ when $ n_i \equal{} i, i \equal{} 1,2,3..,9$. Suppose another configuration $ m_i$ is such that $ \sum_{i} \frac {1}{m_i} > \sum_{i} \frac {1}{n_i}$ with the same constraints, and that the largest $ m_i$ contains no more than $ k \ge 2$ digits. Further more let $ m_i$ be that sequence with largest $ \sum_{i} \frac {1}{m_i}$. (If there are more than one then pick an abritary)

Let $ m_r$ be the largest number, and hence the number with the most digits.(Obviously $ m_r\ge10$) Let $ t \equal{} \lfloor \frac {m_r}{10} \rfloor$. Then there is at most $ 10$ numbers from $ m_i$(Including $ m_r$) on the form $ \overline{td}$, where $ d$ is a digit. The sum of the reciprocicals of these are strichtly less than $ \frac {1}{t}$. Hence there exist another sequence where all the numbers on the form $ \overline{td}$ is replaced with $ t$, which still contains no element with more than $ k$ digits, but whos sum of reciprocicals is strictly bigger than $ m_i$'s. Since there are only finitly many numbers with $ k$ digits - less than $ 10^{k \plus{} 1}$ at least - we obtain a contradiction. Hence there doesn't exist such a sequence.

Now we have proved the problem when $ n_i$ has finitly many elements. How can we expand it to infinitly many elements?
Further more: The proof get's a bit clumsy.. Using $ \frac {1}{n} \ge \frac {1}{\overline{n0}} \plus{} \frac {1}{\overline{n2}} \plus{} ... \plus{} \frac {1}{\overline{n9}}$ would do the job, but would be hard to argument with, if there were infinitly many elements..[/quote]thx ... nice solution :D

[quote="Mathias_DK"][quote="long14893"]sr for asking but I searched the solution of this 4th problem(4th round) in this linkhttp://www.mathlinks.ro/Forum/resources.php?c=80&cid=101&year=1998 but have not found ... can anyone solve it >"< ?[/quote]
A lemma:
$ \frac {1}{n} > \frac {1}{10n \plus{} 0} \plus{} \frac1{10n \plus{} 1} \plus{} \frac {1}{10n \plus{} 2} \plus{} ... \plus{} \frac1{10n \plus{} 9}$ (Which is quite obvious...)

$ \sum_{i} \frac {1}{n_i}$ can reach $ 1 \plus{} \frac12 \plus{} .. \plus{} \frac19$ when $ n_i \equal{} i, i \equal{} 1,2,3..,9$. Suppose another configuration $ m_i$ is such that $ \sum_{i} \frac {1}{m_i} > \sum_{i} \frac {1}{n_i}$ with the same constraints, and that the largest $ m_i$ contains no more than $ k \ge 2$ digits. Further more let $ m_i$ be that sequence with largest $ \sum_{i} \frac {1}{m_i}$. (If there are more than one then pick an abritary)

Let $ m_r$ be the largest number, and hence the number with the most digits.(Obviously $ m_r\ge10$) Let $ t \equal{} \lfloor \frac {m_r}{10} \rfloor$. Then there is at most $ 10$ numbers from $ m_i$(Including $ m_r$) on the form $ \overline{td}$, where $ d$ is a digit. The sum of the reciprocicals of these are strichtly less than $ \frac {1}{t}$. Hence there exist another sequence where all the numbers on the form $ \overline{td}$ is replaced with $ t$, which still contains no element with more than $ k$ digits, but whos sum of reciprocicals is strictly bigger than $ m_i$'s. Since there are only finitly many numbers with $ k$ digits - less than $ 10^{k \plus{} 1}$ at least - we obtain a contradiction. Hence there doesn't exist such a sequence.

Now we have proved the problem when $ n_i$ has finitly many elements. How can we expand it to infinitly many elements?
Further more: The proof get's a bit clumsy.. Using $ \frac {1}{n} \ge \frac {1}{\overline{n0}} \plus{} \frac {1}{\overline{n2}} \plus{} ... \plus{} \frac {1}{\overline{n9}}$ would do the job, but would be hard to argument with, if there were infinitly many elements..[/quote]

By the way , can u help me the  2nd problem of Day 1 of the VMO 1997 :http://www.mathlinks.ro/resources.php?c=186&cid=40&year=1997   I searched the solution and could'nt fine it as well ...............>"<


function
parameterization
analytic geometry
real analysis
real analysis unsolved

Is there a differentiable function $f:[0,1]^2->R^3$ whose image is a Moebius band?

Hi, Harazi,

Here is the parametric function (I first saw it in the Thorpe's book, but it's quite straightforward): (t, u) -> ( (1 + t*cos(u/2))*cos(u), (1 + t*cos(u/2))*sin(u), t*sin(u/2)), -1/4 < t < 1/4. To understand geometrical meaning you can fix one of the parameters and see what coordinate lines.
I bet it can't be represented as a graph of a regular smooth function.

Stan


function
induction
algebra proposed
algebra

[hide="Solution"]Put in $ x \equal{} y \equal{} 1$ to get that $ f(1\plus{}f(1)) \equal{} 1 \plus{} f(1)$. So we know a fixed point of $ f$ exists. Call it $ c$. Putting in $ x \equal{} 1, y \equal{} c$ gives $ f(1\plus{}c) \equal{} f(1) \plus{} c$. By induction, $ f(k\plus{}c) \equal{} kf(1) \plus{} c$ for all positive integers $ k$. Now we get that $ a^2f(1) \plus{} c \equal{} f(a^2 \plus{} c) \equal{} af(a) \plus{} c$, and so $ af(1) \equal{} f(a)$. By this identity $ f(x^2 \plus{} f(y)) \equal{} x^2f(1) \plus{} yf(1)^2$ and $ xf(x) \plus{} y \equal{} x^2f(1) \plus{} y$. Setting these two equal gets $ yf(1)^2 \equal{} y$ so $ f(1) \equal{} 1$ and $ f(x) \equal{} x$ for all naturals $ x$.[/hide]

[quote="MellowMelon"][hide="Solution"]Put in $ x \equal{} y \equal{} 1$ to get that $ f(1 \plus{} f(1)) \equal{} 1 \plus{} f(1)$. So we know a fixed point of $ f$ exists. Call it $ c$. Putting in $ x \equal{} 1, y \equal{} c$ gives $ f(1 \plus{} c) \equal{} f(1) \plus{} c$. By induction, $ f(k \plus{} c) \equal{} kf(1) \plus{} c$ for all positive integers $ k$. Now we get that $ a^2f(1) \plus{} c \equal{} f(a^2 \plus{} c) \equal{} af(a) \plus{} c$, and so $ af(1) \equal{} f(a)$. By this identity $ f(x^2 \plus{} f(y)) \equal{} x^2f(1) \plus{} yf(1)^2$ and $ xf(x) \plus{} y \equal{} x^2f(1) \plus{} y$. Setting these two equal gets $ yf(1)^2 \equal{} y$ so $ f(1) \equal{} 1$ and $ f(x) \equal{} x$ for all naturals $ x$.[/hide][/quote]
How do you induct ?  :wink: .
Here is my solution .
Take $ y \equal{} x \equal{} 1$ then $ f(f(1) \plus{} 1) \equal{} f(1) \plus{} 1$
$ a \equal{} f(1) \plus{} 1$
Take $ x \equal{} 1$ then
$ f(f(y) \plus{} 1) \equal{} y \plus{} f(1)$
Take $ y\to f(y) \plus{} 1$ then
$ f(y \plus{} a) \equal{} f(y) \plus{} a$
Induction we have $ f(x \plus{} ka) \equal{} f(x) \plus{} ka$
Because $ f(a) \equal{} a$ so $ f(ka) \equal{} ka,\forall k\in N$
Now take $ y \equal{} a$ then
$ f(x^2) \equal{} xf(x)$
Take $ x\to a \plus{} 1$ then
$ f(a^2 \plus{} 2a \plus{} 1) \equal{} (a \plus{} 1)f(a \plus{} 1)$
$ \Leftrightarrow a^2 \plus{} 2a \plus{} f(1) \equal{} (a \plus{} 1)(a \plus{} f(1))$
$ \Leftrightarrow f(1) \equal{} 1$
 So $ a \equal{} 2$
We have $ f(x \plus{} 2) \equal{} f(x) \plus{} 2$
and $ f(2) \equal{} 2,f(1) \equal{} 1$
Induction we have $ f(n) \equal{} n,\forall n\in N$
Solution of this function equation :$ f(n)\equiv n$

[quote=Ahiles]Find functions $ f: \mathbb{N} \rightarrow \mathbb{N}$, such that $ f(x^2 \plus{} f(y)) \equal{} xf(x) \plus{} y$, for $ x,y \in \mathbb{N}$.[/quote]

P(0,y) => $f(f(y))=y$(*)
P(f(x),y) with (*)=>$f(f(x)^2+f(y))=xf(x)+y$
=>$f(f(x)^2+f(y))=f(x^2+f(y))$
And using (*) => $f(x)^2=x^2=>f(x)=x$


At 12:00 noon, Anne, Beth and Carmen begin running laps around a circular track of length $300$ meters, all starting from the same point on the track. Each jogger maintains a constant speed in one of the two possible directions for an indefinite period of time. Show that if Anne's speed is different from the other two speeds, then at some later time Anne will be at least $100$ meters from each of the other runners.  (Here, distance is measured along the  shorter of the two arcs separating two runners.)

[quote]
Let $M=300$ and let us define by $f: [0, \infty) \rightarrow [0, \frac{M}{2}]$ the distance between Anne and Beth and $g: [0, \infty) \rightarrow [0, \frac{M}{2}]$ the distance between Anne and Carmen. $f$ and $g$ are not constant functions by hypothesis. $f$ and $g$ are periodic functions and let $S$ and $T$  be their periods respectively. Since $f$ and $g$ are not constant, $S>0, T>0$. Suppose $S\leq T$.

Moreover, by hypothesis $f$ and $g$ are linear functions on any interval of the form $[k\frac{S}{2}, (k+1)\frac{S}{2}]$ and $[k\frac{T}{2}, (k+1)\frac{T}{2}]$, $k\geq0$ respectively. Hence $f$ and $g$ are two saw functions of the following form:
$f(x)=\{ \begin{array}{c}{11} f_{2k}(x), x\in[2k\frac{S}{2}, (2k+1)\frac{S}{2}]\ f_{2k+1}(x), x\in[(2k+1)\frac{S}{2}, (2k+2)\frac{S}{2}]\ \end{array} $
, $k\geq0$
$g(x)=\{ \begin{array}{c}{11} g_{2k}(x), x\in[2k\frac{T}{2}, (2k+1)\frac{T}{2}]\ g_{2k+1}(x), x\in[(2k+1)\frac{T}{2}, (2k+2)\frac{T}{2}]\ \end{array} $
, $k\geq0$.

We have to prove that $(\exists) \alpha\geq0$ s.t. $f(\alpha)\geq\frac{M}{3}$ and $g(\alpha)\geq\frac{M}{3}$.

We shall prove that by analyzing how $g_0$ and $g_1$ intersect different $f_k$.
Let $k_0$ be the maximum $k\geq0$ for which $g_0\cap f_{k_0}\neq\oslash$. Such a $k_0$ exists because we supposed that $S\leq T$. This obviously implies that $g_1\cap f_{k_0+1}\neq\oslash$. Let ${P_0(x_0, y_0)=g_0\cap f_{k_0}}$ and $P_1(x_1, y_1)=g_1\cap f_{k_0+1}$. Let us notice that if $y_0\geq\frac{M}{3}$ or $y_1\geq\frac{M}{3}$ than $x_0$ or $x_1$ are the searched $\alpha$ so suppose that (1) $y_0<\frac{M}{3}$ and $y_1<\frac{M}{3}$.
After solving the two equations ${P_0(x_0, y_0)=g_0\cap f_{k_0}}$ and $P_1(x_1, y_1)=g_1\cap f_{k_0+1}$ we get that $P_0$ and $P_1$ are defined by $P_0((k_0+1)\frac{ST}{S+T}, (k_0+1)\frac{SM}{S+T})$ and $P_1((k_0+2)\frac{ST}{S+T}, M\frac{T-(k_0+1)S}{S+T})$.
From this and (1) we get that
$\{ \begin{array}{c}{11} (k_0+1)\frac{SM}{S+T}<\frac{M}{3}\ M\frac{T-(k_0+1)S}{S+T})<\frac{M}{3}\ \end{array}  \Leftrightarrow \{ \begin{array}{11} k_0+1<\frac{S+T}{3S}\ k_0+1>\frac{2T-S}{3S}\ \end{array}  \Leftrightarrow \{ \begin{array}{11} k_0<\frac{T-2S}{3S}\ k_0>\frac{2T-4S}{3S}\ \end{array} $ (3).
From this we get that $\frac{2T-4S}{3S}<\frac{T-2S}{3S}$ which is equivalent with (4) $T<2S$.
But (3) and (4) imply that $k_0<0$ which is a contradiction and this completes the proof.

[/quote]

[quote=Altheman]For a simpler proof, consider the frame of reference of Anne, and do cases for Beth and Carmen.[/quote]

I have also solved using the topic of reference point.Seems ur a physics student


you know in defining piece wise function you do stuff like:

$ |x| = \{_{-x, x < 0}^{ x, x \ge 0}$

But of course that first bracket looks ugly.  I tried to use \left which fixes it, but then I need a \right.  I can just put the \right and ignore the closing delimiter, and it works fine, but gives me an annoying error message.  Any other cleaner and smarter ways?

edit: hmm I know the words look tiny small too, but for my tex adding \displaystyle fixes it, but I guess on AoPS they automatically remove those stuffs??

Use \phantom

$ |x| = \{\begin{array}{l}x,\ x\geq0\\-x,\ x<0\end{array}\phantom$

Better yet, use the \cases environment (you need the amsmath package)

$ |x|=\begin{cases}x&\text{if }x\ge0\\-x&\text{if }x<0 \end{cases}$
[code]$|x|=\begin{cases}
x&\text{if }x\ge0\\
-x&\text{if }x<0
\end{cases}$[/code]

[quote="pkerichang"]I tried to use \left which fixes it, but then I need a \right.  I can just put the \right and ignore the closing delimiter, and it works fine, but gives me an annoying error message.  Any other cleaner and smarter ways?[/quote]Do follow roadnottaken's advice for such functions but, in another situation, if you ever need a left bracket but no right bracket then use[color=red]\right.[/color] with the [b][color=red].[/color][/b] at the end. So you can have, for example, $ \left[\frac{1}{2}\right.$ and using \left. you get $ \left.\frac{1}{2}\right)$
[quote]edit: hmm I know the words look tiny small too, but for my tex adding \displaystyle fixes it, but I guess on AoPS they automatically remove those stuffs??[/quote]In fact the forum adds \displaystyle to your formula so you don't need to use it.

also you can try the following-gives a slightly improved look for inline(the one using cases seems to be taller):[code]
\left|x\right| =
\left\{
\begin{matrix}
   x & \text{if } x \geq 0  \\
   -x & \text{if } x < 0
\end{matrix}
\right.[/code]

\left|x\right| =
\left\{
\begin{matrix}
   x & \text{if } x \geq 0  \\
   -x & \text{if } x < 0
\end{matrix}
\right.

(Testing how to get the forum to render \LaTeX code?????

Ok how is latex code rendered inline...is there a special tag???

Use dollar signs like you would in a tex document. [code]$|x|=\begin{cases}
x&\text{if }x\ge0\\
-x&\text{if }x<0
\end{cases}$[/code]$ |x|=\begin{cases}x&\text{if }x\ge0\\-x&\text{if }x<0 \end{cases}$


Se considera multimea
\[A=\{x\in\mathbb{N}| x=(p^{2}-q)(q^{2}-p), p, q\in\mathbb{N}\}. \]
Sa se arate ca:
a) $2006\in{A}$;
b) $A$ contine o infinitate de patrate perfecte;
c) nu exista nicio multime nevida $B$ astfel incat $|B|=p^{2}-q$; $|P(B)|=q^{2}-p$ cu $p,q\in\mathbb{N}$.

[quote="Marius Damian"]Se considera multimea
\[A=\{x\in\mathbb{N}| x=2(p^{2}-q)(q^{2}-p), p, q\in\mathbb{N}\}. \]
Sa se arate ca:
a) $2006\in{A}$;
b) $A$ contine o infinitate de patrate perfecte;
c) nu exista nicio multime nevida $B$ astfel incat $|B|=p^{2}-q$; $|P(B)|=q^{2}-p$ cu $p,q\in\mathbb{N}$.[/quote]

[color=blue]Am corectat enuntul.[/color]


Put $ a;b;c \ge 0$ and $ a\plus{}b\plus{}c\equal{}1$. Prove that:
$ 4 \plus{} 15abc \plus{} (a \minus{} b)(b \minus{} c)(c \minus{} a) \ge\ 5(a^2b \plus{} b^2c \plus{} c^2a) \plus{} 12(ab \plus{} bc \plus{} ca)$

[quote="onlylove_math"]Put $ a;b;c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 1$. Prove that:
$ 4 \plus{} 15abc \plus{} (a \minus{} b)(b \minus{} c)(c \minus{} a) \ge\ 5(a^2b \plus{} b^2c \plus{} c^2a) \plus{} 12(ab \plus{} bc \plus{} ca)$[/quote]
It is really easy ,my friend.This ineq is the mixing of two ineq
The first $ 2\sum a^3 \plus{} 3abc\geq 3(a^2b \plus{} b^2c \plus{} c^2a)$  (vey well-know)
And $ ab^2 \plus{} bc^2 \plus{} ca^2\geq 3abc$
It is nice

[quote="Allnames"][quote="onlylove_math"]Put $ a;b;c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 1$. Prove that:
$ 4 \plus{} 15abc \plus{} (a \minus{} b)(b \minus{} c)(c \minus{} a) \ge\ 5(a^2b \plus{} b^2c \plus{} c^2a) \plus{} 12(ab \plus{} bc \plus{} ca)$[/quote]
It is really easy ,my friend.This ineq is the mixing of two ineq
The first $ 2(\sum a^3 \plus{} 3abc\geq 3(a^2b \plus{} b^2c \plus{} c^2a)$  (vey well-know)
And $ ab^2 \plus{} bc^2 \plus{} ca^2\geq 3abc$
It is nice[/quote]
I don't understand your proof
Can you rewrite detail of your proof? please  :)

[quote="bunhiacovski"][quote="Allnames"][quote="onlylove_math"]Put $ a;b;c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 1$. Prove that:
$ 4 \plus{} 15abc \plus{} (a \minus{} b)(b \minus{} c)(c \minus{} a) \ge\ 5(a^2b \plus{} b^2c \plus{} c^2a) \plus{} 12(ab \plus{} bc \plus{} ca)$[/quote]
It is really easy ,my friend.This ineq is the mixing of two ineq
The first(1) $ 2\sum a^3 \plus{} 3abc\geq 3(a^2b \plus{} b^2c \plus{} c^2a)$  (vey well-know)
And $ ab^2 \plus{} bc^2 \plus{} ca^2\geq 3abc$
It is nice[/quote]
I don't understand your proof
Can you rewrite detail of your proof? please  :)[/quote]
It is quite simple.onlylove_math's ineq <=>$ 4\sum a^3 \plus{} 3abc\plus{}ab^2 \plus{} bc^2 \plus{} ca^2\geq 6(a^2b \plus{} b^2c \plus{} c^2a)$
and use two ineq that i said we have a proof ((1) is a ineq in Secrets in ineq of PKH  or you can prove it by SS or pqr)

[quote="Allnames"][/quote][quote="bunhiacovski"][quote="Allnames"][quote="onlylove_math"]Put $ a;b;c \ge 0$ and $ a + b + c = 1$. Prove that:
$ 4 + 15abc + (a - b)(b - c)(c - a) \ge\ 5(a^2b + b^2c + c^2a) + 12(ab + bc + ca)$[/quote]
It is really easy ,my friend.This ineq is the mixing of two ineq
The first(1) $ 2\sum a^3 + 3abc\geq 3(a^2b + b^2c + c^2a)$  (vey well-know)
And $ ab^2 + bc^2 + ca^2\geq 3abc$
It is nice[/quote]
I don't understand your proof
Can you rewrite detail of your proof? please  :)[/quote][quote="Allnames"]
It is quite simple.onlylove_math's ineq <=>$ 4\sum a^3 + 3abc + ab^2 + bc^2 + ca^2\geq 6(a^2b + b^2c + c^2a)$
and use two ineq that i said we have a proof ((1) is a ineq in Secrets in ineq of PKH  or you can prove it by SS or pqr)[/quote]
Thank Allnames; It is my proof.
But with my proof i dont use $ ab^2 + bc^2 + ca^2\geq 3abc$ . I have used
$ a^3 + b^3 + c^3 + ab^2 + bc^2 + ca^2 \ge\ 2(a^2b + b^2c + c^2a)$


Today I just won my 100th game. That means you get the 5th win thingy (click on the user and see how many win thingies they have). Just wondering, has anyone had trouble getting that?

The 5th win badge is many orders of magnitude easier to earn than the 5th rating badge, and infinite orders of magnitude easier to earn than the 5th game badge. Needless to say, most users have it.

@cheze___

Whenever you win a certain amount of games, you are awarded with badges.  When you win enough games, you will have 5 badges when someone(even yourself) clicks on your name on the left bar.  You also get to win not only win badges, but also rating and game badges.  There are a total of 14 badges that you can have: 5 ratings, 5 wins, and 4 games.

I don't know when you get the win and game badges, but after the rating badges in order are when you have achieved a "record" rating of 1200, 1350, 1500, 1750, and 2000(not very many people have this...i only know jongy, tcoas, aime15, chaus, and 1337rox)


trigonometry
geometry
geometric transformation
rotation
dilation
ratio
circumcircle

Give sharp triangle $ ABC$,erect on the outside of $ BC,CA,AB$ in turn the squares with center are $ M,N,P$.
Prove that:

$ \bigtriangleup{BMN}\equal{}\bigtriangleup{MCP}$

Please help me with this problem,it's hard with me :( .Thank you!

Let $ BC\equal{}a$, $ AC\equal{}b$, $ AB\equal{}c$, $ \angle A\equal{}\alpha$, $ \angle B\equal{}\beta$,
$ \angle C\equal{}\gamma$. Let us prove that $ BN\equal{}PM$. Considering
$ \triangle BAN$ we get
\[ BN^{2}\equal{}c^2\plus{}\frac{b^2}{2}\minus{}bc\sqrt{2}\cos(\alpha\plus{}\frac{\pi}{4}) \equal{}
c^2 \plus{}\frac{b^2}{2} \minus{}bc(\cos\alpha \minus{}\sin\alpha).\] Considering
$ \triangle PBM$ we get \[ PM^{2}\equal{}\frac{c^2}{2}
\plus{}\frac{a^2}{2}\plus{}ac\sin\beta .\] Since $ b\sin\alpha \equal{} a\sin \beta$
we have $ bc\cos\alpha \equal{} bc\cos\alpha \minus{}bc\sin\alpha \plus{}ac\sin\beta$,
whence $ \frac{c^2}{2}\plus{}\frac{b^2}{2}\minus{}\frac{a^2}{2}\equal{}bc(\cos\alpha
\minus{}\sin\alpha)\plus{}ac\sin\beta ,$ and thus
\[ c^2 \plus{}\frac{b^2}{2}\minus{}bc(\cos\alpha \minus{}\sin\alpha) \equal{} \frac{c^2}{2}\plus{}
\frac{a^2}{2} \plus{}ac\sin\beta .\] So we have proved that $ BN\equal{}PM$.
Analogously $ NM\equal{}CP.$ Obviously $ BM\equal{}MC.$

So $ \triangle BMN\equal{}\triangle MCP. \     \Box$

Well, unlike other people here, I do not like boring calculations. My goal is to prove that AM and PN are equal and perpendicular, similarly for the other 2 pairs.
So, let's note D - midpoint of BC, whilst AE and AF being sides of the squares, E and F the opposite vertices of C and B respectively.
A $ 90^\circ$ rotation about A will map B to F and E to C, hence BE and CF are equal and perpendicular; the same are PD and DN, as midlines in the triangles BCF and BCE, consequently $ PN \equal{} PD \cdot \sqrt {2}$ ( 1 ).
A $ 45^\circ$ rotation about B followed by a dilation of $ \sqrt 2$ ratio will map P to A and D to M, while a $ 45^\circ$ rotation about C followed by a dilation of $ \sqrt 2$ ratio will map N to A and D to M, hence $ AM \equal{} PD \cdot \sqrt 2$ ( 2 ) and $ \angle (PD, AM) \equal{} 45^\circ$ ( 3 ).
From (1) and (2) we get AM = PN, and, with (3) we get $ AM \perp PN$.
Similarly, BN = MP, CP = BN and, with BM = MC, $ \triangle BMN \equal{} \triangle MCP$, q.e.d.

Best regards,
sunken rock

Proving that $ AM$ and $ PN$ are perpendicular will suffice . (Since after that, we can prove that the two triangles are similar and since $ BM \equal{} MC$, we'll be done)
[img]http://img41.imageshack.us/img41/1326/10561676.png[/img]

The circumcircles of two of the erected squares are shown. This is equivalent to showing that $ M$ lies on the radical axis of the circles.
Let $ MB$ and $ MC$ intersect the circles at $ X$ and $ Y$. $ BU$ and $ CV$ are diameters. Observe that $ \angle BUX \equal{} B$ and $ \angle CVY \equal{} C$. We easily see that $ BX \equal{} CY$ ($ BUsinB \equal{} CVsinC$ iff $ ABsinB \equal{} ACsinC$)
So,  $ MC \equal{} MB$ and $ MX \equal{} MY$ $ \rightarrow MB.MX \equal{} MC.MY$. So, power of $ M$ w.r.t. the two circles are equal and we're done.


vector
topology
real analysis
real analysis unsolved

[quote="senouy"]E is a normed vector space , and M is a submanifold of finite dimension of E;
I am facing the following confusion: since M is of finite dimension , it is topologically isomorphic to Rn , which is complete, so M is cmplete
but since the interior of M is empty,so it contains no spheroid, and then M is of first Baire category which contrdict its cmpleteness.
Thanks.[/quote]The same argument could be made to argue that $ \mathbb R \sim  \mathbb R \times \{ 0 \} \subset \mathbb R^2$ is not complete. Think about it.

thanks , but i can not understand what you read

Basically, your argument is nonsense. That's not what the Baire theorem says, and the completeness of $ M$ depends on the choice of metric, not just a topology. Is the open unit disk complete? It's homeomorphic to $ \mathbb{R}^2$.

[quote]the interior of M is empty,so it contains no spheroid, and then M is of first Baire category which contrdict its cmpleteness[/quote]
The Baire category theorem asserts that a complete metric space is of second category [b]in itself[/b]. This is not related to the fact that $ M$ is of first category in $ E$.


floor function
algebra
functional equation
algebra proposed

Find $f:\mathbb{N}\to\mathbb{N}$ satisfying :
   \[f(m+f(n))= f(f(m)) +f(n)\]

This one was problem 3 in IMO 1996.

I'll assume that $0\in \mathbb{N}$ (as IMO 1996). Substituting $m=n=0$ we get $f(0)=0$. Plugging in $n=0$, one gets $f(m) = f(f(m))$. One solution is $f(n) = 0$ for all $n\in \mathbb{n}$.

The functional equation becomes $f(m+f(n)) = f(m) + f(n)$.

Assume that there is $n$ such that $f(n)\neq 0$. Thus there is at least one nonzero fixed point ($f(n)$, for example). Let $a>0$ be the minimum fixed point of $f$ and let $f(i) = a_i$ for $0 < i < a$. Since $f(m+a) = f(m+f(a)) = f(m) + f(a) = f(m) + a$ inductively we have $f(m+qa) = f(m) + qa$ for all nonnegative integers $q$. Thus if $n=qa+r$, $q,r$ integers and $0\leq r<a$ then $f(n) = qa + f(r) = qa + a_r$. So, defined $f(0)=0,f(1),\ldots,f(a)=a$, we have $f(n) = \left\lfloor\frac{n}{a}\right\rfloor a + f(n\bmod{a})$.

I didn't think about the problem if $0\notin \mathbb{N}$, but it seems to be harder...


SMO 2007, Open #17

In $\Delta ABC$, $AC > AB$, the internal angle bisector of $\angle A$ meets $BC$ at $D$, and $E$ is the foot of the perpendicular from $B$ onto $AD$. Suppose $AB = 5$, $BE = 4$ and $AE = 3$. Find the value of the expression \[(\frac{AC+AB}{AC-AB})ED\].

We have $AE=c\cos{A/2}\longrightarrow \cos{A/2}=\frac{3}{5}$
then we know that $AD=\frac{2bc\cos{A/2}}{b+c}=\frac{6b}{b+5}$ $\longrightarrow ED=AD-3=\frac{3b-15}{b+5}$ and
now :$(\frac{b+c}{b-c})ED=3$

[quote="tjhance"][quote="khashi70"]$AD=\frac{2bc\cos{A/2}}{b+c}$[/quote]

How did you get this?[/quote]
with $\sin$ law in $\triangle ABD$ we get that $\frac{AD}{\sin B}=\frac{BD}{\sin{\frac{A}{2}}}$
but we know that $\frac{BD}{DC}=\frac{AB}{AC}$ so we get that $\frac{BD}{a}=\frac{c}{b+c}$ so we get that $BD=\frac{ac}{b+c}$ so we get that $\frac{AD}{\sin B}=\frac{ac}{(b+c)\sin{\frac{A}{2}}}$
but we know that $\frac{b}{\sin B}=\frac{a}{\sin A}$ so we get that:
$AD=\frac{bc\sin A}{(b+c)\sin{\frac{A}{2}}}$
but with $\sin A=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}$ we get that:
$AD=\frac{2bc\cos{\frac{A}{2}}}{b+c}$...


conics
ellipse
linear algebra
matrix
ratio
integration
parabola

[color=darkred]First of all, I'd like to say sorry for making this urgent as it is despised in this board. However, I'm having maths board exams 14th and I'd be very grateful if you can try to help me in the following questions ASAP. Those questions are very important ones, taken from last years papers....

Thank you in advance.

Guys wish me luck for exams!
[/color]

[b][u]Questions which I don't know how to solve:[/u][/b]

1. Find the equation of tangent to the ellipse $ \frac {x^2}{25} + \frac {y^2}{16} = 1$ on the point $ (\frac {5}{2},2\sqrt {3})$

[b]✓[/b] [hide]2. Solve the equation sin2x+cosx=0 for general solution. [/hide] [b]Got this, ty![/b]

3. If $ C_0, C_1, C_2 ... C_n$ are binomial coefficients in the expanion of $ (1 + x)^n$, prove that $ C_0 + C_2 + C_4 + ... = 2^n^ - ^1$

4. Find the sum of the following infinite series $ 1 + \frac {1}{3}(\frac {1}{2})^2 + \frac {1}{5}(\frac {1}{2})^4 + ...$

5. If $ e^x^ + ^y = xy$, show that $ \frac {dy}{dx} = \frac {y(1 - x)}{x(y - 1)}$

6. Using properties of determinants, prove that$ \lvert A\rvert = \begin{vmatrix} a - b - c & 2a & 2a \\
2b & b - c - a & 2b \\
2c & 2c & c - a - b \end{vmatrix} = (a + b + c)^3$

7. Find $ n$ if $ {}^2{}^nC_3: ^nC_3 = 44: 3$

8. Prove that $ (2n)! = 2^n.n![1.3.5....(2n - 1)]$

9. Find the sum to n terms of the following series: $ \frac {1}{1.3} + \frac {1}{3.5} + \frac {1}{5.7} + ....$             [color=olive][PS: (dot) is the multiplication and not the decimal][/color]

10. If $ C_0, C_1, C_2,...C_n$ are the binomial coefficients in the expansion of $ (1 + x)^n$. Prove that $ C_0+C_1+C_2+C_3+...+C_n=2^n$

11. Find the ratio in which the line joingin $ ( - 5, 1)$ and $ (1, - 3)$ divides the join of the points $ (3, 4)$ and $ (7, 8)$.

12. Evaluate $ \int{\frac {3x^2}{\sqrt {9 - 16x^6}}}dx$

[b]✓[/b] [hide]13. Solve $ 2cos^2\theta + 3sin\theta = 0$.[/hide] [b]Got this, ty![/b]

14. Find the vertex, axis, focus and latus rectum of the parabola $ y^2 + 4x - 2y + 3 = 0$

15. Show that the derivative of $ tan^ - ^1{\frac {\sqrt {1 + x^} - 1}{x}} w.r.t. sin^ - ^1{\frac {2x}{1 + x^2}} is \frac {1}{4}.$

16. Evaluate: $ \int^\pi_0{\frac {xsinx}{1 + cos^2x}}dx$
Ans: $ \int^\pi_0{\frac {xsinx}{1 + cos^2x}}dx$=$ \int^\pi_0{\frac {xsinx}{sin^x}}dx$
$ = \int^\pi_0{\frac {x}{sinx}}dx$? [color=olive][Just gotta know how to integrate this atleast, then I know how to put the pi and 0 values there][/color]

17. Solve the equation cos 3x=sin2x for x.
My work: $ 4cos^3x-3cosx=2cosxsinx$
$ 4cos^2x-3-2sinx=0$
$ 4(1-sin^2x)-3-2sinx=0$
$ 4sin^2x+2sinx-1=0$
$ 42sinx(2sinx+1)-1=0$ [color=olive][<<<What to do next??][/color]

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

[b][u]For correction on my work:[/u][/b]

[color=darkred]Following are my works....which I'm not sure if it's correct... so..plz check the following works and point out the fault if any....[/color]

A. If $ y = \frac {e{}^2{}^x.cosx}{xsinx}$, find $ \frac {dy}{dx}$
Ans: $ \frac {dy}{dx} = \frac {\frac {d(e{}^2{}^x.cosx)}{dx}xsinx - \frac {d(xsinx)}{dx}(e{}^2{}^x.cosx)}{x{}^2sin{}^2x}$
$ = \frac {2x e{}^2{}^xcosxsinx - xe{}^2{}^xsin{}^2x - xe{}^2{}^xcos{}^2x - e{}^2{}^xsinxcosx}{x{}^2sin{}^2x}$
$ = \frac {2x - xe{}^2{}^x}{x{}^2sin{}^2x}$

B. Evaluate: $ \lim_{x \to - 1}(\frac {x + 1}{\sqrt {6x^2 + 3} + 3x})$
Ans: $ \lim_{x \to - 1}(\frac {x + 1}{\sqrt {6x^2 + 3} + 3x}) = \lim_{x\to - 1}\frac {x + 1}{x\sqrt {6 + 3} + 3x}$
$ = \lim_{x \to - 1}\frac {x + 1}{6x}$
$ = 0$

C.  Solve the differential equation $ \frac {dy}{dx} = 1 + x + y + xy$ given that y=0 when x=0.
Ans: $ \frac {dy}{dx} = 1 + 0 + 0 + 0.0$
$ \int{dy} = \int{dx}$
$ y = x + C$
$ = 0 + C = C$

[b]✓ Understood this 1, ty![/b]
[hide]D. Evaluate $ \sqrt {13 - 20\sqrt {3}i}$
Ans: Let $ \sqrt {13 - 20\sqrt {3}i} = x + yi$
$ 13 - 20\sqrt {3}i = (x + yi)^2 = x^2 - y^2 + 2xyi$

$ 13 = x^2 - y^2$                                      .............(1)
$ - 20\sqrt {3}i = 2xyi$

$ (x^2 + y{}^2){}^2 = (x^2 - y{}^2){}^2 + (2xy)^2$
$ (x^2 + y{}^2){}^2 = 169 - 20\sqrt {3}$
$ x^2 + y^2 = \sqrt {169 - 20\sqrt {3}}$                     ...................(2)

Subtracting (1) and (2), I get:
$ y = \sqrt {\frac {\sqrt {169 - 20\sqrt {3}} - 13}{2}}$

Also, $ x = \sqrt {\frac {\sqrt {169 - 20\sqrt {3}} - 13}{2}}$

Therefore, $ \sqrt {13 - 20\sqrt {3}i}$=$ \sqrt {\frac {\sqrt {169 - 20\sqrt {3}} - 13}{2}} + \sqrt {\frac {\sqrt {169 - 20\sqrt {3}} - 13}{2}}i$
=$ \sqrt {\frac {\sqrt {169 - 20\sqrt {3}} - 13}{2}} - \sqrt {\frac {\sqrt {169 - 20\sqrt {3}} - 13}{2}}$=0[/hide]

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[b][u]For Explanation:[/u][/b]

[color=darkred]Need your explanation for the following questions with solution from the text books....also plz let me know whatever formula is used....[/color]

[b]✓ Understood these hiddens....ty![/b]
[hide]a. The equation of the chord of the circle $ x^2 + y^2 - 2ax = 0$ is y=mx. Find the equation of the circle described on this chord as diameter.
Ans: The coordinates of the pints of intersection of the circle and the given chord are, (0, 0) and $ (\frac {2a}{1 + m^2},\frac {2ma}{1 + m^2})$    [color=olive]<<<How did they get this??[/color]
[color=darkblue][size=75]------rest of the solution censored....sorry no time to type out------[/size][/color][/hide]

[hide]b. Evaluate: $ \lim_{x \to 0}(\frac {1 - cos x}{2x^2})$
Ans: $ \lim_{x \to 0}(\frac {1 - cos x}{2x^2}) = \lim_{x \to 0}(\frac {2sin\frac {x}{2}}{2x})^2$[color=olive]<<<How did they get this??
[/color]
[color=darkblue][size=75]------rest of the solution censored------[/size][/color][/hide]

c. Find the Mean Deviation (MD) and Standard Deviation (SD) of the AP: a, a+d, a+2d,...,a+2nd.
Ans: The number of items in the A.P. is (n+1)
$ %Error. "mean" is a bad command.
=a+nd$
Mean deviation about the mean
$ =\frac{1}{(2n+1)}\displaystyle\sum_{k=1}^n{2n}{r=0} |(a+rd)-(a+nd)|$
$ =\frac{1}{(2n+1)}.2[nd+(n-1)d+ .... +d]$
$ =\frac{2}{(2n+1)}[1+2+ ... +(n-1)+n]d$
$ =\frac{2n(n+1)}{(2n+1)2}.d=\frac{n(n+1)d}{(2n+1)}$
$ \sqrt{Variance}=\frac{1}{(2n+1)}\displaystyle\sum_{k=1}^n[(a+rd)-(a+nd)]^2$
$ =\frac{2n^2}{(2n+1)}[n^2+(n-1)^2+...+2^2+1^2]$
$ =\frac{2d^2}{(2n+1)}\frac{n(n+1)(2n+1)}{6}$
$ =\frac{n(n+1)d^2}{3}$
$ variance=d.\sqrt{(\frac{n(n+1)}{3})}$

[hide="#1"]Differentiating implicitly we have
$ \frac{dy}{dx}\equal{}\minus{}\frac{16x}{25y}\equal{}\minus{}\frac{4}{5\sqrt{3}}$
$ \Leftrightarrow \ell_T : 4x\plus{}5\sqrt{3}y\equal{}40$[/hide]
[hide="#2"]$ \sin{2x}\plus{}\cos{x}\equal{}0$
$ \displaystyle \sin{2x}\equal{}\sin{\left(\minus{}\frac{\pi}{2}\minus{}x\right)}$
$ x\equal{}\minus{}\frac{\pi}{6}\pm 2\pi n$

$ 2\sin{x}\cos{x}\equal{}\minus{}\cos{x}$
$ \sin{x}\equal{}\minus{}\frac{1}{2}$
$ x\equal{}\frac{7\pi}{6}\pm2\pi n$
Not 100% on those, but oh well.
[/hide]

#3
we know that..
$ (1 \plus{} x)^n \equal{} \sum_{i \equal{} 0}^n C_i^n 1^i x^{n \minus{} i}$

thus we can easily put $ x \equal{} 1$ and $ x \equal{} \minus{} 1$ to get the "part" $ x^{n \minus{} i}$ always $ 1$
$ (1 \plus{} 1)^n \equal{} C_0^n \plus{} C_1^n \plus{} C_2^n \plus{} C_3^n \plus{} C_4^n.....$
$ (1 \minus{} 1)^n \equal{} C_0^n \minus{} C_1^n \plus{} C_2^n \minus{} C_3^n \plus{} C_4^n.....$

add the 2 parts.. we have
$ 2^n \equal{} 2(C_0^n \plus{} C_2^n \plus{} C_4^n....)$

that's why $ C_0^n \plus{} C_2^n \plus{} C_4^n.... \equal{} 2^{n \minus{} 1}$

[hide="5"]
Take the derivative of both sides:
$ (1\plus{}y')e^{(x\plus{}y)} \equal{} y\plus{}xy'$, where the chain rule was used on the left side and the product rule was used on the right side.
We know $ e^{x\plus{}y}\equal{}xy$, so:
$ (1\plus{}y')xy \equal{} y\plus{}xy'$
$ xy\plus{}xyy' \equal{} y\plus{}xy'$
$ xyy'\minus{}xy' \equal{} y\minus{}xy$
$ x(y\minus{}1)y' \equal{} y(1\minus{}x)$
$ y' \equal{} \frac{y(1\minus{}x)}{x(y\minus{}1)}$[/hide]

[hide="12"]
Let $ u\equal{}4x^3 \implies \frac{du}{dx} \equal{} 12x^2 \implies du \equal{} 12x^2 dx$

Then, $ \frac{1}{4} \int \frac{12x^2 dx}{\sqrt{9 \minus{} 16 x^6}} \equal{} \frac{1}{4} \int \frac{du}{\sqrt{9 \minus{} u^2}}$, which you may recognize as $ \frac{1}{4} \sin^{\minus{}1} \frac{4u}{3} \plus{}C \equal{} \frac{1}{4} \sin ^{\minus{}1} \frac{16 x^3}{3} \plus{}C$[/hide]

[hide="C"]They do not tell you that $ y\equal{}0$ and $ x\equal{}0$. They tell you that $ y(0)\equal{}0$, where $ y$ is treated as a function of $ x$.

The correct approach is to factor the equation as $ \frac{dy}{dx} \equal{} (x\plus{}1)(y\plus{}1)$, and then use separation of variables to get $ \frac{1}{y\plus{}1} dy \equal{} (x\plus{}1)dx$. Integrate to get $ \ln |y\plus{}1| \equal{} \frac{x^2}{2} \plus{} x \plus{} C$, and then exponentiate to get $ |y\plus{}1| \equal{} e^{\frac{x^2}{2} \plus{} x \plus{} C} \equal{} e^c e^{\frac{x^2}{2} \plus{} x}$.

$ e^C$ is positive, but we get the $ \pm$ from the absolute value, so the general solution is $ y \equal{} K e^{\frac{x^2}{2} \plus{} x}\minus{}1$
for some $ K$.

Now, plug in $ (x,y) \equal{} (0,0)$, which is just one point on the function, to get $ 0\equal{}K\minus{}1 \implies K\equal{}1$, so $ y \equal{} e^{\frac{x^2}{2} \plus{} x}\minus{}1$[/hide]

[hide="D"]If you squareroot, you must assume positive and negative values, and then find the right one.

$ x^2\minus{}y^2 \equal{} 13$ and $ xy \equal{} \minus{} 10 \sqrt{3} \implies x^2y^2 \equal{} 300$

Let $ (u,v) \equal{} (x^2,y^2)$. Then, $ u\minus{}v \equal{} 13$ and $ uv \equal{} 300$, thus $ u(u\minus{}13) \equal{} 300 \implies u \equal{} \minus{}12 \; \; or \; \; u \equal{} 25$

Clearly, $ u^2 > 0$ so $ u \equal{} 25 \implies x \equal{} \pm 5$

Substituting, $ u\minus{}v \equal{} 13 \implies v \equal{} 12 \implies y \equal{} \pm 2 \sqrt{3}$

Finally, you have that the simplified answer must be $ \pm 5 \pm 2 \sqrt{3} i$.

Note that if the signs are the same, then you get $ 13 \plus{} 20 \sqrt{3} i$ on the inside of the squareroot, so they must be different. And graphing in the complex plane makes it clear that we have $ 5 \minus{} 2 \sqrt{3} i$ [/hide]

[hide="a"]$ x^2 \plus{} m^2 x^2 \minus{} 2ax \equal{} 0$ by direct substitution of $ y\equal{}mx$.

Now, we may either have $ x\equal{}0$, in which case $ y\equal{}0$ as well, giving $ (0,0)$

Otherwise, we may divide by $ x$ to get $ x\plus{}m^2 x \minus{} 2a \equal{} 0 \implies (m^2\plus{}1) x \equal{} 2a \implies x \equal{} \frac{2a}{m^2\plus{}1}$

Since $ y\equal{}mx$, we have the associated $ y$ value of $ \frac{2ma}{m^2\plus{}1}$, thus $ \left( \frac{2a}{m^2\plus{}1}, \; \frac{2ma}{m^2\plus{}1} \right)$ is the other intersection point.
[/hide]

Q9

[hide]\[ \displaystyle\sum_{k\equal{}1}^n \frac{1}{k(k\plus{}2)}\equal{}\displaystyle\sum_{k\equal{}1}^n \frac{1}{2k}\minus{}\displaystyle\sum_{k\equal{}1}^n \frac{1}{2(k\plus{}2)}\] Start writing out terms and it telescopes.[/hide]

Your Q10 doesn't seem to be complete.

for part b it's just the half-angle identity $ \sin^2(\frac{x}2)\equal{}\frac{1\minus{}\cos(x)}2.$

[hide="Problem 13"]
$ 2cos^2 \theta \plus{}3 sin \theta \equal{} 0 \Leftrightarrow 2(1\minus{}sin^2 \theta)\plus{}3sin \theta\equal{}0 \Leftrightarrow 2sin^2 \theta\minus{}3sin \theta\minus{}2\equal{}0$
$ \star \ \ sin\theta \equal{}2$, impossible
$ \star \ \ sin \theta\equal{}\minus{}0.5 \Rightarrow \theta\equal{}\frac{\minus{}\pi}{6}\plus{}k\cdot 2\pi$ or $ \theta\equal{}\frac{7\pi}{6}\plus{}k\cdot 2\pi$ where $ k \in Z$
[/hide]

[hide="Problem 16"]
Set $ x \equal{} \pi \minus{} t \Rightarrow dx \equal{} \minus{} dt$
$ I \equal{} \int_{0}^{\pi}{\frac {x sin x}{1 \plus{} cos^2 x}dx} \equal{} \int_{ \minus{} \pi}^{0}{\frac {(\pi \minus{} t) sin t}{1 \plus{} cos^2 t}( \minus{} dt)} \equal{} \int_{0}^{\pi}{\frac {(\pi \minus{} t) sin t}{1 \plus{} cos^2 t}dt}$
$ \Rightarrow I \equal{} \frac {\pi}{2}\int_{0}^{\pi}{\frac {sin t}{1 \plus{} cos^2 t}dt} \equal{} \frac {\pi}{2}\int_{0}^{\pi}{\frac {d_{cos t}}{1 \plus{} cos^2 t}} \equal{} \frac { \minus{} \pi}{2}.\left[arctan\ \cos t\right]_{0}^{\pi} \equal{} \left(\frac {\pi}{2}\right)^2$
[/hide]

Ty for your help!

Now what about the other questions?
[b]✓-ed[/b]\hidden ones means I have understood it...


Prove that: f(n)= n2 + 3n is O(n^2)

This what I have:
n^2 + 3n <= n^2 + 3n^2 <= 4n^2 <= n^2

This really belongs in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=296]Calculus Computations and Tutorials[/url] section of the forum, you'll have more luck there.

For $ n \geq 3$, we have $ n^2 \leq f(n) \leq n^2\plus{}3n^2\equal{}4n^2 \implies 1 \leq \frac{f(n)}{n^2} \leq 4$

So, as $ n$ gets arbitrarily large, $ \frac{f(n)}{n^2}$ is bounded.