I believe this is a problem from rudin, but I am having a hard time with it.

if $X$ is complete and $M$ ios a closed subspace of $X$, then $X/M$ is complete.

uhm...
i think i've got it: take a cauchy sequence $(x_i) \subset X' = X/M$.
now we've got two cases:
(a) there exist $\varepsilon > 0, N\in\mathbb{N}$ such that $d'(x_i, M') > \varepsilon \; \forall i > N$.
(b) for all $\varepsilon > 0$ there exist $N$ such that $d'(x_i, M') < \varepsilon$,
where $d'$ is the distance on $X'$, and $M' = [m]$, where $m\in M$.

now, case (a): this is trivial, since that means that definitely $x_i = [y_i] = {y_i}$, where $y_i \in X$.
the distance $(X' - {M'},d')$ is isometric to $(X-M,d)$. so $(y_i)$ is still a cauchy sequence, and converges to $y$, and $x_i \rightarrow [y]$.

case (b): we have that there exist a subsequence $(x_{n(k)}) \subset (x_k)$ converging to $M'$. but sequence is a cauchy sequence, so we have that every subsequence converge to the same limit, if one converges, and that's done.

A "different" way to tell the things:
Take a Cauchy sequence in $X$ \ $M$.It converges in $X$ to a limit $a$.If $a\in M$ then an entire neighborehood of $a$ should belong($M$ is closed) to $M$,which includes the chosen sequence(after a certain term).This is false...which means $a\in X$\ $M$.So $X$ \ $M$ is complete.


When four standard six-sided dice are rolled, what is the most likely sum of the four numbers?

The average roll will be $ \frac{1\plus{}2\plus{}3\plus{}4\plus{}5\plus{}6}{6}\equal{}3.5$, so the average sum of four rolls will be $ 3.5 \times 4\equal{}\boxed{14}$.

Although that is correct you can easily see that when rolling two die together, as in a board game, the easiest sum to produce is 7 due to the fact that:

$  6+1=7, 3+4=7. and 5+2=7  $

So with four die the easiest sum to produce is $  7(2)= \box{14}  $


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I can't be the only one following it closely!

Second matchday today. Yesterday went as expected. Today I think:
Austria - Croatia: 2
Germany - Poland: 1

Discuss!

Btw: I've created a Mathlinks league at http://en.fantasy.euro2008.uefa.com
Join it, the league code is 12699-47365. Let's find out which mathlinker is the best.  :lol:

[quote="sabyasachi"]It seems each team in this competition is bent upon fighting till the end. I predict, today it's going to be Austria 1 Croatia 2.Gemany will have the Poles close at heels and a scoreline of 1-0 can be expected. But football can be quite unpredictible at times because a few moments of excellance can change the game. Good matches can be expected today.[/quote]Usually Austria are pretty bad while Croatia have a very strong defence, but with the crowd in their back, Austria might be able to give them a match. Although, I doubt it.

If the second match goes as expected, Germany will crush Poland. I would say 3 - 0.  Ballack is probably hungry after the CL loss. Lehmann -> Lahm -> Ballack -> Klose, that would be scary.  :ninja: Btw, Isn't Klose born in Poland?

Edit: According to Wikipedia he is.

Darn. Germany beat Poland 2-0. The ironic thing is that Poles scored both of Germany's goals. :)

Today:
Romania - France: 2
[b]Netherlands - Italy[/b]: x

I see that Spain is leading the poll. Spain is the team with most star players and they have a good chance of winning, but do they ever succeed when it comes to the big tournaments? Maybe they'll break the trend....

stuck: And the other Pole stood for both assists.  :)

hey dude, it's the SECOND poll about Euro 2008 in the Other Topics forum!!

Romania - France was such a boring game. I was expecting Romania to play really well - they're vastly underrated over here, but this was just dull. I didn't even watch the second half, and I'm basically glued in front of the TV watching the games and all the discussion surrounding them this month. I suppose the tactic of slowing down the game in order to be able to compete is also how Sweden plays, but I've got patriotic stake in those.

I'm interested in what other countries had to say about the offside-goal in the Holland - Italy game. Here they consulted a bunch of swedish referees, and they said technically the goal counted; it was not offside. A player is apparently not allowed to exit the field, even if they're hurt - they should tell the referee they're getting off. It's a pretty ridiculous ruling since Panucci was obviously not purposely leaving the field.

I'm guessing Italy is pretty pissed off with the referee. In the Italian Serie A they give penalties like ten times more often than in the UEFA rules that were applied here. Myself I'm pretty tired of the fact that finding penalties has become such a huge part of the game.

Italian Referee Association's chairman Gussoni gave an objective opinion:

[quote]The president of the Italian referee association (AIA), Cesare Gussoni, believes that Ruud Van Nistelrooy’s opener was legitimate despite a massive claim for offside.

Replay’s can clearly show the player in an offside position but Italy defender Christian Panucci was deemed by the linesman to have played Van Nistelrooy onside even though the stopper was off the pitch and down on the touchline with what seemed like an injury.

Nevertheless, the goal stood and Gussoni believes it was legitimate despite the whole of Italy going berzerk when it was allowed to stand.

“The linesman applied UEFA’s ruling 100 per cent,” said Gussoni.

“He evaluated Panucci’s position as the player who kept Van Nistelrooy onside.

“I thought it was offside but the replay showed Panucci off the pitch.”

“The sense of the rule is to avoid players leaving the pitch on purpose in an attempt to gain an advantage from playing the other player offside,” said Gussoni.[/quote]

His colleague from Austria (their national chairman) Gerhard Kapl gave the same reply, quoting article 11.4.1 of the refereeing code.

P. S. UEFA also backed the decision: [url]http://www.euro2008.uefa.com/news/kind=1/newsid=711341.html#uefa+supports+dutch+goal+decision[/url]

Even if it was an offside, the Netherladns really played better.
This Euro 2008 is not interesting at all. There is not a team that is 100 winner, but also, a there will not be a surprise like Greece on last Euro.
I think that you can not guess a winner, yet.

Well, the fact that there's not a sole title favourite makes the competition even more interesting. Spaniards and the Dutch have shown they can go to the end, being efficient, and playing attractive football (although the beauty of the play isn't proportional to the success... just remember Greece in 2004, or Glasgow Rangers in this year's UEFA cup). So did Portugal. But we still cannot say that one of the sleeping giants of the first round (Italy, France, even Croatia) won't be the European champions.

(Ah, the first round is not over yet, I'm still waiting to see Ibra and co. tonight against Cup holders...)

P.S. My favourites are the same as in WC 2008...

[b]ALLEZ LES BLEUS! [/b]   :winner_first:

[quote="Kalle"]I don't know if you guys get this information, but right now there is chaos in Sweden. Ibrahimovic was exchanged due to his knee injury, Niclas Alexandersson was also switched out due to cramp, Henrik Larsson is tired and wants out (he hasn't played any non-swedish league game in a couple of years), and Wilhelmsson just had a serious injury and the dirty greek refuse to get the ball out of play ....

The whole team is so brittle. Like half of the players on the field are playing with injuries, and they even asked a retired player to play.

Man, that greek with the pony tail is playing so dirty. He should've had at least two yellow cards by now. Also easily missed because they didn't show any replays, but there was a pretty clear penalty in the goal area for sweden earlier. Wilhelmson was tackled when the ball was relatively far away from him, but stupid as he is he didn't fall but instead continued to stand up.

edit: The same thing just happened to greece now in the 86th minute. A greek player could easily have let himself be tackled without touching the ball in the goal area, but he stood up as well.[/quote]
Greece, the only team being able to make a euro match truly boring. At least Sweden won.  :)

[quote="Pink Floyd"]I think that you can not guess a winner, yet.[/quote]
My guess is that in the final we'll have Germany vs. Spain. A match that Spain will win

I have to say that Greeks played very hard on the Swedes tonight, some of the tackles were quite dangerous... Nevertheless, the match was good, Swedes have shown their good form, but how long will it last, considering the fitness and doctor's report after tonight...

Ibrahimović finally scored in the national team...

Austria - Poland was a weird game. The Poland goal was an extremely obvious offside. I can't believe the Euro 2008 article does not mention this.

Star is born: Arshavin. Simply brilliant.

Firm favourites in these three quarterfinals so far go home... Seems that there's some kind of "first place in the group" curse this year on EURO.

wait, Russia isn't at the poll... can someone edit it?

I agree that Russia should be in the poll... After yesterday's game I hope they win Euro 2008. This was really unexpected, I mean Holland had won every game before with scores that made the team very confient...
I think that this is the main problem for teams that reach this level without any difficulties. See Croatia or Portugal and there are many other examples of teams that in the past years were - apparently unfair - eliminated and didn't make the semi-finals.

Russia has young and fast players and all of them have a high level technique...
The team has in my opinion the best coach...
And don;t forget the bonus will have the players .(600.000 euros olny for Guus Hiddink :!: )
For these reasons i think Russia will win the cup

Germany vs Spain in the final. My vote in the poll (which had a time limit set to the day before the quarter-finals) went to Spain and I stick with it.  :lol:

Germany did not impress on Turkey while Spain played a fantastic match against Russia. But then again, one of the tactics of Spain is to make the opponent tired by holding the ball in the team, not taking too many chances. It now just happens to be that Germany is very organized and not so offensive, so I'm not sure how well that tactic will work. But Spain should (on paper) be too fast for Germany. They'll never know what (who) hit them until they see the name "Fabregas".  :wink:

Dude, I stayed up till 2 here watching that game, I was so happy! GO SPAIN

It's too bad they don't have Telemundo in India, soccer is so much more fun on the Spanish channel =(

And I'm going to be on the f[b][color=red]X[/color][/b]ing plane during the finals; lame

[quote="Anaxerzia"]I think I was the only one to vote for Germany. Well see how this plays out xD[/quote]
that was me, and it was before any of the matches were played :) so even if they are only 2nd it's not a bad call by me I think. I am still hoping for them.. I agree they haven't played too well, especially in the semi, almost a disgrace that they won with such a performance. But they have some great players and I still like them :D

My vote was for Germany too, but I didn't make it here in time to actually vote in the poll!  Too busy marching around the Ku'damm with an enormous German flag, yelling with the other revelers. :P   I wasn't a football fan before I came to Berlin, but the excitement was infectious....you should have seen the party after we won against Turkey.  Too bad we lost, but it did make it a lot easier for me to get home afterwards!

OMG Spain won that's so sick!  :D  :D  :D

I wanted them to win so badly.

My friend tivoed (yes, that's a verb) the match, so I got to watch it too =)

Spain deserved it, in the end. After all, they haven't won much since like, what the 60s?

Agreed.
I think most people think that Turkey stole a victory from the Czech Republic but actually Turkey  finished up with more possession, more shots and the best runners and passers.

Russia also had a great Euro Cup.
I think it would be very interesting to see Turkey and Russia face off for a third place title.

Nice clips...
http://www.youtube.com/watch?v=a_vkUiFjtlw
http://www.youtube.com/watch?v=kiaEw9bXXts


Source: Bulgarian MO 1980
6. Prove that if the squares of edge lengths of a rectangular parallelepiped are intgers and its main diagonal is $\sqrt{73}$, then its volume does not exceed $120$.

[hide="solution"]
I thought this was easy pbm....(maybe im wrong....)
Let $x$, $y$, and $z$ be the sides of the box.
Then $\sqrt{x^2+y^2+z^2}=\sqrt{73}$.
or $x^2+y^2+z^2=73$.
Now apply AM-GM inequal on the above equation.
Basically, I got $\frac{73\sqrt{219}}{9}\geq xyz$.
And i estimated the RHS and got around 121...
Is this right?? :?

[/hide]

Your proof isn't complete.  You haven't used the condition that $x, y, z$ are integers to show that the maximum is [b]exactly[/b] $120$, which the LHS isn't.  (It's slightly larger.)

there are not that many positive integer solutions to

$x^2+y^2+z^2=73$

since $1\le x,y,z\le 8$

wlog $x\ge y\ge z$, then

$x^2\ge\frac{73}{3}$

so try $x=5,6,7,8$ and solve


[quote="sheepwarrior"]what defines the line between liking and loving? the difference between true love and just a crush? post ur thots...[/quote]A few days ago I read this quote, and I liked it a lot, even when probably some people will disagree with me! (or with the phrase!)

[quote="Antonio Gala"]Love is friendship with some erotic moments.[/quote]

Love. Love... Love is so different from a crush. It comes on you in a different way. A crush comes on quickly, and heavily, and with a passion that seems that it cannot be broken- it is almost an obsession that you cannot control. Love? Love is slower, but warmer. And, and David said, at the beginning and the end- past all things- the person you love is your friend.
Love. Love changes things. True, you may not always get along with the person you love, that never happens, but with someone you love... you WANT to fix things. For example, I have a reputation for being hard-headed and not conceding, even when I know I'm wrong. With the person I love, I'm not ashamed to admit my wrongs, because I don't WANT things to come between us.
And, as David said, there's also the physical attraction. You can appreciate the bond of friendship with someone from now until doomsday, but unless you think they are beautiful, unless you can look at them and smile and smile and smile, [and unless all that pheremone stuff...]... it's hard, if not impossible, to love someone you are not physically, as well as intellectually and emotionally, attracted to.

I don't know what others think but I think liking and love are both closely related.

Like you might have this good looking friend you've known for years. You liker her/him as a friend right? But one day you saw that friend again and suddenly, you feel something is different. You feel as even you knew him/her for long time, it seems like you want to know about other side of him/her, not part where you're friend.

So, I guess you can like someone then maybe at some point (not sure how it comes), it turns into love. I'm not saying all of them do. At least some do.

For crush and love, they are absolutely different. For crush, it's more like instant feeling of human being. Like you might have crush for the "Miss Popular" but that doesn't mean you would risk your life or sacrifice something important for you to her. But for love, you do. It changes you. I don't know how it happens or why it happens. If you're in love, you just begin to see the things differently. Maybe the stuff on the love poem where "O' all the worlds turn into Heaven when I met you." Well, maybe not that much :D but it does make things optimistic. And pessimistic when you break up.

For David's comment about love being a friend first, I highly agree.  I don't think anyone (unless you're thinking of dating someone at the beginning) would come to a girl because he just wants to date. Like it may happen but I think most cases are where you know he/she as a friend and then you know, get to know her/him better.

[quote="probability1.01"]I wouldn't know anything, but historically, our concept of "love" didn't always exist.  So in some ways we made it up, although some of it I believe must be natural.[/quote]Well, it depends highly in your definition of love. The phisiological reaction to the phenylethylamine in our bodies is not that different (probably not different at all) to the one our ancestors had.

Nevertheless, the romantic love such as we conceive it nowdays is for sure not the same that people have had along the centuries. First than all, because it is not a natural phenomenon, it is a social phenomenon, and different societies have different schemes of values for it, and different visions for it.

As far as I remember, I read somewhere that the ancient egiptians and the ancient romans had a very different vision of it. Any way, I am too lazy right now to look for that on the web!  ;)

[quote="djimenez"][quote="mcalderbank"]This may sound bollucks because I am not in love, but here goes:

Two people are in love if they can just be together by themselves and just appreciate each other without having to say a word, or feel obligated to say anything. This means that they are most comfortable with each other, and have no need for any extras.

 :|[/quote]Michael,

Theoretically, what you say sounds very romantic and platonic, but unrealistic, at least in the general setting, and during adulthood. At least I am assuming that the post goes over the [i]romantic[/i] or [i]erotic[/i] love. What you describe is, in my opinion, called a deep, good and strong friendship.

Love use to have a very vage definition, as many of the human emotions and feelings, mainly because they use to be a little beyond the words. Who says that you cannot feel love for your closest friends, it is just not a romantic love. The truth is that my wife is, before and after every other thing, my best friend. That is complemented with the fact that is a person that I know in the deepest ways and a woman for whom I feel a strong attraction, and yes, that carries with it a lot of [i]extras[/i], as you call them, but mainly, she is my friend! (I am just giving a particular example, that in some sence justify the quote I posted)

There is something that I suppose you implicitely tell and I agree, love and phisical attraction (to keep the vocabulary rated G) are two completely different things, even though they are good complements, but that does not mean that you cannot have one and not the other!

Anyway, just a few comments![/quote]

This is way too complicated for me, I will just stick to math.  :P

[quote="mcalderbank"]This is way too complicated for me, I will just stick to math.  :P[/quote] :what?: not only bread needs the man to live...

But why is that so complicated?

[sarcasm]
My grandmother is the hottest person alive. I LOVE HER WITH A PASSION!
[/sarcasm]

It's a different kind of love, a parent-child bond, silly :P .

I suggest reading Four Loves by C.S. Lewis.

That's quite book of philosophy on love. I enjoyed it even I didn't quite get much.

On chapter about love between man and woman, it does contain the things that David said.

[quote="stupidkid"][quote="djimenez"]A few days ago I read this quote, and I liked it a lot, even when probably some people will disagree with me! (or with the phrase!)

[quote="Antonio Gala"]Love is friendship with some erotic moments.[/quote][/quote]
But what about loving your parents? Or do you not call that love?[/quote]

That was my whole point in the second post. Is that other kind of love? what is the difference between love and friendship? if you love your father or your mather, aren't they your friends too?

I think the main thing is that you can love someone and not be attracted to them.

For me, love is more of a deep care and concern for a person and their general welfare. Passion is more what lovers can feel for one another.

I always stood by, that if I love someone, that means I'm willing to put my life on the line for that person without regard.

Infatuation is what you feel when you are extremely attracted to someone. For example, I am infactuated by Aishwarya Rai's good looks. She's hot!

[quote="ankur87"]For me, love is more of a deep care and concern for a person and their general welfare. Passion is more what lovers can feel for one another.[/quote]This is a very interesting affirmation, because by definition, the lover is that who loves, not necessarly that how feels passion! (would it be Passioner or passionate instead :P ). If we go for what the word really means and not for the sence it is sometimes used, then Love is what lovers can feel for one another!

Theoretically, what you say sounds very romantic and platonic, but unrealistic, at least in the general setting, and during adulthood. At least I am assuming that the post goes over the romantic or erotic love. What you describe is, in my opinion, called a deep, good and strong friendship.

There is something that I suppose you implicitely tell and I agree, love and phisical attraction (to keep the vocabulary rated G) are two completely different things, even though they are good complements, but that does not mean that you cannot have one and not the other!

[quote="4everwise"]Theoretically, what you say sounds very romantic and platonic, but unrealistic, at least in the general setting, and during adulthood. At least I am assuming that the post goes over the romantic or erotic love. What you describe is, in my opinion, called a deep, good and strong friendship.[/quote]No, I am not talking about an opinion, I am talking about the correct etymological meaning of the word lover. If you look in the dictionary you'll find (approximately)

[b]Love[/b]: ([i]Noun[/i]) A deep, tender, ineffable feeling of affection and solicitude toward a person, such as that arising from kinship, recognition of attractive qualities, or a sense of underlying oneness.  ([i]verb[/i])To have a deep, tender, ineffable feeling of affection and solicitude toward (a person)

Now, the suffix -er or -r at the end of a verb transform the word in the noun of who either receive or perform the action, therefore we would say that

[b]Lover[/b]: A person who loves or is loved.

Now, of course, as I said, that is the correct etymological meaning of the word, but english is a live language, therefore it is modified by its users each and every day, therefore I understand that most of the people understand,

[b]Lover[/b]:  A sexual partner, a significant other to whom you are not related by marriage.

Now, my question for you is, what is the difference between your [i]deep, good and strong friendship[/i] and the [i][u]non-romantic[/u] love[/i]?

Love is just a word until you find someone to give the definition to  :D .

I think a few studies have argued that evolutionarily,  love and lust have different roots.  Lust, which has always been there, makes us want to have sex, thereby ensuring the survival of the species.  Love, on the other hand, came afterwards and makes us want to take care of those we love.  Thus, families and eventually communities can be built.  Often, the two go together, which makes it so difficult to separate the two.  These days, I can't really imagine anyone loving someone else and not being physically attracted to him/her.  Indeed, a cynic may argue that love doesn't exist; it is just a more benign form of lust.  This is a crude picture.  Someone who knows more about studies like these can post more.

Polymorphic,

That depends on what you define by love, because the regular definition of love, or at least romantic love, is something that hasn't been present on all the eras or societies, and I am not talking just about prehistorical societies, but also about important societies of ancient times (Egyptians, Greeks, Romans to name a few) at least not as a social value.

Nevertheless, it is right that sex drive and love (or infatuation) are evolutionary very different. One comes from testosteron and the other from fenyletylamin, or at least that is what I remember.

You have to take into consideration that many people (including antropologist) consider that love, or the idea of romantic love ( :love: ) as we considered nowdays does exists, but it is an invention of mankind (even some argue that it is an invention of the [i]womankind[/i]) and the arguments are not that crazy as they may look. The main facts to take into consideration is that there is no primate is naturally monogamous, and actually monogamy, if on one hand it solves some social problems, it (partially) attempts against one of the most basic principles of evolution: Genetic variance. You see, if each female has children from several males and viseversa, there is a greater genetic mixure. That happens most of the species.

Now, what happened on the begining, I pressume. Well, females are pregnate for nine months, and it is not healthy for a female to deliver and inmediately (within a few months of the delivery) to get pregnate again, that means that there should be at least 18 months between one kid and the other (and for mothern standars, that is too few), while a male can have as much decendance as females he can find. That, plus the fact that on prehistorical times females took care of the decendence while the male went to hunt or collect food made the males more prone to die, therefore there would be more females per male, what probably made quite natural the polygamy where males could have several sexual partners, but not females (presumly, a male wouldn't like to share freely his hunt with others' children). Now, eventually females realized that if the males have just one sexual partners, their children would have a better chance to survive, because then the male does not have to share his resources with his children with other females.

The last parragraph is just my personal theory!

Bumping this because it's a great topic, and many posts here, I will edit this comment and insert my opinion here real soon.


topology
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I'm sorry if this is a stupid question (I just started teaching myself topology) but is there a method for proving the mean value theorem using topology?

The proof of the extreme value theorem, which is needed to prove Rolle's theorem (and consequently MVT), can be done topologically; it is the combination of the facts that a continuous image of a compact set is compact and that a compact set in $ \mathbb{R}$ is closed and bounded.

Thanks for the reply. So there is no drect topological proof then. How about a topologoical interpretation?

What do you mean by a "direct topological proof"?  Both of the statements Nukular cited are topological facts - one about general topology and one about the topology of $ \mathbb{R}$.

I should probably say is there a more general version of MVT.

That depends on what you mean.  Derivatives aren't a topological notion, so you're not going to get any meaningful generalization to all topological spaces.  The MVT, interpreted naively, doesn't even hold in higher-dimensional Euclidean spaces.

The EVT, on the other hand, has a clear generalization, which is precisely the first fact Nukular stated.

Thanks for the clarification. Thats what I was wondering.

Let me try to interpret MVT so that it may hold in higher dimensions:

Theorem*: [i]Let $ f\colon [a,b]\to\mathbb R^n$ (where $ n\ge 2$) be a continuous function such that $ f'(t)$ exists (as a vector in $ \mathbb R^n$) for all $ t\in (a,b)$. Then there exist points $ t_1,\dots,t_{n \minus{} 1}$ in $ (a,b)$ such that $ f(b) \minus{} f(a)$ belongs to the linear span of $ \{f'(t_j)\}$.
[/i]

 The special case $ n \equal{} 2$ of the above can be restated as follows, writing $ f$ in components $ f \equal{} (f_1,f_2)$: there is $ t\in (a,b)$ such that $ f_1'(t) (f_2(b) \minus{} f_2(a)) \equal{} f_2'(t) (f_1(b) \minus{} f_1(a))$. This is a generalized version of MVT, which in turn reduces to classical MVT be setting $ f_1(t) \equal{} t$.

The asterisk * is because I haven't actually proved this. Proofs (of counterexamples) are welcome.  :)


i was wondering if someone could post a solution using algebraic number theory to the following problem:

K > L > M > N are positive integers such that KM + LN = (K + L - M + N)(-K + L + M + N). Prove that KL + MN is composite.

(this is problem three of day two from the 2001 IMO)

thanks

This isn't exactly algebraic number theory, but...

Hint: [hide] Try a geometric approach with quadrilaterals [/hide]

Ah. I remember this problem. There was 2 minutes left of the exam, and I had nothing to do, so I decided to write down all the pairs of things that I could find that were coprime. Or not coprime. I forget. It got me a point :D


Inspired by http://www.mathlinks.ro/viewtopic.php?t=221169.

The terms in the sequence for $ n\equal{}1,2,3,...$ are
$ \frac{2}{1},\frac{8}{3},\frac{26}{7},\frac{80}{15},...$
Taking the differences of the top over the bottom we get
$ \frac{6}{2},\frac{18}{4},\frac{54}{8}$
And taking the differences again
$ \frac{12}{2},\frac{36}{4}$.

Now the two rationals in the 3rd sequence are all natural numbers. Is it true the values (this third sequence) will always be a natural number?

The very next term in the third sequence is $ \frac{108}{8}$.

Hahaha, I guess I should have checked just one more case :P

How about creating a 4th sequence? It would have $ \frac{24}{2}, \frac{72}{4}$

If this doesn't yield a natural number, would the 3rd term always be a non-natural value?

An explicit formula for the $ k^{th}$ sequence is possible; find it.


quadratics
algebra
number theory open
number theory

if a,b,c bea possitive integer and abc+bc+c=a^3 find a,b,c[/code][/quote]

Here are some solutions: $a=n, b=n-1, c=n$ for any integer $n\geq 2.$

From $abc+bc+c=a^{3}$, we have, $(a+1)(b+1)c=a(a^{2}+c)$ ...(1)
Since $(a+1,a)=1$, therefore, $a+1|a^{2}+c$. That is, $a+1|c-a$
It follows that $c=a$ or $c=k(a+1)+a$ where $k$ is a positive integer (if $c<a$ then $0<|c-a|<|a+1|$)
Case(i) $c=a$, substitute this in (1) and solve for $b$ yields the solution $(a,b,c)=(n+1,n,n+1)$ where $n$ is in $\mathbb{N}$
Case(ii) $c=k(a+1)+a$, substitute this in (1) and simplify the eqaution, we have
$(b+1)(k(a+1)+a)=a(a+k)$
$a^{2}-(1+b+bk)a-(k+bk)=0$
Let $a_{1}$ denotes the other solution of the above quadratic equation $x^{2}-(1+b+bk)x-(k+bk)=0$
We have $a+a_{1}=1+b+bk$  ---(2) and
$aa_{1}=-(k+bk)$ ---(3)
From (2),(3) and the fact that $a,b,k$ are positive integers, it follows that $a_{1}$ is a negative integer
Next, if $a_{1}\neq-1$, then, $|1+a_{1}|\geq 1$, and we have the following contradiction
$2+b+bk=|2+b+bk|\leq |1+a|\leq |(1+a)(1+a_{1})|=|1+(1+b+bk)-(k+kb)|=|2+b-k|<2+b+bk$
Thus, $a_{1}=-1$, substitute this in (2),(3) yields $k=b+2, a=(b+2)(b+1)$
Also, $c=k(a+1)+a=(b+2)((b+2)(b+1)+1))+(b+2)(b+1)=(b+2)^{3}$
That is, $(a,b,c)=((n+2)(n+1),n,(n+2)^{3})$ where $n$ is in $\mathbb{N}$


[quote="FlaME_HiRo"]$Q(y)=9g-gx+18-2x/gx-9g+2x-18$

I got 1, the book got -1, I looked over and over and -1 seems wrong..

$Q(y)=m^3-mn^2/mn^2+m^2n-2m^3$
my answer
 $Q(y)=m^2+n^2/-2m+n$
books
 $Q(y)=-(m+n/2m+n)$

I dont know how they got those answers, anyone know?[/quote]
I also don't know how you got your answers.  :?  :huh:
$\frac{9g-gx+18-2x}{gx-9g+2x-18}=\frac{-(gx-9g+2x-18)}{gx-9g+2x-18}=-1$ provided that $g\neq -2$ and $x\neq 9$.

$\frac{m^3-mn^2}{mn^2+m^2n-2m^3}=\frac{m(m-n)(m+n)}{-m(m-n)(2m+n)}=-(\frac{m+n}{2m+n})$ when $m\neq 0$ and $m\neq n$.
Next time post these problems in high school forum. ;)

But I'm in college, ah doesnt matter I will..

Main reason I dont know is because this guy needs to speak proper english,  :rotfl:


two part question:

a).  Derive the operations  $+, -, \times, \div $ from $-$ and reciprocal

b).  Invent a single binary operation from which $+, -, \times, \div $ can be derived

[quote="Mikey"]I don't understand the question.[/quote]
For example:
[b]1)[/b] $a+b=a-(-b)$;
2) $2a=a+a$;
3) $a^2=1+2\cdot\frac{1}{1/(a-1)-1/(a+1)}$;
4) $a/2=\frac{1}{1/a+1/a}$;
[b]5)[/b] $a\times b=((a+b/2)^2-(a-b/2)^2)/2$.
So we are able to obtain $a+b$ and $a\times b$ using only two operations $-$ and $\frac{1}{x}$.

well, I meant this as a challenge problem, but part b is still up for grabs, any takers?

A binary operation is a function that takes two numbers as input and outputs one number.  For example, addition and subtraction are binary operations, whereas square root is not.

For this problem, you get to invent any such operation (function) you like.

I wouldn't expect Myth giving answers in the Intermediate Topics Forum...  :D  :D  :D


Find the coefficient of x^99 in the expansion of

(1+2x+3x^2+4x^3+........+100x^99)^2.

[hide="Solution"]
Interpret it as the square of the first $ 100$ terms of the power series \[ 1\plus{}2x\plus{}3x^2\plus{}\dots.\] This power series can also be represented as \[ (1\plus{}x\plus{}x^2\plus{}x^3\plus{}\dots)^2,\] so we really want to find the coefficient of $ x^{99}$ in the expansion of \[ (1\plus{}x\plus{}x^2\plus{}x^3\plus{}\dots)^4.\]

[hide="Trick"]
The coefficient of $ x^{k}$ in the expansion of \[ (1\plus{}x\plus{}x^2\plus{}x^3\plus{}\dots)^{n}\] is \[ \binom{n\plus{}k\minus{}1}{n\minus{}1}.\][/hide]

Applying the "trick" the coefficient of $ x^{99}$ is \[ \binom{4\plus{}99\minus{}1}{4\minus{}1}\equal{}\binom{102}{3}\equal{}\boxed{171700}.\][/hide]

Sorry if that was a little hard to understand.

How do u show that  that given expression can be reduced to
\[ (1\plus{}x\plus{}x^{2}\plus{}x^{3}\plus{}\dots)^{4}.\] ?

let $ y\equal{}1\plus{}2x\plus{}3x^2\plus{}...$

$ y\minus{}yx\equal{}1\plus{}x\plus{}x^2\plus{}x^3\plus{}...$

$ y\equal{}(1\plus{}x\plus{}x^2\plus{}x^3...)\frac{1}{1\minus{}x}$

note$ \frac{1}{1\minus{}x}\equal{}(1\plus{}x\plus{}x^2\plus{}x^3\plus{}...)$

therefore $ y\equal{}(1\plus{}x\plus{}x^2\plus{}x^3\plus{}...)^2$

Could anyone please explain this result intuitively?

Hello, I think you might be assuming that the given is an infinite series. The last term of the series is 100x^99.

$ y\equal{}1\plus{}2x\plus{}3x^{2}\plus{}... \plus{}100x^{99}$
$ y\minus{}yx\equal{}1\plus{}x\plus{}x^{2}\plus{}x^{3}\plus{}...\plus{}99x^{100}\plus{}100x^{100}$
$ y\equal{}(1\plus{}x\plus{}x^{2}\plus{}x^{3}...\plus{}x^{99}\minus{}100x^{100})\frac{1}{1\minus{}x}$

I got stuck here.

But certainly you can use an infinite series. The coefficient of $ x^{99}$ is the same.

I get the point. Since you wont be able to get any coefficient of x^99 in x^100.

That's correct.

Read chapter 14 in Intermediate Counting+Probability if this is still unclear.


geometry
3D geometry
sphere
parameterization

A glass sphere of volume 7litres contains air at 300K. It is connected to a pipe filled with mercury. At the start, the mercury meniscus is level with the bottom of the sphere on both arms of the pipe. The air in the sphere is now heated and the mercury level on the right arm rises by 5mm. If the cross sectional area of the pipe is 10cm^2, what is the new temperature of the air ?

Figure: http://img403.imageshack.us/img403/6254/thermqd2.jpg

In the horizontal arm of the tube. Pressure on both ends must be equal, after and before heating. So the change in Pressure inside the sphere equal to $ \rho gh$. Using this find the temperature. Volume of the sphere/gas remains a constant. Done.

Same as I thought. But why do you think volume of the sphere and the area of cross section is given?

Dunno. I find the area of the pipe a useless fact. And I don't find anything wrong with the argument. Answer does not match?

I posted the question in another forum too.. Check the link.
http://www.goiit.com/posts/list/heat-thermodynamics-thermodynamics-fluids-question-52286.htm

I will treat air as an ideal gas...
Let $ p_0$, $ V_0$, $ T_0$ be the initial parameters of the air.
After the heating is done, the mercury level on the right rises by $ x$.
Hence the pressure of the air should be:
\[ p_1 \equal{} p_0\plus{}g\rho 2x
\]
Here $ \rho$ is the mass density of mercury
The volume of the air is now:
\[ V_1 \equal{} V_0 \plus{} xA
\]
Where $ A$ is the cross sectional area of the pipe.
So using the below law:
\[ \dfrac{p_0V_0}{T_0} \equal{} \dfrac{p_1V_1}{T_1}
\]
We get:
\[ T_1 \equal{} \dfrac{(V_0 \plus{} xA)(p_0\plus{}2x\rho g)}{p_0V_0}T_0
\]


abstract algebra
search
superior algebra
superior algebra unsolved

Prove that every projective module $ P$ has a free complement.

* A complement of a projective module $ P$ is a module $ K$ such that $ P \oplus K \cong F$ for some free module $ F$.

What definition of projective are you using?

Presumably we're given the existence of a complement, and we need to show that we can take it to be free.  In fact, we can ensure $ K\cong F$.  See [url=http://en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle]here[/url].

Projectives (more or less by definition) have projective complements.  If the original projective is finitely generated, then you can choose the projective complement to be finitely generated too.  However, if you want the complement to be free, then it may not be (and often is not) the case that you can choose the free complement to be finitely generated.

However, there is a silly choice of free complement that always works.  If X=P+Q is free, then so is a direct sum of countably infinitely many copies of X.  However P+X = P+(Q+P+Q+P+...) = (P+Q)+(P+Q)+.... = X is free.

This sort of thing is a nice comparison to some of the Krull-Schmidt results out there that I think Steve looked at some time ago.  X is free (so nice), but cannot be cancelled.

Another direction this is taken is usually called "stably free"; this is where you require the free complement to be finitely generated (and so, it often does not exist).  I think McConnel-Robson has a nice chapter on this.  "Finite free resolutions" are another good search term for this sort of thing.


I recently bought a new iMac that's totally awesome in every way except one.

I just downloaded Texshop,  but i don't know how to use asymptote on it.  can anyone help me?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200956]Asymptote compilation - Mac[/url]. If that doesn't help and undefined117 doesn't respond to this thread, try sending them a PM as they seem to be the expert with this topic.

Uhh steve, when you say "they", does that mean undefined117 is more than one person?  Oh yeah, I PM'ed them and the installation seems to be working ok....for now

It means that stevem doesn't know whether undefined117 is male or female, and isn't guessing as you just did.

[quote="worthawholebean"]Really, all you should need to do is compile it. Then, you'll need to fiddle around with the commands TexShop runs. What I do is I have a script in /usr/local/texlive/2007/bin/i386-darwin/ called asypdftex that contains the following code:
[code]#!/bin/bash

pdflatex --shell-escape $1
echo "running asymptote on $1"
/sft/bin/asy "`basename -s .tex $1`"
pdflatex --shell-escape $1[/code]Then, go into TeXShop preferences->Engine. Change the Latex field under pdfTeX to "asypdftex".[/quote]How do I navigate to "/usr/local/texlive/2007/bin/i386-darwin/".  Can I create the script file in any texteditor?  What extension do I use?

Thanks!

I'm not sure about how this stuff works on Mac, but here are some (possibly wrong) answers to those questions:
1. If you give the file execute permissions (probably by typing chmod u+x asypdftex in Terminal), you should be able to put the file anywhere specified by your PATH. (To set your PATH, edit the .bash_profile file in your home directory (by typing nano .bash_profile in Terminal), and add the line [code=bash]export PATH=$PATH:<directory you put the code in>[/code] at the end.
2. Yes.
3. None. Bash scripts usually don't have extensions.
Edit: The last post on this thread was in 2009, but I'm answering this anyways because there's a link on the Wiki.


Hello,

I don't know if this is the best place to post this, but I have been reading some of the geometry problems, and I have notice that some people (specially Romanians) mention frequently Math Gazette, but when I go to Google, it seems like every university has a Math Gazette, that is, mainly an anouncement media for seminars an talks. When I go to the library it says that it is a magazine that was published from 1894 to 1970, edited by a brittish editorial.

Are you making reference to that magazine, or is that in your countries there is a magazine with that name? I ask because it seems to be an intersting magazine!

Best regards,

Yes it's the main [b]high-school[/b] publication in Romania (it's over 100 years old) and has monthly issues. However there is no english version (sometimes some articles appear in English, but very rare) and there's really nothing that important about it. Usually people select problems from it and post them on the forum :)

djimenez,

"Math Gazette" is a semi-common term in Anglophone countries.  I believe the seminar announcement mailing list at MIT/Harvard is called the Mathematical Gazette, for example, and there is a British undergraduate/recreational math magazine with the same name (still in publication).

The Romanians are referring specifically to their high-school magazine,  [i]Gazeta Matematica[/i],  which has some dozens of problems in each issue and appears every 1-3 months.   It is an analogue of Kvant, Komal, Alpha, Matematika v Shkole, etc, with a problem-solving contest for students.   It is more specifically competition-oriented than the other East European journals, with numerous contest problems (including some IMO and BMO shortlisted material) and articles where people analyze and generalize various competition problems.  So it is, perhaps, more like a high school version of Crux Mathematicorum.   On the last 1-2 pages they translate the IMO-style problems into English.

I haven't seen a copy of this journal in a long time.  Very few libraries in the USA seem to  subscribe to it, but it would be inexpensive to purchase a personal subscription.

Also, Valentin --- was there not a second series of Gazeta containing professional math/physics research articles?   Does it still exist?

[quote="fleeting_guest"]Also, Valentin --- was there not a second series of Gazeta containing professional math/physics research articles?   Does it still exist?[/quote]There is, but it's also more problem solving orientated than research (of course some good research articles appear, but not too often). Also this one has only 4 issues per year.

Another question for Valentin ---

can a subforum be added for Mathematical Journals?
For problems and other discussions pertaining to journals.

I just want to note that the German math journal Alpha is out of print.

[quote="fleeting_guest"]can a subforum be added for Mathematical Journals?
For problems and other discussions pertaining to journals.[/quote]It's done.

[quote="hbreiter"]Two journals with good problem sections that are accessible to high schoolers are Mathematics and Informatics Quarterly  and Pi Mu Epsilon Journal.
Sadly, neither is availbale online.
hbreiter[/quote]

What are those journals? I mean, the name, because I have some known people in Pi Mu Epsilon, and they could find them for me.

[quote="fanaticsm"][quote="djimenez"] What are those journals? I mean, the name, because I have some known people in Pi Mu Epsilon, and they could find them for me.[/quote]

I am not sure what you are asking because the names of those journals are Mathematics and Informatics Quarterly  and Pi Mu Epsilon Journal respectively. If you are looking for more information, [url=http://olympiads.win.tue.nl/ioi/misc/miq.html]here[/url] is some more info on M & IQ and  [url=http://www.pme-math.org/journal/overview.html]here[/url] is some more info on the Pi Mu Epsilon journal.[/quote]

Sorry, I feel so stupid. I just misread, I misunderstood. I appologize  :blush: :oops:


what is the area of a square inscribed in an ellipse $16x^{2}+9y^{2}=144$?

[hide="Useful principle"] A point on an ellipse has the property that the sum of the distances to the foci are constant.  Calculate the foci and the rest is easy. [/hide]

[hide]
Since the center of the ellipse is the origin, the center of the square is the origin, too. Consider the vertex $(x,y)$ in the first quadrant. It must have $x = y$ to be aligned with the elipse. So

$16x^{2}+9y^{2}= 16x^{2}+9x^{2}= 25x^{2}= 144 \Rightarrow x = \frac{12}{5}$.

Then the area would just be $\left(\frac{24}{5}\right)^{2}= \frac{576}{25}$.[/hide]

Denote the length $L$ and area $S$ of the required square $\Longrightarrow$ $\{\begin{array}{c}16x^{2}+9y^{2}=144\\\ x=y\end{array}$ $\Longleftrightarrow$ $\frac{L}{2}=x=y=\frac{12}{5}$ $\Longrightarrow$ $S=(\frac{24}{5})^{2}<\pi\cdot 4\cdot 3\ .$


A and B play a game on a $ nXn$ board ($ n>1$). A starts putting a checker on the lower left corner. In each turn, the corresponding player puts a checker on an empty square that is adjacent to the square where the previous checker was placed. (Two squares are adjacent if they have a common side). The player who cannot play on his/her turn loses. Find for all n which player has a winning strategy.

I will post the [url=http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=68983]jestrada[/url]'s solution.  :D
1- If $n$ is even, you can fill the board with dominoes, (without overlapping of course), and so the second player wins cos he can always play in the correspondent cell of the domino where was the last move of the first player.
2- If $n$ is odd, reading the first case you can easily get that the first player wins, (filling the board with dominoes, except the lower left corner).


Is the problem that you aren't sure how to use Newton's sums?  What part is confusing?

[quote="kcn2rivers"]Let the roots of the function $ f(x) \equal{} x^4 \minus{} 7x^3 \plus{} 17x^2 \minus{} 17x \plus{} 6$ be $ a, b, c, d$. Find the value of $ a^7 \plus{} b^7 \plus{} c^7 \plus{} d^7$.

I'd like to see a solution using Newton's sums  :)[/quote]

Newton's sums? Why?

$ f(x)\equal{}(x\minus{}1)^2(x\minus{}2)(x\minus{}3)$, so $ a^7\plus{}b^7\plus{}c^7\plus{}d^7\equal{}1^7\plus{}1^7\plus{}2^7\plus{}3^7\equal{}2317$

Well I expanded the polynomial after making up that factorization so I know what the answer should be.

I would just like to see Newton's sums because I am new to the subject and want to see how to calculate $ s_n$ when $ n>\deg(f)$. The wiki doesn't show how.

[quote="kcn2rivers"]want to see how to calculate $ s_n$ when $ n > \deg(f)$. The wiki doesn't show how.[/quote]
All that happens is that the $ ka_{n \minus{} k}$ term disappears.  There is a very good reason for this in terms of generating functions; if you're interested, you should try learning about [url=http://en.wikipedia.org/wiki/Recurrence_relation#Linear_homogeneous_recurrence_relations_with_constant_coefficients]linear recurrences[/url] in general.

If you're having trouble with degree, note that you can multiply by $x^3$ to get $x^7 - 7x^6 + 17x^5 - 17x^4 + 6x^3$ without changing the sum of the roots to the seventh power.


trigonometry
calculus
calculus computations

hello, making the ansatz $ r(t)=e^{\lambda t}$ we get $ \lambda=\pm\sqrt{b}i$ and your solution is
$ r(t)=C_1e^{\lambda_1t}+C_2e^{\lambda_2t}$
this can also be written in the form
$ r(t)=C_1\sin(\sqrt{b}t)+C_2\cos(\sqrt{b}t)$
Sonnhard.

I have found  what u have posted.

I want a relation in x and y involving r0 , v0 and b.

Debanik


function
algebra
polynomial
real analysis
real analysis unsolved

Let $ \Omega \subseteq R^{n}$ open and bounded, $ x_{0} \in R^{n}$ and $ D : C^{1}(\Omega) \rightarrow R$ be a function satisfying conditions:
1. $ D(f\plus{}g)\equal{}Df\plus{}Dg$
2. $ D(fg)\equal{}f(x_{0})Dg\plus{}g(x_{0})Df$
3. $ D(limf_{n})\equal{}limD(f_{n})$.

Prove that $ Df \equal{} \frac{\partial f}{\partial v}$ for some $ v \in R^{n}$.

It's easy to check that $ v \equal{} (D(x_1),D(x_2),\dots, D(x_n))$ gives $ Df \equal{} \frac {\partial f}{\partial v}$ for all polynomials. And polynomials are dense in $ C^1(\Omega)$.

Ok, but how do you prove that polynomials are dense in $ \Omega$ if $ \Omega$ is not compact, as in this case (Stone-Weierstrass can't be applied)?


calculus
integration
trigonometry
calculus computations

what is the definite integral of x^(3)  (x^2 + 1)^(-1/2) dx on the interval [0,1]

You don't even need integration by parts.

Why don't you set $ u^2\equal{}x^2\plus{}1$?

[quote="kunny"]$ I_2\equiv\int_0^1 \frac {x^3}{\sqrt {x^2 \plus{} 1}}\ \mathrm {dx} \equal{}$ $ \int_0^1 \left(x\sqrt {x^2 \plus{} 1} \minus{} \frac {x}{\sqrt {x^2 \plus{} 1}}\right)\ \mathrm {dx} \equal{}$

$ \left[\frac {1}{3}(x^2 \plus{} 1)^{\frac {3}{2}} \minus{} \sqrt {x^2 \plus{} 1}\right]_0^1$ $ \implies\boxed {I_2 \equal{} \frac {2 \minus{} \sqrt {2}}{3}}\ .$[/quote]
[color=darkblue][b]Kunny[/b], I like the your nice proof because [b]it is without integration by parts
and without substitutions[/b], i.e. it is [b]a integration by "binocular"[/b] ! Thank you !

Can show analogously that $ I_n\equiv\boxed {\int_0^1\frac {x^{2n \minus{} 1}}{\sqrt {x^n \plus{} 1}}\ \mathrm {dx} \equal{} \frac {2(2 \minus{} \sqrt 2)}{3n}}\ !$

Indeed, $ \frac {x^{2n \minus{} 1}}{\sqrt {x^n \plus{} 1}} \equal{}$ $ x^{n \minus{} 1}\left(\sqrt {x^n \plus{} 1} \minus{} \frac {1}{\sqrt {x^n \plus{} 1}}\right) \equal{}$

$ \frac 1n\left[\left(x^n \plus{} 1\right)^{\frac 12} \minus{} \left(x^n \plus{} 1\right)^{ \minus{} \frac 12}\right]\left(x^n \plus{} 1\right)'$ $ \implies$ $ \frac 1n\left[\frac 23\left(x^n \plus{} 1\right)^{\frac 32} \minus{} 2\left(x^n \plus{} 1\right)^{\frac 12}\right]_0^1$ $ \implies$

$ I_n \equal{} \frac 1n\left[\frac 23\left(2\sqrt 2 \minus{} 1\right) \minus{} 2\left(\sqrt 2 \minus{} 1\right)\right]$ $ \implies$  $ I_n\equiv\boxed {\int_0^1\frac {x^{2n \minus{} 1}}{\sqrt {x^n \plus{} 1}}\ \mathrm {dx} \equal{} \frac {2(2 \minus{} \sqrt 2)}{3n}}\ .$

[b]Remark.[/b] Can show analogously more generally that

$ I_{n,p}\equiv$ $ \boxed {\int_0^1\frac {x^{2n \minus{} 1}}{\sqrt [p] {x^n \plus{} 1}}\ \mathrm {dx} \equal{}\frac {p\left(2 \minus{} \sqrt [p] {2^{p\minus{}1}}\right)}{n(p\minus{}1)(2p\minus{}1)}}\ !$[/color]

Thank you, Virgil Nicula. :)
Your generalization seems to hold.

How could we integrate with interation by parts.

I imagine that the substition of $ x\equal{}\tan \theta$ will be O.K.

[quote="Virgil Nicula"][color=darkred][b]Remark.[/b] Can show analogously more generally that

$ I_{n,p}\equiv$ $ \boxed {\int_0^1\frac {x^{2n - 1}}{\sqrt [p] {x^n + 1}}\ \mathrm {dx} = \frac {p(2 - \sqrt [p] {2^{p - 1}})}{n(p - 1)(2p - 1)}}\ !$[/color][/quote]
We can use the substitution $ \{\begin{array}{c}
t = \phi (x) = x^n + 1\\\\
\phi '(x)=nx^{n-1}\end{array}$ , i.e. $ \int_0^1\frac {x^{2n - 1}}{\sqrt [p] {x^n + 1}}\ \mathrm {dx} =$

$ \frac 1n\int_0^1\frac {(x^n + 1) - 1}{\sqrt [p] {x^n + 1}}(x^n + 1)'\ \mathrm {dx} =$ $ \frac 1n\int_1^2\frac {t - 1}{\sqrt [p] t}\ \mathrm {dt}$ a.s.o.


i posted this on intermediate topics, but I figure this is mathcounts level.

prove that sum of digits of 1981^n (where n>=1) is more than or equal to 19.

Gee, Mathcounts? I'm not sure whether I can prove this one cleanly. Anyone have a nice solution?

I think there is a solution in the following thread: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8590[/url]


AMC
USA(J)MO
USAMO
inequalities unsolved
inequalities

Let $p,q,r$ be positive real numbers and $n \in N$. Show that if $pqr=1$, then :

$\frac{1}{p^n+q^n+1}+\frac{1}{q^n+r^n+1}+\frac{1}{r^n+p^n+1}=<1$

Take $a^3 = p^n$, $b^3 = q^n$ and $c^3 = r^n$ such that $abc = 1$. We need to prove ...

\[\frac{abc}{a^3 + b^3 + abc} + \frac{abc}{b^3 + c^3 + abc} + \frac{abc}{c^3 + a^3 + abc} \leq 1\] which is USAMO 1997, posted at least 100 times before.

This was some Baltic Way problem, last year or maybe a few years ago.

Are you going to Sweden too, Labiliau?

There is a test this saturday to choose two among us to go to Sweden.

I see. The Flemish contestants have been chosen already. They are

Guolong Li (obviously)
Clara Mertens
Wim van Rie

Good luck with the test.


trigonometry
number theory
relatively prime
trig identities

If (sin 10)*(sin 50)*(sin 70) is expressed in the form m/n, where m and n are relatively prime positive integers, find m+n. (The angles are in degrees.)

That is correct. I did it almost the same way, except from

$\displaystyle 1/4\sin(70)(-\cos(140)-2(\cos^2(70))\\$

I got
1/4 (cos20cos40 - 2sin70cos70cos70)
1/4 (cos20cos40 - sin140sin20)
1/4 (cos20cos40 - sin40sin20)
1/4 (cos60)
1/8

BTW this was a NYCIML Problem (3 sets of problems, 2 problems per set, 10 minutes per set).

Here's how I would do it:

\begin{eqnarray*}
\sin 10 \sin 50 \sin 70  &=& \cos 80 \cos 40 \cos 20\\
                                    &=& \cos 80 \cos 40 (2\cos 20 \sin 20)\left(\frac{1}{2\sin 20}\right) \\
                                     &=& \left(\frac{1}{2\sin 20}\right)\cos 80 \cos 40 \sin 40 \\
                                     &=& \left(\frac{1}{4\sin 20}\right)\cos 80 \sin 80\\
                                      &=& \left(\frac{1}{8\sin 20}\right)\sin 160\\
                                      &=& \dfrac{\sin 20}{8\sin 20}\\
                                      &=& \dfrac 18
\end{eqnarray*}

Just to make sure, joml88's solution is based on the double angle formula for sin(2 :theta: ), right?

Yeah here's a summary of the steps I took:

1st step:  use the identity $\sin \theta= \cos(90^\circ -\theta).$

2-5 steps:  use the identity $\sin(2\theta)=2\sin \theta \cos \theta.$

6th step:  use the identity $\sin \theta=\sin(180^\circ -\theta).$

Nice solution Joml. This problem also appeared on a USC (University of South Carolina) Math Contest a few years ago.

All of their past contests and solutions can be found here: http://www.math.sc.edu/contest/problems.html The level of the problems provide good practice for the AMC-12 contests.

[quote="metru"]It is true?
$4\sin \left( {60^0  - \alpha } \right) \cdot \sin \alpha  \cdot \sin \left( {60^0  + \alpha } \right) = \sin 3\alpha ,\forall \alpha  \in \mathbb{R}$[/quote]

[hide]Yes, in fact there is a nice identity posted by Gyan in the Trig Identities section that says: $\sin{(nx)} = 2^{n-1} \cdot \prod^{n-1}_{k=0} \sin{\Big(x + \frac{k\pi}{n}\Big)}$. It would be a very good exercise for students to try to prove this.[/hide]


modular arithmetic
number theory unsolved
number theory
IMO Shortlist

Let $1=d_1<d_2<....<d_k=n$ be all different divisors of positive integer n written in ascending order. Determine all n such that:
\[d_6^{2} +d_7^{2} - 1=n\]

$ n \equal{} 144$ and $ n \equal{} 1984$ are the only solutions.

It is clear that $ d_6$ and $ d_7$ are co-prime, implying that the total number of divisors of $ d_6$ and $ d_7$ does not exceed 8. Consider the following cases:

1) $ d_7 \equal{} p$ is prime. Then $ d_6^2 \equiv 1\pmod{p}$, implying that $ d_6 \equal{} p \minus{} 1$ and $ n \equal{} (p \minus{} 1)^2 \plus{} p^2 \minus{} 1 \equal{} 2p(p \minus{} 1)$, where the number of divisors of $ 2(p \minus{} 1)$ is $ 7$, i.e., $ 2(p \minus{} 1) \equal{} q^6$ where $ q$ is prime. Since the l.h.s. is even, we have $ q \equal{} 2$ and $ p \equal{} 33$, which is not prime, a contradiction.

2) $ d_6 \equal{} p$ is prime (it is clear that $ p > 5$). Then $ d_7^2 \equiv 1\pmod{p},$ implying that $ d_7 \equal{} kp \pm 1$. We also have $ d_7$ divides $ p^2 \minus{} 1$ and thus $ p d_7 \minus{} k (p^2 \minus{} 1) \equal{} k\pm p$, implying that $ d_7 \equal{} p \plus{} 1$. Therefore, $ n \equal{} p^2 \plus{} (p \plus{} 1)^2 \minus{} 1 \equal{} 2p(p \plus{} 1)$, where the number of divisors of $ 2(p \plus{} 1)$ is 7, implying (as before) that $ 2(p \plus{} 1) \equal{} 2^6$, $ p \equal{} 31$, and $ n \equal{} 1984$.

3) Neither of $ d_6$ and $ d_7$ is prime. In this case the product of their smallest prime divisors represents yet another divisor of $ n$ among the smallest five, hence the total number of divisors of $ d_6$ and $ d_7$ does not exceed 7. So, either both $ d_6$ and $ d_7$ are squares of primes, or one is square while the other is the product of two distinct primes or a third power power of a prime.

Let $ p$ and $ q$ are the smallest prime divisors of $ d_6$ and $ d_7$ respectively. One of them must equal 2, since otherwise $ \{1, p, q, 2p, 2q, pq \}$ would be among the first five divisors of $ n$, a contradiction.
Therefore, the case when both $ d_6$ and $ d_7$ are squares of primes is not possible. The remaining two cases are:

3.1) If $ d_6$ and $ d_7$ are a square of prime and the product of two distinct primes, then they are $ p^2$ and $ 2q$ (or vice versa), where $ p$ and $ q$ are odd primes. However, in this case $ n \equal{} p^4 \plus{} (2q)^2 \minus{} 1$ is a multiple of $ 4$ and $ 1, 2, 4, p, q, 2p, 2q, p^2$ must be among smallest 7 divisors of $ n$, a contradiction.

3.2) If $ d_6$ and $ d_7$ are a square of prime and a third power of prime then they are $ 2^3$ and $ p^2$, implying that $ n \equal{} p^4 \plus{} 63$. Therefore, $ p^2$ divides $ 63$, implying that $ p \equal{} 3$ and $ n \equal{} 144$.


I need to know a procedure to correctly factor expressions with negative fractional exponents.

Here are a couple of problems and I just kind of guessed at the answer, because I don't know the proper procedure.

1. ((x + 4) ^ (-1/5) - (x + 4) ^ (-6/5))

I came up with this :

(x + 6) / (x + 4) ^ (6/5)

2. (6x ^ (-5/3)  + 12x ^ (1/3))

I came up with this :

((6(1+(2x)^2)/ x ^ (5/3))

Any help on this would be greatly appreciated.  I cannot find any "clear" procedure anyway to solve these types of problems.

thanks

The way I like to think about these problems is to make a substitution that makes everything clearer.  For example, in the first problem the substitution $ y \equal{} (x\plus{}4)^{\minus{}\frac{1}{5}}$ allows us to rewrite the expression as

$ y\minus{}y^{6}\equal{} y(1\minus{}y^{5})$

in accordance with hjbrasch's factoring, but now the algebra behind the steps are made that much clearer.  :)

when i was learning it, u just remember to factor out common expressions/values with the smallest exponent on them. e.g. $ (x\plus{}1)^{\minus{}\frac{1}{5}}\plus{}(x\plus{}1)^{\minus{}\frac{6}{5}}\Longrightarrow (x\plus{}1)^{\minus{}\frac{6}{5}}\cdot(x^{1}\plus{}(x\plus{}1))$

Thanks for the help, but I still don't fully understand the process.  Can someone point me to some web resources with directions and many examples?

thanks

i don't feel like finding a web source although there might be something on the AoPS wiki.

regardless,
think about it in non-fractional exponents:
if u were asked to factor $ (x\plus{}1)^{3}\plus{}(x\plus{}1)^{2}$, you would factor out a $ (x\plus{}1)^{2}$ out of both of them, right, because u take factor of that expression with the smallest exponent.

think about it that way. i hope that helps a little.  :lol:


The statement " There are exactly 4 integer values of n for which (2n+y)/(n-2) is itself an integer" is true for certain values of y only. For how many values of y in the rangw 1<=y<=20 is the statement true?

[hide="Solution"]
Using long division, we find that $ \frac{2n\plus{}y}{n\minus{}2}\equal{} 2\plus{}\frac{4\plus{}y}{n\minus{}2}$. Looking at the denominator of the second half of the right hand side, we see that it can attain any integral value other than 0. Thus, if $ 4\plus{}y$ is divisible by x, it's also divisible by -x. This means that we only have to look for 2 positive values of the denominator, as we automatically get 2 more. Also, the fraction is always an integer if the denominator equals 1, so we only have to look for 1 more positive value. The only possible numerators that are only divisible by $ 1$ and another number, and their negatives, are primes. Thus $ y$ can take on $ 7$ values, as their are 7 primes between 5 and 24.
[/hide]


There are $ 4$ large boxes. There are $ 4 \times 3\equal{}12$ small boxes. There are $ 12 \times 2 \equal{} 24$ smaller boxes. Thus, there is a total of $ 4\plus{}12\plus{}24\equal{}40$ boxes

[quote=GameBot]There are 4 separate large boxes, and inside each large box there are 3
separate small boxes, and inside each of these small boxes there are 2
separate smaller boxes. How many boxes, counting all sizes, are there
altogether?[/quote]

#1 Who the heck did this?!?! You will pull out so many boxes and I swear the next problem we see is what is the probability my teddy bear is in there or something.
#2 DA SOLUTION!!!!! :)   :)   :)
[hide=solution step 1] So, $4$ large boxes, then, that means there is $12$ Small boxes inside the Large boxes, then, there is $24$ smaller boxes.[/hide]
[hide=solution step 2] Add them up. $4+12+24=40$[/hide]
[hide=further explanation] Let's use pictures and colors to help imagine it. [color=#f00]|____| |____| |____| |____|[/color] Those are the Large boxes. Now, [color=#f00]|[/color] [color=#00f]|__|[/color] [color=#f00]| |[/color][color=#00f]|___|[/color] [color=#f00]| |[/color][color=#00f]|___|[/color][color=#f00]|[/color] Now, let continue. I'm sure you know what to do, so, I'll leave that to youl Add all your little boxes up. [/hide]
[hide=add them! :)] Like the first time, we get $4+12+24=40$[/hide]


linear algebra
matrix
vector
linear algebra theorems

Let W be the space of $ n\times n$ matrices over the field F, and let $ W_0$ be the subspace spanned by the matrices C of the form  C=AB-BA. Prove that $ W_0$ is exactly the subspace of matrices which have trace zero. (Hint: What is the dimension of the space of matrices of trace zero? Use the matrix 'units', i.e., matrices with exactly one non-zero entry, to construct enough linearly independent matrices of the form  AB-BA.)

Since the dimension of $ n\times n$ matrices is $ n^2$ and the trace functional maps to the one-dimensional space F, by the dimension theorem dim range Trace + dim null Trace = $ n^2$, we obtain, the answer for the first question is $ n^2\minus{}1$. But how can I construct enough linearly independent matrices of the required form?

Let $ U_{jk}$ be the matrix with a $ 1$ in the $ (j,k)$ position and zeros elsewhere.

Suppose $ j\ne\ell.$ Choose any $ k.$ Then $ U_{jk}U_{k\ell}\minus{}U_{k\ell}U_{jk}\equal{}U_{j\ell}.$

We can do that for every off-diagonal location, which gives us $ n^2\minus{}n$ independent matrices of trace zero. Next, we focus on the main diagonal:

$ U_{jk}U_{kj}\minus{}U_{kj}U_{jk}\equal{}U_{jj}\minus{}U_{kk},$ which is a diagonal matrix with a $ 1$ and a $ \minus{}1$ in two different places on the main diagonal. (Watch as we proceed below, but it turns out that we don't need to know whether $ 1\equal{}\minus{}1$ or not. In other words, it will be OK even if $ \mathbb{F}$ has characteristic $ 2.$)

Just consider the $ n\minus{}1$ cases for which $ k\equal{}j\plus{}1,j\equal{}1,2,\dots,n\minus{}1.$ I claim that these $ n\minus{}1$ matrices are independent. We can see that by writing each such diagonal matrix as a row vector in $ \mathbb{F}^n,$ then stacking these rows together into an $ (n\minus{}1)\times n$ matrix which is already in row echelon form. These diagonal matrices are also clearly independent of the off-diagonal matrices from the previous step. Add our sets together and we have $ n^2\minus{}1$ independent matrices of the form $ AB\minus{}BA.$

Note that we have not shown that all matrices of trace zero can be written as $ AB\minus{}BA.$ Rather we have shown that all matrices of trace zero can be written as linear combinations of such matrices.


1. A sheep is tied to the edge of a circular patch of grass of radius ‘r’. The length of the rope used to tie the sheep is ‘l’. Find the length of the rope if the sheep eats away exactly half of the grass.

2.The overgrown grass patch is in the shape of an equilateral triangle. John is assigned the work of cutting down the grass and for that; he is given a machine which has a long blade whose length is equal to half the height of the triangular region. That is, he can trim off a sector of grass, in one swipe, with the radius of the sector being half the height of the triangle.
    John starts from one of the vertices. He has very little time left with him. Find the path that John has to follow so that he travels the least distance and still trims the entire grass patch?

please solve this problems as soon as possible

Thank u!!!

[quote="msg_1989"]1. A sheep is tied to the edge of a circular patch of grass of radius ‘r’. The length of the rope used to tie the sheep is ‘l’. Find the length of the rope if the sheep eats away exactly half of the grass.

please solve this problems as soon as possible

Thank u!!![/quote]

http://www.mathlinks.ro/viewtopic.php?t=249993

I don't know if I understood the second...

http://www.mathlinks.ro/viewtopic.php?search_id=1983298318&t=58587


Let $ a, b, c$-positive real numbers, $ a, b, c>0$
Show that: $ a\plus{}b\plus{}c\plus{}\frac{3}{\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}}>\equal{}4\sqrt[3] {abc}$.  Thanx  :wink:

Nice!
[hide]Let abc=1, then we have
$ a \plus{} b \plus{} c \plus{} \frac {3}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}} \ge 4$
But $ a \plus{} b \plus{} c \plus{} \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \ge 6$ by AM-GM
So for a fixed $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$, a+b+c is minimized when it is 6 minus it.
So we need to show $ 6 \minus{} x \plus{} \frac {3}{x} \ge 4$
$ (x \minus{} 3)(x \plus{} 1) \ge 0$
Clearly true, as $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c} \ge 3$ by AM-GM.[/hide]

[quote="dgreenb801"]$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \ge 3$ by AM-GM.[/quote]

$ a\equal{}b\equal{}c\equal{}2$

He scaled so that $ abc \equal{} 1$, but isn't the last inequality backwards? $ 6 \minus{} x \plus{} \frac 3x \ge 4 \Longleftrightarrow x^2 \minus{} 2x \minus{} 3 \le 0$.

Edit: Oh, so "But $ a \plus{} b \plus{} c \plus{} \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \ge 6$ by AM-GM" equality holds only when $ a \equal{} b \equal{} c \equal{} 1$.  :(

Wait, the problem isn't true. Take $ a\equal{}b\equal{}10$ and $ c\equal{}\frac{1}{100}$

Hmm no I think that works   :wink: $ LHS > 10 > 4 \equal{} RHS$

Let abc=1, then
$ a\plus{}b\plus{}c\plus{} \frac{3}{ab\plus{}bc\plus{}ac} \ge 4$
But $ (ab\plus{}bc\plus{}ac) \le (a\plus{}b\plus{}c)^2/3$, so we need to show
$ x\plus{}\frac{9}{x^2} \ge 4$ for all $ x \ge 3$, easy to verify.


the mean of six positive integers is 5, and the median is 6. What is the largest the mode could be, given that the mode is unique?

I got 8.  If the six numbers are 1, 1, 4, 8, 8, 8, we get a mean of 5 and a median of 6.


trigonometry
inequalities proposed
inequalities

Show that there exists an $x$ such that $\begin{equation*} \tan \left( \frac{\pi \sin x}{4\sin \alpha }\right) +\tan \left( \frac{\pi \cos x}{4\cos \alpha }\right) \leq 2. \end{equation*}$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally.  :)


calculus
integration
limit
logarithms
real analysis
real analysis unsolved

Please compute $ \lim_{x\to1}f(x)$, where $ f(x) \equal{} \int^{x^2}_{x} \frac{dt}{\ln{t}}$ and $ x>0$, $ x\neq 1$.

Substitute $ t \equal{} e^u$, use the power series expansion of $ x \mapsto \frac{e^x}{x}$ and integrate.


Source: USSR
(a) Find all integers with initial digit $6$ which have the following property, that if this initial digit is deleted, the resulting number is reduced to $\frac{1}{25}$ of its original value. And prove that you have found all such numbers.

(b) Prove that there does not exist any integer with the property that if its first digit is deleted, the resulting number is $\frac{1}{35}$ of the original number.

It's been posted before.

[hide="Solution"]
(a)  Let the number after deletion be $a$.  We have

$a + 6 \times 10^k = 25a$
$6 \times 10^{k} = 24a$
$10^{k} = 4a$

Hence $a = 25 \times 10^{m}$ and the numbers we want are of the form $625 \times 10^{m}$.

(b)  We have $d \times 10^{k} = 34a$.  But the LHS cannot have a prime factor greater than $7$.  Contradiction.  (The generalization is pretty obvious.)[/hide]

hmm for part b) i had like

[hide]

let the number be $x$, with first digit $a$ and has $n$ digits, then we have:

$35(x - a\cdot 10^{n-1}) = x$

this then gives:

$34x = 35a\cdot 10^{n-1}$

which cant be possible as the RHS does not contain a factor of $17$ and cannot contain a factor of $17$

[/hide]


In G proteins, does the GDP aquire one phosphate (to turn into GTP) or is the entire molecule of GDP removed from the protein and GTP re-inserted?


Let $ b$ be a prime. If $ j<b$, it seems the numerator won't have any factors of $ b$ in it, so the problem would be wrong.

[quote="caubetoanhoc94"]Let a;b;j are positive integer number such that:$ j \le a \plus{} 1$.Prove that:
$ {b}^{a \minus{} j \plus{} 1}| \frac {({b}^{a} \minus{} 1)({b}^{a} \minus{} 2)..({b}^{a} \minus{} j \plus{} 1)}{j!}$[/quote]

I think that it is not hard :arrow:

[quote="caubetoanhoc94"]Let a;b;j are positive integer number such that:$ j \le a \plus{} 1$.Prove that:
$ {b}^{a \minus{} j \plus{} 1}| \frac {({b}^{a} \minus{} 1)({b}^{a} \minus{} 2)..({b}^{a} \minus{} j \plus{} 1)}{j!}$[/quote]

You should at least check your problems with trivial values before posting :

Let $ (a,b,j) \equal{} (2,3,2)$ and you asked us to prove that $ 3|4$

I guess this is wrong :( (as dgreenb801 clearly said previously, and although shortlist seems to have an undisclosed easy solution)


superior algebra
superior algebra theorems

1.) Find the index of $<3>$ in the group $Z_{24}$?
2.) Let $A = (1 2 5 4) (2 3)$ in $S_{5}$. Find the index of $A$ in $S_{5}$?

1) The definition of index is just $\frac{|\mathbb{Z}_{24}|}{|<3>|}$. It shouldn't be hard to find the orders of these. $\mathbb{Z}_{24}=\{0,1,2,3,...\}$ and $<3> = \{0,3,6,9,...\}$.

2) Find the order of $S_{5}$ (should be defined in your book) and divide by the order of $(1254)(23)$. You could find the order of that element by composing it with itself until you get the identity. The number of times you do that is its order.


Hi everyone,
I'm currently the math club president, so I set up a summer training programs in this summer for the members. Do you think 25 problems a week with AMC level (all multiple choice) and 2 monthly problems a month with AIME level (show work) is enough in order to pass the AMC and get at least a 5 on AIME next year? How many problems a week/day (with what level) would you reccomend in order to pass the AMC and get at least a 5 on AIME ?
Thanks in advance.

There is no set amount of practice that will get some score. It all depends upon a myriad of factors, including but not limited to prior knowledge, ability to focus, intelligence, value of the practice, and other mathematical learning that goes on. Frankly, I did three AIME practice tests and qualified for USAMO. Someone else could have done none and qualified for it, while someone else could have done 20 practice tests and not even made AIME. If you want results, practice as much as you can. In particular, see what the scores are like, and, if you aren't satisfied, practice some more. If 25 problems a week seems like a reasonable amount to you, sure, do it. As well, I [i]greatly[/i] recommend that you look into AoPS books and classes for improvement, especially for people who are struggling to even qualify for the AIME.

thanks for reply, calc rulz
My problem is: since I wanna assign the problems for the math club members, so I dont know how many are good and effective enough to do well in the AMC/AIME. I agree with you that we should study as much as possible, but they might like Math but be lazy, or they are not agressive enough, ... So I want to like force them to study in the beginning, to make them self-aware in studying, know how to study on their own in the long run, especially when the AMC/AIME comes up next year. My school team didnt do very well in AMC/AIME this year, since we have no preparation and most of us dont even know what AMC/AIME are and their importance until we are told to take it.

I'm the president of my math team too and I don't know how useful assigning problem, however many, over the summer would be.  From my experience, it is a lot easier to motivate people and help them learn in group settings rather than assigning individual work.  Your method might work well if everyone on your team is extremely motivated and devoted to math, but the fact is that most people aren't.  This website represents the minority.  Your purpose might be better served by having everyone get together somewhere twice a month for a few hours and do the problems there.  That way, people can get help with questions that they don't understand.


quadratics algorithm calculus integration algebra function domain

Proove that $\sqrt{p}$, with $p$ a Prime number, is a irrational number.

Note: You don't know that all numbers can be rewrite as a product of prime numbers

[hide]Suppose $\sqrt{p}= \frac{m}{n}$, with $m$ and $n$ coprime positive integers. Then $m^{2}= n^{2}p$. So $p | m^{2}$, which is only possible if $p | m$. (Because if $p | ab$ then $p | a$ and/or $p | b$, for all integers $a,b$ and primes $p$.) Therefore, $m = kp$ for a certain integer $k$. Then $n^{2}p = m^{2}= k^{2}p^{2}$, so $n^{2}= k^{2}p$, i.e. $p | n^{2}$, from which it follows once again that $p | n$, making $1 = \gcd(m,n) \geq p$. Contradiction.[/hide]

[hide="Generalization"] For a non-square positive integer $d$, the quantity $\sqrt{d}$ is irrational.

Proof:  $\sqrt{d}$ is a root of the quadratic $x^{2}-d = 0$, which by Rational Root Theorem cannot have rational root $\sqrt{d}$ as $d$ is not square. [/hide]

you can't say "coprime positive integers"

because if you said that you are saying "m,n are a irreducible fraction"

since that, you are using the fact that a number can be write as a product of prime numbers..

[quote="yagaron"]since that, you are using the fact that a number can be write as a product of prime numbers..[/quote]

Not true.  $(m,n) = 1 \Leftrightarrow \not \exists 1 < d < \text{min}(m, n) : d | m, d | n$, which only requires the existence of a division algorithm.

if you dont know the fact that numbers can be write as a product of prime numbers. there is not word called "factor"

divisor, etc... this dont have meaning here.

so using rational root theorem is not a proof

[quote="yagaron"]divisor, etc... this dont have meaning here.[/quote]

Not true.  Again, the word "divisor" has meaning whenever there is a division algorithm.

http://en.wikipedia.org/wiki/Division_algorithm

Edit:  And if we are not allowed to consider the division algorithm, there is not very much we can do with the integers!

I have a point...

You don't know that All numbers can be write as a product of numbers. for more precision, primes.

so. how can you divide?

there is no sense in division. because there is nothing proving that division is possible.

$4/2$ is not $2$ because we can't make 4 in factors since we dont knwo that all numbers can be rewrite as a product.

please... don't wast your forth post talking about that.

i'm not stupid to make a problem that can be easily solved if we know the nature of prime numbers...

but i have already said... oh jeez

You don't need prime factorization. The expression $\alpha =\frac{m}{n}$ means that $n\cdot \alpha =m$, and we can talk about division by saying that $a$ divides $b$ if $ad=b$ for [i]some[/i] integer $d$.

If you don't want to talk about division at all, then talking about prime numbers ([i]defined[/i] in terms of division) is meaningless, and so is this topic.

[quote="yagaron"]there is no sense in division. because there is nothing proving that division is possible.[/quote]

Did you read the link or not?

[b]Division algorithm:[/b]  Given two positive integers $a, b$, there exist unique non-negative integers $q$ and $r < b$ such that

$a = bq+r$

Proof:  Consider the set

$\mathbb{S}= \{ R \in \mathbb{Z}: \exists Q \in \mathbb{Z}: R = a-bQ \}$

By the Well-Ordering Principle, $\mathbb{S}$ has a smallest non-negative member; call it $r$.  Now, $r-b \in \mathbb{S}$, but since we assumed that $r$ was the smallest non-negative member it follows that $r-b$ is negative; in other words, $r < b$.  QED.

[b]Definition:[/b]  $b | a \Leftrightarrow \exists q \in \mathbb{Z}: a = bq+0$

Edit:  On the other hand, the proof of the Rational Root Theorem probably involves prime factorization.  There is nothing wrong with Kurt's proof, though.

[quote="yagaron"]you guys have a simple way of think

toro.. kurt proof have a error.

let's make it in simple words

No prime-Factors ->
a/b = m does not hold, because how we can divide since we dont know that numbers can be multiplied such as we can get all numbers ->
or in anothers words $p_{1}^{a}.p_{2}^{b}\cdots=m$ is not true for all n, but you are using that is true. i know that this is true, but since we dont know that is true, so it's false->
Let's divide 15 by 3.  15/3 = 5*3/3 = 5 . just see? to divide we need to make a number in factors.[/quote]
Knowing that $5\cdot 3=15$ doesn't mean we know prime factorization works in general. If we're not allowed to multiply two numbers, what's the point of talking about rational numbers? (i.e $2/3$ is a number that $2/3$ [i]multiplied[/i] by $3$ gives $2$. We can't multiply, so I guess we can't talk about $2/3$.)

Yagaron, you're basically talking nonsense -- that some numbers can be written as a product of other numbers follows from [i]multiplication[/i] -- that $15 = 5 \cdot 3$ is a rule of multiplication, so for purposes of their proofs, the "division" $\frac{15}{3}= 5$ is literally defined to mean the multiplication $15 = 5 \cdot 3$.  Most significantly, this requires no assumptions about unique factorization or anything of the sort: that $15 = 3 \cdot 5$ does not automatically preclude $15 = 2 \cdot 7$, for example.

What you should probably do, if you want this thread to be interesting, is actually suggest the axioms of the system you want us to use.  For example, are you claiming that your statement is true in any integral domain?  In principle ideal domains?  In totally ordered commutative rings?

[i]Edit: this post has some duplication with Scorpius'.  Sorry about that.[/i]

the thing i'm saying is

1 - we know multiplication: we know that $5.3=15$ we can multiply $a.b$ with a,b being integer numbers.

2 - we don't know division. or in anothers words we dont know the meaning of $\frac{15}{3}$. or in another words. there is no algorithm of division. or in another words. we know that some numbers when multiplied give another number, but we canno generalize it and say that "if we multiply some numbers we can have all possible numbers". do you understand? just because it holds for a few numbers we cannot generalize.

Of corse we can divide some numbers, or given 2 non-coprime numbers, it cannot be rigth as a division of coprime numbers. this is false, but i'm assuming this is true, because we dont know that is true.

understand? since we dont know one thing we cannot use this thing, it's the same thing that use a conjecture to proove a problem,

suppose the problem is the opposite, so i can solve it like that

"p_1^a.p_2^b \cdots" is irrational if $a,b,\cdots$ is not divissible by 2

to use this problem you know even numbers, odd numbers, but you dont know prime numbers.

integers division is factor in prime numbers and take the common factors

but multiplication is not just that. to make multiplication we just need to know that

1.1 =1
1.2=2
...
9.9=81
etc

and use decimal numbers.

[quote="yagaron"]2 - we don't know division. [/quote]

Then how are you supposed to talk about prime numbers or irrationality at all?

[quote="yagaron"]integers division is factor in prime numbers and take the common factors[/quote]

Integer division is repeated subtraction, and the sooner you realize this the sooner this thread might take a better turn.

[quote="I"]You don't need prime factorization. The expression $\alpha =\frac{m}{n}$ means that $n\cdot \alpha =m$...[/quote]

There's division without prime factorization. Because $5\cdot 3=15$, we know that $\frac{15}{3}=5$. At the very least, we need to use division to have rational numbers. Otherwise, what is $\frac{15}{3}$? If we can't divide, we don't know its value. Then I could say something like, "Define $\frac{15}{3}$ to be the result of raising 2 to power of the square root of 2." Of course, this is silly, since $2^\sqrt{2}$ is obviously not equal to 5 or any integer. But what if I denote $2^\sqrt{2}$ by the Greek letter $\gamma$? Sure enough, the letter $\gamma$ is often used to represent a certain constant, and one obviously not equal to $2^\sqrt{2}$. But for those who don't know about $\gamma$, it makes sense to use $\gamma$ like a variable, like $x$. Why not use $\frac{15}{3}$ like a variable if it has no meaning?

I can't really understand much of what yagaron is saying, but let me see if I can restate his question.

Let R be an integral domain - meaning it is a set of elements with
1. additive closure
2. a zero element (0)
3. commutative addition
4. associative addition
5. additive inverses for each element
6. multiplicative closure
7. an identity (or one) element (1)
8. commutative multiplication
9. associative multiplication
10. the distributive law of multiplication over addition
11. for a and b in R if ab = 0 then at least one of a or b is zero

(Note: an integral domain may or may not have unique factorization and the euclidean algorithm does not necessarily hold.)

Let Q be the field of fractions of R - meaning it is the set of ordered pairs (a,b) such that:

1. (a,b) = (c,d) iff ad = bc
2. (a,b) + (c,d) = (ad+bc,bd)
3. (a,b)(c,d) = (ac,bd)
for any a,b,c,d in R.

Prove that if p is irreducible in R (meaning if p = ab then at least one of a or b is a *unit* in R), then $ (p,1) \neq x^{2}$ for any x in Q.

(*Note: a unit is an element u in R such that a there exists an element v in R with uv = 1).

This very much remembers me to http://www.mathlinks.ro/Forum/viewtopic.php?t=150038 ...
This question is, as is the one in the link, just strange. And especially the reactions to answers are _not_ ok. I suggest locking again...


function
algebra
polynomial
algebra theorems

Could someone post here Rouche's theorem (the one with polynoms)

Rouche's theorem:
Let $f,g$ be analytical functions (e.g. polynomials) defined on (some simple connected (so connected and it has no holes) open) $D \subset \mathbb{C}$. Let $\alpha$ be a closed curve in $D$ that goes exactly one time around every point in it's interior (so is a 'circle').
When now also $|g(z)| < |f(z)|$ for all $z \in \alpha$, then have $f$ and $f+g$ as functions the same number of zeros in the interior of $\alpha$ (counted with multiplicativity).

Since this is a theorem of complex analysis, any proof will be at the advanced undergraduate or early graduate level.

Since it's going to be in the next lecture of a course I'm teaching ...

In J.B. Conway's textbook, he gives the following general and flexible statement of the theorem, and attributes it to Irving Glicksberg ([i]Amer. Math. Monthly,[/i] [b]83[/b] (1976), 186-187):

----

Suppose $f$ and $g$ are meromorphic in a neighborhood of $\overline{B}(a;R)$ with no zeros or poles on the circle $\gamma=\{z: |z-a|=R\}.$ If $Z_f,Z_g\,(P_f,P_g)$ are the number of zeros (poles) of $f$ and $g$ inside $\gamma$ counted according the their multiplicities and if
\[ |f(z)+g(z)|<|f(z)|+|g(z)| \]
on $\gamma,$ then
\[ Z_f-P_f=Z_g-P_g \]
----

If "meromorphic functions" sounds too general for your tastes, you could apply it to rational functions.


geometry
circumcircle
Euler
analytic geometry
geometry solved

Suppose that G is the centroid of triangle ABC, and that
$\angle{GAB}+\angle{GBC}+\angle{GCA}=90^{\circ}$
Characterize triangle ABC

This problem is from CRUX 1999 but I don't heve its solution in CRUX 2000.Somebody helps me!Thanks a lot! :omighty:

Can you tell me the number of the problem or page ?

  Babis

It is Crux Mathematicorum problem #2427, solved in Crux Mathematicorum 26 (2000) nr. 3, p. 182-183. I attach the solution as a PDF file.

Another solution is straightforward if you are familiar with the McCay cubic of a triangle; in fact, the equation < GAB + < GBC + < GCA = 90° shows that the point G lies on the McCay cubic of triangle ABC; thus, the point G, its isogonal conjugate - the symmedian point K of triangle ABC -, and the circumcenter O of triangle ABC are collinear; in other words, the symmedian point K of triangle ABC lies on the Euler line GO, and this easily yields (e. g., using barycentric coordinates) that the triangle ABC is isosceles. However, I fear you are not used to this notions and thus I didn't post this solution until I located the Crux solution (which is elementary).

By the way, next time, please do tell us the number of the problem in Crux. This makes searching for it much easier.

  Darij


analytic geometry
ratio
Pythagorean Theorem
geometry
similar triangles

Okay, I've tried these types of problems but I can't figure out how to do them.

Point C is on the segment AB which has endpoints A(2,-1) and B(11,5).  Point C is twice as far from Point A as it is form point B.  What are the coordinates of point C?

Thankx!

[quote="007math"]
Point C is on the segment AB which has endpoints A(2,-1) and B(11,5).  Point C is twice as far from Point A as it is form point B.  What are the coordinates of point C?
[/quote]

Let's find the distance between A and B. Using the distance formula(AKA Pythagorean Theorem), we find it is $ 3\sqrt{13}$. C is twice as far from point A as it is from point B. This is eerily reminiscent of ratio problems! This distance from A to C compared to C to B is 2:1. So we set up the equation $ 3x \equal{} 3\sqrt{13}$, $ x \equal{} \sqrt{13}$. We multiply by 2 to get $ 2\sqrt{13}$, which is the distance from C to A.

We now have the equation $ \sqrt{(x\minus{}(\minus{}1))^2 \plus{} (y\minus{}2)^2} \equal{} 2\sqrt{13}$, where (x,y) is Point C. Now solve this equation to get your answer.

[quote="mathemagician1729"]Twelve zaps weigh the same as 9 yaps. Seven xaps weigh as much as six yaps. How
many xaps weigh as much as 32 zaps?

I always set up messy equations to deal with these, but is their a method that is "countdown ready"?[/quote]

12 zaps=9 yaps or yap=4/3 zap.

6 yaps=8 zaps.

7 xaps=8 zaps

28 xaps=32 zaps.

28.

[quote="mewto55555"]Except that not only gets nasty, it has 2 variables, so solving that gets the locus of all points that satisfy this condition not on the line.

Lets look at the coordinates seperately.

X: Its 2x as far from 2 as it is from 11. Clearly it is 8.

Y: Similar logic shows 3

Thus its (11,3)

Note: We could solve this problem izzys way, but we would have to include the equation of the line through A and B.[/quote]

Note that mewto55555's method can be proven easily with similar triangles--draw a diagram, and extend lines parallel to the axes from the points.


inequalities
calculus
integration
LaTeX
inequalities unsolved

It is clear that the sum is increasing, so since we have equality on the left for $n=1$, the left part is proven.

The right part I can solve with some calculus: the sum is ascending, and in the limit for n to infinity, we replace the sum by integral: $\int^n_1 \frac1{n+x} dx$, which has limit ln(2), and since the sum is increasing for increasing n, each instance of the sum should be below ln(2).

Now all we should prove is that $\ln(2) < \frac{\sqrt{2}}{2}$... this is of course easy to find out it's true, but I don't really see an elementary way... do you have one? My best attempt would be taylor...

[quote="blahblahblah"]The taylor series for $\ln(1+x)$ converges at $x=1$.

Hence $\ln(2)=\frac {1}{1}-\frac {1}{2}+\frac {1}{3}-\frac{1}{4}+\ldots$

Hence $\ln(2)<\frac {1}{1}-\frac {1}{2}+\frac {1}{3}-\frac{1}{4}=\frac {7}{12}<\frac{\sqrt{2}}{2}$

Alternately, $2<1+\frac {\sqrt{2}}{2}+\frac {1}{2}\left(\frac {\sqrt{2}}{2}\right)^2+\frac {1}{6}\left(\frac {\sqrt{2}}{2}\right)^2<e^{\frac {\sqrt{2}}{2}}$[/quote]

Hmm... sorry, but... $\ln(2) > \frac{7}{12}$ :)
The second path was what I meant with taylor, but I could not find the simple rigorous argument for $2<1+\frac {\sqrt{2}}{2}+\frac {1}{2}\left(\frac {\sqrt{2}}{2}\right)^2+...$ :?

For your 4 terms it is already 2.016032346, but it seems a very close bound to show with the hand... how do you show it without calculator?

I see it's larger than 1 + 0.7 + 0.25 + 0.028 but this leads me nowhere... stuck on about 1.98

The first approach was a bad one - we'll ignore it from now on :)

For the second one, it's easy to expand the four terms taylor expansion into

$\frac {13\sqrt{2}}{24}+\frac {5}{4}$

I claim that this is $>2$.  Move the irrational and rational terms onto separate sides, we need to show that

$\frac {13\sqrt{2}}{24}>\frac {3}{4}$

Squaring, we get

$\frac {169}{288}>\frac {9}{16}$

Subtract them, and we get

$\frac {7}{288}$ :)

Ok, that solves the problem then. :) thanks.

However I guess there must be a solution without this 'calculation', socrates?

I doubt it - partial sums of the harmonic series are notoriously hard to calculate exactly.

[quote="dblues"]Checking for small cases $n=1, 2$, we can assume $n \geq 2$. Pairing up the summands, first with the last, second with the second last and so on, the sum becomes $\sum_{k=1}^n \frac{1}{n+k} = (3n+1)(\frac{1}{(n+1)(2n)} + \frac{1}{(n+2)(2n-1)} + \ldots )$
$\leq (3n+1)(\frac{1}{(n+1)(2n)} + \frac{n-2}{2} (\frac{1}{(n+2)(2n-1)}))$,

which after a whole mess of algebraic manipulation, is $< \frac{1}{\sqrt{2}}$ for sufficiently large $n$, thus we only need to check a finite number of cases for small values of $n$.[/quote]I don't understand you dblues. For sufficiently large $n$ (e.g. for the limit of $n\rightarrow\infty$) your term goes to $\frac34 > \frac{1}{\sqrt{2}}$.. right?

This is not a difficult one,I suppose?
Indeed,there is a solution without 'calculation'.
By Cauchy Ineq,
$\displaystyle \sum\limits_{k=1}^n \dfrac{1}{n+k} \leq \sqrt{n\sum\limits_{k=1}^n \dfrac{1}{(n+k)^2}}$
Then,
$\displaystyle \sum\limits_{k=1}^n \dfrac{1}{(n+k)^2} < \sum\limits_{k=1}^n \dfrac{1}{(n+k)(n+k-1)}$
What is more,
$\displaystyle \sum\limits_{k=1}^n \dfrac{1}{(n+k)(n+k-1)} = \sum\limits_{k=1}^n \dfrac{1}{n+k-1}-\sum\limits_{k=1}^n \dfrac{1}{n+k} = \dfrac {1}{n}-\dfrac{1}{2n} = \dfrac {1}{2n}$
Thus,we have
$\displaystyle \sum\limits_{k=1}^n \dfrac{1}{n+k} < \sqrt{\dfrac{n}{2n}}=\dfrac{\sqrt{2}}{2}$

[quote="Forward"]What has happened to the LATeX system?[/quote]Nothing, your post looks fine here...

Peter, there was a big problem with LaTeX on the forum today.


integration
calculus
trigonometry
calculus computations


If a, b, and c are real numbers such that a^2 +2b=7, b^2 +4c=-7, and c^2 +6a = -14, find a^2 +b^2+c^2. The only way I could think of how to do this problem is by actually substituting, but that is too long. Is there a quicker way?

[hide]add them up

get

a :^2:  + b :^2:  + c :^2:  + 6a + 2b + 4c + 14 = 0

(a+3) :^2:  + (b+1) :^2:  + (c+2) :^2:  = 0

each one of them have to equal zero

so... a = -3, b = -1, c=-2

the sum of their squares is 14[/hide]

Yeah. I figured that a few minutes after posting this. Thanks a lot anyways. Is there a trick to looking for clever factorizations, or is it all just practice?

I dunno. I just do as many as you can. And when you come across one, u can figure out what to do.

Then there are ways to get the factorization, like adding x, then subtracting it. Im not pro at this, but factoring is fun.

Main thing: do as many different factorings as possible. You'll be thankful when you come across them again.

Well this isn't as much a factorisation, the main thing here is that a sum of squares mustn't be negative.. that is definitely something that pops up all over the place. in fact thats where most (all?) of the more complicated inequalities come from in the first place..

well rearrangement is just based on ordering instead of the trivial inequality, but that's probably not complicated from your viewpoint...

oh yeah and then we have cos:theta::le:1 as in dot-product proof of cauchy


trigonometry
quadratics
trig identities
Law of Cosines
geometry
Law of Sines

If the diameter of a circle is $ 10\sqrt {2}$ and $ \overline{AB} \equal{} 2$, $ \overline{BC} \equal{} 10$ are two chords of this circle, find $ \overline{AC}$.

$ 6\sqrt{2}$.There really has to be a darn better way.....
[hide]Draw the diameter $ CD$ through $ C$. So,$ \angle CDB\equal{}\angle CAB\equal{}45$.Put law of cosines in $ \Delta ABC$ to get $ AC$[/hide]

[quote="LoboSolitario"][quote="hell_ever"]$ 6\sqrt {2}$[/quote]

I think you are missing something.[/quote]

Must be calclation mistake. my approach is the same.u'll sympathise with me if u tried out that quadratic without a calculator. :|

English isn't my first language, so i'm not sure if i did a good translation. The figure below will make it clear (i hope).

[quote="hell_ever"]$ 6\sqrt {2}$.There really has to be a darn better way.....
[hide]Draw the diameter $ CD$ through $ C$. So,$ \angle CDB \equal{} \angle CAB \equal{} 45$.Put law of cosines in $ \Delta ABC$ to get $ AC$[/hide][/quote]
[hide]From what you've written, I think your solution should be $ 8\sqrt2$. (You had the quadratic $ x^2\minus{}2\sqrt2 x\minus{}96\equal{}0$ ?) But $ C$ can also be at $ D$, because $ BC\equal{}BD\equal{}10$. So there are 2 solutions. It is easy to find the second solution, because there is a right triangle inscribed in the circle with hypotenuse $ 10\sqrt2$ and one leg $ 8\sqrt2$. The other leg is $ \boxed{6\sqrt2}$.[/hide]

[quote="LoboSolitario"]If the diameter of a circle is $ 10\sqrt {2}$ and $ \overline{AB} \equal{} 2$ and $ \overline{BC} \equal{} 10$ are two chords of this circle, find $ \overline{AC}$.[/quote]
I'll assume that $ B$ is between $ A$ and $ C$.
[hide="Alternate Solution"]
Pick $ P$ diametrically opposite to $ B$.  So, by Ptotemy's, we have that $ (PA)(BC) \plus{} (PC)(AB) \equal{} (PB)(AC)$.  A bit of substitution gives us that $ PA(10) \plus{} PC(2) \equal{} AC(PB) \equal{} AC(10\sqrt {2})$.  Now, $ PB$ is the diameter, so $ \angle PAB \equal{} 90 \equal{} \angle PCB$.  Thus, from Pythagoras on $ \triangle PAB$ and $ \triangle PBC$, we get that $ AB^2 \plus{} AP^2 \equal{} PB^2\implies 4 \plus{} PA^2 \equal{} 200\implies PA \equal{} 14$ and $ BC^2 \plus{} PC^2 \equal{} 200\implies 100 \plus{} PC^2 \equal{} 200\implies PC \equal{} 10$.  Now, we plug $ PC$ and $ PA$ back in to obtain that $ AC(10\sqrt {2}) \equal{} (2)(10) \plus{} (10)(14) \equal{} 160\implies AC \equal{} \boxed{8\sqrt {2}}$.  [/hide]

[hide="Another Approach"]Using the relation between arcs and angles, you can work with some trig finding easily $ \sin A\widehat{O}Q\equal{}\dfrac{1}{5\sqrt{2}}\equal{}\sin A\widehat{C}B$ (check file/figure attached above) and $ \sin C \widehat{O} P \equal{}\dfrac{\sqrt2}{2} \Longrightarrow C\widehat{O}P\equal{}45^{\circ}\equal{}D\widehat{A}B$. Hence, calculate $ \sin (A\widehat{C}B\plus{}45^{\circ})\equal{}\dfrac{4}{5}$. Note that $ A\widehat{B}C\equal{}180^{\circ}\minus{}(A\widehat{C}B\plus{}D\widehat{A}B).$
Now by law of sines on triangle $ ABC$, $ AC\equal{}2R\cdot\sin (A\widehat{C}B\plus{}45^{\circ})\equal{}8\sqrt2$.[/hide]


How do we define a function on a point? more over how do we define the sum of 2  or more functions on points because I noticed that in section 3.2 they had something like $ f(x_1) \plus{} f(x_2) \plus{} ... \plus{} f(x_k) \equal{} 0$ what does this mean? Has anyone read that part of the book?

You'll have to be more specific about what you mean by "point."  Do you mean a point in $ \mathbb{R}^n$?


analytic geometry
geometry
USAMTS
arithmetic sequence
probability
expected value
complex numbers

Hi  My teacher assigned me with these problems but i cant manage to solve them :( so can anyone help me with these?

Problem 1
An increasing arithmetic sequence with infinitely many terms is determined as follows.
A single die is thrown and the number that appears is taken as the first term. The die is thrown again and the second number that appears is taken as the common difference between each pair of consecutive terms. Determine with proof how many of the 36 possible sequences formed in this way contain at least one perfect square.

Problem 2
George has six ropes. He chooses two of the twelve loose ends at random (possibly from the same rope), and ties them together, leaving ten loose ends. He again chooses two loose ends at random and joins them, and so on, until there are no loose ends. Find, with proof, the expected value of the number of loops George ends up with.

Problem 3
Let r be a nonzero real number. The values of z which satisfy the equation
R^4z^4 + (10r^6 - 2r^2)z^2 - 16r^5z + (9r^8 + 10r^4 + 1) = 0 are plotted on the complex plane (i.e. using the real part of each root as the x-coordinate and the imaginary part as the y-coordinate). Show that the area of the convex quadrilateral with these points as vertices is independent of r, and find this area.

Problem 4
Homer gives mathematicians Patty and Selma each a di_erent integer, not known to the other or to you. Homer tells them, within each other’s hearing, that the number given to Patty is the product ab of the positive integers a and b, and that the number given to Selma is the sum a + b of the same numbers a and b, where b > a > 1. He doesn’t, however, tell Patty or Selma the numbers a and b. The following (honest) conversation then takes place:
Patty: “I can’t tell what numbers a and b are.”
Selma: “I knew before that you couldn’t tell.”
Patty: “In that case, I now know what a and b are.”
Selma: “Now I also know what a and b are.”
Supposing that Homer tells you (but neither Patty nor Selma) that neither a nor b is greater
than 20, find a and b, and prove your answer can result in the conversation above.

Problem 5
Given triangle ABC, let M be the midpoint of side AB and N be the midpoint of
side AC. A circle is inscribed inside quadrilateral NMBC, tangent to all four sides, and that circle touches MN at point X. The circle inscribed in triangle AMN touches MN at point Y , with Y between X and N. If XY = 1 and BC = 12, find, with proof, the lengths of the sides AB and AC.

Thanks alot

 :lol:  :(  :(  :(  :(  :)  :lol:  :(  :(

These are the problems from the current [url=http://www.usamts.org/]USAMTS[/url].  Needless to say, discussion of these are not allowed before the deadline.