real analysis
function
real analysis unsolved

Define a sequence $ ( f_n )_{n\equal{}1}^{\infty}$ of functions on $ [0,1]$ by $ f_n(t)\equal{}t^n$.

(1) Does $ (f_n)_{n\equal{}1}^{\infty}$ converge in $ (C[0,1],\| \|_{\infty})$?
(2) Does $ (f_n)_{n\equal{}1}^{\infty}$ converge in $ (L^2[0,1],\| \|_{2})$?.

Note that convergence in $ C[0,1]$ under $ \left\|\cdot\right\|_{\infty}$ is equivalent to uniform convergence.  Does $ f_{n}(t)$ converge uniformly on $ [0,1]$?  (Hint: Consider $ t\equal{}1$.)

For the second one, $ f_{n}\to 0$ pointwise a.e. in $ [0,1]$, the exception being $ \{1\}$.  Then $ \left\|f_{n}\minus{}f\right\|_{2}\to 0$ if and only if $ \left\|f_{n}\right\|_{2}\to\left\|f\right\|_{2}$.

so the first one doesn't converge uniformly because a series of 1s (t=1) is not uniformly convergent?
Am i correct?
I also think that the second is convergent.

[quote]because a series of 1s (t=1) is not uniformly convergent? [/quote]
That doesn't mean anything. If you restrict to a single point, all you have is a sequence of numbers, which can either converge or diverge - and this one converges. "Uniform" makes no sense at a single point. A sequence is (or isn't) uniformly convergent on some whole [i]set[/i].

JRav's hint about considering $ t \equal{} 1$ may really a reference to a possible contrapositive of the following theorem: if a sequence of continuous functions $ f_n$ converges uniformly on $ S$ to the limit function $ f,$ then $ f$ is continuous on $ S.$ (Or for short, "uniform limits of continuous functions are continuous.")

Also: I'm moving this topic to the analysis forum, where it fits better.

What happens with the second example? does this equence converge? if a function is convergent pointwise then is it convergent?

That doesn't make any sense-it converges pointwise hence it converges.  There are different modes of convergence, however, and what you want is convergence in $ L^{2}$.  I'm telling you that you've got pointwise convergence to 0 a.e., so use the result I gave you.  Or explicitly compute it, they're equally easy. (Or better, prove the result!)

The first step in determining whether or not a sequence converges uniformly is to understand the definition.  Uniform convergence says that if you go far enough out in your sequence, you can make [b]every[/b] point arbitrarily close to the limit.  Why does this create problems for $ t^{n}$ on $ [0,1]$?


Αν $ a,b,c > 0$ , να αποδείξετε ότι :

  $ \frac {2a^3}{a^6 \plus{} bc} \plus{} \frac {2b^3}{b^6 \plus{} ca} \plus{} \frac {2c^3}{c^6 \plus{} ab} \leq \frac {a}{bc} \plus{} \frac {b}{ca} \plus{} \frac {c}{ab}$

[b]ΣΧΟΛΙΟ[/b]

 Στο περιοδικό δίνεται ως πρώτο ερώτημα η ανισότητα :

 $ \frac {a}{a^2 \plus{} bc} \leq \frac {1}{4} ( \frac {1}{b} \plus{} \frac {1}{c}$

 Μπάμπης

[quote="maniopas"]Το δύσκολο θα ήταν χωρίς το πρώτο υποερώτημα.

$ \frac {a}{a^2 \plus{} bc} \le \frac {1}{4}(\frac {1}{b} \plus{} \frac {1}{c}) \Leftrightarrow$

$ 4a \le b \plus{} c \plus{} \frac {a^2}{b} \plus{} \frac {a^2}{c}$, που ισχύει γιατί

$ b \plus{} c \plus{} \frac {a^2}{b} \plus{} \frac {a^2}{c} \ge$ (AM - GM) $ 4\sqrt [4]{\frac {a^2a^2bc}{bc}} \equal{} 4|a| \equal{} 4a$

άρα. βάζοντας στο υποερώτημα όπου $ a$ το $ a^3$ είναι:

$ \frac {a^3}{a^6 \plus{} bc} \le \frac {1}{4}(\frac {1}{b} \plus{} \frac {1}{c}) \equal{} \frac {1}{4}(\frac {a}{ab} \plus{} \frac {b}{bc})$

εφαρμόζοντας το ίδιο και για τα b, c και προσθέτοντας κατά μέλη προκύπτει

$ \frac {1}{2}LHS \le \frac {1}{4}(2\frac {a}{ab} \plus{} 2\frac {b}{bc} \plus{} 2\frac {c}{ca}) \le$
(από ανισότητα αναδιάταξης) $ \le 2\frac {1}{4}RHS \Leftrightarrow LHS \le RHS$ q.e.d. :D[/quote]

[color=red][b]Πoλύ ωραία λύση ! [/b][/color]

 Σας έδωσα  το α΄ερώτημα , για να μη σας βγάλω από τη μελέτη σας για πολύ ώρα.Σημασία έχει  να δείτε τη διάσπαση (δηλαδή το α΄ερώτημα) και να προβληματιστείτε πώς θα μπορούσε κάποιος να την επινοήσει !.

 Τώρα ,αν θέλει κάποιος πιο απλά πράγματα , βλέπω ότι μπορεί να γράψει :

 $ b \plus{} c \plus{} \frac {a^2}{b} \plus{} \frac {a^2}{c} \equal{}( b\plus{}\frac {a^2}{b})\plus{}(c \plus{} \frac {a^2}{c}) \geq 2 \sqrt {b \cdot \frac {a^2}{b}} \plus{} 2 \sqrt {c \cdot \frac {a^2}{c}} \equal{} 2a\plus{}2a \equal{} 4a$

 Αλλά και η  αναδιάταξη μπορεί να αποφευχθεί(εμένα μου αρέσει πολύ!) , αρκεί να χρησιμοποιήσουμε την :
   $ x^2\plus{}y^2\plus{}z^2 \geq xy\plus{}yz\plus{}zx$ , με $ x,y,z$ τους όρους(τετρ. ρίζες) της αποδεικτέας(β΄μέλους).

 [i]Η άσκηση είναι ωραία εφαρμογή στη μέθοδο της διάσπασης -  παρεμβολής.[/i]

΄Το ερώτημά μου κύριε Μπάμπη είναι αν μπορούμε να καταλάβουμε βλέποντας μια τέτοια ανίσωση , αν λύνεται με τη μέθοδο της παρεμβολής - διάσπασης.
Αν κάποιος πέρναγε και μας το σφύριζε " έει ψιτ ...  λύσε το με διάσπαση "
τότε θα προσπαθούσαμε να βρούμε τη βασική και μετά η κλασική προσθεση των ανισοτήτων που έχουν προκύψει από κυκλική εναλλαγή των γραμμάτων ..
αλλά στην πραγματικότητα τα πράγματα είναι κάπως διαφορετικά!!!
Θέλω να πώ ότι πολλές φορές κάποιες ανισότητες μπορούν να σου φανε μια μέρα και να μη το βρείς :mad: τι θέλουν. Πιστεύετε ότι είναι θέμα εμπειρίας, διαίσθησης ή υπάρχει και κάτι άλλο ( πιό τεχνικό ίσως ) που σε βοηθάει να αποφεύγεις τις κακοτοπιές ?

[quote="vangelismak"]΄Το ερώτημά μου κύριε Μπάμπη είναι αν μπορούμε να καταλάβουμε βλέποντας μια τέτοια ανίσωση , αν λύνετε με τη μέθοδο της παρεμβολής - διάσπασης.
Αν κάποιος πέρναγε και μας το σφύριζε " έει ψιτ ...  λύσε το με διάσπαση "
τότε θα προσπαθούσαμε να βρούμε τη βασική και μετά η κλασική προσθεση των ανισοτήτων που έχουν προκύψει από κυκλική εναλλαγή των γραμμάτων ..
αλλά στην πραγματικότητα τα πράγματα είναι κάπως διαφορετικά!!!
Θέλω να πώ ότι πολλές φορές κάποιες ανισότητες μπορούν να σου φανε μια μέρα και να μη το βρείς :mad: τι θέλουν. Πιστεύετε ότι είναι θέμα εμπειρίας, διαίσθησης ή υπάρχει και κάτι άλλο ( πιό τεχνικό ίσως ) που σε βοηθάει να αποφεύγεις τις κακοτοπιές ?[/quote]

Βαγγέλη,
  για αυτόν ακριβώς το λόγο που αναφέρεις βάζουν ανισότητες και στην ΙΜΟ !
 Επειδή και μια απλή ανισότητα μπορεί να σου φάει μια βδομάδα , μέχρι να βρεις το τέχνασμα.
 Εδώ χρειάζεται και εμπειρία  και ευφυία και τύχη ! Όλα μαζί !

 Μπάμπης

Μια άλλη λύση (αφιερωμένη στον vangelismak)

Είναι $ \frac{2a^3}{a^6\plus{}bc}\leq\frac{2a^3}{2a^3\sqrt{bc}}\equal{}\frac{1}{\sqrt{bc}}$

Θέτω $ a\equal{}x^2$ κτλ  γιατί βαριέμαι τις ρίζες  :D   και κάνοντας ομόνυμα αρκεί νδο

$ \frac{x\plus{}y\plus{}z}{xyz}\leq\frac{x^4\plus{}y^4\plus{}z^4}{(xyz)^2}$ ή $ x^4\plus{}y^4\plus{}z^4\geq xyz(x\plus{}y\plus{}z)$

Όμως $ x^4\plus{}y^4\plus{}z^4\geq x^2y^2\plus{}y^2z^2\plus{}z^2x^2\geq xyz(x\plus{}y\plus{}z)$   qed  :)

Siluan Μια μέρα θα σε φτάσω  :P


Hi,

I've got an topological problem. Define $ \mathbb{RP}^n$ as the set of all 1-dimensional linear subspaces of $ \mathbb{R}^{n\plus{}1}$. We give it the quotient topology determined by the map $ \pi: \mathbb{R}^{n\plus{}1}\backslash\{0\}\to\mathbb{RP}^n$ sending each vector to its linear span.
I have to prove that $ \mathbb{RP}^n$ is second-countable. Obviously it is sufficient to show that the projection $ \pi$ is open, because then $ \pi$ maps a countable base of $ \mathbb{R}^{n\plus{}1}\backslash\{0\}$ to a countable base of $ \mathbb{RP}^n$.
To prove that $ \pi$ is open it is sufficient to show that $ \pi$ maps sets of the type $ \left((a_1,b_1)\times (a_2,b_2)\times\dots (a_{n\plus{}1},b_{n\plus{}1})\right)\cap\left(\mathbb{R}^{n\plus{}1}\backslash\{0\}\right)$ (that are basis elements of $ \mathbb{R}^{n\plus{}1}\backslash\{0\}$) to open sets. But for any set $ U\subseteq\mathbb{R}^{n\plus{}1}$ the sets $ U$ and $ \pi^{\minus{}1}\pi U$ have the same image under the map $ \pi$, but $ \pi^{\minus{}1}\pi U$ is saturated. So it is sufficient to prove that $ \pi^{\minus{}1}\pi U$ is open whereas $ U\equal{}\left((a_1,b_1)\times (a_2,b_2)\times\dots (a_{n\plus{}1},b_{n\plus{}1})\right)\cap\left(\mathbb{R}^{n\plus{}1}\backslash\{0\}\right)$.

1. If $ (a_1,b_1)\times (a_2,b_2)\times\dots (a_{n\plus{}1},b_{n\plus{}1})$ contains $ 0$ then it also contains an $ \epsilon$-ball around $ 0$ and $ U$ contains an $ \epsilon$-ball around $ 0$ (except for point $ 0$ of course). But this implies that $ \pi^{\minus{}1}\pi U$ is the entire $ \mathbb{R}^{n\plus{}1}\backslash\{0\}$ which is open.

2. If $ (a_1,b_1)\times (a_2,b_2)\times\dots (a_{n\plus{}1},b_{n\plus{}1})$ doesn't contain $ 0$ then $ 0\notin (a_1,b_1)$ and you can write $ U\equal{}\bigcup_{x\in (a_1,b_1)}x\times (a_2,b_2)\times\dots\times (a_{n\plus{}1},b_{n\plus{}1})$. So it is sufficient to prove that $ \pi^{\minus{}1}\pi\left(x\times (a_2,b_2)\times\dots\times (a_{n\plus{}1},b_{n\plus{}1})\right)$ is open for $ x\neq 0$. But now I have no idea how to go ahead.

Of course I would appreciate any other proof for the second-countability of $ \mathbb{RP}^n$.

Thanks,
Mark

Let $ X\equal{}\mathbb{R}^{n_1}\setminus 0$. Define $ \phi_t(x)\equal{}tx$. Then this is a homeomorphism, since we have the inverse $ \phi^{\minus{}1}_t\equal{}\phi_{1/t}$. Let $ U\subset X$ be open, then $ \pi(U)\equal{}[U]\equal{}\cup\phi_t(U)$ (which unions over all $ t\neq 0$. But since each $ \phi_t(U)$ is open, $ \pi(U)$ is open, thus $ \pi$ is open. (I have to run, I'll answer questions on details if needed when asked).

I think you meant $ \pi^{\minus{}1}\pi(U)\equal{}\bigcup_{t\neq 0}\phi_t(U)$, didn't you? Then it's clear. Thank you very much.

Mark


The problem is that I have access to Engel but not ACoPs and won't have access to ACoPs for about another 2 months. Can I just go ahead to engel or should I complete ACoPs before that?

Engel is much harder than ACoPS. ACoPS is for people just learning to do proof style olympiad/putnam stuff. Engel is for experienced problem solvers looking to practice on gobs of problems.

Can u score high consistently on the AIME? Don't get ahead of yourself.

I tried to do the same thing, believe me, it isn't a good idea. Read ACoPS. Check how well you do on AIME. Then proceed to Engel.

One question: why do people call it Engel instead of PSS? Or ACoPS instead of Zeitz?

Engel has plenty of (relatively) easy problems, and ACoPS has plenty of quite hard ones. Don't lose too much sleep over this - just do problems.

I never got much out of ACoPS, but I loved Engel's PSS.  PSS is singlehandedly the reason I was able to transition into Olympiad math.  To give you an idea of when it was useful to me, I started using it when I could get about a 10 on the AIME, and I stopped using it when I started to be able to solve the easiest USAMO problems.  However, PSS's problems tend to be the type of problem where there's one trick that will solve the problem, and you need to figure out what that trick is.  USAMO problems usually can't be solved by a single trick, but that's not to say that all the proof techniques and tricks you'll learn through PSS won't be very useful on USAMO problems.

ACoPS's main draw is that it attempts to teach general problem-solving technique, but I found these attempts to be not helpful at all.  They'll set you up with a problem or two where their specific problem-solving technique is gold, but you're still lost when they're not there to hand-pick the problems for you.  I think the best way to learn problem-solving is to solve problems (duh), and you've already done a lot of that, so your problem-solving ability is most likely pretty good already.

Many people may disagree with me, though - ACoPS is pretty highly regarded around here, as you seem to know.


Prove or Disprove that:
[color=red]
$ \frac {a \plus{} b^2}{b^2 \plus{} c} \plus{} \frac {b \plus{} c^2}{c^2 \plus{} a} \plus{} \frac {c \plus{} a^2}{a^2 \plus{} b} \geq 3$[/color]

Thanks !

EDIT: $ a,b,c  \geq 0$

ooops i forgot to put a, b, c are greater than 0.

[quote="dgreenb801"]It's wrong. Take a=5 b=1 c=10.[/quote]

Thanks...  :blush:

I will change it.

[color=blue]If $ a,b,c$ are sides of the triangle.[/color]
[color=red]Prove or disprove that:[/color]

$ \frac {a \plus{} b^2}{b^2 \plus{} c} \plus{} \frac {b \plus{} c^2}{c^2 \plus{} a} \plus{} \frac {c \plus{} a^2}{a^2 \plus{} b} \geq 3$

[quote="dgreenb801"]Still wrong. Take a=5 b=2 c=6[/quote]

thanks again!

Let $a,b,c>0 $ and $a+b+c=3.$ Prove that$$\frac{a + b^2}{b^2 + c} + \frac {b + c^2}{c^2 + a} + \frac {c + a^2}{a^2 + b} \geq3 $$
Equality holds when $a=b=c=1.$
Let $a,b,c>0 $ and $a+kb+c=3.$ Prove that$$\frac{a + b^2}{b^2 + c} + \frac {b + c^2}{c^2 + a} + \frac {c + a^2}{a^2 + b} \geq3 $$
Where $k\in  N^+.$Equality holds when $a=b=c=\frac{3}{k+2}.$
Let $a,b,c>0 $ and $a+k_1b+k_2c=3.$ Prove that$$\frac{a + b^2}{b^2 + c} + \frac {b + c^2}{c^2 + a} + \frac {c + a^2}{a^2 + b} \geq3 $$
Where $k_1,k_2 \in  N^+.$ Equality holds when $a=b=c=\frac{3}{k_1+k_2+1}.$


Quite often, one may want to create an internal reference to an article whose name is slightly different from the highlighted text that signifies the link. A typical example is "where $\mathbb R$ is the set of [b]real numbers[/b]" with the reference words "real numbers" (plural, no capitalization) that one wants to point to an article that is actually called "Real number" (singular, with capitalization). How can this be done?

By the way, we already have several separate articles on the same topic like [b]Prime number[/b] and [b]Prime[/b] and, I'm afraid, without setting some clear policy on naming the topics, such pairs are destined to grow like mushrooms after a rain. What should one do with them? Just move the content to one topic and set the #REDIRECT links to all other ones?

I think REDIRECT can solve both of the problems. For instance on wikipedia we see that both Primes and Prime_numbers redirect to Prime_number (Prime alone goes to a disambiguation page).

Edit: For the first question, you can also do a piped link. http://en.wikipedia.org/wiki/Wikipedia:How_to_edit_a_page#Links_and_URLs

[[real number|real numbers]] links to real_number but displays real numbers.

Edit 2: From the same link, endings are 'blended' in to a link so that [[real number]]s would look right but direct to real number. (And it says this is the preferred style)

[quote="MCrawford"]As for the "prime vs. prime number" problem, we should definitely give the most formal names possible to articles, so "prime number" is best.  However, "prime" should redirect to "prime number" because it's so often used in writing.[/quote]
 Prime is also used in other contexts. A disambig should be used instead in my opinion.

You can just write [[real number]]s and it will show up just how you want it.


I found this Vista theme on the net.  I posted the same thing on the Round Table, but then I changed it and tried to get a mod to delete it, afraid that it would be classified as spam.

See the attachment (it's a screenshot) PDF.


quadratics
conics
parabola
rotation
algebra
quadratic formula

Well, I had my math mid-term exam today and there was one question I was not sure about. This is what I recall:
I can't remember the situation or what was being thrown but I think it was a ball. So the equation that describes the movement of the ball is: [b]h=5t 2 -5t+1[/b], where h is the height in metres of the ball and t is the time in seconds it takes.
a) How long will it take the ball to be at a height of 16 metres?
b) When the ball hit the ground?

 ;) Try it

a) [hide]You have
$16=5t^2-5t+1$
$5t^2-5t-15=0$
$t^2-t-3=0$
Quadratic formula yields $\frac{1\pm\sqrt{13}}{2}$. The negative solution doesn't work because time can't be negative. Final answer: $\frac{1+\sqrt{13}}{2}$ seconds.
[/hide]
b) [hide]The ball hits the ground when h=0. So
5t^2-5t+1=0
Quadratic formula gives $\frac{5\pm\sqrt{5}}{10}$, and both those solutions are positive. But the ball only hits the ground once, right? So what am I doing wrong?
[/hide]

I think the formula in the question is a bit wrong.. a ball usually moves in an upsidedown parabola (negative something x squared), not the other way round! (the ball is going down then getting higher and higher faster and faster :))

:lol:  a positively accelerating ball under negative gravitation!!! That ball must have been repelled by gravity. What a weird planet.

[b]
Not really.
We always think that the positive direction is upward(the arrow on y-axis always points upward), but if we rotate it 180, the positive direction became downward. That is the arrow on y-axis point downward.Then the equation becomes perfect.[/b]

[b]
Hope you understand what i say.
My English is really really bad.[/b]

I'm afraid you just can't do that :)
And anyway, 'height' is how high the ball is from the ground.

Hey people
Well we got the exam back today and it WAS the teacher's mistake so never mind the question. It makes no sense... :D


Post 69 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1356495#1356495]here[/url], my bold tag at the bottom isnt working.

[b]test[/b]

Well that seems to work...

There is an unclosed tag in your last hidden part. Still the effect is somewhat unwanted (see below for a simpler example).

[hide="Nonsense"][b]Blah[/hide]

[b]Blub[/b]


ARML
MATHCOUNTS
AMC
USA(J)MO
USAMO
induction
pigeonhole principle

The eloquent insight and sheer amazing beauty recorded by the previous poster has left me dumbfounded and unable to formulate a proper response. I was taken aback by the third and fourth repetitions of the 18th letter of the alphabet, simply a WONDERFUL twist of plot. Finally, when I reached the end, I was nearly in tears over the disappearance of the main character and the shady romantic interest developed.

Thank you for enlightening my day, bookaholic. I'm now able to put off my english homework because I can claim that I have just read the most perfect thing ever to be written on the face of the planet.

*all sarcasm aside, what should we be agreeing with you on?*

Me. I had to suffer in school today because I didn't make *****.

[color=darkred]Mod edit: Dont say 01110101 01110011 01100001 01101101 01101111. No discussion until May 1st.  :lol: [/color]

Aaaaagony.....  But no one from my school took it.  SO I'm okay.
I agree anyway.  And while we're on the topic of ARML, I don't think we're allowed to discuss the 18th letter of the alphabet until school gets out in July.

For one, I agree with any and all "Grrrr"'s today...

[quote]*all sarcasm aside, what should we be agreeing with you on?*[/quote]
Something which we technically shouldn't be discussing, but as long as we stay away from details, I don't see the problem. It's all over Facebook anyway.

Also, it's not about who should agree, but who does agree. There are many people who have no right to agree with me.

Dude, one word: ****. Totally agree. (EDIT: Apologies for profanity.)

[color=darkred]Mod edit: Now that word wasn't the U-word, but it was inappropriate. Keep in mind the age level of kids that are going on here.[/color]
[color=green]Mod 2 edit: Yes, I agree the age level must be quite low on this forum.  Think of the children, guys.[/color]

[quote="bowei"]Ditto.

Btw, shame on you, bookaholic! You're a mod and you created this topic.[/quote]
I think it's useful. Some people need every chance to whine and complain that they can get. Like me. Besides, there will be actual useful discussion as soon as it's allowed.

Grrr I didn't understand some parts of the stuff so I wrote randoms stuff for last part of the stuffs.

Oh, when you say stuff, do you mean like stuff stuff or the other stuff?

to arch: YESH

and to David...
blahh too many four letter words
dangit
I came too late to see what the word was

Well, according to some people (okay, one person), doing this year's #1 using induction and not saying that x^2+x+1 and x^2-x+1 are relatively prime may merit a 3-5, since proving that there are infinite x^2+x+1's could be considered one part of the problem and proving that they're relatively prime the other part, and also the fact that the official solution doesn't prove it but merely states it in a sentence.

I hope that's the case, for my own sake, though it doesn't matter since I didn't make red MOP or anything either way :|

[quote="The QuattoMaster 6000"][quote="serialk11r"][quote="davidyko"]Well, I beg to differ on the "failing on USAMO = not making it" statement, as it's clearly false, but I will agree on giving oneself a low estimate, and I agree with your descriptions of the scores. Still not sure whether I got 2 or 6 on #4 though, so I'll go with 2 and give myself an estimate of 3. :P[/quote]
Okay maybe, but like if I can solve some USAMO problems, but didn't make it, I think of myself as better than people who fail the USAMO after making it, since IMO is the ultimate reason for USAMO, and IMO is proof stuff :P[/quote]
dude you can't say that everyone who did badly on this year's USAMO sucks
the USAMO was easy but a bunch of people messed on it
even like beashtly tesht takersh like Ubey meshed on it....
so like if you get like a ballin score in practice, it doesn't mean that you'll pwn the actual USAMO (yeah I have this through experience)
but you hopefully will :)
depends on how pressured you are

to bookaholic: I don't think you get a point for the equality...
because like
davidlizeng did the equality and he got nothing..
unless like
he got one point on 4 and one point on 6 dunno[/quote]
No no no, but what I mean is that most people getting 2s would not get any more than that in practice.

[quote]huh how do you know[/quote]
Pigeonhole Principle. I got 8 points' worth of partial credit somehow...

[quote="moogra"][quote="The QuattoMaster 6000"][quote="serialk11r"]
No no no, but what I mean is that most people getting 2s would not get any more than that in practice.[/quote]
uhh most of the people get less than 5 (points) anyways on the USAMO every year...
what's the difference for this year?[/quote]

Median was like 15 last year.[/quote]
What??? So the person ranked 200th scored 15? MOP cutoff was like 35?

Mop cutoff 35...hm...I sense sarcasm...

Is the conversation about like the difference between practice and real or something, cause I can't tell. From my experience, I do better in competition settings for some reason...There isn't much of a difference anyways in terms of score.

[quote="bookaholic"]Moogra is confused. The median was around 10.[/quote]
Hmm are you serious? Weren't there about 400 USAMO takers? 10 seems pretty high for a median score...well I don't know, I just see tons of people claiming they got 0s so...

There weren't THAT many 0s. See [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2007-ua/2007usamoscores.shtml]here[/url]; a 10 was the lowest score in the 48-75th percentile level, so we can assume that the median was about 10-11.

[quote="bookaholic"]Moogra is confused. The median was around 10.[/quote]

Remeber that list I showed you? It was 12 (at least the average)

Uh... did you actually calculate it... cause that seems improbable. The median was definitely 10, but don't forget the skewed scores: that means there are over 200 people with 10s and under, probably around 400 people with 15s and under, and then the rest of the score were spread out. The mean should be even less than the median.

I think someone did and the average or median was around that. ANYWAYS NO POINT OF FUTURE ARGUEMENT. WE HAVE ESTABLISHED THE FACT THAT THE MEDIAN IS AROUND 10.

oh dang the median is around zero

although I guess I don't really know what you're talking about

The USAMO scores from last year? The scores are on the internet.


calculus
integration
trigonometry
calculus computations

yes... but it' standart integral which easly canbe calculated by substitution:
$ \tan x \equal{} y$
$ y\plus{}2\equal{}z$
$ \sqrt z \equal{}t$

Just a question: shouldn't it be 2 - y = z?


trigonometry
ARML
geometry
analytic geometry
graphing lines
slope
Duke

AMC is coming less than two weeks so I'm going through large number of problems. Instead of just doing the problems, I decided to do them by topic by topic and share the problems here. Hopefully you guys do these as well!  :)

1. (ARML 1988) Compute the number of degrees in the smallest positive angle $ x$ such that $ 8 \sin x \cos^5 x \minus{} 8 \sin^5 x \cos x \equal{} 1$.

2. (AMC 12 2004 #21) If $ \sum^{\infty}_{n\equal{}0} \cos^{2n} \theta \equal{} 5,$ what is the value of $ \cos 2 \theta$?
A) 1/5
B) 2/5
C) $ \sqrt{5} / 5$
D) 3/5
E) 4/5

3. (AMC 12 2003 #23) The number of $ x$-intercepts on the graph of $ y \equal{} \sin\left(\frac{1}{x}\right)$ in the interval (0.0001,0.001) is closed to
A) 2900
B) 3000
C) 3100
D) 3200
E) 3300

4. (ARML 1997 I6) Compute the remainder when $ \cot^{1997} \left(\frac{\pi}{12}\right) \plus{} \tan^{1997} \left(\frac{\pi}{12}\right)$ is divided by 9.

5. (ARML 1998 I4) The sides of a non-right isosceles $ \triangle ABC$ are $ \sin x, \cos x,$ and $ \tan x$. Compute $ \sin x$.

6. (ARML 2000 I7) The measure of the vertex angle of isosceles triangle $ ABC$ is $ \theta$ and the sides of the triangle are $ \sin \theta, \sqrt{\sin \theta},$ and $ \sqrt{\sin \theta}$. Compute the area of $ \triangle ABC$.

7. (Myself) If a line makes an angle $ \theta$ with horizontal axis, express the slope in terms of $ \theta$.

8. (Myself) Prove that $ \tan \theta \equal{} \frac{m_2 \minus{} m_1}{1 \plus{} m_2 m_1}$ where $ \theta$ is the angle between the lines $ l_1$ and $ l_2,$ and $ m_1$ and $ m_2$ each denote the slopes of lines $ l_1$ and $ l_2$.

Both 7 and 8 are found in AoPS Volume II (but it wasn't any contest problem so I put "myself" as source).

9. (ARML 2001 I5) For $ m>0, f(x) \equal{} mx$. The acute angle $ \theta$ formed by the graphs of $ f$ and $ f^{\minus{}1}$ is such that $ \tan \theta \equal{} \frac{5}{12}.$ Compute the sum of the slopes of $ f$ and $ f^{\minus{}1}$.

10. (AHSME 1999 #27) In triangle $ ABC, 3 \sin A \plus{} 4 \cos B \equal{} 6$ and $ 4 \sin B \plus{} 3 \cos A \equal{} 1.$ Then $ \angle C$ in degrees is
A) 30
B) 60
C) 90
D) 120
E) 150

11. (Duke Math Meet 2002 T3) Evaluate
$ \displaystyle \frac{\tan \frac{\pi}{4}}{4} \plus{} \frac{\tan \frac{3 \pi}{4}}{8} \plus{} \frac{\tan \frac{5 \pi}{4}}{16} \plus{} \frac{\tan {7 \pi}{4}}{32} \plus{} \cdots$

12. (ARML 2004 T3) $ ABCDE$ is inscribed in a circle with $ AB \equal{} BC \equal{} CD \equal{} DE \equal{} 4$ and $ AE \equal{} 1.$ Compute $ (1 \minus{} \cos \angle B)(1 \minus{} \cos \angle ACE)$.

13. (ARML 2005 T6) For $ x$ and $ y$ in radian measure with $ 0 \leq x,y \leq 2 \pi,$ compute the largest possible value of the sum $ x\plus{}y$ if $ 2 \sin x \cos y \plus{} \sin x \plus{} \cos y \equal{} \minus{} \frac{1}{2}$.

14. (NYSML 1998 I7) If $ \sec(x) \plus{} \cos(x) \equal{} 3,$ then compute $ \frac{\sec^{10} x \plus{} 1}{\sec^5 x}$.

Since this is the first post in this thread, I have the liberty of posting the solution to the easiest questions!

[hide="1"]

$ 8\sin x\cos^5 x \equal{} 8\sin x \cos x (\cos^4 x \minus{} \sin^4 x) \equal{} 4\sin 2x (cos^2 x \minus{} sin^2 x)(cos^2 x \plus{} sin^2 x) \equal{} 4\sin 2x \cos 2x \equal{} 2 \sin 4x \equal{} \frac {1}{2}$

$ \boxed{x \equal{} \frac {15}{2}}$[/hide]

[hide="2"]

We have an infinite geometric series, so the sum is equal to $ \frac{1}{1\minus{}\cos^2 x} \equal{} 5 \rightarrow \sin^2 x \equal{} \frac{1}{5}$.  Finding $ \cos^2 x$ using a triangle, we get the answer $ \cos^2 x \minus{} \sin^2 x \equal{} \boxed{\frac{3}{5}}$[/hide]

[hide="#3"]This is the same as the number of intercepts to $ y \equal{} \sin{x}$ on $ (1000,10000)$
Every time $ x$ is an integer multiplied by $ \pi$, the RHS equals $ 0$, so there are about $ \frac {10000 \minus{} 1000}{\pi}$ solutions, which is closest to $ 2900$, so the answer is A.[/hide]

[hide="7"]Well to find the slope of the line, you would do rise/run and since the rise/run in the triangle created is simply the opposite/adjacent sides, which would make the slope, by definition $ \boxed{\tan{\theta}}$  :blush:, i dunno.[/hide]

[hide="5"]
If $ \sin\equal{}\cos$, then both are $ \frac{\sqrt{2}}{2}$, and $ \tan$ is $ 1$, making the triangle right.

If $ \sin\equal{}\tan$, then $ \sin\equal{}\frac{\sin}{\cos}$, so $ \cos$ is $ 1$ and $ \sin$ is $ 0$, which is impossible in a triangle.

If $ \cos\equal{}\tan$, then $ \cos\equal{}\frac{\sin}{\cos}$, and $ \cos^2\equal{}\sin$, so $ \sin^2\plus{}\sin\minus{}1\equal{}0$.  The solution is $ \sin\equal{}\frac{\sqrt{5}\minus{}1}{2}$.
 [/hide]

I get number 12 to be 1/64 but I dont get what the point of it inscribed in a circle is so I think I did it wrong.

I just used law of cosines..

number 9 i guess  you can use that tan forumla in problem 7 but theres a lot easier way to do it with triangles-i get like 13/6.

number 6 is laws of cosines and i get like 8/25..

number 10 is the backwords of the sum to product formulas.

i should be posting solutions but i have to eat lunch..ill try later if i can.

[hide="6"]Well we do know that the are of a triangle, with sides a and b and included angle $ \theta$, is $ \frac{1}{2}$ $ ab$ ${ \sin{\theta}}$. so if we substitute the given information, we can say that the area of the triangle is

$ \frac{1}{2}$ $ {\sqrt{\sin{\theta}}}^2$ $ \times{\sin{\theta}}$

= $ {\boxed{1/2 {\sin^2(\theta)}}}$, im probly wrong, was it supposed to be a numerical answer, if so, someone can finish it from here. [/hide]

[hide="4"]
$ \tan\frac {\pi}{12} \equal{} 2 \minus{} \sqrt {3}$, $ \cot\frac {\pi}{12} \equal{} 2 \plus{} \sqrt {3}$.
Let $ x \equal{} (2 \minus{} \sqrt {3})^{1997}$, and let $ y \equal{} (2 \plus{} \sqrt {3})^{1997}$.
$ x \equal{} 2^{1997} \minus{} 1997(2^{1996})(\sqrt {3}) \plus{} \binom{1997}{2}(2^{1995})(3) \minus{} \binom{1997}{3}(2^{1994})(3\sqrt {3}) \plus{} \ldots$.  The rest of the terms are of the form $ 9n$, so we can ignore them.
$ y \equal{} 2^{1997} \plus{} 1997(2^{1996})(\sqrt {3}) \plus{} \binom{1997}{2}(2^{1995})(3) \plus{} \binom{1997}{3}(2^{1994})(3\sqrt {3}) \plus{} \ldots$.  The rest of the terms are of the form $ 9n$, so we can ignore them.

Note that when we add $ x \plus{} y$, the terms with $ \sqrt {3}$ cancel out.  So we are left with $ 2\left(2^{1997} \plus{} \binom{1997}{2}(3)(2^{1995})\right)$.  Noting that the residues of $ 2^n$ in $ \mod{9}$ are 2-4-8-7-5-1..., we see that $ 2^{1997}\equiv 5 \pmod{9}$ since $ 1997\equiv 5\pmod{6}$.  Similarly, $ 2^{1995}\equiv 8\pmod{9}$ since $ 1995\equiv 3\pmod{6}$.

Next, $ \binom{1997}{2} \equal{} \frac {1997\cdot1996}{2}\equiv\frac {8\cdot 7}{2} \equal{} 28\equiv 1\pmod{9}$.

Therefore $ 2\left(2^{1997} \plus{} \binom{1997}{2}(3)(2^{1995})\right)\equiv 2\left(5 \plus{} (1)(3)(8)\right) \equal{} 2(29) \equal{} 58\equiv \boxed{4}\pmod{9}.$[/hide]

[hide="14"]Let $ n \equal{} \sec x$.  We want to find $ \frac {n^10 \plus{} 1}{n^5} \equal{} \frac {n^10}{n^5} \plus{} \frac {1}{n^5} \equal{} n^5 \plus{} n^{ \minus{} 5}$.
We are told that $ n \plus{} n^{ \minus{} 1} \equal{} 3$.
$ (n \plus{} n^{ \minus{} 1})^3 \equal{} n^3 \plus{} 3n \plus{} 3n^{ \minus{} 1} \plus{} n^{ \minus{} 3} \equal{} n^3 \plus{} 3(n \plus{} n^{ \minus{} 1}) \plus{} n^{ \minus{} 3} \equal{} n^3 \plus{} n^{ \minus{} 3} \plus{} 3(3) \equal{} 27\Rightarrow n^3 \plus{} n^{ \minus{} 3} \equal{} 18$.
$ (n \plus{} n^{ \minus{} 1})^5 \equal{} n^5 \plus{} 5n^3 \plus{} 10n \plus{} 10n^{ \minus{} 1} \plus{} 5n^{ \minus{} 3} \plus{} n^{ \minus{} 5} \equal{} n^5 \plus{} 10(n \plus{} n^{ \minus{} 1}) \plus{} 5(n^3 \plus{} n^{ \minus{} 3}) \plus{} n^{ \minus{} 5} \equal{} n^5 \plus{} n^{ \minus{} 5} \plus{} 10(3) \plus{} 5(18) \equal{} 243\Rightarrow n^5 \plus{} n^{ \minus{} 5} \equal{} \boxed{123}$.[/hide]

[quote="cognos599"][hide="6"]Well we do know that the are of a triangle, with sides a and b and included angle $ \theta$, is $ \frac {1}{2}$ $ ab$ ${ \sin{\theta}}$. so if we substitute the given information, we can say that the area of the triangle is

$ \frac {1}{2}$ $ {\sqrt {\sin{\theta}}}^2$ $ \times{\sin{\theta}}$

= $ {\boxed{1/2 {\sin^2(\theta)}}}$, im probly wrong, was it supposed to be a numerical answer, if so, someone can finish it from here. [/hide][/quote]

so then 2sin(theta)-2(sin(theta))(cos(theta))=sin^2(theta).  use that cos^2+sin^2=1, substitute that in and we get sin^2(theta)-2sin(theta)=-2sin(theta)(sqrt(1-cos^2(theta))).  cancel out a sin(theta) since theta is not going to be 0, we get sin(theta)-2=-2sqrt(1-sin^2(theta)).  then square both sides we end up getting after combiniting terms 5sin^2(theta)=sin(theta), again cancel out a sin(theta) to get that sin(theta)=4/5.  Make sure that works in your initial equation, it does so its using your forumla 1/2(16/25)=8/25...i need to learn latex...just try working it out with laws of cosines since this solution is really hard to read.

[hide="14"]Really messy, well with the given equation, we can say that $ \cos{x} + \frac{1}{\cos{x}}$ = $ 3$, now if we let $ \cos{x} = a$, then we get a simple quadratic equation, $ a^2 - 3a + 1 = 0$, now if we solved that equation, we get $ a$ = ${{ (\frac{3 +\sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}}})$.

now $ \frac{\frac{1}{a^{10}} + 1}{a^5}$ = $ \frac{a^{10} + 1}{a^5}$,

so now you substitute, and you get $ (\frac{3 + \sqrt{5}}{2})^{10} +1$, for now i will leave out the denominator so i can simplify this.

now i used a calculator to plug-n-chug. So for the numerator i get, $ \frac{6765\sqrt{5} + 15127}{2}$, and when i divide that by $ (\frac{3 + \sqrt{5}}{2})^5$, i get $ \fbox{123}$, lingomac's method was probably more elegant, seeing as this problem probably doesn't require the use of a calcuator.[/hide]

Is the one with tangents 1/6 (#11)?  IOf it is ill post a solution

You were not allowed to use a calculator for that problem.

[hide="9"]
Let $ \alpha$ be the angle formed by $ f(x)$ and the $ x$-axis.  Therefore $ m=\tan\alpha$.  It follows that this is also the angle formed by $ f^{-1}(x)$ and the $ y$-axis.  Using the fact that $ f^{-1}(x)=\frac{1}{m}x$, We have:
\begin{eqnarray*}
\frac{\pi}{2}-2\alpha &=& \theta \\
\tan\left(\frac{\pi}{2}-2\alpha\right) &=& \tan\theta \\
\cot 2\alpha &=& \frac{5}{12} \\
\tan 2\alpha &=& \frac{12}{5} \\
\frac{2\tan\alpha}{1-\tan^2\alpha} &=& \frac{12}{5} \\
\frac{m}{1-m^2} &=& \frac{6}{5} \\
5m &=& 6-6m^2 \\
6m^2+5m-6 &=& 0 \\
m &=& \frac{2}{3} \\
m+\frac{1}{m} &=& \boxed{\frac{13}{6}}
\end{eqnarray*}
[/hide]

[hide="9"]
By $ \tan \theta = \frac{m_1 - m_2}{1 + m_1m_2}$, we have \[ \tan \theta = \frac{m - \frac 1m}{1 + 1} = \frac{1}{2}\left(\frac{m^2-1}{m}\right) = \frac {5}{12} \Longrightarrow 6m^2 - 5m - 6 = (2m-3)(3m+2) = 0\]. Thus the absolute value of the answer is $ \frac{13}{6}$ as above.
[/hide]

[hide="10"]
Squaring,
\begin{align*} 9\sin^2 A + 16 \cos^2 B + 24\sin A \cos B & = 36 \\
9\cos^2 A + 16 \sin^2 B + 24\sin B \cos A & = 1 \end{align*}
Summing and using the Pythagorean identity gives $ 24(\sin A \cos B + \sin B \cos A) = 37 - 25 = 12,$ and $ \sin (A + B) = \sin C = \frac 12$. Thus $ C = 30$ or $ 150$.

Unfortunately, both are choices, but notice that if $ C = 150 \Longrightarrow A + B = 30$, then $ 3\sin A + 4\cos B < 3\sin (30) + 4\cos(0) = 5.5$ (using $ \sin$ strictly increasing, $ \cos$ strictly decreasing on the interval $ 0 \le \theta \le 30$), contradiction. Hence the answer is $ C = 30 \mathrm{(A)}$.
[/hide]

[hide="11"]
I’m surprised that nobody has posted this yet; it is just
\[ \frac {1}{4} - \frac {1}{8} + \frac {1}{16} + \cdots = \frac {1/4}{1 + \frac {1}{2}} = \frac {1}{6}
\]
[/hide]

[hide="12"]
By Law of Cosines on $ \triangle ABC, ACE$ (note $ AC = CE$), we have
\begin{align*} AC^2 & = 4^2 + 4^2 - 2 \cdot 4 \cdot 4 \cos \angle B \\
1 & = AC^2 + AC^2 - 2 \cdot AC \cdot AC \cos \angle ACE \end{align*}
Solving, we have $ \cos \angle B = 1 - \frac {AC^2}{32}$ and $ \cos \angle ACE = 1 - \frac {1}{2AC^2}$. Thus the product $ (1 - \cos B)(1 - \cos ACE) = \frac {1}{64}$.
[/hide]

[hide="13"]
After multiplying out by $ 2$, this is factorable as
\[ (2\sin x + 1)(2\cos y + 1) = 0
\]
We just need one quantity to be zero. The maximum possible value of $ x$ such that $ \sin x = \frac { - 1}{2}$ is $ \frac {11\pi}{6}$, where as for $ y$ it is $ \frac {2\pi}{3}$. Since $ \frac {11\pi}6 > \frac {2\pi}{3}$, we let $ x = \frac {11\pi}{6}, y = 2\pi$ to get the maximal sum of $ \frac {23\pi}{6}$.
[/hide]

[hide="8"]
Simple. Use tangent angle addition formula.

Let the angle between $ l_1$ and the horizontal be $ \theta_1$, and similarly for $ l_2$.

WLOG, let $ \theta_2 > \theta_1$.
Then, since $ \theta \equal{} \theta_2 \minus{} \theta_1$,

$ \tan(\theta \pm \varphi) \equal{} \frac{\tan \theta \pm \tan \varphi}{1 \mp \tan \theta \tan \varphi}$
Substitute in our variables,
$ \tan\theta \equal{} \tan(\theta_2 \minus{} \theta_1) \equal{} \frac{\tan \theta_2 \minus{} \tan \theta_1}{1 \plus{} \tan \theta_2 \tan \theta_1}$
and then use $ \tan\theta_1 \equal{} m_1$ and likewise for $ \theta_1$ and $ m_2$ to get
$ \tan \theta \equal{} \tan(\theta_2 \minus{} \theta_1) \equal{} \frac{m_2 \minus{} m_1}{1 \plus{} m_2 m_1}$
[/hide]

I didn't look through all of your solutions but it seems that you guys had most of them right.

[hide="Comment for lingomaniac88 and his Number 4"]
For those who might not understand how he came up with $ \tan \frac{\pi}{12}$ and $ \cot \frac{\pi}{12},$ just use tangent addition/subtraction formula with $ \frac{\pi}{3}$ and $ \frac{\pi}{4}$. [/hide]

And yes, all of problems in any AMC Practice should be solved WITHOUT calculator because that's the whole point of it.


Let a,b,c>0
prove that
\[ \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \ge 3\sqrt {\frac{{\mathop a\nolimits^2  \plus{} \mathop b\nolimits^2  \plus{} \mathop c\nolimits^2 }}{{ab \plus{} bc \plus{} ca}}}
\]

after clearing denominators it's equivalent to prove
$ \sum_{cyc}a^5bc^2\plus{}\sum_{cyc}a^4bc^3\plus{}\sum_{cyc}a^5c^3\plus{}3\sum_{cyc}a^4b^3c\plus{}2\sum_{cyc}a^3b^3c^2\plus{}2\sum_{cyc}a^4b^2c^2 \geq 9\sum_{cyc}a^4b^2c^2$
but it's obviously true by am-gm:$ a^5bc^2\plus{}a^4bc^3\plus{}a^5c^3\plus{}2a^4b^3c\plus{}2a^3b^3c^2\plus{}2a^4b^2c^2 \geq 9a^4b^2c^2$

Another solution:squaring both side;It's equivalent to prove that
\[ \sum_{cyc}\frac{a^2}{b^2}\plus{}2\sum_{cyc}\frac{a}{c} \geq \frac{9(a^2\plus{}b^2\plus{}c^2)}{ab\plus{}bc\plus{}ca}\]
\[ \leftrightarrow (ab\plus{}bc\plus{}ca)(\sum_{cyc}a^4c^2\plus{}2\sum_{cyc}a^3b^2c) \geq 9(a^2b^2c^2)(a^2\plus{}b^2\plus{}c^2)\]
which is true by $ ab\plus{}bc\plus{}ca \geq 3\sqrt[3]{a^2b^2c^2}$ and $ \sum_{cyc}a^4c^2\plus{}2\sum_{cyc}a^3b^2c \geq 3\sqrt[3]{a^4b^4c^4}(a^2\plus{}b^2\plus{}c^2)$


Suppose that $n$ is a positive integer and let \[d_{1}<d_{2}<d_{3}<d_{4}\] be the four smallest positive integer divisors of $n$. Find all integers $n$ such that \[n={d_{1}}^{2}+{d_{2}}^{2}+{d_{3}}^{2}+{d_{4}}^{2}.\]

If n is odd we get contradition $ d_1^2\plus{}d_2^2\plus{}d_3^2\plus{}d_4^2$ - is even. Therefore $ d_1\equal{}1,d_2\equal{}2$.
Because $ n>d_4^2$ n had $ k\ge 8$ divisors and $ d_{k\plus{}1\minus{}4}\equal{}\frac{d}{d_4}\equal{}d_4\plus{}\frac{d_3^2\plus{}5}{d_4}\le 3d_4\minus{}5.$ Therefore $ n\le d_4(3d_4\minus{}5).$
If $ m$ number of prime divisors equal 1 $ d_{k\minus{}3}$ is not integer, therefore $ m\ge 2$.
If $ m\equal{}p_1^{k_1}p_2^{k_2}$. It is easy to chek, that for small prime $ p_1<p_2$ degree $ k_1\ge 2$.
If $ p_2<p_1^2$ we get $ d_2\equal{}p_1,d_3\equal{}p_2,d_4\equal{}p_1^2|p_2^2\plus{}1\to p_1>2.$ It give contradition ($ 1\plus{}d_1^2\plus{}d_2^3\plus{}d_4^2$ is even). If $ p_2>p_1^2$ we get $ d_2\equal{}p_1,d_3\equal{}p_1^2,d_4\equal{}p_2|1\plus{}p_1^2\plus{}p_1^4\equal{}(p_1^2\plus{}1\minus{}p_1)(p_1^2\plus{}p_1\plus{}1).$
Therefore $ p_2\equal{}p_1^2\plus{}p_1\plus{}1$ and $ n\equal{}1\plus{}p_1^2\plus{}p_1^4\plus{}p_2^2\equal{}p_2(p_2\plus{}p_1^2\minus{}p_1\plus{}1)\equal{}2p_2(p_1^2\plus{}1)$ - contradition. Therefore $ m\ge 3$. If $ m>3$ we get contradition $ p_1p_2p_3p_4\ge n\equal{}d_1^2\plus{}d_2^2\plus{}d_3^2\plus{}d_4^2\le 1\plus{}p_1^2\plus{}p_2^2\plus{}p_3^2<3p_3^2$.
If $ n\equal{}2^{k_1}p_2^{k_2}p_3^{k_3}$ by these whay we get $ k_3\equal{}1$. Chek by mod 4 give, that $ k_1\equal{}1$.
Therefore $ n\equal{}2*p^kq$ p<q odd prime divisors. If $ q>2p$, then $ n\equal{}1\plus{}2^2\plus{}p^2\plus{}(2p)^2\equal{}5(p^2\plus{}1)\to p\equal{}5,q\equal{}13$ solution unique $ n\equal{}130$, because if $ p<q<2p$ from $ 2p^kq\equal{}5\plus{}p^2\plus{}q^2<5p^2$ had not solution.


geometry
angle bisector
geometry proposed

Given a triangle $ ABC$ with $ B$ obtuse and $ A < 60$ degrees. $ P$ is located on $ AC$ such that angle $ CPB \equal{} 60$ degrees. Also , the internal angle bisector of $ A$ passes through $ CP$ at a point $ D$ such that angle $ CBD\equal{}30$ degrees.. Prove that $ CR$ trisects angle $ ACB$.


Consider the smallest number in each of the $^nC_r$ subsets of size $r$ of S = {1,2,3...n}. Show that the arithmetic mean of the numbers so obtained is $\frac{n+1}{r+1}$.

[hide]Let $i$ be a natural number, the number of subsets of size $r$ which smallest element is $i$(ie.the number of times that $i$ appear) is $^1C_1 \cdot ^{n-i}C_{r-1}$ where $(n-i)$ is the number of elements that are greater then $i$, consequently, the sum of all smallest elements of each subsets of size $r$ of S = {1,2,3...n} for all $1 \le i \le n-r+1$ is given by,

$\sum_{i=1}^{n-r+1}i \cdot ^{n-i}C_{r-1}=\sum_{j=1}^{n-r+1} \sum_{i=j}^{n-r+1}\ ^{n-i}C_{r-1}$ but Stiffel's theorem state that $\sum_{i=n}^{n+m}^{i}C_{n}=^{n+m+1}C_{n+1}$ so,

$\sum_{j=1}^{n-r+1} \sum_{i=j}^{n-r+1}\ ^{n-i}C_{r-1}=\sum_{j=1}^{n-r+1}^{n-j+1}C_{r}=^{n+1}C_{r+1}$ therefore the arithmetic mean is,

$\frac{\sum_{i=1}^{n-r+1}i \cdot ^{n-i}C_{r-1}}{\sum_{i=1}^{n-r+1} ^{n-i}C_{r-1}}=\frac{^{n+1}C_{r+1}}{^{n}C_{r}}=\frac{n+1}{r+1}$[/hide]


Solve in $ \mathbb{N}$ this equation :

$ (x^2\plus{}2)(y^2\plus{}3)(z^2\plus{}4)\equal{}60xyz$

Already posted here:

http://www.mathlinks.ro/viewtopic.php?p=839754#839754


function
topology
Ring Theory
superior algebra
superior algebra unsolved

Suppose that $ R$ is the ring of continuous real-valued functions on $ [0,1]$.
1)What are maximal ideals of $ R$?
2)What are prime ideals of $ R$?

First thought: The ideals I of $ C([0,1])$ are in bijection with the filters $ \phi$ over $ [0,1]$. The inverse maps should be:
$ \phi\mapsto (\chi_{A}| A\in\phi)$ and $ I\mapsto\{A\subseteq [0,1] | \exists f\in I: A=f^{-1}(A)\}$.

These maps are inclusion perserving, so the maximal ideals correspond to the ultrafilters. Not a very good description of the maximal ideals, but I think there's no really good way to do it, since there are some very "ugly" ultrafilters which are constructed with the AC (and only with the AC, the situation is similar to the theorem "every vectorspace has a basis")

The description of the prime ideals might indeed run into such trouble, but this is not the case for maximal ideals: they are precisely the sets of functions vanishing in $ x$, for various points $ x\in[0,1]$. In fact, this extends to rings of continuous functions on any compact Hausdorff space.

[quote="grobber"] they are precisely the sets of functions vanishing in $ x$, for various points $ x\in[0,1]$. [/quote]
Why? Please post a proof. Thanks!

obviously, these ideals are maximal, because evaluation at $ x$ shows that their factor ring is isomorphic to $ \mathbb{R}$.

for the converse, take a maximal ideal $ I$ of $ C(X)$ where $ X$ is a compact space. if for every $ x \in X$ there is some $ f \in I$ s.t. $ f(x) \neq 0$, then the $ \{x : f(x) \neq 0\}$ yield an open cover of $ X$. thus there are $ f_{1}, ... , f_{n}\in I$, s.t. $ X$ is covered by the $ \{x : f_{i}\neq 0\}$. but then $ f_{1}^{2}+...+f_{n}^{2}$ lies in $ I$ and has no zeroes, so that $ 1 \in I$, a contradiction.
so there is a $ x \in X$ s.t. $ f(x)=0$ for all $ f \in I$. then $ I$ lies in the maximal ideal of functions vanishing at $ x$. because of maximality, they are equal. qed.

by the way, one can also show that the [i]closed[/i] ideals of $ C(X)$ are in one-to-one correspondence with the closed subsets of $ X$. I've also read (some time ago with google) that there's a complete classification of the prime ideals, but I think it's very complicated.

@gockel: $ C(X)$ is completely different from $ \mathbb{R}^{X}$ ...

[quote="-oo-"]obviously, these ideals are maximal, because evaluation at $ x$ shows that their factor ring is isomorphic to $ \mathbb{R}$.

for the converse, take a maximal ideal $ I$ of $ C(X)$ where $ X$ is a compact space. if for every $ x \in X$ there is some $ f \in I$ s.t. $ f(x) \neq 0$, then the $ \{x : f(x) \neq 0\}$ yield an open cover of $ X$. thus there are $ f_{1}, ... , f_{n}\in I$, s.t. $ X$ is covered by the $ \{x : f_{i}\neq 0\}$. but then $ f_{1}^{2}+...+f_{n}^{2}$ lies in $ I$ and has no zeroes, so that $ 1 \in I$, a contradiction.
so there is a $ x \in X$ s.t. $ f(x)=0$ for all $ f \in I$. then $ I$ lies in the maximal ideal of functions vanishing at $ x$. because of maximality, they are equal. qed.

[/quote]
You are true, thank you very much :) I am reading a proof of 1) in a lecture of Greenberg, Ralph . When I see at 1) I think to 2) .


hi :)

a,b,c are positif real numbers, prove or give a counterexemple:

$ \frac{a}{{\sqrt {b^2  \plus{} c^2 } }} \plus{} \frac{b}{{\sqrt {a^2  \plus{} c^2 } }} \plus{} \frac{c}{{\sqrt {a^2 \plus{}b^2 } }} \geq \frac{{3\sqrt 2 }}
{2}$

Put a^2=x,b^2=y, c^2=z and wlog x+y+z = 1

The LHS reduces to: sum(sqrt(x/1-x))

[quote="arqady"][quote="anas"]hi :)

a,b,c are positif real numbers, prove or give a counterexemple:

$ \frac {a}{{\sqrt {b^2 \plus{} c^2 } }} \plus{} \frac {b}{{\sqrt {a^2 \plus{} c^2 } }} \plus{} \frac {c}{{\sqrt {a^2 \plus{} b^2 } }} \geq \frac {{3\sqrt 2 }} {2}$[/quote]
Try $ a \equal{} b \equal{} 1$ and $ c \equal{} 0.2.$ :wink:[/quote]

thank you arqady

I think it must be like this:
For all $x,y,z>0$
$ \frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{x^{2}+z^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}}>2 $
It's proof is there http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=421480&p=2381196#p2381196


After a roughly 50% chance of a safe landing, the Phoenix spacecraft touched down successfully on the surface of Mars. It landed in the northern polar region and will search the soil and permafrost there for signs of life in the past or the present.

[url=http://www.cnn.com/2008/TECH/space/05/25/mars.lander/index.html]Here[/url] is the full story on CNN.com.

Personally, I hope the lander finds signs of life there, at least past life. Not only would it be something cool, but it would also be a huge paradigm shift for people as a whole. Just knowing that there's the possibility of something else out there right NOW could make us "Earthlings" look at ourselves in a different light.

Awesome thread!  :lol:

Agree that if life is found people will start to look at their "position" in the universe differently.  Regardless if some psychopath creates a cult worshiping these lifeforms or not, it will has an effect, especially for the scientific organizations.

Speaking of Martian life, what do you guys think of the evidence for life on Martian meteorites?  Since the SEM blobs on Martian meteorites are smaller than normal bacteria, the arguments revolve around whether you believe that the "nanobacteria" researchers claim to have found are really life.

It seems ironic to me that we are sending probes to Mars to look for the existence of life when we can't agree on whether the closest Earth analogues to Martian life are really alive!

If you look at the traveling with robots NASA webpage, they got a picture of it coming down with the parachute deployed.  Not a great picture, but amazing

Update: The Phoenix mission has been extended until September 30th. Phoenix's laboratory tests have identified water in a soil sample! :D

[url=http://phoenix.lpl.arizona.edu/]Here's[/url] the full story from the Phoenix Mars Mission's website.

hey anyone joining the carl sagan contest in exobiology?
and also astrochemistry?


AMC
AIME
Pythagorean Theorem
geometry
power of a point

14.  A unicorn is tethered by a rope length 20 to the base of a cylindrical tower. The rope is attached to the tower at ground level and to the unicorn at height 4 and pulled tight. The unicorn's end of the rope is a distance 4 from the nearest point of the tower. Find the length of the rope which is in contact with the tower.

Does anyone know how to start this problem...

A hint: [hide]Think helix -- try (a) drawing your situation and (b) unwrapping the cylinder.[/hide]

is there a certain formula involving helix?

What is easier to deal with other than the helix?

Spoiler: [hide]Can't you just find the length of the rope thats not touching the tower using power of a point and then the pythagorean theorem?[/hide]

Ian -- no, because the rope is at an angle, and you don't know how much of an angle.  Or, at least, what you said could work, but there needs to be plenty of other stuff done first.

Correct me if I'm wrong, but this question seems not to have enough information as stated.  Changing the radius of the cylinder should alter the result, should it not?

The problem has all the information that are needed to solve the problem. By the way how does radius = 8?

Since the radius is given, just pythag a few things first. :-)

I think it's a lot easier with a radius given...no?

(A solution is in the latest Math Jam transcript)

Easier in that it is possible to find an answer, yes   The question isn't solvable without it.  (Well, I suppose you could solve it with r as a variable, but whatever.)  Anyhow, the observation to make is that you are [hide]wrapping a triangle around a cylinder[/hide].

I don't understand what's actually happening in this question.  Anyone want to describe the picture to me?

Imagine a triangle being wrapped around a cylinder.  The rope is taught, and part of it is against the cylinder and part of it isn't.


I beat you 8 out of 10 at Trivial Pursuit at TIP.  That's 80%, in case you couldn't figure that out.  What's your AoPS screename?

He doesn't, except he probably saw that testimonial Neal Wu gave about the aops books.

[quote="joml88"]The one smack dab in the middle of [url=http://www.artofproblemsolving.com/Books/AoPS_B_About.php]this[/url] page.[/quote]

Wow, what a testimonial it is. Right up and center on the AoPS books page. It'll probably help them sell a lot, too :P .

[quote="math92"]if you want to find out if he has a username then email him or talk to him, or call him on the phone.[/quote]

How are we supposed to know his email or phone number?

[quote="noneoftheabove"]i have his email but im not sure he will want me to give it out...ill ask him[/quote]

(not that I really care)

But, it would probably be best if you just tell him to come to this thread..

Neal Wu's e-mail was part of the long list that was emailed by solafidefarms to most MC competitors.

It was?! I don't remember getting his... Oh yeah, that was at breakfast the last day...  :oops:


this is a fairly easy problem from IRMO.
if $a=10^{2*1376}-10^{1376}+1$ then which one of the following is the value for the $[2\sqrt{a}]$?
($[x]$ is the biggest integer, which is equal or less than $x$)
$A:2*10^{1376}-2,        B:2*10^{1376}-1,        C:2*10^{1376},    D:2*10^{1376}+1,     E:\text{none of the above}$

i'm not gonna say whether it is right or not, :D  cause it is multiple choice, but why don't you show some proof.. :lol:

I meant in the original equation, guess not.

no the original equation for $a$, is fine.. :D

[hide="solution"]
[b]B
[/b]
let 101376=x
x2-x+1=a
a=(x-1\2)2+3/4
:sqrt:a =x-1\2 plus a very small and insignificant number.
2:sqrt:a =2x-1 plus a very small and insignificant number.
greatest integer of this value is 2*101376-1
[/hide]

[hide]
I'm leaning towards b, although my solution isn't good

$2\sqrt a$ is same as $\sqrt{4a}$, and completing the square, we get
$\sqrt{4(x-\frac 1 2 )^2 + 3}$ where $x=10^{1376}$.
Because x is such a ridiculously large number, I kinda disregarded the 3, and got
$2x-1$ or $2*10^{1376}-1$
[/hide]


calculus
function
trigonometry
calculus computations

$ z \equal{} \frac {(x\plus{}y)(1\minus{}xy)}{(1 \plus{} x^2)(1 \plus{} y^2)}$

Therefore find maximum possible vale of $ z$.

This is a multi-valued function so take the gradient.  Then set each of the components of the gradient and set them equal to 0 to determine any possible critical points.  Then use the Hessian to determine the nature of these points.  The function is ugly, but taking the gradient is no different from taking an ordinary derivative.

hello, we have $ \frac{(x\plus{}y)(1\minus{}xy)}{(1\plus{}x^2)(1\plus{}y^2)}\le \frac{1}{2} \Leftrightarrow \frac{(\minus{}1\plus{}x\plus{}y\plus{}xy)^2}{2(1\plus{}x^2)(1\plus{}y^2)}\ge 0$.
Sonnhard.

Sorry,
You could do it in a lot easier way than the mile long solution you suggested.
It goes thus with the convenient substitution:
$ x \equal{} \tan \alpha$
$ y \equal{} \tan \beta$
Now proceed and you'll get $ z \epsilon [\minus{}\frac {1}{2} , \frac {1}{2}]$


In a pyramid with base $B$ and apex $A$, where $B$ is any convex polygon, do the centroids of the all triangles that make up the sides of the pyramid lie in the same plane?

If the polygon is regular, then I believe they would. If not, I don't think it's guaranteed.


ratio
geometry
geometric transformation
rotation
geometry solved

Let $ P$ be a point inside a triangle $ \triangle ABC$ such that $ \angle PBA \equal{} \angle PCA$. Let $ E$, and $ F$ be the feet of the perpendiculars from $ P$ to $ AB$ and $ AC$ respectively. Let $ D$ be the midpoint of the side $ BC$. Show that $ \triangle EDF$ is isosceles with $ DE \equal{} DF$.

Yes, some transformations help.

$F$ is taken to $P$ by the spiral similarity centered at $C$, of angle $\angle PCA$ and ratio $\frac{PC}{FC}$. After this, $P$ is taken to $E$ by the spiral similarity centered at $B$, of angle $\angle PBA=\angle PCA$ and ratio $\frac{EB}{PB}=\frac 1{\frac{PC}{FC}}$. When you compose the second transformation with the first one, you get a rotation, because the ratios cancel out. It means that $F$ is taken to $E$ by a rotation having as a center the fixed point of the composition of the two transformations mentioned above. Now show that $D$ is a fixed point, so it must be the center of that rotation.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=16761 .

  darij

Wow, thanks a lot, very nice solution grobber  :D

Let the midpoints of $BP$ and $CP$ be $U$ and $V$. Then: $\triangle{FVD}\cong \triangle{EUD}$.


algebra
polynomial
geometric sequence
number theory unsolved
number theory

Hey Mathlinkers,

A problem that all of you know too well (first from Kvant, then Bulgarian 2003 MO, and then again from the MathLinks [url=http://www.mathlinks.ro/Forum/contest/4th%20Edition/4th%20edition%20-%20round%201.pdf]contest[/url] in a somewhat stronger form) is this: if $f \in \mathbb{Z}[x]$ is an integer polynomial such that the set $f(\mathbb{Z}) := \{f(n) \mid n \in \mathbb{Z}\}$ contains an infinite geometric progression, then $f(x)$ is the power of a linear polynomial. Here is a generalization of this. I can prove it using the Chebotarev Density theorem, but I put it in the 'Unsolved' section because I don't know an elementary solution.

[b]Problem. [/b] Let $f \in \mathbb{Z}[x]$ be an integer polynomial. Assume that for every $n \in \mathbb{N}$ there exists an integer $x_n \in \mathbb{Z}$ such that $f(x_n) $ is a perfect $n$th power different from $0$ and $\pm 1$. Prove that $f(x) = (ax+b)^k$ for some $a,b \in \mathbb{Z}, k \in \mathbb{N}$.

--Vesselin

there is similar problem on forum
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=i+like+it&t=6554


A sequence is defined as follows:
$ a_1 \equal{} 1$, $ a_2 \equal{} 2$ ($ a_i$ is the $ i^{th}$ term, of the sequence)
Each term of the sequence is equal to the average of the previous two terms. (for third term onwards)

As $ n$ tends to infinity, $ a_n$ tends to _  ?

[hide="P.S."]
how to type a subscript n-1 etc. in latex?[/hide]

Assuming i did not do any stupid mistakes

The formula for the recursion is $ a_n \equal{} 5/3 \plus{} (\minus{}1/2)^n$

So if $ n$ tends to infinity then $ a_n$ tends to $ 5/3$.

[hide]I knew 60% it was u keyree but i had a doubt of 40% so i was scared :rotfl: .

Good to see u and arjun on this forum. [/hide]

Please explain the recursion thing  :huh:

The result of $ a_n$ is incorrect, but the limit is correct.

MERCY! Explain before my poor head bursts!  :P

$ T_z \equal{} (T_x \plus{} T_y)/2$
(y=x+1, z=y+1)
$ T_z \minus{} T_y \equal{} \minus{} (T_y \minus{} T_x)/2$
so each diff. is -1/2 times previous difference of terms.
so the seq. goes like 1, 2, 2 - 1/2 (=3/2), 3/2 + 1/4 (=7/4), prev. term - 1/8... and so on.
we can use infinite GP summation formula to find the value to which the sequence converges.

[hide]
I still dunno how to type A subscript (n-1)
PLEASE HELP ME !!!!!  :(
[/hide]

Subscript $ n\minus{}1$ for $ a$ :
Type a_{n-1} within latex.
Now on typing it comes like :
$ a_{n\minus{}1}$.

OOPs i just posted the answer wrongly y mistake.

For linear recursions of the form

$ x_n \equal{} ax_{n\minus{}1} \plus{} bx_{n\minus{}2}$

solve the quadratic $ x^2 \minus{} ax \minus{} b$ to get the roots

If l and m are the roots then we can write

$ x_n \equal{} s* l^n \plus{} t*m^n$

where s and t are real number constants.


[quote="czczczczczcz1991"]LET [b]a[/b],[b]b[/b],[b]c[/b]>0 and [b]a[/b]^4+[b]b[/b]^4+[b]c[/b]^4=3   :maybe:  prove the inequality:
  1/(4-[b]ab[/b])+1/(4-[b]bc[/b])+1/(4-[b]ac[/b])[b]<=[/b]1[/b][/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=158834


algebra
polynomial
quadratics
search
algebra unsolved

Let [b]P(x)=ax^2+bx+c[/b]
suppose:  for every [b]x[/b] in [b]Z[/b],[b]P(x)[/b] is a number square
prove that:
[b]P(x)[/b]=[b](dx+e)^2[/b] with [b]d,e[/b] in[b]Z[/b]

This is true in the general case of any polynomial with integer coefficients (not necessarly quadratic ones). I'm sure that this appeared before on this forum, with other generalisations (in particular, with restriction on the $x$ for which $P(x)$ is a square, a$x$ is a power of $2$ etc...).
Use the search function...

Pierre.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=polynomial+square&t=15102
The post of Peter Scholze.

Pierre.


Find the area of shaded region in the figure below.

Please note:
a) The square is of unit length
b) The curves joining the opposite corners of square are quarter of a circle.

Do mention the method in detail  :)  :)  :)

[quote="sen"]hint**
[hide]
separate the diagram into 3 different parts, the centre (shaded area) c, and the other two different areas, a and b

equilateral triangle.

initially have 4a+4b+c=1

you just need to find a+b

my final answer was : $1-\sqrt{3}+\frac{\pi}{3}$
[/hide][/quote]

4a+4b+c=1 --- this is correct, but when I use the same logic, I get a different answer !

Can you tell the values of a and b.

I get a = 3*sqrt(3)/4 - pi/6
and b = 1 - 3*sqrt(3)/2 + pi/12

Something is wrong here...

[quote="sen"]my a is the thin bit on each of the sides, my b is the other bit[/quote]

My "a" and "b" are as follows:

a = area(DHC)
b = area(GHC)

are these same as yours?

[quote="sen"][hide]
$4a+4b+c=1$

$2a+3b+c=\frac{\pi}{4}$

To find $a+2b+c$ WLOG look at area $BEC$. Draw in $\triangle BEC$

Now we see that it is equilateral because $BE=BC$ and $BC=EB$ quite obvious [radii of circle]

$\implies area |\triangle BEC| = \sin60*1*1*\frac{1}{2} = \frac{\sqrt{3}}{4}$

$\implies a+2b+c = \frac{\sqrt{3}}{4} + 2(\frac{\pi}{6}-\frac{\sqrt{3}}{4})$

after simplification...

$\implies a+2b+c = \frac{4\pi-3\sqrt{3}}{12}$

SUbtract this result from $2a+3b+c=\frac{\pi}{4}$ and we obtain:

$\implies a+b = \frac{3\sqrt{3}-\pi}{12}$

$\implies 4a+4b = \frac{3\sqrt{3}-\pi}{3}$

Subtract this result from $4a+4b+c=1$

we obtain $c=1-\sqrt{3}+\frac{\pi}{3}$[/hide][/quote]

Thanks... I can see where I made the mistake...

Firstly I tried like this only and now realise what I missed...

$area |\triangle BEC| = \sin60*1*1*\frac{1}{2} = \frac{\sqrt{3}}{4}$

I forgot the $\frac{1}{2}$ term.

 :(  :(  :(  :(  :(

[quote="sen"][hide]
$4a+4b+c=1$

$2a+3b+c=\frac{\pi}{4}$

To find $a+2b+c$ WLOG look at area $BEC$. Draw in $\triangle BEC$

Now we see that it is equilateral because $BE=BC$ and $BC=EB$ quite obvious [radii of circle]

$\implies area |\triangle BEC| = \sin60*1*1*\frac{1}{2} = \frac{\sqrt{3}}{4}$

$\implies a+2b+c = \frac{\sqrt{3}}{4} + 2(\frac{\pi}{6}-\frac{\sqrt{3}}{4})$

after simplification...

$\implies a+2b+c = \frac{4\pi-3\sqrt{3}}{12}$

SUbtract this result from $2a+3b+c=\frac{\pi}{4}$ and we obtain:

$\implies a+b = \frac{3\sqrt{3}-\pi}{12}$

$\implies 4a+4b = \frac{3\sqrt{3}-\pi}{3}$

Subtract this result from $4a+4b+c=1$

we obtain $c=1-\sqrt{3}+\frac{\pi}{3}$[/hide][/quote]

should it not be ...
$\implies a+2b+c = \frac{\sqrt{3}}{4} + 2(\frac{\pi}{6}-\frac{\sqrt{3}}{8})$

rest all terms I can match... may be I am making some mistake in definite integral :( :?:  :?:  :?:

[quote="sen"]no i dont think thats right. its simply the sector minus the triangle to give the segment[/quote]

I see now... again missed the $\frac{1}{2}$

just another bad day... :mad:  :mad:  :mad:


Does the code compile and give the page?

[quote="aufha"]In the bottom box of TeXnicCenter after building, it said "There's no line here to edit." What does that mean? How can I get to the line where the problem is?[/quote]I think you mean it says "There's no line here to end".
That means you've used something like \\ on a blank line. \\ can only be used at the end of a line with something in it. Press ! or F9 will show you where it is.


Consider the polynomial equation $q_{1}z^{4}+iq_{3}z^{3}+q_{2}z^{2}+iq_{1}z+q_{0}=0$

If $z=a+bi$ is a solution to this equation, where a and b are real constants and $i^{2}=-1$, which of the following must also be a solution?

A. $-a-bi$
B. $a-bi$
C. $-a+bi$
D. $b+ai$
E. None of these

If $f(z)$ is the function above, we have $g(z) = f(iz) = q_{4}z^{4}+q_{3}z^{3}-q_{2}z^{2}-q_{1}z+q_{0}$, which has integer coefficients. So if $x+yi$ is a root of $g$, we know $x-yi$ is a root as well.

But $\frac{a+bi}{i}= b-ai$, so $b+ai$ is a root of $g$. Then $i(b+ai) =-a+bi$ is a root of $f$.

aha! that makes sense, I think.

We have $g(z)=f(zi)=0$, or $g(\frac{z}{i})=f(z)=0$

Since $g$ has integer coefficients, if $x+yi$ is a root, so is $x-yi$.

Thus, $f(\frac{x+yi}{i})=f(\frac{x-yi}{i})=0=f(y+xi)=f(-y+xi)$

Hence if $a+bi$ is a root of $f$, so is $-a+bi$.

thanks.


When you are anywhere in the "Alcumus" part of the website, and you hover over the left side tabs, the following occur, different from the normal.

Over the "Community" tab, there is a subtab for "T-Shirts".

Over the "For The Win" tab, there is no subtab for "Test Version".

For the former, there isn't supposed to be one, and the latter there should be.

That's strange, because for me everything is how it should be.

This has been posted about before. Just [url=http://www.artofproblemsolving.com/Forum/search.php]search[/url].

EDIT: My bad.

Whoops, forgot to include that the first of the two glitches has already been mentioned. However, nothing was done and I found another one. Whatever, the glitch is so trivial.


logarithms
function
trigonometry
integration
calculus
calculus computations

waza ?  :D

can you show

$ \sum_{k \equal{} 1}^\infty\;\dfrac{1}{k^3\cdot \dbinom{2k}{k}}\; \equal{} \;\boxed{\dfrac{\pi}{6\sqrt 3}\left\{\psi^{(1)}\,\left(\dfrac{1}{3}\right) \minus{} \psi^{(1)}\,\left(\dfrac{2}{3}\right)\right\} \minus{} \dfrac{4}{3}\zeta(3)}$

where $ \psi^{(1)}\,(x) \equiv \dfrac{d^2}{dx^2}\,\left(\ln\,\left[\Gamma(x)\right]\right),\quad\Gamma(x)$ is the gamma function.

you may  leave the answer of the question in terms of  the [b]clausen[/b] function, particularly $ \textbf{Cl}_2\left(\dfrac{2\pi}{3}\right)\;\equal{}\;\sum_{j\equal{}1}^\infty\;\dfrac{\sin\,\left(\dfrac{2\,j\,\pi}{3}\right)}{j^2}$

We can use the following expansion:

$ \arcsin^{2}(x)\equal{}\frac{1}{2}\sum\limits_{k\equal{}1}^{\infty}\frac{(2x)^{2k}}{k^{2}\binom{2k}{k}},$

from which it follows that

$ 2\int\limits_{0}^{1/2}\frac{\arcsin(x)^{2}}{x}dx\equal{}\sum\limits_{k\equal{}1}^{\infty}\frac{1}{k^{3}\binom{2k}{k}}\equal{}S$.

Use the substitution $ x\equal{}\sin y$, and integrate by parts to get that

$ \frac{\pi^{2}}{18}\ln \frac{1}{2}\minus{}2\int\limits_{0}^{\pi/6}x\ln \sin(x)dx\equal{}S$.

I prefer to leave the answer in this form but if you wish to express it in terms of the Clausen function simply integrate one more time by parts to have it rearranged in a different form.

I hope I made no mistakes.


dar mostatile ABCD faseleie noghteie P az se rase A,B,C betartib 10,3,4 mibashad.
faseleie P az rase D cheghadr ast?
 :?:  :(

rooye yecki az rasa, kolle shekl ro davarane 90 daraje, bede.

PS.albate, man in joori hal kardam, shayad rahe behtari ham bashe

khob agar az noghteye p be azla'e mostatil amood konid va rabeteye fisaghoores ra dar harkodam az mosallas haye be vojood amade benevisid andazeye dp niz be dast miayad ingoone ke agar h1 ,h2,h3,h4 betartib paye amood az p bar rooye ab,bc,cd bashand va be dalile mostatil bnoodan abcd darim ah1=dh3=y,bh1=ch3=x,bh2=ah4=z,ch2=dh4=w va tebghe fisaghoores:
y^2+z^2=100
x^2+z^2=9
w^2+x^2=16
w^2+y^2=dp^2=100-16+9=107 --->dp=radikal 107


modular arithmetic
Miscellaneous Problems

Show that if $a$ and $b$ are positive integers, then \[\left( a+\frac{1}{2}\right)^{n}+\left( b+\frac{1}{2}\right)^{n}\] is an integer for only finitely many positive integer $n$.

Suppose $ \left(a\plus{}\dfrac12\right)^n\plus{}\left(b\plus{}\dfrac12\right)^n\equal{}\dfrac{(2a\plus{}1)^n\plus{}(2b\plus{}1)^n}{2^n}\in\mathbb{Z}$. Let $ 2a\plus{}1\equal{}A$, $ 2b\plus{}1\equal{}B$. If $ n\ge2$, then $ n$ must be odd, otherwise $ A^n\plus{}B^n\equiv2\pmod 4$. Then $ A^n\plus{}B^n\equal{}(A\plus{}B)(A^{n\minus{}1}\minus{}A^{n\minus{}2}B\plus{}A^{n\minus{}3}B^2\minus{}\ldots\minus{}AB^{n\minus{}2}\plus{}B^{n\minus{}1})$. The second paranthesis consists of an odd ($ n$) number of odd terms, so it is odd. Then $ 2^n|A\plus{}B$, which holds for finitely many $ n$.


Let positive real numbers $ a,b,c$ such that $ 5a^2\plus{}4b^2\plus{}3c^2\plus{}2abc\equal{}60$. Prove that $ a\plus{}b\plus{}c\le 6$

[quote="Mikel2008"]Let positive real numbers $ a,b,c$ such that $ 5a^2 \plus{} 4b^2 \plus{} 3c^2 \plus{} 2abc \equal{} 60$. Prove that $ a \plus{} b \plus{} c\le 6$[/quote]
No!.No Mikel2008.It is in  $ 45^\text {th}$ anniversary contest.It also is one of the most  important contests of Mathematics  and Youth magazine in our country. So I don't think it is   suitable for us to discuss it here
 :)
P/s:The generalization is here
\[ ax^2 \plus{} by^2 \plus{} cz^2 \equal{} a \plus{} b \plus{} c
\]
Prove that
\[ x \plus{} y \plus{} z\le a \plus{} b \plus{} c
\]
Regards

[quote="Allnames"][quote="Mikel2008"]Let positive real numbers $ a,b,c$ such that $ 5a^2 \plus{} 4b^2 \plus{} 3c^2 \plus{} 2abc \equal{} 60$. Prove that $ a \plus{} b \plus{} c\le 6$[/quote]
No!.No Mikel2008.It is in  $ 45^\text {th}$ anniversary contest.It also is one of the most  important contests of Mathematics  and Youth magazine in our country. So I don't think it is   suitable for us to discuss it here
 :)

Regards[/quote]
Dear friends,
Why isn't it suitable for discussion here? I'm a new member and I think Mathlinks is the best forum to discuss all math problems from the world with others.
[quote="Allnames"]
P/s:The generalization is here
\[ ax^2 \plus{} by^2 \plus{} cz^2 \equal{} a \plus{} b \plus{} c
\]
Prove that
\[ x \plus{} y \plus{} z\le a \plus{} b \plus{} c
\][/quote]
Sorry, but yours is wrong.


[quote="NkMAsTeR"]Let a,b,c be the positive real number such that $a^{2}+b^{2}+c^{2}=1$
Prove that
$9xyz+x+y+z \ge 4(xy+yz+zx)$[/quote]
Maybe:

Let $x,y,z$ be the positive real number such that $x^{2}+y^{2}+z^{2}=1$
Prove that
$9xyz+x+y+z \geqslant 4(xy+yz+zx)$.

[quote="NkMAsTeR"]I'm sorry, there is some mistake. It'w wrong. Can you prove it with $x^{2}+y^{2}+z^{2}=3$?[/quote]

I think that's wrong too. Try $\frac{17}{10}, \frac{1}{10}, \frac{1}{\sqrt{10}}$


email
AMC
AIME
USA(J)MO
USAMO
summer program
Mathcamp

Anyone know what the acceptance rate to SIMUW is?

Also when do they send out acceptance letters?

I'm sure the acceptance rate is fairly high, since it doesn't seem to be that well publicized. They told me that there were "many people willing to take my spot."

You should know if you've been accepted, but they haven't sent any follow up materials yet.

tey already have sent the acceptence letters..

what do you mean? Have you received an actual, physical letter? I just got an email saying I got in.

yeah, i recieved my letter in mail, like couple days ago. :lol:

yeah i know i just did too. I guess I'll see you there.

yeah, i'm %90 sure i'm goin....
c u there. :lol:

Ah, so I'll see you both there too.  One of my classmates got in too (female), but she only answered 4 of the problems and doesn't even have Pre-Calculus. They're really trying to recruit girls. And SIMUW waitlisted a SSP alumni who alredy had HL Math for here.

I'll also be doing research with a UW Astronomy prof on top of that.

From what I heard, SIMUW has a pretty relaxed atmosphere.

So as for roster we know so far: (it'll be interesting to see how many people we know are in)

 Me
The girl in my school who got in
 jeffrey
amirhtlusa

So as for any SIMUWers interested in getting to know others, I'll give some info about myself:

I'm currently a sophomore at Inglemoor but am going to the UW Academy next year. I'm not terribly great in math or on other words, I qualified for AIME but didn't do great in it. I'm not huge into math (yet) and am more into science research but I hope that the program will spark enthusiasm into it.

people get in so easy, i'm one too.
i wonder if simuw would actually have 24 people apply.
anyways, i'll see you guys soon. :starwars:

SIMUW's over now. *sigh*

I wasn't the most social person but I can definitely say that there were some fun moments and this program was really what it said - an opportunity for HS students to expose themselves to math. I wasn't one of the strongest students but I can say that I did learn quite a lot and gained a lot of motivation to study math (heh, I hang around the math section of the library more than the physics section).

i agree.
it was fun.
although i was expecting to see stronger people (i mean mathematically) but anyways it was a fun experience for me.
i hope it helps me to do better in competitions in the upcoming year.

amir

Hmm, I thought that SIMUW did a good job of recruiting some of the best of the state's students, though there definitely was a wide variation in skill level. There weren't any USAMO qualifiers (Hesterberg went MOSP and MathCamp, Young Jeong considered SIMUW, then decided to go to MathCamp IIRC, Jeffrey Wang went PROMYS last year then didn't do anything this year[EDIT:  he went to MAO this year], and Will Johnson from my school isn't ambitious), but there were quite a few people who had high AIME scores (5 and above) and some other very strong students who screwed up on the AIME or didn't have the opportunity to take AMCs. IIRC, there were at 6 people who got AIME scores of 5 and above at SIMUW and there aren't very many of them in our state)

I think it may be a good idea to post some of SIMUW's problems on AoPS(McGovern's and Cohn's, specifically). Of course, my thoughts reflect my strong support of the notion that all information is for everyone, not to be contained within any program.

If anyone wants to play Mafia, I'll mod.

the thing about the classes was that i already knew 90% of the material in them;
and the problem was that i was expecting to see classes that would be more useful for HS students; for example i found the last two classes (elliptic curves and Combinatorics), way more usefull than the knot theory.

but alex; if you want to play mafia. i'm in :D

[quote="Simfish"]Hmm, I thought that SIMUW did a good job of recruiting some of the best of the state's students, though there definitely was a wide variation in skill level. There weren't any USAMO qualifiers (Hesterberg went MOSP and MathCamp, Young Jeong considered SIMUW, then decided to go to MathCamp IIRC, Jeffrey Wang went PROMYS last year then didn't do anything this year, and Will Johnson from my school isn't ambitious), but there were quite a few people who had high AIME scores (5 and above) and some other very strong students who screwed up on the AIME or didn't have the opportunity to take AMCs. IIRC, there were at 6 people who got AIME scores of 5 and above at SIMUW and there aren't very many of them in our state)

I think it may be a good idea to post some of SIMUW's problems on AoPS(McGovern's and Cohn's, specifically). Of course, my thoughts reflect my strong support of the notion that all information is for everyone, not to be contained within any program.

If anyone wants to play Mafia, I'll mod.[/quote]

Young and I were at MAO nationals in HAWAII. Ha. Beat that. A friend of mine, Edward, went though. He said it was pretty good, too. Seems like a decent program.

PS. I know Mr. Cohn :P

[quote]Young and I were at MAO nationals in HAWAII. Ha. Beat that. A friend of mine, Edward, went though. He said it was pretty good, too. Seems like a decent program. [/quote]

Ok, I stand corrected. I thought Young was considering not doing MAO since he told me it wasn't really problem solving. I think I recall a post from you saying that you couldn't do PROMYS this summer - but umm yeah - why did I  say that you didn't do anything?!? I guess I was just tired..

Now that I'm out of SIMUW, I'm missing it so much now.. :( My favorite class: elliptic curves. It's so awesome to learn a specialized math topic because I won't have to review it later again for a while, even though I'm going to UW next year. Many of SIMUW's topics are about very specializied math topics.

I think knot theory was a good class - while not intrinsically useful for competitions, it's another specialized math topic that one won't see again in a while and it's a very active area of research (there are several REUs in knot theory)/

I had to do MAO because I have made a verbal commitment before i applied to MathCamp and others. I heard that some of the ARMLers went to SIMUW ----- noticeably Brian Basham and Rachel Hinman. I will consider it next yr, although i would 1. Apply to RSI or 2. Return to MAO, since I met a girl there this year that I would hate to not meet next yr.

yeah i would recommend it to you.
it's a fun program.
the best part is the FOOD. :D
they gave us \$1000 to spend on anything we wanted.

I saw a SIMUW thread.. and holy poop, my name was in it.  On top of that, from someone who... wasn't from SIMUW! Anyway, SIMUW was awesome.  My only complaint is there were only four girls this year :( Hah, and a dance party.  Funny :D Anyway, I'm off to bed.  I'll be AOPSing more this year.. need to get serious about my math if I'm going to compete with the likes of Amir... YEAH, BABY!  Ciao guys.
~Basham

heh.
brian dont worry you don't have to go against me :D  (since you'll be a sophomore and ill be a senior.) :)
but if you had, then you didn't have much chance :D  :D
just kidding. :)

but  i'll be looking forward to see you next year in Vancouver, and other competitions.

amir

There were more girls last year - most of who were Chinese. For some reason, this year had no East Asian girls. They definitely were trying to recruit more girls this year and even mailed all girls accepted by the UW Academy (early entrance program for 10th graders). However, that plan was unsuccessful - as I turned out to be the only UW Academy student who was also SIMUW 2005.

But anyways, we somewhat compensated for the lack of girls with other relationships.

I didn't mind it that much, as I don't want to be distracted by a girlfriend and wish to stay a virgin through college.

Oh, and there's one game in SIMUW I've never heard about before that's really fun Assasins, which could be introduced to another summer program. ;)

You silly, Simfish.  Man, I hang out with more girls on a regular basis than there were at the camp total.  Including the TACs. Which is probably why I'm so bad at math... :O lol :)

And when you said assassins... did you mean Assassins, or AssAssins? :D

$\mathsf{(math skill)} \varpropto  \frac1{\mathsf{(\# girlfriends)}}$

Anyways, I'll probably apply to SIMUW next year (I'm going into 10th grade).  It shouldn't be too hard to get in, I looked at the questions for admission.

Does that sign thingy mean proportional to?  If so, it must be true, cuz I didn't know :(


integration
linear algebra
matrix
calculus
calculus computations

$ \int_{0}^{2}\int_{\frac{x}{2}}^{\frac{x}{2}+1}x^{5}(2y-x){e^{\left(2y-x\right)}^{2}}dy dx$

Using change of variables, $ u=x$ and $ v=2y-x$ so ultimately it becomes $ \frac{1}{2}\int_{0}^{2}\int_{0}^{2}u^{5}v{e{}^{v}{}^{2}}du dv$.  I can do everything and solve it but I can't figure out why the Jacobian is $ \frac{1}{2}$.  I got $ 1$.  Any help?

Since $ \begin{pmatrix}u\\v\end{pmatrix}\equal{}\begin{pmatrix}1 & 0\\\minus{}1 & 2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$, taking inverse, $ \begin{pmatrix}x\\y\end{pmatrix}\equal{}\begin{pmatrix}1 & 0\\ \frac{1}{2}&\frac{1}{2}\end{pmatrix}\begin{pmatrix}u\\v\end{pmatrix}$ and the Jacobian is given by $ \left|\begin{matrix}1 & 0\\ \frac{1}{2}&\frac{1}{2}\end{matrix}\right| \equal{}\frac{1}{2}$.


The perimeter of an isosceles triangle 36 cm, and the altitude to its base is 12 cm.  What is the number of square centimeters in the area of the triangle?

:? Is there enough information to complete this problem? I'll give it a shot first.

well since it's an isosceles, i guessed and checked
the sides are 13,10,13
the area is 60units sq.

[quote="4everwise"];)  Yeah, I just got that a little while ago. With two right triangles, test with the pythagorean theorem.[/quote]
that's what i did


function
number theory proposed
number theory

Determine for which integers $k>0$ exists some positive integers $m,n>0$ such that:
$n^2$ , $n^2+m$ , $n^2+4m$ , $n^2+9m$ , ..., $n^2+k^2 m$
are simultaneously squares of integers?

[quote="RobertuY"]Very very very obvious.
I am teacher at university.

______
RY TVCM[/quote]

If this is very obvious, then which is your answer?

I thought I did it, but I was wrong!
Can you show that for any integer $k \geq 3$ there are no positive integers $m,n>0$ such that:
$n^2$ , $n^2+m$ , $n^2+4m$ , $n^2+9m$ , ..., $n^2+k^2 m$
are simultaneously squares of integers?

[quote="harazi"]Just ignore him, he already said enough while "solving" other problems.[/quote]

 :lol:  :lol: He reminds me of a certain Canadian IMO team member..

The wrong proof I gave uses strong results involving sum of squares function:
http://mathworld.wolfram.com/SumofSquaresFunction.html
The number of representations of some $N$ by $2$ squares ignoring order and signs is $r'_2 (N)$ as the number of distinct solutions of:
$x^2+y^2=N$, $(x,y) \in N^2$, $0 \leq x \leq y$
For any $k$ arbitrarily large, we may take some $N$ such that $r'_2 (N)$=$k$.
If $N$=$2^{a_0} p_{1}^{2a_1} p_{2}^{2a_2} ... p_{r}^{2a_r} q_{1}^{b_1} q_{2}^{b_2} ... q_{s}^{b_s} $, where $p_i$ are primes of the form $4k+3$, $q_j$ primes of the form $4k+1$ and all $a_i, b_j$ integers and if we denote:
$B=(b_1 +1)(b_2 +1) ... (b_s +1)$, then:
$r'_2 (N)$=$\frac{B}{2}$, for $B$-even
$r'_2 (N)$=$\frac{B-(-1)^{a_0}}{2}$, for $B$-odd
I am still looking for a nice elementary proof!!


topology
real analysis
real analysis unsolved

a space $E$ is said polish if there is a distance $d$ on $E$ such that:
*$(E,d)$ is complet
* there exists a countable dense subset of $E$.

1/Prove that an open subset of a polish space is a polish space.
2/ prove that $\mathbb{R}-\mathbb{Q}$ is a polish space

We don't need to be concerned with separability here: in our problem, all the spaces are separable, so it suffices to prove this (a result due to Alexandrov, I think):

Given a complete metric space $(X,d)$ and a $G_\delta$ subset $M$ of $X,\ M$ can be remetrized into a complete metric space.

Let $M=\bigcap_{n\ge 1}G_n$, where $(G_n)_n$ is a descending chain of open subsets of $X$. For each $n$, we define a distance $d_n$ on $G_n$ as follows: let $d_n(x,y)=\left|\frac 1{d(x,X\setminus G_n)}-\frac 1{d(y,X\setminus G_n)}\right|+d(x,y)$, and $\rho_n(x,y)=\frac{d_n(x,y)}{1+d_n(x,y)}$. Checking that $(G_n,\rho_n)$ is a complete metric space is a simple exercise. Finally, define the distance $\rho$ on $M$ by $\rho(x,y)=\sum_{n\ge 1}\frac 1{2^n}\rho_n(x,y)$. Again, I'll leave out the actual checking of the fact that $(M,\rho)$ is a complete metric space :).


Regularly it is AMC score + 10 * AIME score but what if we qualified by the USAMTS instead? Is the USAMTS score converted into some kind of equivalent score?
If we took both the AMC and USAMTS, is the higher of the AMC score and the equivalent USAMTS score used?

Thanks in advance.

[quote="BOGTRO"]You recieve 100 points for your AMC score, unless you have a better score actually taking it.[/quote]

That is, you will not make USAMO unless

1) You beast at AIME getting 10+
2) You get high score taking AMC B

Has anyone ever qualified for USAMO via USAMTS?

[quote="georgiamathdude"]I doubt it, since someone good enough to get 10+ on AIME will most likely score much higher than 100 on the AMC.[/quote]

what if someone is smart enough to take AIME, but didn't take AMCs. its possible.

It's possible that they just got very unlucky and repeatedly got sick on both days, so they have to rely on it.

Last year I recall someone said their school didn't do the AMCs, and they were 1 point away from the AIME cutoff by USAMTS. I don't know if they protested and got that point they needed.


function
algebra
polynomial
geometry
geometric transformation
reflection
quadratics

let $ T$ be the complex t-plane, and let $ F \equal{} \mathbb{C}(t)$ be the field of rational functions in $ t$.  For an extension $ K$ over $ F$, we choose a primitive element $ \alpha$ generated by an irreducible polynomial $ f(x)$ such that $ K \equal{} F[x]/(f)$.  then, the locus of points $ (t,x)$ in $ \mathbb{C}^2$ such that $ f(t,x) \equal{} 0$ defines a riemann surface.  suppose $ f$ has degree $ n$ in $ x$.  Then, the surface generated by $ f$ is an n-sheeted branched covering of $ T$.

we analyze this surface by considering the branch points (i.e. the points $ t$ for which there are where there are fewer than $ n$ points of the surface).  in particular, we can categorize riemann surfaces into isomorphism classes that reflect the topological structure of the surface as follows: cut-and-paste the sheets of the riemann surface and associate each branch point with a permutation on the sheets.  the permutation is induced by traveling in small circles in t-space around each branch point, recording which sheets we visit and in which order we visit them before returning to where we started.  this set of permutations and branch points defines an isomorphism class, where equivalent representations of the surface are related by conjugating the permutations by some element $ \sigma \in S_n$.

there is a difficulty, however: suppose we restrict our attention to quadratic extensions of $ F$.  then, the extensions generated by $ f \equal{} q(t)x^2 \minus{} p(t)$ and $ g \equal{} p(t)x^2 \minus{} q(t)$ ought to be the same since $ \sqrt\frac {p}{q}$ is the reciprocal of $ \sqrt\frac {q}{p}$, and fields contain inverses.  however, the all branch points of the extension generated by $ f$ must be some subset of the roots of $ p$, while all the branch points of the extension generated by $ g$ should be some subset of the roots of $ q$.

on one hand, it seems like we should be able to exclude those points of the riemann surface generated by $ f$ where $ q \equal{} 0$ (i.e. where poles exist), since rational functions are completely determined by any infinite set of points.  removing some finite number of poles should do nothing.  on the other hand, there is no reason, a priori, to assume that $ p$ and $ q$ share any roots at all, and yet they must represent the same surface.  if this is the case, then most quadratic field extensions are trivial, which is clearly false.  this suggests we cannot ignore the poles, but then what are we supposed to do with them?

clearly i am confused somewhere.  set me straight!

edit: crap. this is the wrong forum.  sorry, everyone.  someone move this into superior algebra or complex analysis or something.

Riemann Surfaces should be compact. By taking finite covers of the complex affine line (your complex t-plane), you are missing what happens over the point at infinity of $ \mathbb P^1(\mathbb C)$ which is your basic Riemann surface $ \mathbb S^2$ of genus $ 0$.

You should view the example you gave as the picture of the same Riemann Surface (a double cover of $ \mathbb P^1$) in the two different standard charts that cover $ \mathbb P^1$. The projective equation in coordinates X,Y,Z would be $ p(X)Y^2\minus{}q(X)Z^2\equal{}0$. The charts are $ Y\neq 0$ and $ Z\neq 0$. You should also make new coordinates of $ Z/Y$ and $ Y/Z$ respectively.


search
combinatorics unsolved
combinatorics

How an earth do you solve this problem? I just don't even know where to start.

[u]Two integers [/u]

I'm thinking of two integer numbers, each if them is more than 1 and their sum is less than 100.
I tell my friend Serge the sum of these two numbers, and another friend, Polly, product of these two numbers.
Then such a dialog took place:
P: I can't determine what are these numbers.
S: Ah, i knew you wouldn't be able to do this.
P: Oh, then i know what they are!
S: Oh, then i know them too!

Can you determine the numbers?

Thanks.

Let the sum be $S$ and the product $P$.

Consider the information you have. You know that $S$ tells you that $P$ will not uniquely determine the pair. That means that for every possible pair of numbers adding up to $S$, $P$ will also be a product of another possible pair.

This can only happen for $S\in\{5, 7, 9, 10, 11, 13, 15, 16, 17, 19, 21, 23, 25, 27, 29, 31, 35, 37, 41, 43,45, 47, 49, 53\}$. (1)

Now, for each of these sums there is a finite set of possible products. From knowing (i) the value of $P$ and (ii) that $S$ is in the above set, Polly can determine the pair. This means that $P$ can only be expressed in the form $i(S-i)$ for one of the $S$ in the above set. Moreover, from knowing $S$ and that $P$ uniquely determines $S$, Serge can determine the pair - this means that, for the given $S$, there must be exactly one possible product which is unique to that sum.

Now, the only values of $S$ which have exactly one product of the form $i(S-i)$ which is not expressible in this form for any of the other $S$ are 5, 9, 15 and 19, corresponding to the products 4, 8, 56 and 88 respectively. (2)

This means that the only possible pairs are $\{1,4\}$, $\{1,8\}$, $\{7,8\}$ and $\{8,11\}$.

I believe that any of these are possible solutions, which would meet the above conditions.

Incidentally, I was able to solve this problem in reasonable time only by doing the searches on a computer (using Mathematica) - but to do this by hand, you'd either need a lot of time on your hands, or a much less labour intensive stategy.

(1) I proved this by an exhaustive search. Given $S$, the possible values of $P$ are $i(S-i)$ for $i\in\{1,2,\cdots,S-1\}$. For any $S$, if any one of these values can be written as a product of a possible solution pair other than $\{i,S-i\}$ then $S$ does not have the required property.

(2) Again, I proved this by an exhaustive search - for each $S$, I comiled the list of possible products, and then found the values of $S$ which had exactly one product not to occur in any other list.

I might have misunderstood this problem, but I got the sum to be $17$, the product to be $52$, and thus the two integers to be $4$ and $13$.

First let me mention this to make the solution easier to comprehend:

Suppose i tell the sum to Sam and the product to Paul.  Now I can post my solution for this problem later if you want tracer-bullet, but it lies heavily on the fact that if Paul does not initially figure out the two numbers, and Sam can tell Paul he knew that he wouldn't figure it out, then

$S\in\{11,17,23,27,29,35,37,41,47,51,\dots\}$

Note that if $S=p_{1}+p_{2}$ where $p_{1}$ and $p_{2}$ are primes, then Paul could determine the solution immediately (both integers are greater than $1$).  Now, to generate this set, we reason that if $S$ is even, then the Goldbach conjecture tells us that there are two primes with sum $S$ (I know it's not proved yet, but it has definitely been checked up to $100$) and as long as $S$ is odd and $S=p+2$ where $p$ is a prime number, then $S$ cannot work.  The rest of the solution relies on making sure each person has exactly only one solution in integers to their equations...

For example, since $52=2\times 26=4\times 13$, Paul cannot be sure on the two integers; however, the two possible sums that Sam could have are $17$ and $28$.  However, since $5+23=28$, and $5$ and $23$ are prime, Sam could not have said he was certain Paul could not do this problem if the sum was $28$.  Thus Paul knows that the sum is $17$, and accordingly the two integers.

Now, if Sam writes down the possible products that Paul could have, he comes up with the list $30,42,52,60,66,70,72$.  Now for all of the integers but $52$, there are at least two pairs of integers that have the same product of one of these numbers and also two sums that are not the sum of two primes.

For instance, if $P=30$, then the possible sums would be $11,13$ and $17$.  But since Sam could have been given the sum of  $11$ or $17$ and still said "Paul couldn't do it," Paul would not have been able to come up with the sum.  This happens for all the numbers but $52$.   Hence if Paul was able to figure it out, Sam knows that the product was $52$.  Whence Sam deduces the two integers, $4$ and $13$.

How did you find the list $30,42,52,60,66,70,72$ ?
Because, by inspection would be too tedious and gloomy for an Olympic solution , I think...

I used the method I described above on the possibility that the sum was $11$, and this did not work out.  So next I tried $17$ as the sum, and the list comes from the products of two integers that have a sum of $17$ and are both greater than $1$.

I didn't know that this was an Olympiad problem (is it really?) but if someone has a shorter solution, it would be worth seeing.

I’m not sure, either, if this is an Olympiad problem, but by “ the name of the game”…

Anyhow why did you stop at $17$  as the sum?

How about $23$ that gives a product list like $42,60,76,90,102,112,120,126,130,132.$, where, actually, we can find at least $76$ and $112$ satisfying the requisite, and so, the last assertion wouldn’t be possible.

Even so, it should to be tested.  Did you do this boring work for all the allowed sums, or there is a better way to do it?


Define $ x \heartsuit y$ to be $ |x\minus{}y|$ for all real numbers $ x$ and $ y$. Which of the following statements is [b]not[/b] true?

$\textbf{(A)}\ x \heartsuit y \equal{} y \heartsuit x \text{ for all } x \text{ and } y$
$\textbf{(B)}\ 2(x \heartsuit y) \equal{} (2x) \heartsuit (2y) \text{ for all } x \text{ and } y$
$\textbf{(C)}\ x \heartsuit 0 \equal{} x \text{ for all } x$
$\textbf{(D)}\ x \heartsuit x \equal{} 0 \text{ for all } x$
$\textbf{(E)}\ x \heartsuit y > 0 \text{ if } x \ne y$

$ \text{C is not true because it is true only for positive numbers for negative numbers it is wrong because for positive numbers is}\\ x \heartsuit 0\equal{}|x|\equal{}x \\\text{and for negative numbers is}\\x\heartsuit0\equal{}|x|\equal{}\minus{}x$

So I got this one right, because C is obviously wrong, but I wasn't sure about B. I guess it's true, but can someone explain why.

If $x$ is negative, the absolute value functions makes it positive. So, $\boxed{C}$ is incorrect. :love:

[quote=Zanderman][quote=JustinLee2017]2(x heart y) = 2|x-y| = |2x-2y| = (2x) heart (2y)
:love:[/quote]

Wait you can factor out of an absolute value?[/quote]

Well |5x| = 5|x|, as an example

You can only factor positive numbers out of absolute value, for example, $|-5x| \neq -5|x|$

[quote=tyseo]this is from an amc8 right[/quote]

I think this is from the AMC 10...

Ah, okay. Thanks @below.

[quote=michaelchang1][quote=tyseo]this is from an amc8 right[/quote]

I think this is from the AMC 10...[/quote]

2003 12 a #6 or so