integration
function
calculus
calculus computations

Suppose that a sequence $ \{a_n\}$ is defined as $ a_n\equal{}\int^{n\plus{}1}_nf(x)dx$ ,$ n\in\mathbb{N}$ . If we wish to show that $ a_n$ is monotonic, can we just show that $ f(x)$ is monotonic?

[hide="original question"]let $ a_n\equal{}\int^{n\plus{}1}_n\frac{1}{xe^x}dx$, for $ n\geq 1$ . Show that $ a_n$ is decreasing [/hide]

Well, yes.  $ a_{n\plus{}1} \minus{} a_n \equal{} \int_n^{n\plus{}1} \left( f(x \plus{} 1) \minus{} f(x) \right) \, dx$.

(Although of course the converse is not true -- there are non-monotonic functions for which that sequence is monotonic.)


induction
combinatorics unsolved
combinatorics

There are $37$ political leaders. They all know that some of them are being spied. They make a deal that every day have a lunch all together and talk about spying. Each one of them talks with each of the others.

Rules:
1) All of them will tell the truth (what they know) about who's being spied, but not who spies;
2) If one of the two knew that the other one was being spied, he would [b]not[/b] say it to him (so, this is an exception to rule 1);
3) As soon as somebody find himself being spied, he immediately moves to some other office.

After the first and the second lunch nothing happened. But after the third lunch some of the leaders has moved.

Question: how many leaders has moved?

I've seen this kind of idea before. In general, if nobody leaves on the first k days and somebody leaves on the k+1 th day, then k+1 people leave then (so 3 people in our case). We can prove by induction the assertion: if a person has been told that m people are spied, and nobody leaves after the first m days, he will realise that he is spied and will leave on the m+1 th day.

For 0 it is trivial; now assume it true for m-1 and let us prove it for m. Person X has been told that m people are spied; then it can either be that m others are spied, or m other and himself is spied. But if it were that only m others are spied, then one of those (say Y) would be told that m-1 others are spied. So by the induction hypothesis, it would mean that he would get up by the m-th day. But our assumption is that nobody leaves in the first m days, so our X must also be spied. He will thus leave on the m+1 th day (he cannot leave before that, because if no one leaves, he cannot discern in the first m days whether there are m other people spied or m others and himself)

So if no one leaves in the first k days, and people leave on the k+1 th, then those people who leave must be the ones who are told that k people are spied. So in truth, there are k+1 people spied, and all those are told that k people are spied. So these k+1 must leave.

Your solution looks nice, but now I see that there is a problem. I'll try to prove that somebody must move after the first lunch.

Let person $A_1$ be one of the leaders.

Suppose that he heard that nobody is spied. That means that he should move.

Suppose that he heard that person $A_2$ is spied. If that person doesn't move it means that person $A_2$ heard that somebody is spied, and that must be person $A_1$, so he should move.

Suppose that he heard that persons $A_2$ and $A_3$ are spied. If nobody of them moves that means that persons $A_2$ and $A_3$ heard that two persons are spied, so $A_1$ is spied and he will move.

Continuing this discussion, I don't see that there is a possibility that nobody will move after the first lunch.

If you (or somebody else) knew this problem, could you please post the exact statement, cause I just heard it and there may be a mistake.

Bojan, The problem seems to be in rule 3). Let suppose that two men are spied - A_1 and A_2. A_1(he is 'faster' thinker than A_2) sees that A_2 hasn't left, so he will leave. When A_2 sees that A_1 is leaving he won't know what to do :blush:
If rule 3) means:
When somebody understand that he is being spied he won't appear on next lunch.
Then if on fourth(this is after third) lunch somebody is missing(these men had realized that they are spied during third lunch) than it would be 3 people.
And the solution of RHS is OK.


[quote="thaithuan_GC"]Given a positive integer $ n$. Prove that there exists an infinite number of intergers $ x,y,z,t$ satisfying $ x^3 \plus{} y^3 \plus{} z^3 \plus{} t^3 \equal{} 2(\frac {n(n \plus{} 1)}{2})^3 \minus{} 1$.[/quote]
This problem has some ways to approach , my solution is only one way . Try to find out some other methods . Let consider :
$ x \equal{} \frac {a \plus{} b}{2},y \equal{} \frac {a \minus{} b}{2},z \equal{} \frac {u \plus{} v}{2},t \equal{} \frac {u \minus{} v}{2}$
Then the equation becomes :
\[ a^3 \plus{} 3ab^2 \plus{} u^3 \plus{} 3uv^2 \equal{} 8m^3 \minus{} 4
\]
where $ m \equal{} \frac {n(n \plus{} 1)}{2}$
Choose $ a \equal{} 2m,u \equal{} \minus{} 1$ and consider equation :
\[ 6mb^2 \minus{} 3u^2 \equal{} \minus{} 3
\]
It is equivalent to equation :
\[ v^2\minus{}2mb^2\equal{}1
\]
It is clear $ 2m$ is not a perfect square and this equation has infinite solution $ (v_n,b_n)$ . Because $ b_0 \equal{} 0$ so the sequence $ \{b_n\}$ contains infinite even number , and it clear that all term of sequence $ \{v_n\}$ is odd ,thus $ x,y,z,t$ are integer . Proof completes .


So here are the problems, answers are due by Sunday, December 11th, 2005, 8:00 PM EST. Have fun!

PM or IM me your answers.

Okay, so far only 2 people have submitted answers. Yeah I realized this is a horrible time in the year for something like this. Therefore, the deadline will be extended to [b]Sunday, December 11th, 2005, 8:00 PM EST[/b].

I'm definitely going to take the exam - I just haven't gotten time yet; probably I'll do it tomorrow.

[quote="zanttrang"][quote="Altheman"]i want to take it, but i usually don't have a 3 hour block of free time[/quote]
Then take it in smaller blocks, and just make sure that you only work on it for a total of three hours.[/quote]

Yeah do that if it works better for you, right now I just need people sending in answers,  :lol: .

Who was the liar who voted easy? :rotfl:

[quote="JesusFreak197"]Nevermind, I'm not taking it. I looked over the first few problems, and it's way harder than the standard AIME, not to mention a huge overabundance of geometry problems. I'm not going to spend my time doing something like this. It took me 20 minutes to get #1. :dry:[/quote]

Yeah but 6 geometry problems is not exactly too many, check out 2004 AIME1
 [url]http://www.kalva.demon.co.uk/aime/aime04a.html[/url]. I agree that the first few problems are much harder than something on a regular AIME, sorry about that :( .

But what's wrong with it being too difficult (wouldn't it be a good practice then)?

[quote="Mildorf"]My philosophy is that you should train, in part, for problems more difficult than the ones you will encounter, so I have made these mock contests extremely difficult.  The idea is that, once you become acclimated to them, the real AIMEs will seem easier, and you will approach them with justifiable confidence[/quote]

[quote="beta"]

Yeah but 6 geometry problems is not exactly too many[/quote]
Well, I was mostly looking at the first half or so of the test, and was a little frustrated by the abundance of geometry, which is by far my worst subject.

[quote="beta"]But what's wrong with it being too difficult (wouldn't it be a good practice then)?

[quote="Mildorf"]My philosophy is that you should train, in part, for problems more difficult than the ones you will encounter, so I have made these mock contests extremely difficult.  The idea is that, once you become acclimated to them, the real AIMEs will seem easier, and you will approach them with justifiable confidence[/quote][/quote]

I know I should practice the harder problems; I might do some off-and-on at some point. I don't know if I'm an unusual mathy person here, but I don't like spending a long time on problems when I'm not actually in a competition. On the previous mock AIME, I only spent one hour and fifty minutes on it because three hours is just too long to spend straight. I wasn't even going to do it all straight, but I'm not a big fan of individual problems in the first five that take 20 minutes to solve. :| Maybe I'll take a look at some of the later ones, except those are probably going to be even harder, anyway... =\

Thomas made an important point that it was important to work on the hard problems so the actual one will be a lot easier. This is how Barron's books are for SAT (I'm pretty positive that they're much harder than real SAT) but does that mean everyone who reads Barron's books get good grades on SAT? Not necessarily. This is same with Mock AMC. Although harder problems can benefit people, it may not benefit [b]everyone[/b].

Just for sidenote, don't worry too much about beta's geometry problems. They are tough problems but that's because he's good at geometry. If you can manage to solve all of them by yourself (which I can't, sadly), I'm pretty sure that you will be able to do most of the geometry problems on AIME by yourself.

beta, nice set of problems by the way!  :)

you should want to challenge yourself jesusfreak, and also, you should practice for the real thing by working the full 3 hours

[quote="JesusFreak197"]I very rarely have three free hours straight, and when I do, I'd rather do something other than math.[/quote]

Hmmm then why do you do math competitions if you don't like doing math?

I never said that. I enjoy doing math, but I don't like spending long periods of time straight on it.

[quote="JesusFreak197"]...I'd rather do something other than math.[/quote]
 :what?: like what?  :P

Yay! Now I can actually finish them! (too much homework this week....)

[quote="Altheman"]wow, i just finished this, and it was really difficult, and on number 6, i kept getting an answer that did not fit the form of a,b,c, but oh well, and number 1 was a lot like the number 14 on the 2005 aime A? talk about difficult...[/quote]

I don't want to give too much away but #1 is much simpler than 2005 AIME #14...

wow p13 :blush:

only thing bad abt this test is that 7/15 are geo

just did a couple of problems
and def recommend

[quote=MockSolutionsCompiler]I know I'm bumping a thread after 17 years, but this mock hasn't been listed anywhere or talked about, so I think it is severely underappreciated.

Do not be fooled by this test's age. The difficulty is still on par with the harder real AIMEs from recent years. If you don't believe me, just try out the first 7-8 problems and you'll see.

And yeah, I wrote solutions, since I was in the mood to and I thoroughly enjoyed solving these:
[url]https://www.dropbox.com/s/n434y3tiytfrzki/beta_Mock_AIME_Solutions.pdf?dl=0[/url][/quote]

attached those solutions

(problems are to be posted in separate threads in HSM in a while)

all problems have been posted in separate threads in HSM,  I just created the related post collection

[url=https://artofproblemsolving.com/community/c3741459_2005_beta_mock_aime]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)

19 years, and @beta’s legacy lives.

GO AIME people!!!!!!

Gosh this thread was made when I was less than two months old.

this thread was made when i was -4 years old

By the way, there is a results thread, which contains answers and solve rates:

[url]https://artofproblemsolving.com/community/c5h65552[/url]


For the USAMTS problems, can we use programming knowledge to write a program to solve the problem for us?  If so, are there any rules or guidelines in doing so?

Yes.  Please see [url=http://www.usamts.org/TipsFAQ/U_FAQ.php#computer]this item in our FAQ list[/url].


function
real analysis
real analysis solved

Suppose $E\subset \mathbb{R}^1$ and the cardinality of $E$ is smaller than $c$. Show that there exists real number $a$ such that $E+\{a\} = \{x+a: x\in E\}\subset \mathbb{R}^1\setminus \mathbb{Q}$.

If not, then there are functions f and g, both from the reals into the rationals, respective E such that $ f(x)=x+g(x) $. Since R is not numerable, there exists u such that $ f^{-1}(u) $ is not numerable. But then $ u-f^{-1}{u} $ is in E, and hence E is not numerable. I assumed here that any set with cardinality smaller than c is at most numerable.

I was thinking of taking a Hamel basis of $R$ (a basis of $R$ over $Q$) which includes $1$. Let $a_i,\ i\in I$ be the members of this basis. $I$ must have continuum cardinality $c$ and the difference of each two different $a_i$'s is not rational because otherwise it wouldn't be a basis (because th basis contains $1$ and then the $a_i$'s wouldn't be independent). Then if we assume that for each $a_i$ there is an $x_i$ s.t. $x_i+a_i$ is rational, the $x_i$'s must be distinct, so the cardinality of $E$ would be at least the cardinality of $(a_i)_{i\in I}$, which is $c$, and we have a contradiction.

[edited :D: Hammel$\rightarrow$ Hamel]

[quote="grobber"]I was thinking of taking a Hammel basis of $R$ (a basis of $R$ over $Q$) ... [/quote]
Hamel basis  ;)

it is obviously that $a\not\in \mathbb{Q}-E$, so we need only to prove that the  cardinality of $\mathbb{Q}-E$ is $<c$, but how to prove it without assuming that continuum hypothesis is correct?


geometry
geometric transformation
reflection

i want to have aconversation about movies.i mean talk about the actors of a movie,the editors,...of it.
let's start with $A$ $walk$ $to$ $remember$.
for the first time i saw a movie that has no famous actor and all members of it were new,but the film was incridible.ok now if somebody saw this movie,join. :D

is there any body home?
this forum is look like a dead forum,so where are the art lovers? :?:

Maybe no one has seen that particular movie. (I haven't seen it.) There are some film lovers here, as I recall an earlier thread asking about favorite film directors.

Most people who post here are more into music than into visual art, and that may reflect the self-selected population of mostly math-liking people who post here.

Do you play any musical instruments? (I used to play flute, badly, but haven't played it for years. My wife and children play piano.)

no i don't play any musical instrument but i really like to hear a beautiful song,because i think songs are alive and they are telling us some truth,if we can get them we will enjoy alot.
and i love movies because i think a movie can tell the truth and we can get it easier that songs,and even a movie can show something that in our world is not fact,and i think with a movie (or song) we can travel anywhere we want,easily. :D

well, some good films i've seen recently: YES, Crash, Nostalghia (old classic by Tarkovsky), Maria full of Grace ( it is about colombian drug dealing).

My favorite film is Persona by the famous director Ingman Berman.

thanks i will see these moviesif i can. :D

Maybe no one has seen that particular movie.
so lets have a marathon about actors and actresses.
lets start with Robert Deniru(if i type worng plz correct it.)
i think he is a good actor in HOLLYWOOD and he has a lot of good films like A FEW GOOD MEN,... .
i love his playing so much. :D

A good De Niro character: the guerrilla plumber in Terry Gilliam's [i]Brazil.[/i]

i think most of the time he play good and you enjoy watching his movies,except few movies of him,that it is not for genral people... :D

are you refferring to Jack Nickelson? Di Nero wasnt in a few good men.  I like di nero in casino.


ratio Asymptote trigonometry geometry inradius

[hide]
Draw $ OP$ and the altitude from $ P$ to $ OB$
this creates a right triangle with angle $ 30^{\circ}$
using right triangle trig

$ \sin(30) \equal{} \frac{r}{R\minus{}r}$

$ \frac{1}{2} \equal{} \frac{r}{R\minus{}r}$

$ 2 \equal{} \frac{R\minus{}r}{r} \equal{} \frac{R}{r} \minus{} 1$

$ \boxed{\frac{R}{r} \equal{} 3}$
[/hide]

[quote="i_like_pie"]Circle $ P$, whose radius is $ r$, is inscribed in a $ 60^{\circ}$ sector of circle $ O$, whose radius is $ R$. Find the ratio $ R: r$.
[asy]import cse5; size(150); pen p=fontsize(10); pair O=(0,0), A=O+dir(60), B=O+dir(0), P=O+2/3*dir(30); D(CR(O,1)); D(MP("A",D(A),NNE,p)--MP("O",D(O),WSW,p)--MP("B",D(B),ENE,p)); D(CR(MP("P",D(P),N,p),1/3));[/asy][/quote]
[hide="solution"]
Connect one radius each of circle $ P$ to the radii of circle $ O$ that are tangent to circle $ P$.
The angles formed by the connections are right angles. Connect another line segment from the center of $ O$ to the center of $ P$. Now there are two right triangles.
Since the angle of the two radii of $ O$ is 60 degrees, each smallest angle in the right triangles are 30 degrees each since the 60 degree angle got bisected. The length of the hypoteneuse of both triangles is $ r$ less than $ R$. This can by shown by extending it by drawing another radius of circle $ P$ to touch circle $ O$. The lengths of the opposite sides of the smallest angles (The shortest sides) of both right triangles are $ r$.
In a $ 30 \minus{} 60 \minus{} 90$ triangle, the smallest leg and the hypoteneuse are in a $ 1: 2$ ratio in length. The hypoteneuse is $ 2r$, so $ R \equal{} 2r \plus{} r$.
$ R \equal{} 3r$
$ R: r \equal{} 3: 1$
[/hide]

[quote="ProtestanT"][quote="Farenhajt"][hide="No fuss"]Draw the common tangent and extend $ OA, OB$ to the respective intersections with it. You obtain a bigger equilateral triangle whose height is $ R$ and inradius $ r$, therefore $ R: r \equal{} 3: 1$.[/hide]

[hide="Also no fuss"]Since "there's an answer hidden in every question", we just click on the initial pic and analyze i_like_pie's code to obtain the same result ;)[/hide][/quote]
Can you show that the bigger triangle is equilateral? I can't see why...it's a fuss to me...so it's not really "no fuss"[/quote]

The common tangent is parallel to $ AB$, being perpendicular to $ OP$, so the bigger triangle is similar to $ \triangle OAB$.


calculus
integration
trigonometry
calculus computations

$ \int {cos(t^2 / 2)}dt$

How do you integrate this?

Maybe you made a typo, if not, the answer would be $ x\cos ^ 2 \frac{t}{2}\plus{}C$.

[quote="kunny"]Maybe you made a typo, if not, the answer would be $ x\cos ^ 2 \frac {t}{2} \plus{} C$.[/quote]

How exactly? I tried $ u \equal{} t^2 /2$  but I don't know how to evaluate $ \int { cos u du/t }$

Sorry I made typo.

What's the variable for the integral?

If it were $ x$, the answer is $ x\cos \frac{t^2}{2}\plus{}C$.

Sorry, I fixed my original post. Could you show your work for how to integrate that, though?

Thank you for quick the correction.

In case of $ t\neq 0$, Using integral by parts,

$ \int \cos \frac {t^2}{2}dt \equal{} \int \frac {1}{t}\left(\sin \frac {t^2}{2}\right)'dt$

$ \equal{} \frac {1}{t}\sin \frac {t^2}{2} \plus{} \int \frac {1}{t^2}\sin \frac {t^2}{2}dt$, I'm stuck here.

Maybe the integral is related to Fresnel Integration.

Naturally, there is no elementary antiderivative. Perhaps the original poster meant

$ \int _0^{\plus{} \infty} \cos(t^2/2)\,dt$.


Let $p$ be a prime number. Prove that there are at least $\frac{\varphi(p-1)}{2}$ numbers such as $g$ such that$0<g<p$ and $g$ be a [b]primitive root[/b] of $p^K$ for every natural $K$.

Just a sketch:

It's well-known that if a number is a primitive root modulo $p^2$, it's a primitive root modulo $p^k$ for all $k\in\mathbb N$, so we only need to look for primitive roots of $p^2$ in $(0,p)$.

It's easy to prove that those $x\in(0,p)$ which are primitive roots of $p$ but not of $p^2$ are precisely those for which $p^2|x^{p-1}-1$. If $x$ satisfies this, then $p-x$ does not, so there are at least as many primitive roots of $p^2$ in $(0,p)$ as there are primitive roots of $p$ but not of $p^2$. Since their total number is $\varphi(p-1)$, the desired result follows.

[quote="grobber"]for which $p^2|x^{p-1}-1$. If $x$ satisfies this, then $p-x$ does not[/quote]
Are you sure !!!!!!!!!

Of course, man :).

Suppose you have $p^2|x^{p-1}-1$. Then $(x-p)^{p-1}-1$ equals $x^{p-1}-1-(p-1)px^{p-2}$ plus some multiple of $p^2$. Since $p^2|x^{p-1}-1$, the necessary and sufficient condition for $p^2|(x-p)^{p-1}-1$ is that $p^2|(p-1)px^{p-2}$, which is clearly false.

Anything wrong here? :?


analytic geometry
function
inequalities
inequalities proposed

[quote="Litlle 1000t"]$ a_1,a_2,....,a_n$ be positive reals. Prove that
 $ \frac {a_1}{a_2 \plus{} a_3} \plus{} \frac {a_2}{a_3 \plus{} a_4} \plus{} ...... \plus{} \frac {a_{n \minus{} 1}}{a_n \plus{} a_1} \plus{} \frac {a_n}{a_1 \plus{} a_2} > 0.4577996$[/quote]

It's evidence! Because $ LHF \ge \frac 3{2}$ for $ n \ge 3$  :wink:

However,we have the result (in Tran Phuong'book)
$ \frac {a_1}{a_2 \plus{} a_3} \plus{} \frac {a_2}{a_3 \plus{} a_4} \plus{} ...... \plus{} \frac {a_{n \minus{} 1}}{a_n \plus{} a_1} \plus{} \frac {a_n}{a_1 \plus{} a_2} > 0.4577996n$
 :)

i'm sorry, i forgot $ n$. I edited  :wink:

The maximal  result here ( V.Drinfeld, 1969 ):
$ \frac{a_1}{a_2\plus{}a_3}\plus{}\frac{a_2}{a_3\plus{}a_4}\plus{}......\plus{}\frac{a_{n\minus{}1}}{a_n\plus{}a_1}\plus{}\frac{a_n}{a_1\plus{}a_2}\geq\gamma\cdot\frac{n}{2},$
where $ \gamma$ is $ y$- coordinate  of a crossing point of general tangent line to functions graphs $ y\equal{}e^{\minus{}x}$ and $ y\equal{}\frac{2}{e^x\plus{}e^{\frac{x}{2}}}$ with $ y$ -axis. $ \gamma\equal{}0.989...$

[quote="dduclam"]

However,we have the result (in Tran Phuong'book)
$ \frac {a_1}{a_2 \plus{} a_3} \plus{} \frac {a_2}{a_3 \plus{} a_4} \plus{} ...... \plus{} \frac {a_{n \minus{} 1}}{a_n \plus{} a_1} \plus{} \frac {a_n}{a_1 \plus{} a_2} > 0.4577996n$
 :)[/quote]

Could you show this book's link or solution?

[quote="dduclam"][quote="arqady"]The maximal  result here ( V.Drinfeld, 1969 ):
$ \frac {a_1}{a_2 \plus{} a_3} \plus{} \frac {a_2}{a_3 \plus{} a_4} \plus{} ...... \plus{} \frac {a_{n \minus{} 1}}{a_n \plus{} a_1} \plus{} \frac {a_n}{a_1 \plus{} a_2}\geq\gamma\cdot\frac {n}{2},$
where $ \gamma$ is $ y$- coordinate  of a crossing point of general tangent line to functions graphs $ y \equal{} e^{ \minus{} x}$ and $ y \equal{} \frac {2}{e^x \plus{} e^{\frac {x}{2}}}$ with $ y$ -axis. $ \gamma \equal{} 0.989...$[/quote]

The stronger result is $ RHS \equal{} 0.4577996n$ (by Tran Phuong-2004)  :)[/quote]
Are you sure that the Tran Phuong's inequality is stronger? :wink:

[quote="arqady"]
The stronger result is $ RHS \equal{} 0.4577996n$ (by Tran Phuong-2004)  :)[/quote]
Are you sure that the Tran Phuong's inequality is stronger? :wink:[/quote]

He tell me,so. But that result is not the best result  :wink: The best result (which i know,althought i don't know solution) is $ RHS\equal{}(\sqrt2\minus{}1)n$ (1991-Kavant Magazine)

[quote="Litlle 1000t"]
Could you show this book's link or solution?[/quote]

That solution is very long and hard! The Tran Phuong'book is [b]Diamonds in Mathematical Inequalites[/b] (published in Vietnam)

[quote="dduclam"][quote="dduclam"]
The stronger result is $ RHS \equal{} 0.4577996n$ (by Tran Phuong-2004)  :)[/quote]
[quote="arqady"]Are you sure that the Tran Phuong's inequality is stronger? :wink:[/quote]

He tell me,so. But that result is not the best result  :wink: The best result (which i know,althought i don't know solution) is $ RHS \equal{} (\sqrt2 \minus{} 1)n$ (1991-Kavant Magazine)
[/quote]
I don't think so. See here:
http://kvant.mccme.ru/1991/04/istoriya_odnogo_neravenstva.htm
Are you think that the Drinfeld's proof is wrong?

[quote="arqady"]
I don't think so. See here:
http://kvant.mccme.ru/1991/04/istoriya_odnogo_neravenstva.htm
Are you think that the Drinfeld's proof is wrong?[/quote]

Thanks arqady! I will read and consider,again. But not now,i'm too busy    :(

Wow. It looks like the proof is elementary although I don't know Russian.
Can anyone translate into english?


Arrange the following in increasing order of magnitude:

$ 2^{100}, 3^{75}, 5^{50}$.

Actually, it's $ \boxed{5^{30}, 2^{100}, 3^{75}}$, if I'm not mistaken

[hide] We can make comparisons among the $ 25$th roots of those numbers. Their $ 25$th roots are $ 2^4$, $ 3^3$, and $ 5^2$. Clearly, the order of these is $ 2^4$, $ 5^2$, $ 3^3$. Thus the order in increasing magnitude of the original problem is $ \boxed{2^{100},5^{50},3^{75}}$.[/hide]

@AIME15USAMO The problem says $ 5^{50}$, not $ 5^{30}$. (Then again, maybe it was just a typo on your part).

Oh whoops, misread the problem :blush:
Thank you for correcting me.

except how can we figure this out without a calculator?

[quote="gaussintraining"]We can make comparisons among the $ 25$th roots of those numbers. Their $ 25$th roots are $ 2^4$, $ 3^3$, and $ 5^2$.[/quote]
The values of $ 2^4$, $ 3^3$, and $ 5^2$ shouldn't be too hard to compute:

$ 2^4 \equal{} 2 \cdot 2 \cdot 2 \cdot 2\equal{} 4 \cdot 2 \cdot 2\equal{}8 \cdot 2\equal{}\mathbf{16}$
$ 3^3\equal{}3 \cdot 3 \cdot 3\equal{}9 \cdot 3 \equal{} \mathbf{27}$
$ 5^2\equal{}5 \cdot 5 \equal{} \mathbf{25}$

And we order this as $ 16$, $ 25$, and $ 27$.

Well, we would actually order them as $ 2^{100}, 5^{50}, 3^{75}$ because it asks to arrange the original numbers, not the $ 25th$ roots of the numbers.


algebra
polynomial
function
ratio
induction
algebra proposed

The problem is about real polynomial functions, denoted by $f$, of degree $\deg f$.

a) Prove that a polynomial function $f$ can`t be wrriten as sum of at most $\deg f$ periodic functions.

b) Show that if a polynomial function of degree $1$ is written as sum of two periodic functions, then they are unbounded on every interval (thus, they are "wild").

c) Show that every polynomial function of degree $1$ can be written as sum of two periodic functions.

d) Show that every polynomial function $f$ can be written as sum of $\deg f+1$ periodic functions.

e) Give an example of a function that can`t be written as a finite sum of periodic functions.

[i]Dan Schwarz[/i]

a) By induction. Let $f$ be a polynomial that can be written as sum of $\deg f = n$ periodic functions $f_{1},f_{2},\ldots,f_{n}$ with periods $t_{1},t_{2},\ldots,t_{n}$.
We have:
$f(x) = f_{1}(x)+\ldots+f_{n}(x)$
Since $f_{1}(x)-f_{1}(x+t_{1}) = 0$, we get:
$f(x)-f(x+t_{1}) = f_{2}(x)-f_{2}(x+t_{1})+\ldots+f_{n}(x)-f_{n}(x+t_{1})$
Note that $f(x)-f(x+t_{1})$ is a polynomial of degree $n-1$, and it's written as sum of $n-1$ periodic functions.
So we can "go down" until a polynomial of degree 1 is written as sum of only one periodic function... but a polynomial of degree 1 is not periodic!  :)

b) Since $f(x)$ is periodic if and only if $af(x)+b$ is, with a,b nonzero constants, and the same for boundedness, we can suppose the polynomial to be x.
Let $f_{1}(x)+f_{2}(x) = x$ for all x, and the period of $f_{1},f_{2}$ be $t_{1},t_{2}$.
Without loss of generality we suppose $f_{1}(0) = 0$. Then $f_{1}(nt_{1}) = 0$ for all integers n, and $f_{2}(nt_{1}) = kt_{1}$ for all integers n, and $f_{2}(nt_{1}+mt_{2}) = nt_{1}$ for all integers n,m. If the ratio $\frac{t_{1}}{t_{2}}$ is rational, than  $f_{1}(x)+f_{2}(x)$ is a periodic function, but this is impossible. Therefore $\frac{t_{1}}{t_{2}}$ is irrational, and by a well-known theorem $nt_{1}+mt_{2}$ can go as near as we want to every real.
For each $\epsilon > 0$, there are integers n,m such that $0<nt_{1}+mt_{2}<\epsilon$, and we can also make $nt_{1}$ very large by going nearer to 0, and then multiplying n,m by a large factor (if $\epsilon < |t_{2}|$ we must have $n \ge 1$).
Then we can also go near to each real a with a very large $nt_{1}$, to which corresponds a very large value of $f_{2}$ which can't be bounded in an interval. The same for $f_{1}$.

c) I think that here we need the Axiom of Choice.
Take two nonzero reals $t_{1},t_{2}$ such that $\frac{t_{1}}{t_{2}}$ is irrational. Let $A = \{nt_{1}+mt_{2}| n,m \in \mathbb{Z}\}$. $a-b \in A$ is an equivalence relation among the reals. Let $[a]$ be the equivalence class of a, and let $c: \{[a]: a \in \mathbb{R}\}\rightarrow \mathbb{R}$ a choice function of the classes.

Using the definition of our equivalence relation and the fact that $\frac{t_{1}}{t_{2}}$ is irrational, we show that for each $x \in \mathbb{R}$ there exist unique integers n,m such that $x = c([x])+nt_{1}+mt_{2}$. Then we define:
$f_{1}(x) = f_{1}(c([x])+nt_{1}+mt_{2}) = c([x])+mt_{2}$
$f_{2}(x) = f_{2}(c([x])+nt_{1}+mt_{2}) = nt_{1}$

$f_{1},f_{2}$ are periodic, since obviously $f_{1}(x+t_{1})= f_{1}(c([x])+(n+1)t_{1}+mt_{2})= c([x])+mt_{2}= f_{1}(x)$ and the same for $f_{2}$.
And also $f_{1}(x)+f_{2}(x) = c([x])+mt_{2}+nt_{1}= x$.

d) This can be done by induction, using c). Suppose the statament is true for all polynomial with deg < n, and that you can choose freely the periods of the functions of which the polynomial is sum. (not completely freely, their  ratio is irrational).
Take the polynomial f with $\deg f = n$. Choose a number $t_{n+1}$ as you want.
Consider $f(x)-f(x-t_{n+1})$. It is a polynomial of degree n-1. Then there exists periodic functions such that
$f(x)-f(x-t_{n+1})= g_{1}(x)+\ldots+g_{n}(x)$.

We first prove the lemma: if $g$ is a periodic function of period t, and $\frac{t}{t_{1}}$ is irrational, then there exists a function $g'$ such that $g(x) = g'(x)-g'(x-t_{1})$ and $g'$ is still periodic of period t.
The proof is somehow the same of the point c)... with the same terminology, define:
$g'(x) = g'(c([x])+nt+m t_{1}) = \sum_{i=0}^{m}g(c([x])+nt+it_{1})$ if $m \ge 0$, 0 is m = -1 and $\sum_{i=1}^{-m-1}g(c([x])+nt+-it)$ if $m<-1$.
Then it's easily seen that g' satisfies $g'(x)-g'(x-t_{1}) = g(x)$ and $g'(x) = g'(x+t)$.

Using the lemma, we get that there are periodic functions such that:
$f(x)-f(x-t_{n+1}) = f_{1}(x)-f_{1}(x-t_{n+1})+\ldots+f_{n}(x)-f_{n}(x-t_{n+1})$
Now we have just to add the suitable function $f_{n+1}$.
For each $0\le x < t_{n+1}$ define:
$f_{n+1}(x) = f(x)-f(x-t_{n+1})-f_{1}(x)+f_{1}(x-t_{n+1})+\ldots-f_{n}(x)+f_{n}(x-t_{n+1})$
And then extend $f_{n+1}$ to the reals using periodicity, and check that
$f(x) = f_{1}(x)+f_{2}(x)+\ldots+f_{n}(x)$ for all real x.  :)

my solution to c)
[hide="solution"]
let $B$ be a Hamel-basis of $\mathbb{R}$ over $\mathbb{Q}$ (as a vectorial space), and let $B_{1},B_{2}$ be a partition of it.
now, take functions $f_{1},f_{2}$, define them over $B$, then extend them as linear maps :
if $x\in B_{1}$, then $f_{1}(x)=x$ and $f_{2}(x)=0$
if $x\in B_{2}$, then $f_{1}(x)=0$ and $f_{2}(x)=x$.
now, for $y\in \mathbb{R}$, $y=\sum a_{i}x_{i}$ with $x_{i}\in B$ and $a_{i}\in \mathbb{Q}$.
$f_{1}(y)+f_{2}(y)=\sum_{x_{i}\in B_{1}}a_{i}f_{1}(x_{i})+\sum_{x_{i}\in B_{2}}a_{i}f_{2}(x_{i}) = \sum a_{i}x_{i}= y$.
also it is easy to check that $f_{1}$ is periodic (with period any $x\in B_{1}$) and $f_{2}$ is periodic also (with period any $x\in B_{2}$).
[/hide]

edit : oops, i just said a biiiiiiiiiiiiiig stupid thing.  :blush:

[quote="maky"]my solution to c) (without AC)
let $B$ be a Hamel-basis of $\mathbb{R}$ over $\mathbb{Q}$ (as a vectorial space), ...
[/quote]

But this needs AC !

[quote="maky"] e) give an example of a function that can`t be written as a finite sum of periodic functions.[/quote]

Take $f(x)=e^{x}$

Demo:

If $e^{x}$ is sum of $n$ periodic functions $e^{x}=\sum_{i=1}^{n}f_{i}(x)$, with $f_{i}(x+t_{i})=f(x)$ and $t_{i}>0$, then :

$e^{x+t_{n}}=\sum_{i=1}^{n-1}f_{i}(x+t_{n})+f_{n}(x+t_{n})=\sum_{i=1}^{n-1}f_{i}(x+t_{n})+f_{n}(x)$

$e^{x+t_{n}}-e^{x}= \sum_{i=1}^{n-1}(f_{i}(x+t_{n})-f_{i}(x))$

$e^{x}=\sum_{i=1}^{n-1}\frac{f_{i}(x+t_{n})-f_{i}(x)}{e^{t_{n}}-1}$

$e^{x}=\sum_{i=1}^{n-1}h_{i}(x)$, with $h_{i}(x)=\frac{f_{i}(x+t_{n})-f_{i}(x)}{e^{t_{n}}-1}$, $h_{i}(x+t_{i})=h_{i}(x)$ and $t_{i}>0$

So, if $e^{x}$ is sum of $n$ periodic functions, $e^{x}$ is sum of $n-1$ periodic functions, which implies $e^{x}$ is sum of $1$ periodic function, which is false.

So $e^{x}$ can't be sum of a finite number of periodic functions.

Nice!
Here is my proof:
Suppose a function $f$ is sum of n periodic functions $f_{1},\ldots,f_{n}$ with periods $t_{1},\ldots,t_{n}$, $t_{i}\neq 0$.
Let $T = \{t_{1},\ldots,t_{n}\}$. Then you can see by induction that, for all real x,

$\sum_{X \subset T}(-1)^{|X|}f\left( x+\sum_{t \in X}t \right) = 0$.

I think that, with some conditions on the $t_{i}$, this can become a sufficient condition.

If we put $f(x) = e^{x}$, we can factor the condition as:
$e^{x}(e^{t_{1}}-1)(e^{t_{2}}-1)\cdots (e^{t_{n}}-1) = 0$

But this is nonzero because all $t_{i}$ are different from 0  :D


geometry
perimeter
circumcircle
trigonometry
perpendicular bisector
trig identities
Law of Sines

The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?

$ \textbf{(A)}\ \frac{3\sqrt2}{\pi} \qquad
\textbf{(B)}\ \frac{3\sqrt3}{\pi} \qquad
\textbf{(C)}\ \sqrt3 \qquad
\textbf{(D)}\ \frac{6}{\pi} \qquad
\textbf{(E)}\ \sqrt3\pi$

hello, we have $ p\equal{}3a$ where $ a$ is the sidelength of the given triangle, further we get $ R\equal{}\frac{a}{2\cos(30^{\circ})}$, from here we get $ 3a\equal{}\left(\frac{a}{2\cos(30^{\circ})}\right)^2 \Leftrightarrow a\equal{}\frac{9}{\pi}$, so our radius is $ R\equal{}\frac{9}{\pi\sqrt{3}}\equal{}\frac{3\sqrt{3}}{\pi}$, the answer is $ \boxed{B}$.
Sonnhard.

Let the radius of the circle be $ r$. The area of the circle is obviously pi$ r^2$. The height of the equilateral triangle be $ h$. Let the side length of the triangle be $ s$. Using the pythagorean theorum we find that;

$ (s/2)^2 \plus{} h^2 \equal{} s^2$
$ (s^2)/4 \plus{} h^2 \equal{} (4s^2)/4$
$ h^2 \equal{} (3/4)(s^2)$
$ (4/3)h^2 \equal{} s^2$
Take the square root to get;
$ 2h/\sqrt{3} \equal{} s$
So the perimeter of the triangle is
$ 6h/\sqrt{3}$
Now if we can find a way to compare $ h$ to $ r$, we can find an equation to solve for $ r$.
If you break the triangle into a bunch of 30-60-90 triangles, you should find that $ h$ turns out to be $ 3/2$ of $ r$.

So if $ h \equal{} (3/2)r$, and pi$ r^2 \equal{} 6h/\sqrt{3}$
Plugging $ (3/2)r$ for $ h$, we get;

pi$ r^2 \equal{} 6(3/2)r/\sqrt{3}$
pi$ r^2 \equal{} 9r/\sqrt{3}$
Multiplying the right side of the equation by $ \sqrt{3}/\sqrt{3}$, we get;
pi$ r^2 \equal{} 3r\sqrt{3}$
pi$ r \equal{} 3\sqrt{3}$
So the radius of the circle equals $ (3\sqrt{3})/$pi, or B.

[hide] Let the side of the equilateral triangle be 2s. Consider the circumcenter and draw a perpendicular bisector down to one of the sides. Also, connect the circumcenter to one of the endpoints of the side. We now have a 30-60-90 triangle. The radius has length $ \frac{2s\sqrt{3}}{3}$.

Using the fact that the perimeter of the triangle is the same as the area of the circle, we find $ 6s\equal{}\frac{4s^2\pi}{3}$ which gives $ s\equal{}\frac{9}{2\pi}$ That means the radius is $ \textbf{(B)}\ \frac{3\sqrt3}{\pi}$.[geogebra]6c3f3927e40fbc3f3c043709da452f1b927b5242[/geogebra] [/hide]

[hide]

Let the side length of the equilateral triangle be $ s$. By the Extended Law of Sines, we have $ R\equal{}\frac{s}{2\sin 60^\circ}\equal{}\frac{s}{\sqrt{3}}$. The area of this, $ \pi R^2$, is $ \frac{s^2\pi}{3}$. This is equal to $ 3s$. So, $ 3s\equal{}\frac{s^2\pi}{3}$. Hence, $ s\equal{}9/\pi$. Finishing, the radius is $ \frac{9}{\pi\sqrt3}\equal{}\frac{3\sqrt3}{\pi}\Longrightarrow\mathrm{ (B) \ }$

[/hide]

is the circle inside or outside the triangle? iam confused on the vocab there


farz konid $f: \mathbb{Z}\longrightarrow \mathbb{Z}$ va $f$ chandjomlei nabashad ke be ezaye har $a,b,m\in \mathbb{Z}$ agar $a\equiv b (\mod m)$ on vaght  $f(a)\equiv f(b) (\mod m)$.sabet konid ke be ezaye har $k$

$\lim_{n\rightarrow \infty}\frac{|f(n)|}{n^k}=\infty$

in soal ke ghalate masalan tebe hamani sharayete masale ra darad amma an had mosavi binahaiat nist.

Ooooops!!!!

ye sharte dige ham dare ineke $f(x)$ chandjomlei nist.


USAMTS calculus logarithms inequalities trigonometry AoPSwiki Asymptote

[quote="AMC37.5"]I could not find a comprehensive rules page on the USAMTS web site (only the brief [url]http://www.usamts.org/About/U_AbRules.php[/url]). I have some questions:

1. Is there a more detailed rules page that I missed?
2. Would the following types of solutions get credit:
- brute-force computer programs, if they run in a reasonable amount of time
- calculus-based solutions
3. Can solutions to previous USAMTS questions (in the same year) be referenced or used as justifications in subsequent questions? If so, will you get credit for an otherwise correct solution to a subsequent question that references an invalid proof for a previous question?
4. Are e-mails to solutions@usamts.org read by humans?[/quote]

Look at the FAQ section. from what i know:

1. the FAQ section and also a print out in pdf form comes with your entry form.
2. computer programs: yes, but you must include an outline of the program and your output.
calculus: idk, i dont see why not.
3. idk
4. yes

3. Probably, but why don't you just include the proof? The solutions are already on the internet already, it doesn't take much work, so why take the chance?

[quote="AMC37.5"]1. Is there a more detailed rules page that I missed?[/quote]There is also the [url=http://www.usamts.org/TipsFAQ/U_FAQ.php]FAQ page[/url].
[quote]2. Would the following types of solutions get credit:
- brute-force computer programs, if they run in a reasonable amount of time[/quote]Yes, provided that you prove that your program is correct and solves the problem.
[quote]- calculus-based solutions[/quote]Yes, but all USAMTS problems can be solved without calculus.
[quote]3. Can solutions to previous USAMTS questions (in the same year) be referenced or used as justifications in subsequent questions? If so, will you get credit for an otherwise correct solution to a subsequent question that references an invalid proof for a previous question?[/quote]Anything that you find on the internet (or anywhere else) can be referenced (if properly cited).  If the stuff you're referencing is not "well-known" and/or from a reputable source, you should justify any claimed assertions.  It's difficult to precisely define this -- you will have to use your judgment.  If you reference something invalid, you will not receive full credit.  I don't understand the last part of your question -- there should never be a need for a subsequent question to reference a previous question; all of the questions are independent.  (Unless you mean parts (a) and (b) of the same question, in which case it depends on the grading rubric used, but usually you can get some partial credit if, for example, you assume the result of part (a) of a question to correctly prove part (b), even if your solution to part (a) was incorrect; it depends on the particular circumstance.)
[quote]4. Are e-mails to solutions@usamts.org read by humans?[/quote]Yes and no.  A human processes them.  I'm not sure why you're asking this.

Are we allowed to cite obvious calculator computation (eg $ 0.99 <\log\pi^{2}< 1$) or something of the sort?

[quote="AMC37.5"]2. Would the following types of solutions get credit:
- brute-force computer programs, if they run in a reasonable amount of time[/quote]
We would even accept programs that run in an unreasonable amount of time. We aren't going to run them ourselves. To verify that the program runs, we want an explanation of why you think that the program will give the correct answer. For example, suppose you have already proven that the answer is an integer between 1 and $ 10^{20}$. Then, "I wrote a program that tested all of those integers. The only answers that passed the test were 3 and 5," would be an acceptable though inelegant conclusion to your solution.

[quote]3. Can solutions to previous USAMTS questions (in the same year) be referenced or used as justifications in subsequent questions? If so, will you get credit for an otherwise correct solution to a subsequent question that references an invalid proof for a previous question?[/quote]
We NSA graders don't have access to the questions or solutions to previous USAMTS problem sets during our grading sessions. Even if we did, we wouldn't have time to look anything up. Your solutions should be self-contained, except for referencing well-known theorems. If you reference an obscure theorem, please state the full theorem.

[quote="ccy"]Are we allowed to cite obvious calculator computation (eg 0.99 <\log\pi^{2}< 1) or something of the sort?[/quote]
Yes. However all such calculations could have rounding error and your proof must take the possibility of rounding error into account. Your inequality example does that, so it is just fine.

Erin Schram
USAMTS grader

[quote="roadnottaken"][quote="Erin J. Schram"][quote="ccy"]Are we allowed to cite obvious calculator computation (eg 0.99 <\log\pi^{2}< 1) or something of the sort?[/quote]
Yes. However all such calculations could have rounding error and your proof must take the possibility of rounding error into account. Your inequality example does that, so it is just fine.[/quote]

What?  I distinctly remember two questions from last year, in which the rubric specified that using a calculator to "solve" complicated trig expressions was not a true solution, and was grounds for deducting a point.[/quote]
solving complication trig stuff $ \ne$obvious calculator computation

[quote="buzzer11"][quote="rrusczyk"][quote="Erin J. Schram"]Then, "I wrote a program that tested all of those integers. The only answers that passed the test were 3 and 5," would be an acceptable though inelegant conclusion to your solution.
[/quote]

Actually, students need to do a little more than this: they should show us their program and explain why it works.

In other words, you have to prove that your program does indeed perform a correct test.  (Sometimes, it's not so trivial to test an assertion for certain integers!)[/quote]

How many points would you get for "I left the program running for a long long long time and the only solutions that turned up were _____."?[/quote]

That's an impossible question to answer, but I'll note that this definitely doesn't constitute 'proof'!

[quote="13375P34K43V312"]You know, your (by your, everyone's) time would be better spent trying to find an elegant non-programming solution than to ask all these questions.[/quote]He asked 1 question...

[quote="rrusczyk"][quote="Erin J. Schram"]Then, "I wrote a program that tested all of those integers. The only answers that passed the test were 3 and 5," would be an acceptable though inelegant conclusion to your solution.
[/quote]

Actually, students need to do a little more than this: they should show us their program and explain why it works.

In other words, you have to prove that your program does indeed perform a correct test.  (Sometimes, it's not so trivial to test an assertion for certain integers!)[/quote]
Sorry about oversimplifying my example. We graders do like short solutions, but not as short as what I wrote. I left out too many details because I would have had to invent a problem to add them.

[quote="13375P34K43V312"]You know, your (by your, everyone's) time would be better spent trying to find an elegant non-programming solution...[/quote]
That is our not-so-hidden agenda. We prefer the elegant non-programming solution, because the solver discovers more mathematics. However, we would rather have a programming solution than have anyone give up.

[quote="roadnottaken"]What? I distinctly remember two questions from last year, in which the rubric specified that using a calculator to "solve" complicated trig expressions was not a true solution, and was grounds for deducting a point.[/quote]
That is due to the rounding error difficulty that I mentioned. Consider the problem, "Prove that $ \cos^{2}(36^{\circ})\plus{}\cos^{2}(54^{\circ})\equal{}1$." I could claim that $ \cos(36^{\circ}) \equal{} 0.809$ and $ \cos(54^{\circ}) \equal{} 0.588$, so the sum of their squares is $ 1.000$. But that is not a proof, because my calculator actually said $ 1.000225$, which is not $ 1$. Sure, when I keep all the digits of the cosines in my calculator's memory, it says the sum of the squares is $ 1.000000$, but that just means that the rounding error is less that $ 10^{\minus{}6}$, not that there is no rounding error. The rounding error could cause no visible error, but so long as the answer could have rounding error, it does not count as a proof.

Erin Schram
USAMTS grader

[quote="Altheman"]5. If I have finished my solutions, would there be any benefit to submitting them early?[/quote]

There is not a competitive benefit to doing so (i.e., there aren't extra points or anything like that).

[quote="Altheman"]I was more thinking along the lines of: would I get results earlier? But based on the idea of grading sessions, I suppose that wouldn't happpen. I answered my own question...lol[/quote]
You get your checkmark earlier, which provides a secure feeling :P

Wondering where can I get Asymptote/Geo's Sketch Pad if I wanna turn in my solutions w/ diagrams?

The [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_%28Vector_Graphics_Language%29]AoPSWiki[/url] provides an excellant tutorial on installing and using asymptote.

Thank you, Altheman.  I have deleted the offending posts.  Do not discuss anything about the problems - not how you created diagrams, not how long it took you, nothing.  (It is OK to ask things like 'How do I install Asymptote?')

If we use a programming solution, should we state what programming language was used?  (What programing language is used in the TI-83 series calculators?)

I will repeat: DO NOT DISCUSS SPECIFIC PROBLEMS WHILE A ROUND IS IN PROGRESS!

I have again removed offending posts, one of which was immediately after my previous statement to not discuss specific problems.

So, I'll say it again: DO NOT DISCUSS SPECIFIC PROBLEMS WHILE A ROUND IS IN PROGRESS!

I don't exactly understand your question.  If you're saying that you used a calculator on one of the problems to find a number(which looks kind of hard and time-consuming for Round 3), then you should just say, "I plugged something into my calculator, and it spit out ******".  If you're saying you used a calculator to get some insight on how to solve a problem but then proved it without the calculator, you shouldn't "cite" the calculator because a calculator can't really qualify as a source, and you didn't use it in the proof.  Only cite stuff that comes from a source and is not too well known.

[quote="variable"]Should I, please, cite calculator-use, if I use a calculator to number-crunch and provide hints at how a proof might run, even if a resulting proof does not, in fact, use any of that number-crunching directly?[/quote]

A proof doesn't have to give any insight as to how you came about the proof. It just has to be a correct logically.

Thanks.  I didn't know if it came under the same category as citing collaboration on an AoPS Class Problem Set (and yes, I do realize that collaboration is illegal for the USAMTS!).

Again, thanks, and Happy New Year!

variable


Hello all. I am thinking about buying Pokemon Diamond or Pearl with money I get for my birthday this year. Should I buy this game if I played and enjoyed the previous ones, or should I save myself from disappointment?

Thanks in advance and hoping I put this in the right forum,
dynamo729.

I found it very fun, especially with a new generation of pokemon. PLus, judging by your avatar, chansey is a beast and pretty ez to get.

I would definitely recommend it.  For more information on the topic of Pokemon, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=803926712&t=201270]see here[/url].

I know that it's kinda off topic, but in the link that cf249 posted, some people mentioned some crazy (I U) Pikachu thing. What is that?

And dynamo, get pearl, because it has SLOWPOKE!!!

Although I haven't played this one yet, but it should be pretty similar. However, release date for english is as of yet unannounced, so you might have to wait a while if you want this:

http://en.wikipedia.org/wiki/Pok%C3%A9mon_Platinum

[quote="(^_^)"]pokemon sucks.[/quote]

ditto

[quote]The message contains only quote and code tags. Please add some text of your own.[/quote]

er... no it doesnt

@erabel:  In combinatorics, $ \binom{6}{2}$ can be read as "6 choose 2".  So, $ \binom{I}{u} Pikachu$ is a mathematical way of saying "I choose you, Pikachu!"

Thanks for replying guys. However, ernie and (^_^) (and a few others, I'm just picking on these two), I'm looking for people who have played the previous games and Diamond or Pearl and have an opinion backed with reasons, and your previous were basically spam, but if you have an actual opinion to contribute backed with some reasons, I'd love to have you comment.

I personally think the best pokemon games were gold, silver and crystal. But I would wait for Pokemon Platinum

Hm... why do you think trhe best were g/s/c and why would you wait for platinum

....i kno nothing about pokemon and am just posting for the heck of it.  :P

no wait. cross that. i am a genious at the OLD pokemon, the original 151. but i dont like any of the new stuff... the pokemon aren't cute anymor!  :mad: and they have lame names!  :mad:

For one thing, in g/s/c the story line included the Johto and Kanto Regions. All the others were only one region (Hoenn, Kanto, Orre, Sinnoh). Also I think there is too much pokemon/ moves to memorize. (400+pokemon, 250+moves).

Dreambold: Relevant to your interests.

http://www.vgcats.com/comics/images/080519.jpg

...ok just to tell you, I'm 12.  :wink:

was that comic supposed to be an insult??? im not sure.

poor lucario...  :(  he's ok. he still looks good.

they're rite, there ARE too many legendaries.

I SAY IT ENDS WITH MEW.

Um sure whatever.
Ask a mod to move this to Games, from fun factory, if you ever want to be able to see your posts again without having to scroll down and down and down.

Hey guys, I've decided to buy Pokemon Diamond when I get the money. (Which means I'll also need a DS :D) Thanks for your input and when I get it I'll let you know it will most likely be in about 2 and a half weeks. :)

Thanks.

dynamo729


Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \frac{3}{\sqrt{2}}\cdot \sqrt{\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}} \le \sqrt{\frac{a}{a\plus{}b}}\plus{}\sqrt{\frac{b}{b\plus{}c}}\plus{}\sqrt{\frac{c}{c\plus{}a}} \le \frac{3}{\sqrt{2}}\]
:)

Yeah :D . I don't know how to prove them : :(  =))
$ (\sum \sqrt {\frac {a}{a \plus{} b}})^2 \leq 2(a \plus{} b \plus{} c)( \sum \frac {a}{(a \plus{} b)(a \plus{} c)} ) \\ \equal{} 4\frac { (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} abc} \leq \frac {9}{2}$

[quote=luofangxiang]The following inequality is also true
[img]http://s15.sinaimg.cn/middle/0018zOAxgy71y4fXBJY4e&690[/img][/quote]
$$\sum_{cyc}\sqrt{\frac{a}{a+b}}\geq\sqrt{\frac{\sum\limits_{cyc}(a^2+ab)}{a^2+b^2+c^2}}$$ is also true.

The following inequality is also true!

Let $a,b,c>0$,prove that
\[\sum{\sqrt{\frac{a}{a+b}}}\geq{\sqrt{1+\frac{7}{2}\cdot\frac{ab+bc+ca}{a^2+b^2+c^2}}}\]

By the way,we have

\[\sum{\sqrt{\frac{a}{a+b}}}\leq{\sqrt{1+\frac{7}{2}\cdot\frac{a^2+b^2+c^2}{ab+bc+ca}}}\]

Hence,if $a,b,c>0$,then
\[\sqrt{1+\frac{7}{2}\cdot\frac{ab+bc+ca}{a^2+b^2+c^2}}\leq{\sum{\sqrt{\frac{a}{a+b}}}}\leq{\sqrt{1+\frac{7}{2}\cdot\frac{a^2+b^2+c^2}{ab+bc+ca}}}\]


A gumball machine contains 12 red, 6 blue, 1 white and 7 green gumballs. What is the least number of gumballs Cho must buy to guarantee having 3 gumballs of the same color?

Simply writing down those numbers give no insight as to how you got them.

Let us conisder the worst-case scenario, that is, how many gumballs can we get and still not have 3 of the same color. This will occur with 2 red, 2 blue, 1 white, and 2 green, or 7 total gumballs. No matter what color the 8th gumball is, Cho will have 3 of the same color. So our answer is indeed 8.


function
integration
probability and stats

It has a density function of $ f(y)\equal{}  \frac{my^{m\minus{}1}}{\alpha}e^{{\minus{}y^m}/\alpha}$ where $ 0\leq y \ll \infty$ and $ \alpha, m \gneqq 0$
Find the mean and variance for a Weibull distributed random variable with $ m\equal{}2$

I know the answers are $ E[Y]\equal{} \alpha\Gamma(1\plus{}\frac{1}{m})\equal{}\frac{\alpha\sqrt{\pi}}{2}$ and
$ V[Y]\equal{} \alpha^{2}\Gamma(1\plus{}\frac{2}{m})\minus{} {E}^{2}[Y]\equal{}\alpha(1\minus{}\frac{\pi}{4})$

but I'm having trouble integrating $ E[Y]\equal{}\int^\infty_0 y\frac{2y}{\alpha}e^{{\minus{}y^{2}}/\alpha} dy$ to fit those answers.

are you sure about the density of the Weibull distribution: http://en.wikipedia.org/wiki/Weibull_distribution

It's similar to the one posted on this website.
[url]http://www76.wolframalpha.com/input/?i=weibull[/url]


Given an inscribed circle with center S inside the sector AOB with AO=OB=R. Also angle(AOB)=2 :theta: .
Find angle(ASB) if cos^3 (:theta:) +cos^2 (:theta:) +cos (:theta:) =1.

[quote="hardsoul"]Given an inscribed circle with center S inside the sector AOB with AO=OB=R.Also angle(AOB)=2 :theta: .
Find angle(ASB) if cos^3 (:theta:) +cos^2 (:theta:) +cos (:theta:) =1.[/quote]

You should have said that $AOB$ is a [b]circular[/b] sector with radius $R = OA = OB$.

Assume a circle $(S)$ is touching the circle $(O)$ with radius $R$ internally at a point $T$ and $AB$ is a chord of $(O)$ with the midpoint $S$. Assume that the radius $r$ of the circle $(S)$ is such that its tangents form the points $A,\ B$ intersect at the center $O$ of the circle $(O)$, i.e., $AO = BO = R$. The question is, what is the angle $\theta = \angle AOB$?

Let $P,\ Q$ be the tangency points of the circle $(S)$ with the lines $AO,\ BO$.

$OS = R\ \cos\theta$
$r = ST = SP = OS \sin\theta$
$R = OS + ST = OS(1 + \sin\theta) = R \cos\theta (1 + \sin\theta)$
$1 = \cos\theta(1 + \sin\theta)$

Rearranging the equation for $\cos\theta$ and squaring it to get rid of $\sin\theta$,

$1 - \cos^2{\theta} = \frac{1 - 2 \cos\theta + cos^2{\theta}}{\cos^2{\theta}}$
$\cos^4{\theta} - 2 \cos\theta + 1 = 0$

Obviously, this quartic equation for $\cos\theta$ has one root equal to 1. Factoring it out:

$\cos^4{\theta} - 2 \cos\theta + 1 = (\cos\theta - 1)(\cos^3{\theta} + \cos^2{\theta} + \cos\theta - 1)$

If the remaining cubic factor is zero, we get exactly the equation that the angle $\theta$ is supposed to satisfy. Consequently, the angle $\angle ASB = \pi$.

Did you work backwards: if angle(ASB)=pi then cos^3 ( :theta: ) +cos^2 ( :theta: ) +cos ( :theta: ) =1,and because there's only one  :theta:  satisfying the equation there is no other posibilities.

I cheated, numerically solving the equation

$\cos^3{\theta} + \cos^2{\theta} + \cos\theta = 1$

(about 8 iteration). The solution gave the angle $\theta \doteq 57.065^o$. Then I draw 2 radii of the circle $(O)$ approximately forming this angle and inscribed a circle $(S)$ in the sector $AOB$. It appeared that the angle $\angle ASB = \pi$ and I looked for a reason why.

yetti


Hi,

I'll be taking SAT next year and to be honest, I'm not confident to get a good grade on the writing/vocab. I was wondering if anyone can offer me a good book(s).

Also, I saw this online course in collegeboard.com and it looks good but I don't want to do it without knowing how hard it is and also, how useful that is.

Thanks for all your help.

---> I don't think I'll have problem in math.  :D

[quote="blahblahblah"]It's worth it to brush up on the math anyways.  I knew a very bright kid who qualified for the CMO (and is currently going to an Ivy League school) who didn't score higher than 760 on the math.  So a bit of practice can make sure that you get the best results.[/quote]

I'll do that for sure on math. I'm going to be really careful and working to get perfect on math section which won't be too hard for me.

Anyone knows good SAT books?

I've tried almost everything in the market (do NOT do this), and nothing beats College Board's own material; they have fewer errors and less contrived questions than the rest. Their new blue book, the old 10 Real SATs book if you can find it (it's still helpful for the reading comprehension bit), and the SAT online course, if you like, will give you plenty of *official* practice. Oh, and if you know people who signed up for the Question-and Answer service, then ask to borrow their copy of the test.

Good luck.

BTW, there are a couple other posts on this, u might want to check those out.


limit
real analysis
real analysis unsolved

find the limit of sum of sequence as n-> infinity of 9n/((2n-1)(2n+1))

Well, sure, if that's what the O.P. meant. (To be honest, I don't know for sure what the O.P. meant.)

The limit comparison test settles this quickly. That is, when I look at $ \sum_{n\equal{}1}^{\infty}\frac{9n}{(2n\minus{}1)(2n\plus{}1)},$ I see essentially a constant times $ \sum_{n\equal{}1}^{\infty}\frac1n.$

All the careful arithmetic that kenn4000 did is just overkill; you don't need that elaborate a proof.

[quote="mathwizarddude"][quote="kenn4000"]i got infinity:

${ lim_{n\to\infty}\sum^{n}_{k = 1}\frac {9k}{(2k - 1)(2k + 1)} = lim_{n\to\infty}\sum^{n}_{k = 1}(\frac {\frac {9}{4}}{2k - 1} + \frac {\frac {9}{4}}{2k + 1}}) =$

$ lim_{n\to\infty}\sum^{n}_{k = 1}( \frac {\frac {9}{2}}{2k - 1}) + \frac {9}{4} ( \frac {1}{2n + 1} - 1) =$

$ lim_{n\to\infty}\frac {9}{2}(H_{2n} - \frac {1}{2}H_n) - \frac {9}{4} ( 1 - \frac {1}{2n + 1}) =$
$ lim_{n\to\infty}\frac {9}{4}(H_{2n} + (H_{2n} - H_n - 1 + \frac {1}{2n + 1}))\ge$

$ lim_{n\to\infty}\frac {9}{4}H_{2n} = \infty$

anyone convinced?[/quote]

but how did you get to the third line; i.e., "
\[ lim_{n\to\infty}\sum^{n}_{k = 1}( \frac {\frac {9}{2}}{2k - 1}) + \frac {9}{4} ( \frac {1}{2n + 1} - 1)
\]
"[/quote]

Oh, thank you; now I see it.  But where does the H come from?

$ H_n$ denotes the $ n^{th}$ [url=http://mathworld.wolfram.com/HarmonicNumber.html]harmonic number[/url] $ \sum_{k\equal{}1}^{n} \frac{1}{k}$.


IMO Shortlist
geometry
circumcircle
reflection
trigonometry
rotation

A circle $ C$ with center $ O.$ and a line $ L$ which does not touch circle $ C.$ $ OQ$ is perpendicular to $ L,$ $ Q$ is on $ L.$ $ P$ is on $ L,$ draw two tangents $ L_1, L_2$ to circle $ C.$  $ QA, QB$ are perpendicular to $ L_1, L_2$ respectively. ($ A$ on $ L_1,$ $ B$ on $ L_2$). Prove that, line $ AB$ intersect $ QO$ at a fixed point.

[i]Original formulation:[/i]

A line $ l$ does not meet a circle $ \omega$ with center $ O.$ $ E$ is the point on $ l$ such that $ OE$ is perpendicular to $ l.$ $ M$ is any point on $ l$ other than $ E.$ The tangents from $ M$ to $ \omega$ touch it at $ A$ and $ B.$ $ C$ is the point on $ MA$ such that $ EC$ is perpendicular to $ MA.$ $ D$ is the point on $ MB$ such that $ ED$ is perpendicular to $ MB.$ The line $ CD$ cuts $ OE$ at $ F.$ Prove that the location of $ F$ is independent of that of $ M.$

I'll give you some steps (it's a beautiful problem, but I seem to have some trouble with writing full solutions :)):

(1) Let L1 and L2 touch C at X and Y. Show that XY and OQ intersect in a fixed point S;

(2) Show that If D and E are the intersection points of L1 and L2 with OQ then (S, Q; D, E)=-1 (harmonic cross-ratio);

(3) Let R be the intersection point of AB and OQ for a fixed position of P; show that RD*RE=RA*RB=RQ2;

(4) From (2) and (3) derive th fact that R is the midpoint of QS (meaning that it's fixed, because both Q and S are fixed; I think this is the only part which requires a bit of computation.

The problem is very nice, by the way.

(2) Show that If D and E are the intersection points of L1 and L2 with OQ then (S, Q; D, E)=-1 (harmonic cross-ratio);

Well,
This is so cool！ :D
I just need to know property(2) then can complete the proof.
Thank you very much~

Chao

I have a different solution of the problem, using angle chasing and some properties of orthogonal circles.

First I rewrite the problem with modified notations (I find the namings L, $L_1$, $L_2$ pretty useless):

[color=blue][b]Theorem 1.[/b] Let k be a circle with center O, let Q be a point not lying on this circle k, and let P be a point on the perpendicular to the line OQ at the point Q. Assume that the point P lies outside the circle k. Let the two tangents from the point P to the circle k touch this circle k at the points X and Y. Denote by A and B the orthogonal projections of the point Q on the lines PX and PY. Let the line XY meet the line OQ at a point S. Let R be the midpoint of the segment QS.

[b](a)[/b] The point S is the image of the point Q in the inversion with respect to the circle k.
[b](b)[/b] The point R lies on the line AB.
[b](c)[/b] The point R lies on the radical axis of the circle k and the circle with diameter PQ.
[b](d)[/b] If the line OQ meets the circle k at two points D and E, then the points A, B, D and E are concyclic.[/color]

Note that parts [b](a)[/b] and [b](b)[/b] of this theorem yield Chaogold's initial problem (in fact, since the point Q and the circle k are fixed, according to part [b](a)[/b] the point S must be fixed, too, so the midpoint R of the segment QS is fixed; but according to part [b](b)[/b], the point R lies on the line AB, so the line AB intersects the line QO at a fixed point - namely, at the point R). Part [b](d)[/b] yields the additional assertion that ABDE are cyclic.

[i]Proof of Theorem 1.[/i] [b](a)[/b] This is a trivial consequence of poles and polars, but here is an elementary proof: Since the tangents from the point P to the circle k (whose center is O) touch this circle k at the points X and Y, we have $PX\perp OX$ and $PY\perp OY$, so that < OXP = 90° and < OYP = 90°. Also, < OQP = 90° (since the point P lies on the perpendicular to the line OQ at the point Q). Hence, the points X, Y and Q lie on the circle with diameter OP. Thus, < OPX = < OQX.

Now, since the points X and Y are the points where the tangents from the point P to the circle k (with center O) touch this circle k, these points X and Y are symmetric to each other with respect to the line OP. Thus, $XY\perp OP$, so that < (XY; OP) = 90°. On the other hand, $PX\perp OX$ yields < (OX; PX) = 90°. Thus,

  < SXO = < (XY; OX) = < (XY; OP) + < (OP; PX) - < (OX; PX) = 90° + < (OP; PX) - 90°
              = < (OP; PX) = < OPX = < OQX = - < XQO.

Also, trivially, < SOX = - < XOQ. Thus, the triangles SXO and XQO are inversely similar, so that OS : OX = OX : OQ. In other words, $OS\cdot OQ=OX^2$. Since O is the center and OX is the radius of the circle k, this equation yields that the point S is the image of the point Q in the inversion with respect to the circle k. Thus, Theorem 1 [b](a)[/b] is proven.

[b](b)[/b] Let W be the orthogonal projection of the point Q on the line XY. Then, < QWX = 90°. On the other hand, < QAX = 90°. Hence, the points W and A lie on the circle with diameter QX; consequently, < QWA = < QXA. But < QXA = < QXP, and since the points X, Y and Q lie on the circle with diameter OP, we have < QXP = < QOP. Thus, < QWA = < QOP. Since $XY\perp OP$ and $QW\perp XY$, we have QW || OP, and thus < (QO; OP) = < (QO; QW), what rewrites as < QOP = < RQW. Hence, < QWA = < QOP becomes < QWA = < RQW. On the other hand, since the triangle QWS is right-angled at its vertex W, the point R, being the midpoint of its hypotenuse QS, must be the circumcenter of this triangle QWS; hence, RQ = RW, so that the triangle WRQ is isosceles, and thus < RQW = < QWR. Hence, we get < QWA = < QWR. But this shows that the point A lies on the line WR. Similarly, the point B lies on the line WR. Thus, the four points A, B, W and R lie on one line, what yields that the point R lies on the line AB, and Theorem 1 [b](b)[/b] is proven.

[b](c)[/b] The point R, being the midpoint of the segment QS, is the center of the circle with diameter QS. On the other hand, if we call U the midpoint of the segment PQ, then this point U is the center of the circle with diameter PQ. Since $UQ\perp RQ$ (this is just a rephrase of $PQ\perp OQ$), and the point Q is a common point of the circles with diameters QS and PQ, it follows that the circle with diameter QS is orthogonal to the circle with diameter PQ. On the other hand, since the point S is the image of the point Q in the inversion with respect to the circle k, the circle with diameter QS is orthogonal to the circle k (in fact, any circle through two distinct mutually inverse points with respect to a circle is orthogonal to this circle). Hence, the circle with diameter QS is orthogonal to both the circle k and the circle with diameter PQ. Thus, the center R of this circle with diameter QS lies on the radical axis of the circle k and the circle with diameter PQ. This proves Theorem 1 [b](c)[/b].

[b](d)[/b] Since < PAQ = 90° and < PBQ = 90°, the points A and B lie on the circle with diameter PQ. Since the point R lies on the line AB, it follows that the power of the point R with respect to the circle with diameter PQ equals $RA\cdot RB$. On the other hand, the power of the point R with respect to the circle k equals $RD\cdot RE$. Now, since the point R lies on the radical axis of the circle k and the circle with diameter PQ, its powers with respect to the circle with diameter PQ and the circle k are equal; hence, $RA\cdot RB=RD\cdot RE$. By the converse of the intersecting chords theorem, this yields that the points A, B, D and E are concyclic, and Theorem 1 [b](d)[/b] is proven. $\blacksquare$

  Darij

Approach by [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=26]grobber[/url]:

There are going to be quite a few definitions:

Fix $ P$, and let $ M \equal{} PA\cap QB,N \equal{} PB\cap QA,\ell \equal{} MN$. Let $ S \equal{} AB\cap PQ,T \equal{} AB\cap\ell$, and let $ X,Y$ be the intersections between $ AB$ and the perpendiculars to $ PQ$ in $ Q$ and $ P$, respectively.

If we move a point $ U$ on $ \ell$, then $ QU\cap AB\mapsto PU\cap AB$ is an involution on $ AB$. Through this involution, $ A$ corresponds to $ B$, $ X$ corresponds to $ Y$, while $ S,T$ are fixed points. This means that $ (S,T;A,B) \equal{} (S,T;X,Y) \equal{} \minus{} 1\ (*)$.

From $ (PS,PT;PA,PB) \equal{} \minus{} 1$ we find that $ PT$ passes through the pole $ P'$ of $ PQ$ wrt the circle (the initial circle, centered at $ O$). Now, since $ PT \equal{} PP',PS \equal{} PQ$, the second equality in $ (*)$ can be written $ (PQ,PP';PX,PY) \equal{} \minus{} 1\ (**)$. Since $ PY\|QP',\ (**)$ shows that $ X$ is, in fact, the midpoint of $ QP'$. $ Q$ and $ P'$ do not depend on $ P$, so $ X \equal{} AB\cap OQ$ is fixed.

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

[i]A semicircle is drawn on one side of a straight line $ l.$ $ C$ and $ D$ are points on the semicircle. The tangents to the semicircle at $ C$ and $ D$ meet $ l$ at $ B$ and $ A$ respectively, with the centre of the semicircle between them. Let $ E$ be the point of intersection of $ AC$ and $ BD,$ and $ F$ be the point on $ l$ so $ EF$ is perpendicular to $ l.$:[/i]

Let P be the intersection of the 2 tangents, R the radius of the semicircle  and O its center.

What we need first is to prove that the altitude from P in triangle ABP is concurrent with AC and BD. This is done by proving that BC/CP*PD/DA=tgA/tgB (we want to use the reciprocal of Ceva), where A and B are angles of the triangle ABP. This is done easily since PC=PD and BC=OC/tgB=R/tgB and AD=OD/tgA=R/tgA. By replacing these we get exactly what we want. Thisa means that line EF and line PF are, in fact, one and the same.

Now to prove the actual result:
We draw the reflection of the entire figure with respect to line l. Let the image of any pt X be named X'. It's obvious that P, F, P' are collinear (F is the middle of PP'). We know by Newton's thm that PP', AB, CD', DC' are concurrent (it states that for a circumscribed quadrilateral the diagonals and the 2 lines joining the 2 pts of tangency on 2 opposite sides are all concurrent). This means that the (C, F, D') and (D, F, C') are 2 triplets of        collinear pts.

angle PFD=angle P'FD' because the figure is symmetric with respect to l. Angle CFP=angle P'FD' because C, F, D' are collinear and so are P, F, F'. From these 2 equalities we get angle PFD=angle CFP, which means that EF=PF is the bisector of angle CFD Q.E.D.

I love writing that (QED)  :D

Prove EF bisects angle CFD.

A line $ l$ does not meet a circle $ \omega$ with center $ O.$ $ E$ is the point on $ l$ such that $ OE$ is perpendicular to $ l.$ $ M$ is any point on $ l$ other than $ E.$ The tangents from $ M$ to $ \omega$ touch it at $ A$ and $ B.$ $ C$ is the point on $ MA$ such that $ EC$ is perpendicular to $ MA.$ $ D$ is the point on $ MB$ such that $ ED$ is perpendicular to $ MB.$ The line $ CD$ cuts $ OE$ at $ F.$ Prove that the location of $ F$ is independent of that of $ M.$

Approach by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=37447]mr.danh[/url]:

Four points M,E,A,B all lie on the circle diametered OM. Let I be the orthogonal projection of E on AB.
E lies on the circumcircle (AMB), so I,C,D lie on Simson line.
OE meets line (ICD) at F and line (IAB) at K. F is the midpoint of EK because of the righ-angle EIK.
To prove that F is a fixed point, show that K is a fixed point. It is easy since $ OK.OE \equal{} ON.OM \equal{} R^2$ (where N is the intersection of AB and OM)

[quote="orl"]F is the midpoint of EK because of the righ-angle EIK.[/quote]Whoops I failed to realize this even though I constructed $I$...

It's easy to see that $CEMD$ and $OAEMB$ are both cyclic by right angles, so $\alpha=\angle{EDF}=\angle{EMA}=\angle{EOA}$. Since $FE\perp EM$ and $EM$ is the diameter of the circle passing through $CEMD$, $\angle{FEC}=\angle{EDF}=\alpha$ and
\[\angle{FCE}=\pi-\angle{ECD}=\angle{EMD}=\angle{EDF}+2\angle{AMO}=\alpha+2\beta,\]where $\beta=\angle{AMO}=\angle{BMO}=\angle{AEO}$, using the fact that $MO$ is the angle bisector of $\angle{AMB}$. Thus
\[\angle{EFC}=\pi-\angle{FEC}-\angle{FCE}=\pi-2\alpha-2\beta.\]Now the Law of Sines on $\triangle{EDF}$ gives us
\begin{align*}
EF=\frac{DE\sin\angle{EDF}}{\sin\angle{EFC}}=\frac{ME\sin\angle{EMD}\sin\alpha}{\sin2(\alpha+\beta)} &= \frac{ME\sin((\alpha+\beta)+\beta))\sin\alpha}{2\sin(\alpha+\beta)\cos(\alpha+\beta)}\\
&= \frac{ME(\sin(\alpha+\beta)\cos\beta+\cos(\alpha+\beta)\sin\beta)\sin\alpha}{2\sin(\alpha+\beta)\cos(\alpha+\beta)}\\
&= \frac{ME\left(\frac{OE}{OM}\cdot\frac{BM}{OM}+\frac{ME}{OM}\cdot\frac{BO}{OM}\right)\sin\alpha}{2\cdot\frac{OE}{OM}\cdot\frac{ME}{OM}}\\
&= \frac{ME(OE\cdot BM+ME\cdot BO)\sin\alpha}{2\cdot OE\cdot ME}\\
&= \frac{(OM\cdot BE)\sin\alpha}{2\cdot OE},
\end{align*}where we have used Ptolemy's Theorem on $OEMB$ in the last step. Now by the Law of Sines on $\triangle{AOE}$, we have
\[\sin\alpha=\sin\angle{AOE}=\frac{AE\sin\angle{AEO}}{AO}=\frac{AE\sin\beta}{AO}=\frac{AE\cdot\frac{AO}{OM}}{AO}=\frac{AE}{OM}.\]Thus
\[EF=\frac{(OM\cdot BE)\sin\alpha}{2\cdot OE}=\frac{AE\cdot BE}{2\cdot OE}.\]Finally, since $\angle{AEO}=\angle{BEO}=\beta$, we can reflect $B$ over $EO$ to $B'$ so that $E,A,B'$ are collinear with $B'$ on $\omega$. This tells us that $AE\cdot BE=AE\cdot B'E$ is the power of $E$ with respect to $\omega$, which is clearly constant as $M$ varies. But $OE$ is constant as well, so we're done.

Wait,

I do not understand why F is the midpoint of KE. I do understand that angle KIE is a right angle, but why does that force F to be the midpoint?

At least one of the equalities $\angle{FEI}=\angle{FIE}$ and $\angle{FKI}=\angle{FIK}$ should be easy to angle chase directly.

Let $AB$ meet $OM$ at $G$ and $OE$ at $H$. Then $\angle MEH = \angle MGH = 90$, so $GHEM$ is cyclic, so $OH\cdot OE = OG\cdot OM$. Triangles $AOG$ and $MOA$ are similar, so $OG\cdot OM = OA^2$. Hence $OH = OA^2/OE$, which is constant.

Let the lines $AB$ and $CD$ meet at $K$. Angles $OAM, OEM, OBM$ are all $90$, so $EAOBM$ is cyclic. Similarly $\angle ECM = \angle EDM = 90$, so $ECDM$ is cyclic. Hence $\angle EAK = \angle EMB = \angle EMD = \angle ECK$, so $EKAC$ is cylic. Hence $\angle EKA = 90$. So $EKBD$ is cyclic. So $\angle EKF = \angle EKD = \angle EBD = \angle EBM = \angle EOM$ ($EAOBM$ is cyclic)$ = \angle FEK (OM \parallel EK)$. Hence $EF = FK$. But $\angle EKH = 90$, so $EF = FH$. So $F$ is the midpoint of $EH$. But $E$ and $H$ are fixed, so $F$ is fixed also. $\Box$

Recall a few basic complex numbers things. First \[ \frac{a-b}{\overline{a-b}} = \frac{a-c}{\overline{a-c}} \implies A, B, C \] are collinear and
 \[ \frac{a-b}{\overline{a-b}} = -\frac{c-d}{\overline{c-d}} \implies AB \perp CD \]
WLOG, let $\omega$ be the unit circle, and the line containing $E$ be parallel to the real axis. (We can assume this because we can rotate it).
Let the points be represented by their lowercase equivalents. Let $a, b$ be points on the unit circle, and let $m$ be the intersection of the tangents. We can easily find that $m = \frac{2ab}{a+b}$.
Note that since we assumed that $ME$ was parallel to the real axis, $E$ lies on the complex axis $\implies e = -\bar{e}$.
Also $EM \perp OE$, so
\[ \frac{e}{\overline{e}} = -\frac{e-m}{\bar{e}-\bar{m}} \]
\[ e-\frac{2ab}{a+b} = -e-\frac{2}{a+b} \implies e = \frac{ab-1}{a+b} \]
We can now get a system of equations for $c$.
\[ \frac{c-a}{\bar{c} - \frac{1}{a}} = \frac{m-a}{\overline{m-a}} = -a^2, \frac{c-e}{\bar{c}-\bar{e}} = -\frac{m-a}
{\overline{m-a}} = a^2 \]
\[c = -a^2\bar{c}+2a,  c = a^2\bar{c}-a^2\bar{e}+e\], so we can solve to get that
\[c = \frac{2a+e-a^2\bar{e}}{2} = \frac{2a+e+a^2e}{2} =  \frac{2a+\frac{ab-1}{a+b}+a^2\frac{ab-1}{a+b}}{2} = \frac{a^2+3ab+a^3b-1}{2(a+b)}\] Similarly, $d = \frac{2b+e+b^2e}{2}$.
Now we proceed to find $f$. Note that since $f$ lies on the complex axis, $f = -\bar{f}$.
Also, \[\frac{f-c}{\bar{f}-\bar{c}} = \frac{c-d}{\overline{c-d}} = \frac{\frac{2a+e+a^2e}{2}-\frac{2b+e+b^2e}{2}}{\overline{\frac{2a+e+a^2e}{2}-\frac{2b+e+b^2e}{2}}} = \frac{((a+b)e+2)(a-b)}{\overline{((a+b)e+2)(a-b)}}
= \frac{(ab+1)(a-b)}{\frac{ab+1}{ab}\frac{b-a}{ab}} = -a^2b^2\]
So \[f = \frac{c+a^2b^2\bar{c}}{1-a^2b^2} = \frac{\frac{a^2+3ab+a^3b-1}{2(a+b)}+a^2b^2\overline{\frac{a^2+3ab+a^3b-1}{2(a+b)}}}{1-a^2b^2} = \frac{a^3b^3-a^3b-ab^3-3a^2b^2-a^2-b^2-3ab+1}{2(a^2b^2-1)(a+b)} = \frac{(1+ab)(a^2b^2-a^2-b^2-4ab+1)}{2(1+ab)(ab-1)(a+b)} = \frac{a^2b^2-a^2-b^2-4ab+1}{2(ab-1)(a+b)}\].
We will show that $F$ is the midpoint of $E$ and the pole of $E$ wrt $\omega$. Note that the pole of $E$ would be the point $\frac{e}{\vert e \vert^2} = \frac{e}{e\bar{e}} = -\frac{1}{e}$.
Therefore the midpoint is \[\frac{e - \frac{1}{e}}{2} = \frac{\frac{ab-1}{a+b} - \frac{a+b}{ab-1}}{2} = \frac{a^2b^2-a^2-b^2-4ab+1}{2(a+b)(ab-1)} = f\], as required.

What an EXCELLENT problem.

[hide="Solution"]Let $R, S=L_1,L_2\cap C$, $T=RS\cap QO$, $X=AB\cap QO$. As $T\in RS$, the polar of $P$, it follows that $QP$ is the polar of $T$, whence $T$ is fixed. Observe that quadrilateral $PQRS$ is cyclic, whence $U\in AB$ by Simson, where $U$ is the projection of $Q$ onto $RS$. Angle chasing yields $X$ is the circumcenter of right triangle $QTU$, whence $X$ is the midpoint of $QT$ and consequently fixed. $\blacksquare$[/hide]

Let $PA$ touch $C$ at $D$, $PB$ at $E$. Let $G$ be the polar of $L$ wrt $C$. Let $C$ be the foot from $Q$ to $GD$. Then by Simson line on $\triangle PDE$, it follows that $C$ lies on $AB$.
Note that $CDQA$ is cyclic.

[b][color=red]Claim:[/color][/b] $\triangle ODG \sim \triangle OQD$.
[i]Proof.[/i] Follows by being a polar. $\blacksquare$
As such, it now follows that \[ \measuredangle GCA = \measuredangle DQA = \measuredangle DQA = \measuredangle DQG + \measuredangle GQA = \measuredangle ODG + GOD = \measuredangle OGD = \measuredangle QGC \] which gives the result.

This took longer than I'd like to admit


[color=red]If
$x,y,z > 0$
then:
$3^3 \left( {x + y} \right)^4 \left( {y + z} \right)^4 \left( {z + x} \right)^4  \ge 2^{12} x^3 y^3 z^3 \left( {x + y + z} \right)^3 $[/color]

By popoviciu's inequality, we can obtain a better inequality $3^6 (x+y)^4(y+z)^4(z+x)^4 \ge 2^{12} x^2y^2z^2(x+y+z)^6$.
so your inequality is true.

[quote="Viet Math"][quote="metru"]Thanks, siuhochung.
You are the best.[/quote]

I think sinhochung will be very happy if he read it.  :rotfl:[/quote]
I did read it, but i am not very happy, because metru is dishonest...  :D

Metru,your inequality it follows from the well known

      $ (x+y)(y+z)(z+x)>=8/9. (a+b+c)(ab+bc+ca)$

" metru is dishonest ..."
Dear siuhochung,
This is FALSE.
Metru.

[quote="metru"]Thanks, siuhochung.
You are the best.[/quote]
this shows that you are not really honest...  :rotfl:
anyway, i am just joking that you are dishonest, don't take it serious  :lol:


A point $ M$ is taken on the perpendicular bisector of the side $ AC$ of an acute triangle $ ABC$ so that $ M,B$ are on the same side of $ AC$. If $ \angle BAC=\angle MCB$ and $ \angle ABC+\angle MBC=180$, find $ \angle BAC$.

Are you sure the problem is correct?
see the following figure, if we have the condition $ \angle ABC+\angle MBC=180^{\circ}$ and $ M$ and $ B$ are on the same side of $ AC$ then $ M$ must be on the line passing through side $ AB$, but then the triangle $ AMC$ will be isosceles and its impossible that we have: $ \angle BAC=\angle MAC=\angle BCM$ !!!!! :wink:

OK, it's wrong. And I don't know what's the correction. Sorry  :(


At each stage, a new square is drawn on each side of the perimeter of the figure in the previous stage. How many unit squares will be in stage 10?

[asy]unitsize(.25cm);
label("stage 1",(0,0),S); label("stage 2",(8,0),S);
label("stage 3",(20,0),S); label("stage 4",(36,0),S);
for(int k = 1; k < 5; ++k){
pair A=(2*k*(k-1)+4(k-1),7);

for(int c = 0; c < k; ++c){
draw( A+(1,1) + 2(k-1-c,c) -- A+(1,-1) + 2(k-1-c,-c) -- A+(-1,-1) - 2(k-1-c,c) -- A+(-1,1) - 2(k-1-c,-c)--cycle);
}
}[/asy]

Looking at the pattern, we see that a formula can be derived:
The pattern is basically a square with 4 polygons removed from the corners...
For the $ n^{th}$ stage, the number of squares is $ 2n^2 \minus{} 2n \plus{} 1$
So the number of squares in the $ 10^{th}$ stage is $ 2 * 100 \minus{} 2 * 10 \plus{} 1 \equal{} \boxed{181}$


superior algebra
superior algebra unsolved

[quote="Jumbler"]How many elements of order $ 2$ does the symmetric group $ S_4$ contain?[/quote]

It's a combinatorial problem...

(12), (13), (14), (23), (24), (34)

Can you find the general formulae?

And of course $ (12)(34)$, $ (13)(24)$ and $ (14)(23)$.  See my post at:
http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=2968.0001.0001.0001

Steve


logarithms integration function LaTeX calculus derivative symmetry

$ \log x$ is defined to be $ \int_1^x 1/t\,dt$ for $ x > 0$.

(a) Prove that log is a [i]smooth function[/i].

(b) Prove that $ \log xy \equal{} \log x \plus{} \log y$ $ \forall x, y > 0$. [Hint: fix $ y$ and define $ f(x) \equal{} \log (xy) \minus{} \log x \minus{} \log y$. show that $ f(x)\equiv 0$.]

(c) Prove log is [i]strictly monotone increasing[/i] in $ \mathbb{R}$

I cleaned up your post some. If you want text italics, use the bbcode tags (bracket)i(bracket) and (bracket)/i(bracket), not $ \text{\LaTeX}.$ There's a perfectly good looking "log" in $ \text{\LaTeX}$: it's \log. Just be sure to leave a space after it: \log x. And putting bbcode boldface around $ \text{\LaTeX}$ doesn't do anything. You could have tried \mathbf{R}, but what you really wanted there was \mathbb{R} instead.

Now, as to the substance: You do know the Fundamental Theorem of Calculus, right? And you do know how to make substitutions in integrals, right? Can you show us what you've done so far?

Thanks Kent for correcting my errors. It was my first attempt at using \LaTeX and I am glad I didn't make even more mistakes.

as for the question...

I know F.T.C:
(1)  $ F(x) \equal{} \int_a^b f(t) dt$ where $ F'(x) \equal{} f(x)$
(2)  $ \int_a^b f(x) dt \equal{} g(b) \minus{} g(a)$ where $ g$ is the antiderivative

a) I need to show that all derivatives of $ \log$ exist OR it is $ C^ \infty$. can you give me a hint?

b) this part "seems" very easy but I cannot figure it out. I know substituting in the definition of log will give me: $ f(x) \equal{} \int_1^{xy} {1/t} dt \minus{} \int_1^x 1/t dt \minus{} \int_1^y 1/t dt$  but I cannot go any further.

c) don't know how to apply the definition of strictly monotone increasing to this question.

That's a misquote of the F.T.C.

Here's a correct version, or versions (there are two sides to the theorem):

(1) Derivative of the integral theorem: if we define $ F(x)\equal{}\int_a^xf(t)\,dt$ and $ f$ is continuous (on an interval containing $ a$), then $ F$ is differentiable on that interval and $ F'(x)\equal{}f(x).$

(2) Integral of the derivative theorem: if $ g$ is differentiable and $ g'$ is integrable, then $ \int_a^bg'(x)\,dx\equal{}g(b)\minus{}g(a).$

The one you want here is the derivative of the integral theorem. What does it say about $ \frac{d}{dx}\log x?$

OK
so, $ \frac {d}{dx}\log x \equal{} \frac {1}{x}$
we know $ \frac {1}{x}$ is well defined and continuous when $ x > 0.$
We also know that $ \frac {1}{x}$ is infinitely differentiable
$ \therefore \log x$ is smooth.

How about part b and c? would you give me more hints on them please?

C is easy.  $ \frac{d}{dx} \log x \equal{} \frac{1}{x} > 0$ for $ x>0$, which is where the logarithm is defined.  Since the derivative of log is always positive, log is monotone increasing.

b) $ y>0$ is a constant.  Let $ f(x) \equal{} log xy \minus{} log x \minus{} log y$.  Then $ f'(x) \equal{} \frac{y}{xy} \minus{} \frac{1}{x} \equal{} 0$.  Thus $ f(x)$ is a constant function.  By symmetry, we can also conclude that changing $ y$ has no effect on the function (or we could define f as a function of x and y and show both partial derivatives are 0).  In any case it remains only to evaluate $ f(x)$ at some point, since the function is constant.  For simplicity, choose $ x\equal{}1$.  From the definition of log it is clear that $ \log 1 \equal{} 0$ so $ f(1) \equal{} \log y \minus{} \log 1 \minus{} \log y \equal{} 0$.  Thus $ f(x) \equiv 0$.  So we conclude $ \log xy \equal{} \log x \plus{} \log y$.

[quote="facis"]C is easy.  $ \frac {d}{dx} \log x \equal{} \frac {1}{x} > 0$ for $ x > 0$, which is where the logarithm is defined.  Since the derivative of log is always positive, log is monotone increasing.

...[/quote]

Thanks facis for your response. Very helpful.


Hi,

I have discovered something very interesting and I do not know how to prove it (for now).
Checked up to 103.

Here is the picture describing the statement.

Thank you for any comment.

I'm sorry for this post.
It is a Fermat theorem ....

But it gave me another idea.

Hi,

I have found that the numbers failing the test [u]are equal to 1 for other values less than t.[/u]
For example :

170^10 mod 341=1
170^20 mod 341=1
and so on

At the contrary for the certified primes[u] it does not work before t^t[/u]
So I have to [u]add a second test [/u]to be sure that it is a certified prime.
But the difficulty hold on how to know which power of t is giving 1 as output.

Maybe [u]someone can check or find how to implement the second test.[/u]
Thank you

Youpie!!!!!!!

I have found why!!!

For example

p=341

You have to factorize 340=2*2*5*17

170^10 mod 341=1
and for every 170^(10*k) mod 341=1

Why 10 and not 2,4,5 or 17? Still searching.

Help please.
Maybe I have found the solution.

Hi,

I tried without success to find a composite number c such as :

c=2t+1

t^t mod c = 2t

So if such composite does not exist then we can state that if the ouput is 2t the number 2t+1 is a [u]certified prime.[/u]
I do not know if we can prove it.

Thank you for any help.

Hi,

Let p=2t +1

o the odd part of 2t

Example : p=19

2t=18

o = 9

9^9 mod 19 = 1

More generally :

2t=o*(2^k)

If p=2t+1 is prime then

o^o mod p = 1 or p-1

(some composite numbers pass the test, I know it but I have found another way to sieve those numbers and I will publish it soon)

------------------------------
The odd part is the last part obtained by dividing by 2.

If 2t=1260
we divide by 2
1260/2=630 (is still even)
we divide by 2
630/2=315 is odd then o(1260)=315

2t=315*(2^2)

It does not work with some primes like 41

o=5

5^5 mod 41 = 9

But a question still remain : why it works with a lot of values?

The odd part works only for some primes.
I tried to find some specific divisor of 2t.
I do not know how to name it.
Let us call it d*

d*^d* mod 2t +1 = 1 or 2t only if 2t+1 is prime.

How to identify it?
I tried some simulation

for example p=41
5^5 mod 41 = 9
but 10^10 mod 41 = 1
for example p=97
o=3

3^3 mod 97=27
but
6^6 mod 97 = 96

the composite numbers like 9

o=1
1^1 mod 9=1
but 4^4 mod 9 = 4

There is something I do not understand.
Why it works for some values and not for others.
If we find d* then it will be easy to test the primality of any number.

[quote="SAPOSTO"]Have you ever heard about Fermat's little theorem. Look also [url=http://en.wikipedia.org/wiki/Carmichael_number]here[/url].[/quote]

Thank you for your comment.
Did you read what is written above? I mean all the posts.


calculus
integration
logarithms
number theory unsolved
number theory

We are going to be using the "American" definition for the logarithmic integral:

If $ x \ge 2$, let

Li(x) = $ \int_2^x \frac{dt}{\log{t}}$

a) Prove that

Li(x) = $ \frac{x}{\log{x}} \plus{} \int_2^x \frac{dt}{\log^2 t} \minus{} \frac{2}{\log 2}$

and that, more generally,

Li(x)= $ \frac{x}{\log x} (1 \plus{} \sum_{k\equal{}1}^{n\minus{}1} \frac{k!}{\log^k x}) \plus{} n!\int_2^x \frac{dt}{\log ^{n\plus{}1} t} \plus{} C_n$,

where $ C_n$ is independent of $ x$

b) If $ x \ge 2$ prove that

$ \int_2^x \frac{dt}{\log^n t} \equal{} O(\frac{x}{\log ^n x})$


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2\plus{}b^2\plus{}c^2\neq0.$ Prove that
\[ \frac {a}{a^3 \plus{} (b \plus{} c)^3} \plus{} \frac {b}{b^3 \plus{} (a \plus{} c)^3} \plus{} \frac {c}{c^3 \plus{} (a \plus{} b)^3}\geq\frac {a \plus{} b \plus{} c}{a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc}
\]

Yes it is nice!
Using Cauchy-Schwarz:

$ \frac {a}{a^3 \plus{} (b \plus{} c)^3} \plus{} \frac {b}{b^3 \plus{} (a \plus{} c)^3} \plus{} \frac {c}{c^3 \plus{} (a \plus{} b)^3}$

$ \geq \frac {(a \plus{} b \plus{} c)^2}{a^4 \plus{} b^4 \plus{} c^4 \plus{} a(b \plus{} c)^3 \plus{} b(c \plus{} a)^3 \plus{} c(a \plus{} b)^3} \equal{} \frac {a \plus{} b \plus{} c}{a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc}$

[hide="Because"]$ \left(\sum_{cyc} a \right) \cdot \left(\sum_{cyc} a^3 \plus{} 6abc \right) \equal{} \left(\sum_{cyc} a^4 \plus{} \sum_{cyc}a(b \plus{} c)^3\right)$

Both are equivalent to

$ \sum_{cyc} (a^4 \plus{} a^3b \plus{} a^3c \plus{} 6a^2bc)$[/hide]

I meant to $ \sum_{cyc}\frac {a}{a^3 \plus{} (b \plus{} c)^3} \equal{} \sum_{cyc}\frac {a^2}{a(a \plus{} b \plus{} c)(a^2 \minus{} a(b \plus{} c) \plus{} (b \plus{} c)^2)}\geq$
$ \geq\frac {(a \plus{} b \plus{} c)^2}{(a \plus{} b \plus{} c)\sum(a^3 \minus{} a^2(b \plus{} c) \plus{} a(b^2 \plus{} c^2) \plus{} 2abc)} \equal{} \frac {a \plus{} b \plus{} c}{a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc}.$


Anna purchased 1 pound of fusilli pasta for $ \$$3.40 and a 12-ounce package of spaghetti for $ \$$2.20. There are 16 ounces in 1 pound. What was the average price per ounce, in cents, for her total purchase?

She purchased a total of 28 ounces for a total of $ \$$5.60.  Therefore, the average price per ounce is 20 cents.


function AMC AIME calculus derivative integration Asymptote

Are even and odd functions ever needed on AIME?

I don't know if you explicitly need to use the definition, but you probably want the works with the even/odd functions

Basic properties from Wikipedia:
The only function which is both even and odd is the constant function which is identically zero (i.e., f(x) = 0 for all x).
The sum of an even and odd function is neither even nor odd, unless one of the functions is identically zero.
The sum of two even functions is even, and any constant multiple of an even function is even.
The sum of two odd functions is odd, and any constant multiple of an odd function is odd.
The product of two even functions is an even function.
The product of two odd functions is an even function.
The product of an even function and an odd function is an odd function.
The quotient of two even functions is an even function.
The quotient of two odd functions is an even function.
The quotient of an even function and an odd function is an odd function.
The derivative of an even function is odd.
The derivative of an odd function is even.
The composition of two even functions is even, and the composition of two odd functions is odd.
The composition of an even function and an odd function is even.
The composition of any function with an even function is even (but not vice versa).
The integral of an odd function from −A to +A is zero (where A is finite, and the function has no vertical asymptotes between −A and A).
The integral of an even function from −A to +A is twice the integral from 0 to +A (where A is finite, and the function has no vertical asymptotes between −A and A).


Find all integer solutions to $\frac{ab+bc+ac}{abc}=\frac{1}{2}$ where a, b, and c are integers $\geq 3$.

This problem is related to how 3 kinds of tiles can be arranged without over lapping.


Ha, welcome to my life ;)  Good luck on the English paper, but no good advice here.

[quote="Fierytycoon"]I need help with my addiction. I often cannot seem to concentrate on homework anymore. I always feel compelled to go and study more math and do more problems, but I know that I cannot because I have so much homework (ie my english paper that I should be writing right now). As a consequence, I have trouble efficiently accomplishing either of the tasks. Does anyone have any suggestions on how I can help myself?[/quote]

Visualize your hard work on your English paper paying off in the future when you win a prize for mathematical exposition when you write a paper on one of your famous theorems.

First of all, take it as a personal challenge to do the best at whatever you do. Then try to find something interesting in your paper - something that you can't wait to tell the world about! Something that's an expression of your own unique ideas! Then it will be much more fun to write.

Also, organize your workspace in a way that you have fewer distractions; i.e. no tv, no internet, no (gasp) math books, etc. until you complete your task. This is always essential to me when I have something important to do.

you should be like me, give up on math.
on a more serious note, take a break.  go finish your papers first.

i'm going through the exact same thing now. i just feel detached from school and doing the work therein. particularly history. i just hate it. i wish i could strike a happy medium between school and mathematical study, until i do though, i'll stick to doing problems. some things just don't interest me...

it's definitely a matter of will power. it would be so much easier without that awful earthlink icon though..

[quote="mathfanatic"]What I usually do to cure any addiction is set an extremely strict schedule.  (I'll tell myself something like "turn off the computer and work on USAMTS for three hours.  Then turn on the computer and do your history section reviews.  Then turn off the computer and learn maths for two hours.  Then turn it back on and outline a biology chapter.  Under no circumstances should you move your mouse toward the internet explorer icon.  If you do, I will punish you."  It's funny how I talk to myself in the second person.)

Eventually, you get used to working on something and shutting everything else out.  It was effective in abolishing video games from my life forever.  I'm not sure how effective it has been at forcing me to study for AIME, but we'll see.[/quote]

relax... at least for a few days right before

The best solution is not to destroy your addiction. Get a good schedule. I spend most of my extra time on this web site or doing math elsewhere. I wake up at 6:00. Since school has a ton of free time, I try to finish everything to do with school (i find that finishing no brainer stuff in school is efficient) because there are often five mintue periods during a boring teacher lecture and during breaks. Then I come home do swimming or math circle or aops class. Then I do math. At 8:30 I see what homework I have and I see that I have only one hour left till I go to sleep so I work like a beast on that homework. I probably don't finish. Then I wake up at 6:00 and work like a beast again to finish my homework, and usually I'm done with homework and I do math in the morning. YEAH!


prove that the fibonacci numbers cannot be expressed as a polynomial with integer coefficients plus a^b where b is the bth term of the sequence and a is a constant

Not very formal, but here it goes.

$f(x) =1,1,2,3,5,8,13,21,34,55,89,144,233.... $
Change of f(x) $= 0,1,1,2,3,5,8, 13......$
Change in Change of f(x) $= 0,0,1,1,2,3,5,8,13.......$
nth Changeth of f(x) = strings of length n of 0's, $1,1,2,3,5......$

Basically, since the F_i+1 term of a Fibonacci sequence is the sum of all terms below it, and the F_i term of the Fibonacci sequence is the sum of all terms below it it. F_i+1 - F_i = F_ i-1. So we will have the above sequence. So we'll always have n 0's, then the Fibonaci sequence, which will always be increasing, since the Fibonacci sequence continues on infinitely. The terms of the nth Changeth in f(x) must be 0 if the function is a nth degree polynomial. But we just showed the terms of the nth Changeth in f are never all 0's, so f(x), the fibonacci sequence can never be represented by a nth degree polynomial. QED.

Wow, that was an awful proof. Sorry for making up mathematical terms (nth changeth in f) :lol:


induction
linear algebra
matrix
combinatorics unsolved
combinatorics

We write different positive integers into each of the squares of an $n$x$n$ chessboard. In each column we color red the square with the greatest number. We call a set of $n$ unit squares [i]good[/i] if no two of them lie in the same column or row. Prove that the good set with maximal sum contains at least one red square.

hi

i think that the q can be done by induction :!:

its obvious that there must exist atleast 1 row which doesn't have a red square, othetwise the max sum of good set wud be the set of red squares.

if we denote the board by an array \[ a{}_i{}_j \] where i denotes the row no. and j the col no. (i,j frm o to n-1)

then the condition for good squares is that all the js must be diff
so doesn't  matter if we switch 2 rows, or shift the position of a row

so we bring the any one row which doesn't have a red square to the 1st row position

we choose any 1 element say\[ a{}_0{}_j \]  frm that row and omit that row and the column in which the element is present

so now we r left with a \[ (n-1)X(n-1) \] matrix

by induction for max sum of good set this array we must have atleast 1 red square

so for any value of j a red square is present

hence out of these sums the max one will also satisfy the condition :)

I don't think that works. When you choose a good combination from that $(n-1)\times(n-1)$ aray, by the induction hypothesis, after the addition of $a_{0j}$, you may get a good combination with sum less than maximal. I think this should work though:

Choose a good set of squares $a_{\sigma(j)j},\ j=\overline{1,n}$, and assume it contains no red squares. We'll show that we can find another good combination with (strictly) greater sum.

Change $a_{\sigma(1)1}$ with the red square $a_{\tau(1)1}$. Then change the square in our initial combination which was on line $\tau(1)$ with the red square on the same column (this is a non-trivial change, because the initial square was not red), and so on, until we first meet a situation when the red square $a_{\tau(j_2)j_2}$ that we've just added to our selection lies on the same row as a red square $a_{\tau(j_1)j_1}$ we've selected a few steps back. Then move all squares selected before $a_{\tau(j_1)j_1}$ (including this one) back to their original positions. If, on the other hand, we never meet such a situation (i.e. all selected red squares lie on different rows), the process must end naturally, without our intervention. Either way, in the end we'll have the desired good combination with sum strictly larger than the initial one.

[quote="grobber"]I don't think that works. When you choose a good combination from that $(n-1)\times(n-1)$ aray, by the induction hypothesis, after the addition of $a_{0j}$, you may get a good combination with sum less than maximal. [/quote]

No, the first proof was correct (maybe the part you misunderstood could be better written): The argument that the smaller square contains a red square works for any $a_{0j}$, $j=0,\dots,n-1$ and there is some $a_{0j}$ in the optimal solution for the big square.


I am having touble with this problem, please help out!  :) :

          The sum of three different positive unit fractions is 6/7. What is the least number that could be the sum of the denominators of these fractions?

Here is what I think:
[hide]2/7 + 2/7 + 2/7 = 21? that doesn't seem right, but i can't find any other combination less.  1/2 + 1/4 + 3/28 doesn't work, 1/2 + 1/3 + 1/42 doesn't, 1/2 + 1/6 + 4/21 doesn't, and 1/3 + 1/4 + 23/84 doesn't work. I can't find anything else!  :? Sorry.[/hide]

I think unit fractions are in the form $\frac 1n$?
[hide]
The most obvious choice is that one of them is $\frac 12$.
That leaves us with $\frac{5}{14}$
From here, we see that $\frac{5}{14}-\frac{1}{3}=\frac{1}{42}$
So the answer is $2+3+42=47$[/hide]

oh, three different.

biffandoc:

[hide]1/2 plus 1/4 plus 3/28 is 6/7 and the denominator total is 34.[/hide]

I believe this problem has been posted.
[url]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=74297[/url] should get you there, thanks!

This has been locked.


geometry
inradius
inequalities
logarithms
inequalities proposed

Given triangle $ ABC$, altitude $ h_{a},h_{b},h_{c}$ inradius $ r$, prove that
\[ h_{a}^{a}h_{b}^{b}h_{c}^{c}\ge (3r)^{a+b+c}\]

I think $ h_{a}^{a}h_{b}^{b}h_{c}^{c}\le (3r)^{a+b+c}$, and here's the proof:
$ h_{a}=\frac{2S}{a},h_{b}=\frac{2S}{b},h_{c}=\frac{2S}{c}$ and $ 3r=\frac{2S}{\frac{a+b+c}{3}}$, so the original inequality is equivalent to $ \frac{a+b+c}{3}\le a^{\frac{a}{a+b+c}}b^{\frac{b}{a+b+c}}c^{\frac{c}{a+b+c}}$. If we assume that $ a+b+c=1$, we have to prove that $ \log \frac{1}{3}\le a \log a+b \log b+c \log c$, and since $ f(x)=x \log x$ is concave upward in $ (0,\inf)$, $ f(x) \ge f'(\frac{1}{3})(x-\frac{1}{3})+f(\frac{1}{3})$, which indicates that $ a \log{a}+b \log{b}+c \log{c}\ge f'(\frac{1}{3})(a+b+c-1)+3f(\frac{1}{3})=\log{\frac{1}{3}}$.


Let $ a$ be a real number. Show that the area of the triangle with the sides $ \sqrt{a^2\minus{}a\plus{}1}$ , $ \sqrt{a^2\plus{}a\plus{}1}$ , $ \sqrt{4a^2\plus{}3}$ doesn't depends on $ a$.

[quote="sai00"]Let $ a$ be a real number. Show that the area of the triangle with the sides $ \sqrt {a^2 \minus{} a \plus{} 1}$ , $ \sqrt {a^2 \plus{} a \plus{} 1}$ , $ \sqrt {4a^2 \plus{} 3}$ doesn't depends on $ a$.[/quote]

It's easy to show that the area $ S$ of a triangle whose sides are $ a,b,c$ is $ \frac {\sqrt {2a^2b^2 \plus{} 2a^2c^2 \plus{} 2b^2c^2 \minus{} a^4 \minus{} b^4 \minus{} c^4}}4$

Applying this here, we get $ S \equal{} \frac {\sqrt 3}4$