[hide="Is the answer..."]
65?
[/hide]
If it's right I'll post my solution.

can you post your solution? on the test, 65 was one of the choices, but i duno if it's right since I don't have the answer key.

[hide="hint"]Let AB=10x+y[/hide]
[hide="another hint"]CD=10y+x[/hide]
[hide="another hint"]connect C and the center of the circle[/hide]
[hide="last hint"]You have a right triangle. For what values of x and y result in a rational number?[/hide]

[hide="Full Solution"]
Call the diameter 10x + y. The chord will be 10y + x. Now call the distance from H to the center a. Using a right triangle with legs a and (10y + x)/2 and hypotenuse (10x + y)/2 we find that a = $\frac{3\sqrt{11(a-b)(a+b)}}{2}$. a + b must be 11 and a - b must be a perfect square. The only value that works is a = 6 and b = 5, so the answer is $65$.[/hide]

[quote="now a ranger"]Diameter AB of a circle has length a 2 digit integer in base 10. Reversing the digits gives the length of the perpendicular chord CD. The distance from the intersection point H to the center ) is a positive rational number. Determine the length  of AB.[/quote]
This is so an old AIME problem, it isn't even funny.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=371260#p371260
It's particularly easy for a number 12...


Yes , I think that : We can download it  :wink:

[quote="surshu"]It is possible to download A. Engel: Problem-solving strategies for free?[/quote]

If yes, if would be unlegal.

But you can order it eg here:

http://www.amazon.com/Problem-Solving-Strategies-Problem-Books-Mathematics/dp/0387982191/ref=pd_bbs_sr_1/002-5056188-1487200?ie=UTF8&s=books&qid=1183389978&sr=8-1

conceive


probability
real analysis
real analysis unsolved

Let $I = [0,1]$ and $S \subset I$. Suppose that the statements $m(S) = 0$ and $m(I\backslash S) = 0$ are not true, and that the points of $S$ are uniformly distributed in $I$. Prove that $S$ is not measurable.

If this is a little unclear, I could also state it this way: If $S \subset I$ and $S$ is uniformly distributed in $I$, then $S$ measurable implies $m(S) = 0$ or $m(S) = 1$.

I think this is interesting because it is a proof that there are nonmeasurable sets that is simple compared to that weird equivalence class construction that I still have a hard time visualizing (blah axoim of choice).

Even it's true, you still have to build such a set ... which is not necessarily easier than the "equivalence classes" thing. I say that, but I haven't given it a thought yet.

Ah, you're right! Second part of the question -- prove that such non-measurable sets exist. (I don't know how to do this one at the moment myself.)

edit: for those interested, it's been proven that you need to use the axiom of choice on an uncountable set in order to construct a non-measurable set.

Whay exactly does "uniformly distributed" mean here? My intuition tells me that it should mean, for measurable sets $S$, something like $m(S\cap U)=\alpha\ell(U)$ for every interval $U\subset I$ for some fixed $1>\alpha>0$, where $\ell(\cdot)$ denotes the length of the interval, but I'm not sure.  And how do you define the notion for unmeasurable $S$?

I suppose you could get around it by saying that any purely random countable sequence of points in $S$ is uniformly distributed in $I$ (in the normal way, say using Weyl's Criterion). But I'm still a little unsure of that, and I should be wary of talking about random things when I haven't really studied probability theory at all.

You're right, this is a tricky problem to state the way I want to. This is pretty much the same as a question I had on homework, but it was worded much differently (it was really just an exercise on points of Lebesgue density). I noticed that the result we proved actually said something about "uniformly" distributed subsets of intervals, so I thought I'd post it here. Perhaps I should have thought about it a bit more carefully rather than toss it into the ring so casually.

But I my meaning is clear, at least -- that you can't make a measurable set that's uniformly distributed as grobber defined it unless its measure is 0 or differs by a set of measure 0 from the smallest interval containing it.

Well, in that case, it follows immediately from the fact that a measurable set can be approximated arbitrarily well with open sets. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=poorly&t=68528]this[/url], for example.

Edit:

Sorry; I erased that stupid comment of mine.

[quote]you can always choose a convergent subsequence[/quote]

I won't challenge that, but if you did, then it wouldn't be random, would it? Or are you saying that there are enough convergent subsequences that if I looked at a random sequence it might be a convergent one? That seems unlikely to me. (heh.)

But on the other hand, does the fact that such sequences exist already screw me up, even if the probability of such a sequence being chosen is infinitesimal?


I have a question. In this website, you have your username, and on the bottom of it it says your rank in stars. How do I improve my rank, since my rank says New Member and no stars?

Well, you're a P versus NP now. :D You can improve your rank by the number of posts you have (I think 50 will get you the next higher rank, Hodge Conjecture).
btw, this belongs in games and fun factory or something.

Correction for easyasPI... according to my sources after 15 posts you gain your first rank, but you can ask an admin to be sure.  But if you don,t want to take advice from a new user, thats okay. :starwars:

Use the search function, Luke... :yoda:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Rank+Stars&t=79143[/url]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Rank+Stars&t=73553[/url]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Rank+Stars&t=28375[/url]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Rank+Stars&t=18966[/url]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Rank+Stars&t=13045[/url]

Okay, I'll stop.

[quote="dogseatcheese"]You should try searching for these kinds of things before posting. I posted this question before and the post got locked :blush:  ;)[/quote]Exactly :)


Hello,

I hope this will not be considered spam. I found this trick (that worked) for Hotmail. Since a few months ago they are offering 250 MB, but that only work for the new accounts, and they start first with 25 and afterwards 250. But all of us who are old member don't have that nice feature and still have our simple 2MB of storage (what is ridiculus if you compare with 1GB of Yahoo and 2+GB of Gmail). Then, here there is a trick to get first your 25 and afterwards the other 250 (after ten days)

Login to your account
Go to Option
Click My Profile
Change Country to United States
Wait for browser to Load US Settings
Change state to Florida and Zip Code to 33332 (I tried with mine, Georgia and 30332 and didn't work, but with that one it does work)
Click Update
Click Continue

Go to Language and Make sure it is English. Paste this link in Same Browser: http://by17fd.bay17.hotmail.msn.com/cgi-bin/AccountClose

Wait until the screen says "Your Hotmail account is closed and ready to be deleted." Click Close Account.

Go Back to Login and relogin to ur Account. Just reactivate ur account. No email will be deleted from ur account and now ur account size limit increase to 25MB. (Really!)

Now, it is important to do it inmediately. It shouldn't be that you do half of it one day and the other half the next day, because as you see, YOUR ACCOUNT IS CLOSED AND SCHEDULED TO DELETION, then, if you wait a day, may be it will be deleted!

Any way, I hope it helps!

[quote="furious"]who still uses hotmail?  :o
gmail is way better. i have 50 invites if anyone wants one.[/quote]

I use hotmail... and gmail... I actually like hotmail more, but that's just me.

[quote="furious"]who still uses hotmail?  :o
gmail is way better. i have 50 invites if anyone wants one.[/quote]

i use yahoo for regular email.
i use gmail to send myself projects and things.
i use hotmail for livejournal stuff, and when i fill out college things for standardized tests.

yeah im with paladin, i use both hotmail and gmail, but im just more used to hotmail. so yeah. i use hotmail more.

i like hotmail too :) didn't even consider getting gmail (i guess it's also the fact that i don't exchange that many messages) this trick was very cool though thanks alot djimenez

GMAIL>any thing else.
Gmail is the only free email that doesn't attach a stupid auto generated signature at the end of each email.
That gets really annoying some times.

Gmail owns you.

Recently they changed their style of displaying their ad's in Gmail.

Look at the spam section for some recipes.  :)

OR another way to boost your HOTMAIL MB up:

1. Find a friend who has MSN Internet(Broadband)
2. Tell them to add their address to their network of registered users.
3. you should have it

[quote="themonster"][quote="djimenez"]
Since a few months ago they are offering 250 MB, but that only work for the new accounts
[/quote]

I started my account ~5 years ago, and it automatically got upgraded to 250 MB.[/quote]You are lucky, most of the people didn't got that (well, at least most of the people who opened hotmail from out of the country).

Any way, I use several emails for different things. Hotmail mainly for email lists, walla for the replys from ML only, yahoo for personal emails, and recently gmail too for personal emails, but not too many people have the gmail account!

[quote="xxreddevilzxx"][quote="djimenez"]and recently gmail too for personal emails, but not too many people have the gmail account![/quote]

actually that is not true. With all the account holders having 50 invites, and refilled each time they use some, there are more than you might expect. I mean a lot more.[/quote]Well, let me rephrase though. Not too many people who use to write me have my gmail address. But for example, I usually send all my invitations to a place where they give to several people, and twice the receiver has emailed me saying... "Hey, do I know you?". Usually, the people who have my gmail account is the people who usually send me files attached.

[quote="bleumoose"]THAT would explain why mine didn't auto-upgrade. I put my country of origin as Zimbabwe.[/quote]May be this could help to clarify this,

[quote="Hotmail Main Page"]250MB inbox [b]available only in the 50 United States, District of Columbia, and Puerto Rico[/b]. Eligible Hotmail users will first receive 25MB at sign-up. Please allow at least 30 days for activation of your 250MB storage to verify your e-mail account and help prevent abuse. Microsoft Corporation reserves the right to provide 250MB inbox to free Hotmail accounts at its discretion.[/quote]That is written in very small letter, and at the end. That means that non-american people didn't have that automatic update.

wow. didn't know so many ppl liked hotmail.
also, when i had a hotmail account, it automatically updated to 250MB. but since i like gmail better, i let the account expire. and in hotmail i was getting constant junk email. it was messages with just symbols in it. like "$%(@#(*~\". that annoyed me greatly.

[quote="lightspeed"]What is G-Mail and how do I get it?
Thanks[/quote]

search the words "gmail" and "invite" on aops.

Or go here: http://isnoop.net/gmail/  :)

[quote="lightspeed"]What is G-Mail and how do I get it?
Thanks[/quote]

gmail is a mail started by google.

[quote="lightspeed"]What is G-Mail and how do I get it?
Thanks[/quote]Give me an email address and I can send you an invitation.

[quote="djimenez"][quote="shobber"]ok, i've already got the 25mb account. And for 250's, do i have to do it again after ten days?[/quote]No, in a few days it will automatically upgrade itself. In my case that took two days![/quote]
That's cool!
Thanks a lot!


calculus
integration
logarithms
function
calculus computations

Evaluate $ \int_0^1\frac {\ln\left(x\right)}{1 \minus{} x^2}\,dx$

There are three ways that I have done on my own. I don't want to do anymore for I didn't come up with them on my own.

[hide]

[b]Way one[/b]

We know that

$ \forall{x}\in(0,1)\quad\frac {1}{1 - x^2} = \sum_{n = 0}^{\infty}x^{2n}$

$ \therefore\qua\int_0^1\frac {\ln(x)}{1 - x^2}\,dx$

$ = \int_0^1\ln(x)\sum_{n = 0}^{\infty}x^{2n}\,dx$

Now assuming uniform convergence

$ = \sum_{n = 0}^{\infty}\int_0^1\ln(x)x^{2n}\,dx$

$ = - \sum_{n = 0}^{\infty}\frac {1}{(2n + 1)^2}$

$ = \frac { - \pi^2}{8}$

[b]Way two[/b]

We know that

$ \ln(x) = \int_1^x\frac {dy}{y}$

So we can see that

$ \int_0^1\frac {\ln(x)}{1 - x^2}\,dx$

$ = \int_0^1\int_1^x\frac {dy}{y(1 - x^2)}\,dx$

Now switching the integration order

$ = - \int_0^1\int_0^y\frac {dx}{y(1 - x^2)}\,dy$

Evaluation on the inner integral leads to

$ = - \int_0^1\frac {\tanh^{ - 1}(y)}{y}\,dy$

Now we know that

$ \forall{y}\in(0,1)\quad\tanh^{ - 1}(y) = \sum_{n = 0}^{\infty}\frac {y^{2n + 1}}{2n + 1}$

So

$ - \int_0^1\frac {\tanh^{ - 1}(y)}{y}\,dy$

$ = - \int_0^1\frac {1}{y}\sum_{n = 0}^{\infty}\frac {y^{2n + 1}}{2n + 1}\,dt$

$ = - \int_0^1\sum_{n = 0}^{\infty}\frac {y^{2n}}{2n + 1}$

Now assuming uniform convergence

$ = - \sum_{n = 0}^{\infty}\frac {1}{2n + 1}\int_0^1{y^{2n}}\,dy$

$ = - \sum_{n = 0}^{\infty}\frac {1}{(2n + 1)^2}$

$ = \frac { - \pi^2}{8}$

[b]Way three[/b]

$ \int_0^1\frac {\ln(x)}{1 - x^2}\,dx$

$ = \frac {1}{2}\int_0^1\frac {\ln(x)}{1 - x}\,d + \int_0^1\frac {\ln(x)}{1 + x}$

Now we know that

$ \forall{x}\in(0,1)\quad\frac {1}{1 + x} = \sum_{n = 0}^{\infty}( - 1)^nx^n\,\wedge\, \frac {1}{1 - x} = \sum_{n = 0}^{\infty}x^n$

So we have that

$ = \frac {1}{2}\int_0^1\frac {\ln(x)}{1 - x}\,d + \int_0^1\frac {\ln(x)}{1 + x}$

$ = \frac {1}{2}\int_0^1\ln(x)\sum_{n = 0}^{\infty}( - 1)^nx^n\,dx + \frac {1}{2}\int_0^1\ln(x)\sum_{n = 0}^{\infty}x^n\,dx$

Now assuming uniform convergence

$ = \frac {1}{2}\sum_{n = 0}^{\infty}( - 1)^n\int_0^1\ln(x)x^n\,dx + \frac {1}{2}\sum_{n = 0}^{\infty}\int_0^1\ln(x)x^n\,dx$

$ = - \frac {1}{2}\sum_{n = 0}^{\infty}\frac {( - 1)^n}{(n + 1)^2} - \frac {1}{2}\sum_{n = 0}^{\infty}\frac {1}{(n + 1)^2}$

$ = - \frac {1}{2}\sum_{n = 1}^{\infty}\frac {( - 1)^{n - 1}}{n^2} - \frac {1}{2}\sum_{n = 1}^{\infty}\frac {1}{n^2}$

$ = \frac { - 1}{2}\left(1 - 2^{1 - 2}\right)\zeta(2) - \frac {1}{2}\zeta(2)$

$ = \frac { - \pi^2}{24} - \frac {\pi^2}{12}$

$ = \frac { - \pi^2}{8}$

By any means

$ \boxed{\int_0^1\frac {\ln(x)}{1 - x^2}\,dx = \frac { - \pi^2}{8}}\,\,\blacksquare$[/hide]

[quote="Mathstud28"][quote="vidyamanohar"]Thanks.
I too have solved the problem using series.  I was looking for any other method if possible.

Thanks again[/quote]
Just out of curiosity, which of the three (if any) did you use?[/quote]

The first one.

Sorry for reviving the thread, but I wanted to show another way of solving the integral (inspired by one of Kouichi Nakagawa's posts)

Using the substitution $ x \equal{} e^{t}$, we obtain:

$ \minus{} \int_{0}^{\infty}\frac {te^tdt}{e^2t \minus{} 1}$

Now, lets substitute t with $ \frac {u}{2}$:

$ \minus{} \int_{0}^{\infty}\frac {\frac {u}{4}e^{u/2}du}{e^u \minus{} 1}$

which is $ \minus{} \int_{0}^{\infty}\frac {u}{4}e^{u/2}\sum_{n \equal{} 1}^{\infty}e^{ \minus{} nu}du$

As the interchange of sum-integral order can be justified, we have

$ \minus{} \sum_{n \equal{} 1}^{\infty}\int_{0}^{\infty}\frac {u}{4}e^{u/2}e^{ \minus{} nu}du$

or

$ \minus{} \sum_{n \equal{} 1}^{\infty}\mathcal{L}\left[\frac {u}{4}e^{u/2}\right](n)$

where $ \mathcal{L}[f(x)](z)$ denotes the Laplace Transform of the function $ f$. Finding the transform is easy, and we have:

$ \minus{} \sum_{n \equal{} 1}^{\infty}\frac {1}{(2n \minus{} 1)^2} \equal{} \minus{} \frac {\pi^2}{8}$.

Or something in between:

$ \int^1_0 \frac{x ^ \alpha}{1\minus{}x^2}\,dx\equal{}\int^1_0 x ^ \alpha \sum ^\infty_{n\equal{}0} x^{2n}\,dx\equal{}\int^1_0 \sum ^\infty_{n\equal{}0} x^{2n\plus{}\alpha}\,dx\equal{}$

$ \equal{} \sum ^\infty_{n\equal{}0} \int^1_0x^{2n\plus{}\alpha}\,dx \equal{} \sum ^\infty_{n\equal{}0} \frac{x^{2n\plus{}\alpha\plus{}1}}{2n\plus{}\alpha\plus{}1}\Big|^1_0\equal{}\sum ^\infty_{n\equal{}0} \frac{1}{2n\plus{}\alpha\plus{}1}\equal{}F(\alpha)$

$ I\equal{}\frac{d F}{d\alpha}\Big|_{\alpha \equal{} 0} \equal{}\minus{}\sum ^\infty_{n\equal{}0} \frac{1}{(2n\plus{}1)^2}\equal{}\minus{}\frac{\pi}{8}$


ok so the problem is (10n-1)(10n+1)= 999,999,999,999
and the answer is 6, which doesn't make any sense at all.

I got 100,000 as the answer doing 100n^2-1= 999,999,999,999 and then dividing by 100 and then taking the square root. Could anybody corroborate this answer or tell me what's wrong?
Thanks in advance :lol:

i agree with you
go report this question

if it said how many digits, then 6  is correct

Someone entered the problem incorrectly.  It is now fixed.  Thanks for the report.


geometry
circumcircle
trigonometry
trig identities
Law of Cosines

A triangle inscribed in a circle of radius 5 has two sides of length 5 and 6. Find the two possible lengths of the third side of the triangle.

Call the triangle $ \triangle ABC$, with $ AB \equal{} 5$ and $ BC \equal{} 6$.  Then we want to find $ AC$.  Let the circumcenter of the triangle be point $ E$.  Then we have $ \triangle ABE$ is regular, so $ \angle ABE \equal{} 60^\circ$.  In $ \triangle BCE$, we use the law of cosines to find $ \angle EBC \equal{} \cos^{ \minus{} 1}\bigg(\frac {5^2 \minus{} 5^2 \minus{} 6^2}{ \minus{} 2*5*6}\bigg) \equal{} \cos^{ \minus{} 1} \frac {3}{5}$.  Then $ \angle ABC \equal{} 60^\circ \pm \cos^{ \minus{} 1}\frac {3}{5}$.  Now, by law of cosines:

$ AC^2 \equal{} BC^2 \plus{} AB^2 \minus{} 2AB*BC*\cos \angle ABC$
$ \implies \equal{} 5^2 \plus{} 6^2 \minus{} 2*5*6\cos \bigg(60 \pm \cos^{ \minus{} 1}\frac {3}{5}\bigg)$
$ AC \equal{} \sqrt {61 \minus{} 60 \bigg(\frac {3}{5}\cos 60 \mp \frac {4}{5}\sin 60\bigg)}$
$ \implies \equal{} \sqrt {61 \minus{} 60 (3/10 \mp 4/10 \sqrt{3})}$
$ \implies \equal{} \sqrt {43 \pm 24 \sqrt {3}}$

So $ AC \approx 9.196152423$ or $ AC \approx 1.196152423$


How many positive integers $ N$ less than $ 1000$ are there such that the equation $ x^{\lfloor x\rfloor} \equal{} N$ has a solution for $ x$? (The notation $ \lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $ x$.)

[hide="solution"] We divide this problem into some cases.

Case 1: x is an integer-

Clearly, this can only be $ 1^1, 2^2, 3^3$, and $ 4^4$, so four here.

Case 2: $ \lfloor{x}\rfloor\equal{}2$

Then let $ x\equal{}\sqrt{a}$ and a ranges from 5 to 8, inclusive, so four more.

Case 3: $ \lfloor{x}\rfloor\equal{}3$

Then let $ x\equal{}\sqrt[3]{a}$. a ranges from 28 to 63, so 36 choices.

Case 4: $ \lfloor{x}\rfloor\equal{}4$

Then let $ x\equal{}\sqrt[4]{a}$. a ranges from 257 to 624, which gives 368.

After this, it is easy to see that $ \lfloor{x}\rfloor\equal{}5$ will not work as then N would not be in the given range.

In all, we have 412.[/hide]

Yes, in fact it does.

Take some positive integer $ k$. Then for any $ j$ in the interval $ [k, k \plus{} 1)$, we have $ \lfloor j \rfloor \equal{} k$, so $ j^k \equal{} N$. The interval for $ N$ is $ [k^k, (k \plus{} 1)^k)$. Thus we have $ (k \plus{} 1)^k \minus{} k^k$.

 :| I think my one-line solution explains it better.

[quote="AIME15"]Which totally explains how it works :roll:[/quote]

It does, but if you were retarded like me, you would say $ 4^4\equal{}64$ and get the wrong answer.

Yeah, but my solution prevents the retardedness. You see, you would've calculated $ 4^3\equal{}64$ before $ 4^4$, and if you got $ 4^4\equal{}64$, then you would know something was amiss since you already know $ 4^3\equal{}64$.

[hide="i got this one wrong"]
I put 788 =(
Let's graph this thing.

So we have 1 as a solution, right? So that's 1.

Then case 2: floor x = 2.  We know $ x^2$ ranges from 4 to 9, including 4 but not 9...

Then case 3: floor x = 3 => similar method, [27, 64)

Case 4: floor x = 4 => similar method, [256, 625)

1 + 5 + 37 + 369 = 412[/hide]

[quote="Yongyi781"]One line:
\[ (2^1 \minus{} 1^1) \plus{} (3^2 \minus{} 2^2) \plus{} (4^3 \minus{} 3^3) \plus{} (5^4 \minus{} 4^4) \equal{} 412
\]
[/quote]
I did it the same way, but since I can't write, I almost thought $ 2^1 \equal{} 2^2$, $ 3^2 \equal{} 3^3$, and $ 4^3 \equal{} 4^4$ and calculated $ 5^4 \minus{} 1^1 \equal{} 624$ before I did a

SANITY CHECK

[quote="abacadaea"][quote="AIME15"]Which totally explains how it works :roll:[/quote]

It does, but if you were retarded like me, you would say $ 4^4 \equal{} 64$ and get the wrong answer.[/quote]

You know this is very interesting: I made the [b]exact[/b] same mistake the first time I attempted the problem. Luckily I checked my work and caught it (motivated by the similar $ 2^6$ mistake on the AMC 12A), but it's just amazing that all around the country people were making the exact same mistake.  :o

I got 413 because I counted when N = 1 twice. I found the same thing as what Yongyi said first, but I also added the case when it's like (1/2)^0...which is the same as 1^1. I was kind of annoyed when I realized afterwards.

[quote="brightzhu"][hide="epic failure"]
I added wrong at he end and got 410.  :(
[/hide][/quote]

I failed more. I added correctly to get 412. I even wrote it on my scantron. Then while checking my work I added wrong and changed it to 422.

[quote="Walk Around The River"][quote="abacadaea"][quote="AIME15"]Which totally explains how it works :roll:[/quote]

It does, but if you were retarded like me, you would say $ 4^4 \equal{} 64$ and get the wrong answer.[/quote]

That makes 2 of us.
I triple-checked all my arithmetic on that one too.
 :([/quote]

Heh.

Um, I kinda failed by adding wrong to get 411.  A friend of mine added badly and got 410.  Fail?

I also checked my arithmetic...

I can't complain though. I decided to check my work, and in doing so I changed 6's right answer to a wrong answer but I also changed 8's wrong answer to the right one. Funny, huh?

no not really... =(

Somehow I ended up with 410. Is this explainable?

during the actual contest, i had no idea what to do with this question

afterwards, I play with some numbers on the calculator..and realize you just have to sum

(n+1)^n - n^n

cool problem anyway

Why is our first case $x>0$? In other words why can’t we have negative $x$ that satisfy $x^{\lfloor x\rfloor} = N$ for an integer $N$ that satisfies the conditions of the problem?

Also HMMO Team #1

This problem was pretty nice but my team and I didnt notice in time that $x$ doesn't have to be an integer  :wallbash:

Good problem overall :)

We take cases on $\lfloor x \rfloor$. If $\lfloor x \rfloor = 1$, we can only have $N = 1$, for $1$ way. If $\lfloor x \rfloor = 2$, we can have $N$ ranging from $2^2$ to $3^2-1$, for $3^2-2^2$ ways. Similarly, $\lfloor x \rfloor = 3$ gives $4^3 - 3^3$ ways and $\lfloor x \rfloor = 4$ gives $5^4 - 4^4$ ways. But $\lfloor x \rfloor = 5$ is not possible, so the answer is $1+3^2-2^2+4^3-3^3+5^4-4^4 = \boxed{412}$.


function
trigonometry
algebra unsolved
algebra

Hi guys,

I would like to know the solution of

x^x = k, where k is a defined constant

I've thought about it for a long time, but... nothing to do  :(

Thanks in advance,

Gabriele  :)

hello, here we have $ x\equal{}\frac{\ln(k)}{\text{LambertW}(\ln(k))}$.
Sonnhard.

Thank you very much, Sonnhar...

I watched what LambertW is...

It is the Lambert W function... but... really... what is it???

I didn't understand anything, except that it is the inverse function of $ f(x)\equal{}we^w$...
Is it a sort of "sine", a transcendental function? Otherwise, how can I calculate its values?

Thank you again
Gabriele  :)

hello, here you will found some informations about this function
http://mathworld.wolfram.com/LambertW-Function.html
Sonnhard.


What would be a good book to help me get something better than 4 or 5 on an AIME?

AoPS I & II & Subject books
Art and Craft of Problem Solving
Old AIME's

I heard of a book called Engel. Would that help?

What kind of topics does Art and Craft of Problem Solving cover?

[quote="game.guy"]I heard of a book called Engel. Would that help? [/quote]
I think you want, problem solving strategies by Engel :wink: . I am studying from that book right now, and it 'can' help you with  the harder questions on the AIME, but it is mostly at an Olympiad level. On the few practice AIMES i did recently i got 6,7.. and still Engel's book is hard to comprehend... Maybe its just me


probability
function
conditional probability
\/closed

I was wondering if subforums could be created for each class so that the problem sets would be more organized.

For example, in the Intermediate Counting/Probability class, make subforums by week or by subject. (Week is less work :D)

Week 1, Week 2, Week 3,...., Week 12

or

Conditional Probability, 1-1 Correspondences, PIE, Generating Functions, etc

Anyway, just a suggestion.

Not going to happen I'm afraid. That would mean creating too many subforums, and making proper permissions for each.


function
inequalities
integration
calculus
calculus computations

Suppose f is a differentiable real function such that f(x) + f'(x)<=1 for all x, and
f(0) = 0. What is the largest possible value of f(1)?

Faced with a differential inequality, the first place to look would be techniques from differential equations. In particular, the multiplying factor method for first order linear equations suggests that we multiply both sides of the inequality by $ e^x.$ Try that, and see what develops.

[hide="Short answer."]$ f(1)\le 1\minus{}e^{\minus{}1}$[/hide]

Let $ h(x)\equal{}f(x)\plus{}f'(x)\le 1$. Then $ f$ is a solution of $ y'(x)\plus{}y(x)\equal{}h(x)$. This can be written $ \left(e^xy(x)\right)'\equal{}h(x)e^x$ and so $ f(x)\equal{}e^{\minus{}x}\int_{0}^xe^th(t)dt$. for $ x\equal{}1$ we have $ f(1)\equal{}e^{\minus{}1}\int_0^1e^th(t)dt\le e^{\minus{}1}\int_0^1e^tdt\equal{}1\minus{}e^{\minus{}1}$. Equality is obtained when $ h(t)\equiv 1$, i.e for $ f(x)\equal{}1\minus{}e^{\minus{}x}$.

Solution...
[hide]
Let $ f(x) \equal{} e^{ \minus{} x}v(x)$, $ v(0) \equal{} 0$. We have $ e^{ \minus{} x}v(x) \minus{} e^{ \minus{} x}v(x) \plus{} e^{ \minus{} x}v'(x)\leq 1$, so $ v'(x)\leq e^{x}$.

Taking integral, we will have $ \int_0^x v'(t) dt \leq \int_0^x e^{t} dt$, or $ v(x) \minus{} v(0)\leq e^x \minus{} 1$. Putting x=1: $ v(1)\leq e \minus{} 1$, so $ f(1)\leq 1 \minus{} e^{ \minus{} 1}$, and $ f(1) \equal{} 1 \minus{} e^{ \minus{} 1}$ if $ v(x) \equal{} e^x \minus{} 1$, or $ f(x) \equal{} 1 \minus{} e^{ \minus{} x}$.

Ottem, sorry, my solution is same to your's...
[/hide]

Ottem's solution, Kirill's solution, and the solution I hinted at but didn't write up differ in organization and notation, but are all essentially the same solution, and are motivated by the same idea.


USAMTS
geometry
geometric transformation
rotation
linear algebra
matrix
analytic geometry

Post scores for USAMTS here and your grade. For example (here are my statistics):

Score: 20, Grade: 7

You may also post what you got off on. For example (here is what I got off on):

I got a 2 on problem 4 for not getting the 40 case.
I got a 3 on problem 5 for my proof of part b being incomplete.

25.

I'm glad my ridiculous "1489+3=1502" was excused...

5/5/5/4/5.

On #4 for whatever reason I didn't justify that my constructions implied convexity.

Lol 17. Which is pretty good considering how much time+effort I actually put into it. Yay this means I don't have to try for gold anymore and can be content with 15+  :)

Score: 17 Grade: 7

Didn't put in a lot of work, 5/5/2/0/5

#4 had no idea what to do (I don't like working on hard geometry)

#3 my PPW was rejected...

5/5/5/5/5

Comment for #3:

$ 1488\plus{}4 \neq 1490$
Otherwise nice.

 :rotfl:

5/5/5/3/4 Grade 11
I wonder how I got full points for 3, I needlessly complicated the question with a rotation matrix in order to avoid slope arguments.
I'm mad at myself for number 4. I considered including an argument about being able to construct both the 40 and 60 (or whatever they were), but decided against it because it would increase my page count and wanted the beautiful short proof, not the legit long one.
Number 5 I am also mad with myself because I lost a point for cutting out an algebra step in part (b)
GL to everyone on round 2.

5/5/5/5/5.  Apparently I satisfied their criteria for number 4, even though I didn't show that the hexagons were constructable.  Also, my proof for #3 was not as concise as the grader would have liked.  Hi five for that! :lol:

I think it's funny how of the 15 people on the leaderboard who got 25's, 10 of them are in 11th grade.

In 8th grade, and it was my first time writing formal proofs. I got a 3/5/2/2/2=14, which I am satisfied with.

I lost 2 points for screwing up casework and getting 72. Also, I didn't notice the slick hexagon thing in the posted solution.

This was easy for me, and I got full credit.

Only got 8 points, and failed to show that they were the only ones. I can't understand the connection to number theory in this problem. Lost 3 points.

At the last second, I noticed the flipping over. I didn't see the 40 case, and forgot to prove that they could be constructed (didn't use the fact that the hexagon was convex...). Lost 3 points

I assumed something when doing (a), and my proof wasn't rigorous in part (b).

I learned a lot though. I typed these in Word, but the ugliness motivated me to learn $ \text{\LaTeX}$ for this round. Also, it was a very good experience on proofs, since this was my first time. After really understanding the problems and seeing other people's solution, I learned a lot.[hide=Index][/hide]

3/5/5/0/4 - 17  :lol:
which is good pretty for me, since i got 7 last year :(

Oh well seeing that everyone did better than me um I am not going to post mine.  I never wrote proofs and I kinda didn't write them right.

Blegh. I've never done proofs before...but I still did so horrible...

3/3/0/1/0=7.  :(  :(  :(

25 for me: I guess the limited rigor of my hexagon constructability argument was enough.  (I thought I would get 4 on #4)

3/4/5/2/5 = 19.

Not as good as I hoped but I didn't spend much time on it.
I got a point off on 2 for not considering 1*11111 as a possible factorization, but I figured it was clear that it wouldn't work in that case... should I protest it?  :(

EDIT: looking at the rubric, I did earn a 4... :( too bad. I thought I wouldn't have to explain that case. oh well.

I'm so proud that three of my solutions were programmed :) way too lazy to solve them by hand.

5/4/5/5/5=24, even though i didn't prove existence for either case on #4 (maybe just because my diagram was so good). But i still don't know why i lost a point on #2...my grader said i needed either to check or dismiss the case x+y=11111, x-y=1 in the 5-digit part (where $ (x^2, y^2)$ is the pair), except i clearly said that in this case, x=5556 and so $ x^2$ has more than 5 digits, and therefore this case is clearly absurd. Um.

5/5/0/0/0
Basically I was like FIDDLESTICKS, I DONT SEE AN IMMEDIATELY TRIVIAL AND STRAIGHTFORWARD SOLUTION TO #3-5 and didn't spend the time figure out more. Maybe I'll try harder on round 2?

2/0/0/0/0...  LOL.  I'm in 8th grade and this is my first time writing proofs.  And I only had a week to work on the problems.  I managed to screw up the casework on the only one I did, though. >.<

I'm definitely not going to protest these scores that I really don't deserve because the solutions were written on the weekend before (along with college apps...).

1: 3 - missed cases
2: 5 - yay!
3: 4 - yeah over counted, like I expected, but only one point off??
4: 1 - I went nowhere
5: 5 - really? I didn't know I could write good proofs!

Total: 18

5/5/5/2/5 can only  miss one more point to get a gold >_<

9th grade
I got 5/5/5/3/5 cuz i apparently failed to show that the hexagon was convex...

[quote="dnkywin"]Darn I got a 22... missed 3 points on #4 for forgetting about the 40 case and not constructing the 60 case... oh well  :([/quote]
good job, and now you are going to score worse than last year... :P

3/5/5/2/5 = 20. 9th grade

Overcounted/Messed up casework for #1. Completely forgot about 40 case for #4. I'm really surprised that I got full credit for 5b, since I made up a lot of stuff about sequences for my proof...

3/5/4/4/5=21

hmmm maybe i should not be up til 4am the night before doing these, but i am unfortunately procrastinating on round 2 more so than i did i round one as of right now.

somehow got 120 on number 1, was too sleepy at 4am to finish number 3 ( i am very surprised that my claim of clearly the answer is 1488 (which i guessed instead of 1492) in the end got a 4), and wasn't sure if constructing the diagrams was necessary on number 4 and decided to keep it short.

i guess i'll have to try for gold next year.

8th grade...like others, this was my first real attempt at writing proofs.  Luckily, USAMTS graders are not very harsh...

5/5/4/0/5=19

Eh, only 4 grade 8 or younger on the leaderboard, so I don't feel too bad about this.

For #3 I just stated the answer and said to use analytic geometry, and somehow I got a 4...?

#4 I trig-bashed unsuccessfully because I had an error in, I think, the third line in a three page solution.

On pace for silver :)

Voila.
5/4/2/0/1 = 12 (10th grade)
A list of excuses for my peers  :| :
Very first USAMTS. Very first time I wrote formal proofs, ever. Spent too much time figuring out LATEX. I overslept.
#2: one small, small error in counting cases and I lost a point. Guess that's what happens when you write it the night before.  :!:  #3: thought it was only one point for putting the answer only? Apparently the scoring criterion modified that. #5: WHAT?? No points for second part?
Anyways, enough viewing the forums for me. Time to get back to finishing Round 2!

5/5/5/5/5 11th grade

a Bit surprised I got full points on 4 as I did not prove both were constructable

5/4/5/2/5 = 21 (8th)

#2: I found a solution for the 6-digit case. Go me.
#4: Same as a lot of other people, omitted the 40 case.

9th grade,
5/5/5/3/5

Ugh I did them all 11 o'clock at night before the due date.
I didn't know you had to construct the hexagons  :(


A piece of cardboard has the shape of a pentagon $ABCDE$ in which $BCDE$ is a square and $ABE$ is an isosceles triangle with a right angle at $A$. Prove that the pentagon can be divided in two different ways in three parts that can be rearranged in order to recompose a right isosceles triangle.

Let a = BC be the square side. Area $|ABCDE| = \frac{5a^2}{4}.$ Let x be the leg of the isosceles right triangle with the area $\frac{x^2}{2} = \frac 5 4 a^2.$ Then $x = a\ \sqrt{\frac 5 2.}$ Since $AC = AD = a\ \sqrt{\frac 1 4 + \frac 9 4} = a\ \sqrt{\frac 5 2}$, one of these is the 1st cut, say AD. Rotate the triangle $\triangle ADE$ by $90^\circ$ around A, so that AE coincides with AB and D goes into $D' \in BE$. Let M be the midpoint of BC. Since the triangles $\triangle MCD \cong \triangle MBD'$ are congruent, DM is the 2nd cut.


i had some free time so i looked up how many people actually post on AOPS. out of 19361, only 7625 have ever posted anything. that's about 40%. that also includes the people that posted 1 thing, or 2 things. that's not very good. :(  does anyone have any opinions on this?

There are lots of people that are guests or just look around, reading threads.

Maybe some people join just to take classes.

You'll usually get that with most online forums.  Only about 10% of the people are active.

Well, for one, there are a lot of people, like my dad, who don't have very many things to post about, but read through the forums, and do some things on AoPSWiki.

Besides, I'm sure that a lot of people are searching for forums and join lots of them, but find that they aren't that interested in math after all, so they stop coming.

Well i know alot of people at my school who buy the AoPS books to learn and have an account in the fourm. But they don't really like math that much and find the fourm quite boring.

Some people sign up because they thing this forum is really nice and all, but they get bored really fast and do not visit the  website  for the rest of their lives.

Also, some might sign up only for the Classes, or to browse Resources and such. I think the level of activity here is pretty astounding and 40% is not low at all.

[quote="drunner2007"]Yeah, I've signed up to tons of websites but I end up not posting. I wonder if any of the moderators or administrators have any histograms that would give us a good idea of member participation.[/quote]We do, but they are not for public use ;)

in an other forum members that had 0 posts for a year were deleted... but it was only forum

[quote="rem"]Another reason why some ppl have so little posts is because they want to register to see various ppl's profiles or talk to them by private messaging; they think that the opportunity of posting something so that many ppl around the world will see it is really cool, but then they either get bored or have nothing to post.
I still think that this aops -mathlinks forum is much more active and has much more posts than any other forum where you need to be a regusterd user to post. :!:[/quote]
That's true. Some users don't post because they rather just read or they don't know what to post.


Anybody got thei PSAT scores yet? If you have, post it here...if you wish

got them today, 235: 80v, 77m, 78w, -1, -1, -3, respectively

impressive...

what grade are you in?

shame that they deduct so many points for so few errors on the math...no?

that would be "sir," im in the 11th grade, Tuesday test, from last year, verbal up 3 points, math down 3, writing up 10; overall: up 10 points from my score.

you should get national merit with ease, but then you have to write an essay

did the rest of you get scores yet?  i have not, but i think we should within the week, or maybe next week even.

My friends got theirs today, but I'm sick so I'll get my Mom to pick them up tomorrow.

I got mine, but...yeah...they're not as good as I wanted (nowhere near *******'s) :D. But it's ok, cuz I'm a sophomore. *Frantically buys SAT books off ebay*

i got mine back, since my counselor was so nice to tell me early. my score is 232, with 77 writing, 75 math, 80 writing. a little ironic, since i dont do anything for writing, but i spend a lot of time on math. even so, im in 10th grade, so im pretty happy with my results.

Do they mail the scores to you?

-interesting_move

[quote="mathnovice"]Are the cutoff scores in the Princeton Review the cutoff for Commended or Semi-finalist?[/quote]
The cutoff scores change each year. Last year, the cutoff for semifinalist was 209, and the cutoff for commended was 200 or 201.

222, Am worried that the cutoff for semifinalist will be 223 or something like that, though...

221 --- I should have had higher, but I when I boxed in my answer I wrote "92" instead of "192." Note to self- never do that again. : ( I hope I make the cutoff though.

mathnovice arent you from MA?  I'll pray that it's 221 too.  That'd be horrible to be so close and not make it.

Matt -- I'm pretty sure it varies from state to state.  MA probably has a higher cutoff than most places, although I strongly doubt it will be 220's.

I did so badly...[hide]191[/hide].

-interesting_move

i'm positive that it's been ~222 meaning =/- 1 or 2 points for many years now.  My guidance counselor says that in his ~10 years it's gone from 220 to 222, and I know for sure that last year's cutoff was 222.  And it is the highest in the nation...  I read that NJ's was 221 for this year's seniors, not positive that it's true though.

it is right, oklahoma's was like 207 last year, when other states were in the 220's and suches

[quote="confuted"]Wait... the cutoff for NATIONAL merit varies between states? That doesn't seem right.[/quote]

The statement is correct. And I think the statement that MA is usually one of the two or three most competitive states is also correct. I was comfortably above the cutoff, decades ago, in Minnesota. The states of the upper Midwest, where many high school students score well on all manner of standardized tests, historically have NOT been very competitive for getting above the National Merit Scholarship cutoff, because many students in that region self-select to take the ACT rather than the SAT and its associated PSAT/NMSQT. What makes a state have a high cutoff score is BOTH lots of well-prepared students AND a high percentage of the well-prepared students showing up to take the PSAT/NMSQT, which is, after all, a voluntary test.

I suppose the policy underlying the differing cutoff scores around the county is to ensure that young people from all states of the Union have a chance at getting a [url=http://www.nationalmerit.org/entering.html]National Merit Scholarship[/url]. (After edit: yes, that's it, and the link is below.)

http://www.nationalmerit.org/nmsp.html#semifinalists

"To ensure that academically able young people from all parts of the United States are included in this talent pool, Semifinalists are designated on a state representational basis."

It's not a perfect system (the scholarships are terribly underfunded, as they long have been), but it's a decades-old first attempt at helping bright young people from families of modest means go to college. State university systems with low list prices have undoubtedly had a much bigger impact on access to higher education than has the National Merit Scholarship program--the former worked better for me than the latter, given my family circumstances at the time.

Yeah, my mother was one of the top scorers in Maine back in the day -- odds are not good that she would have made it in New York or Massachusetts.  Many of the scholarships are distributed with a large helping of arbitraryness, though -- look at the business-backed scholarships (one of which my mom recieved, because my grandmother worked for Worldbook at the time, who happened to sponsor a National Merit Scholarship.)

[quote="mathnovice"]When do you find out if you qualified for semi-finalist? If I had gotten one more question right, I probably would not have to worry. : ([/quote]
You'll find out at the beginning of your senior year.  It'll be a long wait.

227(80W 77M 70V) but i 1590ed the SAT so its all good : )

I'm a sophomore, I got my scores a while back, 206(61V, 78M, 67W) This was my first time taking the PSATs.

240 but i took them before as a sophomore also and got 220 (80 M, 72 V, 68 W)


Find all functions $ f: \mathbb{R}\to\mathbb{R}$ that satisfy $ f((x\minus{}y)^2)\equal{}f^2(x)\minus{}2xf(y)\plus{}y^2,\forall x,y\in\mathbb{R}$

[hide]
Plug in $ x\equal{}y\equal{}0$ to get $ f(0)\equal{}f(0)^2$, so $ f(0)$ is either 0 or 1.

Plug in $ x\equal{}y$ to get $ f(0)\equal{}f(x)^2\minus{}2xf(x)\plus{}x^2\equal{}(f(x)\minus{}x)^2$. If $ f(0)\equal{}0$, then $ f(x)\equal{}x$ for all $ x$. This function works.

If $ f(0)\equal{}1$, then for each $ x$, $ f(x)\equal{}x\plus{}1$ or $ f(x)\equal{}x\minus{}1$. Suppose there exists $ a$ with $ f(a)\equal{}a\minus{}1$ Then $ x\equal{}a,y\equal{}0$ gives
\[ a^2\pm 1\equal{}f(a^2)\equal{}(a\minus{}1)^2\minus{}2a\equal{}a^2\minus{}4a\plus{}1\]
If $ f(a^2)\equal{}a^2\plus{}1$, then $ a\equal{}0$, which we know is false. Otherwise, $ a\equal{}1/2$ but then $ f(1/4)\equal{}1/4\minus{}1$, repeating the argument gives $ 1/4\equal{}1/2$, a contradiction.

So the only other solution is $ f(x)\equal{}x\plus{}1$ for all $ x$. This solution turns out to work, too, since
\[ (x\minus{}y)^2\plus{}1\equal{}(x\plus{}1)^2\minus{}2x(y\plus{}1)\plus{}y^2\]
for all $ x,y$.
[/hide]

$ x\equal{}0$ gives $ f(y^2)\equal{}f(0)\plus{}y^2$ so $ \forall x\ge 0: \ f(x)\equal{}x\plus{}a\ (f(0)\equal{}a)$
so if $ x\ge 0$ gives $ (x\minus{}y)^2\plus{}a\equal{}(x\plus{}a)^2\minus{}2xf(y)\plus{}y^2\iff 2xy\minus{}a\equal{}\minus{}2x(a\minus{}f(y))\minus{}a^2$
then $ \forall y\in R: \ f(y)\equal{}y\plus{}a$ because $ x\equal{}y\equal{}0$ gives $ a^2\equal{}a$

$ f((x\minus{}y)^2)\equal{}x^2\plus{}y^2\minus{}2xy\plus{}a$
$ f^2(x)\plus{}y^2\minus{}2xf(y)\equal{}x^2\plus{}y^2\minus{}2xy\plus{}a^2$

then $ f(x)\equal{}x$ or $ f(x)\equal{}x\plus{}1$

[quote="scorpius119"][hide]

If $ f(0) \equal{} 1$, then for each $ x$, $ f(x) \equal{} x \plus{} 1$ or $ f(x) \equal{} x \minus{} 1$. Suppose there exists $ a$ with $ f(a) \equal{} a \minus{} 1$ Then $ x \equal{} a,y \equal{} 0$ gives
\[ a^2\pm 1 \equal{} f(a^2) \equal{} (a \minus{} 1)^2 \minus{} 2a \equal{} a^2 \minus{} 4a \plus{} 1
\]
If $ f(a^2) \equal{} a^2 \plus{} 1$, then $ a \equal{} 0$, which we know is false. Otherwise, $ a \equal{} 1/2$ but then $ f(1/4) \equal{} 1/4 \minus{} 1$, repeating the argument gives $ 1/4 \equal{} 1/2$, a contradiction.

[/quote]

In my opinion, there is a intricacy in above argument. Because there exist [b]some[/b] values $ a$ such that $ f(a) \equal{} a \minus{} 1$. With $ a \equal{} \frac{1}{2}$ then $ f(1/2)\equal{}1/2\minus{}1$ and $ f(1/4)\equal{}1/4\minus{}1$, we can't conclude $ \frac{1}{2}\equal{}\frac{1}{4}$

Well, I was able to conclude that if there is an $ a$ that satisfies $ f(a)\equal{}a\minus{}1$, then $ a$ must equal $ 1/2$. But if we let $ b\equal{}1/4$, we must also have $ f(b)\equal{}b\minus{}1$, and then $ b$ must equal $ 1/2$. Since $ b\equal{}1/4$, this is not possible.

Alternatively, plug in $ x\equal{}1/4,y\equal{}0$ to get that $ f(1/16)$ is something other than $ (1/16)\pm 1$.


trigonometry
geometry
trig identities
Law of Sines
Law of Cosines

The skyhawk in Cedar point?
http://www.cedarpoint.com/public/park/rides/thrill/skyhawk/index.cfm

I just need to make trig questions like something that could be put on a worksheet with the skyhawk. Does anyone have any ideas.

Just putting out random ideas:

1) The length of the skyhawk arm is ______ ft and they form an isosceles triangle with angle ____degrees.  Find the length of the third side.

2) Find the area of the triangle spanned by the arms of the skyhawk.

If you can create some sort of triangle, you can make a trig problem revolving around the Law of Sines or Law of Cosines (or anything else if I'm leaving them out).


The area of triangle $ ABC$ is $ 1$, $ AC \equal{} 2BC$, point $ K$ is middle of $ AC$. The circle with center $ K$ crosses side $ AB$ in points $ M$ and $ N$, where  $ AM \equal{} MN \equal{} NB$. Find the area of that part of triangle $ ABC$, which is inside the circle.

[quote="hell_ever"]wat do u mean by circle with circumcentre K? K is the circumcentre of which triangle? Please be clear.[/quote]

Sorry for my bad english. I mean K- is center of circle

Meatman, your problems are too easy for the olympiad level. If you continue to post easy problems here, you will probably not get useful replies. I am moving this one to the high school section as well.

Regards, yetti.
________________________________________

Denote $ b \equal{} AC, c \equal{} AB.$ $ AK \equal{} \frac {b}{2},$ $ AM \equal{} \frac {c}{3},$ $ AN \equal{} \frac {2c}{3}.$ By the cosine theorem for the $ \triangle AKM, \triangle AKN,$

$ KM^2 \equal{} AK^2 \plus{} AM^2 \minus{} 2 AK \cdot AM \cos A \equal{} \frac {b^2}{4} \plus{} \frac {c^2}{9} \minus{} \frac {bc}{3}\cos A$

$ KN^2 \equal{} AK^2 \plus{} AN^2 \minus{} 2 AK \cdot AN \cos A \equal{} \frac {b^2}{4} \plus{} \frac {4c^2}{9} \minus{} \frac {2bc}{3}\cos A$

From $ KM \equal{} KN,$

$ KN^2 \minus{} KM^2 \equal{} \frac {c^2}{3} \minus{} \frac {bc}{3} \cos A \equal{} \frac {c}{3} \left(c \minus{} b \cos A \right) \equal{} 0$

It follows that $ \cos A \equal{} \frac {c}{b},$ which means that the angle $ \angle B \equal{} 90^\circ$ is right, $ B$ is on a circle $ (O)$ with diameter $ AC.$ From the condition $ AC \equal{} 2 BC,$ $ B$ is also on a circle $ (C)$ centered at $ C$ and with radius $ CB \equal{} \frac {AC}{2},$ congruent to the circle $ (O).$ $ B$ is one of the equivalent intersections of the circles $ (O), (C)\ \Longrightarrow$ the right angle $ \triangle ABC$ has angles $ \angle B \equal{} 90^\circ, \angle C \equal{} 60^\circ, \angle A \equal{} 30^\circ$ and $ \cos A \equal{} \frac {c}{b} \equal{} \frac {\sqrt 3}{2}.$ Substituting $ b \equal{} \frac {2c}{\sqrt 3}, b^2 \equal{} \frac {4c^2}{3}$ into the cosine formula for $ KM$ yields $ KM^2 \equal{} \frac {c^2}{9}, KN \equal{} KM \equal{} \frac {c}{3} \equal{} MN,$ the $ \triangle KMN$ is equilateral. The desired area is then, using $ KM^2 \equal{} \frac {c^2}{9} \equal{} \frac {b^2}{12},$

$ S \equal{} \frac {\pi \cdot KM^2}{3} \plus{} \frac {KM^2 \sqrt 3}{4} \equal{} b^2\ \frac {4 \pi \plus{} 3 \sqrt 3}{144}.$

Oh, sorry :blush:  I won't post such kind of problems there.
Thanks for solution.


I found this problem in Harold Reiter's mathcount's workshops: http://www.math.uncc.edu/~hbreiter/problems/HawaiiIndex.htm , but I have no clue where to even begin. Can someone help? Thanks.

How many rectangles are defined by the grid lines below? How many of them are squares?

Uh look at this

[hide]You know the largest rectangle can at most have a width of 2 squares.  So break it up into 4, 6*2, rectangles count how many squares/rectangles in each one and times by 4.  I count 63 total rectanges 17 of which are squares... that means 252 rectanles and 68 squares.  There are more rectangles however.

Now you also have to account for rectangles that overlap the boundaries that you created.  There are only 4 squares total that cross those borders.  I count 24 extra rectangles on each segment *4=96.  Now just add them all together...

I get 352 total squares and 72 of those are squares.  I'm quite sure about that answer...  I used to know a formula that is applied to counting rectangles... but I can't remember it.[/hide]

hope that helps.

[quote="Treething"][color=darkred]Don't forget to count rectangles with non-integral side lengths...[/color][/quote]

But those aren't defined by the gridlines.

Oh. The gridlines. Sorry, I thought it meant intersections :blush: . I gotsta learn how to read...

[quote] You know the largest rectangle can at most have a width of 2 squares. [/quote]

Sorry if I  am not reading the problem carefully .. but there are rectangles  ( may be with "holes" in the middle , but formed by lines of the grid) which are larger...

I think your best bet may be to determine the number of rectangles if there were grid lines where the 'hole' is, then subtract out the ones that don't exist as a result of the hole.

[hide="Doing so yields..."]If there were no hole, there are $\binom{9}{2}$ ways to choose the two vertical sides and $\binom{9}{2}$ ways to choose the two horizontal sides, so a total of 36^2.

The rectangle will not exist as a result of the hole iff it has at least one vertical and one horizontal side that passes through the hole.  So how many ways are there to choose two vertical lines such that at least one passes through the hole?  There are 3 choices for the line that will pass through the hole, times 8 choices for the second line, minus 3 choices for repetitions (if we number the lines from 1 to 9, then 45, 46, and 56 are each counted twice).  So 21 ways.  Similarly, there are 21 ways for the horizontal lines.  So in all 21^2 are bad.

Thus the answer is 36^2 - 21^2 = 855.[/hide]

As for how many squares there are,
[hide]Just counting them may work best.  There are 48 1x1s, 24 2x2s, 1 4x4, 16 5x5s, 9 6x6s, 4 7x7s, and 1 8x8 (note that for the 5x5s through 8x8s, there are the usual (8-n)^2 n x ns because the hole is too small to take any away.  In all, 103.[/hide]

Or, you can actually count the rectangles directly.  I did it by breaking them down into height and witdth and making a chart:

[code]
\W   1   2   3   4   5   6   7   8
H\
1   48  36  24  20  16  12   8   4
2   36  24  12  10   8   6   4   2
3   24  12   0   0   0   0   0   0
4   20  10   0   1   2   3   2   1
5   16   8   0   2   4   6   4   2
6   12   6   0   3   6   9   6   3
7    8   4   0   2   4   6   4   2
8    4   2   0   1   2   3   2   1[/code]
Adding up all of the entries, I get 477 rectangles and 91 squares -- anyone want to chime in on why this is so much smaller than Keone's answer?  I notice that he made a definite mistake in counting the number of 5x5 squares -- did he do something similar on the rectangles as a whole, or is it my mistake?

Actually, I do see one mistake in his work -- you don't necessarily need a vertical and a horizontal side to cross the hole.  Consider, for example, some 3x8 rectangles.

I get 477 total rectangles.  91 of these are square.

There are 6 full-height vertical lines and 6 full-width horizontal lines.  Choose one pair of horizontal lines plus a pair of vertical lines:  6C2x6C2=225.

Consider the top section directly above the empty square.  There are 3 full-width horizontal lines and 3 short vertical lines.  One pair from each:  3C2x3C2=9.

Continuing the entire top section above the empty square (and extending the width of the large square).  There are 3 full-width horizontal lines.  Choose 2.  There are 6 full-height vertical lines.  Choose 1.  There are 3 short vertical lines.  Choose 1.  3C2x6x3=54.

Repeat for the section below, left, and right of empty square.

Total:  $225+4(9+54)=225+252=477$

[quote="JBL"]Adding up all of the entries, I get 447 rectangles and 91 squares [/quote]
JBL:  I think your table is correct, so there may be an arithmetic error.

Fixed -- no arithmetic error, actually, just a typo.  Thanks.


geometry
perpendicular bisector
geometric transformation
geometry unsolved

On the exterior of a triangle  $ABC$  we draw two similar right triangles $ABD , ACE$ , with right angles at $D , E$ and $\angle BAD = \angle CAE$. If $M$ is the midpoint of $BC$ , prove that  $MD = ME$.

  [u]Babis[/u]

  [i]Mery Cristmas !!![/i]

That's easy.Just let F is the midpoint of AB,and G is the midpoint of AC,DF=AB/2=MG,GE=AC/2=FM,and ∠DFB=∠EGC,∠BFM=∠A=∠MGC,so ∠DFM=∠MGE,so triangle DFM and MGE is congruent,so MD=ME. :)

Two additional solutions

M1) Take $K$ and $L$ the midpoints of $AB$ and $AC$, see that $\triangle DKM \equiv \triangle MLE$ (s.a.s), done.

M2) Take $P$ on the perpendicular bisector of $BC$ so that $\angle BCP = \angle ABD$, then see that there is a spiral similarity centered at $B$, which takes $P$ to $M$ and $A$ to $D$, and a spiral similarity centered at $C$, taking $P$ to $M$ and $A$ to $E$, hence $\frac{DM}{AP}=\frac{ME}{AP}$ , consequently $ME=MD$ and, as it has been seen, $\widehat {(DM, AP)} = \widehat {(ME, AP)} = \angle ABD$, so $\angle DME = 2\angle ABD$ .

Best regards,
sunken rock


trigonometry
function
limit
symmetry
algebra proposed
algebra

1.  Suppose $ f: \left[ \minus{} \frac {\pi}{2},\frac {\pi}{2}\right ]\to \mathbb{R}$ is a continuous function.  Prove that $ f(x) \minus{} \tan x \equal{} 0$ has at least one solution in $ \left( \minus{} \frac {\pi}{2},\frac {\pi}{2}\right)$.

2. Suppose $ f: [ \minus{} \pi,\pi]\to \mathbb{R}$ is a continuous function and $ f(x) \equal{} f( \minus{} x)$.  Prove that $ f(x) \minus{} \tan x \equal{} 0$ has at least two solutions in $ \left[ \minus{} \pi, \pi\right]$

[quote="v.dinesh"]the first one is obvious throug graph  :) but dont know how to do it rigorously :oops:[/quote]

It is obvious throug the graph and it also follows from the Bolzano Theorem, since g(x) = f(x) - tanx is continious in [m, n] for $ m,n \in (\minus{}\frac{\pi}{2}, \frac{\pi}{2})$ with g(m) < 0 and g(n) > 0 (these m,n exists because $ \lim_{x \rightarrow \frac {\pi}{2}}g(x) \equal{} \plus{}\infty$, and $ \lim_{x \rightarrow \minus{}\frac {\pi}{2}}g(x) \equal{} \minus{}\infty$, because f is continious in $ [\minus{}\frac{\pi}{2}, \frac{\pi}{2}]$ and so the limits exists)

Another way for 1 is to use the fact that continuous functions have maxima and minima on closed and bounded intervals. Take the maximum $ M$ of $ f$, and note that if $ \tan x > M$ (which it is somewhere on $ (\minus{}\pi/2,\pi/2)$, then $ f(x) \minus{} \tan x < 0$. Likewise, the same reasoning on the minimum $ m$ gives an $ x$ such that $ f(x) \minus{} \tan x > 0$. Then the intermediate value theorem finishes.

For 2, reducing it showing that $ |f(x)| \equal{} \tan x$ has a root in $ [0,\pi/2)$. Suppose WLOG that $ f(0) > 0$, and then use the maximum argument to show there is an $ x$ such that $ f(x) < \tan x$.

[quote="MellowMelon"]For 2, reducing it showing that $ |f(x)| \equal{} \tan x$ has a root in $ [0,\pi/2)$. Suppose WLOG that $ f(0) > 0$, and then use the maximum argument to show there is an $ x$ such that $ f(x) < \tan x$.[/quote]How does this give us two roots in $ [ \minus{} \pi ,\pi]$?  Note the function satisfies $ f(x)\equal{}f(\minus{}x)$, not $ f(x)\equal{}\minus{}f(\minus{}x)$.

In $ [0,\pi]$, if $ f(x) \equal{} \tan x$ we have a solution on that interval, obviously. If $ \minus{}f(x) \equal{} \tan x$, then $ \minus{}f(\minus{}x) \equal{} \tan x \Rightarrow f(\minus{}x) \equal{} \tan \minus{}x$. Then we can reduce that to $ [0,\pi/2]$ by symmetry and find one solution there, giving two on the larger one.

I still must not be understanding.
[quote="MellowMelon"]In $ [0,\pi]$, if $ f(x) \equal{} \tan x$ we have a solution on that interval, obviously. If $ \minus{} f(x) \equal{} \tan x$, then $ \minus{} f( \minus{} x) \equal{} \tan x \Rightarrow f( \minus{} x) \equal{} \tan \minus{} x$.[/quote]So far you have proven that the existence of one solution to $ f(x) \equal{} \tan x$ in $ [ \minus{} \pi,\pi]$ is equivalent to the existence of one solution of $ |f(x)| \equal{} \tan x$ in $ [0,\pi]$.  I agree, at least one such solution certainly exists.[quote="MellowMellon"]Then we can reduce that to $ [0,\pi/2]$ by symmetry and find one solution there, giving two on the larger one.[/quote]Reduce what? Find one solution where?

I really don't think symmetry is important here.  For example, if I were to replace $ \tan x$ by something non-symmetric like, say,  $ \tan x \plus{} x^2 \plus{} 7x \plus{} 1$, the conclusion that $ f(x) \equal{} \tan x \plus{} x^2 \plus{} 7x \plus{} 1$  has at least two roots in $ [ \minus{} \pi, \pi]$ for any continuous even $ f$ would still hold.  And in fact it would still hold for a much wider class of functions than just continuous even ones.

(fix: when I talked about $ |f(x)| \equal{} \tan x$, I probably meant $ |f(x)| \equal{} |\tan x|$, which changes things when going to other intervals)

The symmetry is just a way of simplifying the solution. If it were $ \tan x$ plus some asymmetric function, you'd probably be right that the conclusion would remain the same, and I admit it would take a stronger proof than mine.

The symmetry involved is that $ \tan x \equal{} \minus{}\tan (\pi \minus{} x)$. If any $ f(x)$ defined on $ [0,\pi/2$ has a solution to $ |f(x)| \equal{} |\tan x|$, then any $ f(x)$ defined on $ [\pi/2,\pi]$ has a solution to that equation because we let $ g(x) \equal{} f(\pi \minus{} x)$ be defined on $ [0,\pi/2]$, and the solution we know exists for $ |g(x)| \equal{} |\tan x|$ implies that \[ |g(x)| \equal{} |\tan (x)|\]
\[ |f(\pi \minus{} x)| \equal{} |\minus{}\tan (\pi \minus{} x)|\]
\[ |f(\pi \minus{} x)| \equal{} |\tan ( \pi \minus{} x )|\]


function probability calculus integration quadratics

The new wave function for a particle is made by adding two functions of particle in states $ a$ and $ b$ ,such that the function is so:

$ \psi\equal{}C_a\psi_a\plus{}C_b\psi_b$

In this equation,$ C_a$ and $ C_b$ are constants.$ C_a\equal{}\frac{1}{2}$ is given.What is the probability of particle being in state b?

Don't we just evaluate the integral $ \int_{\minus{}L}^{L}\Psi\Psi^{*}\ \text{d}x$?

I need the solution for checking whether  mine is correct or not  :) .

I could be wrong, but I'd say that the probability of the particle being in state b is just 3/4, due to the normalization condition for wavefunctions.  Namely, I think $ C_b \equal{} \sqrt(3)/2$, since $ |C_a|^2 \plus{} |C_b|^2 \equal{} 1$.

I conclude this because we already know that
\[ \int_{ \minus{} \infty}^\infty |\psi_a|^2 d(\mbox{whatever}) \equal{} \int_{ \minus{} \infty}^\infty |\psi_b|^2 d(\mbox{whatever}) \equal{} 1.
\]
Oh wow, I'm a "Hodge Conjecture" now.

Oh wait, nevermind, it's not so simple.  That is, we want to find $ |C_b|^2$ given that
\[ \int_{ \minus{} \infty}^\infty |\psi|^2 d(\mbox{whatever}) \equal{} \int_{ \minus{} \infty}^\infty |C_a\psi_a \plus{} C_b\psi_b|^2 d(\mbox{whatever}) \equal{} 1.
\]
Then
\[ 1 \equal{} \int_{ \minus{} \infty}^\infty |C_a\psi_a \plus{} C_b\psi_b|^2 \equal{} \int_{ \minus{} \infty}^\infty (C_a\psi_a \plus{} C_b\psi_b)(C_a\psi_a^* \plus{} C_b\psi_b^*) \equal{} \\
C_a^2 \int_{ \minus{} \infty}^\infty |\psi_a|^2 \plus{} C_a C_b \int_{ \minus{} \infty}^\infty \psi_a \psi_b^* \plus{} C_a C_b \int_{ \minus{} \infty}^\infty \psi_b \psi_a^* \plus{} C_b^2 \int_{ \minus{} \infty}^\infty |\psi_b|^2 \equal{} \\
C_a^2 \plus{} C_b^2 \plus{} C_a C_b \int_{ \minus{} \infty}^\infty \psi_a \psi_b^* \plus{} \psi_b \psi_a^*.
\]
And that's a quadratic in $ C_b$, assuming you know the last integral.

The answer is $ \frac{3}{4}$ if $ \psi_1$ and $ \psi_2$ are orthogonal.  If they
are not, the answer may be different, as genericme pointed out.


Find $y_{max}$ and $y_{min}$ of $y=\frac{x^{2}-3x+4}{x^{2}+3x+4}$. (hint - do not use calculus -- theres a faster way with inequality  :) )

You can use the discriminant and A.M-G.M. :D

Why can you use discriminant immediately?
You should answer the values of $x,\ y$ for which the given equation is minimized or maximized.

I consider function
$f(x)=x^{2}(y-1)+x(3x+3)+4y-4$ This function has at least one root,so it's discriminant is greater  that zero or equal

You should classify the case of $y=1$ and $y\neq 1.$


$ (\exists )\ f: \mathcal R\rightarrow\mathcal R$ so that $ f(1)\equal{}2$ and $ (\forall )\ x\in\mathcal R$ , $ f(f(x))\equal{}x^2\minus{}2x\plus{}2\ \ ?$

No.  Put x=1 in the equation given to get $ f(2) \equal{} 1$.  Since $ f(1) \equal{} 2$, by continuity there exists $ 1<y<2$ with $ f(y) \equal{} y$.  So $ f(f(y)) \equal{} y^2 \minus{} 2y \plus{}2$ implying $ (y\minus{}1)(y\minus{}2) \equal{} 0$.  Contradiction.


function
search
logarithms
complex analysis

I have just read the topic posted below and I am curious bout one thing

[quote]Show that $ u(x,y) \equal{} \log{\sqrt {x^2 \plus{} y^2}}$ is harmonic but that it has no harmonic conjugate on C\{0}.[/quote]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1992478425&t=216875]Click Here for the solution[/url]

The solution given by Prof Merryfield in the above link seems fine to me but I don't understand why $ v(x,y)\equal{}Arg(z)$ can't be its conjugate harmonic function. Since it seems pretty natural to me that $ f(z)\equal{}\ln z$ satisfy the above equation.

Can anyone clear my doubt? Thanks  :)


We only need slove this problem with $x+y=1$, in fact if $x+y=k\geq 1$ then $\frac{x}{k}+\frac{y}{k}=1$ and therefore $\frac{\frac{x}{k}}{\sqrt{\frac{x^{2}}{k^{2}}+3\frac{y}{k}}}+\frac{\frac{y}{k}}{\sqrt{\frac{y^{2}}{k^{2}}+3\frac{x}{k}}}\geq 1\Rightarrow \frac{x}{\sqrt{x^{2}+3ky}}+\frac{y}{\sqrt{y^{2}+3kx}}\geq 1\Rightarrow\frac{x}{\sqrt{x^{2}+3y}}+\frac{y}{\sqrt{y^{2}+3x}}\geq 1$ (because $k\geq 1$).

[quote="chien than"]Prove that:
$\frac{x}{\sqrt{x^{2}+3y}}+\frac{y}{\sqrt{y^{2}+3x}}\geq 1$
where x;y are non-negative satisfying $x+y \geq 1$[/quote]
I think, it's wrong. Try $x=y=\frac{1}{2}.$ :wink:
Maybe $\frac{x}{\sqrt{x^{2}+1.5y}}+\frac{y}{\sqrt{y^{2}+1.5x}}\geq 1$ $?$


LaTeX
real analysis
real analysis unsolved

Hi, it's me again.
I'm having hard time trying to prove the following:

Let $ A$ be an (associative) algebra with identity $ e$. Let $ \| \cdot \|$ be a norm on $ A$ and suppose that $ (A,\| \cdot \|)$ is a Banach space. Suppose that there exists an $ M \geq 1$ such that $ \|xy\| \leq M \|x\| \|y\|$ for all $ x,y \in A$. (We do not assume that $ M \equal{} 1$ nor that $ \|e\| \equal{} 1$.) Define $ \mathcal{A} \equal{} B(A)$. For all $ x \in A$ define $ U_x,V_x \in \mathcal{A}$ by
$ U_xy \equal{} xy$ and $ V_xy \equal{} yx$ $ (y \in A)$
Define $ \parallel{}|\cdot\parallel{}|: A \longrightarrow [0,\infty)$ by
$ \parallel{}|x\parallel{}| \equal{} \|U_x\| (x \in A)$

Prove that $ \parallel{}|xy\parallel{}| \leq \parallel{}|x\parallel{}| \parallel{}|y\parallel{}|$ for all $ x,y \in A$

gosh that triple bar looks horrible. does anyone know the proper latex symbol for that?

I don't know why you defined $ V_x$ here. If I understand correctly, $ x\mapsto U_x$ is a homomorphism of $ A$ into $ B(A)$. And of course, $ B(A)$ is a Banach algebra.

yes you are right, $ V_x$ is used for different parts of the question. For this part however it's irrelevant as you said.

It seems that here we have to relate the following two sets, in an analagous manner to my previous posting Question realted to Banach Algebras -PART 2-.

That is,
$ \inf\{M \geq 0 \forall z \in A [\|xyz\| \leq M \|z\|]\}$

and the product
$ \inf\{M' \geq 0 \forall z \in A [\|xz\| \leq M' \|z\|]\}\inf\{M'' \geq 0 \forall z \in A [\|yz\| \leq M'' \|z\|]\}$

recall that in part2 of my posting roughly speaking we related infimum of a set with the 'SUM' of two infimums, here we have to relate the infimunm of a set with the 'PRODUCT' of two infimums.


This website explains trig alot better than most trig books I've seen and has a good study card with all of the trig formulas on it- http://www.triglessons.com- hope this helps  :lol:

yeah it is good (maybe better than aops books as well), but it is only basic trig. it doesn't really help in high school? also you might want to get rid of the - on the end of the link, because you can't click on the link otherwise (it doesn't take you to the site)


geometry
3D geometry
octahedron
MATHCOUNTS
pyramid

[quote="SplashD"][hide]$2+4+8=\boxed{14}$

with some scissors and tape, anything is possible  :P  :wink: [/hide][/quote]

But you're not helping her solve it.
You're not allowed scissors/tape at mathcounts, how would you solve this problem if doing it during a test?

I'm really bad at these problems also but the fact that an octahedron is made up of two pyramids sometimes helps when trying to visualize it.

[hide="hints"]1, 2, 3, and 4 make up the top pyramid.

Note how 4 matches up with 5 and 3 matches up with 6. [/hide]

I can see how the 2 and the 4 are connected to the 1 but how do you really know about the 8 w/o cutting it apart etc.

[hide]Hint:
Try to eliminate faces that aren't possible.[/hide]

There are 3 sides to the triangle, so we need to add up 3 numbers. Obviously, since the triangle with the 2 is next to the triangle with the 1, one of the numbers is 2. Also, note that 4 will be next to the 1, and so will the 8. The answer is
$2+4+8=14$

These sort of problems require that you 'visualize' the octahedron.

[quote="Ignite168"]Klebian, you rip it and fold it, I ripped a piece of paper and folded it for a problem during a state competition this year.[/quote]

That's usually what I end up doing....  :roll:

[quote="SplashD"][quote="Ignite168"]Klebian, you rip it and fold it, I ripped a piece of paper and folded it for a problem during a state competition this year.[/quote]

was it the one with 8 numbers in a row?[/quote]
THat one wasn't that hard.

[quote="dodobird"][quote="nonie"]Look at the attatchment please!

I'm terrible at this type of problem. I can do them, but it takes me a really long time to figure it out.

Does anyone have any tips on how to get better at them?

Thanks, any help is appreciated.[/quote]

You can easily see that the 2 and the 4 will connect. Next if you follow the lines to the 8 you can see that they will line up perfectly if you try to fold it that way. a tip might be to imagine it and if that doesnt work try variations of the problem and create a scale model of it using paper. Eventually you will recognize particular patterns and figure out tricks to solving the problem. Everybody learns differently therefore everybody sees things in different ways. So everybody will have tricks of their own that make it easier for themselves to do it. Saying i might have a trick but when you use it, it might become even more confusing.[/quote] Well said.

[hide]
There can be only three other numbers that share an edge with 1.
2 and 4 share an edge, and if you imagine or create the octahedron, you will find that the other number is 8.
$2+4+8=\boxed{14}$[/hide]

[quote="nonie"]Look at the attatchment please!

I'm terrible at this type of problem. I can do them, but it takes me a really long time to figure it out.

Does anyone have any tips on how to get better at them?

Thanks, any help is appreciated.[/quote]
[hide]
you have to visualize this one.
It is quite obvious that 2 and 4 are next to it. Then think about it. It helps to have good logical skills. You will notice 8 to be right.
2+4+8=14 which I believe is the answer[/hide]


function
limit
induction
complex analysis
complex analysis unsolved

Suppose $f$ is analytic on the closed unit disk except at the point $z=1$, where $f$ instead has a pole. We let $f(z) = \sum_{n=0}^\infty a_{n}z^{n}$ be the power series for $f$ around 0. Show that the sum does not converge for any point on the unit circle.

I have given this problem quite some thought and I believe I am completely at the wrong track.

Subtract off the pole at 1; write $f(z)=g(z)+\frac{a}{1-z}$, where $a$ is a constant and $g$ is analytic on the closed unit disk.
Then the power series for $g$ converges uniformly on the closed disk, and the power series $\sum az^{n}$ for $\frac{a}{1-z}$ diverges everywhere on the circle because its terms don't go to zero. The power series for $f$ is the sum of these, and it diverges- $a_{n}\to a$ as $n\to\infty$.

Continuity of $g$ on the boundary is enough to make this work- the coefficients of the power series are the Fourier coefficients of the boundary values.

Well I did this, in a slightly different (worse) way. For a simple pole $g(z) = (z-1)f(z)$ has the series $-a_{0}+\sum_{n=1}^{\infty}(a_{n-1}-a_{n})z^{n}$ which has radius of convergence > 1. In particular we have convergence for $z=1$. But the sum telescopes and we get $\lim_{n}a_{n}= g(1) \neq 0$.

This can easily be extended by induction to show that $a_{n}$ never converges to 0 (for any order pole). I don't know why I didn't realize this when I posted originally.

Does it matter that it was a pole and not some other kind of singularity? Yes.

$f(z)=\sum_{n=1}^{\infty}\frac{z^{n}}n=-\log(1-z)$ has a branch point at $1.$ The power series converges everywhere else on the unit circle.

Note that for this branch point, you cannot employ jmerry's proof; you can't subtract off the singularity.

(jmerry didn't address the issue of poles of higher order, but you would simply subtract off the principal part of the Laurent expansion about 1, and it should work out the same.)

One thing I kept thinking about in doing this problem is this: We must have that $\sum a_{n}$ diverges because otherwise we would by Abel's theorem have that the sum is continuous on the ray between 0 and 1 which would contradict having a pole. However, is it possible to extend this to say that the partial sums diverge to infinity?

If we instead had an essential singularity at 1 I don't see how the continuity is contradictory. Is it?


function
group theory
abstract algebra
number theory
prime numbers
superior algebra
superior algebra solved

Prove that the in group of bijections (one to one functions) from $\mathbb{N}$ to $\mathbb{N}$, denote it by $(G,\circ )$, there are three subgroups, each isomorphic with one of the the three folowing groups:
a) $(\mathbb{Z},+)$
b) $(\mathbb{Q},+)$
c) $(\mathbb{Q}_+^*,\cdot)$

You can get every countably infinite group as a subgroup: given a countably infinite group $G$, identify the countable set $\mathbb{N}$ with the set of elements of $G$, and take the bijections on this set induced by left action of the group ($g(x)=gx$). That's obviously isomorphic to $G$. Since each of the groups we care about is a countably infinite group, we can do this for them.

The original solution was quite ugly. They constructed some particular examples of izomophimes for every desired group.
-for the first they chosen a bijections of infinite order
-for the third they partitioned $\mathbb{N}$ into a infinite number of infinite subsets. and on each subset they find a bijection ,$f_{p_i}$, of infinite order. Then using the fact that $(\mathbb{Q}_+^*,\cdot)$ is generated by prime numbers $p_i$, the conclusion folows.

Thinking about this I got to the same aproach as jmerry. thank you for your time.


I recently figured out something. (maybe some one has figured it out before me)

I was looking at the animation on the side that said The sum of the first n odd natural numbers is n^2. Then I though what the sum of the first n even natural numbers are. After about 30 seconds, I figured out what it is. The sum of the first n even natural numbers is n(n+1). How would you prove this though?

If you accept that the sum of the first $ n$ odd natural numbers is $ n^2$, then you get the sum of the first $ n$ even natural numbers by adding $ 1$ to each of the first $ n$ odd natural numbers; in other words, you get the sum $ n^2 \plus{} n \equal{} n(n \plus{} 1)$.  :)

This also allows you to prove that the sum of the first $ n$ natural numbers (which is half the sum of the first $ n$ even natural numbers) is $ \frac {n(n \plus{} 1)}{2}$.  This is the $ n^{th}$ triangular number $ T_n$.

There are well-known techniques for a lot of different summations; in particular, it is possible to find formulas for the sum of the first $ n$ squares, cubes, fourth powers, ...

We prove $ 2\plus{}4\plus{}6\plus{}...\plus{}2n\equal{}n(n\plus{}1)$ using induction:

[b]Base Case ($ n\equal{}1$)[/b]:
The sum of the first $ 1$ even natural number is $ 2$, and the formula yields $ 1(1\plus{}1)\equal{}2$, so our base case is proven.

[b]Show that $ n$ implies $ n\plus{}1$[/b]:
Assume that our statement is true for $ n\equal{}m$:
$ 2\plus{}4\plus{}6\plus{}...\plus{}2m\equal{}m(m\plus{}1)$
Clearly, since the next even natural number is $ 2(m\plus{}1)$, we add the quantity to both sides:
$ 2\plus{}4\plus{}6\plus{}...\plus{}2m\plus{}2(m\plus{}1)\equal{}m(m\plus{}1)\plus{}2(m\plus{}1)$
By algebraically manipulating the RHS, we get:
$ 2\plus{}4\plus{}6\plus{}...\plus{}2(m\plus{}1)\equal{}(m\plus{}1)[(m\plus{}1)\plus{}1]$

And this looks just like our original formula, so we are done.


1) Let $ A,B \in S_n(R)$  slove matrix equations :
               $ \begin{cases} e^A \equal{} I_n \plus{} B \\
e^B \equal{} I_n \plus{} A \end{cases}$
  $ I_n$ is identity matrix
 2) Let $ A,B \in S^{ \plus{} }_n(R)$ slove matrix equations  :
              $ \begin{cases} Ae^A \equal{} Be^B \\
A^3 \equal{} B \plus{} 6I_n\end{cases}$
 My solution 1) ..2) the same
 From 1) we have $ e^A \plus{} A \equal{} e^B \plus{} B$
 By $ A \in S_n(R)$ the exists $ T \in O_n(R)$ ,$ D \equal{} diag(\lambda_1,\lambda_2,...,\lambda_n) \in D_n(R)$ ,$ A \equal{} TDT^{ \minus{} 1}$ then
              $ e^A \plus{} A \equal{} T(e^D \plus{} D)T^{ \minus{} 1}$
By $ \lambda_1,\lambda_2,...,\lambda_n \in R$ ,$ \forall (i,j) \in$ {1,2,...,n} $ (e^{\lambda_i} \plus{} \lambda_i \equal{} e^{\lambda_j} \plus{} \lambda_j$ then $ \lambda_i \equal{} \lambda_j$ for all $ 1 \leq i,j \leq n$
 Show that the exists $ P \in R[X]$ Such that : $ \forall i \in$ {1,2,...,n} $ P(e^{\lambda_i} \plus{} \lambda_i) \equal{} \lambda_i$
 (by interpolation for $ e^{\lambda_i}\plus{}\lambda_i$ distinct ) then We have :
 $ Tdiag(P(e^{\lambda_1} \plus{} \lambda_1),...,P(e^{\lambda_n} \plus{} \lambda_n))T^{ \minus{} 1} \equal{} Tdiag(\lambda_1,...,\lambda_n)T^{ \minus{} 1}$ Or $ P(e^{A} \plus{} A) \equal{} A$ the same the exists $ Q \in R[X]$ such that $ Q(e^B \plus{} B) \equal{} B$
and $ AB \equal{} P(e^{A} \plus{} A)Q(e^{B} \plus{} B) \equal{} P(e^{A} \plus{} A)Q(e^{A} \plus{} A) \equal{} Q(e^A \plus{} A)P(e^{A} \plus{} A) \equal{} Q(e^{B} \plus{} B)P(e^{A} \plus{} A) \equal{} BA$
  Then the exists $ \omega \in O_n(R)$ Such that :
       $ \omega^{ \minus{} 1}A \omega \equal{} diag(a_1,...,a_n)$ ({$ a_1,...,a_n$}={$ \lambda_1,...,\lambda_n$})
       $ \omega^{ \minus{} 1}B \omega \equal{} diag(b_1,...,b_n)$
       $ \omega^{ \minus{} 1}(e^A \plus{} A) \omega \equal{} diag(e^{a_1} \plus{} a_1,...,e^{a_n} \plus{} a_n)$
       $ \omega^{ \minus{} 1}(e^B \plus{} B) \omega \equal{} diag(e^{b_1} \plus{} b_1,...,e^{b_n} \plus{} b_n)$
 Since $ e^A \plus{} A \equal{} e^B \plus{} B$ then $ e^{a_i} \plus{} a_i \equal{} e^{b_i} \plus{} b_i$ for all $ 1 \leq i \leq n$ then $ a_i \equal{} b_i$ Or $ A \equal{} B$
and need solution $ e^A \equal{} A \plus{} I_n$ eassy slove $ e^{a_i} \equal{} a_i \plus{} 1$ have unique root $ a_i \equal{} 0$ for all $ 1 \leq i \leq n$ and $ A \equal{} B \equal{} 0$ :P
 :)

hi...[b]QuyBac[/b]
 You poof that  $ AB\equal{}BA$ is complicated too . Imples as : $ B\equal{}e^{A}\minus{}I_n$ then $ AB\equal{}A(e^A\minus{}I_n)\equal{}(e^A\minus{}I_n)A\equal{}BA$
 Imples for equation : $ e^{A}\plus{}A\equal{}e^B\plus{}B$


http://www.UploadYourImages.com/view/597759induction.jpg

pretty sure i got the second part wrong but i do not understand why, anyone help?

Yes, that's correct, but you lack a base case. :lol:

Well just do n=1 and it's obvious so yeah.

[hide]
say k=4.  Then, let k+1=u.  u=5.  Then, substituting u for k, we know that it is true.
It's kind of a deep intuition that all of us have.[/hide]


Let try this one. :D (I thinks it is very weak but anyway it is nice)

$a,b,c$ are positive number such that $a+b+c=3$ then
$\sum \frac{a^3(2a^2+bc)}{(b+c)^2} \geq \frac32 \sum \frac{a^4}{b^2+c^2}$
Have fun

Use the ineq $(b+c)^2 \leq 2(b^2+c^2)$ , we need to prove :
 $\sum\frac{a^5+a^3bc-a^4b-a^4c}{b^2+c^2} \geq 0$
$\iff \sum\frac{a^3(a-b)(a-c)}{b^2+c^2} \geq 0$ , which is right by Vornicu-Schur .

Wow, it is very nice. But I don't know what is Vornicu-Schur. I have looked it up in Mathlink and found that $\sum f(a)(a-b)(a-c) \geq 0$ But what is a condition of $f(a)$ in order to satisfy this inequality is a little difficult for me to understand. Can anyone explain it for me please? I think it is a very interesting equality.
   By the way, from here, we can find that $\sum \frac{a^3(2a^2+bc)}{(b+c)^2} \geq \frac{9}{4}$ easily .Right?


AMC
USA(J)MO
USAMO
email
AMC 10
geometry
trapezoid

Hey. . .

I got invited to take the APMO, you guys got any info on it? I'd appreciate any help you may give me.

Thanks.

I think all of the people who went to MOP last year, plus anyone not at MOP who got a perfect score on the AMC-12 this year, were invited to do APMO. So maybe 20-30 people.

Everyone who takes it submits their papers to AMC, which grades them all and then chooses the top 10 (by their grading) to be the official US entries. People don't actually travel anywhere to compete in this.

[quote="Idiot_without_a_village"]Yeah I think that dudes right. . . but like, the problems don't seem to be as hard as the USAMO, I dunno though.[/quote]

Wait..you went to MOP last year?

[quote="Idiot_without_a_village"]Naw dawg, I only got a perfect score on dis yer's AMC 12. . . Mebbe MOP dis yer, but I dun think it'll happen man.[/quote]

So you're one of the 12 who got the perfect score? Nice!

Wait, are you kidding?  Is this the first year they've invited perfect AMCer's?  I feel so cheated! :)

Yeah, I got some email or something. . . I dunno, it all happened so fast, my math teacher dude was like, you wanna do this APMO shiznit, and I was like, fo shizzle. . . so, it's all good.

Got in through the AMC 10A Perfect score.

AMC10? That's pretty tite homie. . . yeah man, I guess we just need to plow through some of their old problems or something, the early ones were really easy and then a lil' later ya know they got tougher. . . aight man, I dunno about this thing but I guess we'll see what happens, it's all about the AIME/USAMO this year mon.

Did anyone here take this? How did it go?

#1,#2, #4 are almost trivial... #3 and #5 just seems impossible

[quote="ComplexZeta"]Are the problems posted somewhere? If not everyone has taken it yet (I don't know how this works), please don't post them here.[/quote]

Yes, please wait until the problems are posted on http://cms.math.ca/Competitions/APMO/ to discuss them publicly.  In past years, the offical results have been nullified because the problems were posted online before all participants had taken the exam.  The organizers are rather slow to post the official problems, but please respect their rules.

The questions are finally up.  http://www.cms.math.ca/Competitions/APMO/exam/apmo2004.html

The only one I'm 100% sure about my answer is #1.

I got:

[hide]

{2}

[/hide]

#1, not quite complete: [hide]assume you have at least 2 numbers: if you have two relatively prime numbers, you lose, where winning is defined as having finitely many terms.  Thus, all terms must have a greatest common divisor, k.  So our numbers can be written as k*a1, ..., k*an.  Then, using our law, k | ai + aj for all i, j.  That means that all of the ai are congruent to 0 mod k, or all divisible by k.  This is a contradiction unless k = 1, which also makes us lose.  Otherwise, we must have either (a) k even and all numbers congruent to k/2, in which case we lose unless k = 2, from which we get the solution {2} by going back and reapplying the process or (b) we only have two numbers.  I didn't work these last two cases through at all, but I'm sure that the first leads to a contradiction if the solution isn't {2}, and the second leads to a contradiction.[/hide]

Problem 5 I can reduce to proving (abc)^2 + 2 :ge: ab + bc + ca, but I don't actually know if that is true or not.

Number 2 I can do uglily with abSinC/2 area formula.  Looking for a nicer way.

I think number 3 can be solved by somehow coloring the plane or setting up an isomorphism (I don't know if this is exactly the right word) between the points and sections of the plane, or something.  Basically, I know that it's easy to two-color a plane cut up by any number of lines so that each section of the plane borders only different-colored sections.  I'm wondering if there's some way to use that fact to do this question.

Oh, and I think I just got number 4:

[hide]Cases:  1) n = 1, 2, 3, 4 : the result follows by calculation.

From now on, n > 4.  2) n and n + 1 are composite.  In that case, we have that they are both > 4, and as such, all of their prime factors are represented in the numerator of the fraction.  There are clearly more than enough 2's to go around.  Thus, it is even.

3) n odd prime > 3.  (n - 1)! :equiv: -1 (mod n), and (n + 1) | (n - 1)! because, as above, there are plenty of factors to go around.  (The only problem could possibly be n + 1 = 8, but even there, we're fine.)  (n + 1) :equiv: 1 (mod n), thus (n - 1)! = (n + 1)*(k*n - 1).  We get that k must be odd, since (n - 1)!/(n + 1) is even and n is odd, so our expression is just less than an odd number and thus floor of it is even.

4) (n + 1) odd prime > 5.  Everything basically works the same way.  The n disappears into the (n - 1)!, leaving us with k*(n + 1) - 1, and k must be odd for the same reasons, so everything again works.

[/hide]

This was kind of sketchy, but I'm confident of it.

number 1 is kinda easy, but it might be a slightly more interesting problem if i is not allowed to equal j.

If i is not allowed to equal j, I think any proof will still work with the only added affect of removing the one solution.

right, you just get slightly more cases to examine

Demon... Which one are you on the AMC website?

[quote="Demon"]Then, you'll get infinitely many sets starting with two member sets we get (a+b)/(a,b)=a. It follows that a and b aren't relatilevly prime, the solution I have for 2 element sets is
{a, a(a-1)}
Most likely you'll be able to getb infinitely many sets for any given number of elements.[/quote]

actually i have proved that if i is not allowed equal to j, then |S|<3.  so S={a,a(a+1)} or S={a}.  The proof is quiet complicated and involve some case work so i am interested in seeing a shorter proof.
this would make a nice question 4 and the original question four can be bumped to 1.

JBL, for number 2, since OH is consistent in all three triangles, you gotta prove the average of the two heights (WLOG A to OH and B to OH) is equal to C to OH.  but the average of the heights is equilvalent to the midpoint height of the right trapezoid, which is the perpendicular bisector of AB.  Then using the fact that the centroid lies on the Euler line and dividing the median into the ratio of 1:2, we complete our geometric proof

2004APMO #2 can be solved by vector easily.

Let X be the vector from O to X. Then H=A+B+C. The cross product
0=HxH=(A+B+C)xH=AxH+BxH+CxH. Note AxH,BxH,CxH are perpendicular
to the plane of ABC and their magnitudes are twice the area of triangles AOH,
BOH,COH. So the magnitude part of the equation AxH+BxH+CxH=0 gives
the desired conclusion.


topology
real analysis
real analysis unsolved

Let S be a subset of R. Show that S is an inverval if and only if for all a,b in S, we have [a,b] contained in S.

How, in particular, are you defining interval?  Some books take it to be only sets of the form $ \{a\leq x\leq b\}$, others allow open, and half-open.  And are we allowed to take $ a\equal{}\minus{}\infty$, $ b\equal{}\infty$?

we're allowing open and closed and unbounded (infinitis). But he actually said it doesn't matter because you can prove it for any form.

Still, what is you definition of 'interval'? In our analysis course, the definition was exactly what you want to prove.

This result only states that the only connected sets in $ \mathbb{R}$ with the usual topology (since here connected = connected by arcs) are the intervals. One implication is trivial. For the other one, embed $ \mathbb{R}$ in $ \overline{\mathbb{R}}$ and for that set $ S$ take $ \alpha \equal{} \inf S$ and $ \beta \equal{} \sup S$. It follows that $ (\alpha, \beta) \subseteq S$ and we are done.


Yup. He's my current professor for discrete mathematics.

what should I ask him...he's so approachable yet..he also shines in a way too bright for me.

That depends on what you are interested in yourself and how you are doing in his course. If you like the subject and are getting straight A's, it may be a good idea just to come and say that you want some additional extracurricular reading to do and proceed from there but if you are not interested in his area in the slightest or if your typical midterm grade is between C and D, then all you'll be really able to ask is questions about homeworks.

As for being "too bright", that shouldn't prevent you from talking to him. Bright professors usually do not bite their students :P.


i have no idea what that "proof without words" flash animation explains?
can someone explain to me how they prove that the sum of squares equal that equation?

Thats a good question.  I will try to figure it out over the next day so check back in a few.

At the beginning of the animation, we see all the squares laid out. Then they use all of the squares three times to make a rectangle. The height of the rectangle is 2n+1. The width of the rectangle is 1+2+3+...+n=(n)(n+1)/2, which is proven in another animation. So we have $1^2+2^2+3^2+\cdots+n^2=\frac{(2n+1)\frac{n(n+1)}{2}}{3}=\frac{n(n+1)(2n+1)}{6}$


geometry
incenter
geometric transformation
homothety
ratio
radical axis
angle bisector

Prove that in triangle $ABC$, radical center of its excircles lies on line $GI$, which $G$ is Centroid of triangle $ABC$, and $I$ is the incenter.

According to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=20029#20029]http://www.mathlinks.ro/Forum/viewtopic.php?t=5958 post #2[/url], the radical center of the excircles of triangle ABC is the Spieker point of triangle ABC, that is, the incenter of triangle A'B'C', where A', B', C' are the midpoints of the sides BC, CA, AB of triangle ABC. Thus, it remains to show that the incenter of triangle A'B'C' lies on the line GI. Now, it is well-known that the homothety with center G and ratio $-\frac12$ maps the points A, B, C to the points A', B', C'. Hence, this homothety maps the incenter I of triangle ABC to the incenter of triangle A'B'C'. Since the center of the homothety is G, this yields that the incenter of triangle A'B'C' lies on the line GI. Proof complete.

Actually, this problem is way too well-known for an exam imho...

  darij

this was posted also on the italian forum [url=http://olimpiadi.dm.unipi.it/oliForum/viewtopic.php?t=6470]here[/url], this is the edriv's solution:

Let the A-excircle touch AB on L an the B-excircle touch AB on K. Then as well-know the radical axes of $ \Gamma_a$ and $ \Gamma_b$ pass through the point M that is the midpoint of LK that is the midpoint of AB. Also we have that the radical axes is perpendicular to $ I_aI_b$ that is parallel to the internal angle bisector of $ \angle CAB$ and the radical axes  of $ \Gamma_a$ and $ \Gamma_b$ is the internal M-angle bisector of the medial triangle and equally for the others, from which the radical center P of the excircles is the incenter of the medial triangle (the Spieker point of ABC) so for an homothety with center G it lies on IG with  $ IG \equal{} 2 GP$.

ok It's the same solution of darij   :|

[quote="Omid Hatami"]Prove that in triangle $ ABC$, radical center of its excircles lies on line $ GI$, which $ G$ is Centroid of triangle $ ABC$, and $ I$ is the incenter.[/quote]
I have a solution  :D  Maybe it is same of you  :D
[img]http://i209.photobucket.com/albums/bb172/Ineq/excircles.jpg[/img]
Let $ D$ is midpoint of $ AC$
$ a$ is radical axis of $ O_1)$ and $ (O_3)$
Easy to prove $ I$ is orthocenter of $ O_1O_2O_3$
and $ AE\equal{}CF$
So $ DE\equal{}DF \Rightarrow P_{D|(O_3)}\equal{}P_{D|(O_1)}$
so $ D \in a$ and $ a // BI$
$ K\equal{}GI\cap a$ then $ 2GK\equal{}IG$
Similar we have desire


Give a, b, c>0, and:∑(a).^2=3 ,Find The minimum of this product:
                             (1-3(a.^3)*b)(1-3(b.^3)*c)(1-3(c.^3)*a)

                                            Have a good time.

[quote="navid"]Give $a,\,b,\,c>0$ and $\sum a^{2}=3$. Find the minimum of product: \[\left(1-3a^{3}b\right)\left(1-3b^{3}c\right)\left(1-3c^{3}a\right)\]
[/quote]
[b]navid[/b], Is it true?

Yes it is tru and is there any one who solve it?


function algebra polynomial induction LaTeX quadratics Putnam

A certain 2005-degree polynomial f(x) has this property:
For any n, the numbers    n, f(n), f(f(n)), f(f(f(n))), ...    are all relatively prime.
Find f(x).

does it have integer coefficients?  i think that might make it easier...
f(x)=1 works but it is not exactly of degree 2005. :lol:

is natural number 1 considered 'relatively prime' to every integer?

Quite frankly, I don't know the answer.  I think 1 is relatively prime to itself, but is 0 relatively prime to itself?  And how does relatively prime work with negative numbers?

I thought that this function was impossible at first, but it may be possible afterall.

By the way, n must be an integer.

[quote="Palytoxin"][quote="Wumbate"]...is 0 relatively prime to itself?  And how does relatively prime work with negative numbers?

...
By the way, n must be an integer.[/quote]

0 has no common factor with any integer becuz any number times 0 is 0, which can't possibly be relatively prime to any number, maybe to itself since 1*0=0. but im not sure if they have a convention for 1. yes 'relatively prime' works with negative numbers because it is just a matter of sign, it doesn't affect the divisors. so $n$ must be a [b]nonzero[/b] integer  :D[/quote]

Ah, I never knew that negative integers could be relatively prime to each other.
According to Mathworld, two integers that are relatively prime to each other if their greatest common divisor is 1. So, in fact, 1 is relatively prime to any nonzero integer.

Here's the link: http://mathworld.wolfram.com/RelativelyPrime.html

1 is relatively prime to all other integers.  0 is [i]not[/i] relatively prime with any integer other than 1 (or -1, if we include the negatives), as for any $p \geq 2$, $p | p$ and $p | 0$.  Thus, the greatest common divisior of 0 and anything else is that thing (or its absolute value, if we include negatives), with the exception of 0 and 0, for which there is no greatest common division (as every integer divides both 0 and 0).

For the purposes of this problem, however, it is best to assume (1) that n is a positive number and (2) that none of the numbers f(n), f(f(n)), ... are 0.

my guess is... $f(x)=x^{2005}+1$ but i dunno how to prove it  :(   :wallbash:

That is wrong, since f(2)=2^2005+1=odd
and f(odd)=odd^2005+1=even, so 2 and the even number share 2 as a factor.

I'm trying to work considerations like that into a general polynomial.

hmm i see your point

how about $f(x)=2x^{2005}+1$ ?  :D

Wumbate, how do you know such a polynomial exists?

Paly, you should test before you guess.  For any multiple of 3, that second polynomial fails after 2 interations.

I don't know it exists, in fact my hunch is that it doesn't.
However, this problem is one from a pack of problems I got, without the answers, so it would be weird if this problem was actually a trick question, since all the others weren't.
I don't know where the problem comes from originally.

Well, my challenge then:  solve the problem where the degree of f is replaced by (a) 0 and (b) 1.  Then let's think about 2005.

Actually, there's almost certainly a modular argument of some sort.  I just don't know what it would be :-)

How about degree 2?  Let $f(x) = x^2 - x + 1$.  It's well known that the sequence will be relatively prime in that case.  Can you guys prove it?

For degree 0:  f(x)=1
For degree 1:  I feel like I'm missing something obvious.  Impossible?

Proof for degree 2:
$f(x)=x^2-x+1=(x)(x-1)+1$

Let $x_0$ be the first of the iteration.

First, note that $x_{n+1}-1=(x_n)*(x_n-1)$ which, by induction backwards, is simply the product of all the previous n's times $x_0-1$ (don't know how to latex this nicely).

Then $f(x_{n+1})=(x_n)*(x_n-1)+1$.  The value $(x_n-1)$, by the above, has every previous value multiplied into it.  So taking this mod any previous x will = 1, i.e. they are relatively prime.

This function allows me to solve the 2005 degree one:
$f(x)=x^{2004}*(x-1)+1=x^{2005}-x^{2004}+1$
The proof would be the same as above.

Pretty good.  The quadratic problem was half of a Putnam problem from 1956.

Let's generalize.  Let $f$ be an integer polynomial such that $f(0) = 1$ and $f(1) = 1$.  Then, believe it or not, the given sequence will be relatively prime.  One way to prove it is to use your "product" method.  Another way (perhaps slicker) is to use mods (JBL was right).

:? and what makes this problem intermediate topics? am i missing something?  :help:

Yes, the 2005-degree polynomial I gave isn't the only one.  It can be any polynomial of the form:
$f(x)=(x)^a(x-1)^b+1$
where a+b=2005 and a, b>0

I thought it was intermediate because it came from a set of problems which were easier than this one.

Even more general than that: $f(x) = 1 + x(x-1) g(x)$, where $g$ is an arbitrary polynomial (of degree say 2003).

Oh, that's right, thanks for the correction.

:help:  intermediate topics too hard for me, i better stay in getting started...  :help:  :oops_sign:

Is this the only polynomial like that? I somehow doubt it

Here are 6 general forms that I can think of:

1. $f(x) = x(x-1) g(x) + 1$.
2. $f(x) = x(x+ 1) g(x) - 1$.
3. $f(x) = (x-1) x (x + 1) g(x) - 2x^2 + 1$.
4. $f(x) = (x-1) x (x+1) g(x) + 2x^2  - 1$.
5. $f(x) = (x - 1) x (x + 1) g(x) - x (x + 1) + 1$.
6. $f(x) = (x - 1) x (x + 1) g(x) + x(x - 1) - 1$.

I think I can prove that those 6 forms cover all such polynomials, but I would have to think some more to be sure.

Those 6 forms may look mysterious, but they can be stated more simply by looking at their values:

1. $f(0) = 1$, $f(1) = 1$.
2. $f(0) = -1$, $f(-1) = -1$.
3. $f(0) =  1$, $f(1) = -1$, $f(-1) = -1$.
4. $f(0) = -1$, $f(-1) = 1$, $f(1) = 1$.
5. $f(0) = 1$, $f(1) = -1$, $f(-1) = 1$.
6. $f(0) = -1$, $f(-1) = 1$, $f(1) = -1$.

In each, we get a cycle of values starting from 0 and going through -1 or 1.

(By the way, notice that even the quadratic example gives a infinite sequence of relatively prime numbers.  As a corollary, we get an amusing proof that there are infinitely many primes.)

[quote]How about degree 2?  Let $f(x) = x^2 - x + 1$.  It's well known that the sequence will be relatively prime in that case.[/quote]

Are you thinking of $x^2 - 2x + 2$, which generates the sequence $1 + a^{2^k}$ with $a=x-1$?

No, I meant $x^2 - x + 1$.  The function $x^2 - 2x + 2$, while interesting, doesn't work for this problem.  (Just set $n = 4$ say.)

Perhaps "well known" was an exaggeration.  What I really meant was "I have seen it before" (probably from the Putnam 1956 problem).